Brian C. Hall Lie Groups, Lie Algebras, and Representationsgxyau.github.io/pdf-documents/ETHZ/Autumn...

453
Graduate Texts in Mathematics Brian C. Hall Lie Groups, Lie Algebras, and Representations An Elementary Introduction Second Edition

Transcript of Brian C. Hall Lie Groups, Lie Algebras, and Representationsgxyau.github.io/pdf-documents/ETHZ/Autumn...

Graduate Texts in Mathematics

Brian C. Hall

Lie Groups, Lie Algebras, and RepresentationsAn Elementary Introduction

Second Edition

Graduate Texts in Mathematics 222

Graduate Texts in Mathematics

Series Editors:

Sheldon AxlerSan Francisco State University, San Francisco, CA, USA

Kenneth RibetUniversity of California, Berkeley, CA, USA

Advisory Board:

Alejandro Adem, University of British ColumbiaDavid Jerison, University of California Berkeley & MSRIIrene M. Gamba, The University of Texas at AustinJeffrey C. Lagarias, University of MichiganKen Ono, Emory UniversityJeremy Quastel, University of TorontoFadil Santosa, University of MinnesotaBarry Simon, California Institute of Technology

Graduate Texts in Mathematics bridge the gap between passive study andcreative understanding, offering graduate-level introductions to advanced topicsin mathematics. The volumes are carefully written as teaching aids and highlightcharacteristic features of the theory. Although these books are frequently used astextbooks in graduate courses, they are also suitable for individual study.

More information about this series at http://www.springer.com/series/136

Brian C. Hall

Lie Groups, Lie Algebras,and RepresentationsAn Elementary Introduction

Second Edition

123

Brian C. HallDepartment of MathematicsUniversity of Notre DameNotre Dame, IN, USA

ISSN 0072-5285 ISSN 2197-5612 (electronic)Graduate Texts in MathematicsISBN 978-3-319-13466-6 ISBN 978-3-319-13467-3 (eBook)DOI 10.1007/978-3-319-13467-3

Library of Congress Control Number: 2015935277

Springer Cham Heidelberg New York Dordrecht London© Springer International Publishing Switzerland 2003, 2015This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part ofthe material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation,broadcasting, reproduction on microfilms or in any other physical way, and transmission or informationstorage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodologynow known or hereafter developed.The use of general descriptive names, registered names, trademarks, service marks, etc. in this publicationdoes not imply, even in the absence of a specific statement, that such names are exempt from the relevantprotective laws and regulations and therefore free for general use.The publisher, the authors and the editors are safe to assume that the advice and information in this bookare believed to be true and accurate at the date of publication. Neither the publisher nor the authors orthe editors give a warranty, express or implied, with respect to the material contained herein or for anyerrors or omissions that may have been made.

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For Carla

Contents

Part I General Theory

1 Matrix Lie Groups. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Topological Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.4 Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.5 Lie Groups .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2 The Matrix Exponential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312.1 The Exponential of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312.2 Computing the Exponential .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342.3 The Matrix Logarithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362.4 Further Properties of the Exponential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402.5 The Polar Decomposition .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

3 Lie Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.1 Definitions and First Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.2 Simple, Solvable, and Nilpotent Lie Algebras . . . . . . . . . . . . . . . . . . . . . . 533.3 The Lie Algebra of a Matrix Lie Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573.5 Lie Group and Lie Algebra Homomorphisms . . . . . . . . . . . . . . . . . . . . . . 603.6 The Complexification of a Real Lie Algebra . . . . . . . . . . . . . . . . . . . . . . . 653.7 The Exponential Map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 673.8 Consequences of Theorem 3.42 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 703.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

4 Basic Representation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 774.1 Representations .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 774.2 Examples of Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 814.3 New Representations from Old . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

vii

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4.4 Complete Reducibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 904.5 Schur’s Lemma .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 944.6 Representations of sl.2IC/ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 964.7 Group Versus Lie Algebra Representations .. . . . . . . . . . . . . . . . . . . . . . . . 1014.8 A Nonmatrix Lie Group.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1034.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

5 The Baker–Campbell–Hausdorff Formula and Its Consequences . . . . 1095.1 The “Hard” Questions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1095.2 An Illustrative Example .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1105.3 The Baker–Campbell–Hausdorff Formula . . . . . . . . . . . . . . . . . . . . . . . . . . 1135.4 The Derivative of the Exponential Map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1145.5 Proof of the BCH Formula .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1175.6 The Series Form of the BCH Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1185.7 Group Versus Lie Algebra Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . 1195.8 Universal Covers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1265.9 Subgroups and Subalgebras. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1285.10 Lie’s Third Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1355.11 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

Part II Semisimple Lie Algebras

6 The Representations of sl.3IC/ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1416.1 Preliminaries. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1416.2 Weights and Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1426.3 The Theorem of the Highest Weight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1466.4 Proof of the Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1486.5 An Example: Highest Weight .1; 1/ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1536.6 The Weyl Group .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1546.7 Weight Diagrams .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1586.8 Further Properties of the Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . 1596.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

7 Semisimple Lie Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1697.1 Semisimple and Reductive Lie Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1697.2 Cartan Subalgebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1747.3 Roots and Root Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1767.4 The Weyl Group .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1827.5 Root Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1837.6 Simple Lie Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1857.7 The Root Systems of the Classical Lie Algebras . . . . . . . . . . . . . . . . . . . 1887.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193

8 Root Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1978.1 Abstract Root Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1978.2 Examples in Rank Two. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2018.3 Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204

Contents ix

8.4 Bases and Weyl Chambers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2068.5 Weyl Chambers and the Weyl Group .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2128.6 Dynkin Diagrams. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2168.7 Integral and Dominant Integral Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . 2188.8 The Partial Ordering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2218.9 Examples in Rank Three .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2288.10 The Classical Root Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2328.11 The Classification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2368.12 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238

9 Representations of Semisimple Lie Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2419.1 Weights of Representations .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2429.2 Introduction to Verma Modules. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2449.3 Universal Enveloping Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2469.4 Proof of the PBW Theorem .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2519.5 Construction of Verma Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2559.6 Irreducible Quotient Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2589.7 Finite-Dimensional Quotient Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2619.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264

10 Further Properties of the Representations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26710.1 The Structure of the Weights . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26710.2 The Casimir Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27110.3 Complete Reducibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27510.4 The Weyl Character Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27710.5 The Weyl Dimension Formula .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28310.6 The Kostant Multiplicity Formula .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28910.7 The Character Formula for Verma Modules . . . . . . . . . . . . . . . . . . . . . . . . 29610.8 Proof of the Character Formula .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29710.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305

Part III Compact Lie Groups

11 Compact Lie Groups and Maximal Tori . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30911.1 Tori . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31011.2 Maximal Tori and the Weyl Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31411.3 Mapping Degrees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31711.4 Quotient Manifolds. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32311.5 Proof of the Torus Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32811.6 The Weyl Integral Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33211.7 Roots and the Structure of the Weyl Group . . . . . . . . . . . . . . . . . . . . . . . . . 33511.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341

12 The Compact Group Approach to Representation Theory . . . . . . . . . . . . 34512.1 Representations .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34512.2 Analytically Integral Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34812.3 Orthonormality and Completeness for Characters . . . . . . . . . . . . . . . . . . 353

x Contents

12.4 The Analytic Proof of the Weyl Character Formula . . . . . . . . . . . . . . . . 35912.5 Constructing the Representations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36312.6 The Case in Which ı is Not Analytically Integral . . . . . . . . . . . . . . . . . . 36812.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371

13 Fundamental Groups of Compact Lie Groups . . . . . . . . . . . . . . . . . . . . . . . . . . 37313.1 The Fundamental Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37313.2 Fundamental Groups of Compact Classical Groups .. . . . . . . . . . . . . . . 37513.3 Fundamental Groups of Noncompact Classical Groups .. . . . . . . . . . . 37913.4 The Fundamental Groups of K and T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37913.5 Regular Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38513.6 The Stiefel Diagram .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39113.7 Proofs of the Main Theorems .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39613.8 The Center of K . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40113.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405

A Linear Algebra Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409A.1 Eigenvectors and Eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409A.2 Diagonalization .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 410A.3 Generalized Eigenvectors and the SN Decomposition .. . . . . . . . . . . . 411A.4 The Jordan Canonical Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413A.5 The Trace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413A.6 Inner Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414A.7 Dual Spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416A.8 Simultaneous Diagonalization .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417

B Differential Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 421

C Clebsch–Gordan Theory and the Wigner–Eckart Theorem . . . . . . . . . . . 427C.1 Tensor Products of sl.2IC/ Representations . . . . . . . . . . . . . . . . . . . . . . . . 427C.2 The Wigner–Eckart Theorem .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 430C.3 More on Vector Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434

D Peter–Weyl Theorem and Completeness of Characters . . . . . . . . . . . . . . . . 437

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447

Erratum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . E1

Preface

This text treats Lie groups, Lie algebras, and their representations. My pedagogicalgoals are twofold. First, I strive to develop the theory of Lie groups in an elementaryfashion, with minimal prerequisites. In particular, in Part I, I develop the theoryof (matrix) Lie groups and their Lie algebras using only linear algebra, withoutrequiring any knowledge of manifold theory. Second, I strive to provide moremotivation and intuition for the proofs, often using a figure, than in some of theclassic texts on the subject. At the same time, I aim to be fully rigorous; anexplanation or figure is a supplement to and not a replacement for a traditionalproof.

Although Lie theory is widely used in both mathematics and physics, there isoften a wide gulf between the presentations of the subject in the two disciplines:Physics books get down to business quickly but are often imprecise in definitionsand statements of theorems, whereas math books are more rigorous but oftenhave a high barrier to entry. It is my hope that this book will be useful to bothmathematicians and physicists. In particular, the matrix approach in Part I allowsfor definitions that are precise but comprehensible. Although I do not delve into thedetails of how Lie algebras are used in particle theory, I do include an extendeddiscussion of the representations of SU.3/, which has obvious applications to thatfield. (My recent book, Quantum Theory for Mathematicians [Hall], also aimsto bridge a gap between the mathematics and physics literatures, and it containssome discussion of Lie-theoretic issues in quantum mechanics. The emphasis there,however, is on nonrelativistic quantum mechanics and not on quantum field theory.)

Content of the Book Part I of the text covers the general theory of matrix Liegroups (i.e., closed subgroups of GL.nIC/) and their Lie algebras. Chapter 1introduces numerous examples of matrix Lie groups and examines their topologicalproperties. After discussing the matrix exponential in Chapter 2, I turn to Liealgebras in Chapter 3, examining both abstract Lie algebras and Lie algebrasassociated with matrix Lie groups. Chapter 3 shows, among other things, that everymatrix Lie group is an embedded submanifold of GL.nIC/ and, thus, a Lie group.In Chapter 4, I consider elementary representation theory. Finally, Chapter 5 covers

xi

xii Preface

the Baker–Campbell–Hausdorff formula and its consequences. I use this formula(in place of the more traditional Frobenius theorem) to establish some of the deeperresults about the relationship between Lie groups and Lie algebras.

Part II of the text covers semisimple Lie algebras and their representations. Ibegin with an entire chapter on the representation theory of sl.3IC/, that is, thecomplexification of the Lie algebra of the group SU.3/. On the one hand, thisexample can be treated in an elementary way, simply by writing down a basis andcalculating. On the other hand, this example allows the reader to see the machineryof roots, weights, and the Weyl group in action in a simple example, thus motivatingthe general version of these structures. For the general case, I use an unconventionaldefinition of “semisimple,” namely that a complex Lie algebra is semisimple if it hastrivial center and is the complexification of the Lie algebra of a compact group. Ishow that every such Lie algebra decomposes as a direct sum of simple algebras, andis thus semisimple in the conventional sense. Actually, every complex Lie algebrathat is semisimple in the conventional sense has a “compact real form,” so that mydefinition of semisimple is equivalent to the standard one—but I do not prove thisclaim. As with the choice to consider matrix Lie groups in Part I, this (apparent)reduction in scope allows for a rapid development of the structure of semisimple Liealgebras. After developing the necessary properties of root systems in Chapter 8, Igive the classification of representations in Chapter 9, as expressed in the theoremof the highest weight. Finally, Chapter 10 gives several additional properties of therepresentations, including complete reducibility, the Weyl character formula, andthe Kostant multiplicity formula.

Finally, Part III of the book presents the compact-group approach to represen-tation theory. Chapter 11 gives a proof of the torus theorem and establishes theequivalence between the Lie-group and Lie-algebra definitions of the Weyl group.This chapter does, however, make use of some of the manifold theory that I avoidedpreviously. The reader who is unfamiliar with manifold theory but willing to takea few things on faith should be able to proceed on to Chapter 12, where I developthe Weyl character formula and the theorem of the highest weight from the compactgroup point of view. In particular, Chapter 12 gives a self-contained constructionof the representations, independent of the Lie-algebraic argument in Chapter 9.Lastly, in Chapter 13, I examine the fundamental group of a compact group fromtwo different perspectives, one that treats the classical groups by induction on thedimension and one that is based on the torus theorem and uses the structure of theroot system. This chapter shows, among other things, that for a simply connectedcompact group, the integral elements from the group point of view coincide withthe integral elements from the Lie algebra point of view. This result shows that forsimply connected compact groups, the theorem of the highest weight for the groupis equivalent to the theorem of the highest weight for the Lie algebra.

The first four chapters of the book cover elementary Lie theory and could be usedfor an undergraduate course. At the graduate level, one could pass quickly throughPart I and then cover either Part II or Part III, depending on the interests of theinstructor. Although I have tried to explain and motivate the results in Parts II and IIIof the book, using figures whenever possible, the material there is unquestionably

Preface xiii

more challenging than in Part I. Nevertheless, I hope that the explicit working out ofthe case of the Lie algebra sl.3IC/ (or, equivalently, the group SU.3/) in Chapter 6will give the reader a good sense of the flavor of the results in the subsequentchapters.

In recent years, there have been several other books on Lie theory that use thematrix-group approach. Of these, the book of Rossmann [Ross] is most similar instyle to my own. The first three chapters of [Ross] cover much of the same materialas the first four chapters of this book. Although the organization of my book is,I believe, substantially different from that of other books on the subject, I makeno claim to originality in any of the proofs. I myself learned most of the materialhere from the books of Bröcker and tom Dieck [BtD], Humphreys [Hum], andMiller [Mill].

New Features of Second Edition This second edition of the book is substantiallyexpanded from the first edition. Part I has been reorganized but covers mostly thesame material as in the first edition. In Part II, however, at least half of the material isnew. Chapter 8 now provides a complete derivation of all relevant properties of rootsystems. In Chapter 9, the construction of the finite-dimensional representations ofa semisimple Lie algebra has been fleshed out, with the definition of the universalenveloping algebra, a proof of the Poincaré–Birkhoff–Witt theorem, and a proof ofthe existence of Verma modules. Chapter 10 is mostly new and includes completeproofs of the Weyl character formula, the Weyl dimension formula, and the Kostantmultiplicity formula. Part III, on the structure and representation theory of compactgroups, is new in this edition.

I have also included many more figures in the second edition. The black-and-white images were created in Mathematica, while the color images in Sect. 8.9were modeled in the Zometool system (www.zometool.com) and rendered in ScottVorthmann’s vZome program (vzome.com). I thank Paul Hildebrandt for assistingme with construction of the Zometool models and Scott Vorthmann for going aboveand beyond in assisting me with use of vZome.

Acknowledgments I am grateful for the input of many people on various versions of this text,which has improved it immensely. Contributors to the first printing of the first edition include EdBueler, Wesley Calvert, Tom Goebeler, Ruth Gornet, Keith Hubbard, Wicharn Lewkeeratiyutkul,Jeffrey Mitchell, Ambar Sengupta, and Erdinch Tatar. For the second printing of the first edition,contributors include Moshe Adrian, Kamthorn Chailuek, Paul Gibson, Keith Hubbard, DennisMuhonen, Jason Quinn, Rebecca Weber, and Reed Wickner. Additional corrections to the firstedition since the second printing appeared are due to Kate Brenneman, Edward Burkard, MoritzFirsching, Nathan Gray, Ishan Mata, Jean-Renaud Pycke, and Jason Quinn. Contributors to thesecond edition include Matt Cecil, Alexander Diaz-Lopez, Todd Kemp, Ben Lewis, GeorgeMcNinch, and Ambar Sengupta. Thanks to Jonathan Conder, Christopher Gilbreth, Ian Iscoe,Benjamin Lewis, Brian Stoyell-Mulholland, and Reed Wickner for additional input on the secondprinting of the second edition. Please write to me with questions or corrections at [email protected] further information, click on the “Book” tab of my web site: www.nd.edu/~bhall/.

Notre Dame, IN Brian C. HallJanuary 2015

Part IGeneral Theory

Chapter 1Matrix Lie Groups

1.1 Definitions

A Lie group is, roughly speaking, a continuous group, that is, a group describedby several real parameters. In this book, we consider matrix Lie groups, which areLie groups realized as groups of matrices. As an example, consider the set of all2 � 2 real matrices with determinant 1, customarily denoted SL.2IR/. Since thedeterminant of a product is the product of the determinants, this set forms a groupunder the operation of matrix multiplication. If we think of the set of all 2 � 2

matrices, with entries a; b; c; d , as R4, then SL.2IR/ is the set of points in R4 forwhich the smooth function ad � bc has the value 1.

Suppose f is a smooth function on Rk and we consider the set E where f .x/equals some constant value c. If, at each point x0 in E , at least one of the partialderivatives of f is nonzero, then the implicit function theorem tells us that we cansolve the equation f .x/ D c near x0 for one of the variables as a function of theother k � 1 variables. Thus, E is a smooth “surface” (or embedded submanifold) inRk of dimension k � 1. In the case of SL.2IR/ inside R4, we note that the partialderivatives of ad�bc with respect to a, b, c, and d are d , �c, �b, and a, respectively.Thus, at each point where ad � bc D 1, at least one of these partial derivatives isnonzero, and we conclude that SL.2IR/ is a smooth surface of dimension 3. Thus,SL.2IR/ is a Lie group of dimension 3.

For other groups of matrices (such as the ones we will encounter later inthis section), one could use a similar approach. The analysis is, however, morecomplicated because most of the groups are defined by setting several different

A previous version of this book was inadvertently published without the middle initial of theauthor’s name as “Brian Hall”. For this reason an erratum has been published, correcting themistake in the previous version and showing the correct name as Brian C. Hall (see DOI http://dx.doi.org/10.1007/978-3-319-13467-3_14). The version readers currently see is the correctedversion. The Publisher would like to apologize for the earlier mistake.

© Springer International Publishing Switzerland 2015B.C. Hall, Lie Groups, Lie Algebras, and Representations, GraduateTexts in Mathematics 222, DOI 10.1007/978-3-319-13467-3_1

3

4 1 Matrix Lie Groups

smooth functions equal to constants. One therefore has to check that these functionsare “independent” in the sense of the implicit function theorem, which means thattheir gradient vectors have to be linearly independent at each point in the group.

We will use an alternative approach that makes all such analysis unnecessary.We consider groups G of matrices that are closed in the sense of Definition 1.4. Toeach suchG, we will associate in Chapter 3 a “Lie algebra” g, which is a real vectorspace. A general result (Corollary 3.45) will then show thatG is a smooth manifoldwhose dimension is equal the dimension of g as a vector space.

This chapter makes use of various standard results from linear algebra, which aresummarized in Appendix A.

Definition 1.1. The general linear group over the real numbers, denotedGL.nIR/, is the group of all n � n invertible matrices with real entries. Thegeneral linear group over the complex numbers, denoted GL.nIC/, is the group ofall n � n invertible matrices with complex entries.

Definition 1.2. Let Mn.C/ denote the space of all n � n matrices with complexentries.

We may identifyMn.C/ with Cn2

and use the standard notion of convergence inCn

2: Explicitly, this means the following.

Definition 1.3. Let Am be a sequence of complex matrices in Mn.C/. We say thatAm converges to a matrix A if each entry of Am converges (as m ! 1) to thecorresponding entry of A (i.e., if .Am/jk converges to Ajk for all 1 � j; k � n).

We now consider subgroups of GL.nIC/, that is, subsets G of GL.nIC/ suchthat the identity matrix is in G and such that for all A and B in G, the matrices ABand A�1 are also in G.

Definition 1.4. A matrix Lie group is a subgroup G of GL.nIC/ with thefollowing property: If Am is any sequence of matrices in G, and Am converges tosome matrix A, then either A is in G or A is not invertible.

The condition on G amounts to saying that G is a closed subset of GL.nIC/.(This does not necessarily mean that G is closed in Mn.C/.) Thus, Definition 1.4is equivalent to saying that a matrix Lie group is a closed subgroup of GL.nIC/.Throughout the book, all topological properties of a matrix Lie group G will beconsidered with respect to the topologyG inherits as a subset ofMn.C/ Š Cn

2.

The condition that G be a closed subgroup, as opposed to merely a subgroup,should be regarded as a technicality, in that most of the interesting subgroups ofGL.nIC/ have this property. Most of the matrix Lie groups G we will considerhave the stronger property that if Am is any sequence of matrices in G, and Amconverges to some matrix A, then A 2 G (i.e., that G is closed in Mn.C/).

An example of a subgroup of GL.nIC/ which is not closed (and hence is not amatrix Lie group) is the set of all n�n invertible matrices with rational entries. Thisset is, in fact, a subgroup of GL.nIC/, but not a closed subgroup. That is, one can

1.2 Examples 5

Fig. 1.1 A small portion of the group G inside NG (left) and a larger portion (right)

(easily) have a sequence of invertible matrices with rational entries converging to aninvertible matrix with some irrational entries. (In fact, every real invertible matrix isthe limit of some sequence of invertible matrices with rational entries.)

Another example of a group of matrices which is not a matrix Lie group is thefollowing subgroup of GL.2IC/. Let a be an irrational real number and let

G D� �

eit 0

0 eita

�ˇˇ t 2 R

�: (1.1)

Clearly,G is a subgroup of GL.2IC/. According to Exercise 10, the closure of G isthe group

NG D� �

ei� 0

0 ei�

�ˇˇ �; � 2 R

�:

The groupG inside NG is known as an “irrational line in a torus”; see Figure 1.1.

1.2 Examples

Mastering the subject of Lie groups involves not only learning the general theorybut also familiarizing oneself with examples. In this section, we introduce someof the most important examples of (matrix) Lie groups. Among these are theclassical groups, consisting of the general and special linear groups, the unitaryand orthogonal groups, and the symplectic groups. The classical groups, and theirassociated Lie algebras, will be key examples in Parts II and III of the book.

6 1 Matrix Lie Groups

1.2.1 General and Special Linear Groups

The general linear groups (over R or C) are themselves matrix Lie groups. Ofcourse, GL.nIC/ is a subgroup of itself. Furthermore, if Am is a sequence ofmatrices in GL.nIC/ and Am converges to A, then by the definition of GL.nIC/,either A is in GL.nIC/, or A is not invertible.

Moreover, GL.nIR/ is a subgroup of GL.nIC/, and if Am 2 GL.nIR/ and Amconverges to A, then the entries of A are real. Thus, either A is not invertible orA 2 GL.nIR/.

The special linear group (over R or C) is the group of n � n invertible matrices(with real or complex entries) having determinant one. Both of these are subgroupsof GL.nIC/. Furthermore, if An is a sequence of matrices with determinant one andAn converges to A, then A also has determinant one, because the determinant is acontinuous function. Thus, SL.nIR/ and SL .nIC/ are matrix Lie groups.

1.2.2 Unitary and Orthogonal Groups

An n � n complex matrix A is said to be unitary if the column vectors of A areorthonormal, that is, if

nXlD1

AljAlk D ıjk: (1.2)

We may rewrite (1.2) as

nXlD1.A�/jlAlk D ıjk; (1.3)

where ıjk is the Kronecker delta, equal to 1 if j D k and equal to zero if j ¤ k.Here A� is the adjoint of A, defined by

.A�/jk D Akj:

Equation (1.3) says that A�A D I ; thus, we see that A is unitary if and only ifA� D A�1. In particular, every unitary matrix is invertible.

The adjoint operation on matrices satisfies .AB/� D B�A�. From this, we cansee that if A and B are unitary, then

.AB/�.AB/ D B�A�AB D B�1A�1AB D I;

showing that AB is also unitary. Furthermore, since .AA�1/� D I� D I , we see that.A�1/�A� D I , which shows that .A�1/� D .A�/�1. Thus, if A is unitary, we have

1.2 Examples 7

.A�1/�A�1 D .A�/�1A�1 D .AA�/�1 D I;

showing that A�1 is again unitary.Thus, the collection of unitary matrices is a subgroup of GL.nIC/. We call this

group the unitary group and we denote it by U.n/. We may also define the specialunitary group SU.n/, the subgroup of U.n/ consisting of unitary matrices withdeterminant 1. It is easy to check that both U.n/ and SU.n/ are closed subgroups ofGL.nIC/ and thus matrix Lie groups.

Meanwhile, let h�; �i denote the standard inner product on Cn, given by

hx; yi DXj

xj yj :

(Note that we put the conjugate on the first factor in the inner product.) ByProposition A.8, we have

hx;Ayi D hA�x; yi

for all x; y 2 Cn. Thus,

hAx;Ayi D hA�Ax; yi ;

from which we can see that if A is unitary, then A preserves the inner product onCn, that is,

hAx;Ayi D hx; yi

for all x and y. Conversely, if A preserves the inner product, we must havehA�Ax; yi D hx; yi for all x; y. It is not hard to see that this condition holds onlyif A�A D I . Thus, an equivalent characterization of unitarity is that A is unitary ifand only if A preserves the standard inner product on Cn.

Finally, for any matrix A, we have that detA� D detA. Thus, if A is unitary, wehave

det.A�A/ D jdetAj2 D det I D 1:

Hence, for all unitary matrices A, we have jdetAj D 1.In a similar fashion, an n � n real matrix A is said to be orthogonal if the

column vectors ofA are orthonormal. As in the unitary case, we may give equivalentversions of this condition. The only difference is that if A is real, A� is the same asthe transpose Atr of A, given by

.Atr/jk D Akj:

8 1 Matrix Lie Groups

Thus, A is orthogonal if and only if Atr D A�1, and this holds if and only if Apreserves the inner product on Rn. Since det.Atr/ D detA, if A is orthogonal, wehave

det.AtrA/ D det.A/2 D det.I / D 1;

so that det.A/ D ˙1. The collection of all orthogonal matrices forms a closedsubgroup of GL.nIC/, which we call the orthogonal group and denote by O.n/.The set of n�n orthogonal matrices with determinant one is the special orthogonalgroup, denoted SO.n/. Geometrically, elements of SO.n/ are rotations, while theelements of O.n/ are either rotations or combinations of rotations and reflections.

Consider now the bilinear form .�; �/ on Cn defined by

.x; y/ DXj

xj yj : (1.4)

This form is not an inner product (Sect. A.6) because, for example, it is symmetricrather than conjugate-symmetric. The set of all n � n complex matrices A whichpreserve this form (i.e., such that .Ax;Ay/ D .x; y/ for all x; y 2 Cn) is the complexorthogonal group O.nIC/, and it is a subgroup of GL.nIC/. Since there are noconjugates in the definition of the form .�; �/, we have

.x;Ay/ D .Atrx; y/;

for all x; y 2 Cn, where on the right-hand side of the above relation, we haveAtr rather than A�. Repeating the arguments for the case of O.n/, but now allowingcomplex entries in our matrices, we find that an n�n complex matrixA is in O.nIC/if and only if AtrA D I , that O.nIC/ is a matrix Lie group, and that detA D ˙1for all A in O.nIC/. Note that O.nIC/ is not the same as the unitary group U.n/.The group SO.nIC/ is defined to be the set of all A in O.nIC/ with detA D 1 andit is also a matrix Lie group.

1.2.3 Generalized Orthogonal and Lorentz Groups

Let n and k be positive integers, and consider RnCk . Define a symmetric bilinearform Œ�; ��n;k on RnCk by the formula

Œx; y�n;k D x1y1 C � � � C xnyn � xnC1ynC1 � � � � � xnCkynCk (1.5)

The set of .nCk/� .nCk/ real matricesA which preserve this form (i.e., such thatŒAx;Ay�n;k D Œx; y�n;k for all x; y 2 RnCk) is the generalized orthogonal groupO.nI k/. It is a subgroup of GL.nC kIR/ and a matrix Lie group (Exercise 1). Ofparticular interest in physics is the Lorentz group O.3I 1/. We also define SO.nI k/to be the subgroup of O.nI k/ consisting of elements with determinant 1.

1.2 Examples 9

If A is an .n C k/ � .n C k/ real matrix, let A.j / denote the j th column vectorof A, that is,

A.j / D

0B@

A1;j:::

AnCk;j

1CA :

Note that A.j / is equal to Aej , that is, the result of applying A to the j th standardbasis element ej . Then A will belong to O.nI k/ if and only if ŒAej ;Ael � D Œej ; el �

for all 1 � j ,l � n C k. Explicitly, this means that A 2 O.nI k/ if and only if thefollowing conditions are satisfied:

�A.j /; A.l/

�n;k

D 0 j ¤ l;�A.j /; A.j /

�n;k

D 1 1 � j � n;�A.j /; A.j /

�n;k

D �1 nC 1 � j � nC k:

(1.6)

Let g denote the .nCk/�.nCk/ diagonal matrix with ones in the first n diagonalentries and minus ones in the last k diagonal entries:

g D

0BBBBBBBBB@

1: : :

1

�1: : :

�1

1CCCCCCCCCA:

Then A is in O.nI k/ if and only if AtrgA D g (Exercise 1). Taking the determinantof this equation gives .detA/2 detg D detg, or (detA/2 D 1. Thus, for any A inO.nI k/, detA D ˙1.

1.2.4 Symplectic Groups

Consider the skew-symmetric bilinear form B on R2n defined as follows:

!.x; y/ DnX

jD1.xj ynCj � xnCj yj /: (1.7)

The set of all 2n� 2n real matrices A which preserve ! (i.e., such that !.Ax;Ay/ D!.x; y/ for all x; y 2 R2n) is the real symplectic group Sp.nIR/, and it is a closedsubgroup of GL.2nIR/. (Some authors refer to the group we have just defined asSp.2nIR/ rather than Sp.nIR/.) If � is the 2n � 2n matrix

10 1 Matrix Lie Groups

� D�0 I

�I 0�; (1.8)

then

!.x; y/ D hx;�yi :From this, it is not hard to show that a 2n� 2n real matrix A belongs to Sp.nIR/ ifand only if

��Atr� D A�1: (1.9)

(See Exercise 2.) Taking the determinant of this identity gives detA D .detA/�1,i.e., .detA/2 D 1. This shows that detA D ˙1, for all A 2 Sp.nIR/. In fact,detA D 1 for all A 2 Sp.nIR/, although this is not obvious.

One can define a bilinear form ! on C2n by the same formula as in (1.7) (withno conjugates). Over C, we have the relation

!.z;w/ D .z; �w/;

where .�; �/ is the complex bilinear form in (1.4). The set of 2n�2n complex matriceswhich preserve this form is the complex symplectic group Sp.nIC/. A 2n � 2n

complex matrix A is in Sp.nIC/ if and only if (1.9) holds. (Note: This conditioninvolves Atr, not A�.) Again, we can easily show that each A 2 Sp.nIC/ satisfiesdetA D ˙1 and, again, it is actually the case that detA D 1. Finally, we have thecompact symplectic group Sp.n/ defined as

Sp.n/ D Sp .nIC/ \ U.2n/:

That is to say, Sp.n/ is the group of 2n � 2n matrices that preserve both the innerproduct and the bilinear form !. For more information about Sp.n/, see Sect. 1.2.8.

1.2.5 The Euclidean and Poincaré Groups

The Euclidean group E.n/ is the group of all transformations of Rn that can beexpressed as a composition of a translation and an orthogonal linear transformation.We write elements of E.n/ as pairs fx;Rg with x 2 Rn and R 2 O.n/, and we letfx;Rg act on Rn by the formula

fx;Rg y D Ry C x:

Since

fx1;R1gfx2;R2gy D R1.R2y C x2/C x1 D R1R2y C .x1 CR1x2/;

1.2 Examples 11

the product operation for E.n/ is the following:

fx1;R1gfx2;R2g D fx1 CR1x2;R1R2g: (1.10)

The inverse of an element of E.n/ is given by

fx;Rg�1 D f�R�1x;R�1g:

The group E.n/ is not a subgroup of GL.nIR/, since translations are not linearmaps. However, E.n/ is isomorphic to the (closed) subgroup of GL.n C 1IR/consisting of matrices of the form

0BBB@

x1

R:::

xn0 � � � 0 1

1CCCA ; (1.11)

with R 2 O.n/. (The reader may easily verify that matrices of the form (1.11)multiply according to the formula in (1.10).)

We similarly define the Poincaré group P.nI 1/ (also known as the inhomoge-neous Lorentz group) to be the group of all transformations of RnC1 of the form

T D TxA

with x 2 RnC1 and A 2 O.nI 1/. This group is isomorphic to the group of .nC2/�

.nC 2/ matrices of the form

0BBB@

x1

A:::

xnC10 � � � 0 1

1CCCA (1.12)

with A 2 O.nI 1/.

1.2.6 The Heisenberg Group

The set of all 3 � 3 real matrices A of the form

A D0@1 a b0 1 c

0 0 1

1A ; (1.13)

12 1 Matrix Lie Groups

where a, b, and c are arbitrary real numbers, is the Heisenberg group. It is easy tocheck that the product of two matrices of the form (1.13) is again of that form, and,clearly, the identity matrix is of the form (1.13). Furthermore, direct computationshows that if A is as in (1.13), then

A�1 D0@1 �a ac � b

0 1 �c0 0 1

1A :

Thus, H is a subgroup of GL.3IR/. The Heisenberg group is a model for theHeisenberg–Weyl commutation relations in physics and also serves as a illumi-nating example for the Baker–Campbell–Hausdorff formula (Sect. 5.2). See alsoExercise 8.

1.2.7 The Groups R�, C�, S1, R, and Rn

Several important groups which are not defined as groups of matrices can bethought of as such. The group R� of non-zero real numbers under multiplication isisomorphic to GL.1IR/. Similarly, the groupC� of nonzero complex numbers undermultiplication is isomorphic to GL.1IC/ and the group S1 of complex numbers withabsolute value one is isomorphic to U.1/.

The group R under addition is isomorphic to GL.1IR/C (1�1 real matrices withpositive determinant) via the map x ! Œex�. The group Rn (with vector addition) isisomorphic to the group of diagonal real matrices with positive diagonal entries, viathe map

.x1; : : : ; xn/ !

0B@ex1 0

: : :

0 exn

1CA .

1.2.8 The Compact Symplectic Group

Of the groups introduced in the preceding subsections, the compact symplecticgroup Sp.n/ WD Sp.nIC/ \ U.2n/ is the most mysterious. In this section, weattempt to understand the structure of Sp.n/ and to show that it can be understoodas being the “unitary group over the quaternions.”

Since the definition of Sp.n/ involves unitarity, it is convenient to express thebilinear form ! on C2n in terms of the inner product h�; �i, rather than in terms of thebilinear form .�; �/, as we did in Sect. 1.2.4. To this end, define a conjugate-linearmap J W C2n ! C2n by

J.˛; ˇ/ D .� N; N /;

1.2 Examples 13

where ˛ and ˇ are in Cn and .˛; ˇ/ is in C2n. We can easily check that for allz;w 2 C2n, we have

!.z;w/ D hJ z;wi :

Recall that we take our inner product to be conjugate linear in the first factor; sinceJ is also conjugate linear, hJ z;wi is actually linear in z. We may easily check that

hJ z;wi D �hz; Jwi D � hJw; zi

for all z;w 2 C2n and that

J 2 D �I:

Proposition 1.5. If U belongs to U.2n/ then U belongs to Sp.n/ if and only if Ucommutes with J .

Proof. Fix some U in U.2n/. Then for z and w in C2n, we have, on the one hand,

!.Uz;Uw/ D hJUz;Uwi D hU �JUz;wi D ˝U�1JUz;w

˛;

and, on the other hand,

!.z;w/ D hJz;wi :

From this it is easy to check that U preserves ! if and only if

U�1JU D J;

which is equivalent to JU D UJ. utThe preceding result can be used to give a different perspective on the definition

of Sp.n/, as follows. The quaternion algebraH is the four-dimensional associativealgebra over R spanned by elements 1 (the identity), i, j, and k satisfying

i2 D j2 D k2 D �1

and

ij D kI ji D �kIjk D iI kj D �iIki D jI ik D �j:

14 1 Matrix Lie Groups

We may realize the quaternion algebra inside M2.C/ by identifying 1 with theidentity matrix and setting

i D�i 0

0 �i�

I j D�

0 1

�1 0�

I k D�0 i

i 0

�:

The algebra H is then the space of real linear combinations of I , i, j, and k.Now, since J is conjugate linear, we have

J.iz/ D �iJ.z/

for all z 2 C2n; that is, iJ D �J i . Thus, if we define K to be iJ, we have

K2 D iJiJ D �J.i/2J D J 2 D �I;

and one can easily check that iI, J , and K satisfy the same commutation relationsas i, j, and k. We can therefore make C

2n into a “vector space” over thenoncommutative algebra H by setting

i � z D iz

j � z D Jz

k � z D iJz:

Now, if U belongs to Sp.n/, then U commutes with multiplication by i and withJ (Proposition 1.5) and thus, also, with K WD iJ. Thus, U is actually “quaternionlinear.” A 2n�2nmatrixU therefore belongs to Sp.n/ if and only ifU is quaternionlinear and preserves the norm. Thus, we may think of Sp.n/ as the “unitary groupover the quaternions.” The compact symplectic group then fits naturally with theorthogonal groups (norm-preserving maps over R) and the unitary groups (norm-preserving maps over C).

Every U 2 U.2n/ has an orthonormal basis of eigenvectors, with eigenvalueshaving absolute value 1. We now determine the additional properties the eigenvec-tors and eigenvalues must satisfy in order forU to be in Sp.n/ D U.2n/\Sp.nIC/.Theorem 1.6. If U 2 Sp.n/, then there exists an orthonormal basis u1; : : : ; un,v1; : : : vn for C2n such that the following properties hold: First, J uj D vj ; second,for some real numbers �1; : : : ; �n, we have

U uj D ei�j uj

Uvj D e�i�j vj I

1.2 Examples 15

and third,

!.uj ; uk/ D !.vj ; vk/ D 0

!.uj ; vk/ D ıjk:

Conversely, if there exists an orthonormal basis with these properties, U belongsto Sp.n/.

Lemma 1.7. Suppose V is a complex subspace of C2n that is invariant under theconjugate-linear map J . Then the orthogonal complement V ? of V (with respectto the inner product h�; �i) is also invariant under J . Furthermore, V and V ? areorthogonal with respect to !; that is,

!.z;w/ D 0

for all z 2 V and w 2 V ?.

Proof. If w 2 V ? then for all z 2 V , we have

hJw; zi D � hJz;wi D 0;

because J z is again in V . Thus, V ? is invariant under J . Then if z 2 V and w 2 V ?,we have

!.z;w/ D hJ z;wi D 0;

because J z is again in V . utProof of Theorem 1.6. Consider U in Sp.nIC/\ U.2n/, choose an eigenvector forU , normalized to be a unit vector, and call it u1. Since U preserves the norms ofvectors, the eigenvalue �1 for u1 must be of the form ei�1 , for some �1 2 R. If we setv1 D J u1, then since J is conjugate linear and commutes with U (Proposition 1.5),we have

Uv1 D J.Uu1/ D J.ei�1u1/ D e�i�1v1:

That is to say, v1 is an eigenvector for U with eigenvalue e�i�1 . Furthermore,

hv1; u1i D hJ u1; u1i D !.u1; u1/ D 0;

since ! is a skew-symmetric form. On the other hand,

!.u1; v1/ D hJ u1; v1i D hJu1; Ju1i D 1;

since J preserves the magnitude of vectors.Now, since J 2 D �I , we can easily check that the span V of u1 and v1 D J u1

is invariant under J . Thus, by Lemma 1.7, V ? is also invariant under J and is

16 1 Matrix Lie Groups

!-orthogonal to V . Meanwhile, V is invariant under both U and U � D U�1.Thus, by Proposition A.10, V ? is invariant under both U �� D U and U �. SinceU preserves V ?, the restriction of U to V ? will have an eigenvector, which wecan normalize to be a unit vector and call u2. If we let v2 D J u2, then wehave all the same properties for u2 and v2 as for u1 and v1. Furthermore, u2 andv2 are orthogonal—with respect to both h�; �i and !.�; �/—to u1 and v1. We canthen proceed on in a similar fashion to obtain the full set of vectors u1; : : : ; unand v1; : : : ; vn. (If u1; : : : ; uk and v1; : : : ; vk have been chosen, we take ukC1 andvkC1 WD J ukC1 in the orthogonal complement of the span of u1; : : : ; uk andv1; : : : ; vk .)

The other direction of the theorem is left to the reader (Exercise 6). ut

1.3 Topological Properties

In this section, we investigate three important topological properties of matrix Liegroups, each of which is satisfied by some groups but not others.

1.3.1 Compactness

The first property we consider is compactness.

Definition 1.8. A matrix Lie group G � GL.nIC/ is said to be compact if it iscompact in the usual topological sense as a subset of Mn.C/ Š R

2n2 .

In light of the Heine–Borel theorem (Theorem 2.41 in [Rud1]), a matrix Liegroup G is compact if and only if it is closed (as a subset of Mn.C/, not just as asubset of GL.nIC/) and bounded. Explicitly, this means that G is compact if andonly if (1) whenever Am 2 G and Am ! A, then A is in G, and (2) there exists aconstant C such that for all A 2 G, we have

ˇAjk

ˇ � C for all 1 � j; k � n.The following groups are compact: O.n/ and SO.n/, U.n/ and SU.n/, and

Sp.n/. Each of these groups is easily seen to be closed in Mn.C/ and each satisfiesthe bound

ˇAjk

ˇ � 1, since in each case, the columns of A 2 G are required tobe unit vectors. Most of the other groups we have considered are noncompact. Thespecial linear group SL.nIR/, for example, is unbounded (except in the trivial casen D 1), because for all m ¤ 0, the matrix

Am D

0BBBBB@

m1m

1: : :

1

1CCCCCA

has determinant one.

1.3 Topological Properties 17

1.3.2 Connectedness

The second property we consider is connectedness.

Definition 1.9. A matrix Lie group G is said to be connected if for all A and B inG, there exists a continuous path A.t/, a � t � b, lying in G with A.a/ D A andA.b/ D B . For any matrix Lie groupG, the identity component of G, denotedG0,is the set of A 2 G for which there exists a continuous path A.t/, a � t � b, lyingin G with A.a/ D I and A.b/ D A.

The property we have called “connected” in Definition 1.9 is what is calledpath connected in topology, which is not (in general) the same as connected.However, we will eventually prove that a matrix Lie group is connected (in theusual topological sense) if and only if it is path-connected. Thus, in a slight abuseof terminology, we shall continue to refer to the above property as connectedness.(See the remarks following Corollary 3.45.)

To show that a matrix Lie group G is connected, it suffices to show that eachA 2 G can be connected to the identity by a continuous path lying in G.

Proposition 1.10. If G is a matrix Lie group, the identity component G0 of G is anormal subgroup of G.

We will see in Sect. 3.7 that G0 is closed and hence a matrix Lie group.

Proof. If A and B are any two elements ofG0, then there are continuous pathsA.t/and B.t/ connecting I to A and to B in G. Then the path A.t/B.t/ is a continuouspath connecting I to AB in G, and .A.t//�1 is a continuous path connecting I toA�1 inG. Thus, both AB andA�1 belong toG0, showing thatG0 is a subgroup ofG.Now suppose A is in G0 and B is any element ofG. Then there is a continuous pathA.t/ connecting I to A in G, and the path BA.t/B�1 connects I to BAB�1 in G.Thus, BAB�1 2 G0, showing that G0 is normal. ut

Note that because matrix multiplication and matrix inversion are continuous onGL.nIC/, it follows that if A.t/ and B.t/ are continuous, then so are A.t/B.t/and A.t/�1. The continuity of the matrix product is obvious. The continuity ofthe inverse follows from the formula for the inverse in terms of cofactors; thisformula is continuous as long as we remain in the set of invertible matrices wherethe determinant in the denominator is nonzero.

Proposition 1.11. The group GL.nIC/ is connected for all n � 1.

Proof. We make use of the result that every matrix is similar to an upper triangularmatrix (Theorem A.4). That is to say, we can express any A 2 Mn.C/ in the formA D CBC�1, where

B D

0B@�1 �: : :

0 �n

1CA :

18 1 Matrix Lie Groups

If A is invertible, each �j must be nonzero. Let B.t/ be obtained by multiplying thepart of B above the diagonal by .1 � t/, for 0 � t � 1, and let A.t/ D CB.t/C�1.Then A.t/ is a continuous path lying in GL.nIC/ which starts at A and ends atCDC�1, where D is the diagonal matrix with diagonal entries �1; : : : ; �n. We cannow define paths �j .t/ connecting �j to 1 in C� as t goes from 1 to 2, and we candefine A.t/ on the interval 1 � t � 2 by

A.t/ D C

0B@�1.t/ 0

: : :

0 �n.t/

1CAC�1:

Then A.t/, 0 � t � 2, is a continuous path in GL.nIC/ connecting A to I . utAn alternative proof of this result is given in Exercise 12.

Proposition 1.12. The group SL.nIC/ is connected for all n � 1.

Proof. The proof is almost the same as for GL.nIC/, except that we must makesure our path connecting A 2 SL.nIC/ to I lies entirely in SL.nIC/. We canensure this by choosing �n.t/, in the second part of the preceding proof, to be equalto .�1.t/ � � ��n�1.t//�1. utProposition 1.13. The groups U.n/ and SU.n/ are connected, for all n � 1.

Proof. By Theorem A.3, every unitary matrix has an orthonormal basis of eigen-vectors, with eigenvalues having absolute value 1. Thus, each U 2 U.n/ can bewritten as U1DU�1

1 , where U1 2 U.n/ and D is diagonal with diagonal entriesei�1; : : : ; ei�n . We may then define

U.t/ D U1

[email protected]�t /�1 0

: : :

0 ei.1�t /�n

1CAU�1

1 ; 0 � t � 1:

It is easy to see that U.t/ is in U.n/ for all t , and U.t/ connects U to I . A slightmodification of this argument, as in the proof of Proposition 1.12, shows that SU.n/is connected. ut

The group SO.n/ is also connected; see Exercise 13.

1.3.3 Simple Connectedness

The last topological property we consider is simple connectedness.

1.3 Topological Properties 19

Definition 1.14. A matrix Lie group G is said to be simply connected if it isconnected and, in addition, every loop in G can be shrunk continuously to a pointin G.

More precisely, assume that G is connected. Then G is simply connected if forevery continuous path A.t/, 0 � t � 1, lying in G and with A.0/ D A.1/, thereexists a continuous function A.s; t/, 0 � s; t � 1, taking values in G and havingthe following properties: (1) A.s; 0/ D A.s; 1/ for all s, (2) A.0; t/ D A.t/, and (3)A.1; t/ D A.1; 0/ for all t .

One should think ofA.t/ as a loop andA.s; t/ as a family of loops, parameterizedby the variable s which shrinks A.t/ to a point. Condition 1 says that for eachvalue of the parameter s, we have a loop; Condition 2 says that when s D 0 theloop is the specified loop A.t/; and Condition 3 says that when s D 1 our loopis a point. The condition of simple connectedness is important because for simplyconnected groups, there is a particularly close relationship between the group andthe Lie algebra. (See Sect. 5.7.)

Proposition 1.15. The group SU.2/ is simply connected.

Proof. Exercise 5 shows that SU.2/ may be thought of (topologically) as thethree-dimensional sphere S3 sitting inside R4. It is well known that S3 is simplyconnected; see, for example, Proposition 1.14 in [Hat]. ut

If a matrix Lie group G is not simply connected, the degree to which it fails tobe simply connected is encoded in the fundamental group of G. (See Sect. 13.1.)Sections 13.2 and 13.3 analyze several additional examples. It is shown there, forexample, that SU.n/ is simply connected for all n.

1.3.4 The Topology of SO.3/

We conclude this section with an analysis of the topological structure of the groupSO.3/. We begin by describing real projective spaces.

Definition 1.16. The real projective space of dimension n, denoted RPn, is theset of lines through the origin in R

nC1. Since each line through the origin intersectsthe unit sphere exactly twice, we may think of RPn as the unit sphere Sn with“antipodal” points u and �u identified.

Using the second description, we think of points in RPn as pairs fu;�ug, withu 2 Sn. There is a natural map � W Sn ! RPn, given by

�.u/ D fu;�ug:

20 1 Matrix Lie Groups

We may define a distance function on RPn by defining

d.fu;�ug; fv;�vg/ D min.d.u; v/; d.u;�v/; d.�u; v/; d.�u;�v//D min.d.u; v/; d.u;�v//:

(The second equality holds because d.x; y/ D d.�x;�y/.) With this metric, RPn

is locally isometric to Sn, since if u and v are nearby points in Sn, we haved.fu;�ug; fv;�vg/ D d.u; v/.

It is known that RPn is not simply connected. (See, for example, Example 1.43in [Hat].) Indeed, suppose u is any unit vector in RnC1 and B.t/ is any path in Sn

connecting u to �u. Then

A.t/ WD �.B.t//

is a loop in RPn, and this loop cannot be shrunk continuously to a point in RPn.To prove this claim, suppose that a map A.s; t/ as in Definition 1.14 exists. ThenA.s; t/ can be “lifted” to a continuous map B.s; t/ into Sn such that B.0; t/ DB.t/ and such that A.s; t/ D �.B.s; t//. (See Proposition 1.30 in [Hat].) SinceA.s; 0/ D A.s; 1/ for all s, we must have B.s; 0/ D ˙B.s; 1/. But by construction,B.0; 0/ D �B.0; 1/. If order for B.s; t/ to be continuous in s, we must then haveB.s; 0/ D �B.s; 1/ for all s. It follows that B.1; t/ is a nonconstant path in Sn. It isthen easily verified that A.1; t/ D �.B.1; t// cannot be constant, contradicting ourassumption about A.s; t/.

LetDn denote the closed upper hemisphere in Sn, that is, the set of points u 2 Snwith unC1 � 0. Then � maps Dn onto RPn, since at least one of u and �u is inDn. The restriction of � to Dn is injective except on the equator, that is, the set ofu 2 Sn with unC1 D 0. If u is in the equator, then �u is also in the equator, and�.�u/ D �.u/. Thus, we may also think of RPn as the upper hemisphereDn, withantipodal points on the equator identified (Figure 1.2).

We may now make one last identification using the projection P of RnC1 ontoRn. (That is to say, P is the map sending .x1; : : : ; xn; xnC1/ to .x1; : : : ; xn/.) Therestriction of P toDn is a continuous bijection betweenDn and the closed unit ballBn in Rn, with the equator in Dn mapping to the boundary of the ball. Thus, our

u

−u

Fig. 1.2 The space RPn is the upper hemisphere with antipodal points on the equator identified.The indicated path from u to �u corresponds to a loop in RPn that cannot be shrunk to a point

1.4 Homomorphisms 21

last model of RPn is the closed unit ball Bn � Rn, with antipodal points on theboundary of Bn identified.

We now turn to a topological analysis of SO.3/.

Proposition 1.17. There is a continuous bijection between SO.3/ and RP3.

Since RP3 is not simply connected, it follows that SO.3/ is not simplyconnected, either.

Proof. If v is a unit vector in R3, let Rv;� be the element of SO.3/ consisting ofa “right-handed” rotation by angle � in the plane orthogonal to v. That is to say,let v? denote the plane orthogonal to v and choose an orthonormal basis .u1; u2/for v? in such a way that the linear map taking the orthonormal basis .u1; u2; v/to the standard basis .e1; e2; e3/ has positive determinant. We use the basis .u1; u2/to identify v? with R2, and the rotation is then in the counterclockwise directionin R2. It is easily seen that R�v;� is the same as Rv;�� . It is also not hard to show(Exercise 14) that every element of SO.3/ can be expressed as Rv;� , for some v and� with �� � � � � . Furthermore, we can arrange that 0 � � � � by replacing vwith �v if necessary.

IfR D I , thenR D Rv;0 for any unit vector v. IfR is a rotation by angle � aboutsome axis v, thenR can be expressed both as Rv;� and as R�v;� . It is not hard to seethat ifR ¤ I andR is not a rotation by angle � , thenR has a unique representationas Rv;� with 0 < � < � .

Now let B3 denote the closed ball of radius � in R3 and consider the map ˆ WB3 ! SO.3/ given by

ˆ.u/ D ROu;kuk; u ¤ 0;

ˆ.0/ D I:

Here, Ou D u= kuk is the unit vector in the u-direction. The mapˆ is continuous, evenat I , since Rv;� approaches the identity as � approaches zero, regardless of how v

is behaving. The discussion in the preceding paragraph shows that ˆ maps B3 ontoSO.3/. The mapˆ is injective except that “antipodal” points on the boundary ofB3

have the same image: Rv;� D R�v;� . Thus, ˆ descends to a continuous, injectivemap of RP3 onto SO.3/. Since both RP3 and SO.3/ are compact, Theorem 4.17in [Rud1] tells us that the inverse map is also continuous, meaning that SO.3/ ishomeomorphic to RP3. ut

For a different approach to proving Proposition 1.17, see the discussion followingProposition 1.19.

1.4 Homomorphisms

We now look at the notion of homomorphisms for matrix Lie groups.

22 1 Matrix Lie Groups

Definition 1.18. Let G and H be matrix Lie groups. A map ˆ from G to H iscalled a Lie group homomorphism if (1) ˆ is a group homomorphism and (2) ˆis continuous. If, in addition, ˆ is one-to-one and onto and the inverse map ˆ�1 iscontinuous, then ˆ is called a Lie group isomorphism.

The condition thatˆ be continuous should be regarded as a technicality, in that itis very difficult to give an example of a group homomorphism between two matrixLie groups which is not continuous. In fact, if G D R andH D C�, then any grouphomomorphism from G to H which is even measurable (a very weak condition)must be continuous. (See Exercise 17 in Chapter 9 of [Rud2].)

Note that the inverse of a Lie group isomorphism is continuous (by definition)and a group homomorphism (by elementary group theory), and thus a Lie groupisomorphism. If G and H are matrix Lie groups and there exists a Lie groupisomorphism from G to H , then G and H are said to be isomorphic, and we writeG Š H .

The simplest interesting example of a Lie group homomorphism is the determi-nant, which is a homomorphism of GL.nIC/ into C�. Another simple example isthe map ˆ W R ! SO.2/ given by

ˆ.�/ D�

cos � � sin �sin � cos �

�:

This map is clearly continuous, and calculation (using standard trigonometricidentities) shows that it is a homomorphism.

An important topic for us will be the relationship between the groups SU.2/ andSO.3/, which are almost, but not quite, isomorphic. Specifically, we now show thatthere exists a Lie group homomorphismˆ W SU.2/ ! SO.3/ that is two-to-one andonto. Consider the space V of all 2 � 2 complex matrices X which are self-adjoint(i.e., X� D X ) and have trace zero. Elements of V are precisely the matrices of theform

X D�

x1 x2 C ix3x2 � ix3 �x1

�; (1.14)

with x1; x2; x3 2 R. If we identify V with R3 by means of the coordinates x1, x2,

and x3 in (1.14), then the standard inner product on R3 can be computed as

hX1;X2i D 1

2trace.X1X2/:

That is to say,

1

2trace

��x1 x2 C ix3

x2 � ix3 �x1��

x01 x0

2 C ix03

x02 � ix0

3 �x01

��

D x1x01 C x2x

02 C x3x

03;

as one may easily check by direct calculation.

1.4 Homomorphisms 23

For each U 2 SU.2/, define a linear mapˆU W V ! V by

ˆU .X/ D UXU�1:

Since U is unitary,

.UXU�1/� D .U�1/�XU� D UXU�1;

showing that UXU�1 is again in V .It is easy to see that ˆU1U2 D ˆU1ˆU2 . Furthermore,

1

2trace..UX1U

�1/.UX2U�1// D 1

2trace.UX1X2U

�1/

D 1

2trace.X1X2/;

since the trace is invariant under conjugation. Thus, each ˆU preserves the innerproduct trace.X1X2/=2 on V . It follows that the map U 7! ˆU is a homomorphismof SU.2/ into the group of orthogonal linear transformations of V Š R3, that is, intoO.3/. Since SU.2/ is connected (Proposition 1.13), ˆU must actually lie in SO.3/for all U 2 SU.2/. Thus, ˆ (i.e., the map U 7! ˆU ) is a homomorphism of SU.2/into SO.3/, which is easily seen to be continuous. Since .�I /X.�I /�1 D X , wesee that ˆ�I is the identity element of SO.3/.

Suppose, for example, that U is the matrix

U D�ei�=2 0

0 e�i�=2�:

Then by direct calculation, we obtain

U

�x1 x2 C ix3

x2 � ix3 �x1�U�1 D

�x01 x0

2 C ix03

x02 � ix0

3 �x01

�; (1.15)

where x01 D x1 and

x02 C ix0

3 D ei� .x2 C ix3/

D .x2 cos � � x3 sin �/C i.x2 sin � C x3 cos �/: (1.16)

In this case, then, ˆU is a rotation by angle � in the .x2; x3/-plane. Note that eventhough the diagonal entries of U are e˙i�=2, the map ˆU is a rotation by angle � ,not �=2.

24 1 Matrix Lie Groups

Proposition 1.19. The map U 7! ˆU is a 2-1 and onto map of SU.2/ to SO.3/,with kernel equal to fI;�I g.

Since SU.2/ is homeomorphic to S3, the proposition gives another way of seeingthat SO.3/ is homeomorphic to RP3, that is, S3 with antipodal points identified.This result was obtained in a different way in Proposition 1.17.

It is not hard to show that ˆ is a “covering map” in the topological sense(Section 1.3 of [Hat]). Since SU.2/ is simply connected (Proposition 1.15) andthe map is 2-1, it follows by the theory of covering maps (e.g., Theorem 1.38 in[Hat]) that SO.3/ cannot be simply connected and, indeed, it must have fundamentalgroup Z=2. See Chapter 13 for general information about fundamental groups andfor a computation of the fundamental group of SO.n/, n > 2.

Proof. Exercise 16 shows that the kernel of ˆ is precisely the set fI;�I g. To seethatˆmaps onto SO.3/, letR be a rotation of V Š R3. By Exercise 14, there existsan “axis” X 2 V such that R is a rotation by some angle � in the plane orthogonalto X . If we express X in the form

X D U0

�x1 0

0 �x1�U�10

with U0 2 U.2/, then the plane orthogonal to X in V is the space of matrices of theform

X 0 D U0

�0 x2 C ix3

x2 � ix3 0

�U�10 : (1.17)

If we now take

U D U0

�ei�=2 0

0 e�i�=2�U�10 ;

we can easily see that UXU�1 D X . On the other hand, the calculations in (1.15)and (1.16) show that UX0U�1 is of the same form as in (1.17), but with .x2; x3/rotated by angle � . Thus, ˆU is a rotation by angle � in the plane perpendicular toX , showing that ˆU coincides with R. ut

It is possible, if not terribly useful, to calculate ˆ explicitly. If you write anelement of SU.2/ as in Exercise 5, you (or your computer) may calculate

�˛ � Nˇ N

��x1 x2 C ix3

x2 � ix3 �x1�� N N

�ˇ ˛�

D�

x01 x0

2 C ix03

x02 � ix0

3 �x03

1.5 Lie Groups 25

explicitly. Then .x01; x

02; x

03/ will depend linearly on .x1; x2; x3/ and you can express

.x01; x

02; x

03/ as a matrix applied to .x1; x2; x3/, with the result that

ˆU D0@ j˛j2 � jˇj2 �2Re.˛ˇ/ 2 Im.˛ˇ/2Re.˛ N/ Re.˛2 � ˇ2/ Im.ˇ2 � ˛2/2 Im.˛ N/ Im.˛2 C ˇ2/ Re.˛2 C ˇ2/

1A :

If we take ˛ D ei�=2 and ˇ D 0, we may see directly that ˆU is a rotation by angle� in the .x2; x3/-plane, as we saw already in (1.15) and (1.16).

1.5 Lie Groups

A Lie group is a smooth manifold equipped with a group structure such that theoperations of group multiplication and inversion are smooth. As the terminologysuggests, every matrix Lie group is a Lie group. (See Corollary 3.45 in Chapter 3.)The reverse is not true: Not every Lie group is isomorphic to a matrix Lie group.Nevertheless, we have restricted our attention in this book to matrix Lie groups,in order to minimize prerequisites and keep the discussion as concrete as possible.Most of the interesting examples of Lie groups are, in any case, matrix Lie groups.

A manifold is an object M that looks locally like a piece of Rn. More precisely,an n-dimensional manifold is a second-countable, Hausdorff topological space withthe property that eachm 2 M has a neighborhood that is homeomorphic to an opensubset of Rn. A two-dimensional torus, for example, looks locally but not globallylike R2 and is, thus, a two-dimensional manifold. A smooth manifold is a manifoldM together with a collection of local coordinates coveringM such that the change-of-coordinates map between two overlapping coordinate systems is smooth.

Definition 1.20. A Lie group is a smooth manifold G which is also a group andsuch that the group product

G �G ! G

and the inverse map G ! G are smooth.

Example 1.21. Let

G D R � R � S1 D ˚.x; y; u/jx 2 R; y 2 R; u 2 S1 � C

�;

equipped with the group product given by

.x1; y1; u1/ � .x2; y2; u2/ D .x1 C x2; y1 C y2; eix1y2u1u2/:

Then G is a Lie group.

26 1 Matrix Lie Groups

Proof. It is easily checked that this operation is associative; the product of threeelements with either grouping is

.x1 C x2 C x3; y1 C y2 C y3; ei.x1y2Cx1y3Cx2y3/u1u2u3/:

There is an identity element in G, namely e D .0; 0; 1/ and each element .x; y; u/has an inverse given by .�x;�y; eixyu�1/. Thus,G is, in fact, a group. Furthermore,both the group product and the map that sends each element to its inverse are clearlysmooth, showing that G is a Lie group. ut

Although there is nothing about matrices in the definition of the group G inExample 1.21, we may still ask whetherG is isomorphic to some matrix Lie group.This turns out to be false. As shown in Sect. 4.8, there is no continuous, injectivehomomorphism ofG into any GL.nIC/. We conclude, then, that not every Lie groupis isomorphic to a matrix Lie group. Nevertheless, most of the interesting examplesof Lie groups are matrix Lie groups.

Let us now think briefly about how we might show that every matrix Lie group isa Lie group. We will prove in Sect. 3.7 that every matrix Lie group is an “embeddedsubmanifold” of Mn.C/ Š R

2n2 . The operations of matrix multiplication andinversion are smooth on Mn.C/ (after restricting to the open subset of invertiblematrices in the case of inversion). Thus, the restriction of these operations to a matrixLie group G � Mn.C/ is also smooth, making G into a Lie group.

It is customary to call a map ˆ between two Lie groups a Lie group homo-morphism if ˆ is a group homomorphism and ˆ is smooth, whereas we have(in Definition 1.18) required only that ˆ be continuous. We will show, however,that every continuous homomorphism between matrix Lie groups is automaticallysmooth, so that there is no conflict of terminology. See Corollary 3.50 to Theo-rem 3.42. Finally, we note that since every matrix Lie group G is a manifold, Gmust be locally path connected. It then follows by a standard topological argumentthat G is connected if and only if it is path connected.

1.6 Exercises

1. Let Œ�; ��n;k be the symmetric bilinear form on RnCk defined in (1.5). Let g bethe .nC k/� .nC k/ diagonal matrix with first n diagonal entries equal to oneand last k diagonal entries equal to minus one:

g D�In 0

0 �Ik�:

Show that for all x; y 2 RnCk,

Œx; y�n;k D hx; gyi :Show that a .nC k/ � .nC k/ real matrix A belongs to O.nI k/ if and only ifgAtrg D A�1.

1.6 Exercises 27

2. Let ! be the skew-symmetric bilinear form on R2n given by (1.7). Let� be the2n � 2n matrix

� D�

0 I

�I 0�:

Show that for all x; y 2 R2n, we have

!.x; y/ D hx;�yi :

Show that a 2n � 2n matrix A belongs to Sp.nIR/ if and only if ��Atr� DA�1.Note: A similar analysis applies to Sp.nIC/.

3. Show that the symplectic group Sp.1IR/ � GL.2IR/ is equal to SL.2IR/.Show that Sp.1IC/ D SL.2IC/ and that Sp.1/ D SU.2/.

4. Show that a matrix R belongs to SO.2/ if and only if it can be expressed in theform

�cos � � sin �sin � cos �

for some � 2 R. Show that a matrix R belongs to O.2/ if and only if it is ofone of the two forms:

A D�

cos � � sin �sin � cos �

�or A D

�cos � sin �sin � � cos �

�:

Hint: Recall that for A to be in O.2/, the columns of A must be orthonormal.5. Show that if ˛ and ˇ are arbitrary complex numbers satisfying j˛j2 Cjˇj2 D 1,

then the matrix

A D�˛ �ˇˇ ˛

is in SU.2/. Show that every A 2 SU.2/ can be expressed in this form for aunique pair .˛; ˇ/ satisfying j˛j2 C jˇj2 D 1.

6. Suppose U belongs toM2n.C/ and U has an orthonormal basis of eigenvectorssatisfying the conditions in Theorem 1.6. Show that U belongs to Sp.n/.Hint: Start by showing that U is unitary. Then show that !.Uz;Uw/ D !.z;w/if z and w belong to the basis u1; : : : ; un, v1; : : : ; vn.

7. Using Theorem 1.6, show that Sp.n/ is connected and that every element ofSp.n/ has determinant 1.

28 1 Matrix Lie Groups

8. Determine the centerZ.H/ of the Heisenberg groupH . Show that the quotientgroupH=Z.H/ is commutative.

9. Suppose a is an irrational real number. Show that the set Ea of numbers of theform e2�ina, n 2 Z, is dense in the unit circle S1.Hint: Show that if we divide S1 into N equally sized “bins” of length 2�=N ,there is at least one bin that contains infinitely many elements of Ea. Then usethe fact that Ea is a subgroup of S1.

10. Let a be an irrational real number and let G be the following subgroup ofGL.2IC/:

G D� �

eit 0

0 eita

�ˇˇ t 2 R

�:

Show that

G D� �

ei� 0

0 ei�

�ˇˇ �; � 2 R

�;

where G denotes the closure of the set G inside the space of 2 � 2 matrices.Hint: Use Exercise 9.

11. A subset E of a matrix Lie group G is called discrete if for each A in E thereis a neighborhood U of A in G such that U contains no point in E except forA. Suppose that G is a connected matrix Lie group and N is a discrete normalsubgroup of G. Show that N is contained in the center of G.

12. This problem gives an alternative proof of Proposition 1.11, namely thatGL.nIC/ is connected. Suppose A and B are invertible n � n matrices.Show that there are only finitely many complex numbers � for whichdet .�AC .1 � �/B/ D 0. Show that there exists a continuous path A.t/

of the form A.t/ D �.t/AC .1��.t//B connectingA to B and such that A.t/lies in GL.nIC/. Here, �.t/ is a continuous path in the plane with �.0/ D 0

and �.1/ D 1.13. Show that SO.n/ is connected, using the following outline.

For the case n D 1, there is nothing to show, since a 1 � 1 matrix withdeterminant one must be Œ1�. Assume, then, that n � 2. Let e1 denote the unitvector with entries 1; 0; : : : ; 0 in Rn. For every unit vector v 2 Rn, show thatthere exists a continuous path R.t/ in SO.n/ with R.0/ D I and R.1/v D e1.(Thus, any unit vector can be “continuously rotated” to e1.)Now, show that any element R of SO.n/ can be connected to a block-diagonalmatrix of the form

�1

R1

with R1 2 SO.n � 1/ and proceed by induction.

1.6 Exercises 29

14. If R is an element of SO.3/, show that R must have an eigenvector v witheigenvalue 1. Show that R maps the plane orthogonal to v into itself. Concludethat R is a rotation by some angle � around the “axis” v.Hint: Since SO.3/ � SU.3/, every (real or complex) eigenvalue of R musthave absolute value 1. Since, also,R is real, any nonreal eigenvalues ofR comein conjugate pairs.

15. Let R be an element of SO.n/.

(a) Suppose v 2 Cn is an eigenvector forR with eigenvalue � 2 C, and suppose� is not real. Let V � Rn be the two-dimensional span of .v C Nv/=2 and.v � Nv/=.2i/. Show that V is invariant under R and that the restriction of Rto V has determinant 1.

(b) Suppose that a subspace V � Rn is invariant under both R and R�1. Showthat the orthogonal complement V ? of V is also invariant under both Rand R�1.

(c) Show that if n D 2k, there exists S 2 SO.n/ such that

R D S

0BBBBB@

cos �1 � sin �1sin �1 cos �1

: : :

cos �k � sin �ksin �k cos �k

1CCCCCAS�1

and that if n D 2k C 1, there exists S 2 SO.n/ such that

R D S

0BBBBBBBB@

cos �1 � sin �1sin �1 cos �1

: : :

cos �k � sin �ksin �k cos �k

1

1CCCCCCCCAS�1:

That is, in a suitable orthonormal basis,R is block diagonal with 2�2 blocksof the indicated form, with a single 1 � 1 block if n is odd.

Hint: Show that the number of eigenvalues of R equal to �1 is even.16. (a) Show that if a matrix A commutes with every matrix X of the form (1.14),

then A commutes with every element of M2.C/. Conclude that A must bea multiple of the identity.

(b) Show that the kernel of the map U 7! ˆU in Proposition 1.19 is preciselythe set fI;�I g.

30 1 Matrix Lie Groups

17. Suppose G � GL.n1IC/ and H � GL.n2IC/ are matrix Lie groups and thatˆ W G ! H is a Lie group homomorphism. Then the image of G under ˆ is asubgroup ofH and thus of GL.n2IC/. Is the image of G underˆ necessarily amatrix Lie group? Prove or give a counter-example.

18. Show that every continuous homomorphism ˆ from R to S1 is of the formˆ.x/ D eiax for some a 2 R.Hint: Sinceˆ is continuous, there is some " > 0 such that if jxj < ", thenˆ.x/belongs to the right half of the unit circle.

Chapter 2The Matrix Exponential

2.1 The Exponential of a Matrix

The exponential of a matrix plays a crucial role in the theory of Lie groups. Theexponential enters into the definition of the Lie algebra of a matrix Lie group(Sect. 3.3) and is the mechanism for passing information from the Lie algebra tothe Lie group.

If X is an n � n matrix, we define the exponential of X , denoted eX or expX ,by the usual power series

eX D1XmD0

Xm

mŠ, (2.1)

whereX0 is defined to be the identity matrix I and whereXm is the repeated matrixproduct of X with itself.

Proposition 2.1. The series (2.1) converges for all X 2 Mn.C/ and eX is acontinuous function of X .

Our proof will use the notion of the norm of a matrix X 2 Mn.C/, which wedefine by thinking of Mn.C/ as Cn

2.

A previous version of this book was inadvertently published without the middle initial of theauthor’s name as “Brian Hall”. For this reason an erratum has been published, correcting themistake in the previous version and showing the correct name as Brian C. Hall (see DOI http://dx.doi.org/10.1007/978-3-319-13467-3_14). The version readers currently see is the correctedversion. The Publisher would like to apologize for the earlier mistake.

© Springer International Publishing Switzerland 2015B.C. Hall, Lie Groups, Lie Algebras, and Representations, GraduateTexts in Mathematics 222, DOI 10.1007/978-3-319-13467-3_2

31

32 2 The Matrix Exponential

Definition 2.2. For any X 2 Mn.C/, we define

kXk D0@ nXj;kD1

ˇXjk

ˇ21A1=2

:

The quantity kXk is called the Hilbert–Schmidt norm of X .

The Hilbert–Schmidt norm may be computed in a basis-independent way as

kXk D .trace.X�X//1=2: (2.2)

(See Sect. A.6.) This norm satisfies the inequalities

kX C Y k � kXk C kY k ; (2.3)

kXYk � kXk kY k (2.4)

for all X; Y 2 Mn.C/. The first of these inequalities is the triangle inequality forCn2 and the second follows from the Cauchy–Schwarz inequality (Exercise 1). If

Xm is a sequence of matrices, then it is easy to see that Xm converges to a matrix Xin the sense of Definition 1.3 if and only if kXm � Xk ! 0 as m ! 1.

Proof of Proposition 2.1. In light of (2.4), we see that

kXmk � kXkm

for all m � 1, and, hence,

1XmD0

Xm

� kIk C1XmD1

kXkmmŠ

< 1.

Thus, the series (2.1) converges absolutely.To show continuity, note that since Xm is a continuous function of X , the partial

sums of (2.1) are continuous. By the Weierstrass M-test, the series (2.1) convergesuniformly on each set of the form fkXk � Rg. Thus, eX is continuous on each suchset, and, thus, continuous on all ofMn.C/. ut

We now list some elementary properties of the matrix exponential.

Proposition 2.3. Let X and Y be arbitrary n � n matrices. Then we have thefollowing:

1. e0 D I .2.eX�� D eX

.

3. eX is invertible andeX��1 D e�X .

4. e.˛Cˇ/X D e˛XeˇX for all ˛ and ˇ in C.5. If XY D YX, then eXCY D eXeY D eY eX .6. If C is in GL.N IC/, then eCXC

�1 D CeXC�1.

2.1 The Exponential of a Matrix 33

Although eXCY D eXeY when X and Y commute, this identity fails in general.This is an important point, which we will return to in the Lie product formula inSect. 2.4 and the Baker–Campbell–Hausdorff formula in Chapter 5.

Proof. Point 1 is obvious and Point 2 follows from taking term-by-term adjointsof the series for eX . Points 3 and 4 are special cases of Point 5. To verify Point 5,we simply multiply the two power series term by term, which is permitted becauseboth series converge absolutely. Multiplying out eXeY and collecting terms wherethe power of X plus the power of Y equalsm, we obtain

eXeY D1XmD0

mXkD0

Xk

Y m�k

.m � k/Š D1XmD0

1

mXkD0

kŠ.m � k/ŠXkY m�k . (2.5)

Now, because (and only because) X and Y commute,

.X C Y /m DmXkD0

kŠ.m � k/ŠXkY m�k ,

and, thus, (2.5) becomes

eXeY D1XmD0

1

mŠ.X C Y /m D eXCY .

To prove Point 6, simply note that

CXC�1�m D CXmC�1

and, thus, the two sides of Point 6 are equal term by term. utProposition 2.4. Let X be a n � n complex matrix. Then etX is a smooth curve inMn.C/ and

d

dtetX D XetX D etXX .

In particular,

d

dtetX

ˇˇtD0

D X .

Results that hold for the exponential of numbers may or may not hold for thematrix exponential. Although Proposition 2.4 is what one would expect from thescalar case, it should be noted that, in general, the derivative of eXCtY is not equalto eXCtYY . See Sect. 5.4.

34 2 The Matrix Exponential

Proof. Differentiate the power series for etX term by term. This is permitted because,for each j and k,

etX�

jkis given by a convergent power series in t , and one can

differentiate a power series term by term inside its radius of convergence (e.g.,Theorem 12 in Chapter 4 of [Pugh]). ut

2.2 Computing the Exponential

We consider here methods for exponentiating general matrices. A special methodfor exponentiating 2 � 2 matrices is described in Exercises 6 and 7. Suppose thatX 2 Mn.C/ has n linearly independent eigenvectors v1; : : : ; vn with eigenvalues�1; : : : ; �n. Let C be the n � n matrix whose columns are v1; : : : ; vn and let Dbe the diagonal matrix with diagonal entries �1; : : : ; �n. Then X D CDC�1. It iseasily verified that eD is the diagonal matrix with diagonal entries e�1 ; : : : ; e�n , andthus, by Proposition 2.3, we have

eX D C

0B@e�1 0

: : :

0 e�n

1CAC�1.

Meanwhile, if X is nilpotent (i.e., Xk D 0 for some k), then the series thatdefines eX terminates. Finally, according to Theorem A.6, every matrix X can bewritten (uniquely) in the formX D SCN , with S diagonalizable,N nilpotent, andSN D NS . Then, since N and S commute,

eX D eSCN D eSeN :

Example 2.5. Consider the matrices

X1 D�0 �aa 0

�I X2 D

0@0 a b0 0 c

0 0 0

1A I X3 D

�a b

0 a

�:

Then

eX1 D�

cos a � sin asin a cosa

and

eX2 D0@1 a b C ac=20 1 c

0 0 1

1A

2.2 Computing the Exponential 35

and

eX3 D�ea eab

0 ea

�:

Proof. The eigenvectors of X1 are .1; i/ and .i; 1/, with eigenvalues �ia and ia,respectively. Thus,

eX1 D�1 i

i 1

��e�ia 0

0 eia

��1=2 �i=2

�i=2 1=2

�;

which simplifies to the claimed result. Meanwhile,X22 has the value ac in the upper

right-hand corner and all other entries equal to zero, whereas X32 D 0. Thus, eX2 D

1CX2 CX22 =2, which reduces to the claimed result. Finally,

X3 D�a 0

0 a

�C�0 b

0 0

�;

where the two terms clearly commute and the second term is nilpotent. Thus, weobtain

eX3 D�ea 0

0 ea

��1 b

0 1

�;

which reduces to the claimed result. utThe matrix exponential is used in the elementary theory of differential equations,

to solve systems of linear equations. Consider a first-order differential equation ofthe form

dvdt

D Xv;

v.0/ D v0;

where v.t/ 2 Rn andX is a fixed n�nmatrix. The (unique) solution of this equation

is given by

v.t/ D etXv0;

as may be easily verified using Proposition 2.4. Curves of the form t 7! etXv0, withv0 fixed, trace out the flow along the vector field v 7! Xv.

Let us consider the two matrices

X4 D�

1 2

�2 1�

I X5 D�1 2

2 1

�: (2.6)

Figure 2.1 plots several curves of this form for each matrix (see Exercise 8).

36 2 The Matrix Exponential

Fig. 2.1 Curves of the form t 7! etX4v0 (left) and t 7! etX5v0 (right)

2.3 The Matrix Logarithm

We now wish to define a matrix logarithm, which should be an inverse function(to the extent possible) to the matrix exponential. Let us recall the situation for thelogarithm of complex numbers, in order to see what is reasonable to expect in thematrix case. Since ez is never zero, only nonzero numbers can have a logarithm.Every nonzero complex number can be written as ez for some z, but the z is notunique and cannot be defined continuously onC�. In the matrix case, eX is invertiblefor all X 2 Mn.C/. We will see (Theorem 2.10) that every invertible matrix can bewritten as eX , for some X 2 Mn.C/, but the X is not unique.

The simplest way to define the matrix logarithm is by a power series. We recallhow this works in the complex case.

Lemma 2.6. The function

log z D1XmD1

.�1/mC1 .z � 1/mm

(2.7)

is defined and holomorphic in a circle of radius 1 about z D 1.For all z with jz � 1j < 1,

elog z D z.

For all u with juj < log 2, we have jeu � 1j < 1 and

log eu D u.

2.3 The Matrix Logarithm 37

Proof. The usual logarithm for real, positive numbers satisfies

d

dxlog.1 � x/ D �1

1 � xD � 1C x C x2 C � � � �

for jxj < 1. Integrating term by term and noting that log1 D 0 gives

log.1 � x/ D ��x C x2

2C x3

3C � � �

.

Taking z D 1 � x (so that x D 1 � z), we have

log z D ��.1 � z/C .1 � z/2

2C .1 � z/3

3C � � �

D1XmD1

.�1/mC1 .z � 1/m

m. (2.8)

The series (2.8) has radius of convergence 1 and defines a holomorphic functionon the set fjz � 1j < 1g, which coincides with the usual logarithm for real z in theinterval .0; 2/. Now, exp.log z/ D z for z 2 .0; 2/ and since both sides of this identityare holomorphic in z, the identity continues to hold on the whole set fjz � 1j < 1g.

On the other hand, if juj < log 2, then

jeu � 1j Dˇˇu C u2

2ŠC � � �

ˇˇ � juj C juj2

2ŠC � � � D ejuj � 1 < 1:

Thus, log.exp u/ makes sense for all such u. Since log.exp u/ D u for real u withjuj < log 2, it follows by holomorphicity that log.exp u/ D u for all complexnumbers with juj < log 2. utDefinition 2.7. For an n � n matrix A, define logA by

logA D1XmD1

.�1/mC1 .A� I /m

m(2.9)

whenever the series converges.

Since the complex-valued series (2.7) has radius of convergence 1 and sincek.A � I /mk � kA � Ikm for m � 1, the matrix-valued series (2.9) will convergeif kA� Ik < 1. Even if kA � Ik > 1, the series might converge, for example, ifA� I is nilpotent (see Exercise 9).

38 2 The Matrix Exponential

Theorem 2.8. The function

logA D1XmD1

.�1/mC1 .A� I /m

m

is defined and continuous on the set of all n � n complex matrices A

with kA� Ik < 1.For all A 2 Mn.C/ with kA� Ik < 1,

elogA D A.

For all X 2 Mn.C/ with kXk < log 2,eX � I

< 1 and

log eX D X .

Although it might seem plausible that log.eX/ should be equal to X wheneverthe series for the logarithm is convergent, this claim is false (even over C). If, forexample, X D 2�iI , then eX D e2�i I D I . Then eX � I D 0, so that log.eX/is defined and equal to 0. In this case, log.eX/ is defined but not equal to X . Thus,the assumption that kXk < log 2 cannot be replaced by, say, the assumption thateX � I

< 1.The proof of the theorem will establish a variant of the result, as follows. If A

is diagonalizable and all the eigenvalues � of A satisfy j� � 1j < 1, then logA isdefined and elogA D A. Similarly, if X is diagonalizable and all the eigenvalues �of X satisfy j�j < log 2, then log eX is defined and log eX D X .

Proof. Since k.A � I /mk � k.A� I /km and since the series (2.7) has radius ofconvergence 1, the series (2.9) converges absolutely for all A with kA � Ik < 1.The proof of continuity is essentially the same as for the exponential.

Suppose now that A satisfies kA � Ik < 1. If A is diagonalizable witheigenvalues z1; : : : ; zn, then we can expressA in the form CDC�1 withD diagonal,in which case

.A � I /m D C

[email protected] � 1/m 0

: : :

0 .zn � 1/m

1CAC�1.

Since kA � Ik < 1, each eigenvalue zj of A must satisfyˇzj � 1

ˇ< 1 (Exercise 2).

Thus,

1XmD1

.�1/mC1 .A� I /m

mD C

0B@

log z1 0: : :

0 log zn

1CAC�1;

2.3 The Matrix Logarithm 39

and by Lemma 2.6,

elogA D C

0B@elog z1 0

: : :

0 elog zn

1CAC�1 D A.

IfA is not diagonalizable, we approximateA by a sequenceAm of diagonalizablematrices (Exercise 4) and appeal to the continuity of the logarithm and exponentialfunctions. Thus, exp.logA/ D A for all A with kA� Ik < 1.

Now, the same argument as in the complex case shows that if kXk < log 2, theneX � I < 1. The proof that log.eX/ D X is then very similar to the proof that

exp.logA/ D A. utProposition 2.9. There exists a constant c such that for all n � n matrices B withkBk < 1=2, we have

klog.I C B/ � Bk � c kBk2 .

Proof. Note that

log.I C B/� B D1XmD2

.�1/mC1Bm

mD B2

1XmD2

.�1/mC1Bm�2

m

so that if kBk < 1=2; we have

klog.I C B/� Bk � kBk21XmD2

12

�m�2

m,

which is an estimate of the desired form. utWe may restate the proposition in a more concise way by saying that

log.I C B/ D B CO.kBk2/;

where O.kBk2/ denotes a quantity of order kBk2 (i.e., a quantity that is boundedby a constant times kBk2 for all sufficiently small values of kBk).

We conclude this section with a result that, although we will not use it elsewhere,is worth recording. The proof is sketched in Exercises 9 and 10.

Theorem 2.10. Every invertible n � n matrix can be expressed as eX for someX 2 Mn.C/.

40 2 The Matrix Exponential

2.4 Further Properties of the Exponential

In this section, we give several additional results involving the exponential of amatrix that will be important in our study of Lie algebras.

Theorem 2.11 (Lie Product Formula). For all X; Y 2 Mn.C/, we have

eXCY D limm!1

�eXm e

Ym

m.

There is a version of this result, known as the Trotter product formula, whichholds for suitable unbounded operators on an infinite-dimensional Hilbert space.See, for example, Theorem 20.1 in [Hall].

Proof. If we multiply the power series for eXm and e

Ym , all but three of the terms will

involve 1=m2 or higher powers of 1=m. Thus,

eXm e

Ym D I C X

mC Y

mCO

�1

m2

�:

Now, since eXm e

Ym ! I as m ! 1, e

Xm e

Ym is in the domain of the logarithm for all

sufficiently largem. By Proposition 2.9,

log�eXm e

Ym

D log

�I C X

mC Y

mCO

�1

m2

��

D X

mC Y

mCO

Xm C Y

mCO

�1

m2

�2!

D X

mC Y

mCO

�1

m2

�:

Exponentiating the logarithm then gives

eXm e

Ym D exp

�X

mC Y

mCO

�1

m2

��

and, therefore,

�eXm e

Ym

m D exp

�X C Y CO

�1

m

��:

Thus, by the continuity of the exponential, we conclude that

limm!1

�eXm e

Ym

m D exp .X C Y / ;

which is the Lie product formula. ut

2.4 Further Properties of the Exponential 41

Recall (Sect. A.5) that the trace of matrix is defined as the sum of its diagonalentries and that similar matrices have the same trace.

Theorem 2.12. For any X 2 Mn.C/, we have

deteX� D etrace.X/.

Proof. IfX is diagonalizable with eigenvalues�1; : : : ; �n, then eX is diagonalizablewith eigenvalues e�1 ; : : : ; e�n . Thus, trace.X/ D P

j �j and

det.eX/ D e�1 � � � e�n D e�1C����n D etrace.X/:

IfX is not diagonalizable, we can approximate it by matrices that are diagonalizable(Exercise 4). utDefinition 2.13. A function A W R ! GL.nIC/ is called a one-parametersubgroup of GL.nIC/ if

1. A is continuous,2. A.0/ D I ,3. A.t C s/ D A.t/A.s/ for all t; s 2 R.

Theorem 2.14 (One-Parameter Subgroups). IfA.�/ is a one-parameter subgroupof GL.nIC/, there exists a unique n � n complex matrix X such that

A.t/ D etX.

By taking n D 1, and noting that GL.1IC/ Š C�, this theorem provides amethod of solving Exercise 18 in Chapter 1.

Lemma 2.15. Fix some " with " < log 2. Let B"=2 be the ball of radius "=2 aroundthe origin in Mn.C/, and let U D exp.B"=2/. Then every B 2 U has a uniquesquare root C in U , given by C D exp. 1

2logB/.

Proof. It is evident that C is a square root of B and that C is in U . To establishuniqueness, suppose C 0 2 U satisfies .C 0/2 D B . Let Y D logC 0; then exp.Y / DC 0 and

exp.2Y / D .C 0/2 D B D exp.logB/:

We have that Y 2 B"=2 and, thus, 2Y 2 B", and also that logB 2 B"=2 � B". Since,by Theorem 2.8, exp is injective on B" and exp.2Y / D exp.logB/, we must have2Y D logB . Thus, C 0 D exp. 1

2logB/ D C . ut

Proof of Theorem 2.14. The uniqueness is immediate, since if there is such an X ,then X D d

dtA.t/ˇtD0. To prove existence, let U be as in Lemma 2.15, which is an

open set in GL.nIC/. The continuity of A guarantees that there exists t0 > 0 suchthat A.t/ 2 U for all t with jt j � t0. Define

42 2 The Matrix Exponential

X D 1

t0log.A.t0//;

so that t0X D log.A.t0//. Then t0X 2 B"=2 and

et0X D A.t0/:

Now, A.t0=2/ is again in U and A.t0=2/2 D A.t0/. But by Lemma 2.15, A.t0/ hasa unique square root in U , and that unique square root is exp.t0X=2/. Thus,

A.t0=2/ D exp.t0X=2/:

Applying this argument repeatedly, we conclude that

A.t0=2k/ D exp.t0X=2k/

for all positive integers k. Then for any integerm, we have

A.mt0=2k/ D A.t0=2

k/m D exp.mt0X=2k/:

It follows that A.t/ D exp.tX/ for all real numbers t of the form t D mt0=2k, andthe set of such t’s is dense in R. Since both exp.tX/ and A.t/ are continuous, itfollows that A.t/ D exp.tX/ for all real numbers t . utProposition 2.16. The exponential map is an infinitely differentiable map ofMn.C/

into Mn.C/.

We will compute the derivative of the matrix exponential in Chapter 5.

Proof. Note that for each j and k, the quantity .Xm/jk is a homogeneous polynomialof degreem in the entries ofX . Thus, the series for the function .Xm/jk has the formof a multivariable power series onMn.C/ Š R2n

2. Since the series converges on all

of R2n2, it is permissible to differentiate the series term by term as many times as

we like. (Apply Theorem 12 in Chapter 4 of [Pugh] in each of the n2 variables withthe other variables fixed.) ut

2.5 The Polar Decomposition

The polar decomposition for a nonzero complex number z states that z can be writtenuniquely as z D up, where juj D 1 and p is real and positive. (If z D 0, thedecomposition still exists, with p D 0, but u is not unique.) Since p is real andpositive, it can be written as p D ex for a unique real number x. This gives anunconventional form of the polar decomposition for z, namely

2.5 The Polar Decomposition 43

z D uex; (2.10)

with x 2 R and juj D 1. Although it is customary to leave p as a positive realnumber and to write u as u D ei� , the decomposition in (2.10) is more convenientfor us because x, unlike � , is unique.

We wish to establish a similar polar decomposition first for GL.nIC/ and thenfor various subgroups thereof. If P is a self-adjoint n�nmatrix (i.e., P � D P ), wesay that P is positive if hv; Pvi > 0 for all nonzero v 2 Cn. It is easy to check thata self-adjoint matrixP is positive if and only if all the eigenvalues of P are positive.Suppose now that A is an invertible n � n matrix. We wish to write A as A D UP

where U is unitary and P is self-adjoint and positive. We will then write the self-adjoint, positive matrix P as P D eX where X is self-adjoint but not necessarilypositive.

Theorem 2.17.

1. Every A 2 GL.nIC/ can be written uniquely in the form

A D UP

where U is unitary and P is self-adjoint and positive.2. Every self-adjoint positive matrix P can be written uniquely in the form

P D eX

with X self-adjoint. Conversely, if X is self-adjoint, then eX is self-adjoint andpositive.

3. If we decompose each A 2 GL.nIC/ (uniquely) as

A D UeX

with U unitary and X self-adjoint, then U and X depend continuously on A.

Lemma 2.18. If Q is a self-adjoint, positive matrix, then Q has a unique positive,self-adjoint square root.

Proof. Since Q has an orthonormal basis of eigenvectors,Q can be written as

Q D U

0B@�1: : :

�n

1CAU�1

with U unitary. Since Q is self-adjoint and positive, each �j is positive. Thus, wecan construct a square root of Q as

44 2 The Matrix Exponential

Q1=2 D U

0B@�1=21

: : :

�1=2n

1CAU�1; (2.11)

and Q1=2 will still be self-adjoint and positive, establishing the existence of thesquare root.

If P is a self-adjoint, positive matrix, the eigenspaces of P2 are precisely thesame as the eigenspaces of P , with the eigenvalues of P2 being, of course, thesquares of the eigenvalues of P . The point here is that because the function x 7! x2

is injective on positive real numbers, eigenspaces with distinct eigenvalues remainwith distinct eigenvalues after squaring. Looking at this claim the other way around,if a positive, self-adjoint matrix Q is to have a positive self-adjoint square root P ,the eigenspaces of P must be the same as the eigenspaces ofQ, and the eigenvaluesof P must be the positive square roots of the eigenvalues ofQ. Thus, P is uniquelydetermined by Q. utProof of Theorem 2.17. For the existence of the decomposition in Point 1, note thatif A D UP , then A�A D PU �UP D P2. Now, for any matrix A, the matrix A�Ais self-adjoint. If, in addition,A is invertible, then for all nonzero v 2 Cn, we have

hv;A�Avi D hAv;Avi > 0;showing that A�A is positive. For all invertible A, then, let us define P by

P D .A�A/1=2;

where .�/1=2 is the unique self-adjoint positive square root of Lemma 2.18. We thendefine

U D AP�1 D AŒ.A�A/1=2��1:

Since P is, by construction, self-adjoint and positive, and since A D UP by thedefinition of U , it remains only to check that U is unitary. To that end, we checkthat

U �U D Œ.A�A/1=2��1A�AŒ.A�A/1=2��1;

since the inverse of a positive self-adjoint matrix is self-adjoint. Since A�A is thesquare of .A�A/1=2, we see that U �U D I , showing that U is unitary.

For the uniqueness of the decomposition, we have already noted that if A D UP ,then P2 D A�A, where A�A is self-adjoint and positive. Thus, the uniqueness ofP follows from the uniqueness in Lemma 2.18. The uniqueness of U then follows,since if A D UP, then U D AP�1.

The existence and uniqueness of the decomposition in Point 2 are proved inprecisely the same way as in Lemma 2.18, with the logarithm function (which isa bijection between .0;1/ and R) replacing the square root function. The same sort

2.5 The Polar Decomposition 45

of reasoning shows that for any self-adjoint X , the matrix eX is self-adjoint andpositive. We refer to the matrix X as the “logarithm” of P; even thoughX is not, ingeneral, computable by the series in Definition 2.7.

Finally, we address the continuity claim in Point 3. We first show that thelogarithmX of a self-adjoint, positive matrix P depends continuously on P: To seethis, note that if the eigenvalues of P are between 0 and 2; then by the remarks justbefore the proof of Theorem 2.8, the power series for logP will converge to X; inwhich case, continuity follows by the same argument as in the proof of Proposition2.1. In general, fix some positive, self-adjoint matrix P0: Fix a small neighborhoodV of P0 and choose a large positive number a: For P 2 V; write P D ea.e�aP /:Then the unique self-adjoint logarithmX of P may be computed as

X D aI C log.e�aP /:

If a is large enough, then for all P 2 V; the (positive) eigenvalues of e�aP will allbe less than 2; and the series for log.e�aP / will converge and depend continuouslyon P; showing that X depends continuously on P:

We then note that the square root operation on self-adjoint, positive matricesP isalso continuous, since P1=2 may be computed as exp.log.P /=2/: It is then evidentfrom the formulas for U and P that these quantities depend continuously on A: ut

We now establish polar decompositions for GL.nIR/, SL.nIC/, and SL.nIR/.Proposition 2.19.

1. Every A 2 GL.nIR/ can be written uniquely as

A D ReX;

where R is in O.n/ and X is real and symmetric.2. Every A 2 SL.nIC/ can be written uniquely as

A D UeX ;

where U is in SU.n/ and X is self-adjoint with trace zero.3. Every A 2 SL.nIR/ can be written uniquely as

A D ReX;

where R is in SO.n/ and X is real and symmetric and has trace zero.

Proof. If A is real, then A�A is real and symmetric. Now, a real, symmetric matrixcan be diagonalized over R. Thus, P , which is the unique self-adjoint positivesquare root of A�A (constructed as in (2.11)), is real. Then U D AP�1 is realand unitary, hence in O.n/.

Meanwhile, if A 2 SL.nIC/ and we write A D UeX with U 2 U.n/ andX self-adjoint, then det.A/ D det.U /etrace.X/. Now, jdet.U /j D 1, and etrace.X/ is real andpositive. Thus, by the uniqueness of the polar decomposition for nonzero complexnumbers, we must have det.U / D 1 and trace.X/ D 0. The case of A 2 SL.nIR/follows by combining the results of the two previous cases. ut

46 2 The Matrix Exponential

2.6 Exercises

1. The Cauchy–Schwarz inequality from elementary analysis tells us that for allu D .u1; : : : ; un/ and v D .v1; : : : ; vn/ in C

n, we have

ju1v1 C � � � C unvnj2 �0@ nXjD1

ˇujˇ21A

nXkD1

jvkj2!:

Use this to verify that kXYk � kXk kY k for all X; Y 2 Mn.C/, where k�k isthe Hilbert–Schmidt norm in Definition 2.2.

2. Show that for X 2 Mn.C/ and any orthonormal basis fu1; : : : ; ung of Cn,kXk2 D Pn

j;kD1ˇ˝

uj ; Xuk˛ˇ2

, where kXk is as in Definition 2.2. Now showthat if v is an eigenvector for X with eigenvalue �, then j�j � kXk.

3. The product rule. Recall that a matrix-valued functionA.t/ is said to be smoothif each Ajk.t/ is smooth. The derivative of such a function is defined as

�dA

dt

�jk

D dAjk

dt

or, equivalently,

d

dtA.t/ D lim

h!0

A.t C h/� A.t/

h.

Let A.t/ and B.t/ be two such functions. Prove that A.t/B.t/ is again smoothand that

d

dtŒA.t/B.t/� D dA

dtB.t/C A.t/

dB

dt:

4. Using Theorem A.4, show that every n � n complex matrix A is the limit of asequence of diagonalizable matrices.Hint: If an n � n matrix has n distinct eigenvalues, it is necessarilydiagonalizable.

5. For any a and d in C, define the expression .ea � ed /=.a � d/ in the obviousway for a ¤ d and by means of the limit

lima!d

ea � ed

a � dD ea

when a D d . Show that for any a; b; d 2 C, we have

exp

�a b

0 d

�D ea b e

a�eda�d

0 ed

!:

2.6 Exercises 47

Hint: Show that if a ¤ d , then

�a b

0 d

�mD�am b a

m�dma�d

0 bm

for every positive integerm.6. Show that every 2 � 2 matrix X with trace.X/ D 0 satisfies

X2 D � det.X/I:

If X is 2 � 2 with trace zero, show by direct calculation using the power seriesfor the exponential that

eX D cos�p

detX I C sin

pdetXp

detXX; (2.12)

wherep

detX is either of the two (possibly complex) square roots of detX . Usethis to give an alternative computation of the exponential eX1 in Example 2.5.Note: The value of the coefficient of X in (2.12) is to be interpreted as 1 whendetX D 0, in accordance with the limit lim�!0 sin �=� D 1.

7. Use the result of Exercise 6 to compute the exponential of the matrix

X D�

4 3

�1 2�:

Hint: Reduce the calculation to the trace-zero case.8. Consider the two matrices X4 and X5 in (2.6). Compute etX4 and etX5 either by

diagonalization or by the method in Exercises 6 and 7. Show that curves of theform t 7! etX4v0, with v0 ¤ 0, spiral out to infinity. Show that for v0 outside ofa certain one-dimensional subspace of R2, curves of the form t 7! etX5v0 tendto infinity in the direction of .1; 1/ or the direction of .�1;�1/.

9. A matrix A is said to be unipotent if A � I is nilpotent (i.e., if A is of theform A D I CN , with N nilpotent). Note that logA is defined whenever A isunipotent, because the series in Definition 2.7 terminates.

(a) Show that if A is unipotent, then logA is nilpotent.(b) Show that if X is nilpotent, then eX is unipotent.(c) Show that if A is unipotent, then exp.logA/ D A and that if X is nilpotent,

then log.expX/ D X .

Hint: LetA.t/ D ICt.A�I /. Show that exp.log.A.t/// depends polynomiallyon t and that exp.log.A.t/// D A.t/ for all sufficiently small t .

48 2 The Matrix Exponential

10. Show that every invertible n � n matrix A can be written as A D eX for someX 2 Mn.C/.Hint: Theorem A.5 implies thatA is similar to a block-diagonal matrix in whicheach block is of the form �I CN�, with N� being nilpotent. Use this result andExercise 9.

11. Show that for all X 2 Mn.C/, we have

limm!1

�I C X

m

�mD eX :

Hint: Use the matrix logarithm.

Chapter 3Lie Algebras

3.1 Definitions and First Examples

We now introduce the “abstract” notion of a Lie algebra. In Sect. 3.3, we willassociate to each matrix Lie group a Lie algebra. It is customary to use lowercaseGothic (Fraktur) characters such as g and h to refer to Lie algebras.

Definition 3.1. A finite-dimensional real or complex Lie algebra is a finite-dimensional real or complex vector space g, together with a map Œ�; �� from g � ginto g, with the following properties:

1. Œ�; �� is bilinear.2. Œ�; �� is skew symmetric: ŒX; Y � D � ŒY;X� for all X; Y 2 g.3. The Jacobi identity holds:

ŒX; ŒY;Z��C ŒY; ŒZ;X��C ŒZ; ŒX; Y �� D 0

for all X; Y;Z 2 g.

Two elementsX and Y of a Lie algebra g commute if ŒX; Y � D 0. A Lie algebrag is commutative if ŒX; Y � D 0 for all X; Y 2 g.

The map Œ�; �� is referred to as the bracket operation on g. Note also thatCondition 2 implies that ŒX;X� D 0 for all X 2 g. The bracket operation on aLie algebra is not, in general associative; nevertheless, the Jacobi identity can beviewed as a substitute for associativity.

A previous version of this book was inadvertently published without the middle initial of theauthor’s name as “Brian Hall”. For this reason an erratum has been published, correcting themistake in the previous version and showing the correct name as Brian C. Hall (see DOI http://dx.doi.org/10.1007/978-3-319-13467-3_14). The version readers currently see is the correctedversion. The Publisher would like to apologize for the earlier mistake.

© Springer International Publishing Switzerland 2015B.C. Hall, Lie Groups, Lie Algebras, and Representations, GraduateTexts in Mathematics 222, DOI 10.1007/978-3-319-13467-3_3

49

50 3 Lie Algebras

Example 3.2. Let g D R3 and let Œ�; �� W R3 � R3 ! R3 be given by

Œx; y� D x � y;where x � y is the cross product (or vector product). Then g is a Lie algebra.

Proof. Bilinearity and skew symmetry are standard properties of the cross product.To verify the Jacobi identity, it suffices (by bilinearity) to verify it when x D ej ,y D ek , and z D el , where e1, e2, and e3 are the standard basis elements for R3.If j , k, and l are all equal, each term in the Jacobi identity is zero. If j , k, and lare all different, the cross product of any two of ej , ek , and el is equal to a multipleof the third, so again, each term in the Jacobi identity is zero. It remains to considerthe case in which two of j; k; l are equal and the third is different. By re-orderingthe terms in the Jacobi identity as necessary, it suffices to verify the identity

Œej ; Œej ; ek��C Œej ; Œek; ej ��C Œek; Œej ; ej �� D 0: (3.1)

The first two terms in (3.1) are negatives of each other and the third is zero. utExample 3.3. Let A be an associative algebra and let g be a subspace of A such thatXY�YX 2 g for allX; Y 2 g. Then g is a Lie algebra with bracket operation given by

ŒX; Y � D XY � YX:

Proof. The bilinearity and skew symmetry of the bracket are evident. To verify theJacobi identity, note that each double bracket generates four terms, for a total of 12terms. It is left to the reader to verify that the product of X , Y , and Z in each of thesix possible orderings occurs twice, once with a plus sign and once with a minussign. ut

If we look carefully at the proof of the Jacobi identity, we see that the twooccurrences of, say, XYZ occur with different groupings, once as X.YZ/ and onceas .XY/Z. Thus, associativity of the algebra A is essential. For any Lie algebra, theJacobi identity means that the bracket operation behaves as if it were XY � YX insome associative algebra, even if it is not actually defined this way. Indeed, we willprove in Chapter 9 that every Lie algebra g can be embedded into an associativealgebra A in such a way that the bracket becomes XY � YX. (This claim followsfrom Theorem 9.9, the Poincaré–Birkhoff–Witt theorem.)

Of particular interest to us is the case in which A is the spaceMn.C/ of all n�ncomplex matrices.

Example 3.4. Let sl.nIC/ denote the space of all X 2 Mn.C/ for whichtrace.X/ D 0. Then sl.nIC/ is a Lie algebra with bracket ŒX; Y � D XY � YX.

Proof. For any X and Y in Mn.C/, we have

trace.XY � YX/ D trace.XY/� trace.YX/ D 0:

This holds, in particular, if X and Y have trace zero. Thus, Example 3.3 applies. ut

3.1 Definitions and First Examples 51

Definition 3.5. A subalgebra of a real or complex Lie algebra g is a subspace h ofg such that ŒH1;H2� 2 h for all H1 and H2 2 h. If g is a complex Lie algebra andh is a real subspace of g which is closed under brackets, then h is said to be a realsubalgebra of g.

A subalgebra h of a Lie algebra g is said to be an ideal in g if ŒX;H� 2 h for allX in g and H in h.

The center of a Lie algebra g is the set of all X 2 g for which ŒX; Y � D 0 for allY 2 g.

Definition 3.6. If g and h are Lie algebras, then a linear map � W g ! h is called aLie algebra homomorphism if � .ŒX; Y �/ D Œ�.X/; �.Y /� for all X; Y 2 g. If, inaddition, � is one-to-one and onto, then � is called a Lie algebra isomorphism.A Lie algebra isomorphism of a Lie algebra with itself is called a Lie algebraautomorphism.

Definition 3.7. If g is a Lie algebra and X is an element of g, define a linear mapadX W g ! g by

adX.Y / D ŒX; Y �.

The map X 7! adX is the adjoint map or adjoint representation.

Although adX.Y / is just ŒX; Y �, the alternative “ad” notation can be useful. Forexample, instead of writing

ŒX; ŒX; ŒX; ŒX; Y ����;

we can now write

.adX/4 .Y /.

This sort of notation will be essential in Chapter 5. We can view ad (that is, the mapX 7! adX ) as a linear map of g into End.g/, the space of linear operators on g.The Jacobi identity is then equivalent to the assertion that adX is a derivation of thebracket:

adX.ŒY;Z�/ D ŒadX.Y /;Z�C ŒY; adX.Z/�: (3.2)

Proposition 3.8. If g is a Lie algebra, then

adŒX;Y � D adXadY � adY adX D ŒadX ; adY �;

that is, adW g ! End.g/ is a Lie algebra homomorphism.

Proof. Observe that

adŒX;Y �.Z/ D ŒŒX; Y �; Z�;

52 3 Lie Algebras

whereas

ŒadX; adY �.Z/ D ŒX; ŒY;Z�� � ŒY; ŒX;Z��.

Thus, we want to show that

ŒŒX; Y �; Z� D ŒX; ŒY;Z�� � ŒY; ŒX;Z��;

which is equivalent to the Jacobi identity. utDefinition 3.9. If g1 and g2 are Lie algebras, the direct sum of g1 and g2 is thevector space direct sum of g1 and g2, with bracket given by

Œ.X1;X2/; .Y1; Y2/� D .ŒX1; Y1�; ŒX2; Y2�/: (3.3)

If g is a Lie algebra and g1 and g2 are subalgebras, we say that g decomposes asthe Lie algebra direct sum of g1 and g2 if g is the direct sum of g1 and g2 as vectorspaces and ŒX1;X2� D 0 for all X1 2 g1 and X2 2 g2.

It is straightforward to verify that the bracket in (3.3) makes g1 ˚ g2 into a Liealgebra. If g decomposes as a Lie algebra direct sum of subalgebras g1 and g2, it iseasy to check that g is isomorphic as a Lie algebra to the “abstract” direct sum of g1and g2. (This would not be the case without the assumption that every element of g1commutes with every element of g2.)

Definition 3.10. Let g be a finite-dimensional real or complex Lie algebra, and letX1; : : : ; XN be a basis for g (as a vector space). Then the unique constants cjkl suchthat

ŒXj ;Xk� DNXlD1

cjklXl (3.4)

are called the structure constants of g (with respect to the chosen basis).

Although we will not have much occasion to use them, structure constantsdo appear frequently in the physics literature. The structure constants satisfy thefollowing two conditions:

cjkl C ckjl D 0;Xn

.cjkncnlm C cklncnjm C cljncnkm/ D 0

for all j; k; l;m. The first of these conditions comes from the skew symmetry of thebracket, and the second comes from the Jacobi identity.

3.2 Simple, Solvable, and Nilpotent Lie Algebras 53

3.2 Simple, Solvable, and Nilpotent Lie Algebras

In this section, we consider various special types of Lie algebras. Recall fromDefinition 3.5 the notion of an ideal in a Lie algebra.

Definition 3.11. A Lie algebra g is called irreducible if the only ideals in g are gand f0g. A Lie algebra g is called simple if it is irreducible and dim g � 2.

A one-dimensional Lie algebra is certainly irreducible, since it is has nonontrivial subspaces and therefore no nontrivial subalgebras and no nontrivialideals. Nevertheless, such a Lie algebra is, by definition, not considered simple.

Note that a one-dimensional Lie algebra g is necessarily commutative, sinceŒaX; bX� D 0 for any X 2 g and any scalars a and b. On the other hand,if g is commutative, then any subspace of g is an ideal. Thus, the only way acommutative Lie algebra can be irreducible is if it is one dimensional. Thus, anequivalent definition of “simple” is that a Lie algebra is simple if it is irreducibleand noncommutative.

There is an analogy between groups and Lie algebras, in which the role ofsubgroups is played by subalgebras and the role of normal subgroups is played byideals. (For example, the kernel of a Lie algebra homomorphism is always an ideal,just as the kernel of a Lie group homomorphism is always a normal subgroup.) Thereis, however, an inconsistency in the terminology in the two fields. On the group side,any group with no nontrivial normal subgroups is called simple, including the mostobvious example, a cyclic group of prime order. On the Lie algebra side, by contrast,the most obvious example of an algebra with no nontrivial ideals—namely, a one-dimensional algebra—is not called simple.

We will eventually see many examples of simple Lie algebras, but for nowwe content ourselves with a single example. Recall the Lie algebra sl.nIC/ inExample 3.4.

Proposition 3.12. The Lie algebra sl.2IC/ is simple.

Proof. We use the following basis for sl.2IC/:

X D�0 1

0 0

�I Y D

�0 0

1 0

�; H D

�1 0

0 �1�:

Direct calculation shows that these basis elements have the following commutationrelations: ŒX; Y � D H , ŒH;X� D 2X , and ŒH; Y � D �2Y . Suppose h is an ideal insl.2IC/ and that h contains an element Z D aX C bH C cY , where a, b, and c arenot all zero. We will show, then, that h D sl.2IC/. Suppose first that c ¤ 0. Thenthe element

ŒX; ŒX;Z�� D ŒX; Œ�2bX C cH�� D �2cX

54 3 Lie Algebras

is a nonzero multiple of X . Since h is an ideal, we conclude that X 2 h. But ŒY;X�is a nonzero multiple of H and ŒY; ŒY;X�� is a nonzero multiple of Y , showing thatY andH also belong to h, from which we conclude that h D sl.2IC/.

Suppose next that c D 0 but b ¤ 0. Then ŒX;Z� is a nonzero multiple of Xand we may then apply the same argument in the previous paragraph to show thath D sl.2IC/. Finally, if c D 0 and b D 0 but a ¤ 0, then Z itself is a nonzeromultiple of X and we again conclude that h D sl.2IC/. utDefinition 3.13. If g is a Lie algebra, then the commutator ideal in g, denotedŒg; g�, is the space of linear combinations of commutators, that is, the space ofelements Z in g that can be expressed as

Z D c1ŒX1; Y1�C � � � C cmŒXm; Ym�

for some constants cj and vectorsXj ; Yj 2 g.

For anyX and Y in g, the commutator ŒX; Y � is in Œg; g�. This holds, in particular,if X is in Œg; g�, showing that Œg; g� is an ideal in g.

Definition 3.14. For any Lie algebra g, we define a sequence of subalgebrasg0; g1; g2; : : : of g inductively as follows: g0 D g, g1 D Œg0; g0�, g2 D Œg1; g1�,etc. These subalgebras are called the derived series of g. A Lie algebra g is calledsolvable if gj D f0g for some j .

It is not hard to show, using the Jacobi identity and induction on j; that each gjis an ideal in g.

Definition 3.15. For any Lie algebra g, we define a sequence of ideals gj in ginductively as follows. We set g0 D g and then define gjC1 to be the space oflinear combinations of commutators of the form ŒX; Y � with X 2 g and Y 2 gj .These algebras are called the upper central series of g. A Lie algebra g is said tobe nilpotent if gj D f0g for some j .

Equivalently, gj is the space spanned by all j th-order commutators,

ŒX1; ŒX2; ŒX3; : : : ŒXj ;XjC1� : : :���:

Note that every j th-order commutator is also a .j � 1/th-order commutator, bysetting QXj D ŒXj ;XjC1�. Thus, gj�1 gj . For every X 2 g and Y 2 gj , we haveŒX; Y � 2 gjC1 � gj , showing that gj is an ideal in g. Furthermore, it is clear thatgj � gj for all j ; thus, if g is nilpotent, g is also solvable.

Proposition 3.16. If g � M3.R/ denotes the space of 3 � 3 upper triangularmatrices with zeros on the diagonal, then g satisfies the assumptions of Example 3.3.The Lie algebra g is a nilpotent Lie algebra.

3.3 The Lie Algebra of a Matrix Lie Group 55

Proof. We will use the following basis for g,

X D0@0 1 00 0 0

0 0 0

1A I Y D

0@0 0 00 0 1

0 0 0

1A I Z D

0@0 0 10 0 0

0 0 0

1A : (3.5)

Direct calculation then establishes the following commutation relations: ŒX; Y � DZ and ŒX;Z� D ŒY;Z� D 0. In particular, the bracket of two elements of g is againin g, so that g is a Lie algebra. Then Œg; g� is the span of Z and Œg; Œg; g�� D 0,showing that g is nilpotent. utProposition 3.17. If g � M2.C/ denotes the space of 2 � 2 matrices of the form

�a b

0 c

with a, b, and c in C, then g satisfies the assumptions of Example 3.3. The Liealgebra g is solvable but not nilpotent.

Proof. Direct calculation shows that

��a b

0 c

�;

�d e

0 f

��D�0 h

0 0

�; (3.6)

where h D aeCbf �bd �ce, showing that g is a Lie subalgebra ofM2.C/. Further-more, the commutator ideal Œg; g� is one dimensional and hence commutative. Thus,g2 D f0g, showing that g is solvable. On the other hand, consider the followingelements of g:

H D�1 0

0 �1�

I X D�0 1

0 0

�:

Using (3.6), we can see that ŒH;X� D 2X , and thus that

ŒH; ŒH; ŒH; � � � ŒH;X� � � � ���

is a nonzero multiple of X , showing that gj ¤ f0g for all j . ut

3.3 The Lie Algebra of a Matrix Lie Group

In this section, we associate to each matrix Lie group G a Lie algebra g. Manyquestions involving a group can be studied by transferring them to the Lie algebra,where we can use tools of linear algebra. We begin by defining g as a set, and thenproceed to give g the structure of a Lie algebra.

56 3 Lie Algebras

Definition 3.18. Let G be a matrix Lie group. The Lie algebra of G, denoted g, isthe set of all matrices X such that etX is in G for all real numbers t .

Equivalently, X is in g if and only if the entire one-parameter subgroup(Definition 2.13) generated by X lies in G. Note that merely having eX in Gdoes not guarantee that X is in g. Even though G is a subgroup of GL.nIC/ (andnot necessarily of GL.nIR/), we do not require that etX be in G for all complexnumbers t , but only for all real numbers t . We will show in Sect. 3.7 that everymatrix Lie group is an embedded submanifold of GL.nIC/. We will then show(Corollary 3.46) that g is the tangent space to G at the identity.

We will now establish various basic properties of the Lie algebra g of a matrixLie group G. In particular, we will see that there is a bracket operation on g thatmakes g into a Lie algebra in the sense of Definition 3.1.

Proposition 3.19. LetG be a matrix Lie group, andX an element of its Lie algebra.Then eX is an element of the identity componentG0 of G.

Proof. By definition of the Lie algebra, etX lies in G for all real t . However, as tvaries from 0 to 1, etX is a continuous path connecting the identity to eX . utTheorem 3.20. Let G be a matrix Lie group with Lie algebra g. If X and Y areelements of g, the following results hold.

1. AXA�1 2 g for all A 2 G.2. sX 2 g for all real numbers s.3. X C Y 2 g.4. XY � YX 2 g.

It follows from this result and Example 3.3 that the Lie algebra of a matrix Liegroup is a real Lie algebra, with bracket given by ŒX; Y � D XY � YX. For X and Yin g, we refer to ŒX; Y � D XY � YX 2 g as the bracket or commutator of X and Y .

Proof. For Point 1, we observe that, by Proposition 2.3,

et.AXA�1/ D AetXA�1 2 G

for all t , showing that AXA�1 is in g. For Point 2, we observe that et.sX/ D e.ts/X ,which must be in G for all t 2 R if X is in g. For Point 3 we use the Lie productformula, which says that

et.XCY / D limm!1

etX=metY=m

�m.

Now,etX=metY=m

�mis in G for all m. Since G is closed, the limit (which is

invertible) must be again in G. This shows that X C Y is again in g:Finally, for Point 4, we use the product rule (Exercise 3 in Chapter 2) and

Proposition 2.4 to compute

3.4 Examples 57

d

dt

etXYe�tX

�ˇˇtD0

D .XY/e0 C .e0Y /.�X/

D XY � YX.

Now, by Point 1, etXYe�tX is in g for all t . Furthermore, by Points 2 and 3, g is a realsubspace of Mn.C/, from which it follows that g is a (topologically) closed subsetof Mn.C/. Thus,

XY � YX D limh!0

ehXYe�hX � Yh

belongs to g. utNote that even if the elements of G have complex entries, the Lie algebra g of

G is not necessarily a complex vector space, since Point 2 holds, in general, onlyfor s 2 R. Nevertheless, it may happen in certain cases that g is a complex vectorspace.

Definition 3.21. A matrix Lie group G is said to be complex if its Lie algebra g isa complex subspace of Mn.C/, that is, if iX 2 g for all X 2 g.

Examples of complex groups are GL.nIC/, SL.nIC/, SO.nIC/, and Sp.nIC/,as the calculations in Sect. 3.4 will show.

Proposition 3.22. If G is commutative then g is commutative.

We will see in Sect. 3.7 that if G is connected and g is commutative, G must becommutative.

Proof. For any two matrices X; Y 2 Mn.C/, the commutator of X and Y may becomputed as

ŒX; Y � D d

dt

�d

dsetXesY e�tX

ˇˇsD0

�ˇˇtD0

: (3.7)

If G is commutative and X and Y belong to g, then etX commutes with esY and theexpression in parentheses on the right hand side of (3.7) is independent of t , so thatŒX; Y � D 0. ut

3.4 Examples

Physicists are accustomed to using the map t 7! eitX rather than t 7! etX . Thus,the physicists’ expressions for the Lie algebras of matrix Lie groups will differ by afactor of i from the expressions we now derive.

58 3 Lie Algebras

Proposition 3.23. The Lie algebra of GL.nIC/ is the space Mn.C/ of all n � n

matrices with complex entries. Similarly, the Lie algebra of GL.nIR/ is equal toMn.R/. The Lie algebra of SL.nIC/ consists of all n � n complex matrices withtrace zero, and the Lie algebra of SL.nIR/ consists of all n � n real matrices withtrace zero.

We denote the Lie algebras of these groups as gl.nIC/, gl.nIR/, sl.nIC/, andsl.nIR/, respectively.

Proof. If X 2 Mn.C/, then etX is invertible, so that X belongs to the Lie algebraof GL.nIC/. If X 2 Mn.R/, then etX is invertible and real, so that X is inthe Lie algebra of GL.nIR/. Conversely, if etX is real for all real t , then X Dd.etX/=dt

ˇtD0 must also real. If X 2 Mn.C/ has trace zero, then by Theorem 2.12,

det.etX/ D 1, showing that X is in the Lie algebra of SL.nIC/. Conversely, ifdet.etX/ D et trace.X/ D 1 for all real t , then

trace.X/ D d

dtet trace.X/

ˇˇtD0

D 0:

Finally, if X is real and has trace zero, then etX is real and has determinant 1 for allreal t , showing that X is in the Lie algebra of SL.nIR/. Conversely, if etX is realand has determinant 1 for all real t , the preceding arguments show that X must bereal and have trace zero. utProposition 3.24. The Lie algebra of U.n/ consists of all complex matricessatisfying X� D �X and the Lie algebra of SU.n/ consists of all complex matricessatisfying X� D �X and trace.X/ D 0. The Lie algebra of the orthogonal groupO.n/ consists of all real matrices X satisfying X tr D �X and the Lie algebra ofSO.n/ is the same as that of O.n/.

The Lie algebras of U.n/ and SU.n/ are denoted u.n/ and su.n/, respectively.The Lie algebra of SO.n/ (which is the same as that of O.n/) is denoted so.n/.

Proof. A matrix U is unitary if and only if U � D U�1. Thus, etX is unitary if andonly if

etX�� D

etX��1 D e�tX . (3.8)

By Point 2 of Proposition 2.3,etX�� D etX�

, and so (3.8) becomes

etX� D e�tX . (3.9)

The condition (3.9) holds for all real t if and only if X� D �X . Thus, the Liealgebra of U.n/ consists precisely of matrices X such that X� D �X . As in theproof of Proposition 3.23, adding the “determinant 1” condition at the group leveladds the “trace 0” condition at the Lie algebra level.

3.4 Examples 59

An exactly similar argument over R shows that a real matrixX belongs to the Liealgebra of O.n/ if and only if X tr D �X . Since any such matrix has trace.X/ D 0

(since the diagonal entries of X are all zero), we see that every element of the Liealgebra of O.n/ is also in the Lie algebra of SO.n/. utProposition 3.25. If g is the matrix in Exercise 1 of Chapter 1, then the Lie algebraof O.nI k/ consists precisely of those real matrices X such that

gX trg D �X;

and the Lie algebra of SO.nI k/ is the same as that of O.nI k/. If � is thematrix (1.8), then the Lie algebra of Sp.nIR/ consists precisely of those realmatrices X such that

�X tr� D X;

and the Lie algebra of Sp.nIC/ consists precisely of those complex matrices Xsatisfying the same condition. The Lie algebra of Sp.n/ consists precisely of thosecomplex matrices X such that �X tr� D X and X� D �X .

The verification of Proposition 3.25 is similar to our previous computations andis omitted. The Lie algebra of SO.nI k/ (which is the same as that of O.nI k/) isdenoted so.nI k/, whereas the Lie algebras of the symplectic groups are denotedsp.nIR/, sp.nIC/, and sp.n/.

Proposition 3.26. The Lie algebra of the Heisenberg group H in Sect. 1.2.6 is thespace of all matrices of the form

X D0@0 a b0 0 c

0 0 0

1A ; (3.10)

with a; b; c 2 R.

Proof. If X is strictly upper triangular, it is easy to verify that Xm will be strictlyupper triangular for all positive integers m. Thus, for X as in (3.10), we will haveetX D I C B with B strictly upper triangular, showing that etX 2 H . Conversely, ifetX belongs toH for all real t , then all of the entries of etX on or below the diagonalare independent of t . Thus, X D d.etX/=dt

ˇtD0 will be of the form in (3.10). ut

We leave it as an exercise to determine the Lie algebras of the Euclidean andPoincaré groups.

Example 3.27. The following elements form a basis for the Lie algebra su.2/:

E1 D 12

�i 0

0 �i�

I E2 D 12

�0 i

i 0

�I E3 D 1

2

�0 �11 0

�:

60 3 Lie Algebras

These elements satisfy the commutation relations ŒE1; E2� D E3, ŒE2; E3� D E1,and ŒE3; E1� D E2. The following elements form a basis for the Lie algebra so.3/:

F1 D0@0 0 0

0 0 �10 1 0

1A ; F2 D

0@ 0 0 1

0 0 0

�1 0 0

1A ; F3 D

0@0 �1 01 0 0

0 0 0

1A ;

These elements satisfy the commutation relations ŒF1; F2� D F3, ŒF2; F3� D F1, andŒF3; F1� D F2.

Note that the listed relations completely determine all commutation relationsamong, say, E1, E2, and E3, since by the skew symmetry of the bracket, we musthave ŒE1; E1� D 0, ŒE2; E1� D �E3, and so on. Since E1, E2, and E3 satisfy thesame commutation relations as F1, F2, and F3, the two Lie algebras are isomorphic.

Proof. Direct calculation from Proposition 3.24. ut

3.5 Lie Group and Lie Algebra Homomorphisms

The following theorem tells us that a Lie group homomorphism between two Liegroups gives rise in a natural way to a map between the corresponding Lie algebras.It will follow (Exercise 8) that isomorphic Lie groups have isomorphic Lie algebras.

Theorem 3.28. Let G and H be matrix Lie groups, with Lie algebras g and h,respectively. Suppose that ˆ W G ! H is a Lie group homomorphism. Then thereexists a unique real-linear map � W g ! h such that

ˆ.eX/ D e�.X/ (3.11)

for all X 2 g. The map � has following additional properties:

1. �AXA�1� D ˆ.A/�.X/ˆ.A/�1, for all X 2 g, A 2 G.

2. �.ŒX; Y �/ D Œ�.X/; �.Y /�, for all X; Y 2 g.3. �.X/ D d

dtˆ.etX/ˇtD0, for all X 2 g.

In practice, given a Lie group homomorphism ˆ, the way one goes aboutcomputing � is by using Property 3. In the language of manifolds, Property 3 saysthat � is the derivative (or differential) of ˆ at the identity. By Point 2, � W g ! h isa Lie algebra homomorphism. Thus, every Lie group homomorphism gives rise to aLie algebra homomorphism. In Chapter 5, we will investigate the reverse question:If � is a homomorphism between the Lie algebras of two Lie groups, is there anassociated Lie group homomorphismˆ?

Proof. The proof is similar to the proof of Theorem 3.20. Since ˆ is a continuousgroup homomorphism, ˆ.etX/ will be a one-parameter subgroup of H , for eachX 2 g. Thus, by Theorem 2.14, there is a unique matrix Z such that

3.5 Lie Group and Lie Algebra Homomorphisms 61

ˆ.etX/ D etZ (3.12)

for all t 2 R. We define �.X/ D Z and check that � has the required properties.First, by putting t D 1 in (3.12), we see that ˆ.eX/ D e�.X/ for all X 2 g. Next, ifˆ.etX/ D etZ for all t , then ˆ.etsX/ D etsZ , showing that �.sX/ D s�.X/. Usingthe Lie product formula and the continuity of ˆ, we then compute that

et�.XCY / D ˆ�

limm!1

etX=metY=m

�m

D limm!1

.etX=m/ˆ.etY=m/

�m.

Thus,

et�.XCY / D limm!1

et�.X/=met�.Y /=m

�m D et.�.X/C�.Y //.

Differentiating this result at t D 0 shows that �.X C Y / D �.X/C �.Y /.We have thus obtained a real-linear map � satisfying (3.11). If there were another

real-linear map �0 with this property, we would have

et�.X/ D et�0.X/ D ˆ.etX/

for all t 2 R. Differentiating this result at t D 0 shows that �.X/ D �0.X/.We now verify the remaining claimed properties of �. For any A 2 G, we have

et�.AXA�1/ D e�.tAXA

�1/ D ˆ.etAXA�1

/.

Thus,

et�.AXA�1/ D ˆ.A/ˆ.etX/ˆ.A/�1

D ˆ.A/et�.X/ˆ.A/�1.

Differentiating this identity at t D 0 gives Point 1.Meanwhile, for any X and Y in g, we have, as in the proof of Theorem 3.20,

� .ŒX; Y �/ D �

�d

dtetXYe�tX

ˇˇtD0

D d

dt�etXYe�tX

�ˇˇtD0

;

where we have used the fact that a derivative commutes with a linear transformation.Thus,

62 3 Lie Algebras

� .ŒX; Y �/ D d

dtˆ.etX/�.Y /ˆ.e�tX/

ˇˇtD0

D d

dtet�.X/�.Y /e�t�.X/

ˇˇtD0

D Œ�.X/; �.Y /� ,

establishing Point 2. Finally, sinceˆ.etX/ D e�.tX/ D et�.X/, we can compute �.X/as in Point 3. utExample 3.29. Let ˆ W SU.2/ ! SO.3/ be the homomorphism in Proposi-tion 1.19. Then the associated Lie algebra homomorphism � W su.2/ ! so.3/satisfies

�.Ej / D Fj ; j D 1; 2; 3;

where fE1;E2;E3g and fF1; F2; F3g are the bases for su.2/ and so.3/, respectively,given in Example 3.27.

Since � maps a basis for su.2/ to a basis for so.3/, we see that � is aLie algebra isomorphism, even though ˆ is not a Lie group isomorphism (sinceker.ˆ/ D fI;�I g).

Proof. If X is in su.2/ and Y is in the space V in (1.14), then

d

dtˆ.etX/Y

ˇˇtD0

D d

dtetXYe�tX

ˇˇtD0

D ŒX; Y �:

Thus, �.X/ is the linear map of V Š R3 to itself given by Y 7! ŒX; Y �. If, say,X D E1, then direct computation shows that

�E1;

�x1 x2 C ix3

x2 � ix3 �x1��

D�

x01 x0

2 C ix03

x02 � ix0

3 �x01

�;

where .x01; x

02; x

03/ D .0;�x3; x2/. Since

0@ 0

�x3x2

1A D

0@0 0 0

0 0 �10 1 0

1A0@x1x2x3

1A ; (3.13)

we conclude that �.E1/ is the 3�3matrix appearing on the right-hand side of (3.13),which is precisely F1. The computation of �.E2/ and �.E3/ is similar and is left tothe reader. ut

3.5 Lie Group and Lie Algebra Homomorphisms 63

Proposition 3.30. Suppose that G, H , and K are matrix Lie groups andˆ W H ! K and ‰ W G ! H are Lie group homomorphisms. Let ƒ W G ! K bethe composition ofˆ and‰ and let �, , and � be the Lie algebra maps associatedto ˆ, ‰, andƒ, respectively. Then we have

� D � ı .

Proof. For any X 2 g,

ƒ.etX/ D .‰.etX// D .et .X// D et�. .X//.

Thus, �.X/ D �. .X//. utProposition 3.31. If ˆ W G ! H is a Lie group homomorphism and � W g ! h isthe associated Lie algebra homomorphism, then the kernel of ˆ is a closed, normalsubgroup of G and the Lie algebra of the kernel is given by

Lie.ker.ˆ// D ker.�/:

Proof. The usual algebraic argument shows that ker.ˆ/ is normal subgroup of G.Since, also, ˆ is continuous, ker.ˆ/ is closed. If X 2 ker.�/, then

ˆ.etX/ D et�.X/ D I;

for all t 2 R, showing that X is in the Lie algebra of ker.ˆ/. In the other direction,if etX lies in ker.ˆ/ for all t 2 R, then

et�.X/ D ˆ.etX/ D I

for all t . Differentiating this relation with respect to t at t D 0 gives that �.X/ D 0,showing that X 2 ker.�/. utDefinition 3.32 (The Adjoint Map). LetG be a matrix Lie group, with Lie algebrag. Then for each A 2 G, define a linear map AdA W g ! g by the formula

AdA.X/ D AXA�1.

Proposition 3.33. Let G be a matrix Lie group, with Lie algebra g. Let GL.g/denote the group of all invertible linear transformations of g. Then the map A !AdA is a homomorphism of G into GL.g/. Furthermore, for each A 2 G, AdAsatisfies AdA.ŒX; Y �/ D ŒAdA.X/;AdA.Y /� for all X; Y 2 g.

Proof. Easy. Note that Point 1 of Theorem 3.20 guarantees that AdA.X/ is actuallyin g for all X 2 g. ut

Since g is a real vector space with some dimension k, GL.g/ is essentially thesame as GL.kIR/. Thus, we will regard GL.g/ as a matrix Lie group. It is easy to

64 3 Lie Algebras

show that Ad W G ! GL.g/ is continuous, and so is a Lie group homomorphism.By Theorem 3.28, there is an associated real linear map X ! adX from the Liealgebra of G to the Lie algebra of GL.g/ (i.e., from g to gl.g/), with the propertythat

eadX D AdeX .

Here, gl.g/ is the Lie algebra of GL.g/, namely the space of all linear maps of g toitself.

Proposition 3.34. Let G be a matrix Lie group, let g be its Lie algebra, and letAd W G ! GL.g/ be as in Proposition 3.33. Let ad W g ! gl.g/ be the associatedLie algebra map. Then for all X; Y 2 g

adX.Y / D ŒX; Y �. (3.14)

The proposition shows that our usage of the notation adX in this section isconsistent with that in Definition 3.7.

Proof. By Point 3 of Theorem 3.28, ad can be computed as follows:

adX D d

dtAdetX

ˇˇtD0

.

Thus,

adX.Y / D d

dtetXYe�tX

ˇˇtD0

D ŒX; Y �;

as claimed. utWe have proved, as a consequence of Theorem 3.28 and Proposition 3.34, the

following result, which we will make use of later.

Proposition 3.35. For any X in Mn.C/, let adX W Mn.C/ ! Mn.C/ be given byadXY D ŒX; Y �. Then for any Y in Mn.C/, we have

eXYe�X D AdeX .Y / D eadX .Y /;

where

eadX .Y / D Y C ŒX; Y �C 1

2ŒX; ŒX; Y ��C � � � :

This result can also be proved by direct calculation—see Exercise 14.

3.6 The Complexification of a Real Lie Algebra 65

3.6 The Complexification of a Real Lie Algebra

In studying the representations of a matrix Lie group G (as we will do in laterchapters), it is often useful to pass to the Lie algebra g of G, which is, in general,only a real Lie algebra. It is then often useful to pass to an associated complex Liealgebra, called the complexification of g.

Definition 3.36. If V is a finite-dimensional real vector space, then thecomplexification of V , denoted VC, is the space of formal linear combinations

v1 C iv2;

with v1; v2 2 V . This becomes a real vector space in the obvious way and becomesa complex vector space if we define

i.v1 C iv2/ D �v2 C iv1.

We could more pedantically define VC to be the space of ordered pairs .v1; v2/with v1; v2 2 V , but this is notationally cumbersome. It is straightforward to verifythat the above definition really makes VC into a complex vector space. We willregard V as a real subspace of VC in the obvious way.

Proposition 3.37. Let g be a finite-dimensional real Lie algebra and gC itscomplexification. Then the bracket operation on g has a unique extension to gCthat makes gC into a complex Lie algebra. The complex Lie algebra gC is called thecomplexification of the real Lie algebra g.

Proof. The uniqueness of the extension is obvious, since if the bracket operation ongC is to be bilinear, then it must be given by

ŒX1 C iX2; Y1 C iY2� D .ŒX1; Y1� � ŒX2; Y2�/C i .ŒX1; Y2�C ŒX2; Y1�/ . (3.15)

To show existence, we must now check that (3.15) is really bilinear and skewsymmetric and that it satisfies the Jacobi identity. It is clear that (3.15) is realbilinear, and skew-symmetric. The skew symmetry means that if (3.15) is complexlinear in the first factor, it is also complex linear in the second factor. Thus, we needonly show that

Œi.X1 C iX2/; Y1 C iY2� D i ŒX1 C iX2; Y1 C iY2� . (3.16)

The left-hand side of (3.16) is

Œ�X2 C iX1; Y1 C iY2� D .� ŒX2; Y1� � ŒX1; Y2�/C i .ŒX1; Y1� � ŒX2; Y2�/ ;

66 3 Lie Algebras

whereas the right-hand side of (3.16) is

i f.ŒX1; Y1� � ŒX2; Y2�/C i .ŒX2; Y1�C ŒX1; Y2�/gD .� ŒX2; Y1� � ŒX1; Y2�/C i .ŒX1; Y1� � ŒX2; Y2�/ ,

and, indeed, these expressions are equal.It remains to check the Jacobi identity. Of course, the Jacobi identity holds if

X; Y , and Z are in g. Furthermore, for all X; Y;Z 2 gC, the expression

ŒX; ŒY;Z��C ŒY; ŒZ;X�� C ŒZ; ŒX; Y ��

is complex-linear in X with Y and Z fixed. Thus, the Jacobi identity continues tohold if X is in gC and Y and Z are in g. The same argument then shows that theJacobi identity holds whenX and Y are in gC andZ is in g. Applying this argumentone more time establishes the Jacobi identity for gC in general. utProposition 3.38. Suppose that g � Mn.C/ is a real Lie algebra and that for allnonzero X in g, the element iX is not in g. Then the “abstract”complexification gCof g in Definition 3.36 is isomorphic to the set of matrices in Mn.C/ that can beexpressed in the form X C iY with X and Y in g.

Proof. Consider the map from gC into Mn.C/ sending the formal linear combina-tion X C iY to the linear combination X C iY of matrices. This map is easily seento be a complex Lie algebra homomorphism. If g satisfies the assumption in thestatement of the proposition, this map is also injective and thus an isomorphism ofgC with g C ig � Mn.C/. ut

Using the proposition, we easily obtain the following list of isomorphisms:

gl.nIR/C Š gl.nIC/;u.n/C Š gl.nIC/;su.n/C Š sl.nIC/;

sl.nIR/C Š sl.nIC/;so.n/C Š so.nIC/;

sp.nIR/C Š sp.nIC/;sp.n/C Š sp.nIC/.

(3.17)

Let us verify just one example, that of u.n/. IfX� D �X , then .iX/� D iX. Thus,Xand iX cannot both be in u.n/ unlessX is zero. Furthermore, everyX inMn.C/ canbe expressed asX D X1C iX2, whereX1 D .X�X�/=2 andX2 D .XCX�/=.2i/are both in u.n/. This shows that u.n/C Š gl.nIC/.

Although both su.2/C and sl.2IR/C are isomorphic to sl.2IC/, the Lie algebrasu.2/ is not isomorphic to sl.2IR/. See Exercise 11.

3.7 The Exponential Map 67

Proposition 3.39. Let g be a real Lie algebra, gC its complexification, and h anarbitrary complex Lie algebra. Then every real Lie algebra homomorphism of ginto h extends uniquely to a complex Lie algebra homomorphism of gC into h.

This result is the universal property of the complexification of a real Liealgebra.

Proof. The unique extension is given by �.XCiY/ D �.X/Ci�.Y / for allX; Y 2g. It is easy to check that this map is, indeed, a homomorphism of complex Liealgebras. ut

3.7 The Exponential Map

Definition 3.40. IfG is a matrix Lie group with Lie algebra g, then the exponentialmap for G is the map

exp W g ! G.

That is to say, the exponential map for G is the matrix exponential restrictedto the Lie algebra g of G. We have shown (Theorem 2.10) that every matrix inGL.nIC/ is the exponential of some n�n matrix. Nevertheless, ifG � GL.nIC/ isa closed subgroup, there may exist A in G such that there is noX in the Lie algebrag of G with expX D A.

Example 3.41. There does not exist a matrix X 2 sl.2IC/ with

eX D��1 1

0 �1�; (3.18)

even though the matrix on the right-hand side of (3.18) is in SL.2IC/.Proof. If X 2 sl.2IC/ has distinct eigenvalues, then X is diagonalizable and eX

will also be diagonalizable, unlike the matrix on the right-hand side of (3.18). IfX 2 sl.2IC/ has a repeated eigenvalue, this eigenvalue must be 0 or the trace of Xwould not be zero. Thus, there is a nonzero vector v with Xv D 0, from which itfollows that eXv D e0v D v. We conclude that eX has 1 as an eigenvalue, unlikethe matrix on the right-hand side of (3.18). ut

We see, then, that the exponential map for a matrix Lie group G does notnecessarily map g onto G. Furthermore, the exponential map may not be one-to-one on g, as may be seen, for example, from the case g D su.2/. Nevertheless, itprovides a crucial mechanism for passing information between the group and theLie algebra. Indeed, we will see (Corollary 3.44) that the exponential map is locallyone-to-one and onto, a result that will be essential later.

68 3 Lie Algebras

Theorem 3.42. For 0 < " < log 2, let U" D fX 2 Mn.C/ jkXk < "g and letV" D exp.U"/. Suppose G � GL.nIC/ is a matrix Lie group with Lie algebra g.Then there exists " 2 .0; log 2/ such that for all A 2 V", A is in G if and only iflogA is in g.

The condition " < log 2 guarantees (Theorem 2.8) that for all X 2 U", log.eX/is defined and equal to X . Note that if X D logA is in g, then A D eX is in G.Thus, the content of the theorem is that for some ", having A in V" \G implies thatlogA must be in g. See Figure 3.1.

We begin with a lemma.

Lemma 3.43. Suppose Bm are elements of G and that Bm ! I . Let Ym D logBm,which is defined for all sufficiently large m. Suppose that Ym is nonzero for all mand that Ym= kYmk ! Y 2 Mn.C/. Then Y is in g.

Proof. For any t 2 R, we have .t= kYmk/ Ym ! tY . Note that since Bm ! I , wehave kYmk ! 0. Thus, we can find integers km such that km kYmk ! t . We have,then,

ekmYm D exp

�.km kYmk/ YmkYmk

�! etY :

Fig. 3.1 If A 2 V" belongs to G, then logA belongs to g

3.7 The Exponential Map 69

Fig. 3.2 The points kmYm are converging to tY

However,

ekmYm D .eYm/km D .Bm/km 2 G

and G is closed, and we conclude that etY 2 G. This shows that Y 2 g. (SeeFigure 3.2.) ut

Proof of Theorem 3.42. Let us think of Mn.C/ as Cn2 Š R2n

2and let D denote

the orthogonal complement of g with respect to the usual inner product on R2n2.

Consider the map ˆ W Mn.C/ ! Mn.C/ given by

ˆ.Z/ D eXeY ,

whereZ D XCY withX 2 g and Y 2 D. Since (Proposition 2.16) the exponentialis continuously differentiable, the mapˆ is also continuously differentiable, and wemay compute that

d

dtˆ.tX; 0/

ˇˇtD0

D X;

d

dtˆ.0; tY/

ˇˇtD0

D Y .

70 3 Lie Algebras

This calculation shows that the derivative of ˆ at the point 0 2 R2n2

is theidentity. (Recall that the derivative at a point of a function from R2n

2to itself is

a linear map of R2n2

to itself.) Since the derivative of ˆ at the origin is invertible,the inverse function theorem says that ˆ has a continuous local inverse, defined ina neighborhood of I .

We need to prove that for some ", if A 2 V" \G, then logA 2 g. If this were notthe case, we could find a sequenceAm in G such that Am ! I asm ! 1 and suchthat for all m, logAm … g. Using the local inverse of the map ˆ, we can write Am(for all sufficiently largem) as

Am D eXmeYm ; Xm 2 g; Ym 2 D;with Xm and Ym tending to zero as m tends to infinity. We must have Ym ¤ 0, sinceotherwise we would have logAm D Xm 2 g. Since eXm and Am are in G, we seethat

Bm WD e�XmAm D eYm

is in G.Since the unit sphere in D is compact, we can choose a subsequence of the Ym’s

(still called Ym) so that Ym= kYmk converges to some Y 2 D, with kY k D 1. Then,by the lemma, Y 2 g. This is a contradiction, because D is the orthogonal comple-ment of g. Thus, there must be some " such that logA 2 g for all A in V" \G. ut

3.8 Consequences of Theorem 3.42

In this section, we derive several consequences of the main result of the last section,Theorem 3.42.

Corollary 3.44. If G is a matrix Lie group with Lie algebra g, there exists aneighborhoodU of 0 in g and a neighborhoodV of I inG such that the exponentialmap takes U homeomorphically onto V .

Proof. Let " be such that Theorem 3.42 holds and set U D U" \ g and V DV" \ G. The theorem implies that exp takes U onto V . Furthermore, exp is ahomeomorphism of U onto V , since there is a continuous inverse map, namely,the restriction of the matrix logarithm to V . utCorollary 3.45. Let G be a matrix Lie group with Lie algebra g and let k be thedimension of g as a real vector space. Then G is a smooth embedded submanifoldof Mn.C/ of dimension k and hence a Lie group.

It follows from the corollary that G is locally path connected: every point inG has a neighborhood U that is homeomorphic to a ball in R

k and hence pathconnected. It then follows thatG is connected (in the usual topological sense) if andonly if it is path connected. (See, for example, Proposition 3.4.25 of [Run].)

3.8 Consequences of Theorem 3.42 71

Proof. Let " 2 .0; log 2/ be such that Theorem 3.42 holds. Then for any A0 2 G,consider the neighborhood A0V" of A0 in Mn.C/. Note that A 2 A0V" if and onlyif A�1

0 A 2 V". Define a local coordinate system on A0V" by writing each A 2 A0V"as A D A0e

X , for X 2 U" � Mn.C/. It follows from Theorem 3.42 that (forA 2 A0V") A 2 G if and only if X 2 g. Thus, in this local coordinate systemdefined near A0, the groupG looks like the subspace g ofMn.C/. Since we can findsuch local coordinates near any point A0 in G, we conclude that G is an embeddedsubmanifold of Mn.C/.

Now, the operation of matrix multiplication is clearly smooth. Furthermore, bythe formula for the inverse of a matrix in terms of cofactors, the map A 7! A�1 isalso smooth on GL.nIC/. The restrictions of these maps to G are then also smooth,showing that G is a Lie group. utCorollary 3.46. Suppose G � GL.nIC/ is a matrix Lie group with Lie algebra g.Then a matrix X is in g if and only if there exists a smooth curve � in Mn.C/ with�.t/ 2 G for all t and such that �.0/ D I and d�=dtjtD0 D X . Thus, g is thetangent space at the identity to G.

This result is illustrated in Figure 3.1.

Proof. If X is in g, then we may take �.t/ D etX and then �.0/ D I andd�=dtjtD0 D X . In the other direction, suppose that �.t/ is a smooth curve in Gwith �.0/ D I . For all sufficiently small t , we can write �.t/ D eı.t/, where ı is asmooth curve in g. Now, the derivative of ı.t/ at t D 0 is the same as the derivativeof t 7! tı0.0/ at t D 0. Thus, by the chain rule, we have

� 0.0/ D d

dteı.t/

ˇˇtD0

D d

dtetı

0.0/

ˇˇtD0

D ı0.0/:

Since ı.t/ belongs to g for all sufficiently small t , we conclude (as in the proof ofTheorem 3.20) that ı0.0/ D � 0.0/ belongs to g. utCorollary 3.47. IfG is a connected matrix Lie group, every element A ofG can bewritten in the form

A D eX1eX2 � � � eXm (3.19)

for some X1;X2; : : : ; Xm in g.

Even if G is connected, it is in general not possible to write every A 2 G assingle exponential, A D expX , with X 2 g. (See Example 3.41.) We begin with asimple analytic lemma.

Lemma 3.48. Suppose A W Œa; b� ! Gl.nIC/ is a continuous map. Then for all" > 0 there exists ı > 0 such that if s; t 2 Œa; b� satisfy js � t j < ı, then

A.s/A.t/�1 � I < ":

72 3 Lie Algebras

Proof. We note that

A.s/A.t/�1 � I D .A.s/� A.t//A.t/�1

� kA.s/� A.t/k A.t/�1 : (3.20)

Since Œa; b� is compact and the map t 7! A.t/�1 is continuous, there is a constantC such that

A.t/�1 � C for all t 2 Œa; b�. Furthermore, since Œa; b� is compact,Theorem 4.19 in [Rud1] tells us that the map A is actually uniformly continuous onŒa; b�. Thus, for any " > 0, there exists ı > 0 such that when js � t j < ı, we havekA.s/ �A.t/k < "=C . Thus, in light of (3.20), we have the desired ı. utProof of Corollary 3.47. Let V" be as in Theorem 3.42. For any A 2 G, choose acontinuous path A W Œ0; 1� ! G with A.0/ D I and A.1/ D A. By Lemma 3.48,we can find some ı > 0 such that if js � t j < ı, then A.s/A.t/�1 2 V". Nowdivide Œ0; 1� into m pieces, where 1=m < ı. Then for j D 1; : : : ; m, we see thatA..j � 1/=m/�1A.j=m/ belongs to V", so that

A..j � 1/=m/�1A.j=m/ D eXj

for some elements X1; : : : ; Xm of g: Thus,

A D A.0/�1A.1/

D A.0/�1A.1=m/A.1=m/�1A.2=m/ � � �A..m � 1/=m/�1A.1/

D eX1eX2 � � � eXm;

as claimed. utCorollary 3.49. Let G and H be matrix Lie groups with Lie algebras g andh, respectively, and assume G is connected. Suppose ˆ1 and ˆ2 are Lie grouphomomorphisms of G into H and that �1 and �2 be the associated Lie algebrahomomorphisms of g into h. Then if �1 D �2, we have ˆ1 D ˆ2.

Proof. Let g be any element ofG. SinceG is connected, Corollary 3.47 tells us thatevery element of G can be written as eX1eX2 � � � eXm , with Xj 2 g. Thus,

ˆ1.eX1 � � � eXm/ D e�1.X1/ � � � e�1.Xm/

D e�2.X1/ � � � e�2.Xm/

D ˆ2.eX1 � � � eXm/;

as claimed. utCorollary 3.50. Every continuous homomorphism between two matrix Lie groupsis smooth.

3.9 Exercises 73

Proof. For all g 2 G, we write nearby elements h 2 G as h D geX , with X 2 g, sothat

ˆ.h/ D ˆ.g/ˆ.eX / D ˆ.g/e�.X/:

This relation says that in the exponential coordinates near g, the map ˆ is acomposition of a linear map, the exponential map, and multiplication on the leftby ˆ.g/, all of which are smooth. This shows that ˆ is smooth near g. utCorollary 3.51. IfG is a connected matrix Lie group and the Lie algebra g ofG iscommutative, then G is commutative.

This result is a partial converse to Proposition 3.22.

Proof. Since g is commutative, any two elements of G, when written as inCorollary 3.47, will commute. utCorollary 3.52. If G � Mn.C/ is a matrix Lie group, the identity component G0of G is a closed subgroup of GL.N IC/ and thus a matrix Lie group. Furthermore,the Lie algebra of G0 is the same as the Lie algebra of G.

Proof. Suppose that hAmi is a sequence in G0 converging to some A 2 GL.nIC/.Then certainly A 2 G, since G is closed. Furthermore, AmA�1 lies in G for allm and AmA�1 ! I as m ! 1. If m is large enough, Theorem 3.42 tells us thatAmA

�1 D eX for some X 2 g, so that A D e�XAm. Since Am 2 G0, there is a pathjoining I to Am in G. But we can then join Am to e�XAm D A by the path e�tXAm,0 � t � 1. By combining these two paths, we can join I to A in G, showing that Abelongs to G0.

Now, since G0 � G, the Lie algebra of G0 is contained in the Lie algebra of G.In the other direction, if etX lies in G for all t , then it actually lies in G0, since anypoint et0X on the curve etX can be connected to I inG, using the curve etX itself, for0 � t � t0. ut

3.9 Exercises

1. (a) If g is a Lie algebra, show that the center of g is an ideal in g.(b) If g and h are Lie algebras and � W g ! h is a Lie algebra homomorphism,

show that the kernel of � is an ideal in g.2. Classify up to isomorphism all one-dimensional and two-dimensional real Lie

algebras. There is one isomorphism class of one-dimensional algebras and twoisomorphism classes of two-dimensional algebras.

3. Let g denote the space of n � n upper triangular matrices with zeros on thediagonal. Show that g is a nilpotent Lie algebra under the bracket given byŒX; Y � D XY � YX.

4. Give an example of a matrix Lie group G and a matrix X such that eX 2 G,but X … g.

74 3 Lie Algebras

5. If G1 � GL.n1IC/ and G2 � GL.n2IC/ are matrix Lie groups and G1 �G2 istheir direct product (regarded as a subgroup of GL.n1 C n2IC/ in the obviousway), show that the Lie algebra of G1 �G2 is isomorphic to g1 ˚ g2.

6. LetG andH be matrix Lie groups withH � G, so that the Lie algebra h ofHis a subalgebra of the Lie algebra g of G.

(a) Show that if H is a normal subgroup of G, then h is an ideal in g.(b) Show that if G and H are connected and h is an ideal in g, then H is a

normal subgroup of G.

7. Suppose G � GL.nIC/ is a matrix Lie group with Lie algebra g. Suppose thatA is inG and that kA� Ik < 1, so that the power series for logA is convergent.Is it necessarily the case that logA is in g? Prove or give a counterexample.

8. Show that two isomorphic matrix Lie groups have isomorphic Lie algebras.9. Write out explicitly the general form of a 4 � 4 real matrix in so.3I 1/ (see

Proposition 3.25).10. Show that there is an invertible linear map � W su.2/ ! R3 such that

�.ŒX; Y �/ D �.X/ � �.Y / for all X; Y 2 su.2/, where � denotes the crossproduct (vector product) on R3.

11. Show that su.2/ and sl.2IR/ are not isomorphic Lie algebras, even thoughsu.2/C Š sl.2IR/C Š sl.2IC/.

12. Show the groups SU.2/ and SO.3/ are not isomorphic, even though theassociated Lie algebras su.2/ and so.3/ are isomorphic.

13. Let G be a matrix Lie group and let g be its Lie algebra. For each A 2 G, showthat AdA is a Lie algebra automorphism of g.

14. Let X and Y be n � n matrices. Show by induction that

.adX/m .Y / D

mXkD0

m

k

!XkY.�X/m�k;

where

.adX/m .Y / D ŒX; � � � ŒX; ŒX; Y �� � � � �„ ƒ‚ …

m

:

Now, show by direct computation that

eadX .Y / D AdeX .Y / D eXYe�X .

Assume it is legal to multiply power series term by term. (This result wasobtained indirectly in Proposition 3.35.)Hint: Use Pascal’s triangle.

15. If G is a matrix Lie group, a component ofG is the collection of all points thatcan be connected to a fixed A 2 G by a continuous path in G. Show that if Gis compact, G has only finitely many components.

3.9 Exercises 75

Hint: Suppose G had infinitely many components and consider a sequenceAj with each element of the sequence in a different component. Extract aconvergent subsequence and Bk D Ajk and consider B�1

k BkC1.16. Suppose that G is a connected, commutative matrix Lie group with Lie

algebra g. Show that the exponential map for G maps g onto G.17. Suppose G is a connected matrix Lie group with Lie algebra g and that A is an

element of G. Show that A belongs to the center of G if and only if AdA.X/ DX for all X 2 g.

18. Show that the exponential map from the Lie algebra of the Heisenberg group tothe Heisenberg group is one-to-one and onto.

19. Show that the exponential map from u.n/ to U.n/ is onto, but not one-to-one.Hint: Every unitary matrix has an orthonormal basis of eigenvectors.

20. Suppose X is a 2 � 2 real matrix with trace zero, and assume X has a nonrealeigenvalue.

(a) Show that the eigenvalues ofX must be of the form ia;�ia with a a nonzeroreal number.

(b) Show that the corresponding eigenvectors of X can be chosen to becomplex conjugates of each other, say, v and Nv.

(c) Show that there exists an invertible real matrix C such that

X D C

�0 a

�a 0�C�1:

Hint: Use v and Nv to construct a real basis for R2, and determine the matrix Xin this basis.

21. Suppose A is a 2 � 2 real matrix with determinant one, and assume A has anonreal eigenvalue. Show that there exists a real number � that is not an integermultiple of � and an invertible real matrix C such that

A D C

�cos � sin �

� sin � cos �

�C�1:

22. Show that the image of the exponential map for SL.2IR/ consists of preciselythose matrices A 2 SL.2IR/ such that trace .A/ > �2, together with thematrix �I (which has trace �2). To do this, consider the possibilities for theeigenvalues of a matrix in the Lie algebra sl.2IR/ and in the group SL.2IR/. Inthe Lie algebra, show that the eigenvalues are of the form .�;��/ or .i�;�i�/,with � real. In the group, show that the eigenvalues are of the form .a; 1=a/ or.�a;�1=a/, with a real and positive, or of the form .ei� ; e�i� /, with � real. Thecase of a repeated eigenvalue (.0; 0/ in the Lie algebra and .1; 1/ or .�1;�1/in the group) will have to be treated separately using the Jordan canonical form(Sect. A.4).Hint: You may assume that if a real matrix X has real eigenvalues, then X issimilar over the reals to its Jordan canonical form. Then use the two previousexercises.

Chapter 4Basic Representation Theory

4.1 Representations

If V is a finite-dimensional real or complex vector space, let GL.V / denote thegroup of invertible linear transformations of V . If we choose a basis for V , we canidentify GL.V / with GL.nIR/ or GL.nIC/. Any such identification gives rise to atopology on GL.V /, which is easily seen to be independent of the choice of basis.With this discussion in mind, we think of GL.V / as a matrix Lie group. Similarly,we let gl.V / D End.V / denote the space of all linear operators from V to itself,which forms a Lie algebra under the bracket ŒX; Y � D XY � YX.

Definition 4.1. Let G be a matrix Lie group. A finite-dimensional complexrepresentation of G is a Lie group homomorphism

… W G ! GL.V /;

where V is a finite-dimensional complex vector space (with dim.V / � 1). A finite-dimensional real representation of G is a Lie group homomorphism… of G intoGL.V /, where V is a finite-dimensional real vector space.

If g is a real or complex Lie algebra, then a finite-dimensional complexrepresentation of g is a Lie algebra homomorphism � of g into gl.V /, whereV is a finite-dimensional complex vector space. If g is a real Lie algebra, then afinite-dimensional real representation of g is a Lie algebra homomorphism � ofg into gl.V /, where V is a finite-dimensional real vector space.

A previous version of this book was inadvertently published without the middle initial of theauthor’s name as “Brian Hall”. For this reason an erratum has been published, correcting themistake in the previous version and showing the correct name as Brian C. Hall (see DOI http://dx.doi.org/10.1007/978-3-319-13467-3_14). The version readers currently see is the correctedversion. The Publisher would like to apologize for the earlier mistake.

© Springer International Publishing Switzerland 2015B.C. Hall, Lie Groups, Lie Algebras, and Representations, GraduateTexts in Mathematics 222, DOI 10.1007/978-3-319-13467-3_4

77

78 4 Basic Representation Theory

If … or � is a one-to-one homomorphism, the representation is called faithful.

One should think of a representation as a linear action of a group or Lie algebraon a vector space (since, say, to every g 2 G, there is associated an operator ….g/,which acts on the vector space V ). If the homomorphism… W G ! GL.V / is fixed,we will occasionally use the alternative notation

g � v (4.1)

in place ….g/v. We will often use terminology such as “Let … be a representationof G acting on the space V .”

If a representation … is a faithful representation of a matrix Lie group G, thenf….A/ jA 2 G g is a group of matrices that is isomorphic to the original group G.Thus, … allows us to represent G as a group of matrices. This is the motivationfor the term “representation.” (Of course, we still call … a representation even ifit is not faithful.) Despite the origin of the term, the goal of representation theoryis not simply to represent a group as a group of matrices. After all, the groups westudy in this book are already matrix groups! Rather, the goal is to determine (up toisomorphism) all the ways a fixed group can act as a group of matrices.

Linear actions of groups on vector spaces arise naturally in many branches ofboth mathematics and physics. A typical example would be a linear differentialequation in three-dimensional space which has rotational symmetry, such as theequations that describe the energy states of a hydrogen atom in quantum mechanics.Since the equation is rotationally invariant, the space of solutions is invariant underrotations and thus constitutes a representation of the rotation group SO.3/. Therepresentation theory of SO.3/ (or of its Lie algebra) is very helpful in narrowingdown what the space of solutions can be. See, for example, Chapter 18 in [Hall].

Definition 4.2. Let … be a finite-dimensional real or complex representation of amatrix Lie group G, acting on a space V . A subspace W of V is called invariantif ….A/w 2 W for all w 2 W and all A 2 G. An invariant subspace W is callednontrivial if W ¤ f0g and W ¤ V . A representation with no nontrivial invariantsubspaces is called irreducible.

The terms invariant, nontrivial, and irreducible are defined analogously forrepresentations of Lie algebras.

Even if g is a real Lie algebra, we will consider mainly complex representations ofg. It should be emphasized that if we are speaking about complex representationsof a real Lie algebra g acting on a complex subspace V , an invariant subspaceW isrequired to be a complex subspace of V .

Definition 4.3. Let G be a matrix Lie group, let … be a representation of G actingon the space V , and let † be a representation of G acting on the space W . A linearmap � W V ! W is called an intertwining map of representations if

�.….A/v/ D †.A/�.v/

for all A 2 G and all v 2 V . The analogous property defines intertwining maps ofrepresentations of a Lie algebra.

4.1 Representations 79

If � is an intertwining map of representations and, in addition, � is invertible,then � is said to be an isomorphism of representations. If there exists anisomorphism between V andW , then the representations are said to be isomorphic.

If we use the “action” notation of (4.1), the defining property of an intertwiningmap may be written as

�.A � v/ D A � �.v/

for all A 2 G and v 2 V . That is to say, � should commute with the action of G.A typical problem in representation theory is to determine, up to isomorphism, allof the irreducible representations of a particular group or Lie algebra. In Sect. 4.6,we will determine all the finite-dimensional complex irreducible representations ofthe Lie algebra sl.2IC/.

After identifying GL.V / with GL.nIR/ or GL.nIC/, Theorem 3.28 has thefollowing consequence.

Proposition 4.4. Let G be a matrix Lie group with Lie algebra g and let … be a(finite-dimensional real or complex) representation of G, acting on the space V .Then there is a unique representation � of g acting on the same space such that

….eX/ D e�.X/

for all X 2 g. The representation � can be computed as

�.X/ D d

dt….etX/

ˇˇtD0

and satisfies

�.AXA�1/ D ….A/�.X/….A/�1

for all X 2 g and all A 2 G.

Given a matrix Lie group G with Lie algebra g, we may ask whether everyrepresentation � of g comes from a representation … of G. As it turns out, this isnot true in general, but is true if G is simply connected. See Sect. 4.7 for examplesof this phenomenon and Sect. 5.7 for the general result.

Proposition 4.5.

1. Let G be a connected matrix Lie group with Lie algebra g. Let … be arepresentation of G and � the associated representation of g. Then … isirreducible if and only if � is irreducible.

2. Let G be a connected matrix Lie group, let …1 and …2 be representations of G,and let �1 and �2 be the associated Lie algebra representations. Then �1 and �2are isomorphic if and only if …1 and…2 are isomorphic.

80 4 Basic Representation Theory

Proof. For Point 1, suppose first that … is irreducible. We then want to show that� is irreducible. So, let W be a subspace of V that is invariant under �.X/ for allX 2 g. We want to show that W is either f0g or V . Now, suppose A is an elementof G. Since G is assumed connected, Corollary 3.47 tells us that A can be writtenas A D eX1 � � � eXm for some X1; : : : ; Xm in g. Since W is invariant under �.Xj /it will also be invariant under exp.�.Xj // D I C �.Xj / C �.Xj /

2=2 C � � � and,hence, under

….A/ D ….eX1 � � � eXm/ D ….eX1/ � � �….eXm/D e�.X1/ � � � e�.Xm/:

Since … is irreducible and W is invariant under each ….A/, W must be either f0gor V . This shows that � is irreducible.

In the other direction, assume that � is irreducible and that W is an invariantsubspace for …. Then W is invariant under ….exp tX/ for all X 2 g and, hence,under

�.X/ D d

dt….etX/

ˇˇtD0

:

Thus, since � is irreducible,W is f0g or V , and we conclude that … is irreducible.This establishes Point 1 of the proposition.

Point 2 of the proposition is similar and is left as an exercise to the reader(Exercise 1). utProposition 4.6. Let g be a real Lie algebra and gC its complexification. Thenevery finite-dimensional complex representation � of g has a unique extension to acomplex-linear representation of gC, also denoted � . Furthermore, � is irreducibleas a representation of gC if and only if it is irreducible as a representation of g.

Of course, the extension of � to gC is given by �.X C iY/ D �.X/C i�.Y / forall X; Y 2 g.

Proof. The existence and uniqueness of the extension follow from Proposition 3.39.The claim about irreducibility holds because a complex subspace W of V isinvariant under �.X C iY/, with X and Y in g, if and only if it is invariant underthe operators �.X/ and �.Y /. Thus, the representation of g and its extension to gChave precisely the same invariant subspaces. utDefinition 4.7. If V is a finite-dimensional inner product space and G is a matrixLie group, a representation … W G ! GL.V / is unitary if ….A/ is a unitaryoperator on V for every A 2 G.

Proposition 4.8. Suppose G is a matrix Lie group with Lie algebra g: Suppose Vis a finite-dimensional inner product space,… is a representation ofG acting on V ,and � is the associated representation of g. If … is unitary, then �.X/ is skew self-adjoint for all X 2 g. Conversely, if G is connected and �.X/ is skew self-adjointfor all X 2 g, then … is unitary.

4.2 Examples of Representations 81

In a slight abuse of notation, we will say that a representation � of a real Liealgebra g acting on a finite-dimensional inner product space is unitary if �.X/ isskew self-adjoint for all X 2 g.

Proof. The proof is similar to the computation of the Lie algebra of the unitarygroup U.n/. If … is unitary, then for all X 2 g we have

.et�.X//� D ….etX/� D ….etX/�1 D e�t�.X/; t 2 R;

so that et�.X/� D e�t�.X/. Differentiating this relation with respect to t at t D 0

gives �.X/� D ��.X/. In the other direction, if �.X/� D ��.X/, then theabove calculation shows that ….etX/ D et�.X/ is unitary. If G is connected, thenby Corollary 3.47, every A 2 G is a product of exponentials, showing that ….A/ isunitary. ut

4.2 Examples of Representations

A matrix Lie group G is, by definition, a subset of some GL.nIC/. The inclusionmap of G into GL.nIC/ (i.e., the map ….A/ D A) is a representation ofG, called the standard representation of G. If G happens to be contained inGL.nIR/ � GL.nIC/, then we can also think of the standard representation asa real representation. Thus, for example, the standard representation of SO.3/ is theone in which SO.3/ acts in the usual way on R

3 and the standard representationof SU.2/ is the one in which SU.2/ acts on C

2 in the usual way. Similarly, ifg � Mn.C/ is a Lie algebra of matrices, the map �.X/ D X is called the standardrepresentation of g.

Consider the one-dimensional complex vector space C. For any matrix Lie groupG, we can define the trivial representation,… W G ! GL.1IC/, by the formula

….A/ D I

for all A 2 G. Of course, this is an irreducible representation, since C has nonontrivial subspaces, let alone nontrivial invariant subspaces. If g is a Lie algebra,we can also define the trivial representation of g, � W g ! gl.1IC/, by

�.X/ D 0

for all X 2 g. This is an irreducible representation.Recall the adjoint map of a group or Lie algebra, described in Definitions 3.32

and 3.7.

Definition 4.9. IfG is a matrix Lie group with Lie algebra g, the adjoint represen-tation of G is the map Ad W G ! GL.g/ given by A 7! AdA. Similarly, the adjoint

82 4 Basic Representation Theory

representation of a finite-dimensional Lie algebra g is the map ad W g ! gl.g/given by X 7! adX .

If G is a matrix Lie group with Lie algebra g, then by Proposition 3.34, the Liealgebra representation associated to the adjoint representation of G is the adjointrepresentation of g. Note that in the case of SO.3/, the standard representation andthe adjoint representation are both three-dimensional real representations. In fact,these two representations are isomorphic (Exercise 2).

Example 4.10. Let Vm denote the space of homogeneous polynomials of degreem in two complex variables. For each U 2 SU.2/, define a linear transformation…m.U / on the space Vm by the formula

Œ…m.U /f � .z/ D f .U�1z/; z 2 C2: (4.2)

Then…m is a representation of SU.2/.

Elements of Vm have the form

f .z1; z2/ D a0zm1 C a1z

m�11 z2 C a2z

m�21 z22 C � � � C amzm2 (4.3)

with z1; z2 2 C and the aj ’s arbitrary complex constants, from which we see thatdim.Vm/ D mC 1. Explicitly, if f is as in (4.3), then

Œ…m.U /f �.z1; z2/ DmXkD0

ak.U�111 z1 C U�1

12 z2/m�k.U�1

21 z1 C U�122 z2/

k.

By expanding out the right-hand side of this formula, we see that…m.U /f is againa homogeneous polynomial of degreem. Thus, …m.U / actually maps Vm into Vm.

To see that …m is actually a representation, compute that

…m.U1/Œ…m.U2/f �.z/ D Œ…m.U2/f �.U�11 z/ D f .U�1

2 U�11 z/

D …m.U1U2/f .z/.

The inverse on the right-hand side of (4.2) is necessary in order to make …m arepresentation. We will see in Proposition 4.11 that each …m is irreducible andwe will see in Sect. 4.6 that every finite-dimensional irreducible representation ofSU.2/ is isomorphic to one (and only one) of the …m’s. (Of course, no two of the…m’s are isomorphic, since they do not even have the same dimension.)

The associated representation �m of su.2/ can be computed as

.�m.X/f /.z/ D d

dtf .e�tXz/

ˇˇtD0

:

4.2 Examples of Representations 83

Now, let z.t/ D .z1.t/; z2.t// be the curve in C2 defined as z.t/ D e�tXz. By thechain rule, we have

�m.X/f D @f

@z1

d z1dt

ˇˇtD0

C @f

@z2

d z2dt

ˇˇtD0

.

Since d z=dtjtD0 D �Xz, so we obtain

�m.X/f D � @f@z1

.X11z1 CX12z2/ � @f

@z2.X21z1 CX22z2/. (4.4)

We may then take the unique complex-linear extension of � to sl.2IC/ Š su.2/C,as in Proposition 3.39. This extension is given by the same formula, but with X 2sl.2IC/.

If X , Y , andH are the following basis elements for sl.2IC/:

H D�1 0

0 �1�

I X D�0 1

0 0

�I Y D

�0 0

1 0

�;

then applying formula (4.4) gives

�m.H/ D �z1@

@z1C z2

@

@z2

�m.X/ D �z2@

@z1

�m.Y / D �z1@

@z2:

Applying these operators to a basis element zm�k1 zk2 for Vm gives

�m.H/.zm�k1 zk2/ D .�mC 2k/zm�k

1 zk2

�m.X/.zm�k1 zk2/ D �.m � k/ zm�k�1

1 zkC12 ;

�m.Y /.zm�k1 zk2/ D �kzm�kC1

1 zk�12 . (4.5)

Thus, zm�k1 zk2 is an eigenvector for �m.H/ with eigenvalue �mC 2k, while �m.X/

and �m.Y / have the effect of shifting the exponent k of z2 up or down by one. Notethat since �m.X/ increases the value of k, this operator increases the eigenvalue of�m.H/ by 2, whereas �m.Y / decreases the eigenvalue of �m.H/ by 2.

Fig. 4.1 The black dotsindicate the nonzero terms fora vector w in the space V6.Applying �6.X/4 to w gives anonzero multiple of z62 z1

6 z15z2 z1

4 z22 z1

3z23 z1

2z24 z1z2

5 z26

6 X 4

84 4 Basic Representation Theory

Proposition 4.11. For eachm � 0, the representation �m is irreducible.

Proof. It suffices to show that every nonzero invariant subspace of Vm is equal toVm. So, let W be such a space and let w be a nonzero element of W . Then w can bewritten in the form

w D a0zm1 C a1z

m�11 z2 C a2z

m�21 z22 C � � � C amzm2

with at least one of the ak’s being nonzero. Let k0 be the smallest value of k forwhich ak ¤ 0 and consider

�m.X/m�k0w.

(See Figure 4.1.)Since each application of �m.X/ raises the power of z2 by 1, �m.X/m�k0 will

kill all the terms in w except the ak0zm�k01 zk02 term. On the other hand, since

�m.X/.zm�k1 zk2/ is zero only if k D m, we see that �m.X/m�k0w is a nonzero

multiple of zm2 . Since W is assumed invariant, W must contain this multiple of zm2and so also zm2 itself. Now, for 0 � k � m, it follows from (4.5) that �m.Y /kzm2is a nonzero multiple of zk1zm�k

2 . Therefore, W must also contain zk1zm�k2 for all

0 � k � m. Since these elements form a basis for Vm, we see that W D Vm, asdesired. ut

4.3 New Representations from Old

One way of generating representations is to take some representations one knowsand combine them in some fashion. In this section, we will consider threestandard methods of obtaining new representations from old, namely direct sumsof representations, tensor products of representations, and dual representations.

4.3.1 Direct Sums

Definition 4.12. Let G be a matrix Lie group and let …1;…2; : : : ;…m be repre-sentations of G acting on vector spaces V1; V2; : : : ; Vm. Then the direct sum of…1;…2; : : : ;…m is a representation …1 ˚ � � � ˚ …m of G acting on the spaceV1 ˚ � � � ˚ Vm, defined by

Œ…1 ˚ � � � ˚…m.A/� .v1; : : : ; vm/ D .…1.A/v1; : : : ;…m.A/vm/

for all A 2 G.

4.3 New Representations from Old 85

Similarly, if g is a Lie algebra, and �1; �2; : : : ; �m are representations of g actingon V1; V2; : : : ; Vm, then we define the direct sum of �1; �2; : : : ; �m, acting on V1 ˚� � � ˚ Vm by

Œ�1 ˚ � � � ˚ �m.X/� .v1; : : : ; vm/ D .�1.X/v1; : : : ; �m.X/vm/

for all X 2 g.

It is straightforward to check that, say, …1 ˚ � � � ˚…m is really a representationof G.

4.3.2 Tensor Products

Let U and V be finite-dimensional real or complex vector spaces. We wish to definethe tensor product of U and V , which will be a new vector space U˝V “built” outof U and V . We will discuss the idea of this first and then give the precise definition.

We wish to consider a formal “product” of an element u of U with an element vof V , denoted u ˝ v. The space U ˝ V is then the space of linear combinations ofsuch products, that is, the space of elements of the form

a1u1 ˝ v1 C a2u2 ˝ v2 C � � � C anun ˝ vn. (4.6)

Of course, if “˝” is to be interpreted as a product, then it should be bilinear:

.u1 C au2/˝ v D u1 ˝ v C au2 ˝ v;

u ˝ .v1 C av2/ D u ˝ v1 C au ˝ v2.

We do not assume that the product is commutative. That is to say, even if U D V sothat U ˝ V and V ˝ U are the same space, u ˝ v will not, in general, equal v ˝ u.

Now, if e1; e2; : : : ; en is a basis for U and f1; f2; : : : ; fm is a basis for V , then,using bilinearity, it is easy to see that any element of the form (4.6) can be written asa linear combination of the elements ej ˝ fk . In fact, it seems reasonable to expectthat fej ˝fk j1 � j � n; 1 � k � mg should be a basis for the space U ˝V . This,in fact, turns out to be the case.

Definition 4.13. If U and V are finite-dimensional real or complex vector spaces,then a tensor product of U with V is a vector space W , together with a bilinearmap � W U � V ! W with the following property: If is any bilinear map ofU � V into a vector space X , there exists a unique linear map Q of W into X suchthat the following diagram commutes:

86 4 Basic Representation Theory

U � V �! W

& . Q X

.

Note that the bilinear map from U � V into X turns into the linear map Q ofW intoX . This is one of the points of tensor products: Bilinear maps onU �V turninto linear maps onW .

Theorem 4.14. If U and V are any finite-dimensional real or complex vectorspaces, then a tensor product .W; �/ exists. Furthermore, .W; �/ is unique up tocanonical isomorphism. That is, if .W1; �1/ and .W2; �2/ are two tensor products,then there exists a unique vector space isomorphism ˆ W W1 ! W2 such that thefollowing diagram commutes:

U � V �1! W1

�2 & . ˆ

W2

.

Suppose that .W; �/ is a tensor product and that e1; e2; : : : ; en is a basis for Uand f1; f2; : : : ; fm is a basis for V . Then

f�.ej ; fk/ j1 � j � n; 1 � k � mg

is a basis forW .

In particular, dim .U ˝ V / D .dimU / .dimV /.

Proof. Exercise 7. utNotation 4.15 Since the tensor product of U and V is essentially unique, we willlet U ˝V denote an arbitrary tensor product space and we will write u ˝ v insteadof �.u; v/. In this notation, Theorem 4.14 says that

fej ˝ fk j1 � j � n; 1 � k � m g

is a basis for U ˝ V , as expected.

The defining property of U ˝ V is called the universal property of tensorproducts. To understand the significance of this property, suppose we want to definea linear map T from U ˝ V into some other space. We could try to define T usingbases for U and V , but then we would have to worry about whether T depends onthe choice of basis. Instead, we could try to define T on elements of the form u ˝ v,with u 2 U and v 2 V . While it follows from Theorem 4.14 that elements of thisform spanU˝V , the decomposition of an element ofU˝V as a linear combinationof elements of the form u ˝ v is far from unique. Thus, if we wish to define T onsuch elements, we have to worry whether T is well defined.

4.3 New Representations from Old 87

This is where the universal property comes in. If .u; v/ is any bilinearexpression in .u; v/, the universal property says precisely that there is a unique linearmap T (D Q ) such that

T .u ˝ v/ D .u; v/.

Thus, we can construct a well-defined linear map T on U ˝ V simply by definingit on elements of the form u ˝ v, provided that our definition of T .u ˝ v/ is bilinearin u and v. The following result is an application of this line of reasoning.

Proposition 4.16. Let U and V be finite-dimensional real or complex vectorspaces. Let A W U ! U and B W V ! V be linear operators. Then there exists aunique linear operator from U ˝ V to U ˝ V , denoted A˝ B , such that

.A˝ B/.u ˝ v/ D .Au/˝ .Bv/

for all u 2 U and v 2 V . If A1 and A2 are linear operators on U and B1 and B2are linear operators on V , then

.A1 ˝ B1/.A2 ˝ B2/ D .A1A2/˝ .B1B2/. (4.7)

Proof. Define a map from U � V into U ˝ V by

.u; v/ D .Au/˝ .Bv/.

Since A and B are linear and since ˝ is bilinear, is a bilinear map of U � V intoU ˝ V . By the universal property, there is an associated linear map Q W U ˝ V !U ˝ V such that

Q .u ˝ v/ D .u; v/ D .Au/˝ .Bv/.

Thus, Q is the desired map A ˝ B . An elementary calculation shows that theidentity (4.7) holds on elements of the form u ˝ v. Since, by Theorem 4.14, suchelements span U ˝ V , the identity holds in general. ut

We are now ready to define tensor products of representations. There are twodifferent approaches to this, both of which are important. The first approach startswith a representation of a group G acting on a space U and a representation ofanother group H acting on a space V and produces a representation of the productgroup G � H acting on the space U ˝ V . The second approach starts with twodifferent representations of the same group G, acting on spaces U and V , andproduces a representation of G acting on U ˝ V . Both of these approaches canbe adapted to apply to Lie algebras.

Definition 4.17. Let G and H be matrix Lie groups. Let …1 be a representation ofG acting on a space U and let …2 be a representation of H acting on a space V .

88 4 Basic Representation Theory

Then the tensor product of …1 and …2 is a representation …1 ˝ …2 of G � H

acting on U ˝ V defined by

.…1 ˝…2/.A;B/ D …1.A/˝…2.B/

for all A 2 G and B 2 H .

Using Proposition 4.16, it is easy to check that …1 ˝ …2 is, in fact, arepresentation of G �H .

Now, ifG andH are matrix Lie groups (i.e.,G is a closed subgroup of GL.nIC/andH is a closed subgroup of GL.mIC/), thenG�H can be regarded in an obviousway as a closed subgroup of GL.nCmIC/. Thus, the direct product of matrix Liegroups can be regarded as a matrix Lie group. It is easy to check that the Lie algebraofG�H is isomorphic to the direct sum of the Lie algebra ofG and the Lie algebraof H .

Proposition 4.18. Let G and H be matrix Lie groups with Lie algebras g andh, respectively. Let …1 and …2 be representations of G and H , respectively, andconsider the representation …1 ˝…2 of G �H . If �1 ˝ �2 denotes the associatedrepresentation of g ˚ h, then

.�1 ˝ �2/.X; Y / D �1.X/˝ I C I ˝ �2.Y /

for all X 2 g and Y 2 h.

Proof. Suppose that u.t/ is a smooth curve in U and v.t/ is a smooth curve in V .Then, by repeating the proof of the product rule for scalar-valued functions (or bycalculating everything in a basis), we have

d

dt.u.t/˝ v.t// D du

dt˝ v.t/C u.t/˝ dv

dt.

This being the case, we compute as follows:

.�1 ˝ �2/.X; Y /.u ˝ v/

D d

dt…1.e

tX/u ˝…2.etY/v

ˇˇtD0

D�d

dt…1.e

tX/u

ˇˇtD0

�˝ v C u ˝

�d

dt…2.e

tY/v

ˇˇtD0

�.

This establishes the claimed form of .�1˝�2/.X; Y / on elements of the form u˝v,which span U ˝ V . ut

The proposition motivates the following definition.

4.3 New Representations from Old 89

Definition 4.19. Let g and h be Lie algebras and let �1 and �2 be representations ofg and h, acting on spaces U and V . Then the tensor product of �1 and �2, denoted�1 ˝ �2, is a representation of g ˚ h acting on U ˝ V , given by

.�1 ˝ �2/.X; Y / D �1.X/˝ I C I ˝ �2.Y /

for all X 2 g and Y 2 h.

It is easy to check that this indeed defines a representation of g ˚ h. Note that ifwe defined .�1 ˝ �2/.X; Y / D �1.X/˝ �2.Y /, this would not be a representationof g ˚ h, since this is expression is not linear in .X; Y /.

We now define a variant of the above definitions in which we take the tensorproduct of two representations of the same group G and regard the result as arepresentation of G rather than of G �G.

Definition 4.20. LetG be a matrix Lie group and let…1 and…2 be representationsof G, acting on spaces V1 and V2. Then the tensor product representation of G,acting on V1 ˝ V2, is defined by

.…1 ˝…2/.A/ D …1.A/˝…2.A/

for all A 2 G. Similarly, if �1 and �2 are representations of a Lie algebra g, wedefine a tensor product representation of g on V1 ˝ V2 by

.�1 ˝ �2/.X/ D �1.X/˝ I C I ˝ �2.X/:

It is easy to check that …1 ˝ …2 and �1 ˝ �2 are actually representations ofG and g, respectively. The notation is, unfortunately, ambiguous, since if …1 and…2 are representations of the same group G, we can regard …1 ˝ …2 either as arepresentation of G or as a representation of G �G. We must, therefore, be carefulto specify which way we are thinking about…1 ˝…2.

If …1 and …2 are irreducible representations of a group G, then …1 ˝ …2 willtypically not be irreducible when viewed as a representation of G. One can, then,attempt to decompose…1 ˝…2 as a direct sum of irreducible representations. Thisprocess is called the Clebsch–Gordan theory or, in the physics literature, “additionof angular momentum.” See Exercise 12 and Appendix C for more informationabout this topic.

4.3.3 Dual Representations

Suppose that � is a representation of a Lie algebra g acting on a finite-dimensionalvector space V . Let V � denote the dual space of V , that is, the space of linear

90 4 Basic Representation Theory

functionals on V (Sect. A.7). If A is a linear operator on V , let Atr denote the dualor transpose operator on V � , given by

.Atr�/.v/ D �.Av/

for � 2 V �, v 2 V . If v1; : : : ; vn is a basis for V , then there is a naturally associated“dual basis” �1; : : : ; �n with the property that �j .vk/ D ıjk. The matrix for Atr

in the dual basis is then simply the transpose (not the conjugate transpose!) of thematrix of A in the original basis. If A and B are linear operators on V , it is easilyverified that

.AB/tr D B trAtr: (4.8)

Definition 4.21. Suppose G is a matrix Lie group and … is a representation of Gacting on a finite-dimensional vector space V . Then the dual representation…� to… is the representation of G acting on V � and given by

…�.g/ D Œ….g�1/�tr: (4.9)

If � is a representation of a Lie algebra g acting on a finite-dimensional vector spaceV , then �� is the representation of g acting on V � and given by

��.X/ D ��.X/tr: (4.10)

Using (4.8), it is easy to check that both …� and �� are actually representations.(Here the inverse on the right-hand side of (4.9) and the minus sign on the right-handside of (4.10) are essential.) The dual representation is also called contragredientrepresentation.

Proposition 4.22. If … is a representation of a matrix Lie group G, then (1) …� isirreducible if and only if… is irreducible and (2) .…�/� is isomorphic to…. Similarstatements apply to Lie algebra representations.

Proof. See Exercise 6. ut

4.4 Complete Reducibility

Much of representation theory is concerned with studying irreducible represen-tations of a group or Lie algebra. In favorable cases, knowing the irreduciblerepresentations leads to a description of all representations.

Definition 4.23. A finite-dimensional representation of a group or Lie algebra issaid to be completely reducible if it is isomorphic to a direct sum of a finite numberof irreducible representations.

4.4 Complete Reducibility 91

Definition 4.24. A group or Lie algebra is said to have the complete reducibilityproperty if every finite-dimensional representation of it is completely reducible.

As it turns out, most groups and Lie algebras do not have the complete reducibil-ity property. Nevertheless, many interesting example groups and Lie algebras dohave this property, as we will see in this section and Sect. 10.3.

Example 4.25. Let … W R ! GL.2IC/ be given by

….x/ D�1 x

0 1

�:

Then… is not completely reducible.

Proof. Direct calculation shows that … is, in fact, a representation of R. If fe1; e2gis the standard basis for C2, then clearly the span of e1 is an invariant subspace.We now claim that he1i is the only nontrivial invariant subspace for …. To see this,suppose V is a nonzero invariant subspace and suppose V contains a vector not inthe span of e1, say, v D ae1 C be2 with b ¤ 0. Then

….1/v � v D be1

also belongs to V . Thus, e1 and e2 D .v � ae1/=b belong to V , showing thatV D C2. We conclude, then, that C2 does not decompose as a direct sum ofirreducible invariant subspaces. utProposition 4.26. If V is a completely reducible representation of a group or Liealgebra, then the following properties hold.

1. For every invariant subspace U of V , there is another invariant subspace Wsuch that V is the direct sum of U andW .

2. Every invariant subspace of V is completely reducible.

Proof. For Point 1, suppose that V decomposes as

V D U1 ˚ U2 ˚ � � � ˚ Uk;

where the Uj ’s are irreducible invariant subspaces, and that U is any invariantsubspace of V . If U is all of V , then we can take W D f0g and we are done. IfU ¤ V , there must be some j1 such that Uj1 is not contained in U . Since Uj1 isirreducible, it follows that the invariant subspaceUj1 \U must be f0g. Suppose nowthat U C Uj1 D V . If so, the sum is direct (since Uj1 \ U D f0g) and we are done.

If U C Uj1 ¤ V , there is some j2 such that U C Uj1 does not contain Uj2 ,in which case, .U C Uj1/ \ Uj2 D f0g. Proceeding on in the same way, we musteventually obtain some family j1; j2; : : : ; jl such that U CUj1 C� � �CUjl D V andthe sum is direct. ThenW WD Uj1 C � � � C Ujl is the desired complement to U .

For Point 2, suppose U is an invariant subspace of V . We first establish that Uhas the “invariant complement property” in Point 1. Suppose, then, thatX is anotherinvariant subspace of V with X � U . By Point 1, we can find invariant subspace Y

92 4 Basic Representation Theory

such that V D X ˚ Y . Let Z D Y \ U , which is then an invariant subspace. Wewant to show that U D X ˚Z. For all u 2 U , we can write u D xC y with x 2 Xand y 2 Y . But since X � U , we have x 2 U and therefore y D u � x 2 U . Thus,y 2 Z D Y \U . We have shown, then, that every u 2 U can be written as the sumof an element of X and an element of Z. Furthermore,X \Z � X \ Y D f0g, soactually U is the direct sum of X and Z.

We may now easily show that U is completely reducible. If U is irreducible, weare done. If not, U has a nontrivial invariant subspace X and thus U decomposesas U D X ˚ Z for some invariant subspace Z. If X and Z are irreducible, weare done, and if not, we proceed on in the same way. Since U is finite dimensional,this process must eventually terminate with U being decomposed as a direct sum ofirreducibles. utProposition 4.27. If G is a matrix Lie group and … is a finite-dimensional unitaryrepresentation of G, then … is completely reducible. Similarly, if g is a real Liealgebra and � is a finite-dimensional “unitary”representation of g (meaning that�.X/� D ��.X/ for all X 2 g), then � is completely reducible.

Proof. Let V denote the Hilbert space on which … acts and let h�; �i denote theinner product on V . If W � V is an invariant subspace, let W ? be the orthogonalcomplement of W , so that V is the direct sum of W and W ?. We claim that W ? isalso an invariant subspace for … or � .

To see this, note that since … is unitary, ….A/� D ….A/�1 D ….A�1/ for allA 2 G. Then, for any w 2 W and any v 2 W ?, we have

h….A/v;wi D hv;….A/�wi D ˝v;….A�1/w

˛D ˝v;w0˛ D 0:

In the last step, we have used that w0 D ….A�1/w is in W , since W is invariant.This shows that ….A/v is orthogonal to every element of W , as claimed. A similarargument, with….A�1/ replaced by ��.X/, shows that the orthogonal complementof an invariant subspace for � is also invariant.

We have established, then, that for a unitary representation, the orthogonalcomplement of an invariant subspace is again invariant. Suppose now that V is notirreducible. Then we can find an invariant subspaceW that is neither f0g nor V , andwe decompose V as W ˚W ?. Then W and W ? are both invariant subspaces andthus unitary representations of G in their own right. ThenW is either irreducible orit splits as an orthogonal direct sum of invariant subspaces, and similarly for W ?.We continue this process, and since V is finite dimensional, it cannot go on forever,and we eventually arrive at a decomposition of V as a direct sum of irreducibleinvariant subspaces. utTheorem 4.28. If G is a compact matrix Lie group, every finite-dimensionalrepresentation of G is completely reducible.

See also Sect. 10.3 for a similar result for semisimple Lie algebras. The argumentbelow is sometimes called “Weyl’s unitarian trick” for the role of unitarity in the

4.4 Complete Reducibility 93

proof. We require a notion of integration over matrix Lie groups that is invariantunder the right action of the group. One way to construct such a right-invariantintegral is to construct a right-invariant measure on G, known as a Haar measure.It is, however, simpler to introduce the integral by means of a right-invariantdifferential form on G. (See Appendix B for a quick introduction to the notion ofdifferential forms.)

IfG � Mn.C/ is a matrix Lie group, then the tangent space toG at the identity isthe Lie algebra g of G (Corollary 3.46). It is then easy to see that the tangent spaceTAG at any point A 2 G is the space of vectors of the form XA with X 2 g. Ifthe dimension of g as a real vector space is k, choose a nonzero k-linear, alternatingform ˛I W gk ! R. (Such a form exists and is unique up to multiplication by aconstant.) Then we may define a k-linear, alternating form ˛A W .TAG/k ! R bysetting

˛A.Y1; : : : ; Yk/ D ˛I .Y1A�1; : : : ; YkA�1/

for all Y1; : : : ; Yk 2 TAG. That is to say, we define ˛ in an arbitrary nonzero fashionat the identity, and we then use the right action of G to “transport” ˛ to every otherpoint in G. The resulting family of functionals is a k-form on G.

Once such an ˛ has been constructed, we can use it to construct an orientationon G, by decreeing that an ordered basis Y1; : : : ; Yk for TAG is positively orientedif ˛A.Y1; : : : ; Yk/ > 0. If f W G ! R is a smooth function, we can integrate thek-form f ˛ over nice domains in G. If G is compact, we may f ˛ integrate over allof G, leading to a notion of integration, which we denote as

ZG

f .A/˛.A/:

Since the orientation on G was defined in terms of the k-form ˛ itself, it is nothard to see that if f .A/ > 0 for all A, then

RGf ˛ > 0. Furthermore, since the form

˛ was constructed using the right action of G, it is easily seen to be invariant underthat action. As a result, the notion of integration of a function over a compact groupG is invariant under the right action of G: For all B 2 G, we have

ZG

f .AB/˛.A/ DZG

f .A/˛.A/:

Proof of Theorem 4.28. Choose an arbitrary inner product h�; �i on V , and thendefine a map h�; �iG W V � V ! C by the formula

hv;wiG DZG

h….A/v;….A/wi ˛.A/:

It is easy to check that h�; �iG is an inner product; in particular, the positivity of h�; �iGholds because h….A/v;….A/vi > 0 for all A if v ¤ 0. We now compute that foreach B 2 G, we have

94 4 Basic Representation Theory

h….B/v;….B/wiG DZG

h….A/….B/v;….A/….B/wi˛.A/

DZG

h….AB/v;….AB/wi ˛.A/

DZG

h….A/v;….A/wi ˛.A/

D hv;wiG ;

where we have used the right invariance of the integral in the third equality. Thiscomputation shows that for each B 2 G, the operator….B/ is unitary with respectto h�; �iG . Thus, by Proposition 4.27,… is completely reducible. ut

Note that compactness of the group G is needed to ensure that the integraldefining h�; �iG is convergent.

4.5 Schur’s Lemma

It is desirable to be able to state Schur’s lemma simultaneously for groups and Liealgebras. In order to do so, we need to indulge in a common abuse of notation.If, say, … is a representation of G acting on a space V , we will refer to V as therepresentation, without explicit reference to ….

Theorem 4.29 (Schur’s Lemma).

1. Let V and W be irreducible real or complex representations of a group or Liealgebra and let � W V ! W be an intertwining map. Then either � D 0 or � isan isomorphism.

2. Let V be an irreducible complex representation of a group or Lie algebra andlet � W V ! V be an intertwining map of V with itself. Then � D �I , for some� 2 C.

3. Let V and W be irreducible complex representations of a group or Lie algebraand let �1; �2 W V ! W be nonzero intertwining maps. Then �1 D ��2, forsome � 2 C.

It is important to note that the last two points in the theorem hold only over C (orsome other algebraically closed field) and not over R. See Exercise 8.

Before proving Schur’s lemma, we obtain two corollaries of it.

Corollary 4.30. Let … be an irreducible complex representation of a matrix Liegroup G. If A is in the center of G, then ….A/ D �I , for some � 2 C. Similarly,if � is an irreducible complex representation of a Lie algebra g and if X is in thecenter of g, then �.X/ D �I .

4.5 Schur’s Lemma 95

Proof. We prove the group case; the proof of the Lie algebra case is similar. If A isin the center of G, then for all B 2 G,

….A/….B/ D ….AB/ D ….BA/ D ….B/….A/.

However, this says exactly that….A/ is an intertwining map of the space with itself.Thus, by Point 2 of Schur’s lemma, ….A/ is a multiple of the identity. ut

Corollary 4.31. An irreducible complex representation of a commutative group orLie algebra is one dimensional.

Proof. Again, we prove only the group case. If G is commutative, the center ofG is all of G, so by the previous corollary ….A/ is a multiple of the identity foreach A 2 G. However, this means that every subspace of V is invariant! Thus,the only way that V can fail to have a nontrivial invariant subspace is if it is onedimensional. ut

We now provide the proof of Schur’s lemma.

Proof of Theorem 4.29. As usual, we will prove just the group case; the proof ofthe Lie algebra case requires only the obvious notational changes. For Point 1, ifv 2 ker.�/, then

�.….A/v/ D †.A/�.v/ D 0.

This shows that ker� is an invariant subspace of V . Since V is irreducible, we musthave ker� D 0 or ker� D V . Thus, � is either one-to-one or zero.

Suppose � is one-to-one. Then the image of � is a nonzero subspace of W . Onthe other hand, the image of � is invariant, for if w 2 W is of the form �.v/ forsome v 2 V , then

†.A/w D †.A/�.v/ D �.….A/v/.

Since W is irreducible and image.V / is nonzero and invariant, we must haveimage.V / D W . Thus, � is either zero or one-to-one and onto.

For Point 2, suppose V is an irreducible complex representation and that � WV ! V is an intertwining map of V to itself, that is that �….A/ D ….A/� forall A 2 G. Since we are working over an algebraically closed field, � must haveat least one eigenvalue � 2 C. If U denotes the corresponding eigenspace for �,then Proposition A.2 tells us that each….A/ maps U to itself, meaning that U is aninvariant subspace. Since � is an eigenvalue, U ¤ 0, and so we must have U D V ,which means that � D �I on all of V .

For Point 3, if �2 ¤ 0, then by Point 1, �2 is an isomorphism. Then �1 ı ��12

is an intertwining map of W with itself. Thus, by Point 2, �1 ı ��12 D �I , whence

�1 D ��2. ut

96 4 Basic Representation Theory

4.6 Representations of sl.2IC/

In this section, we will compute (up to isomorphism) all of the finite-dimensionalirreducible complex representations of the Lie algebra sl.2IC/. This computation isimportant for several reasons. First, sl.2IC/ is the complexification of su.2/, whichin turn is isomorphic to so.3/, and the representations of so.3/ are of physicalsignificance. Indeed, the computation we will perform in the proof of Theorem 4.32is found in every standard textbook on quantum mechanics, under the heading“angular momentum.” Second, the representation theory of su.2/ is an illuminatingexample of how one uses commutation relations to determine the representations ofa Lie algebra. Third, in determining the representations of a semisimple Lie algebrag (Chapters 6 and 7), we will make frequent use of the representation theory ofsl.2IC/, applying it to various subalgebras of g that are isomorphic to sl.2IC/.

We use the following basis for sl.2IC/:

X D�0 1

0 0

�I Y D

�0 0

1 0

�I H D

�1 0

0 �1�;

which have the commutation relations

ŒH;X� D 2X;

ŒH; Y � D �2Y;ŒX; Y � D H:

(4.11)

If V is a finite-dimensional complex vector space and A, B , and C are operators onV satisfying ŒA;B� D 2B , ŒA; C � D �2C , and ŒB; C � D A, then because of theskew symmetry and bilinearity of brackets, the unique linear map � W sl.2IC/ !gl.V / satisfying

�.H/ D A; �.X/ D B; �.Y / D C

will be a representation of sl.2IC/.Theorem 4.32. For each integer m � 0, there is an irreducible complex represen-tation of sl.2IC/with dimensionmC1. Any two irreducible complex representationsof sl.2IC/ with the same dimension are isomorphic. If � is an irreducible complexrepresentation of sl.2IC/ with dimension m C 1, then � is isomorphic to therepresentation �m described in Sect. 4.2.

Our goal is to show that any finite-dimensional irreducible representation ofsl.2IC/ “looks like” one of the representations �m coming from Example 4.10. Inthat example, the space Vm is spanned by eigenvectors for �m.H/ and the operators�m.X/ and �m.Y / act by shifting the eigenvalues up or down in increments of 2.We now introduce a simple but crucial lemma that allows us to develop a similarstructure in an arbitrary irreducible representation of sl.2IC/.

4.6 Representations of sl.2IC/ 97

Lemma 4.33. Let u be an eigenvector of �.H/ with eigenvalue ˛ 2 C. Then wehave

�.H/�.X/u D .˛ C 2/�.X/u.

Thus, either �.X/u D 0 or �.X/u is an eigenvector for �.H/ with eigenvalue˛ C 2. Similarly,

�.H/�.Y /u D .˛ � 2/�.Y /u;

so that either �.Y /u D 0 or �.Y /u is an eigenvector for �.H/ with eigenvalue˛ � 2.

Proof. We know that Œ�.H/; �.X/� D � .ŒH;X�/ D 2�.X/. Thus,

�.H/�.X/u D �.X/�.H/u C 2�.X/u

D �.X/ .˛u/C 2�.X/u

D .˛ C 2/�.X/u:

The argument with �.X/ replaced by �.Y / is similar. utProof of Theorem 4.32. Let � be an irreducible representation of sl.2IC/ actingon a finite-dimensional complex vector space V . Our strategy is to diagonalize theoperator �.H/. Since we are working over C, the operator �.H/ must have at leastone eigenvector. Let u be an eigenvector for �.H/ with eigenvalue ˛. ApplyingLemma 4.33 repeatedly, we see that

�.H/�.X/ku D .˛ C 2k/�.X/ku:

Since operator on a finite-dimensional space can have only finitely many eigenval-ues, the �.X/ku’s cannot all be nonzero. Thus, there is some N � 0 such that

�.X/N u ¤ 0

but

�.X/NC1u D 0.

If we set u0 D �.X/N u and � D ˛ C 2N , then,

�.H/u0 D �u0; (4.12)

�.X/u0 D 0: (4.13)

98 4 Basic Representation Theory

Let us then define

uk D �.Y /ku0

for k � 0. By Lemma 4.33, we have

�.H/uk D .� � 2k/ uk . (4.14)

Now, it is easily verified by induction (Exercise 3) that

�.X/uk D kŒ� � .k � 1/�uk�1 .k � 1/: (4.15)

Furthermore, since �.H/ can have only finitely many eigenvalues, the uk’s cannotall be nonzero. There must, therefore, be a non-negative integerm such that

uk D �.Y /ku0 ¤ 0

for all k � m, but

umC1 D �.Y /mC1u0 D 0.

If umC1 D 0, then �.X/umC1 D 0 and so, by (4.15),

0 D �.X/umC1 D .mC 1/.��m/um:

Since um and mC 1 are nonzero, we must have � �m D 0. Thus, � must coincidewith the non-negative integerm.

Thus, for every irreducible representation .�; V /, there exists an integer m � 0

and nonzero vectors u0; : : : ; um such that

�.H/uk D .m � 2k/uk

�.Y /uk D�

ukC1 if k < m0 if k D m

�.X/uk D�k.m � .k � 1//uk�1 if k > 0

0 if k D 0: (4.16)

The vectors u0; : : : ; um must be linearly independent, since they are eigenvectorsof �.H/ with distinct eigenvalues (Proposition A.1). Moreover, the .m C 1/-dimensional span of u0; : : : ; um is explicitly invariant under �.H/, �.X/, and �.Y /and, hence, under �.Z/ for all Z 2 sl.2IC/. Since � is irreducible, this space mustbe all of V .

We have shown that every irreducible representation of sl.2IC/ is of theform (4.16). Conversely, if we define �.H/, �.X/, and �.Y / by (4.16) (wherethe uk’s are basis elements for some .m C 1/-dimensional vector space), it is

4.6 Representations of sl.2IC/ 99

not hard to check that operators defined as in (4.16) really do satisfy the sl.2IC/commutation relations (Exercise 4). Furthermore, we may prove irreducibility ofthis representation in the same way as in the proof of Proposition 4.11.

The preceding analysis shows that every irreducible representation of dimensionmC 1 must have the form in (4.16), which shows that any two such representationsare isomorphic. In particular, the .mC 1/-dimensional representation �m describedin Sect. 4.2 must be isomorphic to (4.16).

This completes the proof of Theorem 4.32. utAs mentioned earlier in this section, the representation theory of sl.2IC/ plays a

key role in the representation theory of other Lie algebras, such as sl.3IC/, becausethese Lie algebras contain subalgebras isomorphic to sl.2IC/. For such applications,we need a few results about finite-dimensional representations of sl.2IC/ that arenot necessarily irreducible.

Theorem 4.34. If .�; V / is a finite-dimensional representation of sl.2IC/, notnecessarily irreducible, the following results hold.

1. Every eigenvalue of �.H/ is an integer. Furthermore, if v is an eigenvector for�.H/ with eigenvalue � and �.X/v D 0, then � is a non-negative integer.

2. The operators �.X/ and �.Y / are nilpotent.3. If we define S W V ! V by

S D e�.X/e��.Y /e�.X/;

then S satisfies

S�.H/S�1 D ��.H/:

4. If an integer k is an eigenvalue for �.H/, so is each of the numbers

� jkj ;� jkj C 2; : : : ; jkj � 2; jkj :

Since SU.2/ is simply connected, Theorem 5.6 will tell us that the repre-sentations of sl.2IC/ Š su.2/C are in one-to-one correspondence with therepresentations of SU.2/. Since SU.2/ is compact, Theorem 4.28 then tells usthat every representation of sl.2IC/ is completely reducible. If we decompose Vas a direct sum of irreducibles, it is easy to prove the theorem for each summandseparately. It is, however, preferable to give a proof of the theorem that does not relyon Theorem 5.6, which in turn relies on the Baker–Campbell–Hausdorff formula.

See also Exercise 13 for a different approach to the first part of Point 1, andExercise 14 for a different approach to Point 3.

Proof. For Point 1, suppose v is an eigenvector of �.H/ with eigenvalue �. Thenthere is some N � 0 such that �.X/N v ¤ 0 but �.X/NC1v D 0, where �.X/Nvis an eigenvector of �.H/ with eigenvalue � C 2N . The proof of Theorem 4.32shows that m WD �C 2N must be a non-negative integer, so that � is an integer. If�.X/v D 0 then we take N D 0 and � D m is non-negative.

100 4 Basic Representation Theory

For Point 2, it follows from the SN decomposition (Sect. A.3) that �.H/ has abasis of generalized eigenvectors, that is, vectors v for which .�.H/ � �I/kv D 0

for some positive integer k. But, using the commutation relation ŒH;X� D 2X andinduction on k, we can see that

Œ�.H/ � .�C 2/I �k�.X/ D �.X/Œ�.H/ � �I �k:

Thus, if v is a generalized eigenvector for �.H/ with eigenvalue �, then �.X/vis either zero or a generalized eigenvector with eigenvalue � C 2. Applying �.X/repeatedly to a generalized eigenvector for �.H/ must eventually give zero, since�.H/ can have only finitely many generalized eigenvalues. Thus, �.X/ is nilpotent.A similar argument applies to �.Y /.

For Point 3, we note that

S�.H/S�1 D e�.X/e��.Y /e�.X/�.H/e��.X/e�.Y /e��.X/: (4.17)

By Proposition 3.35, we have

e�.X/�.H/e��.X/ D Ade�.X/ .�.H// D ead�.X/ .�.H//

and similarly for the remaining products in (4.17).Now, adX.X/ D 0, adX.H/ D �2X and adX.Y / D H , and similarly with �

applied to each Lie algebra element. Thus,

ead�.X/ .�.H// D �.H/ � 2�.X/:

Meanwhile, adY .X/ D �H , adY .H/ D 2Y , and adY .Y / D 0. Thus,

e�ad�.Y / .�.H/ � 2�.X//

D �.H/ � 2�.X/� 2�.Y /� 2�.H/C 1

24�.Y /

D ��.H/ � 2�.X/:

Finally,

ead�.X/ .��.H/� 2�.X// D ��.H/ � 2�.X/C 2�.X/

D ��.H/;

which establishes Point 3.For Point 4, assume first that k is non-negative and let v be an eigenvector for

�.H/ with eigenvalue k. Then as in Point 1, there is then another eigenvector v0for �.H/ with eigenvalue m WD k C 2N � k and such that �.X/v0 D 0. Thenby the proof of Theorem 4.32, we obtain a chain of eigenvectors v0; v1; : : : ; vm for

4.7 Group Versus Lie Algebra Representations 101

�.H/with eigenvalues ranging fromm to �m in increments of 2. These eigenvaluesinclude all of the numbers k; k � 2; : : : ;�k. If k is negative and v is an eigenvectorfor �.H/ with eigenvalue k, then Sv is an eigenvector for �.H/ with eigenvaluejkj. Hence, by the preceding argument, each of the numbers from jkj to � jkj inincrements of 2 is an eigenvalue. ut

4.7 Group Versus Lie Algebra Representations

We know from Chapter 3 (Theorem 3.28) that every Lie group homomorphism givesrise to a Lie algebra homomorphism. In particular, every representation of a matrixLie group gives rise to a representation of the associated Lie algebra. In Chapter 5,we will prove a partial converse to this result: If G is a simply connected matrixLie group with Lie algebra g, every representation of g comes from a representationof G. (See Theorem 5.6). Thus, for a simply connected matrix Lie group G, thereis a natural one-to-one correspondence between the representations of G and therepresentations of g.

It is instructive to see how this general theory works out in the case of SU.2/(which is simply connected) and SO.3/ (which is not). For every irreduciblerepresentation � of su.2/, the complex-linear extension of � to sl.2IC/ mustbe isomorphic (Theorem 4.32) to one of the representations �m described inSect. 4.2. Since those representations are constructed from representations of thegroup SU.2/, we can see directly (without appealing to Theorem 5.6) that everyirreducible representation of su.2/ comes from a representation of SU.2/.

Now, by Example 3.27, there is a Lie algebra isomorphism � W su.2/ ! so.3/such that �.Ej / D Fj , j D 1; 2; 3, where fE1;E2;E3g and fF1; F2; F3g arethe bases listed in the example. Thus, the irreducible representations of so.3/ areprecisely of the form m D �m ı ��1. We wish to determine, for a particularm, whether or not there is a representation †m of the group SO.3/ such that†m.expX/ D exp.m.X// for all X in so.3/.

Proposition 4.35. Let m D �m ı ��1 be an irreducible complex representation ofthe Lie algebra so.3/ (m � 0). If m is even, there is a representation †m of thegroup SO.3/ such that†m.eX/ D em.X/ for all X in so.3/. Ifm is odd, there is nosuch representation of SO.3/.

Note that the condition that m be even is equivalent to the condition thatdimVm D m C 1 be odd. Thus, it is the odd-dimensional representations of theLie algebra so.3/ which come from group representations. In the physics literature,the representations of su.2/ Š so.3/ are labeled by the parameter l D m=2. Interms of this notation, a representation of so.3/ comes from a representation ofSO.3/ if and only if l is an integer. The representations with l an integer are called“integer spin”; the others are called “half-integer spin.”

For any m, one could attempt to construct †m by the construction in the proofof Theorem 5.6. The construction is based on defining †m.A/ along a path joining

102 4 Basic Representation Theory

I toA and then proving that the value of†m.A/ is independent of the choice of path.The construction of †m along a path goes through without change. Since SO.3/ isnot simply connected, however, two paths in SO.3/ with the same endpoint arenot necessarily homotopic with endpoints fixed and the proof of independence ofthe path breaks down. One can join the identity to itself, for example, either bythe constant path or by the path consisting of rotations by angle 2�t in the .y; z/-plane, 0 � t � 1. If one defines †m along the constant path, one gets the value†m.I / D I . If m is odd, however, and one defines †m along the path of rotationsin the .y; z/-plane, then one gets the value †m.I / D �I , as the calculations in theproof of Proposition 4.35 will show. This strongly suggests (and Proposition 4.35confirms) that when m is odd, there is no way to define †m as a “single-valued”representation of SO.3/.

An electron, for example, is a “spin one-half” particle, which means that itis described in quantum mechanics in a way that involves the two-dimensionalrepresentation 1 of so.3/. In the physics literature, one finds statements to the effectthat applying a 360ı rotation to the wave function of the electron gives back thenegative of the original wave function. This statement reflects that if one attempts toconstruct the nonexistent representation†1 of SO.3/, then when defining†1 alonga path of rotations in the .x; y/-plane, one gets that †1.I / D �I .

Proof. Suppose, first, thatm is odd and suppose that such a†m existed. Computingas in Sect. 2.2, we see that

e2�F1 D0@1 0 0

0 cos 2� � sin 2�0 sin 2� cos 2�

1A D I:

Meanwhile, m.F1/ D �m.��1.F1// D �m.E1/, withE1 D iH=2, where, as usual,

H is the diagonal matrix with diagonal entries .1;�1/. We know that there is abasis u0; u1; : : : ; um for Vm such that uj is an eigenvector for �m.H/with eigenvaluem � 2j . This means that uj is also an eigenvector for m.F1/ D i�m.H/=2, witheigenvalue i.m� 2j /=2. Thus, in the basis

˚uj�, we have

m.F1/ D

0BBB@

i2m

i2.m � 2/

: : :i2.�m/

1CCCA :

Since we are assuming m is odd, m � 2j is an odd integer. Thus, e2�m.F1/ haseigenvalues e2�i.m�2j /=2 D �1 in the basis

˚uj�, showing that e2�m.F1/ D �I .

Thus, on the one hand,

†m.e2�F1 / D †m.I / D I;

4.8 A Nonmatrix Lie Group 103

whereas, on the other hand,

†m.e2�F1 / D e2�m.F1/ D �I:

Thus, there can be no such group representation†m.Suppose now that m is even. Recall that the Lie algebra isomorphism � comes

from the surjective group homomorphism ˆ in Proposition 1.19, where ker.ˆ/ DfI;�I g. Let …m be the representation of SU.2/ in Example 4.10. Now, e2�E1 D�I , and, thus,

…m.�I / D …m.e2�E1/ D e�m.2�E1/:

If, howeverm is even, then e�m.2�E1/ is diagonal in the basis fuj g with eigenvaluese2�i.m�2j /=2 D 1, showing that …m.�I / D e�m.2�E1/ D I .

Now, for each R 2 SO.3/, there is a unique pair of elements fU;�U g such thatˆ.U / D ˆ.�U / D R. Since …m.�I / D I , we see that …m.U / D …m.�U /. Itthus makes sense to define

†m.R/ D …m.U /:

It is easy to see that†m is a homomorphism. To see that†m is continuous, the readermay first verify that if a homomorphism between matrix Lie groups is continuousat the identity, it is continuous everywhere. Thus, it suffices to observe that for Rnear the identity, U may be chosen to depend continuously on R; by setting U Dexp.��1.logR//. Now, by construction, we have …m D †m ı ˆ. Thus, the Liealgebra representation m associated to†m satisfies �m D mı� or m D �mı��1;showing that †m is the desired representation of SO.3/: ut

4.8 A Nonmatrix Lie Group

In this section, we will show that the Lie group introduced in Sect. 1.5 is notisomorphic to a matrix Lie group. (The universal cover of SL.2IR/ is anotherexample of a Lie group that is not a matrix Lie group; see Sect. 5.8.) The groupin question is G D R � R � S1, with the group product defined by

.x1; y1; u1/ � .x2; y2; u2/ D .x1 C x2; y1 C y2; eix1y2u1u2/:

Meanwhile, let H be the Heisenberg group and consider the map ˆ W H ! G

given by

ˆ

0@1 a b0 1 c

0 0 1

1A D .a; c; e2�ib/:

104 4 Basic Representation Theory

Direct computation shows that ˆ is a homomorphism. The kernel of ˆ is thediscrete normal subgroupN of H given by

N D8<:0@1 0 n0 1 0

0 0 1

1Aˇˇˇn 2 Z

9=; :

Now, suppose that † is any finite-dimensional representation of G. Then we candefine an associated representation … of H by … D † ı ˆ. Clearly, the kernelof any such representation of H must include the kernel N of ˆ. Now, let Z.H/denote the center of H , which is easily shown to be

Z.H/ D8<:0@1 0 b0 1 0

0 0 1

1Aˇˇˇ b 2 R

9=; :

Theorem 4.36. Let … be any finite-dimensional representation of H . If ker… N , then ker… Z.H/.

Once this is established, we will be able to conclude that there are no faithfulfinite-dimensional representations of G. After all, if † is any finite-dimensionalrepresentation of G, then the kernel of… D † ıˆ will containN and, thus,Z.H/,by the theorem. Thus, for all b 2 R,

0@1 0 b0 1 0

0 0 1

1A D †.0; 0; e2�ib/ D I:

This means that the kernel of † contains all elements of the form .0; 0; u/ and † isnot faithful. Thus, we obtain the following result.

Corollary 4.37. The Lie group G has no faithful finite-dimensional representa-tions. In particular, G is not isomorphic to any matrix Lie group.

We now begin the proof of Theorem 4.36.

Lemma 4.38. If X is a nilpotent matrix and etX D I for some nonzero t , thenX D 0.

Proof. SinceX is nilpotent, the power series for etX terminates after a finite numberof terms. Thus, each entry of etX depends polynomially on t ; that is, there existpolynomials pjk.t/ such that .etX/jk D pjk.t/. If etX D I for some nonzero t , thenentX D I for all n 2 Z, showing that pjk.nt/ D ıjk for all n. However, a polynomialthat takes on a certain value infinitely many times must be constant. Thus, actually,etX D I for all t , from which it follows that X D 0. ut

4.9 Exercises 105

Proof of Theorem 4.36. Let � be the associated representation of the Lie algebra hof H . Let fX; Y;Zg be the following basis for h:

X D0@0 1 00 0 0

0 0 0

1A ; Y D

0@0 0 00 0 1

0 0 0

1A ; Z D

0@0 0 10 0 0

0 0 0

1A : (4.18)

These satisfy the commutation relations ŒX; Y � D Z and ŒX;Z� D ŒY;Z� D 0.We now claim that �.Z/ must be nilpotent, or, equivalently, that all of the

eigenvalues of �.Z/ are zero. Let � be an eigenvalue for �.Z/ and let V� be theassociated eigenspace. Then V� is certainly invariant under �.Z/. Furthermore,since �.X/ and �.Y / commute with �.Z/, Proposition A.2 tells us that V� isinvariant under �.X/ and �.Y /. Thus, the restriction of �.Z/ to V�—namely, �I—is the commutator of the restrictions to V� of �.X/ and �.Y /. Since the trace of acommutator is zero, we have 0 D � dim.V�/ and � must be zero.

Now, direct calculation shows that enZ belongs to N for all integers n. Thus, if… is a representation of H and ker… N , we have ….enZ/ D I for all n. Since�.Z/ is nilpotent, Lemma 4.38 tells us that �.Z/ is zero and thus that ….etZ/ Det�.Z/ D I for all t 2 R. Since every element of Z.H/ is of the form etZ for somet , we have the desired conclusion. ut

4.9 Exercises

1. Prove Point 2 of Proposition 4.5.2. Show that the adjoint representation and the standard representation are

isomorphic representations of the Lie algebra so.3/. Show that the adjoint andstandard representations of the group SO.3/ are isomorphic.

3. Using the commutation relation Œ�.X/; �.Y /� D �.H/ and induction on k,verify the relation (4.15).

4. Define a vector space with basis u0; u1; : : : ; um. Now, define operators �.H/,�.X/, and �.Y / by formula (4.16). Verify by direct computation that the oper-ators defined by (4.16) satisfy the commutation relations (4.11) for sl.2IC/.Hint: When dealing with �.Y /, treat the case of uk, k < m, separately from thecase of um, and similarly for �.X/.

5. Consider the standard representation of the Heisenberg group, acting on C3.Determine all subspaces of C3 which are invariant under the action of theHeisenberg group. Is this representation completely reducible?

6. Prove Proposition 4.22.Hint: There is a one-to-one correspondence between subspaces of V andsubspaces of V � as follows: For any subspaceW of V , the annihilator ofW isthe subspace of all � in V � such that � is zero on W . See Sect. A.7.

7. Prove Theorem 4.14.

106 4 Basic Representation Theory

Hints: For existence, choose bases˚ej�

and ffkg for U and V . Thendefine a space W which has as a basis fwjk j1 � j � n; 1 � k � mg. Define�.ej ; fk/ D wjk and extend by bilinearity. For uniqueness, use the universalproperty.

8. Let SO.2/ act on R2 in the obvious way. Show that R

2 is an irreduciblereal representation under this action, but that Point 2 of Schur’s lemma(Theorem 4.29) fails.

9. Suppose V is a finite-dimensional representation of a group or Lie algebra andthat W is a nonzero invariant subspace of V . Show that there exists a nonzeroirreducible invariant subspace for V that is contained in W .

10. Suppose that V1 and V2 are nonisomorphic irreducible representations of agroup or Lie algebra, and consider the associated representation V1 ˚ V2.Regard V1 and V2 as subspaces of V1 ˚ V2 in the obvious way. Following theoutline below, show that V1 and V2 are the only nontrivial invariant subspacesof V1 ˚ V2.

(a) First assume that U is a nontrivial irreducible invariant subspace. Let P1 WV1 ˚ V2 ! V1 be the projection onto the first factor and let P2 be theprojection onto the second factor. Show that P1 and P2 are intertwiningmaps. Show that U D V1 or U D V2.

(b) Using Exercise 9, show that V1 and V2 are the only nontrivial invariantsubspaces of V1 ˚ V2.

11. Suppose that V is an irreducible finite-dimensional representation of a group orLie algebra over C, and consider the associated representation V ˚ V . Showthat every nontrivial invariant subspace U of V ˚ V is isomorphic to V and isof the form

U D f.�1v; �2v/jv 2 V g

for some constants �1 and �2, not both zero.12. Recall the spaces Vm introduced in Sect. 4.2, viewed as representations of

the Lie algebra sl.2IC/. In particular, consider the space V1 (which hasdimension 2).

(a) Regard V1 ˝ V1 as a representation of sl.2IC/, as in Definition 4.20. Showthat this representation is not irreducible.

(b) Now, view V1 ˝ V1 as a representation of sl.2IC/ ˚ sl.2IC/, as inDefinition 4.19. Show that this representation is irreducible.

13. Let .…; V / be a finite-dimensional representation of SU.2/with associated rep-resentation � of su.2/, which extends by complex linearity to sl.2IC/. (SinceSU.2/ is simply connected, Theorem 5.6 will show that every representation� of sl.2IC/ arise in this way.) If H is the diagonal matrix with diagonalentries .1;�1/, show that e2�iH D I and use this to prove (independently ofTheorem 4.34) that every eigenvalue of �.H/ is an integer.

4.9 Exercises 107

14. Let .…; V / be a finite-dimensional representation of SU.2/ with associatedrepresentation � of su.2/, which extends by complex linearity to sl.2IC/. IfX , Y , and H are the usual basis element for sl.2IC/, compute eXe�Y eX andshow that

eXe�Y eXH.eXe�Y eX/�1 D �H:

Use this result to give a different proof of Point 3 of Theorem 4.34.

Chapter 5The Baker–Campbell–Hausdorff Formulaand Its Consequences

5.1 The “Hard” Questions

Consider three elementary results from the preceding chapters of this book: (1)Every matrix Lie group G has a Lie algebra g. (2) A continuous homomorphismˆ between matrix Lie groups G and H gives rise to a Lie algebra homomorphism� W g ! h. (3) If G and H are matrix Lie groups and H is a subgroup of G,then the Lie algebra h ofH is a subalgebra of the Lie algebra g of G. Each of theseresults goes in the “easy” direction, from a group notion to an associated Lie algebranotion. In this chapter, we attempt to go in the “hard” direction, from the Lie algebrato the Lie group. We will investigate three questions relating to the preceding threetheorems.

• Question 1: Is every finite-dimensional, real Lie algebra the Lie algebra of somematrix Lie group?

• Question 2: Let G and H be matrix Lie groups with Lie algebras g and h,respectively, and let � W g ! h be a Lie algebra homomorphism. Does thereexists a Lie group homomorphismˆ W G ! H such that ˆ.eX/ D e�.X/ for allX 2 g?

• Question 3: If G is a matrix Lie group with Lie algebra g and h is a subalgebraof g, is there a matrix Lie groupH � G whose Lie algebra is h?

The answer to Question 1 is yes; see Sect. 5.10. The answer to Question 2 is,in general, no, but is yes if G is simply connected; see Sect. 5.7. The answer to

A previous version of this book was inadvertently published without the middle initial of theauthor’s name as “Brian Hall”. For this reason an erratum has been published, correcting themistake in the previous version and showing the correct name as Brian C. Hall (see DOI http://dx.doi.org/10.1007/978-3-319-13467-3_14). The version readers currently see is the correctedversion. The Publisher would like to apologize for the earlier mistake.

© Springer International Publishing Switzerland 2015B.C. Hall, Lie Groups, Lie Algebras, and Representations, GraduateTexts in Mathematics 222, DOI 10.1007/978-3-319-13467-3_5

109

110 5 The Baker–Campbell–Hausdorff Formula and Its Consequences

Question 3 is no, in general, but is yes if we allow H to be a “connected Liesubgroup” that is not necessarily closed; see Sect. 5.9.

Our tool for investigating these questions is the Baker–Campbell–Hausdorffformula, which expresses log.eXeY /, where X and Y are sufficiently small n � n

matrices, in Lie-algebraic terms, that is, in terms of iterated commutators involvingX and Y . The formula implies that all information about the product operation ona matrix Lie group, at least near the identity, is encoded in the Lie algebra. In thecase of Questions 2 and 3 in the preceding paragraph, we will give a complete proofof the theorem that answers the question. In the case of Question 1, we will need toassume Ado’s theorem, which asserts that every finite-dimensional real Lie algebrais isomorphic to an algebra of matrices.

5.2 An Illustrative Example

In this section, we prove one of the main theorems alluded to above (the answerto Question 2), in the case of the Heisenberg group. We introduce the problem ina general setting and then specialize to the Heisenberg case. Suppose G and Hare matrix Lie groups with Lie algebras g and h, respectively, and suppose � Wg ! h is a Lie algebra homomorphism. We would like to construct a Lie grouphomomorphism ˆ W G ! H such that ˆ.eX/ D e�.X/ for all X in g. In light ofTheorem 3.42, we can define a map ˆ from a neighborhood U of the identity in Ginto H as follows:

ˆ.A/ D e�.logA/; (5.1)

so that

ˆ.eX/ D e�.X/; (5.2)

at least for sufficiently small X .A key issue is then to show thatˆ, as defined near the identity by (5.1) or (5.2), is

a “local homomorphism.” Suppose A D eX and B D eY , whereX; Y 2 g are smallthat eX , eY , and eXeY are all in the domain of ˆ. To compute ˆ.AB/ D ˆ.eXeY /,we need to express eXeY in the form eZ , so that ˆ.eXeY / will equal e�.Z/. TheBaker–Campbell–Hausdorff formula states that for sufficiently small X and Y , wehave

Z D log.eXeY /

D X C Y C 12ŒX; Y �C 1

12ŒX; ŒX; Y �� � 1

12ŒY; ŒX; Y ��C � � � , (5.3)

where the “� � � ” refers to additional terms involving iterated brackets ofX and Y . (Aprecise statement of and a proof of the formula will be given in subsequent sections.)If � is a Lie algebra homomorphism, then

5.2 An Illustrative Example 111

�log.eXeY /

� D �.X/C �.Y /C 12Œ�.X/; �.Y /�

C 112Œ�.X/; Œ�.X/; �.Y /�� � 1

12Œ�.Y /; Œ�.X/; �.Y /��C � � �

D log.e�.X/e�.Y //: (5.4)

It then follows that ˆ.eXeY / D e�.X/e�.Y / D ˆ.eX/ˆ.eY /.In the general case, it requires considerable effort to prove the Baker–Campbell–

Hausdorff formula and then to prove that, when G is simply connected, ˆ canbe extended to all of G. In the case of the Heisenberg group (which is simplyconnected), the argument can be greatly simplified.

Theorem 5.1. Suppose X and Y are n � n complex matrices, and that X and Ycommute with their commutator:

ŒX; ŒX; Y �� D ŒY; ŒX; Y �� D 0. (5.5)

Then we have

eXeY D eXCYC 12ŒX;Y �.

This is the special case of (5.3) in which the series terminates after the ŒX; Y �term. See Exercise 5 for another special case of the Baker–Campbell–Hausdorffformula.

Proof. Consider X and Y in Mn.C/ satisfying (5.5). We will prove that

etXetY D exp

�tX C tY C t2

2ŒX; Y �

�,

which reduces to the desired result in the case t D 1. Since, by assumption, ŒX; Y �commutes with X and Y , the above relation is equivalent to

etXetYe� t2

2ŒX;Y � D et.XCY /. (5.6)

Let us denote byA.t/ the left-hand side of (5.6) and byB.t/ the right-hand side. Ourstrategy will be to show that A .t/ and B .t/ satisfy the same differential equation,with the same initial conditions. We can see immediately that

dB

dtD B.t/.X C Y /:

On the other hand, differentiating A.t/ by means of the product rule gives

dA

dtD etXXetYe� t2

2ŒX;Y � C etXetYYe� t2

2ŒX;Y �

C etXetYe� t2

2ŒX;Y �.�t ŒX; Y �/. (5.7)

112 5 The Baker–Campbell–Hausdorff Formula and Its Consequences

(The correctness of the last term may be verified by differentiating e�t 2ŒX;Y �=2 termby term.) Now, since Y commutes with ŒX; Y �, it also commute with e�t 2ŒX;Y �=2.Thus, the second term on the right in (5.7) can be rewritten as

etXetYe� t2

2ŒX;Y �Y .

For the first term on the right-hand side of (5.7), we compute, using Proposition 3.35,that

XetY D etYe�tYXetY

D etYAde�tY .X/

D etYe�tadY .X/:

Since ŒY; ŒY;X�� D � ŒY; ŒX; Y �� D 0, we have

e�tadY .X/ D X � t ŒY;X� D X C t ŒX; Y � ;

with all higher terms being zero. We may then simplify (5.7) to

dA

dtD etXetYe� t2

2ŒX;Y �.X C Y / D A.t/.X C Y /.

We see, then, that A.t/ and B.t/ satisfy the same differential equation, with thesame initial condition A.0/ D B.0/ D I . Thus, by standard uniqueness results for(linear) ordinary differential equations, A.t/ D B.t/ for all t . utTheorem 5.2. Let H denote the Heisenberg group and h its Lie algebra. Let Gbe a matrix Lie group with Lie algebra g and let � W h ! g be a Lie algebrahomomorphism. Then there exists a unique Lie group homomorphism ˆ W H ! G

such that

ˆ.eX/ D e�.X/

for all X 2 h.

Proof. Recall (Exercise 18 in Chapter 3) that the Heisenberg group has the specialproperty that its exponential map is one-to-one and onto. Let “log” denote theinverse of this map and defineˆ W H ! G by the formula

ˆ.A/ D e�.logA/.

We will show that ˆ is a group homomorphism.If X and Y are in the Lie algebra of the Heisenberg group (3 � 3 strictly upper

triangular matrices), direct computation shows that every entry of ŒX; Y � is zeroexcept possibly for the entry in the upper right corner. It is then easily seen that

5.3 The Baker–Campbell–Hausdorff Formula 113

ŒX; Y � commutes with both X and Y . Since � is a Lie algebra homomorphism,� .X/ and � .Y / will also commute with their commutator. Thus, by Theorem 5.1,for any X and Y in the Lie algebra of the Heisenberg group, we have

ˆeXeY

� D ˆ�eXCYC 1

2ŒX;Y �

D e�.X/C�.Y /C 12 Œ�.X/;�.Y /�

D e�.X/e�.Y /

D ˆ.eX/ˆ.eY /:

Thus, ˆ is a group homomorphism, which is continuous because each of exp, log,and � is continuous. ut

5.3 The Baker–Campbell–Hausdorff Formula

The goal of the Baker–Campbell–Hausdorff formula (BCH formula) is to computelog.eXeY /. One may well ask, “Why do we not simply expand both exponentialsand the logarithm in power series and multiply everything out?” While it is possibleto do this, what is not clear is why the answer is expressible in terms of commutators.Consider, for example, the quadratic term in the expression for log.eXeY /, whichwill be a linear combination of X2, Y 2, XY, and YX. For this term to be expressiblein terms of commutators, it must be a multiple of XY � YX. Although directcomputation verifies that this is, indeed, the case, it is far from obvious how toprove that a similar result occurs for all the higher terms.

We will actually state and prove an integral form of the BCH formula, ratherthan the series form (5.3). The integral version of the formula, along with theargument we present in Sect. 5.5, is actually due to Poincaré. (See [Poin1, Poin2]and Section 1.1.2.2 of [BF].)

Consider the function

g.z/ D log z

1 � 1z

;

which is defined and holomorphic in the disk fjz � 1j < 1g. Thus, g.z/ can beexpressed as a series

g.z/ D1XmD0

am.z � 1/m,

with radius of convergence one. If V is a finite-dimensional vector space, we mayidentify V with C

n by means of an arbitrary basis, so that the Hilbert–Schmidt norm

114 5 The Baker–Campbell–Hausdorff Formula and Its Consequences

(Definition 2.2) of a linear operator on V can be defined. For any operator A on Vwith kA� Ik < 1, we can define

g.A/ D1XmD0

am.A� I /m.

We are now ready to state the integral form of the BCH formula.

Theorem 5.3 (Baker–Campbell–Hausdorff). For all n � n complex matrices Xand Y with kXk and kY k sufficiently small, we have

log.eXeY / D X CZ 1

0

g.eadX etadY /.Y / dt. (5.8)

The proof of this theorem is given in Sect. 5.5 of this chapter. Note that eadX etadY

and, hence, also g.eadX etadY / are linear operators on the space Mn.C/ of all n � ncomplex matrices. In (5.8), this operator is being applied to the matrix Y . The factthat X and Y are assumed small guarantees that eadX etadY is close to the identityoperator on Mn.C/ for 0 � t � 1, so that g.eadX etadY / is well defined. Althoughthe right-hand side of (5.8) is rather complicated to compute explicitly, we are notso much interested in the details of the formula but in the fact that it expresseslog.eXeY / (and hence eXeY ) in terms of the Lie-algebraic quantities adX and adY .

5.4 The Derivative of the Exponential Map

In this section we prove a result that is useful in its own right and will play a keyrole in our proof of the BCH formula. Consider the directional derivative of exp ata point X in the direction of Y :

d

dteXCtY

ˇˇtD0

. (5.9)

Unless X and Y commute, this derivative may not equal eXY . Nevertheless, sinceexp is continuously differentiable (Proposition 2.16), the directional derivativesin (5.9) will depend linearly on Y with X fixed. Now, the function .1 � e�z/=zis an entire function of z (with a removable singularity at the origin) and is given bythe power series

1 � e�z

zD

1XkD0.�1/k zk

.k C 1/Š;

which has infinite radius of convergence. It thus makes sense to replace z by anarbitrary linear operator A on a finite-dimensional vector space.

5.4 The Derivative of the Exponential Map 115

Theorem 5.4 (Derivative of Exponential). For all X; Y 2 Mn.C/, we have

d

dteXCtY

ˇˇtD0

D eX�I � e�adX

adX.Y /

D eX�Y � ŒX; Y �

2ŠC ŒX; ŒX; Y ��

3Š� � � �

�. (5.10)

More generally, if X .t/ is a smooth matrix-valued function, then

d

dteX.t/ D eX.t/

�I � e�adX.t/

adX.t/

�dX

dt

��. (5.11)

Our proof follows [Tuy].

Lemma 5.5. If Z is a linear operator on a finite-dimensional vector space, then

limm!1

1

m

m�1XkD0

.e�Z=m/k D 1 � e�Z

Z: (5.12)

Proof. If we formally applied the formula for the sum of a finite geometric series toe�Z=m; we would get

1

m

m�1XkD0

.e�Z=m/k D 1

m

1 � e�Z

1 � e�Z=m m ! 1�����!1 � e�Z

Z:

To give a rigorous argument, we observe that

1 � e�x

xDZ 1

0

e�tx dt;

from which it follows that

1 � e�Z

ZDZ 1

0

e�tZ dt: (5.13)

(The reader may check, using term-by-term integration of the series expansion ofe�tZ , that this formula for .1 � e�Z/=Z agrees with our earlier definition.)

Since .e�Z=m/k D e�kZ=m; the left-hand side of (5.12) is a Riemann-sumapproximation to the matrix-valued integral on the right-hand side of (5.13). TheseRiemann sums converge to the integral of e�tZ—which is a continuous function oft—establishing (5.12). ut

116 5 The Baker–Campbell–Hausdorff Formula and Its Consequences

Proof of Theorem 5.4. The formula (5.11) follows from (5.10) by applying thechain rule to the composition of exp and X.t/. Thus, it suffices to prove (5.10).For any n � n matrices X and Y , set

.X; Y / D d

dteXCtY

ˇˇtD0

:

Since (Proposition 2.16) exp is a continuously differentiable map,.X; Y / is jointlycontinuous in X and Y and is linear in Y for each fixed X .

Now, for every positive integerm, we have

eXCtY D�

exp

�X

mC t

Y

m

��m: (5.14)

Applying the product rule, we will get m terms, where in each term, m � 1 ofthe factors in (5.14) are simply evaluated at t D 0 and the remaining factor isdifferentiated at t D 0. Thus,

d

dteXCtY

ˇˇtD0

Dm�1XkD0

.eX=m/m�k�1�d

dtexp

�X

mC t

Y

m

�ˇˇtD0

�.eX=m/k

D e.m�1/X=mm�1XkD0

.eX=m/�k�X

m;Y

m

�.eX=m/k

D e.m�1/X=m 1m

m�1XkD0

exp

��adXm

�k �

�X

m; Y

��: (5.15)

In the third equality, we have used the linearity of.X; Y / in Y and the relationshipbetween Ad and ad (Proposition 3.35).

We now wish to let m tend to infinity in (5.15). The factor in front tendsto exp.X/. Since .X; Y / is jointly continuous in X and Y , the expression.X=m; Y / tends to .0; Y /, where it is easily verified that .0; Y / D Y . Finally,applying Lemma 5.5 with Z D adX , we see that

limm!1

1

m

m�1XkD0

exp

��adXm

�kD 1 � e�adX

adX:

Thus, by letting m tend to infinity in (5.15), we obtain the desired result. ut

5.5 Proof of the BCH Formula 117

5.5 Proof of the BCH Formula

We now turn to the proof of Theorem 5.3. For sufficiently small X and Y inMn.C/,let

Z.t/ D log.eXetY /

for 0 � t � 1. Our goal is to compute Z.1/. Since eZ.t/ D eXetY , we have

e�Z.t/ ddteZ.t/ D

eXetY��1

eXetYY D Y .

On the other hand, by Theorem 5.4,

e�Z.t/ ddteZ.t/ D

�I � e�adZ.t/

adZ.t/

� �dZ

dt

�.

Hence,

�I � e�adZ.t/

adZ.t/

� �dZ

dt

�D Y .

Now, if X and Y are small enough, Z.t/ will also be small, so that ŒI �e�adZ.t/ �=adZ.t/ will be close to the identity and thus invertible. In that case, weobtain

dZ

dtD�I � e�adZ.t/

adZ.t/

��1.Y /. (5.16)

Meanwhile, if we apply the homomorphism “Ad” to the equation eZ.t/ D eXetY ,use the relationship between “Ad” and “ad,” and take a logarithm, we obtain thefollowing relations:

AdeZ.t/ D AdeX AdetY

eadZ.t/ D eadX etadY

adZ.t/ D log.eadX etadY /:

Plugging the last two of these relations into (5.16) gives

dZ

dtD�I � .eadX etadY /�1

log.eadX etadY /

��1.Y /. (5.17)

118 5 The Baker–Campbell–Hausdorff Formula and Its Consequences

Now, observe that

g.z/ D�1 � z�1

log z

��1

so that (5.17) is the same as

dZ

dtD g.eadX etadY /.Y /. (5.18)

Noting that Z.0/ D X and integrating (5.18) gives

log.eXeY / D Z.1/ D X CZ 1

0

g.eadX etadY /.Y / dt;

which is the Baker–Campbell–Hausdorff formula.

5.6 The Series Form of the BCH Formula

Let us see how to get the first few terms of the series form of the BCH formula fromthe integral form in Theorem 5.3. Using the Taylor series (2.7) for the logarithm, wemay easily compute that

g .z/ D 1C 1

2.z � 1/� 1

6.z � 1/2 C 1

12.z � 1/3 � � � � :

Meanwhile,

eadX et adY � I

D�I C adX C .adX/2

2C � � �

��I C t adY C t2.adY /2

2C � � �

�� I

D adX C t adY C t adX adY C .adX/2

2C t2.adY /2

2C � � � .

Since eadX et adY � I has no zeroth-order term, .eadX etadY � I /m will contributeonly terms of degree m or higher in adX and/or adY . Computing up to degree 2 inadX and adY gives

g.eadX et adY /

D I C 1

2

�adX C t adY C t adX adY C .adX/2

2C t2.adY /2

2

5.7 Group Versus Lie Algebra Homomorphisms 119

� 1

6

�.adX/2 C t2.adY /2 C t adX adY C t adY adX

C higher-order terms.

We now apply geadX etadY

�to Y and integrate. Computing to second order and

noting that any term with adY acting first is zero, we obtain:

log.eXeY /

X CZ 1

0

�Y C 1

2ŒX; Y �C 1

4ŒX; ŒX; Y �� � 1

6ŒX; ŒX; Y �� � t

6ŒY; ŒX; Y ��

�dt

X C Y C 1

2ŒX; Y �C 1

12ŒX; ŒX; Y �� � 1

12ŒY; ŒX; Y �� ;

which is the expression in (5.3).

5.7 Group Versus Lie Algebra Homomorphisms

Recall Theorem 3.28, which says that given matrix Lie groups G and H and a Liegroup homomorphism ˆ W G ! H , we can find a Lie algebra homomorphism� W g ! h such that ˆ.eX/ D e�.X/ for all X 2 g. In this section, we prove aconverse to this result in the case that G is simply connected.

Theorem 5.6. Let G and H be matrix Lie groups with Lie algebras g and h;respectively, and let � W g ! h be a Lie algebra homomorphism. If G is simplyconnected, there exists a unique Lie group homomorphism ˆ W G ! H such thatˆ.eX/ D e�.X/ for all X 2 g.

This result has the following corollary.

Corollary 5.7. SupposeG andH are simply connected matrix Lie groups with Liealgebras g and h, respectively. If g is isomorphic to h; then G is isomorphic to H .

Proof. Let � W g ! h be a Lie algebra isomorphism. By Theorem 5.6, there existsan associated Lie group homomorphism ˆ W G ! H . Since WD ��1 is alsoa Lie algebra homomorphism, there is a corresponding Lie group homomorphism‰ W H ! G. We want to show that ˆ and ‰ are inverses of each other.

Now, the Lie algebra homomorphism associated toˆı‰ is, by Proposition 3.30,equal to � ı D I , and similarly for ‰ ı ˆ. Thus, by Corollary 3.49, both ˆ ı ‰and ‰ ıˆ are equal to the identity maps on H and G, respectively. ut

We now proceed with the proof of Theorem 5.6. The first step is to construct a“local homomorphism” from �. This step is the only place in the argument in whichwe use the BCH formula.

120 5 The Baker–Campbell–Hausdorff Formula and Its Consequences

Definition 5.8. If G and H are matrix Lie groups, a local homomorphism of Gto H is pair .U; f / where U is a path-connected neighborhood of the identity in Gand f W U ! H is a continuous map such that f .AB/ D f .A/f .B/ whenever A,B , and AB all belong to U .

The definition says that f is as much of a homomorphism as it makes sense tobe, given that U is not necessarily a subgroup of G.

Proposition 5.9. Let G and H be matrix Lie groups with Lie algebras g and h,respectively, and let � W g ! h be a Lie algebra homomorphism. Define U" � G by

U" D fA 2 Gj kA� Ik < 1 and klogAk < "g:

Then there exists some " > 0 such that the map f W U" ! H given by

f .A/ D e�.logA/

is a local homomorphism.

Note that by Theorem 3.42, if " is small enough, logAwill be in g for allA 2 U",so that ˆ makes sense.

Proof. Choose " small enough that Theorem 3.42 applies and small enough that forall A;B 2 U", the BCH formula applies to X WD logA and Y WD logB and also to�.X/ and �.Y /. Then if AB happens to be in U", we have

f .AB/ D f .eXeY / D e�.log.eX eY //:

We now compute log.eXeY / by the BCH formula and then apply �. Since � is a Liealgebra homomorphism, it will change all the Lie-algebraic quantities involving Xand Y in the BCH formula into the analogous quantities involving �.X/ and �.Y /.Thus, as in (5.4), we have

�Œlog.eXeY /� D �.X/CZ 1

0

1XmD0

am.ead�.X/etad�.Y / � I /m.�.Y // dt

D log.e�.X/e�.Y //;

We obtain, then,

f .AB/ D exp˚log.e�.X/e�.Y //

�D e�.X/e�.Y /

D f .A/f .B/;

as claimed. ut

5.7 Group Versus Lie Algebra Homomorphisms 121

Theorem 5.10. Let G and H be matrix Lie groups, with G simply connected. If.U; f / is a local homomorphism of G into H , there exists a unique (global) Liegroup homomorphismˆ W G ! H such that ˆ agrees with f on U .

Proof. Step 1: Define ˆ along a path. Since G is simply connected and thusconnected, for any A 2 G, there exists a path A.t/ 2 G with A.0/ D I andA.1/ D A. Let us call a partition 0 D t0 < t1 < t2 � � � < tm D 1 of Œ0; 1� a goodpartition if for all s and t belonging to the same subinterval of the partition, wehave

A.t/A.s/�1 2 U: (5.19)

Lemma 3.48 guarantees that good partitions exist. If a partition is good, then,in particular, since t0 D 0 and A.0/ D I , we have A.t1/ 2 U . Choose a goodpartition and write A as

A D ŒA.1/A.tm�1/�1�ŒA.tm�1/A.tm�2/�1� � � � ŒA.t2/A.t1/�1�A.t1/:

Since ˆ is supposed to be a homomorphism and is supposed to agree with fnear the identity it is reasonable to “define”ˆ.A/ by

ˆ.A/ D f .A.1/A.tm�1/�1/ � � �f .A.t2/A.t1/�1/f .A.t1//: (5.20)

In the next two steps, we will prove that ˆ.A/ is independent of the choice ofpartition for a fixed path and independent of the choice of path.Step 2: Prove independence of the partition. For any good partition, if we insertan extra partition point s between tj and tjC1, the result is easily seen to beanother good partition. This change in the partition has the effect of replacing thefactor f .A.tjC1/A.tj /�1/ in (5.20) by

f .A.tjC1/A.s/�1/f .A.s/A.tj /�1/:

Since s is between tj and tjC1, the condition (5.19) on the original partitionguarantees that A.tjC1/A.s/�1, A.s/A.tj /�1 and A.tjC1/A.tj /�1 are all in U .Thus, since f is a local homomorphism, we have

f .A.tjC1/A.tj /�1/ D f .A.tjC1/A.s/�1/f .A.s/A.tj /�1/;

showing that the value ofˆ.A/ is unchanged by the addition of the extra partitionpoint.By repeating this argument, we see that the value of ˆ.A/ does not change bythe addition of any finite number of points to the partition. Now, any two goodpartitions have a common refinement, namely their union, which is also a goodpartition. The above argument shows that the value of ˆ.A/ computed from thefirst partition is the same as for the common refinement, which is the same as forthe second partition.

122 5 The Baker–Campbell–Hausdorff Formula and Its Consequences

Step 3: Prove independence of the path. It is in this step that we use thesimple connectedness of G. Suppose A0.t/ and A1.t/ are two paths joiningthe identity to some A 2 G. Then, since G is simply connected, a standardtopological argument (e.g., Proposition 1.6 in [Hat]) shows that A0 and A1 arehomotopic with endpoints fixed. This means that there exists a continuous mapA W Œ0; 1� � Œ0; 1� ! G with

A.0; t/ D A0.t/; A.1; t/ D A1.t/

for all t 2 Œ0; 1� and also

A.s; 0/ D I; A.s; 1/ D A

for all s 2 Œ0; 1�.As in the proof of Lemma 3.48, there exists an integer N such that for all .s; t/and .s0; t 0/ in Œ0; 1� � Œ0; 1� with js � s0j < 2=N and jt � t 0j < 2=N , we have

A.s; t/A.s0; t 0/�1 2 U:

We now deform A0 “a little bit at a time” into A1. This means that we define asequence Bk;l of paths, with k D 0; : : : ; N � 1 and l D 0; : : : ; N . We definethese paths so that Bk;l .t/ coincides with A..k C 1/=N; t/ for t between 0 and.l � 1/=N , and Bk;l .t/ coincides with A.k=N; t/ for t between l=N and 1. Fort between .l � 1/=N and l=N , we define Bk;l .t/ to coincide with the valuesof A.�; �/ on the path that goes “diagonally” in the .s; t/-plane, as indicated inFigure 5.1. When l D 0, there are no t-values between 0 and .l � 1/=N , soBk;0.t/ D A.k=N; t/ for all t 2 Œ0; 1�. In particular, B0;0.t/ D A0.t/.

Fig. 5.1 The path in the.s; t / plane defining Bk;l .t /

1kN

k 1N

s

l 1N

lN

1t

5.7 Group Versus Lie Algebra Homomorphisms 123

We now deform A0 D B0;0 into B0;1 and then into B0;2, B0;3, and so on until wereach B0;N , which we then deform into B1;0 and so on until we reach BN�1;N ,which we finally deform into A1. We claim that the value of ˆ.A/ is the same ateach stage of this deformation. Note that for k < l , Bk;l .t/ and Bk;lC1.t/ are thesame except for t’s in the interval

Œ.l � 1/=N; .l C 1/=N �:

Furthermore, by Step 2, we are free to choose any good partition we like tocomputeˆ.A/. For both Bk;l and Bk;lC1, we choose the partition points to be

0;1

N; : : : ;

l � 1

N;l C 1

N;l C 2

N; : : : ; 1;

which gives a good partition by the way N was chosen.Now, from (5.20), the value ofˆ.A/ depends only on the values of the path at thepartition points. Since we have chosen our partition in such a way that the valuesof Bk;l and Bk;lC1 are identical at all the partition points, the value of ˆ.A/ isthe same for these two paths. (See Figure 5.2.) A similar argument shows thatthe value of ˆ.A/ computed along Bk;N is the same as along BkC1;0. Thus, thevalue of ˆ.A/ is the same for each path from A0 D B0;0 all the way to BN�1;Nand then the same as A1.

Fig. 5.2 The paths Bk;l andBk;lC1 agree at each partitionpoint. In the figure, sincreases as we move fromthe top toward the bottom

A0(t)

A1(t)

Bk,l (t)

Bk,l 1(t)

124 5 The Baker–Campbell–Hausdorff Formula and Its Consequences

Step 4: Prove thatˆ is a homomorphism and agrees with f on U . The proof thatˆ is a homomorphism is a straightforward unpacking of the definition of ˆ andis left to the reader; see Exercise 7. To show that ˆ agrees with f on U , chooseA 2 U . Since U is path connected, we can find a path A.t/, 0 � t � 1, lying inU joining I to A. Choose a good partition ftj gmjD1 for A.t/, and we then claimthat for all j , we have ˆ.A.tj // D f .A.tj //.Note that A.t/, 0 � t � tj is a path joining I to A.tj / and that ft0; t1; : : : ; tj g isa good partition of this path. (Technically, we should reparameterize this path sothat the time interval is Œ0; 1�.) Hence,

ˆ.A.tj // D f .A.tj /A.tj�1/�1/ � � �f .A.t2/A.t1/�1/f .A.t1//;

for all j . In particular,

ˆ.A.t1// D f .A.t1//:

Now assume that ˆ.A.tj // D f .A.tj //, and compute that

ˆ.A.tjC1// D f .A.tjC1/A.tj /�1/f .A.tj /A.tj�1/�1/ � � �f .A.t1//D f .A.tjC1/A.tj /�1/ˆ.A.tj //

D f .A.tjC1/A.tj /�1/f .A.tj //

D f .A.tj //:

The last equality holds because f is a local homomorphism and becauseA.tjC1/A.tj /�1, A.tj /, and their product all lie in U . Thus, by induction,ˆ.A.tj // D f .A.tj // for all j ; when j D m, we obtain ˆ.A/ D f .A/. utIt is important to note that when proving independence of the path (Step 3 of

the proof), it is essential to know already that the value of ˆ.A/ is independent ofthe choice of good partition (Step 2 of the proof). Specifically, when we move fromBk;l to Bk;lC1, we use one partition for Bk;lC1, but when we move from Bk;lC1 toBk;lC2, we use a different partition for Bk;lC1. Essentially, the proof proceeds bydeforming the path between partition points—which clearly does not change thevalue of ˆ.A/—then picking a new partition and doing the same thing again. Notealso that it is in proving independence of the partition that we use the assumptionthat f is a local homomorphism.

Proof of Theorem 5.6. For the existence part of the proof, let f be the localhomomorphism in Proposition 5.9 and let ˆ be the global homomorphism inTheorem 5.10. Then for anyX 2 g, the element eX=m will be inU for all sufficientlylargem, showing that

ˆ.eX=m/ D f .eX=m/ D e�.X/=m:

5.7 Group Versus Lie Algebra Homomorphisms 125

Since ˆ is a homomorphism, we have

ˆ.eX/ D ˆ.eX=m/m D e�.X/;

as required.For the uniqueness part of the proof, supposeˆ1 andˆ2 are two homomorphisms

related in the desired way to �. Then for any A 2 G, we express A as eX1 � � � eXNwith Xj 2 g, as in Corollary 3.47, and observe that

ˆ1.A/ D ˆ2.A/ D e�.X1/ � � � e�.XN /;

so that ˆ1 agrees with ˆ2. utWe conclude this section with a typical application of Theorem 5.6.

Theorem 5.11. Suppose that G is a simply connected matrix Lie group and thatthe Lie algebra g ofG decomposes as a Lie algebra direct sum g D h1˚h2, for twosubalgebras h1 and h2 of g. Then there exists closed, simply connected subgroupsH1 and H2 of G whose Lie algebras are h1 and h2, respectively. Furthermore, G isisomorphic to the direct product of H1 andH2.

Proof. Consider the Lie algebra homomorphism � W g ! g that sends X C Y

to X , where X 2 h1 and Y 2 h2. Since G is simply connected, there is a Liegroup homomorphism ˆ W G ! G associated to �, and the Lie algebra of thekernel of ˆ is the kernel of � (Proposition 3.31), which is h2. LetH2 be the identitycomponent of kerˆ. Since ˆ is continuous, kerˆ is closed, and so is its identitycomponent H2 (Corollary 3.52). Thus, H2 is a closed, connected subgroup of Gwith Lie algebra h2. By a similar argument, we may construct a closed, connectedsubgroupH1 of G whose Lie algebra is h1.

Suppose now that A.t/ is a loop in H1. Since G is simply connected, there isa homotopy A.s; t/ shrinking A.t/ to a point in G. Now, � is the identity on h1,from which it follows that ˆ is the identity on H1. Thus, if we define B.s; t/ Dˆ.A.s; t//, we see that

B.0; t/ D ˆ.A.t// D A.t/:

Furthermore, since � maps G into h1, we see that ˆ maps G into H1. We concludethat B is a homotopy of A.t/ to a point in H1. Thus, H1 is simply connected, and,by a similar argument, so is H2.

Finally, since g is the Lie algebra direct sum of h1 and h2, elements of h1commutes with elements of h2. It follows that elements ofH1 (which are all productof exponentials of elements of h1) commute with elements of H2. Thus, we have aLie group homomorphism ‰ W H1 � H2 ! G given by ‰.A;B/ D AB. Theassociated Lie algebra homomorphism is then just the original isomorphism ofh1 ˚ h2 with g. Since G is simply connected, there is a homomorphism � W G !H1 �H2 for which the associated Lie algebra homomorphism is �1. By the proofof Corollary 5.7, � and ‰ are inverses of each other, showing that G is isomorphicto H1 �H2. ut

126 5 The Baker–Campbell–Hausdorff Formula and Its Consequences

5.8 Universal Covers

Theorem 5.6 says that if G is simply connected, every homomorphism of the Liealgebra g of G can be exponentiated to a homomorphism of G. If G is not simplyconnected, we may look for another group QG that has the same Lie algebra as G butsuch that QG is simply connected.

Definition 5.12. Let G be a connected matrix Lie group. Then a universal coverof G is a simply connected matrix Lie group H together with a Lie grouphomomorphism ˆ W H ! G such that the associated Lie algebra homomorphism� W h ! g is a Lie algebra isomorphism. The homomorphism ˆ is called thecovering map.

If a universal cover of G exists, it is unique up to “canonical isomorphism,” asfollows.

Proposition 5.13. IfG is a connected matrix Lie group and .H1;ˆ1/ and .H2;ˆ2/

are universal covers ofG, then there exists a Lie group isomorphism‰ W H1 ! H2

such thatˆ2 ı‰ D ˆ1.

Proof. See Exercise 9. utSince a connected matrix Lie group has at most one universal cover (up to

canonical isomorphism), it is reasonable to speak of the universal cover . QG;ˆ/ ofG. Furthermore, if H is a simply connected Lie group and � W h ! g is a Liealgebra isomorphism, then by Theorem 5.6, we can construct an associated Liegroup homomorphism ˆ W H ! G, so that .H;ˆ/ is a universal cover of G.Since � is an isomorphism, we can use � to identify Qg with g. Thus, in slightly lessformal terms, we may define the notion of universal cover as follows: The universalcover of a matrix Lie group G is a simply connected matrix Lie group QG such thatthe Lie algebra of QG is equal to the Lie algebra of G. With this perspective, wehave the following immediate corollary of Theorem 5.6.

Corollary 5.14. Let G be a connected matrix Lie group and let QG be the universalcover of G, where we think of G and QG as having the same Lie algebra g. If H is amatrix Lie group with Lie algebra h and � W g ! h is a Lie algebra homomorphism,there exists a unique homomorphism ˆ W QG ! H such that ˆ.eX/ D e�.X/ for allX 2 g.

An example of importance in physics is the universal cover of SO.3/.

Example 5.15. The universal cover of SO.3/ is SU.2/.

Proof. The group SU.2/ is simply connected by Proposition 1.15. Proposition 1.19and Example 3.29 then provide the desired covering map. ut

The topic of universal covers is one place where we pay a price for our decisionto consider only matrix Lie groups: a matrix Lie group may not have a universalcover that is a matrix Lie group. It is not hard to show that every Lie group G

5.8 Universal Covers 127

has a universal cover in the class of (not necessarily matrix) Lie groups. Indeed,G has a universal cover in the topological sense (Definition 13.1), and this covercan be given a group structure in such a way that the covering map is a Lie grouphomomorphism. It turns out, however, that the universal cover of a matrix Lie groupmay not be a matrix group.

We now show that the group SL.2IR/ does not have a universal cover in the classof matrix Lie groups. We begin by showing that SL.2IR/ is not simply connected.By Theorem 2.17 and Proposition 2.19, SL.2IR/, as a manifold, is homeomorphicto SO.2/ � V , where V is the space of 2 � 2 real, symmetric matrices with tracezero. Now, V , being a vector space, is certainly simply connected, but SO.2/, whichis homeomorphic to the unit circle S1, is not. Thus, SL.2IR/ itself is not simplyconnected. By contrast, the group SL.2IC/ decomposes as SU.2/�W , whereW isthe space of 2 � 2 self-adjoint, complex matrices with trace zero. Since both SU.2/andW are simply connected, SL.2IC/ is simply connected. (See Appendix 13.3 formore information on this type of calculation.)

Proposition 5.16. Let G � GL.nIC/ be a connected matrix Lie group with Liealgebra g. Suppose ˆ W G ! SL.2IR/ is a Lie group homomorphism for whichthe associated Lie algebra � W g ! sl.2IR/ is a Lie algebra isomorphism. Thenˆ is a Lie group isomorphism and, therefore, G cannot be simply connected. Thus,SL.2IR/ has no universal cover in the class of matrix Lie groups.

The result relies essentially on the assumption that G is a matrix Lie group—or,more precisely, on the assumption that G is contained in a group whose Lie algebrais complex.

Lemma 5.17. Suppose W sl.2IR/ ! gl.nIC/ is a Lie algebra homomorphism.Then there exists a Lie group homomorphism ˆ W SL.2IR/ ! GL.nIC/ such thatˆ.eX/ D e�.X/ for all X 2 sl.2IR/.

The significance of the lemma is that the result holds even though SL.2IR/ isnot simply connected.

Proof. Let C W sl.2IC/ ! gl.nIC/ be the complex-linear extension of tosl.2IC/ Š sl.2IR/C, which is a Lie algebra homomorphism (Proposition 3.39).Since SL.2IC/ is simply connected, there exists a Lie group homomorphism‰C W SL.2IC/ ! GL.nIC/ such that ‰C.e

X/ D e C.X/ for all X 2 sl.2IC/. If welet ‰ be the restriction of ‰C to SL.2IR/, then ‰ is a Lie group homomorphismwhich satisfies ‰.eX/ D e .X/ for X 2 sl.2IR/. utProof of Proposition 5.16. Since � is a Lie algebra isomorphism, the inverse map W sl.2IR/ ! g is a Lie algebra homomorphism. Thus, by the lemma, there is aLie group homomorphism ‰ W SL.2IR/ ! G corresponding to . Since � and are inverses of each other, it follows from Proposition 3.30 and Corollary 3.49 thatˆ and ‰ are also inverses of each other. ut

128 5 The Baker–Campbell–Hausdorff Formula and Its Consequences

5.9 Subgroups and Subalgebras

In this section, we address Question 3 from Sect. 5.1: IfG is a matrix Lie group withLie algebra g and h is a subalgebra of g, does there exist a matrix Lie groupH � G

whose Lie algebra is H? If the exponential map for G were a homeomorphismbetween g and G and if the BCH formula worked globally instead of locally, theanswer would be yes, since we could simply define H to be the set of elements ofthe form eX , X 2 g, and the BCH formula would show that H is a subgroup.

In reality, the answer to Question 3, as stated, is no. Suppose, for example, thatG D GL .2IC/ and

h D� �

it 00 ita

�ˇˇ t 2 R

�, (5.21)

where a is irrational. If there is going to be a matrix Lie group H with Lie algebrah, thenH would have to contain the closure of the group

H0 D� �

eit 0

0 eita

�ˇˇ t 2 R

�; (5.22)

which is (Exercise 10 in Chapter 1) is the group

H1 D� �

ei� 0

0 ei�

�ˇˇ �; � 2 R

�.

But then the Lie algebra of H would have to contain the Lie algebra of H1, whichis two dimensional!

Fortunately, all is not lost. We can still get a subgroup H for each subalgebra hif we weaken the condition that H be a matrix Lie group. In the above example, thesubgroup we want is H0, despite the fact that H0 is not closed.

Definition 5.18. If H is any subgroup of GL .nIC/, the Lie algebra h of H is theset of all matrices X such that

etX 2 H

for all real t .

It is possible to prove that for any subgroupH of GL.nIC/, the Lie algebra h ofH is actually a Lie algebra, that is, a real vector space—possibly zero dimensional—and closed under brackets. (See Proposition 1 and Corollary 7 in Chapter 2 of[Ross].) This result is not, however, directly relevant to our goal in this section,which is to construct, for each subalgebra h of gl.nIC/ a subgroup with Lie algebrah. Note, however, that if h is at least a real subspace of gl.nIC/, then the proof ofPoint 4 of Theorem 3.20 shows that h is also closed under brackets.

5.9 Subgroups and Subalgebras 129

Definition 5.19. If G is a matrix Lie group with Lie algebra g, then H � G is aconnected Lie subgroup of G if the following conditions are satisfied:

1. H is a subgroup of G.2. The Lie algebra h of H is a Lie subalgebra of g.3. Every element of H can be written in the form eX1eX2 � � � eXm , withX1; : : : ; Xm 2 h.

Connected Lie subgroups are also called analytic subgroups. Note that anygroup H as in the definition is path connected, since each element of H can beconnected to the identity in H by a path of the form

t 7! e.1�t /X1e.1�t /X2 � � � e.1�t /Xm:

The groupH0 in (5.22) is a connected Lie subgroup of GL.2IC/ whose Lie algebrais the algebra h in (5.21).

We are now ready to state the main result of this section, which is our secondmajor application of the Baker–Campbell–Hausdorff formula.

Theorem 5.20. Let G be a matrix Lie group with Lie algebra g and let h be a Liesubalgebra of g. Then there exists a unique connected Lie subgroup H of G withLie algebra h.

If h is the subalgebra of gl.2IC/ in (5.21), then the connected Lie subgroup His the group H0 in (5.22), which is not closed. In practice, Theorem 5.20 is mostuseful in those cases where the connected Lie subgroup H turns out to be closed.See Proposition 5.24 and Exercises 10, 13, and 14 for conditions under which thisis the case.

We now begin working toward the proof of Theorem 5.20. Since G is assumedto be a matrix Lie group, we may as well assume thatG D GL.nIC/. After all, ifGis a closed subgroup of GL.nIC/ and H is a connected Lie subgroup of GL.nIC/whose Lie algebra h is contained in g, then H is also a connected Lie subgroupof G. We now let

H D feX1eX2 � � � eXN ˇX1; : : : ; XN 2 hg; (5.23)

which is a subgroup of G. The key issue is to prove that the Lie algebra of H ,in the sense of Definition 5.18, is h. Once we know that Lie.H/ D h, we willimmediately conclude that H is a connected Lie subgroup with Lie algebra h, theremaining properties in Definition 5.19 being true by definition. Note that for theclaim Lie.H/ D h to be true, it essential that h be a subalgebra of gl.nIC/, and notmerely a subspace; compare Exercise 11.

As in the proof of Theorem 3.42, we think of gl.nIC/ as R2n2

and we decomposegl.nIC/ as the direct sum of h and D, where D is the orthogonal complement ofh with respect to the usual inner product on R2n

2. Then, as shown in the proof

of Theorem 3.42, there exist neighborhoods U and V of the origin in h and D,

130 5 The Baker–Campbell–Hausdorff Formula and Its Consequences

respectively, and a neighborhoodW of I in GL.nIC/ with the following properties:Each A 2 W can be written uniquely as

A D eXeY ; X 2 U; Y 2 V; (5.24)

in such a way that X and Y depend continuously on A. We think of thedecomposition in (5.24) as our local coordinates in a neighborhood of the identityin GL.nIC/.

If X is a small element of h, the decomposition of eX is just eXe0. If we take theproduct of two elements of the form eX1eX2 , with X1 and X2 small elements of h,then since h is a subalgebra, if we combine the exponentials as eX1eX2 D eX3 bymeans of the Baker–Campbell–Hausdorff formula, X3 will again be in h. Thus, ifwe take a small number of products as in (5.23) with the Xj ’s being small elementsof h, we will move from the identity in theX -direction in the decomposition (5.24).Globally, however, H may wind around and come back to points in W of theform (5.24) with Y ¤ 0. (See Figure 5.3.) Indeed, as the example of the “irrationalline” in (5.22) shows, there may be elements of H in W with arbitrarily smallnonzero values of Y . Nevertheless, we will see that the set of Y values that occursis at most countable.

Lemma 5.21. Decompose gl.nIC/ as h ˚ D and let V be a neighborhood of theorigin in D as in (5.24). If E � V is defined by

E D fY 2 V ˇeY 2 H g;

then E is at most countable.

Assuming the lemma, we may now prove Theorem 5.20.

eU

eV

Fig. 5.3 The black lines indicate the portion of H in the set W . The group H intersects eV in atmost countably many points

5.9 Subgroups and Subalgebras 131

Proof of Theorem 5.20. As we have already observed, it suffices to show that theLie algebra of H is h. Let h0 be the Lie algebra of H , which clearly contains h. ForZ 2 h0, we may write, for all sufficiently small t ,

etZ D eX.t/eY.t/;

where X.t/ 2 U � h and Y.t/ 2 V � D and where X.t/ and Y.t/ are continuousfunctions of t . Since Z is in the Lie algebra ofH , we have etZ 2 H for all t . Since,also, eX.t/ is in the group H , we conclude that eY.t/ is in H for all sufficientlysmall t . If Y.t/ were not constant, then it would take on uncountably many values,which would mean that E is uncountable, violating Lemma 5.21. So, Y.t/ must beconstant, and since Y.0/ D 0, this means that Y.t/ is identically equal to zero. Thus,for small t , we have etZ D eX.t/ and, therefore, tZ D X.t/ 2 h. This means Z 2 hand we conclude that h0 � h. ut

Before proving Lemma 5.21, we prove another lemma.

Lemma 5.22. Pick a basis for h and call an element of h rational if its coefficientswith respect to this basis are rational. Then for every ı > 0 and every A 2 H , thereexist rational elements R1; : : : ; Rm of h such that

A D eR1eR2 � � � eRmeX;where X is in h and kXk < ı.

Suppose we take ı small enough that the ball of radius ı in h is contained in U .Then since there are only countably many m-tuples of the form .R1; : : : ; Rm/ withRj rational, the lemma tells us thatH can be covered by countably many translatesof the set eU .

Proof. Choose " > 0 so that for all X; Y 2 h with kXk < " and kY k < ", theBaker–Campbell–Hausdorff holds for X and Y . Let C.�; �/ denote the right-handside of the formula, so that

eXeY D eC.X;Y /

whenever kXk ; kY k < ". It is not hard to see thatC.�; �/ is a continuous. Now, if thelemma holds for some ı, it also holds for any ı0 > ı. Thus, it is harmless to assumeı is less than " and small enough that if kXk ; kY k < ı, we have kC.X; Y /k < ".

Since eX D .eX=k/k , every element A of H can be written as

A D eX1 � � � eXN (5.25)

with Xj 2 h andXj < ı. We now proceed by induction on N . If N D 0, then

A D I D e0, and there is nothing to prove. Assume the lemma for A’s that can beexpressed as in (5.25) for some integer N , and consider A of the form

A D eX1 � � � eXN eXNC1 (5.26)

132 5 The Baker–Campbell–Hausdorff Formula and Its Consequences

with Xj 2 h andXj < ı. Applying our induction hypothesis to eX1 � � � eXN , we

obtain

A D eR1 � � � eRmeXeXNC1

D eR1 � � � eRmeC.X;XNC1/:

where the Rj ’s are rational and kC.X;XNC1/k < ". Since h is a subalgebra ofgl.nIC/, the element C.X;XNC1/ is again in h, but may not have norm less than ı.

Now choose a rational element RmC1 of h that is very close to C.X;XNC1/ andsuch that kRmC1k < ". We then have

A D eR1 � � � eRmeRmC1e�RmC1eC.X;XNC1/

D eR1 � � � eRmeRmC1eX0

;

where

X 0 D C.�RmC1; C.X;XNC1//:

Then X 0 will be in h, and by choosing RmC1 sufficiently close to C.X;XNC1/, wecan make kX 0k < ı. After all, since C.�Z;Z/ D log.e�ZeZ/ D 0 for all small Z,if Z0 is close to Z, then C.�Z0; Z/ will be small. ut

We now supply the proof of Lemma 5.21.

Proof of Lemma 5.21. Fix ı so that for all X and Y with kXk ; kY k < ı, thequantity C.X; Y / is defined and contained in U . We then claim that for eachsequence R1; : : : ; Rm of rational elements in h, there is at most one X 2 h withkXk < ı such that the element

eR1eR2 � � � eRmeX (5.27)

belongs to eV . After all, if we have

eR1eR2 � � � eRmeX1 D eY1 ; (5.28)

eR1eR2 � � � eRmeX2 D eY2 (5.29)

with Y1; Y2 2 V , then

e�X1eX2 D e�Y1eY2

and so

e�Y1 D e�X1eX2e�Y2 D eC.�X1;X2/e�Y2 ;

5.9 Subgroups and Subalgebras 133

with C.�X1;X2/ 2 U . However, each element of eU eV has a unique representationas eY eX with X 2 U and Y 2 V . Thus, we must have �Y2 D �Y1 and, by (5.28)and (5.29), eX1 D eX2 and X1 D X2.

By Lemma 5.22, every element of H can be expressed in the form (5.27) withkXk < ı. Now, there are only countably many rational elements in h and thus onlycountably many expressions of the form eR1 � � � eRm , each of which produces at mostone element of the form (5.27) that belongs to eV . Thus, the set E in Lemma 5.21is at most countable. ut

This completes the proof of Theorem 5.20.If a connected Lie subgroup H of GL.nIC/ is not closed, the topology H

inherits from GL.nIC/ may be pathological, e.g., not locally connected. (CompareFigure 1.1.) Nevertheless, we can giveH a new topology that is much nicer.

Theorem 5.23. LetH be a connected Lie subgroup of GL.nIC/ with Lie algebra h.ThenH can be given the structure of a smooth manifold in such a way that the groupoperations on H are smooth and the inclusion map ofH into GL.nIC/ is smooth.

Thus, every connected Lie subgroup of GL.nIC/ can be made into a Lie group. Inthe case of the groupH0 in (5.22), the new topology onH0 is obtained by identifyingH0 with R by means of the parameter t in the definition of H0.

Proof. For any A 2 H and any " > 0, define

UA;" D fAeXˇX 2 h and kXk < "g:

Now define a topology on H as follows: A set U � H is open if for each A 2 U

there exists " > 0 such that UA;" � U . (See Figure 5.4.) In this topology, twoelements A and B of H are “close” if we can express B as B D AeX with X 2 hand kXk small. This topology is finer than the topologyH inherits from GI that is,if A and B are close in this new topology, they are certainly close in the ordinarysense in G, but not vice versa.

It is easy to check that this topology is Hausdorff, and using Lemma 5.22, it isnot hard to see that the topology is second countable. Furthermore, in this topology,H is locally homeomorphic to RN , where N D dim h, by identifying each UA;"with the ball of radius " in h.

We may define a smooth structure onH by using theUA;"’s, with " less than somesmall number "0, as our “atlas.” If two of these sets overlap, then some elementC ofH can be written as C D AeX D BeY for some A;B 2 H andX; Y 2 h. It followsthat B D AeXe�Y , which means (since kXk and kY k are less than "0) that A andB are close. The change-of-coordinates map is then Y D log.B�1AeX/. Since Aand B are close and kXk is small, we will have that

B�1AeX � I < 1, so thatB�1AeX is in the domain where the matrix logarithm is defined and smooth. Thus,the change-of-coordinates map is smooth as function of X . Finally, in any of thecoordinate neighborhoods UA;", the inclusion of H into G is given by X 7! AeX ,which is smooth as a function of X . ut

134 5 The Baker–Campbell–Hausdorff Formula and Its Consequences

Fig. 5.4 The set U inH isopen in the new topology butnot in the topology inheritedfrom GL.2IC/. The elementB is close to A in GL.2IC/but not in the new topologyon H

AB

U

2p

2pf

q

As we have already noted, Theorem 5.20 is most useful in cases where theconnected Lie subgroup H is actually closed. The following result gives onecondition under which this is guaranteed to be the case. See also Exercises 10, 13,and 14.

Proposition 5.24. Suppose G � GL.nIC/ is a matrix Lie group with Lie algebrag and that h is a maximal commutative subalgebra of g, meaning that h iscommutative and h is not contained in any larger commutative subalgebra of g.Then the connected Lie subgroupH of G with Lie algebra h is closed.

Proof. Since h is commutative, H is also commutative, since every element of His a product of exponentials of elements of h. It easily follows that the closure NH ofH in GL.nIC/ is also commutative. We now claim that NH is connected. To see this,take A 2 NH , so that A is in G (since G is closed) and A is the limit of a sequenceAm in H . Since NH is closed, Theorem 3.42 applies. Thus, for all sufficiently largem, the element AA�1

m is expressible as AA�1m D eX , for some X in the Lie algebra h0

of NH . Thus, A D eXAm, which means that A can be connected to Am by the pathA.t/ D e.1�t /XAm, 0 � t � 1, in NH . Since Am can be connected to the identity inH � NH , we see that A can be connected to the identity in NH .

Now, since NH is commutative, its Lie algebra h0 is also commutative. But sinceh was maximal commutative, we must have h0 D h. Since, also, NH is connected, weconclude that NH D H , showing that H is closed. ut

5.11 Exercises 135

5.10 Lie’s Third Theorem

Lie’s third theorem (in its modern, global form) says that for every finite-dimensional, real Lie algebra g, there exists a Lie group G with Lie algebra g.We will construct G as a connected Lie subgroup of GL.nIC/.Theorem 5.25. If g is any finite-dimensional, real Lie algebra, there exists aconnected Lie subgroup G of GL.nIC/ whose Lie algebra is isomorphic to g.

Our proof assumes Ado’s theorem, which asserts that every finite-dimensionalreal or complex Lie algebra is isomorphic to an algebra of matrices. (See, forexample, Theorem 3.17.7 in [Var].)

Proof. By Ado’s theorem, we may identify g with a real subalgebra of gl.nIC/.Then by Theorem 5.20, there is a connected Lie subgroup of GL.nIC/ with Liealgebra g. ut

It is actually possible to choose the subgroup G in Theorem 5.25 to be closed.Indeed, according to Theorem 9 on p. 105 of [Got], if a connected Lie group Gcan be embedded into some GL.nIC/ as a connected Lie subgroup, then G can beembedded into some other GL.n0IC/ as a closed subgroup. Assuming this result,we may reach the following conclusion.

Conclusion 5.26. Every finite-dimensional, real Lie algebra is isomorphic to theLie algebra of some matrix Lie group.

This result does not, however, mean that every Lie group is isomorphic to amatrix Lie group, since there can be several nonisomorphic Lie groups with thesame Lie algebra. See, for example, Sect. 4.8.

5.11 Exercises

1. LetX be a linear transformation on a finite-dimensional real or complex vectorspace. Show that

I � e�X

X

is invertible if and only if none of the eigenvalues of X (over C) is of the form2�in, with n an nonzero integer.

Remark. This exercise, combined with the formula in Theorem 5.4, gives thefollowing result (in the language of differentiable manifolds): The exponentialmap exp W g ! G is a local diffeomorphism near X 2 g if and only if adX Wg ! g has no eigenvalue of the form 2�in, with n a nonzero integer.

136 5 The Baker–Campbell–Hausdorff Formula and Its Consequences

2. Show that for any X and Y in Mn.C/, even if X and Y do not commute,

d

dttrace.eXCtY/

ˇˇtD0

D trace.eXY /:

3. Compute log.eXeY / through third order in X and Y by calculating directlywith the power series for the exponential and the logarithm. Show this gives thesame answer as the Baker–Campbell–Hausdorff formula.

4. Suppose that X and Y are upper triangular matrices with zeros on the diagonal.Show that the power series for log.eXeY / is convergent. What happens to theseries form of the Baker–Campbell–Hausdorff formula in this case?

5. SupposeX and Y are n�n complex matrices satisfying ŒX; Y � D ˛Y for somecomplex number ˛. Suppose further that there is no nonzero integer n such that˛ D 2�in. Show that

eXeY D expnX C ˛

1 � e�˛ Yo:

Hint: Let A.t/ D eXetY and let

B.t/ D expnX C ˛

1 � e�˛ tYo:

Using Theorem 5.4, show that A.t/ and B.t/ satisfy the same differentialequation with the same value at t D 0.

6. Give an example of matrices X and Y in sl.2IC/ such that ŒX; Y � D 2�iYbut such that there does not exist any Z in sl.2IC/ with eXeY D eZ . UseExample 3.41 and compare Exercise 5.

7. Complete Step 4 in the proof of Theorem 5.6 by showing that ˆ is ahomomorphism. For all A;B 2 G, choose a path A.t/ connecting I to A anda path B.t/ connecting I to B . Then define a path C connecting I to AB bysetting C.t/ D B.2t/ for 0 � t � 1=2 and setting C.t/ D A.2t � 1/B for1=2 � t � 1. If t0; : : : ; tm is a good partition for A.t/ and s0; : : : ; sM is a goodpartition for B.t/, show that

s0

2; � � � ; sM

2;1

2C t0

2; � � � ; 1

2C tm

2

is a good partition forC.t/. Now, computeˆ.A/,ˆ.B/, andˆ.AB/ using thesepaths and partitions and show that ˆ.AB/ D ˆ.A/ˆ.B/.

8. If QG is a universal cover of a connected group G with projection map ˆ, showthat ˆ maps QG onto G.

9. Prove the uniqueness of the universal cover, as stated in Proposition 5.13.10. Let a be a subalgebra of the Lie algebra of the Heisenberg group. Show that

exp.a/ is a connected Lie subgroup of the Heisenberg group and that thissubgroup is closed.

5.11 Exercises 137

11. Consider the Lie algebra h of the Heisenberg group H , as computed inProposition 3.26. Let X , Y , and Z be the basis elements for h in (4.18), whichsatisfy ŒX; Y � D Z and ŒX;Z� D ŒY;Z� D 0. Let V be the subspace of hspanned by X and Y (which is not a subalgebra of h) and let K denote thesubgroup of H consisting of products of exponential of elements of V . Showthat K D H and, thus, that the Lie algebra of K is not equal to V .Hint: Use Theorem 5.1 and the surjectivity of the exponential map for H(Exercise 18 in Chapter 3).

12. Show that every connected Lie subgroup of SU.2/ is closed. Show that this isnot the case for SU.3/.

13. Let G be a matrix Lie group with Lie algebra g, let h be a subalgebra of g, andlet H be the unique connected Lie subgroup of G with Lie algebra h. Supposethat there exists a simply connected, compact matrix Lie groupK such that theLie algebra of K is isomorphic to h. Show that H is closed. Is H necessarilyisomorphic to K?

14. This exercise asks you to prove, assuming Ado’s theorem (Sect. 5.10), thefollowing result: If G is a simply connected matrix Lie group with Lie algebrag and h is an ideal in g, then the connected Lie subgroupH with Lie algebra his closed.

(a) Show that there exists a Lie algebra homomorphism � W g ! gl.N IC/with ker.�/ D h.Hint: Since h is an ideal in g, the quotient space g=h has a natural Liealgebra structure.

(b) Since G is simply connected, there exists a Lie group homomorphism ˆ WG ! gl.N IC/ for which the associated Lie algebra homomorphism is �.Show that the identity component of the kernel of ˆ is a closed subgroupof G whose Lie algebra is h.

(c) Show that the result fails if the assumption that G be simply connected isomitted.

Part IISemisimple Lie Algebras

Chapter 6The Representations of sl.3IC/

6.1 Preliminaries

In this chapter, we investigate the representations of the Lie algebra sl.3IC/,which is the complexification of the Lie algebra of the group SU.3/. The mainresult of this chapter is Theorem 6.7, which states that an irreducible finite-dimensional representation of sl.3IC/ can be classified in terms of its “highestweight.” This result is analogous to the results of Sect. 4.6, in which we classifythe irreducible representations by the largest eigenvalue of �.H/, namely the non-negative integerm.

The results of this chapter are special cases of the general theory of repre-sentations of semisimple Lie algebras (Chapters 7 and 9) and of the theory ofrepresentations of compact Lie groups (Chapters 11 and 12). It is nevertheless usefulto consider this case separately, in part because of the importance of SU.3/ inphysical applications but mainly because seeing roots, weights, and the Weyl group“in action” in a simple example motivates the introduction of these structures laterin a more general setting.

Every finite-dimensional representation of SU.3/ (over a complex vector space)gives rise to a representation of su.3/, which can then be extended by complexlinearity to sl.3IC/ Š su.3/C. Since SU.3/ is simply connected, we can go in theopposite direction by restricting any representation of sl.3IC/ to su.3/ and thenapplying Theorem 5.6 to obtain a representation of SU.3/. Propositions 4.5 and 4.6tell us that a representation of SU.3/ is irreducible if and only if the associated rep-resentation of sl.3IC/ is irreducible, thus establishing a one-to-one correspondence

A previous version of this book was inadvertently published without the middle initial of theauthor’s name as “Brian Hall”. For this reason an erratum has been published, correcting themistake in the previous version and showing the correct name as Brian C. Hall (see DOI http://dx.doi.org/10.1007/978-3-319-13467-3_14). The version readers currently see is the correctedversion. The Publisher would like to apologize for the earlier mistake.

© Springer International Publishing Switzerland 2015B.C. Hall, Lie Groups, Lie Algebras, and Representations, GraduateTexts in Mathematics 222, DOI 10.1007/978-3-319-13467-3_6

141

142 6 The Representations of sl.3IC/

between the irreducible representations of SU.3/ and the irreducible representationsof sl.3IC/. Furthermore, since SU.3/ is compact, Theorem 4.28 then tells us thatall finite-dimensional representations of SU.3/—and thus, also, of sl.3IC/—arecompletely reducible.

It is desirable, however, to avoid relying unnecessarily on Theorem 5.6, whichin turn relies on the Baker–Campbell–Hausdorff formula. If we look the repre-sentations from the Lie algebra point of view, we can classify the irreduciblerepresentations of sl.3IC/ without knowing that they come from representationsof SU.3/. Of course, classifying the irreducible representations of sl.3IC/ doesnot tell one what a general representation of sl.3IC/ looks like, unless one knowscomplete reducibility. Nevertheless, it is possible to give an algebraic proof ofcomplete reducibility, without referring to the group SU.3/. This proof is givenin the setting of general semisimple Lie algebras in Sect. 10.3, but it should be fairlyeasy to specialize the argument to the sl.3IC/ case.

Meanwhile, if we look at the representations from the group point of view,we can construct the irreducible representations of SU.3/ without knowing thatevery representation of sl.3IC/ gives rise to a representation of SU.3/. Indeed, theirreducible representations of SU.3/ are constructed as subspaces of tensor productsof several copies of the standard representation with several copies of the dual of thestandard representation. Since the standard representation and its dual are defineddirectly at the level of the group SU.3/, there is no need to appeal to Theorem 5.6.

In short, this chapter provides a self-contained classification of the irreduciblerepresentations of both SU.3/ and sl.3IC/, without needing to know the resultsof Chapter 5. We establish results for sl.3IC/ first, and then pass to SU.3/(Theorem 6.8).

6.2 Weights and Roots

We will use the following basis for sl.3IC/:

H1 D0@1 0 0

0 �1 00 0 0

1A ; H2 D

0@0 0 0

0 1 0

0 0 �1

1A ;

X1 D0@0 1 00 0 0

0 0 0

1A ; X2 D

0@ 0 0 00 0 1

0 0 0

1A ; X3 D

0@0 0 10 0 0

0 0 0

1A ;

Y1 D0@0 0 01 0 0

0 0 0

1A ; Y2 D

0@0 0 00 0 0

0 1 0

1A ; Y3 D

0@0 0 00 0 0

1 0 0

1A .

6.2 Weights and Roots 143

Note that the span hH1;X1; Y1i of H1, X1, and Y1 is a subalgebra of sl.3IC/isomorphic to sl.2IC/, as can be seen by ignoring the third row and the thirdcolumn in each matrix. The subalgebra hH2;X2; Y2i is also, similarly, isomorphicto sl.2IC/. Thus, we have the following commutation relations:

ŒH1;X1� D 2X1; ŒH2;X2� D 2X2;

ŒH1; Y1� D �2Y1; ŒH2; Y2� D �2Y2;ŒX1; Y1� D H1; ŒX2; Y2� D H2.

We now list all of the commutation relations among the basis elements whichinvolve at least one of H1 and H2. (This includes some repetitions of the abovecommutation relations.)

ŒH1;H2� D 0I

ŒH1;X1� D 2X1; ŒH1; Y1� D �2Y1;ŒH2;X1� D �X1; ŒH2; Y1� D Y1I

ŒH1;X2� D �X2; ŒH1; Y2� D Y2;

ŒH2;X2� D 2X2; ŒH2; Y2� D �2Y2I

ŒH1;X3� D X3; ŒH1; Y3� D �Y3;ŒH2;X3� D X3; ŒH2; Y3� D �Y3:

(6.1)

Finally, we list all of the remaining commutation relations.

ŒX1; Y1� D H1;

ŒX2; Y2� D H2;

ŒX3; Y3� D H1 CH2I

ŒX1;X2� D X3; ŒY1; Y2� D �Y3;ŒX1; Y2� D 0; ŒX2; Y1� D 0I

ŒX1;X3� D 0; ŒY1; Y3� D 0;

ŒX2;X3� D 0; ŒY2; Y3� D 0I

ŒX2; Y3� D Y1; ŒX3; Y2� D X1;

ŒX1; Y3� D �Y2; ŒX3; Y1� D �X2:

All of our analysis of the representations of sl.3IC/ will be in terms of the abovebasis. From now on, all representations of sl.3IC/ will be assumed to be finitedimensional and complex linear.

Our basic strategy in classifying the representations of sl.3IC/ is to simultane-ously diagonalize�.H1/ and �.H2/. (See Sect. A.8 for information on simultaneous

144 6 The Representations of sl.3IC/

diagonalization.) Since H1 and H2 commute, �.H1/ and �.H2/ will also commute(for any representation �) and so there is at least a chance that �.H1/ and �.H2/

can be simultaneously diagonalized. (Compare Proposition A.16.)

Definition 6.1. If .�; V / is a representation of sl.3IC/, then an ordered pair� D .m1;m2/ 2 C2 is called a weight for � if there exists v ¤ 0 in V such that

�.H1/v D m1v;

�.H2/v D m2v. (6.2)

A nonzero vector v satisfying (6.2) is called a weight vector corresponding to theweight�. If� D .m1;m2/ is a weight, then the space of all vectors v satisfying (6.2)is the weight space corresponding to the weight �. The multiplicity of a weight isthe dimension of the corresponding weight space.

Thus, a weight is simply a pair of simultaneous eigenvalues for �.H1/ and�.H2/. It is easily shown that isomorphic representations have the same weightsand multiplicities.

Proposition 6.2. Every representation of sl.3IC/ has at least one weight.

Proof. Since we are working over the complex numbers, �.H1/ has at least oneeigenvalue m1 2 C. Let W � V be the eigenspace for �.H1/ with eigenvalue m1.Since ŒH1;H2� D 0, �.H2/ commutes with �.H1/, and, so, by Proposition A.2,�.H2/must mapW into itself. Then the restriction of�.H2/ toW must have at leastone eigenvector w with eigenvalue m2 2 C, which means that w is a simultaneouseigenvector for �.H1/ and �.H2/ with eigenvaluesm1 and m2. ut

Every representation � of sl.3IC/ can be viewed, by restriction, as a representa-tion of the subalgebras hH1;X1; Y1i and hH2;X2; Y2i, both of which are isomorphicto sl.2IC/.Proposition 6.3. If .�; V / is a representation of sl.3IC/ and � D .m1;m2/ is aweight of V , then both m1 andm2 are integers.

Proof. Apply Point 1 of Theorem 4.34 to the restriction of � to hH1;X1; Y1i and tothe restriction of � to hH2;X2; Y2i. ut

Our strategy now is to begin with one simultaneous eigenvector for �.H1/ and�.H2/ and then to apply �.Xj / or �.Yj / and see what the effect is. The followingdefinition is relevant in this context.

Definition 6.4. An ordered pair ˛ D .a1; a2/ 2 C2 is called a root if

1. a1 and a2 are not both zero, and2. there exists a nonzero Z 2 sl.3IC/ such that

ŒH1;Z� D a1Z;

ŒH2;Z� D a2Z.

The element Z is called a root vector corresponding to the root ˛.

6.2 Weights and Roots 145

Condition 2 in the definition says that Z is a simultaneous eigenvector for adH1and adH2 . This means that Z is a weight vector for the adjoint representation withweight .a1; a2/. Thus, taking into account Condition 1, we may say that the rootsare precisely the nonzero weights of the adjoint representation. The commutationrelations (6.1) tell us that we have the following six roots for sl.3IC/:

˛ Z.2;�1/ X1.�1; 2/ X2.1; 1/ X3

˛ Z.�2; 1/ Y1.1;�2/ Y2.�1;�1/ Y3

: (6.3)

Note thatH1 andH2 are also simultaneous eigenvectors for adH1 and adH2 , but theyare not root vectors because the simultaneous eigenvalues are both zero. Since thevectors in (6.3), together with H1 and H2, form a basis for sl.3IC/, it is not hardto show that the roots listed in (6.3) are the only roots (Exercise 1). These six rootsform a “root system,” conventionally called A2. (For much more information aboutroot systems, see Chapter 8.)

It is convenient to single out the two roots corresponding to X1 and X2:

˛1 D .2;�1/I ˛2 D .�1; 2/; (6.4)

which we call the positive simple roots. They have the property that all of the rootscan be expressed as linear combinations of ˛1 and ˛2 with integer coefficients, andthese coefficients are (for each root) either all greater than or equal to zero or all lessthan or equal to zero. This is verified by direct computation:

.2;�1/ D ˛1I .�1; 2/ D ˛2I .1; 1/ D ˛1 C ˛2;

with the remaining three roots being the negatives of the ones above. The decisionto designate ˛1 and ˛2 as the positive simple roots is arbitrary; any other pair ofroots with similar properties would do just as well.

The significance of the roots for the representation theory of sl.3IC/ is containedin the following lemma, which is the analog of Lemma 4.33 in the sl.2IC/ case.

Lemma 6.5. Let ˛ D .a1; a2/ be a root and let Z˛ 2 sl.3IC/ be a correspondingroot vector. Let � be a representation of sl.3IC/, let � D .m1;m2/ be a weight for� , and let v ¤ 0 be a corresponding weight vector. Then we have

�.H1/�.Z˛/v D .m1 C a1/�.Z˛/v;

�.H2/�.Z˛/v D .m2 C a2/�.Z˛/v.

Thus, either �.Z˛/v D 0 or �.Z˛/v is a new weight vector with weight

�C ˛ D .m1 C a1;m2 C a2/.

146 6 The Representations of sl.3IC/

Proof. By the definition of a root, we have the commutation relation ŒH1;Z˛� Da1Z˛ . Thus,

�.H1/�.Z˛/v D .�.Z˛/�.H1/C a1�.Z˛// v

D �.Z˛/.m1v/C a1�.Z˛/v

D .m1 C a1/�.Z˛/v.

A similar argument allows us to compute �.H2/�.Z˛/v. ut

6.3 The Theorem of the Highest Weight

If we have a representation with a weight � D .m1;m2/, then by applying the rootvectors X1; X2; X3; Y1; Y2, and Y3, we obtain new weights of the form � C ˛,where ˛ is the root. Of course, some of the time, �.Z˛/v will be zero, in whichcase � C ˛ is not necessarily a weight. In fact, since our representation is finitedimensional, there can be only finitely many weights, so we must get zero quiteoften. By analogy to the classification of the representations of sl.2IC/, we wouldlike to single out in each representation a “highest” weight and then work fromthere. The following definition gives the “right” notion of highest.

Definition 6.6. Let ˛1 D .2;�1/ and ˛2 D .�1; 2/ be the roots introduced in (6.4).Let �1 and �2 be two weights. Then �1 is higher than �2 (or, equivalently, �2 islower than �1) if �1 � �2 can be written in the form

�1 � �2 D a˛1 C b˛2 (6.5)

with a � 0 and b � 0. This relationship is written as �1 � �2 or �2 � �1.If � is a representation of sl.3IC/, then a weight �0 for � is said to be a highest

weight if for all weights � of � , � � �0.

Note that the relation of “higher” is only a partial ordering; for example, ˛1 �˛2is neither higher nor lower than 0. In particular, a finite set of weights need not havea highest element. Note also that the coefficients a and b in (6.5) do not have to beintegers, even if both �1 and �2 have integer entries. For example, .1; 0/ is higherthan .0; 0/ since .1; 0/ D 2

3˛1 C 1

3˛2.

We are now ready to state the main theorem regarding the irreducible represen-tations of sl .3IC/, the theorem of the highest weight. The proof of the theorem isfound in Sect. 6.4.

Theorem 6.7. 1. Every irreducible representation � of sl.3IC/ is the direct sumof its weight spaces.

2. Every irreducible representation of sl.3IC/ has a unique highest weight �.

6.3 The Theorem of the Highest Weight 147

3. Two irreducible representations of sl.3IC/ with the same highest weight areisomorphic.

4. The highest weight � of an irreducible representation must be of the form

� D .m1;m2/;

where m1 and m2 are non-negative integers.5. For every pair .m1;m2/ of non-negative integers, there exists an irreducible

representation of sl.3IC/ with highest weight .m1;m2/.

We will also prove (without appealing to Theorem 5.6) a similar result for thegroup SU.3/. Since every irreducible representation of SU.3/ gives rise to anirreducible representation of sl.3IC/ Š su.3/C, the only nontrivial matter is toprove Point 5 for SU.3/.

Theorem 6.8. For every pair .m1;m2/ of non-negative integers, there exists anirreducible representation … of SU.3/ such that the associated representation �of sl.3IC/ has highest weight .m1;m2/.

One might naturally attempt to construct representations of SU.3/ by a methodsimilar to that used in Example 4.10, acting on spaces of homogeneous polynomialson C3. This is, indeed, possible and the resulting representations of SU.3/ turn outto be irreducible. Not every irreducible representation of SU.3/, however, arises inthis way, but only those with highest weight of the form .0;m/. See Exercise 8.

For � D .m1;m2/ 2 C2, we may say that � is an integral element if m1 andm2 are integers and that � is dominant if m1 and m2 are real and non-negative.Thus, the set of possible highest weights in Theorem 6.7 are the dominant integralelements. Figure 6.1 shows the roots and dominant integral elements for sl.3IC/.

Fig. 6.1 The roots (arrows)and dominant integralelements (black dots), shownin the obvious basis

a2

a1

148 6 The Representations of sl.3IC/

This picture is made using the obvious basis for the space of weights; that is, thex-coordinate is the eigenvalue of H1 and the y-coordinate is the eigenvalue of H2.Once we have introduced the Weyl group (Sect. 6.6), we will see the same picturerendered using a Weyl-invariant inner product, which will give a more symmetricview of the situation.

Note the parallels between this result and the classification of the irreduciblerepresentations of sl.2IC/: In each irreducible representation of sl.2IC/, �.H/is diagonalizable, and there is a largest eigenvalue of �.H/. Two irreduciblerepresentations of sl.2IC/ with the same largest eigenvalue are isomorphic. Thehighest eigenvalue is always a non-negative integer and every non-negative integeris the highest weight of some irreducible representation.

6.4 Proof of the Theorem

The proof consists of a series of propositions.

Proposition 6.9. In every irreducible representation .�; V / of sl.3IC/, the opera-tors �.H1/ and �.H2/ can be simultaneously diagonalized; that is, V is the directsum of its weight spaces.

Proof. LetW be the sum of the weight spaces in V . Equivalently,W is the space ofall vectors w 2 V such that w can be written as a linear combination of simultaneouseigenvectors for�.H1/ and �.H2/. Since (Proposition 6.2)� always has at least oneweight,W ¤ f0g.

On the other hand, Lemma 6.5 tells us that if Z˛ is a root vector correspondingto the root ˛, then �.Z˛/maps the weight space corresponding to � into the weightspace corresponding to �C ˛. Thus, W is invariant under the action of each of theroot vectors,X1;X2;X3; Y1; Y2, and Y3. SinceW is certainly also invariant under theaction of H1 and H2, W is invariant under all of sl.3IC/. Thus, by irreducibility,W D V . Finally, since, by Proposition A.17, weight vectors with distinct weightsare independent, V is actually the direct sum of its weight spaces. utDefinition 6.10. A representation .�; V / of sl.3IC/ is said to be a highest weightcyclic representation with weight � D .m1;m2/ if there exists v ¤ 0 in V suchthat

1. v is a weight vector with weight �,2. �.Xj /v D 0, for j D 1; 2; 3,3. the smallest invariant subspace of V containing v is all of V .

Proposition 6.11. Let .�; V / be a highest weight cyclic representation of sl.3IC/with weight �. Then the following results hold.

1. The representation � has highest weight �.2. The weight space corresponding to the weight � is one dimensional.

6.4 Proof of the Theorem 149

Before turning to the proof of this proposition, let us record a simple lemmathat applies to arbitrary Lie algebras and which will be useful also in the setting ofgeneral semisimple Lie algebras.

Lemma 6.12 (Reordering Lemma). Suppose that g is any Lie algebra and that� is a representation of g. Suppose that X1; : : : ; Xm is an ordered basis for g as avector space. Then any expression of the form

�.Xj1/�.Xj2/ � � ��.XjN /; (6.6)

can be expressed as a linear combination of terms of the form

�.Xm/km�.Xm�1/km�1 � � ��.X1/k1 (6.7)

where each kl is a non-negative integer and where k1 C k2 C � � � C km � N .

Proof. The idea is to use the commutation relations of g to re-order the factorsinto the desired order, at the expense of generating terms with one fewer factors,which then be handled by the same method. To be more formal, we use inductionon N . If N D 1, there is nothing to do: Any expression of the form �.Xj / is ofthe form (6.7) with kj D 1 and all the other kl ’s equal to zero. Assume, then, thatthe result holds for a product of at mostN factors, and consider an expression of theform (6.6) with N C 1 factors. By induction, we can assume that the last N factorsare in the desired form, giving an expression of the form

�.Xj /�.Xm/km�.Xm�1/km�1 � � ��.X1/k1

with k1 C � � � C km D N .We now move the factor of �.Xj / to the right one step at a time until it is in the

right spot. Each time we have �.Xj /�.Xk/ somewhere in the expression we use therelation

�.Xj /�.Xk/ D �.Xk/�.Xj /C �.ŒXj ;Xk�/

D �.Xk/�.Xj /CXl

cjkl�.Xl/;

where the constants cjkl are the structure constants for the basis fXj g (Definition3.10). Each commutator term has at most at most N factors. Thus, we ultimatelyobtain several terms with N factors, which can be handled by induction, and oneterm with N factors that is of the desired form (once �.Xj / finally gets to the rightspot). ut

We now proceed with the proof of Proposition 6.11.

150 6 The Representations of sl.3IC/

Proof. Let v be as in the definition. Consider the subspace W of V spanned byelements of the form

w D �.Yj1/�.Yj2/ � � ��.YjN /v (6.8)

with each jl equal to 1, 2, or 3 and N � 0. (If N D 0, then w D v.) We now claimthat W is invariant. We take as our basis for sl.3IC/ the elements X1, X2, X3, H1,H2, Y1, Y2, and Y3, in that order. If we apply a basis element to w, the lemma tellsus that we can rewrite the resulting vector as a linear combination of terms in whichthe �.Xj /’s act first, the �.Hj /’s act second, and the �.Yj /’s act last, and all ofthese are applied to the vector v. Since v is annihilated by each �.Xj /, any termhaving a positive power of anyXj is simply zero. Since v is an eigenvector for each�.Hj /, any factors of �.Hj / acting on v can be replaced by constants. That leavesonly factors of �.Yj / applied to v, which means that we have a linear combinationof vectors of the form (6.8). Thus,W is invariant and contains v, so W D V .

Now, Y1, Y2, and Y3 are root vectors with roots �˛1, �˛2, and �˛1 � ˛2,respectively. Thus, by Lemma 6.5, each element of the form (6.8) with N > 0

is a weight vector with weight lower than �. Thus, the only weight vectors withweight � are multiples of v. utProposition 6.13. Every irreducible representation of sl.3IC/ is a highest weightcyclic representation, with a unique highest weight �.

Proof. We have already shown that every irreducible representation � is the directsum of its weight spaces. Since the representation is finite dimensional, there can beonly finitely many weights, so there must be a maximal weight �, that is, such thatthere is no weight strictly higher than �. Thus, for any nonzero weight vector v withweight �, we must have

�.Xj /v D 0; j D 1; 2; 3:

Since � is irreducible, the smallest invariant subspace containing v must be thewhole space; therefore, the representation is highest weight cyclic. utProposition 6.14. Suppose .�; V / is a completely reducible representation ofsl.3IC/ that is also highest weight cyclic. Then � is irreducible.

As it turns out, every finite-dimensional representation of sl.3IC/ is completelyreducible. This claim can be verified analytically (by passing to the simply con-nected group SU.3/ and using Theorem 4.28) or algebraically (as in Sect. 10.3). Wedo not, however, require this result here, since we will only apply Proposition 6.14to representations that are manifestly completely reducible.

Meanwhile, it is tempting to think that any representation with a cyclic vector(that is, a vector satisfying Point 3 of Definition 6.10) must be irreducible, but thisis false. (What is true is that if every nonzero vector in a representation is cyclic,then the representation is irreducible.) Thus, Proposition 6.14 relies on the specialform of the cyclic vector in Definition 6.10.

6.4 Proof of the Theorem 151

Proof. Let .�; V / be a highest weight cyclic representation with highest weight �and let v be a weight vector with weight �. By assumption, V decomposes as adirect sum of irreducible representations

V ŠMj

Vj . (6.9)

By Proposition 6.9, each of the Vj ’s is the direct sum of its weight spaces. Sincethe weight � occurs in V , it must occur in some Vj (compare the last part ofProposition A.17). But by Proposition 6.11, v is (up to a constant) the only vector inV with weight �. Thus, Vj is an invariant subspace containing v, which means thatVj D V . There is, therefore, only one term in the sum (6.9), and V is irreducible. utProposition 6.15. Two irreducible representations of sl.3IC/ with the same high-est weight are isomorphic.

Proof. Suppose .�; V / and .;W / are irreducible representations with the samehighest weight � and let v and w be the highest weight vectors for V and W ,respectively. Consider the representation V ˚ W and let U be smallest invariantsubspace of V ˚ W which contains the vector .v;w/. Then U is a highest weightcyclic representation. Furthermore, since V ˚ W is, by definition, completelyreducible, it follows from Proposition 4.26 that U is completely reducible. Thus,by Proposition 6.14, U is irreducible.

Consider now the two “projection” maps P1 and P2, mapping V ˚W to V andW , respectively, and given by

P1.v;w/ D vI P2.v;w/ D w:

Since P1 and P2 are easily seen to be intertwining maps, their restrictions toU � V ˚ W are also intertwining maps. Now, neither P1jU nor P2jU is the zeromap, since both are nonzero on .v;w/. Moreover, U , V , and W are all irreducible.Therefore, by Schur’s lemma, P1jU is an isomorphism of U with V and P2jU is anisomorphism of U with W , showing that V Š U Š W . utProposition 6.16. If � is an irreducible representation of sl.3IC/ with highestweight � D .m1;m2/, then m1 andm2 non-negative integers.

Proof. By Proposition 6.3, m1 and m2 are integers. If v is a weight vectorwith weight �, then �.X1/v and �.X2/v must be zero, or � would not be thehighest weight for � . Thus, if we then apply Point 1 of Theorem 4.34 to therestrictions of � to hH1;X1; Y1i and hH2;X2; Y2i, we conclude that m1 and m2

are non-negative. utProposition 6.17. If m1 and m2 are non-negative integers, then there exists anirreducible representation of sl.3IC/ with highest weight � D .m1;m2/.

152 6 The Representations of sl.3IC/

Proof. Since the trivial representation is an irreducible representation with highestweight .0; 0/, we need only construct representations with at least one ofm1 andm2

positive.First, we construct two irreducible representations, with highest weights .1; 0/

and .0; 1/, which we call the fundamental representations. The standard represen-tation of sl.3IC/, acting on C3 in the obvious say, is easily seen to be irreducible. Ithas weight vectors e1; e2, and e3, with corresponding weights .1; 0/; .�1; 1/, and.0;�1/, and with highest weight is .1; 0/. The dual of the standard representation,given by

�.Z/ D �Ztr (6.10)

for allZ 2 sl.3IC/, is also irreducible. It also has weight vectors e1, e2, and e3, withcorresponding weights .�1; 0/, .1;�1/, and .0; 1/ and with highest weight .0; 1/.

Let .�1; V1/ and .�2; V2/ be the standard representation and its dual, respectively,and let v1 D e1 and v2 D e3 be the respective highest weight vectors. Now, considerthe representation �m1;m2 given by

.V1 ˝ � � � ˝ V1/˝ .V2 ˝ � � � ˝ V2/; (6.11)

where V1 occurs m1 times and V2 occurs m2 times. The action of sl.3IC/ on thisspace is given by the obvious extension of Definition 4.20 to multiple factors. It theneasy to check that the vector

vm1;m2 D v1 ˝ v1 � � � ˝ v1 ˝ v2 ˝ v2 � � � ˝ v2

is a weight vector with weight .m1;m2/ and that vm1;m2 is annihilated by�m1;m2.Xj /, j D 1; 2; 3.

Now let W be the smallest invariant subspace containing vm1;m2 . Assumingthat �m1;m2 is completely reducible, W will also be completely reducible andProposition 6.14 will tell us that W is the desired irreducible representation withhighest weight .m1;m2/.

It remains only to establish complete reducibility. Note first that both the standardrepresentation and its dual are “unitary” for the action of su.3/, meaning that�.X/� D ��.X/ for all X 2 su.3/. Meanwhile, it is easy to verify (Exercise 5)that if V and W are inner product spaces, then there is a unique inner product onV ˝W for which

hv1 ˝ w1; v2 ˝ w2i D hv1; v2i hw1;w2i

for all v1; v2 2 V and w1;w2 2 W . Extending this construction to tensor productsof several vector spaces, use the standard inner product on C3 to construct an innerproduct on the space in (6.11). It is then easy to check that �m1;m2 is also unitary forthe action of su.3/. Thus, by Proposition 4.27, �m1;m2 is completely reducible underthe action of su.3/ and thus, also, under the action of sl.3IC/ Š su.3/C. ut

6.5 An Example: Highest Weight .1; 1/ 153

We have now completed the proof of Theorem 6.7.

Proof of Theorem 6.8. The standard representation �1 of sl.3IC/ comes from thestandard representation …1 of SU.3/, and similarly for the dual of the standardrepresentation. By taking tensor products, we see that there is a representation…m1;m2 corresponding to the representation �m1;m2 of sl.3IC/. The irreducibleinvariant subspaceW in the proof of Proposition 6.17 is then also invariant under theaction of SU.3/, so that the restriction of …m1;m2 to W is the desired representationof SU.3/. ut

6.5 An Example: Highest Weight .1; 1/

To obtain the irreducible representation with highest weight .1; 1/, we take thetensor product of the standard representation and its dual, take the highest weightvector in the tensor product, and then consider the space obtained by repeatedapplications of the operators �1;1.Yj /, j D 1; 2; 3. Since, however, Y3 D �ŒY1; Y2�,it suffices to apply only �1;1.Y1/ and �1;1.Y2/.

Now, the standard representation has highest weight e1 and the action of theoperators �.Y1/ D Y1 and �.Y2/ D Y2 is given by

Y1e1 D e2 Y1e2 D 0 Y1e3 D 0

Y2e1 D 0 Y2e2 D e3 Y2e3 D 0:

For the dual of the standard representation, let use the notation Z D �Ztr, so that�.Z/ D Z. If we introduce the new basis

f1 D e3I f2 D �e2I f3 D e1;

then the highest weight is f1 and we have

Y1f1 D 0 Y1f2 D f3 Y1f3 D 0

Y2f1 D f2 Y2f2 D 0 Y2f3 D 0:

We must now repeatedly apply the operators

�1;1.Y1/ D Y1 ˝ I C I ˝ Y1

�1;1.Y2/ D Y2 ˝ I C I ˝ Y2 (6.12)

until we get zero. This calculation is contained in the following chart. Here, there aretwo arrows coming out of each vector. Of these, the left arrow indicates the actionof Y1 ˝ I C I ˝ Y1 and the right arrow indicates the action of Y2 ˝ I C I ˝ Y2. Tosave space, we omit the tensor product symbol, writing, for example, e2f2 insteadof e2 ˝ f2.

154 6 The Representations of sl.3IC/

e1f1. &

e2f1 e1f2. # # &

0 e3f1 C e2f2 e2f2 C e1f3 0

. # # &e2f3 2e3f2 2e2f3 e3f2

. # # & . # # &0 e3f3 2e3f3 0 2e3f3 e3f3 0

A basis for the space spanned by these vectors is e1f1, e2f1, e1f2, e3f1 C e2f2,e2f2 C e1f3, e2f3, e3f2, and e3f3. Thus, the dimension of this representation is8; it is (isomorphic to) the adjoint representation. Now, e1, e2, and e3 have weights.1; 0/, .�1; 1/, and .0;�1/, respectively, whereas f1, f2, and f3 have weights .0; 1/,.1;�1/, and .�1; 0/, respectively. From (6.12), we can see that the weight for ej˝fkis just the sum of the weight for ej and the weight for fk . Thus, the weights forthe basis elements listed above are .1; 1/, .�1; 2/, .2;�1/, .0; 0/ (twice), .1;�2/,.�2; 1/, and .�1;�1/. Each weight has multiplicity 1 except for .0; 0/, which hasmultiplicity 2. See the first image in Figure 6.4.

6.6 The Weyl Group

This section describes an important symmetry of the representations of SU.3/,involving something called the Weyl group. Our discussion follows the compact-group approach to the Weyl group. See Sect. 7.4 for the Lie algebra approach, in thecontext of general semisimple Lie algebras.

Definition 6.18. Let h be the two-dimensional subspace of sl.3IC/ spanned byH1

and H2. Let N be the subgroup of SU.3/ consisting of those A 2 SU.3/ suchthat AdA.H/ is an element of h for all H in h. Let Z be the subgroup of SU.3/consisting of those A 2 SU.3/ such that AdA.H/ D H for all H 2 h.

The space h is a Cartan subalgebra of sl.3IC/. It is a straightforward exercise(Exercise 9) to verify that Z and N are subgroups of SU.3/ and that Z is a normalsubgroup of N . This leads us to the definition of the Weyl group.

Definition 6.19. The Weyl group of SU.3/, denoted W , is the quotientgroupN=Z.

The primary significance of W for the representation theory of SU.3/ is thatit gives rise to a symmetry of the weights occurring in a fixed representation; seeTheorem 6.22. We can define an action of W on h as follows. For each element wof W , choose an element A of the corresponding coset in N . Then for H in h wedefine the action w �H of w on H by

w �H D AdA.H/:

6.6 The Weyl Group 155

To see that this action is well defined, suppose B is an element of the same cosetas A. Then B D AC with C 2 Z and, thus,

AdB.H/ D AdA.AdC .H// D AdA.H/;

by the definition ofZ. Note that by definition, if w �H D H for allH 2 h, then w isthe identity element ofW (that is, the associated A 2 N is actually in Z). Thus, wemay identifyW with the group of linear transformations of h that can be expressedin the form H 7! w �H for some w 2 W .

Proposition 6.20. The group Z consists precisely of the diagonal matrices insideSU.3/, namely the diagonal matrices with diagonal entries .ei� ; ei�; e�i.�C�// with�; � 2 R. The group N consists of precisely those matrices A 2 SU.3/ such thatfor each j D 1; 2; 3, there exist kj 2 f1; 2; 3g and �j 2 R such that Aej D ei�j ekj .Here, e1;e2; e3 is the standard basis for C3.

The Weyl group W D N=Z is isomorphic to the permutation group on threeelements.

Proof. Suppose A is in Z, which means that A commutes with all elements of h,includingH1, which has eigenvectors e1, e2, and e3, with corresponding eigenvalues1, �1, and 0. SinceA commutes withH1, it must preserve each of these eigenspaces(Proposition A.2). Thus, Aej must be a multiple of ej for each j , meaning that A isdiagonal. Conversely, any diagonal matrix in SU.3/ does indeed commute not onlywith H1 but also with H2 and, thus, with every element of h.

Suppose, now, that A is in N . Then AH1A�1 must be in h and therefore must

be diagonal, meaning that e1, e2, and e3 are eigenvectors for AH1A�1, with the

same eigenvalues 1;�1; 0 asH1, but not necessarily in the same order. On the otherhand, the eigenvectors of AH1A

�1 must be Ae1, Ae2, and Ae3. Thus, Aej must bea multiple of some ekj , and the constant must have absolute value 1 if A is unitary.Conversely, if Aej is a multiple of ekj for each j , then for any (diagonal) matrixHin h, the matrix AHA�1 will again be diagonal and thus in h.

Finally, if A maps each ej to a multiple of ekj , for some kj depending on j ,then for each diagonal matrix H , the matrix AHA�1 will be diagonal with diagonalentries rearranged by the permutation j 7! kj . For any permutation, we can choosethe constants to that the map taking ej to ei�j ekj has determinant 1, showing thatevery permutation actually arises in this way. Thus, W—which we think of as thegroup of linear transformations of h of the form AdA, A 2 N—is isomorphic to thepermutation group on three elements. ut

We want to show that the Weyl group is a symmetry of the weights of any finite-dimensional representation of sl.3IC/. To understand this, we need to adopt a lessbasis-dependent view of the weights. We have defined a weight as a pair .m1;m2/

of simultaneous eigenvalues for �.H1/ and �.H2/. However, if a vector v is aneigenvector for �.H1/ and �.H2/ then it is also an eigenvector for �.H/ for anyelement H of the space h spanned by H1 and H2, and the eigenvalues will dependlinearly on H in h. Thus, we may think of a weight not as a pair of numbers but asa linear functional on h.

156 6 The Representations of sl.3IC/

It is then convenient to use an inner product on h to identity linear functionals onh with elements of h itself. We define the inner product of H and H 0 in h by

˝H;H 0˛ D trace.H�H 0/; (6.13)

or, explicitly,

hdiag.a; b; c/; diag.d; e; f /i D Nad C Nbe C Ncf;where diag.�; �; �/ is the diagonal matrix with the indicated diagonal entries. If � isa linear functional on h, there is (Proposition A.11) a unique vector � in h suchthat � may be represented as �.H/ D h�;H i for all H 2 h. If we represent thelinear functional in the previous paragraph in this way, we arrive at a new, basis-independent notion of a weight.

Definition 6.21. Let h be the subspace of sl.3IC/ spanned by H1 and H2 and let.�; V / be a representation of sl.3IC/. An element � of h is called a weight for � ifthere exists a nonzero vector v in V such that

�.H/v D h�;H i vfor all H in h. Such a vector v is called a weight vector with weight �.

If � is a weight in our new sense, the ordered pair .m1;m2/ in Definition 6.1 isgiven by

m1 D h�;H1i I m2 D h�;H2i :

It is easy to check that for all U 2 N , the adjoint action of U on h preservesthe inner product in (6.13). Thus, the action of the Weyl group on h is unitary:hw �H;w �H 0i D hH;H 0i. Since the roots are just the nonzero weights of theadjoint representation, we now also think of the roots as elements of h:

Theorem 6.22. Suppose that .…; V / is a finite-dimensional representation ofSU.3/ with associated representation .�; V / of sl.3IC/. If � 2 h is a weight forV then w � � is also a weight of V with the same multiplicity. In particular, the rootsare invariant under the action of the Weyl group.

Proof. Suppose that � is a weight for V with weight vector v. Then for all U 2 NandH 2 h, we have

�.H/….U /v D ….U /.….U /�1�.H/….U //v

D ….U /�.U�1HU/v

D ˝�;U�1HU

˛….U /v:

Here, we have used that U is in N , which guarantees that U�1HU is, again, in h.Thus, if w is the Weyl group element represented by U , we have

6.6 The Weyl Group 157

�.H/….U /v D ˝�;w�1 �H ˛….U /v D hw � �;H i….U /v:

We conclude that ….U /v is a weight vector with weight w � �.The same sort of reasoning shows that ….U / is an invertible map of the weight

space with weight � onto the weight space with weight w � �, whose inverse is….U /�1. This means that the two weights have the same multiplicity. ut

To represent the basic weights, .1; 0/ and .0; 1/, in our new approach, we lookfor diagonal, trace-zero matrices �1 and �2 such that

h�1;H1i D 1; h�1;H2i D 0

h�2;H1i D 0; h�2;H2i D 1:

These are easily found as

�1 D diag.2=3;�1=3;�1=3/I �2 D diag.1=3; 1=3;�2=3/:

The positive simple roots .2;�1/ and .�1; 2/ are then represented as

˛1 D 2�1 � �2 D diag.1;�1; 0/I˛2 D ��1 C 2�2 D diag.0; 1;�1/: (6.14)

Note that both ˛1 and ˛2 have lengthp2 and h˛1; ˛2i D �1. Thus, the angle �

between them satisfies cos � D �1=2, so that � D 2�=3.Figure 6.2 shows the same information as Figure 6.1, namely, the roots and

the dominant integral elements, but now drawn relative to the Weyl-invariant innerproduct in (6.13). We draw only the two-dimensional real subspace of h consisting

Fig. 6.2 The roots anddominant integral elementsfor sl.3IC/, computedrelative to a Weyl-invariantinner product

a2

a1

m1

m2

158 6 The Representations of sl.3IC/

Fig. 6.3 The Weyl group isthe symmetry group of theindicated equilateral triangle

a2

a1

of those elements � such that h�;H1i and h�;H2i are real, since all the roots andweights have this property. Let w.1;2;3/ denote the Weyl group element that acts bycyclically permuting the diagonal entries of eachH 2 h. Then w.1;2;3/ takes ˛1 to ˛2and ˛2 to �.˛1 C ˛2/, which is a counterclockwise rotation by 2�=3 in Figure 6.2.Similarly, if w.1;2/ the element that interchanges the first two diagonal entries ofH 2 h, then w.1;2/ maps ˛1 to �˛1 and ˛2 to ˛1 C ˛2. Thus, w.1;2/ is the reflectionacross the line perpendicular to ˛1. The reader is invited to compute the action of theremaining elements of the Weyl group and to verify that it is the symmetry group ofthe equilateral triangle in Figure 6.3.

We previously defined a pair .m1;m2/ to be integral ifm1 andm2 are integers anddominant if m1 � 0 and m2 � 0. These concepts translate into our new languageas follows. If � 2 h, then � is integral if h�;H1i and h�;H2i are integers and �is dominant if h�;H1i � 0 and h�;H2i � 0. Geometrically, the set of dominantelements is a sector spanning an angle of �=3.

6.7 Weight Diagrams

In this section, we display the weights and multiplicities for several irreduciblerepresentations of sl.3IC/. Figure 6.4 covers the irreducible representations withhighest weighs .1; 1/, .1; 2/, .0; 4/, and .2; 2/. The first of these examples wasanalyzed in Sect. 6.5, and the other examples can be analyzed by the same method.In each part of the figure, the arrows indicate the roots, the two black lines indicatethe boundary of the set of dominant elements, and the dashed lines indicate theboundary of the set of points lower than the highest weight. Each weight of

6.8 Further Properties of the Representations 159

2

22

22

22

2

2

2

2

2

2

333333

Fig. 6.4 Weight diagrams for representations with highest weights .1; 1/, .1; 2/, .0; 4/, and .2; 2/

a particular representation is indicated by a black dot, with a number next toa dot indicating its multiplicity. A dot without a number indicates a weight ofmultiplicity 1.

Our last example is the representation with highest weight .9; 2/ (Figure 6.5),which cannot feasibly be analyzed using the method of Sect. 6.5. Instead, theweights are determined by the results of Sect. 6.8 and the multiplicities are computedusing the Kostant multiplicity formula. (See Figure 10.8 in Sect. 10.6.) See alsoExercises 11 and 12 for another approach to computing multiplicities.

6.8 Further Properties of the Representations

Although we now have a classification of the irreducible representations of sl.3IC/by means of their highest weights, there are other things we might like to knowabout the representations, such as (1) the other weights that occur, besides thehighest weight, (2) the multiplicities of those weights, and (3) the dimension of therepresentation. In this section, we establish which weights occur and state withoutproof the formula for the dimension. A formula for the multiplicities and a proof ofthe dimension formula are given in Chapter 10 in the setting of general semisimpleLie algebras.

160 6 The Representations of sl.3IC/

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

22

22

22

3

33

33

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

33

33

33

3

33

33

3

3

3

3

3

3

3

3

33

33

3

Fig. 6.5 Weight diagram for the irreducible representation with highest weight .9; 2/

Definition 6.23. If v1; : : : ; vN are elements of a real or complex vector space, theconvex hull of v1; : : : ; vN is the set of all vectors of the form

c1v1 C c2v2 C � � � C cN vN

where the cj ’s are non-negative real numbers satisfying c1 C c2 C � � � C cN D 1.

Equivalently, the convex hull of v1; : : : ; vN is the smallest convex set thatcontains all of the vj ’s.

Theorem 6.24. Let � be a dominant integral element and let V� be the irreduciblerepresentation with highest weight �. If � is a weight of V�, then � satisfies thefollowing two conditions: (1) � � � can be expressed as an integer combination ofroots, and (2) � belongs to the convex hull of W � �, the orbit of � under the actionof W .

Proof. According to the proof of Proposition 6.11, V� is spanned by vectors ofthe form in (6.8). These vectors are weight vectors with weights of the form � WD� � ˛j1 � � � � � ˛jN . Thus, every weight of V� satisfies the first property in thetheorem.

6.8 Further Properties of the Representations 161

w

Fig. 6.6 The integral element � is outside the convex hull of the orbit of �, and the element w � �is not lower than �

The second property in the theorem is based on the following idea: If � is aweight of V�, then w � � is also a weight for all w 2 W , which means that w � � islower than �. We can now argue “pictorially” that if � were not in the convex hullofW ��, there would be some w 2 W for which w � � is not lower than �, so that �could not be a weight of V�. See Figure 6.6.

We can give a more formal argument as follows. For any weight � of V�, wecan, by Exercise 10, find some w 2 W so that �0 WD w � � is dominant. Since �0is also a weight of V�, we must have �0 � �. Thus, �0 is in the quadrilateral Q�

consisting of dominant elements that are lower than � (Figure 6.7). We now arguethat the vertices of Q� are all in the convex hull. First, it is easy to see that for any�, the average of w �� over all w 2 W is zero, which means that 0 is in E�. Second,the vertices marked v1 and v2 in the figure are expressible as follows:

v1 D 1

2�C 1

2s˛1 � �

v2 D 1

2�C 1

2s˛2 � �;

where s˛1 and s˛2 are the Weyl group elements given by reflecting about the linesorthogonal to ˛1 and ˛2. Thus, all the vertices ofQ� are inE�, from which it followsthat Q� itself is contained in E�.

Now, W � � is clearly W -invariant, which means that E� is also W -invariant.Since �0 2 Q� � E�, we have � D w�1�0 2 E� as well. ut

162 6 The Representations of sl.3IC/

v1

v2

1

2

Fig. 6.7 The shaded quadrilateral is the set of all points that are dominant and lower than �

Theorem 6.25. Suppose V� is an irreducible representation with highest weight �and that � is an integral element satisfying the two conditions in Theorem 6.24.Then � is a weight of V�.

Theorem 6.25 says, in effect, that there are no unexpected holes in the set ofweights of V�. The key to the proof is the “no holes” result (Point 4 of Theorem 4.34)we previously established for sl.2IC/.Lemma 6.26. Let � be a weight of V�, let ˛ be a root, and let s˛ 2 W be thereflection about the line orthogonal to ˛. Suppose � is a point on the line segmentjoining � to s˛ � � with the property that � � � is an integer multiple of ˛. Then � isalso a weight of V�.

See Figure 6.8 for an example. Note from Figure 6.3 that for each root ˛, thereflection s˛ is an element of the Weyl group.

Proof. Since the reflections associated to ˛ and �˛ are the same, it suffices toconsider the roots ˛1, ˛2, and ˛3 WD ˛1 C ˛2. If we let H3 D H1 C H2, thenfor j D 1; 2; 3we have a subalgebra sj D ˝

Xj ; Yj ;Hj

˛isomorphic to sl.2IC/ such

that Xj is a root vector with root ˛j and Yj is a root vector with root �˛j . Since

ŒHj ;Xj � D 2Xj D ˝˛j ;Hj

˛Xj ;

we have˝˛j ;Hj

˛ D 2 for each j .Let us now fix a weight � of V� and let U be the span of all the weight vectors in

V� whose weights are of the form � C k˛j for some real number k. (These weightsare circled in Figure 6.8.) Since, by Lemma 6.5, �.Xj / and �.Yj / shift weights

6.8 Further Properties of the Representations 163

s

Fig. 6.8 Since � is a weight of V�, each of the elements � � ˛; � � 2˛; : : : ; s˛ � � must also be aweight of V�

by ˙˛j , we see that U is invariant under sj and thus constitutes a representationof sj (not necessarily irreducible). With our new perspective that roots are elementsof h, we can verify from (6.14) that for each j , we have ˛j D Hj , from whichit follows that s˛j � Hj D �Hj . Thus, if u and v are weight vectors with weights� and s˛ � � , respectively, u and v are in U and are eigenvectors for �.Hj / witheigenvalues

˝�;Hj

˛and

˝s˛ � �;Hj

˛ D ˝�; s˛ �Hj

˛ D � ˝�;Hj

˛;

respectively.If � is on the line segment joining � to s˛ �� , we see that

˝�;Hj

˛is between

˝�;Hj

˛and

˝s˛ � �;Hj

˛ D � ˝�;Hj

˛. If, in addition, � differs from � by an integer multiple

of ˛j , then˝�;Hj

˛differs from

˝�;Hj

˛by an integer multiple of

˝˛j ;Hj

˛ D 2.Thus, by applying Point 4 of Theorem 4.34 to the action of sj on U , there must bean eigenvector w for �.Hj / in U with eigenvalue l D ˝

�;Hj

˛. Since the unique

weight of the form � C k˛j for which˝� C k˛j ;Hj

˛ D ˝�;Hj

˛is the one where

� C k˛j D �, we conclude that � is a weight of V�. ut

Proof of Theorem 6.25. Suppose that � satisfies the two conditions in the theorem,and write � D � � n1˛1 � n2˛2. Consider first the case n1 � n2, so that

� D � � .n1 � n2/˛1 � n2.˛1 C ˛2/

D � � .n1 � n2/˛1 � n2˛3;

164 6 The Representations of sl.3IC/

1

2

3

s 3

s 1

Fig. 6.9 By applying Lemma 6.26 twice, we can see that � and � must be weights of V�

where ˛3 D ˛1 C˛2. If we start at � and travel in the direction of ˛3, we will hit theboundary of E� at the point

� WD � � .n1 � n2/˛1:

(See Figure 6.9.) Thus, � is inE� and must therefore be between� and s˛1 ��. Sincealso � differs from � by an integer multiple of ˛1 (namely n1 � n2) Lemma 6.26says that � is a weight of V . Meanwhile, � is between � and s˛3 � � (see, again,Figure 6.9) and differs from � by an integer multiple of ˛3 (namely n2). Thus, thelemma tells us that � must be a weight of V , as claimed. If n1 � n2, we can use asimilar argument with the roles of ˛1 and ˛2 reversed. ut

We close this section by stating the formula for the dimension of an irreduciblerepresentation of sl.3IC/. We will prove the result in Chapter 10 as a special caseof the Weyl dimension formula.

Theorem 6.27. The dimension of the irreducible representation with highestweight .m1;m2/ is

1

2.m1 C 1/.m2 C 1/.m1 Cm2 C 2/.

The reader is invited to verify this formula by direct computation in therepresentations depicted in Figure 6.4.

6.9 Exercises 165

6.9 Exercises

1. Show that the roots listed in (6.3) are the only roots.2. Let � be an irreducible finite-dimensional representation of sl.3IC/ acting on a

space V and let �� be the dual representation to � , acting on V �, as defined inSect. 4.3.3. Show that the weights of �� are the negatives of the weights of � .Hint: Choose a basis for V in which both �.H1/ and �.H2/ are diagonal.

3. Let � be an irreducible representation of sl.3IC/ with highest weight �.

(a) Let ˛3 D ˛1 C˛2 and let s˛3 denote the reflection about the line orthogonalto ˛3. Show the lowest weight for � is s˛3 � �.

(b) Show that the highest weight for the dual representation �� to � is theweight

�0 WD �s˛3 � �:

(c) Let �1 and �2 be the fundamental weights, as in Figure 6.2. If � Dm1�1Cm2�2, show that�0 D m2�1Cm1�2. That is to say, the dual to therepresentation with highest weight .m1;m2/ has highest weight .m2;m1/.

4. Consider the adjoint representation of sl.3IC/ as a representation of sl.2IC/by restricting the adjoint representation to the subalgebra spanned by X1; Y1,andH1. Decompose this representation as a direct sum of irreducible represen-tations of sl.2IC/. Which representations occur and with what multiplicity?

5. Suppose that V and W are finite-dimensional inner product spaces over C.Show that there exists a unique inner product on V ˝W such that

˝v ˝ w; v0 ˝ w0˛ D ˝

v; v0˛ ˝w;w0˛

for all v; v0 2 V and w;w0 2 W .Hint: Let fej g and ffkg be orthonormal bases for V and W , respectively. Takethe inner product on V ˝W for which fej ˝ fkg is an orthonormal basis.

6. Following the method of Sect. 6.5, work out the representation of sl.3IC/ withhighest weight .2; 0/, acting on a subspace of C3 ˝ C3. Determine all theweights of this representation and their multiplicity (i.e., the dimension of thecorresponding weight space). Verify that the dimension formula (Theorem 6.27)holds in this case.

7. Consider the nine-dimensional representation of sl.3IC/ considered inSect. 6.5, namely the tensor product of the representations with highestweights .1; 0/ and .0; 1/. Decompose this representation as a direct sum ofirreducibles. Do the same for the tensor product of two copies of the irreduciblerepresentation with highest weight .1; 0/. (Compare Exercise 6.)

166 6 The Representations of sl.3IC/

8. Let Wm denote the space of homogeneous polynomials on C3 of degree m. LetSU.3/ act onWm by the obvious generalization of the action in Example 4.10.

(a) Show that the associated representation of sl.3IC/ contains a highestweight cyclic representation with highest weight .0;m/ and highest weightvector zm3 .

(b) By imitating the proof of Proposition 4.11, show that any nonzero invariantsubspace ofWm must contain zm3 .

(c) Conclude that Wm is irreducible with highest weight .0;m/.

9. Show that Z and N (defined in Definition 6.18) are subgroups of SU.3/. Showthat Z is a normal subgroup of N .

10. Suppose � is an integral element, that is, one of the triangular lattice points inFigure 6.2. Show that there is an element w of the Weyl group such that w � � isdominant integral, that is, one of the black dots in Figure 6.2.Hint: Recall that the Weyl group is the symmetry group of the triangle inFigure 6.3.

(a) Regard the Weyl group as a group of linear transformations of h. Show that�I is not an element of the Weyl group.

(b) Which irreducible representations of sl.3IC/ have the property that theirweights are invariant under �I ?

11. Suppose .�; V / is an irreducible representation of sl.3IC/ with highest weight� and highest weight vector v0. Show that the weight space with weight� � ˛1 � ˛2 has multiplicity at most 2 and is spanned by the vectors

�.Y1/�.Y2/v0; �.Y2/�.Y1/v0:

12. Let .�; V / be the irreducible representation with highest weight .m1;m2/. As inthe proof of Proposition 6.17, choose an inner product on V such that �.X/� D��.X/ for all X 2 su.3/ � sl.3IC/. Let v0 be a highest weight vector for V ,normalized to be a unit vector, and define vectors u1 and u2 in V as

u1 D �.Y1/�.Y2/v0I u2 D �.Y2/�.Y1/v0:

Each of these vectors is either zero or a weight vector with weight��˛1�˛2.(a) Using and the commutation relations among the basis elements of sl.3IC/,

show that

hu1; u1i D m2.m1 C 1/

hu2; u2i D m1.m2 C 1/

hu1; u2i D m1m2:

Hint: Show that �.Xj /� D �.Yj / for j D 1; 2; 3.

6.9 Exercises 167

(b) Show that if m1 � 1 and m2 � 1 then u1 and u2 are linearly independent.Conclude that the weight � � ˛1 � ˛2 has multiplicity 2.

(c) Show that if m1 D 0 and m2 � 1 or m1 � 1 and m2 D 0, then the weight� � ˛1 � ˛2 has multiplicity 1.

Note: The reader may verify the results of this exercise in the representationsdepicted in Figure 6.4.

Chapter 7Semisimple Lie Algebras

In this chapter we introduce a class of Lie algebras, the semisimple algebras, forwhich we can classify the irreducible representations using a strategy similar tothe one we used for sl.3IC/. In this chapter, we develop the relevant structuresof semisimple Lie algebras. In Chapter 8, we look into the properties of the setof roots. Then in Chapter 9, we construct and classify the irreducible, finite-dimensional representations of semisimple Lie algebras. Finally, in Chapter 10,we consider several additional properties of the representations constructed inChapter 9. Meanwhile, in Chapters 11 and 12, we consider representation theoryfrom the closely related viewpoint of compact Lie groups.

7.1 Semisimple and Reductive Lie Algebras

We begin by defining the term semisimple. There are many equivalent characteriza-tions of semisimple Lie algebras. It is not, however, always easy to prove that twoof these various characterizations are equivalent. We will use an atypical definition,which allows for a rapid development of the structure of semisimple Lie algebras.Recall from Sect. 3.6 the notion of the complexification of a real Lie algebra.

Definition 7.1. A complex Lie algebra g is reductive if there exists a compactmatrix Lie groupK such that

g Š kC:

A previous version of this book was inadvertently published without the middle initial of theauthor’s name as “Brian Hall”. For this reason an erratum has been published, correcting themistake in the previous version and showing the correct name as Brian C. Hall (see DOI http://dx.doi.org/10.1007/978-3-319-13467-3_14). The version readers currently see is the correctedversion. The Publisher would like to apologize for the earlier mistake.

© Springer International Publishing Switzerland 2015B.C. Hall, Lie Groups, Lie Algebras, and Representations, GraduateTexts in Mathematics 222, DOI 10.1007/978-3-319-13467-3_7

169

170 7 Semisimple Lie Algebras

A complex Lie algebra g is semisimple if it is reductive and the center of g is trivial.

Definition 7.2. If g is a semisimple Lie algebra, a real subalgebra k of g is acompact real form of g if k is isomorphic to the Lie algebra of some compact matrixLie group and every element Z of g can be expressed uniquely as Z D X C iY ,with X; Y 2 k.

On the one hand, using Definition 7.1 gives an easy method of constructingCartan subalgebras and fits naturally with our study of compact Lie groups inPart III. On the other hand, this definition covers an apparently smaller class ofLie algebras than some of the more standard definitions. That is to say, we willprove (Theorem 7.8 and Exercise 6) that the condition in Definition 7.1 impliestwo of the standard definitions of “semisimple,” but we will not prove the reverseimplications. These reverse implications are, in fact, true, so that our definition ofsemisimplicity is ultimately equivalent to any other definition. But it is not possibleto prove the reverse implications without giving up the gains in efficiency that gowith Definition 7.1. The reader who wishes to see a development of the theorystarting from a more traditional definition of semisimplicity may consult ChapterII (along with the first several sections of Chapter I) of [Kna2].

The only time we use the compact group in Definition 7.1 is to construct the innerproduct in Proposition 7.4. In the standard treatment of semisimple Lie algebras, theKilling form (Exercise 6) is used in place of this inner product. Our use of an innerproduct in place of the bilinear Killing form substantially simplifies some of thearguments. Notably, in our construction of Cartan subalgebras (Proposition 7.11),we use that a skew self-adjoint operator is always diagonalizable. By contrast,an operator that is skew symmetric with respect to a nondegenerate bilinear formneed not be diagonalizable. Thus, the construction of Cartan subalgebras in theconventional approach is substantially more involved than in our approach.

For a complex semisimple Lie algebra g, we will always assume we have chosena compact real form k of g, so that g D kC.

Example 7.3. The following Lie algebras are semisimple:

sl.nIC/; n � 2

so.nIC/; n � 3

sp.nIC/; n � 1:

The Lie algebras gl.nIC/ and so.2IC/ are reductive but not semisimple.

Proof. It is easy to see that the listed Lie algebras are reductive, with the correspond-ing compact groupsK being SU.n/, SO.n/, Sp.n/, U.n/, and SO.2/, respectively.(Compare (3.17) in Sect. 3.6.) The Lie algebra gl.nIC/ has a nontrivial center,consisting of scalar multiples of the identity, while the Lie algebra so.2IC/ iscommutative. It remains only to show that the centers of sl.nIC/, so.nIC/, andsp.nIC/ are trivial for the indicated values of n.

7.1 Semisimple and Reductive Lie Algebras 171

Consider first the case of sl.nIC/ and let X be an element of the center ofsl.nIC/. For any 1 � j; k � n, let Ejk be the matrix with a 1 in the .j; k/ spotand zeros elsewhere. Consider the matrixHjk 2 sl.nIC/ given by

Hjk D Ejj � Ekk;

for j < k. Then we may easily calculate that

0 D ŒHjk; X� D 2XjkEjk � 2XkjEkj:

Since Ejk and Ekj are linearly independent for j < k, we conclude thatXjk D Xkj D 0. Since this holds for all j < k, we see that X must be diagonal.

Once X is known to be diagonal, we may compute that for j ¤ k,

0 D ŒX;Ejk� D .Xjj �Xkk/Ejk:

Thus, all the diagonal entries ofX must be equal. But since, also, trace.X/ D 0, weconclude that X must be zero.

For the remaining semisimple Lie algebras in Example 7.3, the calculations inSect. 7.7 will allow us to carry out a similar argument. It is proved there that thecenter of so.2nIC/ is trivial for n � 2, and a similar analysis shows that the centersof so.2nC 1IC/ and sp.nIC/ are also trivial. utProposition 7.4. Let g WD kC be a reductive Lie algebra. Then there exists an innerproduct on g that is real valued on k and such that the adjoint action of k on g is“unitary,” meaning that

hadX.Y /;Zi D � hY; adX.Z/i (7.1)

for all X 2 k and all X; Y 2 g. If we define a operation X 7! X� on g by theformula

.X1 C iX2/� D �X1 C iX2 (7.2)

for X1;X2 2 k, then any inner product satisfying (7.1) also satisfies

hadX.Y /;Zi D hY; adX�.Z/i (7.3)

for all X , Y , and Z in g.

The motivation for the definition of X� is that g D gl.nIC/ and k D u.n/, thenX� is the usual matrix adjoint of X .

Proof. By the proof of Theorem 4.28, there is a (real-valued) inner product on kthat is invariant under the adjoint action of K . This inner product then extends to acomplex inner product on g D kC for which the adjoint action ofK is unitary. Thus,

172 7 Semisimple Lie Algebras

by Proposition 4.8, the adjoint action of k on g is unitary in the sense of (7.1). It isthen a simple matter to check the relation (7.3). ut

Recall from Definition 3.9 what it means for a Lie algebra to decompose as a Liealgebra direct sum of subalgebras.

Proposition 7.5. Suppose g D kC is a reductive Lie algebra. Choose an innerproduct on g as in Proposition 7.4. Then if h is an ideal in g, the orthogonalcomplement of h is also an ideal, and g decomposes as the Lie algebra direct sumof h and h?.

Proof. An ideal in g is nothing but an invariant subspace for the adjoint action of gon itself. If h is a complex subspace of g and h is invariant under the adjoint actionof k, it will also be invariant under adjoint action of g D kC. Thus, if h is an idealin g, then by Proposition 4.27, h? is invariant under the adjoint action of k and is,therefore, also an ideal.

Now, g decomposes as a vector space as g D h ˚ h?. But since both h and h?are ideals, for all X 2 h and Y 2 h?, we have

ŒX; Y � 2 h \ h? D f0g:

Thus, g is actually the Lie algebra direct sum of h and h?. utProposition 7.6. Every reductive Lie algebra g over C decomposes as a Liealgebra direct sum g D g1˚z, where z is the center of g and where g1 is semisimple.

Proof. Since z is an ideal in g, Proposition 7.5 shows that g1 WD z? is also an idealin g and that g decomposes as a Lie algebra direct sum g D g1˚z. It remains only toshow that g1 is semisimple. It is apparent that g1 has trivial center, since any centralelement Z of g1 would also be central in g D g1 ˚ z, in which case, Z must be inz \ g1 D f0g. Thus, we must only construct a compact real form of g1.

It is easy to see that Z 2 g belongs to z if and only if Z commutes with everyelement of k. From this, it is easily seen that z is invariant under “conjugation,”that is, under the map X C iY 7! X � iY, where X; Y 2 k. Since z is invariantunder conjugation, its orthogonal complement g1 is also closed under conjugation. Itfollows that z is the complexification of z0 WD z\k and that g1 is the complexificationof k1 WD g1 \ k.

We will now show that k1 is a compact real form of g1. LetK 0 be the adjoint groupof K , that is, the image in GL.k/ of the adjoint map. Since K is compact and theadjoint map is continuous,K 0 is compact and thus closed. Now, by Proposition 3.34,the Lie algebra map associated to the map A 7! AdA is the map X 7! adX , thekernel of which is the center z0 of k. Thus, the image of ad is isomorphic to k=z Š k1,showing that k1 is isomorphic to the Lie algebra of the image of Ad, namelyK 0. utProposition 7.7. If K is a simply connected compact matrix Lie group with Liealgebra k, then g WD kC is semisimple.

Proof. As in the proof of Proposition 7.6, k decomposes as a Lie algebra direct sumk D k1 ˚ z0, where z0 is the center of k and where g1 WD .k1/C is semisimple. Then

7.1 Semisimple and Reductive Lie Algebras 173

by Theorem 5.11, K decomposes as a direct product of closed, simply connectedsubgroups K1 and Z0. Now, since z0 is commutative, it is isomorphic to the Liealgebra of the commutative Lie group Rn, for some n. Since both Z0 and Rn aresimply connected, Corollary 5.7 tells us that Z0 Š Rn. On the other hand, since Z0is a closed subgroup of K , we see that Z0 Š Rn is compact, which is possible onlyif n D 0. Thus, z0 and z D z0 C iz0 are zero dimensional, meaning that g D g1 issemisimple. ut

Recall (Definition 3.11) that a Lie algebra g is said to be simple if g has nonontrivial ideals and dim g is at least 2. That is to say, a one-dimensional Lie algebrais, by decree, not simple, even though it clearly has no nontrivial ideals.

Theorem 7.8. Suppose that g is semisimple in the sense of Definition 7.1. Then gdecomposes as a Lie algebra direct sum

g DmMjD1

gj ; (7.4)

where each gj � g is a simple Lie algebra.

We will see in Sect. 7.6 that most of our examples of semisimple Lie algebras areactually simple, meaning that there is only one term in the decomposition in (7.4).The converse of Theorem 7.8 is also true; if a complex Lie algebra g decomposes asa direct sum of simple algebras, then g is semisimple in the sense of Definition 7.1.(See Theorem 6.11 in [Kna2].)

Proof. If g has a nontrivial ideal h, then by Proposition 7.5, g decomposes as theLie algebra direct sum g D h˚h?, where h? is also an ideal in g. Suppose that, say,h has a nontrivial ideal h0. Since every element of h commutes with every elementof h?, we see that h0 is actually an ideal in g. Thus, h00 WD .h0/? \ h is an ideal in gand we can decompose g as h0 ˚ h00 ˚ h?. Proceeding on in the same way, we candecompose g as a direct sum of ideals gj , where each gj has no nontrivial ideals.It remains only to show that each gj has dimension at least 2. Suppose, toward acontradiction, that dim gj D 1 for some j . Then gj is necessarily commutative,which means (since elements of gj commute with elements of gk for j ¤ k) thatgj is contained in the center Z.g/ of g, contradicting the assumption that the centerof g is trivial. utProposition 7.9. If g is a complex semisimple Lie algebra, then the subalgebras gjappearing in the decomposition (7.8) are unique up to order.

To be more precise, suppose g is isomorphic to a Lie algebra direct sum of simplesubalgebras g1; : : : ; gl and also to a Lie algebra direct sum of simple subalgebrash1; : : : ; hm. Then the proposition asserts that each hj is actually equal to (not justisomorphic to) gk for some k. The proposition depends crucially on the fact that thesummands gj in (7.8) are simple (hence dim gj � 2) and not just that gj has nonontrivial ideals. Indeed, if g is a two-dimensional commutative Lie algebra, theng can be decomposed as a direct sum of one-dimensional commutative algebras inmany different ways.

174 7 Semisimple Lie Algebras

Proof. If h is an ideal in g, then h can be viewed as a representation of g bymeans of the map X 7! .adX/jh. If we decompose g as the direct sum of simplesubalgebra g1; : : : ; gm, then each gj is irreducible as a representation of g, since anyinvariant subspace would be an ideal in g and thus an ideal in gj . Furthermore, theserepresentations are nonisomorphic, since action of gj on gj is nontrivial (since gjis not commutative), whereas the action of each gk , k ¤ j , on gj is trivial. Supposenow that h is an ideal in g that is simple as a Lie algebra and, thus, irreducibleunder the adjoint action of g. Now, for any j , the projection map �j W g ! gj isan intertwining map of representations of g. Thus, for each j , the restriction of �jto h must be either 0 or an isomorphism. There must be some j for which �j

ˇh

isnonzero and hence an isomorphism. But since the various gj ’s are nonisomorphicrepresentations of g, we must have �k jh D 0 for all k ¤ j . Thus, actually,h D gj . ut

Before getting into the details of semisimple Lie algebras, let us briefly outlinewhat our strategy will be in classifying their representations and what structureswe will need to carry out this strategy. We will look for commuting elementsH1; : : : ;Hr in our Lie algebra that we will try to simultaneously diagonalizein each representation. We should find as many such elements as possible, andif they are going to be simultaneously diagonalizable in every representation,they must certainly be diagonalizable in the adjoint representation. This leads(in basis-independent language) to the definition of a Cartan subalgebra. Thenonzero sets of simultaneous eigenvalues for adH1; : : : ; adHr are called roots andthe corresponding simultaneous eigenvectors are called root vectors. The rootvectors will serve to raise and lower the eigenvalues of �.H1/; : : : ; �.Hr/ ineach representation � . We will also have the Weyl group, which is an importantsymmetry of the roots and also of the weights in each representation.

One crucial part of the structure of semisimple Lie algebras is the existence ofcertain special subalgebras isomorphic to sl.2IC/. Several times over the course ofthis chapter and the subsequent ones, we will make use of our knowledge of therepresentations of sl.2IC/, applied to these subalgebras. If X , Y , and H are theusual basis elements for sl.2IC/, then of particular importance is the fact that in anyfinite-dimensional representation � of sl.2IC/ (not necessarily irreducible), everyeigenvalue of �.H/ must be an integer. (Compare Point 1 of Theorem 4.34 andExercise 13 in Chapter 4.)

7.2 Cartan Subalgebras

Our first task is to identify certain special sorts of commutative subalgebras, calledCartan subalgebras.

Definition 7.10. If g is a complex semisimple Lie algebra, then a Cartan sub-algebra of g is a complex subspace h of g with the following properties:

7.2 Cartan Subalgebras 175

1. For all H1 and H2 in h, ŒH1;H2� D 0.2. If, for some X 2 g, we have ŒH;X� D 0 for all H in h, then X is in h.3. For all H in h, adH is diagonalizable.

Condition 1 says that h is a commutative subalgebra of g. Condition 2 says that his a maximal commutative subalgebra (i.e., not contained in any larger commutativesubalgebra). Condition 3 says that each adH (H 2 h) is diagonalizable. Sincethe H ’s in h commute, the adH ’s also commute, and thus they are simultaneouslydiagonalizable (Proposition A.16).

Of course, the definition of a Cartan subalgebra makes sense in any Lie algebra,semisimple or not. However, if g is not semisimple, then g may not have any Cartansubalgebras in the sense of Definition 7.10; see Exercise 1. (Sometimes a differentdefinition of “Cartan subalgebra” is used, one that allows every complex Lie algebrato have a Cartan subalgebra. This other definition is equivalent to Definition 7.10when g is semisimple but not in general.) Even in the semisimple case we mustprove that a Cartan subalgebra exists.

Proposition 7.11. Let g D kC be a complex semisimple Lie algebra and let t be anymaximal commutative subalgebra of k. Define h � g by

h D tC D t C i t:

Then h is a Cartan subalgebra of g:

Note that k (or any other finite-dimensional Lie algebra) contains a maximalcommutative subalgebra. After all, any one-dimensional subalgebra t1 of k is com-mutative. If t1 is maximal, then we are done; if not, then we choose somecommutative subalgebra t2 properly containing t1, and so on.

Proof of Proposition 7.11. It is clear that h is a commutative subalgebra of g. Wemust first show that h is maximal commutative. So, suppose that X 2 g commuteswith every element of h, which certainly means that it commutes with every elementof t. If we write X D X1 C iX2 with X1 and X2 in k, then forH in t, we have

ŒH;X1 C iX2� D ŒH;X1�C i ŒH;X2� D 0;

where ŒH;X1� and ŒH;X2� are in k. However, since every element of g has a uniquedecomposition as an element of k plus an element of ik, we see that ŒH;X1� andŒH;X2�must separately be zero. Since this holds for allH in t and since t is maximalcommutative, we must have X1 and X2 in t, which means that X D X1 C iX2 isin h. This shows that h is maximal commutative.

If h�; �i is an inner product on g as in Proposition 7.4, then for each X in k, theoperator adX W g ! g is skew self-adjoint. In particular, each adH , H 2 t, is skewself-adjoint and thus diagonalizable (Theorem A.3). Finally, if H is any element ofh D t C i t, thenH D H1 C iH2, with H1 and H2 in t. Since H1 and H2 commute,adH1 and adH2 also commute and, thus, by Proposition A.16, adH1 and adH2 aresimultaneously diagonalizable. It follows that adH is diagonalizable. ut

176 7 Semisimple Lie Algebras

Throughout this chapter and the subsequent chapters, we consider only Cartansubalgebras of the form h D tC, where t is a maximal commutative subalgebraof some compact real form k of g. It is true, but not obvious, that every Cartansubalgebra arises in this way. Indeed, for any two Cartan subalgebras h1 and h2 ofg, there exists an automorphism of g mapping h1 to h2. (See Proposition 2.13 andTheorem 2.15 in Chapter II of [Kna2].) While we will not prove this result, we willprove in Chapter 11 that for a fixed k, the maximal commutative subalgebra t � kis unique up to an automorphism of k. (See Proposition 11.7 and Theorem 11.9.)In light of the uniqueness of Cartan subalgebras up to automorphism, the followingdefinition is meaningful.

Definition 7.12. If g is a complex semisimple Lie algebra, the rank of g is thedimension of any Cartan subalgebra.

7.3 Roots and Root Spaces

For the rest of this chapter, we assume that we have chosen a compact real formk of g and a maximal commutative subalgebra t of k, and we consider the Cartansubalgebra h D t C i t. We assume also that we have chosen an inner product on gthat is real on k and invariant under the adjoint action ofK (Proposition 7.4).

Since the operators adH , H 2 h, commute, and each such adH is diagonalizable,Proposition A.16 tell us that the adH ’s with H 2 h are simultaneously diagonal-izable. If X 2 g is a simultaneous eigenvalue for each adH , H 2 h, then thecorresponding eigenvalues depend linearly on H 2 h. If this linear functional isnonzero, we call it a root. As in Sect. 6.6, it is convenient to express this linearfunctional as H 7! h˛;H i for some ˛ 2 h. The preceding discussion leads to thefollowing definition.

Definition 7.13. A nonzero element ˛ of h is a root (for g relative to h) if thereexists a nonzeroX 2 g such that

ŒH;X� D h˛;H iX

for all H in h. The set of all roots is denotedR.

Note that we follow the convention that an inner product be linear in the secondfactor, so that h˛;H i depends linearly on H for a fixed ˛.

Proposition 7.14. Each root ˛ belongs to i t � h.

Proof. As we have already noted, each adH , H 2 t, is a skew self-adjoint operatoron h, which means that adH has pure imaginary eigenvalues. It follows that if ˛ isa root, h˛;H i must be pure imaginary for H 2 t, which (since our inner product isreal on t � k) can only happen if ˛ is in i t. ut

7.3 Roots and Root Spaces 177

Definition 7.15. If ˛ is a root, then the root space g˛ is the space of all X in g forwhich ŒH;X� D h˛;H iX for all H in h: A nonzero element of g˛ is called a rootvector for ˛.

More generally, if ˛ is any element of h, we define g˛ to be the space of all X ing for which ŒH;X� D h˛;H iX for all H in h (but we do not call g˛ a root spaceunless ˛ is actually a root).

Taking ˛ D 0, we see that g0 is the set of all elements of g that commute withevery element of h. Since h is a maximal commutative subalgebra, we conclude thatg0 D h. If ˛ is not zero and not a root, then g˛ D f0g.

As we have noted, the operators adH ,H 2 h, are simultaneously diagonalizable.As a result, g can be decomposed as the sum of h and the root spaces g˛. Actually,by Proposition A.17, the sum is direct and we have established the following result.

Proposition 7.16. The Lie algebra g can be decomposed as a direct sum of vectorspaces as follows:

g D h ˚M˛2R

g˛:

That is to say, every element of g can be written uniquely as a sum of an elementof h and one element from each root space g˛ . Note that the decomposition isnot a Lie algebra direct sum, since, for example, elements of h do not, in general,commute with elements of g˛.

Proposition 7.17. For any ˛ and ˇ in h, we have

Œg˛; gˇ� � g˛Cˇ:

In particular, if X is in g˛ and Y is in g�˛ , then ŒX; Y � is in h. Furthermore, if Xis in g˛ , Y is in gˇ, and ˛ C ˇ is neither zero nor a root, then ŒX; Y � D 0.

Proof. It follows from the Jacobi identity that adH is a derivation:

ŒH; ŒX; Y �� D ŒŒH;X�; Y �C ŒX; ŒH; Y ��:

Thus, if X is in g˛ and Y is in gˇ , we have

ŒH; ŒX; Y �� D Œh˛;H iX; Y �C ŒX; hˇ;H i Y �D h˛ C ˇ;H i ŒX; Y �:

for all H 2 h, showing that ŒX; Y � is in g˛Cˇ. utProposition 7.18. 1. If ˛ 2 h is a root, so is �˛. Specifically, if X is in g˛, thenX� is in g�˛, where X� is defined by (7.2) in Proposition 7.4.

2. The roots span h.

178 7 Semisimple Lie Algebras

Proof. For Point 1, if X D X1 C iX2 with X1;X2 2 k, let

NX D X1 � iX2: (7.5)

Since k is closed under brackets, if H 2 t � k and X 2 g, we have

ŒH;X� D ŒH;X1�� i ŒH;X2� D ŒH; NX�:In particular, if X is a root vector with root ˛ 2 i t, then for all H 2 h, we have

ŒH; NX� D ŒH;X� D h˛;H iX D � h˛;H i NX; (7.6)

since h˛;H i is pure imaginary forH 2 t. Extending (7.6) by linearity inH , we seethat ŒH; NX� D � h˛;H i NX for all H 2 h. Thus, NX is a root vector corresponding tothe root �˛. It follows that X� D � NX belongs to g�˛ .

For Point 2, suppose that the roots did not span h. Then there would exist anonzero H 2 h such that h˛;H i D 0 for all ˛ 2 R. Then we would haveŒH;H 0� D 0 for all H 0 in h, and also

ŒH;X� D h˛;H iX D 0

forX in g˛ . Thus, by Proposition 7.16,H would be in the center of g, contradictingthe definition of a semisimple Lie algebra. ut

We now develop a key tool in the study of a semisimple Lie algebra g, theexistence of certain subalgebras of g isomorphic to sl.2IC/.Theorem 7.19. For each root ˛, we can find linearly independent elements X˛ ing˛, Y˛ in g�˛ , and H˛ in h such that H˛ is a multiple of ˛ and such that

ŒH˛;X˛� D 2X˛;

ŒH˛; Y˛� D �2Y˛;ŒX˛; Y˛� D H˛: (7.7)

Furthermore, Y˛ can be chosen to equal X �.

If X˛; Y˛;H˛ are as in the theorem, then on the one hand, ŒH˛;X˛� D 2X˛,while on the other hand, ŒH˛;X˛� D h˛;H˛iX˛. Thus,

h˛;X˛i D 2: (7.8)

Meanwhile, H˛ is a multiple of ˛ and the unique such multiple compatiblewith (7.8) is

H˛ D 2˛

h˛; ˛i : (7.9)

For use in Part III, we record the following consequence of Theorem 7.19.

7.3 Roots and Root Spaces 179

Corollary 7.20. For each root ˛, let X˛ , Y˛ , and H˛ be as in Theorem 7.19, withY˛ D X �. Then the elements

E˛1 WD i

2H˛I E˛

2 WD i

2.X˛ C Y˛/I E˛

3 WD 1

2.Y˛ � X˛/

are linearly independent elements of k and satisfy the commutation relations

ŒE˛1 ; E

˛2 � D E˛

3 I ŒE˛2 ; E

˛3 � D E˛

1 I ŒE˛3 ; E

˛1 � D E˛

2 :

Thus, by Example 3.27, the span ofE˛1 ,E˛

2 , andE˛3 is a subalgebra of k isomorphic

to su.2/.

Proof. Since ˛ belongs to i t and H˛ is, by (7.9), a real multiple of ˛, the element.i=2/H˛ will belong to t � k. Meanwhile, we may check that .E˛

2 /� D �E˛

2 and.E˛

3 /� D �E˛

3 . From the way the map X 7! X� was defined (Proposition 7.4),it follows that E˛

2 and E˛3 also belong to k. Furthermore, since X˛, Y˛ , and

H˛ are linearly independent, E˛1 , E˛

2 , and E˛3 are also independent. Finally,

direct computation with the commutation relations in (7.7) confirms the claimedcommutation relations for the E˛

j ’s. utDefinition 7.21. The element H˛ D 2˛= h˛; ˛i in Theorem 7.19 is the corootassociated to the root ˛.

We begin the proof of Theorem 7.19 with a lemma.

Lemma 7.22. Suppose that X is in g˛, that Y is in g�˛, and that H is in h. ThenŒX; Y � is in h and

hŒX; Y �;H i D h˛;H i hY;X�i ; (7.10)

where X� is as in Proposition 7.4.

In the proof of Theorem 7.19, we will need to know not just that ŒX; Y � 2 h, butwhere in h the element ŒX; Y � lies. This information is obtained by computing theinner product of ŒX; Y � with each element of H , as we have done in (7.10).

Proof. That ŒX; Y � is in h follows from Proposition 7.17. Then, using Proposition7.4, we compute that

hŒX; Y �;H i D hadX.Y /;H i D hY; adX�.H/i D � hY; ŒH;X��i : (7.11)

Since X is in g˛, Proposition 7.18 tells us that the element X� is in g�˛, so thatŒH;X�� D � h˛;H iX� and (7.11) reduces to the desired result. (Recall that wetake inner products to be linear in the second factor.) utProof of Theorem 7.19. Choose a nonzero X in g˛, so that X� D � NX is a nonzeroelement of g�˛ . Applying Lemma 7.22 with Y D X� gives

hŒX;X��;H i D h˛;H i hX�; X�i : (7.12)

180 7 Semisimple Lie Algebras

From (7.12), we see that ŒX;X�� 2 h is orthogonal to everyH 2 h that is orthogonalto ˛, which happens only if ŒX;X�� is a multiple of ˛. Furthermore, if we chooseH so that h˛;H i ¤ 0, we see that hŒX;X��;H i ¤ 0, so that ŒX;X�� ¤ 0. Now, ifwe evaluate (7.12) with H D ŒX;X��, we obtain

hŒX;X��; ŒX;X��i D h˛; ŒX;X��i hX�; X�i :Since ŒX;X�� ¤ 0, we conclude that h˛; ŒX;X��i is real and strictly positive.

Let H D ŒX;X�� and define elements of g as follows:

H˛ D 2

h˛;H iH;

X˛ Ds

2

h˛;H iX;

Y˛ Ds

2

h˛;H iY:

Then h˛;H˛i D 2, from which it follows that ŒH˛;X˛� D 2X˛ and ŒH˛; Y˛� D�2Y˛. Furthermore, ŒX˛; Y˛� D 2ŒX; Y �= h˛;H i D H˛ . Thus, these elementssatisfy the relations claimed in the theorem, and Y˛ D X �. Furthermore, since theseelements are eigenvectors for adH with distinct eigenvalues 2, �2, and 0, they mustbe linearly independent (Proposition A.1). ut

We now make use of the subalgebras in Theorem 7.19 to obtain results aboutroots and root spaces. Note that

s˛ WD hX˛; Y˛;H˛i (7.13)

acts on g by (the restriction to s˛ of) the adjoint representation.

Theorem 7.23. 1. For each root ˛, the only multiples of ˛ that are roots are ˛and �˛.

2. For each root ˛, the root space g˛ is one dimensional.

Point 1 of Theorem 7.23 should be contrasted with the results of Exercise 8 inthe case of a Lie algebra that is not semisimple.

Lemma 7.24. If ˛ and c˛ are both roots with jcj > 1, then c D ˙2.

Proof. Let s˛ be as in (7.13). If ˇ D c˛ is also a root and X is a nonzero elementof gˇ, then by (7.8), we have

ŒH˛;X� D hˇ;H˛iX D Nc h˛;H˛iX D 2 NcX:Thus, 2 Nc is an eigenvalue for the adjoint action of H˛ 2 s˛ on g. By Point 1 ofTheorem 4.34, any such eigenvalue must be an integer, meaning that c is an integermultiple of 1=2. But by reversing the roles of ˛ and ˇ in the argument, we see that1=c must also be an integer multiple of 1=2.

7.3 Roots and Root Spaces 181

Fig. 7.1 If ˇ D 2˛ were aroot, the orthogonalcomplement of s˛ in V ˛

would contain an element ofh orthogonal toH˛

V

Now, suppose c D n=2 for some integer n and that 1=c D 2=n is an integermultiple of 1=2, so that 2=c D 4=n is an integer. Then n D ˙1, ˙2, or ˙4. Thus,c D ˙1=2, ˙1, or ˙2, of which only ˙2 have jcj > 1. utProof of Theorem 7.23. Since there are only finitely many roots, there is somesmallest positive multiple of ˛ that is a root. Replacing ˛ by this smallest multiple,we can assume that if c˛ is a root and c˛ ¤ ˙˛, then jcj > 1. By Lemma 7.24, wemust then have c D ˙2.

Let s˛ be as in (7.13), with Y˛ chosen to equal X �. Let V ˛ be the subspace ofg spanned by H˛ and all the root spaces gˇ for which ˇ is a multiple of ˛. (SeeFigure 7.1.) We claim that V ˛ is a subalgebra of g. To verify this claim, observe firstthat if a root ˇ is a multiple of ˛ then by Lemma 7.22, every element of Œgˇ; g�ˇ�is in h and is orthogonal to every element of h that is orthogonal to ˇ. Thus, everyelement of Œgˇ; g�ˇ� is a multiple of ˇ, which is a multiple of ˛, which is a multipleofH˛ . Observe next that ifX 2 gˇ, then ŒH˛;X� is a multiple ofX . Observe, finally,that if ˇ and ˇ0 are roots that are multiples of ˛ with ˇ0 ¤ �ˇ, then Œgˇ; gˇ0 � �gˇCˇ0 , where ˇ C ˇ0 ¤ 0 is again a multiple of ˛.

Since V ˛ is a subalgebra of g, it is certainly invariant under the adjoint action ofs˛ � V ˛ . Note also that s˛ itself is an invariant subspace for the adjoint action of s˛

on V ˛ . Now, since Y˛ D X � and H˛ is a positive multiple of ˛ 2 i t � ik, we seethat X� 2 s˛ for every X 2 s˛. Thus, by Propositions 7.4 and 4.27, the orthogonalcomplement U˛ of s˛ in V ˛ will also be an invariant under the adjoint action of s˛ .

Now, h˛;H˛i D 2 and, by Lemma 7.24, any multiples ˇ of ˛ that are rootsare of the form ˇ D ˙˛ or ˇ D ˙2˛. Thus, the eigenvalues of adH˛ in V ˛ are0, ˙2, and, possibly, ˙4. If U˛ ¤ f0g, then U˛ will contain an eigenvector foradH˛ , with an eigenvalue � 2 f0;˙2;˙4g. Since � is even, it follows from Point 4

182 7 Semisimple Lie Algebras

of Theorem 4.34 that 0 must also be an eigenvalue for adH˛ in U˛ . But this isimpossible, since H˛ is the only eigenvector for adH˛ in V ˛ with eigenvalue 0, andU˛ is orthogonal toH˛ 2 s˛ . Thus, actually,U˛ D f0g, which means that V ˛ D s˛ .Thus, the only multiples of ˛ that are roots are ˙˛ and the root spaces with roots˙˛ are one dimensional. ut

Figure 7.1 shows a configuration of “roots” consistent with Lemma 7.24, butwhich cannot actually be the root system of a semisimple Lie algebra.

7.4 The Weyl Group

We now introduce an important symmetry of the set R of roots, known as the Weylgroup. In this section, we follow the Lie algebra approach to the Weyl group, asopposed to the Lie group approach we followed in Sect. 6.6 in the SU.3/ case. Wewill return to the Lie group approach to the Weyl group in Chapter 11 and show(Sect. 11.7) that the two approaches are equivalent.

Definition 7.25. For each root ˛ 2 R, define a linear map s˛ W h ! h by theformula

s˛ �H D H � 2 h˛;H ih˛; ˛i ˛: (7.14)

The Weyl group of R, denoted W , is then the subgroup of GL.h/ generated by allthe s˛’s with ˛ 2 R.

Note that since each root ˛ is in i t and our inner product is real on t, if H is ini t, then s˛ � H is also in i t. As a map of i t to itself, s˛ is the reflection about thehyperplane orthogonal to ˛. That is to say, s˛ �H D H wheneverH is orthogonalto ˛, and s˛ � ˛ D �˛. Since each reflection is an orthogonal linear transformation,we see that W is a subgroup of the orthogonal group O.i t/.

Theorem 7.26. The action of W on i t preserves R. That is to say, if ˛ is a root,then w � ˛ is a root for all w 2 W .

Proof. For each ˛ 2 R, consider the invertible linear operator S˛ on g given by

S˛ D eadX˛ e�adY˛ eadX˛ :

Now, if H 2 h satisfies h˛;H i D 0, then ŒH;X˛� D h˛;H iX˛ D 0. Thus, H andX˛ commute, which means that adH and adX˛ also commute, and similarly for adHand adY˛ . Thus, if h˛;H i D 0, the operator S˛ will commute with adH , so that

S˛adHS�1˛ D adH ; h˛;H i D 0: (7.15)

7.5 Root Systems 183

On the other hand, if we apply Point 3 of Theorem 4.34 to the adjoint action ofs˛ Š sl.2IC/ on g, we see that

S˛adH˛S�1˛ D �adH˛ : (7.16)

By combining (7.15) and (7.16), we see that for all H 2 h, we have

S˛adHS�1˛ D ads˛ �H : (7.17)

Now if ˇ is any root and X is an associated root vector, consider the vectorS�1˛ .X/ 2 g. We compute that

adH.S�1˛ .X// D S�1

˛ .S˛adHS�1˛ /.X/

D S�1˛ ads˛ �H.X/

D hˇ; s˛ �H iS�1˛ .X/

D ˝s�1˛ � ˇ;H ˛S�1

˛ .X/:

Thus, S�1˛ .X/ is a root vector with root s�1

˛ � ˇ D s˛ � ˇ. This shows that the set ofroots is invariant under each s˛ and, thus, underW . ut

Actually, since s˛ � s˛ � ˇ D ˇ, each reflection maps R onto R. It follows thateach w 2 W also maps R onto R.

Corollary 7.27. The Weyl group is finite.

Proof. Since the roots span h, each w 2 W is determined by its action on R. Since,also, w maps R onto R, we see that W may be thought of as a subgroup of thepermutation group on the roots. ut

7.5 Root Systems

In this section, we record several important properties of the roots, using results fromthe two previous sections. Recall that for each root ˛, we have an element H˛ of hcontained in Œg˛; g�˛� as in Theorem 7.19. As we saw in (7.8) and (7.9),H˛ satisfiesh˛;H˛i D 2 and is related to ˛ by the formula H˛ D 2˛= h˛; ˛i. In particular, theelementH˛ is independent of the choice of X˛ and Y˛ in Theorem 7.19.

Definition 7.28. For each root ˛, the elementH˛ 2 h given by

H˛ D 2˛

h˛; ˛iis the coroot associated to the root ˛.

184 7 Semisimple Lie Algebras

Proposition 7.29. For all roots ˛ and ˇ, we have that

hˇ;H˛i D 2h˛; ˇih˛; ˛i (7.18)

is an integer.

We have actually already made use of this result in the proof of Lemma 7.24.

Proof. If s˛ D hX˛; Y˛;H˛i is as in Theorem 7.19 andX is a root vector associatedto the root ˇ, then ŒH˛;X� D hˇ;H˛iX . Thus, hˇ;H˛i is an eigenvalue for theadjoint action of s˛ Š sl.2IC/ on g. Point 1 of Theorem 4.34 then shows thathˇ;H˛i must be an integer. ut

Recall from elementary linear algebra that if ˛ and ˇ are elements of an innerproduct space, the orthogonal projection of ˇ onto ˛ is given by

h˛; ˇih˛; ˛i˛:

The quantity on the right-hand side of (7.18) is thus twice the coefficient of ˛in the projection of ˇ onto ˛. We may therefore interpret the integrality result inProposition 7.29 in the following geometric way:

If ˛ and ˇ are roots, the orthogonal projection of ˛ onto ˇ must be an integer or half-integermultiple of ˇ.

Alternatively, we may think about Proposition 7.29 as saying that ˇ and s˛ � ˇ mustdiffer by an integer multiple of ˛ [compare (7.14)].

If we think of the set R of roots as a subset of the real inner product spaceE WD i t, we may summarize the properties of R as follows.

Theorem 7.30. The set R of roots is a finite set of nonzero elements of a real innerproduct space E , and R has the following additional properties.

1. The roots span E .2. If ˛ 2 R, then �˛ 2 R and the only multiples of ˛ in R are ˛ and �˛.3. If ˛ and ˇ are in R, so is s˛ � ˇ, where s˛ is the reflection defined by (7.14).4. For all ˛ and ˇ in R, the quantity

2h˛; ˇih˛; ˛i

is an integer.

Any such collection of vectors is called a root system. We will look in detail atthe properties of root systems in Chapter 8.

7.6 Simple Lie Algebras 185

7.6 Simple Lie Algebras

Every semisimple Lie algebra decomposes as a direct sum of simple algebras(Theorem 7.8). In this section, we give a criterion for a semisimple Lie algebrato be simple. We will eventually see (Sect. 8.11) that most of the familiar examplesof semisimple Lie algebras are actually simple.

Proposition 7.31. Suppose g is a real Lie algebra and that the complexification gCof g is simple. Then g is also simple.

Proof. Since gC is simple, the dimension of gC over C is at least 2, so that thedimension of g over R is also at least 2. If g had a nontrivial ideal h, then thecomplexification hC of h would be a nontrivial ideal in g. ut

The converse of Proposition 7.31 is false in general. The Lie algebra so.3I 1/, forexample, is simple as a real Lie algebra, and yet its complexification is isomorphicto so.4IC/, which in turn is isomorphic to sl.2IC/˚ sl.2IC/. See Exercise 14.

Theorem 7.32. Suppose K is a compact matrix Lie group whose Lie algebra k issimple as a real Lie algebra. Then the complexification g WD kC of k is simple as acomplex Lie algebra.

For more results about simple algebras over R, see Exercises 12 and 13. Beforeproving Theorem 7.32 result, we introduce a definition.

Definition 7.33. If g is a real Lie algebra, g admits a complex structure if thereexists a “multiplication by i” map J W g ! g that makes g into a complex vectorspace in such a way that the bracket map Œ�; �� W g � g ! g is complex bilinear.

Here, bilinearity of the bracket means, in particular, that ŒJX; Y � D J ŒX; Y � forall X; Y 2 k. Equivalently, g admits a complex structure if there exists a complexLie algebra h and a real-linear map � W h ! g that is one-to-one and onto andsatisfies �.ŒX; Y �/ D Œ�.X/; �.Y /� for all X; Y 2 h.

Lemma 7.34. Suppose that K is a compact matrix Lie group whose Lie algebra kis noncommutative. Then k does not admit a complex structure.

Proof. Suppose, toward a contradiction, that k does admit a complex structure J .By Proposition 7.4, there exists a real inner product on k with respect to which adHis skew symmetric for all H 2 k. If we choose H not in the center of k then adHis nonzero and skew symmetric, hence diagonalizable in kC with pure-imaginaryeigenvalues, not all of which are zero. In particular, adH is not nilpotent.

On the other hand, since J is complex bilinear, if we view k as a complex vectorspace with respect to the map J , then adH is complex linear. Since adH is notnilpotent, it has a nonzero eigenvalue � D a C ib 2 C. Thus, there is a nonzeroX 2 k such that

ŒH;X� D .a C ib/ �X D aX C bJX:

186 7 Semisimple Lie Algebras

If we then consider element

H 0 WD N� �H D aH � bJH;

we may compute, using the linearity of the bracket with respect to J and the identityJ 2 D �I , that

ŒH 0; X� D j�j2 X D .a2 C b2/X:

But adH 0 is also skew symmetric, and, thus,

j�j2 hX;Xi D hadH 0.X/;Xi D � hX; adH 0.X/i D � j�j2 hX;Xi ;which is impossible if � and X are both nonzero. utProof of Theorem 7.32. Suppose to the contrary that g were not simple, so that itdecomposes as a sum of at least two simple algebras gj . By Proposition 7.9, thedecomposition of a semisimple algebra Lie algebra into a sum of simple algebras isunique up to ordering of the summands. On the other hand, if g decomposes as thesum of the gj ’s, it also decomposes as the sum of the gj ’s, where the map X 7! NXis as in (7.5). Thus, for each j , we must have gj D gk for some k.

Suppose there is some j for which gj D gj . Then for all X 2 gj , the elementXC NX is in gj \k, from which it follows that gj \k is a nonzero ideal in k. But gj \kcannot be all of k, or else gj would be all of g. Thus, gj \ k would be a nontrivialideal in k, contradicting our assumption that k is simple. On the other hand, if wepick some j with gj D gk , k ¤ j , then .gj ˚ gk/\ k is a nonzero ideal in k, whichmust be all of k. We conclude, then, that there must be exactly two summands g1and g2 in the decomposition of g, satisfying g1 D g2.

Let us then define a real-linear map � W g1 ! k by �.X/ D X C NX . Note thatfor anyX 2 g1, the element NX is in g2, so that ŒX; NX� D 0. From this, we can easilycheck that � satisfies �.ŒX; Y �/ D Œ�.X/; �.Y /� for all X; Y 2 g1. Furthermore, �is injective because, for any X 2 g1, we have NX 2 g2 and thus NX cannot equal �XunlessX D 0. Finally, by counting real dimensions, we see that � is also surjective.Since g1 is a complex Lie algebra and � is a real Lie algebra isomorphism, we seethat k admits a complex structure, contradicting Lemma 7.34. utTheorem 7.35. Let g Š kC be a complex semisimple Lie algebra, let t be a maximalcommutative subalgebra of k, and let h D tC be the associated Cartan subalgebraof g. Let R � h be the root system for g relative to h. If g is not simple then hdecomposes as an orthogonal direct sum of nonzero subspaces h1 and h2 in such away that every element of R is either in h1 or in h2. Conversely, if h decomposes inthis way, then g is not simple.

We may restate the first part of the theorem in contrapositive form: If there doesnot exist an orthogonal decomposition of h as h D h1 ˚ h2 (with dim h1 > 0 anddim h2 > 0) such that every root is either in h1 or h2, then g is simple. In Sect. 8.6,we will show that if the “Dynkin diagram” of the root system is connected, thenno such decomposition of h exists. We will then be able to check that most of ourexamples of semisimple Lie algebras are actually simple.

7.6 Simple Lie Algebras 187

Proof of Theorem 7.35, Part 1. Assume first that g is not simple, so that, byTheorem 7.32, k is not simple either. Thus, k has a nontrivial ideal k1. Now, idealsin k are precisely invariant subspaces for the adjoint action of k on itself, which arethe same as the invariant subspaces for the adjoint action of K on k. If we choosean inner product on k is Ad-K-invariant, the orthogonal complement of k1 is also anideal. Thus, k decomposes as the Lie algebra direct sum k1 ˚ k2, where k2 D .k1/

?,in which case g decomposes as g1 ˚ g2, where g1 D .k1/C and g2 D .k2/C.

Let t be any maximal commutative subalgebra of k and h D tC the associatedCartan subalgebra of g. We claim that t must decompose as t1 ˚ t2, where t1 � k1and t2 � k2. Suppose H and H 0 are two elements of t, with H D X1 C X2 andH 0 D Y1 C Y2, with X1; Y1 2 k1 and X2; Y2 2 k2. Then

0 D ŒH;H 0� D ŒX1; Y1�C ŒX2; Y2�;

with ŒX1; Y1� 2 k1 and ŒX2; Y2� 2 k2, which can happen only if ŒX1; Y1� DŒX2; Y2� D 0. Thus,

ŒX1;H0� D ŒX1; Y1� D 0;

showing that X1 commutes with every element of h. Since h is maximal commuta-tive, X1 must actually be in h, and similarly for X2. That is to say, for every X 2 t,the k1- and k2-components of X also belong to t. From this observation, it followseasily that h is the direct sum of the subalgebras

t1 WD t \ k1; t2 WD t \ k2:

The algebra h then splits as h1 ˚ h2, where h1 D .t1/C and h2 D .t2/C. Let R1and R2 be the roots for g1 relative to h1 and g2 relative to h2, respectively. If ˛ is anelement of R1 and X 2 g1 is an associated root vector, then consider any elementH D H1 CH2 of h, where H1 2 h1 and H2 2 h2. We have

ŒH;X� D ŒH1;X�C ŒH2;X� D h˛;H1iX;since H2 2 g2 and X 2 g1. Thus, X is also a root vector for g relative to h, withthe associated root being ˛. Similarly, every root ˇ 2 R2 is also a root for g relativeto h.

We now claim that every root ˛ for g relative to h is either an element of R1 oran element of R2. If X D X1 CX2 is a root vector associated to the root ˛, then forH1 2 h1 we have

ŒH1;X� D ŒH1;X1 CX2� D ŒH1;X1� D h˛;H1iX1:On the other hand, ŒH1;X� must be a multiple of X . Thus, either X2 D 0 or ˛ isorthogonal to h1. If ˛ is orthogonal to h1 then it belongs to h2 and so ˛ is a root forg2 (i.e., ˛ 2 R2). If, on the other hand, X2 D 0, then 0 D ŒH2;X� D h˛;H2iX forall H2 2 h2, so that ˛ is orthogonal to h2. In that case, ˛ 2 h1 and ˛ must belongto R1. ut

188 7 Semisimple Lie Algebras

Proof of Theorem 7.35, Part 2. Suppose now h splits as h D h1 ˚ h2, with h1 andh2 being nonzero, orthogonal subspaces of h, in such a way that every root is eitherin h1 or h2. Let Rj D R \ hj , j D 1; 2, and define subspaces gj of g as

gj D hj ˚M˛2Rj

g˛; j D 1; 2:

Then g decomposes as a vector space as g1 ˚ g2. But h1 commutes with each g˛with ˛ 2 R2, because ˛ is in h2, which is orthogonal to h1. Similarly, h2 commuteswith g˛, ˛ 2 R1, and, of course, with h1. Finally, if ˛ 2 R1 and ˇ 2 R2, thenŒg˛; gˇ� D f0g, because ˛ C ˇ is not a root. Thus, actually, g is the Lie algebradirect sum of g1 and g2, showing that g is not simple. ut

7.7 The Root Systems of the Classical Lie Algebras

In this section, we look at how the structures described in this chapter work out in thecase of the “classical” semisimple Lie algebras, that is, the special linear, orthogonal,and symplectic algebras over C. We label each of our Lie algebras in such a waythat the rank is n, and we split our analysis of the orthogonal algebras into the evenand odd cases. For each of the classical Lie algebras, we use a constant multiple ofthe Hilbert–Schmidt inner product hX; Y i D trace.X�Y /, which is invariant underthe adjoint action of the corresponding compact group.

7.7.1 The Special Linear Algebras sl.nC 1IC/

We work with the compact real form k D su.nC1/ and the commutative subalgebrat which is the intersection of the set of diagonal matrices with su.nC 1/; that is,

t D

8<:

0B@

ia1: : :

ianC1

1CAˇˇˇ aj 2 R; a1 C � � � C anC1 D 0

9>=>; : (7.19)

We also consider h WD tC, which is described as follows:

h D

8<:

0B@�1: : :

�nC1

1CAˇˇˇ�j 2 C; �1 C � � � C �nC1 D 0

9>=>; : (7.20)

7.7 The Root Systems of the Classical Lie Algebras 189

If a matrix X commutes with each element of t, it will also commute with eachelement of h. But then, as in the proof of Example 7.3,X would have to be diagonal,and if X 2 su.nC 1/, then X would have to be in t. Thus, t is actually a maximalcommutative subalgebra of su.nC 1/.

Now, let Ejk denote the matrix that has a one in the j th row and kth column andhas zeros elsewhere. A simple calculation shows that if H 2 h is as in (7.20), thenHEjk D �jEjk and EjkH D �kEjk. Thus,

ŒH;Ejk� D .�j � �k/Ejk: (7.21)

If j ¤ k, then Ejk is in sl.n C 1IC/ and (7.21) shows that Ejk is a simultaneouseigenvector for each adH with H in h, with eigenvalue �j � �k . Note that everyelement X of sl.n C 1IC/ can be written uniquely as an element of the Cartansubalgebra (the diagonal entries of X ) plus a linear combination of the Ejk’s withj ¤ k (the off-diagonal entries of X ).

If we think at first of the roots as elements of h�, then [according to (7.21)]the roots are the linear functionals ˛jk that associate to each H 2 h, as in (7.20), thequantity �j ��k. We identify h with the subspace of CnC1 consisting vectors whosecomponents sum to zero. The inner product hX; Y i D trace.X�Y / on h is just therestriction to this subspace of the usual inner product on CnC1. If we use this innerproduct to transfer the roots from h� to h, we obtain the vectors

˛jk D ej � ek .j ¤ k/:

The roots of sl.n C 1IC/ form a root system that is conventionally called An,with the subscript n indicating that the rank of sl.nC1IC/ (i.e., the dimension of h)is n. We see that each root has length

p2 and

˝˛jk; ˛j 0k0

˛

has the value 0, ˙1, or ˙2, depending on whether fj; kg and fj 0; k0g have zero, one,or two elements in common. Thus

2

˝˛jk; ˛j 0k0

˛˝˛jk; ˛jk

˛ 2 f0;˙1;˙2g:

If ˛ and ˇ are roots and ˛ ¤ ˇ and ˛ ¤ �ˇ, then the angle between ˛ and ˇ iseither �=3, �=2, or 2�=3, depending on whether h˛; ˇi has the value 1, 0, or �1.

It is easy to see that for any j and k, the reflection s˛jk acts on CnC1 byinterchanging the j th and kth entries of each vector. It follows that the Weyl groupof the An root system is the permutation group on nC 1 elements.

190 7 Semisimple Lie Algebras

7.7.2 The Orthogonal Algebras so.2nIC/

The root system for so.2nIC/ is denoted Dn. We consider the compact real formso.2n/ of so.2nIC/, and we consider in so.2n/ the commutative subalgebra tconsisting of 2 � 2 block-diagonal matrices in which the j th diagonal block is ofthe form

�0 aj

�aj 0

�; (7.22)

for some aj 2 R. We then consider h D tC i t of so.2nIC/, which consists of 2�2block-diagonal matrices in which the j th diagonal block is of the form (7.22) withaj 2 C. As we will show below, t is actually a maximal commutative subalgebra ofso.2n/, so that h is a Cartan subalgebra of so.2nIC/.

The root vectors are now 2 � 2 block matrices having a 2 � 2 matrix C in the.j; k/ block (j < k), the matrix �C tr in the .k; j / block, and zero in all otherblocks, where C is one of the four matrices

C1 D�1 i

i �1�; C2 D

�1 �i

�i �1�;

C3 D�1 �ii 1

�; C4 D

�1 i

�i 1�:

Direct calculation shows that these are, indeed, root vectors and that the correspond-ing roots are the linear functionals on h given by i.aj Cak/, �i.aj Cak/, i.aj �ak/,and �i.aj � ak/, respectively.

Let us use on h the inner product hX; Y i D trace.X�Y /=2. If we then identify hwith Cn by means of the map

H 7! i.a1; : : : ; an/;

our inner product on h will correspond to the standard inner product on Cn. Theroots as elements of Cn are then the vectors

˙ ej ˙ ek; j < k; (7.23)

where fej g is the standard basis for Rn.We now demonstrate that t is a maximal commutative subalgebra of k, and also

that the center of so.2nIC/ is trivial for n � 2, as claimed in Example 7.3. Itis easy to check that every X 2 g can be expresses as an element of h plus alinear combination of the root vectors described above. If X commutes with everyelement of t, thenX also commutes with every element of h. Since each of the linearfunctionals i.˙aj ˙ak/, j < k, is nonzero on h, the coefficients of the root vectorsin the expansion of X must be zero; that is, X must be in h. If, also, X is in k, thenX must be in h \ k D t. This shows that t is maximal commutative in k.

7.7 The Root Systems of the Classical Lie Algebras 191

Meanwhile, if X is in the center of so.2nIC/, then as shown in the previousparagraph, X must belong to h. But then for each root vector X˛ with root ˛, wemust have

0 D ŒX;X˛� D h˛;XiX˛;so that h˛;Xi D 0. It is then easy to check that if n � 2, the roots in (7.23) spanh Š Cn and we conclude that X must be zero. (If, on the other hand, n D 1, thenthere are no roots and so.2IC/ D h is commutative.)

7.7.3 The Orthogonal Algebras so.2nC 1IC/

The root system for so.2n C 1IC/ is denoted Bn. We consider the compact realform so.2nC1/ of so.2nC1IC/, and we consider in so.2nC1/ the commutativesubalgebra t consisting of block diagonal matrices with n blocks of size 2 � 2

followed by one block of size 1 � 1. We take the 2 � 2 blocks to be of the sameform as in so.2n/ and we take the 1 � 1 block to be zero. Then h WD tC consists ofthose matrices in so.2nC1IC/ of the same form as in t except that the off-diagonalelements of the 2 � 2 blocks are permitted to be complex. The same argument asin the case of so.2nIC/, based on the calculations below, shows that t is maximalcommutative, so that h is a Cartan subalgebra.

The Cartan subalgebra in so.2n C 1IC/ is identifiable in an obvious way withthe Cartan subalgebra in so.2nIC/. In particular, both so.2nIC/ and so.2nC1IC/have rank n. With this identification of the Cartan subalgebras, every root forso.2nIC/ is also a root for so.2n C 1IC/. There are 2n additional roots forso.2nC 1IC/. These have the matrices having the following 2 � 1 block in entries.2k; 2nC 1/ and .2k C 1; 2nC 1/ as their root vectors:

B1 D�1

i

and having �B tr1 in entries .2n C 1; 2k/ and .2n C 1; 2k C 1/, together with the

matrices having

B2 D�1

�i�

in entries .2k; 2nC 1/ and .2k C 1; 2nC 1/ and �B tr2 in entries .2nC 1; 2k/ and

.2nC 1; 2k C 1/. The corresponding roots, viewed as elements of h�, are given byiak and �iak .

192 7 Semisimple Lie Algebras

If we use the inner product to identify the roots with elements of h and thenidentify h with Cn in the same way as in the previous subsection, the roots forso.2n C 1IC/ consist of the roots ˙ej ˙ ek , j < k, for so.2nIC/, together withadditional roots of the form

˙ej ; j D 1; : : : ; n:

These additional roots are shorter by a factor ofp2 than the roots ˙ej ˙ ek for

so.2nIC/.

7.7.4 The Symplectic Algebras sp.nIC/

The root system for sp.nIC/ is denoted Cn. We consider sp.nIC/, the space of2n � 2n complex matrices X satisfying �X tr� D X , where � is the 2n � 2n

matrix

� D�

0 I

�I 0�:

(Compare Proposition 3.25.) Explicitly, the elements of sp.nIC/ are matrices of theform

X D�A B

C �Atr

�;

where A is an arbitrary n � n matrix and B and C are arbitrary symmetricmatrices. We consider the compact real form sp.n/ D sp.nIC/\ u.2n/. (CompareSect. 1.2.8.)

We consider the commutative subalgebra t of sp.n/ consisting of matrices of theform 0

BBBBBBBBB@

a1: : :

an

�a1: : :

�an

1CCCCCCCCCA;

where each aj is pure imaginary. We then consider the subalgebra h D t C i t ofsp.nIC/, which consists of matrices of the same form but where each aj is now anarbitrary complex number. As in previous subsections, the calculations below willshow that t is maximal commutative, so that h is a Cartan subalgebra.

7.8 Exercises 193

The 2n � 2n matrices of the block form�0 Ejk C Ekj

0 0

�;

�0 0

Ejk C Ekj 0

�(7.24)

(j ¤ k) are root vectors for which the corresponding roots are .aj C ak/ and�.aj C ak/, respectively. Next, matrices of the block form

�Ejk 0

0 �Ekj

�; (7.25)

(j ¤ k) are root vectors for which the corresponding roots are .ak � al /. Finally,matrices of the block form �

0 Ejj

0 0

�;

�0 0

Ejj 0

�(7.26)

are root vectors for which the corresponding roots are 2aj and �2aj .We use on h the inner product hX; Y i D trace.X�Y /=2. If we then identify h

with Cn by means of the map

H 7! .a1; : : : ; an/;

our inner product on h will correspond to the standard inner product on Cn. Theroots are then the vectors of the form

˙ej ˙ ek; j < k

and of the form

˙2ej ; j D 1; : : : ; n:

This root system is the same as that for so.2n C 1IC/, except that instead of ˙ejwe now have ˙2ej .

7.8 Exercises

1. Let hC denote the complexification of the Lie algebra of the Heisenberg group,namely the space of all complex 3 � 3 upper triangular matrices with zeros onthe diagonal.

(a) Show the maximal commutative subalgebras of hC are precisely the two-dimensional subalgebras of hC that contain the center of hC.

(b) Show that hC does not have any Cartan subalgebras (in the sense ofDefinition 7.10).

194 7 Semisimple Lie Algebras

2. Give an example of a maximal commutative subalgebra of sl.2IC/ that is not aCartan subalgebra.

3. Show that the Hilbert–Schmidt inner product on sl.nIC/, given by hX; Y i Dtrace.X�Y /, is invariant under the adjoint action of SU.n/ and is real on su.n/.

4. Using the root space decomposition in Sect. 7.7.2, show that the Lie algebraso.4IC/ is isomorphic to the Lie algebra direct sum sl.2IC/˚ sl.2IC/. Thenshow that so.4/ is isomorphic to su.2/˚ su.2/.

5. Suppose g is a Lie subalgebra ofMn.C/ and assume that for all X 2 g, we haveX� 2 g, where X� is the usual matrix adjoint of X . Let k D g \ u.n/.

(a) Show that k is a real subalgebra of g and that g Š kC.(b) Define an inner product on g by

hX; Y i D trace.X�Y /:

Show that this inner product satisfies all the properties in Proposition 7.4.

6. For any Lie algebra g, let the Killing form be the symmetric bilinear form B

on g defined by

B.X; Y / D trace.adXadY /:

(a) Show that for any X 2 g, the operator adX W g ! g is skew-symmetricwith respect to B , meaning that

B.adX.Y /;Z/ D �B.Y; adX.Z//

for all Y;Z 2 g.(b) Suppose g D kC is semisimple. Show that B is nondegenerate, meaning

that for all nonzero X 2 g, there is some Y 2 g for which B.X; Y / ¤ 0.

Hint: Use Proposition 7.4.7. Let g be a Lie algebra and let B be the Killing form on g (Exercise 6). Show

that if g is a nilpotent Lie algebra (Definition 3.15), then B.X; Y / D 0 for allX; Y 2 g.

8. Let g denote the vector space of 3 � 3 complex matrices of the form

�A B

0 0

�;

where A is a 2� 2 matrix with trace zero and B is an arbitrary 2� 1 matrix.

(a) Show that g is a subalgebra of M3.C/.(b) Let X , Y , H , e1, and e2 be the following basis for g. We let X , Y , and

H be the usual sl.2IC/ basis in the “A” slot, with B D 0. We let e1 and

7.8 Exercises 195

e2 be the matrices with A D 0 and with .1 0/tr and .0 1/tr in the “B”slot, respectively. Compute the commutation relations among these basiselements.

(c) Show that g has precisely one nontrivial ideal, namely the span of e1 ande2. Conclude that g is not semisimple.Hint: First, determine the subspaces of g that are invariant under the adjointaction of the sl.2IC/ algebra spanned by X , Y , andH , and then determinewhich of these subspaces are also invariant under the adjoint action of e1and e2. In determining the sl.2IC/-invariant subspaces, use Exercise 10 ofChapter 4.

(d) Let h be the one-dimensional subspace of g spanned by the element H .Show that h is a maximal commutative subalgebra of g and that adH isdiagonalizable. Show that the eigenvalues of adH are 0, ˙1, and ˙2.Note: The “roots” for h are thus the numbers ˙1 and ˙2, which wouldnot be possible if h were a Cartan subalgebra of the form h D tC in asemisimple Lie algebra.

(e) Let g1 and g�1 denote the eigenspaces of adH with eigenvalues 1 and �1,respectively. Show that Œg1; g�1� D f0g.Note: This result should be contrasted with the semisimple case, whereŒg˛; g�˛� is always a one-dimensional subspace of h, so that g˛, g�˛ , andŒg˛; g�˛� span a three-dimensional subalgebra of g isomorphic to sl.2IC/.

9. Using Theorem 7.35 and the calculations in Sect. 7.7.3, show that the Liealgebra so.5IC/ is simple.

10. Let E D Rn, n � 2, and consider the Dn root system, consisting of the vectorsof the form ˙ej ˙ek , with j < k. Show that the Weyl group ofDn is the groupof transformations of Rn expressible as a composition of a permutation of theentries and an even number of sign changes.

11. Let E D Rn and consider the Bn root system, consisting of the vectors ofthe form ˙ej ˙ ek , with j < k, together with the vectors of the form ˙ej ,j D 1; : : : ; n. Show that the Weyl group of Bn is the group of transformationsexpressible as a composition of a permutation of the entries and an arbitrarynumber of sign changes.Note: Since each root in Cn is a multiple of a root in Bn, and vice versa, thereflections for Cn are the same as the reflections for Bn. Thus, the Weyl groupof Cn is the same as that of Bn.

12. Let g be a complex simple Lie algebra, with complex structure denoted byJ . Let gR denote the Lie algebra g viewed as a real Lie algebra of twice thedimension. Now let g0 be the complexification of gR, with the complex structureon g0 denoted by i .

(a) Show that g0 decomposes as a Lie algebra direct sum g0 D g1 ˚ g2, whereboth g1 and g2 are isomorphic to g.Hint: Consider element of g0 of the formX C iJX and of the formX � iJX,with X 2 g.

196 7 Semisimple Lie Algebras

(b) Show that gR is simple as a real Lie algebra. (That is to say, there is nonontrivial real subspace h of g such that ŒX; Y � 2 h for all X 2 g andY 2 h.)

13. Let h be a real simple Lie algebra, and assume that hC is not simple. Show thatthere is a complex simple Lie algebra g such that h Š gR, where the notation isas in Exercise 12.Hint: Imitate the proof of Theorem 7.32.

14. Show that the real Lie algebra so.3I 1/ is isomorphic to sl.2IC/R, where thenotation is as in Exercise 12. Conclude that so.3I 1/ is simple as a real Liealgebra, but that so.3I 1/C is not simple and is isomorphic to sl.2IC/˚sl.2IC/.Hint: First show that so.3I 1/C Š sl.2IC/ ˚ sl.2IC/ and then show that thetwo copies of sl.2IC/ are conjugates of each other.

Chapter 8Root Systems

In this chapter, we consider root systems apart from their origins in semisimpleLie algebras. We establish numerous “factoids” about root systems, which willbe used extensively in subsequent chapters. Here is one example of how resultsabout root systems will be used. In Chapter 9, we construct each finite-dimensional,irreducible representation of a semisimple Lie algebra as a quotient of an infinite-dimensional representation known as a Verma module. To prove that the quotientrepresentation is finite-dimensional, we prove that the weights of the quotient areinvariant under the action of the Weyl group, that is, the group generated by thereflections about the hyperplanes orthogonal to the roots. It is not possible, however,to prove directly that the weights are invariant under all reflections, but only underreflections coming from a special subset of the root system known as the base. Tocomplete the argument, then, we need to know that the Weyl group is actuallygenerated by the reflections associated to the roots in the base. This claim is thecontent of Proposition 8.24.

8.1 Abstract Root Systems

A root system is any collection of vectors having the properties satisfied by theroots (viewed as a subset of i t � h) of a semisimple Lie algebra, as encoded in thefollowing definition.

A previous version of this book was inadvertently published without the middle initial of theauthor’s name as “Brian Hall”. For this reason an erratum has been published, correcting themistake in the previous version and showing the correct name as Brian C. Hall (see DOI http://dx.doi.org/10.1007/978-3-319-13467-3_14). The version readers currently see is the correctedversion. The Publisher would like to apologize for the earlier mistake.

© Springer International Publishing Switzerland 2015B.C. Hall, Lie Groups, Lie Algebras, and Representations, GraduateTexts in Mathematics 222, DOI 10.1007/978-3-319-13467-3_8

197

198 8 Root Systems

Definition 8.1. A root system .E;R/ is a finite-dimensional real vector space Ewith an inner product h�; �i, together with a finite collection R of nonzero vectors inE satisfying the following properties:

1. The vectors in R span E .2. If ˛ is in R and c 2 R, then c˛ is in R only if c D ˙1.3. If ˛ and ˇ are in R, then so is s˛ � ˇ, where s˛ is the linear transformation of E

defined by

s˛ � ˇ D ˇ � 2 hˇ; ˛ih˛; ˛i˛; ˇ 2 E:

4. For all ˛ and ˇ in R, the quantity

2hˇ; ˛ih˛; ˛i

is an integer.

The dimension of E is called the rank of the root system and the elements of Rare called roots.

Note that since s˛ �˛ D �˛, we have that �˛ 2 R whenever ˛ 2 R. In the theoryof symmetric spaces, there arise systems satisfying Conditions 1, 3, and 4, but notCondition 2. These are called “nonreduced” root systems. In the theory of Coxetergroups, there arise systems satisfying Conditions 1, 2, and 3, but not Property 4.These are called “noncrystallographic” or “nonintegral” root systems. In this book,we consider only root systems satisfying all of the conditions in Definition 8.1.

The map s˛ is the reflection about the hyperplane orthogonal to ˛; that is, s˛ �˛ D�˛ and s˛ � ˇ D ˇ for all ˇ that are orthogonal to ˛, as is easily verified from theformula for s˛ . From this description, it should be evident that s˛ is an orthogonaltransformation of E with determinant �1.

We can interpret Property 4 geometrically in one of two ways. In light of theformula for s˛ , Property 4 is equivalent to saying that s˛ � ˇ should differ from ˇ

by an integer multiple of ˛. Alternatively, if we recall that the orthogonal projectionof ˇ onto ˛ is given by .hˇ; ˛i=h˛; ˛i/˛, we note that the quantity in Property 4 istwice the coefficient of ˛ in this projection. Thus, Property 4 is equivalent to sayingthat the projection of ˇ onto ˛ is an integer or half-integer multiple of ˛.

We have shown that one can associate a root system to every complex semisimpleLie algebra. It turns out that every root system arises in this way, although this is farfrom obvious—see Sect. 8.11.

Definition 8.2. If .E;R/ is a root system, the Weyl groupW of R is the subgroupof the orthogonal group of E generated by the reflections s˛ , ˛ 2 R.

By assumption, each s˛ maps R into itself, indeed onto itself, since each ˇ 2 Rsatisfies ˇ D s˛ � .s˛ � ˇ/. It follows that every element of W maps R onto itself.Since the roots span E , a linear transformation of E is determined by its action

8.1 Abstract Root Systems 199

on R. Thus, the Weyl group is a finite subgroup of O.E/ and may be regarded as asubgroup of the permutation group onR. We denote the action of w 2 W onH 2 Eby w �H .

Proposition 8.3. Suppose .E;R/ and .F; S/ are root systems. Consider the vectorspace E ˚ F , with the natural inner product determined by the inner products onE and F . Then R[ S is a root system in E ˚F , called the direct sum of R and S .

Here, we are identifyingE with the subspace of E ˚ F consisting of all vectorsof the form .e; 0/ with e in E , and similarly for F . Thus, more precisely, R [ S

means the elements of the form .˛; 0/ with ˛ in R together with elements of theform .0; ˇ/ with ˇ in S . (Elements of the form .˛; ˇ/ with ˛ 2 R and ˇ 2 S arenot in R [ S .)

Proof. If R spans E and S spans F , then R [ S spans E ˚ F , so Condition 1is satisfied. Condition 2 holds because R and S are root systems in E and F ,respectively. For Condition 3, if ˛ and ˇ are both in R or both in S , then s˛ � ˇ 2R [ S because R and S are root systems. If ˛ 2 R and ˇ 2 S or vice versa, thenh˛; ˇi D 0, so that

s˛ � ˇ D ˇ 2 R [ S:

Similarly, if ˛ and ˇ are both in R or both in S , then 2 h˛; ˇi = h˛; ˛i is an integerbecause R and S are root systems, and if ˛ 2 R and ˇ 2 S or vice versa, then2 h˛; ˇi = h˛; ˛i D 0. Thus, Condition 4 holds for R [ S . utDefinition 8.4. A root system .E;R/ is called reducible if there exists an orthogo-nal decompositionE D E1 ˚ E2 with dimE1 > 0 and dimE2 > 0 such that everyelement of R is either in E1 or in E2. If no such decomposition exists, .E;R/ iscalled irreducible.

If .E;R/ is reducible, then it is not hard to see that the part of R in E1 is a rootsystem in E1 and the part of R in E2 is a root system in E2. Thus, a root system isreducible precisely if it can be realized as a direct sum of two other root systems.In the Lie algebra setting, the root system associated to a complex semisimple Liealgebra g is irreducible precisely if g is simple (Theorem 7.35).

Definition 8.5. Two root systems .E;R/ and .F; S/ are said to be isomorphic ifthere exists an invertible linear transformationA W E ! F such thatAmapsR ontoS and such that for all ˛ 2 R and ˇ 2 E , we have

A .s˛ � ˇ/ D sA˛ � .Aˇ/:

A map A with this property is called an isomorphism.

Note that the linear map A is not required to preserve inner products, but only topreserve the reflections about the roots. If, for example, F D E and S consists ofelements of the form c˛ with ˛ 2 R, then .F; S/ is isomorphic to .E;R/, with theisomorphism being the map A D cI .

200 8 Root Systems

3

2

24

2

6

2

3 2

Fig. 8.1 The basic acute angles and length ratios

We now establish a basic result limiting the possible angles and length ratiosoccurring in a root system.

Proposition 8.6. Suppose ˛ and ˇ are roots, ˛ is not a multiple of ˇ, and h˛; ˛i �hˇ; ˇi. Then one of the following holds:

1. h˛; ˇi D 0

2. h˛; ˛i D hˇ; ˇi and the angle between ˛ and ˇ is �=3 or 2�=33. h˛; ˛i D 2 hˇ; ˇi and the angle between ˛ and ˇ is �=4 or 3�=44. h˛; ˛i D 3 hˇ; ˇi and the angle between ˛ and ˇ is �=6 or 5�=6

Figure 8.1 shows the allowed angles and length ratios, for the case of an acuteangle. In each case, 2 h˛; ˇi=h˛; ˛i D 1, whereas 2 hˇ; ˛i=hˇ; ˇi takes the values1, 2, and 3 in the three successive cases. Section 8.2 shows that each of the anglesand length ratios permitted by Proposition 8.6 actually occurs in some root system.

Proof. Suppose that ˛ and ˇ are roots and let m1 D 2h˛; ˇi=h˛; ˛i and m2 D2hˇ; ˛i=hˇ; ˇi, so that m1 and m2 are integers. Assume h˛; ˛i � hˇ; ˇi and notethat

m1m2 D 4h˛; ˇi2

h˛; ˛i hˇ; ˇi D 4 cos2 �; (8.1)

where � is the angle between ˛ and ˇ, and that

m2

m1

D h˛; ˛ihˇ; ˇi � 1 (8.2)

8.2 Examples in Rank Two 201

22

s

Fig. 8.2 The projection of ˇ onto ˛ equals ˛=2 and s˛ � ˇ equals ˇ � ˛

whenever h˛; ˇi ¤ 0. From (8.1), we conclude that 0 � m1m2 � 4. If m1m2 D 0,then cos � D 0, so ˛ and ˇ are orthogonal. If m1m2 D 4, then cos2 � D 1, whichmeans that ˛ and ˇ are multiples of one another.

The remaining possible values for m1m2 are 1, 2, and 3. If m1m2 D 1, thencos2 � D 1=4, so � is �=3 or 2�=3. Since m1 and m2 are both integers, we musthave m1 D 1 and m2 D 1 or m1 D �1 and m2 D �1. In the first case, h˛; ˇi > 0

and we have � D �=3 and in the second case, h˛; ˇi < 0 and we have � D 2�=3.In either case, (8.2) tells us that ˛ and ˇ have the same length, establishing Case 2of the proposition.

If m1m2 D 2, then cos2 � D 1=2, so that � is �=4 or 3�=4. Since m1 andm2 areintegers and jm2j � jm1j by (8.2), we must have m1 D 1 and m2 D 2 or m1 D �1andm2 D �2. In the first case, we have � D �=4 and in the second case, � D 3�=4.In either case, (8.2) tells us that ˛ is longer than ˇ by a factor of

p2. The analysis

of the case m1m2 D 3 is similar. utCorollary 8.7. Suppose ˛ and ˇ are roots. If the angle between ˛ and ˇ is strictlyobtuse (i.e., strictly between �=2 and �), then ˛ C ˇ is a root. If the angle between˛ and ˇ is strictly acute (i.e., strictly between 0 and �=2), then ˛�ˇ and ˇ�˛ arealso roots.

See Figure 8.2.

Proof. The proof is by examining each of the three obtuse angles and each of thethree acute angles allowed by Proposition 8.6. Consider first the acute case andadjust the labeling so that h˛; ˛i � hˇ; ˇi. An examination of Cases 2, 3, and 4 inthe proposition (see Figure 8.1) shows that in each of these cases, the projection ofˇ onto ˛ is equal to ˛=2. Thus, s˛ � ˇ D ˇ � ˛ is again a root—and so, therefore,is �.ˇ � ˛/ D ˛ � ˇ. In the obtuse case (with h˛; ˛i � hˇ; ˇi), the projection of ˇonto ˛ equals �˛=2, and, thus, s˛ � ˇ D ˛ C ˇ is again a root. ut

8.2 Examples in Rank Two

If the rank of the root system is one, then there is only one possibility: R mustconsist of a pair f˛;�˛g, where ˛ is a nonzero element of E . In rank two, there arefour possibilities, pictured in Figure 8.3 with their conventional names. In the case

202 8 Root Systems

A1 A1 A2

B2 G2

Fig. 8.3 The rank-two root systems

of A1 � A1, the lengths of the horizontal roots are unrelated to the lengths of thevertical roots. In A2, all roots have the same length; in B2, the length of the longerroots is

p2 times the length of the shorter roots; inG2, the length of the longer roots

isp3 times the length of the shorter roots. The angle between successive roots is

�=2 for A1 � A1, �=3 for A2, �=4 for B2, and �=6 for G2. It is easy to checkthat each of these systems is actually a root system; in particular, Proposition 8.6 issatisfied for each pair of roots.

Proposition 8.8. Every rank-two root system is isomorphic to one of the systems inFigure 8.3.

Proof. It is harmless to assume E D R2; thus, let R � R2 be a root system. Let �be the smallest angle occurring between any two vectors inR. Since the elements ofR span R2, we can find two linearly independent vectors ˛ and ˇ in R. If the anglebetween ˛ and ˇ is greater than �=2, then the angle between ˛ and �ˇ is less than�=2; thus, the minimum angle � is at most �=2. Then, according to Proposition 8.6,� must be one of the following: �=2, �=3, �=4, �=6.

Let ˛ and ˇ be two elements of R such that the angle between them is theminimum angle � . Then the vector �sˇ � ˛ will be a vector that is at angle � toˇ but on the opposite side of ˇ from ˛, as shown in Figure 8.4. Thus, �sˇ � ˛ is at

8.2 Examples in Rank Two 203

Fig. 8.4 The root �sˇ � ˛ isat angle 2� from ˛

s

angle 2� to ˛. Similarly, �ssˇ �˛ � ˇ is at angle 3� to ˛. Continuing in the same way,we can obtain vectors at angle n� to ˛ for all n. These vectors are unique since anontrivial positive multiple of a root is not allowed to be a root. Now, since eachof the allowed values of � evenly divides 2� , we will eventually come around to ˛again. Furthermore, there cannot be any other vectors besides those at angles n� to˛, or else there would be an angle smaller than � .

Thus, R must consist of n equally spaced vectors, with consecutive vectorsseparated by angle � , where � is one of the acute angles in Proposition 8.6. If,say, � D �=4, then in order to satisfy the length requirement in Proposition 8.6, theroots must alternate between a shorter length and a second length that is greater bya factor of

p2. Thus, our root system must be isomorphic to B2. Similar reasoning

shows that all remaining values of � yield one of the root systems in Figure 8.3. utUsing the results of Sect. 7.7, we can see that the root systems A1 � A1, A2, and

B2 arise as root systems of classical Lie algebras as follows. The root systemA1�A1is the root system of so.4IC/, which is isomorphic to sl.2IC/˚ sl.2IC/; A2 is theroot system of sl.3IC/; and B2 is the root system of so.5IC/, which is isomorphicto the root system of sp.2IC/. The root system G2, meanwhile, is the root systemof an “exceptional” Lie algebra, which is also referred to as G2 (Sect. 8.11).

Proposition 8.9. If R is a rank-two root system with minimum angle � D 2�=n,n D 4; 6; 8; 12, then the Weyl group of R is the symmetry group of a regularn=2-gon.

Proof. The group W will contain at least n=2 reflections, one for each pair ˙˛ ofroots. If ˛ and ˇ are roots with some angle � between them, then the compositionof the reflections s˛ and sˇ will be a rotation by angle ˙2�, with the direction ofthe rotation depending on the order of the composition. To see this, note that s˛

204 8 Root Systems

and sˇ both have determinant �1 and so s˛sˇ has determinant 1 and is, therefore,a rotation by some angle. To determine the angle, it suffices to apply sˇs˛ to anynonzero vector, for example, ˛. However, s˛ � ˛ D �˛ and sˇ � .�˛/ is at angle 2�to ˛, as in Figure 8.4.

Now, since � can be any integer multiple of � , we obtain all rotations by integermultiples of 2� . Meanwhile, the composition of a reflection s˛ and a rotation byangle 2� will be another reflection sˇ , where ˇ is a root at angle � to ˛, as thereader may verify. Thus, the set of n=2 reflections together with rotations by integermultiples of 2� form a group; this is the Weyl group of the rank-two root system.But this group is also the symmetry group of a regular n=2-gon, also known as thedihedral group on n=2 elements. ut

Note that in the case of A2, the Weyl group consists of three reflections togetherwith three rotations (by multiples of 2�=3). In this case, the Weyl group is not thefull symmetry group of the root system: Rotations by �=3 map R onto itself but arenot elements of the Weyl group.

8.3 Duality

In this section, we introduce an important duality operation on root systems.

Definition 8.10. If .E;R/ is a root system, then for each root ˛ 2 R, the corootH˛ is the vector given by

H˛ D 2˛

h˛; ˛i :

The set of all coroots is denoted R_ and is called the dual root system to R.

This definition is consistent with the use of the term “coroot” in Chapter 7; seeDefinition 7.28. Point 4 in the definition of a root system may be restated as sayingthat hˇ;H˛i should be an integer for all roots ˛ and ˇ.

Proposition 8.11. If R is a root system, then R_ is also a root system and the Weylgroup for R_ is the same as the Weyl group for R. Furthermore, .R_/_ D R.

Proof. We compute that

hH˛;H˛i D 4h˛; ˛ih˛; ˛i2 D 4

h˛; ˛i

and, therefore,

2H˛

hH˛;H˛i D 2

�2˛

h˛; ˛i� h˛; ˛i

4D ˛: (8.3)

8.3 Duality 205

If we take the inner product of (8.3) with Hˇ , we see that

2

˝H˛;Hˇ

˛hH˛;H˛i D ˝

˛;Hˇ

˛ D 2h˛; ˇihˇ; ˇi ; (8.4)

which means that the left-hand side of (8.4) is an integer.Furthermore, since H˛ is a multiple of ˛, it is evident that s˛ D sH˛ . Since, also,

s˛ is an orthogonal transformation, we have

s˛ �Hˇ D 2s˛ � ˇhˇ; ˇi D 2

s˛ � ˇhs˛ � ˇ; s˛ � ˇi D Hs˛ �ˇ:

Thus, the set of coroots is invariant under each reflection s˛ (D sH˛ ). Thisobservation, together with (8.4), shows that R_ is again a root system, with theremaining properties of root systems for R_ following immediately from thecorresponding properties of R.

Since s˛ D sH˛ , we see that R and R_ have the same Weyl group. Finally, (8.3)says that the formula for ˛ in terms ofH˛ is the same as the formula forH˛ in termsof ˛. Thus, HH˛ D ˛ and .R_/_ D R. utNote from (8.4) that the integer associated to the pair .H˛;Hˇ/ in R_ is the sameas the integer associated to the pair .ˇ; ˛/ (not .˛; ˇ/) in R. If all the roots in Rhave the same length,R_ is isomorphic toR. Even if not all the roots have the samelength, R and R_ could be isomorphic. In the case of B2, for example, the dualroot system R_ can be converted back to R by a �=4 rotation and a scaling. (SeeFigure 8.5.) On the other hand, the rank-three root systems B3 and C3 (Sect. 8.9)are dual to each other but not isomorphic to each other.

Figure 8.5 shows the dual root system for the root system B2. On the left-handside of the figure, the long roots have been normalized to have length

p2. Thus, for

each long root we haveH˛ D ˛ and for each short root we haveH˛ D 2˛, yieldingthe root system on the right-hand side of the figure.

Fig. 8.5 The B2 root system and its dual

206 8 Root Systems

8.4 Bases and Weyl Chambers

We now introduce the notion of a base, or a system of positive simple roots, for aroot system.

Definition 8.12. If .E;R/ is a root system, a subset of R is called a base if thefollowing conditions hold:

1. is a basis for E as a vector space.2. Each root ˛ 2 R can be expressed as a linear combination of elements of

with integer coefficients and in such a way that the coefficients are either allnon-negative or all nonpositive.

The roots for which the coefficients are non-negative are called positive rootsand the others are called negative roots (relative to the base ). The set of positiveroots relative to a fixed base is denotedRC and the set of negative roots is denotedR�. The elements of are called the positive simple roots.

Note that since is a basis for E , each ˛ can be expressed uniquely as a linearcombination the elements of . We require that be such that the coefficients inthe expansion of each ˛ 2 R be integers and such that all the nonzero coefficientshave the same sign.

Proposition 8.13. If ˛ and ˇ are distinct elements of a base for R, thenh˛; ˇi � 0.

Geometrically, this means that either ˛ and ˇ are orthogonal or the angle betweenthem is obtuse.

Proof. Since ˛ ¤ ˇ, if we had h˛; ˇi > 0, then the angle between ˛ and ˇ would bestrictly between 0 and �=2. Then, by Corollary 8.7, ˛�ˇ would be an element ofR.Since the elements of form a basis forE as a vector space, each element ofR hasa unique expansion in terms of elements of, and the coefficients of that expansionare supposed to be either all non-negative or all nonpositive. The expansion of ˛�ˇ,however, has one positive and one negative coefficient. Thus, ˛�ˇ cannot be a root,which means that h˛; ˇi � 0. ut

The reader is invited to find a base for each of the rank-two root systems inFigure 8.3. We now show that every root system has a base.

Proposition 8.14. There exists a hyperplane V through the origin in E that doesnot contain any element of R.

Proof. For each ˛ 2 R, let V˛ denote the hyperplane

V˛ D fH 2 Ej h˛;H i D 0g:

Since there are only finitely many of these hyperplanes, there exists H 2 E not inany V˛ . (See Exercise 2.) Let V be the hyperplane through the origin orthogonal to

8.4 Bases and Weyl Chambers 207

H . SinceH is not in any V˛ , we see thatH is not orthogonal to any ˛, which meansthat no ˛ is in V . utDefinition 8.15. Let .E;R/ be a root system. Let V be a hyperplane through theorigin in E such that V does not contain any root. Choose one “side” of V and letRC denote the set of roots on this side of V . An element ˛ of RC is decomposableif there exist ˇ and � in RC such that ˛ D ˇ C � ; if no such elements exist, ˛ isindecomposable.

The “sides” of V can be defined as the connected components of the set E n V .Alternatively, if H is a nonzero vector orthogonal to V , then V is the set of � 2 Efor which h�;H i D 0. The two “sides” of V are then the sets f� 2 Ej h�;H i > 0gand f� 2 Ej h�;H i < 0g.

Theorem 8.16. Suppose .E;R/ is a root system, V is a hyperplane through theorigin in E not containing any element of R, and RC is the set of roots lying on afixed side of V . Then the set of indecomposable elements of RC is a base for R.

This construction of a base motivates the term “positive simple root” for theelements of a base. We first define the positive roots (RC) as the roots on one side ofV and then define the positive simple roots (the base) as the set of indecomposable(or “simple”) elements of RC.

Proof. Let denote the set of indecomposable elements in RC. Choose a nonzerovectorH orthogonal to V so that the chosen side of V is the set of � 2 E for whichhH;�i > 0.

Step 1: Every ˛ 2 RC can be expressed as a linear combination of elements of with non-negative integer coefficients. If not, then among all of the elementsof RC that cannot be expressed in this way, choose ˛ so that hH;˛i is as smallas possible. Certainly ˛ cannot be an element of, so ˛ must be decomposable,˛ D ˇ1 C ˇ2, with ˇ1; ˇ2 2 RC. Now, ˇ1 and ˇ2 cannot both be expressible aslinear combinations of elements of with non-negative integer coefficients, orelse ˛ would be expressible in this way. However, hH;˛i D hH;ˇ1i C hH;ˇ2i,and since the numbers hH;ˇ1i and hH;ˇ2i are both positive, they must besmaller than hH;˛i, contradicting the minimality of ˛.Step 2: If ˛ and ˇ are distinct elements of , then h˛; ˇi � 0. Note that since is not yet known to be a base, Proposition 8.13 does not apply. Nevertheless,if we had h˛; ˇi > 0, then by Corollary 8.7, ˛ � ˇ and ˇ � ˛ would both beroots, one of which would have to be inRC. If ˛�ˇ were inRC, we would have˛ D .˛�ˇ/Cˇ and ˛ would be decomposable. If, on the other hand, ˇ�˛ werein RC, we would have ˇ D .ˇ � ˛/C ˛, and ˛ would, again, be decomposable.Since ˛ and ˇ are assumed indecomposable, we must have h˛; ˇi � 0.Step 3: The elements of are linearly independent. Suppose we have

X˛2

c˛˛ D 0 (8.5)

208 8 Root Systems

for some collection of constants c˛ . Separate the sum into those terms wherec˛ � 0 and those where c˛ D �d˛ < 0, so that

Xc˛˛ D

Xdˇˇ (8.6)

where the sums range over disjoint subsets of . If u denotes the vector in (8.6),we have

hu; ui DDX

c˛˛;X

dˇˇE

DXX

c˛dˇh˛; ˇi:

However, c˛ and d˛ are non-negative and (by Step 2) h˛; ˇi � 0. Thus, hu; ui � 0

and u must be zero.Now, if u D 0, then hH; ui D P

c˛ hH;˛i D 0, which implies that all the c˛’sare zero since c˛ � 0 and hH;˛i > 0. The same argument then shows that allthe d˛’s are zero as well.Step 4: is a base. We have shown that is linearly independent and that allof the elements of RC can be expressed as linear combinations of elements of with non-negative integer coefficients. The remaining elements of R, namely theelements of R�, are simply the negatives of the elements of RC, and so they canbe expressed as linear combinations of elements of with nonpositive integercoefficients. Since the elements of R span E , then must also span E and it isa base. utFigure 8.6 illustrates Theorem 8.16 in the case of the G2 root system, with ˛1

and ˛2 being the indecomposable roots on one side of the dashed line.

Theorem 8.17. For any base for R, there exists a hyperplane V and a side of Vsuch that arises as in Theorem 8.16.

Proof. If D f˛1; : : : ; ˛r g is a base for R, then is a basis for E in the vectorspace sense. Then, by elementary linear algebra, for any sequence of numbersc1; : : : ; cr there exists a unique � 2 E with

˝�; ˛j

˛ D cj , j D 1; : : : ; r . In particular,we can choose � so that

˝�; ˛j

˛> 0 for all j . Then if RC denotes the positive

roots with respect to , we will have h�; ˛i > 0 for all ˛ 2 RC, since ˛ is alinear combination of elements of with non-negative coefficients. Thus, all of theelements of RC lie on the same side of the hyperplane V D fH 2 E jh�;H i D 0g.

Suppose now that ˛ is an element of and that ˛ were expressible as a sumof at least two elements of RC. Then at least one of these elements would bedistinct from ˛ and, thus, not a multiple of ˛. Thus, ˛ would be expressible as alinear combination of the elements of with non-negative coefficients, where thecoefficient of some ˇ ¤ ˛ would have to be nonzero. This would contradict theindependence of the elements of . We conclude, then, that every ˛ 2 is anindecomposable element ofRC. Thus, is contained in the base associated to V as

8.4 Bases and Weyl Chambers 209

1

2

2 1

2 2 1

2 3 1

2 2 3 1

Fig. 8.6 The roots ˛1 and ˛2 form a base for the root system G2

in Theorem 8.16. However, the number of elements in a base must equal dimE , so,actually, is the base associated to V . utProposition 8.18. If is a base for R, then the set of all coroots H˛ , ˛ 2 , is abase for the dual root system R_.

Figure 8.7 illustrates Proposition 8.18 in the case of the root system B2.

Lemma 8.19. Let be a base, RC the associated set of positive roots, and ˛ anelement of . Then ˛ cannot be expressed as a linear combination of elements ofRC n f˛g with non-negative real coefficients.

Proof. Let ˛1 D ˛ and let ˛2; : : : ; ˛r be the remaining elements of . Suppose˛1 is a linear combination elements ˇ ¤ ˛1 in RC with non-negative coefficients.Each such ˇ can then be expanded in terms of ˛1; : : : ; ˛r with non-negative (integer)coefficients. Thus, we end up with

˛1 D c1˛1 C c2˛2 C � � � C cr˛r ;

210 8 Root Systems

1

2

1 2

2 1 2

H 1

H 2 H1

2H2

H 1 H 2

Fig. 8.7 Bases for B2 and its dual

with each cj � 0. Since ˛1; : : : ; ˛r are independent, we must have c1 D 1 andall other cj ’s equal 0. But since each ˇ in the expansion of ˛ has non-negativecoefficients in the basis ˛1; : : : ; ˛r , each ˇ would have to be a multiple of ˛1,and thus actually equal to ˛1, since the only multiple of ˛1 in RC is ˛1. But thiscontradicts the assumption that ˇ was different from ˛1. utProof of Proposition 8.18. Choose a hyperplaneV such that the base forR arisesas in Theorem 8.16, and call the side of V on which lies the positive side. LetRCdenote the set of positive roots in R relative to the base . Then the coroots H˛ ,˛ 2 RC, also lie on the positive side of V , and all the remaining coroots lie on thenegative side of V . Thus, applying Theorem 8.16 to R_, there exists a base _ forR_ such that the positive roots associated to are precisely theH˛’s with ˛ 2 RC.

Now, if ˛ 2 RC but ˛ … , then ˛ is a linear combination of ˛1; : : : ; ˛r withnon-negative integer coefficients, at least two of which are nonzero. Thus, H˛ is alinear combination ofH˛1 ; : : : ;H˛r with non-negative real coefficients, at least twoof which are nonzero. Since, by Lemma 8.19, such an H˛ cannot be in _, the relements of _ must be precisely H˛1; : : : ;H˛r . ut

Definition 8.20. The open Weyl chambers for a root system .E;R/ are theconnected components of

E �[˛2R

V˛;

where V˛ is the hyperplane through the origin orthogonal to ˛. If D f˛1; : : : ; ˛rgis a base for R, then the open fundamental Weyl chamber in E (relative to ) isthe set of all H in E such that

˝˛j ;H

˛> 0 for all j D 1; : : : ; r .

Figure 8.8 shows that open fundamental Weyl chamber C associated to aparticular base for the A2 root system. Note that the boundary of C is made upof portions of the lines orthogonal to the roots (not the lines through the roots).

Since the elements of a base form a basis for E as a vector space, elementarylinear algebra shows that the open fundamental Weyl chamber is convex, henceconnected, and nonempty. Since the only way one can exit a fundamental Weyl

8.4 Bases and Weyl Chambers 211

C

1

2

Fig. 8.8 The shaded region C is the open fundamental Weyl chamber associated to the basef˛1; ˛2g for A2

chamber is by passing through a pointH where˝˛j ;H

˛ D 0, the open fundamentalWeyl chamber is, indeed, an open Weyl chamber. Note, also, that if

˝˛j ;H

˛> 0 for

all j D 1; : : : ; r , then h˛;H i > 0 for all ˛ 2 RC, since ˛ is a linear combinationof ˛1; : : : ; ˛r with non-negative coefficients.

Each w 2 W is an orthogonal linear transformation that maps R to R and, thus,maps the set of hyperplanes orthogonal to the roots to itself. It then easily followsthat for each open Weyl chamber C , the set w � C is another open Weyl chamber.

For any base and the associated set RC of positive roots, we have defined thefundamental Weyl chamber to be the set of those elements having positive innerproduct with each element of , and therefore also with each element of RC. Ournext result says that we can reverse this process.

Proposition 8.21. For each open Weyl chamber C , there exists a unique base C

for R such that C is the open fundamental Weyl chamber associated to C . Thepositive roots with respect to C are precisely those elements ˛ of R such that ˛has positive inner product with each element of C .

Thus, there is a one-to-one correspondence between bases and Weyl chambers.

Proof. Let H be any element of C and let V be the hyperplane orthogonal to H .Since H is contained in an open chamber, H is not orthogonal to any root, and,thus, V does not contain any root. Thus, by Theorem 8.16, there exists a base D f˛1; : : : ; ˛r g lying on the same side of V as H . Since

˝˛j ;H

˛has constant

sign on C , we see that every element of C has positive inner product with each

212 8 Root Systems

element of and thus with every element of the associated setRC of positive roots.Thus, C must be the fundamental Weyl chamber associated to . We have justsaid that every ˛ 2 RC has positive inner product with every element of C , whichmeans that each ˛ 2 R� has negative inner product with each element of C . Thus,RC consists precisely of those roots having positive inner product with C .

Finally, if 0 is any base whose fundamental chamber is C , then each elementof 0 has positive inner product with H 2 C , meaning that 0 lies entirely on thesame side of V as . Thus, 0 has the same positive roots as and therefore alsothe same set of positive simple (i.e., indecomposable) roots as . That is to say,0 D . utProposition 8.22. Every root is an element of some base.

Since C is the set of H 2 E for which˝˛j ;H

˛> 0 for j D 1; : : : ; r ,

the codimension-one pieces of the boundary of C will consist of portions of thehyperplanes V˛j orthogonal to the elements of . Thus, to prove the proposition,we merely need to show that for every root ˛, the hyperplane V˛ contains acodimension-one piece of the boundary of some C .

Proof. Let ˛ be a root and V˛ the hyperplane orthogonal to ˛. If we apply Exercise 2to V˛ we see that there exists some H 2 V˛ such that H does not belong to Vˇ forany root ˇ other than ˇ D ˙˛. Then for small positive ", the elementH C "˛ willbe in a open Weyl chamber C .

We now claim that ˛ must be a member of the base C D f˛1; : : : ; ˛rg inProposition 8.21. Since ˛ has positive inner product with H C "˛ 2 C , wemust at least have ˛ 2 RC, the set of positive roots associated to C . Write˛ D Pr

jD1 cj ˛j , with cj � 0, so that

0 D h˛;H i DrX

jD1cj˝˛j ;H

˛:

SinceH is clearly in the closure of C , we must have˝H;˛j

˛ � 0, which means that˝˛j ;H

˛must be zero whenever cj ¤ 0. If more than one of the cj ’s were nonzero,

H would be orthogonal to two distinct roots in , contradicting our choice of H .Thus, actually, ˛ is in . ut

8.5 Weyl Chambers and the Weyl Group

In this section, we establish several results about how the action of the Weyl grouprelates to the Weyl chambers.

8.5 Weyl Chambers and the Weyl Group 213

Proposition 8.23. The Weyl group acts transitively on the set of open Weylchambers.

Proposition 8.24. If is a base, then W is generated by the reflections s˛ with˛ 2 .

Proof of Propositions 8.23 and 8.24. Fix a base, let C be the fundamental cham-ber associated to , and let W 0 be the subgroup of W generated by the s˛’s with˛ 2 . Let D be any other chamber and let H and H 0 be fixed elements of C andD, respectively. We wish to show that there exists w 2 W 0 so that w �H 0 belongs toC . To this end, choose w 2 W 0 so that jw �H 0 �H j is minimized, which is possiblebecauseW 0 � W is finite. If w �H 0 were not in C , there would be some ˛ 2 suchthat h˛;w �H 0i < 0. In that case, direct calculation would show that

ˇw �H 0 �H ˇ2 � ˇ

s˛ � w �H 0 �Hˇ2 D �4 h˛;w �H 0i

h˛; ˛i h˛;H i > 0;

which means that s˛ �w �H 0 is closer toH than w �H 0 is (Figure 8.9). Since s˛ 2 W 0,this situation would contradict the minimality of w. Thus, actually, w � H 0 2 C ,showing that D can be mapped to C by an element of W 0. Since any chambercan be mapped to C by an element of W 0, any chamber can be mapped to anyother chamber by an element of W 0. We conclude that W 0, and thus also W , actstransitively on the chambers, proving Proposition 8.23.

H

w H

s 1 w H

1

2

Fig. 8.9 If w � H 0 and H are on opposite sides of the line orthogonal to ˛1, then s˛1 � w � H 0 iscloser to H than w �H 0 is

214 8 Root Systems

To prove Proposition 8.24, we must show that W 0 D W . Let s˛ be the reflectionassociated to an arbitrary root ˛. By Proposition 8.22, ˛ belongs to D for somechamber D. If w 2 W 0 is chosen so that w � D D C , then w � ˛ will belong to .Now, it is easily seen that

sw�˛ D ws˛w�1;

so that s˛ D w�1sw�˛w. But since w � ˛ 2 , both w and sw�˛ belong to W 0. Thus,s˛ belongs to W 0 for every root ˛, which means that W (the group generated by thes˛’s) equalsW 0. utProposition 8.25. Let C be a Weyl chamber and let H and H 0 be elements of NC ,the closure of C . If w �H D H 0 for some w 2 W , then H D H 0.

That is to say, two distinct elements of NC cannot be in the same orbit of W .By Proposition 8.24, each w 2 W can be written as a product of reflections from

. If k � 0 is the smallest number of reflections from needed to express w, thenany expression for w as the product of k reflections from is called a minimalexpression for w. Such a minimal expression need not be unique.

The following technical lemma is the key to the proof of Proposition 8.25.

Lemma 8.26. Let be a base for R and let C be the associated fundamental Weylchamber. Let w be an element of W with w ¤ I and let w D s˛1s˛2 � � � s˛k , with˛j 2 , be a minimal expression for w. Then C and w � C lie on opposite sides ofthe hyperplane V˛1 orthogonal to ˛1.

Proof. Since w ¤ I , we must have k � 1. If k D 1, then w D s˛1 , so thatw �C D s˛1 �C is on the opposite side of V˛1 from C . Assume, inductively, that theresult holds for u 2 W where the minimal number of reflections needed to expressu is k � 1. Then consider w 2 W having a minimal expression of the form w Ds˛1 � � � s˛k . If we let u D s˛1 � � � s˛k�1

, then s˛1 � � � s˛k�1must be a minimal expression

for u, since any shorter expression for u would result in a shorter expression forw D us˛k . Thus, by induction, u � C and C must lie on opposite sides of V˛1 .Suppose, toward a contradiction, that w � C lies on the same side of V˛1 as C . Thenu � C and w � C D us˛k � C lie on opposite sides of V˛1 , which implies that C ands˛k � C lie on opposite sides of u�1V˛1 D Vu�1�˛1 .

We now claim that there is only one hyperplane Vˇ, with ˇ 2 R, such that Cand s˛k � C lie on opposite sides of Vˇ , namely, Vˇ D V˛k . After all, as in theproof of Proposition 8.22, we can choose H in the boundary of C so that H lies inV˛k but in no other hyperplane orthogonal to a root. We may then pass from C tos˛k � C along a line segment of the form H C t˛k , �" < t < ", and we will passthrough no hyperplane orthogonal to a root, other than V˛k . We conclude, then, thatVu�1�˛1 D V˛k .

Now, if Vu�1 �˛1 D V˛k , it follows that su�1�˛1 D s˛k , so that

s˛k D su�1 �˛1 D u�1s˛1u: (8.7)

8.5 Weyl Chambers and the Weyl Group 215

Substituting (8.7) into the formula w D us˛k gives

w D s˛1u D s2˛1 s˛2 � � � s˛k�1:

Since s2˛ D I , we conclude that w D s˛2 � � � s˛k�1, which contradicts the assumption

that s˛1 � � � s˛k was a minimal expression for w. utProof of Proposition 8.25. We proceed by induction on the minimal number ofreflections fromC needed to express w. If the minimal number is zero, then w D I

and the result holds. If the minimal number is greater than zero, let w D s˛1 � � � s˛kbe a minimal expression for w. By Lemma 8.26, C and w � C lie on opposite sidesof the hyperplane V˛1 orthogonal to ˛1. Thus,

.w � NC/ \ NC � V˛1;

which means that w �H D H 0 must lie in V˛1 . It follows that

s˛1 � w �H D s˛1 �H 0 D H 0:

That is to say, the Weyl group element w0 WD s˛1w also mapsH to H 0. But

w0 D s˛1w D s2˛1s˛2 � � � s˛k D s˛2 � � � s˛kis a product of fewer than k reflections from C . Thus, by induction, we haveH 0 D H . utProposition 8.27. The Weyl group acts freely on the set of open Weyl chambers. IfH belongs to an open chamber C and w �H D H for some w 2 W , then w D I .

Proof. Let C be any chamber and let be the associated base (Proposition 8.21).If w 2 W and W ¤ I; Lemma 8.26 tells us that w � C and C lie on opposite sidesof a hyperplane, so that w � C cannot equal C .

Meanwhile, ifH belongs to an open chamber C and w �H D H , then w �C mustequal C , so that w D I . utProposition 8.28. For any two bases1 and2 forR, there exists a unique w 2 Wsuch that w �1 D 2.

Proof. By Proposition 8.21, there is a bijective correspondence between bases andopen Weyl chambers, and this correspondence is easily seen to respect the action ofthe Weyl group. Since, by Propositions 8.23 and 8.27,W acts freely and transitivelyon the chambers, the same is true for bases. utProposition 8.29. Let C be a Weyl chamber and H an element of E . Then thereexists exactly one point in the W -orbit of H that lies in the closure NC of C .

216 8 Root Systems

We are not saying that there is a unique w such that w � H 2 NC , but rather thatthere exists a unique point H 0 2 NC such that H 0 can be expressed (not necessarilyuniquely) as H 0 D w �H .

Proof. If U is any neighborhood of H , then by the argument in Exercise 2, thehyperplanes V˛, ˛ 2 R, cannot fill up U , which means U contains points in someopen Weyl chamber. It follows that H belongs to ND for some chamber D. ByProposition 8.23, there exists w 2 W such that w � D D C and, thus, w � ND D NC .Thus,H 0 WD w�H is in NC . Meanwhile, ifH 00 is a point in theW -orbit ofH such thatH 00 2 NC , thenH 0 andH 00 lie in the same W -orbit, which means (Proposition 8.25)that H 0 D H 00. utProposition 8.30. Let be a base for R, let RC be the associated set of positiveroots, and let ˛ be an element of . If ˇ 2 RC and ˇ ¤ ˛, then s˛ � ˇ 2 RC. Thatis to say, if ˛ 2 , then s˛ permutes the positive roots different from ˛.

Proof. Write ˇ D P�2 c�˛� with c� � 0. Since ˇ ¤ ˛, there is some c� with

� ¤ ˛ and c� > 0. Now, since s˛ �ˇ D ˇ� n˛ for some integer n, in the expansionof s˛ �ˇ, only the coefficient of ˛ has changed compared to the expansion of ˇ. Thus,the coefficient of � in the expansion of s˛ �ˇ remains positive. But if one coefficientof s˛ � ˇ is positive, all the other coefficients must be non-negative, showing thats˛ � ˇ is a positive root. ut

8.6 Dynkin Diagrams

A Dynkin diagram is a convenient graphical way of encoding the structure of a basefor a root system R, and thus also of R itself.

Definition 8.31. If D f˛1; : : : ; ˛rg is a base for a root system R, the Dynkindiagram for R is a graph having vertices v1; : : : ; vr . Between two distinct verticesvj and vk , we place zero, one, two, or three edges according to whether the anglebetween ˛j and ˛k is �=2, 2�=3, 3�=4, or 5�=6. In addition, if ˛j and ˛k are notorthogonal and have different lengths, we decorate the edges between vj and vkwith an arrow pointing from the vertex associated to the longer root to the vertexassociated to the shorter root.

Note that by Proposition 8.6, angles of 2�=3, 3�=4, or 5�=6 correspond tolength ratios of 1;

p2,

p3, respectively. Thus, the number of edges between vertices

corresponding to two nonorthogonal roots is 1, 2, or 3 according to whether thelength ratio (of the longer to the shorter) is 1,

p2, or

p3. Thinking of the arrow

decorating the edges as a “greater than” sign helps one to recall which way thearrow should go.

Two Dynkin diagrams are said to be isomorphic if there is a one-to-one, ontomap of the vertices of one to the vertices of the other that preserves the number ofbonds and the direction of the arrows. By Proposition 8.28, any two bases 1 and

8.6 Dynkin Diagrams 217

A2

B2 G2

A1 A1

Fig. 8.10 The Dynkin diagrams for the rank-two root systems

2 for a fixed root system are related by the action of a unique Weyl group elementw. Since w preserves angles and lengths, the Dynkin diagrams associated to twodifferent bases for the same root system are isomorphic.

In the case of the root system G2, for example, a base consists of two roots atangle of 5�=6, with a length ratio of

p3 (Figure 8.6). Thus, the Dynkin diagram

consists of two vertices connected by three edges, with an arrow pointing from thelonger root (˛2) to the shorter (˛1). One can similarly read off the Dynkin diagramforA2 from Figure 6.2 and the diagram forB2 from Figure 8.7. Finally, forA1�A1,the two elements of the base are orthogonal, yielding the results in Figure 8.10.

Proposition 8.32. 1. A root system is irreducible if and only if its Dynkin diagramis connected.

2. If the Dynkin diagrams of two root systems R1 and R2 are isomorphic, then R1and R2 themselves are isomorphic.

Proof. For Point 1, if a root system R decomposes as the direct sum of two rootsystems R1 and R2, then we can obtain a base for R as D 1 [2, where 1

and 2 are bases for R1 and R2, respectively. Since elements of R1 are orthogonalto elements R2, each element of 1 is orthogonal to each element of 2. Thus, theDynkin diagram associated to is disconnected.

Conversely, suppose the Dynkin diagram of R is disconnected. Then the base decomposes into two pieces 1 and 2 where (since there are no edges connectingthe pieces), each element of1 is orthogonal to each element of 2. Thus, E is theorthogonal direct sum of E1 WD span.1/ and E2 WD span.2/. If R1 D R \ E1and R2 D R \ E2, then it easy to check that R1 and R2 are root systems in E1 andE2, respectively, and that 1 is a base for R1 and 2 is a base for R2.

Now, for each ˛ 2 1, the reflection s˛ will act as the identity onE2 and similarlyfor ˛ 2 2. Since the Weyl groups ofR,R1, andR2 are generated by the reflectionsfrom , 1, and 2, respectively, we see that the Weyl group W of R is the directproduct of the Weyl groups W1 and W2 of R1 and R2, with W1 acting only on E1and W2 acting only on E2. Since W acts transitively on the bases of R and everyelement ofR is part of some base, we see thatW � D R. But sinceW D W1�W2,we have W � D .W1 �1/ [ .W2 �2/. Thus, every element of R is either in E1or in E2, meaning that R is the direct sum of R1 and R2.

218 8 Root Systems

For Point 2, using Point 1, we can reduce the problem to the case in whichR1 andR2 are irreducible and the Dynkin diagrams of R1 and R2 are connected. Let 1 Df˛1; : : : ; ˛rg and 2 D fˇ1; : : : ; ˇrg be bases for R1 and R2, respectively, orderedso that the isomorphism of the Dynkin diagrams maps the vertex associated to ˛j tothe vertex associated to ˇj . We may rescale the inner product on R1 so that k˛1k Dkˇ1k. Since the Dynkin diagrams are connected and the diagram determines thelength ratios between vertices joined by an edge, it follows that

˝˛j ; ˛k

˛ D ˝ˇj ; ˇk

˛for all j and k. It is then easy to check that the unique linear map A W E1 ! E2such that A˛j D ˇj , j D 1; : : : ; r is an isometry. We then have that As˛j D sˇj A

for all j .Now, if ˛ is any element of R1, then, since W1 is generated by the reflections

from1 andW1 � D R1, we see that

˛ D s˛j1 � � � s˛jN ˛kfor some indices j1; : : : ; jN and k. Thus,

A˛ D sˇj1 � � � sˇjN ˇk

is an element of R2. The same reasoning shows that A�1ˇ 2 R1 for all ˇ 2 R2.Thus,A is an isometry mappingR1 ontoR2, which implies thatA is an isomorphismof R1 with R2. utCorollary 8.33. Let g D kC be a semisimple Lie algebra, let h D tC be a Cartansubalgebra of g, and letR � i t be the root system of g relative to h. Then g is simpleif and only if the Dynkin diagram of R is connected.

Proof. According to Theorem 7.35, g is simple if and only if R is irreducible. Butaccording to Point 1 of Proposition 8.32, R is irreducible if and only if the Dynkindiagram of R is connected. ut

8.7 Integral and Dominant Integral Elements

We now introduce a notion of integrality for elements of E . In the setting of therepresentations of a semisimple Lie algebra g, the weights of a finite-dimensionalrepresentation of g are always integral elements. Recall from Definition 8.10 thenotion of the corootH˛ associated to a root ˛.

Definition 8.34. An element � of E is an integral element if for all ˛ in R, thequantity

h�;H˛i D 2h�; ˛ih˛; ˛i

8.7 Integral and Dominant Integral Elements 219

is an integer. If is a base for R, an element � of E is dominant (relative to ) if

h˛;�i � 0

for all ˛ 2 and strictly dominant if

h˛;�i > 0

for all ˛ 2 .

See Figure 6.2 for the integral and dominant integral elements in the case of A2.Additional examples will be given shortly.

A point � 2 E is strictly dominant relative to if and only if � is containedin the open fundamental Weyl chamber associated to , and � is dominant ifand only if � is contained in the closure of the open fundamental Weyl chamber.Proposition 8.29 therefore implies the following result: For all � 2 E , there existsw 2 W such that w � � is dominant.

Note that by the definition of a root system, every root is an integral element.Thus, every integer linear combination of roots is also an integral element. In mostcases, however, there exist integral elements that are not expressible as an integercombination of roots. In the case of A2, for example, the elements labeled �1 and�2 in Figure 6.2 are integral, but their expansions in terms of ˛1 and ˛2 are �1 D2˛1=3C ˛2=3 and �2 D ˛1=3C 2˛2=3. Since ˛1 and ˛2 form a base for A2, if �1or �2 were an integer combination of roots, it would also be an integer combinationof ˛1 and ˛2.

Proposition 8.35. If � 2 E has the property that

2h�; ˛ih˛; ˛i

is an integer for all ˛ 2 , then the same holds for all ˛ 2 R and, thus, � is anintegral element.

Proof. An element � is integral if and only if h�;H˛i is an integer for all ˛ 2 R.By Proposition 8.18, eachH˛ with ˛ 2 R can be expressed as a linear combinationof the H˛’s with ˛ 2 , with integer coefficients. Thus, if h�;H˛i 2 Z for ˛ 2 ,then same is true for ˛ 2 R, showing that � is integral. utDefinition 8.36. Let D f˛1; : : : ; ˛rg be a base. Then the fundamental weights(relative to ) are the elements �1; : : : ; �r with the property that

2

˝�j ; ˛k

˛h˛k; ˛ki D ıjk; j; k D 1; : : : ; r: (8.8)

Elementary linear algebra shows there exists a unique set of integral elementssatisfying (8.8). Geometrically, the j th fundamental weight is the unique element

220 8 Root Systems

of E that is orthogonal to each ˛j , j ¤ k, and whose orthogonal projection onto˛j is one-half of ˛j . Note that the set of dominant integral elements is precisely theset of linear combinations of the fundamental weights with non-negative integercoefficients; the set of all integral elements is the set of linear combinations offundamental weights with arbitrary integer coefficients.

Definition 8.37. Let be a base for R and RC the associated set of positive roots.We then let ı denote half the sum of the positive roots:

ı D 1

2

X˛2RC

˛:

The element ı plays a key role in many of the developments in subsequentchapters. It appears, for example, in the statement of the Weyl character formulaand the Weyl integral formula. Figure 8.11 shows the integral and dominant integralelements forB2 andG2. In each case, the base f˛1; ˛2g and the element ı are labeled,and the fundamental weights are circled. The background square lattice (B2 case) ortriangular lattice (G2 case) indicates the set of all integral elements. The root systemG2 is unusual in that the fundamental weights are roots, which means that everyintegral element is expressible as an integer combination of roots.

Proposition 8.38. The element ı is a strictly dominant integral element; indeed,

2hˇ; ıihˇ; ˇi D 1

for each ˇ 2 .

Proof. If ˇ 2 , then by Proposition 8.30, sˇ permutes the elements ofRC differentfrom ˇ. Thus, we can decompose RC n fˇg as E1 [ E2, where elements of E1 areorthogonal to ˇ and elements of E2 are not. Then the elements of E2 split up intopairs f˛; sˇ � ˛g, where

˝sˇ � ˛; ˇ˛ D ˝

˛; sˇ � ˇ˛ D � h˛; ˇi :

Thus, when we compute the inner product of ˇ with half the sum of the positiveroots, the roots inE1 do not contribute and the contributions from roots inE2 cancelin pairs. Thus, only the contribution from ˇ itself remains:

2hˇ; ıihˇ; ˇi D 2

˝ˇ; 1

2ˇ˛

hˇ; ˇi D 1;

as claimed. ut

8.8 The Partial Ordering 221

Fig. 8.11 Integral anddominant integral elementsfor B2 (top) and G2 (bottom).The black dots indicatedominant integral elements,while the background latticeindicates the set of all integralelements

1

2

1

2

1

2

1

2

8.8 The Partial Ordering

We now introduce a partial ordering on the set of integral elements, which willbe used to formulate the theorem of the highest weight for representations of asemisimple Lie algebra.

222 8 Root Systems

Fig. 8.12 Points that arehigher than zero (light anddark gray) and points that aredominant (dark gray) for B2

1

2

Definition 8.39. If D f˛1; : : : ; ˛rg is a base, an element � 2 E is said to behigher than � 2 E (relative to ) if �� � can be expressed as

�� � D c1˛1 C � � � C cr˛r ;

where each cj is a non-negative real number. We equivalently say that � is lowerthan � and we write this relation as � � � or � � �.

The relation � defines a partial ordering on E . We now develop various usefulproperties of this relation. For the rest of the section, we assume a base for R hasbeen chosen, and that the notions of higher, lower, and dominant are defined relativeto .

Proposition 8.40. If � 2 E is dominant, then � � 0.

See Figure 8.12 for an example of the proposition.For any basis fv1; : : : ; vrg of E , we can form a basis for the dual space E� by

considering the linear functionals j given by j .vk/ D ıjk. We can then find unique

vectors v�j 2 E such that j .u/ D

Dv�j ; u

E, so that

Dv�j ; vk

ED ıjk:

The basis fv�1 ; : : : ; v

�r g for E is called the dual basis to fv1; : : : ; vrg.

8.8 The Partial Ordering 223

Lemma 8.41. Suppose fv1; : : : ; vrg is an obtuse basis for E , meaning that˝vj ; vk

˛ � 0 for all j ¤ k. Then fv�1 ; : : : ; v

�r g is an acute basis for E , meaning thatD

v�j ; v

�k

E� 0 for all j; k.

The proof of this elementary lemma is deferred until the end of this section.

Proof of Proposition 8.40. Any vector u can be expanded as u D Pj cj ˛j and the

coefficients may be computed as cj DD˛�j ; u

E, where f˛�

j g is the dual basis to f˛j g.

Applying this with u D ˛�j gives

˛�j D

rXkD1

D˛�k ; ˛

�j

E˛k:

Now, if � is dominant, the coefficients in the expansion � D Pj cj ˛j are given by

cj DD˛�j ; �

EDXk

D˛�k ; ˛

�j

Eh˛k; �i :

Since � is dominant, h˛k; �i � 0. Furthermore, the ˛j ’s form an obtuse basis for

E (Proposition 8.13) and thus by Lemma 8.41, we haveD˛�k ; ˛

�j

E� 0 for all j; k.

Thus, cj � 0 for all j , which shows that � is higher than zero. utProposition 8.42. If � is dominant, then w � � � � for all w 2 W .

Proof. Let O be the Weyl-group orbit of �. Since O is a finite set, it contains amaximal element �, i.e., one such that there is no �0 ¤ � in O that is higher than �.Then for all ˛ 2 , we must have h˛; �i � 0, since if h˛; �i were negative, then

s˛ � � D � � 2 h�; ˛ih˛; ˛i˛

would be higher than �. Thus, � is dominant. But since, by Proposition 8.29, � isthe unique dominant element of O , we must have � D �. Thus, � is the uniquemaximal element of O .

We now claim that every element of O is lower than �. Certainly, no elementof O can be higher than �. Let O 0 be the set of � 2 O that are neither higher norlower than �. If O 0 is not empty, it contains a maximal element �. We now arguethat � is actually maximal in O . For any � 2 O , if � 2 O 0, then certainly � cannotbe higher than �, which is maximal in O 0. On the other hand, if � 2 O n O 0, then� is lower than �, in which case, � cannot be higher than �, or else we would have� � � � �, so that � would not be in O 0. We conclude that � is maximal in O , notjust inO 0, which means that �must equal�, the unique maximal element ofO . Butthis contradicts the assumption that � 2 O 0. Thus, O 0 must actually be empty, and� is the highest element of O . ut

224 8 Root Systems

Proposition 8.43. If � is a strictly dominant integral element, then � � ı, where ıis as in Definition 8.37.

Proof. Since � is strictly dominant, � � ı will still be dominant in light ofProposition 8.38. Thus, by Proposition 8.40, � � ı � 0, which is equivalent to� � ı. ut

Recall from Definition 6.23 the notion of the convex hull of a finite collectionof vectors in E . We let W � � denote the Weyl-group orbit of � 2 E and we letConv.W � �/ denote the convex hull of W � �.

Proposition 8.44. 1. If � and � are dominant, then � belongs to Conv.W � �/ ifand only if � � �.

2. Let � and � be elements of E with � dominant. Then � belongs to Conv.W � �/if and only if w � � � � for all w 2 W .

Figure 8.13 illustrates Point 2 of the proposition in the case of B2. In the figure,the shaded region represents the set of points that are lower than �. The point �1 isinside Conv.W � �/ and w � �1 is lower than � for all w. By contrast, �2 is outsideConv.W � �/ and there is some w for which w � �2 is not lower than �.

Since Conv.W � �/ is convex and Weyl invariant, we see that if � belongs toConv.W � �/, then every point in Conv.W � �/ also belongs to Conv.W � �/. Thus,Point 1 of the proposition may be restated as follows:

If � and � are dominant, then � � � if and only if

Conv.W � �/ � Conv.W � �/:

We establish two lemmas that will lead to a proof of Proposition 8.44.

Lemma 8.45. Suppose K is a compact, convex subset of E and � is an element ofE that it is not in K . Then there is an element � of E such that for all � 2 K , wehave

h�; �i > h�; �i :

If we let V be the hyperplane (not necessarily through the origin) given by

V D f� 2 Ej h�; �i D h�; �i � "g

for some small ", then K and � lie on opposite sides of V . Lemma 8.45 is a specialcase of the hyperplane separation theorem in the theory of convex sets.

Proof. Since K is compact, we can choose an element �0 of K that minimizes thedistance to �. Set � D � � �0, so that

h�; � � �0i D h� � �0; � � �0i > 0;

and, thus, h�; �i > h�; �0i.

8.8 The Partial Ordering 225

1

2

w 2

1

2

Fig. 8.13 The element w � �2 is not lower than �

Now, for any � 2 K , the vector �0 C s.� � �0/ belongs to K for 0 � s � 1, andwe compute that

d.�; �0 C s.�� �0//2 D h� � �0; � � �0i � 2s h� � �0; � � �0i

C s2 h� � �0; �� �0i :

The only way this quantity can be greater than or equal to h� � �0; � � �0i Dd.�; �0/

2 for small positive s is if

h� � �0; �� �0i D h�; � � �0i � 0:

Thus,

h�; �i � h�; �0i < h�; �i ;which is what we wanted to prove. utLemma 8.46. If � and � are dominant and � … Conv.W � �/, there exists adominant element � 2 E such that

h�; �i > h�;w � �i (8.9)

for all w 2 W .

226 8 Root Systems

Proof. By Lemma 8.45, we can find some � in E , not necessarily dominant, suchthat h�; �i > h�; �i for all � 2 Conv.W � �/. In particular,

h�; �i > h�;w � �i

for all w 2 W . Choose some w0 so that � 0 WD w0 � � is dominant. We will show thatreplacing � by � 0 makes h�; �i bigger while permuting the values of h�;w � �i.

By Proposition 8.42, � � � 0, meaning that � 0 equals � plus a non-negative linearcombination of positive simple roots. But since � is dominant, it has non-negativeinner product with each positive simple root, and we see that h� 0; �i � h�; �i. Thus,

˝� 0; �

˛ � h�; �i > h�;w � �i

for all w. But

h�;w � �i D ˝w�10 � � 0;w � �˛ D ˝

� 0; .w0w/ � �˛ :Thus, as w ranges overW , the values of h�;w � �i and h� 0; .w0w/ � �i range throughthe same set of real numbers. Thus, h� 0; �i > h�0;w � �i for all w, as claimed. ut

The proof of Lemma 8.46 is illustrated in Figure 8.14. The dominant element �is not in Conv.W ��/ and is separated from Conv.W ��/ by a line with orthogonalvector � . The element � 0 WD s˛2 � � is dominant and � is also separated from

1

2

Fig. 8.14 The element � is separated from Conv.W � �/ first by a line orthogonal to � and thenby a line orthogonal to the dominant element � 0

8.8 The Partial Ordering 227

Conv.W ��/ by a line with orthogonal vector � 0. The existence of such a line meansthat � cannot be lower than �.

Proof of Proposition 8.44. For Point 1, let � and � be dominant. Assume first that� is in Conv.W � �/. By Proposition 8.42, every element of the form w � � is lowerthan �. But the set E of elements lower than � is easily seen to be convex, andso E must contain Conv.W � �/ and, in particular, �. Next, assume � � � andsuppose, toward a contradiction, that � … Conv.W � �/. Let � be a dominantelement as in Lemma 8.46. Then � � � is a non-negative linear combination ofpositive simple roots, and � , being dominant, has non-negative inner product witheach positive simple root. Thus, h�; � � �i � 0 and, hence, h�; �i � h�; �i, whichcontradicts (8.9). Thus, � must actually belong to Conv.W � �/.

For Point 2, assume first that w �� � � for all w 2 W , and choose w so that w � �is dominant. Since, w � � � �, Point 1 tells us that w � � belongs to Conv.W � �/,which implies that � also belongs to Conv.W � �/. In the other direction, assume� 2 Conv.W ��/ so that w �� 2 Conv.W ��/ for all w 2 W . Using Proposition 8.42we can easily see that every element of Conv.W ��/ is lower than �. Thus, w �� � �

for all w. utIt remains only to supply the proof of Lemma 8.41.

Proof of Lemma 8.41. We proceed by induction on the dimension r of E . Whenr D 1 the result is trivial. When r D 2, the result should be geometrically obvious,but we give an algebraic proof. The Gram matrix of a basis is the collection of innerproducts, Gjk WD ˝

vj ; vk˛. It is an elementary exercise (Exercise 3) to show that the

Gram matrix of the dual basis is the inverse of the Gram matrix of the original basis.Thus, in the r D 2 case, we have

� ˝v�1 ; v

�1

˛ ˝v�1 ; v

�2

˛˝v�1 ; v

�2

˛ ˝v�2 ; v

�2

˛�

D 1

.hv1; v1i hv2; v2i � hv1; v2i2/� hv2; v2i � hv1; v2i

� hv1; v2i hv1; v1i�: (8.10)

Since v1 is not a multiple of v2, the Cauchy–Schwarz inequality tells us that thedenominator on the right-hand side of (8.10) is positive, which means that

˝v�1 ; v

�2

˛has the opposite sign of hv1; v2i.

Assume now that the result holds in dimension r � 2 and consider the case ofdimension r C 1. Fix any index m and let P be the orthogonal projection onto theorthogonal complement of vm, which is given by

P.u/ D u � hvm; uihvm; vmivm:

The operator P is easily seen to be self-adjoint, meaning that hu; P vi D hP u; vifor all u; v.

228 8 Root Systems

We now claim that Pv1; : : : ;bPvm; : : : ;PvrC1 is an obtuse basis for hvmi?, wherethe notation bPvm indicates that Pvm is omitted. Indeed, a little algebra shows that

˝Pvj ;Pvk

˛ D ˝vj ; vk

˛ �˝vm; vj

˛ hvm; vkihvm; vmi � 0;

since˝vj ; vk

˛,˝vm; vj

˛, and hvm; vki are all less than or equal to zero. Furthermore,

for j and k different fromm, we have

Dv�j ;Pvk

EDDPv�

j ; vk

EDDv�j ; vk

ED ıjk

since v�j is orthogonal to vm. Thus, the dual basis to Pv1; : : : ;bPvm; : : : ;PvrC1

consists simply of the vectors v�1 ; : : : ;

cv�m; : : : ; v

�rC1 (all of which are orthogonal

to vm).Now fix any two distinct indices j and k. Since r C 1 � 3, we can choose some

other indexm distinct from both j and k. Applying our induction hypothesis to the

basis Pv1; : : : ;bPvm; : : : ;PvrC1 for hvmi?, we conclude thatDv�j ; v

�k

E� 0, which is

what we are trying to prove. ut

8.9 Examples in Rank Three

In rank three, we can have a reducible root system, which must be a direct sum ofA1 with one of the rank-two root systems described in the previous section. In thissection, we will consider only the irreducible root systems of rank three. There are,up to isomorphism, three irreducible root systems in rank three, customarily denotedA3, B3, and C3. They arise from the Lie algebras sl.4IC/, so.7IC/, and sp.3IC/,respectively, as described in Sect. 7.7.

The models in this section can be constructed using the Zometool system,available at www.zometool.com. The reader is encouraged to obtain some Zometoolpieces and build the rank-three root systems for him- or herself. The models requirethe green lines, which are not part of the basic Zometool kits. The models of theC3 root system use half-length greens, although one can alternatively use wholegreens together with double (end to end) blue pieces. The images shown herewere rendered in Scott Vorthmann’s vZome software, available at vzome.com. Fordetailed instructions on how to build rank-three root systems using Zometool, clickon the “Book” tab of the author’s web site: www.nd.edu/~bhall/.

Figure 8.15 shows the A3 root system, with a base highlighted. The elementsof A3 form the vertices of a polyhedron known as a cuboctahedron, which has sixsquare faces and eight triangular faces, as shown in Figure 8.16. The points in A3can also be visualized as the midpoints of the edges of a cube, as in Figure 8.17.Algebraically, we can describe A3 as the set of 12 vectors in R3 of the form

8.9 Examples in Rank Three 229

Fig. 8.15 The A3 rootsystem, with the elements ofthe base in dark gray

Fig. 8.16 The roots in A3make up the vertices of acuboctahedron

.˙1;˙1; 0/, .˙1; 0;˙1/, and .0;˙1;˙1/. (This set of vectors actually correspondsto the conventional description of the D3 root system, as in Sect. 7.7.2, which turnsout to be isomorphic to A3.) It is then a simple exercise to check that this collectionof vectors is, in fact, a root system. A base for this system is given by the vectors.1;�1; 0/, .0; 1;�1/, and .0; 1; 1/.

The Weyl group W for A3 is the symmetry group of the tetrahedron pictured inFigure 8.18. The group W is the full permutation group on the four vertices of thetetrahedron. As in the A2 case, the Weyl group of A3 is not the full symmetry groupof the root system, since �I is not an element of W .

The B3 root system is obtained from the A3 root system by adding six additionalvectors, consisting of three mutually orthogonal pairs. Each of the new roots

230 8 Root Systems

Fig. 8.17 The roots in A3 lieat the midpoints of the edgesof a cube

Fig. 8.18 The Weyl group ofA3 is the symmetry group ofa regular tetrahedron

is shorter than the original roots by a factor ofp2, as shown in Figure 8.19.

Algebraically, B3 consists of the twelve vectors .˙1;˙1; 0/, .˙1; 0;˙1/, and.0;˙1;˙1/ of A3, together with the six vectors .˙1; 0; 0/, .0;˙1; 0/, and.0; 0;˙1/.

The C3 root system, meanwhile, is obtained from A3 by adding six new vectors,as in the case of B3, except that this time the new roots are longer than the originalroots by a factor of

p2, as in Figure 8.20. That is to say, the new roots are the

six vectors .˙2; 0; 0/, .0;˙2; 0/, and .0; 0;˙2/. The C3 root system is the dual ofB3, in the sense of Definition 8.10. The elements of C3 make up the vertices of anoctahedron, together with the midpoints of the edges of the octahedron, as shown inFigure 8.21.

8.9 Examples in Rank Three 231

Fig. 8.19 The B3 root system, with the elements of the base in dark gray

Fig. 8.20 The C3 root system, with the elements of the base in dark gray

232 8 Root Systems

Fig. 8.21 TheC3 root system consists of the vertices of an octahedron, together with the midpointsof the edges of the octahedron

The root systems B3 and C3 have the same Weyl group, which is the symmetrygroup of the cube in Figure 8.17. In both cases, the Weyl group is the full symmetrygroup of the root system.

8.10 The Classical Root Systems

We now return to the root systems of the classical semisimple Lie algebras, ascomputed in Sect. 7.7. For each of these root systems, we describe a base anddetermine the associated Dynkin diagram.

8.10.1 The An Root System

TheAn root system is associated to the Lie algebra sl.nC1IC/. For this root system,E is the subspace of RnC1 consisting of vectors whose entries sum to zero. The rootsare the vectors of the form

ej � ek; j ¤ k;

8.10 The Classical Root Systems 233

where fej g is the standard basis for RnC1. As a base, we may take the vectors

e1 � e2; e2 � e3; : : : ; en � enC1

Note that for j < k,

ej � ek D .ej � ejC1/C .ejC1 � ejC2/C � � � C .ek�1 � ek/;

so that every root is a sum of elements of the base, or the negative thereof.All roots in the base have the same length, two consecutive roots are at an angle

of 2�=3 to one another, and nonconsecutive roots are orthogonal.

8.10.2 The Dn Root System

The Dn root system is associated to the Lie algebra so.2nIC/, n � 2. For this rootsystem, E D Rn and the roots are the vectors of the form

˙ej ˙ ek; j < k:

As a base, we may take the n � 1 roots

e1 � e2; e2 � e3; � � � ; ; en�2 � en�1; en�1 � en; (8.11)

together with the one additional root,

en�1 C en: (8.12)

Note that for j < k, we have the following formulas:

ej � ek D .ej � ejC1/C .ejC1 � ejC2/C � � � C .ek�1 � ek/;ej C en D .ej � en�1/C .en�1 C en/;

ej C ek D .ej C en/C .ek � en/: (8.13)

This shows that every root of the form ej � ek or ej C ek (j < k) can be writtenas a linear combination of the base in (8.11) and (8.12) with non-negative integercoefficients. The roots of this form are the positive roots, and the remaining rootsare the negatives of these roots.

Two consecutive roots in the list (8.11) have an angle of 2�=3 and twononconsecutive roots in the list (8.11) are orthogonal. The angle between the rootin (8.12) and the second-to-last element in the list (8.11) is 2�=3; the root in (8.12)is orthogonal to all the other roots in (8.11).

234 8 Root Systems

8.10.3 The Bn Root System

The Bn root system is associated to the Lie algebra so.2n C 1IC/. For this rootsystem, E D Rn and the roots are the vectors of the form

˙ej ˙ ek; j < k;

and of the form

˙ej ; j D 1; : : : ; n:

As a base for our root system, we may take the n � 1 roots

e1 � e2; e2 � e3; : : : ; en�1 � en; (8.14)

(exactly as in the so.2nIC/ case) together with the one additional root,

en: (8.15)

The positive roots are those of the form ej C ek or ej � ek (j < k) and those of theform ej (1 � j � n). To expand every positive root in terms of the base, we use theformulas in (8.13), except with the second line replaced by

ej C en D .ej � en/C 2en; (8.16)

and with the additional relation

ej D .ej � en/C en: (8.17)

As in the so.2nIC/ case, consecutive roots in the list (8.14) have an angle of2�=3, whereas nonconsecutive roots on the list (8.14) are orthogonal. Meanwhile,the root in (8.15) has an angle of 3�=4 with the last root in (8.14) and is orthogonalto the remaining roots in (8.14).

In Sect. 8.2, we have pictured the B2 root system rotated by �=4 relative tothe n D 2 case of the root system described in this subsection. The pictures inSect. 8.2 actually correspond to the conventional description of the C2 root system(Sect. 8.10.4), which is isomorphic to B2.

8.10.4 The Cn Root System

The Cn root system is associated to the Lie algebra sp.nIC/. For this root system,E D Rn and the roots are the vectors of the form

˙ej ˙ ek; j < k

8.10 The Classical Root Systems 235

and of the form

˙2ej ; j D 1; : : : ; n:

As a base, we may take the n � 1 roots

e1 � e2; e2 � e3; : : : ; en�1 � en (8.18)

(as in the two preceding subsections), together with the root 2en. We use the sameformula for expanding roots in terms of the base as in the case of so.2n C 1IC/,except that (8.17) is rewritten as

2ej D 2.ej � en/C .2en/:

The angle between two consecutive roots in (8.18) is 2�=3I nonconsecutive rootsin (8.18) are orthogonal. The angle between 2en and the last root in (8.18) is 3�=4;the root 2en is orthogonal to the other roots in (8.18).

8.10.5 The Classical Dynkin Diagrams

From the calculations in the previous subsections, we can read off the Dynkindiagram for the root systems An, Bn, Cn, and Dn; the results are recorded inFigure 8.22. We can see that certain special things happen for small values of n.First, the Dynkin diagram for Dn does not make sense when n D 1, since thediagram always has at least two vertices. This observation reflects the fact thatthe Lie algebra so.2IC/ is not semisimple. Second, the Dynkin diagram for D2

is not connected, which means (Corollary 8.33) that the Lie algebra so.4IC/ issemisimple but not simple. (Compare Exercise 4 in Chapter 7.) Third, the Dynkindiagrams forA1,B1, andC1 are isomorphic, reflecting that the rank-one Lie algebrassl.2IC/, so.3IC/, and sp.1IC/ are isomorphic. Last, we have an isomorphismbetween the diagrams for B2 and C2 and an isomorphism between the diagrams for

An

Cn

Bn

Dn

Fig. 8.22 The Dynkin diagrams of the classical Lie algebras

236 8 Root Systems

A3 and D3, which reflects (Sect. 8.11) an isomorphism between the correspondingLie algebras.

Corollary 8.47. The following semisimple Lie algebras are simple: the speciallinear algebras sl.n C 1IC/, n � 1; the odd orthogonal algebras so.2n C 1IC/,n � 1; the even orthogonal algebras so.2nIC/, n � 3; and the symplectic algebrassp.nIC/, n � 1.

Proof. The Dynkin diagrams for An, Bn, and Cn are always connected, whereas theDynkin diagram forDn is connected for n � 3. Thus, Corollary 8.33 shows that theclaimed Lie algebras are simple. ut

8.11 The Classification

In this section, we describe, without proof, the classification of irreducible rootsystems and of simple Lie algebras. Recall (Corollary 8.33) that a semisimple Liealgebra is simple if and only if its Dynkin diagram is connected.

Every irreducible root system is either the root system of a classical Lie algebra(types An, Bn, Cn, and Dn) or one of five exceptional root systems. We begin bylisting the Dynkin diagrams of the exceptional root systems.

Theorem 8.48. For each of the graphs in Figure 8.23, there exists a root systemhaving that graph as its Dynkin diagram.

We have already described the root system G2 in Figure 8.3. Although it ispossible to write down the remaining exceptional root systems explicitly, it is notterribly useful to do so, since there is no comparably easy way to construct the Liealgebras associated to these root systems.

E6

E7

E8

F4

G2

Fig. 8.23 The exceptional Dynkin diagrams

8.11 The Classification 237

Theorem 8.49. Every irreducible root system is isomorphic to exactly one of thefollowing:

• An, n � 1

• Bn, n � 2

• Cn, n � 3

• Dn, n � 4

• One of the exceptional root systems G2, F4, E6, E7, and E8.

The restrictions on n are to avoid the case of D2, which is not irreducible, andto avoid repetitions. The Dynkin diagram for D3, for example, is isomorphic to theDynkin diagram for A3, which means (Proposition 8.32) that the A3 and D3 rootsystems are also isomorphic. Similarly, all root systems in rank one are isomorphic,and B2 is isomorphic to C2.

Theorem 8.50. 1. If g is a complex semisimple Lie algebra and h1 and h2 areCartan subalgebra of g, there exists an automorphism � W g ! g such that�.h1/ D �.h2/.

2. Suppose g1 and g2 are semisimple Lie algebras with Cartan subalgebras h1and h2, respectively. If the root systems associated to .g1; h1/ and .g2; h2/ areisomorphic, then g1 and g2 are isomorphic.

3. For every root system R, there exists a semisimple Lie algebra g and a Cartansubalgebra h of g such that the root system of g relative to h is isomorphic to R.

Point 1 of the theorem says that there is only one root system associated to eachsemisimple Lie algebra g. Since all bases of a fixed root system are equivalent underW , it follows that there is only one Dynkin diagram associated to each g. Points 2and 3 then tell us that there is a one-to-one correspondence between isomorphismclasses of semisimple Lie algebras and isomorphism classes of root systems. Thus,the classification of irreducible root systems also gives rise to a classification ofsimple Lie algebras.

For Point 1 of the theorem see Section 16.4 of [Hum] and for Point 2 see Section14.2 of [Hum]. For Point 3, one can proceed on a case-by-case basis, where the LiealgebrasAn,Bn,Cn, andDn have already been constructed as classical Lie algebras.The exceptional Lie algebras can then be constructed by special methods, as in [Jac]or [Baez]. Alternatively, one can construct all of the simple Lie algebras by a unifiedmethod, as in Section 18 of [Hum].

In particular, the isomorphisms among root systems of small rank translateinto isomorphisms among the associated semisimple Lie algebras. In rank one,for example, sl.2IC/, so.3IC/, and sp.1IC/ are isomorphic. In rank two, theisomorphism between B2 and C2 reflects an isomorphism of the Lie algebrasso.5IC/ and sp.2IC/. In rank three, the isomorphism between A3 and D3 reflectsan isomorphism of the Lie algebras sl.4IC/ and so.6IC/.

By combining the classification of irreducible root systems in Theorem 8.49with Proposition 8.32 and Theorem 8.50, we arrive at the following classificationof simple Lie algebras over the field of complex numbers.

238 8 Root Systems

Theorem 8.51. Every simple Lie algebra over C is isomorphic to precisely onealgebra from the following list:

1. sl.nC 1IC/, n � 1

2. so.2nC 1IC/, n � 2

3. sp.nIC/, n � 3

4. so.2nIC/, n � 4

5. The exceptional Lie algebras G2, F4, E6, E7, and E8

A semisimple Lie algebra is then determined up to isomorphism by specifyingwhich simple summands occur and how many times each one occurs; seeProposition 7.9. It is also possible to classify simple Lie algebras over R. Aswe showed in Sect. 7.6, every such algebra is either a complex simple Lie algebra,viewed as a real Lie algebra, or a real form of a complex simple Lie algebra. Realforms of complex Lie algebras can then be enumerated using the Dynkin diagramof the complex Lie algebra as a starting point. See Section VI.10 and Appendix Cof [Kna2] for a description of this enumeration.

8.12 Exercises

Unless otherwise noted, the notation in the exercises is as follows: .E;R/ is a rootsystem with Weyl group W , D f˛1; : : : ; ˛rg is a fixed base for R, RC is theassociated set of positive roots, and C is the open fundamental chamber associatedto .

1. (a) Suppose that ˛ and ˇ are linearly independent elements of R and that forsome positive integer k, the vector ˛C kˇ belongs to R. Show that ˛C lˇ

also belongs to R for all integers l with 0 < l < k.Hint: If F D span.˛; ˇ/, then R \ F is a rank-two root system in F .

(b) A collection of roots of the form ˛; ˛ C ˇ; : : : ; ˛ C kˇ for which neither˛�ˇ nor ˛C.kC1/ˇ is a root is called a root string. What is the maximumnumber of roots that can occur in a root string?

2. Let E be a finite-dimensional real inner product space and let V1; : : : ; Vk besubspaces of E of codimension at least one. Show that the union of the Vk’s isnot all of E .Hint: Show by induction on k that the complement of the union the Vk’s is anonempty open subset of E .

3. Let E be a finite-dimensional real inner product space, let fv1; : : : ; vr g be basis

forE , and let fv�1 ; : : : ; v

�r g the dual basis, satisfying

Dv�j ; vk

ED ıjk for all j and

k. Let G and H be the Gram matrices for these two bases: Gjk D ˝vj ; vk

˛and

Hjk DDv�j ; v

�k

E. Show that G and H are inverses of each other.

8.12 Exercises 239

Hint: First show that for any u 2 E , we have u D Pj

Dv�j ; u

Evj . Then apply

this result to the vector u D v�k .

4. Show that Lemma 8.26 fails (even with u D e) if ˇ is not assumed to be anelement of C .

5. Show that if R is an irreducible root system, thenW acts irreducibly on E .Hint: Suppose V � E is a W -invariant subspace. Show that every element ofR is either in V or in the orthogonal complement of E .

6. Let .E;R/ be an irreducible root system and let h�; �i be the inner product onE .Using Exercise 5, show that if h�; �i1 is a W -invariant inner product on E , thereis some constant c such that hH;H 0i1 D c hH;H 0i for all H;H 0 2 E .Hint: Consider the unique linear operator A W E ! E that is symmetric withrespect to h�; �i and that satisfies

˝H;H 0˛

1D ˝H;AH0˛

for all H;H 0 2 E . Then imitate the proof of Schur’s lemma, noting that theeigenvalues of A are real.

7. Suppose .E;R/ and .F; S/ are irreducible root systems and that A W E ! F isan isomorphism ofR with S . Show thatA is a constant multiple of an isometry.

8. Using the outline below, prove the following result: For all �; � 2 E , we have� � � if and only if h�� �;H i � 0 for all H 2 C .

(a) Show that if � � �, then h� � �;H i � 0 for all H 2 C .

(b) Let f˛�1 ; : : : ; ˛

�r g be the dual basis to , satisfying

D˛�j ; ˛k

ED ıjk for all j

and k. Show that if h�;H i � 0 for all H 2 C , thenD�; ˛�

j

E� 0 for all j .

(c) Show that if h� � �;H i � 0 for all H 2 C , then � � �.

9. Let P W E ! R be the function given by

P.H/ DY˛2RC

h˛;H i :

Show that P satisfies

P.w �H/ D det.w/P.H/

for all w 2 W and all H 2 E .10. Show that if �I is not in the Weyl of R, the Dynkin diagram of R must have a

nontrivial automorphism.Hint: By Proposition 8.23, there exists an element w of W mapping �C to C .Consider the map H 7! �w �H , which maps C to itself.

11. For which rank-two root systems is �I an element of the Weyl group?12. Show that the Weyl group of the An root system, described in Sect. 7.7.1, does

not contain �I , except when n D 1.

240 8 Root Systems

13. Let E D Rn and let R denote the collection of vectors of the following threeforms:

˙ej ˙ ek j < k

˙ej j D 1; : : : ; n

˙2ej j D 1; : : : ; n

:

Show thatR satisfies all the properties of a root system in Definition 8.1 exceptCondition 2. The collection R is a “nonreduced root system” and is known asBCn, since it is the union of Bn and Cn. (Compare Figure 7.1 in the n D 2

case.)14. Determine which of the Dynkin diagrams in Figures 8.22 and 8.23 have a

nontrivial automorphism. Show that only the Dynkin diagram of D4 has anautomorphism group with more than two elements.

Chapter 9Representations of Semisimple Lie Algebras

In this chapter, we prove the theorem of the highest weight for irreducible, finite-dimensional representations of a complex semisimple Lie algebra g: We firstprove that every such representation has a highest weight, that two irreduciblerepresentations with the same highest weight are isomorphic, and that the highestweight of an irreducible representation must be dominant integral. This part ofthe theorem is established in precisely the same way as in the case of sl.3IC/in Chapter 6. It then remains to prove that every dominant integral element is, infact, the highest weight of some irreducible representation. In the sl.3IC/ case,we did this by first constructing the representations whose highest weights werethe fundamental weights .1; 0/ and .0; 1/; and then taking tensor products of theserepresentations. For a general semisimple Lie algebra g; however, there is no simpleway to construct the representations whose highest weights are the fundamentalweights in Definition 8.36. Thus, we require a new method of constructing theirreducible representation of g with a given dominant integral highest weight. Thisconstruction will be the main topic of the present chapter.

In Chapter 10, we will derive several additional properties of the irreduciblerepresentations, including the structure of the set of weights, the multiplicities ofthe weights, and the dimensions of the representations. In that chapter, we willalso prove complete reducibility for representations of g; that is, that every finite-dimensional representation of g decomposes as a direct sum of irreducibles.

A previous version of this book was inadvertently published without the middle initial of theauthor’s name as “Brian Hall”. For this reason an erratum has been published, correcting themistake in the previous version and showing the correct name as Brian C. Hall (see DOI http://dx.doi.org/10.1007/978-3-319-13467-3_14). The version readers currently see is the correctedversion. The Publisher would like to apologize for the earlier mistake.

© Springer International Publishing Switzerland 2015B.C. Hall, Lie Groups, Lie Algebras, and Representations, GraduateTexts in Mathematics 222, DOI 10.1007/978-3-319-13467-3_9

241

242 9 Representations of Semisimple Lie Algebras

9.1 Weights of Representations

Throughout the chapter, we assume that g D kC is a complex semisimple Lie algebraand that h D tC is a fixed Cartan subalgebra of g (compare Proposition 7.11). We fixon g an inner product that is real on k and that is invariant under the adjoint actionof k; as in Proposition 7.4. We let R � i t denote the set of roots of g relative to h,we let be a fixed base for R; and we let RC and R� be the set of positive andnegative roots relative to ; respectively. For each root ˛; we consider the corootH˛ 2 h given by

H˛ D 2˛

h˛; ˛i :

We also consider the Weyl group W; that is, the group of linear transformationsof h generated by the reflections about the hyperplanes orthogonal to the roots.Finally, we consider the notions of integral and dominant integral elements, as inDefinition 8.34.

We now introduce the notion of a weight of a representation, as in the sl.3IC/case.

Definition 9.1. Let .�; V / be a representation of g; possibly infinite dimensional.An element � of h is a weight of � if there exists a nonzero vector v 2 V such that

�.H/v D h�;H i v (9.1)

for all H 2 h: The weight space corresponding to � is the set of all v 2 V

satisfying (9.1) and the multiplicity of � is the dimension of the correspondingweight space.

Throughout the chapter, we will use, without comment, Proposition A.17, whichsays that weight vectors with distinct weights are linearly independent.

Proposition 9.2. If .�; V / is a finite-dimensional representation of g, every weightof � is an integral element.

Proof. For each root ˛; let s˛ D hX˛; Y˛;H˛i Š sl.2IC/ be the subalgebra of g inTheorem 7.19. If v is a weight vector with weight �; then

�.H˛/v D h�;H˛i v:Thus, by applying Point 1 of Theorem 4.34 to the restriction of � to s˛; we see thath�;H˛i must be an integer, showing that � is integral. utTheorem 9.3. If .�; V / is a finite-dimensional representation of g; the weights of� and their multiplicities are invariant under the action ofW on H:

Proof. For each ˛ 2 R; we may construct the operator

S˛ WD e�.X˛/e��.Y˛/e�.X˛/:

9.1 Weights of Representations 243

If h˛;H i D 0; then H will commute with both X˛ and Y˛ and thus with S˛: Onthe other hand, by Point 3 of Theorem 4.34, we have S˛�.H˛/S

�1˛ D ��.H˛/:We

see, then, that

S˛�.H/S�1˛ D �.s˛ �H/

for all H 2 h:Suppose now that v is a weight vector with some weight �: Then

�.H/S�1˛ v D S�1

˛ �.s˛ �H/vD S�1

˛ h�; s˛ �H i vD ˝s�1˛ � �;H ˛S�1

˛ v;

showing that S�1˛ v is a weight vector with weight s�1

˛ � �: Thus, S�1˛ maps the

weight space with weight � into the weight space with weight s�1˛ � �: Meanwhile,

essentially the same argument shows that S˛ maps the weight space with weight s�1˛ �

� into the weight space with weight �; showing that the two spaces are isomorphic.Thus, s�1

˛ � � is again a weight with the same multiplicity as �: Thus, the weightsand multiplicities are invariant under each s�1

˛ D s˛ and, thus, underW: utWe now state the “easy” part of the theorem of the highest weight for represen-

tations of g:

Theorem 9.4. 1. Every irreducible, finite-dimensional representation of g has ahighest weight.

2. Two irreducible, finite-dimensional representations of g with the same highestweight are isomorphic.

3. If .�; V / is an irreducible, finite-dimensional representation of g with highestweight �; then � is dominant integral.

Proof. Enumerate the positive roots as ˛1; : : : ; ˛N : Choose a basis for g consistingof elements X1; : : : ; XN with Xj 2 g˛j ; elements Y1; : : : ; YN with Yj 2 g�˛j ; anda basisH1; : : : ;Hr for h: Then the proof of Proposition 6.11 from Chapter 6 carriesover to the present setting, with only the obvious notational changes, showing thatevery irreducible, finite-dimensional representation of g has a highest weight.

The proofs of Propositions 6.14 and 6.15 then also go through without change toshow that two irreducible, finite-dimensional representations with the same highestweight are isomorphic. Finally, using the sl.2IC/-subalgebras in Theorem 7.19, wemay follow the proof of Proposition 6.16 to show that the highest weight of a finite-dimensional, irreducible representation must be dominant integral. ut

We now come to the “hard” part of the theorem of the highest weight.

Theorem 9.5. If� is a dominant integral element, there exists an irreducible, finite-dimensional representation of g with highest weight �:

As we have noted, the method of proof of Proposition 6.17, from the sl.3IC/case, does not readily extend to general semisimple Lie algebras. The proof ofTheorem 9.5 will occupy the remainder of this chapter.

244 9 Representations of Semisimple Lie Algebras

9.2 Introduction to Verma Modules

Our goal is to construct, for each dominant integral element � 2 h; a finite-dimensional, irreducible representation of g with highest weight�:Our constructionwill proceed in two stages. The first stage consists of constructing an infinite-dimensional representation V� of g; known as a Verma module. This representationwill not be irreducible, but will be a highest weight cyclic representation withhighest weight �: We will construct V� as a quotient of the so-called universalenveloping algebra U.g/ of g: In order to show that the highest weight vectorin V� is nonzero, we will need to develop a structure result for U.g/ known asthe Poincaré–Birkhoff–Witt theorem (Theorem 9.10). Unlike the finite-dimensionalrepresentations of g; the weights of the Verma module are not invariant under theaction of the Weyl group.

The second stage in our construction consists of showing that when � isdominant integral, V� has an invariant subspace W� for which the quotient spaceV�=W� is finite dimensional and irreducible, but not zero. To establish the finitedimensionality of the quotient, we will show that when � is dominant integral,the weights of V�=W�; unlike those of V�; are invariant under the action of theWeyl group. Thus, each weight � of V�=W� is integral and satisfies w � � � � forall w in the Weyl group. It turns out that there are only finitely many �’s with thisproperty. Since each weight � has finite multiplicity (even in V�), it follows thatV�=W� is finite dimensional.

Definition 9.6. A (possibly infinite-dimensional) representation .�; V / of g ishighest weight cyclic with highest weight � 2 h if there exists a nonzero vectorv 2 V such that (1) �.H/v D h�;H i v for all H 2 h; (2) �.X/v D 0 for allX 2 g˛ with ˛ 2 RC; (3) the smallest invariant subspace containing v is V:

Note that � 2 h is not required to be integral. Although it will turn out thatall finite-dimensional highest weight cyclic representations are irreducible, thisis not the case in infinite dimensions. Furthermore, two highest weight cyclicrepresentations with the same highest weight may not be isomorphic, unless bothof them are finite dimensional. In this chapter, we will construct, for any � 2 h;a particular highest weight cyclic representation V� with highest weight �; knownas a Verma module. The Verma module is the “maximal” highest weight cyclicrepresentation with a particular highest weight, and it is always infinite dimensional,even when � is dominant integral.

In the sl.2IC/ case, Verma modules can be constructed explicitly as follows.For any complex number �; construct an infinite-dimensional vector space V� withbasis v0; v1; : : : : (The elements of V� are finite linear combinations of the vj ’s.) Wedefine an action of sl.2IC/ on V� by the same formulas as in Sect. 4.6:

��.Y /vj D vjC1��.H/vj D .� � 2j /vj

9.2 Introduction to Verma Modules 245

��.X/v0 D 0

��.X/vj D j.� � .j � 1//vj�1:

Note that in Sect. 4.6, the vectors vj equaled zero for large j; whereas here the vj ’sare, by definition, linearly independent. Direct calculation shows that these formulasdo, in fact, define a representation of sl.2IC/:

When � is a non-negative integerm, the spaceW� spanned by vmC1; vmC2; : : : isinvariant under the action of sl.2IC/: After all, this space is clearly invariant underthe action of ��.Y / and ��.H/; and it is invariant under ��.X/ because (with� D m) we have

��.X/vmC1 D k.m �m/vm D 0:

Since W� is invariant, the quotient vector space V�=W� inherits a natural actionof sl.2IC/: This quotient space is then the unique finite-dimensional irreduciblerepresentation of sl.2IC/ with highest weight �:

In the case of general semisimple Lie algebra g; we would like to do somethingsimilar. Pick an basis consisting of vectors

Y1; : : : ; YN ; H1; : : : ;Hr ; X1; : : : ; XN ; (9.2)

as in the proof of Theorem 9.4. If V� is any highest weight cyclic representationwith highest weight � and highest weight vector v0; then V� is spanned by productsof the basis elements applied to v0: By the reordering lemma (Lemma 6.12), we canreorder any such product as a linear combination of terms in which the elements arein the order listed in (9.2). Once the basis elements are in this order, any term thatcontains any Xj ’s will give zero when applied to v0. Furthermore, in any term thatdoes not have any Xj ’s, any factors of Hk will simply give �.Hj / when appliedto v0: Thus, V� must be spanned by elements of the form

��.Y1/n1��.Y2/

n2 � � ���.YN /nN v0: (9.3)

The idea of a Verma module is that we should proceed as in the sl.2IC/ caseand simply decree that the vectors in (9.3) form a basis for our Verma module. Theweights of the Verma module will the consist of all elements of the form

� � n1˛1 � � � � � nN˛N ;

where each nj is a non-negative integer. (See Figure 9.1.) If we do this, then thereis only one possible way that the Lie algebra g can act. After all, if we apply someLie algebra element �.Z/ to a vector as in (9.3), we can reorder the elements as inthe previous paragraph until they are in the order of (9.2). Then, as we have alreadynoted, any factors of ��.Xj / give zero and any factors of ��.Hk/ give constants.We will, thus, eventually get back a linear combination of elements of the form (9.3).

246 9 Representations of Semisimple Lie Algebras

Fig. 9.1 The weights of the Verma module with highest weight �

The difficulty with the above description of the Verma module is that it doesnot provide any reasonable method for checking that �� actually constitutes arepresentation of g: After all, unless g D sl.2IC/; it is impossible to writedown an explicit description of how the various basis elements act, and it is thusimpossible to verify directly that these elements satisfy the correct commutationrelations. Nevertheless, we will eventually prove (Sect. 9.5) that there is a well-defined representation of g having the elements (9.3) as a basis and in which g actsin the way described above. In the case in which � is dominant and integral, we willthen construct an invariant subspace of the Verma module for which the quotient isfinite dimensional and irreducible.

9.3 Universal Enveloping Algebras

In Sect. 9.5, we will construct each Verma module V� as a quotient of somethingcalled the universal enveloping algebra of a Lie algebra g: If g is a Lie algebra, wemay try to embed g as a subspace of some associative algebra A in such a way thatthe bracket on g may be computed as ŒX; Y � D XY � YX; where XY and YX arecomputed in A: If g is the Lie algebra of a matrix Lie group G � Gl.nIC/; then gis a subspace of the associative algebraMn.C/ and the bracket on g is indeed givenas ŒX; Y � D XY � YX: There may be, however, many other ways to embed g intoan associative algebra A: For example, if g D sl.2IC/; then for each m � 1; theirreducible representation �m of g of dimension m C 1 allows us to embed g intoMmC1.C/:

9.3 Universal Enveloping Algebras 247

Let us now give a useful but slightly imprecise definition of the universalenveloping algebra of g; denoted U.g/ For any Lie algebra g; the universalenveloping algebra of g will be an associative algebra A with identity with thefollowing properties. (1) The Lie algebra g embeds into A in such a way thatŒX; Y � D XY � YX: (2) The algebra A is generated by elements of g; meaningthat the smallest subalgebra with identity of A containing g is all of A: (3) Thealgebra A is maximal among all algebras A with the two previous properties.The maximality property of A will be explained more precisely in the discussionfollowing Theorem 9.7.

Consider, for example, the case of a one-dimensional Lie algebra g spanned bya single nonzero element X; which of course satisfies ŒX;X� D 0: Then U.g/should be an associative algebra with identity generated a single element X; inwhich case, U.g/ must also be commutative. Now, any associative algebra A withidentity generated by a single nonzero element X will satisfy Properties 1 and 2in the definition of the enveloping algebra. But for A to be maximal, there shouldbe no relations between the different powers of X; meaning that p.X/ should be anonzero element of A for every nonzero polynomial p: In this case, then, we maytake U.g/ to be the algebra of polynomials in a single variable.

Suppose the algebra g in the previous paragraph is a matrix algebra, meaningthat X is a single n � n matrix. Contrary to what we might, at first, expect, theenveloping algebra U.g/ will not coincide with the associative algebra with identityA generated by X inside Mn.C/: After all, for any X 2 Mn.C/; the Cayley–Hamilton theorem implies that there exists a nonzero polynomial p (namely, thecharacteristic polynomial of X ) for which p.X/ D 0: In U.g/; by contrast, we havesaid that p.X/ should be nonzero for all nonzero polynomials p:

Actually, it follows from the PBW theorem (Theorem 9.10) that the universalenveloping algebra of any nonzero Lie algebra g is infinite dimensional. In fact, ifXis any nonzero element of g; the elements 1; X; X2; : : : will be linearly independentin U.g/: Thus, even if g is an algebra of matrices, U.g/ cannot be isomorphic to asubalgebra of an algebra of matrices.

We now give the formal definition of a universal enveloping algebra.

Theorem 9.7. For any Lie algebra g; there exists an associative algebra withidentity, denoted U.g/; together with a linear map i W g ! U.g/ such that thefollowing properties hold. (1) For all X; Y 2 g; we have

i.ŒX; Y �/ D i.X/i.Y / � i.Y /i.X/: (9.4)

(2) The algebra U.g/ is generated by elements of the form i.X/; X 2 g; meaningthat the smallest subalgebra with identity ofU.g/ containing every i.X/ is U.g/: (3)Suppose A is an associative algebra with identity and j W g ! A is a linear mapsuch that j.ŒX; Y �/ coincides with j.X/j.Y / � j.Y /j.X/ for all X; Y 2 g. Thenthere exists a unique algebra homomorphism � W U.g/ ! A such that �.1/ D 1

and such that �.i.X// D j.X/ for all X 2 g:A pair .U.g/; i/ with the preceding properties is called a universal enveloping

algebra for g:

248 9 Representations of Semisimple Lie Algebras

A simple argument (Exercise 1) shows that any two universal enveloping algebrasfor a fixed Lie algebra g are “canonically” isomorphic.

Let us define an enveloping algebra of g to be an associative algebra A togetherwith a linear map j W g ! A as in Point 3 of the theorem, with the additionalproperty that A is generated by elements of the form j.X/; X 2 g: In this case, thehomomorphism � W U.g/ ! A as in Theorem 9.7 is surjective and A is isomorphicto the quotient algebra U.g/= ker.�/: Thus, the universal enveloping algebra U.g/of g has the property that every other enveloping algebra of g is a quotient of U.g/:This property of U.g/ is a more precise formulation of the maximality condition wediscussed in the second paragraph of this subsection.

Example 9.8. Let g D sl.2IC/ with the usual basis fX; Y;H g: The universalenveloping algebra of g is then the associative algebra with identity generated bythree elements x; y; and h; subject to the relations

hx � xh D 2x

hy � yh D �2yxy � yx D h;

and no other relations. The map i W g ! U.g/ is then the unique linear map suchthat i.X/ D x; i.Y / D y; and i.H/ D h and � W U.g/ ! A is the uniquehomomorphism such that �.x/ D j.X/; �.y/ D j.Y /; and �.h/ D j.H/:

The meaning of the phrase “no other relations” will be made more precise inthe proof of Theorem 9.7, in which J is the smallest two-sided ideal containing allelements of the form X ˝ Y � Y ˝ X � ŒX; Y �: The fact that U.g/ has no otherrelations guarantees that the homomorphism � is well defined.

The construction of U.g/ is in some sense easy or “soft.” But for U.g/ to beuseful in practice (for example, in constructing Verma modules), we need a structuretheorem for it known as the Poincaré–Birkhoff–Witt theorem. The Poincaré–Birkhoff–Witt theorem will, in particular, show that the map i in Theorem 9.7 isactually injective. (See also Exercise 2.) Once this is established, we will be able toidentify g with its image under i and thus think of g as embedded into U.g/:

Proof of Theorem 9.7. The operation of tensor product on vector spaces is associa-tive, in the sense that U ˝ .V ˝ W / is canonically isomorphic to .U ˝ V / ˝ W;

with the isomorphism taking u ˝ .v ˝ w/ to .u ˝ v/ ˝ w for each u 2 U; v 2 V;

and w 2 W: We may thus drop the parentheses and write simply U ˝ V ˝ W andu˝v˝w; and similarly for the tensor product of any finite number of vector spaces.In particular, we will let V ˝k denote the k-fold tensor product V ˝ � � � ˝ V: The0-fold tensor product V ˝0 is defined to be C (or whatever field we are workingover).

For a Lie algebra g; let us first define the tensor algebra T .g/ over g; which isdefined as

T .g/ D1MkD0

g˝k:

9.3 Universal Enveloping Algebras 249

In the direct sum, each element of T .g/ is required to be a finite linear combinationsof elements g˝k for different values of k: We can make T .g/ into an associativealgebra with identity by defining

.u1 ˝ u2 ˝ � � � ˝ uk/ � .v1 ˝ v2 ˝ � � � ˝ vl/

D u1 ˝ u2 ˝ � � � ˝ uk ˝ v1 ˝ v2 ˝ � � � ˝ vl (9.5)

and then extending the product by linearity. That is to say, the product operationis the unique bilinear map of T .g/ � T .g/ into T .g/ that coincides with thetensor product (9.5) on g˝k � g˝l : Since C ˝ g˝k is naturally isomorphic to g˝k;the identity element 1 2 C D g˝0 is the multiplicative identity for T .g/: Theassociativity of the tensor product assures that T .g/ is an associative algebra.

We now claim that the algebra T .g/ has the following property: If A is anyassociative algebra with identity and j W g ! A is any linear map, there exists analgebra homomorphism W T .g/ ! A such that .1/ D 1 and .X/ D j.X/

for all X 2 g � T .g/: Note that this property differs from the desired property ofU.g/ in that j is an arbitrary linear map and does not have to have any particularrelationship to the algebra structure of A: To construct ; we require that therestriction of to g˝k to be the unique linear map of g˝� into A such that

.X1 ˝ � � � ˝Xk/ D j.X1/ � � � j.Xk/ (9.6)

for all X1; : : : ; Xk in g: (Here we are using the natural k-fold extension of theuniversal property of tensor products in Definition 4.13.) It is then simple tocheck that is an algebra homomorphism. Furthermore, if is to be an algebrahomomorphism that agrees with j on g; then must have the form in (9.6).

We now proceed to construct U.g/ as a quotient of T .g/: A two-sided ideal inT .g/ is a subspace J of T .g/ such that for all ˛ 2 T .g/ and ˇ 2 J , the elements ˛ˇand ˇ˛ belong to J:We now let J be the smallest two-sided ideal in T .g/ containingall elements of the form

X ˝ Y � Y ˝X � ŒX; Y �; X; Y 2 g: (9.7)

That is to say, J is the intersection of all two-sided ideals in T .g/ containing allsuch elements, which is, again, a two-sided ideal containing these elements. Moreconcretely, J can be constructed as the space of elements of the form

NXjD1

˛j .Xj ˝ Yj � Yj ˝Xj � ŒXj ; Yj �/ˇj ;

with Xj and Yj in g and ˛j and ˇj being arbitrary elements of T .g/:We now form the quotient vector space T .g/=J; which is an algebra. If

j W g ! A is any linear map, we can form the algebra homomorphism W T .g/ ! A as above. If j satisfies j.ŒX; Y �/ D j.X/j.Y / � j.Y /j.X/;

then the kernel of will contain all elements of the formX ˝Y �Y ˝X � ŒX; Y �:

250 9 Representations of Semisimple Lie Algebras

Furthermore, the kernel of an algebra homomorphism is always a two-sided ideal.Thus, ker. / contains J: It follows that the map W T .g/ ! A factors throughU.g/ WD T .g/=J; giving the desired homomorphism � of U.g/ into A: Since U.g/is spanned by products of elements of g; there can be at most one map � with thedesired property, establishing the claimed uniqueness of �: utProposition 9.9. If � W g ! End.V / is a representation of a Lie algebra g(not necessarily finite dimensional), there is a unique algebra homomorphismQ� W U.g/ ! End.V / such that Q�.1/ D I and Q�.X/ D �.X/ for all X 2 g � U.g/:

Proof. Apply Theorem 9.7 with A D End.V / and j.X/ D �.X/: utWe now state the Poincaré–Birkhoff–Witt theorem—or PBW theorem, for

short—which is the key structure result for universal enveloping algebras. Althoughthe result holds even for infinite-dimensional Lie algebras, we state it here in thefinite-dimensional case, for notational simplicity. The proof of the PBW theorem isin Sect. 9.4.

Theorem 9.10 (PBW Theorem). If g is a finite-dimensional Lie algebra with basisX1; : : : ; Xk; then elements of the form

i.X1/n1 i.X2/

n2 � � � i.Xk/nk ; (9.8)

where each nk is a non-negative integer, span U.g/ and are linearly independent.In particular, the elements i.X1/; : : : ; i.Xk/ are linearly independent, meaning thatthe map i W g ! U.g/ is injective.

In (9.8), we interpret i.Xj /nj as 1 if nj D 0: Since, actually, i is injective, wewill henceforth identify g with its image under i and thus regard g as a subspace ofU.g/: Thus, we may now write X in place of i.X/: In our new notation, we maywrite (9.4) as

ŒX; Y � D XY � YX

and we may write the basis elements (9.8) as

Xn11 X

n22 � � �Xnk

k : (9.9)

It is straightforward to show that the elements in (9.8) span U.g/; the hard part is toprove they are linearly independent.

Corollary 9.11. If g is a Lie algebra and h is a subalgebra of g; then there is anatural injection of U.h/ into U.g/ given by mapping any product X1X2 � � �XN ofelements of h to the same product in U.g/:

Proof. The inclusion of h into g induces an algebra homomorphism of � W U.h/ !U.g/: Let us choose a basis X1; : : : ; Xk for h and extend it to a basis X1; : : : ; XNfor g: By the PBW theorem for h; the elements Xn1

1 � � �Xnkk form a basis for U.h/:

Then by the PBW theorem for g; the corresponding elements of U.g/ are linearlyindependent, showing that � is injective. ut

9.4 Proof of the PBW Theorem 251

9.4 Proof of the PBW Theorem

It is notationally convenient to write the elements of the claimed basis for U.g/ as

i.Xj1/i.Xj2/ � � � i.XjN /; (9.10)

with j1 � j2 � � � � � jN ; where we interpret the above expression as 1 ifN D 0: The easy part of the PBW theorem is to show that these elements spanU.g/: The proof of this claim is essentially the same as the proof of the reorderinglemma (Lemma 6.12). Every element of the tensor algebra T .g/, and hence also ofthe universal enveloping algebra U.g/; is a linear combination of products of Liealgebra elements. Expanding each Lie algebra element in our basis shows that everyelement of U.g/ is a linear combination of products of basis elements, but not (sofar) necessarily in nondecreasing order. But using the relation XY � YX D ŒX; Y �;

we can reorder any product of basis elements into the desired order, at the expenseof introducing several terms that are products of one fewer basis elements. Thesesmaller products can then, inductively, be rewritten as a linear combination of termsthat are in the correct order.

It may seem “obvious” that elements of the form (9.10) are linearly independent.Note, however, that any proof of independence of these elements must make use ofthe Jacobi identity. After all, if g is a vector space with any skew-symmetric, bilinear“bracket” operation, we can still construct a “universal enveloping algebra” by theconstruction in Sect. 9.3, and the elements of the form (9.10) will still span thisenveloping algebra. If, however, the bracket does not satisfy the Jacobi identity, theelements in (9.10) will not be linearly independent. If they were, then, in particular,the map i W g ! U.g/ would be injective. We could then identify g with its imageunder i; which means that the bracket on g would be given by ŒX; Y � D XY � YX;where XY and YX are computed in the associative algebra U.g/: But any bracketof this form does satisfy the Jacobi identity.

We now proceed with the proof of the independence of the elements in (9.10).The reader is encouraged to note the role of the Jacobi identity in our proof. Let Dbe any vector space having a basis

fv.j1;��� ;jN /g:indexed by all nondecreasing tuples .j1; : : : ; jN /. We wish to construct a linear map� W U.g/ ! D with the property that

�.i.Xj1/i.Xj2/ � � � i.XjN // D v.j1;��� ;jN /

for each nondecreasing tuple .j1; : : : ; jN /: Since the elements v.j1;��� ;jN / are,by construction, linearly independent, if such a map � exists, the elementsXj1Xj2 � � �XjN must be linearly independent as well. (Any linear relation amongthe Xj1Xj2 � � �XjN ’s would translate under � into a linear relation among thev.j1;��� ;jN /’s.)

252 9 Representations of Semisimple Lie Algebras

Instead of directly constructing �; we will construct a linear map ı W T .g/ ! D

with the properties (1) that

ı.Xj1 ˝Xj2 ˝ � � � ˝XjN / D v.j1;��� ;jN / (9.11)

for all nondecreasing tuples .j1; : : : ; jN /; and (2) that ı is zero on the two-sidedideal J . Since ı is zero on J; it gives rise to a map � of U.g/ WD T .g/=J into Dwith the analogous property.

To keep our notation compact, we will now omit the tensor product symbol formultiplication in T .g/: Since all computations in the remainder of the section arein T .g/; there will be no confusion. Suppose we can construct ı in such a waythat (9.11) holds for nondecreasing tuples and that for all tuples .j1; : : : ; jN /, wehave

ı.Xj1 � � �XjkXjkC1� � �XjN /

D ı.Xj1 � � �XjkC1Xjk � � �XjN /C ı.Xj1 � � � ŒXjk ; XjkC1

� � � �XjN /: (9.12)

Then ı will indeed by zero on J: After all, J is spanned by elements of the form

˛.XY � YX � ŒX; Y �/ˇ:After moving all terms in (9.12) to the other side and taking linear combinations,we can see that ı will be zero on every such element.

Define the degree of a monomial Xj1Xj2 � � �XjN to be the number N and theindex of the monomial to be the number of pairs l < k for which jl > jk: We willconstruct ı inductively, first on the degree of the monomial, and then on the indexof the monomial for a given degree, verifying (9.12) as we proceed. If N D 0, weset ı.1/ D v.0;:::;0/ and if N D 1; we set ı.Xj / D v.0;:::;1;:::;0/: In both cases, (9.11)holds by construction and (9.12) holds vacuously.

For a fixed N � 2 and p; we now assume that ı has been defined on the spanof all monomials of degree less than N; and also on all monomials of degree Nand index less than p. If p D 0; this means simply that ı has been defined on thespan of all monomials of degree less than N: Our induction hypothesis is that ı; asdefined up to this point, satisfies (9.12) whenever all the terms in (9.12) have beendefined. That is to say, we assume (9.12) holds whenever both the monomials on theleft-hand side of (9.12) have degree less than N or degreeN and index less than p:(Under these assumptions, the argument of ı on the right-hand side of (9.12) will bea linear combination of monomials of degree less than N; so that the left-hand sideof (9.12) has been defined.)

We now need to show that we can extend the definition of ı to monomials ofdegreeN and index p in such a way that (9.12) continues to hold. If p D 0; the newmonomials we have to consider are the nondecreasing ones, in which case (9.11)requires us to set

ı.Xj1 � � �XjN / D v.j1;��� ;jN /:

9.4 Proof of the PBW Theorem 253

Now, the only way both the monomials on the left-hand side of (9.12) can havedegreeN and index zero is if jkC1 D jk; in which case, both sides of (9.12) will bezero. It remains, then, to consider the case p > 0:

Let us consider an example that illustrates the most important part of theargument. Suppose N D 3 and p D 3; meaning that we have defined a mapı satisfying (9.12) on all monomials of degree less than 3 and all monomials ofdegree 3 and index less than 3. We now attempt to define ı on monomials of degree3 and index 3 and verify that (9.12) still holds. A representative such monomialwould be X3X2X1: Since we want (9.12) to hold, we may attempt to use (9.12) asour definition of ı.X3X2X1/: But this strategy gives two possible ways of definingı.X3X2X1/; either

ı.X3X2X1/ D ı.X2X3X1/C ı.ŒX3;X2�X1/ (9.13)

or

ı.X3X2X1/ D ı.X3X1X2/C ı.X3ŒX2;X1�/: (9.14)

Note that the monomials on the right-hand sides of (9.13) and (9.14) have degree 2or degree 3 and index 2, so that ı has already been defined on these monomials. Wenow verify that these two expression for ı.X3X2X1/ agree.

Since ı has already been defined for the terms on the right-hand side of (9.13),we may apply our induction hypothesis to these terms. Using induction twice, wemay simplify the right-hand side of (9.13) until we obtain a term in the correct PBWorder of X1X2X3; plus commutator terms:

ı.X2X3X1/C ı.ŒX3;X2�X1/

D ı.X2X1X3/C ı.X2ŒX3;X1�/C ı.ŒX3;X2�X1/

D ı.X1X2X3/C ı.ŒX2;X1�X3/

C ı.X2ŒX3;X1�/C ı.ŒX3;X2�X1/:

Similarly, the other candidate (9.14) for ı.X3X2X1/ may be computed by ourinduction hypothesis as

ı.X3X1X2/C ı.X3ŒX2;X1�/

D ı.X1X3X2/C ı.ŒX3;X1�X2/C ı.X3ŒX2;X1�/

D ı.X1X2X3/C ı.X1ŒX3;X2�/

C ı.ŒX3;X1�X2/C ı.X3ŒX2;X1�/:

254 9 Representations of Semisimple Lie Algebras

Subtracting the two expressions gives the quantity

ı.ŒX2;X1�X3/� ı.X3ŒX2;X1�/

C ı.X2ŒX3;X1�/ � ı.ŒX3;X1�X2/C ı.ŒX3;X2�X1/ � ı.X1ŒX3;X2�/:

Since all terms are of degree 2, we can use our induction hypothesis to reduce thisquantity to

ı.ŒŒX2;X1�; X3�C ŒX2; ŒX3;X1��C ŒŒX3;X2�; X1�/

D ı.ŒX3; ŒX1;X2��C ŒX2; ŒX3;X1��C ŒX1; ŒX2;X3��/

D 0;

by the Jacobi identity.Thus, the two apparently different definitions of ı.X3X2X1/ in (9.13) and (9.14)

agree. Using this result, it should be apparent that (9.12) holds when we extend thedomain of definition of ı to include the monomialX3X2X1 of degree 3 and index 3.

We now proceed with the general induction step in the construction of ı;meaningthat we assume ı has been constructed on monomials of degree less than N and onmonomials of degree N and index less than p; in such a way that (9.12) holdswhenever both monomials on the left-hand side of (9.12) are in the current domainof definition of ı: Since we have already addressed the p D 0 case, we assumep > 0: We now consider a monomial Xj1Xj2 � � �XjN of index p � 1: Since theindex of the monomial is at least 1, the monomial is not weakly increasing and theremust be some jk with jk > jkC1: Pick such a k and “define” ı on the monomial by

ı.Xj1 � � �XjkXjkC1� � �XjN / D ı.Xj1 � � �XjkC1

Xjk � � �XjN /C ı.Xj1 � � � ŒXjk ; XjkC1

� � � �XjN /: (9.15)

Note that the first term on the right-hand side of (9.15) has index p � 1 and thesecond term on the right-hand side has degree N � 1; which means that both ofthese terms have been previously defined.

The crux of the matter is to show that the value of ı on a monomial of index pis independent of the choice of k in (9.15). Suppose, then, that there is some l < k

such that jl > jlC1 and jk > jkC1: We now proceed to check that the value of theright-hand side of (9.15) is unchanged if we replace k by l:

Case 1: l � k � 2: In this case, the numbers l; l C 1; k; kC 1 are all distinct. Letus consider the two apparently different ways of calculating ı: If we use l; thenwe have

ı.� � �XlXlC1 � � �XkXkC1 � � � /D ı.� � �XlC1Xl � � �XkXkC1 � � � /C ı.� � � ŒXl ; XlC1� � � �XkXkC1 � � � /: (9.16)

9.5 Construction of Verma Modules 255

Now, the second term on the right-hand side of (9.16) has degreeN �1: The firstterm has index p � 1, and if we reverse Xk and XkC1 we obtain a term of indexp � 2: Thus, we can apply our induction hypothesis to reverse the order of Xkand XkC1 in both terms on the right-hand side of (9.16), giving

ı.� � �XlXlC1 � � �XkXkC1 � � � /D ı.� � �XlC1Xl � � �XkC1Xk � � � /C ı.� � �XlC1Xl � � � ŒXk;XkC1� � � � /C ı.� � � ŒXl ; XlC1� � � �XkC1Xk � � � /C ı.� � � ŒXl ; XlC1� � � � ŒXk;XkC1� � � � /:

(9.17)

(Note that on the right-hand side of (9.17), all terms have both Xl and XlC1 andXk and XkC1 back in their correct PBW order, with XlC1 to the left of Xl andXkC1 to the left of Xk:) Since the right-hand side of (9.17) is symmetric in k andl; we would get the same result if we started with k instead of l:

Case 2: l D k�1. In this case, the indices jl ; jlC1 D jk; and jlC2 D jkC1 are inthe completely wrong order, jl > jlC1 > jlC2: Let us use the notation X D Xl;

Y D XlC1, and Z D XlC2: We wish to show that the value of ı.� � �XYZ � � � /is the same whether we use l (that is, interchanging X and Y ) or we use k (thatis, interchanging Y andZ). But the argument is then precisely the same as in thespecial case of ı.X3X2X1/ considered at the beginning of our proof, with someextra factors, indicated by � � � ; tagging along for the ride.

Once we have verified that the value of ı is independent of the choice of k in (9.15),it should be clear that (9.12) holds, since we have used (9.12) with respect to agiven pair of indices as our “definition” of ı: We have, therefore, completed theconstruction of ı and the proof of the PBW theorem.

9.5 Construction of Verma Modules

The proof will make use of the following definition: A subspace I of U.g/ is calleda left ideal if ˛ˇ 2 I for all ˛ 2 U.g/ and all ˇ 2 I: For any collection of vectorsf˛j g in U.g/; we may form the left ideal I “generated by” these vectors, that is, thesmallest left ideal in U.g/ containing each ˛j : The left ideal I is precisely the spaceof elements of the form X

j

ˇj ˛j

with ˇj being arbitrary elements of U.g/:Let I� denote the left ideal in U.g/ generated by elements of the form

H � h�;H i 1; H 2 h (9.18)

256 9 Representations of Semisimple Lie Algebras

and of the form

X 2 g˛; ˛ 2 RC: (9.19)

We now let W� denote the quotient vector space

W� D U.g/=I�;

and we let Œ˛� denote the image of ˛ 2 U.g/ in the quotient space.We may define a representation �� of U.g/ acting on W� by setting

��.˛/.Œˇ�/ D Œ˛ˇ� (9.20)

for all ˛ and ˇ in U.g/: To verify that ��.˛/ is well defined, note that if ˇ0 isanother representative of the equivalence class Œˇ�; then ˇ0 D ˇ C � for some � inI�: But then ˛ˇ0 D ˛ˇC˛�; and ˛� belongs to I�; because I� is a left ideal. Thus,Œ˛ˇ0� D Œ˛ˇ�: We may check that �� is a homomorphism by noting that ��.˛ˇ/Œ��and ��.˛/��.ˇ/Œ�� both equal Œ˛ˇ��; by the associativity of U.g/: The restrictionof �� to g constitutes a representation of g acting on W�:

Definition 9.12. The Verma module with highest weight �; denoted W�; is thequotient space U.g/=I�, where I� is the left ideal in U.g/ generated by elements ofthe form (9.18) and (9.19).

Theorem 9.13. The vector v0 WD Œ1� is a nonzero element of W� and W� isa highest weight cyclic representation with highest weight � and highest weightvector v0:

The hard part of the proof is establishing that v0 is nonzero; this amounts toshowing that the element 1 of U.g/ is not in I�: For purposes of constructing theirreducible, finite-dimensional representations of g; Theorem 9.13 is sufficient. Ourmethod of proof, however, gives more information about the structure ofW�; whichwe will make use of in Chapter 10.

Let nC denote the span of the root vectors X˛ 2 g˛ with ˛ 2 RC; and letn� denote the span of the root vectors Y˛ 2 g�˛ with ˛ 2 RC: Because Œg˛; gˇ� �g˛Cˇ; both nC and n� are subalgebras of g:

Theorem 9.14. If Y1; : : : ; Yk form a basis for n�; then the elements

��.Y1/n1��.Y2/

n2 � � ���.Yk/nkv0; (9.21)

where each nj is a non-negative integer, form a basis for W�:

The theorem, together with the PBW theorem, tells us that there is a vector spaceisomorphism between U.n�/ and W� given by ˛ 7! ��.˛/v0; where �� is theaction of U.g/ on W�; given by (9.20).

Lemma 9.15. Let J� denote the left ideal in U.b/ � U.g/ generated by elementsof the form (9.18) and (9.19). Then 1 does not belong to J�.

9.5 Construction of Verma Modules 257

Proof. Let b be the direct sum (as a vector space) of nC and h; which is easily seento be a subalgebra of g: Let us define a one-dimensional representation � of b;acting on C; by the formula

�.X CH/ D h�;H i :

(That is to say, �.X C H/ is the 1 � 1 matrix with entry h�;H i :) To see that �is actually a representation, we note that all 1 � 1 matrices commute. On the otherhand, the commutator of two elements Z1 and Z2 of b will lie in nC; and � isdefined to be zero on n: Thus, �.ŒZ1;Z2�/ and Œ�.Z1/; �.Z2/� are both zero.

By Proposition 9.9, the representation � of b extends to a representation Q�of U.b/ satisfying Q�.1/ D 1: Now, the kernel of Q� is easily seen to be a leftideal in U.b/; and by construction, the kernel of Q� will contain all elements ofthe form (9.18) and (9.19). But since Q�.1/ D 1; the element 1 does not belong tothe kernel of Q�: Thus, ker. Q�/ is a left ideal in U.b/ containing all elements of theform (9.18) and (9.19), which means that ker. Q�/ contains J�: Since ker. Q�/ doesnot contain 1; neither does J�: utProof of Theorems 9.13 and 9.14. Note that for any H 2 h; we have

.��.H � h�;H i 1//v0 D ŒH � h�;H i 1� D 0;

because H � h�;H i 1 belongs to I�: Thus, ��.H/v0 D h�;H i v0: Similarly, forany ˛ 2 RC; we have

��.X˛/v0 D ŒX˛� D 0:

Since elements of U.g/ are linear combinations of products of elements of g; anyinvariant subspace for the action of g onW� is also invariant under the action ofU.g/on W�: Suppose, then, that U is an invariant subspace of W� containing v0 D Œ1�:

Then for any Œ˛� 2 W�; we have ��.˛/v0 D Œ˛�; and so U D W�: Thus, v0 is acyclic vector forW�: To prove Theorem 9.13, it remains only to show that v0 ¤ 0:

The Lie algebra g decomposes as a vector space direct sum of n� and b; wheren� is the span of the root spaces corresponding to roots in R� and where b is thespan of h and the root spaces corresponding to roots in RC: Let us choose a basisY1; : : : ; Yk; Z1; : : : ; Zl for g consisting of a basis Y1; : : : ; Yk for n� together with abasis Z1; : : : ; Zl for b: By applying the PBW theorem to this basis, we can easilysee (Exercise 4) that every element ˛ of U.g/ can be expressed uniquely in the form

˛ D1X

n1;:::;nkD0Yn11 Y

n22 � � �Y nkk an1;:::;nk ; (9.22)

where each an1;:::;nk belongs to U.b/ � U.g/: (For each ˛ 2 U.g/; only finitelymany of the an1;:::;nk ’s will be nonzero.)

258 9 Representations of Semisimple Lie Algebras

Suppose now that ˛ belongs to I�; which means that ˛ is a linear combinationof terms of the form ˇ.H � h�;H i 1/ and ˇX˛; with ˇ in U.g/; H in h; and X˛ ing˛; ˛ 2 RC: By writing each ˇ as in (9.22), we see that ˛ is a linear combinationof terms of the form

Yn11 Y

n22 � � �Y nkk bn1;:::;nk .H � h�;H i 1/

and

Yn11 Y

n22 � � �Y nkk bn1;:::;nkX˛;

with bn1;:::;nk in U.b/: Note that bn1;:::;nk .H � h�;H i 1/ and bn1;:::;nkX˛ belongto the left ideal J� � U.b/: Thus, if ˛ is in I�; each an1;:::;nk in the (unique)expansion (9.22) of ˛ must belong to J�.

Now, by the uniqueness of the expansion, the only way the element ˛ in (9.22)can equal 1 is if there is only one term, the one with n1 D � � � D nk D 0; and ifa0;:::;0 D 1: On the other hand, for ˛ to be in I�; each an1;:::;nk must belong to J�:Since (Lemma 9.15) 1 is not in J�; we see that 1 is not in I�; so that v0 D Œ1� isnonzero in U.g/=I�:

We now argue that the vectors in (9.21) are linearly independent in U.g/=I�:Suppose, then, that a linear combination of these vectors, with coefficients cn1;:::;nkequals zero. Then the corresponding linear combination of elements inU.g/; namely

˛ WD1X

n1;:::;nkD0Yn11 Y

n22 : : : Y

nkk cn1;:::;nk ;

belongs to I�: But as shown above, for ˛ to be in I�; each of the constants cn1;:::;nkmust be in J�: Thus, by Lemma 9.15, each of the constants cn1;:::;nk is zero. ut

The main role of the PBW theorem in the preceding proof is to establish theuniqueness of the expansion (9.22).

9.6 Irreducible Quotient Modules

In this section, we show that every Verma module has a largest proper invariantsubspace U� and that the quotient space V� WD W�=U� is irreducible with highestweight �: In the next section, we will show that if � is dominant integral, thisquotient space is finite dimensional.

It is easy to see (Exercise 6) that the Verma module W� is the direct sum of itsweight spaces. It therefore makes sense to talk about the component of a vectorv 2 W� in the one-dimensional subspace spanned by v0; which we refer to as thev0-component of v: As in the previous section, we let nC denote the subalgebra ofg spanned by weight vectors X˛ 2 g˛; with ˛ 2 RC:

9.6 Irreducible Quotient Modules 259

Definition 9.16. For any Verma module W�; let U� be the subspace of W�

consisting of all vectors v such that the v0-component of v is zero and such thatthe v0-component of

��.X1/ � � ���.XN /vis also zero for any collection of vectorsX1; : : : ; XN in nC:

That is to say, a vector v belongs to U� if we cannot “get to” v0 from v byapplying “raising operators” X 2 g˛; ˛ 2 RC: Certainly the zero vector is in U�Ifor some g’s and �’s, it happens that U� D f0g:Proposition 9.17. The space U� is an invariant subspace for the action of g:

Proof. Suppose that v is in U� and that Z is some element of g: We want to showthat ��.Z/v is also in U�: Thus, we consider

��.X1/ � � ���.Xl/��.Z/v (9.23)

and we must show that the v0-component of this vector is zero. Using the reorderinglemma (Lemma 6.12), we may rewrite the vector in (9.23) as a linear combinationof vectors of the form

��.Y1/ � � ���.Y j /��.H1/ � � ���.Hk/��. QX1/ � � ���. QXm/v; (9.24)

where the Y ’s are in n�; the H ’s are in h; and the QX ’s are in nC: However, since vis in U�; the v0-component of

��. QX1/ � � ���. QXm/v (9.25)

is zero; thus, this vector is a linear combination of weight vectors with weight lowerthan �: Then applying elements of h and n� to the vector in (9.25) will only keepthe weights the same or lower them. Thus, the v0-component of the vector in (9.24),and hence also the v0-component of the vector in (9.23), is zero. This shows that��.Z/v is, again, in U�: ut

Since U� is an invariant subspace of W�; the quotient vectors space W�=U�carries a natural action of g and thus constitutes a representation of g:

Proposition 9.18. The quotient space V� WD W�=U� is an irreducible representa-tion of g:

Proof. A simple argument shows that the invariant subspaces of the representationW�=U� are in one-to-one correspondence with the invariant subspaces of W� thatcontain U�: Thus, proving that V� is irreducible is equivalent to showing that anyinvariant subspace of W� that contains U� is either U� or W�: Suppose, then, thatX is an invariant subspace that contains U� and at least one vector v that is notin U�: This means that X also contains a vector u D ��.X1/ � � ���.Xk/v whosev0-component is nonzero.

260 9 Representations of Semisimple Lie Algebras

We now claim that X must contain v0 itself. To see this, we decompose u asa nonzero multiple of v0 plus a sum of weight vectors corresponding to weights� ¤ �: Since � ¤ �; we can find H in h with h�;H i ¤ h�;H i and then wemay apply to u the operator ��.H/ � h�;H i I: This operator will keep us in Xand will “kill” the component of u that is in the weight space corresponding tothe weight � while leaving the v0-component of u nonzero. We can then continueapplying operators of this form until we have killed all the components of u inweight spaces different from �; giving us a nonzero multiple of v0: We conclude,then, that X contains v0 and, therefore, all of W�: Thus, any invariant subspace ofW� that properly contains U� must be equal to W�: ut

Since for each u 2 U� the v0-component of u is zero, the vector v0 is notinU�: Thus, the quotient spaceW�=U� is still a highest weight cyclic representationwith highest weight � and with highest weight vector being the image of v0 in thequotient.

Example 9.19. Let ˛ be an element of and let s˛ D hX˛; Y˛;H˛i be as inTheorem 7.19. If h�;H˛i is a non-negative integer m; then the vector v WD�.Y˛/

mC1v0 belongs to U�:

This result is illustrated in Figure 9.2.

Proof. By the argument in proof of Theorem 4.32, the analog of (4.15) will holdhere:

�.X˛/�.Y˛/j v0 D j.m � .j � 1//�.Y˛/

j�1v0:

v

2

1

Fig. 9.2 Example 9.19 in the case ˛ D ˛2 and m D 2: The vector v belongs to U�

9.7 Finite-Dimensional Quotient Modules 261

Thus,

�.X˛/v D m.m�m/�.Y˛/mv0 D 0:

Meanwhile, if ˇ 2 RC with ˇ ¤ ˛; then for Xˇ 2 gˇ; if �.Xˇ/v were nonzero, itwould be a weight vector with weight

� D � � .mC 1/˛ C ˇ:

Since � is not lower than �; we must have �.Xˇ/v D 0: Thus, the v0-component ofv is zero and �.X�/v D 0 for all � 2 RC; which implies that v is in U�: ut

9.7 Finite-Dimensional Quotient Modules

Throughout this section, we assume that � is a dominant integral element. We willnow show that, in this case, the irreducible quotient space V� WD W�=U� is finitedimensional. Our strategy is to show that the set of weights for V� is invariantunder the action of the Weyl group on h: Now, if � is dominant integral, then everyweight � of W�—and thus also of V�—must be integral, since � � � is an integercombination of roots. Hence, if the weights of V� are invariant under W; we musthave w � � � � for all w 2 W: But it is not hard to show that there are only finitelymany integral elements this property. We will conclude, then, that there are onlyfinitely many weights in V�: Since (even in the Verma module) each weight hasfinite multiplicity, this will show that W�=U� is finite dimensional.

How, then, do we construct an action of the Weyl group on V�? If we attempt tofollow the proof of Theorem 9.3, we must contend with the fact that V� is not yetknown to be finite dimensional. Thus, we need a method of exponentiating operatorson a possibly infinite-dimensional space.

Definition 9.20. A linear operatorX on a vector space V is locally nilpotent if foreach v 2 V; there exists a positive integer k such that Xkv D 0:

If V is finite dimensional, then a locally nilpotent operator must actually benilpotent, that is, there must exist a single k such that Xkv D 0 for all v: Inthe infinite-dimensional case, the value of k depends on v and there may be nosingle value of k that works for all v: If X is locally nilpotent, then we define eX tobe the operator satisfying

eXv D1XkD0

Xk

kŠv; (9.26)

where for each v 2 V; the series on the right terminates.

262 9 Representations of Semisimple Lie Algebras

Proposition 9.21. For each ˛ 2 ; let s˛ D hX˛; Y˛;H˛i be as in Theorem 7.19.If � is dominant integral, then X˛ and Y˛ act in a locally nilpotent fashion on thequotient space V�:

Proof. For any X 2 g; we use QX as an abbreviation for the action of X on thequotient space V�: We say that a vector in V� is s˛-finite if it is contained in afinite-dimensional, s˛-invariant subspace. Let

m D h�;H˛i ;which is a non-negative integer because � is dominant integral. Let Qv0 denote theimage in V� of the highest vector v0 2 W�, and consider the vectors

Qvk WD QY k˛ Qv0; k D 0; 1; 2; : : : :

By the calculations in Sect. 4.6, the span of the Qvk’s is invariant under the actionof s˛: On the other hand, Example 9.19 shows that �.Y˛/mC1v0 is in U�; whichmeans that QvmC1 D QY mC1

˛ Qv0 is the zero element of V�: Thus, h Qv0; : : : ; Qvmi is afinite-dimensional, s˛-invariant subspace (Figure 9.3). In particular, there exists anonzero, s˛-finite vector in V�:

Now let T˛ � V� be the space of all s˛-finite vectors, which we have just shownto be nonzero. We now claim that T˛ is invariant under the action of g: To seethis, fix a vector v in T˛ and an element X of g: Let S be a finite-dimensional,s˛-invariant subspace containing v and let S 0 be the span of all vectors of the form

v

2

1

Fig. 9.3 Since the vector v is in U� (Figure 9.2), the circled weights span a s˛2 -invariant subspaceof W�=U�

9.7 Finite-Dimensional Quotient Modules 263

QYw with Y 2 g and w 2 S: Then S 0 is finite dimensional, having dimension at most.dim g/.dimS/: Furthermore, if Z 2 s˛; then for all w 2 S; we have

QZ QYw D QY QZw C AŒZ; Y �w;

which belongs to S 0; because QZw is again in S: Thus, S 0 is also invariant underthe action of s˛: We see, then, that QXv is contained in the finite-dimensional, s˛-invariant subspace S 0; that is, QXv 2 T˛: Since V� is irreducible and T˛ is nonzeroand invariant under the action of g, we have T˛ D V�:

We conclude that every v 2 V� is contained in a finite-dimensional, s˛-invariantsubspace. It then follows from Point 2 of Theorem 4.34 that . QX˛/kv D . QY˛/kv D 0

for some k, showing that QX˛ and QY˛ are locally nilpotent. utProposition 9.22. If � is dominant integral, the set of weights for V� is invariantunder the action of the Weyl group on h:

Proof. We continue the notation from the proof of Proposition 9.21. Since (Propo-sition 8.24)W is generated by the reflections s˛ with w 2 ; it suffices to show thatthe weights ofW�=U� are invariant under each such reflection. By Proposition 9.21,QX˛ and QY˛ are locally nilpotent, and thus it makes sense to define operators S˛ by

S˛ D eQX˛e� QY˛ e QX˛ :

We may now imitate the proof of Theorem 9.3 as follows. If H 2 h satisfiesh˛;H i D 0; then ŒH;X˛� D ŒH; Y˛� D 0; which means that QH commutes withQX˛ and QY˛ and, thus, with S˛: Meanwhile, for any v 2 V�; we may find a finite-

dimensional, s˛-invariant subspace S containing v: In the space S; we may applyPoint 3 of Theorem 4.34 to show that

S˛ QH˛S�1˛ v D � QH˛v:

We conclude that for all H 2 h; we have

S˛ QHS�1˛ D s˛ � QH:

From this point on, the proof of Theorem 9.3 applies without change. utFigure 9.4 illustrates the result of Proposition 9.22 in the case of sl.2IC/ and

highest weight 3: If v is a weight vector with weight �5; then �.X/v D 0;

by (4.15) in the proof of Theorem 4.32. Thus, the span of the weight vectors withweights l � �5 is invariant. The quotient space has weights ranging from �3 to 3in increments of 2 and is, thus, invariant under the action of W D fI;�I g:

We are now ready for the proof of the existence of finite-dimensional, irreduciblerepresentations.

264 9 Representations of Semisimple Lie Algebras

1 37 5 3 1

0 0

Fig. 9.4 In the Verma module for sl.2IC/ with highest weight 3; the span of the weight vectorswith weights �5; �7; : : : ; is invariant

Proof of Theorem 9.5. The quotient space V� WD W�=U� is irreducible and hashighest weight �: Every weight � of V� is integral and satisfies � � �: ByProposition 9.22, the weights are also invariant under the action ofW; which meansthat every weight � satisfies w � � � � for all w 2 W: Thus, by Proposition 8.44, �must be in the convex hull of the W -orbit of �; which implies that k�k � k�k :Since there are only finitely many integral elements � with this property, weconclude that V� has only finitely many weights.

Now, V� has at least one weight, namely �: Since V� is irreducible, it must bethe direct sum of its weight spaces. Now, since the elements in (9.21) form a basisfor the Verma module W�; the corresponding elements of V� certainly span V�:But for a given weight �; there are only finitely many choices of the exponentsn1; : : : ; nk in (9.21) that give a weight vector with weight �: (After all, if any of thenj ’s is large, the weight of the vector in (9.21) will be much lower than �:) Thus,each weight of V� has finite multiplicity. Since, also, there are only finitely manyweights, V� is finite dimensional. ut

9.8 Exercises

1. Suppose .i; U.g// and .i 0; U 0.g// are algebras as in Theorem 9.7. Show that thereis an isomorphismˆ W U.g/ ! U 0.g/ such that ˆ.1/ D 1 and such that

ˆ.i.X// D i 0.X/

for all X 2 g:Hint: Use the defining property of U.g/ to constructˆ:

2. Suppose that g � Mn.C/ is a Lie algebra of matrices (with bracket given byXY � YX). Prove, without appealing to the PBW theorem, that the map i W g !U.g/ in Theorem 9.7 is injective.

3. Suppose g is a Lie algebra and h is a subalgebra. Apply Theorem 9.7 to the Liealgebra h with A D U.g/ and with j being the inclusion of h into g � U.g/:If � W U.h/ ! U.g/ is the associated algebra homomorphism, show that � isinjective.Hint: Use the PBW theorem.

9.8 Exercises 265

4. Using the PBW theorem for b (applied to the basisZ1; : : : ; Zl ) and for g (appliedto the basis Y1; : : : ; Yk; Z1; : : : ; Zl ), establish first the existence and then theuniqueness of the expansion in (9.22).Hint: For the uniqueness result, first prove that if ˛ is a nonzero element of U.b/;then Y n11 � � �Y nkk ˛ is a nonzero element of U.g/; for any sequence n1; : : : ; nk ofnon-negative integers. Then prove that a linear combination as in (9.22) cannotbe zero unless each of the elements an1;:::nk in U.b/ is zero.

5. Let � be any element of h and let W� WD U.g/=I� be the Verma module withhighest weight �: Now let � be any other highest weight cyclic representationof g with highest weight �; acting on a vector space W�: Show that there is asurjective intertwining map � of V� ontoW�:

Note: It follows that W� is isomorphic to the quotient space V�= ker.�/: Thus,V� is maximal among highest weight cyclic representations with highest weight�; in the sense that every other such representation is a quotient of V�:Hint: If Q� is the extension of � to U.g/, as in Proposition 9.9, construct a map W U.g/ ! W� by mapping ˛ 2 U.g/ to Q�.˛/w0, where w0 is a highest weightvector for W�:

6. Let W� WD U.g/=I� be the Verma module with highest weight � and highestweight vector v0: Let X be the subspace of W� consisting of all those vectorsthat can be expressed as finite linear combinations of weight vectors. Show thatX contains v0 and is invariant under the action of g on W�: Conclude that W� isthe direct sum of its weight spaces.

7. Let � be any element of h and let W� be the associated Verma module. Suppose� 2 h can be expressed in the form

� D � � n1˛1 � � � � � nk˛k (9.27)

where ˛1; : : : ; ˛k are the positive roots and n1; : : : ; nk are non-negative integers.Show that the multiplicity of � in W� is equal to the number of ways that � � �

can be expressed as a linear combination of positive roots with non-negativeinteger coefficients. That is to say, the multiplicity of � is the number of k-tuplesof non-negative integers .n1; : : : ; nk/ for which (9.27) holds.

Chapter 10Further Properties of the Representations

In this chapter we derive several important properties of the representations weconstructed in the previous chapter. Throughout the chapter, g D kC denotes acomplex semisimple Lie algebra, h D tC denotes a fixed Cartan subalgebra of g;and R denotes the set of roots for g relative to h: We let W denote the Weyl group,we let denote a fixed base for R; and we let RC and R� denote the positive andnegative roots with respect to ; respectively.

10.1 The Structure of the Weights

In this section, we establish the general version of Theorems 6.24 and 6.25 in thecase of sl.3IC/; which tells us which integral elements appear as weights of afixed finite-dimensional irreducible representation. In Sect. 10.6, we will establish aformula for the multiplicities of the weights, as a consequence of the Weyl characterformula.

Recall that the weights of a finite-dimensional representation of g are integralelements (Proposition 9.2) and that the weights and their multiplicities are invariantunder the action of W (Theorem 9.3). We now determine which weights occur inthe representation with highest weight �: Recall from Definition 6.23 the notion ofthe convex hull of a collection of vectors.

A previous version of this book was inadvertently published without the middle initial of theauthor’s name as “Brian Hall”. For this reason an erratum has been published, correcting themistake in the previous version and showing the correct name as Brian C. Hall (see DOI http://dx.doi.org/10.1007/978-3-319-13467-3_14). The version readers currently see is the correctedversion. The Publisher would like to apologize for the earlier mistake.

© Springer International Publishing Switzerland 2015B.C. Hall, Lie Groups, Lie Algebras, and Representations, GraduateTexts in Mathematics 222, DOI 10.1007/978-3-319-13467-3_10

267

268 10 Further Properties of the Representations

2

2

2

2

2

2

2

2

3

3

3

3

3

3

3

3

4

4

4

44

4

4

4

5

5

5

5

5

5

5

5

6

6

6

66

6

6

6

Fig. 10.1 Typical weight diagram for the Lie algebra so.5IC/

Theorem 10.1. Let .�; V�/ be an irreducible finite-dimensional representation ofg with highest weight �: An integral element � is a weight of V� if and only if thefollowing two conditions are satisfied.

1. � belongs to the convex hull of the Weyl-group orbit of �:2. � � � can be expressed as an integer combination of roots.

Figure 10.1 shows a typical example for the Lie algebra so.5IC/: In the figure,the square lattice indicates the set of integral elements, the highest weight is circled,and the black dots indicate the weights of the representation. A number next toa dot indicates the multiplicity, with an unnumbered dot representing a weight ofmultiplicity 1. The multiplicities can be calculated using the Kostant multiplicityformula (Sect. 10.6). Note that the multiplicities do not have the sort of simplepattern that we saw in Sect. 6.7 for the case of sl.3IC/; that is, the multiplicitiesfor so.5IC/ are not constant on the “rings” in the weight diagram.

We will use the notation W � � to denote the Weyl group orbit of an element �;and the notation Conv.E/ to denote the convex hull of E: The following result isthe key step on the way to proving Theorem 10.1.

Proposition 10.2. Let � be a dominant integral element. Suppose � is dominant, �is lower than �; and � � � can be expressed as an integer combination of roots.Then � is a weight of the irreducible representation with highest weight �:

10.1 The Structure of the Weights 269

Lemma 10.3 (“No Holes” Lemma). Suppose .�; V / is a finite-dimensional repre-sentation of g: Suppose that � is a weight of V and that h�; ˛i > 0 for some root ˛:If we define j by

j D h�;H˛i D 2h�; ˛ih˛; ˛i ;

then ��k˛ is a weight of � for every integer k with 0 � k � j: In particular, ��˛is a weight of V:

Note that since � is integral, j must be a (positive) integer.

Proof. Let s˛ D hX˛; Y˛;H˛i be the copy of sl.2IC/ corresponding to the weight ˛(Theorem 7.19). LetU be the subspace of V spanned by weight spaces with weights� of the form � D � � k˛ for k 2 Z: Since X˛ and Y˛ shift weights by ˙˛; thespace U is invariant under s˛: Note that since h˛;H˛i D 2; we have

h� � k˛;H˛i D j � 2k:That is, the weight space corresponding to weight ��k˛ is precisely the eigenspacefor �.H˛/ inside U corresponding to the eigenvalue j � 2k:

By Point 4 of Theorem 4.34, if j > 0 is an eigenvalue for �.H˛/ inside U; thenall of the integers j � 2k; 0 � k � j; must also be eigenvalues for �.H˛/ inside U:Thus, � � k˛ must be a weight of � for 0 � k � j: Since j is a positive integer, jmust be at least 1, and, thus, � � ˛ must be a weight of �: ut

Figure 10.2 illustrates the results of the “no holes” lemma. For the indicatedweight �, the orthogonal projection of � onto ˛ equals .3=2/˛; so that j D 3: Thus,� � ˛; � � 2˛; and � � 3˛ must also be weights.

Proof of Proposition 10.2. Since � � � is an integer combination of roots, � � �

is also an integer combination of the positive simple roots ˛1; : : : ; ˛r : Since, also,� � �; we have

� D �CrX

jD1kj ˛j

for some non-negative integers k1; : : : ; kr : Consider now the following set P ofintegral elements,

P D8<:� D �C

rXjD1

lj ˛j

ˇˇˇ 0 � lj � kj

9=; : (10.1)

The elements of P form a discrete parallelepiped.We do not claim that every element of P is a weight of �; which is not, in

general, true (Figure 10.3). Rather, we will show that if � ¤ � is a weight of � in P;then there is another weight of V� in P that is “closer” to �: Specifically, for each

270 10 Further Properties of the Representations

Fig. 10.2 Since � is a weight, each of the circled elements must also be a weight

element � of P as in (10.1), let L.�/ D Pj lj : Starting from �; we will construct a

sequence of weights of V� in P with decreasing values of L.�/; until we reach onewith L.�/ D 0; which means that � D �:

Suppose then that � is a weight of � in P with L.�/ > 0: In that case, the secondterm in the formula for � is nonzero, and, thus,

*rX

jD1lj ˛j ;

rXkD1

lk˛k

+D

rXkD1

lk

*rX

jD1lj ˛j ; ˛k

+> 0:

Since each lk is non-negative, there must be some ˛k for which lk > 0 and for which

*rX

jD1lj ˛j ; ˛k

+> 0:

On the other hand, since � is dominant, h�; ˛ki � 0; and we conclude that h�; ˛ki >0: Thus, by the “no holes” lemma, � � ˛k must also be a weight of �:

Now, since lk is positive, lk � 1 is non-negative, meaning that � � ˛k is stillin P , where all the lj ’s are unchanged except that lk is replaced by lk � 1: Thus,L.� � ˛k/ D L.�/ � 1: We can then repeat the process starting with � � ˛k andobtain a sequence of weights of � with successively smaller values of L; until wereach L D 0; which corresponds to � D �: ut

Figure 10.3 illustrates the proof of Proposition 10.2. Starting at �; we look fora sequence of weights of � in P: Each weight in the sequence has positive inner

10.2 The Casimir Element 271

P

2

1

Fig. 10.3 The thick line indicates a path of weights in P connecting � to �

product either with ˛1 or with ˛2; allowing us to move in the direction of �˛1 or�˛2 to another weight of � in P; until we reach �:

Proof of Theorem 10.1. Let X � V� denote the span of all weight vectors whoseweights differ from � by a linear combination of roots. Then X is easily seen to beinvariant under the action of g and X contains v0; so X D V�: Thus, every weightof � must satisfy Point 2 of Theorem 10.1. Furthermore, if � is a weight of �; thenw � � � � for all w 2 W: Thus, by Proposition 8.44, � 2 Conv.W � �/:

Conversely, suppose that � satisfies the two conditions of Theorem 10.1. Wecan choose w 2 W so that �0 WD w � � is dominant. Clearly, �0 still belongs toConv.W � �/: Furthermore, since � is integral, w � � � � is an element of the rootlattice. After all, the definition of integrality implies that s˛ � � � � is an integermultiple of ˛; since the s˛’s generate W; the result holds for all w 2 W: Thus,� � �0 D � � � C � � �0 is an element of the root lattice, which means that �0also satisfies the two conditions of Theorem 10.1. Thus, by Proposition 10.2, �0 is aweight of �; which means that � D w�1 � �0 is also a weight. ut

10.2 The Casimir Element

In this section, we construct an element of U.g/ known as the Casimir, whichbelongs to the center of U.g/: The Casimir element is important in its own rightand also plays a crucial role in the proof of complete reducibility (Sect. 10.3) and ofthe Weyl character formula (Sect. 10.8).

272 10 Further Properties of the Representations

Definition 10.4. Let Xj be an orthonormal basis for k: Then the Casimir elementC of U.g/ is given by

C D �Xj

X2j :

Proposition 10.5. 1. The value of C is independent of the choice of orthonormalbasis for k.

2. The element C is in the center of U.g/:

Proof. If fXj g and fYj g are two different orthonormal bases for k; then there is anorthogonal matrix R such that

Yj DXk

RkjXk:

Then

Xj

Y 2j DXj;k;l

RkjXkRljXl

DXk;l

Xj

Rkj.Rtr/jlXkXl

DXk;l

ıklXkXl

DXk

X2k :

This shows that C is independent of the choice of basis.Meanwhile, if fXj g is an orthonormal basis, let cjkl be the associated structure

constants:

ŒXj ;Xk� DXl

cjklXl:

Note that for a fixed j; the matrix Aj given by .Aj /kl D cjlk is the matrixrepresenting the operator adXj in the chosen basis. Since the inner product on kis Ad-K-invariant, adXj is a skew operator, which means that cjkl is skew symmetricin k and l for a fixed j: If we compute the commutator of some Xj with C in U.g/;we obtain

ŒXj ; C � DXk

ŒXj ;X2k �

DXk

.ŒXj ;Xk�Xk CXkŒXj ;Xk�/

DXk;l

cjklXlXk CXk;l

cjklXkXl : (10.2)

10.2 The Casimir Element 273

In the first sum in the last line of (10.2), we may reverse the labeling of thesummation variables and use the skew symmetry of cjkl in k and l to obtain

ŒXj ; C � DXk;l

.�cjkl C cjkl/XlXk D 0:

Thus, C commutes the each Xj : But since U.g/ is generated by elements of g; wesee that C actually commutes with every element of U.g/: ut

Let � be a finite-dimensional, irreducible representation of g:By Proposition 9.9,we can extend � to a representation of U.g/; which we also denote by � . Wenow show �.C / is a constant multiple of the identity. The formula for the constantinvolves the element ı (Definition 8.37), equal to half the sum of the positive roots.The element ı also arises in our discussion of the Weyl character formula, the Weyldimension formula, and the Kostant multiplicity formula.

Proposition 10.6. Let .�; V / be a finite-dimensional irreducible representation ofg (extended to U.g/) with highest weight �: Then we have

�.C / D �Xj

�.Xj /2 D c�I;

where c� is a constant given by

c� D h�C ı; �C ıi � hı; ıi :Furthermore, c� � 0 with c� D 0 only if � D 0:

Lemma 10.7. LetX 2 g˛ be a unit vector, so thatX� 2 g�˛: Then under our usualidentification of h with h�; we have

ŒX;X�� D ˛:

Proof. According to Lemma 7.22, we have

hŒX;X��;H˛i D h˛;H˛i hX�; X�i D h˛;H˛i ; (10.3)

since X (and thus, also, X�) is a unit vector. On the other hand, we know that thecommutator of any element of g˛ with any element of g�˛ is a multiple ofH˛;whichis (under our identification of h with h�) a multiple of ˛:But if ŒX;X�� D c˛; (10.3)tells us that c must equal 1. utProof of Proposition 10.6. Since C is in the center of the universal envelopingalgebra, �.C / commutes with each �.X/;X 2 g: Thus, by Schur’s lemma, �.C /must act as a constant multiple c� of the identity operator.

To compute the constant c�; we choose an orthonormal basis for k as follows.Take an orthonormal basis H1; : : : ;Hj for t: Then for each ˛ 2 RC; choose a unitvector X˛ in g˛; so that X � is a unit vector in g�˛: Then the elements

Y˛ WD .X˛ CX �/=.p2i/I Z˛ WD .X˛ �X �/=

p2;

274 10 Further Properties of the Representations

satisfy Y � D �Y˛ and Z� D �Z˛; which shows that these elements belong to k:Since g˛ is orthogonal to g�˛; it is easy to see that these vectors are also unit vectorsand orthogonal to each other. The set of vectors of the form Hj ; j D 1; : : : ; r; andY˛; ˛ 2 RC; and Z˛; ˛ 2 RC; form an orthonormal basis for k:

We compute that

Y 2˛ D �12.X2

˛ CX˛X� CX �X˛ C .X �/2/

Z2˛ D 1

2.X2

˛ �X˛X � � X �X˛ C .X �/2/;

so that

�Y 2˛ �Z2˛ D X˛X

� CX �X˛D 2X �X˛ C ŒX˛;X

��:

Thus, the Casimir element C may be computed as

C DX˛2RC

.2X �X˛ C ŒX˛;X��/ �

rXjD1

H2j :

Suppose now that v is a highest weight vector, and compute that

�.C /v D �rX

jD1�.Hj /

2v C 2X˛2RC

�.X �/�.X˛/v

CX˛2RC

�.ŒX˛;X��/v:

Since v is a highest weight vector, �.X˛/v D 0; and since Hj 2 t � h; we

have �.Hj /2v D ˝

�;Hj

˛2v: Now, since the roots live in i t and h�; ˛i is real for

all roots ˛; we see that � also lives in i t: Thus,˝�;Hj

˛is pure imaginary and˝

�;Hj

˛2 D � ˇ˝�;Hj

˛ˇ2: Using Lemma 10.7, we then see that

�.C /v D0@ rXjD1

ˇ˝�;Hj

˛ˇ2 CX˛2RC

h�; ˛i1A v; (10.4)

where the coefficient of v on the right-hand side of (10.4) must equal c�:Now, since fHj grjD1 is an orthonormal basis for t; the first term in the coefficient

of v equals h�;�i. Moving the sum over ˛ inside the inner product in the secondterm gives

c� D h�;�i C h�; 2ıi ; (10.5)

10.3 Complete Reducibility 275

which is the same as h�C ı; �C ıi � hı; ıi : Finally, we note that since � isdominant, h�; ˛i � 0 for every positive root, from which it follows that h�; 2ıi DP

˛2RC h�; ˛i is non-negative. Thus, c� � 0 for all � and c� > 0 if � ¤ 0: ut

10.3 Complete Reducibility

Let g be a complex semisimple Lie algebra. Then g is isomorphic to the complexi-fication of a the Lie algebra of a compact matrix Lie group K: (This is, for us, trueby definition; see Definition 7.1.) Actually, it is possible to show that there existsa simply connected compact Lie group K with Lie algebra k such that g Š kC:(This claim follows from Theorems 4.11.6 and 4.11.10 of [Var].) Assuming thisresult, we can see that every finite-dimensional representation � of g gives riseto a representation of K by restricting � to k and then applying Theorem 5.6.Theorem 4.28 then tells us that every finite-dimensional representation of g iscompletely reducible; compare Corollary 4.11.11 in [Var].

Rather than relying on the existence of a simply connected K; we now givean algebraic proof of complete reducibility. Our proof makes use of the Casimirelement and, in particular, the fact that the eigenvalue of the Casimir is nonzero ineach nontrivial irreducible representation.

Proposition 10.8. If .�; V / is a one-dimensional representation of g; then�.X/ D 0 for all X 2 g:

Proof. By Theorem 7.8, g decomposes as a Lie algebra direct sum of simplealgebras gj : Since the kernel of �jgj is a ideal, the restriction of � to gj must beeither zero or injective. But since dim gj � 2 and dim.End.V // D 1; this restrictioncannot be injective, so it must be zero for each j: utTheorem 10.9. Every finite-dimensional representation of a semisimple Lie alge-bra is completely reducible.

We begin by considering what appears to be a very special case of the theorem.

Lemma 10.10. Suppose .�; V / is a finite-dimensional representation of g and thatW is an invariant subspace of V of codimension 1. Then V decomposes asW ˚ U

for some invariant subspace U of V:

Proof. We consider first the case in which W is irreducible. If W is one-dimensional, then by Proposition 10.8, the restriction of � to W is zero. SinceW isone-dimensional and has codimension 1, V must be two dimensional. The space oflinear operators on V that are zero on W then has dimension 2. (Pick a basis for Vconsisting of a nonzero element of W and another linearly independent vector; inthis basis, any such operator will have first column equal to zero.) On the other hand,each of the simple summands gj in the decomposition of g has dimension at least3, since an algebra of dimension 2 cannot be simple. (An algebra of dimension 2 iseither commutative or has a commutator ideal of dimension 1.) Thus, the restriction

276 10 Further Properties of the Representations

of � to gj cannot be injective and thus must be zero. We conclude, then, that � isidentically zero, and we may take U to be any subspace of V complementary to W:

Assume now that W is irreducible and nontrivial. Let C be the Casimir elementof U.g/ and let �.C / denote the action of C on V; by means of the extension of �to U.g/: By Proposition 10.6, the restriction of �.C / to W is a nonzero multiple cof the identity. On the other hand, since V=W is one dimensional, the action of gon V=W is trivial, by Proposition 10.8. Thus, the action of �.C / on V=W is zero,from which it follows that �.C / must have a nonzero kernel. (If we pick a basis forV consisting of a basis for W together with one other vector, the bottom row of thematrix of �.C / in this basis be identically zero.) Because �.C / commutes with each�.X/; this kernel is an invariant subspace of V: Furthermore, ker.�.C //\W D f0g;because �.C / acts as a nonzero scalar on W: Thus, U WD ker.�.C // is the desiredinvariant complement to W:

We consider next the case in which W has a nontrivial invariant subspace W 0;which is, of course, also an invariant subspace of V: ThenW=W 0 is a codimension-one invariant subspace of V=W 0: Thus, by induction on the dimension of W; wemay assume that W=W 0 has a one-dimensional invariant complement, say Y=W 0:Then W 0 is a codimension-one invariant subspace of Y: Since dimW 0 < dimW;we may apply induction again to find a one-dimensional invariant complementU toW 0 in Y; so that Y D W 0 ˚ U: Now, Y \W D W 0 and U \W 0 D 0, from whichit follows that U \W D f0g: Thus, U is the desired complement to W in V: utProof of Theorem 10.9. Let .�; V / be a finite-dimensional representation of gand let W be nontrivial invariant subspace of V: We now look for an invariantcomplement to W; that is, an invariant subspace U such that V D W ˚ U: If wecan always find such an U; then we may proceed by induction on the dimension toestablish complete reducibility. If A W V ! W is an intertwining map, the kernelof A will be an invariant subspace of V: If, in addition, the restriction of A to W isinjective, then ker.A/ \W D f0g and, by a dimension count, V will decompose asW ˚ ker.A/: Now, the simplest way to ensure that A is injective onW is to assumethat AjW is a nonzero multiple of the identity (If, for example, the restriction ofA to W is the identity, we may think of A as a “projection” of V onto W:) If Wis irreducible, nontrivial, and of codimension 1, we may take A D �.C / as in theproof of Lemma 10.10.

To construct A in general, we proceed as follows. Let Hom.V;W / denote thespace of linear maps of V to W (not necessarily intertwining maps). This space canbe viewed as a representation of g by means of the action

X � A D �.X/A� A�.X/ (10.6)

for all X 2 g and A 2 Hom.V;W /: Let V denote the subspace of Hom.V;W /consisting of those maps A whose restriction to W is a scalar multiple cA of theidentity and let W denote the subspace of Hom.V;W / consisting of those mapswhose restriction to W is zero. The map A 7! cA is a linear functional on V whichis easily seen not to be identically zero. The space W is the kernel of this linearfunctional and is, thus, a codimension-one subspace of V :

10.4 The Weyl Character Formula 277

We now claim that both V and W are invariant subspaces of Hom.V;W /: To seethis, suppose A 2 Hom.V;W / is equal to cI on W: Then for w 2 W; we have

.X � A/w D �.X/Aw �A�.X/wD �.X/cw � c�.X/w

D 0;

because �.X/w is again inW; showing that X �A is actually in W : Thus, the actionof g maps V into W ; showing that both V and W are invariant.

Since W has codimension 1 in V ; we are in the situation of Lemma 10.10. Thus,there exists an invariant complementU to W in V : LetA be a nonzero element of U :SinceA is not in W ; the restriction ofA toW is a nonzero scalar. Furthermore, sinceU is one dimensional, Proposition 10.8 tells us that the action of g on A is zero,meaning that A commutes with each �.X/: That is to say, A is an intertwining mapof V to W: Thus, A is precisely the operator we were looking for: an intertwiningmap of V to W whose restriction to W is a nonzero multiple of the identity. Now,A maps V into W; and actually maps onto W; since A acts as a nonzero scalar onW itself. Thus, dim.ker.A// D dimV � dimW and since ker.A/ \W D f0g; weconclude that U WD ker.A/ is an invariant complement to W:

We conclude, then, that every nontrivial invariant subspace W of V has aninvariant complement U . By induction on the dimension, we may then assumethat both W and U decompose as direct sums of irreducible invariant subspaces,in which case, V also has such a decomposition. ut

10.4 The Weyl Character Formula

The character formula is a major result in the structure of the irreducible represen-tations of g: Its consequences include a formula for the dimension of an irreduciblerepresentation (Sect. 10.5) and a formula for the multiplicities of the weights in anirreducible representation (Sect. 10.6).

We now introduce the notion of the character of a finite-dimensional representa-tion of a group.

Definition 10.11. Suppose .�; V / is a finite-dimensional representation of a com-plex semisimple Lie algebra g: The character of � is the function �� W g ! C

given by

��.X/ D trace.e�.X//:

It turns out that the character of � encodes many interesting properties of �: Wewill give a formula for the character of an irreducible representation of a semisimpleLie algebra in terms of the highest weight of the representation. This section givesthe statement of the character formula, Sects. 10.5 and 10.6 give consequences of it,and Sect. 10.8 gives the proof.

278 10 Further Properties of the Representations

If the representation � of g D kC comes from a representation… of the compactgroupK; then forX 2 k; we have ��.X/ D trace.….eX//: In Chapter 12, the groupversion of a character, namely, the function given by x 7! trace.….x//; x 2 K;

plays a key role in the compact group approach to representation theory.

Proposition 10.12. If .�; V / is a finite-dimensional representation of g; we havethe following results.

1. The dimension of V is equal to the value of �� at the origin:

dim.V / D ��.0/:

2. Suppose V decomposes as a direct sum of weight spaces V� with multiplicitymult.�/: Then forH 2 h we have

��.H/ DX�

mult.�/eh�;H i: (10.7)

Proof. The first point holds because trace.I / D dimV . Meanwhile, since �.H/acts as h�;H i I in each weight space V�; the second point follows from thedefinition of ��: utExample 10.13. Let � denote the irreducible representation of sl.2IC/ of dimen-sion mC 1 and let

H D�1 0

0 �1�:

Then

�� .aH/ D ��

��a 0

0 �a��

D ema C e.m�2/a C � � � C e�ma: (10.8)

We may also compute �� as

��.aH/ D sinh..mC 1/a/

sinh.a/(10.9)

whenever a not an integer multiple of i�:

Proof. The eigenvalues of�m.H/ arem;m�2; : : : ;�m; from which (10.8) follows.To obtain (10.9), we note that

.ea � e�a/��.aH/

D e.mC1/a C e.m�1/a C � � � C e�.m�1/a

� e.m�1/a � � � � � e�.m�1/a � e�.mC1/a

D e.mC1/a � e�.mC1/a; (10.10)

10.4 The Weyl Character Formula 279

so that

��.aH/ D e.mC1/a � e�.mC1/a

ea � e�a D sinh..mC 1/a/

sinh.a/;

as claimed. utNote that in deriving (10.9) from (10.8), we multiplied the character by a cleverly

chosen combination of exponentials (ea � e�a), leading to a large cancellation, sothat only two terms remain in (10.10). The Weyl character formula asserts thatwe can perform a similar trick for the characters of irreducible representations ofarbitrary semisimple Lie algebras.

Recall that each element w of the Weyl group W acts as an orthogonallinear transformation of i t � h: We let det.w/ denote the determinant of thistransformation, so that det.w/ D ˙1: Recall from Definition 8.37 that ı denoteshalf the sum of the positive roots. We are now ready to state the main result of thischapter.

Theorem 10.14 (Weyl Character Formula). If .�; V�/ is an irreducible represen-tation of g with highest weight �; then

��.H/ DP

w2W det.w/ehw�.�Cı/;H iPw2W det.w/ehw�ı;H i (10.11)

for all H 2 h for which the denominator is nonzero.

Since we will have frequent occasion to refer to the function in the denominatorin (10.11), we give it a name.

Definition 10.15. Let q W h ! C be the function given by

q.H/ DXw2W

det.w/ehw�ı;H i:

The function q is called the Weyl denominator.

The character formula may also be written as

q.H/��.H/ DXw2W

det.w/ehw�.�Cı/;H i: (10.12)

Let us pause for a moment to reflect on what is going on in (10.12). The Weyldenominator q.H/ is a sum of jW j exponentials with coefficients equal to ˙1:Meanwhile, the character��.H/ is a large sum of exponentials with positive integercoefficients, as in (10.7). When we multiply these two functions, we seeminglyobtain an even larger sum of exponentials of the form

ehw�ıC�;H i;

for w 2 W and � a weight of �; with integer coefficients.

280 10 Further Properties of the Representations

The character formula, however, asserts that most of these terms are not actuallypresent. Specifically, the only exponentials that actually appear are those of the formehw�.�Cı/;H i; which occurs with a coefficient of det.w/: The point is that, in mostcases, if a weight � can be written in the form � D w � ı C �; with � a weight of�; then � can be written in this form in more than one way. The character formulaasserts that unless � is in the Weyl-group orbit of ı C �; the coefficient of eh�;H i;after all the different contributions are taken into account, ends up being zero.

By contrast, the weight � D ıC� only occurs once, since it corresponds to takingthe highest weight occurring in q (namely ı) and the highest weight occurring in ��(namely the highest weight � of �). (Note that by Propositions 8.38 and 8.27, theelements of the form w � ı are all distinct. Then by Proposition 8.42, w � ı � ı for allw:) Furthermore, since the weights occurring in both q and �� are Weyl invariant,the weight w � .ı C �/ also occurs only once. The Weyl character formula, then,can be expressed as stating that if we compute the product q�� ; a huge cancellationoccurs: Every exponential in the product ends canceling out to zero, except for thosethat occur only once, namely those of the form w � .ı C �/:

In the case of sl.2IC/; we have already observed how this cancellation occurs,in (10.10). The Weyl denominator is equal to ea�e�a in this case. Each exponentialela occurring in the product .ea�e�a/��.aH/will occur once with a plus sign (fromeae.l�1/a) and once with a minus sign (from e�ae.lC1/a), except for the extremecases l D ˙.mC1/: This cancellation occurs because the multiplicity of the weightl � 1 equals the multiplicity of the weight l C 1; namely 1.

Figure 10.4, meanwhile, illustrates the case of the irreducible representation ofsl.3IC/ with highest weight .1; 2/: The top part of the figure indicates the sixexponentials occurring in the Weyl denominator, with alternating signs. The middlepart of the figure indicates the exponentials in the character of the representationwith highest weight .1; 2/: The bottom part of the figure shows the product of theWeyl denominator and the character, in which only the six exponentials indicatedby black dots survive. The white dots in the bottom part of the figure indicateexponentials that occur at least once in the product, but which end up with acoefficient of zero.

The cancellation inherent in the Weyl character formula reflects a very specialstructure to the multiplicities of the various weights that occur. For an integralelement �; each product of the form ehw�ı;H ieh�;H i; where � D � � w � ı; makesa contribution of det.w/mult.�/ to the coefficient of eh�;H i: Thus, if � is not in theWeyl orbit of �C ı; the Weyl character formula implies that

Xw2W

det.w/mult.� � w � ı/ D 0; � … W � .�C ı/: (10.13)

(In (10.13), some of elements of the form � � w � ı may not actually be weights of�; in which case, the multiplicity should be considered to be zero.) In the case ofsl.3IC/; the weights of the form � � w � ı form a small hexagon around the weight�: Figure 10.5 illustrates how the alternating sum of multiplicities around one suchhexagon equals zero.

10.4 The Weyl Character Formula 281

Fig. 10.4 The product of theWeyl denominator (top) andthe character of arepresentation (middle)produces an alternating sumof exponentials (bottom). Thewhite dots indicateexponentials that occur atleast once in the product butthat end up with a coefficientof zero

1

1

1

1

1

1

22

22

22

1

1

1

1

1

1

We will see in Sect. 10.6 that the character formula leads to a formula for themultiplicities of all the weights occurring in a particular irreducible representation.

Before concluding this section, we establish a technical result that we will use inthe remainder of the chapter.

Proposition 10.16. The exponential functions H 7! eh�;H i; with � 2 h, arelinearly independent in C1.h/:

282 10 Further Properties of the Representations

22

22

22

Fig. 10.5 We compute the alternating sum of multiplicities around the hexagon enclosing thecircled weight, beginning in the fundamental Weyl chamber and proceeding counterclockwise.The result is 1� 1C 1� 2C 2� 1 D 0

The proposition means that if a function f 2 C1.h/ can be expressed as a finitelinear combination of exponentials, it has a unique such expression.

Proof. We need to show that if the function f W h ! C given by

f .H/ D c1eh�1;H i C � � � C cne

h�n;H i

is identically zero, where �1; : : : ; �n are distinct elements of h; then c1 D � � � Dcn D 0: If n D 1;we evaluate atH D 0 and conclude that c1 must be zero. If n > 1;we choose, for each k D 2; : : : ; n; some Hk 2 h such that h�1;Hki ¤ h�k;Hki :Since f is identically zero, so is the function

g WD .DH2 � h�2;H2i/ � � � .DHn � h�n;Hni/f

whereDX denotes the directional derivative in the direction of X :

.DXf /.H/ D d

dtf .H C tX/

ˇˇtD0

: (10.14)

10.5 The Weyl Dimension Formula 283

Direct calculation then shows that

g.H/ DnX

jD1cj

nY

kD2.˝�j ;Hk

˛ � h�k;Hki/!eh�j ;Hi

D c1

nY

kD2.h�1;Hki � h�k;Hki/

!eh�1;H i: (10.15)

By evaluating at H D 0 and noting that, by construction, the product in the secondline of (10.15) is nonzero, we conclude that c1 D 0: An entirely similar argumentthen shows that each cj D 0 as well. utCorollary 10.17. Suppose f 2 C1.h/ can be expressed as a finite linearcombination of exponentials eh�;H i; with � integral,

f .H/ DX�

c�eh�;H i:

If f satisfies f .w �H/ D det.w/f .H/; then

cw�� D det.w/c�

for each � occurring in the expansion of f:

Proof. On the one hand,

f .w �H/ DX�

c�ehw�1��;Hi D

X�

cw��eh�;H i: (10.16)

On the other hand,

f .w �H/ D det.w/f .H/ DX�

det.w/c�eh�;H i: (10.17)

By the linear independence of the exponentials, the only way the expansionsin (10.16) and (10.17) can agree is if cw�� D det.w/c� for all �: ut

10.5 The Weyl Dimension Formula

Before coming to the proof of the Weyl character formula, we derive two importantconsequences of it, the Weyl dimension formula (described in this section) and theKostant multiplicity formula (described in the next section).

The dimension of a representation is equal to the value of the character at theidentity (Proposition 10.12). In the Weyl character formula, however, both thenumerator and the denominator are equal to zero when H D 0: In the case ofsl.2IC/ case, for example, the character formula reads

284 10 Further Properties of the Representations

��.aH/ D sinh..mC 1/a/

sinh a:

The limit of this expression as � tends to zero may be computed by l’Hospital’s ruleto be mC 1; which is, of course, the dimension of Vm:

In the general case, we will expand both numerator and denominator of thecharacter formula in a power series. We will see that in both numerator anddenominator, the first nonzero term has degree k; where

k D the number of positive roots.

To evaluate the limit of this expression at the origin, we will develop a version ofl’Hospital’s rule. The limit is then computed as the ratio of a certain k-fold derivativeof the numerator and the corresponding k-fold derivative of the denominator,evaluated at the origin. The result of this analysis is expressed in the followingtheorem.

Theorem 10.18. If .��; V�/ is the irreducible representation of g with highestweight �; then the dimension of V� may be computed as

dim.V�/ DQ˛2RC h˛;�C ıiQ

˛2RC h˛; ıi :

Note that both �C ı and ı are strictly dominant elements, so that all the factorsin both the numerator and the denominator are nonzero.

A function P on h is called a polynomial if for every basis of h; the functionP is a polynomial in the coordinates z1; : : : ; zr associated to that basis. That is tosay, P should be expressible as a finite linear combination of terms of the formzn11 zn22 � � � znrr ; where n1; : : : ; nr are non-negative integers. It is easy to see that if Pis a polynomial in any one basis, then it is also a polynomial in every other basis aswell. A polynomial P is said to be homogeneous of degree l if P.cH/ D clP.H/

for all constants c and all H 2 h:

Definition 10.19. Let P W h ! C be the function given by

P.H/ DY˛2RC

h˛;H i :

Note that P is a product of k linear functions and is, thus, a homogeneouspolynomial of degree k: The dimension formula may be restated in terms of P as

dim.V�/ D P.�C ı/

P.ı/:

A key property of the polynomial P is its behavior under the action of the Weylgroup.

10.5 The Weyl Dimension Formula 285

Definition 10.20. A function f W h ! C is said to be Weyl alternating if

f .w �H/ D det.w/f .H/

for all w 2 W andH 2 h:

It is easy to see, for example, that the Weyl denominator q is Weyl alternating(Exercise 3).

Proposition 10.21. 1. The function P is Weyl alternating.2. If f W h ! C is a Weyl-alternating polynomial, there is a polynomial g W h ! C

such that

f .H/ D P.H/g.H/:

In particular, if f is homogeneous of degree l < k, then f must be identicallyzero, and if f is homogeneous of degree k; then f must be a constant multipleof P:

Proof. For any w 2 W; consider the collection of roots of the form w�1 � ˛ for˛ 2 RC: Since RC contains exactly one element out of each pair ˙˛ of roots, thesame is true of the collection of w�1 �˛’s, with w fixed and ˛ varying overRC: Thus,

P.w �H/ DY˛2RC

h˛;w �H i

DY˛2RC

˝w�1 � ˛;H ˛

D .�1/jY˛2RC

h˛;H i ;

where j is the number of negative roots in the collection fw�1 � ˛g˛2RC :

Suppose first that w D w�1 D s˛; where ˛ is a positive simple root. Accordingto Proposition 8.30, s˛ permutes the positive roots different from ˛; whereass˛ � ˛ D �˛. Thus, j D 1 in this case, and so

P.w �H/ D �P.H/ D det.w/P.H/;

since the determinant of a reflection is �1: By Proposition 8.24, every element w ofW is a product of reflections associated to positive simple roots, and so

P.w �H/ D P.s˛j1 � � � s˛jN �H/D .�1/NP.H/D det.s˛j1 � � � s˛jN /P.H/;

showing that P is alternating.

286 10 Further Properties of the Representations

Suppose now that f is any Weyl-alternating polynomial. Then for any positiveroot ˛; if h˛;H i D 0; we will have

f .H/ D f .s˛ �H/ D �f .H/;

since det.s˛/ D �1: Thus, f must vanish on the hyperplane orthogonal to ˛; whichwe denote as V˛: It is then not hard to show (Exercise 4) that f is divisible in thespace of polynomials by the linear function h˛;H i ; that is, f .H/ D h˛;H i f 0.H/for some polynomial f 0: Now, if ˇ is any positive root different from ˛; thepolynomial f 0 must vanish at least on the portion of Vˇ not contained in V˛: Butsince ˇ is not a multiple of ˛; Vˇ is distinct from V˛; so that Vˇ \ V˛ is a subspaceof dimension r � 2: Thus, Vˇ � .V˛ \ Vˇ/ is dense in Vˇ: Since f 0 is continuous, itmust actually vanish on all of Vˇ:

It follows that f 0 is divisible in the space of polynomials by hˇ;H i ; so that

f .H/ D h˛;H i hˇ;H i f 00.H/

for some polynomial f 00: Proceeding on in the same way, we see that f contains afactor of h˛;H i for each positive root ˛; meaning that

f .H/ D0@ Y˛2RC

h˛;H i1A g.H/

D P.H/g.H/

for some polynomial g; as claimed. utRecall the notion of directional derivativeDX , defined in (10.14).

Lemma 10.22. Let A denote the differential operator

A DY˛2RC

D˛:

For any � 2 h; let f� W h ! C be the function given by

f�.H/ DXw2W

det.w/ehw��;H i:

Then f� is Weyl alternating and is given by a convergent power series of the form

f�.H/ D c�P.H/C terms of degree at least k C 1 (10.18)

10.5 The Weyl Dimension Formula 287

for some constant c�: Furthermore, .AP/.0/ ¤ 0 and the constant c� may becomputed as

c� D .Af�/.0/.AP/.0/ ;

where

.Af�/.0/ D jW jP.�/:

Proof. The proof that f� is Weyl alternating is elementary. Since f� is a sum ofexponentials, it is a real-analytic function, meaning that it can be expanded in aconvergent power series in the coordinates x1; : : : ; xr associated to any basis forh: In the power-series expansion of f�; we collect together all the terms that arehomogeneous of degree l: Thus, f� is expressible as the sum of homogeneouspolynomials q�;l of degree l: Since f� is Weyl-alternating, it is not hard to show(Exercise 5) that each of the polynomials q�;l is also Weyl-alternating. Thus, byProposition 10.21, all the polynomials q�;l with l < k must zero, and the polynomialq�;k must be a constant multiple of P.H/: This establishes the claimed form of theseries for f�:

On the one hand, applying A to a homogeneous term of degree l > k gives ahomogeneous term of degree l � k > 0; which will evaluate to zero at the origin.Thus,

.Af�/.0/ D c�.AP/.0/:On the other hand, by directly differentiating the exponentials in the definition off�, we get

.Af�/.0/ DXw2W

det.w/Y˛2RC

hw � �; ˛i

DXw2W

det.w/P.w � �/

D jW jP.�/:

This shows that

c�.AP/.0/ D jW jP.�/: (10.19)

Now, if � is strictly dominant, each factor in the formula for P.�/ is nonzero, sothat P.�/ ¤ 0: Applying (10.19) in such a case shows that .AP/.0/ ¤ 0: We canthus solve (10.19) for c� to obtain the claimed formula. ut

288 10 Further Properties of the Representations

Proof of the Weyl Dimension Formula. As we have noted already, P.� C ı/ andP.ı/ are nonzero, so that c�Cı and cı are also nonzero. For anyH 2 h; we have

��.tH/ D f�Cı.tH/fı.tH/

D c�CıtkP.H/CO.tkC1/cıtkP.H/CO.tkC1/

D c�CıP.H/CO.t/

cıP.H/CO.t/(10.20)

for any t for which the numerator of the last expression is nonzero.Now, we know from the definition of a character, that �� is a continuous function.

To determine the value of �� at the identity, we choose H to be in the openfundamental Weyl chamber, so that P.H/ ¤ 0; in which case the denominatorin (10.20) is nonzero for all sufficiently small nonzero t: Thus,

dim.V�/ D limt!0

��.tH/

D c�Cıcı

D P.�C ı/

P.ı/;

where we have used the formula for c� in Lemma 10.22. Recalling the definition ofP gives the dimension formula as stated in Theorem 10.18. utExample 10.23. If �1 and �2 denote the two fundamental weights for sl.3IC/ thenthe dimension of the representation with highest weightm1�1 Cm2�2 is given by

1

2.m1 C 1/.m2 C 1/.m1 Cm2 C 2/:

See Exercises 7 and 8 for the analogous formulas for B2 and G2:

Proof. Note that scaling the inner product on h by a constant does not affect theright-hand side of the dimension formula, since the inner product occurs an equalnumber of times in the numerator and denominator. Let us then normalize the innerproduct so that all roots ˛ satisfy h˛; ˛i D 2: With this normalization,H˛ D ˛ andwe have

m1 D h�;H1i D h˛1; �im2 D h�;H2i D h˛2; �i :

10.6 The Kostant Multiplicity Formula 289

Letting ˛3 D ˛1 C ˛2; we have ı D 12.˛1 C ˛2 C ˛3/ D ˛3: We then note that

h˛1; ıi D 1; h˛2; ıi D 1; and h˛3; ıi D 2: Thus, the numerator in the dimensionformula is

.h˛1; �i C h˛1; ıi/ .h˛2; �i C h˛2; ıi/ .h˛3; �i C h˛3; ıi/D .m1 C 1/.m2 C 1/.m1 Cm2 C 2/

and the denominator is .1/.1/.2/: ut

10.6 The Kostant Multiplicity Formula

We will obtain Kostant’s multiplicity formula from the Weyl character formula bydeveloping a method for dividing by the Weyl denominator q:We now illustrate thismethod in the case of sl.2IC/; where the character formula takes the form

�m.aH/ D e.mC1/a � e�.mC1/a

ea � e�a : (10.21)

One way to divide the numerator of (10.21) by the denominator is to use thegeometric series 1=.1 � x/ D 1 C x C x2 C � � � : Applying this formally withx D e�2a gives the following result:

1

ea � e�a D 1

ea.1 � e�2a/D e�a.1C e�2a C e�4a C � � � /: (10.22)

If the real part of a is negative, the series will not converge in the ordinarysense. Nevertheless, if we treat the right-hand side (10.22) as simply a formal series,then (10.22) holds in the sense that if we multiply the right-hand side by ea � e�a;we get 1. Using (10.22), we have

e.mC1/a � e�.mC1/a

ea � e�a

D e�ae.mC1/a.1C e�2a C � � � / � e�ae�.mC1/a.1C e�2a C � � � /D .ema C e.m�2/a C � � � /� .e�.mC2/a C e�.mC4/a C � � � /D ema C e.m�2/a C � � � C e�ma:

This last expression is, indeed, the character of Vm: From this formula, we can readoff that each weight of Vm has multiplicity 1 (Proposition 10.12).

We now develop a similar method for dividing by the Weyl denominator in thegeneral case.

290 10 Further Properties of the Representations

Definition 10.24. A formal exponential series is a formal series of the form

X�

c�eh�;H i;

where each � is an integral element and where the c�’s are complex numbers.A finite exponential series is a series of the same form where all but finitely manyof the c�’s are equal to zero.

Since we make no restrictions on the coefficients c�; a formal exponential seriesmay not converge. Thus, the series should properly be thought of not as a functionof H but simply as a list of coefficients c�: If f is a formal exponential series withcoefficients c� and g is a finite exponential series with coefficients d�, the productof f and g is a well-defined formal exponential series with coefficients e� given by

e� DX�0

c���0d�0 : (10.23)

Note that only finitely many of the terms in (10.23) are nonzero, since g is a finiteexponential series. (The product of two formal exponential series is, in general, notdefined, because the sum defining the coefficients in the product might be divergent.)

Definition 10.25. If � is an integral element, we let p.�/ denote the number ofways (possibly zero) that � can be expressed as a non-negative integer combinationof positive roots. The function p is known as the Kostant partition function.

More explicitly, if the positive roots are ˛1; : : : ; ˛k; then p.�/ is the number ofk-tuples of non-negative integers .n1; : : : ; nk/ such that n1˛1 C � � � C nk˛k D �:

Example 10.26. If ˛1 and ˛2 are the two positive simple roots for sl.3IC/; then forany two non-negative integersm1 andm2; we have

p.m1˛1 Cm2˛2/ D 1C min.m1;m2/:

If � is not a non-negative integer combination of ˛1 and ˛2; then p.�/ D 0:

The result of Example 10.26 is shown graphically in Figure 10.6.

Proof. We have the two positive simple roots ˛1 and ˛2; together with one otherpositive root ˛3 D ˛1 C ˛2: Thus, if � can be expressed as a non-negative integercombination of ˛1 C ˛2 C ˛3, we can rewrite ˛3 as ˛1 C ˛2 to express � as � Dm1˛1 Cm2˛2; with m1;m2 � 0: Every expression for � is then of the form

� D .m1 � k/˛1 C .m2 � k/˛2 C k˛3;

for 0 � k � min.m1;m2/: utWe are now ready to explain how to invert the Weyl denominator.

10.6 The Kostant Multiplicity Formula 291

1

1

1

1

1

1

1

1

2

2

2

2

2

2

3

3

3

3

3

3

4

Fig. 10.6 The Kostant partition function p for A2: Unlabeled points � have p.�/ D 0

Proposition 10.27 (Formal Reciprocal of the Weyl Denominator). At the levelof formal exponential series, we have

1

q.H/DX��0

p.�/e�h�Cı;H i: (10.24)

Here the sum is nominally over all integral elements � with � � 0; but p.�/ D 0

unless � is an integer combination of roots.

The proposition means, more precisely, that the product of q.H/ and the formalexponential series on the right-hand side of (10.24) is equal to 1 (i.e., to e0). Toprove Proposition 10.24, we first rewrite q as a product.

Lemma 10.28. The Weyl denominator may be computed as

q.H/ DY˛2RC

.eh˛;H i=2 � e�h˛;H i=2/: (10.25)

See Exercise 9 for the explicit form of this identity in the case g D sl.nC 1IC/:Proof. Let Qq denote the product on the right-hand side of (10.25). If we expand outthe product in the definition of Qq; there will be a term equal to

Y˛2RC

eh˛;H i=2 D ehı;H i;

292 10 Further Properties of the Representations

where ı is half the sum of the positive roots. Note that even though ˛=2 is notnecessarily integral, the element ı is integral (Proposition 8.38). We now claim thatevery other exponential in the expansion will be of the form ˙eh�;H i with � integraland strictly lower than ı: To see this, note that every time we take e�h˛;H i=2 insteadof eh˛;H i=2; we lower the exponent by ˛: Thus, any � appearing will be of the form

� D ı �X˛2E

˛;

for some subset E of RC: Such a � is integral and lower than ı:Meanwhile, by precisely the same argument as in the proof of Point 1 of

Proposition 10.21, the function Qq is alternating with respect to the action of W:Thus, if we write

Qq.H/ DX�

a�eh�;H i; (10.26)

then by Corollary 10.17, the coefficients a� must satisfy

aw�� D det.w/a�: (10.27)

Since the exponential ehı;H i occurs in the expansion (10.26) with a coefficient of1, ehw�ı;H i must occur with a coefficient of det.w�1/ D det.w/: Thus, it remains onlyto show that no other exponentials can occur in (10.26). To see this, note that if anyexponential eh�;H i occurs for which � is not in the W -orbit of ı; then by (10.27),another exponential eh�0 ;H i must appear with �0 dominant but strictly lower thanı: Since ı is the minimal strictly dominant integral element, �0 cannot be strictlydominant (see Proposition 8.43) and thus must be orthogonal to one of the positivesimple roots ˛j : Thus, s˛j ��0 D �0; where det.s˛j / D �1: Applying (10.27) to this

case shows that the coefficient of eh�0;H i must be zero. utFigure 10.7 illustrates the proof of Lemma 10.28 for the root system G2. The

white dots indicate weights of exponentials that do not, in fact, occur in theexpansion of Qq: Each of these white dots lies on the line orthogonal to some root,which means that the corresponding exponential cannot occur in the expansion of aWeyl-alternating function.

Proof of Proposition 10.27. As in the sl.2IC/ example considered at the beginningof this section, we have

1

eh˛;H i=2 � e�h˛;H i=2 D e�h˛;H i=2.1C e�h˛;H i C e�2h˛;H i C � � � /;

at the level of formal exponential series. Taking a product over ˛ 2 RC gives

1

q.H/D e�hı;H i Y

˛2RC

.1C e�h˛;H i C e�2h˛;H i C � � � /:

10.6 The Kostant Multiplicity Formula 293

2

1

Fig. 10.7 The black dots indicate the W -orbit of ı for G2: The white dots indicate weights ofexponentials that do not occur in the expansion of Qq

In the product, a term of the form e�h�;H i will occur precisely as many times thereare ways to express � as a non-negative integer combination of the ˛’s, namely,p.�/ times. utTheorem 10.29 (Kostant’s Multiplicty Formula). Suppose � is a dominant inte-gral element and V� is the finite-dimensional irreducible representation with highestweight �: Then if � is a weight of V�; the multiplicity of � is given by

mult.�/ DXw2W

det.w/p.w � .�C ı/ � .�C ı//:

Proof. By the Weyl character formula and Proposition 10.27, we have

��.H/ D0@X��0

p.�/e�h�Cı;H i1A X

w2Wdet.w/ehw�.�Cı/;H i

!:

For a fixed weight �; the coefficient of eh�;H i in the character �� is just themultiplicity of � in V� (Proposition 10.12). This coefficient is the sum of thequantity

p.�/ det.w/ (10.28)

294 10 Further Properties of the Representations

over all pairs .�;w/ for which

�� � ı C w � .�C ı/ D �;

or

� D w � .�C ı/ � .�C ı/: (10.29)

Substituting (10.29) into (10.28) and summing overW gives Kostant’s formula. utIf � is well in the interior of the fundamental Weyl chamber and � is very close to

�; then for all nontrivial elements ofW;w � .�C ı/ will fail to be higher than �C ı:

In those cases, there is only one nonzero term in the formula and the multiplicity of� will be simply p.���/: (By Exercise 7 in Chapter 9, p.���/ is the multiplicityof � in the Verma module W�:) In general, it suffices to compute the multiplicitiesof dominant weights �: For any dominant �; there will be many elements w of Wfor which w � .� C ı/ will fail to be higher than � C ı: Nevertheless, in high-rankexamples, the order of the Weyl group is very large and the number of nonzero termsin Kostant’s formula can be large, even if it is not as large as the order of W:

Figure 10.8 carries out the multiplicity calculation for sl.3IC/ in the irreduciblerepresentation with highest weight .2; 9/: (These multiplicities were presentedwithout proof in Figure 6.5.) The calculation is done only for the weights inthe fundamental Weyl chamber; all other multiplicities are determined by Weylinvariance. The term involving w 2 W makes a nonzero contribution to themultiplicity of � only if � C ı is lower than w � .� C ı/; or equivalently if � islower than w � .� C ı/ � ı: For most weights � in the fundamental chamber, onlythe w D 1 term makes a nonzero contribution. In those cases, the multiplicity of �is simply p.� � �/.

Now, by Example 10.26 and Figure 10.6, p.� � �/ increases by 1 each time �moves from one “ring” of weights to the ring immediately inside. For the weightsindicated by white dots, however, there are two nonzero terms, the second being theone in which w is the reflection about the vertical root ˛1: Since the determinant ofthe reflection is �1; the second term enters with a minus sign. On the medium-sizedtriangle, the first term is 4 and the second term is �1; while on the small triangle,the two terms are 5 and �2. Thus, in the end, all of the weights in all three of thetriangles end up with a multiplicity of 3.

It is not hard to see that the pattern in Figure 10.8 holds in general forrepresentations of sl.3IC/: As we move inward from one “ring” of weights to thenext, the multiplicities increase by 1 at each step, until the rings become triangles, atwhich point the multiplicities become constant. (There is an increase in multiplicityfrom the last hexagon to the first triangle, but not from the first triangle to anyof the subsequent triangles.) In particular, if the highest weight is on the edge ofthe fundamental Weyl chamber, all of the rings will be triangles and, thus, all ofthe weights have multiplicity 1. A small piece of this pattern of multiplicities wasdetermined in a more elementary way in Exercises 11 and 12 in Chapter 6.

10.6 The Kostant Multiplicity Formula 295

1

1

11

11

22

22

2

33

33

4 1

4 14 1

5 2

w

w

Fig. 10.8 Multiplicity calculation for the representation with highest weight .2; 9/

For other low-rank Lie algebras, the pattern of multiplicities is more complicated,but can, in principle, be computed explicitly. In particular, the Kostant partitionfunction can computed explicitly for B2 and G2; at which point one only needsto work out which terms in the multiplicity formula contribute, for weights in thefundamental chamber. (See, for example, [Tar] or [Cap] and compare [CT], whichgives an elegant graphical method of computing the multiplicities in the B2 Š C2case.) For Lie algebras of rank higher than two, [BBCV] gives efficient algorithmsfor computing the partition function for the classical Lie algebras, either numericallyor symbolically. If the order of the Weyl group is not too large, one can thenimplement Kostant’s formula to read off the multiplicities.

For higher-rank cases, the order of the Weyl group can be very large, in whichcase it is not feasible to use Kostant’s formula, even if the partition function isknown. Freudenthal’s formula (Sect. 22.3 in [Hum]) gives an alternative methodof computing the multiplicities, which is preferable in these cases, because it doesnot involve a sum over the Weyl group.

296 10 Further Properties of the Representations

10.7 The Character Formula for Verma Modules

Before coming to the proof of the Weyl character formula, we consider a “warm upcase,” that of the character of a Verma module. In the character formula for Vermamodules, there is an even larger cancellation than in the Weyl character formula. Theproduct of the Weyl denominator and the character would appear to be an infinitesum of exponentials, and yet all but one of these exponentials cancels out! The proofof this character formula follows a very natural course, consisting of writing downexplicitly the multiplicities for the various weight spaces and then using the formulafor the multiplicities to establish the desired cancellation. The character formula forVerma modules is a key ingredient in the proof of the character formula for finite-dimensional representations.

Since a Verma module .�; V�/ is infinite dimensional, the operator e�.X/ maynot have a well-defined trace. On the other hand, Point 2 of Proposition 10.12 givesan expression for the character of a finite-dimensional representation, evaluated ata point in h; in terms of the weights for the representation. This observation allowsus to define the character of any representation that decomposes as a direct sum ofweight spaces, as a formal exponential series on h: (Recall Definition 10.24.)

Definition 10.30. Let V be any representation (possibly infinite dimensional) thatdecomposes as a direct sum of integral weight spaces of finite multiplicity. We definethe formal character of V by the formula

QV .H/ DX�

mult.�/eh�;H i; H 2 h;

where the QV is interpreted as a formal exponential series.

Proposition 10.31 (Character Formula for Verma Modules). For any integralelement �; the formal character of the Verma module is given by

QW�.H/ DX��0

p.�/eh���;H i; (10.30)

where p is the Kostant partition function in Definition 10.25. This formal charactermay also be expressed as

QW�.H/ D eh�Cı;H i�

1

q.H/

�(10.31)

where 1=q.H/ is the formal reciprocal of the Weyl denominator given inProposition 10.27.

The formula (10.31) is similar to the Weyl character formula, except that inthe numerator we have only a single exponential rather than an alternating sumof exponentials. Of course, (10.31) implies that

q.H/QW�.H/ D eh�Cı;H i: (10.32)

10.8 Proof of the Character Formula 297

This result means that when we multiply the Weyl denominator (an alternating finitesum of exponentials) and the formal characterQW� (an infinite sum of exponentials),the result is just a single exponential, with all the other terms cancelling out. Asusual, it is easy to see this cancellation explicitly in the case of sl.nIC/: There, ifm is any integer (possibly negative) and H is the usual diagonal basis element, wehave

QWm.aH/ D ema C e.m�2/a C e.m�4/a C � � � :

Since q.aH/ D ea C e�a; we obtain

q.aH/QWm.aH/ D e.mC1/a C e.m�1/a C e.m�3/a C � � �� e.m�1/a � e.m�3/a � � � �

D e.mC1/a:

Proof. Enumerate the positive roots as ˛1; : : : ; ˛k and let fYj g be nonzero elementsof g�˛j ; j D 1; : : : ; k: By Theorem 9.14, the elements of the form

�.Y1/n1 � � ��.Yk/nkv0 (10.33)

form a basis for the Verma module. An element of the form (10.33) is a weightvector with weight

D � � n1˛1 � � � � � nk˛k:

The number of times the weight will occur is the number of ways that � � canbe written as a non-negative integer combinations of the positive roots. Thus, weobtain the first expression for QW�: The second expression the follows easily fromthe first expression and the formula (Proposition 10.27) for the formal inverse of theWeyl denominator. [Compare (10.30) to (10.24).] ut

10.8 Proof of the Character Formula

Our strategy in proving the character formula is as follows. We will show first thatthe (formal) character of a Verma module W� can be expressed as a finite linearcombination of characters of irreducible highest weight cyclic representations V�:(The V�’s may be infinite dimensional.) By using the action of the Casimir element,we will see that only the only �’s appearing in this decomposition are thosesatisfying j�C �j2 D j�C �j2 :We will then invert this relationship and express thecharacter of an irreducible representation V� as a linear combination of charactersof Verma modules W�; with � again satisfying j�C �j2 D j�C �j2 : We will then

298 10 Further Properties of the Representations

specialize to the case in which� is dominant integral, where V� is finite dimensionaland its character �� is a finite sum of exponentials. By the character formula forVerma modules from Sect. 10.7, we obtain the following conclusion: The productq.H/��.H/ is a finite linear combination of exponentials eih�;H i; where each� D � C � satisfies j�j2 D j�C �j2 : From this point, it is a short step to showthat only �’s of the form � D w � .� C �/ occur and then to prove the characterformula.

We know from general principles that the product q.H/��.H/ is a finite sum ofexponentials. We need to show that the only exponential that actually occur in thisproduct (with a nonzero coefficient) are those of the form ehw�.�Cı/;H i; and that suchexponentials occur with a coefficient of det.w/: We begin with a simple observationthat limits which exponentials could possibly occur in the product.

Proposition 10.32. If an exponential eh�;H i occurs with nonzero coefficient in theproduct q.H/��.H/; then � must be in the convex hull of theW -orbit of �C ı and� must differ from �C ı by an element of the root lattice.

In the last image in Figure 10.4, the white and black circles indicate weightsconsistent with Proposition 10.32. Only a small fraction of these weights (the blackcircles) actually occur in q.H/��.H/:

Proof. If � is an integral element, then for each root ˛;

s˛ � � D � � 2 h�; ˛ih˛; ˛i˛

will differ from � by an integer multiple of ˛: Since roots are integral, we concludethat s˛ �� is, again, integral. It follows that for any w 2 W; the element w�� is integraland differs from � by an integer combination of roots. Thus, by Proposition 8.38,w � ı is integral and differs from ı by an integer combination of roots. Similarly,each weight of � is integral and differs from � by an integer combination ofroots (Theorem 10.1). Thus, for each exponential eh�;H i occurring in the productq.H/��.H/; the element � will be integral and will differ from ıC� by an integercombination of roots.

Meanwhile, since each exponential in q is in the Weyl-orbit of ı and eachexponential in �� is in the convex hull of the Weyl-orbit of � (Theorem 10.1), eachexponential in the product will be in the convex hull of the Weyl-orbit of �C ı; asclaimed. ut

Our next result is the key to the proof of the character formula. In fact, we willsee that it, in conjunction with Proposition 10.32, limits the exponentials that canoccur in q.H/��.H/ to only those whose weights are in the Weyl orbit of �C ı:

Proposition 10.33. If an exponential eh�;H i occurs with nonzero coefficient in theproduct q.H/��.h/; then � must satisfy

h�; �i D h�C ı; �C ıi : (10.34)

10.8 Proof of the Character Formula 299

Fig. 10.9 The weights indicated by white dots satisfy the condition in Proposition 10.33 but notthe condition in Proposition 10.32. The weights indicated by black dots satisfy both conditions

In the last image in Figure 10.4, for example, the exponentials � representedby the white dots do not satisfy (10.34) and, thus, cannot occur in the productq.H/��.h/: Much of the rest of this section will be occupied with the proof ofProposition 10.33. Before turning to this task, however, let us see how Proposi-tions 10.32 and 10.33 together imply the character formula. The claim is that theonly weights satisfying the conditions in both of these propositions are the ones inthe W -orbit of �C ı: See Figure 10.9.

Proof of Character Formula Assuming Proposition 10.33. Suppose v1; : : : ; vm aredistinct elements of a real inner product space, all of which have the same norm, S .The convex hull of these elements is the space of vectors of the form

mXjD1

aj vj

with 0 � aj � 1 andP

j aj D 1: It is then not hard to show by inductionon m that the only way such a convex combination can have norm S is if oneof the aj ’s is equal to 1 and all the others are zero. Applying this with the vj ’sequal to w � .� C ı/;w 2 W; shows that the only exponentials eh�;H i satisfyingboth Propositions 10.32 and 10.33 are those of the form � D w � .� C ı/:

Thus, the only exponentials appearing in the product q�� are those of the formcehw�.�Cı/;H i;w 2 W:

Now, the exponential eh�Cı;H i occurs in the product exactly once, since itcorresponds to taking the highest weight from q (i.e., ı) and the highest weight

300 10 Further Properties of the Representations

from �� (i.e., �) and the coefficient of eh�Cı;H i is 1. (Note that since ı is strictlydominant, w �ı is strictly lower than ı for all nontrivial w 2 W; by Proposition 8.42.)Since the character is Weyl invariant and the Weyl denominator is Weyl alternating,their product is Weyl alternating. Thus, by Corollary 10.17, the coefficients in theexpansion of ��q are alternating; that is, the coefficient of ehw�.�Cı/;H i must equaldet.w/. ut

We now begin the process of proving Proposition 10.33. A crucial role in theproof is played by the Casimir element, which we have already introduced inSect. 10.2. (See Definition 10.4.) We now extend the result of Proposition 10.6to highest-weight cyclic representations that may not be irreducible or finitedimensional.

Proposition 10.34. Let .�; V / be a highest-weight cyclic representation of g(possibly infinite dimensional) with highest weight � and let Q� be the extensionof � to U.g/: Then Q�.C / D c�I; where

c� D h�C ı; �C ıi � hı; ıi :

Proposition 10.34 applies, in particular, to the Verma module W� and to theirreducible representation V�: A key consequence of the proposition is that iftwo highest weight cyclic representations, with highest weights �1 and �2; havethe same eigenvalue of the Casimir, then h�1 C ı; �1 C ıi must coincide withh�2 C ı; �2 C ıi :Proof. The same argument as in the proof of Proposition 10.6 shows if v0 is thehighest weight vector for V; then Q�.C /v0 D c�v0: Now let U be the space of allv 2 V for which Q�.C /v D c�v: Since C is in the center of U.g/ (Proposition 10.5),we see that if v 2 U , then

Q�.C /.�.X/v/ D �.X/ Q�.C /v D c��.X/v;

showing that �.X/v is again in U: Thus,U is an invariant subspace of V containingv0; which means that U D V; that is, that Q�.C /v D c�v for all v 2 V: utDefinition 10.35. Let ƒ denote the set of integral elements. If � is a dominantintegral element, define a set S� � ƒ as follows:

S� D f� 2 ƒj h�C ı; �C ıi D h�C ı; �C ıig :

Note that S� is the intersection of the integral lattice ƒ with sphere of radiusk�C ık centered at �ı; see Figure 10.10. Since there are only finitely manyelements of ƒ in any bounded region, the set S� is finite, for any fixed �: Ourstrategy for the proof of Proposition 10.33 is as discussed at the beginning of thissection. We will decompose the formal character of each Verma module W� with� 2 S� as a finite sum of formal characters of irreducible representations V� ;with each � also belonging to S�: This expansion turns out to be of an “upper

10.8 Proof of the Character Formula 301

1

2

Fig. 10.10 The set S� (black dots) consists of integral elements � for which k�C ık D k�C ık

triangular with ones on the diagonal” form, allowing us to invert the expansionto express the formal character of irreducible representations in terms of formalcharacters of Verma modules. In particular, the character of the finite-dimensionalrepresentation V� will be expressed as a linear combination of formal characters ofVerma modules W� with � 2 S�: When we multiply both sides of this formulaby the Weyl denominator and use the character formula for the Verma module(Proposition 10.31), we obtain the claimed form for q��:

Of course, a key point in the argument is to verify that whenever the characterof an irreducible representation V� appears in the expansion of the characterof W�; � 2 S�; the highest weight � is also in S�: This claim holds becauseany subrepresentation V� occurring in the decomposition of W� must have thesame eigenvalue of the Casimir as W�: In light of Proposition 10.34, this meansthat h� C ı; � C ıi must equal h�C ı; �C ıi ; which is assumed to be equal toh�C ı; �C ıi :Proposition 10.36. For each � in S�; the formal character of the Verma moduleW� can be expressed as a linear combination of formal characters of irreduciblerepresentations V� with � in S� and � � �:

QW� DX�2S����

a��QV� (10.35)

Furthermore, the coefficient a�� of QV� in this decomposition is equal to 1.

See Figure 10.11.

302 10 Further Properties of the Representations

Fig. 10.11 For � 2 S�; the character of W� is a linear combination of characters of irreduciblerepresentations V� with highest weights � � � in S�

Lemma 10.37. Let .�; V / be a representation of g; possibly infinite dimensional,that decomposes as direct sum of weight spaces of finite multiplicity, and let U be anonzero invariant subspace of V: Then both U and the quotient representation V=Udecompose as a direct sum of weight spaces. Furthermore, the multiplicity of anyweight in V is the sum of its multiplicity in U and its multiplicity in V=U:

Proof. Let u be an element of U: By assumption, we can decompose u as u Dv1C� � �Cvj ; where the vk’s belong to weight spaces in V corresponding to distinctweights �1; : : : ; �j : We wish to show that each vk actually belongs to U: If j D 1;

there is nothing to prove. If j > 1; then �j ¤ �1; which means that there issome H 2 h for which

˝�j ;H

˛ ¤ h�1;H i : Then apply to u the operator �.H/ �h�1;H i I

.�.H/ � h�1;H i I /u DjXkD1

.h�k;H i � h�1;H i/vk: (10.36)

Since the coefficient of v1 is zero, the vector in (10.36) is the sum of fewer than jweight vectors. Thus, by induction on j; we can assume that each term on the right-hand side of (10.36) belongs to U: In particular, a nonzero multiple of vj belongsto U; which means vj itself belongs to U: Now, if vj is in U; then u � vj D v1 C� � �Cvj�1 is also in U: Thus, using induction again, we see that each of v1; : : : ; vj�1belongs to U:

We conclude that the sum of the weight spaces in U is all of U: Since weightvectors with distinct weights are linearly independent (Proposition A.17), the sum

10.8 Proof of the Character Formula 303

must be direct. We turn, then, to the quotient space V=U: It is evident that the imagesof the weight spaces in V are weight spaces in V=U with the same weight. Thus,the sum of the weight spaces in V=U is all of V=U and, again, the sum must bedirect.

Finally, consider a fixed weight � occurring in V; and let V� be the associatedweight space. Let q� be the restriction to V� of the quotient map q W V ! V=U: Thekernel of q� consists precisely of the weight vectors with weight � in U: Thus, thedimension of the image of q�; which is the weight space in V=U with weight �; isequal to dimV� � dim.V� \ U /: The claim about multiplicities in V;U; and V=Ufollows. utProof of Proposition 10.36. We actually prove a stronger result, that the formalcharacter of any highest-weight cyclic representationU� with � 2 S� can be decom-posed as in (10.35). As the proof of Proposition 6.11, any such U� decomposes as adirect sum of weight spaces with weights lower than � and with the multiplicity ofthe weight � being 1. For any such U�; let

M DX�2S�

mult.�/:

Our proof will be by induction onM:We first argue that if M D 1; then U� must be irreducible. If not, U� would

have a nontrivial invariant subspace X; and this subspace would, by Lemma 10.37,decompose as a direct sum of weight spaces, all of which are lower than �: Thus, Xwould have to contain a weight vector w that is annihilated by each raising operator�.X˛/; ˛ 2 RC: Thus, X would contain a highest weight cyclic subspace X 0 withsome highest weight �: By Proposition 10.34, the Casimir would act as the scalarh� C ı; � C ıi � hı; ıi in X 0: On the other hand, since X 0 is contained in U�; theCasimir has to act as h�C ı; �C ıi � hı; ıi in X 0; which, since � 2 S�; is equalto h�C ı; �C ıi � hı; ıi : Thus, � must belong to S�: Meanwhile, � cannot equal� or else X 0 would be all of U�: Thus, both � and � would have to have positivemultiplicities andM would have to be at least 2.

Thus, when M D 1; the representation U� is irreducible, in which case,Proposition 10.36 holds trivially. Assume now that the proposition holds for highestweight cyclic representations with M � M0; and consider a representation U� withM D M0C1: IfU� is irreducible, there is nothing to prove. If not, then as we arguedin the previous paragraph,U� must contain a nontrivial invariant subspaceX 0 that ishighest weight cyclic with some highest weight � that belongs to S� and is strictlylower than �:We can then form the quotient vector space U�=X 0; which will still behighest weight cyclic with highest weight �: By Lemma 10.37, the multiplicity of in U� is the sum of the multiplicities of in X 0 and in U�=X 0: Thus,

QU� D QX 0 CQU�=X 0 :

304 10 Further Properties of the Representations

Now, both X 0 and U�=X0 contain at least one weight in S� with nonzero

multiplicity, namely � forX 0 and � forU�=X 0: Thus, both of these spaces must haveM � M0 and we may assume, by induction, that their formal characters decomposeas a sum of characters of irreducible representations with highest weights in S�:These highest weights will be lower than �, in fact lower than � in the case of X 0.Thus, the character of V� will not occur in the expansion of QX 0 . But since U�=X 0still has highest weight �; we may assume by induction that the character of V�will occur exactly once in the expansion of QU�=X 0 ; and, thus, exactly once in theexpansion of QU�: utProposition 10.38. If we enumerate the elements of S� in nondecreasing order asS� D f�1; : : : ; �lg; then the matrix Ajk WD a

�j�k is upper triangular with ones on the

diagonal. Thus, A is invertible, and the inverse of A is also upper triangular withones on the diagonal. It follows that we can invert the decomposition in (10.35) toa decomposition of the form

QV� DX�2S�

b��QW� ; (10.37)

where b�� D 1:

It is easy to check (using, say, the formula for the inverse of a matrix in terms ofcofactors) that the inverse of an upper triangular matrix with ones on the diagonal isagain upper triangular with ones on the diagonal.

Proof. For any finite partially ordered set, it is possible (Exercise 6) to enumeratethe elements in nondecreasing order. In our case, this means that we can enumeratethe elements of S� as �1; : : : ; �l in such a way that if �j � �k then j � k: If weexpandQW�k

in terms ofQV�jas in Proposition 10.38, the only nonzero coefficients

are those with �j � �k; which means that j � k: Thus, the matrix is uppertriangular. Since, also, the coefficient of QV�k

in the expansion of QW�kis 1, the

expansion has ones on the diagonal. utProof of Proposition 10.33. We apply (10.37) with � D �; so that QV� is the char-acter of ��.H/ of the finite-dimensional, irreducible representation with highestweight�:We then multiply both sides of (10.37) by the Weyl denominator q: Usingthe character formula for Verma modules [Proposition 10.31 and Eq. (10.32)], weobtain

q.H/��.H/ DX�2S�

b�� eh�Cı;H i: (10.38)

Since each � belongs to S�; each weight � WD � C ı occurring on the right-handside of (10.38) satisfies

h�; �i D h� C ı; � C ıi D h�C ı; �C ıi :

10.9 Exercises 305

Thus, we have expressed q.H/��.H/ as a linear combination of exponentials withweights � satisfying Proposition 10.33. Since any such decomposition is uniqueby Proposition 10.16, it must be the one obtained by multiplying together theexponentials in q and ��: ut

10.9 Exercises

1. Suppose g is a complex Lie algebra that is reductive (Definition 7.1) but notsemisimple. Show that there exists a finite-dimensional representation of g thatis not completely reducible. (Compare Theorem 10.9 in the semisimple case.)

2. Suppose g is a complex semisimple Lie algebra with the property that everyfinite-dimensional representation of g is completely reducible. Show that gdecomposes as a direct sum of simple algebras.Hint: Consider the adjoint representation of g:

3. Using Proposition 10.16, show that the Weyl denominator is a Weyl-alternatingfunction.

4. Suppose that f W h ! C is a polynomial and that f .H/ D 0 wheneverh˛;H i D 0: Show that f is divisible in the space of polynomial functionsby h˛;H i :Hint: Choose coordinates z1; : : : ; zr on h for which h˛;H i D z1:

5. Suppose f is an analytic function on h; meaning that f can be expressed incoordinates in a globally convergent power series. Collect together all the termsin this power series that are homogeneous of degree k; so that f is the sum ofhomogeneous polynomials pk of degree k: Show that if f is alternating withrespect to the action of W; so is each of the polynomials pk:Hint: Show that the composition of a homogeneous polynomial with a lineartransformation is again a homogeneous polynomial.

6. Show that any finite partially ordered setE can be enumerated in nondecreasingorder.Hint: Every finite partially ordered set has a minimal element, that is, an elementx 2 E such that no y ¤ x in E is smaller than x:

7. Let f˛1; ˛2g be a base for the B2 root system, with ˛1 being the shorter root, asin the left-hand side of Figure 8.7. Let �1 and �2 be the associated fundamentalweights (Definition 8.36), so that every dominant integral element is uniquelyexpressible as

� D m1�1 Cm2�2;

with m1 and m2 being non-negative integers. Show that the dimension of theirreducible representation with highest weight � is

1

6.m1 C 1/.m2 C 1/.m1 Cm2 C 2/.m1 C 2m2 C 3/:

Hint: Imitate the calculations in Example 10.23.

306 10 Further Properties of the Representations

8. Let the notation be as in Exercise 7, but with B2 replaced by G2: (SeeFigures 8.6 and 8.11.) Show that the dimension of the irreducible representationwith highest weight � is

1

5Š.m1 C 1/.m2 C 1/.m1 Cm2 C 2/

� .m1 C 2m2 C 3/.m1 C 3m2 C 4/.2m1 C 3m2 C 5/:

Show that the smallest nontrivial irreducible representation of G2 has dimen-sion 7.

9. According to Lemma 10.28, we have the identity

Xw2W

det.w/eihw�ı;H i DY˛2RC

.eh˛;H i=2 � e�h˛;H i=2/: (10.39)

Work out the explicit form of this identity for the case of the Lie algebrasl.n C 1IC/; using the Cartan subalgebra h described in Sect. 7.7.1 and thesystem of positive roots described in Sect. 8.10.1. If a typical element of hhas the form .a0; : : : ; an/; introduce the variables zj D eaj : Show that aftermultiplying both sides by .z0 � � � zn/n=2; the identity (10.39) takes the form of aVandermonde determinant:

det

0BBB@

zn0 zn�10 � � � 1

zn1 zn�11 � � � 1

::::::

:::

znn zn�1n � � � 1

1CCCA D

Yj<k

.zj � zk/:

10. Let � be an irreducible, finite-dimensional representation of g with highestweight �; and let �� be the dual representation to �:

(a) Show that the weights of �� are the negative of the weights of �:(b) Let w0 be the unique element of W that maps the fundamental Weyl

chamber C to �C: Show that the highest weight �� of �� may becomputed as

�� D �w0 � �:

(Compare Exercises 2 and 3 in Chapter 6.)(c) Show that if �I is an element of the Weyl group, then every representation

of g is isomorphic to its dual.

Part IIICompact Lie Groups

Chapter 11Compact Lie Groups and Maximal Tori

In this chapter and Chapter 12 we develop the representation theory of a connected,compact matrix Lie groupK: The main result is a “theorem of the highest weight,”which is very similar to our main results for semisimple Lie algebras. If we letk be the Lie algebra of K and we let g be the complexification of k; then g isreductive, which means (Proposition 7.6) that g is the direct sum of a semisimplealgebra and a commutative algebra. We can, therefore, draw on our structure resultsfor semisimple Lie algebras to introduce the notions of roots, weights, and theWeyl group. We will, however, give a completely different proof of the theoremof the highest weight. In particular, our proof of the hard part of the theorem, theexistence of a irreducible representation for each weight of the appropriate sort, willbe based on decomposing the space of functions onK under the left and right actionof K: This argument is independent of the Lie-algebraic construction using Vermamodules.

In the present chapter, we develop the structures needed to formulate a theoremof the highest weight for K; and we develop some key tools that will aid is in theproof of the theorem. The representations themselves will appear in the next chapter.The key results of this chapter are the torus theorem and the Weyl integral formula.Although parts of the chapter assume familiarity with the theory of manifolds anddifferential forms, the reader who is not familiar with that theory can still follow thestatements of the key results. Furthermore, the torus theorem can easily be provedby hand in the case of SU.n/: The reader who is willing to take the results of thischapter on faith can proceed on to Chapter 12, where they are applied to provethe compact-group versions of the Weyl character formula and the theorem of the

A previous version of this book was inadvertently published without the middle initial of theauthor’s name as “Brian Hall”. For this reason an erratum has been published, correcting themistake in the previous version and showing the correct name as Brian C. Hall (see DOI http://dx.doi.org/10.1007/978-3-319-13467-3_14). The version readers currently see is the correctedversion. The Publisher would like to apologize for the earlier mistake.

© Springer International Publishing Switzerland 2015B.C. Hall, Lie Groups, Lie Algebras, and Representations, GraduateTexts in Mathematics 222, DOI 10.1007/978-3-319-13467-3_11

309

310 11 Compact Lie Groups and Maximal Tori

highest weight. Finally, in Chapter 13, we will take a close look at the fundamentalgroup of K: We will prove, among other things, that when K is simply connected,the notion of “dominant integral element” forK coincides with the analogous notionfor the Lie algebra g:

Throughout the chapter, we assume that K is a connected, compact matrix Liegroup with Lie algebra k: We allow k; and thus also g WD kC; to have a nontrivialcenter, which means that g is reductive but not necessarily semisimple. As inProposition 7.4, we fix on g an inner product that is real on k and invariant under theadjoint action of K .

11.1 Tori

In this section, we consider tori, that is, groups isomorphic to a product of copies ofS1: In the rest of the chapter, we will be interested in tori that arise as subgroups ofconnected compact groups.

Definition 11.1. A matrix Lie group T is a torus if T is isomorphic to the directproduct of k copies of the group S1 Š U.1/; for some k:

Consider, for example, the group T of diagonal, unitary n � n matrices withdeterminant 1. Every element of T can be expressed uniquely as

diag.u1; : : : ; un�1; .u1 � � � un�1/�1/

for some complex numbers u1; : : : ; un�1 with absolute value 1. Thus, T is isomor-phic to n � 1 copies of S1:

Theorem 11.2. Every compact, connected, commutative matrix Lie group is atorus.

Recall that a subset E of a topological space is discrete if every element e of Ehas a neighborhood containing no points of E other than e:

Lemma 11.3. Let V be a finite-dimensional inner product space over R; viewed asa group under vector addition, and let � be a discrete subgroup of V: Then thereexist linearly independent vectors v1; : : : ; vk in V such that � is precisely the set ofvectors of the form

m1v1 C � � � Cmkvk;

with eachmj 2 Z:

Proof. Since � is discrete, there is some " > 0 such that the only point � in � withk�k < " is � D 0: For any �; � 0 2 �; if k� 0 � �k < "; then since � 0 � � is also in�; we must have � 0 D �: It then follows easily that there can be only finitely manypoints in � in any bounded region of V: If � D f0g; the result holds with k D 0:

Otherwise, we can find some nonzero �0 2 � with such that k�0k is minimal among

11.1 Tori 311

0

n 0

Fig. 11.1 If � is very close to the line through �0; then k� � n�0k is smaller than k�0k for some n

nonzero elements of �: Let W denote the orthogonal complement of the span of �0;let P denote the orthogonal projection of V onto W; and let � 0 denote the image of� under P: Since P is linear, � 0 will be a subgroup of W: We now claim that � 0 isdiscrete in W:

Suppose, toward a contradiction that � 0 is not discrete. Then for every " > 0;

there must exist ı ¤ ı0 2 � 0 with kı0 � ık < ": Thus, ı0 � ı is a nonzero elementof � 0 with norm less than ": There then exists some � 2 � with P.�/ D ı0 � ı:

Since kı0 � ık < "; the distance from � to the span of �0 is less than ": Now let ˇbe the orthogonal projection of � onto the span of �0; which is the point closest to� in line through �0: Then ˇ lies between m�0 and .m C 1/�0 for some integer m:By taking n D m or n D mC 1; we can assume that the distance between n�0 andˇ is at most half the length of �0: Meanwhile, the distance from ˇ to � is at most ":Thus, the distance from n�0 to � is at most "C k�0k =2; which is less than k�0k ; if" is small enough. But then � � n�0 is a nonzero element of � with norm less thanthe norm of �0, contradicting the minimality of �0: (See Figure 11.1.)

Now that � 0 is known to be discrete, we can apply induction on the dimensionof V with the base case corresponding to dimension zero. Thus, there exist linearlyindependent vectors u1; : : : ; uk�1 in � 0 such that � 0 is precisely the set of integerlinear combinations of u1; : : : ; uk�1: Let us then choose vectors v1; : : : ; vk�1 in �such that P.vj / D uj : Since P.v1/; : : : ; P.vk�1/ are linearly independent in W , itis easy to see that v1; : : : ; vk�1 and �0 are linearly independent in V: For any � 2 �;the element P.�/ is of the form m1u1 C � � � Cmk�1uk�1: Thus, � must be equal tom1v1 C � � � Cmk�1vk�1 C ; where 2 � satisfies P./ D 0; meaning that is amultiple of �0: But then must be an integer multiple of �0, or else, by an argumentsimilar to the one in the previous paragraph, there would be an element of � in thespan of �0 with norm less than k�0k : We conclude, then that

� D m1v1 C � � � Cmk�1vk�1 Cmk�0;

establishing the desired form of �: ut

312 11 Compact Lie Groups and Maximal Tori

Proof of Theorem 11.2. Let T be compact, connected, and compact, and let t bethe Lie algebra of T: Then t is also commutative (Proposition 3.22), in whichcase Corollary 3.47 tells us that the exponential map exp W t ! T is surjective.Furthermore, the exponential map for T is a Lie group homomorphism and itskernel � must be discrete, since the exponential map is injective in a neighborhoodof the origin. Thus, by Lemma 11.3, � is the set of integer linear combinationsof independent vectors v1; : : : ; vk: If dim t D n; then exp descends to a bijectivehomomorphism of t=� Š .S1/k�Rn�k with T:Now, the Lie algebra map associatedto this homomorphism is invertible (since the Lie algebra of t is t), which means thatthe inverse map is also continuous. Thus, T is homeomorphic to .S1/k�Rn�k: SinceT is compact, this can only happen if k D n; in which case, T is the torus .S1/n: ut

At various points in the developments in later chapters, it will be useful toconsider elements t in a torus T for which the subgroup of S generated by s isdense in s as Proposition 11.4 in The following result guarantees the existence ofsuch elements.

Proposition 11.4. Let T D .S1/k and let t D .e2��1 ; : : : ; e2��k / be an element ofT: Then t generates a dense subgroup of T if and only if the numbers

1; �1; : : : ; �k

are linearly independent over the field Q of rational numbers.

The k D 1 case of this result is Exercise 9 in Chapter 1. In particular, if x isany transcendental real number and we define �j D xj ; j D 1; : : : ; k; then t willgenerate a dense subgroup of T: See Figure 11.2 for an example of an element thatgenerates a dense subgroup of S1 � S1:Lemma 11.5. If T is a torus and t is an element of T; then the subgroup generatedby t is not dense in T if and only if there exists a nonconstant homomorphism ˆ WT ! S1 such that ˆ.t/ D 1:

Proof. Suppose first that there exists a nonconstant homomorphism ˆ W T ! S1

with T .t/ D 1: Then ker.ˆ/ is a closed subgroup of T that contains t and thus thegroup generated by t: But since ˆ is nonconstant, ker.ˆ/ ¤ T; which means thatthe closure of the group generated by t is not all of T:

In the other direction, let S be the closure of the group generated by t and supposethat S is not all of T: We will proceed by describing the preimage of S under theexponential map, using an extension of Lemma 11.3. Thus, let ƒ be the set of allH 2 t such that e2�H 2 S: Since S is a closed subgroup of T; the set ƒ will be aclosed subgroup of the additive group t: Now let ƒ0 be the identity component ofƒ; which must be a subspace of t: (Indeed, by Corollaries 3.47 and 3.52, ƒ0 mustbe equal to the Lie algebra ofƒ:) Since S is not all of T; the subspaceƒ0 cannot beall of t:

The entire groupƒ now decomposes as ƒ0 �ƒ1, where

ƒ1 WD ƒ \ .ƒ0/?

11.1 Tori 313

t

t2

t3

t4

t5

t6

2

2

Fig. 11.2 A portion of the dense subgroup generated by t in S1 � S1

is a closed subgroup of .ƒ0/?: Furthermore, the identity component of ƒ1 must be

trivial, which means that the Lie algebra ofƒ1 must be f0g: Thus, by Theorem 3.42,ƒ1 is discrete. Let us now define a homomorphism � W t ! S1 by setting

�.H/ D e2�i .H/

for some linear functional on t: By Lemma 11.3,ƒ1 is the integer span of linearlyindependent vectors v1; : : : ; vl . Since ƒ0 ¤ t; we can arrange things so that iszero on ƒ0; the values of on v1; : : : ; vl are integers, but is not identically zero.

Then ker.�/ will containƒ; and, in particular, the kernel of the mapH 7! e2�H :

Thus, there is a nonconstant, continuous homomorphismˆ W T ! S1 satisfying

ˆ.e2�H / D �.H/

for all H 2 t: If we choose H so that e2�H D t 2 S; then H 2 ƒ; which meansthat

ˆ.t/ D ˆ.e2�H / D 1;

but ˆ is not constant. ut

314 11 Compact Lie Groups and Maximal Tori

Proof of Proposition 11.4. In light of Lemma 11.5, we may reformulate the propo-sition as follows: The numbers 1; �1; : : : ; �k are linearly dependent over Q ifand only if there exists a nonconstant homomorphism ˆ W T ! S1 with.e2�i�1; : : : ; e2�i�k / 2 ker.ˆ/: Suppose first that there is a dependence relationamong 1; �1; : : : ; �k overQ: Then after clearing the denominators from this relation,we find that there exist integersm1; : : : ; mk; not all zero, such that

m1�1 C � � � Cmk�k 2 Z:

Thus, we may define a nonconstantˆ W T ! S1 by

ˆ.u1; : : : ; uk/ D um11 � � � umkk (11.1)

and the kernel of ˆ will contain .e2�i�1 ; : : : ; e2�i�k /:In the other direction, Exercise 2 tells us that every continuous homomorphism

ˆ W T ! S1 is of the form (11.1) for some set of integersm1; : : : ; mk: Furthermore,if ˆ is nonconstant, these integers cannot all be zero. Thus, if

1 D ˆ.e2�i�1 ; : : : ; e2�i�k / D e2�i.m1�1C���Cmk�k/;

we must have m1�1 C � � � C mk�k D n for some integer n; which implies that1; �1; : : : ; �k are linearly dependent over Q: ut

11.2 Maximal Tori and the Weyl Group

In this section, we introduce a the concept of a maximal torus, which plays the samerole in the compact group approach to representation theory as the Cartan subalgebraplays in the Lie algebra approach.

Definition 11.6. A subgroup T of K is a torus if T is isomorphic to .S1/k forsome k: A subgroup T of K is a maximal torus if it is a torus and is not properlycontained in any other torus in K .

If K D SU.n/; we may consider

T D

8ˆ<ˆ:

0BBB@ei�1

: : :

ei�n�1

e�i.�1C���C�n�1/

1CCCA

ˇˇˇˇ�j 2 R

9>>>=>>>;;

which is a torus of dimension n� 1: If T is contained in another torus S � SU.n/;then every element s of S would commute with every element t of T: If we choose

11.2 Maximal Tori and the Weyl Group 315

t to have distinct eigenvalues, then by Proposition A.2, s would have to be diagonalin the standard basis, meaning that s 2 T: Thus, T is actually a maximal torus.

Proposition 11.7. If T is a maximal torus, the Lie algebra t of T is a maximalcommutative subalgebra of k: Conversely, if t is a maximal commutative subalgebraof k; the connected Lie subgroup T of k with Lie algebra t is a maximal torus.

Proof. If T is a maximal torus, it is commutative, which means that its Lie algebra tis also commutative (Proposition 3.22). Suppose t is contained in a commutativesubalgebra s: Then it is also contained in a maximal commutative subalgebras0 containing s: The connected Lie subgroup S 0 with Lie algebra s0 must becommutative (since S 0 is generated by exponentials of elements of s0) and closed(Proposition 5.24) and hence compact. Thus, by Theorem 11.2, S 0 is a torus. SinceT is a maximal torus, we must have S 0 D T and thus s0 D s D t; showing that t ismaximal commutative.

In the other direction, if t is maximal commutative, the connected Lie subgroupT with Lie algebra t is closed (Proposition 5.24), hence compact. But T is alsocommutative and connected, hence a torus, by Theorem 11.2. If T is contained in atorusS; then t is contained in the commutative Lie algebra s of S: Since t is maximalcommutative, we have s D t and since S is connected, S D T; showing that T is amaximal torus. utDefinition 11.8. If T is a maximal torus in K; then the normalizer of T , denotedN.T /; is the group of elements x 2 K such that xTx�1 D T: The quotient group

W WD N.T /=T

is the Weyl group of T:

Note that T is, almost by definition, a normal subgroup of N.T /: If w is anelement of W represented by x 2 N.T /; then w acts on T by the formula

w � t D xtx�1; t 2 T:

If x 2 N.T /; the conjugation action of x maps T onto T: It follows that Adx mapsthe Lie algebra t of T into itself. We define an action of W on t by

w �H D Adx.H/; H 2 t: (11.2)

Since our inner product is invariant under the adjoint action of K; the action of Won t is by orthogonal linear transformations.

We will see in Sect. 11.7 that the centralizer of T—that is, the group of thosex 2 K such that xtx�1 D t for all t 2 T—is equal to T: It follows that W actseffectively on T; meaning that if w � t D t for all t 2 T; then w is the identityelement of W: It then follows from Corollary 3.49 (with G D H D T ) that Walso acts effectively on t: Thus, W may be identified with the group of orthogonallinear transformations of t of the formH 7! w �H . We will also show in Sect. 11.7

316 11 Compact Lie Groups and Maximal Tori

that this group of linear transformations coincides with the group generated by thereflections through the hyperplanes orthogonal to the roots. Thus, the Weyl groupas defined in Definition 11.8 is naturally isomorphic to the Weyl group associatedto the Lie algebra g WD kC in Sect. 7.4. Exercise 3, meanwhile, asks the reader toverify directly in the case of SU.n/ that the centralizer of T is T and that N.T /=Tis the permutation group on n entries, thus agreeing with the Weyl group for sl.nIC/computed in Sect. 7.7.1.

The following “torus theorem” is a key result that underlies many of thedevelopments in this chapter and the next two chapters.

Theorem 11.9 (Torus Theorem). If K is a connected, compact matrix Lie group,the following results hold.

1. If S and T are maximal tori in K; there exists an element x of K such thatT D xSx�1:

2. Every element ofK is contained in some maximal torus.

The torus theorem has many important consequences; we will mention just twoof these now.

Corollary 11.10. If K is a connected, compact matrix Lie group, the exponentialmap forK is surjective.

Proof. For any x 2 K; choose a maximal torus T containing x: Since theexponential map for T Š .S1/k is surjective, x can be expressed as the exponentialof an element of the Lie algebra of T: utCorollary 11.11. Let K be a connected, compact matrix Lie group and let x anarbitrary element of K: Then x belongs to the center of K if and only if x belongsto every maximal torus in K:

Proof. Assume first that x belongs to the center Z.K/ of K; and let T be anymaximal torus in K: By the torus theorem, x is contained in a maximal torus S;and this torus is conjugate to T: Thus, there is some y 2 K such that S D yTy�1:Since x 2 S; we have x D yty�1 for some t 2 T; and thus t D y�1xy: But we areassuming that x is central, and so, actually, t D x; showing that x belongs to T:

In the other direction, assume x belongs to every maximal torus in K: Then forany y 2 K; we can find some torus T containing y; and this torus will also containx: Since T is commutative, we conclude that x and y commute, showing that x isin Z.K/: ut

The torus theorem follows from the following result.

Lemma 11.12. Let T be a fixed maximal torus in K: Then every y 2 K can bewritten in the form

y D xtx�1

for some x 2 K and t 2 T:

11.3 Mapping Degrees 317

IfK D SU.n/ and T is the diagonal subgroup ofK; Lemma 11.12 follows easilyfrom the fact that every unitary matrix has an orthonormal basis of eigenvectors.The proof of the general case of Lemma 11.12 requires substantial preparation andis given in Sect. 11.5.

Proof of torus theorem assuming Lemma 11.12. Since each y 2 K can be writtenas y D xtx�1; we see that y belongs to the maximal torus xTx�1:

Next we show that every maximal torus S in K is conjugate to T; from whichit follows any two maximal tori S1 and S2 are conjugate to each other. Suppose sis an element of S such that the subgroup of S generated by s is dense in S , as inProposition 11.4. Then we can choose some x 2 K and t 2 T such that s D xtx�1and t D x�1sx: Thus, x�1skx D tk 2 T for all integers k: Since the set of elementsof the form sk is dense in S; we must have x�1Sx � T . But since S is maximal, weactually have x�1Sx D T: ut

11.3 Mapping Degrees

We now introduce a method that we will use in proving the torus theorem inSect. 11.5. The current section requires greater familiarity with manifolds thanelsewhere in the book. In addition to differential forms (Appendix B), we makeuse of the exterior derivative (“d”), the pullback of a form by a smooth map, andStoke’s theorem. See Chapters 14 and 16 in [Lee] for more information. For ourpurposes, the main result of this section is Corollary 11.17, which gives a conditionguaranteeing that a map between two manifolds of the same dimension is surjective.The reader who is not familiar with manifold theory should still be able to get anidea of what is going on from the example in Figures 11.3 and 11.4.

If V is a finite-dimensional vector space over R; we may define an equivalencerelation on ordered bases of V by declaring two ordered bases .v1; : : : ; vn/ and.v01; : : : ; v

0n/ to be equivalent if the unique linear transformationLmapping vj to v0

j

has positive determinant. The set of ordered bases for V then consists of exactlytwo equivalence classes. An orientation for V is a choice of one of these twoequivalence classes. Once an orientation of V has been fixed, an ordered basis for Vis said to be oriented if it belongs to the chosen equivalence class. An orientationon a smooth manifoldM is then a continuous choice of orientation on each tangentspace to M: A smooth manifold together with a choice of orientation is called anoriented manifold.

We consider manifolds that are closed—that is, compact, connected, and withoutboundary—and oriented. We will be interested in smooth maps between two closed,oriented manifolds of the same dimension.

Definition 11.13. Let X and Y be closed, oriented manifolds of dimension n � 1

and let f W X ! Y be a smooth map. A point y 2 Y is a regular value of f iffor all x 2 X such that f .x/ D y; the differential f�.x/ of f at x is invertible.

318 11 Compact Lie Groups and Maximal Tori

A point y 2 Y is a singular value of f if there exists some x 2 X such thatf .x/ D y and the differential f�.x/ of f at x is not invertible.

It is important to note that if y is not in the range of f; then y is a regular value.After all, if there is no x with f .x/ D y; then it is vacuously true that for every xwith f .x/ D y; the differential f�.x/ is invertible.

Proposition 11.14. Let X; Y; and f be as in Definition 11.13. If y is a regularvalue of f; then y has only finitely many preimages under f:

Proof. If f �1.fyg/ were infinite, the set would have to have an accumulation pointx0, by the assumed compactness of X . Then by continuity, we would have f .x0/ Dy: Since y is a regular value of f; then f�.x0/ would be invertible. But then theinverse function theorem would say that f is injective in a neighborhood of x0;which is impossible since every neighborhood of x0 contains infinitely many pointsx with f .x/ D y: ut

Saying that X and Y are oriented means that we have chosen a consistentorientation on each tangent space to X and to Y: If f W X ! Y is smooth andthe differential f�.x/ of f at x is invertible, then f�.x/ is either an orientationpreserving or an orientation reversing map of Tx.X/ to Tf.x/.Y /: Since f 0 isassumed to be continuous, if f� is invertible and orientation preserving at x; itis invertible and orientation preserving in a neighborhood of x; and similarly iff� is invertible and orientation reversing at x:

Definition 11.15. If y is a regular value of f; let the signed number of preimagesof y denote the number of preimages, where x 2 f �1.fyg/ counted with a plussign if f�.x/ is orientation preserving and with a minus sign if f�.x/ is orientationreversing.

The main result of the section is the following.

Theorem 11.16. Let X and Y be closed, oriented manifolds of dimension n � 1

and let f W X ! Y be a smooth map. Then there exists an integer k such that forevery regular value y of f; the signed number of preimages of y is equal to k:

If there are, in fact, any regular values of f; the integer k is unique and is calledthe mapping degree of f: (Actually, Sard’s theorem guarantees that every such fhas a nonempty set of regular values, but we do not need to know this, since inour application of Theorem 11.16 in Sect. 11.5, we will find regular values of therelevant map “by hand.”) See the “Degree Theory” section in Chapter 17 of [Lee]for more information.

Corollary 11.17. Let X; Y; and f be as in Theorem 11.16. If there exists a regularvalue y of f for which the signed number of preimages is nonzero, then f mustmap onto Y:

Proof. If there existed some y0 that is not in the range of f; then y0 would be aregular value and the (signed) number of preimages of y0 would be zero. This wouldcontradict Theorem 11.16. ut

11.3 Mapping Degrees 319

Fig. 11.3 The graph of amap f from S1 to S1: Thesigned number of preimagesof each regular value is 1

a

b

y

y

2

2

Figure 11.3 illustrates Theorem 11.16. The figure shows the graph of a map ffrom S1 (which we think of as Œ0; 2�� with ends identified) to itself. The singularvalues are the points marked a and b on the y-axis; all other values are regular. Fortwo different values y and y0; we compute the signed number of preimages. Thepoint y has three preimages, but f 0 is negative at one of these, so that the signednumber of preimages is 1 � 1C 1 D 1: Meanwhile, the point y0 has one preimage,at which f 0 is positive so that the signed number of preimages is 1. The mappingdegree of f is 1 and, consistent with Corollary 11.17, f is surjective. Meanwhile,Figure 11.4 shows the same map in a more geometric way, as a map between twomanifoldsX and Y; each of which is diffeomorphic to S1:

We now turn to the proof of Theorem 11.16; see also Theorem 17.35 in[Lee]. Using the inverse function theorem, it is not hard to show that the signednumber of preimages and the unsigned number of preimages are both constant in aneighborhood of any regular value y: This result, however, does not really help us,because the set of regular values may be disconnected. (See Figure 11.3.) Indeed,Figure 11.3 shows that the unsigned number of preimages may not be constant; weneed a creative method to show that the signed number of preimages is constant onthe set of regular values.

Our tool for proving this result is that of differential forms. If f W X ! Y isan orientation-preserving diffeomorphism, then for any n-form ˛ on Y; the integralof f �.˛/ over X will equal the integral of ˛ over Y: If, on the other hand, f is anorientation-reversing diffeomorphism, the integral of f �.˛/ will be the negative ofthe integral ˛: Suppose now that f W X ! Y is a smooth map, not necessarily adiffeomorphism. Suppose that y is a regular value of f and that x1; : : : ; xN are the

320 11 Compact Lie Groups and Maximal Tori

X

Y

ba

Fig. 11.4 The map indicated by the arrows is the same one as in Figure 11.3, but shown moregeometrically

elements of f �1.fyg/: Then we can find a neighborhoodV of y such that f �1.V / isa disjoint union of neighborhoodsU1; : : : ; UN of x1; : : : ; xN and such that f mapseach Uj diffeomorphically onto V: Furthermore, by shrinking V if necessary, wecan assume that for each j; the differential f�.x/ is either orientation preserving atevery point of Uj or orientation reversing at every point of Uj : Then for any n-form˛ supported in V; we see that

ZX

f �.˛/ D k

ZY

˛; (11.3)

where k is the signed number of preimages of y:If ˛ is chosen so that

RY˛ ¤ 0; then (11.3) becomes

k DRXf �.˛/RY˛

: (11.4)

The right-hand side of (11.4) gives us an analytic method for determining the signednumber of preimages. Our goal is to use (11.4) to show that k is constant on the setof regular values. To this end, we will establish a key result: The value of the right-hand side of (11.4) is unchanged if ˛ is “deformed” by pulling it back by a familyof diffeomorphisms of Y:

Suppose, then, that y and y0 are regular values of f with the signed number ofpreimages being k and k0; respectively. We will construct a family ˛t ; 0 � t � 1; ofn-forms on Y such that ˛0 is supported near y and ˛1 is supported near y0: For allt; the expression

RX f

�.˛t /RY˛t

(11.5)

11.3 Mapping Degrees 321

makes sense, even if the support of ˛t contains singular values of f: Furthermore,the values of both integrals in (11.5) are—as we will show—independent of t: Thus,we will conclude that

k DRXf �.˛0/RY˛0

DRXf �.˛1/RY˛1

D k0;

as claimed.

Lemma 11.18. Suppose ‰t is a continuous, piecewise-smooth family oforientation-preserving diffeomorphisms of Y; with ‰0 being the identity map.For any n-form ˛ on Y; let ˛t D ‰�

t .˛/: Then for all t; we have

ZY

˛t DZY

˛ (11.6)

andZX

f �.˛t / DZX

f �.˛/: (11.7)

Proof. Saying that ‰t is piecewise smooth means that it we can divide Œ0; T � intofinitely many subintervals on each of which ‰t.x/ is smooth in x and t: Since ‰t iscontinuous, it suffices to prove the result on each of the subintervals, that is, in thecase where‰t.x/ is smooth, which we now assume. The result (11.6) holds because‰t is an orientation preserving diffeomorphism.

To establish (11.7), we show that the left-hand side of (11.7) is independent of t:Note that

f �.˛t / D f �.‰�t .˛// D .‰t ı f /�.˛/:

Thus, if we define g W X � Œ0; T � ! Y by

g.x; t/ D ‰t.f .x//;

we haveZX

f �.˛T / �ZX

f .˛0/ [email protected]�Œ0;T �/

g�.˛/:

Using Stoke’s theorem and a standard result relating pullbacks and exterior deriva-tives, we obtain

322 11 Compact Lie Groups and Maximal Tori

ZX

f �.˛T / �ZX

f .˛0/ DZX�Œ0;T �

d.g�.˛//

DZX�Œ0;T �

g�.d˛/:

But since ˛ is a top-degree form on Y; we must have d˛ D 0; showing thatRXf �.˛T / D R

Xf .˛0/: ut

Proof of Theorem 11.16. To complete the proof of Theorem 11.16, it remains onlyto address the existence of a continuous, piecewise smooth, orientation-preservingfamily ‰t of diffeomorphisms of Y such that ‰0 is the identity and such that‰1.y

0/ D y: (Thus, if ˛ is supported near y; then ‰�1 .˛/ will be supported near

y0:) We actually only require this in the case that Y is a compact Lie group, inwhich case, the diffeomorphisms can easily be constructed using the group structureon Y: Nevertheless, we will outline an argument for the general result. Let U be aneighborhood of y0 that is a rectangle in some local coordinate system around y0:Then it is not hard to construct a family of diffeomorphisms of Y that are the identityon Y n U and that map y0 to any desired point of U: (See Exercise 7.)

If y 2 U; we are done. If not, we consider the set E of points z 2 Y such thaty0 can be moved to z by a family of diffeomorphisms of the desired sort. If z 2 E;

we have, by assumption, a family moving y0 to z: We can then use the argument inthe preceding paragraph to move z to any point z0 in a neighborhood of z: Thus, Eis open and contains y0: We now claim that E is closed. If z is a limit point of E;then in any neighborhood V of z; there is an element z0 of E: Thus, by the argumentin the preceding paragraph, we can move z0 to z by a family of diffeomorphisms.Since E is both open and closed and nonempty (because it contains y), E must beall of Y . utProposition 11.19. Let X; Y; and f be as in Theorem 11.16, and suppose f hasmapping degree k: Then for every n-form ˛ on Y; we have

ZX

f �.˛/ D k

ZY

˛: (11.8)

In Figure 11.4, for example, the map f indicated by the arrows has mappingdegree 1. Thus, for every form ˛ on Y; the integral of f �.˛/ over X is the sameas the integral of ˛ over Y; even though f is not a diffeomorphism. When pullingback the part of ˛ between a and b; we get three separate integrals on X; but one ofthese integrals occurs with a minus sign, because f� is orientation reversing on themiddle of the three intervals over Œa; b�:

Proof. We have already noted in (11.3) that if y is a regular value of f; there isa neighborhood U of y such that (11.8) holds for all ˛ supported in U: By thedeformation argument in the proof of Theorem 11.16, the same result holds for anyy in Y: Since Y is compact, we can then cover Y by a finite number of open sets Uj

11.4 Quotient Manifolds 323

such that (11.8) holds whenever ˛ is supported in Uj : Using a partition of unity, wecan express any form ˛ as a sum of forms ˛j such that ˛j is supported on Uj ; andthe general result follows. ut

11.4 Quotient Manifolds

Before coming the proof of the torus theorem, we require one more preparatoryconcept, that of the quotient of a matrix Lie group by a closed subgroup. Throughoutthis section, we assume that G is a matrix Lie group with Lie algebra g and that His a closed subgroup of G with Lie algebra h: Even if H is not normal, we can stillconsider the quotient G=H as a set (the set of left cosets gH of H ). We now showthat G=H has the structure of a smooth manifold. We will let Œg� denote the cosetgH of H in G and we will let Q W G ! G=H denote the quotient map. Recall thata topological structure on a set E is Hausdorff if for every pair of distinct pointsx; y 2 E; there exist disjoint open sets U and V in E with x 2 U and y 2 V:Lemma 11.20. Define a topology on G=H by decreeing that a set U in G=H isopen if and only if Q�1.U / is open in G: Then G=H is Hausdorff with respect tothis topology. Furthermore, G=H has a countable dense subset.

If we did not assume that H is closed, the Hausdorff condition for G=H would,in general, fail.

Proof. We begin by noting that the quotient map Q W G ! G=H is open; that is,Q.U / is open in G=H whenever U is open in G: To establish this, we must showthat Q�1.Q.U // is open in G: But

Q�1.Q.U // D[h2H

Uh;

whereUh D fuh ju 2 U g : But since right-multiplication by h is a homeomorphismof G with itself, each Uh is open, so that Q�1.Q.U // is also open. Next, let � WG �G ! G be the continuous map given by �.x; y/ D x�1y: Note that Œx� D Œy�

if and only if �.x; y/ 2 H: Suppose now that Œx� and Œy� are distinct points in G=H;so that �.x; y/ … H:

Then since H is closed, G n H is open, so that ��1.G n H/ is open in G � G.Thus, we can find an open rectangle U � V in ��1.G n H/, where U and V areopen sets in G containing x and y; respectively. For all u 2 U and v 2 V; we have�.u; v/ … H; meaning that Œu� ¤ Œv�: It follows that Q.U / and Q.V / are disjointopen sets in G=H containing Œx� and Œy�; respectively.

Finally, since G inherits its topology from the separable metric space Mn.C/ ŠR2n

2; it follows that G has a countable dense subset E . Furthermore, the quotient

map Q W G ! G=H is surjective and (by the definition of the topology on G=H )continuous. Thus,Q.E/ will be dense in G=H: ut

324 11 Compact Lie Groups and Maximal Tori

Fig. 11.5 The gray regionindicates the set of points ofthe form gh with g 2 exp.U /and h 2 H: This set isdiffeomorphic to exp.U /�H

H

expU

Lemma 11.21 (Slice Lemma). Let G be a matrix Lie group with Lie algebra g;let H be a closed subgroup of G; and let h be the Lie algebra of H: Decomposeg as a vector space as g D h ˚ f for some subspace f of g and define a mapƒ W f �H ! G by

ƒ.X; h/ D eXh:

Then there is a neighborhood U of 0 in f such that ƒ is injective on U �H and ƒmaps U � H diffeomorphically onto an open subset of G: In particular, if X1 andX2 are distinct elements of U; then eX1 and eX2 belong to distinct cosets of G=H:

The term “slice lemma” refers to the fact that the map sending X 2 U to eX

slices across the different cosets ofH: Figure 11.5 illustrates the slice lemma in thecase in which G D S1 � S1 andH is the subgroup consisting of points of the form.ei� ; e3i� /:

Proof. We identify the tangent spaces at the identity to both f�H andG with f˚h:If X.t/ is a smooth curve in f passing through 0 at t D 0; we have

d

dtƒ.X.t/; I /

ˇˇtD0

D d

dteX.t/

ˇˇtD0

D X 0.0/:

Meanwhile, if h.t/ is a smooth curve in H passing through I at t D 0; we have

d

dtƒ.0; h.t//

ˇˇtD0

D d

dth.t/

ˇˇtD0

D h0.0/:

11.4 Quotient Manifolds 325

From this calculation, and the linearity of the differential, we can see that thedifferentialƒ� ofƒ at .0; I / is the identity map of f˚h to itself. Thus, by continuity,ƒ�.X; e/ is invertible for X in some neighborhoodU of 0 in f.

Meanwhile, the mapƒ commutes with the right action of H :

ƒ.X; hh0/ D ƒ.X; h/h0:

From this, it is easy to see that if ƒ�.X; I / is invertible, then ƒ�.X; h/ is invertiblefor all h: We conclude, then, that ƒ�.X; h/ is invertible for all .X; h/ 2 U �H: Bythe inverse function theorem, then, ƒ maps a small neighborhood of each point inU �H injectively onto an open set in G: In particular, the image of U � H underƒ is an open subset of G:

We must now show that by shrinking U as necessary, we can ensure thatƒ is globally injective on U � H: By the inverse function theorem, there areneighborhoods U 0 of 0 in f and V of I in H such that ƒ maps U 0 � V injectivelyintoG. If we choose a small subneighborhoodU 00 of 0 in f andX andX 0 are in U 00;then e�X 0

eX will be close to the identity in G: Indeed, if U 00 is small enough, thenwhenever e�X 0

eX happens to be in H; the element e�X 0

eX will actually be in V:We now claim that ƒ is injective on U 00 �H: To see this, suppose eXh D eX

0

h0with X;X 0 2 U 00 and h; h0 2 H: Then

h0h�1 D e�X 0

eX 2 V;

by our choice of U 00: But then

eX D eX0

.h0h�1/;

and since h0h�1 2 V; we must have X D X 0 and I D h0h�1; by the injectivity ofƒon U 0 � V: Thus, actually, X D X 0 and h D h0; establishing the desired injectivity.

utIt is instructive to contemplate the role in the preceding proof played by the

assumption that H is closed. It is evident from Figure 1.1 that the slice lemma canfail ifH is not closed. (For example, even for very small nonzeroX 2 f; the elementeX can be inH .) On the other hand, even ifH is not closed, Theorem 5.23 says thatthere is a new topology on H and an atlas of coordinate neighborhoods making Hinto a smooth manifold, in such a way that the inclusion of H into G is smooth. Ifwe use this structure on H; the map ƒ in Lemma 11.21 is smooth and much of theproof proceeds in the same way as whenH is closed. The proof of global injectivityof ƒ; however, breaks down because the new topology on H does not agree withthe topology thatH inherits fromG: Thus, even if e�X 0

eX belongs toH and is veryclose to the identity in G; this element may not be close to the identity in the newtopology onH: (See Figure 5.4.) Thus, we cannot conclude that e�X 0

eX is in V; andthe proof of injectivity fails.

326 11 Compact Lie Groups and Maximal Tori

Theorem 11.22. If G is a matrix Lie group and H a closed subgroup of G; thenG=H can be given the structure of a smooth manifold with

dim.G=H/ D dimG � dimH

in such a say that (1) the quotient mapQ W G ! G=H is smooth, (2) the differentialofQ at the identity maps TI .G/ onto TŒI �.G=H/with kernel h; and (3) the left actionof G on G=H is smooth.

Proof. Let U � f be as in Lemma 11.21. Then for each g 2 G; let ƒg be the mapfrom U �H into G given by

ƒg.X; h/ D geXh:

Combining Lemma 11.21 with a translation by g; we see that ƒg is a diffeomor-phism of U � H onto its image, and that ƒg.X; h/ and ƒg.X

0; h0/ are in distinctcosets of H provided that X ¤ X 0: Let Wg be the (open) image of U � H underƒg and let Vg D Q.Wg/; that is,

Vg D ˚ŒgeX � 2 G=H ˇX 2 U � :

Then Q�1.Vg/ D Wg; showing that Vg is open in G=H: By the above properties ofƒg; the map X 7! ŒgeX � is an injective map of U onto Vg:

We now propose to use the maps X 7! ŒgeX � as local coordinates on G=H;where U � f may be identified with Rk; with k D dim f D dim g � dim h: Bythe way the topology on G=H is defined, the map Q is continuous and a functionf on G=H is continuous if and only if f ı Q is continuous on G: Thus, the mapX 7! Q.geX/ D ŒgeX� is continuous. Furthermore, if we compose the inverse mapŒgeX � 7! X withQ;we obtain the map geXh 7! X: This map is continuous becauseit consists of the inverse of the diffeomorphism .X; h/ 7! geXh; combined with themap .X; h/ 7! X:

Thus, G=H is locally homeomorphic to Rk: Since, also, G=H is Hausdorff andhas a countable dense subset (Lemma 11.20), we see that G=H is a topologicalmanifold.

Now, the coordinate patches ŒgeX � clearly cover G=H: If two such patchesoverlap, the change of coordinates map is the mapX 7! X 0;where ŒgeX� D Œg0eX 0

�:

This map can be computed by mapping X to geX 2 G and then applying .ƒg0/�1to write geX as geX D g0eX 0

h0: Since .ƒg0/�1 is smooth, we see that the changeof coordinates map is smooth. Thus, we may give a smooth structure to G=H usingthese coordinates.

It is now a straightforward matter to check the remaining claims about the smoothstructure onG=H: To see, for example, thatQ is smooth, pick some g inG and writepoints near g as geXh; with X 2 U and h 2 H: Then Q.geXh/ D ŒgeX�: Thus, Qcan be written locally as the inverse of ƒg followed by the map .X; h/ 7! ŒgeX �;which is smooth in our local identification of G=H with U: The remaining claimsare left as an exercise to the reader (Exercise 8). ut

11.4 Quotient Manifolds 327

Proposition 11.23. Suppose there exists an inner product on g that is invariantunder the adjoint action of H; and let V denote the orthogonal complement of hwith respect to this inner product. Then we may identify the tangent space at eachpoint Œg� of G=H with V by writing v 2 TŒg�.G=H/ as

v D d

dtŒgetX�

ˇˇtD0

; X 2 V:

This identification of TŒg�.G=H/ with V is unique up to the adjoint action ofH on V:

Note that since the adjoint action of H on g is orthogonal and preserves h; thisaction also preserves V; by Proposition A.10.

Proof. Since g D h˚V; Point 2 of Theorem 11.22 tells us that every tangent vectorv to G=H at the identity coset can be expressed uniquely as

v D d

dtŒetX �

ˇˇtD0

; X 2 V:

For any Œg� 2 G=H; we identify TŒg�.G=H/ with TŒI �.G=H/ Š V by using the leftaction of g: Thus, each v 2 TŒg�.G=H/ can be written uniquely as

v D d

dtŒgetX �

ˇˇtD0

; X 2 V:

If we use a different element gh; h 2 H; of the same coset, we get a differentidentification of TŒg�.G=H/ with V; as follows:

ŒghetX � D ŒghetXh�1� D ŒgetX0

�; (11.9)

whereX 0 D Adh.X/:Differentiating (11.9) shows that the two identifications differby the adjoint action of H: ut

A volume form on a manifoldM of dimension n is a nowhere-vanishing n-formon M: As we have already discussed in the proof of Theorem 4.28, each matrix Liegroup G has a volume form that is invariant under the right action of G on itself.The same argument shows that G has a volume form invariant under the left actionof G on itself. (For some groups G, it is possible to find a single volume form thatis invariant under both the left and right action of G on itself, but this is not alwaysthe case.) We now address the existence of an invariant volume form on a quotientmanifold.

Proposition 11.24. If G is a matrix Lie group and H is a connected compactsubgroup of G; there exists a volume form on G=H that is invariant under the leftaction of G: This form is unique up to multiplication by a constant.

328 11 Compact Lie Groups and Maximal Tori

In the caseH D fI g;we conclude that there is a left-invariant volume form onGitself and that this form is unique up to a constant. (In this chapter, it is convenientto use a left-invariant volume form on G; rather than a right-invariant form as inSect. 4.4.)

Proof. SinceH is compact, there exists an inner product on g that is invariant underthe adjoint action of H: Let V denote the orthogonal complement of h in g; so thatV is invariant under the adjoint action of H: Since H is connected, the restrictionto V of Adh will actually be in SO.V / for all h 2 H:

Now pick an orientation on V and let ˛ be the standard volume form on V; that is,the unique one for which ˛.e1; : : : ; eN / D 1 whenever .e1; : : : ; eN / is an orientedorthonormal basis for V: Since the action of Adx on V has determinant 1, we have

˛.Adx.v1/; : : : ;Adx.vN // D ˛.v1; : : : ; vN /

for all v1; : : : ; vN 2 V:Now, the tangent space to G=H at the identity coset is identified with V .

We define a form on G=H as follows. At the identity coset, we take it to be ˛:At any other coset Œg�; we use the action of g 2 G to transport ˛ from ŒI � to Œg�:If Œg� D Œg0�; then g0 D xh for h 2 H: The action of h on TŒI �.G=H/ Š V is theadjoint action, which preserves ˛: Thus, the resulting form at Œg� is independent ofthe choice of g:

Finally, to address the uniqueness, note that any two top degree forms on G=Hmust agree up to a constant at the identity coset ŒI �: But the since the left action ofG on G=H is transitive, the value of the form at ŒI � uniquely determines the formeverywhere. Thus, any two left-invariant forms on G=H must be equal up to anoverall constant. ut

11.5 Proof of the Torus Theorem

Having made the required preparations in Sects. 11.3 and 11.4, we are now ready tocomplete the proof of Theorem 11.9. It remains only to prove Lemma 11.12; to thatend, we define a key map.

Definition 11.25. Let T be a fixed maximal torus in K: Let

ˆ W T � .K=T / ! K

be defined by

ˆ.t; Œx�/ D xtx�1; (11.10)

where Œx� denotes the coset xT in K=T:

11.5 Proof of the Torus Theorem 329

Note that if s 2 T; then since T is commutative, we have

.xs/t.xs/�1 D xsts�1x�1 D xtx�1;

showing that ˆ is well defined as a map of T � .K=T / into K: Lemma 11.12amounts to saying that ˆ is surjective. Since T � .K=T / has the same dimensionas K; we may apply Theorem 11.16 and Corollary 11.17. Thus, if there is even oneregular value of ˆ for which the signed number of preimages is nonzero, ˆ mustbe surjective. Our strategy will be to find a certain class of points y 2 K for whichwe can (1) determine all of the preimages of y under ˆ; and (2) verify that ˆ� isinvertible and orientation preserving at each of the preimages.

Lemma 11.26. Let t 2 T be such that the subgroup generated by t is dense in T(Proposition 11.4). Then ˆ�1.ftg/ of t consists precisely of elements of the form.x�1tx; Œx�/ with Œx� belonging to W D N.T /=T . In particular, if xsx�1 D t forsome x 2 K and s 2 T; then s must be of the form s D w�1 � t for some w 2 W:

Note that if x and y in N.T / represent distinct elements of W D N.T /=T; thenŒx� and Œy� are distinct elements of K=T: Thus, the lemma tells us that there is aone-to-one correspondence between ˆ�1.ftg/ and W:

Proof. If x 2 N.T /; then x�1tx 2 T and we can see that ˆ.x�1tx; Œx�/ D t: In theother direction, if xsx�1 D t; then

x�1tmx D sm 2 T

for all integersm; so that x�1Tx � T by our assumption on t: Since x�1Tx is againa maximal torus, we must actually have x�1Tx D T and, thus, T D xTx�1; showingthat x 2 N.T /: Furthermore, since xsx�1 D t; we have s D x�1tx D w�1 � t; wherew D Œx�: ut

We now compute the differential of ˆ: Using Proposition 11.23 with .G;H/equal to .T; fI g/; .K; T /; and .K; fI g/; we identify the tangent space at each pointin T � .K=T / with t ˚ f Š k and the tangent space at each point in K with k:Since we are trying to determine the signed number of preimages of ˆ; we mustchoose orientations on T � .K=T / and on K: To this end, we choose orientationson the vector spaces t and f and use the obvious associated orientation on k Š t˚ f:We then define orientations on T � .K=T / and K using the above identificationsof the tangent spaces with t ˚ f Š k: The identification of the tangent spaces toK=T with f is unique up to the adjoint action of T (Proposition 11.23). Since T isconnected, this action will have positive determinant, showing that the orientationon T � .K=T / is well defined. Recall that the (orthogonal) adjoint action of T on kpreserves t and thus, also, f WD t?:

Proposition 11.27. Let .t; Œx�/ be a fixed point in T � .K=T /. If we identify thetangent spaces to T � .K=T / and to K with t ˚ f Š k; then the differential of ˆ at.t; Œx�/ is represented by the following operator:

330 11 Compact Lie Groups and Maximal Tori

ˆ� D .Adx/

�I 0

0 Ad0t�1

� I

�; (11.11)

where Ad0t�1

denotes the restriction of Adt�1 to f:

Proof. For H 2 t; we compute that

d

d�ˆ.te�H ; Œx�/

ˇˇ�D0

D d

d�xte�Hx�1

ˇˇ�D0

D xtHx�1

D .xtx�1/.Adx.H//:

Since we identify the tangent space to K at xtx�1 with k using the left action ofxtx�1; we see that ˆ�..H; 0// D Adx.H/:

Meanwhile, if X 2 f; we compute that

d

d�ˆ.t; Œxe�X �/

ˇˇ�D0

D d

d�xe�X te��Xx�1

ˇˇ�D0

D xXtx�1 � xtXx�1

D xtx�1.xt�1Xtx�1 � xXx�1/

D .xtx�1/ŒAdx.Adt�1 .X/ �X/�;

so that ˆ�..0;X// D Adx.Adt�1 .X/ � X//: These two calculations, together withthe linearity of the differential, establish the claimed form of ˆ�. ut

We now wish to determine when ˆ�.t; Œx�/ is invertible. Since Adx is invertible,the question becomes whether Ad0

t�1�I is invertible. Whenˆ�.t; Œx�/ is invertible,

we would like to know whether this linear map is orientation preserving ororientation reversing. In light of the way our orientations on T � .K=T / and Khave been chosen, the orientation behavior of ˆ� will be determined by the sign ofthe determinant of ˆ� as a linear map of k to itself. Now, since K is connected andour inner product on k is AdK-invariant, we see that Adx 2 SO.k/ for every x: Thus,det.Adx/ D 1; which means that we only need to calculate the determinant of thesecond factor on the right-hand side of (11.11).

Lemma 11.28. For t 2 T; let Ad0t�1

denote the restriction of Adt�1 to f:

1. If t generates a dense subgroup of T; then Ad0t�1

� I is an invertible lineartransformation of f:

2. For all w 2 W and t 2 T; we have

det.Ad0w�t�1 � I / D det.Ad0

t�1� I /:

11.5 Proof of the Torus Theorem 331

Proof. The operator Ad0t�1

� I is invertible provided that the restriction of Adt�1to f does not have an eigenvalue of 1. Suppose, then, that Adt�1 .X/ D X for someX 2 f: Then for every integerm;we will have Adtm.X/ D X: If t generates a densesubgroup of T; then by taking limits, we conclude that Ads.X/ D X for all s 2 T:

But then for all H 2 t; we have

ŒH;X� D d

d�Ade�H .X/

ˇˇ�D0

D 0:

Since t is maximal commutative (Proposition 11.7), we conclude that X 2 f \ t Df0g: Thus, there is no nonzeroX 2 f for which Adt�1 .X/ D X:

For the second point, if w 2 W is represented by x 2 N.T /; we have

Ad0w�t�1 � I D Ad0

xt�1x�1 � I

D Adx.Ad0t�1

� I /Adx�1 :

Thus, Ad0w�t�1 � I and Ad0

t�1� I are similar and have the same determinant. ut

We are now ready for the proof of Lemma 11.12, which will complete the proofof the torus theorem.

Proof of Lemma 11.12. By Proposition 11.4, we can choose t 2 T so that thesubgroup generated by t is dense in T: Then by Lemma 11.26, the preimagesof t are in one-to-one correspondence with elements of W: Furthermore, byProposition 11.27 and Point 1 of Lemma 11.28, ˆ� is nondegenerate at eachpreimage of t: Finally, by Point 2 of Lemma 11.28, ˆ� has the same orientationbehavior at each point of ˆ�1.ftg/: Thus, t is a regular value of ˆ and the signednumber of preimages of t under ˆ is either jW j or � jW j : It then follows fromCorollary 11.17 that ˆ is surjective, which is the content of Lemma 11.12. utCorollary 11.29. The Weyl group W is finite and the orientations on T � .K=T /andK can be chosen so that the mapping degree of ˆ is jW j ; the order of the Weylgroup.

Proof. If t generates a dense subgroup of T; then by Lemma 11.26, ˆ�1.ftg/ is inone-to-one correspondence withW: Furthermore, Point 1 of Lemma 11.28 then tellsus that such a t is a regular value ofˆ: Thus, by Proposition 11.14,ˆ�1.ftg/ is finite,andW is thus also finite. Meanwhile, we already noted in the proof of Lemma 11.12that ˆ has mapping degree equal to ˙ jW j : By reversing the orientation on K asnecessary, we can ensure that the mapping degree is jW j : ut

332 11 Compact Lie Groups and Maximal Tori

11.6 The Weyl Integral Formula

In this section, we apply Proposition 11.19 to the map ˆ to obtain an integrationformula for functions f on K: Of particular importance will be the special case inwhich f satisfies f .yxy�1/ D f .x/ for all x; y 2 K: This special case of the Weylintegral formula will be a main ingredient in our analytic proof of the Weyl characterformula in Sect. 12.4.

Recall that we have decomposed k as t˚ f; where f is the orthogonal complementof t; and that the adjoint action of T on k preserves both t and f: Define a function� W T ! R by

�.t/ D det.Ad0t�1

� I /; (11.12)

where Ad0t�1

is the restriction of Adt�1 to f: Using Proposition 11.24, we canconstruct volume forms on K;T; and K=T that are invariant under the left actionof K;T; and K; respectively. Since each of these manifolds is compact, the totalvolume is finite, and we can normalize this volume to equal 1.

Theorem 11.30 (Weyl Integral Formula). For all continuous functions f on K;we have

ZK

f .x/ dx D 1

jW jZT

�.t/

ZK=T

f .yty�1/ d Œy� dt; (11.13)

where dx; dt; and dŒy� are the normalized, left-invariant volume forms onK;T; andK=T; respectively and jW j is the order of the Weyl group.

In Sect. 12.4, we will compute � explicitly and relate it to the Weyl denominator.

Proof of Theorem 11.30, up to a constant. Since ˆ has mapping degree jW j,Theorem 11.16 tells us that

jW jZK

f .x/ dx DZT�.K=T /

ˆ�.f .x/ dx/

DZT�.K=T /

.f ıˆ/ ˆ�.dx/; (11.14)

for any smooth function f: Since, by the Stone–Weierstrass theorem (Theorem7.33 in [Rud1]), every continuous function on K can be uniformly approximatedby smooth functions, (11.14) continues to hold when f is continuous. Thus, toestablish (11.13), it suffices to show that ˆ�.dx/ D �.t/ d Œy� ^ dt:

Pick orthonormal basesH1; : : : ;Hr for t andX1; : : : ; XN for f: Then by the proofof Proposition 11.24, we can find invariant volume forms ˛1 on T; ˛2 on K=T , andˇ on K such that at each point, we have

˛1.H1; : : : ;Hr / D ˛2.X1; : : : ; XN / D ˇ.H1; : : : ;Hr ;X1; : : : ; XN / D 1;

11.6 The Weyl Integral Formula 333

so that

.˛1 ^ ˛2/.H1; : : : ;Hr ;X1; : : : ; XN / D 1:

By the uniqueness in Proposition 11.24, ˛1; ˛2; and ˇ will coincide, up tomultiplication by a constant, with the normalized volume forms dt; d Œy�; and dx;respectively.

Now, at each point, the matrix of ˆ�; with respect to the chosen bases for T .T �.K=T // and for T .K/; is given by the matrix in (11.11). Thus, using the definitionof the pulled-back form ˆ�.ˇ/; we have

ˆ�.ˇ/.H1; : : : ;Hr ;X1; : : : ; XN /

D ˇ.ˆ�.H1/; : : : ; ˆ�.Hr/;ˆ�.X1/; : : : ; ˆ�.XN //

D det.ˆ�/ˇ.H1; : : : ;Hr ;X1; : : : ; XN /

D �.t/.˛1 ^ ˛2/.H1; : : : ;Hr ;X1; : : : ; XN /;

where in the third line, we use (B.1) in Appendix B. Since dx coincides with ˇ upto a constant and dŒy� ^ dt coincides with ˛1 ^ ˛2 up to a constant, (11.14) thenbecomes

jW jZK

f .x/ dx D C

ZT�.K=T /

�.t/f .yty�1/ d Œy� dt:

It remains only to show that C D 1: We postpone the proof of this claim untilSect. 12.4. ut

It is possible to verify that C D 1 directly, using Lemma 11.21. According tothat result, if U is a small open set inK=T; then q�1.U / is diffeomorphic to U �T:It is not hard to check that under the diffeomorphism between q�1.U / and U � T ,the volume form ˇ decomposes as the product of ˛2 and ˛2: Thus, for any (niceenough)E � U; the volume of E � T D q�1.E/ is the product of the volume of Eand the volume of T: From this, it is not hard to show that for any (nice enough) setE � K=T; the volume of q�1.E/ equals the volume ofE (with respect to ˛2) timesthe volume of T (with respect to ˛1). In particular, the volume of q�1.K=T / D K

is the product of the volume of K=T and the volume of T: Thus, if we choose ourinner products on t and on f so that the volume forms ˛1 and ˛2 are normalized, thevolume form ˇ will also be normalized. In that case, the above computation holds onthe nose, without any undetermined constant. Since we will offer a different proofof the normalization constant in Sect. 12.4, we omit the details of this argument.

We now consider an important special case of Theorem 11.30.

Definition 11.31. A function f W K ! C is called a class function if f .yxy�1/ Df .x/ for all x; y 2 K:

334 11 Compact Lie Groups and Maximal Tori

That is to say, a function is a class function if it is constant on each conjugacyclass.

Corollary 11.32. If f is a continuous class function on K; then

ZK

f .x/ dx D 1

jW jZT

�.t/f .t/ dt: (11.15)

Proof. If f is a class function, then f .yty�1/ D f .t/ for all y 2 K and t 2 T:

Since the volume form onK=T is normalized, we have

ZK

f .yty�1/ d Œy� D f .t/;

in which case, the Weyl integral formula reduces to (11.15). utExample 11.33. Suppose K D SU.2/ and T is the diagonal subgroup. ThenCorollary 11.32 takes the form

ZSU.2/

f .x/ dx D 1

2

Z �

��f .diag.ei� ; e�i� // 4 sin2.�/

d�

2�; (11.16)

where jW j D 2 and the normalized volume measure on T is d�=.2�/:

Note that if f 1; both sides of (11.16) integrate to 1. See also Exercise 9 inChapter 12 for an explicit version of the Weyl integral formula for U.n/:

Proof. If we use the Hilbert–Schmidt inner product on su.2/; the orthogonalcomplement of t in su.2/ is the space of matrices X of the form

X D�

0 x C iy�x C iy 0

�;

with x; y 2 R: Direct computation then shows that if t D diag.ei� ; e�i� /; then

Adt�1 .X/ D�

0 e�2i� .x C iy/e2i� .�x C iy/ 0

�:

Thus, Adt�1 acts as a rotation by angle �2� in C D R2: It follows that

det.Adt�1 � I / D det

�cos.2�/� 1 sin.2�/� sin.2�/ cos.2�/� 1

�:

This determinant simplifies by elementary trigonometric identities to 4 sin2 �:Finally, since W D fI;�I g; we have jW j D 2: ut

11.7 Roots and the Structure of the Weyl Group 335

Note that the matrix diag.ei� ; e�i� / is conjugate in SU.2/ to the matrixdiag.e�i� ; ei� /:

�e�i� 0

0 ei�

�D�

0 1

�1 0��

ei� 0

0 e�i���

0 �11 0

�:

Thus, if f is a class function on SU.2/; the value of f at diag.e�i� ; ei� / is thesame its value at diag.ei� ; e�i� /: We may therefore rewrite the right-hand sideof (11.16) as

Z �

0

f .diag.ei� ; e�i� // 4 sin2.�/d�

2�: (11.17)

Meanwhile, recall from Exercise 5 in Chapter 1 that SU.2/ can be identifiedwith the unit sphere S3 � C2: By Exercise 9, a function f on SU.2/ is a classfunction if and only if the associated function on S3 depends only on the polarangle. Furthermore, for 0 � � � �; the polar angle associated to diag.ei� ; e�i� /is simply �: With this perspective, (11.17) is simply the formula for integration inspherical coordinates on S3; in the special case in which the function depends onlyon the polar angle. (Apply the m D 4 case of Eq. (21.15) in [Has] to a function thatdepends only on the polar angle.)

11.7 Roots and the Structure of the Weyl Group

In the context of compact Lie groups, it is convenient and customary to redefine thenotion of “root” by a factor of i so that the roots will now live in t rather than in i t:

Definition 11.34. An element ˛ of t is real root of g with respect to t if ˛ ¤ 0 andthere exists a nonzero element X of g such that

ŒH;X� D i h˛;H iX

for all H 2 t: For each real root ˛; we consider also the associated real coroot H˛

given by

H˛ D 2˛

h˛; ˛i :

When working with the groupK and its Lie algebra k; the use of real roots (and,later, real weights for representations) is convenient because it makes the factors ofi explicit, rather than hiding them in the fact that the roots live in i t: If, for example,we wish to compute the complex conjugate of the expression eih˛;H i; where ˛ is areal root andH is in t; the explicit factor of i makes it obvious that the conjugate ise�ih˛;H i:

336 11 Compact Lie Groups and Maximal Tori

If our compact group K is simply connected, then g WD kC is semisimple, byProposition 7.7. In general, k decomposes as k D k1˚z where z is the center of k andwhere g1 WD .k1/C is semisimple. (See the proof of Proposition 7.6.) Furthermore,as in the proof of Theorem 7.35, every maximal commutative subalgebra t of k willbe of the form t1 ˚ z; where t1 is a maximal commutative subalgebra of k1: All theresults from Chapter 7 then apply—with slight modifications to account for the useof real roots—except that the roots may not span t: Nevertheless, the roots form aroot system in their span, namely the space t1. Throughout the section, we will letR � t denote the set of real roots, denote a fixed base for R; and RC denote theassociated set of positive (real) roots.

Now that we have introduced the (real) roots for K; it makes sense to comparethe Weyl group in the compact group sense (the groupN.T /=T ) to the Weyl groupin the Lie algebra sense (the group generated by reflections about the hyperplanesperpendicular to the roots). As it turns out, these two groups are isomorphic. It isnot hard to show that for each reflection there is an associated element of the Weylgroup. The harder part of the proof is to show that these elements generate all ofN.T /=T: This last claim is proved by making a clever use of the torus theorem.

Proposition 11.35. For each ˛ 2 R; there is an element x in N.T / such that

Adx.H˛/ D �H˛

and such that

Adx.H/ D H

for all H 2 t for which h˛;H i D 0: Thus, the adjoint action of x on t is thereflection s˛ about the hyperplane orthogonal to ˛:

Proof. Choose X˛ and Y˛ as in Theorem 7.19, with Y˛ D X �: Then .X˛ � Y˛/� D�.X˛ � Y˛/; from which it follows that X˛ � Y˛ 2 k: Let us define x 2 K by

x D exph�2.X˛ � Y˛/

i

(where � is the number 3:14 � � � ; not a representation). Then by the relationshipbetween Ad and ad (Proposition 3.35), we have

Adx.H/ D exph�2.adX˛ � adY˛ /

i.H/ (11.18)

for all H 2 t: If h˛;H i D 0; then .adX˛ � adY˛ /.H/ D 0; so that AdX.H/ D H:

Consider, then, the case H D H˛: In that case, the entire calculation on theright-hand side of (11.18) taking place in the subalgebra s˛ D hX˛; Y˛;H˛i of g: Ins˛; the elements X˛ � Y˛; iX˛ C iY˛ CH˛; and iX˛ C iY˛ �H˛ are eigenvectorsfor adX˛ � adY˛ with eigenvalues 0; 2i; and �2i; respectively. Since H˛ is half thedifference of the last two vectors, we have

11.7 Roots and the Structure of the Weyl Group 337

exph�2.adX˛ � adY˛ /

i.H˛/

D ei�.iX˛ C iY˛ CH˛/=2� e�i� .iX˛ C iY˛ �H˛/=2

D �H˛: (11.19)

Thus, Adx mapsH˛ to �H˛ and is the identity on the orthogonal complement of ˛:Note that Adx.H/ belongs to t for all H in t: It follows that xtx�1 belongs to T forall t in T; showing that x is in N.T /: ut

See Exercise 10 for an alternative approach to verifying (11.19). We now proceedto show that the Weyl group is generated by the reflections s˛; ˛ 2 R: We let Z.T /denote the centralizer of T; that is

Z.T / D fx 2 Kj xt D tx; for all t 2 T g :Theorem 11.36. If T is a maximal torus in K; the following results hold.

1. Z.T / D T:

2. The Weyl group acts effectively on t and this action is generated by the reflectionss˛; ˛ 2 R; in Proposition 11.35.

Since Z.T / D T; it follows that T is a maximal commutative subgroup of T(i.e., there is no commutative subgroup of K properly containing T ). Nevertheless,there may exist maximal commutative subgroups of K that are not maximal tori;see Exercise 5.

It is not hard to verify Theorem 11.36 directly in the case of SU.n/; seeExercise 3. The following lemma is the key technical result in the proof ofTheorem 11.36 in general.

Lemma 11.37. SupposeS is a connected, commutative subgroup ofK: If x belongsto Z.S/; then there is a maximal torus S 0 containing both S and x:

Proof. Let x be in Z.S/, let B be the subgroup of K generated by S and x; and letNB be the closure of B: We are going to show that there is an element y of NB such

that the group generated by y is dense in NB: The torus theorem will then tell us thatthere is a maximal torus S 0 containing y and, thus, both S and x:

Since NB is compact and commutative, Theorem 11.2 implies that the identitycomponent NB0 of NB is a torus. Since NB is compact, it has only finitely manycomponents (Exercise 15 in Chapter 3), which means that the quotient group NB= NB0is finite.

Now, every element y of NB is the limit of sequence of the form xnk sk for someintegers nk and elements sk 2 S: Thus, for some large k; the element xnk sk willbe in the same component of NB as y: But since S is connected, xnk must also bein the same component of NB as y: It follows that Œy� and Œxnk � represent the sameelement of the quotient group NB= NB0:We conclude, then, that NB= NB0 is a cyclic groupgenerated by Œx�: Since also NB= NB0 is finite, it must be isomorphic to Z=m for somepositive integerm:

338 11 Compact Lie Groups and Maximal Tori

It follows that xm belongs to the torus NB0. Choose some t 2 NB0 such that thesubgroup generated by t is dense in NB0 (Proposition 11.4), and choose g 2 NB0 so thatgm D x�mt . (Since the exponential map for the torus NB0 is surjective, x�mt 2 NB0has an mth root in NB0:) Now set y D gx; so that y is in the same component of NBas x: Since NB is commutative, we have

ym D gmxm D t;

which means that the set of elements of the form ynm D tn is dense in NB0: Now,since NB= NB0 is cyclic with generator Œx�; each component of NB is of the form xk NB0for some k: Furthermore, the set of elements of the form ynmCk D tnxkgk; with kandm fixed and n varying, is dense in NB0:

We see, then, that the group generated by y contains a dense subset of eachcomponent of NB and is, thus, dense in NB: By the torus theorem, there is a maximaltorus S 0 that contains y. It follows that S 0 must contain NB and, in particular, both Sand x: ut

Figure 11.6 illustrates the proof of Lemma 11.37 in the case where NB= NB0 iscyclic of order 5: We choose y in the same component of NB as x in such a way thatt WD y5 generates a dense subgroup of NB0: Then y generates a dense subgroup ofthe whole group NB:

y2y3

y4y5 t

y

B

x

Fig. 11.6 The element y generates a dense subgroup of NB

11.7 Roots and the Structure of the Weyl Group 339

Proof of Point 1 of Theorem 11.36. If we apply Lemma 11.37 with S D T; we seethat any element x of Z.T / must be contained in a maximal torus S 0 that containsT: But since T is also maximal, we must have S 0 D T; so that x 2 T: utLemma 11.38. Suppose C is the fundamental Weyl chamber with respect to andthat w is an element of W that maps C to C: Then w D 1:

For any w 2 W; the action of W constitutes a symmetry of the root system R;

that is, an orthogonal linear transformation that maps R to itself. In some cases,such as the root system B2; there is no nontrivial symmetry of R that maps C to C:In the case of A2; on the other hand, there is a nontrivial symmetry of R that mapsC to C; namely the unique linear transformation that interchanges ˛1 and ˛2: (Thismap is just the reflection about the line through the root ˛3 D ˛1 C ˛2.) The lemmaasserts that although this map is a symmetry of W; it is not given by the action ofa Weyl group element. In the A2 case, of course, we have an explicit description ofthe Weyl group, and we can easily check that there is no w 2 W with w � ˛1 D ˛2and w � ˛2 D ˛1: (See Figure 11.7. where the Weyl group is the symmetry group ofthe indicated triangle.) Nevertheless, we need an argument that works in the generalcase.

The idea of the proof is as follows. We want to show that if x 2 N.T / andAdx.C / � C; then x 2 T . The idea is to show that x must commute with somenonzero H 2 t; and that this H can be chosen to be “nice.” If H could be chosenso that the group exp.tH/ were dense in T; then x would have to commute with

C

3

2

1

Fig. 11.7 The reflection about the line through ˛3 is a symmetry of R that maps C to itself, butthat is not an element of W

340 11 Compact Lie Groups and Maximal Tori

every element of T; so that x would belong to Z.T / D T . Although we cannot, ingeneral, chooseH to be as nice as that, we can chooseH to be in the interior of C;which means that h˛;H i ¤ 0 for all ˛ 2 R: This turns out to be sufficient to showthat x 2 T:Proof. Let x 2 K be a representative of the element w 2 W: Take any H0 in theinterior of C and averageH0 over the action of the (finite) subgroup ofW generatedby w: The resulting vector H is still in the interior of C (which is convex) and isnow fixed by w;meaning that Adx.H/ D H: Thus, x commutes with every elementof the one-parameter subgroup S WD fexp.tH/jt 2 Rg: By Lemma 11.37, there isa maximal torus S 0 containing x and S: Suppose, toward a contradiction, that x isnot in T: Then S 0 cannot equal T; and, since S 0 is maximal, S 0 cannot be containedin T: Thus, there is some X 2 k that is in the Lie algebra s0 of S 0 but not in t; andthis X commutes with exp.tH/; t 2 R; and hence with H:

On the other hand, suppose we decompose X 2 k � g as a sum of an elementof h and elements of the various root spaces g˛: Now, ŒH;X� D 0 and h˛;H i isnonzero for all ˛; since H is in the interior of C: Thus, the component of X ineach g˛ must be zero, meaning that X 2 h \ k D t; which is a contradiction. Thus,actually, x must be in T; which means that w is the identity element of W: utProof of Point 2 of Theorem 11.36. We let W 0 � W be the group generated byreflections. By Proposition 8.23, W 0 acts transitively on the set of Weyl chambers.Thus, for any w 2 W; we can find w0 2 W 0 mapping W.C/ back to C; so that w0wmaps C to C: Then by Lemma 11.38, we have that w0w D 1; which means thatw D .w0/�1 belongs to W 0: Thus, every element of W actually belongs to W 0: utTheorem 11.39. If two elements s and t of T are conjugate in K; then there existsan element w ofW such that w � s D t:

We may restate the theorem equivalently as follows: If t D xsx�1 for some x 2K , then t can also be expressed as t D ysy�1 for some y 2 N.T /:Proof. Suppose s and t are in T and t D xsx�1 for some x 2 K: Let Z.t/ bethe centralizer of t . Since, xux�1 commutes with t D xsx�1 for all u 2 T , we seethat xTx�1 � Z.t/: Thus, both T and xTx�1 are tori in Z.t/: Actually, since T andxTx�1 are connected, they must be contained in the identity component Z.t/0 ofZ.t/: Furthermore, since T and xTx�1 are maximal tori inK; they must be maximaltori in Z.t/0: Thus, we may apply the torus theorem to Z.t/0 to conclude that thereis some z 2 Z.t/0 such that

zxTx�1z�1 D T: (11.20)

Now, (11.20) says that zx is in the normalizer N.T / of T: Furthermore, sincez 2 Z.t/0 commutes with t; we have

.zx/s.zx/�1 D z.xsx�1/z�1 D ztz�1 D t:

Thus, y WD zx is the desired element of N.T / such that t D ysy�1: ut

11.8 Exercises 341

Corollary 11.40. If f is a continuous Weyl-invariant function on T; then f extendsuniquely to a continuous class function on K:

Proof. By the torus theorem, each conjugacy class in K intersects T in at least onepoint. By Theorem 11.39, each conjugacy class intersects T in a single orbit of W:Thus, if f is a W -invariant function on T; we can unambiguously (and uniquely)extend f to a class functionF onK by making F constant on each conjugacy class.

It remains to show that the extended function F is continuous on K: Suppose,then that hxni is a sequence in K converging to some x: We can write each xn asxn D yntny

�1n ; with yn 2 K and tn 2 T: Since both K and T are compact, we

can—after passing to a subsequence—assume that yn converges to some y 2 K

and tn converges to some t 2 T: It follows that

x D limn!1xn D lim

n!1yntny�1n D yty�1:

Now, by our construction of F , we have F.xn/ D f .tn/ and F.x/ D f .t/: Thus,since f is assumed to be continuous on T; we see that

F.xn/ D f .tn/ ! f .t/ D F.x/;

showing that F is continuous onK: ut

11.8 Exercises

1. Let � denote the set of all vectors in R2 that can be expressed in the form

a.1; 1/C b.3; 1/C c.2;�4/;

for a; b; c 2 Z: Then � is a subgroup of R2 and � is discrete, since it iscontained in Z2: Find linearly independent vectors v1 and v2 in � such that� consists precisely of the set of integer linear combinations of v1 and v2:(Compare Lemma 11.3.)

2. (a) Show that every continuous homomorphism from S1 to S1 is of the formu 7! um for some m 2 Z:

Hint: Use Theorem 3.28.(b) Show that every continuous homomorphism from .S1/k ! S1 is of the

form

.u1; : : : ; uk/ 7! um11 � � � umkk

for some integersm1; : : : ; mk:

3. Consider the group SU.n/; with maximal torus T being the intersection ofSU.n/ with the space of diagonal matrices. Prove directly (without appealing

342 11 Compact Lie Groups and Maximal Tori

to Theorem 11.36) that Z.T / D T and that the N.T /=T is isomorphic to theWeyl group of the root system An�1: (Compare Sect. 7.7.1.)Hint: Imitate the calculations in Sect. 6.6.

4. Give an example of closed, oriented manifoldsM andN of the same dimensionand a smooth map f W M ! N such that f has mapping degree zero but f is,nevertheless, surjective.

5. Let K D SO.n/; where n � 3: Let H be the (commutative) subgroup of Kconsisting of the diagonal matrices in SO.n/: (Of course, the diagonal entrieshave to be ˙1 and the number of diagonal entries equal to �1 must be even.)Show that H is a maximal commutative subgroup of K and that H is notcontained in a maximal torus.Hint: Use Proposition A.2.Note: This example shows that in Lemma 11.37, the assumption that S beconnected cannot be omitted. (Otherwise, we could take S D H and x D I

and we would conclude that there is a maximal torus S 0 containingH .)6. Suppose K D SU.n/ and H is any commutative subgroup of K: Show that H

is conjugate to a subgroup of the diagonal subgroup of K and thus that H iscontained in a maximal torus. This result should be contrasted with the resultof Exercise 5.

7. (a) For any interval .a; b/ � R and any x; y 2 .a; b/; show that there exists asmooth family of diffeomorphisms ft W .a; b/ ! .a; b/; 0 � t � 1; withthe following properties. First, f0.z/ D z for all z: Second, there is some" > 0 such that ft .z/ D z for all z 2 .a; a C "/ and for all z 2 .b � "; b/:

Third, f1.x/ D y:

Hint: Take

ft .z/ D z C t

Z z

a

g.u/ du

for some carefully chosen function g:(b) If R � R

n is an open rectangle and x and y belong to R; show that thereis a smooth family of diffeomorphisms‰t W R ! R such that (1) ‰0 is theidentity map, (2) each ‰t is the identity in a neighborhood of @R; and (3)‰1.x/ D y:

8. (a) Show that the left action of G on G=H is smooth with respect tothe collection of coordinate patches on G=H described in the proof ofTheorem 11.22.

(b) Show that the kernel of the differential of the quotient mapQ W G ! G=H

at the identity is precisely h:9. According to Exercise 5 in Chapter 1, each element of SU.2/ can be written

uniquely as

U D�˛ � Nˇ N

�;

11.8 Exercises 343

where .˛; ˇ/ 2 C2 belongs to the unit sphere S3: For each U 2 SU.2/; let vUdenote the corresponding unit vector .˛; ˇ/: Now, the angle � between .˛; ˇ/ 2S3 and the “north pole” .1; 0/ satisfies

cos � D Re.h.˛; ˇ/; .1; 0/i/ D Re.˛/;

and this relation uniquely determines �; if we take 0 � � � �: In sphericalcoordinates on S3; the angle � is the polar angle.

(a) Suppose U 2 SU.2/ has eigenvalues ei� and e�i� . Show that

Re.hvU ; .1; 0/i/ D cos �:

Hint: Use the trace.(b) Conclude that U1 and U2 are conjugate in SU.2/ if and only if vU1 and vU2

have the same polar angle.

10. In this exercise, we give an alternative verification of the identity (11.19). Sincethe left-hand side of (11.19) is expressed in purely Lie-algebraic terms, we maydo the calculation in any Lie algebra isomorphic to hX˛; Y˛;H˛i ; for example,in sl.2IC/ itself. That is to say, it suffices to prove the formula with X˛ DX; Y˛ D Y; and H˛ D H; where X; Y; and H are the usual basis elements forsl.2IC/:

Show that

exph�2.adX � adY /

i.H/ D e

�2 .X�Y /He� �

2 .X�Y / D �H;

as claimed.

Chapter 12The Compact Group Approachto Representation Theory

In this chapter, we follow Hermann Weyl’s original approach to establishing theWeyl character formula and the theorem of the highest weight. Throughout thechapter, we assumeK is a connected, compact matrix Lie group, with Lie algebra k:We fix on k an inner product that is invariant under the adjoint action of K;constructed as in the proof of Theorem 4.28. Throughout the chapter, we let Tdenote a fixed maximal torus in K and we let t denote the Lie algebra of T: Welet R denote the set of real roots for k relative to t (Definition 11.34), we let be a fixed base for R; and we let RC denote the positive real roots relative to .We also let W WD N.T /=T denote the Weyl group for K relative to T: In lightof Theorem 11.36, W is isomorphic to the subgroup of O.t/ generated by thereflections about the hyperplanes orthogonal to the roots.

12.1 Representations

All representations of K considered in this chapter are assumed to be finitedimensional and defined on a vector space overC; unless otherwise stated. Althoughwe are studying in this chapter the representations of the compact group K; it isconvenient to describe the weights of such a representation in terms of the associatedrepresentation of g D kC: Since we are using real roots for g, we will also use realweights for representations of K:

A previous version of this book was inadvertently published without the middle initial of theauthor’s name as “Brian Hall”. For this reason an erratum has been published, correcting themistake in the previous version and showing the correct name as Brian C. Hall (see DOI http://dx.doi.org/10.1007/978-3-319-13467-3_14). The version readers currently see is the correctedversion. The Publisher would like to apologize for the earlier mistake.

© Springer International Publishing Switzerland 2015B.C. Hall, Lie Groups, Lie Algebras, and Representations, GraduateTexts in Mathematics 222, DOI 10.1007/978-3-319-13467-3_12

345

346 12 The Compact Group Approach to Representation Theory

Definition 12.1. Let .…; V / be a finite-dimensional representation of K and � theassociated representation of g: An element � of t is called a real weight of V ifthere exists a nonzero element v of V such that

�.H/v D i h�;H i v (12.1)

for all H 2 t: The weight space with weight � is the set of all v 2 V

satisfying (12.1) and the multiplicity of � is the dimension of the correspondingweight space.

We now consider some elementary properties of the weights of a representation.

Proposition 12.2. If .…; V / is a finite-dimensional representation of K; the realweights for … and their multiplicities are invariant under the action of the Weylgroup.

Proof. Following the proof of Theorem 6.22, we can show that if w 2 W isrepresented by x 2 N.T /; then ….x/ will map the weight space with weight �isomorphically onto the weight space with weight w � �: ut

The weights of representation of K satisfy an integrality condition that does not,in general, coincide with the notion of integrality in Definition 8.34.

Definition 12.3. Let � be the subset of t given by

� D ˚H 2 t

ˇe2�H D I

�:

We refer to � as the kernel of the exponential map for t:

The set � should, more precisely, be referred to as the kernel of the exponentialmap scaled by a factor of 2�: Note that if w 2 W is represented by x 2 N.T /; thenfor all H 2 �; we have

e2�w�H D e2�xHx�1 D xe2�Hx�1 D I:

Thus, � is invariant under the action of W on t:

Definition 12.4. An element of � of t is an analytically integral element if

h�;H i 2 Z

for all H in �: An element � of t is an algebraically integral element if

h�;H˛i D 2h�; ˛ih˛; ˛i 2 Z

for each real root ˛: An element � of t is dominant if

h�; ˛i � 0

for all ˛ 2 : Finally, if D f˛1; : : : ; ˛r g and � and � are two elements of t; wesay that � is higher than � if

12.1 Representations 347

� � � D c1˛1 C � � � C cr˛r

with cj � 0:We denote this relation by � � �:

The notion of an algebraically integral element is essentially the one we used inChapters 8 and 9 in the context of semisimple Lie algebras. Specifically, if g WD kCis semisimple, then � 2 t is algebraically integral if and only if i� is integral in thesense of Definition 8.34. We will see in Sect. 12.2 that every algebraically integralelement is analytically integral, but not vice versa. In Chapter 13, we will show thatwhenK is simply connected, the two notions of integrality coincide.

Proposition 12.5. Let .†; V / be a representation of K and let be the associatedrepresentation of k: If � 2 t is a real weight of ; then � is an analytically integralelement.

Proof. If v is a weight vector with weight � and H is an element of �; then on theone hand,

†.e2�H /v D Iv D v;

while on the other hand,

†.e2�H /v D e2�.H/v D e2�ih�;H iv:

Thus, e2�ih�;H i D 1, which implies that h�;H i 2 Z: utWe are now ready to state the theorem of the highest weight for (finite-

dimensional) representations of a connected compact group.

Theorem 12.6 (Theorem of the Highest Weight). If K is a connected, compactmatrix Lie group and T is a maximal torus in K; the following results hold.

1. Every irreducible representation of K has a highest weight.2. Two irreducible representations of K with the same highest weight are

isomorphic.3. The highest weight of each irreducible representation of K is dominant and

analytically integral.4. If � is a dominant, analytically integral element, there exists an irreducible

representation of K with highest weight �:

Let us suppose now that g WD kC is semisimple. Even in this case, the set ofanalytically integral elements may not coincide with the set of algebraically integralelements, as we will see in several examples in Sect. 12.2. Thus, the theorem ofthe highest weight for g (Theorems 9.4 and 9.5) will, in general, have a differentset of possible highest weights than the theorem of the highest weight for K: Thisdiscrepancy arises because a representation of g may not come from a representationof K; unless K is simply connected. In the simply connected case, there is nosuch discrepancy; according to Corollary 13.20, whenK is simply connected, everyalgebraically integral element is analytically integral.

348 12 The Compact Group Approach to Representation Theory

We will develop the tools for proving Theorem 12.6 in Sects. 12.2–12.4, with theproof itself coming in Sect. 12.5. The hard part of the theorem is Point 4; this willbe established by appealing to a completeness result for characters.

12.2 Analytically Integral Elements

In this section, we establish some elementary properties of the set of analyticallyintegral elements (Definition 12.4) and consider several examples. One additionalkey result that we will establish in Chapter 13 is Corollary 13.20, which says thatwhenK is simply connected, the set of analytically integral elements coincides withthe set of algebraically integral elements.

Proposition 12.7.

1. The set of analytically integral elements is invariant under the action of the Weylgroup.

2. Every analytically integral element is algebraically integral.3. Every real root is analytically integral.

We begin with an important lemma.

Lemma 12.8. If ˛ 2 t is a real root and H˛ D 2˛= h˛; ˛i is the associated realcoroot, we have

e2�H˛ D I

in K: That is to say, H˛ belongs to the kernel � of the exponential map.

Proof. By Corollary 7.20, there is a Lie algebra homomorphism � W su.2/ ! ksuch that the element iH D diag.i;�i/ in su.2/ maps to the real coroot H˛: (Inthe notation of the corollary, H˛ D 2E˛

1 :) Since SU.2/ is simply connected, thereis (Theorem 5.6) a homomorphism ˆ W SU.2/ ! K for which the associated Liealgebra homomorphism is �: Now, the element iH 2 su.2/ satisfies e2� iH D I , andthus

I D ˆ.e2� iH/ D e2��.iH/ D e2�H˛ ;

as claimed. ut

Proof of Proposition 12.7. For Point 1, we have already shown (following Defi-nition 12.3) that � is invariant under the action of W: Thus, if � is analyticallyintegral and w is the Weyl group element represented by x; we have hw � �;H i D˝�;w�1 �H ˛ 2 Z; since w�1 � H is in �: For Point 2, note that for each ˛; we

have H˛ 2 � by Lemma 12.8. Thus, if � is analytically integral, h�;H˛i 2 Z;

showing that � is algebraically integral. For Point 3, we note that the real roots arereal weights of the adjoint representation, which is a representation of the groupK ,and not just its Lie algebra. Thus, by Proposition 12.5, the real roots are analyticallyintegral. ut

12.2 Analytically Integral Elements 349

Proposition 12.9. If � is an analytically integral element, there is a well-definedfunction f� W T ! C such that

f�.eH / D eih�;H i (12.2)

for all H 2 t: Conversely, for any � 2 t; if there is a well-defined function on Tgiven by the right-hand side of (12.2), then � must be analytically integral.

Proof. Replacing H by 2�H; we can equivalently write (12.2) as

f�.e2�H / D e2�ih�;H i; H 2 t: (12.3)

Now, since T is connected and commutative, every t 2 T can be written as t De2�H for some H 2 t: Furthermore, e2�.HCH 0/ D e2�H if and only if e2�H

0 D I;

that is, if and only if H 0 2 �: Thus, the right-hand side of (12.3) defines a functionon T if and only if e2�ih�;HCH 0i D e2�ih�;H i for all H 0 2 �: This happens if andonly if e2�ih�;H 0i D 1 for all H 0 2 �; that is, if and only if h�;H 0i 2 Z for allH 0 2 �: utProposition 12.10. The exponentials f� in (12.2), as � ranges over the set ofanalytically integral elements, are orthonormal with respect to the normalizedvolume form dt on T :

ZT

f�.t/f�0.t/ dt D ı�;�0 : (12.4)

Proof. Let us identify T with .S1/k for some k; so that t is identified with Rk andthe scaled exponential map is given by

.�1; : : : ; �n/ 7! .e2�i�1 ; : : : ; e2�i�k /:

The kernel � of the exponential is the integer lattice inside Rk: The lattice ofanalytically integral elements (points having integer inner product with each elementof �) is then also the integer lattice. Thus, the exponentials in the proposition arethe functions of the form

f�.ei�1 ; : : : ; ei�k / D ei�1�1 � � � ei�k�k ;

with � D .�1; : : : ; �k/ 2 Zk:

Meanwhile, if we use the coordinates �1; : : : ; �k on T , then any k-form on T canbe represented as a density � times d�1^� � �^d�n: Since the volume form dt on T istranslation invariant, the density � must be constant. Thus, the normalized integralin (12.4) becomesZ

T

f�.t/f�0.t/ dt

D .2�/�kZ 2�

0

� � �Z 2�

0

ei�1�1 � � � ei�k�k ei�0

1�1 � � � ei�0

k�k d�1 � � �d�k

350 12 The Compact Group Approach to Representation Theory

Fig. 12.1 Dominant, analytically integral elements (black dots) and dominant, algebraicallyintegral elements (black and white dots) for SO.3/

D .2�/�k�Z 2�

0

ei.�0

1��1/�1d�1�

� � ��Z 2�

0

ei.�0

k��k/�kd�k�:

The claimed orthonormality then follows from direct computation. utWe now calculate the algebraically integral and analytically integral elements

in several examples, with an eye toward clarifying the distinction between thetwo notions. When K is simply connected, Corollary 13.20 shows that the setof analytically integral and algebraically integral elements coincide. Thus, in thesimply connected case, the calculations in Sects. 6.7 and 8.7 provide examples ofthe set of analytically integral elements. We consider now three groups that are notsimply connected.

Example 12.11. Consider the group SO.3/ and let t be the maximal commutativesubalgebra spanned by the element F3 in Example 3.27. Let the unique positive root˛ be chosen so that h˛; F3i D 1: Then � 2 t is dominant and algebraically integralif and only if � D m˛=2; wherem is a non-negative integer, and � 2 t is dominantand analytically integral if and only if � D m˛; where m is a non-negative integer.

See Figure 12.1. Note that in this case, ı (half the sum of the positive roots) isequal to ˛=2; which is not an analytically integral element.

Proof. Following Sect. 7.7, but adjusting for our current convention of using realroots (Definition 11.34), we identify t with R by mapping aF3 to a: The roots arethen the numbers ˙1; where we take 1 as our positive root ˛, so that h˛; F3i D 1:

Then � 2 t Š R is dominant if and only if � � 0: Furthermore, � is algebraicallyintegral if and only if

2 h�; ˛i D 2.�/.1/ 2 Z;

that is, if and only if 2� is an integer. Thus, the dominant, algebraically integralelements are the numbers of the formm=2 D m˛=2:

Now, e2�aF3 D I if and only if a is an integer. Thus, � is analytically integral ifand only if .�/.a/ 2 Z for all a 2 Z; that is, if and only if � is an integer. Thus, thedominant, analytically integral elements are the numbers of the formm D m˛: ut

We consider next the group U.2/, with t consisting of diagonal matrices withpure imaginary diagonal entries. We identify t with R2 by mapping diag.ia; ib/ to.a; b/: The roots are then the elements of the form .1; 1/ and .�1;�1/; and weselect ˛ WD .1; 1/ as our positive root. We now decompose everyH 2 � as a linearcombination of the vectors

˛ D .1; 1/I ˇ D .1;�1/: (12.5)

12.2 Analytically Integral Elements 351

Fig. 12.2 The dominant, analytically integral elements (black dots) and nondominant, analyticallyintegral elements (white dots) for U.2/: The vertical lines indicate the algebraically integralelements

Example 12.12. Let t be the diagonal subalgebra of u.2/, and write every element� 2 t as

� D c˛ C dˇ;

with ˛ and ˇ as in (12.5). Then � is analytically integral if and only if either c and dare both integers or c and d are both of the form integer plus one-half. Furthermore,� is dominant if and only if c � 0: On the other hand, � is algebraically integral ifand only if c is either an integer or an integer plus one-half.

In Figure 12.2, the black dots are the dominant, analytically integral elements andthe white dots are the nondominant, analytically integral elements. All the points inthe vertical lines are algebraically integral.

Proof. If H D diag.ia; ib/; then e2� iH D I if and only if a and b are both integers.Thus, when we identify t with R2; the kernel � of the exponential corresponds to theinteger lattice Z2: The lattice of analytically integral elements is then also identifiedwith Z2:Now, it is straightforward to check that c˛Cdˇ is in Z2 if and only if eitherc and d are both integers or c and d are both integers plus one-half, accounting forthe claimed form of the analytically integral elements. Since ˇ is orthogonal to ˛;an element � D c˛Cdˇ has non-negative inner product with ˛ if and only if c � 0;

accounting for the claimed condition for � to be dominant. Finally, � is algebraicallyintegral if and only if 2 h�; ˛i = h˛; ˛i is an integer, which happens if c is either aninteger or an integer plus one-half, with no restriction on d: ut

352 12 The Compact Group Approach to Representation Theory

1

2

Fig. 12.3 Dominant, analytically integral elements (black dots) and dominant, algebraicallyintegral elements (black and white dots) for SO.5/

Example 12.13. The dominant, analytically integral elements for SO.5/ are asshown in Figure 12.3.

The figure shows the dominant, analytically integral elements (black dots). Thewhite dots are dominant, algebraically integral elements that are not analyticallyintegral. The background square lattice is the set of all analytically integral elements.Note that in Figure 12.3, the B2 root system is rotated by �=4 compared toFigure 8.11. If we rotate Figure 12.3 clockwise by �=4 and then reflect across thex-axis, the set of dominant algebraically integral elements in Figure 12.3 (black andwhite dots) will match the set of “dominant integral” elements in Figure 8.11. Notethat ı (half the sum of the positive roots) is not analytically integral.

Proof. Elements of t are of the form

H D

0BBBBB@

0 a

�a 00 b

�b 00

1CCCCCA; (12.6)

with a; b 2 R: Following Sect. 7.7, but adjusting for our current convention of usingreal roots, we identify t with R

2 by means of the map H 7! .a; b/: The roots are

12.3 Orthonormality and Completeness for Characters 353

then the elements .˙1; 0/; .0;˙1/; and .˙1;˙1/: As a base, we take ˛1 D .1;�1/and ˛2 D .0; 1/: Furthermore, .x; y/ is algebraically integral if

2h.1;�1/; .x; y/i

2D x � y 2 Z

and

2h.0; 1/; .x; y/i

1D 2y 2 Z:

These conditions hold if and only if either x and y are both integers or x and y areboth of the form “integer plus one-half.”

Now, if H is as in (12.6), e2�H D I if and only if a and b are integers. Thus,under our identification of t with R2; the kernel of the exponential map is the set ofelements of the form .a; b/; with a; b 2 Z: Thus, .x; y/ is analytically integral if

h.a; b/; .x; y/i D ax C by 2 Z

for all a; b 2 Z: This condition holds if and only if x and y are both integers.Finally, .x; y/ is dominant if it has non-negative inner product with each of ˛1

and ˛2; which happens if .x; y/ is in the 45-degree sector indicated by dashed linesin Figure 12.3. ut

12.3 Orthonormality and Completeness for Characters

In this section, we show that the characters of irreducible representations form anorthonormal set and that these characters are complete in the space of continuousclass functions on K:

Definition 12.14. Suppose .…; V / is representation of K: Then the character of… is the function �… W K ! C given by

�….x/ D trace.….x//:

Note that we now consider the character as a function on the groupK; rather thanon the Lie algebra g D kC; as in Chapter 10. If � is the associated representationof g; then the character �� of � (Definition 10.11) is related to the character �… of… by

�….eH/ D ��.H/; H 2 k:

Note that each character is a class function on K:

�….yxy�1/ D trace.….y/….x/….y/�1/ D trace.….x//:

354 12 The Compact Group Approach to Representation Theory

The following theorem says that the characters of irreducible representations forman orthonormal set in the space of class functions.

Theorem 12.15. If .…; V / and .†;W / are irreducible representations of K; then

ZK

trace.….x//trace.†.x// dx D�1 if V Š W

0 if V © W;

where dx is the normalized left-invariant volume form on K:

If .…; V / is a representation of K; let V K denote the space given by

V K D fv 2 V j….x/v D v for all x 2 K g :

Lemma 12.16. Suppose .…; V / is a finite-dimensional representation ofK; and letP be the operator on V given by

P DZK

….x/ dx:

Then P is a projection onto V K: That is to say, P maps V into V K and Pv D v forall v 2 V K:

Clearly, V K is an invariant subspace for …: If we pick an inner product on V forwhich… is unitary, then .V K/? is also invariant under each….x/ and thus underP:But since P maps into V K; the map P must be zero on .V K/?; thus, P is actuallythe orthogonal projection onto V K:

Proof. For any y 2 K and v 2 V; we have

….y/Pv D ….y/

�ZK

….x/ dx

�v

D�Z

K

….yx/ dx

�v

D Pv;

by the left-invariance of the form dx: This shows that Pv belongs to V K:Meanwhile,if v 2 V K; then

Pv DZK

….x/v dx

D�Z

K

dx

�v

D v;

by the normalization of the volume form dx: ut

12.3 Orthonormality and Completeness for Characters 355

Note that if V is irreducible and nontrivial, then V K D f0g: In this case, theproposition says that

RK….x/ dx D 0:

Lemma 12.17. For A W V ! V and B W W ! W; we have

trace.A/trace.B/ D trace.A˝ B/;

where A˝ B W V ˝W ! V ˝W is as in Proposition 4.16.

Proof. If fvj g and fwlg are bases for V and W; respectively, then fvj ˝ wlg is abasis for V ˝ W: If Ajk and Blm are the matrices of A and B with respect to fvj gand fwlg; respectively, then the matrix of A˝B with respect to fvj ˝ wlg is easilyseen to be

.A˝ B/.j;l/.k;m/ D AjkBlm:

Thus,

trace.A˝B/ DXj;l

AjjBll D trace.A/trace.B/;

as claimed. utProof of Theorem 12.15. We know that there exists an inner product on V for whicheach ….x/ is unitary. Thus,

trace.….x// D trace.….x/�/ D trace.….x�1//: (12.7)

Recall from Sect. 4.3.3 that for any A W V ! V; we have the transpose operatorAtr W V � ! V �. Since the matrix of Atr with respect to the dual of any basisfvj g of V is the transpose of the matrix of A with respect to fvj g; we see thattrace.Atr/ D trace.A/: Thus,

trace.….x// D trace.….x�1/tr/ D trace.…tr.x//;

where …tr is the dual representation to …. Thus, the complex conjugate of thecharacter of … is the character of the dual representation…tr of …:

Using Lemma 12.17, we then obtain

trace.….x//trace.†.x// D trace.…tr.x/˝†.x//

D trace..…tr ˝†/.x//:

356 12 The Compact Group Approach to Representation Theory

By Lemma 12.16, this becomesZK

trace.….x//trace.†.x// dx DZK

trace..…tr ˝†/.x// dx

D trace

�ZK

.…tr ˝†/.x/ dx

D trace.P /

D dim..V � ˝W /K/ (12.8)

where P is a projection of V � ˝W onto .V � ˝W /K:

Now, for any two finite-dimensional vector spaces V and W; there is a naturalisomorphism between V � ˝ W and End.V;W /; the space of linear maps from V

to W: This isomorphism is actually an intertwining map of representations, wherex 2 K acts on A 2 End.V;W / by

x � A D †.x/A….x/�1:

Finally, under this isomorphism, .V � ˝ W /K maps to the space of intertwiningmaps of V to W: (See Exercises 3 and 4 for the proofs of the preceding claims.)By Schur’s lemma, the space of intertwining maps has dimension 1 if V Š W anddimension 0 otherwise. Thus, (12.8) reduces to the claimed result. ut

Our next result says that the characters form a complete orthonormal set in thespace of class functions on K:

Theorem 12.18. Suppose f is a continuous class function onK and that for everyfinite-dimensional, irreducible representation… of K; the function f is orthogonalto the character of …: Z

K

f .x/trace.….x// dx D 0:

Then f is identically zero.

If K D S1; the irreducible representations are one dimensional and of the form….ei� / D eim�I; m 2 Z; so that �….ei� / D eim� : In this case, the completenessresult for characters reduces to a standard result about Fourier series.

The proof given in this section assumes in an essential way that K is a compactmatrix Lie group. (Of course, we work throughout the book with matrix Lie groups,but most of the proofs we give extend with minor modifications to arbitrary Liegroups.) In Appendix D, we sketch a proof of Theorem 12.18 that does not rely onthe assumption that K is a matrix group. That proof, however, requires a bit morefunctional analysis than the proof given in this section.

We now consider a class of functions called matrix entries, that include as aspecial case the characters of representations. We will prove a completeness resultfor matrix entries and then specialize this result to class functions in order to obtaincompleteness for characters.

12.3 Orthonormality and Completeness for Characters 357

Definition 12.19. If .…; V / is a representation of K and fvj g is a basis for V; thefunctions f W K ! C of the form

f .x/ D .….x//jk (12.9)

are called matrix entries for…. Here .….x//jk denotes the .j; k/ entry of the matrixof ….x/ in the basis fvj g:

In a slight abuse of notation, we will also call f a matrix entry for … if f isexpressible as a linear combination of the functions in (12.9):

f .x/ DXj;k

cjk.….x//jk: (12.10)

We may write functions of the form (12.10) in a basis-independent way as

f .x/ D trace.….x/A/;

where A is the operator on V whose matrix in the basis fvj g is Ajk D ckj: (Thesum over k computes the product of….x/ with A and the sum over j computes thetrace.) Note that if A D I then f is the character of…:

We will actually prove an orthogonality and completeness result for matrixentries, known as the Peter–Weyl theorem, which will imply the desired complete-ness result for characters. According to Exercise 5, matrix entries for nonisomorphicirreducible representations are orthogonal. We will show a completeness result formatrix entries as well: If f is continuous function onK and f is orthogonal to everymatrix entry, then f is identically zero. The Peter–Weyl theorem is an importantresult in the study of functions on a compact group, quite apart from its role in theproof of the theorem of the highest weight.

Lemma 12.20. If .…; V / is an irreducible representation of K; then for eachoperator A on V; we have Z

K

….y/A….y/�1 dy D cI (12.11)

for some constant c.

Proof. Let B denote the operator on the left-hand side of (12.11). By the left-invariance of the integral, we have

….x/B….x/�1 DZK

….xy/A….xy/�1 dy

DZK

….y/A….y/�1 dy

D B:

Thus, B commutes with each ….x/; which, by Schur’s lemma, implies that B is amultiple of the identity. ut

358 12 The Compact Group Approach to Representation Theory

We are now ready for the proof of our completeness result for characters.

Proof of Theorem 12.18. Let A denote the space of continuous functions on Kthat can be expressed as a linear combination of matrix entries, for some finitecollection of representations of K: We claim that A satisfies the hypotheses of thecomplex version of the Stone–Weierstrass theorem, namely, that is an algebra, that itvanishes nowhere, that it is closed under complex conjugation, and that it separatespoints. (See Theorem 7.33 in [Rud1].) First, using Lemma 12.17, we see that theproduct of two matrix entries is a matrix entry for the tensor product representation,which decomposes as a direct sum of irreducibles. Thus, the product of two matrixentries is expressible as a linear combination of matrix entries, showing that Ais an algebra. Second, the matrix entries of the trivial representation are nonzeroconstants, showing that A is nowhere vanishing. Third, by a simple extensionof (12.7), we can show that the complex conjugate of a matrix entry is a matrixentry for the dual representation. Last, since K is a matrix Lie group, it has, bydefinition, a faithful finite-dimensional representation. The matrix entries of anysuch representation separate points in K .

Thus, the complex version of the Stone–Weierstrass theorem applies to A;meaning that if f is continuous, we can find g 2 A such that g is everywherewithin " of f: If f is a class function, then for all x; y 2 K; we have

ˇf .x/ � g.yxy�1/

ˇ D ˇf .yxy�1/ � g.yxy�1/

ˇ< ":

Thus, if h is given by

h.x/ DZK

g.y�1xy/ dy;

then h will also be everywhere within " of f:Now, by assumption, g can be represented as

g.x/ DXj

trace.…j .x/Aj /

for some family of representations .…j ; Vj / of K and some operators Aj 2End.Vj /: We can then easily compute that

h.x/ DXj

trace.…j .x/Bj /;

where

Bj DZK

….y/Aj….y/�1 dy:

But by Lemma 12.20, each Bj is a multiple of the identity, which means that h is alinear combination of characters.

12.4 The Analytic Proof of the Weyl Character Formula 359

We conclude, then, that every continuous class function f can be uniformlyapproximated by a sequence of functions hn;where each hn is a linear combinationsof characters. If f is orthogonal to every character, f is orthogonal to each hn:Then, by letting n tend to infinity, we find that f is orthogonal to itself, meaningthat

RK jf .x/j2 dx D 0: Since f is continuous, this can only happen if f is

identically zero. utWe used the assumption that K is a matrix Lie group to show that the algebra

A separates points in K; which allowed us to prove (using the Stone–Weierstrasstheorem) that A is dense in the space of continuous functions onK: In Appendix D,we sketch a proof of Theorem 12.18 that does not assume ahead of time that K is amatrix group.

12.4 The Analytic Proof of the Weyl Character Formula

In order to simplify certain parts of the analysis, we make the following temporaryassumption concerning ı (half the sum of the real, positive roots).

Assumption 12.21 In this section and in Sect. 12.5, we assume that the element ıis analytically integral.

As we will show in Chapter 13 (Corollary 13.20), ı is analytically integralwhenever K is simply connected. In Sect. 12.6, we describe the modificationsneeded to the arguments when ı is not analytically integral. We will give, in thissection, an an analytic proof of the Weyl character formula, as an alternative to thealgebraic argument in Sect. 10.8. The proof is based on a more-explicit version of theWeyl integral formula, obtained by computing the weight function �.t/ occurring inTheorem 11.30. We will see that the function � is the square of another function,which turns out to be our old friend, the Weyl denominator. This computation willprovide a crucial link between the Weyl integral formula and the Weyl characterformula.

If ı denotes half the sum of the positive real roots, then the Weyl denominatorfunction (Definition 10.15) takes the form

q.H/ DXw2W

det.w/eihw�ı;H i; H 2 t: (12.12)

By Assumption 12.21 and Proposition 12.9, each exponential eihw�ı;H i;w 2 W;

defines a function on T: Thus, there is a unique functionQ W T ! C satisfying

Q.eH / D q.H/; H 2 t: (12.13)

We now state Weyl’s formula for the character (Definition 12.14) �… of …:

Theorem 12.22 (Weyl Character Formula). Suppose .…; V / is an irreduciblerepresentation of K and that � is a maximal weight for …: Then � is dominantand analytically integral, and � is actually the highest weight of …: Furthermore,the character of … is given by the formula

360 12 The Compact Group Approach to Representation Theory

�….eH / D

Pw2W det.w/eihw�.�Cı/;H i

q.H/; (12.14)

at all points eH 2 T for which q.H/ ¤ 0: In particular, every irreduciblerepresentation has a highest weight that is dominant and analytically integral.

The character formula amounts to saying that Q.t/�….t/ is an alternatingsum of exponentials. The right-hand side of (12.14) is the same expression as inTheorem 10.14, adjusted for our current convention of using real roots and realweights.

Example 12.23. Let K D SU.2/, let T be the diagonal subgroup, and let …m bethe irreducible representation for which the largest eigenvalue of �m.H/ ism: Thenthe character formula for…m takes the form

�…m

��ei� 0

0 e�i���

D sin..mC 1/�/

sin �:

This formula may also be obtained from Example 10.13 with a D i�:

Proof. IfH D diag.1;�1/; then we may choose the unique positive root ˛ to satisfyh˛;H i D 2. The highest weight � of the representation then satisfies h�;H i D m:

Note that diag.ei� ; e�i� / D ei�H , that ı satisfies hı;H i D h˛=2;H i D 1; and thatW D f1;�1g: Thus, the character formula reads

�…m.diag.ei� ; e�i� // D ei.mC1/� � e�i.mC1/�

ei� � e�i� ;

which simplifies to the claimed expression. utThe hard part of the proof of the character formula is showing that in the product

q.H/�….eH / of the Weyl denominator and a character, no other exponentials occur

besides the ones of the form eihw�.�Cı/;H i;w 2 W . (Compare Section 10.4.) Now,Theorem 12.15 tells us that the norm of �… is 1:

ZK

j�….x/j2 dx D 1: (12.15)

In the analytic approach, the unwanted exponentials will be ruled out by applyingthe Weyl integral formula to show that if any other exponentials did occur, theintegral on the left-hand side of (12.15) would be greater than 1. To make thisargument work, we must work out the Weyl integral formula in a more explicit form.

Proposition 12.24. If ı is analytically integral, the Weyl integral formula takes theform

ZK

f .x/ dx D 1

jW jZT

jQ.t/j2ZK=T

f .yty�1/ d Œy� dt; (12.16)

12.4 The Analytic Proof of the Weyl Character Formula 361

where dx; d Œy�; and dt are the normalized volume forms on K;K=T; and T;

respectively, and where Q is as in (12.13).

In the proof of this proposition, we also verify the correctness of the normal-ization constant in the Weyl integral formula (Theorem 11.30), a point that wasnot addressed in Sect. 11.6. (We will address the normalization issue when ı is notintegral in Sect. 12.6.) As we have already indicated above, knowing the correctnormalization constant is an essential part of our analytic proof of the Weyl integralformula.

Proof. We may extend Adt�1 to a complex-linear operator on g D kC; where gis the direct sum of h WD tC and fC; where fC is the orthogonal complement ofh in g: Meanwhile, g also decomposes as the direct sum of h and the root spacesg˛: We now claim that each g˛ is orthogonal to h. To see this, choose H 2 t forwhich h˛;H i ¤ 0: Then each H 0 2 h is an eigenvector for adH with eigenvalue 0,whereas X 2 g˛ is an eigenvector for adH with a nonzero eigenvalue. Since adH isskew self-adjoint,H 0 and X must be orthogonal. Thus, fC is actually the direct sumof the g˛’s.

Now, if ˛ is a real root and X belongs to g˛; we have

Ade�H .X/ D e�adH .X/ D e�ih˛;H iX

for X 2 g˛: Thus, letting Ad0e�H denote the restriction of Ade�H to fC; we have

det.Ad0e�H � I / D

Y˛2R

.e�ih˛;H i � 1/

DY˛2RC

.e�ih˛;H i � 1/.eih˛;H i � 1/:

Since .e�i� � 1/.ei� � 1/ D ˇei�=2 � e�i�=2ˇ2 ; we have

det.Ade�H � I / DˇˇˇY˛2RC

.eih˛;H i=2 � e�ih˛;H i=2/

ˇˇˇ2

D jq.H/j2 :

Thus, the Weyl integral formula (Theorem 11.30) takes the claimed form. Here wehave used Lemma 10.28, adapted to our current convention of using real roots.

It remains only to verify the correct normalization constant in the Weyl integralformula. To see that (12.16) is properly normalized, it suffices to consider f 1;

in which case (12.16) says that

1 D 1

jW jZT

jQ.t/j2 dt; (12.17)

362 12 The Compact Group Approach to Representation Theory

where dt is the normalized volume form on T: Now, by Proposition 8.38, ı belongsto the open fundamental Weyl chamber, which means (Proposition 8.27) that theexponentials eihw�ı;H i;w 2 W; are all distinct. Proposition 12.10 then tells us thatthese exponentials are orthonormal, so that the integral of jQ.t/j2 over T equalsjW j ; verifying (12.17). ut

We are now ready for our analytic proof of the Weyl character formula.

Proof of Theorem 12.22. Since … is finite dimensional, it has only finitely manyweights. Thus, there is a maximal weight� (i.e., one such that no other weight of…is higher than �/; with multiplicity mult.�/ � 1: We then claim that in the productq.H/�….e

H /; the exponential eih�Cı;H i occurs only once, by taking the term eihı;H ifrom q and the term mult.�/eih�;H i from �…: To see this, suppose we take eihw�ı;H ifrom q and mult.�/eih�;H i from �…; and that

�C w � ı D �C ı;

so that

� D �C .ı � w � ı/:But by Proposition 8.42, ı � w � ı and, thus, � � �: Since � is maximal, weconclude that � D �; in which case we must also have w D I:

We conclude, then, that eih�Cı;H i occurs with multiplicity exactly mult.�/ inthe product. Now, �….eH / is a Weyl-invariant function, while q.H/ is Weyl-alternating. The product of these two functions is then Weyl alternating. Thus,by Corollary 10.17, each of the exponentials eihw�.�Cı/;H i occurs with multiplicitydet.w/mult.�/:

We now claim that mult.�/ D 1 and that the only exponentials in theproduct q�… are those of the form det.w/eihw�.�Cı/;H i: To see this, recall fromTheorem 12.15 that

ZK

j�….x/j2 dx D 1:

Thus, by Proposition 12.24 (as applied to a class function), we have

ZT

jQ.t/�….t/j2 dt D jW j ; (12.18)

where jW j is the order of the Weyl group.Now, the product Q�… is a sum of exponentials and those exponentials are

orthonormal on T (Proposition 12.10). Since at least jW j exponentials do occur inthe product, namely those of the form det.w/eihw�.�Cı/;H i; these exponentials makea contribution of mult.�/ jW j to the integral in (12.18). By orthonormality, anyremaining exponentials would make a positive contribution to the integral. Thus,the only way (12.18) can hold is if mult.�/ D 1 and there are no exponentials inQ�… other than those of the form det.w/eihw�.�Cı/;H i: Thus, after dividing byQ.t/;we obtain the Weyl character formula.

12.5 Constructing the Representations 363

Since � is a weight of …; it must be analytically integral. It remains to showthat � is dominant and that � is the highest weight of …: If � were not dominant,there would be some w 2 W for which w � � is dominant, in which case (byProposition 8.42), � D w�1 � .w � �/ would be lower than w � �; contradictingthe maximality of �: Meanwhile, if � were not the highest weight of …; we couldchoose a maximal element�0 among the weights of… that are not lower than�; andthis �0 would be another maximal weight for …: As we have just shown, �0 wouldalso have to be dominant, in which case (Proposition 8.29), the Weyl-orbit of �0would be disjoint from the Weyl-orbit of �: But then by the reasoning above, boththe exponentials eihw�.�0Cı/;H i and the exponentials eihw�.�Cı/;H i would occur withnonzero multiplicity in the productQ�…; which would force the integral in (12.18)to be larger than jW j : ut

In Chapter 10, we used the Casimir element and the character formula forVerma modules to show that any exponential eih�;H i that appears in the productq.H/��.H/ must satisfy j�j D j�C ıj : This was the key step in proving thatonly exponentials of the form � D w � .� C ı/ appear in q��: In this chapter,we have instead used the Weyl integral formula to show that if any unwantedexponentials appeared, the integral of j�…j2 overK would be greater than 1, whichwould contradict Theorem 12.15.

Proposition 12.25. Two irreducible representations of K with the same highestweight are isomorphic.

Proof. If … and † both have highest weight �; then the Weyl character formulashows that the characters �… and �† must be equal. But if … and † were not iso-morphic, Theorem 12.15 would imply that �… and �† D �… are orthogonal. Thus,�… would have to be identically zero, which would contradict the normalizationresult in Theorem 12.15. ut

12.5 Constructing the Representations

In our proof of Theorem 12.6, we showed as part of the Weyl character formulathat every irreducible representation of K has a highest weight, and that thishighest weight is dominant and analytically integral. We also showed in Proposi-tion 12.25 that two irreducible representations of K with the same highest weightare isomorphic. Thus, in proving Theorem 12.6, it remains only to prove thatevery dominant, analytically integral element arises as the highest weight of arepresentation. We continue to assume that ı (half the sum of the real, positive roots)is an analytically integral element. This assumption is lifted in Sect. 12.6.

Suppose now that � is a dominant, analytically integral element. We do notyet know that � is the highest weight of an irreducible representation of K:Nevertheless, as we now demonstrate, there is a well-defined function �� on Kwhose restriction to T is given by the right-hand side of the Weyl character formula.

364 12 The Compact Group Approach to Representation Theory

Lemma 12.26. For each dominant, analytically integral element �; there is aunique continuous function �� W T ! C satisfying

��.eH/ D

Pw2W det.w/eihw�.�Cı/;H i

q.H/; H 2 t; (12.19)

whenever q.H/ ¤ 0:

In preparation for the proof of this lemma, recall from Lemma 10.28 (adjustedfor using real roots) that the Weyl denominator can be written as

q.H/ WDXw2W

det.w/eihw�ı;H i D .2i/kY˛2RC

sin.h˛;H i =2/

where k is the number of positive roots. For each ˛ 2 R and each integer n; definea hyperplane (not necessarily through the origin) by

V˛;n D fH 2 t jh˛;H i D 2�n g : (12.20)

Then the zero-set of the Weyl denominator q is the union of all the V˛;n’s, as ˛ rangesover the set of positive real roots and n ranges over the integers. Furthermore, q hasa “simple zero” on each of these hyperplanes. Symmetry properties of the numeratoron the right-hand side of (12.19) will then force this function to be zero each V˛;n,so that the zeros in the numerator cancel the zeros in the denominator.

Proof. By Assumption 12.21, the element ı is analytically integral, which meansthat each of w � .� C ı/ and w � ı;w 2 W; is also analytically integral. Thus, byProposition 12.9, all of the exponentials involved are well-defined functions on T:It follows that the right-hand side of (12.19) is a well-defined function T; outsideof the set where the denominator is zero. We now argue that the right-hand sideof (12.19) extends uniquely to a continuous function on t:

Let �.H/;H 2 t; denote the numerator on the right-hand side of (12.19):

�.H/ DXw2W

det.w/eihw�.�Cı/;H i:

We claim that � vanishes on each of the hyperplanes V˛;n in (12.20). To verify thisclaim, note that each weight � D w � .�Cı/ occurring in � is analytically integral,and thus also algebraically integral, by Proposition 12.7. Thus, h�;H˛i is an integerfor each ˛ 2 R: It follows that each exponential in �—and, thus, � itself—isinvariant under translation by 2�H˛:

�.H C 2�H˛/ D �.H/

for all H 2 t and all ˛ 2 R:

12.5 Constructing the Representations 365

Meanwhile, the coefficients of det.w/ in the formula for � guarantee that � isalternating with respect to the action of W . Finally, if H 2 V˛;n, we have

s˛ �H D H � 2h˛;H ih˛; ˛i ˛ D H � 2�nH˛:

Putting these observations together, we have, for H 2 V˛;n; �.H/ D � �.s˛ �H/ D � �.H � 2�nH˛/ D � �.H/;

which can happen only if �.H/ D 0: (See Figure 12.4, where the gray lines arethe hyperplanes V˛;n with jnj � 2: Compare also Exercise 6.)

Now, for each ˛ and n; let L˛;n W t ! R be given by

L˛;n.H/ D h˛;H i � 2�n;so that the zero-set of L˛;n is precisely V˛;n: Then the function

L˛;n.H/

sin.h˛;H i =2/ (12.21)

extends to a continuous function on a neighborhood of V˛;n in t: On the other hand,we have shown that � vanishes on V˛;n for each ˛ and n: Furthermore, since � isgiven by a globally convergent power series, it is not hard to prove that � can beexpressed as

�.H/ D f .H/L˛;n.H/

H

V ,1

2 H

s H

Fig. 12.4 The function � changes sign under the reflection s˛ and is unchanged under translationby 2�H˛; forcing � to be zero on each V˛;n

366 12 The Compact Group Approach to Representation Theory

for some smooth function f; where f is also given by a globally convergent powerseries and where f still vanishes on each hyperplane Vˇ;m with .ˇ;m/ ¤ .˛; n/:

(See Exercise 7.)Now letH be an arbitrary point in t: Since the hyperplanesV˛;n are disjoint for ˛

fixed and n varying,H is contained in only finitely many of these hyperplanes, sayV˛1;n1 ; : : : ; V˛m;nm : Since � vanishes on each of these hyperplanes, we can showinductively that � can be expressed as

� D L˛1;n1 � � �L˛m;nm g (12.22)

for some smooth function g: Since the function in (12.21) is nonsingular in aneighborhood of V˛;n; we conclude that the factors of L˛j ;nj in (12.22) cancel allthe zeros in q.H/; showing that ��.eH / D �.H/=q.H/ extends to a continuousfunction in a neighborhood of H:

Finally, in local exponential coordinates on T; each point H is contained in atmost finitely many of the hyperplanes on which q vanishes. Thus, H is a limit ofpoints on which q ¤ 0; showing the uniqueness of the continuous extension. utLemma 12.27. Let �� be as in Lemma 12.26 and let ˆ� W K ! C be the uniquecontinuous class function on K such that ˆ�

ˇT

D �� (Corollary 11.40). Then as� ranges over the set of dominant, analytically integral elements, the functions ˆ�form an orthonormal set:

ZK

ˆ�.x/ˆ�0.x/ dx D ı�;�0 :

Note that we do not know, at the moment, that each ˆ� is actually the characterof a representation of K: Thus, we cannot appeal to Theorem 12.15.

Proof. Since the denominator in the definition of �� is the Weyl denominator,Corollary 11.32 to the Weyl integral formula tells us that

ZK

ˆ�.x/ˆ�0.x/ dx

D 1

jW jZT

jQ.t/j2 ��.t/��0.t/ dt

D 1

jW jXw2W

Xw02W

ZT

det.w/eihw�.�Cı/;H i� �det.w0/eihw0�.�0Cı/;H i dH:

(12.23)

Now, since �C ı and �0 C ı are strictly dominant,W acts freely on these elements.If � ¤ �0; the W -orbit of �C ı will be disjoint from the W -orbit of �0 C ı: Thus,the exponentials occurring in �� will be disjoint from those in ��0 ; which means,

12.5 Constructing the Representations 367

by Proposition 12.10, that (12.23) is zero. If, on the other hand, � D �0; we havethe norm-squared of jW j distinct, orthonormal exponentials, so that the right-handside of (12.23) reduces to 1. ut

We are now ready to complete the proof of Theorem 12.6, in the case in which theelement ı is analytically integral. It remains only to prove that for each dominant,analytically integral element �; there is an irreducible representation of K withhighest weight �:

Proof of Theorem 12.6. Here is what we know so far about the characters of irre-ducible representations. First, by Theorem 12.22, the character of each irreduciblerepresentation… must be equal to ˆ�; where � is the highest weight of…: Second,by Lemma 12.26, all theˆ�’s, with � dominant and analytically integral—whetheror not they are characters of a representation—form an orthonormal set. Last, byTheorem 12.18, the characters of the irreducible representations form a completeorthonormal set.

Let �0 be a fixed dominant, analytically integral element and suppose, towarda contradiction, that there did not exist an irreducible representation with highestweight �0: Then the function ˆ�0 would be a continuous class function on K andˆ�0 would be orthogonal to the character of every irreducible representation. (Afterall, every irreducible character is of the form ˆ�; where � is the highest weightof the representation, and we are assuming that �0 is not the highest weight ofany representation.) But Theorem 12.18 says that a continuous class function thatis orthogonal to every irreducible character must be identically zero. Thus, ˆ�0would have to be the zero function, which is impossible, since

ˇˆ�0

ˇ2integrates to

1 overK: utWe may put the argument in a different way as follows. Theorem 12.18 says

that the characters of irreducible representations form a complete orthonormal setinside the space of continuous class functions on K: The Weyl character formula,meanwhile, says that the characters form a subset of the set of ˆ�’s. Finally, theWeyl integral formula says that the collection of all ˆ�’s, with � dominant andanalytically integral, are orthonormal. If there were some such � for which ˆ� wasnot a character, then the set of characters would be a proper subset of an orthonormalset, in which case, the characters could not be complete.

Now, the preceding proof of the theorem of the highest weight is not veryconstructive, in contrast to the Lie-algebraic proof, in which we gave a directconstruction of each finite-dimensional representation as a quotient of a Vermamodule. If one looks carefully at the proof of Theorem 12.18, however, one seesa hint of a more direct description of the representations from the compact groupperspective. Specifically, let C.K/ denote the space of continuous functions on K;and define a left and a right action ofK on C.X/ as follows. For each x 2 K; defineLx and Rx as operators on C.X/ as by

.Lxf /.y/ D f .x�1y/

.Rxf /.y/ D f .yx/:

368 12 The Compact Group Approach to Representation Theory

One may easily check that Lxy D LxLy and Rxy D RxRy: Thus, both L� and R�define representations of K acting on the infinite-dimensional space C.K/:

We now show that each irreducible representation of K occurs as a finite-dimensional subspace of C.K/ that is invariant under the right action ofK: For eachrepresentation irreducible .…; V / of K; we can fix some nonzero vector v0 2 V;

which might, for example, be a highest weight vector. Then for each v 2 V; we canconsider the function fv W K ! C given by

fv.x/ D hv0;….x/vi : (12.24)

(The function fv is a special sort of matrix entry for …, in the sense of the proofof Theorem 12.18.) It is not hard to check that the map v 7! fv is injective; seeExercise 8.

Let W denote the space of all functions of the form fv; with v 2 V: We cancompute that

.Rxfv/.y/ D hv0;….yx/vi D hv0;….y/.….x/v/i ;

which means that

Rxfv D f….x/v:

Thus, W is a finite-dimensional invariant subspace for the right action of K onC.K/: Indeed, the map v 7! fv is a bijective intertwining map between V and W:

Conclusion 12.28. For each irreducible representation .…; V / of K; fix a nonzerovector v0 2 V and let W denote the subspace of C.X/ consisting of functions of theform (12.24), with v 2 V: Then W is invariant under the right action of K and isisomorphic, as a representation of K; to V:

One can pursue this line of analysis further by choosing v0 to be a highestweight vector and then attempting to describe the space W—without referring tothe representation…—by means of its behavior under certain differential operatorson K: See Sect. 4.12 of [DK] for more information.

12.6 The Case in Which ı is Not Analytically Integral

For a general connected compact group K; the element ı may not be analyticallyintegral. (See Sect. 12.2.) If ı is not analytically integral, many of the functionswe have been working with will not be well-defined functions on T: Specifically,exponentials of the form eihw�ı;H i and eihw�.�Cı/;H i; with w 2 W and � a dominant,analytically integral element, no longer define functions on T: Fortunately, all of our

12.6 The Case in Which ı is Not Analytically Integral 369

calculations involve products of such exponentials, and these products turn out to bewell-defined functions on T: Thus, all the arguments in the previous three sectionswill go through with minor modifications.

In all cases, the quantity 2ı D P˛2RC ˛ is analytically integral, by Point 3 of

Proposition 12.7. Thus, for each H 2 t for which e2�H D I; the quantity hı;H imust either be an integer or a half integer. If ı is not analytically integral, there mustexist some H 2 t with e2�H D I for which hı;H i is a half integer (but not aninteger). With this observation in mind, we make the following definition.

Definition 12.29. If ı is not analytically integral, we say that � 2 t is half integralif � � ı is analytically integral.

That is to say, the half integral elements are those of the form � D ı C �0; with�0 being analytically integral.

Proposition 12.30. If � and � are half integral, then �C � is analytically integral.If � is half integral, then �� is also half integral and w � � is half integral for allw 2 W:Proof. If � D ıC �0 and � D ıC �0 are half integral, then �C � D 2ıC �0 C �0 isanalytically integral. If � D ı C �0 is half integral, so is

�� D �ı � �0 D ı � 2ı � �0:

For each w 2 W; the set w � RC will consist of a certain subset S of the positiveroots, together with the negatives of the roots in RC n S: Thus, w � ı will consist ofhalf the sum of the elements of S minus half the sum of the elements of RC n S: Itfollows that

ı � w � ı DX

˛2RCnS˛;

showing that w �ı is again half integral. (Recall that each root is analytically integral,by Proposition 12.7.) More generally, if � D ı C �0 is half integral, so is w � � Dw � ı C w � �0: ut

Note that exponentials of the form eih�;H i; with � being half integral, do notdescend to functions on T: Our next result says that, nevertheless, the productof two such exponentials (possibly conjugated) does descend to T: Furthermore,such exponentials are still “orthonormal on T ,” as in Proposition 12.10 in theintegral case.

Proposition 12.31. If � and � are half integral, there is a well-defined function fon T such that

f .eH / D eih�;H ieih�;H i

370 12 The Compact Group Approach to Representation Theory

andZT

eih�;H ieih�;H i dH D ı�;�:

Proof. If � and � are half integral, then �� is half integral, so that � � � isanalytically integral. Thus, by Proposition 12.9, there is a well-defined functionf W T ! C satisfying

f .eH / D eih�;H ieih�;H i D eih���;H i:

Furthermore, if � D ı C �0 and � D ı C �0 are half integral, we have

ZT

eih�;H ieih�;H i dH DZT

e�ihıC�0 ;H ieihıC�0;H i dH

DZT

e�ih�0 ;H ieih�0;H i dH

D ı�0;ı0 ;

by Proposition 12.10. Since � D � if and only if �0 D ı0; we have the desired“orthonormality” result. ut

We now discuss how the results of Sects. 11.6, 12.4, and 12.5 should be modifiedwhen ı is not analytically integral. In the case of the Weyl integral formula(Theorem 11.30), the function q.H/;H 2 t; does not descend to T when ı is notintegral. That is to say, there is no function Q.t/ on T such that Q.eH/ D q.H/:

Nevertheless, the function jq.H/j2 does descend to T; since jq.H/j2 is a sum ofproducts of half integral exponentials. The Weyl integral formula, with the sameproof, then holds even if ı is not analytically integral, provided that the expressionjQ.t/j2 is interpreted as the function eH 7! jq.H/j2 : We may then consider thecase f 1 and use Proposition 12.31 to verify the correctness of the normalizationin the Weyl integral formula.

In the case of the Weyl character formula, we claim that the right-hand sideof (12.14) descends to a function on T: To see this, note that we can pull a factorof eihı;H i out of each exponential in the numerator and each exponential in thedenominator. After canceling these factors, we are left with exponentials in boththe numerator and denominator that descend to T: Meanwhile, in the proof ofthe character formula, although the function Q.t/�….t/ is not well defined on T;the function jQ.t/�….t/j2 is well defined. The Weyl integral formula (interpretedas in the previous paragraph) tells us that the integral of jQ.t/�….t/j2 over T isequal to jW j ; as in the case where ı is analytically integral. If we then apply theorthonormality result in Proposition 12.31, we see that, just as in the integral case,the only exponentials present in the product q.H/�….eH / are those in the numeratorof the character formula.

12.7 Exercises 371

Finally, we consider the proof that every dominant, analytically integral elementis the highest weight of a representation. If ı is not analytically integral, then neitherthe numerator nor the denominator on the right-hand side of (12.19) descends tofunction on T: Nevertheless, the ratio of these functions does descend to T; by theargument in the preceding paragraph. The argument that �� extends to a continuousfunction on T then goes through without change. (This argument requires only thateach weight � D w � .� C ı/ in � be algebraically integral, which holds evenif ı is not analytically integral, by Proposition 8.38.) Thus, we may apply the half-integral version of the Weyl integral formula to show that the functionsˆ� onK areorthonormal, as � ranges over the set of dominant, analytically integral elements.The rest of the argument then proceeds without change.

12.7 Exercises

1. LetK D SU.2/ and let t be the diagonal subalgebra of su.2/: Prove directly thatevery algebraically integral element is analytically integral.Note: Since SU.2/ is simply connected, this claim also follows from the generalresult in Corollary 13.20.

2. This exercise asks you to use the theory of Fourier series to give a direct proof ofthe completeness result for characters (Theorem 12.18), in the case K D SU.2/:To this end, suppose f is a continuous class function on SU.2/ that f isorthogonal to the character of every representation.

(a) Using the explicit form of the Weyl integral formula for SU.2/ (Exam-ple 11.33) and the explicit form of the characters for SU.2/ (Example 12.23),show that

Z �

��f .diag.ei� ; e�i� //.sin �/ sin..mC 1/�/ d� D 0

for every non-negative integerm:(b) Show that the function � 7! f .diag.ei� ; e�i� //.sin �/ is an odd function

of �:(c) Using standard results from the theory of Fourier series, conclude that f

must be identically zero.

3. Suppose .…; V / and .†;W / are representations of a group G; and letHom.V;W / denote the space of all linear maps from V to W: Let G act onHom.V;W / by

g � A D †.g/A….g/�1; (12.25)

for all g 2 G and A 2 Hom.W; V /: Show that A is an intertwining map of V toW if and only if g � A D A for all g 2 G:

372 12 The Compact Group Approach to Representation Theory

4. If V andW are finite-dimensional vector spaces, letˆ W V � ˝W ! Hom.V;W /be the unique linear map such that for all 2 V � and w 2 W; we have

ˆ. ˝ w/.v/ D .v/w; v 2 V:

(a) Show that ˆ is an isomorphism.(b) Let .…; V / and .†;W / be representations of a group G, let G act on V � as

in Sect. 4.3.3, and let G act on Hom.V;W / as in (12.25). Show that the mapˆ W V � ˝W ! Hom.V;W / in Part (a) is an intertwining map:

5. Suppose f .x/ WD trace.….x/A/ and g.x/ WD trace.†.x/B/ are matrix entriesfor nonisomorphic, irreducible representations… and† ofK (Definition 12.19).Show that f and g are orthogonal:

ZK

f .x/g.x/ dx D 0:

Hint: Imitate the proof of Theorem 12.15.6. Let fV˛;ng denote the collection of hyperplanes in (12.20). If H˛ D 2˛= h˛; ˛i is

the real coroot associated to a real root ˛; show that

V˛;n C �mH˛ D V˛;nCm:

7. Let V be a hyperplane in Rn; not necessarily through the origin, and letL W Rn !

R be an affine function whose zero-set is precisely V: Suppose g W Rn ! C isgiven by a globally convergent power series in n variables and that g vanisheson V:

(a) Show that g can be expressed as g D Lh for some function h; where h isalso given by a globally convergent power series.Hint: Choose a coordinate system y1; : : : ; yn on Rn with origin in L suchthat L.y/ D y1:

(b) Suppose g also vanishes on some hyperplane V 0 distinct from V: Show thatthe function h in Part (a) vanishes on V 0:

8. Let .…; V / be an irreducible representation ofK and let v0 be a nonzero elementof V: For each v 2 V; let fv be the function given in (12.24). Show that if fv isthe zero function, then v D 0:

9. Let K D U.n/ and let T be the diagonal subgroup of K: Show that the density�.�/ in the Weyl integral formula (Theorem 11.30) can be computed explicitly as

�.diag.ei�1 ; : : : ; ei�n// DY

1jkn

ˇei�j � ei�k

ˇ2:

Hint: Use Proposition 12.24, as interpreted in Sect. 12.6 in the case where ı isnot necessarily analytically integral.

Chapter 13Fundamental Groups of Compact Lie Groups

13.1 The Fundamental Group

In this section, we briefly review the notion of the fundamental group of atopological space. For a more detailed treatment, the reader should consult anystandard book on algebraic topology, such as [Hat, Chapter 1]. Let X be anypath-connected Hausdorff topological space and let x0 be a fixed point in X (the“basepoint”). We consider loops inX based at x0 (i.e., continuous maps l W Œ0; 1� !X with the property that l.0/ D l.1/ D x0). The choice of the basepoint makes nosubstantive difference to the constructions that follow. From now on, “based loop”will mean “loop based at x0.” Ultimately, we are interested in the case that X is amatrix Lie group.

If l1 and l2 are two based loops, then we define the concatenation of l1 and l2 tobe the loop l1 � l2 given by

l1 � l2.t/ D�l1.2t/; 0 � t � 1

2

l2.2t � 1/; 12

� t � 1;

that is, l1 � l2 traverses l1 as t goes from 0 to 1=2 and then traverses l2 as t goes from1=2 to 1:

Two based loops l1 and l2 are said to be homotopic if one can be “continuouslydeformed” into the other. More precisely, this means that there exists a continuousmap A W Œ0; 1� � Œ0; 1� ! X such that A.0; t/ D l1.t/ and A.1; t/ D l2.t/ for all

A previous version of this book was inadvertently published without the middle initial of theauthor’s name as “Brian Hall”. For this reason an erratum has been published, correcting themistake in the previous version and showing the correct name as Brian C. Hall (see DOI http://dx.doi.org/10.1007/978-3-319-13467-3_14). The version readers currently see is the correctedversion. The Publisher would like to apologize for the earlier mistake.

© Springer International Publishing Switzerland 2015B.C. Hall, Lie Groups, Lie Algebras, and Representations, GraduateTexts in Mathematics 222, DOI 10.1007/978-3-319-13467-3_13

373

374 13 Fundamental Groups of Compact Lie Groups

t 2 Œ0; 1� and such that A.s; 0/ D A.s; 1/ D x0 for all s 2 Œ0; 1�: One should thinkof A.s; t/ as a family of loops parameterized by s: In some cases, we may use thenotation ls.t/ in place of A.s; t/ to emphasize this point of view.

A based loop is said to be null homotopic if it is homotopic to the constant loop(i.e., the loop l0 for which l0.t/ D x0 for all t 2 Œ0; 1�). If all loops in X based at x0are null homotopic, then X is said to be simply connected. Since we are assumingX is path connected, it is not hard to show that if all loops at one basepoint arenull homotopic, the same it true for every other basepoint. Furthermore, if a loopbased at x0 can be shrunk to a point without fixing the basepoint (i.e., requiring onlythat A.s; 0/ D A.s; 1/), then it can also be shrunk to a point with basepoint fixed(i.e., requiring A.s; 0/ D A.s; 1/ D x0).

The notion of homotopy is an equivalence relation on loops based at x0.The homotopy class of a loop l is then the set of all loops that are homotopicto l , and each loop belongs to one and only one homotopy class. The concatenationoperation “respects homotopy,” meaning that if l1 is homotopic to l2 and m1 ishomotopic to m2; then l1 �m1 is homotopic to l2 �m2: As a result, it makes sense todefine the concatenation operation on equivalence classes.

The operation of concatenation makes the set of homotopy classes of loops basedat x0 into a group, called the fundamental group ofX and denoted�1.X/: To verifyassociativity, we note that although .l1 �l2/�l3 is not the same as l1 �.l2 �l3/; the secondof these two loops is a reparameterization of the first, from which it is not hard to seethat the loops are homotopic. Meanwhile, the identity in �1.X/ is the constant loopl0: This is not an identity at the level of loops but is at the level of homotopy classes;that is, l � l0 and l0 � l are not equal to l; but they are both homotopic to l; since bothare reparameterizations of l: Finally, for inverses, the inverse to a homotopy class Œl �is the homotopy class Œl 0� where l 0.t/ D l.1� t/: (It is not hard to see that both l � l 0and l 0 � l are null homotopic.) A topological space X is simply connected preciselyif its fundamental group is the trivial group.

Some standard examples of fundamental groups are as follows: Rn is simplyconnected for all n; Sn is simply connected for n � 2; and the fundamental groupof S1 is isomorphic to Z:

Definition 13.1. If X and Y are Hausdorff topological space, a continuous map� W Y ! X is a covering map if (1) � maps Y onto X and (2) for each x 2 X;

there is a neighborhoodV of x such that ��1.V / is a disjoint union of open sets U˛;where the restriction of � to each U˛ is a homeomorphism of U˛ onto V: A coverof X is a pair .Y; �/, where � W Y ! X is a covering map. If .Y; �/ is a cover of Xand Y is simply connected, then .Y; �/ is a universal cover of X:

If .Y; �/ is a cover of X and f W Z ! X is a continuous map, then a mapQf W Z ! Y is a lift of f if Qf is continuous and � ı Qf D f:

It is known that every connected manifold (indeed, every reasonably niceconnected topological space) has a universal cover, and that this universal coveris unique up to a “canonical” homeomorphism, that is, one that intertwines thecovering maps. (See, for example, pp. 63–66 in [Hat].) Thus, we may speak about“the” universal cover of any connected manifold. A key property of a covering maps

13.2 Fundamental Groups of Compact Classical Groups 375

� W Y ! X is that lifts of reasonable maps into X always exist, as described in thenext two results. (See Proposition 1.30 in [Hat].)

Proposition 13.2 (Path Lifting Property). Suppose .Y; �/ is a cover of X andthat p W Œ0; 1� ! X is a (continuous) path with p.0/ D x: Then for eachy 2 ��1.fxg/; there is a unique lift Qp of p for which Qp.0/ D y:

Proposition 13.3 (Homotopy Lifting Property). Suppose that l is a loop in X ,that a path p in Y is a lift of l , and that ls is a homotopy of l in X with basepointfixed. Then there is a unique lift of ls to a homotopy ps of p in Y with endpointsfixed.

If we can find a universal cover .Y; �/ of a space X; the cover gives a simplecriterion for determining when a loop in X is null homotopic.

Corollary 13.4. Suppose that .Y; �/ is a universal cover of X , that l is a loop inX; and that p is a lift of l to Y: Then l is null homotopic in X if and only if p is aloop in Y; that is, if and only if p.1/ D p.0/:

Proof. If the lift p of l is a loop, then since Y is simply connected, there is ahomotopy ps of p to a point with basepoint fixed. Then ls WD � ı ps is a homotopyof l to a point in X: In the other direction, if there is a homotopy ls of l to a point inX; then by Proposition 13.3, we can lift this to a homotopy ps with endpoints fixed.Now, if l1 is the constant loop at x0; then p1, which is a lift of l1; must live entirelyin ��1.fx0g/; which is a discrete set. Thus, actually, p1 must be constant, and, inparticular, has equal endpoints. But since ps is a homotopy with endpoints fixed, theendpoints of each ps must be equal. Thus, p D p0 must be a loop in Y: ut

13.2 Fundamental Groups of Compact Classical Groups

In this section, we discuss a method of computing, inductively, the fundamentalgroups of the classical compact groups. The same results can also be obtained byusing the results of Sects. 13.4–13.7; see Exercises 1–4. In all cases, we will findthat �1.K/ is commutative. This is not a coincidence; a general argument showsthat the fundamental group of any Lie group is commutative (Exercise 7). In thecase of a compact matrix Lie group K , the commutativity of �1.K/ also followsfrom Corollary 13.18.

For any nice topological space, one can define higher homotopy groups�k.X/; k D 1; 2; 3; : : : : The group �k.X/ is the set of homotopy classes of maps ofSk intoX;where the notion of homotopy for maps of Sk intoX is analogous to thatfor maps of S1 into X: Although one can define a group structure on �k.X/; thisstructure is not relevant to us. All that is relevant is what it means for �k.X/ to betrivial, which is that every continuous map of the k-sphere Sk into X can be shrunkcontinuously to a point. We will make use of the following standard topologicalresult (e.g., Corollary 4.9 in [Hat]).

376 13 Fundamental Groups of Compact Lie Groups

Proposition 13.5. For a d -sphere Sd ; �k.Sd / is trivial if k < d:

This result is plausible because for k < d; the image of a “typical” continuousmap of Sk into Sd will not be all of Sd : However, if the image of the map omitseven one point in Sd ; then we can remove that point and what is left of the spherecan be contracted continuously to a point.

We now introduce the topological concept underlying all the calculations in thissection, that of a fiber bundle.

Definition 13.6. Suppose that B and F are Hausdorff topological spaces. A fiberbundle with base B and fiber F is a Hausdorff topological space X together witha continuous map p W X ! B; called the projection map, having the followingproperties. First, for each b in B; the preimage p�1.b/ of b in X is homeomorphicto F: Second, for every b in B; there is a neighborhoodU of b such that p�1.U / ishomeomorphic to U �F in such a way that the projection map is simply projectiononto the first factor.

In any fiber bundle, the sets of the form p�1.b/ are called the fibers. The secondcondition in the definition may be stated more pedantically as follows. For eachb 2 B , there should exist a neighborhood U of B and a homeomorphism ˆ

of p�1.U / with U � F having the property that p.x/ D p1.ˆ.x//; wherep1 W U � F ! U is the map p1.u; f / D u:

The simplest sort of fiber bundle is the product space X D B � F; with theprojection map being simply the projection onto the first factor. Such a fiber bundleis called trivial. The second condition in the definition of a fiber bundle is calledlocal triviality and it says that any fiber bundle must look locally like a trivialbundle. In general,X need not be globally homeomorphic to B � F:

IfX were a trivial fiber bundle, then the fundamental group ofX would be simplythe product of the fundamental group of the base B and the fundamental group ofthe fiber F: In particular, if X were a trivial fiber bundle and �1.B/ were trivial,then �1.X/ would be isomorphic to �1.F /: The following result says that if �1.B/and �2.B/ are trivial, then the same conclusion holds, even if X is nontrivial.

Theorem 13.7. Suppose that X is a fiber bundle with base B and fiber F: If �1.B/and �2.B/ are trivial, then �1.X/ is isomorphic to �1.F /:

Proof. According to a standard topological result (e.g., Theorem 4.41 and Proposi-tion 4.48 in [Hat]), there is a long exact sequence of homotopy groups for a fiberbundle. The portion of this sequence relevant to us is the following:

�2.B/ !f�1.F / !

g�1.X/ !

h�1.B/: (13.1)

Saying that the sequence is exact means that each map is a homomorphism andthe image of each map is equal to the kernel of the following map. Since we areassuming �2.B/ is trivial, the image of f is trivial, which means the kernel of g isalso trivial. Since �1.B/ is also trivial, the kernel of hmust be �1.X/; which meansthat the image of g is �1.X/: Thus, g is an isomorphism of �1.F / with �1.X/: ut

13.2 Fundamental Groups of Compact Classical Groups 377

Proposition 13.8. Suppose G is a matrix Lie group andH is a closed subgroup ofG: Then G has the structure of a fiber bundle with base G=H and fiber H; wherethe projection map p W G ! G=H is given by p.x/ D Œx�; with Œx� denoting thecoset xH 2 G=H:Proof. For any coset Œx� in G=H; the preimage of Œx� under p is the set xH � G,which is clearly homeomorphic to H: Meanwhile, the required local trivialityproperty of the bundle follows from Lemma 11.21 and Theorem 11.22. (If we take aopen set U in G=H as in the proof of Theorem 11.22, Lemma 11.21 tells us that thepreimage of U under p is homeomorphic to U �H in such a way that the projectionp is just projection onto the first factor.) utProposition 13.9. Consider the map p W SO.n/ ! Sn�1 given by

p.R/ D Ren; (13.2)

where en D .0; : : : ; 0; 1/: Then .SO.n/; p/ is a fiber bundle with base Sn�1 andfiber SO.n � 1/:Proof. We think of SO.n � 1/ as the (closed) subgroup of SO.n/ consisting ofblock diagonal matrices of the form

R D�R0 00 1

with R0 2 SO.n � 1/: By Proposition 13.8, SO.n/ is a fiber bundle with baseSO.n/=SO.n � 1/ and fiber SO.n � 1/: Now, it is easy to see that SO.n/ actstransitively on the sphere Sn�1: Thus, the map p in (13.2) maps SO.n/ onto Sn�1:Since Ren D en if and onlyR 2 SO.n�1/;we see that p descends to a (continuous)bijection of SO.n/=SO.n� 1/ onto Sn�1: Since both SO.n/=SO.n� 1/ and Sn�1are compact, this map is actually a homeomorphism (Theorem 4.17 in [Rud1]).Thus, SO.n/ is a fiber bundle of the claimed sort. utProposition 13.10. For all n � 3; the fundamental group of SO.n/ is isomorphicto Z=2: Meanwhile, �1.SO.2// Š Z:

Proof. Suppose that n is at least 4, so that n � 1 is at least 3. Then, byProposition 13.5, �1.Sn�1/ and �2.Sn�1/ are trivial and, so, Theorem 13.7 andProposition 13.9 tell us that �1.SO.n// is isomorphic to �1.SO.n � 1//: Thus,�1.SO.n// is isomorphic to �1.SO.3// for all n � 4: It remains to show that�1.SO.3// Š Z=2: This can be done by noting that SO.3/ is homeomorphic toRP

3; as in Proposition 1.17, or by observing that the map ˆ in Proposition 1.19 is atwo-to-one covering map from SU.2/ � S3 onto SO.3/:

Finally, we observe that SO.2/ is homeomorphic to the unit circle S1; so that�1.SO.2// Š Z (Theorem 1.7 in [Hat]). ut

378 13 Fundamental Groups of Compact Lie Groups

If one looks into the proof of the long exact sequence of homotopy groups fora fiber bundle, one finds that the map g in (13.1) is induced by the inclusion of FintoX: Thus, if l is a homotopically nontrivial loop in SO.n/; then after we includeSO.n/ into SO.nC 1/; the loop l is still homotopically nontrivial.

Meanwhile, we may take SU.2/ as the universal cover of SO.n/; with coveringmap being the homomorphism ˆ in Proposition 1.19 (compare Exercise 8). Now,if we take l to be the loop in SO.3/ consisting of rotations by angle � in the.x2; x3/-plane, 0 � � � 2�; the computations in (1.15) and (1.16) show that thelift of l to SU.2/ is not a loop. (Rather, the lift will start at I and end at �I:)Thus, by Corollary 13.4, l is homotopically nontrivial in SO.3/: But this loop l isconjugate in SO.3/ to the loop of rotations in the .x1; x2/-plane, so that loop is alsohomotopically nontrivial. Thus, by the discussion in the previous paragraph, we maysay that, for any n � 3; the one nontrivial homotopy class in SO.n/ is representedby the loop

l.�/ WD

0BBBBB@

cos � � sin �sin � cos �

1: : :

1

1CCCCCA; 0 � � � 2�:

(Compare Exercise 6.)

Proposition 13.11. The group SU.n/ is simply connected for all n � 2: For alln � 1; we have that �1.U.n// Š Z:

Proof. For all n � 3; the group SU.n/ acts transitively on the sphere S2n�1. By asmall modification of the proof of Proposition 13.9, SU.n/ is a fiber bundle withbase S2n�1 and fiber SU.n � 1/: Since 2n � 1 > 3 for all n � 2; Theorem 13.7and Proposition 13.5 tell us that �1.SU.n// Š �1.SU.n � 1//: Since �1.SU.2// Š �1.S

3/ is trivial, we conclude that �1.SU.n// is trivial for all n � 2:

The analysis of the case of U.n/ is similar. The fiber bundle argument showsthat �1.U.n// Š �1.U.n � 1// for all n � 2: Since U.1/ is just the unit circleS1; we have that �1.U.1// Š Z (Theorem 1.7 in [Hat]). Thus, �1.U.1// Š Z forall n � 1: utProposition 13.12. For all n � 1; the compact symplectic group Sp.n/ is simplyconnected.

Proof. Since Sp.n/ is contained in U.2n/; it acts on the unit sphere S4n�1 � C2n: Ifthis action is transitive, then we can imitate the arguments from the cases of SO.n/and SU.n/: Since 4n�1 � 3;we see that �1.S4n�1/ and �2.S4n�1/ are trivial for alln � 1:We conclude, then, that �1.Sp.n// Š �1.Sp.n�1//: Since Sp.1/ D SU.2/;which is simply connected, we see that Sp.n/ is simply connected for all n:

It remains, then, to show that Sp.n/ acts transitively on S4n�1: For this, it sufficesto show that for all unit vectors u 2 C2n; there is some U 2 Sp.n/ with Ue1 D u:

13.4 The Fundamental Groups of K and T 379

To see this, let u1 D u and v1 D Ju; where J is the map in Sect. 1.2.8. Then v1 isorthogonal to u1 (check) and we may consider the space

W WD .spanfu1; v1g/?:

By the calculations in Sect. 1.2.8,W will be invariant under J: Thus, we can choosean arbitrary unit vector u2 in W and let v2 D Ju2: Proceeding on in this way, weeventually obtain an orthonormal family

u1; : : : ; un; v1; : : : ; vn (13.3)

in C2n where Juj D vj : It is then straightforward to check that the matrix U whose

columns are the vectors in (13.3) belongs to Sp.n/; and that Ue1 D u1 D u: ut

13.3 Fundamental Groups of Noncompact Classical Groups

Using the polar decomposition (Sect. 2.5), we can reduce the computation ofthe fundamental group of certain noncompact groups to the computation of thefundamental group of one of the compact groups in Sect. 13.2. Theorem 2.17, forexample, tells us that GL.nIC/ is homeomorphic to U.n/ � V for a certain vectorspace V (the space of n� n self-adjoint matrices). Since V is simply connected, weconclude that �1.GL.nIC// Š Z: Using Proposition 2.19, we can similarly showthat �1.SL.nIC// Š �1.SU.n// and that �1.SL.nIR// Š �1.SO.n//:

Conclusion 13.13. For all n � 1; we have

�1.GL.nIC// Š �1.U.n// Š Z:

For all n � 2; the group SL.nIC/ is simply connected. For n � 3; we have

�1.SL.nIR// Š �1.SO.n// Š Z=2;

whereas

�1.SL.2IR// Š �1.SO.2// Š Z:

13.4 The Fundamental Groups of K and T

In this section and the subsequent ones, we develop a different approach to comput-ing the fundamental group of a compact groupK; based on the torus theorem. In thissection, we state the main results; the proofs will be developed in Sects. 13.5, 13.6,

380 13 Fundamental Groups of Compact Lie Groups

and 13.7. One important consequence of these results will be Corollary 13.20, whichsays that if K is simply connected, then every algebraically integral element isanalytically integral. This claim allows us (in the simply connected case) to matchup our theorem of the highest weight for the compact groupK with the theorem ofthe highest weight for the Lie algebra g WD kC: That is to say, when K is simplyconnected, the set of possible highest weights for the representations of K (namelythe dominant, analytically integral elements) coincides with the set of possiblehighest weights for the representations of g (namely the dominant, algebraicallyintegral elements).

In these sections, we will follow the convention of composing the exponentialmap for T with a factor of 2�; writing an element t of T as

t D e2�H ; H 2 t:

(There are factors of 2� that have to go somewhere in the theory, and this seemsto be the most convenient spot to put them.) Recall (Definition 12.3) that � � tdenotes the kernel of the (scaled) exponential map:

� D ˚H 2 t

ˇe2�H D I

and that � is invariant under the action ofW:We begin with a simple result describing the fundamental group of T:

Proposition 13.14. Every loop in T is homotopic in T to a unique loop of the form

� 7! e2��� ; 0 � � � 1;

with � 2 �: Furthermore, �1.T / is isomorphic to �:

Proof. The main issue is to prove that the (scaled) exponential map H 7! e2�H is acovering map. Since the kernel � of the exponential map is discrete, there is some" > 0 such that every nonzero element � of � has norm at least ": Let B"=2.�/ bethe ball of radius "=2 around a point � 2 �: Now, for any t 2 T; write t as e2�H forsome H 2 t: Then let V � T denote the set of element of the form e2�H

0

; whereH 0 2 B"=2.H/; so that V is a neighborhood of t: The preimage of V under theexponential is the union of the balls

B"=2.H C �/; � 2 �:

By the way " was chosen, these balls are disjoint, and each ball maps homeomor-phically onto V: Since we can do this for each t 2 T; we see that the exponential isa covering map.

Since, also, t is simply connected, t is the universal cover of T: Now, every loop lin T based at the identity has a unique lift to a path Ql in t starting at 0 and ending atsome point � in�: The theory of covering spaces tells us that two loops l1 and l2 in T(based at I ) are homotopic if and only if Ql1.1/ D Ql2.1/:Meanwhile, if Ql.1/ D �; then

13.4 The Fundamental Groups of K and T 381

(since t is simply connected) Ql is homotopic with endpoints fixed to the straight-linepath � 7! ��; showing that l itself is homotopic to � 7! e2��� ; as claimed. Finally,if we compose two loops of the form � 7! e2���1 and � 7! e2���2 ; the lift of thecomposite loop will be the composition of the lifts, where the lift of the second partof the composite loop must be taken to start at �1: Thus, the lift of the compositeloop will go from 0 to �1 and then (shifting the lift of the second loop by �1) from �1to �1 C �2: But any path from 0 to �1 C �2 in t is homotopic with endpoints fixed tothe straight-line path � 7! e2��.�1C�2/: Thus, if we identify elements of �1.T / withelements of �; the composition operator corresponds to addition in �: ut

We now state the first main result of this section; the proof is given in Sect. 13.7.

Theorem 13.15. Every loop in K is homotopic to a loop in T:

The theorem does not mean that �1.K/ is isomorphic to �1.T /; since a loopin T may be null homotopic in K even if it is not null homotopic in T: Indeed,�1.T / (which is isomorphic to �) is often very different from �1.K/ (which is, forexample, trivial when K D SU.n/). Nevertheless, the theorem gives us a usefulway to study �1.K/; because we understand �1.T /. Taking Theorem 13.15 andProposition 13.14 together, we see that every loop in K is homotopic to a loop ofthe form � 7! e2��� ; with � 2 �: Thus, we are faced with a very concrete problemto calculate �1.K/: We must only determine, for each � 2 �; whether the loopt 7! e2�t� is null homotopic in the compact group K: (By Proposition 13.14, sucha loop is null homotopic in the torus T only if � D 0:)

The condition for � 7! e2��� to be null homotopic in K turns out to be relatedto the notion of coroots. Recall that if ˛ is a real root for t; then (Lemma 12.8),the associated real coroot H˛ D 2˛= h˛; ˛i belongs to �: Thus, any integer linearcombination of coroots also belongs to �:

Definition 13.16. The coroot lattice, denoted I; is the set of all integer linearcombinations of real corootsH˛; ˛ 2 R:

We now state the second main result of this section; the proof is also in Sect. 13.7.

Theorem 13.17. For each � 2 �; the loop � 7! e2��� is null homotopic inK if andonly if � belongs to the coroot lattice I:

If we combine this result with Theorem 13.15, we obtain the following descrip-tion of the fundamental group of K:

Corollary 13.18. The fundamental group of K is isomorphic to the quotient group�=I; where the quotient is of commutative groups.

Proof. By Theorem 13.15, every loop in K is homotopic to a loop of the form � 7!e2��� ; with � 2 �: Under this correspondence, composition of loops corresponds toaddition in �: By Theorem 13.17, two loops of the form � 7! e2���1 and � 7! e2���2

are homotopic if and only if �1 � �2 belongs to I: Thus, �1.K/ may be identifiedwith the set of cosets of I in �: ut

382 13 Fundamental Groups of Compact Lie Groups

We now consider three examples. The first two are familiar friends, the groupsSO.5/ and SU.3/: The third is the projective unitary group PSU.3/: In general,PSU.n/ is the quotient of SU.n/ by the subgroup consisting of whichever multiplesof the identity have determinant 1 in U.n/: In the case n D 3; a matrix of the formei�I has determinant 1 if and only if e3i� D 1: Thus,

PSU.3/ D SU.3/=fI; e2�i=3I; e4�i=3I g: (13.4)

We can represent PSU.3/ as a matrix group by using the adjoint representation. Itis easy to check that the center of SU.3/ is the group being divided by on the right-hand side of (13.4); thus, the image SU.3/ under the adjoint representation will be asubgroup of GL.sl.3IC// isomorphic to PSU.3/: Since the center of the Lie algebrasu.3/ is trivial, the Lie algebra version of the adjoint representation is faithful; thus,we may identify the Lie algebra of PSU.3/ with the Lie algebra of SU.3/:

Example 13.19. If K D SO.5/; then �=I Š Z=2: If K D SU.3/; then �=I istrivial. Finally, if K D PSU.3/; then �=I Š Z=3:

The verification of the claims in Example 13.19 is left to the reader (Exercise 5).Corollary 13.18 together with Example 5 give another way of computing the funda-mental groups of SO.5/ (Z=2/ and SU.3/ (trivial), in addition to Proposition 13.10.

Figures 13.1 and 13.2 show � and I in the case of the groups SO.5/ and PSU.3/:The black dots indicate points in I , whereas white dots indicate points in � that are

Fig. 13.1 The coroot lattice (black dots) and the kernel of the exponential mapping (black andwhite dots) for the group SO.5/

13.4 The Fundamental Groups of K and T 383

2

1

Fig. 13.2 The coroot lattice (black dots) and the kernel of the exponential mapping (black andwhite dots) for the group PSU.3/

not in I: In the case of SO.5/; it is easy to see that any two white dots differ by anelement of the coroot lattice, showing that there is exactly one nontrivial element of�=I: In the case of PSU.3/; note that the elements �1 and �2 D 2�1 are not in I; but3�1 is in I; showing that Œ�1� is an element of order 3 in �=I: The reader may verifythat every element of � is either in I; differs from �1 by an element of I; or differsfrom �2 by an element of I: The situation for SU.3/ is similar to that for PSU.3/;the coroot lattice I does not change, but � is now equal to I; so that �=I is trivial.For now, the reader may regard the lines in the figures as merely decorative; theselines will turn out to make up the “Stiefel diagram” for the relevant group. (SeeSect. 13.6.)

Corollary 13.20. If K is simply connected, then every algebraically integralelement is analytically integral.

Proof. In light of Theorem 13.17, K is simply connected if and only if I D �: If� 2 t is algebraically integral, then h�;H˛i 2 Z for all ˛; where H˛ D 2˛= h˛; ˛iis the real coroot associated to ˛: It follows that h�; �i 2 Z for every element of I;the set of integer linear combinations of coroots. Thus, if K is simply connected,h�; �i 2 Z for every element of I D �; which means that � is analyticallyintegral. ut

We may offer a completely different proof of Corollary 13.20, as follows. Wefirst observe that if K is simply connected, then by Proposition 7.7, the complexLie algebra g WD kC is semisimple. Then let � be an algebraically integral element.Since the sets of analytically integral and algebraically integral elements are bothinvariant under the action of W; it is harmless to assume that � is dominant. Thus,

384 13 Fundamental Groups of Compact Lie Groups

by the results of Chapter 9, there is a finite-dimensional irreducible representation.�; V�/ of g WD kC with highest weight �: SinceK is simply connected, Theorem 5.6then tells us that there is an associated representation … of K acting on V�: Thus,� is a weight for a representation of the group K; which implies (Proposition 12.5)that � is analytically integral.

Although it is mathematically correct, the preceding argument may be consideredas “cheating,” since it depends on the whole machinery of Verma modules (toconstruct the representations of g) and on the Baker–Campbell–Hausdorff formula(to prove Theorem 5.6). In the subsequent sections, we will use techniques similarto those in the proof of the torus theorem to prove Theorem 13.17 and thus to give amore direct (but not easy!) proof of Corollary 13.20.

Corollary 13.21. If K is simply connected, the element ı (half the sum of the real,positive roots) is analytically integral.

Proof. According to Proposition 8.38 (translated into the language of real roots), theelement ı is algebraically integral. But by Corollary 13.20, ifK is simply connected,every algebraically integral element is analytically integral. ut

Since it is easy to do so, we will immediately prove one direction of The-orem 13.17, namely that if � is in the coroot lattice I , then � 7! e2��� ishomotopically trivial in K:

Proof of Theorem 13.17, one direction. We assume at first that � D H˛; a singlereal coroot. Then by Corollary 7.20, there is a homomorphism � of su.2/ into ksuch that � maps the element iH D diag.i;�i/ in su.2/ to the real coroot H˛:

Since SU.2/ is simply connected, there exists a homomorphism ˆ W SU.2/ ! K

such that ˆ.eX/ D e�.X/ for all X 2 su.2/:Consider, then, the loop l in SU.2/ given by

l.�/ D e2�� iH D�e2�i� 0

0 e�2�i��; 0 � � � 1:

Observe that ˆ.l.�// D e2��H˛ ; so that the image of l under ˆ is the relevant loopin K: Since SU.2/ is simply connected, there is a family ls of loops connecting l toa constant loop in SU.2/: Thus, the loops

� 7! ˆ.ls.�//

constitute a homotopy of our original loop to a point in K:Now, if � is an integer linear combination of coroots, then by Proposition 13.14,

the loop � 7! e2��� is homotopic (in T and thus in K) to a composition of loopsof the form � 7! e2��H˛ ; for various coroots ˛: Since each of those loops is nullhomotopic in K; so is the loop � 7! e2��� : ut

13.5 Regular Elements 385

13.5 Regular Elements

We will study the topology of K by writing every element x of K as

x D yty�1;

with y 2 K and t 2 T: It is then convenient to write the variable t in the precedingexpression in exponential coordinates. Thus, we may consider the map ‰ W t �.K=T / ! K given by

‰.H; Œy�/ D ye2�Hy�1: (13.5)

(We continue to scale the exponential map for T by a factor of 2�:) The torustheorem tells us that‰ maps ontoK: On the other hand, the behavior of‰ is easiestto understand when the differential‰� is nonsingular.

We now introduce a notion of “regular”elements in K; we will see (Proposi-tion 13.24) that ‰�.H; Œy�/ is invertible if and only if x WD ‰.H; Œy�/ is regular. Itwill turn out that the fundamental group of K is the same as the fundamental groupof the set of regular elements in K:

Definition 13.22. If x 2 K is contained in a unique maximal torus, x is regular;if x is contained in two distinct maximal tori, x is singular. The set of regularelements in K is denotedKreg and the set of singular elements is denotedKsing:

If, for example, t 2 T generates a dense subgroup of T; then the only maximaltorus containing yty�1 is yTy�1: Thus, for such a t; the element yty�1 is regular.As we will see, however, yty�1 can be regular even if t does not generate a densesubgroup of T:

Proposition 13.23. Suppose x 2 K has the form x D yty�1 with t 2 T and y 2 K:If there exists some X 2 k with X … t such that

Adt .X/ D X;

then x is singular. If no such X exists, x is regular.

Proof. The condition of being regular is clearly invariant under conjugation; that is,x is regular if and only if t is regular. Suppose now that there is some X … t withAdt .X/ D X: Then t commutes with e�X for all � 2 R: Applying Lemma 11.37with S D fe�X g�2R; there is a maximal torus S 0 containing both t and fe�X g�2R:But then the Lie algebra s0 of S 0 must contain X; which is not in t; which meansthat S 0 ¤ T: Thus, t (and therefore, also x) is singular.

In the other direction, if x (and therefore, also, t) is singular, then there is amaximal torus S 0 ¤ T containing t: But we cannot have S 0 � T; or else S 0 wouldnot be maximal. Thus, there must be some X in the Lie algebra s0 of S 0 that is notin t: But since t 2 S 0 and S 0 is commutative, we have Adt .X/ D X: ut

386 13 Fundamental Groups of Compact Lie Groups

Proposition 13.24. An element x D ye2�Hy�1 is singular if and only if there issome root ˛ for which

h˛;H i 2 Z:

It follows that x D ye2�Hy�1 is singular if and only if ‰� is singular at the point.H; Œy�/:

Note that for each fixed ˛ 2 R and n 2 Z; the set ofH in t for which h˛;H i D n

is a hyperplane (not necessarily through the origin).

Proof. By Proposition 13.23, x is singular if and only if Ade2�H .X/ D X for someX … t: Now, the dimension of the eigenspace of Ade2�H with eigenvalue 1 2 R isthe same whether we work over R or over C: The eigenvalues of Ade2�H over C are1 (on h) together with the numbers of the form

e2�ih˛;H i; ˛ 2 R;

(from the root space g˛). Thus, the dimension of the 1-eigenspace is greater thandim h if and only if e2�ih˛;H i D 1 for some ˛;which holds if and only if h˛;H i 2 Z:

Meanwhile, the map ‰ is just the map ˆ in Definition 11.25, composed with theexponential map for T: Since T is commutative, the differential of the exponentialmap for T is the identity at each point. Thus, using Proposition 11.27, we see that‰�.H; Œy�/ is singular if and only if the restriction of Ade�2�H to t? (or, equivalently,to .t?/C) has an eigenvalue of 1. Since the eigenvalues of Ade�2�H on .t?/C are thenumbers of the form e�2�ih˛;H i; we see that ‰�.H; Œy�/ is singular if and only ifh˛;H i 2 Z for some ˛: utDefinition 13.25. An element H of t is regular if for all ˛ 2 R; the quantityh˛;H i is not an integer. Otherwise, H is singular. The set of regular elements in tis denoted treg:

A key issue in the proof of Theorem 13.17 is to understand the extent to which themap‰ in (13.5) fails to be injective. There are two obvious sources of noninjectivityfor ‰: The first is the kernel of the exponential map; clearly if � 2 �; then

‰.H C �; Œx�/ D ‰.H; Œx�/:

Meanwhile, if w 2 W and z 2 N.T / represents w; then

‰.w �H; Œxz�1�/ D xz�1e2�.w�H/zx�1

D xe2�Hx�1

D ‰.H; Œx�/;

since z�1 represents w�1:We now demonstrate that if we restrict‰ to treg � .K=T /,these two sources account for all of the noninjectivity of ‰:

13.5 Regular Elements 387

Proposition 13.26. Suppose .H; Œx�/ and .H 0; Œx0�/ belong to treg � .K=T /: Then‰.H; Œx�/ D ‰.H 0; Œx0�/ if and only if there exist some w D Œz� in W and some� 2 � such that

H 0 D w �H C �

Œx0� D Œxz�1�: (13.6)

Here Œz� denotes the coset containing z 2 N.T / in W D N.T /=T: Furthermore,if the elements in (13.6) satisfy H 0 D H and Œx0� D Œx�; then � D 0 and w is theidentity element of W:

Note that if z 2 N.T / and t 2 T; then

xtz�1 D xz�1.ztz�1/;

where ztz�1 2 T: Thus, Œxtz�1� and Œxz�1� are equal in K=T: That is to say, fora fixed z 2 N.T /; the map Œx� 7! Œxz�1� is a well-defined map of K=T to itself.A similar argument shows that this action depends only on the coset of z inN.T /=T:

Proof. IfH 0 and x0 are as in (13.6), then by the calculations preceding the statementof the proposition, we will have ‰.H 0; Œx0�/ D ‰.H; Œx�/: In the other direction, if‰.H; Œx�/ D ‰.H 0; Œx0�/; then

xe2�Hx�1 D x0e2�H 0

.x0/�1;

which means that

e2�H D z�1e2�H 0

z; (13.7)

where z D .x0/�1x:Now, the relation (13.7) implies that e2�H belongs to the torus z�1Tz: SinceH 2

treg; it follows from Proposition 13.24 that z�1Tz D T; that is, that z 2 N.T /: Then,if w D Œz�; we have

e2�H D e2�w�1�H 0

:

From this, we obtain e2�.w�1�H 0�H/ D I; which means that w�1 � H 0 �H belongs

to �: Since � is invariant under the action of W; the element � WD H 0 � w �H alsobelongs to �; and we find that H 0 D w �H C � and x0 D xz�1; as claimed.

Finally, if H 0 D H and Œx0� D Œx�; then x0 and x belong to the same cosetin K=T; which means that z�1 must be in T: Thus, w is the identity element inW D N.T /=T: But once w D e; we see that H 0 D H only if � D 0: ut

We now come to a key result that is essential to the proofs of Theorems 13.15and 13.17.

388 13 Fundamental Groups of Compact Lie Groups

Theorem 13.27. The fundamental groups of K and Kreg are isomorphic. Specif-ically, every loop in K is homotopic to a loop in Kreg and a loop in Kreg is nullhomotopic in K only if it is null homotopic in Kreg:

To prove this result, we will first show that the singular set in K is “small,”meaning that it has codimension at least 3. (In the case K D SU.2/, for example,the singular set is just fI;�I g; so that Ksing has dimension 0, whereas K hasdimension 3.) We will then argue as follows. Let n be the dimension of K andsupposeE and F are subsets ofK of dimension k and l; respectively. If kC l < n;

then “generically”E and F will not intersect. IfE and F do intersect, then (we willshow) it is possible to perturb F slightly so as to be disjoint from E: We first applythis result with E D Ksing and F being a loop in K: Then E has dimension at mostn � 3 while F has dimension 1, so the loop F is homotopic to a loop that does notintersect Ksing: We next apply this result with E D Ksing and F being a homotopyof a loop l � Kreg: ThenE still has dimension at most n�3 while F has dimension2 (the image of a square), so the homotopy can be deformed to a homotopy thatdoes not intersectKsing: In the remainder of this section, we will flesh out the aboveargument.

Lemma 13.28. There exist finitely many smooth compact manifolds M1; : : : ;MN

together with smooth maps fj W Mj ! K such that (1) each Mj has dimension atmost dimK � 3; and (2) each element of Ksing is in the image of fj for some j:

Proof. Since each root ˛ is analytically integral (Proposition 12.7), there exists amap f˛ W T ! S1 such that

f˛.e2�H / D e2�ih˛;H i (13.8)

for all H 2 t: Clearly, f˛ is actually a homomorphism of T into S1: Let T˛ be thekernel of f˛; so that T˛ is a closed subgroup of T: The Lie algebra of T˛ is the setof H 2 t with h˛;H i D 0; thus, T˛ has dimension one less than the dimension ofT: (Note that T˛ may not be connected.) For each H 2 t; we see from (13.8) thate2�H belongs to T˛ if and only if h˛;H i is an integer. Thus, by the torus theoremand Proposition 13.24, each singular element inK is conjugate to an element of T˛;for some ˛:

Now fix a root ˛ and let C.T˛/ denote the centralizer of T˛; that is, the set ofx 2 K such that x commutes with every element of T˛: Then C.T˛/ is a closedsubgroup of K; and the Lie algebra of C.T˛/ consists of those X 2 k such thatAdt .X/ D X for all t 2 T˛: Suppose now that X˛ belongs to the root space g˛ andt D e2�H belongs to T˛: Then

Ade2�H .X˛/ D e2�adH .X˛/ D e2�ih˛;H iX˛ D X˛;

since f .e2�H / D e2�ih˛;H i D 1; by assumption, and similarly with X˛ replaced byY˛ WD X �: Thus, the Lie algebra of C.T˛/ will contain t and at least two additionalelements, X˛ � Y˛ and i.X˛ C Y˛/; that are independent of each other and of t:(Compare Corollary 7.20.) We conclude that the dimension of C.T˛/ is at leastdimT˛ C 2:

13.5 Regular Elements 389

Now, the map g˛ W T˛ �K ! K given by

g˛.t; x/ D xtx�1

descends to a map (still called g˛) of T˛ � .K=C.T˛// into K: Furthermore, wecompute that

dim.T˛ � .K=C.T˛/// D dimT˛ C dimK � dimC.T˛/

� dimT � 1C dimK � dimT � 2

D dimK � 3:

Since, as we have said, every singular element is conjugate to an element of someT˛; we have proved the lemma, with the M ’s being the manifolds T˛ � .K=C.T˛//and the f ’s being the maps g˛: utLemma 13.29. Let M and N be compact manifolds and let D be a compactmanifold with boundary, with

dimM C dimD < dimN:

Let f W M ! N and g W D ! N be smooth maps. Suppose E is a closed subsetof D such that g.E/ is disjoint from f .M/: Then g is homotopic to a map g0 suchthat g0 D g on E and such that g0.D/ is disjoint from f .M/:

Since g.E/ is already disjoint from f .M/; it is plausible that we can deformg without changing its values on E to make the image disjoint from f .M/: Ourproof will make use of the following result: If X and Y are smooth manifolds withdimX < dimY and f W X ! Y is a smooth map, then the image ofX under f is aset of measure zero in Y: (Note that we do not assume f� is injective.) This result isa consequence of Sard’s theorem; see, for example, Corollary 6.11 in [Lee]. We willuse this result to show that g can be moved locally off of f .M/; a finite number ofthese local moves will then produce the desired map g0:

Proof. Step 1: The local move. For x 2 D n E; let us choose a neighborhood Uof g.x/ diffeomorphic to Rn; where n D dimN: We may then define a map

h W f �1.U / � g�1.U / ! U

by

h.m; x/ D f .m/ � g.x/;

where the difference is computed in Rn Š U: By our assumption on thedimensions, the image of h is a set of measure zero in U . Thus, in everyneighborhood of 0; we can find some p that is not in the image of h:

390 13 Fundamental Groups of Compact Lie Groups

Suppose that W is any neighborhood of x in D such that NW is contained ing�1.U / and NW is disjoint from E: Then we can choose a smooth function � onD such that � equals 1 on NW but such that � equals 0 both on E and on thecomplement of g�1.U /: Let us define a family of maps gs W D ! N by setting

gs.x/ D g.x/C s�.x/p; 0 � s � 1

for x 2 g�1.U / and setting gs.x/ D g.x/ for x … g�1.U /:When s D 1; and x 2 NW ; we have

f .m/ � g1.x/ D f .m/ � g.x/ � p D h.m; x/ � p:

Since h never takes the value p; we see that f .m/ does not equal g1.x/; thus,g1. NW / � U is disjoint from f .M/: We conclude that g1 is a map homotopic tog such that g1 D g on E and g1. NW / is disjoint from f .M/: Furthermore, sincep can be chosen to be as small as we want, we can make g1 uniformly as closeto g as we like.

Step 2: The global argument. Choose a neighborhood V of E in D such thatg. NV / is disjoint from f .M/; and letK be the complement of V inD: For each xinK;we can find a neighborhoodU � N of g.x/ such thatU is diffeomorphic toRn:Now choose a neighborhoodW of x so that NW is contained in g�1.U / but NWis disjoint fromE: SinceK is compact, there is some finite collection x1; : : : ; xNof points in K such that the associated open sets W1; : : : ;WN cover K: Thus,each NWj is disjoint from E and is contained in a set of the form g�1.Uj /; withUj � N diffeomorphic to Rn:

By the argument in Step 1, we can find a map g1 homotopic to g such thatg1 D g on E and g1. NW1/ is disjoint from f .M/; but such that g is as close aswe like to g: Since g. NV / and f .M/ are compact, the distance between g. NV / andf .M/ (with respect to an arbitrarily chosen Riemannian metric on N ) achievesa positive minimum. Therefore, if we take g1 close enough to g; then g1. NV /will still be disjoint from f .M/: We can similarly ensure that g1 still maps thecompact set NWj into Uj for j D 2; : : : ; N:

We may now perform a similar perturbation of g1 to a map g2 such that g2 Dg1 D g on E but such that g2. NW2/ is disjoint from f .M/: By making thisperturbation small enough, we can ensure that g2 has the same properties asg1: First, g2. NV / is still disjoint from f .M/; second, g2. NW1/ is still disjoint fromf .M/; and third, g2. NWj / is still contained in Uj for j D 3; : : : ; N: Proceedingon in this fashion, we eventually obtain a map gN homotopic to g such thatgN D g on E and such that gN . NV / and each gN . NWj / are disjoint from f .M/:

Then g0 WD gN is the desired map. utProof of Theorem 13.27. In Lemma 13.28, it is harmless to assume that each Mj

has dimension dimK � 3; since if any Mj has a lower dimension, we can take aproduct of Mj with, say, a sphere dimension of an appropriate dimension, and then

13.6 The Stiefel Diagram 391

make fj independent of the sphere variable. Once this is done, we can letM be thedisjoint union of the Mj ’s and let f be the map equal to fj on Mj : That is to say,the singular set is actually in the image under a smooth map f of single manifoldM of dimension dimK � 3:

If l is any loop in K; it is not hard to prove that l is homotopic to smooth loop.(See, for example, Theorem 6.26 in [Lee].) In fact, this claim can be proved by anargument similar to the proof of Lemma 13.29; locally, any continuous map fromRn to Rm can be approximated by smooth functions (even polynomials), and onecan then patch together these approximations, being careful at each stage not todisrupt the smoothness achieved at the previous stage. We may thus assume that lis smooth and apply Lemma 13.29 with D D S1;N D K; and E D ¿: SincedimS1 C dimM < dimK; the lemma says that we can deform l until it does notintersect f .M/ Ksing:

Next, suppose l is a loop in Kreg; which we can assume to be smooth. If l isnull homotopic in Kreg; it is certainly null homotopic in K: In the other direction,suppose l is null homotopic inK:We think of the homotopy of l to a point as a mapg of a 2-disk D into K; where the boundary of D corresponds to the original loopl and the center of D correspond to the point. After deforming g slightly—by thesame argument as in the previous paragraph—we may assume that g is smooth. Wenow apply Lemma 13.29 with D equal to the 2-disk and E equal to the boundaryof D: The lemma tells us that we can deform g to a map g0 that agrees with g onthe boundary of D but such that g0.D/ is disjoint from f .M/ Ksing: Thus, g0 isa homotopy of l to a point in Kreg: ut

13.6 The Stiefel Diagram

In this section, we look more closely at the structure of the hyperplanes in t whereh˛;H i D n, which appear in Proposition 13.24. The main result of the section isTheorem 13.35, which constructs a cover of the regular set Kreg:

Definition 13.30. For each n 2 Z and ˛ 2 R; consider the hyperplane (notnecessarily through the origin) given by

L˛;n D fH 2 t jh˛;H i D n g :

The union of all such hyperplanes is called the Stiefel diagram. A connectedcomponent of the complement in t of the Stiefel diagram is called an alcove.

In light of Proposition 13.24 and Definition 13.25, the complement of the Stiefeldiagram is just the regular set treg in t: Figures 13.3, 13.4, and 13.5 show the Stiefeldiagrams for A2;B2; and G2; respectively. In each figure, the roots are indicatedwith arrows, with the long roots being normalized to have length

p2. The black

dots in the figure indicate the coroots and one alcove is shaded. Note that since

392 13 Fundamental Groups of Compact Lie Groups

Fig. 13.3 The Stiefeldiagram for A2; with theroots normalized to havelength

p2: The black dots

indicate the coroots

Fig. 13.4 The Stiefel diagram for B2; with the long roots normalized to have lengthp2: The

black dots indicate the coroots

h˛;H˛i D 2; if we start at the origin and travel in the ˛ direction, the coroot H˛

will be located on the second hyperplane orthogonal to ˛:Let us check the correctness of, say, the diagram forG2: Suppose ˛ is a long root,

which we normalize so that h˛; ˛i D 2: Then ˛=2 belongs to the line L˛;1; so thatL˛;1 is the (unique) line orthogonal to ˛ passing through ˛=2: Similarly, L˛;n is the

13.6 The Stiefel Diagram 393

Fig. 13.5 The Stiefel diagram for G2; with the long roots normalized to have lengthp2: The

black dots indicate the coroots

line orthogonal to ˛ passing through n˛=2: Suppose, on the other hand, that ˛ is ashort root, which we must then normalize so that h˛; ˛i D 2=3: Then 3˛=2 belongsto L˛;1, so that L˛;1 is the line orthogonal to ˛ passing through 3˛=2 and L˛;n is theline orthogonal to ˛ passing through 3n˛=2:Meanwhile, for a long root ˛; we haveH˛ D 2˛=2 D ˛, whereas for a short root, we have H˛ D 2˛=.2=3/ D 3˛: Theseresults agree with what we see in Figure 13.5.

It will follow from Proposition 13.34 that every alcove is isometric to every otheralcove. Furthermore, if g WD kC is simple (i.e., if the root system R spans t and isirreducible), each alcove is a “simplex,” that is, a bounded region in a k-dimensionalspace defined by kC 1 linear inequalities. Thus, for rank-two simple algebras, eachalcove is a triangle, as can be seen from Figures 13.3, 13.4, and 13.5. In general,the structure of the alcoves for a simple algebra is described by an extended Dynkindiagram; see Exercises 11, 12, and 13.

Proposition 13.31. The Stiefel diagram is invariant under the action of W andunder translations by elements of �:

Proof. Since W permutes the roots, invariance of the Stiefel diagram is evident.Meanwhile, if � is in the kernel � of the exponential map, the adjoint action of e2��

on g is trivial. Thus, for X 2 g˛; we have

X D Ade2�� .X/ D e2�ih˛;�iX;

which means that h˛; �i is an integer, for all ˛ 2 R: From this we can easily seethat if H is in the Stiefel diagram, so H C �: ut

394 13 Fundamental Groups of Compact Lie Groups

Recall that an affine transformation of a vector space is a transformation thatcan be expressed as a combination of a linear transformation and a translation.

Proposition 13.32. Let � ÌW denote the set of affine transformations of t that canbe expressed in the form

H 7! w �H C �

for some � 2 � and some w 2 W: Then � ÌW forms a group under composition,with the group law given by

.�;w/ � .� 0;w0/ D .� C w � � 0;ww0/:

The group � ÌW is a semidirect product of � and W; with � being the normalfactor.

Proof. We merely compute that

w � .w0 �H C � 0/C � D .ww0/ �H C � C w � � 0;

so that the composition of the affine transformations associated to .�;w/ and .� 0;w0/is the affine transformation associated to .� C w � � 0;ww0/: ut

We now introduce the analog of the Weyl group for the Stiefel diagram. Evenif a hyperplane V � t does not pass through the origin, we can still speak of thereflection s about V; which is the unique affine transformation of t such that

s.H CH 0/ D H �H 0

wheneverH is in V andH 0 is orthogonal to V:

Definition 13.33. The extended Weyl group for K relative to t is the group ofaffine transformations of t generated by reflections about all the hyperplanes in theStiefel diagram of t:

We now establish some key properties of the extended Weyl group. Recall fromDefinition 13.16 the notion of the coroot lattice I:

Proposition 13.34. 1. The extended Weyl group equals I Ì W; the semidirectproduct of the ordinary Weyl group W and the coroot lattice I:

2. The extended Weyl group acts freely and transitively on the set of alcoves.

Note that since I � � is invariant under the action of W; the extended Weylgroup I ÌW is a subgroup of the group � ÌW in Proposition 13.32. That is to say,the group law in I ÌW is the same as in Proposition 13.32, although I ÌW will,in general, be a proper subgroup of � ÌW:

Proof. For Point 1, let V˛; ˛ 2 R; be the hyperplane through the origin orthogonalto ˛: Since h˛;H˛i D 2; we see that the hyperplane L˛;n in Definition 13.30 is the

13.6 The Stiefel Diagram 395

translate of V˛ by nH˛=2: We can then easily verify that the reflection s˛;n aboutL˛;n is given by

s˛;n.H/ D s˛.H/C nH˛; (13.9)

where s˛ is the reflection about V˛ . Thus, s˛;n is a combination of an element s˛ ofW and a translation by an element of I: It follows that the extended Weyl groupis contained in I Ì W: In the other direction, the extended Weyl group certainlycontains W; since the Stiefel diagram contains the hyperplanes through the originorthogonal to the roots. Furthermore, from (13.9), we see that the composition ofs˛;1 and s˛;0 is translation by H˛: Thus, the extended Weyl group contains all ofI ÌW:

For Point 2, we first argue that the orbits of I ÌW in t do not have accumulationpoints. For anyH 2 t; the orbit ofH is the set of all vectors of the form w �H C � ,with w 2 W and � 2 I: SinceW is finite and I contains only finitely many points ineach bounded region, the orbit of H also contains only finitely many points in eachbounded region. Once this observation has been made, we may now repeat, almostword for word, the proof that the ordinary Weyl group acts freely and transitivelyon the set of open Weyl chambers. (Replace the hyperplanes orthogonal to the rootswith the hyperplanes L˛;n and replace the open Weyl chambers with the alcoves.)The only point where a change is necessary is in the proof of the transitivity of theaction (Proposition 8.23). To generalize that argument, we need to know that foreach H and H 0 in t; the orbit .I Ì W / � H 0 contains a point at minimal distancefrom H: Although the extended Weyl group is infinite, since each orbit containsonly finitely many points in each bounded region, the result still holds. The reader isinvited to work through the proofs of Propositions 8.23 and 8.27 with the ordinaryWeyl group replaced by the extended Weyl group, and verify that no other changesare needed. utTheorem 13.35. The map ‰ W treg � .K=T / ! Kreg is a covering map.Furthermore, if A � t is any one alcove, then ‰ maps A � .K=T / onto Kreg and

‰ W A � .K=T / ! Kreg

is also a covering map.

Recall the definition of ‰ in (13.5). We will see in the next section that K=Tis simply connected. Since each alcove A is a convex set, A is contractible andtherefore simply connected. We will thus conclude that A� .K=T / is the universalcover of Kreg:

Proof. Recall the map ˆ W T � .K=T / ! K given by ˆ.t; Œx�/ D xtx�1; andlet Treg D Kreg \ T: Since the exponential map for T is a local diffeomorphism, itfollows from Proposition 13.24 that ˆ is a local diffeomorphism on Treg � .K=T /:Furthermore, by a trivial extension of Proposition 13.26, we have ˆ.t; Œx�/ Dˆ.t 0; Œx0�/ if and only if there is some z 2 N.T / such that t 0 D ztz�1 and x0 D xz�1:Finally, if W D N.T /=T acts on Treg � .K=T / by

396 13 Fundamental Groups of Compact Lie Groups

w � .t; Œx�/ D .ztz�1; Œxz�1�/

for each w D Œz� inW; then this action is free onK=T and thus free on Treg�.K=T /:Fix some y in Kreg and pick some .t; Œx�/ in Treg � .K=T / for which

ˆ.t; Œx�/ D y. Then since W is finite and acts freely, we can easily find aneighborhood U of .t; Œx�/ such that w � U is disjoint from U for all w ¤ e inW: It follows that the sets w � U are pairwise disjoint and thus that ˆ is injectiveon each such set. Now, since ˆ is a local diffeomorphism, ˆ.U / will be open inKreg: The preimage of V is then the disjoint union of the sets w � U and the localdiffeomorphism ˆ will map each w � U homeomorphically onto V: We concludethat ˆ is a covering map of Treg � .K=T / ontoKreg:

Meanwhile, as shown in the proof of Proposition 13.14, the exponential map forT is a covering map. Thus, the map

.H; Œx�/ 7! .e2�H ; Œx�/

is a covering map from treg � .K=T / onto Treg � .K=T /: Since the composition ofcovering maps is a covering map, we conclude that ‰ W treg � .K=T / ontoKreg is acovering map, as claimed.

Finally, Proposition 13.34 shows that each point in treg can be moved into A bythe action of I ÌW � �ÌW: Thus,‰ actually mapsA�.K=T / ontoKreg: For eachy 2 Kreg; choose a neighborhood V of y so that‰�1.V / is a disjoint union of opensets U˛ mapping homeomorphically onto V: By shrinking V if necessary, we canassume V is connected, in which case the U˛’s will also be connected. Thus, eachU˛ is either entirely in A� .K=T / or disjoint from A� .K=T /: Thus, if we restrict‰ to A � .K=T /; the preimage of V will now consist of some subset of the U˛’s,each of which still maps homeomorphically onto V; showing that the restriction of‰ to A � .K=T / is still a covering map. ut

13.7 Proofs of the Main Theorems

We now have all the necessary tools to attack the proofs of our main results. Here isan outline of our strategy in this section. We will first show that the quotientK=T isalways simply connected. On the one hand, the simple connectivity of K=T leadsto a proof of Theorem 13.15, that every loop in K is homotopic to a loop in T: Onthe other hand, the simple connectivity of K=T means that the set A � .K=T / inTheorem 13.35 is actually the universal cover ofKreg: Thus, to determine �1.K/ Š�1.Kreg/; we merely need to determine how close to being injective the coveringmap ‰ W A � .K=T / ! Kreg is.

Now, according to Proposition 13.26, the failure of injectivity for the full map‰ W treg � .K=T / is due to the action of the group � Ì W: If we restrict ‰ to

13.7 Proofs of the Main Theorems 397

A� .K=T /, then by Proposition 13.34, we have eliminated the failure of injectivitydue to the subgroup I ÌW � � Ì W: Thus, the (possible) failure of injectivity of‰ on A � .K=T / will be measured by the extent to which I fails to be all of �:

Proposition 13.36. The quotient manifoldK=T is simply connected.

Proof. Let Œx.�/� be any loop inK=T: LetH be a regular element in T and considerthe loop l in Kreg given by

l.�/ D x.�/e2�Hx.�/�1:

Now let t.s/ be any path in T connecting e2�H to I; and consider the loops

ls.�/ WD x.�/t.s/x.�/�1:

Clearly, ls is a homotopy of l in K to the constant loop at I:Thus, for any loop Œx.�/� in K=T; the corresponding loop l.�/ in Kreg is null

homotopic inK . We now argue that Œx.�/� itself is null homotopic inK=T:As a firststep, we use Theorem 13.27 to deform the homotopy ls to a homotopy l 0s shrinkingl to a point in Kreg: Now, the map ‰ is a covering map and the loop .H; Œx.�/�/is a lift of l to A � .K=T /: Thus, as a second step, we can lift the homotopy l 0sto A � .K=T / to a homotopy l 00s shrinking .H; Œx.�/�/ to a point in A � .K=T /:

(See Proposition 13.3.) Finally, as a third step, we can project the homotopy l 00s fromA � .K=T / onto K=T to obtain a homotopy shrinking Œx.�/� to a point in K=T: utProposition 13.37. Every loop in K is homotopic to a loop in T:

Proof. By Proposition 13.8, the groupK is a fiber bundle with base K=T and fiberT: Suppose now that l is a loop in K and that l 0.�/ WD Œl.�/� is the correspondingloop in K=T: Since K=T is simply connected, there is a homotopy l 0s shrinking l 0to a point in K=T: Furthermore, since K=T is connected, it is harmless to assumethat l 0s shrinks to the point ŒI � in K=T: Now, fiber bundles are known to have thehomotopy lifting property (Proposition 4.48 in [Hat]), which in our case means thatthere is a homotopy ls in K such that l0 D l and such that Œls.�/� D l 0s.�/ for all �and s: Since l 01 is a constant loop at ŒI �; the loop l1 lies in T: utLemma 13.38. Suppose � 2 � but � … I: Then there exist � 0 in I and w inW suchthat the affine transformation

H 7! w � .H C � C � 0/ (13.10)

maps A to itself but is not the identity map of A:

Proof. By Proposition 13.31, translation by � maps the alcove A to some otheralcove A0: Then by Proposition 13.34, there exists an element of the extended Weylgroup that maps A0 back to A: Thus, there exist � 0 2 I and w 2 W such thatthe map (13.10) maps A to itself. If this map were the identity, then translation by� would be (the inverse of) some element of I Ì W; which would mean that �

398 13 Fundamental Groups of Compact Lie Groups

H

H

H

Fig. 13.6 The element � is in the kernel of the exponential mapping for PSU.3/; but is not in thecoroot lattice. If we apply a rotation by 2�=3 to H C �; we obtain an element H 0 in the samealcove as H

would have to be in I; contrary to our assumption. Thus, the map in (13.10) sendsA to itself and is not the identity map on t: But an affine transformation is certainlydetermined by its restriction to any nonempty open set, which means that the mapcannot be the identity on A: ut

Suppose now that � 2 � but � … I: Let us choose � 0 2 I and w 2 W as inLemma 13.38. We may then chooseH 2 A so that H 0 WD w � .H C � C � 0/ lies inA but is distinct from H: (See Figure 13.6 in the case of the group PSU.3/:) Let pdenote the path in A given by

p.�/ D H C �.H 0 �H/; 0 � � � 1:

Now choose x 2 N.T / representing w, let x.�/ be a path connecting I to x in K;and define a path q in A � .K=T / by

q.�/ D .p.�/; Œx.�/�1�/: (13.11)

Note that p.1/ ¤ p.0/ and thus, certainly, q.1/ ¤ q.0/:

Lemma 13.39. Let q.�/ be the path in A� .K=T / given by (13.11). Then the path

� 7! ‰.q.�//

is a loop in Kreg and this loop is homotopic in K to the loop

� 7! e2��� :

13.7 Proofs of the Main Theorems 399

Proof. On the one hand, we have

‰.q.0// D e2�H :

On the other hand, since x�1 represents w�1; we have

‰.q.1// D x�1e2�H 0

x

D e2�.HC�C� 0/

D e2�H ;

since � 2 � and � 0 2 I � �: Thus, ‰ ı q is a loop in Kreg; as claimed.Meanwhile,

‰.q.�// D exp˚2� x.�/�1p.�/x.�/

�;

where x.0/�1p.0/x.0/ D H and x.1/�1p.1/x.1/ D H C � C � 0: Since the vectorspace k is simply connected, the path

� 7! x.�/�1p.�/x.�/

in k is homotopic with endpoints fixed to the straight-line path connecting H toH C � C � 0, namely

� 7! H C �.� C � 0/:

Since the exponential map for K is continuous, we see that ‰ ı q is homotopic inK to the loop

� 7! e2�H e2��.�C� 0/:

We may then continuously deform e2�H to the identity, showing that ‰ ı q ishomotopic in K to the loop � 7! e2��.�C� 0/:

Finally, in light of Proposition 13.14, the loop � 7! e2��.�C� 0/ is homotopic tothe composition (in either order) of the loop � 7! e2��� and the loop � 7! e2���

0

:

But since � 0 belongs to I; we have already shown in Sect. 13.4 that this second loopis null homotopic in K , showing that ‰ ı q is homotopic in K to � 7! e2��� ; asclaimed. ut

It now remains only to assemble the previous results to finish the proof ofTheorem 13.17.

400 13 Fundamental Groups of Compact Lie Groups

Proof of Theorem 13.17, the other direction. We showed in Sect. 13.4 that if � 2 Ithe loop � 7! e2��� is null homotopic in K: In the other direction, suppose that �belongs to � but not to I:We may then construct the path q inA�.K=T / in (13.11).Although this path is not a loop in A � .K=T /; the path ‰ ı q is a loop in Kreg; byLemma 13.39. Since q is a lift of ‰ ı q and q has distinct endpoints, Corollary 13.4tells us that‰ ı q is not null homotopic in Kreg:Meanwhile, since ‰ ı q homotopicin K to � 7! e2��� and �1.Kreg/ is isomorphic to �1.K/ (Theorem 13.27), weconclude that � 7! e2��� is not null homotopic in K: utSummary. It is instructive to summarize the arguments in the proofs ofTheorems 13.15 and 13.17, as described in this section and the previous threesections. We introduced the regular set and the singular set in K and we showedthat the singular set has codimension at least 3. Using this, we proved a key result,that �1.Kreg/ is isomorphic to �1.K/: Next, we introduced the Stiefel diagram andthe local diffeomorphism

‰ W treg � .K=T / ! Kreg:

We found that the failure of injectivity of ‰ on treg � .K=T / is measured by theaction of the group�ÌW; and that the subgroup I ÌW of �ÌW acts transitively onthe alcoves. We concluded that if A is any alcove, the map ‰ W A � .K=T / ! Kreg

is a covering map.We then demonstrated that K=T is simply connected. We did this by mapping

any loop Œx.�/� in K=T to a loop l in K and constructing a homotopy ls of l to apoint inK: Since �1.Kreg/ D �1.K/; we can deform the homotopy ls intoKreg:Wecan then use the covering map ‰ to lift the homotopy ls to A � .K=T / and projectback ontoK=T to obtain a homotopy of Œx.�/� to a point inK=T: After establishingthatK=T is simply connected, we proved our first main result, that every loop inKis homotopic to a loop in T; as follows. We start with a loop l in K; push it down toK=T and then homotope it to the identity coset. We then lift this homotopy to K;giving a homotopy of l into T:

Since �1.T / is easily calculated, we conclude that every loop in K is homotopicto a loop in T of the form � 7! e2��� ; for some � 2 �: When � is a coroot, weshowed that � 7! e2��� is the image under a continuous homomorphism of a loop inthe simply connected group SU.2/; which shows that � 7! e2��� is null homotopic.When � … I; we started with some H in A; translated by �; and then mapped backto some H 0 2 A by the action of I ÌW; where H is chosen so that H 0 ¤ H: Wethen constructed a path q in A � .K=T / with distinct endpoints such that ‰ ı q isa loop in Kreg and is homotopic in K to � 7! e2��� : Since ‰ W A � .K=T / ! Kreg

is a covering map and q has distinct endpoints,‰ ı q is homotopically nontrivial inKreg: Then since �1.Kreg/ D �1.K/;we concluded that � 7! e2��� is homotopicallynontrivial in K:

13.8 The Center of K 401

13.8 The Center of K

In this section, we analyze the center ofK using the tools developed in the previoussections. If T is any one fixed maximal torus in K; Corollary 11.11 tells us that Tcontains the centerZ.K/ ofK:We now give a criterion for an element t D e2�H ofT to be in Z.K/:

Proposition 13.40. If H 2 t, then e2�H belongs to Z.K/ if and only if

h˛;H i 2 Z

for all ˛ 2 R:Proof. By Exercise 17 in Chapter 3, an element x of K is in Z.K/ if and only ifAdx.X/ D X for all X 2 k; or, equivalently, if and only if Adx.X/ D X for allX 2 g D kC: Now, if X 2 g˛; then

Ade2�H .X/ D e2�adH .X/ D e2�h˛;H iX:

Thus, Ade2�H acts as the identity on g˛ if and only if h˛;H i is an integer. Since g isthe direct sum of tC (on which Ade2�H certainly acts trivially) and the g˛’s, we seethat e2�H belongs to Z.K/ if and only if h˛;H i 2 Z for all ˛: utDefinition 13.41. Let ƒ � t denote the root lattice, that is, the set of all integerlinear combinations of roots. Let ƒ� denote the dual of the root lattice, that is,

ƒ� D f� 2 H jh�; �i 2 Z; 8� 2 ƒg :

Note that if � 2 t has the property that h˛; �i 2 Z for every root ˛; then certainlyh�; �i 2 Z whenever � is an integer combination of roots. Thus, Proposition 13.40may be restated as saying that e2�H 2 Z.K/ if and only if H 2 ƒ�: Note also thatif e2�H D I; then certainly e2�H is in the center of K: Thus, the kernel � of theexponential map must be contained in ƒ�:

Proposition 13.42. The map

� 7! e2�� ; � 2 ƒ�;

is a homomorphism of ƒ� onto Z.K/ with kernel equal to �: Thus,

Z.K/ Š ƒ�=�;

where ƒ� is the dual of the root lattice and � is the kernel of the exponential.

402 13 Fundamental Groups of Compact Lie Groups

Proof. As we have noted, Corollary 11.11 implies that Z.K/ � T: Since theexponential map for T is surjective, Proposition 13.40 tells us that the map � 7!e2�� maps ƒ� onto Z.K/: This map is a homomorphism since t is commutative,and the kernel of the map is � � ƒ�: ut

Suppose, for example, that K D T: Then there are no roots, in which case thedual of root lattice is all of t: In this case, we have

Z.K/ D Z.T / Š t=� Š T:

On the other hand, if g is semisimple, both ƒ� and � will be discrete subgroups oft that span t; in which case, ƒ�=� will be finite.

Note that we have several different lattices inside t: Some of these “really” livein t� and only become subsets of t when we use the inner product to identify t�with t: Other lattices naturally live in t itself. The lattices that really live in t�are the root lattice, the lattice of analytically integral elements, and the lattice ofalgebraically integral elements. Meanwhile, the lattices that naturally live in t arethe coroot lattice, the kernel of the exponential map, and the dual of the root lattice.Note that there is a duality relationship between the lattices in t and the lattices in t�:An element is algebraically integral if and only if its inner product with each corootis an integer; thus, the lattice of algebraically integral elements and the coroot latticeare dual to each other. Similarly, the lattice of analytically integral elements is dualto the kernel of the exponential map. Finally, ƒ� is, by definition, dual to the rootlattice ƒ:

The lattices in t� are included in one another as follows:

.root lattice/ � .analytically integral elements/

� .algebraically integral elements/: (13.12)

The dual lattices in t are then included in one another in the reverse order:

.coroot lattice/ � .kernel of exponential/

� .dual of root lattice/: (13.13)

In light of Proposition 13.42 and Corollary 13.18, we have the following isomor-phisms involving quotients of lattices in (13.13):

.kernel of exponential/=.coroot lattice/ Š �1.K/

and

.dual of root lattice/=.kernel of exponential/ Š Z.K/:

13.8 The Center of K 403

Corollary 13.43. Let ƒ� denote the dual of the root lattice and let I denote thecoroot lattice. If K is simply connected, then

Z.K/ Š ƒ�=I:

On the other hand, if Z.K/ is trivial, then

�1.K/ Š ƒ�=I:

Let us define the adjoint group associated to K to be the image of K underthe adjoint representation Ad W K ! GL.k/: (Since K is compact, the adjointgroup of K is compact and thus closed.) If K is semisimple, then the center of kis trivial, which means that the Lie algebra version of the adjoint representation,ad W k ! gl.k/ is faithful. Thus, if g is semisimple, the Lie algebra of the adjointgroup is isomorphic to the Lie algebra k ofK itself. On the other hand, it is not hardto check (still assuming that g is semisimple and thus that the Lie algebra of theadjoint group is isomorphic to k) that the center of the adjoint group is trivial. Thus,whenever g is semisimple, we can construct a new group K 0 where the first part ofthe corollary applies:

�1.K0/ Š ƒ�=I:

Proof. If �1.K/ is trivial, the kernel � of the exponential map must equal the corootlattice I , which means that

Z.K/ Š ƒ�=� D ƒ�=I:

Meanwhile, if Z.K/ is trivial, then the kernel � of the exponential must equal thedual ƒ� of the root lattice, which means that

�1.K/ Š �=I D ƒ�=I;

as claimed. utExample 13.44. If K D SO.4/; then the latticesƒ�; �; and I are as in Figure 13.7and both�1.K/ andZ.K/ are isomorphic to Z=2: Explicitly,Z.SO.4// D fI;�I g:Proof. If we compute as in Sect. 7.7.2, but adjusting for a factor of i to obtain thereal roots and coroots, we find that the coroots are the matrices

0BB@

0 a

�a 00 b

�b 0

1CCA ; (13.14)

404 13 Fundamental Groups of Compact Lie Groups

dual of root lattice

kernel of exponential

coroot lattice

Fig. 13.7 The lattices for SO.4/; with the coroots indicated by arrows

where a D ˙1 and b D ˙1: We identify the coroots with the vectors.a; b/ D .˙1;˙1/ in R2: The coroot lattice I will then consist of pairs .m; n/ 2 Z2

for whichmCn is even. The kernel � of the exponential, meanwhile, is easily seento consist of all pairs .m; n/ 2 Z2: Finally, since the coroots have been normalized tohave length

p2; the roots and coroots coincide. Thus, the dualƒ� of the root lattice

is the set of vectors .x; y/ having integer inner product with .1; 1/ and .1;�1/; thatis, such that x C y and x � y are integers. Thus, either x and y are both integers,or x and y are both integer-plus-one-half. It is then easy to check that both �=I andƒ�=� are isomorphic to Z=2:

If H is any coroot, then H=2 belongs to ƒ� but not to �: Thus, the uniquenontrivial element of Z.SO.4// may be computed as e2�.H=2/. Direct calculationwith (13.14) then shows that this unique nontrivial element is �I 2 SO.4/: ut

Example 13.45. Suppose thatK is a connected, compact matrix Lie group with Liealgebra k, that t is a maximal commutative subalgebra of k; and that the root systemof k relative to t is isomorphic to G2: Then both �1.K/ and Z.K/ are trivial.

It turns out that for compact groups with root systems of type An;Bn; Cn andDn; none of them is simultaneously simply connected and center free. Among theexceptional groups, however, the groups with root systems G2; F4; and E8 all havethis property.

Proof. As we can see from Figure 8.11, each of the fundamental weights for G2 isa root. Thus, every algebraically integral element for G2 (i.e., every integer linearcombination of the fundamental weights) is in the root lattice. Thus, all three ofthe lattices in (13.12) must be equal. By dualizing this observation, we see that allthree of the lattices in (13.13) must also be equal. Thus, both �1.K/ Š �=I andZ.K/ Š ƒ�=� are trivial. ut

13.9 Exercises 405

13.9 Exercises

In Exercises 1–4, we consider the maximal commutative subalgebra t of the relevantLie algebra k given in Sects. 7.7.1–7.7.4. In each case, we identify the Cartansubalgebra h D tC with C

n by the map given in those sections, except that weadjust the map by a factor of i; so that t maps to R

n: We also consider the real rootsand coroots, which differ by a factor of i from the roots and coroots in Chapter 7.

1. For the group SU.n/; n � 1; show that coroot lattice I consists of all integern-tuples .k1; : : : ; kn/ for which k1 C � � � C kn D 0: Show that the kernel � ofthe exponential is the same as the coroot lattice.

2. For the group SO.2n/; n � 2; show that the coroot lattice I is the set of integerlinear combinations of vectors of the form ˙ej ˙ ek; with j ¤ k: Concludethat the coroot lattice consists of all integer n-tuples .k1; k2; : : : ; kn/ for whichk1 C � � � C kn is even. Show that the kernel � of the exponential consists of allinteger n-tuples and that �=I Š Z=2:

3. For the group SO.2n C 1/; n � 1; show that the coroot lattice I is the set ofinteger linear combinations of vectors of the form ˙2ej and ˙ej ˙ ek; withj ¤ k: Conclude that, as for SO.2n/; the coroot lattice consists of all integern-tuples .k1; k2; : : : ; kn/ for which k1C� � �Ckn is even and the kernel � of theexponential consists of all integer n-tuples.

4. For the group Sp.n/; n � 1; show that the coroot lattice I is the set of integerlinear combinations of vectors of the form ˙ej and ˙ej ˙ ek; with j ¤ k:

Conclude that both the coroot lattice and the kernel � of the exponential consistof all integer n-tuples.

5. Verify the claims in Example 13.19.Hints: In the case of SO.5/; make use of the calculations in the proof ofExample 12.13. In the case of PSU.3/; note that X 2 psu.3/ D su.3/exponentiates to the identity in PSU.3/ if and only ifX exponentiates in SU.3/to a constant multiple of the identity.

6. Using Theorems 13.15 and 13.17 and the results of Exercises 2 and 3, showthat every homotopically nontrivial loop in SO.n/; n � 3; is homotopic to theloop

� 7!

0BBBBB@

cos � � sin �sin � cos �

1: : :

1

1CCCCCA; 0 � � � 2�:

(This result also follows from the inductive argument using fiber bundles, asdiscussed following the proof of Proposition 13.10.)

7. Let G be a connected matrix Lie group. Using the following outline, showthat �1.G/ is commutative. Let A.�/ and B.�/ be any two loops in G based at

406 13 Fundamental Groups of Compact Lie Groups

the identity. Construct two families of loops ˛s.�/ (defined in terms of A.�/)and ˇs.�/ (defined in terms of B.�/) with the property that ˛0.t/ˇ0.t/ is theconcatenation of A with B and ˛1.t/ˇ1.t/ is the concatenation of B with A:(Here the product of, say, ˛0.t/ˇ0.t/ is computed in the group G:)

8. Suppose G1 and G2 are connected matrix Lie groups with Lie algebras g1and g2; respectively, and that ˆ W G1 ! G2 is a Lie group homomorphism.Show that if the associated Lie algebra homomorphism � W g1 ! g2 is anisomorphism, then ˆ is a covering map (Definition 13.1).

9. Show that Proposition 13.26 fails if we do not assume thatH 0 andH are in treg:

10. Let E be a real Euclidean space. Suppose V � E is the hyperplane through theorigin orthogonal to a nonzero vector ˛: Now suppose L is the hyperplane (notnecessarily through the origin) obtained by translating V by c˛: Let s W V ! V

be the affine transformation given by

s.v/ D v � 2 h˛; vih˛; ˛i˛ C 2c˛:

Show that if v 2 L and d 2 R; then

s.v C d˛/ D v � d˛:

That is to say, s is the reflection about L:11. Let RC be the set of positive roots associated to a particular base ; and let

˛1; : : : ; ˛N be the positive roots.

(a) Show that for each alcove A; there are integers n1; : : : ; nN such that

A D ˚H 2 t

ˇnj <

˝˛j ;H

˛< nj C 1; j D 1; : : : ; N

�: (13.15)

(b) If n1; : : : ; nN is any sequence of integers, show that if the set A in (13.15)is nonempty, then this set is an alcove.

12. In this exercise, we assume g WD kC is simple. Let RC be the positive rootsassociated to a particular base ; and let C be the fundamental Weyl chamberassociated to RC: Let A be the alcove containing the “bottom” of C; that is,such that all very small elements of C are in A: Let ˇ be the highest root, thatis, the highest weight for the adjoint representation of g; which is irreducibleby assumption.

(a) Show that A may be described as

A D ˚H 2 t

ˇ0 < h˛;H i < 1; 8˛ 2 RC � :

(Compare Exercise 11.)

13.9 Exercises 407

A2 B2 G2

Fig. 13.8 The extended Dynkin diagrams associated to A2; B2; and G2

(b) Show that A may also be described as

A D fH 2 t jhˇ;H i < 1; h˛;H i > 0; 8˛ 2 g :

13. If g D kC is simple, the alcove A in Exercise 12 is a simplex, that is, a boundedregion in t Š Rk defined by k C 1 linear inequalities. By the same argumentas in the proof of Proposition 8.24, the extended Weyl group is generated bythe k C 1 reflections about the walls of A: The structure of A (and thus ofthe extended Weyl group) can be captured in the extended Dynkin diagram,defined as follows. The diagram has k C 1 vertices, representing the elements˛1; : : : ; ˛k and �ˇ: We then define edges and arrows by the same rules as forordinary Dynkin diagrams (Sect. 8.6). (Note that since ˇ is the highest weightof the adjoint representation, ˇ is dominant and, thus, �ˇ is at an obtuse anglewith each element of :)

Verify the extended Dynkin diagrams for A2;B2; and G2 in Figure 13.8,where in each diagram, the black dot indicates the “extra” vertex �ˇ added tothe ordinary Dynkin diagram. (Refer to Figures 13.3, 13.4, and 13.5.)

14. Let K be a connected compact matrix Lie group with Lie algebra k and let t bea maximal commutative subalgebra of k: Let A � t be an alcove and supposeA has no nontrivial symmetries. (That is to say, suppose there is no nontrivialelement of the Euclidean group of t mapping A onto A:)

(a) Show thatK is simply connected.(b) Show that if K 0 is any connected matrix Lie group whose Lie algebra is

isomorphic to k; then K 0 is isomorphic to K:

Note: It is known that there exists a connected compact group K of rank 2whose root system is G2: Since each alcove for G2 is a triangle with 3 distinctedge lengths, Part (b) of the problem shows that any such group K must besimply connected. (Compare Example 13.45.)

15. Consider the quotient space T=W; that is, the set of orbits of W acting on T:Let A � t be any one alcove and let NA be the closure of A in t: Show that if Kis simply connected, then the mapH 7! Œe2�H �; where Œt � denotes the W -orbitof t; is a bijection between NA and T=W:Hint: Imitate the proof of Proposition 8.29 with the Weyl groupW replaced bythe extended Weyl group I ÌW and the chamber C replaced by the alcove A.

Erratum:

Lie Groups, Lie Algebras, and Representations

Brian C. Hall

c Springer International Publishing Switzerland 2015B.C. Hall, Lie Groups, Lie Algebras, and Representations, Graduate Texts in Mathematics 222,DOI 10.1007/978-3-319-13467-3

DOI 10.1007/978-3-319-13467-3_14

The original version of this book was inadvertently published without the middleinitial of the author’s name as “Brian Hall”. The correct name of the author shouldappear as “Brian C. Hall”.

The online version of the original book can be found athttp://dx.doi.org/10.1007/978-3-319-13467-3

© Springer International Publishing Switzerland 2015B.C. Hall, Lie Groups, Lie Algebras, and Representations, GraduateTexts in Mathematics 222, DOI 10.1007/978-3-319-13467-3_14

E1

Appendix ALinear Algebra Review

In this appendix, we review results from linear algebra that are used in the text.The results quoted here are mostly standard, and the proofs are mostly omitted. Formore information, the reader is encouraged to consult such standard linear algebratextbooks as [HK] or [Axl]. Throughout this appendix, we let Mn.C/ denote thespace of n � n matrices with entries in C:

A.1 Eigenvectors and Eigenvalues

For anyA 2 Mn.C/; a nonzero vector v in Cn is called an eigenvector forA if thereis some complex number � such that

Av D �v:

An eigenvalue for A is a complex number � for which there exists a nonzero v 2Cn with Av D �v: Thus, � is an eigenvalue for A if the equation Av D �v or,equivalently, the equation

.A� �I/v D 0;

has a nonzero solution v: This happens precisely when A��I fails to be invertible,which is precisely when det.A � �I/ D 0: For any A 2 Mn.C/; the characteristicpolynomial p of A is given by

A previous version of this book was inadvertently published without the middle initial of theauthor’s name as “Brian Hall”. For this reason an erratum has been published, correcting themistake in the previous version and showing the correct name as Brian C. Hall (see DOI http://dx.doi.org/10.1007/978-3-319-13467-3_14). The version readers currently see is the correctedversion. The Publisher would like to apologize for the earlier mistake.

© Springer International Publishing Switzerland 2015B.C. Hall, Lie Groups, Lie Algebras, and Representations, GraduateTexts in Mathematics 222, DOI 10.1007/978-3-319-13467-3

409

410 A Linear Algebra Review

p.�/ D det.A � �I/; � 2 C:

This polynomial has degree n: In light of the preceding discussion, the eigenvaluesare precisely the zeros of the characteristic polynomial.

We can define, more generally, the notion of eigenvector and eigenvalue for anylinear operator on a vector space. If V is a finite-dimensional vector space overC (orover any algebraically closed field), every linear operator on V will have a least oneeigenvalue. If A is a linear operator on a vector space V and � is an eigenvalue forA; the �-eigenspace forA; denoted V�; is the set of all vectors v 2 V (including thezero vector) that satisfy Av D �v: The �-eigenspace for A is a subspace of V: Thedimension of this space is called the multiplicity of �: (More precisely, this is the“geometric multiplicity” of �: In the finite-dimensional case, there is also a notionof the “algebraic multiplicity” of �; which is the number of times that � occurs as aroot of the characteristic polynomial. The geometric multiplicity of � cannot exceedthe algebraic multiplicity).

Proposition A.1. Suppose that A is a linear operator on a vector space V andv1; : : : ; vk are eigenvectors with distinct eigenvalues �1; : : : ; �k: Then v1; : : : ; vkare linearly independent.

Note that here V does not have to be finite dimensional.

Proposition A.2. Suppose that A and B are linear operators on a finite-dimensional vector space V and suppose that AB D BA: Then for each eigenvalue� of A; the operator B maps the �-eigenspace of A into itself.

Proof. Let � be an eigenvalue of A and let V� be the �-eigenspace of A: Then let vbe an element of V� and consider Bv: Since B commutes with A; we have

A.Bv/ D BAv D �Bv,

showing that Bv is in V�: ut

A.2 Diagonalization

Two matrices A;B 2 Mn.C/ are said to be similar if there exists an invertiblematrix C such that

A D CBC�1:

The operation B ! CBC�1 is called conjugation of B by C: A matrix is saidto be diagonalizable if it is similar to a diagonal matrix. A matrix A 2 Mn.C/ isdiagonalizable if and only if there exist n linearly independent eigenvectors for A:Specifically, if v1; : : : ; vn are linearly independent eigenvectors, let C be the matrixwhose kth column is vk: Then C is invertible and we will have

A.3 Generalized Eigenvectors and the SN Decomposition 411

A D C

0B@�1: : :

�n

1CAC�1; (A.1)

where �1; : : : ; �n are the eigenvalues associated to the eigenvectors v1; : : : ; vn; inthat order. To verify (A.1), note that C maps the standard basis element ej to vj :Thus, C�1 maps vj to ej ; the diagonal matrix on the right-hand side of (A.1) thenmaps ej to �j ej ; and C maps �j ej to �j vj : Thus, both sides of (A.1) map vj to�j vj ; for all j:

IfA 2 Mn.C/ has n distinct eigenvalues (i.e., n distinct roots to the characteristicpolynomial),A is necessarily diagonalizable, by Proposition A.1. If the characteris-tic polynomial of A has repeated roots, A may or may not be diagonalizable.

For A 2 Mn.C/; the adjoint of A; denoted A�; is the conjugate-transpose of A;

.A�/jk D Akj: (A.2)

A matrix A is said to be self-adjoint (or Hermitian) if A� D A: A matrix A issaid to be skew self-adjoint (or skew Hermitian) if A� D �A: A matrix is said tobe unitary if A� D A�1: More generally, A is said to be normal if A commuteswith A�: If A is normal, A is necessarily diagonalizable, and, indeed, it is possibleto find an orthonormal basis of eigenvectors for A: In such cases, the matrix C in(A.1) may be taken to be unitary.

If A is self-adjoint, all of its eigenvalues are real. If A is real and self-adjoint(or, equivalently, real and symmetric), the eigenvectors may be taken to be real aswell, which means that in this case, the matrixC may be taken to be orthogonal. IfAis skew, then its eigenvalues are pure imaginary. If A is unitary, then its eigenvaluesare complex numbers of absolute value 1.

We summarize the results of the previous paragraphs in the following.

Theorem A.3. Suppose that A 2 Mn.C/ has the property that A�A D AA�; (e.g.,if A� D A;A� D A�1; or A� D �A). Then A is diagonalizable and it is possibleto find an orthonormal basis for Cn consisting of eigenvectors for A: If A� D A;

all the eigenvalues of A are real; if A� D �A; all the eigenvalues of A are pureimaginary; and if A� D A�1; all the eigenvalues of A have absolute value 1.

A.3 Generalized Eigenvectors and the SN Decomposition

Not all matrices are diagonalizable, even over C: If, for example,

A D�1 1

0 1

�; (A.3)

412 A Linear Algebra Review

then the only eigenvalue of A is 1, and every eigenvector with eigenvalue 1 is of theform .c; 0/: Thus, we cannot find two linearly independent eigenvectors for A: It isnot hard, however, to prove the following result. Recall that a matrix A is nilpotentif Ak D 0 for some positive integer k:

Theorem A.4. Every matrix is similar to an upper triangular matrix. Everynilpotent matrix is similar to an upper triangular matrix with zeros on the diagonal.

While Theorem A.4 is sufficient for some purposes, we will in general needsomething that comes a bit closer to a diagonal representation. If A 2 Mn.C/

does not have n linearly independent eigenvectors, we may consider the moregeneral concept of generalized eigenvectors. A nonzero vector v 2 C

n is called ageneralized eigenvector forA if there is some complex number� and some positiveinteger k such that

.A � �I/kv D 0: (A.4)

If (A.4) holds for some v ¤ 0, then .A��I/ cannot be invertible. Thus, the number� must be an (ordinary) eigenvalue for A: However, for a fixed eigenvalue �; theremay be generalized eigenvectors v that are not ordinary eigenvectors. In the caseof the matrix A in (A.3), for example, the vector .0; 1/ is a generalized eigenvectorwith eigenvalue 1 (with k D 2).

It can be shown that every A 2 Mn.C/ has a basis of generalized eigenvectors.For any matrix A and any eigenvalue � for A; let W� be the generalized eigenspacewith eigenvalue �:

W� D fv 2 Cnˇ.A� �I/kv D 0 for some k g:

Then Cn decomposes as a direct sum of the W�’s, as � ranges over all theeigenvalues of A: Furthermore, the subspace W� is easily seen to be invariantunder the matrix A. Let A� denote the restriction of A to the subspace W�; andlet N� D A� � �I; so that

A� D �I CN�:

Then N� is nilpotent; that is, Nk� D 0 for some positive integer k: We summarize

the preceding discussion in the following theorem.

Theorem A.5. Let A be an n � n complex matrix. Then there exists a basis for Cn

consisting of generalized eigenvectors for A: Furthermore, Cn is the direct sum ofthe generalized eigenspaces W�: Each W� is invariant under A; and the restrictionof A to W� is of the form �I CN�; where N� is nilpotent.

The preceding result is the basis for the following decomposition.

Theorem A.6. EachA 2 Mn.C/ has a unique decomposition asA D SCN whereS is diagonalizable,N is nilpotent, and SN D NS:

A.5 The Trace 413

The expression A D S C N; with S and N as in the theorem, is called theSN decomposition of A: The existence of an SN decomposition follows from theprevious theorem: We define S to be the operator equal to �I on each generalizedeigenspace W� of A and we set N to be the operator equal to N� on each W�. Forexample, if A is the matrix in (A.3), then we have

S D�1 0

0 1

�; N D

�0 1

0 0

�:

A.4 The Jordan Canonical Form

The Jordan canonical form may be viewed as a refinement of the SN decomposition,based on a further analysis of the nilpotent matrices N� in Theorem A.5.

Theorem A.7. Every A 2 Mn.C/ is similar to a block-diagonal matrix in whicheach block is of the form

0BBBB@

� 1

�: : :

: : : 1

1CCCCA :

Two matrices A andB are similar if and only if they have precisely the same Jordanblocks, up to reordering.

There may be several different Jordan blocks (possibly of different sizes) for thesame value of �: In the case in which A is diagonalizable, each block is 1 � 1; inwhich case, the ones above the diagonal do not appear. Note that each Jordan blockis, in particular, of the form �I CN; where N is nilpotent.

A.5 The Trace

ForA 2 Mn.C/, we define the trace ofA to be the sum of the diagonal entries ofA:

trace.A/ DnX

kD1Akk:

Note that the trace is a linear function of A. For A;B 2 Mn.C/; we note that

trace.AB/ DnX

kD1.AB/kk D

nXkD1

nXlD1

AklBlk: (A.5)

414 A Linear Algebra Review

If we similarly compute trace.BA/; we obtain the same sum with the labels for thesummation variables reversed. Thus,

trace.AB/ D trace.BA/: (A.6)

If C is an invertible matrix and we apply (A.6) to the matrices CA and C�1;we have

trace.CAC�1/ D trace.C�1CA/ D trace.A/;

that is, similar matrices have the same trace.More generally, if A is a linear operator on a finite-dimensional vector space V;

we can define the trace of A by picking a basis and defining the trace of A to be thetrace of the matrix that represents A in that basis. The above calculations show thatthe value of the trace of A is independent of the choice of basis.

A.6 Inner Products

Let h�; �i denote the standard inner product on Cn; defined by

hx; yi DnX

jD1xj yj ;

where we follow the convention of putting the complex-conjugate on the first factor.We have the following basic result relating the inner product to the adjoint of amatrix, as defined in (A.2).

Proposition A.8. For all A 2 Mn.C/; the adjoint A� of A has the property that

hx;Ayi D hA�x; yi (A.7)

for all x; y 2 Cn:

Proof. We compute that

hx;Ayi DnX

jD1xj

nXkD1

Ajkyk

DnX

jD1

nXkD1

Ajkxj yk

DnX

jD1

nXkD1

A�kjxj yk:

This last expression is just the inner product of A�x with y: ut

A.6 Inner Products 415

We may generalize the notion of inner product as follows.

Definition A.9. If V is any vector space over C, an inner product on V is a mapthat associates to any two vectors u and v in V a complex number hu; vi and thathas the following properties:

1. Conjugate symmetry: hv; ui D hu; vi for all u; v 2 V:2. Linearity in the second factor: hu; v1 C av2i D hu; v1i C a hu; v2i ; for all

u; v1; v2 2 V and a 2 C:

3. Positivity: For all v 2 V; the quantity hv; vi is real and satisfies hv; vi � 0; withhv; vi D 0 only if v D 0:

Note that in light of the conjugate-symmetry and the linearity in the secondfactor, an inner product must be conjugate-linear in the first factor:

hv1 C av2; ui D hv1; ui C Na hv2; ui :

(Some authors define an inner product to be linear in the first factor and conjugatelinear in the second factor.) An inner product on a real vector space is defined in thesame way except that conjugate symmetry is replaced by symmetry (hv; ui D hu; vi)and the constant a in Point 2 now takes only real values.

If V is a vector space with inner product, the norm of a vector v 2 V , denotedkvk ; is defined by

kvk D phv; vi:

The positivity condition on the inner product guarantees that kvk is always a non-negative real number and that kvk D 0 only if v D 0: If, for example, V D Mn.C/;

we may define the Hilbert–Schmidt inner product by the formula

hA;Bi D trace.A�B/: (A.8)

It is easy to see check that this expression is conjugate symmetric and linear in thesecond factor. Furthermore, we may compute as in (A.5) that

trace.A�A/ DnX

k;lD1A�

klAlk DnX

k;lD1jAklj2 � 0;

and the sum is zero only if each entry ofA is zero. The associated Hilbert–Schmidtnorm satisfies

kAk2 DnX

k;lD1jAkl j2 :

416 A Linear Algebra Review

Suppose that V is a finite-dimensional vector space with inner product and thatW is a subspace of V: Then the orthogonal complement of W; denoted W ?; isthe set of all vectors v in V such that hw; vi D 0 for all w in W: The space V thendecomposes as the direct sum of W and W ?:

We now introduce the abstract notion of the adjoint of a matrix.

Proposition A.10. Let V be a finite-dimensional vector space with an innerproduct h�; �i : If A is a linear map from V to V; there is a unique operatorA� W V ! V such that

hu;Avi D hA�u; vi

for all u; v 2 V: Furthermore, if W is a subspace of V that is invariant under A;then W ? is invariant under A�:

A.7 Dual Spaces

If V is a vector space over C; a linear functional on V is a linear map of V intoC: If V is finite dimensional with basis v1; : : : ; vn, then for each set of constantsa1; : : : ; an; there is a unique linear functional � such that �.vk/ D ak: If V is acomplex vector space, then the dual space to V; denoted V �; is the set of all linearfunctionals on V: The dual space is also a vector space and its dimension is the sameas that of V: If V is finite dimensional, then V is isomorphic to V �� by the mapsending v 2 V to the “evaluation at v” functional, that is, the map � 7! �.v/;

� 2 V �:If W is a subspace of a vector space V; the annihilator subspace of W; denoted

W ^; is the set of all � in V � such that �.w/ D 0 for all w in W: Then W ^ is asubspace of V �: If V is finite dimensional, then

dimW C dimW ^ D dimV

and the map W ! W ^ provides a one-to-one correspondence between subspacesof V and subspaces of V �:

In general, one should be careful to distinguish between a vector space and itsdual. Nevertheless, when V is finite dimensional and has an inner product, we canproduce an identification between V and V �:

Proposition A.11. Let V be a finite-dimensional inner product space and let � bea linear functional on V: Then there exists a unique w 2 V such that

�.v/ D hw; vifor all v 2 V:

Recall that we follow the convention that inner products are linear in the secondfactor, so that hw; vi is, indeed, linear in v:

A.8 Simultaneous Diagonalization 417

A.8 Simultaneous Diagonalization

We now extend the notion of eigenvectors and diagonalization to families of linearoperators.

Definition A.12. Let V be a vector space and let A be a collection of linearoperators on V: A nonzero vector v 2 V is a simultaneous eigenvector for A iffor all A 2 A; there exists a constant �A such that Av D �Av: The numbers �A arethe simultaneous eigenvalues associated to v:

Consider, for example, the space D of all diagonal n� n matrices. Then for eachk D 1; : : : ; n; the standard basis element ek is a simultaneous eigenvector for D:For each diagonal matrix A; the simultaneous eigenvalue associated to ek is the kthdiagonal entry of A:

Proposition A.13. If A is a commuting family of linear operators on a finite-dimensional complex vector space, then A has at least one simultaneous eigen-vector.

It is essential here that the elements of A commute; noncommuting families ofoperators typically have no simultaneous eigenvectors.

In many cases, the collection A of operators on V is a subspace of End.V /;the space of all linear operators from V to itself. In that case, if v is a simultaneouseigenvector for A, the eigenvalues�A for v depend linearly onA:After all, ifA1v D�1v and A2v D �2v, then

.A1 C cA2/v D .�1 C c�2/v:

The preceding discussion leads to the following definition.

Definition A.14. Suppose that V is a vector space and A is a vector space of linearoperators on V: A weight for A is a linear functional � on A such that there existsa nonzero vector v 2 V satisfying

Av D �.A/v

for all A in A: For a fixed weight �; the set of all vectors v 2 V satisfying Av D�.A/v for all A in A is called the weight space associated to the weight �:

That is to say, a weight is a set of simultaneous eigenvalues for the operators inA: If V is finite dimensional and the elements of A all commute with one another,then there will exist at least one weight for A:

If A is finite dimensional and comes equipped with an inner product, it isconvenient to express the linear functional� in Definition A.14 as the inner productof A with some vector, as in Proposition A.11. From this point of view, we define

418 A Linear Algebra Review

a weight to be an element � of A (not A�) such that there exists a nonzero v inV with

Av D h�;Ai v

for all A 2 A:Definition A.15. Suppose that V is a finite-dimensional vector space and A issome collection of linear operators on V: Then the elements of A are said to besimultaneously diagonalizable if there exists a basis v1; : : : ; vn for V such thateach vk is a simultaneous eigenvector for A:

If A is a vector space of linear operators on V , then saying that the elements of Aare simultaneously diagonalizable is equivalent to saying that V can be decomposedas a direct sum of weight spaces of A:

If a collection A of operators is simultaneously diagonalizable, then the elementsof A must commute, since they commute when applied to each vk: Conversely, ifeach A 2 A is diagonalizable by itself and if the elements of A commute, then(it can be shown), the elements of A are simultaneously diagonalizable. We recordthese results in the following proposition.

Proposition A.16. If A is a commuting collection of linear operators on a finite-dimensional vector space V and each A 2 A is diagonalizable, then the elementsof A are simultaneously diagonalizable.

We close this appendix with an analog of Proposition A.1 for simultaneouseigenvectors.

Proposition A.17. Suppose V is a vector space and A is a vector space of linearoperators on V: Suppose �1; : : : ; �m are distinct weights for A and v1; : : : ; vm areelements of the corresponding weight spaces. If v1 C � � � C vm D 0; then vj D 0 forall j D 1; : : : ; m: Furthermore, if v1 C � � � C vm is a weight vector with weight �,then � D �j for some j and vk D 0 for all k ¤ j:

Since this result is not quite standard, we provide a proof.

Proof. Assume first that v1C� � �Cvm D 0;with vj in the weight space with weight�j : Ifm D 1; then we have v1 D 0; as claimed. Ifm > 1; choose some A 2 A suchthat �1.A/ ¤ �2.A/: If we then apply the operator A � �2.A/I to v1 C � � � C vm;

we obtain

0 DmXjD1

.�j .A/� �2.A//vj : (A.9)

Now, the j D 2 term in (A.9) is zero, so that the sum actually contains at mostm�1nonzero terms. Thus, by induction on m; we can assume that each term in (A.9) iszero. In particular, .�1.A/��2.A//v1 D 0;which implies (by our choice of A) that

A.8 Simultaneous Diagonalization 419

v1 D 0: Once v1 is known to be zero, the original sum v1 C � � � C vm contains atmostm� 1 nonzero terms. Thus, using induction onm again, we see that each termin the sum is zero.

Assume now that v WD v1 C � � � C vm is a (nonzero) weight vector with someweight �; and choose some j for which vj ¤ 0: Then for each A 2 A; we have

0 D Av � �.A/v DmXkD1

.�k.A/� �.A//vk:

Thus, by the first part of the proposition, we must have .�k.A/ � �.A//vk D 0 forall j: Taking k D j; we conclude that �j .A/ � �.A/ D 0: Since this result holdsfor all A 2 A; we see that � D �j : Finally, for any k ¤ j; we can choose A 2 Aso that �k.A/ ¤ �j .A/: With this value of A (and with � D �j ), the fact that.�k.A/ � �j .A//vk D 0 forces vk to be zero. ut

Appendix BDifferential Forms

In this section, we give a very brief outline of the theory of differential forms onmanifolds. Since this is our main requirement, we consider only top-degree forms,that is, k-forms on k-dimensional manifolds. See Chapter 16 in [Lee] for moreinformation. We begin by considering forms at a single point, which is just a topicin linear algebra.

Definition B.1. If V is a k-dimensional real vector space, a map ˛ W V k ! R issaid to be k-linear and alternating if (1) ˛.v1; : : : ; vk/ is linear with respect toeach vj with the other variables fixed, and (2) ˛ changes sign whenever any two ofthe variables are interchanged:

˛.v1; : : : ; vl ; : : : ; vm; : : : vk/ D �˛.v1; : : : ; vm; : : : ; vl ; : : : vk/:

It is a standard result in linear algebra (e.g., Theorem 2 in Section 5.3 of [HK])that every k-dimensional real vector space admits a nonzero k-linear, alternatingform, and that any two such forms differ by multiplication by a constant. If T WV ! V is a linear map and ˛ is a k-linear, alternating form on V; then for anyv1; : : : ; vk 2 V; we have

˛.Tv1; : : : ;Tvk/ D .detT /˛.v1; : : : ; vk/: (B.1)

If v1; : : : ; vk and w1; : : : ;wk are two ordered bases for V; then there is a uniqueinvertible linear transformation T W V ! V such that Tvj D wj :We may divide thecollection of all ordered bases of V into two groups, where two ordered bases belong

A previous version of this book was inadvertently published without the middle initial of theauthor’s name as “Brian Hall”. For this reason an erratum has been published, correcting themistake in the previous version and showing the correct name as Brian C. Hall (see DOI http://dx.doi.org/10.1007/978-3-319-13467-3_14). The version readers currently see is the correctedversion. The Publisher would like to apologize for the earlier mistake.

© Springer International Publishing Switzerland 2015B.C. Hall, Lie Groups, Lie Algebras, and Representations, GraduateTexts in Mathematics 222, DOI 10.1007/978-3-319-13467-3

421

422 B Differential Forms

to the same group if the linear map relating them has positive determinant and thetwo bases belong to different groups if the linear map relating them has negativedeterminant. An orientation of V is then a choice of one of the two groups of bases.Once an orientation of V has been chosen, we say that a basis is positively orientedif it belongs to the chosen group of bases. If ˛ is a nonzero k-linear, alternating formon V; we can define an orientation of V by decreeing an ordered basis v1; : : : ; vk tobe positively oriented if ˛.v1; : : : ; vk/ > 0:

The following example of a k-linear, alternating form on Rk will help motivatethe notion of a k-form. For any vectors v1; : : : ; vk in Rk; define the parallelepipedPv1;:::;vk spanned by these vectors, as follows:

Pv1;:::;vk D fc1v1 C � � � C ckvkj0 � cl � 1g : (B.2)

(If k D 2; then a parallelepiped is just a parallelogram.) Let us use the orientationon Rk in which the standard basis e1; : : : ; ek is positively oriented.

Example B.2. Define a map V W .Rk/k ! R by

V.v1; : : : ; vk/ D ˙Vol.Pv1;:::;vk /; (B.3)

where we take a plus sign if v1; : : : ; vk is a positively oriented basis for Rk and aminus sign if it is a negatively oriented basis. Then V is a k-linear, alternating formon Rk:

Note that the volume of Pv1;:::;vk is zero if v1; : : : ; vk do not form a basis forRk , in which case, we do not have to worry about the sign on the right-hand sideof (B.3). Now, it is known that the volume of Pv1;:::;vk is equal to jdetT j ; where Tis the k � k matrix whose columns are the vectors v1; : : : ; vk: This claim is a veryspecial case of the change-of-variables theorem in multivariate calculus and can beproved by expressing T as a product of elementary matrices. We can then see thatV.v1; : : : ; vk/ is equal to detT (without the absolute value signs). Meanwhile, it is astandard result from linear algebra that the determinant of T is a k-linear, alternatingfunction of its column vectors v1; : : : ; vk:

We now turn to a discussion of top-degree forms on manifolds. If M is a k-dimensional manifold (say, embedded into some RN ), we have the notion of thetangent space to M atm; denoted TmM; which is a k-dimensional subspace of RN :

Definition B.3. Suppose M is a smoothly embedded, k-dimensional submanifoldof RN for some k;N:A k-form ˛ onM is a smoothly varying family ˛m of k-linear,alternating maps on TmM; one for each m 2 M:

To be precise, let us say that a family ˛m of k-linear, alternating forms on eachTmM is “smoothly varying” if the following condition holds. Suppose X1; : : : ; Xkare smooth RN -valued functions on RN with the property that for each m 2 M; thevector Xj .m/ is tangent to M: Then the function ˛m.X1.m/; : : : ; Xk.m//;m 2 M;should be a smooth function on M:

B Differential Forms 423

F

M

v2

v1

Fig. B.1 The integral of ˛ over the small region F � M is approximately equal to ˛.v1; v2/

The “purpose in life” of a k-form ˛ on a k-dimensional manifold M is tobe integrated over regions in M: More precisely, we must assume that M isorientable—meaning that it is possible to choose an orientation of each tangentspace TmM in a way that varies continuously with m—and that we have chosenan orientation on M: Then if E is a “nice” subset of M (to be precise, a compactk-dimensional submanifold with boundary), there is a notion of the integral of ˛over E � M; denoted

ZE

˛:

The value ofRE˛ may be thought of as assigning a sort of (possibly negative)

“volume” to the set E: If ˛ is a k-form on M and f W M ! R is a smoothfunction, then f ˛ is also a k-form on M; which may also be integrated, using thesame orientation we used to integrate ˛.

We may gain an intuitive understanding of the notion of integration of k-formsas follows. For any regionE � M; we may think of choppingE up into very smallsubregions F , each of which is shaped like a parallelepiped (as in (B.2)). Morespecifically, each subregion will look like the parallelepiped spanned by tangentvectors v1; : : : ; vk at some point m 2 E; which we can arrange to be positivelyoriented. The idea is then that the integral of ˛ over each subregion should beapproximately ˛m.v1; : : : ; vk/. (See Figure B.1.) The integral of ˛ over all of Eshould then be the sum of its integrals over the subregions.

If we think ofRE˛ as a sort of volume of the set E; then ˛m.v1; : : : ; vk/

represents the volume (possibly with a minus sign) of a small parallelepiped-shapedsubregion inside E: Example B.2 then makes it natural that we should require˛m.v1; : : : ; vk/ to be k-linear and alternating.

We may give a more precise definition of the integral of a differential form asfollows. We choose a local coordinate system x1; : : : ; xk on our oriented manifoldM; defined in some open set U:We then let @=@x1; : : : ; @=@xk denote the associated

424 B Differential Forms

basis for the tangent space at each point. (In coordinates, @=@xj is the unit vectorin the xj -direction.) We assume the coordinate system is “oriented,” meaning that@=@x1; : : : ; @=@xk is an oriented basis for the tangent space at each point in U:

Definition B.4. Let ˛ be a k-form on an oriented k-dimensional manifold M andsuppose E � M is a compact subset of the domain U of fxj g: We then defineRE˛ as

ZE

˛ DZE

˛

�@

@x1; : : : ;

@

@xk

�dx1 dx2 � � � dxk; (B.4)

where the integral on the right-hand side of (B.4) is an ordinary integral in Euclideanspace.

The integral on the right-hand side of (B.4) may be defined as a Riemann integralor using Lebesgue measure on R

k . A key point in the definition is to verify that thevalue of

RE ˛ is independent of the choice of coordinates. To this end, suppose fykg

is another oriented coordinate system whose domain includes E: Then by the chainrule, we have

@f

@xlDXm

@f

@ym

@ym

@xl

for any smooth function f: That is to say,

@

@xlDXm

@ym

@xl

@

@ym:

Thus, if T is the matrix whose entries are Tlm D @ym=@xl ; we will have, by (B.1),

˛

�@

@x1; : : : ;

@

@xk

�D .detT /˛

�@

@x1; : : : ;

@

@xk

�:

On the other hand, the classical change of variables theorem says that

ZE

f .x1; : : : ; xk/ dx1 dx2 � � � dxk DZE

f .y1; : : : ; yk/ J dy1 dy2 � � � dyk;

where J is the determinant of the matrix f@xm=@ylg: (For example, in the k D 1

case, J is just dx=dy; which is obtained by writing dx D .dx=dy/ dy:) But by thechain rule again, the matrix f@xm=@ylg is the inverse of the matrix f@ym=@xlg: Thus,J is the reciprocal of detT; and we see that

B Differential Forms 425

ZE

˛

�@

@x1; : : : ;

@

@xk

�dx1 dx2 � � � dxk

DZE

˛

�@

@y1; : : : ;

@

@yk

�dy1 dy2 � � � dyk;

as claimed.Note that if we think of the integral in (B.4) as a Riemann integral, we

compute the integral by covering E with small k-dimensional “rectangles,” andthese rectangles may be thought of as being “spanned” by multiples of the vectors@=@x1; : : : ; @=@xk: In the Riemann integral, the integral of ˛ over each smallrectangle is being approximated by ˛.@=@x1; : : : ; @=@xk/ times the volume of therectangle, in agreement with preceding intuitive description of the integral.

If we wish to integrate a k-form ˛ over a general k-dimensional, compact subsetE of M; we use a partition of unity to write ˛ as a sum of forms ˛j , each of whichis supported in a small region in M: For each j; we choose a coordinate systemdefined on a set Uj containing the support of ˛j :We then integrate ˛j overE \Ujand sum over j:

Appendix CClebsch–Gordan Theoryand the Wigner–Eckart Theorem

C.1 Tensor Products of sl.2IC/ Representations

The irreducible representations of SU.2/ (or, equivalently, of sl.2IC/) were classi-fied in Sect. 4.6 and may be realized in spaces of homogeneous polynomials in twocomplex variables as in Example 4.10. For each non-negative integer m; we havean irreducible representation .�m; Vm/ of sl.2IC/ of dimension m C 1; and everyirreducible representation of sl.2IC/ is isomorphic to one of these. We are usinghere the mathematicians’ labeling of the representations; in the physics literature,the representations are labeled by the “spin” l WD m=2:

By the averaging method of Sect. 4.4, we can find on each space Vm an innerproduct that is invariant under the action of the compact group SU.2/: (In the caseof V1 Š C2;we can use the standard inner product onC2 and for anym; it is not hardto describe such an inner product explicitly.) With respect to such an inner product,the orthogonal complement of a subspace invariant under SU.2/ (or, equivalently,under sl.2IC/) is again invariant under SU.2/. Since the elementH D diag.1;�1/of sl.2IC/ is in isu.2/; �m.H/will be self-adjoint with respect to this inner product.Thus, eigenvectors of �m.H/ with distinct eigenvectors must be orthogonal.

Recall from Sect. 4.3.2 the notion of the tensor product of representations of agroup or Lie algebra. We consider this in the case of the irreducible representationsof sl.2IC/: We regard the tensor product Vm ˝ Vn as a representation of sl.2IC/:(Recall that it is also possible to view Vm ˝ Vn as a representation of sl.2IC/ ˚sl.2IC/:) The action of sl.2IC/ on Vm ˝ Vn is given by

A previous version of this book was inadvertently published without the middle initial of theauthor’s name as “Brian Hall”. For this reason an erratum has been published, correcting themistake in the previous version and showing the correct name as Brian C. Hall (see DOI http://dx.doi.org/10.1007/978-3-319-13467-3_14). The version readers currently see is the correctedversion. The Publisher would like to apologize for the earlier mistake.

© Springer International Publishing Switzerland 2015B.C. Hall, Lie Groups, Lie Algebras, and Representations, GraduateTexts in Mathematics 222, DOI 10.1007/978-3-319-13467-3

427

428 C Clebsch–Gordan Theory and the Wigner–Eckart Theorem

.�m ˝ �n/.X/ D �m.X/˝ I C I ˝ �n.X/: (C.1)

We compute in the standard basis fX; Y;H g for sl.2IC/: Once we have chosenSU.2/-invariant inner products on Vm and Vn; there is a unique inner producton Vm ˝ Vn with the property that hu1 ˝ v1; u2 ˝ v2i D hu1; u2i hv1; v2i : (Thisassertion can be proved using the universal property of tensor products.) The innerproduct on Vm ˝ Vn is also invariant under the action of SU.2/: We assume in therest of this section that an inner product of this sort has been chosen on each Vm˝Vn:

In general, Vm ˝ Vn will not be an irreducible representation of sl.2IC/; thegoal of this section is to describe how Vm ˝ Vn decomposes as a direct sum ofirreducible invariant subspaces. This decomposition is referred to as the Clebsch–Gordan theory. Let us consider first the case of V1 ˝ V1; where V1 D C2; thestandard representation of sl.2IC/: If fe1; e2g is the standard basis for C2; thenthe vectors of the form ek ˝ el ; 1 � k; l � 2; form a basis for C2 ˝ C2: Sincee1 and e2 are eigenvalues for �1.H/ with eigenvalues 1 and �1; respectively, then,by (C.1), the basis elements for C2 ˝ C2 are eigenvectors for the action of H witheigenvalues 2; 0; 0; and �2; respectively. Since 2 is the largest eigenvalue forH; thecorresponding eigenvector e1 ˝ e1 must be annihilated by X (i.e., by the operator�1.X/˝ I C I ˝ �1.X/).

If then we apply Y (i.e., by the operator �1.Y /˝ I C I ˝ �1.Y /) repeatedly toe1 ˝ e1; we obtain e1 ˝ e2 C e2 ˝ e1; then 2e2 ˝ e2; and then the zero vector. Thespace spanned by these vectors is invariant under sl.2IC/ and irreducible, and isisomorphic to the three-dimensional representation V2: The orthogonal complementof this space in C2 ˝C2, namely the span of e1 ˝ e2 � e2 ˝ e1; is also invariant, andsl.2IC/ acts trivially on this space. Thus,

C2 ˝ C

2 D spanfe1 ˝ e1; e1 ˝ e2 C e2 ˝ e1; e2 ˝ e2g ˚ spanfe1 ˝ e2 � e2 ˝ e1g:We see, then, that the four-dimensional space V1 ˝ V1 is isomorphic, as an sl.2IC/representation, to V2 ˚ V0:

Theorem C.1. Let m and n be non-negative integers with m � n: If we considerVm ˝ Vn as a representation of sl.2IC/; then

Vm ˝ Vn Š VmCn ˚ VmCn�2 ˚ � � � ˚ Vm�nC2 ˚ Vm�n;

where Š denotes an isomorphism of sl.2IC/ representations.

Note that this theorem is consistent with the special case worked out earlier:V1 ˝ V1 Š V2 ˚ V0: For applications to the Wigner–Eckart theorem, a key propertyof the decomposition in Theorem C.1 is that it is multiplicity free. That is to say,each irreducible representation that occurs in the decomposition of Vm ˝ Vn occursonly once. This is a special feature of the representations of sl.2IC/; the analogousstatement does not hold for tensor products of representations of other Lie algebras.

Proof. Let us take a basis for each of the two spaces that is labeled by theeigenvalues for H: That is to say, we choose a basis um; um�2; : : : ; u�m for Vm and

C.1 Tensor Products of sl.2IC/ Representations 429

vn; vn�2; : : : ; v�n for Vn; with �m.H/uj D juj and �n.H/vk D kvk: Then thevectors of the form uj ˝ vk form a basis for Vm ˝ Vn; and we compute that

Œ�m.H/˝ I C I ˝ �n.H/�uj ˝ vk D .j C k/uj ˝ vk:

Thus, each of our basis elements is an eigenvector for the action ofH on Vm ˝ Vn:

The eigenvalues for the action of H range from mC n to �.mC n/ in incrementsof 2:

The eigenspace with eigenvaluemC n is one dimensional, spanned by um ˝ vn:

If n > 0; then the eigenspace with eigenvaluemC n� 2 has dimension 2; spannedby um�2 ˝ vn and um ˝ vn�2: Each time we decrease the eigenvalue of H by 2we increase the dimension of the corresponding eigenspace by 1, until we reach theeigenvaluem � n; which is spanned by the vectors

um�2n ˝ vn; um�2nC2 ˝ vn�2; : : : ; um ˝ v�n:

This space has dimension nC 1: As we continue to decrease the eigenvalue ofH inincrements of 2, the dimensions remain constant until we reach eigenvalue n �m;

at which point the dimensions begin decreasing by 1 until we reach the eigenvalue�m � n; for which the corresponding eigenspace has dimension one, spanned byu�m˝v�n: This pattern is illustrated by the following table, which lists, for the caseof V4 ˝ V2; each eigenvalue for H and a basis for the corresponding eigenspace.

Eigenvalue forH Basis6 u4 ˝ v2

4 u2 ˝ v2 u4 ˝ v02 u0 ˝ v2 u2 ˝ v0 u4 ˝ v�20 u�2 ˝ v2 u0 ˝ v0 u2 ˝ v�2

�2 u�4 ˝ v2 u�2 ˝ v0 u0 ˝ v�2�4 u�4 ˝ v0 u�2 ˝ v�2�6 u�4 ˝ v�2

Consider now the vector um˝vn;which is annihilated byX and is an eigenvectorforH with eigenvaluemC n: Applying Y repeatedly gives a chain of eigenvectorsfor H with eigenvalues decreasing by 2 until they reach �m � n: By the proofof Theorem 4.32, the span W of these vectors is invariant under sl.2IC/ andirreducible, isomorphic to VmCn: The orthogonal complement ofW is also invariant.Since W contains each of the eigenvalues of H with multiplicity one, eacheigenvalue forH inW ? will have its multiplicity lowered by 1. In particular,mCn

is not an eigenvalue for H in W ?; the largest remaining eigenvalue is m C n � 2

and this eigenvalue has multiplicity one (unless n D 0). Thus, if we start with aneigenvector forH in W ? with eigenvaluemC n� 2; this will be annihilated by Xand will generate an irreducible invariant subspace isomorphic to VmCn�2:

430 C Clebsch–Gordan Theory and the Wigner–Eckart Theorem

We now continue on in the same way, at each stage looking at the orthogonalcomplement of the sum of all the invariant subspaces we have obtained in theprevious stages. Each step reduces the multiplicity of each H -eigenvalue by 1 andthereby reduces the largest remainingH -eigenvalue by 2. This process will continueuntil there is nothing left, which will occur after Vm�n: ut

Theorem C.1 tells us, for example, that the 15-dimensional space V4 ˝ V2; wedecompose as the direct sum of a seven-dimensional invariant subspace isomorphicto V6; a five-dimensional invariant subspace isomorphic to V4; and a three-dimensional invariant subspace isomorphic to V2: By following the arguments inthe proof of the theorem, we could, in principle, compute these subspaces explicitly.

C.2 The Wigner–Eckart Theorem

Recall that the Lie algebras su.2/ and so.3/ are isomorphic. Specifically, we use thebases fE1;E2;E3g for su.2/ and fF1; F2; F3g for so.3/ described in Example 3.27.The unique linear map � W su.2/ ! so.3/ such that �.Ek/ D Fk; k D 1; 2; 3;

is a Lie algebra isomorphism. Thus, the representations of so.3/ are in one-to-onecorrespondence with the representations of su.2/; which, in turn, are in one-to-onecorrespondence with the complex-linear representations of sl.2IC/: In particular,the analysis of the decomposition of tensor products of sl.2IC/ representations inthe previous section applies also to so.3/ representations.

Suppose now that … is a representation of SO.3/ acting on a finite-dimensionalvector space V: Let End.V / denote the space of endomorphisms of V (i.e., the spaceof linear operators of V into itself). Then we can define an associated action ofSO.3/ on End.V / by the formula

R � C D ….R/C….R/�1; (C.2)

for all R 2 SO.3/ and C 2 End.V /: It is easy to check that this action constitutesa representation of SO.3/:

Definition C.2. Let .…; V / be a representation of SO.3/: For any ordered tripleC WD .C1; C2; C3/ of operators on V and any vector v 2 R3; let v � C be the operator

v � C D3X

jD1vj Cj : (C.3)

The triple C is a vector operator if

.Rv/ � C D ….R/.v � C/….R/�1 (C.4)

for all R 2 SO.3/:

C.2 The Wigner–Eckart Theorem 431

That is to say, the triple C is a vector operator if the map v 7! v �C intertwines theobvious action of SO.3/ on R3 with the action of SO.3/ on End.V / given in (C.2).Note that if, say, R 2 SO.3/ maps e1 to e2; then (C.4) implies that

C2 D ….R/C1….R/�1: (C.5)

Equation (C.5) then says that C1 and C2 are “the same operator, up to rotation.”

Example C.3. Let V be the space of smooth functions on R3 and define an actionof SO.3/ on V by

.….R/f /.x/ D f .R�1x/: (C.6)

Define operators X D .X1;X2;X3/ on V by

.Xjf /.x/ D xj f .x/:

Then X is a vector operator.

Note that Xj is the operator of “multiplication by xj .” The operatorsX1;X2 andX3 are called the position operators in the physics literature.

Proof. For any v 2 R3 and R 2 SO.3/; we have

fŒ.Rv/ � X�f g.x/ D ..Rv/ � x/f .x/:

On the other hand, we compute that

Œ.v � X/….R/�1f �.x/ D .v � x/f .Rx/;

so that

Œ….R/.v � X/….R/�1f �.x/ D .v � .R�1x//f .x/

D ..Rv/ � x/f .x/;

as required for a vector operator. utWe are now ready for our first version of the Wigner–Eckart theorem.

Theorem C.4. Let .…; V / be an irreducible finite-dimensional representation ofSO.3/; and let A and B be two vector operators on V , with A being nonzero. Thenthere exists a constant c such that

B D cA:

The computational significance of the theorem is as follows. For each irreduciblerepresentation V; if we can find one single vector operator A acting on V; thenthe action of any other vector operator on V is completely determined by a single

432 C Clebsch–Gordan Theory and the Wigner–Eckart Theorem

constant c: There are two ingredients in the proof. The first is Schur’s lemma and thesecond is Theorem C.1, which implies (as we will see shortly) that when End.V /decomposes as a direct sum of irreducibles, the (complexification of) the standardrepresentation of SO.3/ occurs at most once.

Lemma C.5. Let … be a finite-dimensional, irreducible representation of SO.3/acting on a vector space V; and let SO.3/ act also on End.V / as in (C.2). Then

End.V / Š V ˝ V;

where Š denotes an isomorphism of SO.3/ representations.

Proof. For any finite-dimensional vector space V; there is, by Definition 4.13, aunique linear map from ‰ W V � ˝ V ! End.V / such that for all v 2 V and� 2 V �; we have

‰.� ˝ v/.w/ D �.w/v:

By computing on a basis, it is easy to check that ‰ is an isomorphism of vectorspaces. If, in addition, V is a representation of SO.3/; then ‰ is an isomorphism ofrepresentations, where SO.3/ acts on V � as in Sect. 4.3.3 and acts on End.V / as in(C.2). (Compare Exercises 3 and 4 in Chapter 12.) Thus, End.V / Š V � ˝ V:

Meanwhile, every irreducible representation of SO.3/ is isomorphic to its dual.This can be seen either by noting that there is only one irreducible representationin each dimension, or (more fundamentally) by noting that �I is an element of theWeyl group of the A1 root system. (Compare Exercise 10 in Chapter 10.) Thus,actually, End.V / Š V ˝ V; as claimed. utProof of Theorem C.4. The action of SO.3/ on R3 is irreducible. Indeed, theassociated action of SO.3/ on C3 is irreducible; this is the unique irreduciblerepresentation of SO.3/ of dimension 3. Now, the linear map v 7! v � A extends to acomplex linear map from C3 into End.V /; and this extension is still an intertwiningmap.

Meanwhile, End.V / Š V ˝ V; by the lemma, and V ˝ V decomposes as adirect sum of irreducibles, as in Theorem C.1. In this decomposition, the three-dimensional irreducible representation V2 of SO.3/ occurs exactly once, unless Vis trivial. Thus, by Schur’s lemma, the map v 7! v �A must be zero if V is trivial andmust map into the unique copy of C3 if V is nontrivial. Of course, the same holdsfor the map v 7! v � B: Applying Schur’s lemma a second time, we see that if A isnonzero, B must be a multiple of A: ut

We now turn to a more general form of the Wigner–Eckart theorem, in whichthe space V on which the vector operators act is not assumed irreducible, or evenfinite dimensional. Rather, the theorem describes how vector operators act relativeto a pair of irreducible invariant subspaces of V:

C.2 The Wigner–Eckart Theorem 433

Theorem C.6 (Wigner–Eckart). Let V be an inner product space, possibly infi-nite dimensional. Suppose … is a representation of SO.3/ acting on V in aninner-product-preserving fashion. Let W1 andW2 be finite-dimensional, irreduciblesubspaces of V: Suppose A and B are two vector operators on V and that

˝w; Ajw0˛

is nonzero for some w 2 W1, w0 2 W2; and j 2 f1; 2; 3g: Then there exists aconstant c such that

˝w; Bjw0˛ D c

˝w; Ajw0˛

for all w 2 W1, all w0 2 W2; and all j D 1; 2; 3:

In many applications, the space V is L2.R3/; the space of square-integrablefunctions on R3; and where SO.3/ acts on L2.R3/ by the same formula as in (C.6).The irreducible, SO.3/-invariant subspaces ofL2.R3/ are described in Section 17.7of [Hall]. The computational significance of the theorem is similar to that ofTheorem C.4: For each pair of irreducible subspaces W1 and W2; the “matrixentries” of any vector operator between W1 and W2 (i.e., the quantities

˝w; Ajw0˛

with w 2 W1 and w0 2 W2) are the same, up to a constant. Indeed, these matrixentries really depend only on the isomorphism class of W1 and W2: Thus, if onecan compute the matrix entries for some vector operator once and for all—for eachpair of irreducible representations of SO.3/—the matrix entries for any other vectoroperator are then determined up to the calculation of a single constant.

Proof. Note that the operators Aj and Bj (or more generally, v � A and v � B; forv 2 R3) do not necessarily map W2 into W1: On the other hand, taking the innerproduct of, say, Ajw0 with an element w of W1 has the effect of projecting Ajw0onto W1; since the inner product only depends on the component of Ajw0 in W1:

With this observation in mind, let P1 W V ! W1 be the orthogonal projection ontoW1: (This operator exists even if V is not a Hilbert space and can be constructedusing an orthonormal basis for W1.) Let Hom.W2;W1/ denote the space of linearoperators from W2 to W1 and define a linear map �A W R3 ! Hom.W2;W1/ by

�A.v/.w/ D P1.v � A/.w/

for all w 2 W2:

Now, since bothW1 andW2 are invariant, if C belongs to Hom.W2;W1/; then sodoes the operator

….R/C….R/�1 (C.7)

for all R 2 SO.3/: Under the action (C.7), the space Hom.W2;W1/ becomes arepresentation of SO.3/:We now claim that �A is an intertwining map from R3 intoHom.W2;W1/: To see this, note that since A is a vector operator, we have

�A.Rv/.w/ D P1….R/.v � A/….R/�1.w/: (C.8)

434 C Clebsch–Gordan Theory and the Wigner–Eckart Theorem

But sinceW1 is invariant and the action of SO.3/ preserves the inner product,W ?1 is

also invariant, in which case we can see that P1 commutes with ….R/: Thus, (C.8)becomes

�A.Rv/.w/ D ….R/�A.v/….R/�1.w/;

as claimed.Now, by a simple modification of the proof of Theorem C.4, we have

Hom.W2;W1/ Š W �2 ˝W1 Š W2 ˝W1;

where Š denotes isomorphism of SO.3/ representations. By Theorem C.1, in thedecomposition of W2 ˝W1; the three-dimensional irreducible representation C3 ofSO.3/ occurs at most once. If C3 does not occur, then �A must be identically zero,and similarly for the analogously defined map �B: If C3 does occur, both �A and�B must map into the same irreducible subspace of Hom.W2;W1/; and, by Schur’slemma, they must be equal up to a constant.

Finally, note that the orthogonal projection P1 is self-adjoint on V and is equalto the identity on W1: Thus,

˝w; P1.v � A/w0˛ D ˝

P1w; .v � A/w0˛ D ˝w; .v � A/w0˛ ;

and similarly with A replaced by B: Thus, since �B D c�A; we have

˝w; .v � B/w0˛ D c

˝w; .v � A/w0˛

for all v 2 R3: Specializing to v D ej , j D 1; 2; 3; gives the claimed result. ut

C.3 More on Vector Operators

We now look a bit more closely at the notion of vector operator. We consider first theLie algebra counterpart to Definition C.2. We use the basis fF1; F2; F3g for so.3/from Example 3.27. For j; k; l 2 f1; 2; 3g; define "jkl as follows:

"klm D8<:

0 if any two of j; k; l are equal1 if .j; k; l/ is a cyclic permutation of .1; 2; 3/

�1 if .j; k; l/ is an non-cyclic permutation of .1; 2; 3/:

Thus, for example, "112 D 0 and "132 D �1: The commutation relations amongF1; F2; and F3 may be written as

C.3 More on Vector Operators 435

ŒFj ; Fk� D3XlD1

"jklFl :

Proposition C.7. Let .…; V / be a finite-dimensional representation of SO.3/ andlet � be the associated representation of so.3/: Then a triple C D .C1; C2; C3/ ofoperators is a vector operator if and only if

.Xv/ � C D �.X/.v � C/ � .v � C/�.X/ (C.9)

for all X 2 so.3/: This condition, in turn, holds if and only if C1; C2; and C3 satisfy

Œ�.Fj /; Ck� D3XlD1

"jklCl : (C.10)

In physics terminology, the operators �.Fj / are (up to a factor of i„; where „ isPlanck’s constant) the angular momentum operators. See Section 17.3 of [Hall] formore information.

Proof. If SO.3/ acts on End.V / as R � C D ….R/C….R/�1; the associated actionof so.3/ on End.V / is X � C D �.X/C � C�.X/: The condition (C.9) is just theassertion that the map v 7! v � C is an intertwining map between the action of so.3/on R

3 and its action on End.V /: Since SO.3/ is connected, it is easy to see that thiscondition is equivalent to the intertwining property in Definition C.2.

Meanwhile, (C.9) will hold if and only if it holds for X D Fj and v D ek;

for all j; k D 1; 2; 3: Now, direct calculation with the matrices F1; F2; and F3 inExample 3.27 shows that Fj ek D P3

lD1 "jklel : Putting X D Fj and v D ek in (C.9)gives

3XlD1

"jklCl D Œ�.Fj /; Ck�;

as claimed. utThere is one last aspect of vector operators that should be mentioned. In quantum

physics, it is expected that the vector space of states should carry an action of therotation group SO.3/: This action may not, however, be an ordinary representation,but rather a projective representation. This means that the action is allowed tobe ill defined up to a constant. The reason for allowing this flexibility is that inquantum mechanics, two vectors that differ by a constant are considered the samephysical state. (See Section 16.7.3 of [Hall] for more information on projectiverepresentations.) In particular, the space of states for a “spin one-half” particlecarries a projective representation of SO.3/ that does not come from an ordinaryrepresentation of SO.3/:

436 C Clebsch–Gordan Theory and the Wigner–Eckart Theorem

Suppose, for example, that V carries an action of the group SU.2/; rather thanSO.3/: Suppose, also, that the action of the element �I 2 SU.2/ on V is eitheras I or as �I: If the action of �I 2 SU.2/ on V is as I; then as in the proof ofProposition 4.35, the representation will descend to a representation of SO.3/ ŠSU.2/=fI;�I g on V: Even if the action of �I 2 SU.2/ on V is as �I , we can stillconstruct a representation of SO.3/ that is well defined up to a constant; that is, Vstill carries a projective representation of SO.3/: Furthermore, the associated actionof �I 2 SU.2/ on End.V / will satisfy

.�I / � C D .�I /C.�I /�1 D C:

Thus, the action of SU.2/ on End.V / still descends to an (ordinary) action ofSO.3/: We can, therefore, still define vector operators in the setting of projectiverepresentations of SO.3/; and the proof of the Wigner–Eckart theorem goes throughwith only minor changes.

Appendix DPeter–Weyl Theorem and Completenessof Characters

In this appendix, we sketch a proof of the completeness of characters (Theorem12.18) for an arbitrary compact Lie group, not assumed to be isomorphic to a matrixgroup. The proof requires some functional analytic results, notably the spectraltheorem for compact self-adjoint operators. The needed results from functionalanalysis may be found, for example, in Chapter II of [Kna1].

As in the proof for matrix groups in Chapter 12, we prove completenessfor characters by first proving the Peter–Weyl theorem, which states that thematrix entries for irreducible representations form a complete orthogonal familyof functions. That is to say, matrix entries for nonisomorphic irreducible repre-sentations are orthogonal (Exercise 5 in Chapter 12) and any continuous functionthat is orthogonal to every matrix entry is identically zero. If we do not assumeahead of time that K has a faithful finite-dimensional representation, then it isnot apparent that the matrix entries separate points on K; so we cannot applythe Stone–Weierstrass theorem. Instead, we begin by showing that any finite-dimensional, translation-invariant space of functions on K decomposes in termsof matrix entries. We will then construct such spaces of functions as eigenspaces ofcertain convolution operators.

We consider the normalized left-invariant volume form ˛ on K: If we translate˛ on the right by some x 2 K; the resulting form ˛x is easily seen to be, again, aleft-invariant volume form, which must agree with ˛ up to a constant. On the otherhand, ˛x is still normalized, so it must actually agree with ˛: Similarly, the pullback

A previous version of this book was inadvertently published without the middle initial of theauthor’s name as “Brian Hall”. For this reason an erratum has been published, correcting themistake in the previous version and showing the correct name as Brian C. Hall (see DOI http://dx.doi.org/10.1007/978-3-319-13467-3_14). The version readers currently see is the correctedversion. The Publisher would like to apologize for the earlier mistake.

© Springer International Publishing Switzerland 2015B.C. Hall, Lie Groups, Lie Algebras, and Representations, GraduateTexts in Mathematics 222, DOI 10.1007/978-3-319-13467-3

437

438 D Peter–Weyl Theorem and Completeness of Characters

of ˛ by the map x 7! x�1 is easily seen to be left-invariant and normalized andthus coincides with ˛: Thus, ˛ is invariant under both left and right translations andunder inversions.

Now, integration of a smooth function f against ˛ satisfies

ˇˇZK

f ˛

ˇˇ � sup

K

jf j :

Meanwhile, by the Stone–Weierstrass theorem (Theorem 7.33 in [Rud1]), everycontinuous function on K can be uniformly approximated by smooth functions.Thus, the map f 7! R

Kf ˛ extends by continuity from smooth functions to

continuous functions, and if f is non-negative,RKf ˛ will be non-negative. It then

follows from the Riesz representation theorem that there is a unique measure � onthe Borel -algebra in K such that

ZK

f ˛ DZK

f .x/ d�.x/

for all continuous functions f onK: (See Theorems 2.14 and 2.18 in [Rud2]). Since˛ is normalized and invariant under left and right translations and inversions, thesame is true of �. We refer to � as the (bi-invariant, normalized) Haar measureon K: We consider the Hilbert space L2.K/; the space of (equivalence classes ofalmost-everywhere-equal) square-integrable functions on K with respect to �:

We make use of the left translation and right translation operators, given by

.Lxf /.y/ D f .x�1y/

.Rxf /.y/ D f .xy/:

Both L� and R� constitute representations of K acting on L2.K/: A subspace V �L2.K/ is right invariant, left invariant, or bi-invariant if it is invariant under lefttranslations, right translations, or both left and right translations.

Proposition D.1. Suppose V � L2.K/ is a finite-dimensional, bi-invariant sub-space and that each element of V is continuous. Then each element of V canbe expressed as a finite linear combination of matrix entries for irreduciblerepresentations ofK:

Saying that an element f of K is continuous means, more precisely, that theequivalence class f has a (necessarily unique) continuous representative.

Proof. By complete reducibility, we may decompose V into subspaces Vj that arefinite-dimensional and irreducible under the right action of K: Since the elementsof Vj are continuous, “evaluation at the identity” is a well-defined linear functionalon the finite-dimensional space Vj : Thus, there exists an element �j of Vj such that

f .e/ D ˝�j ; f

˛

D Peter–Weyl Theorem and Completeness of Characters 439

for all f 2 Vj : It follows that for all f 2 Vj ; we have

f .x/ D .Rxf /.e/

D ˝�j ;Rxf

˛D trace.Rx

ˇ�j hf j/;

whereˇ�j hf j is the operator mapping g 2 Vj to hf; gi�j : Thus, each f 2 Vj is

a matrix entry of the irreducible representation .R�; Vj / of K and each f 2 V is alinear combination of such matrix entries. utDefinition D.2. If f and g are inL2.K/; the convolution of f and g is the functionf � g onK given by

.f � g/.x/ DZK

f .xy�1/g.y/ d�.y/: (D.1)

A key property of convolution is that convolution on the left commutes withtranslation on the right, and vice versa. That is to say,

.Lxf / � g D Lx.f � g/ (D.2)

and

f � .Rxg/ D Rx.f � g/: (D.3)

Intuitively, f � g can be viewed as a combination of right-translates of f; weightedby the function g: Thus, say, (D.2) boils down to the fact that right translationcommutes with left translation, which is just a different way of stating thatmultiplication on K is associative. Rigorously, both (D.2) and (D.3) follow easilyfrom the definition of convolution.

Using the Cauchy–Schwarz inequality and the invariance of � under translationand inversion, we see that

j.f � g/.x/j � kf kL2.K/ kgkL2.K/ (D.4)

for all x 2 K: If f and g are continuous, then (since K is compact) f isautomatically uniformly continuous, from which it follows that f �g is continuous.For any f and g in L2.K/; we can approximate f and g in L2.K/ by continuousfunctions and show, with the help of (D.4), that f � g is continuous. We may also“move the norm inside the integral” in (D.1) to obtain the inequality

kf � gkL2.K/ � kf kL2.K/ kgkL1.K/ : (D.5)

Unlike convolution on the real line, convolution on a noncommutative group is,in general, noncommutative. Nevertheless, we have the following result.

440 D Peter–Weyl Theorem and Completeness of Characters

Proposition D.3. If f 2 L2.K/ is a class function, then for all g 2 L2.K/; wehave

f � g D g � f:

Proof. If we make the change of variable z D y�1x; so that y D xz�1 and y�1 Dzx�1 we find that

.f � g/.x/ DZK

f .xzx�1/g.xz�1/ d�.z/:

Since f is a class function, this expression reduces to

.f � g/.x/ DZK

g.xz�1/f .z/ d�.z/ D .g � f /.x/;

as claimed. utWe now introduce the properties of operators that will feature in our version of

the spectral theorem.

Definition D.4. Let H be a Hilbert space and A a bounded linear operator on H:Then A is self-adjoint if

hu;Avi D hAu; vifor all u and v in H; and A is compact if for every bounded set E � H; the imageof E under A has compact closure in H:

Here compactness is understood to be relative to the norm topology onH: IfH isinfinite dimensional, the closed unit ball in H is not compact in the norm topologyand thus, for example, the identity operator onH is not compact.

Proposition D.5. If � 2 L2.K/ is real-valued and invariant under x 7! x�1; theconvolution operator C� given by

C�.f / D � � fis self-adjoint and compact.

Proof. The operator C� is an integral operator with integral kernel k.x; y/ D�.xy�1/: Now, an integral operator is self-adjoint precisely if its kernel satisfiesk.x; y/ D k.y; x/: In the case of C� , this relation holds because

�.xy�1/ D �.yx�1/;

as a consequence of our assumptions on �:Meanwhile, since � is square integrableover K and K has finite measure, the function k.x; y/ D �.xy�1/ is squareintegrable over K � K: It follows that C� is a Hilbert–Schmidt operator, andtherefore compact. (See Theorem 2.4 in Chapter II of [Kna1].) ut

D Peter–Weyl Theorem and Completeness of Characters 441

Since K is compact, we can construct an inner product on the Lie algebra k ofKthat is invariant under the adjoint action of K: Thinking of k as the tangent space toK at the identity, we may then extend this inner product to an inner product on everyother tangent space by using (equivalently) either left or right translations. Thus, weobtain a bi-invariant Riemannian metric onK , which we use in the following result.

Proposition D.6. Let B".I / denote the ball of radius " about I 2 K: There existsa sequence h�ni of non-negative class functions on K such that (1) supp.�n/ �B1=n.I /; (2) �n.x�1/ D �n.x/ for all x 2 K; and (3)

RK�n.x/ d�.x/ D 1: If h�ni

is any such sequence, then

limn!1 kf � �n � f kL2.K/ ! 0

for all f 2 L2.K/:We may think of the functions �n in the proposition as approximating a “ı-

function” at the identity onK:

Proof. Since the metric on K is bi-invariant, each B".I / is invariant under theadjoint action ofK: Thus, if n is any non-negative function with support inB1=n.I /that integrates to 1, we may define

�n.x/ DZK

n.yxy�1/ d�.y/;

and �n will be a class function, still supported in B1=n.I / and still integrating to 1.We may then define

�n.x/ D 1

2.�n.x/C �n.x

�1//

and �n will have the required properties. (Note that d.x�1; I / D d.I; x/ by the leftinvariance of the metric.)

Suppose g is continuous—and thus uniformly continuous—on K: Then if n islarge enough, we will have jg.y/ � g.x/j < "whenever d.y; x/ < 1=n:Now, since� is normalized, we have

.�n � g/.x/ � g.x/ DZK

�n.xy�1/.g.y/ � g.x// d�.y/

and so, for large n;

j.�n � g/.x/ � g.x/j �ZK

�n.xy�1/ jg.y/ � g.x/j d�.y/

� "

ZK

�n.xy�1/ d�.y/

D ":

We conclude that �n � g converges uniformly—and thus, also, in L2.K/—to g.

442 D Peter–Weyl Theorem and Completeness of Characters

For any f 2 L2.K/ is arbitrary, we choose a continuous function g close to fin L2.K/ and observe that

k�n � f � f kL2.K/� k�n � f � �n � gkL2.K/ C k�n � g � gkL2.K/ C kg � f kL2.K/� k�nkL1.K/ kf � gkL2.K/ C k�n � g � gkL2.K/ C kg � f kL2.K/ :

where in the second inequality, we have used (D.5) and Proposition D.3. Since �nis non-negative and integrates to 1, k�nkL1.K/ D 1 for all n: Thus, if we take g withkf � gk < "=3 and then choose N so that k�n � g � gk < "=3 for n � N; we seethat k�n � f � f k < " for n � N: ut

We now appeal to a general functional analytic result, the spectral theorem forcompact self-adjoint operators.

Theorem D.7 (Spectral Theorem for Compact Self-adjoint Operators). Sup-pose H is an infinite-dimensional, separable Hilbert space and A is a compact,self-adjoint operator on H: Then A has an orthonormal basis of eigenvectors withreal eigenvalues that tend to zero.

For a proof, see Section II.2 of [Kna1]. Since the eigenvalues tend to zero, a fixednonzero number can occur only finitely many times as an eigenvalue; that is, eacheigenspace with a nonzero eigenvalue is finite dimensional.

Theorem D.8. If K is any compact Lie group, the space of matrix entries is densein L2.K/:

Proof. Let us say that a function f 2 L2.K/ is K-finite if there exists a finite-dimensional space of continuous functions on K that contains f and is invariantunder both left and right translation. In light of Proposition D.1, it suffices to showthat the space of K-finite functions is dense in L2.K/:

To prove this claim, suppose g 2 L2.K/ is orthogonal to everyK-finite functionf: If h�ni is as in Proposition D.6, then �n � g converges to g in L2.K/: Since �nis a class function, Proposition D.3 and the identities (D.2) and (D.3) tell us that theconvolution operator C�n commutes with both left and right translations. Thus, theeigenspaces of C�n are invariant under both left and right translations. Furthermore,since �n � f is continuous for any f 2 L2.K/; the eigenvectors of C�n withnonzero eigenvalues must be continuous. Finally, since C�n is compact and self-adjoint, the eigenspaces for C�n with nonzero eigenvalues are finite-dimensional.Thus, eigenvectors for C�n with nonzero eigenvalues are K-finite.

We conclude that g must be orthogonal to all the eigenvectors of C�n withnonzero eigenvalues. Thus, by the spectral theorem, g must actually be in theeigenspace for C�n with eigenvalue 0; that is, �n � g D 0 for all n: Letting n tend toinfinity, we conclude that g is the zero function. ut

D Peter–Weyl Theorem and Completeness of Characters 443

We may now prove (a generalization of) Theorem 12.18, without assumingahead of time that K is a matrix group. It is actually not too difficult to prove,using Theorem D.8, that every compact Lie group has a faithful finite-dimensionalrepresentation and is, therefore, isomorphic to a matrix Lie group.

Corollary D.9. If f is a square-integrable class function onK and f is orthogonalto the character of every finite-dimensional, irreducible representation ofK; then fis zero almost everywhere.

Proof. By Theorem D.8, we can find a sequence gn converging in L2.K/ to f;where each gn is a linear combination of matrix entries. Since f is a class function,the L2 distance between f .x/ and gn.y�1xy/ is independent of y: Thus, if wedefine fn by

fn.x/ DZK

gn.y�1xy/ d�.y/;

the sequence fn will also converge to f in L2.K/: But by the proof of Theorem12.18, each fn is a linear combination of characters of irreducible representations.Thus, f must be orthogonal to each fn; and we conclude that

kf k2 D hf; f i D limn!1 hf; fni D 0;

from which the claimed result follows. ut

References

[Axl] Axler, S.: Linear Algebra Done Right, 2nd edn. Undergraduate Texts in Mathematics.Springer, New York (1997)

[Baez] Baez, J.C.: The octonions. Bull. Am. Math. Soc. (N.S.) 39, 145–205 (2002); errata Bull.Am. Math. Soc. (N.S.) 42, 213 (2005)

[BBCV] Baldoni, M.W., Beck, M., Cochet, C., Vergne, M.: Volume computation for polytopesand partition functions for classical root systems. Discret. Comput. Geom. 35, 551–595(2006)

[BF] Bonfiglioli, A., Fulci, R.: Topics in Noncommutative Algebra: The Theorem of Camp-bell, Baker, Hausdorff and Dynkin. Springer, Berlin (2012)

[BtD] Bröcker, T., tom Dieck, T.: Representations of Compact Lie Groups. Graduate Texts inMathematics, vol. 98. Springer, New York (1985)

[CT] Cagliero, L., Tirao, P.: A closed formula for weight multiplicities of representations ofSp2.C/. Manuscripta Math. 115, 417–426 (2004)

[Cap] Capparelli, S.: Computation of the Kostant partition function. (Italian) Boll. Unione Mat.Ital. Sez. B Artic. Ric. Mat. 6(8), 89–110 (2003)

[DK] Duistermaat, J., Kolk, J.: Lie Groups. Universitext. Springer, New York (2000)[Got] Gotô, M.: Faithful representations of Lie groups II. Nagoya Math. J. 1, 91–107 (1950)[Hall] Hall, B.C.: Quantum Theory for Mathematicians. Graduate Texts in Mathematics, vol.

267. Springer, New York (2013)[Has] Hassani, S.: Mathematical Physics: A Modern Introduction to its Foundations, 2nd edn.

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free (and legal!) electronic version of the text is available from the author’s web page atwww.math.cornell.edu/~hatcher/AT/AT.pdf

[HK] Hoffman, K., Kunze, R.: Linear Algebra, 2nd edn. Prentice-Hall, Englewood Cliffs(1971)

[Hum] Humphreys, J.: Introduction to Lie Algebras and Representation Theory. Second print-ing, revised. Graduate Texts in Mathematics, vol. 9. Springer, New York/Berlin (1978)

A previous version of this book was inadvertently published without the middle initial of theauthor’s name as “Brian Hall”. For this reason an erratum has been published, correcting themistake in the previous version and showing the correct name as Brian C. Hall (see DOI http://dx.doi.org/10.1007/978-3-319-13467-3_14). The version readers currently see is the correctedversion. The Publisher would like to apologize for the earlier mistake.

© Springer International Publishing Switzerland 2015B.C. Hall, Lie Groups, Lie Algebras, and Representations, GraduateTexts in Mathematics 222, DOI 10.1007/978-3-319-13467-3

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[Jac] Jacobson, N.: Exceptional Lie Algebras. Lecture Notes in Pure and Applied Mathemat-ics, vol. 1. Marcel Dekker, New York (1971)

[Kna2] Knapp, A.W.: Lie Groups Beyond an Introduction, 2nd edn. Progress in Mathematics,vol. 140. Birkhäuser, Boston (2002)

[Kna1] Knapp, A.W.: Advanced Real Analysis. Birkhäuser, Boston (2005)[Lee] Lee, J.: Introduction to Smooth Manifolds. 2nd edn. Graduate Texts in Mathematics, vol.

218. Springer, New York (2013)[Mill] Miller, W.: Symmetry Groups and Their Applications. Academic, New York (1972)

[Poin1] Poincaré, H.: Sur les groupes continus. Comptes rendus de l’Acad. des Sciences 128,1065–1069 (1899)

[Poin2] Poincaré, H.: Sur les groupes continus. Camb. Philos. Trans. 18, 220–255 (1900)[Pugh] Pugh, C.C.: Real Mathematical Analysis. Springer, New York (2010)[Ross] Rossmann, W.: Lie Groups. An Introduction Through Linear Groups. Oxford Graduate

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and Applied Mathematics. McGraw-Hill, New York-Auckland-Düsseldorf (1976)[Rud2] Rudin, W.: Real and Complex Analysis, 3rd edn. McGraw-Hill, New York (1987)[Run] Runde, V.: A Taste of Topology. Universitext. Springer, New York (2008)[Tar] Tarski, J.: Partition function for certain simple Lie algebras. J. Math. Phys. 4, 569–574

(1963)[Tuy] Tuynman, G.M.: The derivation of the exponential map of matrices. Am. Math. Mon.

102, 818–819 (1995)[Var] Varadarajan, V.S.: Lie Groups, Lie Algebras, and Their Representations. Reprint of the

1974 edn. Graduate Texts in Mathematics, vol. 102. Springer, New York (1984)

Index

AA2 root system, 145, 157, 201A3 root system, 228An root system, 189, 232abelian, see commutativeAdA, 63adjoint

group, 403map, 63of a matrix, 411representation, 51

adX , 51, 64affine transformation, 394alcove, 391, 395, 406algebraically integral element, see integral

element, algebraicanalytic subgroup, see connected Lie subgroupanalytically integral element, see integral

element, analyticangles in root systems, 200angular momentum, 96averaging method, 92

BB2 root system, 201B3 root system, 228

Bn root system, 191, 234Baker–Campbell–Hausdorff formula, 109, 113base of a root system, 206basepoint, 373bilinear form

skew symmetric, 9symmetric, 8

bracket, 56

CC3 root system, 228Cn root system, 192, 234Campbell–Hausdorff formula, see Baker–

Campbell–Hausdorff formulacanonical form, see Jordan canonical formCartan subalgebra, 154, 174Casimir element, 271, 300center

discrete subgroup of, 28of a compact group, 316, 401of a Lie algebra, 51, 94of a matrix Lie group, 94

centralizer, 337chamber, Weyl, see Weyl chambercharacter of a representation, 277, 353, 443characteristic polynomial, 409

A previous version of this book was inadvertently published without the middle initial of theauthor’s name as “Brian Hall”. For this reason an erratum has been published, correcting themistake in the previous version and showing the correct name as Brian C. Hall (see DOI http://dx.doi.org/10.1007/978-3-319-13467-3_14). The version readers currently see is the correctedversion. The Publisher would like to apologize for the earlier mistake.

© Springer International Publishing Switzerland 2015B.C. Hall, Lie Groups, Lie Algebras, and Representations, GraduateTexts in Mathematics 222, DOI 10.1007/978-3-319-13467-3

447

448 Index

class function, 333, 341classical groups and Lie algebras, 188classification

of root systems, 236of simple Lie algebras, 236

Clebsch–Gordan theory, 89, 428closed subgroup, 4, 137commutative

Lie algebra, 49, 57, 73matrix Lie group, 57, 73

commutative grouprepresentations of, 95

commutator, 56commutator ideal, 54compact group

complete reducibility of, 92fundamental group of, 375

compact operator, 440compact real form, 170compact symplectic group, 10, 12

fundamental group of, 378compactness, 16complete reducibility, 90, 275complex

Lie algebra, 49, 57matrix Lie group, 57

complexification of a Lie algebra, 65conjugate linear, 415connected Lie subgroup, 129connectedness, 17, 71contragredient representation, see dual

representationconvergence of a sequence of matrices, 4convex hull, 159, 224, 267convolution, 439coroot, 179, 204

real, 335coroot lattice, 381cover, 374covering group, see universal covercovering map, 374, 395cross product, 50, 74

DDn root system, 190, 233ıjk , 6ı, see half the sum of the positive rootsdense subgroup, 312derivation, 51derivative of exponential mapping, 114derived series, 54diag.�/, 156diagonalization, 410

differential form, 93, 319, 421direct product of matrix Lie groups, 74, 88direct sum

of Lie algebras, 52of representations, 84of root systems, 199

discrete subgroup, 28, 310, 341dominant, 147, 219, 346dual

of root lattice, 401representation, 89, 165root system, 204space, 89, 416

Dynkin diagram, 216, 235, 236extended, 407

Eeigenspace, 410eigenvalue, 409eigenvector, 409Ejk, 189Euclidean group, 10exponential

of a locally nilpotent operator, 261of a matrix, 31

exponential map, 67surjectivity of, 316

extended Dynkin diagram, see Dynkindiagram, extended

extended Weyl group, 394

Ffaithful, 77fiber bundle, 376Freudenthal’s formula, 295fundamental

representations, 152weights, 219Weyl chamber, 210

fundamental group, 373of classical groups, 375, 379of SO.3/, 21, 24

G� , see kernel of exponential mapG2 root system, 201, 208, 220, 404, 407g˛ , 176general linear group, 4, 58generalized eigenvector, 412generalized orthogonal group, 8, 59group versus Lie algebra homomorphisms, 119

Index 449

Hhalf the sum of the positive roots, 220, 359,

368, 384Hausdorff property, 323Heisenberg group, 11, 110higher, 146, 222highest weight, 146, 243highest weight cyclic representation, 148, 244Hilbert–Schmidt

inner product, 188, 194norm, 32, 46

homomorphismof Lie algebras, 51, 60of matrix Lie groups, 22, 60, 72

homotopic, 373homotopy group, 375hyperplane, 198, 206, 207

Iideal, 51, 53, 249, 255identity component of a matrix Lie group, 17,

56indecomposable positive root, 207inhomogeneous Lorentz group, see Poincaré

groupinner product, 414integral element, 144, 147, 218, 242

algebraic, 346, 383analytic, 346, 348, 383

intertwining map, 78invariant subspace, 78irreducible

representation, 78root system, 199

isomorphismof Lie algebras, 51of matrix Lie groups, 22of representations, 78of root systems, 199

Jjoint eigenvector, see simultaneous eigenvectorJordan canonical form, 413

Kkernel

of a Lie group homomorphism, 63of the exponential map, 346, 380

Killing form, 194Kostant multiplicity formula, 293

Kostant partition function, 290Kreg, 385

Llength ratios in root systems, 200Lie algebra

general, 49of a matrix Lie group, 55

Lie group, 25Lie product formula, 40lift of a map, 374linear functional, 416local homomorphism, 119locally nilpotent operator, 261logarithm of a matrix, 36long exact sequence of homotopy groups, 376loop, 373Lorentz group, 8lower, 146, 222

Mmanifold, 25, 70mapping degree, 317, 331matrix entry of a representation, 357, 368, 442matrix Lie group, 4maximal commutative subalgebra, 175, 315maximal torus, 314Mn.C/, 4module, see representationmorphism, see intertwining mapmultiplicity, 144, 346

Nnegative root, 206nilpotent

Lie algebra, 54matrix, 47operator, 412

nonmatrix Lie group, 103nontrivial

ideal, 53invariant subspace, 78

norm of a matrix, 31normalizer, 315N.T /, see normalizernull homotopic, 374

Oone-parameter subgroup, 41, 56orientation, 317

450 Index

orthogonal, 191complement, 416matrix, 7

orthogonal group, 7, 58, 190fundamental group of, 377

orthogonalityof characters, 354of exponentials, 349

orthonormal basis, 411

Ppath connected, 17Peter–Weyl theorem, 357, 437physicists’ convention, 57Poincaré–Birkhoff–Witt theorem, 250Poincaré group, 11polar decomposition, 42, 127polynomials

action of SU.2/ on, 82action of SU.3/ on, 166

positive root, 206positive simple root, 145, 206product rule, 46

Qquotient manifold, 323, 377

Rrank, 176, 198real roots and coroots, 335real weight, 345reflection, 198, 336, 394, 406regular element, 385, 386regular value, 317representation

of a Lie algebra, 77of a matrix Lie group, 77unitary, 80

root, 144, 176lattice, 401real, 335space, 176string, 238system, 184, 197vector, 144, 176beer, 228

rotation, 8RP 3, 21, 377RPn, 19

SS1, 12Sard’s theorem, 318Schur’s lemma, 94Schwarz inequality, 32, 46self-adjoint

matrix, 411operator, 440

semidirect product, 394semisimple, 169simple connectedness, 18, 119, 374simple Lie algebra, 53simple root, see positive simple rootsimply connected, 122simultaneous

diagonalization, 418eigenvector, 417

singular element, 385, 386singular value, 317skew self-adjoint, 411sl.2IC/, 53, 83, 96SL.2IR/, 127sl.3IC/

representations of, 141Weyl group of, 154

slice lemma, 324smooth manifold, 25SN decomposition, 34, 413SO.3/

fundamental group of, 19Lie algebra of, 62representations of, 101universal cover of, 126

solvable Lie algebra, 54, 55special linear group, 6, 58special orthogonal group, 7, 58special unitary group, 6spectral theorem, 442spin, 101Sp.n/, see compact symplectic groupsquare root

of a positive matrix, 43uniqueness of, 41

standard representation, 81Stiefel diagram, 391strictly dominant, 219structure constants, 52SU.2/

Lie algebra of, 62relationship to SO.3/, 22, 101, 126representations of, 82, 96, 101simple connectedness of, 19

Index 451

SU.3/representations of, 141Weyl group of, 154

subalgebra, 51symplectic group, 9, 59, 192

Ttangent space at the identity, 56, 71tensor algebra, 248tensor product

of representations, 87, 106, 427of vector spaces, 85

theorem of the highest weight, 146, 243, 347torus, 310torus theorem, 316, 328trace of a matrix, 413trivial representation, 81Trotter product formula, 40

Uunipotent matrix, 47unitarian trick, 92unitary group, 6, 58

fundamental group of, 378universal cover, 126, 395

of SL.2IR/, 127of SO.3/, 126

universal enveloping algebra, 246universal property of tensor products, 86upper central series, 54upper triangular matrix, 412

VVandermonde determinant, 306vector operator, 430, 434Verma module, 244, 255, 296volume form, 327

Wweight diagram, 158, 268weight of a representation, 144, 242, 345Weyl

chamber, 210, 212character formula, 277, 359denominator, 279, 290, 291dimension formula, 283group, 154, 198, 203, 212, 315, 335integral formula, 332, 360

Weyl-alternating function, 285

ZZometool system, 228Z.T /, see centralizer