The Mathematics Any Physicist Should Know

7/28/2019 The Mathematics Any Physicist Should Know

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The Mathematics any Physicist

Should Know

Thomas Hjortgaard Danielsen


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Contents

Preface 5

I Representation Theory of Groups and Lie Algebras 7

1 Peter-Weyl Theory 91.1 Foundations of Representation Theory . . . . . . . . . . . . . . . 91.2 The Haar Integral . . . . . . . . . . . . . . . . . . . . . . . . . . 151.3 Matrix Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . 181.4 Characters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.5 The Peter-Weyl Theorem . . . . . . . . . . . . . . . . . . . . . . 24

2 Structure Theory for Lie Algebras 29

2.1 Basic Notions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.2 Semisimple Lie Algebras . . . . . . . . . . . . . . . . . . . . . . . 352.3 The Universal Enveloping Algebra . . . . . . . . . . . . . . . . . 42

3 Basic Representation Theory of Lie Algebras 49

3.1 Lie Groups and Lie Algebras . . . . . . . . . . . . . . . . . . . . 493.2 Weyls Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

4 Root Systems 59

4.1 Weights and Roots . . . . . . . . . . . . . . . . . . . . . . . . . . 594.2 Root Systems for Semisimple Lie Algebras . . . . . . . . . . . . . 624.3 Abstract Root Systems . . . . . . . . . . . . . . . . . . . . . . . . 684.4 The Weyl Group . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

5 The Highest Weight Theorem 75

5.1 Highest Weights . . . . . . . . . . . . . . . . . . . . . . . . . . . 755.2 Verma Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . 795.3 The Case sl(3, C) . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

6 Infinite-dimensional Representations 91

6.1 Grding Subspace . . . . . . . . . . . . . . . . . . . . . . . . . . 916.2 Induced Lie Algebra Representations . . . . . . . . . . . . . . . . 956.3 Self-Adjointness . . . . . . . . . . . . . . . . . . . . . . . . . . . . 976.4 Applications to Quantum Mechanics . . . . . . . . . . . . . . . . 102

II Geometric Analysis and Spin Geometry 109

7 Clifford Algebras 111

7.1 Elementary Properties . . . . . . . . . . . . . . . . . . . . . . . . 1117.2 Classification of Clifford Algebras . . . . . . . . . . . . . . . . . . 117

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7.3 Representation Theory . . . . . . . . . . . . . . . . . . . . . . . . 121

8 Spin Groups 125

8.1 The Clifford Group . . . . . . . . . . . . . . . . . . . . . . . . . . 125

8.2 Pin and Spin Groups . . . . . . . . . . . . . . . . . . . . . . . . . 1288.3 Double Coverings . . . . . . . . . . . . . . . . . . . . . . . . . . . 1318.4 Spin Group Representations . . . . . . . . . . . . . . . . . . . . . 135

9 Topological K-Theory 139

9.1 The K-Functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1399.2 The Long Exact Sequence . . . . . . . . . . . . . . . . . . . . . . 1449.3 Exterior Products and Bott Periodicity . . . . . . . . . . . . . . 1499.4 Equivariant K-theory . . . . . . . . . . . . . . . . . . . . . . . . . 1519.5 The Thom Isomorphism . . . . . . . . . . . . . . . . . . . . . . . 155

10 Characteristic Classes 163

10.1 Connections on Vector Bundles . . . . . . . . . . . . . . . . . . . 163

10.2 Connections on Associated Vector Bundles* . . . . . . . . . . . . 16610.3 Pullback Bundles and Pullback Connections . . . . . . . . . . . . 17210. 4 Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17510.5 Metric Connections . . . . . . . . . . . . . . . . . . . . . . . . . . 17810.6 Characteristic Classes . . . . . . . . . . . . . . . . . . . . . . . . 18010.7 Orientation and the Euler Class . . . . . . . . . . . . . . . . . . . 18610.8 Splitting Principle, Multiplicative Sequences . . . . . . . . . . . . 19010.9 The Chern Character . . . . . . . . . . . . . . . . . . . . . . . . . 197

11 Differential Operators 201

11.1 Differential Operators on Manifolds . . . . . . . . . . . . . . . . . 20111.2 The Principal Symbol . . . . . . . . . . . . . . . . . . . . . . . . 20511.3 Dirac Bundles and the Dirac Operator . . . . . . . . . . . . . . . 210

11.4 Sobolev Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22011.5 Elliptic Complexes . . . . . . . . . . . . . . . . . . . . . . . . . . 227

12 The Atiyah-Singer Index Theorem 233

12.1 K-Theoretic Version . . . . . . . . . . . . . . . . . . . . . . . . . 23312.2 Cohomological Version . . . . . . . . . . . . . . . . . . . . . . . . 236

A Table of Clifford Algebras 245

B Calculation of Fundamental Groups 247

Bibliography 251

Index 252


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Preface

When following courses given by Ryszard Nest at the Copenhagen University,you can be almost certain that a reference to the Atiyah-Singer Index Theoremwill appear at least once during the course. Thus it was an obvious project for meto find out what this, apparently great theorem, was all about. However, fromthe beginning I was well aware that this was not an easy task and that it was

necessary for me to delve into a lot of other subjects involved in its formulation,before the goal could be reached. It has never been my intension to actuallyprove the theorem (well except for a few moments of utter over ambitiousness)but merely to pave a road for my own understanding. This road leads throughas various subjects as K-theory, characteristic classes and elliptic theory. I havetried to treat each subject as thoroughly and self-contained as I could, eventhough this meant including stuff which wasnt really necessary for the IndexTheorem.

The starting point is of course my own prerequisites when I began my workhalf a year ago, that is a solid foundation in Riemannian geometry, algebraictopology (notably homology and cohomology) and pseudodifferential calculus onEuclidean space. From here we develop at first, in a systematic way, topologicalK-theory. The approach is via vector bundles as it can be found in for instance

[Atiyah] or [Hatcher], no C-algebras are involved. In the first two sectionsthe basic theory will be outlined and most proofs will be given. In the thirdsection we present the famous Bott-periodicity Theorem, without giving a proof.The last two sections are dedicated to the Thom Isomorphism. To this end weintroduce equivariant K-theory (that is, K-theory involving group actions), aslight generalization of the K-theory treated in the first sections. I follow theoutline given in the classical article by [Segal]. One could argue, that equivariantK-theory could have been introduced from the very beginning, however I havechosen not to, in order not to blur the introductory presentation with too manytechnicalities.

The second chapter deals with the Chern-Weil approach to characteristicclasses of vector bundles. The first four sections are devoted to the study of thebasic theory of connections on vector bundles. From the curvature forms andinvariant polynomials we construct characteristic classes, in particular Chernand Pontrjagin classes and their relationships will be discussed. In the followingsection the Euler class of oriented bundles is defined. I have relied heavily on[Morita] and [Milnor, Stacheff] when working out these sections but also [Mad-sen, Tornehave] has provided valuable inspiration. The chapter ends with adiscussion of certain characteristic classes constructed, not from invariant poly-nomials but from invariant formal power series. Examples of such classes arethe Todd class and the total A-class and the Chern character. No effort hasbeen made to include great theorems, in fact there are really no major resultsin this chapter. It serves as a tool box to be applied to the construction of thetopological index.

The third chapter revolves around differential operators on manifolds. In the

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standard literature on this subject not much care is taken, when transferring thedifferential operators and principal symbols from Euclidean space to manifolds.Ive tried to remedy this, giving a precise and detailed treatment. To this I haveadded a lot of examples of classical differential operators, such as the Lapla-

cian, Hodge-de Rham operators, Dirac operators etc. calculating their formaladjoints and principal symbols. To shed some light on the analytic properties weintroduce Sobolev spaces. Essentially there are two different definitions: in thefirst one, Sobolev spaces are defined in terms of connections, and in the secondthey are defined as the clutching of local Euclidean Sobolev spaces. We provethat the two definitions agree, when the underlying manifold is compact, and weshow how to extend differential operators to continuous operators between theSobolev spaces. The major results such as the Sobolev Embedding Theorem,the Rellich lemma and Elliptic Regularity are given without proofs. We thenmove on to elliptic complexes, which provides us with a link to the K-theorydeveloped in the first chapter.

In the fourth and final chapter the Index Theorem is presented. We constructthe so-called topological index map from the K-group K(T M) to the integersand state the index theorem, which says that the index function when evaluatedon the specific K-class determined from the symbol of an elliptic differential op-erator, is in fact equal to the Fredholm index. I give a short sketch of the proofbased on the original 1968-article by Atiyah and Singer. Then by introducingthe cohomological Thom isomorphism, Thom Defect classes etc. and drawingheavily on the theory developed in the previous chapters we manage to deducethe famous cohomological index formula. To demonstrate the power of the IndexTheorem, we prove two corollaries, namely the generalized Gauss-Bonnet The-orem and the fact that any elliptic differential operator on a compact manifoldof odd dimension has index 0.

I would like to thank Professor Ryszard Nest for his guidance and inspiration,as well as answers to my increasing amount of questions.

Copenhagen, March 2008. Thomas Hjortgaard Danielsen.


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Part I

Representation Theory of

Groups and Lie Algebras

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Chapter 1

Peter-Weyl Theory

1.1 Foundations of Representation Theory

We begin by introducing some basic but fundamental notions and results regard-ing representation theory of topological groups. Soon, however, we shall restrictour focus to compact groups and later to Lie groups and their Lie algebras.

We begin with the basic theory. To define the notion of a representation, let Vdenote a separable Banach space and equip B(V), the space of bounded linearmaps V V, with the strong operator topology i.e. the topology on B(V)generated by the seminorms Ax = Ax. Let Aut(V) B(V) denote thegroup of invertible linear maps and equip it with the subspace topology, whichturns it into a topological group.

Definition 1.1 (Representation). By a continuous representation of a topo-logical group G on a separable Banach space V we understand a continuous

group homomorphism : G Aut(V). We also say that V is given the struc-ture of a G-module. If is an injective homomorphism the representation iscalled faithful.

By the dimension of the representation we mean the dimension of the vectorspace on which it is represented. If V is infinite-dimensional the representationis said to be infinite-dimensional as well.

In what follows a group without further specification will always denote alocally compact topological group, and by a representation we will always un-derstand a continuous representation. The reason why we demand the groupsto be locally compact should be apparent in the next section.

We will distinguish between real and complex representations depending onwhether V is a real or complex Banach space. Without further qualification, the

representations considered will all be complex.The requirement on to be strongly continuous can be a little hard to handle,

so here is an equivalent condition which is more applicable:

Proposition 1.2. Let : G Aut(V) be a group homomorphism. Then thefollowing conditions are equivalent:

1) is continuous w.r.t. the the strong operator topology on Aut(V), i.e. is a continuous representation.

2) The map G V V given by (g, v) (g)v is continuous.

For a proof see [1] Proposition 18.8.


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10 Chapter 1 Peter-Weyl Theory

Example 1.3. The simplest example one can think of is the trivial repre-sentation: Let G be a group and V a Banach space, and consider the mapG g idV. This is obviously a continuous group homomorphism and hencea representation.

Now, let G be a matrix Lie group (i.e. a closed subgroup of GL(n, C)).Choosing a basis for Cn we get an isomorphism Aut(Cn)

GL(n, C), andwe can thus define a representation of G on Cn simply by the inclusion mapG GL(n, C). This is obviously a continuous representation of G, called thedefining representation.

We can form new representations out of old ones. If (1, V1) and (2, V2) arerepresentations of G on Banach spaces we can form their direct sum 1 2to be the representation of G on V1 V2 (which has been given the norm(x, y) = x + y, turning V1 V2 into a Banach space) given by

(1 2)(g)(x, y) = (1(g)x, 2(g)y).If we have a countable family (Hi)i

I of Hilbert spaces we can form the direct

sum Hilbert space iIHi to be the vector space of sequences (xi), xi Hi,satisfying

iIxi2Hi < . Equipped with the inner product (xi), (yi) =

iIxi, yi this is again a Hilbert space. If we have a countable family (i, Hi)of representations such that supiIi(g) < for each g G, then we canform the direct sum of the representations

iIi on

iIHi by

iI

i

(g)(xi) = (i(g)xi).

Finally, if(1, H1) and (2, H2) are representations on Hilbert spaces, we canform the tensor product, namely equip the tensor product vector space H1 H2with the inner product

x1 x2, y1 y2 = x1, y1x2, y2which turns H1 H2 into a Hilbert space, and define the tensor product repre-sentation 1 2 by

(1 2)(g)(x y) = 1(g)x 2(g)y.Definition 1.4 (Unitary Representation). By a unitary representation of agroup G we understand a representation on a Hilbert space H such that (g)is a unitary operator for each g G.

Obviously the trivial representation is a unitary representation. As is thedefining representation of any subgroup of the unitary group U(n). In the nextsection we show unitarity of some more interesting representations.

Definition 1.5 (Intertwiner). Let two representations (1, V1) and (2, V2) ofthe same group G be given. By an intertwiner or an intertwining map between1 and 2 we understand a bounded linear map T : V1 V2 rendering thefollowing diagram commutative

V1

T//

1(g)

V2

2(g)

V1

T// V2

i.e. satisfying T 1(g) = 2(g) T for all g G. The set of all intertwiningmaps is denoted HomG(V1, V2).


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1.1 Foundations of Representation Theory 11

A bijective intertwiner with bounded inverse between two representations iscalled an equivalence of representations and the two representations are said tobe equivalent. This is denoted 1 = 2.

Its easy to see that HomG(V1, V2) is a vector space, and that HomG(V, V) isan algebra. The dimension ofHomG(V1, V2) is called the intertwining number ofthe two representations. If 1 = 2 via an intertwiner T, then we have 2(g) =T1 1(g) T. Since we thus can express the one in terms of the other, foralmost any purpose the two representations can be regarded as the same.

Proposition 1.6. HomG respects direct sum in the sense that

HomG(V1 V2, W) = HomG(V1, W) HomG(V2, W) and (1.1)HomG(V, W1 W2) = HomG(V, W1) HomG(V, W2). (1.2)

Proof. For the first isomorphism we define

: HomG(V1 V2, W) HomG(V1, W) HomG(V2, W)by (T) := (T|V1 , T|V2). It is easy to check that this is indeed an element of thelatter space. It has an inverse 1 given by

1(T1, T2)(v1, v2) := T1(v1) + T2(v2),

and this proves the first isomorphism. The latter can be proved in the sameway.

Definition 1.7. Given a representation (, V) of a group G, we say that alinear subspace U V is -invariant or just invariant if (g)U U for allg G.

If U is a closed invariant subspace for a representation of G on V, weautomatically get a representation ofG on U, simply by restricting all the (g)sto U (U should be a Banach space, and therefore we need U to be closed). Thisis clearly a representation, and we will denote it |U (although we are restrictingthe (g)s to U and not ).

Here is a simple condition to check invariance of a given subspace, at least inthe case of a unitary representation

Lemma 1.8. Let (, H) be a representation of G, let H = U U be a de-composition of H and denote by P : H U the orthogonal projection onto U.If U is -invariant then so is U. Furthermore U is -invariant if and only ifP (g) = (g) P for all g G.Proof. Assume that U is invariant. To show that U is invariant let v

U.

We need to show that (g)v U, i.e. u U : (g)v, u = 0. But thatseasy, exploiting unitarity of (g):

(g)v, u = (g1)((g)v), (g1)u = v, (g1)u

which is 0 since (g1)u U and v U. Thus U is invariantAssume U to be invariant. Then also U is invariant by the above. We split

x H into x = P x + (1 P)x and calculate

P (g)x = P((g)(P x + (1 P)x)) = P (g)P x + P (g)(1 P)x.

The first term is (g)P x, since (g)P x U, and the second term is zero, since(g)(1 P)x U. Thus we have the desired formula.


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Conversely, assume that P (g) = (g) P. Every vector u U is of theform P x for some x H. Since

(g)u = (g)(P x) = P((g)x)

U,

U is an invariant subspace.

For any representation (, V) it is easy to see two obvious invariant subspaces,namely V itself and {0}. We shall focus a lot on representations having noinvariant subspaces except these two:

Definition 1.9. A representation is called irreducible if it has no closed invari-ant subspaces except the trivial ones. The set of equivalence classes of finite-dimensional irreducible representations of a group G is denoted G.

A representation is called completely reducible if it is equivalent to a directsum of finite-dimensional irreducible representations.

Any 1-dimensional representation is obviously irreducible, and if the group is

abelian the converse is actually true. We prove this in Proposition 1.14If (1, V1) and (2, V2) are irreducible representations then the direct sum

1 2 is not irreducible, since V1 is an 1 2-invariant subspace of V1 V2:(1 2)(g)(v, 0) = (1(g)v, 0).

The question is more subtle when considering tensor products of irreduciblerepresentations. Whether or not the tensor product of two irreducible repre-sentations is irreducible and if not, to write is as a direct sum of irreduciblerepresentations is a branch of representation theory known as Clebsch-Gordantheory.

Lemma 1.10. Let(1, V1) and (2, V2) be equivalent representations. Then 1is irreducible if and only if 2 is irreducible.

Proof. Given the symmetry of the problem, it is sufficient to verify that ir-reducibility of 1 implies irreducibility of 2. Let T : V1 V2 denote theintertwiner, which by the Open Mapping Theorem is a linear homeomorphism.Assume that U V2 is a closed invariant subspace. Then T1U V1 is closedand 1-invariant:

1(g)T1U = T12(g)U T1U

But this means that T1U is either 0 or V1, i.e. U is either 0 or V2.

Example 1.11. Consider the group SL(2, C) viewed as a real (hence 6-dimensional)Lie group. We consider the following 4 complex representations of the real Liegroup SL(2, C) on C2:

(A) := A, (A) := A,(A) := (AT)1, (A) := (A)1,where A simply means complex conjugation of all the entries. All four are clearlyirreducible. They are important in physics where they are called spinorial rep-resentations. The physicists have a habit of writing everything in coordinates,thus will usually be written , where = 1, 2 but the exact notation willvary according to which representation we have imposed on C2 (i.e. accordingto how transforms as the physicists say). In other words they view C2 not asa vector space but rather as a SL(2, C)-module. The notations are

C2, C2, C2, C

2.


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1.1 Foundations of Representation Theory 13

The representations are not all mutually inequivalent, actually the map :

C2 C2 given by the matrix

0 11 0

intertwines with

and intertwines

with . On the other hand and are actually inequivalent as we will se in Sec-tion 1.4. These two representations are called the fundamental representationsof SL(2, C).

In short, representation theory has two goals: 1) given a group: find all theirreducible representations and 2) given a representation of this group: split it (ifpossible) into a direct sum of irreducibles. The rest of this chapter deals with thesecond problem (at least for compact groups) and in the end we will achieve somepowerful results (Schur Orthogonality and the Peter-Weyl Theorem). Chapter5 revolves around the first problem of finding irreducible representations.

But already at this stage we are able to state and prove two quite interestingresults. The first result is known as Schurs Lemma. We prove a slightly moregeneral version than is usually seen, allowing the representations to be infinite-dimensional.

Theorem 1.12 (Schurs Lemma). Let (1, H1) and (2, H2) be two irre-ducible unitary representations of a group G, and suppose that F : H1 H2is an intertwiner. Then either F is an equivalence of representations or F is thezero map.

If (, H) is an irreducible unitary representation of G and F B(H) is alinear map which commutes with all (g), then F = idH.

Proof. The proof utilizes a neat result from Gelfand theory: suppose thatA is a commutative unital C*-algebra which is also an integral domain (i.e.ab = 0 implies a = 0 or b = 0), then A = Ce. The proof is rather simple.Gelfands Theorem states that there exists a compact Hausdorff space X suchthat A = C(X). To reach a contradiction, assume that X is not a one-pointset, and pick two distinct points x and y. Then since X is a normal topologicalspace, we can find disjoint open neighborhoods U and V around x and y, andthe Urysohn Lemma gives us two nonzero continuous functions f and g on X,the first one supported in U and the second in V, the product, thus, beingzero. This contradicts the assumption that A = C(X) was an integral domain.Therefore X can contain only one point and thus C(X) = C.

With this result in mente we return to Schurs lemma. F being an intertwinermeans that F 1(g) = 2(g) F, and using unitarity of 1(g) and 2(g) we getthat

F 2(g) = 1(g) Fwhere F is the hermitian adjoint of F. This yields

(F F) 2(g) = F 1(g) F = 2(g) (F F).In the last equality we also used that F intertwines the two representations.Consider the C-algebra A = C(idH2 , F F

), the C-algebra generated by idH2and F F. Its a commutative unital C-algebra, and all the elements are of theform

n=0 an(F F

)n. They commute with 2(g): n=1

an(F F)n

2(g) =

n=1

(an(F F)n2(g)) =

n=1

an(2(g)(F F)n)

= 2(g)n=1

an(F F)n.

We only need to show that A is an integral domain. Assume ST = 0. Since2(g)S = S2(g) its easy to see that ker S is 2-invariant. 2 is irreducible


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so ker S is either H2 or {0}. In the first case S = 0, and we are done, in thesecond case, S is injective, and so T must be the zero map. This means thatA = C idH2 , in particular, there exists a C so that F F = idH2 . Likewise,one shows that FF = idH1 . Thus, we see

F = F(FF) = (F F)F = F

which implies F = 0 or = . In the second case if = = 0 then FF v = 0for all v H1, and hence

0 = v, FF v = F v , F v,

i.e. F = 0. If = and = 0 then it is not hard to see that 12 F is unitary,and that F therefore is an isomorphism.

The second claims is an immediate consequence of the proof of the first.

The content of this can be summed up to the following: If 1 and 2 are irre-

ducible unitary representations of G on H1 and H2, then HomG(H1, H2) = C if1 and 2 are equivalent and HomG(H1, H2) = {0} if1 and 2 are inequivalent.Corollary 1.13. Let (, H1) and (, H2) be finite-dimensional unitary repre-sentations which decompose into irreducibles

=iI

mii and =iI

nii.

Then dim HomG(H1, H2) =

iInimi.

Proof. Denoting the representations spaces of the irreducible representationsby Vi we get from (1.1) and (1.2) that

HomG(H1, H2) = iI

jI

nimj Hom(Vi, Vj),

and by Schurs Lemma the dimension formula now follows.

Now for the promised result on abelian groups

Proposition 1.14. Let G be an abelian group and (, H) be a unitary repre-sentation of G. If is irreducible then is 1-dimensional.

Proof. Since G is abelian we have (g)(h) = (h)(g) i.e. (h) is an in-tertwiner. Since is irreducible, Schurs Lemma says that (h) = (h) idH.Thus, each 1-dimensional subspace of H is invariant, and by irreducibility H is1-dimensional.

Example 1.15. With the previous lemma we are in a position to determine theset of irreducible complex representations of the circle group T = R/Z. Since thisis an abelian group, we have found all the irreducible representations when weknow all the 1-dimensional representations. A 1-dimensional representation is

just a homomorphism R/Z C, so lets find them: It is well-known that theonly continuous homomorphisms R C are those of the form x e2iaxfor some a R. But since we also want it to be periodic with periodicity 1,only integer values of a are allowed. Thus, T consists of the homomorphismsn(x) = e

2inx for n Z.Proposition 1.16. Every finite-dimensional unitary representation is com-pletely reducible.


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1.2 The Haar Integral 15

Proof. If the representation is irreducible then we are done, so assume wehave a unitary representation : G Aut(H) and let {0} = U H be aninvariant subspace. The point is that U is invariant as well cf. Lemma 1.8. Ifboth

|U and

|U are irreducible we are done. If one of them is not, we find

an invariant subspace and perform the above argument once again. Since therepresentation is finite-dimensional and since 1-dimensional representations areirreducible, the argument must stop at some point.

1.2 The Haar Integral

In the representation theory of locally compact groups (also known as harmonicanalysis) the notions of Haar integral and Haar measure play a key role.

Some preliminary definitions: Let X be a locally compact Hausdorff spaceand Cc(X) the space of complex valued functions on X with compact support.By a positive integral on X is understood a linear functional I : Cc(X) Csuch that I(f)

0 iff

0. The Riesz Representation Theorem tells us that to

each such positive integral there exists a unique Radon measure on the Borelalgebra B(X) such that

I(f) =

X

fd.

We say that this measure is associated with the positive integral.Now, let G be a group. For each g0 G we have two maps Lg0 and Rg0 , left

and right translation, on the set of complex-valued functions on G, given by

(Lg0f)(g) = f(g10 g), (Rg0f)(g) = f(gg0).

These obviously satisfy Lg1g2 = Lg1Lg2 and Rg1g2 = Rg1Rg2 .

Definition 1.17 (Haar Measure). Let G be a locally compact group. A

nonzero positive integral I on G is called a left Haar integral if I(Lgf) = I(f)for all g G and f Cc(X). Similarly a nonzero positive integral is called aright Haar integral if I(Rgf) = I(f) for all g G and f Cc(X). An integralwhich is both a left and a right Haar integral is called a Haar integral.

The measures associated with left and right Haar integrals are called left andright Haar measures. The measure associated with a Haar integral is called aHaar measure.

Example 1.18. On (Rn, +) the Lebesgue integral is a Haar integral: it is ob-viously positive, and it is well-known that the Lebesgue integral is translationinvariant:

Rnf(x + a)dx =

Rnf(a + x)dx =

Rnf(x)dx.

The associated Haar measure is of course the Lebesgue measure mn.On the circle group (T, ) we define an integral I by

C(T) f 12

20

f(eit)dt.

As before this is obviously a positive integral and since

I(Leiaf) =1

2

20

f(eiaeit)dt =1

2

20

f(ei(a+t))dt

=1

2

20

f(eit)dt


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again by exploiting translation invariance of the Lebesgue measure, I is a leftHaar integral on T. Likewise one can show that it is a right Haar integral aswell, and hence a Haar integral. The associated Haar measure on T is also calledthe arc measure.

In both cases the groups were abelian and in both cases the left Haar integralswere also right Haar integrals. This is no mere coincidence for if G is an abeliangroup we have Lg0 = Rg10

and thus a positive integral is a left Haar integral if

and only if it is a right Haar integral.The following central theorem attributed to Alfred Haar and acclaimed as

one of the most important mathematical discoveries in the 20th century statesexistence and uniqueness of left and right Haar integrals on locally compactgroups.

Theorem 1.19. Every locally compact group G possesses a left Haar integraland a right Haar integral, and these are unique up to multiplication by a positiveconstant.

If G is compact then the two integrals coincide, and the corresponding Haarmeasure is finite.

It would be far beyond the scope of this thesis to delve into the proof of this.The existence part of the proof is a hard job so we just send some acknowledgingthoughts to Alfred Haar and accept it as a fact of life.

Now we restrict focus to compact groups on which, as we have just seen, wehave a finite Haar measure. The importance of this finiteness is manifested inthe following result:

Theorem 1.20 (Unitarization). Let G be a compact group and (, H) arerepresentation on a Hilbert space (H, , ). Then there exists an inner product, G on H equivalent to , which makes a unitary representation.

Proof. Since the measure is finite, we can integrate all bounded measurablefunctions over G. Let us assume the measure to be normalized, i.e. that (G) =1. For x1, x2 H the map g (g)x1, (g)x2 is continuous (by Proposition1.2), hence bounded and measurable, i.e. integrable. Now define a new innerproduct by

x1, x2G :=G

(g)x1, (g)x2dg. (1.3)

That this is a genuine inner product is not hard to see: it is obviously sesqui-linear by the properties of the integral, it is conjugate-symmetric, as the originalinner product is conjugate-symmetric. Finally, if x = 0 then (g)x = 0 ((g) isinvertible) and thus (g)x > 0 for all g G. Since the map g (g)x2 iscontinuous we have x, xG =

G(g)xdg > 0.

By the translation of the Haar measure we get

(h)x1, (h)x2G =G

(gh)x1, (gh)x2dg

=

G

(g)x1, (g)x2dg= x1, x2G.

Thus, is unitary w.r.t. this new inner product.We just need to show that the two norms and G corresponding to

the two inner products are equivalent, i.e. that there exists a constant C so that CG and G C. To this end, consider the map g (g)x2 forsome x H. Its a continuous map, hence supgG (g)x2 < for all x, and


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1.2 The Haar Integral 17

the Uniform Boundedness Principle now says that C := supgG (g) < .Therefore

x

2 = G x2dg = G (g1)(g)x2dg C2 G (g)x2 = C2x2G.

Conversely we see

x2G =G

(g)x2 G

(g)2x2dg C2G

x2dg = C2x2.

This proves the claim.

If we combine this result with Proposition 1.16 we get

Corollary 1.21. Every finite-dimensional representation of a compact group iscompletely reducible.

The Peter-Weyl Theorem which we prove later in this chapter provides astrong generalization of this result in that it states that every Hilbert spacerepresentation of a compact group is completely reducible.

We end this section by introducing the so-called modular function which is afunction that provides a link between left and right Haar integrals.

Let G be a topological group and I : f G

f(g)dg a left Haar integral.

Let h G and consider the integral Ih : f G f(gh1)dg. This is positiveand satisfies

Ih(Lg0f) = G

f(g10 gh1)dg =

G

f(gh1)dg = Ih(f)i.e. is a left Haar integral. By the uniqueness part of Haars Theorem there existsa positive constant c such that Ih(f) = cI(f). We define the modular function : G R+ by assigning this constant to the group element h i.e.

G

f(gh1)dg = (h)

G

f(g)dg.

It is not hard to see that this is indeed a homomorphism: on one hand we haveG

f(g(hk)1)dg = (hk)

G

f(g)dg,

and on the other hand we have that this equals

Gf(gk1h1)dg = (h)

Gf(gk1)dg = (h)(k)

G.f(g)dg

Since this holds for all integrable functions f we must have (hk) = (h)(k).One can show that this is in fact a continuous group homomorphism and thusin the case of G being a Lie group, a Lie group homomorphism.

If is identically 1, that is if every right Haar integral satisfiesG

f(hg)dg =

G

f(g)dg (1.4)

for all h, then the group G is called unimodular. Eq. (1.4) says that an equivalentcondition for a group to be unimodular is that all right Haar integrals are alsoleft Haar integrals. As we have seen previously in this section abelian groupsand compact groups are unimodular groups.


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1.3 Matrix Coefficients

Definition 1.22 (Matrix Coefficient). Let (, V) be a finite-dimensional rep-resentation of a compact group G. By a matrix coefficient for the representation

we understand a map G C of the formmv,(g) = ((g)v)

for fixed v V and V.If we pick a basis {e1, . . . , en} for V and let {1, . . . , n} denote the correspond-

ing dual basis, then we see that mei,j = j((g)ei) precisely are the entries ofthe matrix-representation of (g), therefore the name matrix coefficient.

IfV comes with an inner product , , then by the Riesz Theorem all matrixcoefficients are of the form mv,w = (g)v, w for fixed v, w V. By Theorem1.20 we can always assume that this is the case.

Denote by C(G) the space of linear combinations of matrix coefficient. Since

a matrix coefficient is obviously a continuous map, C(G) C(G) L2

(G).Thus, we can take the inner product of two functions in C(G). Note, howeverthat the elements of C(G) need not all be matrix coefficients for .

The following technical lemma is an important ingredient in the proof of theSchur Orthogonality Relations which is the main result of this section.

Lemma 1.23. Let (, H) be a finite-dimensional unitary representation of acompact group G. Define the map T : End(H) C(G) by

T(A)(g) = Tr((g) A). (1.5)

Then C(G) = im T.

Proof. Given a matrix coefficient mv,w we should produce a linear map A :H H, such that mv,w = T(A). Consider the map Lv,w : H H definedby Lv,w(u) = u, wv, the claim is that this is the desired map A. To see thiswe need to calculate Tr Lv,w and we claim that the result is v, w. Since Lv,wis sesquilinear in its indices (Lav+bv,w = aLv,w + bLv,w), its enough to checkit on elements of an orthonormal basis {e1, . . . , en} for H.

Tr Lei,ei =n

k=1

Lei,eiek, ek =n

k=1

ek, eiei, ek = 1

while for i = j

Tr Lei,ej =n

k=1Lei,ejek, ek =n

k=1ek, ejei, ek = 0.Thus, Tr Lv,w = v, w. Finally since

Lv,w (g)u = (g)u, wv = u, (g1)wv = Lv,(g1)wu

we see that

T(Lv,w)(g) = Tr((g) Lv,w) = Tr(Lv,w (g)) = v, (g1)w = (g)v, w= mv,w(g).

Conversely, we should show that any map T(A) is a linear combination ofmatrix coefficients. Some linear algebraic manipulations should be enough to


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1.3 Matrix Coefficients 19

convince the reader that we for any A End(H) have A = ni,j=1Aej , eiLei,ejw.r.t some orthonormal basis {e1, . . . , en}. But then we readily see

T(A)(g) = Tn

i,j=1Aej , eiLei,ej(g) =n

i,j=1Aej , eiT(Lei,ej )(g)=

ni,j=1

Aej , eimei,ej (g).

Theorem 1.24 (Schur Orthogonality I). Let (1, H1) and (2, H2) be twounitary, irreducible finite-dimensional representations of a compact group G. If1 and 2 are equivalent, then we have C(G)1 = C(G)2. If they are not, thenC(G)1C(G)2 inside L2(G).

Before the proof, a few remarks on the integral of a vector valued functionwould be in order. Suppose that f : G H is a continuous function intoa finite-dimensional Hilbert space. Choosing a basis {e1, . . . , en} for H we canwrite f in its components f = ni=1 fiei, which are also continuous, and define

G

f(g)dg :=ni=1

G

fi(g)dg ei.

Its a simple change-of-basis calculation to verify that this is independent of thebasis in question. Furthermore, one readily verifies that it is left-invariant andsatisfies

G

f(g)dg,v

=

G

f(g), vdg and AG

f(g)dg =

Af(g)dg

when A End(H).Proof of Theorem 1.24. If1 and 2 are equivalent, there exists an isomor-phism T : H1

H2 such that T 1(g) = 2(g)T. For A

End(H1) we see

that

T2(T AT1)(g) = Tr(2(g)T AT

1) = Tr(T12(g)T A) = Tr(1(g)A)

= T1(A)(g).

Hence the map sending T1(A) to T2(T AT1) is the identity id : C(G)1

C(G)2 proving that the two spaces are equal.Now we show the second claim. Define for fixed w1 H1 and w2 H2 the

map Sw1,w2 : H1 H2 by

Sw1,w2(v) =

G

1(g)v, w12(g1)w2dg.

Sw1,w2 is in HomG(H1, H2) since by left-invariance

Sw1,w21(h)(v) = G1(gh)v, w12(g1)w2dg =

G1(g)v, w12(hg1)w2dg

= 2(h)

G

1(g)v, w12(g1)w2dg= 2(h)Sw1,w2(v).

Assume that we can find two matrix coefficients mv1,w1 and mv2,w2 for 1and 2 that are not orthogonal, i.e. we assume that

0 =G

mv1,w2(g)mv2,w2(g)dg =

G

1(g)v1, w12(g)v2, w2dg

=

G

1(g)v1, w12(g1)w2, v2dg.


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From this we read Sw1,w2v1, v2 = 0, so that Sw1,w2 = 0. Since its an inter-twiner, Schurs Lemma tells us that Sw1,w2 is an isomorphism. By contraposition,the second claim is proved.

In the case of two matrix coefficients for the same representation, we have thefollowing result

Theorem 1.25 (Schur Orthogonality II). Let (, H) be a unitary, finite-dimensional irreducible representation of a compact group G. For two matrixcoefficients mv1,w1 and mv2,w2 we have

mv1,w1 , mv2,w2 =1

dim Hv1, v2w2, w1. (1.6)

Proof. As in the proof of Theorem 1.24 define Sw1,w2 : H H by

Sw1,w2(v) = G1(g)v, w12(g1)w2dg = G (g1)Lw2,w1(g)vdg.We see that

mv1,w1 , mv2,w2 =G

(g)v1, w1(g)v2, w2dg

=

G

(g)v1, w1(g1)w2, v2dg

=

G

(g)v1, w1(g1)w2, v2= Sw1,w2v1, v2.

Furthermore, since Sw1,w2 commutes with (g), Schurs Lemma yields a com-plex number (w1, w2), such that Sw1,w2 = (w1, w2) idH. The operator Sw1,w2is linear in w2 and anti-linear in w1, hence (w1, w2) is a sesquilinear form onH. We now take the trace on both sides of the equation Sw1,w2 = (w1, w2) idH:the right hand side is easy, its just (w1, w2)dim H. For the left hand side wecalculate That is, we get (w1, w2) = (dim H)1w1, w2, and hence

Sw1,w2 = (dim H)1w1, w2 idH .

By substituting this into the equation mv1,w1 , mv2,w2 = Sw1,w2v1, v2 thedesired result follows.

1.4 CharactersDefinition 1.26 (Class Function). For a group G, a class function is a func-tion on G which is constant on conjugacy classes. The set of square-integrableresp. continuous class functions on G are denoted L2(G, class) and C(G, class).

It is not hard to see that the closure ofC(G, class) inside L2(G) is L2(G, class).Thus, L2(G, class) is a Hilbert space. Given an irreducible finite-dimensionalrepresentation the set of continuous class functions inside C(G) is very small:

Lemma 1.27. Let (, H) be a finite-dimensional irreducible unitary represen-tation of a compact group G, then the only class functions inside C(G) arecomplex scalar multiples of T(idH).


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1.4 Characters 21

Proof. To formulate the requirement on a class function, consider the repre-sentation of G on C(G) by ((g)f)(x) = f(g1xg), then in terms of this afunction f is a class function if and only if (g)f = f for all g.

For reasons which will be clear shortly, we introduce another representation

of G on End(H) by

(g)A = (g)A(g1).

Equipping End(H) with the inner product A, B := Tr(BA), it is easy to seethat becomes unitary. The linear map T : End(H) C(G) which weintroduced in Lemma 1.23 is an intertwiner of the representations and :

T((g)A)(x) = Tr

(x)(g)A(g1)

= Tr((g1xg)a)

= (g) Tr((x)A) = (g)T(A)(x).

T was surjective by Lemma 1.23. To show injectivity we define

T :=

dim H T

and show that this is unitary. Since the linear maps Lv,w span End(H) it is

enough to show unitarity on these. But first we need some facts concerningLv,w:

Lv,wx, y = x, wv, y = x, wv, y = x, v, yw= x, y, vw = x, Lw,v

showing that Lv,w = Lw,v. Furthermore

Lw,v Lv,wx = Lw,v(x, wv) = x, wv, vw= v, vx, ww = v, vLw,wx.

With the inner product on End(H) these results now yield

Lv,w, Lv,w = Tr(Lw,v Lv,w) = Tr(v, vLw,w) = v, vw, w.

Since T(Lv,w)(x) = mv,w(x), and using Schur Orthogonality II we see

T(Lv,w), T(Lv,w) = dim Hmv,w, mv,w = v, vw, w = Lv,w, Lv,w.Thus T is unitary and in particular injective.

Now we come to the actual proof: let C(G) be a class function. Tis bijective, so there is a unique A End(H) for which = T(A). That Tintertwines and leads to

(g1xg) = ((g))(x) = (g)

T(A)(x) =

T((g)A)(x) = T((g)A(g

1)),

and since was a class function we get that (g)A(g1) = A, i.e. A intertwines. But was irreducible, which by Schurs Lemma implies A = idH, and hence = T(idH).

In particular there exists a unique class function 0 which is positive on e andwhich has L2-norm 1: namely we have

022 = T(A)22 = A2 = Tr(AA)so if 0 should have norm 1 and be positive on e, then A is forced to be(dim H)1 idH, so that 0 is given by 0(g) = Tr (g). This is a function ofparticular interest:


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Definition 1.28 (Character). Let (, V) be a finite-dimensional representa-tion of a group G. By the character of we mean the function : G Cgiven by

(g) = Tr (g).

If is a character of an irreducible representation, is called an irreduciblecharacter.

The character is a class function, so in the case of two representations 1and 2 being equivalent via the intertwiner T: 2(g) = T 1(g)T

1, we have1 = 2 . Thus, equivalent representations have the same character. Actually,the converse is also true, we show that at the end of the section.

Suppose that G is a topological group, and that H is a Hilbert space withorthonormal basis {e1, . . . , en}. Then we can calculate the trace as

Tr (g) =ni=1

(g)ei, ei

which shows that C(G). In due course we will prove some powerful or-thogonality relations for irreducible characters. But first we will see that thecharacter behaves nicely with respect to direct sum and tensor product opera-tions on representations.

Proposition 1.29. Let (1, V1) and (2, V2) be two finite-dimensional repre-sentations of the group G. The characters of 1 2 and 1 2 are then givenby

12(g) = 1(g) + 2(g) and 12(g) = 1(g)2(g). (1.7)

Proof. Equip V1 and V2 with inner products and pick orthonormal bases (ei)and (fj) for V1 and V2 respectively. Then the vectors (ei, 0), (0, fj) form anorthonormal basis for V1 V2 w.r.t. the inner product

(v1, v2), (w1, w2) := v1, w1 + v2, w2.Thus we see

12(g) = Tr 1 2(g)

=mi=1

1 2(g)(ei, 0), (ei, 0)

+

nj=1

1 2(g)(0, fj), (0, fj)

=

mi=1

1(g)ei, ei +n

j=1

2(g)fj , fj = 1(g) + 2(g).

Likewise, the vectors ei fj constitute an orthonormal basis for V1 V2 w.r.t.the inner product

v1 v2, w1 w2 := v1, w1v2, w2,and hence

12(g) = Tr 1 2(g) =m,ni,j=1

1 2(g)(ei fj), (ei fj)

=

m,ni,j=1

1(g)ei, ei2(g)fj , fj

=mi=1

1(g)ei, ein

j=1

2(g)fj , fj = 1(g)2(g).


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1.4 Characters 23

The following lemma, stating the promised orthogonality relations of char-acters, shows that irreducible characters form an orthonormal set in C(G).The Schur Orthogonality Relations are important ingredients in the proof, thushenceforth we need the groups to be compact.

Lemma 1.30. Let (1, V1) and (2, V2) be two finite-dimensional irreduciblerepresentations of a compact group G. Then the following hold:

1) 1 = 2 implies 1 , 2 = 1.2) 1 2 implies 1 , 2 = 0.

Proof. In the first case, we have a bijective intertwiner T : V1 V2. Choosean inner product on V1 and an orthonormal basis (ei) for V1. Define an innerproduct on V2 by declaring T to be unitary. Then (T ei) is an orthonormal basisfor V2. Let n = dim V1 = dim V2. The expressions 1 =

ni=11(g)ei, ei and

2 =

nj=12(g)T ej , T ej along with (1.6) yield

1 , 2 =n

i,j=1

G

1(g)ei, ei2(g)T ej , T ejdg

=n

i,j=1

G

1(g)ei, eiT 1(g)ej , T ejdg

=

ni,j=1

G

1(g)ei, ei1(g)ej , ejdg

=1

n

ni,j=1

ei, ejei, ej = 1n

ni=1

1 = 1.

In the second case, if 1

and 2

are non-equivalent then by Theorem 1.24we have C(G)1C(G)2 . Since 1 C(G)1 and 2 C(G)2 , the resultfollows.

This leads to the main result on characters:

Theorem 1.31. Let be an finite-dimensional representation of a compactgroup G. Then decomposes according to

=iG

, ii.

Proof. Proposition 1.16 says that =

mii where i is irreducible and mi

is the number of times that i occurs in . From Lemma 1.30 it follows that

= i mii and hence by orthonormality of the irreducible characters thatmi = , i.Example 1.32. A very simple example to illustrate this is the following. Con-sider the 2-dimensional representation of T given by

x 12

e2inx + e2imx e2inx + e2imx

e2inx + e2imx e2inx + e2imx

for n, m Z. It is easily seen to be a continuous homomorphism T Aut(C2)with character (x) = e

2imx + e2inx. But the two terms are irreduciblecharacters for T, cf. Example 1.15, and by Theorem 1.31 we have = n m.


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Corollary 1.33. For finite-dimensional representations 1, 2 and of a com-pact group we have:

1) 1 = 2 if and only if 1 = 2 .2) is irreducible if and only if , = 1.

Proof. For the first statement, the only-if part is true by the remarks followingthe definition of the character. To see the converse, assume that 1 = 2 .Then for each irreducible representation we must have 1 , = 2 , and therefore 1 and 2 are equivalent to the same decomposition of irreduciblerepresentations, hence they are equivalent.

If is irreducible then Lemma 1.30 states that , = 1. Conversely,assume , = 1 and decompose into irreducibles: =

mii. Or-

thonormality of the irreducible characters again gives , =

m2i . Fromthis it is immediate that there is precisely one mi which is 1, while the rest are0, i.e. = i. Therefore is irreducible.

Considering the representations and from Example 1.11 we see that thecorresponding characters satisfy = and since is actually a complexmap, they are certainly not equal. Hence the representations are inequivalent.

1.5 The Peter-Weyl Theorem

The single most important theorem in the representation theory of compacttopological groups is the Peter-Weyl Theorem. It has numerous consequences,some of which we will mention at the end of this section.

Theorem 1.34 (Peter-Weyl I). LetG be a compact group. Then the subspace

M(G) := G

C(G)

of C(G) is dense in L2(G).

In other words the linear span of all matrix coefficients of the finite-dimensionalirreducible representations of G is dense in L2(G).

Proof. We want to show that M(G) = L2(G). We prove it by contradictionand assume that M(G) = 0. Now, suppose that M(G) (which is a closedsubspace ofL2(G) and hence a Hilbert space itself) contains a finite-dimensionalR-invariant subspace W (R is the right-regular representation) such that R|Wis irreducible (we prove below that this is a consequence of the assumptionM(G)

= 0). Then we can pick an finite orthonormal basis (i) for W, and

then for 0 = f Wf(x) =

Ni=1

f, ii(x).

This is a standard result in Hilbert space theory. Then we see that

f(g) = (R|W(g)f)(e) =Ni=1

R|W(g)f, ii(e).

Since R|W is a finite-dimensional irreducible representation, the map g R|W(g), i is a matrix coefficient. But this means that f M(G), hence acontradiction.


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1.5 The Peter-Weyl Theorem 25

Now, lets prove the existence of the finite-dimensional right-invariant sub-space. Let f0 M(G) be nonzero. As C(G) is dense in L2(G) we can find a C(G) such that

, f0 = 0 where

(g) = (g1). Define K C(G G) by

K(x, y) = (xy1) and let T : L2(G)

L2(G) be the integral operator with

K as its kernel:

T f(x) =

G

K(x, y)f(y)dy.

According to functional analysis, this is a well-defined compact operator, and itcommutes with R(g):

T R(g)f(x) =G

K(x, y)R(g)f(y)dy =

G

(xy1)f(yg)dy

=

G

(xgy1)f(y)dy =

G

K(xg,y)f(y)dy

= R(g)(T f)(x).

In the third equation we exploited the invariance of the measure under the righttranslation y yg1.

Since R(g) is unitary, also the adjoint T of T commutes with R(g):

T R(g) = T R(g1) = (R(g1) T) = (T R(g1)) = R(g) T.Thus, the self-adjoint compact operator TT commutes with R(g). The SpectralTheorem for compact operators yields a direct sum decomposition of L2(G):

L2(G) = ker(TT) =0

E

where all the eigenspaces E are finite-dimensional. They are also R-invariant,for if f E then

TT(R(g)f) = R(g)(TT)f = R(g)(f) = (R(g)f) (1.8)

i.e. R(g)f E. Actually M(G) is R-invariant: all functions are of the formni=1 aii(x)i, i and since

R(g)f(x) = f(xg) =ni=1

aii(x)(i(g)i), i

we see that R(g)f M(G). But then also M(G) is invariant. IfP : L2(G) M(G) denotes the orthogonal projection, then by Lemma 1.8, P commuteswith R(g), and a calculation like (1.8) reveals that P E are all R-invariant sub-spaces of M(G). These are very good candidates to the subspace we wanted:they are finite-dimensional and R-invariant, so we can restrict R to a represen-tation on these. We just need to verify that at least one of them is nonzero. Soassume that P E are all 0. This means by definition of P that

E M(G)

and hence that M(G) ( E) = ker TT ker T, where the last inclu-sion follows since f ker TT implies 0 = TT f , f = T f , T f , i.e. T f = 0.But applied to the f0 M(G) we picked at the beginning, we have

T f0(e) =

G

(ey1)f0(y)dy =

G

(y)f0(y)dy = , f0 = 0,and as T f0 is continuous, T f0 = 0 as an L2 function. Thus, we must have atleast one for which P E = 0. IfR restricted to this space is not irreducible, itcontains a nontrivial subspace on which it is. Thus, we have proved the result.


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What we actually have shown in the course of the proof is that we for eachnonzero f can find a finite-dimensional subspace U L2(G) which is R-invariantand, restricted to which, R is irreducible. We can show exactly the same thingfor the left regular representation L, all we need to alter is the definition of K,

which should be K(x, y) = (x1y). This observation will come in useful now,when we prove the promised generalization of Corollary 1.21:

Theorem 1.35 (Peter-Weyl II). Let (, H) be any (possibly infinite-dimen-sional) representation of a compact group G on a Hilbert space H. Then =

i where i is a finite-dimensional irreducible representation of G, i.e. iscompletely reducible.

Proof. By virtue of Theorem 1.20 we can choose a new inner product on Hturning into a unitary representation.

Then we consider the set of collections of finite-dimensional invariant sub-spaces of H restricted to which is irreducible, i.e. an element (Ui)iI in isa collection of subspaces of H satisfying the mentioned properties. We equip

with the ordering defined by (Ui)iI (Uj)jJ ifi Ui j Uj . It is easilyseen that (, ) is inductively ordered, hence Zorns Lemma yields a maximalelement (Vi)iI. To show the desired conclusion, namely that

H =iI

Vi,

we assume that W := (

Vi) = 0. We have a contradiction if we in W can find

a finite-dimensional -invariant subspace on which is irreducible, so thats ourgoal.

First we remark that W is -invariant since its the orthogonal complementto an invariant subspace, thus we can restrict to a representation on W. Now,we will define an intertwiner T : W

L2(G) between

|W and the left regular

representation L. Fix a unit vector x0 H and define (T y)(g) = y, (g)x0.T y : G C is clearly continuous, and since T x0(e) = x0 = 0, T x0 isnonzero in L2(G), hence T is nonzero as a linear map. T is continuous, as theCauchy-Schwartz inequality and unitarity of (g) give

|T y(g)| = |y, (g)x0 | yx0

that is T x0. T is an intertwiner:

(T (h))y(g) = (h)y, (g)x0 = y, (h1g)x0 = L(h) (T y)(g).

The adjoint T : L2(G) W (which is nonzero, as T is) is an intertwiner aswell, for taking the adjoint of the above equation yields (h)

T = T

L(h)

for all h. Using unitarity we get (h1) T = T L(h1), i.e. also T is anintertwiner.

As T is nonzero, there is an f0 L2(G) such that Tf0 = 0. But by theremark following the proof of the first Peter-Weyl Theorem we can find a non-trivial finite-dimensional L-invariant subspace U L2(G) containing f0. ThenTU W is finite-dimensional, nontrivial (it contains Tf0) and -invariant,for ifTf TU, then (h) Tf = T L(h)f TU. Inside TU we can nowfind a subspace on which is irreducible, hence the contradiction.

An immediate corollary of this is:

Corollary 1.36. An irreducible representation of a compact group is automat-ically finite-dimensional.


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1.5 The Peter-Weyl Theorem 27

In particular the second Peter-Weyl Theorem says that the left regular rep-resentation is completely reducible. In many textbooks this is the statement ofthe Peter-Weyl Theorem. The proof of this is not much different from the proofwe gave for the first version of the Peter-Weyl Theorem, and from this it would

also be possible to derive our second version of the Peter-Weyl Theorem. I chosethe version with matrix coefficients since it can be used immediately to provideelegant proofs of some results in Fourier theory, which we now discuss.

Theorem 1.37. Let G be a compact group. The set of irreducible charactersconstitute an orthonormal basis for the Hilbert space L2(G, class). In particularevery square integrable class function f on G can be written

f =G

f, ,

the convergence being L2-convergence.

Proof. Let P : L2

(G) C(G) denote the orthogonal projection ontoC(G). It is not hard to see that P maps class functions to class functions, henceP(L

2(G, class)) C(G) C(G, class), the last space being the 1-dimensionalC by Lemma 1.27. Hence the space

M(G, class) := M(G) C(G, class) = G

C(G) C(G, class)

has as orthonormal basis the set of irreducible characters of G. To see that thecharacters also form an orthonormal basis for the Hilbert space L2(G, class)assume that there exists an f L2(G, class) which is orthogonal to all thecharacters. Then since Pf is just a scalar multiple of we see

P

f =P

f,

=f,

= 0

where in the third equality we exploited self-adjointness of the projection P.Thus we must have f M(G) which by Peter-Weyl I implies f = 0.

Specializing to the circle group T yields the existence of Fourier series. Firstof all, since T is abelian, all functions defined on it are class functions, andfunctions on T are nothing but functions on R with periodicity 1. Specializingthe above theorem to this case then states that the irreducible characters e2inx

constitute an orthonormal basis for L2(T, class) and that we have an expansionof any such square integrable class function

f =

nZcn(f)e

2inx (1.9)

where cn is the nth Fourier coefficient

cn(f) = f, n =1

0

f(x)e2inxdx.

Its important to stress that the convergence in (1.9) is only L2-convergence. Ifwe put some restrictions to f such as differentiability or continuous differentia-bility we can achieve pointwise or uniform convergence of the series. We will nottravel further into this realm of harmonic analysis.


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Chapter 2

Structure Theory for Lie

Algebras

2.1 Basic Notions

Although we succeeded in Chapter 1 to prove some fairly strong results, wemust realize that it is limited how much we can say about topological groups,compact or not. For instance the Peter-Weyl Theorem tells us that every rep-resentation of a compact group is completely reducible, but if we dont knowthe irreducible representations then whats the use? Therefore we change ourfocus to Lie groups. The central difference, when regarding Lie groups, is ofcourse that we have their Lie algebras at our disposal. Often these are mucheasier to handle than the groups themselves, while at the same time sayingquite a lot about the group. Therefore we need to study Lie algebras and their

representation theory.In this section we focus solely on Lie algebras, developing the tools necessary

for the representation theory of the later chapters. We will only consider Liealgebras over the fields R and C (commonly denoted K) although many of theresults in this chapter carry over to arbitrary (possibly algebraically closed)fields of characteristic 0.

Definition 2.1 (Lie Algebra). A Lie algebra g over K is a K-vector space gequipped with a bilinear map [ , ] : g g g satisfying

1) [X, Y] = [Y, X] (antisymmetry)

2) [[X, Y], Z] + [[Y, Z], X] + [[Z, X], Y] = 0 (Jacobi identity).

A Lie subalgebrah ofg is a subspace ofg which is closed under the bracket, i.e.for which [h, h] h. A Lie subalgebra h for which [h, g] h is called an ideal.

In this thesis all Lie algebras will be finite-dimensional unless otherwise spec-ified.

Example 2.2. The first examples of Lie algebras are algebras of matrices. Bygl(n, R) and gl(n, C) we denote the set of real resp. complex n n matricesequipped with the commutator bracket. It is trivial to verify that these areindeed Lie algebras. The list below contains the definition of some of the classicalLie algebras. They are all subalgebras of the two Lie algebras just mentioned.It is a matter of routine calculations to verify that these examples are indeed

29


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30 Chapter 2 Structure Theory for Lie Algebras

closed under the the commutator bracket.

sl(n, R) = {X gl(n, R) | Tr X = 0}sl(n, C) =

{X

gl(n, C)

|Tr X = 0

}so(n) = {X gl(n, R) | X+ Xt = 0}so(m, n) = {X, gl(m + n, R) | XtIm,n + Im,nX = 0}so(n, C) = {X gl(n, C) | X+ Xt = 0}

u(n) = {X gl(n, C) | X+ X = 0}u(m, n) = {X gl(m + n, C) | XIm,n + Im,nX = 0}su(n) = {X gl(n, C) | X+ X = 0, Tr X = 0}

su(m, n) = {X gl(m + n, C) | XIm,n + Im,nX = 0, Tr X = 0}

where Im,n is the block-diagonal matrix whose first m m block is the identityand the last n n block is minus the identity.

Another interesting example is the endomorphism algebra EndK(V) for some

K-vector space V, finite-dimensional or not. Equipped with the commutatorbracket [A, B] = AB BA this becomes a Lie algebra over K, as one can check.To emphasize the Lie algebra structure of this, it is sometimes denoted gl(V).We stick to End(V).

We always have the trivial ideals in g, namely 0 and g itself. If g is a Liealgebra and h is an ideal in g, then we can form the quotient algebra g/h in thefollowing way: The underlying vector space is the vector space g/h and this weequip with the bracket

[X+ h, Y + h] = [X, Y] + h.

Using the ideal-property it is easily checked that this is indeed well-defined and

satisfies the properties of a Lie algebra.

Definition 2.3 (Lie Algebra Homomorphism). Let g and g be Lie algebrasover K. A K-linear map : g g is called a Lie algebra homomorphism if itsatisfies [(X), (Y)] = [X, Y] for all X, Y g. If is bijective it is called aLie algebra isomorphism.

An example of a Lie algebra homomorphism is the canonical map : g g/hmapping X to X+h. It is easy to see that the image of a Lie algebra homomor-phism is a Lie subalgebra of g and that the kernel of a homomorphism is anideal in g. Another interesting example is the so-called adjoint representationad : g End(V) given by ad(X)Y = [X, Y]. We see that ad(X) is linear,hence an endomorphism, and that the map X ad(X) is linear. By virtueof the Jacobi identity it respects the bracket operation and is thus ad is a Liealgebra homomorphism.

In analogy with vector spaces and rings we have the following

Proposition 2.4. Let : g g be a Lie algebra homomorphism andh g anideal which containsker , then there exists a unique Lie algebra homomorphism : g/h g such that = . In the case thath = ker andg = im theinduced map is an isomorphism.

Ifh andk are ideals ing then there exists a natural isomorphism (h+k)/k

h/(h k).Definition 2.5 (Centralizer). Finally, for any element X g we define thecentralizer C(X) ofX to be the set of elements in g which commute with X. Leth be any subalgebra ofg. The centralizer C(h) ofh is the set of all elements of


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2.1 Basic Notions 31

g that commute with all elements ofh. The centralizer ofg is called the centerand is denoted Z(g).

For a subalgebra h ofg we define the normalizer N(h) ofh to be all elementsX

g for which [X, h]

h.

We immediately see that the centralizer of X is just ker ad(X), hence C(X)is an ideal. Furthermore we see that

C(h) =Xh

C(X)

and that Z(g) = ker ad. Hence also the center is an ideal. Finally, a subalgebraofg is an ideal if and only if its normalizer is g.

Now consider the so-called derived algebra: Dg := [g, g] which clearly is anideal. g is called abelian if Dg = 0, i.e. if [X, Y] = 0 for all X, Y g. Every1-dimensional Lie algebra is abelian by antisymmetry of the bracket.

Definition 2.6 (Simple Lie Algebra). A nontrivial Lie algebra is calledindecomposable if the only ideals are the trivial ones: g and 0. A nontrivial Liealgebra is called simple if it is indecomposable and Dg = 0.

Any 1-dimensional Lie algebra is indecomposable and as the next propositionshows, the requirement Dg = 0 is just to get rid of these trivial examples:Proposition 2.7. A Lie algebra is simple if and only if it is indecomposableand dim g 2.Proof. Ifg is simple then it is not abelian, hence we must have dim g 2.

Conversely, assume that g is indecomposable and dim g 2. As Dg is an idealwe can only have Dg = 0 or Dg = g. In the first case, g is abelian and hence allsubspaces are ideals, and since dim g 2, nontrivial ideals exist, contradictingindecomposability. Therefore Dg = g = 0.

Now, lets consider the following sequence of ideals D1g := Dg, D2g :=[Dg, Dg], . . . , Dng := [Dn1g, Dn1g], the so-called derived series. Obviously wehave Dm+ng = Dm(Dng). To see that they are really ideals we use induction:We have already seen that D1g is an ideal, so assume that Dn1g is an ideal.Let X, X Dn1g and let Y g be arbitrary. Then by the Jacobi identity

[[X, X], Y] = [[X, Y], X] [[Y, X], X].

Since Dn1g is an ideal, [X, Y], [Y, X] Dn1g showing that [[X, X], Y] Dng.Definition 2.8 (Solvable Lie Algebra). A Lie algebra is called solvable if

there exists an N such that DNg = 0.Abelian Lie algebras are solvable, since we can take N = 1. On the other

hand, simple Lie algebras are definitely not solvable, for we showed in the proofof Proposition 2.7 that Dg = g which implies that Dng = g for all n.Proposition 2.9. Letg be a Lie algebra.

1) Ifg is solvable, then so are all subalgebras of g.

2) Ifg is solvable and : g g is a Lie algebra homomorphism, then im is solvable.

3) Ifh g is a solvable ideal so that g/h is solvable, theng is solvable.


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4) Ifh and k are solvable ideals of g, then so is h + k.

Proof. 1) It should be clear that Dh Dg. Hence, by induction, Dih Digand since

DNg = 0 for some N, then DNh = 0 as well.

2) Since is a Lie algebra homomorphism, we have D((g)) = (Dg), andagain by induction Di((g)) = (Dig). Thus, DNg = 0 implies DN((g)) = 0.

3) Assume there is an N for which DN(g/h) = 0 and consider the canonicalmap : g g/h. Like above we have Di(g/h) = Di((g)) = (Dig). Thus,since DN(g/h) = 0, we have (DNg) = 0 i.e. DNg h. But h was also solvable,so we can find an M for which DMh = 0. Then

DM+Ng = DM(DNg) DMh = 0

i.e. g is solvable.4) By 3) of this proposition it is enough to prove that (h+ k)/k is solvable. By

Proposition 2.4 there exists an isomorphism (h+k)/k h/(hk), and the right

hand side is solvable since it is the image of the canonical map h h/(hk).The last point of this proposition yields the existence of a maximal solvable

ideal in g, namely if h and k are solvable ideals, then h + k will be a solvableideal containing both. Thus the sum of all solvable ideals is a solvable ideal. Thisworks since the Lie algebra is finite-dimensional. By construction, it is unique.

Definition 2.10 (Radical). The maximal solvable ideal, the existence of whichwe have just verified, is called the radical ofg and is denoted Rad g.

A Lie algebra g is called semisimple if Rad g = 0.

Since all solvable ideals are contained in Rad g another way of formulatingsemisimplicity would be to say that it has no nonzero solvable ideals. In thissense, semisimple Lie algebras are as far as possible from being solvable.

In the next section we prove some equivalent conditions for semisimplicity.

Proposition 2.11. Semisimple Lie algebras have trivial centers.

Proof. The center is an abelian, hence solvable, ideal, and is therefore trivialby definition.

We now consider a concept closely related to solvability. Again we considera sequence of ideals: g0 := g, g1 := Dg, g2 := [g, g1], . . . ,gn := [g, gn1]. Itshouldnt be too hard to see that Dig gi.Definition 2.12 (Nilpotent Lie Algebra). A Lie algebra g is called nilpotentif there exists an N such that gN = 0.

Since Dig gi nilpotency ofg implies solvability ofg. The converse statementis not true in general. So schematically:

abelian nilpotent solvable

in other words, solvability and nilpotency are in some sense generalizations ofbeing abelian.

Here is a proposition analogous to Proposition 2.9

Proposition 2.13. Letg be a Lie algebra.

1) Ifg is nilpotent, then so are all its subalgebras.

2) If g is nilpotent and : g g is a Lie algebra homomorphism, thenim is nilpotent.


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2.1 Basic Notions 33

3) Ifg/Z(g) is nilpotent, then g is nilpotent.4) Ifg is nilpotent, then Z(g) = 0.

Proof.

1) In analogy with the proof of Proposition 2.9 a small induction ar-gument show that if h g is a subalgebra, then hi gi. Thus, gN = 0 implieshN = 0.

2) We have already seen that (g)1 = (Dg). Furthermore(g)2 = [(g), (g)1] = [(g), (Dg)] = ([g, Dg]) = (g2)

and by induction we get (g)i = (gi). Hence nilpotency ofg implies nilpotencyof (g).

3) Letting : g g/Z(g) denote the canonical homomorphism, we see that(g/Z(g))i = ((g))i = (gi) = gi/Z(g).

Thus, if(g/Z(g))N = 0 then gN Z(g). But then gN+1 = [g, gN] [g, Z(g)] =0, hence g is nilpotent.

4) As g is nilpotent there is a smallest n such that gn = 0 and gn+1 = 0. Thismeans that [g, gn] = 0 i.e. everything in gn commutes with all elements of g.Thus, 0 = gn Z(g).Definition 2.14. An element X g is called ad-nilpotent ifad(X) is a nilpotentlinear map, i.e. if there exists an N such that ad(X)N = 0.

If the Lie algebra is a subalgebra of an algebra of endomorphisms (for in-stance End(V)), it makes sense to ask if the elements themselves are nilpotent.In this case nilpotency and ad-nilpotency of an element X need not be the same.For instance in End(V) we have the identity I, which is obviously not nilpo-tent. However, ad(I) = 0, and thus I is ad-nilpotent. The reverse implication,however, is true:

Lemma 2.15. Letg be a Lie algebra of endomorphisms of some vector space.If X g is nilpotent, then it is ad-nilpotent.Proof. We associate to A g two linear maps A, A : End(V) End(V)by A(B) = AB and A(B) = BA. Its easy to see that they commute, andthat ad(A) = A A.

As A is nilpotent, A and A are also nilpotent, so we can find an N for whichNA =

NA = 0. Since they commute, we can use the binomial formula and get

ad(A)2N = (A A)2N =2Nj=0

(1)j

2Nj

2NjA

jA

which is zero since all terms contain either A or A to a power greater thanN.

An equivalent formulation of nilpotency of a Lie algebra is that there exists anN such that ad(X1) ad(XN)Y = 0 for all X1, . . . , X N, Y g. In particular, ifg is nilpotent, then there exists an N such that ad(X)N = 0 for all X g, i.e. Xis ad-nilpotent. Thus, for a nilpotent Lie algebra g, all elements are ad-nilpotent.That the converse is actually true is the statement of Engels Theorem, whichwill be a corollary to the following theorem.

Theorem 2.16. LetV be a finite-dimensional vector space and g End(V) bea subalgebra consisting of nilpotent linear endomorphisms. Then there exists a

nonzero v V which is an eigenvector for all A End(V).


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Proof. We will prove this by induction over the dimension of g. First, assumedim g = 1. Then g = KA for some nonzero A g. As A is nilpotent there is asmallest N such that AN = 0 and AN+1 = 0, i.e. we can find a vector w Vwith ANw

= 0 and A(ANw) = AN+1w = 0. Since all elements of g are scalar

multiples of A the vector ANw will qualify.Now assuming that the theorem holds for all Lie algebras of dimension strictly

less than n, we should prove that it holds for n-dimensional algebras as well.The algebra g consists of nilpotent endomorphisms on V, hence by the previouslemma, all elements are ad-nilpotent. Consider a subalgebra h = g of g whichthus also consists of ad-nilpotent elements. For A h we have that ad(A)h hsince h as a subalgebra is closed under brackets. We can form the vector spaceg/h and define a linear map ad(A) : g/h g/h by

ad(A)(B + h) = (ad(A)B) + h.

This is well defined for if B + h = B + h, then B B h and therefore

ad(A)(B + h) = ad(A)B + h = ad(A)B + ad(A)(B B) + h = ad(A)B + h= ad(A)(B + h).

This map is again nilpotent since ad(A)N(B + h) = (ad(A)NB) + h = h = [0].

So, the situation now is that we have a subalgebra ad(h) of End(g/h) withdimad(h) dimh < dim g = n. Our induction hypothesis then yields an ele-ment 0 = [B0] = B0 + h g/h on which ad(A) is zero for all A h. This meansthat [A, B0] h for all h, i.e. the normalizer N(h) ofh is strictly larger than h.

Now assume that h is any maximal subalgebra h = g. Then since N(h) isa strictly larger subalgebra we must have N(h) = g and consequently h is anideal. Then g/h is a Lie algebra with canonical Lie algebra homomorphism : g

g/h and g/h must have dimension 1 for, assuming otherwise, we could

find a 1-dimensional subalgebra k = g/h in g/h, and then 1(k) = g would bea subalgebra strictly larger that h. This is a contradiction, hence dim g/h = 1and g = h KA0 for some nonzero A0 g \ h.

So far, so good. Now we come to the real proof of the existence of the nonzerovector v V. h was an ideal of dimension n 1 hence the induction hypothesisassures that the subspace W := {v V | B h : Bv = 0} is nonempty. Wewill show that each linear map A g (which maps V V) can be restrictedto a map W W and that it as such a map is still nilpotent. This will, inparticular, hold for A0 which by nilpotency will have the eigenvalue 0 and hencea nonzero eigenvector v W associated to the eigenvalue 0. This will be thedesired vector, for all linear maps in g can according to the decomposition abovebe written as B + A0 for some B h, and Bv = 0 since v was chosen to be inW.

Thus, to finish the proof we only need to see that W is invariant. So let A gbe any map. Since h is an ideal, [A, h] h and hence for w W

B(Aw) = A(Bw) [A, B]w = 0

for all B h. This shows that Aw W and hence that W is invariant. Arestriction of a nilpotent map is clearly nilpotent. This completes the proof.

From this we can prove

Corollary 2.17 (Engels Theorem). A Lie algebra is nilpotent if and only ifall its elements are ad-nilpotent.


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2.2 Semisimple Lie Algebras 35

Proof. We have already showed the only if part. To show the if part weagain invoke induction over the dimension of g. If dim g = 1 then g is abelian,hence nilpotent.

Now set n = dim g and assume that the result holds for all Lie algebras with

dimension strictly less than n. All the elements of g are ad-nilpotent, hencead(g) is a subalgebra ofEnd(g) consisting of nilpotent elements and the previoustheorem yields an element 0 = X g for which ad(Y)(X) = 0 for all Y g i.e.X is contained in the center Z(g) which is therefore a nonzero ideal and g/Z(g)is a Lie algebra whose dimension strictly less than n. Furthermore all elementsofg/Z(g) are ad-nilpotent, since by definition of the quotient bracket

ad([A])[B] = ad(A)B + Z(g)

we have that ad(A)N = 0 implies ad([A])N = 0. Thus g/Z(g) consists solelyof ad-nilpotent elements. Consequently the induction hypothesis assures thatg/Z(g) is nilpotent. Then by Proposition 2.13 g is nilpotent.

2.2 Semisimple Lie Algebras

The primary goal of this section is to reach some equivalent formulations ofsemisimplicity. Our approach to this will be via the so-called Cartan Criterionfor solvability which we will prove shortly.

First we need a quite powerful result from linear algebra regarding advanceddiagonalization:

Theorem 2.18 (SN-Decomposition). Let V be a finite-dimensional vectorspace over K and let A End(V). Then there exist unique commuting linearmaps S, N End(V), S being diagonalizable and N being nilpotent, satisfyingA = S+ N. This is called the SN-decomposition In fact S andN can be realized

as polynomials in A without constant terms.Furthermore, if A = S+ N is the SN-decomposition of A, then ad(S)+ad(N)

is the SN-decomposition of ad(A).

We will not prove this 1.Cartans Criterion gives a sufficient condition for solvability based on the trace

of certain matrices. Therefore the following lemma is necessary.

Lemma 2.19. Let V be a finite-dimensional vector space, W1 and W2 be sub-spaces of End(V) and define M := {B End(V) | ad(A)W1 W2}. If A Msatisfies Tr(AB) = 0 for all B M then A is nilpotent.Proof. Let A M satisfy the required condition, and consider the SN-decom-position of A = S + N. We are done if we can show that S = 0. Well, S is

diagonalizable, so we can find a basis {e1, . . . , en} for V in which S has the formdiag(a1, . . . , an). We will show that all these eigenvalues are 0, and we do so ina curious way: We define E := spanQ{a1, . . . , an} K to be the subspace of Kover the rationals spanned by the eigenvalues. If we can show that this space, orequivalently its dual space E, consisting of Q-linear maps E Q, is 0, thenwe are done.

So, let E be arbitrary. The basis we chose for V readily gives us abasis for End(V), consisting of Eij where Eij is the linear map determined byEijej = ei and Eijek = 0 for k = j. Then we see

(ad(S)Eij)ej = [S, Eij ]ej = SEijej EijSej = Sei ajEijej = (ai aj)ei1For a proof the reader is referred to for instance [5] Section 4.3.


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while [S, Eij]ek = 0 for k = j i.e. ad(S)Eij = (ai aj)Eij .Now, let B End(V) denote the linear map which in the basis {e1, . . . , en} is

diag((a1), . . . , (an)). As with S we have that ad(B)Eij = ((ai) (aj))Eij.There exists a polynomial p = Nn=1 cnXn without constant term which mapsai aj to (ai aj) = (ai) (aj) (its a matter of solving some equationsto find the coefficients cn). Then we have

p(ad S)Eij = cn(ad S)nEij + + c1(ad S)Eij

= cn(ai aj)nEij + + c1(ai aj)Eij= p(ai aj)Eij = ((ai) (aj))Eij

which says that p(ad S) = ad B. A statement in the SN-decomposition was thatad S, being the diagonalizable part of ad A, is itself a polynomial expressionin ad A without constant term, which implies that ad B is a polynomial inad A without constant term. Since A M we have that ad(A)W1 W2, andsince ad(B) was a polynomial expression in ad(A) then also ad(B)W1 W2,i.e. B M, and therefore by assumption Tr(AB) = 0. The trace of AB isthe sum

ni=1 ai(ai) and applying to the equation Tr(AB) = 0 we getn

i=1 (ai)2 = 0, i.e. (ai) = 0 ((ai) are rationals hence (ai)

2 0). Thereforewe must have = 0 which was what we wanted.

Theorem 2.20 (Cartans Criterion). Let V be a finite-dimensional vectorspace andg End(V) a subalgebra. If Tr(AB) = 0 for allA g and allB Dgthen g is solvable.

Proof. As Dng = Dn1(Dg) (Dg)n1 we see that g will be solvable if Dg isnilpotent. To show that Dg is nilpotent we invoke Engels Theorem and Lemma2.15 which combined say that Dg is nilpotent if all X Dg are nilpotent.To this end we use the preceding lemma with W1 = g and W2 = Dg andM =

{B

End(V)

|[B, g]

Dg}

. Notice that g

M. The reverse inclusionneed not hold.

Now, let A Dg be arbitrary, we need to show that it is nilpotent, and byvirtue of the previous lemma it suffices to verify that Tr(AB) = 0 for all B M.A is of the form [X, Y] for X, Y g and we have, in general that

Tr([X, Y]B) = Tr(XY B) Tr(Y XB) = Tr(Y BX) Tr(BY X)= Tr([Y, B]X) = Tr(X[Y, B]). (2.1)

Since B M and Y g we have by construction of M that [Y, B] Dg. Butthen by assumption in the theorem we have that Tr(AB) = Tr([X, Y]B) =0.

With this powerful tool we can prove the promised equivalent conditions fora Lie algebra to be semisimple. One of them involves the so-called Killing form:

Definition 2.21 (Killing Form). By the Killing form for a Lie algebra g overK we understand the bilinear form B : g g K given by

B(X, Y) = Tr(ad(X)ad(Y)).

Proposition 2.22. The Killing form is a symmetric bilinear form satisfying

B([X, Y], Z) = B(X, [Y, Z]). (2.2)

Furthermore, if is any Lie algebra automorphism of g then B((X), (Y)) =B(X, Y).


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2.2 Semisimple Lie Algebras 37

Proof. B is obviously bilinear, and symmetry is a consequence of the propertyof the trace: Tr(AB) = Tr(BA). Eq. (2.2) follows from (2.1). If : g g is a Lie algebra automorphism, then another way of writing the equation[(X), (Y)] = ([X, Y]) is ad((X))

=

ad(X). Therefore

B((X), (Y)) = Tr( ad(X) ad(Y) 1) = Tr(ad(X)ad(Y))= B(X, Y).

Calculating the Killing form directly from the definition is immensely compli-cated. Fortunately, for some of the classical Lie algebras we have a much simplerformula:

B(X, Y) =

2(n + 1) Tr(XY), for X, Y sl(n + 1, K),sp(2n, K)(2n 1) Tr(XY), for X, Y so(2n + 1, K)2(n 1) Tr(XY), for X, Y so(2n, K).

(2.3)

Lemma 2.23. Ifg is a Lie algebra with Killing form B andh g is an ideal,then B|hh is the Killing form ofh.Proof. First a general remark: If : V V is a linear map, and W V is a subspace for which im W, then Tr = Tr(|W): Namely, pick abasis {e1, . . . , ek} for W and extend it to a basis {e1, . . . , ek, . . . , en} for V.Let {1, . . . , n} denote the corresponding dual basis. As (v) W we havek+i((v)) = 0 and hence

Tr =ni=1

i((ei)) =ki=1

i((ei)) = Tr(|W).

Now, let X, Y h, then as h is an ideal: ad(X)g h and ad(Y)g h, whichmeans that the image of ad(X)ad(Y) lies inside h. It should be obvious that

the adjoint representation ofh is just ad(X)|h for X h. ThereforeBh(X, Y) = Tr(ad(X)|h ad(Y)|h) = Tr((ad(X)ad(Y))|h)

= B|hh(X, Y).Theorem 2.24. Ifg is a Lie algebra, then the following are equivalent:

1) g is semisimple i.e. Rad g = 0.

2) g has no nonzero abelian ideals.

3) The Killing form B ofg is non-degenerate.

4) g is a direct sum of simple Lie algebras: g = g1 gn.

Proof.

We first prove that 1 and 2 are equivalent. Ifg

is semisimple, theng

has no nonzero solvable ideals and since abelian ideals are solvable, no nonzeroabelian ideals either. Conversely, if Rad g = 0, then, since Rad g is solvable,there is a smallest N for which DN(Radg) = 0 and DN+1(Rad g) = 0. ThenDN(Radg) and is an abelian ideal hence a solvable ideal. So by contraposition,if no solvable ideals exist, then g is semisimple.

Now we show that 1 implies 3. We consider the so-called radical of the Killingform B namely the subspace

h := {X g | Y g : B(X, Y) = 0}.h is an ideal for if X h and Y g, then for all Z g:

B([X, Y], Z) = B(X, [Y, Z]) = 0


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i.e. [X, Y] h. Obviously B is non-degenerate if and only ifh = 0.Now we assume that Rad g = 0 and want to show that h = 0. We can do

this by showing that h is solvable for then h Rad g. First we use the CartanCriterion on the Lie algebra ad(h) showing that this is solvable: By definition

of h we have that 0 = B(X, Y) = Tr(ad(X)ad(Y)) for all X h and Y g,in particular it holds for all X Dh. In other words we have Tr(AB) = 0 forall A ad(Dh) = D(ad h) and all B ad h. Hence the Cartan Criterion tells usthat ad h is solvable, i.e. 0 = DN(adh) = ad(DNh). This says that DNh Z(g)implying that DN+1h = 0. Thus, h is solvable and consequently equals 0.

Then we prove 3 implies 2. Assume that h = 0, and assume k to be an abelianideal and let X k and Y g. Since the adjoint representations maps accordingto (exploiting the ideal property ofk)

gad(Y) g ad(X) k ad(Y) k ad(X) Dk = 0

we have (ad(X)ad(Y))2 = 0 that is ad(X)ad(Y) is nilpotent. Since nilpotentmatrices have zero trace we see that 0 = Tr(ad(X)ad(Y)) = B(X, Y). Thisimplies X h, i.e. k h and thus the desired conclusion.

We then proceed to show that 1 implies 4. Suppose g is semisimple, andlet h g be any ideal. We consider its orthogonal complement w.r.t. B:h := {X g | Y h : B(X, Y) = 0}. This is again an ideal in g for ifX hand Y g, then for all Z g we have [Y, Z] h and hence

B([X, Y], Z) = B(X, [Y, Z]) = 0

saying that [X, Y] h. To show that we have a decomposition g = h h weneed to show that the ideal h h is zero. We can do this by showing that itis solvable for then semisimplicity forces it to be zero. By some remarks earlierin this proof, solvability ofh h would be a consequence of ad(h h) beingsolvable. To show that ad(h

h) is solvable we invoke the Cartan Criterion: For

X D(hh) hh and Y hh we have Tr(ad(X)ad(Y)) = B(X, Y) = 0since, in particular, X h and Y h. Thus, the Cartan Criterion renderssolvability ofad(hh) implying hh = 0. Hence, hh = 0 and g = hh.

After these preliminary remarks we proceed via induction over the dimensionof g. If dim g = 2, then g is simple, for any nontrivial ideal in g would haveto be 1-dimensional, hence abelian, and such do not exist. Assume now thatdim g = n and that the result is true for Lie algebras of dimension strictly lessthan n. Suppose that g1 is a minimal nonzero ideal in g then g1 is simple sincedim g1 2 and since any nontrivial ideal in g1 would be an ideal in g properlycontained in g1 contradicting minimality. Then we have g = g1 g1 with g1semisimple, for ifk is any abelian ideal in g1 then it is an abelian ideal in g andthese do not exist. Then by the induction hypothesis we have g1 = g2gn, asum of simple Lie algebras, hence g = g1g2 gn, a sum of simple algebras.Finally we show that 4 implies 2. So consider g := g1 gn and let h gbe an abelian ideal. It is not hard to verify that hi := h gi is an abelian idealin gi, thus hi = gi or hi = 0. As hi is abelian and gi is not, we can rule out thefirst possibility, i.e. hi = 0 and hence h = 0.

During the proof we saw that any ideal in a semisimple Lie algebra has acomplementary id

The Mathematics Any Physicist Should Know

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