Abstract - Reductive Grouptesterma/doc/korhonenthesis.pdf · 2018-10-02 · and Zalesski [TZ13]...

Abstract

Let G be a simple linear algebraic group over an algebraically closed �eld Kof characteristic p ≥ 0. In this thesis, we investigate closed connected reductivesubgroups X < G that contain a given distinguished unipotent element u of G.Our main result is the classi�cation of all such X that are maximal among theclosed connected subgroups of G.

When G is simple of exceptional type, the result is easily read from the tablescomputed by Lawther in [Law09]. Our focus is then on the case where G is simpleof classical type, say G = SL(V ), G = Sp(V ), or G = SO(V ). We begin byconsidering the maximal closed connected subgroups X of G which belong to oneof the families of the so-called geometric subgroups. Here the only di�cult caseis the one where X is the stabilizer of a tensor decomposition of V . For p = 2and X = Sp(V1)⊗Sp(V2), we solve the problem with explicit calculations; for theother tensor product subgroups we apply a result of Barry [Bar15].

After the geometric subgroups, the maximal closed connected subgroups thatremain are the X < G such that X is simple and V is an irreducible and tensorindecomposable X-module. The bulk of this thesis is concerned with this case.We determine all triples (X,u, ϕ) where X is a simple algebraic group, u ∈ X isa unipotent element, and ϕ : X → G is a rational irreducible representation suchthat ϕ(u) is a distinguished unipotent element of G. When p = 0, this was donein previous work by Liebeck, Seitz and Testerman [LST15].

In the �nal chapter of the thesis, we consider the more general problem of�nding all connected reductive subgroups X of G that contain a distinguishedunipotent element u of G. This leads us to consider connected reductive overgroupsX of u which are contained in some proper parabolic subgroup of G. Testermanand Zalesski [TZ13] have shown that when u is a regular unipotent element of G,no such X exists. We give several examples which show that their result does notgeneralize to distinguished unipotent elements. As an extension of the Testerman-Zalesski result, we show that except for two known examples which occur in thecase where (G, p) = (C2, 2), a connected reductive overgroup of a distinguishedunipotent element of order p cannot be contained in a proper parabolic subgroupof G.

Keywords: group theory, representation theory, algebraic groups, unipotent ele-ments, classical groups.

i

Résumé

Soit G un groupe algébrique simple sur un corps algébriquement closK de caracté-ristique p ≥ 0. Dans cette thèse, nous nous intéressons aux sous-groupes X < Gfermés réductifs connexes qui contiennent un élément unipotent distingué u de G.Notre principal résultat est une classi�cation de tels X qui sont maximaux parmiles sous-groupes fermés connexes de G.

Quand G est simple de type exceptionnel, le resultat peut facilement êtretrouvé dans les tableaux calculés par Lawther dans [Law09]. L'accent est doncporté sur le cas où G est simple de type classique, disons G = SL(V ), G = Sp(V ),ou G = SO(V ). Nous commençons par examiner les sous-groupes X maximauxparmi les sous-groupes fermés connexes deG, qui appartiennent à l'une des famillesdénommées sous-groupes géométriques. Ici le seul cas di�cile est celui où X est unstabilisateur d'une décomposition tensorielle de V . Pour p = 2 et X = Sp(V1) ⊗Sp(V2), nous résolvons le problème par des calculs explicites; pour les autres sous-groupes tensoriels nous appliquons un résultat de Barry [Bar15].

Aprés les sous-groupes géométriques, les sous-groupes X maximaux parmi lessous-groupes fermés connexes de G restants sont les X < G tels que X est simpleet V est un X-module irréductible et indécomposable en produit tensoriel. Lamajeure partie de cette thèse se rapporte à ce cas. Nous déterminons tous lestriplets (X,u, ϕ) où X est un groupe algébrique simple, u ∈ X est un élémentunipotent, et ϕ : X → G est une représentation rationnelle irréductible telle queϕ(u) est un élément unipotent distingué de G. Dans le cas p = 0, cela a été faitdans des travaux antérieurs de Liebeck, Seitz et Testerman [LST15].

Dans le dernier chapitre de cette thèse, nous considérons un probléme plusgénéral de trouver tous les sous-groupes X < G fermés réductifs connexes quicontiennent un élément unipotent distingué u de G. Cela nous amène à étudierles sous-groupes X qui contiennent u et qui sont contenus dans un sous-groupeparabolique propre de G. Testerman et Zalesski [TZ13] ont montré que si u estrégulier, il n'existe aucun tel X. Nous donnons plusieurs exemples qui démontrentque leur resultat ne se généralise pas aux éléments unipotents distingués. Commeune extension du résultat de Testerman-Zalesski, nous montrons, sauf dans deuxcas où (G, p) = (C2, 2), qu'un sous-groupe fermé réductif connexe qui contientun élément unipotent distingué d'ordre p ne peut pas être contenu dans un sous-groupe parabolique propre de G.

Mots-clefs: théorie des groupes, théorie des représentations, groupes algébriques,éléments unipotents, groupes classiques.

iii

Acknowledgements

The biggest thanks of all are due to my thesis director, Donna Testerman, whointroduced me to the world of algebraic groups and supported me throughoutthis work. She has been very generous with her time, and I thank her for manyfruitful discussions. She read patiently and carefully a countless number of pageswhich eventually became this thesis; I am grateful for her many useful comments,suggestions, and guidance.

I would like to thank Alexander Premet, Alexander Zalesski, and JacquesThévenaz for agreeing to be on the thesis jury, and Kathryn Hess for acting asthe president of the jury.

Special thanks are due to Frank Lübeck, who on request extended the tablesof weight multiplicities on his website, which allowed me to shorten the proof ofLemma 5.1.9 and Lemma 5.1.10. I am thankful to Alastair Litterick and AdamThomas for informing me of their ongoing work. I would also like to thank severalother mathematicians for small comments, discussions, clari�cations, etc. relatedto my thesis work: Michael Barry, Tim Burness, George McNinch, Ross Lawther,Gary Seitz, and Irina Suprunenko.

Many thanks also to past and present colleagues who have made working atEPFL pleasant, especially those within GR-TES: Mikaël Cavallin, Claude Marion,Jay Taylor, Nathan Scheinmann, and Neil Saunders. Finally, to my friends, family,and Fengda: warm thanks for all of your encouragement and support.

v

Contents

Abstract i

Résumé iii

Acknowledgements v

Contents vii

1 Introduction 1

1.1 Main problem and statements of results . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 A more general problem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.3 The structure of this thesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.4 Notation and terminology. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2 Unipotent elements in simple algebraic groups 17

2.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.2 Unipotent elements in SL(V ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.3 Unipotent elements in classical groups (p 6= 2) . . . . . . . . . . . . . . . . . . . . . 192.4 Unipotent elements in classical groups (p = 2) . . . . . . . . . . . . . . . . . . . . . 202.5 Explicit distinguished unipotent elements (p 6= 2) . . . . . . . . . . . . . . . . . 242.6 Unipotent elements and labeled Dynkin diagrams (p good). . . . . . . 252.7 Largest Jordan block size of a unipotent element . . . . . . . . . . . . . . . . . . 282.8 Representatives for unipotent classes in Chevalley groups . . . . . . . . 322.9 Computation of Jordan block sizes in irreducible representations 472.10 Action of triality on unipotent conjugacy classes of D4 . . . . . . . . . . . 48

3 Linear algebra 53

3.1 Basic lemmas about unipotent elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533.2 Jordan blocks in subspaces and quotients . . . . . . . . . . . . . . . . . . . . . . . . . . 543.3 Decomposition of tensor products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 583.4 Jordan decompositions in tensor squares (p 6= 2) . . . . . . . . . . . . . . . . . . 643.5 Symmetric and exterior squares with no repeated blocks (p 6= 2). 693.6 Decomposing tensor products of form modules . . . . . . . . . . . . . . . . . . . . 743.7 Proof of Theorem 1.1.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

4 Some representation theory 91

4.1 Weyl modules and tilting modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 914.2 Irreducible representations of SL2(K). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 924.3 Weyl group and weight orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 954.4 Invariant forms on G-modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

vii

viii Contents

4.5 Weight multiplicities in VG(λ) and LG(λ) . . . . . . . . . . . . . . . . . . . . . . . . . . 1004.6 Representations of SL2(K). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

5 Irreducible overgroups of distinguished unipotent elements 109

5.1 Reduction, unipotent elements of order > p (p 6= 2) . . . . . . . . . . . . . . . 1095.2 Reduction (p = 2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1305.3 Representation LG(ω1 + ωl) for G of type Al (p 6= 2) . . . . . . . . . . . . . 1335.4 Representation LG(ω1 + ωl) for G of type Al (p = 2) . . . . . . . . . . . . . 1335.5 Representation LG(ω2) for G of classical type (p 6= 2). . . . . . . . . . . . . 1355.6 Representation LG(ω2) for G of type Cl (p = 2) . . . . . . . . . . . . . . . . . . . 1415.7 Representation LG(2ω1) for G of classical type (p 6= 2) . . . . . . . . . . . 1445.8 Representation LG(ω3) for G of classical type (p 6= 2). . . . . . . . . . . . . 1505.9 Representation LG(3ω1) for G of classical type (p 6= 2, 3) . . . . . . . . . 1555.10 Spin representations (p 6= 2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1585.11 Spin representations (p = 2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1645.12 Fundamental irreducible representations (p 6= 2) . . . . . . . . . . . . . . . . . . . 1695.13 Unipotent elements of order p (p 6= 2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1765.14 Proof of Theorem 1.1.10 and Theorem 1.1.11 . . . . . . . . . . . . . . . . . . . . . . 193

6 Non-G-completely reducible overgroups 195

6.1 Preliminaries on G-complete reducibility . . . . . . . . . . . . . . . . . . . . . . . . . . . 1956.2 Unipotent elements of order p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1966.3 Unipotent elements of order > p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205

Appendices on simple groups of exceptional type:

A Orders of distinguished unipotent elements 211

B Actions in some irreducible representations (p = 2) 213

C Ordering of the positive roots 219

Bibliography 225

Curriculum vitae 232

Chapter 1

Introduction

This thesis concerns the subgroup structure and representation theory of simplelinear algebraic groups G over an algebraically closed �eld K. Two major openproblems for researchers working in this subject are

(1) Understanding the rational irreducible representations of G;

(2) Classifying all positive-dimensional reductive subgroups of G up to conjugacyin G.

When K is a �eld of characteristic zero, both problems have been well under-stood for a fairly long time. For (1), we have classical results such as the Weylcharacter formula and Freudenthal's formula for weight multiplicities. For (2), thework of Dynkin [Dyn52a] [Dyn52b] gives a full list of all maximal closed connectedsubgroups of G. Furthermore, it follows from a theorem of Mostow [Mos56] thatany maximal connected reductive subgroup of G is either a maximal closed con-nected subgroup, or a Levi subgroup of G. Consequently in characteristic zero, wehave a recursive way of classifying the connected reductive subgroups of G.

In the case where K is a �eld of positive characteristic, considerable progresshas been made towards both problems in the past decades. One key result is in thework of Seitz and Testerman, who considered maximal subgroups for G of classicaltype [Sei87] [Tes88] [Tes89]. Except for the so-called geometric subgroups (e.g.parabolic subgroups, stabilizers of tensor decompositions), any maximal closedconnected subgroup of G is a simple algebraic group X which acts irreducibly andtensor indecomposably on the natural module of G [Sei87, Theorem 3]. In theirwork, Seitz and Testerman proved that apart from a known list of exceptions1,any such X is a maximal closed connected subgroup of G. This is analogous toDynkin's result [Dyn52a] in characteristic zero, but in positive characteristic the�known list of exceptions� is much larger and di�erent techniques are needed forestablishing the list. Later for G simple of exceptional type, Liebeck and Seitz[LS04] found all maximal positive-dimensional closed subgroups of G.

Despite these and plenty of other advances made during the years, many pro-blems still remain open and a complete solution to the two problems (1) and(2) above seems to be unattainable for the foreseeable future. For example, theconsensus among researchers seems to be that even �nding the dimensions of the

1However, we should mention that there is a recently discovered mistake in [Sei87], whichcaused a family of examples to be missing from the �known list of exceptions� given in [Sei87,Table 1]. This mistake is corrected by Cavallin and Testerman in their preprint [CT17].

1

2 Chapter 1.

rational irreducible G-modules will remain an open problem for some time. Henceit is unrealistic to try to classify all possible reductive subgroups of all simplelinear algebraic groups.

In order to get a glimpse of the complete picture, one has to then focus on morespeci�c problems, such as the classi�cation of certain types of reductive subgroups.The main contribution of this thesis is one such classi�cation result: we give acomplete classi�cation of the maximal closed connected reductive subgroups of Gthat contain distinguished unipotent elements (see De�nition 1.1.1 and Problem1.1.2 below). This generalizes previous results found in [Sup95, Theorem 1.9],[SS97], and [LST15].

Roughly, one can think of the main topic of this thesis as a part of a generalapproach or philosophy for studying subgroup structure, where one classi�es thesubgroups containing elements that are �special� in some sense, see [Sax98] fora survey of some results. Besides being helpful for understanding the subgroupstructure of linear groups, solutions to such problems have been motivated byapplications in recognition algorithms for linear groups [NP92] and the InverseGalois Problem in number theory [GM14].

More speci�cally, the problem of classifying overgroups of speci�c classes ofunipotent elements is a topic which many researchers have considered, one exampleis the work of Liebeck and Seitz [LS94] on subgroups containing root elements.It is perhaps slightly surprising that in many cases a full classi�cation is feasible,and just the presence of a single unipotent element of speci�c type in a reductivesubgroup can tell us a lot about its structure. For example, all connected reductivesubgroups of SL(V ) containing a full Jordan block can be classi�ed [SS97] [TZ13].

1.1 Main problem and statements of results

We now introduce the main problem of this thesis and make some preliminaryobversations. Later in this section we state our main results. To begin, we needthe following de�nition.

De�nition 1.1.1. Let u ∈ G be a unipotent element. We say that u ∈ G isdistinguished if CG(u) does not contain a nontrivial torus2.

Basic properties of distinguished unipotent elements that are needed in thiswork will be given in Chapter 2. The main problem we are concerned with (andwhich we will solve in this thesis) is the following.

Problem 1.1.2 (Main problem). Let u ∈ G a distinguished unipotent element.Classify all maximal closed connected subgroups X of G that contain u, up toG-conjugacy.

Recall that as a corollary of the Borel-Tits theorem, any maximal closed con-nected subgroup X of G is either parabolic or reductive. Furthermore, it is wellknown that all parabolic subgroups of G contain a representative of every unipo-tent conjugacy class of G. Therefore Problem 1.1.2 is solved in the case where Xis parabolic.

2Equivalently, we can say that a unipotent element u ∈ G is distinguished if the identitycomponent of CG(u) is unipotent. Another equivalent formulation is that u is not contained inany proper Levi factor of G.

1.1. Main problem and statements of results 3

Suppose then that X is a reductive maximal closed connected subgroup of G.If G is simple of exceptional type, a complete list of such X is given by Liebeckand Seitz in [LS04]. Furthermore, for each such X, Lawther gives in [Law09] foreach unipotent element u ∈ X the conjugacy class of u in G. Thus when G issimple of exceptional type, the solution of Problem 1.1.2 is easily read from thetables in [Law09].

What our work will focus on then is the case where G is simple of classicaltype. It is easy to see (Lemma 2.1.2) that the isogeny type of G will not make anydi�erence in the solution Problem 1.1.2. Therefore we can assume that G = SL(V ),G = Sp(V ), or G = SO(V ), where V is a �nite-dimensional vector space over K.For what follows, set charK = p ≥ 0. The maximal closed connected subgroupsof G can be described with [Sei87, Theorem 3], as follows3:

Theorem 1.1.3. Let G = SL(V ) (dimV ≥ 2), G = Sp(V ) (dimV ≥ 2), orG = SO(V ) (dimV ≥ 5), where V is a �nite-dimensional vector space over K.Let X be a proper closed connected subgroup of G. Then X is maximal among theclosed connected subgroups of G if and only if one of the following holds:

(a) G = SL(V ), and X = Sp(V ) or X = SO(V ) (Exception: X = SO(V ) andp = 2).

(b) G = Sp(V ), p = 2, dimV > 2, and X = SO(V ).

(c) G = SO(V ), p = 2, dimV is even, and X is the stabilizer of a nonsingular1-space.

(d) X is a maximal parabolic subgroup of G.

(e) G = Sp(V ) or G = SO(V ), V = W ⊕ W⊥, where W is a non-degeneratesubspace of V , and X = stabG(W )◦.

(f) V = V1 ⊗ V2 and one of the following holds:

(i) G = SL(V ), and X = SL(V1) ⊗ SL(V2), where dimVi ≥ 2 (Exception:dimV1 = 2 = dimV2).

(ii) G = SO(V ) and X = Sp(V1)⊗ Sp(V2), where dimVi ≥ 2.

(iii) G = SO(V ), p 6= 2, and X = SO(V1)⊗ SO(V2), where dimVi ≥ 3.

(iv) G = Sp(V ), p 6= 2, and X = Sp(V1) ⊗ SO(V2), where dimV1 ≥ 2 anddimV2 ≥ 3.

(g) X is simple, V ↓ X is simple and tensor indecomposable, and there does notexist Y < G with (X,Y, V ) in [Sei87, Table 1] or [CT17, Theorem 1.2].

Our approach will be to solve Problem 1.1.2 for each of the subgroups (a)- (g) in Theorem 1.1.3. The proof will be given in Section 5.14. It will be seen

3Here we have modi�ed the statement of [Sei87, Theorem 3] slightly. The dimension restricti-ons such as dimV ≥ 5 for G = SO(V ) are set because we only consider G simple. Furthermore, asmentioned in footnote 1, it was recently discovered that there is family of examples missing from[Sei87, Table 1]; thus we have included a reference to the correction given by Cavallin and Tester-man. Also, we have corrected a small inaccuracy: the statement of [Sei87, Theorem 3 (f)] excludessome subgroups which are maximal, namely the tensor product subgroups Sp2(K) ⊗ Sp2m(K)in SO4m(K).

4 Chapter 1.

that in cases (a) - (e) of Theorem 1.1.3, the solution of Problem 1.1.2 will followfrom basic results in the literature. In case (f) of Theorem 1.1.3, the solution willfollow with the main result of Section 3.6 (Proposition 3.6.6) and a result of Barry[Bar15]4.

For the rest of this text, we will make the following assumption.

Assume that p > 0.

The solution of Problem 1.1.2 for the subgroups (a) - (f) in Theorem 1.1.3 isgiven by the following result. Below for a unipotent element u ∈ G, the notationVd denotes the indecomposable K[u]-module of dimension d, on which u acts witha single Jordan block. This and other pieces of notation used in this thesis aregiven in 1.4 below.

Theorem 1.1.4. Let G = SL(V ) (dimV ≥ 2), G = Sp(V ) (dimV ≥ 2), orG = SO(V ) (dimV ≥ 5), where V is a �nite-dimensional vector space over K.Fix a distinguished unipotent element u ∈ G. Let X be one of the maximal closedconnected subgroups of G given in (a) - (f) of Theorem 1.1.3. Then the cases whereX contains a G-conjugate of u are precisely the following:

(a) G = SL(V ), X = Sp(V ) (dimV even) or X = SO(V ) (p 6= 2 and dimV odd),and u is a regular unipotent element.

(b) G = Sp(V ), p = 2, dimV > 2, X = SO(V ), and the number of Jordan blocksof u is even.

(c) G = SO(V ), p = 2, dimV is even, X is the stabilizer of a nonsingular 1-space,and u has a Jordan block of size 2.

(d) X is a maximal parabolic subgroup.

(e) G = Sp(V ) or G = SO(V ), V = W ⊕ W⊥ where W is a non-degeneratesubspace of V , X = stabG(W )◦, and V ↓ K[u] ∼=

⊕ti=1 Vdi ⊕

⊕sj=1 Vd′j for

integers di, d′j ≥ 1 such that the following conditions hold:

• dimW =∑t

i=1 di.

• If p = 2 and G = SO(V ), then t ≡ 0 mod 2.

(f) V = V1 ⊗ V2 with dimV1 ≤ dimV2 and one of the following holds:

(i) G = SO(V ), p = 2, dimV1 = 2, X = Sp(V1)⊗Sp(V2), and the orthogonaldecomposition V ↓ K[u] (Proposition 2.4.4) is equal to V (2d1)2 + · · · +V (2dt)

2 for some 1 ≤ d1 < · · · < dt such that di is odd for all i.

(ii) G = Sp(V ) or G = SO(V ), p 6= 2, X is as in Theorem 1.1.3 (ii), (iii) or(iv); and for m = dimV1 and n = dimV2 one of the following conditionshold:

• The pair (m,n) is contained in the set S of De�nition 3.3.12, andV ↓ K[u] ∼=

⊕mi=1 Vm+n−2i+1. (In this case V ↓ K[u] ∼= Vm ⊗ Vn).

4More precisely, the description of the set S that we use in Theorem 1.1.4 (f) is taken from[Bar15, Theorem 2, Theorem 3].

• There exist integers 1 ≤ n1 < n2 < · · · < nt such that∑t

i=1 ni = n,ni ≡ ni′ mod 2 for all 1 ≤ i, i′ ≤ t, ni−ni−1 ≥ 2m for all 2 ≤ i ≤ t,the pair (m,ni) is contained in the set S of De�nition 3.3.12 forall 1 ≤ i ≤ t, and

V ↓ K[u] ∼=min(m,n1)⊕

j=1

Vn1+m−2j+1 ⊕t⊕i=2

m⊕j=1

Vni+m−2j+1.

(In this case V ↓ K[u] ∼= Vm ⊗ (Vn1 ⊕ Vn2 ⊕ · · · ⊕ Vnt)).

Remark 1.1.5. We have assumed that p > 0, but it is easy to see that essentiallythe same result as Theorem 1.1.4 applies when p = 0. Speci�cally, when p = 0,the cases (b), (c) and (f)(i) of Theorem 1.1.4 no longer apply, and in case (f)(ii)the conditions (m,n) ∈ S and (m,ni) ∈ S should be removed.

What remains then is the case of the irreducible subgroups in Theorem 1.1.3(g). The bulk of this thesis is devoted to solving Problem 1.1.2 in this case. To dothis, it will be enough to solve the following problem, which has been solved incharacteristic zero by Liebeck, Seitz and Testerman [LST15].

Problem 1.1.6. Let I (V ) be a connected simple classical group over K (thatis, I (V ) = SL(V ), I (V ) = Sp(V ) or I (V ) = SO(V )). Determine all closedconnected simple subgroups G < I (V ) and unipotent elements u ∈ G such thatthe following hold:

(i) The group G acts on V irreducibly and tensor indecomposably,

(ii) The element u ∈ G is a distinguished unipotent element of I (V ).

This is essentially a problem about the representation theory of simple alge-braic groups. In what follows, we will say that a G-module V (or representationϕ : G → GL(V )) is symplectic if V admits a non-degenerate G-invariant alter-nating bilinear form, and orthogonal if V admits a non-degenerate G-invariantquadratic form. Note that if V is irreducible, then any such form is unique up toa scalar multiple (Lemma 4.4.3 and Lemma 4.4.5). Thus if a rational irreduciblerepresentation ϕ : G → GL(V ) is symplectic or orthogonal, then we can writerespectively ϕ(G) < Sp(V ) or ϕ(G) < SO(V ) without ambiguity.

We will now make the following de�nition to present Problem 1.1.6 in a di�e-rent way.

De�nition 1.1.7. Let u ∈ G be a unipotent element and let ϕ : G → GL(V )be a rational representation. For the G-module V , we say that u acts on V as adistinguished unipotent element, if one of the following holds.

(i) ϕ(u) is a distinguished unipotent element in SL(V ).

(ii) V is symplectic, and ϕ(u) is a distinguished unipotent element in Sp(V ),where Sp(V ) is the stabilizer of a non-degenerate G-invariant alternatingbilinear form.

(iii) V is orthogonal, and ϕ(u) is a distinguished unipotent element in SO(V ),where SO(V ) is the identity component of the stabilizer of a non-degenerateG-invariant quadratic form.

6 Chapter 1.

The following lemma is immediate from Lemma 2.1.2 (ii).

Lemma 1.1.8. Let u ∈ G be a unipotent element. If u acts on a rational G-moduleas a distinguished unipotent element, then u is distinguished in G.

With Lemma 1.1.8, Problem 1.1.6 becomes equivalent to the following problem.

Problem 1.1.9. Let G be connected and simple and let u ∈ G be a distinguishedunipotent element. Find all irreducible, tensor-indecomposable G-modules V suchthat u acts on V as a distinguished unipotent element. Furthermore, for each suchV , determine whether V is symplectic, orthogonal, both, or neither.

Note that by Steinberg's tensor product theorem, it is enough to solve Problem1.1.9 in the case where V is a p-restricted irreducible G-module. The �rst partof the solution to Problem 1.1.9 will follow from the following two theorems. SeeDe�nition 2.7.2 for the de�nition of mu(λ).

Theorem 1.1.10. Assume that p 6= 2. Let G be simple, and let λ be a non-zerop-restricted dominant weight. A unipotent element u ∈ G acts on the irreducibleG-module LG(λ) of highest weight λ as a distinguished unipotent element if andonly if one of the following holds:

(i) (u of order > p) G, λ, p and u occur in Table 1.1 or Table 1.2,

(ii) (u of order p) G, λ and u occur in Table 1.1 or Table 1.2 and p > mu(λ),

(iii) G = G2, λ = ω1 + 2ω2, u is a regular unipotent element, and p = 5.

Theorem 1.1.11. Assume that p = 2. Let G be simple, and let λ be a non-zero2-restricted dominant weight. A unipotent element u ∈ G acts on the irreducibleG-module LG(λ) of highest weight λ as a distinguished unipotent element if andonly if G, λ and u occur in Table 1.3.

For the second part of Problem 1.1.6, we have to determine whether the LG(λ)given in Theorem 1.1.10 and Theorem 1.1.11 are symplectic, orthogonal, both, orneither. In the case where p 6= 2 this is easily done with Lemma 4.4.3 and Table4.1. Consider then p = 2. If λ 6= −w0λ, then LG(λ) is not self-dual and thusis not symplectic nor orthogonal. Suppose then that λ = −w0λ. Then LG(λ) isself-dual and is symplectic by a result observed in [Fon74] (see Lemma 4.4.5). Thequestion of whether LG(λ) is orthogonal is more subtle and in general an openproblem. However, for the λ occurring in Table 1.3, the orthogonality of LG(λ)can be decided from the results in [Kor17]; we have included this information inTable 1.3 below5.

As a �rst step to the solution of Problem 1.1.6, we note the following. It is wellknown that in SL(V ) there is only one class of distinguished unipotent elements,the class of a full Jordan block (Lemma 2.2.2). The irreducible representationsϕ : G→ GL(V ) where ϕ(u) is a full Jordan block have been known for some timealready.

5For many of the entries, this could also done using older results from the literature, see forexample [GW95, Theorem 3.4] and [Gow97, Corollary 4.3].


Theorem 1.1.12 ([Sup95, Theorem 1.9]). Let G be connected simple and letλ ∈ X(T )+ be p-restricted. A unipotent element u ∈ G acts on LG(λ) with asingle Jordan block if and only if u is a regular unipotent element in G and one ofthe following holds:

(i) G = A1.

(ii) G = Al, G = Bl or G = Cl, and λ = ω1.

(iii) G = Al and λ = ωl.

(iv) G = C2 and λ = ω2.

(v) G = G2 and λ = ω1.

By Theorem 1.1.12, we can reduce the proof of Theorem 1.1.10 and Theorem1.1.11 to the case of irreducible representations ϕ : G→ GL(V ) where V is sym-plectic or orthogonal. Equivalently, we reduce to the case where V is self-dual as aG-module (Lemma 4.4.3). When p 6= 2, by the description of distinguished unipo-tent elements in classical groups (Proposition 2.3.4), the problem is then reducedto �nding all distinguished unipotent elements u ∈ G and self-dual irreducibleG-modules V such that ϕ(u) has all Jordan block sizes distinct, and either allblock sizes are even or all block sizes are odd. When p = 2, Jordan block sizesdo not determine whether a unipotent element is distinguished or not, but oneknows that a distinguished unipotent element has all Jordan block sizes even withmultiplicity ≤ 2 (Proposition 2.4.4 (ii)). Our strategy for Theorem 1.1.11 is thento determine the cases where ϕ(u) has all Jordan block sizes even with multiplicity≤ 2, and then determine among these when ϕ(u) is distinguished.

8 Chapter 1.

G λ Class of u u of order > p mu(λ)Al ω1 regular any 2lAl, l ≥ 2 ω1 + ωl regular Proposition 5.3.2 (ii) 2lA1 0 ≤ c ≤ p− 1 regular none cA5 ω3 regular p = 5 6

Bl ω1 any anyB3 101 regular p = 5 12B3 002 regular p = 5 12B3 300 regular none 18Bl, l ≥ 3 2ω1 regular Proposition 5.7.7 (b) (ii) 4lBl, l ≥ 3 ω2 regular Proposition 5.5.5 (b) (ii)-(iii) 4l − 2

Bl, 3 ≤ l ≤ 8 ωn regular Proposition 5.10.1 l(l+1)2

Cl ω1 any anyC2 b0, 1 ≤ b ≤ 5 regular p = 3 3bC2 0b, 1 ≤ b ≤ 5 regular p = 3 4bC2 11 regular p = 3 7C2 12 regular none 11C2 21 regular none 10C3 300 regular none 15C3 ω3 regular p = 3, 5 9C4 ω3 regular none 15C4 ω4 regular none 16C5 ω3 regular none 21C5 ω5 regular none 25Cl, l ≥ 3 2ω1 regular Proposition 5.7.11 (b) (ii) 4l − 2Cl, l ≥ 3 ω2 regular Proposition 5.5.10 (b) (ii)-(iv) 4l − 4

Dl, l ≥ 4 ω1 anyD6 ω6 regular p = 3, 5, 7 15D6 ω6 [3, 9] p = 7 11D8 ω8 regular p = 11 28Dl, l ≥ 4 even 2ω1 regular Proposition 5.7.8 (b) (ii) 4l − 4Dl, l ≥ 4 odd ω2 regular Proposition 5.5.6 (b) (ii)-(iii) 4l − 6

Table 1.1: charK 6= 2: Distinguished actions for irreducible representations ofsimple G of classical type.


G λ Class of u u of order > p mu(λ)G2 10 regular p = 3, 5 6G2 01 regular p = 3, 5 10G2 11 regular none 16G2 20 regular p = 5 12G2 02 regular none 20G2 30 regular none 18

F4 ω1 regular p = 5, 7, 11 22F4 ω4 regular p = 3, 5, 7, 11 16F4 ω4 F4(a1) p = 5, 7 10

E6 ω2 regular p = 5, 7, 11 22

E7 ω1 regular p = 11 34E7 ω7 regular p = 3, 5, 11, 13, 17 27E7 ω7 E7(a1) p = 5, 13 21E7 ω7 E7(a2) p = 5, 7, 11 17

E8 ω8 regular p = 11, 13, 17, 23 58E8 ω8 E8(a1) p = 11 46

Table 1.2: charK 6= 2: Distinguished actions for irreducible representations ofsimple G of exceptional type.

10 Chapter 1.

G λ unipotent class L(λ) ↓ K[u] orthogonal?

Al, l ≥ 1 ω1 regular [l + 1] not self-dualAl, l ≥ 1 ωl regular [l + 1] not self-dualA2 ω1 + ω2 regular [42] yesA3 ω2 regular [2, 4] yesA4 ω1 + ω4 regular [42, 82] yes

Bl, Cl, l ≥ 2 ω1 any distinguished noB2, C2 ω2 regular [4] noB2, C2 ω2 221 [22] noB3, C3 ω2 regular [6, 8] yesB3, C3 ω3 regular [2, 6] yesB4, C4 ω4 regular [82] yesB4, C4 ω4 21, 61 [22, 62] yesB5, C5 ω2 regular [6, 8, 14, 16] yesB5, C5 ω5 regular [2, 6, 10, 14] yesB6, C6 ω2 21, 101 [6, 8, 102, 14, 16] noB6, C6 ω6 21, 101 [22, 62, 102, 142] yes

Dl, l ≥ 4 ω1 any distinguished yesD4 ω3 any distinguished yesD4 ω4 any distinguished yesD6 ω2 regular [6, 8, 102, 14, 16] noD6 ω5 regular [2, 6, 10, 14] yesD6 ω6 regular [2, 6, 10, 14] yes

G2 ω1 regular [6] noG2 ω2 regular [6, 8] yesF4 ω4 regular [10, 16] yesF4 ω1 regular [10, 16] yesE7 ω7 regular [2, 10, 18, 26] yesE7 ω1 regular [8, 10, 16, 18, 22, 26, 32] no

Table 1.3: charK = 2: Distinguished actions for irreducible representations ofsimple G.

1.2. A more general problem 11

1.2 A more general problem

The main problem of this thesis (Problem 1.1.2) is a special case of the followingmore general problem, which remains open.

Problem 1.2.1. Let u ∈ G a distinguished unipotent element. Classify all con-nected reductive subgroups X of G that contain u, up to G-conjugacy.

In the �nal chapter of this thesis, we will present some partial results onProblem 1.2.1. In order to present these results, we will next describe the mostnatural strategy for solving Problem 1.2.1. A �rst step for solving Problem 1.2.1would be to �nd the maximal connected reductive subgroups X of G that contain adistinguished unipotent element u. Then since u must also be distinguished in X,the strategy is to do the same thing with X and proceed in this way inductivelydown the subgroup lattice.

However, in the induction step one encounters a non-trivial problem whichneeds to be dealt with. Let Y < G be a positive-dimensional connected reductivesubgroup containing u. Then Y is contained in some maximal closed connectedsubgroup M of G. As a corollary of the Borel-Tits theorem, we know that Mis parabolic or reductive. If M is connected reductive, then by our solution toProblem 1.1.2, we know the possibilities for M and could apply induction.

But if M is parabolic, it is not obvious what one should do here. In thiscase Y cannot be contained in a Levi factor of M , because u is a distinguishedunipotent element. In the terminology due to Serre [Ser05], this means that Y isa non-G-completely reducible (non-G-cr) subgroup (De�nition 6.1.1). Essentially,the obstacle arising here is that while we have a good understanding of whatreductive maximal connected subgroups look like (by [LS04] and [Sei87]), we knowlittle about subgroups which are maximal connected reductive but not maximalconnected.

Fix u and G in Problem 1.2.1. With the remarks in the previous paragraphsin mind, a natural strategy for solving Problem 1.2.1 proceeds with the followingsteps.

Step I: Find all reductive maximal closed connected subgroups of G containingu.

Step II: Determine if a closed connected reductive subgroup of G containing ucan be contained in a proper parabolic subgroup of G, and if so, �ndthem.

Step III: Use the results in steps I and II to �nd all closed connected reductivesubgroups of G containing u.

The main result of this thesis is the completion of Step I, i.e. the solution ofProblem 1.1.2. It is easy to see (Lemma 6.1.2) that Step II is asking to classify, upto G-conjugacy, all non-G-cr connected reductive subgroups X of G that containu. In previous literature, a result due to Testerman and Zalesski states that nosuch X exists if u is a regular unipotent element [TZ13]. One might hope thatthis would generalize to distinguished unipotent elements, but it turns out thatthis is not the case. Several examples are given in Chapter 6, see e.g. Example6.3.3. However, the general impression is that such examples are quite rare. Thisis seen in the main result of Chapter 6 (Theorem 6.2.12), which gives a complete

12 Chapter 1.

classi�cation in the case where u has order p; the result is that there are only twoexamples, both of which occur in the case (G, p) = (C2, 2).

1.3 The structure of this thesis

The structure of the text is as follows. This introductory chapter will end withSection 1.4 below, where we �x (mostly standard) notation and terminology thatwill be used throughout the text. Chapter 2 contains basic facts about conjugacyclasses of unipotent elements in simple algebraic groups. Most of the results inChapter 2 are well known, proofs are given for some results for which we wereunable to �nd a reference.

Chapter 3 is concerned with the properties of unipotent linear maps on a�nite-dimensional vector space. Most of the chapter concerns the Jordan decom-position of the tensor product, exterior square and symmetric square of unipotentlinear maps. In characteristic zero this would be a trivial topic, but in our setting(p > 0) this becomes more di�cult. We describe results from the literature whichallow one to determine these decompositions, and then apply them to establishvarious facts which will be needed in the sequel. One useful result is Proposition3.5.3, which determines when the symmetric square or the exterior square of aunipotent Jordan block has no repeated block sizes in its Jordan decomposition.Also important for later use will be lemmas 3.4.10, 3.4.11, 3.4.12, and 3.4.13, whichdescribe the smallest Jordan block sizes occurring in the symmetric and exteriorsquare of a unipotent matrix. In Section 3.6, we describe the conjugacy class of aunipotent element u1 ⊗ u2 ∈ Sp(V1)⊗ Sp(V2) in Sp(V1 ⊗ V2) in some small caseswhen p = 2. Chapter 3 �nishes with Section 3.7, where we will prove Theorem1.1.4, one of our main results.

InChapter 4, we consider the representation theory of simple algebraic groupsthat will be needed in the text. Most important result for our main problem isProposition 4.6.8, which is key in our proof of Theorem 1.1.10 in the case where uhas order p. Proposition 4.6.8 gives a classi�cation of SL2(K)-modules on whicha non-identity unipotent element u of SL2(K) acts with ≤ 1 Jordan block of sizep. An important consequence of this will be Proposition 4.6.10, which shows thata self-dual SL2(K)-module is semisimple if u acts on it with ≤ 1 Jordan block ofsize p.

Chapter 5 is devoted to the proof of two of our main results, Theorem 1.1.4and Theorem 1.1.10. Here we begin with Section 5.1, which contains the mainreduction for Theorem 1.1.10 in the case where u has order > p. Essentially, weshow that if u ∈ G has order > p, then it is enough to prove Theorem 1.1.10 fora small number of λ. Section 5.2 contains a similar reduction for Theorem 1.1.11,the main result when p = 2. Sections 5.3-5.12 contain the proof of Theorem 1.1.10and Theorem 1.1.11 for various families of λ that we need to consider. Most of thework is in Section 5.5 and Section 5.7, where we establish Theorem 1.1.10 in thecase where G is of classical type and λ = ω2 or λ = 2ω1. The results established inChapter 3 are key in the proofs of our results. In Section 5.13, we prove Theorem1.1.10 in the case where u has order p. The proof is based on Proposition 4.6.10and Theorem 2.6.8, which allow us to apply the methods of Liebeck, Seitz andTesterman in [LST15]. The proofs of Theorem 1.1.4 and Theorem 1.1.10 are givenin Section 5.14.

In Chapter 6, the �nal chapter of the thesis, we consider connected reductive

1.4. Notation and terminology 13

subgroups that are non-G-cr and contain a distinguished unipotent element u ofG. The main result is Theorem 6.2.12, which is gives a complete classi�cation inthe case where u has order p. In the case where u has order > p, we present someexamples and partial results.

1.4 Notation and terminology

Throughout the whole text, let K be an algebraically closed �eld of characteristicp > 0. Unless otherwise mentioned, G denotes a connected semisimple algebraicgroup over K, and V will be a �nite-dimensional vector space over K. If G issimple, then we say that p is good for G in the following cases:

• G is simple of type Al; for all p > 0,

• G is simple of type Bl, Cl, or Dl; for p > 2,

• G is simple of type G2, F4, E6, or E7; for p > 3,

• G is simple of type E8, for p > 5.

Otherwise we say that p is bad for G. We say that p is very good for G if p isgood for G and in addition p - l+1 if G is of type Al. If G is connected semisimple,then we say that p is good for G if it is good for each simple factor of G, otherwisewe say that p is bad for G.

All groups that we consider are linear algebraic groups over K, and by asubgroup we always mean a closed subgroup. For an algebraic group X, we denoteits identity component by X◦. For an element u ∈ X of �nite order, we denotethe order of u by |u|.

All modules and representations that we consider will be �nite-dimensionaland rational. Throughout we will view G as its group of rational points over K,and often G will studied either as a Chevalley group constructed with the usualChevalley construction (see e.g. [Ste68]), or as a classical group in its naturalrepresentation (i.e. G = SL(V ), G = Sp(V ) or G = SO(V )). We will occasionallydenote G by its type, so notation such as G = Cl means that G is a simplealgebraic group of type Cl.

Let l = rankG. For G we will use the following notation, similarly to thenotation used in [Jan03].

• T : A maximal torus of G.

• X(T ): Character group of T .

• Φ: Root system of G with respect to T . These are elements of X(T ).

• ∆: A system of simple roots for Φ.

• Φ+: The set of positive roots in Φ, with respect to ∆.

• δ: The half-sum of positive roots, i.e. δ = 12

∑α∈Φ+ α.

• �: The usual partial ordering on X(T ), i.e. µ � λ if and only if λ = µ, orλ− µ is a sum of positive roots.

• W : Weyl group of G with respect to T .

14 Chapter 1.

• (−,−): W -invariant inner product on the vector space X(T )⊗Z R.

• α∨ (for α ∈ Φ): the dual root α∨ = 2α(α,α) .

• Φ∨: The dual root system {α∨ : α ∈ Φ}.

• 〈−,−〉: de�ned as 〈λ, µ〉 = 2(λ,µ)(µ,µ) for all λ, µ ∈ X(T ).

• X(T )+: The set of dominant weights for G, with respect to ∆. This is theset of λ ∈ X(T ) which satisfy 〈λ, α〉 ≥ 0 for all α ∈ ∆.

• chV : Character of a G-module V (an element of Z[X(T )]).

• mV (µ): Multiplicity of the weight µ ∈ X(T ) in the G-module V .

• ω1, ω2, . . . , ωl ∈ X(T )+ fundamental dominant weights, with respect to ∆.We use the standard Bourbaki labeling of the simple roots, as given in[Hum72, 11.4, pg. 58].

• mu(λ) (u ∈ G unipotent, λ ∈ X(T )+): See De�nition 2.7.2.

• L(λ), LG(λ): Irreducible G-module with highest weight λ ∈ X(T )+.

• V (λ), VG(λ): Weyl module for G with highest weight λ ∈ X(T )+.

• T (λ), TG(λ): Tilting module for G with highest weight λ ∈ X(T )+.

• radV (λ): The unique maximal submodule of V (λ).

• gG: The G-conjugacy class of g ∈ G.

• L (G): Lie algebra of G.

We will use the notation LG(λ) instead of L(λ) whenever we want to makeclear that LG(λ) is a module for G (similarly VG(λ) and TG(λ)).

For a dominant weight λ ∈ X(T )+, we can write λ =∑l

i=1miωi where mi ∈Z≥0. We say that λ is p-restricted if 0 ≤ mi ≤ p− 1 for all 1 ≤ i ≤ l.

Let u ∈ GL(V ) be unipotent. Suppose that u has Jordan block sizes 1 ≤ d1 <d2 < · · · < dt and let ni ≥ 1 be the number of blocks of size di occurring inthe Jordan form on u. We will often use the notation [dn1

1 , . . . , dntt ] to denote theJordan form of u. Here if ni = 1, we will write di instead of d1

i . We will often saythat u acts on V with Jordan blocks [dn1

1 , . . . , dntt ]. We also say that u acts on Vwith no repeated blocks if ni = 1 for all 1 ≤ i ≤ t. If t = 1 and n1 = 1, we say thatu acts on V with a single Jordan block or as a full Jordan block.

It will often be convenient for us to describe the action of u on a representationin terms of K[u]-modules. Suppose that u has order q = pt. Then there existexactly q indecomposable K[u]-modules V1, V2, . . ., Vq. Here dimVi = i and uacts on Vi as a full Jordan block. We will denote by r · Vn the direct sum Vn ⊕· · · ⊕ Vn, where Vn occurs r times. If V is a K[u]-module with decompositionV = n1Vd1 ⊕ · · · ⊕ ntVdt , where 1 ≤ d1 < d2 < · · · < dt and ni ≥ 1, we willsometimes write this as V = [dn1

1 , . . . , dntt ]. We will say that V has no repeatedblocks if ni = 1 for all 1 ≤ i ≤ t.

A bilinear form b is non-degenerate, if its radical rad b = {v ∈ V : b(v, w) =0 for all w ∈ V } is zero. For a quadratic form Q : V → K on a vector space V , its

1.4. Notation and terminology 15

polarization is the bilinear form bQ de�ned by bQ(v, w) = Q(v+w)−Q(v)−Q(w)for all v, w ∈ V . We say that Q is non-degenerate, if its radical radQ = {v ∈rad bQ : Q(v) = 0} is zero.

For a representation V of G, a bilinear form (−,−) is said to be G-invariantif (gv, gw) = (v, w) for all g ∈ G and v, w ∈ V . We say that V is symplectic ifit has a non-degenerate G-invariant alternating bilinear form, and we say V isorthogonal if it has a non-degenerate G-invariant quadratic form.

For a vector space V over K, we will denote by Sp(V ) the group of linearmaps stabilizing a non-degenerate alternating bilinear form on V , and by O(V )the group of linear maps stabilizing a non-degenerate quadratic form on V . Theidentity component of O(V ) will be denoted by SO(V ). Note that if p 6= 2, thenSO(V ) = O(V )∩SL(V ). If p = 2, then SO(V ) is the kernel of the Dickson invarianton O(V ).

Given a morphism φ : G′ → G of algebraic groups, we can twist representationsof G with φ. That is, if ρ : G → GL(V ) is a representation of G, then ρφ is arepresentation of G′. We denote the corresponding G′-module by V φ. For theFrobenius map F : G→ G, we will denote V Fn = V [n].

If a representation V of G has composition series V = V1 ⊃ V2 ⊃ · · · ⊃ Vt ⊃Vt+1 = 0 with composition factors Wi

∼= Vi/Vi+1, we will occasionally denote thisby V = W1/W2/ · · · /Wt.

We denote by νp the p-adic valuation on the integers, so for a nonzero integern ∈ Z we de�ne νp(n) to be the largest integer k ≥ 0 such that pk divides n.

Chapter 2

Unipotent elements in simplealgebraic groups

2.1 Preliminaries

The purpose of this chapter is to list various basic facts about unipotent elementsin G that will be needed in the sequel. As in the main problem of this thesis,our main interest is in distinguished unipotent elements (De�nition 1.1.1). Oneparticularly important class of distinguished unipotent elements is the class ofregular unipotent elements, which will come up often.

De�nition 2.1.1. Let u ∈ G be a unipotent element. We say that u ∈ G isregular, if dimCG(u) = rankG.

It is well known that a regular unipotent element of G is distinguished [SS70,1.14.(a)]. Furthermore, we know that there exists a unique conjugacy class ofregular unipotent elements in G, and this conjugacy class is dense in the varietyof all unipotent elements of G [SS70, Theorem 1.8]. In the sections that follow, wewill describe the regular and distinguished unipotent conjugacy classes in groupsof classical type.

We will also need the following basic lemma referred to in the introduction,which will allow us to establish our results without being concerned with theisogeny type of G.

Lemma 2.1.2. Let f : G1 → G2 be an isogeny between two simple algebraicgroups G1 and G2. Then

(i) The map f restricts to a bijection between the unipotent varieties of G1 andG2, and f induces a bijection between the unipotent conjugacy classes of G1

and G2.

(ii) Let u ∈ G1 be a unipotent element. Then u is distinguished in G1 if and onlyif f(u) is distinguished in G2.

(iii) The map f induces a bijection between the set of closed connected reductivesubgroups of G1 and G2 via X 7→ f(X).

Proof. Claim (i) follows as in [Car85, Proposition 5.1.1]. Claim (ii) also followswith a standard argument that we give here for completeness. Suppose �rst that

17

18 Chapter 2.

u is centralized by a non-trivial torus S < G1. Then since ker f is �nite, it followsthat f(S) is a non-trivial torus centralizing f(u).

For the other direction of (ii), suppose that f(u) is centralized by a non-trivialtorus S′ < G2. Since f is surjective, we have S′ = f(S) for S = f−1(S′)◦. SinceS is connected and consists of semisimple elements, it is a torus. We proceed toshow that S centralizes u. To this end, note �rst that since S′ centralizes f(u),it follows that [s, u] ∈ ker f for all s ∈ S. On the other hand, the set [S, u] of allcommutators [s, u], s ∈ S, is irreducible, being the image of S under the rationalmap de�ned by s 7→ [s, u]. Since the kernel ker f is �nite, it follows that [S, u] = 1,so S is a non-trivial torus centralizing u.

For (iii), note �rst that images of reductive groups are reductive. Thus wehave a map X 7→ f(X) between the closed connected reductive subgroups of G1

and G2. We show that this map is bijective. If X,Y < G1 are closed connectedreductive subgroups and f(X) = f(Y ), then X ker f = Y ker f . Since ker f is�nite, we have (X ker f)◦ = X and (Y ker f)◦ = Y , which gives X = Y . Thereforethe map in question is injective. To show surjectivity, suppose that Z < G2 isa closed connected reductive subgroup. Since f is surjective, we have f(X) = Zfor X = f−1(Z)◦. What remains is to see that X is reductive, and this followseasily from the fact that ker f is �nite. Indeed, if U is a non-trivial closed connectedunipotent which is normal inX, then f(U) would be a non-trivial closed connectedunipotent subgroup which is normal in Z.

To describe unipotent conjugacy classes for a simple algebraic group with givenroot system Φ, it follows from Lemma 2.1.2 (i) that it is enough to do this forsome �xed isogeny type. For simple groups of classical type, in the sections thatfollow we will give a description of unipotent conjugacy classes in SL(V ), Sp(V ),and SO(V ).

2.2 Unipotent elements in SL(V )

Because we are working over an algebraically closed �eld, the description of theunipotent conjugacy classes in SL(V ) is simply a matter of linear algebra, as shownby the following lemma which is well known.

Lemma 2.2.1. Let x, y ∈ GL(V ). If x and y are conjugate in GL(V ), then x andy are conjugate in SL(V ).

Proof. Suppose that y = g−1xg for some g ∈ GL(V ). Then for any scalar c ∈ K,we have y = h−1xh for h = cg. Since K is algebraically closed, we can choose ascalar c such that det(h) = 1.

Therefore in SL(V ), the conjugacy class of a unipotent element is determinedby its Jordan form. In other words, unipotent conjugacy classes in SL(V ) can belabeled by partitions of dimV .

In SL(V ) there is only one conjugacy class of distinguished unipotent elements,and this is the class of regular unipotent elements. We record this in the followinglemma. For a proof, see for example [LS12, Proposition 3.5].

Lemma 2.2.2. Let u ∈ SL(V ) be unipotent. Then the following statements areequivalent.

(i) u is distinguished in SL(V ).

2.3. Unipotent elements in classical groups (p 6= 2) 19

(ii) u is regular in SL(V ).

(iii) u has only one Jordan block.

2.3 Unipotent elements in classical groups (p 6= 2)

Assume that p 6= 2.

The correspondence between partitions and unipotent conjugacy classes holdsalso in other classical groups in odd characteristic. The following propositions arewell known. See for example Theorem 3.1.(i), Theorem 3.1.(ii) and Proposition3.5 in [LS12].

Proposition 2.3.1. Suppose that G = Sp(V ) or G = O(V ). Then two unipotentelements of G are conjugate in G if and only if they have the same Jordan blockstructure.

Proposition 2.3.2. Let u ∈ SL(V ) be unipotent. Suppose that u has Jordan blocksizes 1 ≤ d1 < d2 < · · · < dt and let ni ≥ 1 be the number of blocks of size dioccurring in the Jordan form on u. Then

(i) A conjugate of u lies in Sp(V ) if and only if ni is even for all i such that diis odd.

(ii) A conjugate of u lies in SO(V ) if and only if ni is even for all i such that diis even.

(iii) Suppose that u ∈ SO(V ). Then the conjugacy class of uO(V ) consists of asingle SO(V )-class, unless di is even for all i, in which case uO(V ) splits intotwo SO(V )-classes.

Proposition 2.3.3. In a simple classical group G with natural module V , theJordan block structure of a regular unipotent element u ∈ G is as follows.

(i) G = SL(V ): single Jordan block of size dimV .

(ii) G = Sp(V ): single Jordan block of size dimV .

(iii) G = SO(V ), dimV odd: single Jordan block of size dimV .

(iv) G = SO(V ), dimV even: [dimV − 1, 1].

Proposition 2.3.4. In a simple classical group G with natural module V , theJordan block structure of a distinguished unipotent element u ∈ G is as follows.

(i) G = SL(V ): single Jordan block of size dimV .

(ii) G = Sp(V ): [d1, d2, . . . , dt] where di are distinct even integers.

(iii) G = SO(V ): [d1, d2, . . . , dt] where di are distinct odd integers.

20 Chapter 2.

2.4 Unipotent elements in classical groups (p = 2)

Assume that p = 2.

We will describe the classi�cation of unipotent conjugacy classes in the classi-cal groups Sp(V ), O(V ) and SO(V ), where dimV is even. For dimV odd, thereexists an isogeny SO(V ) → Sp(V/V ⊥), so with Lemma 2.1.2 we will also get aclassi�cation in this case.

For the rest of this section, let G = Sp(V ), where dimV is even. Let (−,−)be the G-invariant alternating bilinear form on V and let Q : V → K be a non-degenerate quadratic form which has polarization (−,−). We denote by O(V ) thegroup of linear maps on V that stabilize the quadratic form Q. Now O(V ) < G,and the following result shows that a classi�cation of the unipotent conjugacyclasses of G gives a corresponding classi�cation for O(V ).

Theorem 2.4.1. (i) The group O(V ) intersects every conjugacy class of G.

(ii) Let g, g′ ∈ O(V ). Then g and g′ are conjugate in O(V ) if and only if theyare conjugate in G.

(iii) Let u ∈ SO(V ) be a unipotent element. Then u is distinguished in G if andonly if u is distinguished in SO(V ).

Proof. Statements (i) and (ii) follow from [Dye79, Theorem 4, Theorem 5]. State-ment (iii) follows from [LS12, Lemma 6.15 (ii)].

We will be using the classi�cation of unipotent conjugacy classes of G as givenby Hesselink in [Hes79]. Roughly, the result of Hesselink states that the conjugacyclass of a unipotent element of G is determined by its Jordan block sizes on Vand some additional data associated with each Jordan block size. In particular,it turns out (in contrast to the odd characteristic case) that the Jordan blocksizes are not enough to determine whether a unipotent element is distinguishedor not. For example, suppose that dimV = 2d where d is even. Then it followsfrom the results described below that G has exactly two conjugacy classes ofunipotent elements which act on V with two Jordan blocks of size d. One of theclasses consists of distinguished unipotent elements, and the other one consists ofnon-distinguished unipotent elements.

We now describe the main result of Hesselink. Let u ∈ G be unipotent andsuppose that the Jordan block sizes of u acting on V are 1 ≤ d(1) < d(2) < · · · <d(t), with block size d(i) occurring with multiplicity ni ≥ 1. Set X = u − 1. Forn ≥ 0, de�ne a quadratic form αn : V → K by αn(v) = (Xn+1v,Xnv) for allv ∈ V . De�ne the index function of V as the map χV : Z≥0 → Z, where

χV (m) = min{n ≥ 0 : Xmv = 0 implies αn(v) = 0}

for all m ≥ 0. Then by [Hes79, Theorem 3.8] the conjugacy class of u in G isuniquely determined by the d(i), ni and the function χV . In fact, the conjugacyclass of u is determined by the values of χV on the d(i).

Proposition 2.4.2 ([Hes79, 3.9]). The conjugacy class of u in G is uniquelydetermined by the symbol (d(1)n1

χV (d(1)), d(2)n2

χV (d(2)), . . . , d(t)ntχV (d(t))).

To describe when u is a distinguished unipotent element in G, we will use thedescription of Liebeck and Seitz given in [LS12, Chapter 6].

2.4. Unipotent elements in classical groups (p = 2) 21

De�nition 2.4.3. LetW be aK[u]-moduleW with a u-invariant form (−,−). Wesay thatW is orthogonally indecomposable, ifW cannot be written as a orthogonaldirect sum of two non-zero K[u]-submodules.

Now as a K[u]-module, V decomposes into an orthogonal direct sum of ortho-gonally indecomposable modules. In [Hes79], Hesselink classi�es the orthogonallyindecomposable summands that can occur. They fall into two families which arelabeled by integers, we denote them here by V (m) with m ≥ 2 even, and byW (m)with m ≥ 1.

We will not go into any detail in this section on how these indecomposablesare de�ned or classi�ed. For more details and for a construction of V (m) andW (m), see for example [LS12, Chapter 6.1]. For now it will be enough to knowthe following facts.

• The indecomposable V (m) has dimension m and u acts on V (m) with asingle Jordan block of size m.

• The indecomposable W (m) has dimension 2m, and W (m) = W1 ⊕W2 as aK[u]-module, where Wi are totally singular and u acts on Wi with a singleJordan block of size m.

It turns out that the decomposition of V into an orthogonal direct sum ofsubmodules of the form V (m) and W (m) determines the conjugacy class of u inG. More speci�cally, we have the the following:

Proposition 2.4.4 ([LS12, Proposition 6.1, Lemma 6.2, Proposition 6.22]). Letu ∈ G be unipotent. Then there is an orthogonal decomposition

V ↓ K[u] =

t∑i=1

W (mi)ai ⊕

s∑j=1

V (2kj)bj

with bj ≤ 2 for all j and 1 ≤ m1 < · · · < mt, and 1 ≤ k1 < · · · < ks. Furthermore,we have the following:

(i) The summands occurring in the decomposition are unique and determine theconjugacy class of u in G.

(ii) The element u is distinguished in G if and only if V ↓ K[u] =∑s

j=1 V (2kj)bj .

(iii) If u ∈ O(V ), then u ∈ SO(V ) if and only if∑

j bj is even.

(iv) If u ∈ SO(V ), then the conjugacy class uO(V ) consists of a single SO(V )-class, unless V ↓ K[u] =

∑ti=1W (mi)

ai , in which case uO(V ) splits into twoSO(V )-classes.

(v) The element u is regular in G if and only if V ↓ K[u] = V (2k).

(vi) If u ∈ SO(V ), then u is regular in SO(V ) if and only if V ↓ K[u] = V (2k) +V (2).

Finally, we will explain how to translate between the descriptions of the con-jugacy classes given in Proposition 2.4.2 and Proposition 2.4.4.

22 Chapter 2.

Lemma 2.4.5. Let u ∈ G be unipotent, and suppose that

V ↓ K[u] =t∑i=1

W (mi)ai ⊕

s∑j=1

V (2kj)bj

with bj ≤ 2 for all j and 1 ≤ m1 < · · · < mt, and 1 ≤ k1 < · · · < ks. Let1 ≤ d(1) < d(2) < · · · < d(t′) be the Jordan block sizes of u acting on V . Then forall i, we have the following.

(i) If m = d(i) is odd, then χV (m) = m−12 .

(ii) If m = d(i) is even, then χV (m) = m2 if V (m) occurs as an orthogonal direct

summand of V ↓ K[u].

(iii) If m = d(i) is even, then χV (m) = m−22 if V (m) does not occur as an

orthogonal direct summand of V ↓ K[u].

Proof. Write V ↓ K[u] = W1⊕· · ·⊕Wr as an orthogonal direct sum of orthogonallyindecomposable K[u]-modules, so now each summand is equal to some W (mi)(1 ≤ i ≤ t) or V (2kj) (1 ≤ j ≤ s).

Let m = d(i). By [Hes79, 3.5], we have the following:

χV (m)(m) =m

2, if m is even.

χW (m)(m) =m− 2

2, if m is even.

χW (m)(m) =m− 1

2, if m is odd.

Furthermore, by [Hes79, 3.9 (b)] for all 1 ≤ i ≤ r we have 0 ≤ χWi(m) ≤ 12m,

with equality if and only if m is even and Wi = V (m). Now our claim about thevalue of χV (m) follows from the fact that by [Hes79, 3.7 (c)] we have χV (m) =max1≤i≤r{χWi(m)}.

Following [Spa82, I.2.6, pg. 20] we de�ne a map ε on the Jordan block sizes dof a unipotent element u ∈ Sp(V ).

De�nition 2.4.6. Let d be a Jordan block size of u ∈ Sp(V ). We de�ne ε(d) ∈{0, 1} as follows:

• If d is odd, then ε(d) = 0.

• If d is even, then ε(d) = 0 if V (d) does not occur as an orthogonal directsummand of V ↓ K[u]. (By Lemma 2.4.5, this is equivalent to χV (d) = d−2

2 ).

• If d is even, then ε(d) = 1 if V (d) occurs as an orthogonal direct summandof V ↓ K[u]. (By Lemma 2.4.5, this is equivalent to χV (d) = d

2).

As an immediate corollary of Theorem 2.4.2, Lemma 2.4.5, and Proposition2.4.4, we get the following.

Corollary 2.4.7. Let u ∈ G be unipotent and suppose that the Jordan block sizesof u acting on V are 1 ≤ d(1) < d(2) < · · · < d(t), with block size d(i) occurringwith multiplicity ni ≥ 1. Then

2.4. Unipotent elements in classical groups (p = 2) 23

(i) The conjugacy class of u in Sp(V ) is uniquely determined by the symbol(d(1)n1

ε(d(1)), d(2)n2

ε(d(2)), . . . , d(t)ntε(d(t))).

(ii) The element u is distinguished in Sp(V ) if and only if ni ≤ 2 and ε(d(i)) = 1for all i.

In later sections, we will label the conjugacy classes in Sp(V ) by the decompo-sition V ↓ K[u] given by Proposition 2.4.4, or by the symbols given by Corollary2.4.7 (i). By Theorem 2.4.1, we can use the same labeling for unipotent conjugacyclasses of O(V ). Furthermore, by Proposition 2.4.4 (iv), for distinguished SO(V )-classes we can use the same labeling without ambiguity. Note also that by Lemma2.1.2 (i), we can (and will) also use this labeling for any simple group of type Clor Dl.

To end this section, we give a lemma which is useful for computing the valuesε(d(i)) de�ned above.

Lemma 2.4.8. Let u ∈ Sp(V ) be unipotent and suppose that the Jordan blocksizes of u acting on V are 1 ≤ d(1) < d(2) < · · · < d(t), with block size d(i)occurring with multiplicity ni ≥ 1. Fix m = d(i) and set X = u− 1. Then:

(i) If m is odd, then ε(m) = 0.

(ii) Assume that m is even. Let v1, . . . , vr be a basis of KerXm. Then the follo-wing are equivalent:

(a) ε(m) = 0.

(b) (Xm−1v, v) = 0 for all v ∈ KerXm.

(c) (Xm−1vi, vi) = 0 for all 1 ≤ i ≤ r.

Proof. Claim (i) follows from the de�nition of ε.For (ii), suppose that m is even. By Lemma 2.4.5, we have ε(m) = 0 if and

only if χV (m) = m−22 . Since χV (m) ≥ m−2

2 by Lemma 2.4.5, it follows thatε(m) = 0 if and only if αm−2

2(v) = 0 for all v ∈ KerXm. Thus to prove that (a)

and (b) are equivalent, it will be enough to show that (Xm−1v, v) = αm2−1(v) for

all v ∈ KerXm.To this end, let v, w ∈ KerXm. It is a consequence of [Spa82, Lemme II.6.10,

pg. 99] (with ei = Xm−iv and fj = Xm−jw there) that for all 1 ≤ i, j ≤ m − 1such that i+ j = m, we have

(Xi−1v,Xjw) + (Xiv,Xj−1w) = 0. (*)

Applying (*) with i = 1, i = 2, . . ., i = m− 1, we get

i = 1 (v,Xm−1w) = (Xv,Xm−2w)

i = 2 (Xv,Xm−2w) = (X2v,Xm−3w)

......

i = k (Xk−1v,Xm−kw) = (Xkv,Xm−(k+1)w)

......

i = m− 1 (Xm−2v,Xw) = (Xm−1v, w)

24 Chapter 2.

It follows that for all v, w ∈ KerXm we have (Xm−1v, w) = (Xiv,Xm−i−1w)for all 1 ≤ i ≤ m− 1. With i = m/2, this shows that

(Xm−1v, v) = (Xm2 v,X

m2−1v) = αm

2−1(v)

for all v ∈ KerXm. Hence (a) and (b) are equivalent.It is trivial that (b) implies (c). To show that (c) implies (b), �x a basis

v1, . . . , vr of KerXm. Applying (*) with i = m − 1, we get (Xm−1vj , vj′) +(Xm−1vj′ , vj) = 0 for all 1 ≤ j, j′ ≤ r. Thus for v =

∑rj=1 cjvj ∈ KerXm we get

(Xm−1v, v) =∑r

j=1 c2j (X

m−1vj , vj). Hence (Xm−1v, v) = 0 if (Xm−1vj , vj) = 0for all 1 ≤ j ≤ r.

Remark 2.4.9. Following the results above, it is straightforward to implement acomputer program that determines the conjugacy class of u ∈ G, when given thematrix of u and the alternating G-invariant bilinear form (−,−) with respect tosome basis of V . Some of our computer calculations (see Section 2.9) are based onthe use of such a program.

Indeed, calculating with ranks of powers of X = u − 1 allows one to �ndvery quickly the Jordan block sizes of u acting on V (see Lemma 3.1.2). Thenfor each Jordan block size d of u, one needs to calculate ε(d). This is done byapplying Lemma 2.4.8. If d is odd, then ε(d) = 0. If d is even, we compute a basisv1, . . . , vr for KerXd. Then by Lemma 2.4.8 (ii) we have ε(d) = 0 if and only if(Xd−1vi, vi) = 0 for all 1 ≤ i ≤ r.

2.5 Explicit distinguished unipotent elements (p 6= 2)

Assume that p 6= 2.

The purpose of this section is to describe representatives for distinguishedunipotent conjugacy classes in the classical groups Sp(V ) and SO(V ), in terms oftheir action on the natural module V .

Note �rst that it will be enough to construct unipotent elements acting on Vwith a single Jordan block. Indeed, consider the case G = Sp(V ). Then a distin-guished unipotent element u ∈ G should act on V with Jordan block sizes d1, . . . , dtwhere di are distinct even integers such that

∑ti=1 di = dimV (Proposition 2.3.4

(iii)). We can decompose V into a orthogonal direct sum V = W1⊕· · ·⊕Wt, whereeach Wi is a non-degenerate subspace and dimWi = di. Now if ui ∈ Sp(Wi) actsWi with a single Jordan block of size di, then u = u1 ⊕ · · · ⊕ ut ∈ Sp(V ) acts onV with Jordan blocks [d1, . . . , dt]. In precisely the same way, we see that it will beenough to construct (for dimV odd) unipotent elements u ∈ SO(V ) acting on Vwith a single Jordan block.

For the construction of unipotent elements with a single Jordan block, weproceed as in [Jan04, 1.7]. Let d > 0 and let V be a vector space of dimension dwith basis e1, e2, . . . , ed. De�ne a bilinear form (−,−) on V by

(ei, ej) =

{(−1)i if i+ j = d+ 1,

0 otherwise.

Then (−,−) is non-degenerate, and

(−,−) is

{symmetric, if d is odd,

alternating, if d is even.

2.6. Unipotent elements and labeled Dynkin diagrams (p good) 25

De�ne a linear map X : V → V by Xei = −ei−1, where we set ej = 0 forj < 0.

Then X is skew with respect to the form (−,−), that is, (Xv,w)+(v,Xw) = 0for all v, w ∈ V . Now X is a nilpotent element, so the Cayley transform u =(1−X)(1+X)−1 is a unipotent linear map which leaves the form (−,−) invariant[Wey39, Lemma 2.10.A, pg. 57]. That is, (uv, uw) = (v, w) for all v, w ∈ V .Furthermore, the Cayley transform of u is equal to (1 − u)(1 + u)−1 = X, so wehave (u − 1)ei = (u + 1)ei−1 for all 1 ≤ i ≤ n. From this it is straightforward tosee that for all 1 ≤ i ≤ d, we have

uei = ei + 2∑j<i

ej

and that u acts on V with a single Jordan block.Therefore the unipotent element u de�ned by (u− 1)ei = (u+ 1)ei−1 works as

a representative for the unipotent class of a single Jordan block. If d is odd, then(−,−) is symmetric and u ∈ SO(V ), where SO(V ) is the stabilizer of (−,−). If dis even, then (−,−) is alternating and u ∈ Sp(V ), where Sp(V ) is the stabilizerof (−,−).

2.6 Unipotent elements and labeled Dynkin diagrams(p good)

Assume that G is a connected semisimple algebraic group, and that p is good forG.

De�nition 2.6.1. We say that a parabolic subgroup P ≤ G with Levi factor Land unipotent radical Q is distinguished if dimL = dimQ/[Q,Q]. If P = PJ withJ ⊆ ∆, the labeled Dynkin diagram associated with P is de�ned by labeling α ∈ ∆with 0 if α ∈ J and with 2 if α 6∈ J .

De�nition 2.6.2. Let P be a parabolic subgroup of G. We say that a unipotentelement u ∈ Ru(P ) is a Richardson element for P if the P -conjugacy class of u isdense in Ru(P ).

In good characteristic, there is a uniform way of classifying the unipotentconjugacy classes in G. Roughly, this is done by establishing a correspondence be-tween the unipotent elements of G and the Richardson elements of distinguishedparabolic subgroups of Levi factors of G. In this section we will describe this cor-respondence for distinguished unipotent elements of G, and the correspondencebetween distinguished unipotent elements and labeled Dynkin diagrams. In ge-neral, we note that each unipotent conjugacy class (not just distinguished) has alabeled Dynkin diagram associated to it. But for the purposes of the present work,we will only need to know what the labeled Dynkin diagrams of distinguished uni-potent elements look like.

The distinguished parabolic subgroups and their labelings were determined byBala and Carter in [BC76a] and [BC76b]. The following theorem shows that in

26 Chapter 2.

good characteristic there is a bijection between distinguished unipotent classesand distinguished parabolic subgroups6.

Theorem 2.6.3 (Bala-Carter Theorem). Let u ∈ G be a distinguished unipotentelement. Then there exists a unique distinguished parabolic subgroup P < G suchthat u is a Richardson element for P .

In view of Theorem 2.6.3, the following de�nition makes sense.

De�nition 2.6.4. Let u ∈ G be a distinguished unipotent element. Let P = PJ ,J ⊆ ∆ be a distinguished parabolic corresponding to u in Theorem 2.6.3. Thenthe labeled Dynkin diagram associated with u is de�ned to be the labeled Dynkindiagram associated with P .

In good characteristic, the conjugacy classes of distinguished unipotent ele-ments in classical groups (SL(V ), Sp(V ) and SO(V )) are determined by Jordanblock sizes (Propositions 2.3.1, 2.3.4 and 2.3.2). Using the Jordan block sizes, thereis a straightforward method for computing the associated labeled Dynkin diagram.We record this in the following proposition, which is given as in [LS12, Theorem3.18]. Recall that we use the standard Bourbaki labeling of the simple roots, asgiven in [Hum72, 11.4, pg. 58].

Proposition 2.6.5. Let G = SL(V ), Sp(V ), or SO(V ) and let u ∈ G be a dis-tinguished unipotent element. Suppose that u acts on V with Jordan block sizes[d1, . . . , dt], where t ≥ 1 and 1 ≤ d1 < · · · < dt. Then the labeled Dynkin diagramof u is given as follows.

(i) If G = SL(V ) (type A), then t = 1, u is a regular unipotent and the labelingis 22 . . . 2.

(ii) If G = Sp(V ) (type C), the labeling is given as follows. If t = 1 (i.e. u isregular), then the labeling is 22 . . . 2. If t > 1, then starting from the lefthand side of the Dynkin diagram, begin with (dt − dt−1 − 2)/2 labels 2; then(dt−1−dt−2)/2 sequences 20; then (dt−2−dt−3)/2 sequences 200; and so on,until we get to d1/2 sequences 20 . . . 0 (t− 1 zeros); and �nally label 2 on thelast node.

(iii) If G = SO(V ) with dimV odd (type B), the labeling is given as follows. Ift = 1 (i.e. u is regular), then the labeling is 22 . . . 2. If t > 1, then startingfrom the left hand side of the Dynkin diagram, begin with (dt − dt−1 − 2)/2labels 2; then (dt−1 − dt−2)/2 sequences 20; then (dt−2 − dt−3)/2 sequences200; and so on, until we get to (d1 − 1)/2 sequences 20 . . . 0 (t − 1 zeros);and �nally label the rest of the diagram with 20 . . . 0 ((t− 1)/2 zeros).

(iv) If G = SO(V ) with dimV even (type D), the labeling is given as follows.Starting from the left hand side of the Dynkin diagram, begin with (dt −dt−1− 2)/2 labels 2; then (dt−1− dt−2)/2 sequences 20; then (dt−2− dt−3)/2sequences 200; and so on, until we get to (d1 − 1)/2 sequences 20 . . . 0 (t− 1zeros); �nally if t > 2 label the rest of the diagram with 20 . . . 0 (t/2 zeroes);if t = 2 label the last two nodes 22.

6Theorem 2.6.3 was originally proven by Bala and Carter in the case where p = 0 or plarge. In [Pom77] and [Pom80], Pommerening extended the results of Bala and Carter to goodcharacteristic. The approach of Pommerening was based on case-by-case analysis. Later Premet[Pre03] gave the �rst uniform proof, which was simpli�ed by Tsujii in [Tsu08].

2.6. Unipotent elements and labeled Dynkin diagrams (p good) 27

Remark 2.6.6. According to Proposition 2.6.5, for classical groups the labeledDynkin diagrams of distinguished unipotent elements begin with a string of 2s,then some number of 20s, then some number of 200s; and so on, until the lastone or two nodes for which the pattern depends on the type of the group and theunipotent class. We call the string of 2s in the beginning the initial string.

For distinguished unipotent classes in exceptional groups, we name them withthe usual Bala-Carter label of the associated distinguished parabolic subgroup,as given in [BC76b]. For each of these classes, we record the associated labeledDynkin diagrams in the following proposition. See for example [Car85, 5.9, pg.175-177] or [LS12, Table 2.2, pg. 25].

Proposition 2.6.7. Let G be simple of exceptional type and u ∈ G a distinguishedunipotent element. If u is regular, then the labeled Dynkin diagram of u is 22 . . . 2.If u is not regular, then the labeled Dynkin diagram of u is as in Table 2.1.

Type Unipotent class Labeled diagramG2 G2(a1) 02

F4 F4(a1) 2202F4 F4(a2) 0202F4 F4(a3) 0200

E6 E6(a1) 222022E6 E6(a3) 200202

E7 E7(a1) 2220222E7 E7(a2) 2220202E7 E7(a3) 2002022E7 E7(a4) 2002002E7 E7(a5) 0002002

E8 E8(a1) 22202222E8 E8(a2) 22202022E8 E8(a3) 20020222E8 E8(a4) 20020202E8 E8(b4) 20020022E8 E8(a5) 20020020E8 E8(b5) 00020022E8 E8(a6) 00020020E8 E8(b6) 00020002E8 E8(a7) 00002000

Table 2.1: Labeled diagrams of non-regular distinguished unipotent elements forsimple groups of exceptional type.

Following [LT99, pg. 7], we de�ne the labeled diagram of a simple subgroupA < G of type A1 as follows. Let A < G be a simple subgroup of type A1. Nowthere exists a surjective morphism φ : SL2(K)→ A of algebraic groups. Set

TA =

{φ

(µ 00 µ−1

): µ ∈ K∗

}.

28 Chapter 2.

By replacing A with a conjugate, we can assume that TA is contained in the chosenmaximal torus T of G. For each root α ∈ Φ, let cα be the integer such that

α

(φ

(µ 00 µ−1

))= cα.

One can now choose the base ∆ in a way such that cα ≥ 0 for all α ∈ Φ+ [LT99,Proposition 2.3]. We then de�ne a labeled diagram of A to be the Dynkin diagramof G with node α ∈ ∆ labeled by cα.

The following theorem is a key tool for the proof of Theorem 1.1.10 for unipo-tent elements of order p.

Theorem 2.6.8 ([LT99, Theorem 4.2]). Let u ∈ G be a unipotent element oforder p. Then there exists a subgroup A < G of type A1 such that u ∈ A and alabeled diagram of A is that of u.

Now let u ∈ G be a unipotent element of order p and let u ∈ A, where A < Gis as in Theorem 2.6.8. In this setting, one can study the action of u ∈ G on arepresentation V of G by considering the restriction V ↓ A. Indeed, here knowledgeof the labeled diagram of u tells how T -weights restrict to TA-weights. Thereforeone gets information on the composition factors of V ↓ A, which can be used togain information on the action of u on V . This is explored in more detail in Section4.6 and Section 5.13.

2.7 Largest Jordan block size of a unipotent element

Let u ∈ G be a unipotent element and let V be a G-module, a�orded by ϕ :G → GL(V ). Denote by du the largest Jordan block size of ϕ(u). The purposeof this section is to describe various upper bounds which are known for du. Themotivation for this is the fact that if u acts on V as a distinguished unipotentelement, then we get an upper bound on dimV that is a quadratic polynomialin du. This will be key in our solution of Problem 1.1.9, and will be explained insections 5.1 and 5.2.

For basic upper bounds on du, the following elementary observation will beuseful, as found for example in [Sup95, Lemma 2.2, Lemma 2.15].

Lemma 2.7.1. Let u ∈ G be a unipotent element. Then

(i) du ≤ |u|,

(ii) If u′ ∈ G is a regular unipotent element, then du ≤ du′ .

Proof. Let f : V → V be any unipotent linear map on V . For any k ≥ 1, we have(f − 1)p

k= fp

k − 1. Therefore (f − 1)pα

= 0 for |f | = pα, and thus the largestJordan block size of f is ≤ pα. This gives (i).

For (ii), note that if u′ ∈ G is any unipotent element such that the conjugacyclass uG is contained in the Zariski closure of (u′)G, then du ≤ du′ since {f ∈GL(V ) : (f −1)du′ = 0} is Zariski closed in GL(V ). Thus (ii) follows from the factthat the regular unipotent class is dense in the unipotent variety of G.

De�nition 2.7.2. Suppose that p is good for G, let u ∈ G be a unipotent element,and let λ ∈ X(T )+. Write the dominant weight λ as λ =

∑li=1 qiαi, where qi ∈ Q+

2.7. Largest Jordan block size of a unipotent element 29

for all 1 ≤ i ≤ l (this is possible for example by [Hum72, Section 13, Table 1]).De�ne

mu(λ) =

l∑i=1

diqi,

where di is the label of αi in the labeled Dynkin diagram of u.

If u is regular, then mu(λ) = 2∑l

i=1 qi in De�nition 2.7.2. Note also that forany unipotent element u ∈ G, we have mu(λ + µ) = mu(λ) + mu(µ) for allλ, µ ∈ X(T )+.

For the rest of this section we give some properties of mu(λ), mostly in thecase where u is a regular unipotent element.

Lemma 2.7.3. Write λ =∑l

i=1 ciωi, where ωi are the fundamental dominantweights with the usual Bourbaki labeling. For a regular unipotent u, the values ofmu(λ) for each irreducible root system are as in the table below.

Φ mu(λ)

Al∑l

i=1 i(l + 1− i)ci

Bl∑l−1

i=1 i(2l + 1− i)ci + l(l+1)2 cl

Cl∑l

i=1 i(2l − i)ci

Dl∑l−2

i=1 i(2l − i− 1)ci + l(l−1)2 cl−1 + l(l−1)

2 cl

G2 6c1 + 10c2

F4 22c1 + 42c2 + 30c3 + 16c4

E6 16c1 + 22c2 + 30c3 + 42c4 + 30c5 + 16c6

E7 34c1 + 49c2 + 66c3 + 96c4 + 75c5 + 52c6 + 27c7

E8 92c1 + 136c2 + 182c3 + 270c4 + 220c5 + 168c6 + 114c7 + 58c8

Proof. Write λ =∑l

i=1 ciωi. Then mu(λ) =∑l

i=1 cimu(ωi) since mu(λ1 + λ2) =mu(λ1) +mu(λ2). Using the values of the qi for the fundamental weights given in[Hum72, Section 13, Table 1], it is easy to compute mu(ωi) for each irreducibleroot system. We omit this computation.

Lemma 2.7.4. Let u ∈ G be a regular unipotent element. Then u has order p ifand only if p - mu(ωi) for all 1 ≤ i ≤ l.

Proof. Now there exists a numberMG ≥ 0, depending only on the type of G, suchthat a regular unipotent element u ∈ G has order p if and only if p ≥ MG. ForG of type Al, Bl, Cl or Dl we have MG = l + 1, MG = 2l + 1, MG = 2l andMG = 2l − 1 respectively (Proposition 2.3.3).

If G has type Al, we have mu(ωi) = i(l + 1 − i) by Lemma 2.7.3. Thereforep - mu(ωi) for all 1 ≤ i ≤ l if and only if p - i for all 1 ≤ i ≤ l, which is equivalentto p ≥ l + 1. If G has type Bl, Cl, or Dl, a similar application of Lemma 2.7.3shows that p - mu(ωi) for all 1 ≤ i ≤ l if and only if p ≥ 2l + 1, p ≥ 2l andp ≥ 2l − 1, respectively.

30 Chapter 2.

Suppose then that G is simple of exceptional type. If G has type G2, F4, E6,E7, or E8, then we have MG = 7, MG = 13, MG = 13, MG = 19, and MG = 31respectively (Appendix A). For each exceptional type, using Lemma 2.7.3, one caneasily verify that the prime factors occurring in mu(ωi) are precisely all the primes< MG. Therefore p ≥ MG if and only if p - mu(ωi) for all 1 ≤ i ≤ rankG, asdesired. For convenience we have listed the prime factors of the mu(ωi) in Table2.1.

G2 F4 E6 E7 E8

ω1 2, 3 2, 11 2 2, 17 2, 23ω2 2, 5 2, 3, 7 2, 11 7 2, 17ω3 2, 3, 5 2, 3, 5 2, 3, 11 2, 7, 13ω4 2 2, 3, 7 2, 3 2, 3, 5ω5 2, 3, 5 3, 5 2, 5, 11ω6 2 2, 13 2, 3, 7ω7 3 2, 3, 19ω8 2, 29

Table 2.1: Prime factors occurring in mu(ωi), where u is a regular unipotent ele-ment in G of exceptional type.

The following lemma which has some useful implications for the representationtheory of G. Recall that δ ∈ X(T )+ denotes the half-sum of the positive roots inΦ, and that δ =

∑li=1 ωi.

Lemma 2.7.5. Let u ∈ G be a regular unipotent element. Then the inequalitymu(λ) ≥ 〈λ+ δ, α〉 − 1 holds for all λ ∈ X(T )+ and α ∈ Φ+.

Proof. Let λ ∈ X(T )+ and α ∈ Φ+. Write λ =∑l

i=1 ciωi and α =∑l

i=1 kiαi,where ci, ki are nonnegative integers. Then

〈λ+ δ, α〉 =l∑

i=1

(ci + 1)ki2(ωi, αi)

(α, α)=

l∑i=1

(ci + 1)ki(αi, αi)

(α, α)=

l∑i=1

(ci + 1)ti

where ti = ki(αi,αi)(α,α) . Note that α

∨ =∑l

i=1 tiα∨i , so by the equalities above 〈λ+δ, α〉

gets its largest value when α∨ is the highest root of the dual root system Φ∨ withrespect to the base ∆∨ [Hum72, Lemma 10.4A]. In other words, 〈λ+ δ, α〉 gets itslargest value when α is the highest short root β of Φ with respect to the base ∆.Therefore it will be enough to show that mu(λ) ≥ 〈λ+ δ, β〉 − 1.

According to [Ser94, Proposition 5], we have

〈λ+ δ, β〉 ≤ 1 +∑α∈Φ+

〈λ, α〉.

Hence it will be enough to show that∑

α∈Φ+〈λ, α〉 = mu(λ). To this end, write

λ =∑l

i=1 qiαi, where qi ∈ Q+. Then

∑α∈Φ+

〈λ, α〉 =l∑

i=1

qi∑α∈Φ+

〈αi, α〉

2.7. Largest Jordan block size of a unipotent element 31

so it will be enough to prove that∑

α∈Φ+〈αi, α〉 = 2. That is, it su�ces to provethat

∑α∈Φ+−{αi}〈αi, α〉 = 0. Now αi is a simple root, so the re�ection σαi per-

mutes the positive roots other than αi [Hum72, Lemma 10.2B]. Hence∑α∈Φ+−{αi}

〈αi, α〉 =∑

α∈Φ+−{αi}

〈σαi(αi), σαi(α)〉 =∑

α∈Φ+−{αi}

〈−αi, α〉,

which gives∑

α∈Φ+−{αi}〈αi, α〉 = 0.

Now the following corollary of Lemma 2.7.5 is immediate from the Jantzensum formula [Jan03, II.8.19].

Corollary 2.7.6. Let λ ∈ X(T )+ and let u ∈ G be a regular unipotent element.If p > mu(λ), then VG(λ) is irreducible.

We �nish by describing a connection between mu(λ) and simple algebraicgroups in characteristic 0. Let GC be a simple algebraic group over C with the sameroot system Φ as G. Now the Bala-Carter classi�cation of unipotent conjugacyclasses and the results described in 2.6 are still valid for GC. Thus we can makethe following de�nition.

De�nition 2.7.7. Assume that p is good for G and let u ∈ G be a unipotentelement. We denote by uC some representative uC ∈ GC of a unipotent conjugacyclass of GC which has the same labeled diagram as u.

Furthermore, Theorem 2.6.8 holds for GC, so we can �nd a simple subgroupXC < GC of type A1 such that uC ∈ XC and such that XC has the same labeleddiagram as uC.

The following theorem is a straightforward consequence of various results ob-served by Suprunenko [Sup09, Lemma 2.38, Lemma 2.39, Corollary 2.40].

Theorem 2.7.8. Let u ∈ G be a unipotent element and λ ∈ X(T )+. Then u actson LG(λ) with largest Jordan block of size at most mu(λ) + 1.

Proof. Let GC, uC, XC be as above. Now LGC(λ) ↓ XC has highest weight mu(λ),so it follows that uC acts on LGC(λ) with largest Jordan block of size mu(λ) + 1.

Now the claim follows from Lemma 2.38 in [Sup09], once we observe thatLemma 2.39 in [Sup09] holds also for exceptional groups; see for example [LS12,Table 13.3, pg. 172] and [LS12, 18.1, pg. 287].

We will say that the action of u on VG(λ) has the same Jordan block sizes asthe corresponding action in characteristic 0, if the Jordan block sizes of u actingon VG(λ) are same as the Jordan block sizes of uC acting on VGC(λ).

Lemma 2.7.9. Let u ∈ G be a unipotent element and let λ ∈ X(T )+ be nonzero.If p > mu(λ), then the action of u on VG(λ) has the same Jordan block sizes asthe corresponding action in characteristic 0.

Proof. Keep the notation GC, uC,XC as before. Assume that p > mu(λ). It followsnow from Theorem 2.7.8 that u has order p. Thus by Theorem 2.6.8, we can �nd asubgroup X < G of type A1 containing u and having the same labeled diagram asu. The module VG(λ) ↓ X has highest weight mu(λ). Since p > mu(λ), it followsfrom [AJL83, Lemma 2.2] that VG(λ) ↓ X is semisimple.

32 Chapter 2.

Now note that VGC(λ) ↓ XC and VG(λ) ↓ X have the same character, so theyhave the same composition factors. Since both of them are semisimple and allcomposition factors of VG(λ) ↓ X have highest weight strictly less p, the claimfollows.

2.8 Representatives for unipotent classes in Chevalleygroups

For computer calculations, it is often convenient to consider a simple linear alge-braic group G as a Chevalley group, constructed via the Chevalley constructionas in [Ste68]. In this section we will describe how to construct representativesu ∈ G for all unipotent classes of a Chevalley group G. For the purposes of thistext, these representatives are needed only for the implementation of some of ourcomputer calculations, which are described in 2.9.

We begin by recalling the Chevalley construction. Details can be found in[Ste68]. Let Φ be a root system with base ∆ and set of positive roots Φ+. Let gCbe a complex semisimple Lie algebra with root system Φ, and �x some Chevalleybasis {Xα : α ∈ Φ}∪{Hα : α ∈ ∆} of gC. Now for all α, β ∈ Φ with α+β ∈ Φ, wehave [Xα, Xβ] = Nα,βXα+β . We call Nα,β the structure constants of the Chevalleybasis. Here we have Nα,β = ±(k + 1), where k is the largest integer such thatβ − kα ∈ Φ. Let UZ be the Kostant Z-form with respect to this Chevalley basis.That is, UZ is the subring of the universal enveloping algebra of gC generated by

1 and all Xkαk! for α ∈ Φ and k ≥ 1.

Let VC be some irreducible representation of the Lie algebra gC. Then one can�nd a UZ-invariant lattice VZ in VC. De�ne V = VZ ⊗Z K. Since VZ is invariantunder UZ, and since Xα act as nilpotent linear maps on VC, we can de�ne xα(c) =exp(cXα) ∈ GL(V ) for all α ∈ Φ and c ∈ K. Then a Chevalley group over Kinduced by V (which depends on the choice of VC) is the subgroup

G = 〈xα(c) : α ∈ Φ, c ∈ K〉

of GL(V ).One can show that G is a connected semisimple linear algebraic group with

root system Φ. Furthermore, it is well known that any unipotent element of Gis conjugate to an element of the form

∏α∈Φ+ xα(cα), where the product is ta-

ken with respect to some ordering of Φ+. The following useful lemma is due toSteinberg.

Lemma 2.8.1 ([Ste65, Lemma 3.2 (c)]). A unipotent element∏α∈Φ+ xα(cα) of

G is regular if and only if cα 6= 0 for all α ∈ ∆.

Our goal is to �nd representatives of the form∏α∈Φ+ xα(cα) for each unipotent

conjugacy class of G. We will do this in this section, and moreover the represen-tatives that we give will have the property that cα ∈ {−1, 0, 1} for all α ∈ Φ+.Note that by [Ste68, Corollary 5, pg. 44] and Lemma 2.1.2 (i), the conjugacy classof such a representative does not depend on the choice of the representation VCwe choose. However, the conjugacy class of an unipotent element given by theexpression

∏α∈Φ+ xα(cα) might be ambiguous if the structure constants Nα,β are

not speci�ed. We explain next how we will choose the structure constants.We de�ne a total order (as in [Car72, 2.1]) on Φ in the following way. For

α, β ∈ Φ, we de�ne α ≤ β if and only if α = β, or β − α =∑k

i=1 ciαi, for some

2.8. Representatives for unipotent classes in Chevalley groups 33

1 ≤ k ≤ l such that ci ∈ Z for all 1 ≤ i ≤ k, and ck > 0. As in [Car72, 4.2],we will say that a pair (α, β) is special if α, β ∈ Φ+, α ≤ β, and α + β ∈ Φ. Aspecial pair (α, β) will be called extraspecial, if for any special pair (α′, β′) withα′ + β′ = α + β, we have α ≤ α′. As noted in [Car72, 4.2], each γ ∈ Φ+ − ∆corresponds to a unique extraspecial pair (α, β) with α+ β = γ, and this de�nesa bijection between Φ+ −∆ and the set of extraspecial pairs.

It is possible to show [Car72, Proposition 4.2.2] that the structure constantsNα,β of the Chevalley basis are determined by the values of Nα,β for extraspecialpairs (α, β). In what follows, we will always use a Chevalley basis such that thestructure constants for extraspecial pairs are positive. Such a basis can always befound, since by [Car72, Proposition 4.2.2] the signs of the structure constants Nα,β

for extraspecial pairs (α, β) can be chosen arbitrarily. For types Bl, Cl, and Dl, wewill use a Chevalley basis described in [Jan73, 11, pg. 38]. For our purposes, themain reason to use structure constants which are positive for extraspecial pairs isthat these are also the default structure constants used in MAGMA.

De�nition 2.8.2. A subsystem subgroup of G is a connected semisimple subgroupof G which is normalized by a maximal torus of G.

We will see in this section that for most of the unipotent conjugacy classes ofG, we can �nd a class representative in a proper subsystem subgroup of G; thiswill reduce the description of the unipotent class representatives to a relativelysmall number of cases. Typical examples of subsystem subgroups are given byclosed subsets Ψ of Φ.

De�nition 2.8.3. We say that Ψ ⊆ Φ is closed, if both of the following hold:

(i) Ψ = −Ψ.

(ii) For all α, β ∈ Ψ such that α+ β ∈ Φ, we have α+ β ∈ Ψ.

It is well known (see for example [MT11, Theorem 13.6]) that for any closedsubset Ψ ⊆ Φ, we have a subsystem subgroup

〈xα(c) : α ∈ Ψ, c ∈ K〉

of G with root system Ψ.Next we need to make some remarks about graph automorphisms and their

action on unipotent conjugacy classes. The following result is proven in [Ste68,Corollary (b) of Theorem 29].

Lemma 2.8.4. Assume that G is simply connected. Let σ′ : ∆ → ∆ be a graphautomorphism of the Dynkin diagram ∆. Then there exists an isomorphism σ :G → G and signs εα = ±1 such that εα = 1 for all α ∈ ±∆, and σ(xα(c)) =xσ′(α)(εαc) for all α ∈ Φ and c ∈ K.

For a graph automorphism σ′ : ∆ → ∆ of the Dynkin diagram, we call theisomorphism σ in Lemma 2.8.4 the graph automorphism of G induced by σ. Weknow that each graph automorphism of ∆ is either an involution or a triality graphautomorphism of D4. The following two lemmas describe the action of involutorygraph automorphisms on unipotent conjugacy classes.

34 Chapter 2.

Lemma 2.8.5 ([LS12, Corollary 6]). Let G be a simply connected algebraic groupof type Al or E6 and let σ be a graph automorphism of G. Then σ �xes all unipotentclasses of G.

Lemma 2.8.6. Let G be a semisimple algebraic group of type Dl (l ≥ 2) withnatural module V , and let σ be the graph automorphism of G induced by the graphautomorphism swapping the two end nodes of the Dynkin diagram. Then σ �xesall unipotent classes of G, except the following ones:

(i) p 6= 2: For all ci even such that∑t

i=1 ci = l, the automorphism σ swaps thetwo classes of unipotents u which satisfy V ↓ K[u] =

⊕ti=1 2 · Vci .

(ii) p = 2: For all ci even such that∑t

i=1 ci = l, the automorphism σ swaps thetwo classes of unipotents u which satisfy V ↓ K[u] =

⊕ti=1W (ci).

Proof. For l 6= 4 this is [LS12, Corollary 6]. However, the same proof also worksfor l = 4.

To complete the picture, one should still describe the action of a triality graphautomorphism of D4. We postpone a discussion of this to Section 2.10.

We now move on to describing the unipotent class representatives. For simplelinear algebraic groups of exceptional type, explicit representatives for all unipo-tent conjugacy classes have been written down in [Sim13, Tables 3.1-3.9], wherethe representatives were found using [LS12] and computations done by Ross Lawt-her. We list these representatives for the distinguished unipotent classes in tables2.1 - 2.5, where in some cases the representative given depends on the characte-ristic p of K. In the tables we denote xi(c) = xαi(c) for all 1 ≤ i ≤ N , whereN = |Φ+| and Φ+ = {α1, . . . , αN} is ordered with the total order ≤ describedabove. We have given the ordering explicitly in Appendix C. This ordering of theroots is the same one that is used in MAGMA. Note that the ordering of positiveroots used in [Sim13] for type F4 is di�erent to this ordering. We also remark thatin [Sim13] the structure constants used are those given in [GAP16], which areusually di�erent from those used in MAGMA. However, we have veri�ed that therepresentatives given in tables 2.1 - 2.5 work with both sets of structure constants.

For classical types, we will describe representatives for all unipotent conjugacyclasses in the subsections that follow.

Class p RepresentativeG2 any x1(1)x2(1)G2(a1) any x2(1)x5(1)

(A1)3 3 x1(1)x5(1)

Table 2.1: Representatives for distinguished unipotent classes of a Chevalley groupof type G2.

2.8.1 Type Al, l ≥ 1

Let e1, . . . , el, el+1 be a basis for a complex vector space VC, and let VZ be theZ-lattice spanned by this basis. Let sl(VC) be the Lie algebra formed by the linearendomorphisms of VC with trace zero. Then sl(VC) is a semisimple Lie algebraof type Al. Let h be the Cartan subalgebra formed by the diagonal matrices in

Class p RepresentativeF4 any x1(1)x2(1)x3(1)x4(1)F4(a1) any x1(1)x2(1)x6(1)x7(1)F4(a2) 2 x4(1)x7(1)x5(1)x9(1)

6= 2 x4(1)x2(1)x8(1)x9(1)F4(a3) any x2(1)x5(1)x9(1)x18(1)(C3(a1))2 2 x2(1)x4(1)x9(1)x20(1)

(A2A1)2 2 x1(1)x3(1)x4(1)x16(1)

Table 2.2: Representatives for distinguished unipotent classes of a Chevalley groupof type F4.

Class p RepresentativeE6 any x1(1)x2(1)x3(1)x4(1)x5(1)x6(1)E6(a1) any x1(1)x2(1)x9(1)x10(1)x5(1)x6(1)E6(a3) 2 x1(1)x8(1)x9(1)x11(1)x14(1)x19(1)

6= 2 x13(1)x1(1)x15(1)x6(1)x14(1)x4(1)

Table 2.3: Representatives for distinguished unipotent classes of a Chevalley groupof type E6.

Class p RepresentativeE7 any x1(1)x2(1)x3(1)x4(1)x5(1)x6(1)x7(1)E7(a1) any x1(1)x3(1)x10(1)x9(1)x5(1)x6(1)x7(1)E7(a2) any x1(1)x2(1)x3(1)x9(1)x11(1)x12(1)x13(1)E7(a3) 2 x1(1)x7(1)x9(1)x10(1)x12(1)x16(1)x22(1)

6= 2 x15(1)x1(1)x17(1)x6(1)x7(1)x16(1)x4(1)E7(a4) 2 x1(1)x4(1)x7(1)x10(1)x22(1)x23(1)x24(1)

6= 2 x1(1)x4(1)x7(1)x13(1)x15(1)x16(1)x17(1)x23(1)E7(a5) 3 x4(1)x15(1)x16(1)x19(1)x20(1)x21(1)x29(1)

6= 3 x4(1)x7(1)x20(1)x21(1)x22(1)x23(1)x24(1)


sl(VC), with respect to the basis (ei). For 1 ≤ i ≤ n, de�ne maps εi : h → C byεi(h) = hi, where h is a diagonal matrix with diagonal entries (h1, . . . , hl, hl+1).Now Φ = {εi − εj : i 6= j} is the root system for sl(VC) and Φ+ = {εi − εj :i < j} is a system of positive roots. The base ∆ of Φ corresponding to Φ+ is∆ = {α1, . . . , αl}, where αi = εi − εi+1 for all 1 ≤ i ≤ l.

For 1 ≤ i, j ≤ l + 1, let Ei,j be the linear endomorphism on VC such thatEi,j(ej) = ei and Ei,j(ek) = 0 for k 6= j. A Chevalley basis for sl(VC) with positivestructure constants for extraspecial pairs is given by Xεi−εj = Ei,j for all i 6= j.Let UZ be the Kostant Z-form with respect to this Chevalley basis of sl(VC).

Note that VZ is a UZ-invariant lattice in VC. We de�ne V = VZ⊗ZK. Then thefollowing lemma is well known, and amounts to the fact that SL(V ) is generatedby transvections.

Lemma 2.8.7. The Chevalley group of type Al induced by V is equal to G =SL(V ).

36 Chapter 2.

Class p RepresentativeE8 any x1(1)x2(1)x3(1)x4(1)x5(1)x6(1)x7(1)x8(1)E8(a1) any x1(1)x2(1)x10(1)x11(1)x5(1)x6(1)x7(1)x8(1)E8(a2) any x1(1)x2(1)x3(1)x10(1)x12(1)x13(1)x14(1)x8(1)E8(a3) 2 x7(1)x8(1)x9(1)x10(1)x11(1)x12(1)x13(1)x19(1)

6= 2 x1(1)x4(1)x6(1)x7(1)x8(1)x17(1)x18(1)x19(1)E8(a4) 2 x9(1)x10(1)x11(1)x12(1)x13(1)x14(1)x15(1)x19(1)

6= 2 x1(1)x4(1)x6(1)x8(1)x17(1)x18(1)x19(1)x21(1)E8(b4) 2 x7(1)x8(1)x9(1)x17(1)x18(1)x19(1)x20(1)x21(1)





3 x4(1)x17(1)x12(1)x29(1)x30(1)x31(1)x34(1)x35(1)6= 2, 3 x4(1)x8(1)x18(1)x19(1)x23(1)x31(1)x33(1)x39(−1)x41(1)

E8(a7) any x30(1)x33(1)x35(1)x36(1)x40(1)x45(1)x53(1)x58(1)(A7)3 3 x1(1)x3(1)x4(1)x5(1)x6(1)x7(1)x8(1)x67(1)(D7(a1))2 2 x2(1)x3(1)x10(1)x12(1)x6(1)x7(1)x8(1)x83(1)(D5A2)2 2 x1(1)x2(1)x3(1)x4(1)x5(1)x7(1)x8(1)x51(1)


By abuse of notation we will identify the basis (ei⊗1) of V with (ei). Recall thatby Lemma 2.2.1, the conjugacy class of a unipotent element u ∈ G is determinedby its Jordan block sizes. In other words, the unipotent conjugacy classes in G arelabeled by symbols [d1, . . . , dt], where 1 ≤ d1 ≤ · · · ≤ dt and

∑ti=1 di = l + 1. Set

k1 = 1 and ki = 1 +∑i−1

j=1 dj for 1 < i ≤ t. Let Vi be the subspace of V with basisBi = {ej : ki ≤ j ≤ ki + di − 1}. Now V = V1 ⊕ · · · ⊕ Vt, so we have a naturallyembedded subgroup SL(V1)× · · · × SL(Vt) < G, where SL(Vi) is the subgroup ofall g ∈ G such that g(Vi) = Vi and gej = ej for all ej 6∈ Bi.

Note next that a representative of the unipotent conjugacy class labeled by[d1, . . . , dt] is given by u = u1 · · ·ut, where ui ∈ SL(Vi) acts on Vi with a singleJordan block of size di. If di = 1, such an ui is given by ui = 1. Suppose thenthat di > 1. Now by Lemma 2.8.7, the subgroup SL(Vi) is a subsystem subgroupof SL(V ) with root system {±(εj − εj′) : ki ≤ j < j′ ≤ ki + di− 1} of type Adi−1,and a base {αj : ki ≤ j ≤ ki + di − 2}. Thus it follows from Lemma 2.2.2 andLemma 2.8.1 that ui =

∏ki+di−2j=ki

xαj (1) ∈ SL(Vi) acts on Vi with a single Jordanblock of size di. Therefore we have the following result.

Lemma 2.8.8 (Type A). Let G = SL(V ) be a Chevalley group of type Al asde�ned above, where l ≥ 1. Let 1 ≤ d1 ≤ · · · ≤ dt, where

∑ti=1 di = l + 1. Set

k1 = 1 and ki = 1 +∑i−1

j=1 dj for 1 1.

Then u = u1 · · ·ut lies in the unipotent class of G labeled by [d1, . . . , dt].

2.8.2 Type Cl, l ≥ 1

α α =∑l

k=1 ckαi

εi − εj , 1 ≤ i < j ≤ l∑

i≤k≤j−1 αk

εi + εj , 1 ≤ i < j ≤ l∑

i≤k≤j−1 αk +∑

j≤k≤l−1 2αk + αl

2εi, 1 ≤ i ≤ l∑

i≤k≤l−1 2αk + αl

Table 2.6: Type Cl, expressions for roots α ∈ Φ+ in terms of base ∆.

Let e1, . . . , el, e−l, . . . , e−1 be a basis for a complex vector space VC, and let VZbe the Z-lattice spanned by this basis. We de�ne a non-degenerate alternatingbilinear form (−,−) on VC by setting (ei, e−i) = 1 = −(e−i, ei) and (ei, ej) = 0for i 6= −j. Let sp(VC) be the Lie algebra consisting of the linear endomorphismsX of VC satisfying (Xv,w) + (v,Xw) = 0 for all v, w ∈ VC. Then sp(VC) is asemisimple Lie algebra of type Cl. Let h be the Cartan subalgebra formed by thediagonal matrices in sp(VC). Then h = {diag(h1, . . . , hl,−hl, . . . ,−h1) : hi ∈ C}.For 1 ≤ i ≤ l, de�ne maps εi : h→ C by εi(h) = hi where h is a diagonal matrixwith diagonal entries (h1, . . . , hl,−hl, . . . ,−h1). Now Φ = {±(εi ± εj) : 1 ≤ i <j ≤ l} ∪ {±2εi : 1 ≤ i ≤ l} is the root system for sp(VC), and Φ+ = {εi ± εj :1 ≤ i < j ≤ l} ∪ {2εi : 1 ≤ i ≤ l} is a system of positive roots. The base ∆ of Φcorresponding to Φ+ is ∆ = {α1, . . . , αl}, where αi = εi − εi+1 for 1 ≤ i < l andαl = 2εl. We give the expressions of roots α ∈ Φ+ in terms of ∆ in Table 2.6.

For all i, j let Ei,j be the linear endomorphism on VC such that Ei,j(ej) = eiand Ei,j(ek) = 0 for k 6= j. Then a Chevalley basis for sp(VC) with positivestructure constants for extraspecial pairs is given by

Xεi−εj = Ei,j − E−j,−i for all i 6= j,

Xεi+εj = Ej,−i + Ei,−j for all i 6= j,

X−(εi+εj) = E−j,i + E−i,j for all i 6= j,

X2εi = Ei,−i for all i,

X−2εi = E−i,i for all i.

Note that VZ is a UZ-invariant lattice in VC. We de�ne V = VZ ⊗Z K. Now(−,−) also gives a non-degenerate alternating bilinear form on V , and we havethe following result.

Lemma 2.8.9 ([Ree57, 5]). The Chevalley group of type Cl induced by V is equalto the group G = Sp(V ) of invertible linear maps V → V preserving (−,−).

By abuse of notation we identify the basis (ei ⊗ 1) of V with (ei). Let u ∈ Gbe a unipotent element.

Suppose �rst that p 6= 2. Then by Proposition 2.3.1, the conjugacy class of u inG is uniquely determined by the Jordan block sizes of u on V . Furthermore, eachJordan block of odd size must occur with even multiplicity by Proposition 2.3.2

38 Chapter 2.

(i). That is, the conjugacy class of u is uniquely determined by the decomposition

V ↓ K[u] =t⊕i=1

2 · Vci ⊕s⊕j=1

V2dj ,

where 1 ≤ c1 ≤ · · · ≤ ct are odd, and 1 ≤ d1 ≤ · · · ≤ ds, and∑t

i=1 ci+∑s

j=1 dj = l.Consider next p = 2. It follows from Proposition 2.4.4 that the conjugacy class

of u is uniquely determined by the decomposition

V ↓ K[u] = W (c1) + · · ·+W (ct) + V (2d1) + · · ·+ V (2ds)

into an orthogonal direct sum of K[u]-modules, where 1 ≤ c1 ≤ · · · ≤ ct, 1 ≤d1 ≤ · · · ≤ ds, and each V (2k) occurs at most twice as a summand (equivalently,di+2 > di for all i). Furthermore, since u acts on W (ci) with two Jordan blocks ofsize ci and on V (2dj) with a single Jordan block of size 2dj , we have

∑ti=1 ci +∑s

j=1 dj = l.Let p again be arbitrary, and let (ci) and (dj) be the two (possibly empty)

sequences associated with V ↓ K[u] as above. De�ne ct+i = di for all 1 ≤ i ≤ s.Set k1 = 1, and ki = 1 + c1 + · · ·+ ci−1 for 1 < i ≤ t+ s. For all 1 ≤ i ≤ t+ s, letVi be the subspace of V spanned by Bi = {e±j : ki ≤ j ≤ ki + ci − 1}. Now V isan orthogonal direct sum V = V1 ⊕ · · · ⊕ Vt+s, so we have a naturally embeddedsubgroup Sp(V1)× · · · × Sp(Vt+s) < G, where Sp(Vi) is the subgroup of all g ∈ Gsuch that g(Vi) = Vi and gej = ej for all ej 6∈ Bi.

It follows then that a representative of the unipotent class determined by thetwo sequences (ci) and (dj) is given by u1 · · ·ut+s, where ui ∈ Sp(Vi) and we havethe following:

(C1) If 1 ≤ i ≤ t, then ui acts on Vi with two Jordan blocks of size ci. Furthermore,if p = 2, then Vi ↓ K[ui] = W (ci).

(C2) If t+ 1 ≤ i ≤ t+ s, then ui acts on Vi with a single Jordan block of size 2di.

Note that by Lemma 2.8.9, the subgroup Sp(Vi) is a subsystem subgroup ofSp(V ) with root system Φi = {±(εj ± εj′) : ki ≤ j < j′ ≤ ki + ci − 1} ∪ {±2εj :ki ≤ j ≤ ki + ci − 1} of type Cci . Furthermore, the root system Φi has a base∆i = {αj : ki ≤ j ≤ ki + ci − 2} ∪ {2εki+ci−1}.

For 1 ≤ i ≤ t, if ci = 1, it is clear that (C1) is satis�ed by ui = 1. Suppose thenthat ci > 1. Now it is a consequence of the next lemma that ui =

∏ki+ci−2j=ki

xαj (1)satis�es (C1).

Lemma 2.8.10. Let G = Sp(V ) be a Chevalley group of type Cl as de�ned above,where l ≥ 2. Set u = xα1(1) · · ·xαl−1

(1). Then u acts on V with two Jordan blocksizes of l. Furthermore, if p = 2, we have V ↓ K[u] = W (l).

Proof. Now by Lemma 2.8.1, the element u is a regular unipotent element of aLevi factor L ∼= SLl(K) of type Al−1. For p = 2, it is shown in [LS12, 6.1] that wehave V ↓ K[u] = W (l), as desired.

Suppose then that p 6= 2. As noted in [LS12, Proof of Corollary 3.6], we haveV ↓ L ∼= W ⊕W ∗, where W is the natural module of L. Now by Proposition 2.3.3(i), a regular unipotent element of L acts on W (hence also on W ∗) with a singleJordan block of size l, so the claim follows.

Consider then t+1 ≤ i ≤ t+s. It follows from Proposition 2.3.3 and Proposition2.4.4 (v) that (C2) is satis�ed when we take ui to be a regular unipotent elementof Sp(Vi). Thus by Lemma 2.8.1, condition (C2) is satis�ed when we pick ui =x2εki+ci−1

(1) if ci = 1, and ui =∏ki+ci−2j=ki

xαj (1) · x2εki+ci−1(1) if ci > 1.

Putting all of this together, we get the following two lemmas which describerepresentatives for the unipotent classes of G.

Lemma 2.8.11 (Type C (p 6= 2)). Assume that p 6= 2. Let G = Sp(V ) be aChevalley group of type Cl as de�ned above, where l ≥ 1. Let 1 ≤ c1 ≤ · · · ≤ ctbe odd, let 1 ≤ d1 ≤ · · · ≤ ds, where

∑ti=1 ci +

∑sj=1 dj = l. Set ct+i = di for all

1 ≤ i ≤ s, and de�ne k1 = 1, and ki = 1 + c1 + · · ·+ ci−1 for 1 1,

x2εki+ci−1(1), if t+ 1 ≤ i ≤ t+ s and ci = 1,∏ki+ci−2

j=kixαj (1) · x2εki+ci−1

(1), if t+ 1 ≤ i ≤ t+ s and ci > 1.

Then u = u1 · · ·ut+s lies in the unipotent class of G determined by the decompo-

sition V ↓ K[u] =t⊕i=1

2 · Vci ⊕s⊕j=1

V2dj .

Lemma 2.8.12 (Type C (p = 2)). Assume that p = 2. Let G = Sp(V ) be aChevalley group of type Cl as de�ned above, where l ≥ 1. Let 1 ≤ c1 ≤ · · · ≤ ctand 1 ≤ d1 ≤ · · · ≤ ds, where di+2 > di for all i, and

∑ti=1 ci +

∑sj=1 dj = l. Set

ct+i = di for all 1 ≤ i ≤ s, and de�ne k1 = 1, and ki = 1 + c1 + · · · + ci−1 for1 1,

x2εki+ci−1(1), if t+ 1 ≤ i ≤ t+ s and ci = 1,∏ki+ci−2

j=kixαj (1) · x2εki+ci−1

(1), if t+ 1 ≤ i ≤ t+ s and ci > 1.

Then u = u1 · · ·ut+s lies in the unipotent class of G determined by the orthogonaldecomposition V ↓ K[u] = W (c1) + · · ·+W (ct) + V (2d1) + · · ·+ V (2ds).

2.8.3 Type Dl, l ≥ 2

α α =∑l

k=1 ckαi

εi − εj , 1 ≤ i < j ≤ l∑

i≤k≤j−1 αk

εi + εj , 1 ≤ i < j ≤ l − 1∑

i≤k≤j−1 αk +∑

j≤k≤l−2 2αk + αl−1 + αl

εi + εl, 1 ≤ i ≤ l − 1∑

i≤k≤l−2 αk + αl

Table 2.7: Type Dl, expressions for roots α ∈ Φ+ in terms of base ∆.

Let e1, . . . , el, e−l, . . . , e−1 be a basis for a complex vector space VC, and let VZ bethe Z-lattice spanned by this basis. We have a non-degenerate symmetric bilinear

40 Chapter 2.

form (−,−) on VC de�ned by (ei, e−i) = 1 = (e−i, ei) and (ei, ej) = 0 for i 6=−j. Let so(VC) be the Lie algebra formed by the linear endomorphisms X of VCsatisfying (Xv,w) + (v,Xw) = 0 for all v, w ∈ VC. Then so(VC) is a semisimpleLie algebra of type Dl. Note that here type D2 is the same as type A1 ×A1, andD3 is the same as type A3.

Let h be the Cartan subalgebra formed by the diagonal matrices in so(VC).Then h = {diag(h1, . . . , hl,−hl, . . . ,−h1) : hi ∈ C}. For 1 ≤ i ≤ l, de�ne mapsεi : h → C by εi(h) = hi where h is a diagonal matrix with diagonal entries(h1, . . . , hl,−hl, . . . ,−h1). Now Φ = {±(εi ± εj) : 1 ≤ i < j ≤ l} is the rootsystem of so(VC), and Φ+ = {εi ± εj : 1 ≤ i < j ≤ l} is a system of positiveroots. Here the base ∆ of Φ corresponding to Φ+ is ∆ = {α1, . . . , αl}, whereαi = εi − εi+1 for 1 ≤ i < l and αl = εl−1 + εl. We give the expressions of rootsα ∈ Φ+ in terms of ∆ in Table 2.7.

As before, for all i, j let Ei,j be the linear endomorphism on VC such thatEi,j(ej) = ei and Ei,j(ek) = 0 for k 6= j. Then a Chevalley basis for so(VC) withpositive structure constants for extraspecial pairs is given by


Xεi+εj = Ej,−i − Ei,−j for all i < j,

X−(εi+εj) = E−i,j − E−j,i for all i < j.

Now VZ is a UZ-invariant lattice in VC. We de�ne V = VZ ⊗Z K. By abuse ofnotation we identify the basis (ei ⊗ 1) of V with (ei).

Note that (−,−) also de�nes a non-degenerate form on V . Furthermore, thequadratic form qC : VC → C de�ned by qC(v) = 1

2(v, v) restricts to qZ : VZ → Z.Thus we have a quadratic form q = qZ ⊗Z K on V which has the form (−,−)as its polarization. Let G be the Chevalley group of type Dl given by V . Then qis a non-degenerate G-invariant quadratic form on V , and we have the followingresult.

Lemma 2.8.13 ([Ree57, 6], [Hée84, 14.2]). The Chevalley group of type Dl inducedby V is equal to the group G = SO(V ).

Here by SO(V ) we mean the identity component of O(V ), where O(V ) is thesubgroup of all g ∈ GL(V ) which satisfy q(gv) = q(v) for all v ∈ V . Let u ∈ G bea unipotent element.

Suppose �rst that p 6= 2. Considering the action of u on V , by Proposition 2.3.2(ii) each Jordan block of even size must occur with even multiplicity. Furthermore,the dimension dimV is even, so there must be an even number of Jordan blocksof odd size. Thus we have a decomposition

V ↓ K[u] =

t⊕i=1

2 · Vci ⊕t+s⊕j=t+1

(V2dj+1 ⊕ V2d′j+1

),

where 1 < c1 ≤ · · · ≤ ct are even, and 0 ≤ dt+1 ≤ d′t+1 ≤ · · · ≤ dt+s ≤ d′t+s, and∑ti=1 ci +

∑t+sj=t+1(dj + d′j + 1) = l.

By Proposition 2.3.2, this decomposition uniquely determines the conjugacyclass of u in G, except when V ↓ K[u] =

⊕ti=1 2 · Vci , in which case we have two

classes with this decomposition. To �nd representatives for each of these classes,let τ : G → G be the automorphism associated with the involutory graph au-tomorphism of the Dynkin diagram which swaps the two end nodes. By Lemma

2.8.6 (i), if u is some unipotent element with V ↓ K[u] =⊕t

i=1 2 · Vci , then uand τ(u) are not conjugate. Therefore it will su�ce to �nd some u ∈ G with thisdecomposition, as then u and τ(u) give representatives for the two classes.

Consider next p = 2. By Proposition 2.4.4, we have that u acts on V with aneven number of Jordan blocks, and thus there is an orthogonal decomposition

V ↓ K[u] =t⊕i=1

W (ci)⊕t+s⊕j=t+1

(V (2dj) + V (2d′j)

)where 1 ≤ c1 ≤ · · · ≤ ct, and 1 ≤ dt+1 ≤ d′t+1 ≤ · · · ≤ dt+s ≤ d′t+s, and∑t

i=1 ci +∑t+s

j=t+1(dj + d′j) = l.By Proposition 2.4.4 (iv), this decomposition uniquely determines the conju-

gacy class of u in G, except when V ↓ K[u] =⊕t

i=1W (ci), in which case we havetwo classes with this decomposition. As in the p 6= 2 case, it will su�ce to �nd justone u with such a decomposition. Then by Lemma 2.8.6 (ii) the elements u andτ(u) are representatives for the two conjugacy classes with this decomposition.

Let p again be arbitrary. Let (ci), (dj), and (d′j) be the (possibly empty)sequences associated with V ↓ K[u] as above. Now for t + 1 ≤ i ≤ t + s, setci = di+d

′i+1 if p 6= 2 and ci = di+d

′i if p = 2. Set k1 = 1 and ki = 1+c1+· · ·+ci−1

for 1 < i ≤ t+ s. For all 1 ≤ i ≤ t+ s, let Vi be the subspace of spanned by Bi ={e±j : ki ≤ j ≤ ki+ci−1}. Now V is an orthogonal direct sum V = V1⊕· · ·⊕Vt+s,so we have a naturally embedded subgroup SO(V1)×· · ·×SO(Vt+s), where SO(Vi)is the subgroup of all g ∈ SO(V ) such that gej = ej for all ej 6∈ Bi.

It follows then that for a u ∈ G with V ↓ K[u] as given by the sequences (ci),(dj), and (d′j), we can choose u = u1 · · ·ut+s, where ui ∈ SO(Vi) and we have thefollowing:

(D1) If 1 ≤ i ≤ t, then ui acts on Vi with two Jordan blocks of size ci. Furthermore,if p = 2, then Vi ↓ K[ui] = W (ci).

(D2) If t + 1 ≤ i ≤ t + s and p 6= 2, then ui acts on Vi with Jordan blocks[2di + 1, 2d′i + 1].

(D3) If t+ 1 ≤ i ≤ t+ s and p = 2, then Vi ↓ K[ui] = V (2di) + V (2d′i).

Note that by Lemma 2.8.13, the subgroup SO(Vi) is a subsystem subgroupof SO(V ) with root system Φi = {±(εj ± εj′) : ki ≤ j < j′ ≤ ki + ci − 1} oftype Dci . Furthermore, the root system Φi has a base ∆i = {αj : ki ≤ j ≤ki + ci − 2} ∪ {εki+ci−2 + εki+ci−1}.

Consider �rst ui for 1 ≤ i ≤ t. Now it is a consequence of the next lemma that(D1) holds for ui =

∏ki+ci−2j=ki

xαj (1).

Lemma 2.8.14. Let G = SO(V ) be a Chevalley group of type Dl as de�ned above,where l ≥ 2. Set u = xα1(1) · · ·xαl−1

(1). Then u acts on V with two Jordan blocksizes of l. Furthermore, if p = 2, we have V ↓ K[u] = W (l).

Proof. The claim follows with the same proof as Lemma 2.8.10.

For ui with t+1 ≤ i ≤ t+s, consider �rst p 6= 2. If di = d′i = 0, condition (D2)holds for ui = 1. If di = 0 and d′i > 0, then by Proposition 2.3.3 condition (D2)holds when ui is a regular unipotent element of SO(Vi). Thus by Lemma 2.8.1, we

42 Chapter 2.

have (D2) when ui =∏ki+ci−2j=ki

xαj (1) · xεki+ci−2+εki+ci−1(1). If di, d′i > 0, then by

the next lemma, the condition (D2) holds when ui = viv′i, where

vi =

ki+di−2∏j=ki

xαj (1) · xεki+di−1−εki+ci−1(1) · xεki+di−1+εki+ci−1

(1)

and

v′i =

ki+ci−2∏j=ki+di

xαj (1) · xεki+ci−2+εki+ci−1(−1).

Lemma 2.8.15. Assume that p 6= 2. Let G = SO(V ) be a Chevalley group oftype Dl as de�ned above, where l ≥ 3. Let e + f + 1 = l, where e, f > 0. Setu1 =

∏e−1i=1 xαi(1) · xεe−εl(1) · xεe+εl(1) and u2 =

∏l−1i=e+1 xαi(1) · xαl(−1). Then

the unipotent element u = u1u2 acts on V with Jordan blocks [2e+ 1, 2f + 1].

Proof. The claim is proven by Suprunenko in [Sup09, Lemma 2.24], where theexpression with root subgroups has some di�erent signs since the Chevalley basisused in [Sup09] has di�erent structure constants than our Chevalley basis. We givean outline for the proof for completeness. We have V = W ⊕W ′ an orthogonaldirect sum, whereW = 〈e±1, . . . , e±e, el−e−l〉 andW ′ = 〈e±(e+1), . . . , e±(l−1), el+e−l〉. Thus we have a naturally embedded subgroup SO(W )× SO(W ′) < G. NowSO(W ) is generated by the root elements corresponding to simple roots and theirnegatives, and these are

x±αi(c) c ∈ K and 1 ≤ i ≤ e− 1,

xεe−εl(c)xεe+εl(c) c ∈ K,x−εe+εl(c)x−εe−εl(c) c ∈ K.

The corresponding root elements for SO(W ′) are

x±αi(c) c ∈ K and e+ 1 ≤ i ≤ l − 2,

xαl−1(c)xαl(−c) c ∈ K,

x−αl−1(c)x−αl(−c) c ∈ K.

By Lemma 2.8.1 the element u1 is a regular unipotent element of SO(W ) andu2 is a regular unipotent element of SO(W ′), so the claim follows from Proposition2.3.3 (iii).

Finally we explain how to choose ui for t + 1 ≤ i ≤ t + s when p = 2. Ifdi = 1, then by 2.4.4 (vi) we have (D3) when ui is a regular unipotent element ofSO(Vi). It follows then from Lemma 2.8.1 that we can choose ui =

∏ki+ci−2j=ki

xαj (1)·xεki+ci−2+εki+ci−1

(1) and (D3) will hold. Suppose then that di > 1, so now di+d′i ≥

4. It follows from the next lemma (recall that we are assuming d′i ≥ di, so hereci − 2 ≥ d′i ≥

ci2 ) that (D3) holds when

ui =

ki+d′i−2∏

j=ki

xαj (1) · xεki+d′i−1−εki+ci−1(1) · xεki+d′i−1+εki+ci−1

(1) ·ki+ci−2∏j=ki+d′i

xαj (1).

Lemma 2.8.16. Assume that p = 2. Let G = SO(V ) be a Chevalley group oftype Dl as de�ned above, where l ≥ 4. Let l − 2 ≥ e ≥ l

2 . Then for the unipotent

element u =∏e−1i=1 xαi(1)·xεe−εl(1)·xεe+εl(1)·

∏l−1i=e+1 xαi(1), we have an orthogonal

decomposition V ↓ K[u] = V (2e) + V (2l − 2e).

Proof. We compute explicitly the action of u on the basis e±1, . . . , e±l to determinethe decomposition V ↓ K[u]. By looking at the Chevalley basis we are using, wesee that xαi(1) = I+Ei,i+1 +E−(i+1),−i for all 1 ≤ i ≤ l−1, where I is the identityand Ei,j is de�ned as before. Furthermore, we see that xεe−εl(1) = I+Ee,l+E−l,−eand xεe+εl(1) = I +El,−e +Ee,−l. Now it is straightforward to see that u acts onthe basis elements as follows:

e1 7→ e1

(2 ≤ i ≤ e) ei 7→ ei + ei−1 + · · ·+ e1

(1 ≤ i ≤ e− 1) e−i 7→ e−i + e−(i+1)

e−e 7→ el + e−l + e−e + ee + ee−1 + · · ·+ e1

ee+1 7→ ee+1

(e+ 2 ≤ i ≤ l − 1) ei 7→ ei + ei−1 + · · ·+ ee+1

el 7→ el + el−1 + · · ·+ e1

(e+ 1 ≤ i ≤ l − 2) e−i 7→ e−i + e−(i+1)

e−(l−1) 7→ e−(l−1) + e−l + ee + ee−1 + · · ·+ e1

e−l 7→ e−l + ee + ee−1 + · · ·+ e1

A calculation shows that the �xed point space of u has dimension 2, and that itis spanned by e1 and ee+1. Therefore the Jordan normal form of u has two Jordanblocks. To see that the Jordan block sizes are 2e and 2l− 2e, it will be enough toshow that (u− 1)2e = 0 and (u− 1)2e−1 6= 0, as then the largest Jordan block sizein u is 2e. To this end, a calculation shows that (u − 1)2eei = 0 for all the basisvectors ei, and that

(u− 1)2e−1(e−1) =

{e1, if e > l/2,

ee+1 + e1, if e = l/2.

It follows then from Proposition 2.4.4 that either V ↓ K[u] = V (2e)+V (2e−2l)or V ↓ K[u] = W (2e). Our goal is to show that V ↓ K[u] = V (2e) + V (2e − 2l).Now (u− 1)2e = 0, so by Lemma 2.4.8 this is equivalent to �nding a v ∈ V suchthat ((u− 1)2e−1v, v) 6= 0. By the calculation done above, we can choose v = e−1,as then ((u− 1)2e−1v, v) = (e1, e−1) 6= 0.

Now putting all of this together, we get the following two lemmas.

Lemma 2.8.17 (Type D (p 6= 2)). Assume that p 6= 2. Let G = SO(V ) be aChevalley group of type Dl as de�ned above, where l ≥ 2. Let 1 < c1 ≤ · · · ≤ ct beeven, let 0 ≤ dt+1 ≤ d′t+1 ≤ · · · ≤ dt+s ≤ d′t+s such that

∑ti=1 ci+

∑t+sj=t+1(dj+d

′j+

1) = l. Set ci = di+d′i+1 for t+1 ≤ i ≤ t+s, set k1 = 1 and ki = 1+c1+· · ·+ci−1

for 1 0, set

vi =

ki+di−2∏j=ki

xαj (1) · xεki+di−1−εki+ci−1(1) · xεki+di−1+εki+ci−1

(1).

For all t+ 1 ≤ i ≤ t+ s with d′i > 0, set

v′i =

ki+ci−2∏j=ki+di

xαj (1) · xεki+ci−2+εki+ci−1(−1).

44 Chapter 2.

De�ne

ui =

∏ki+ci−2j=ki

xαj (1), if 1 ≤ i ≤ t.1, if t+ 1 ≤ i ≤ t+ s and di = d′i = 0.

v′i, if t+ 1 ≤ i ≤ t+ s and di = 0, d′i > 0.

viv′i if t+ 1 ≤ i ≤ t+ s and di, d′i > 0.

Then u = u1 · · ·ut+s satis�es V ↓ K[u] =t⊕i=1

2 · Vci ⊕t+s⊕j=t+1

(V2dj+1 ⊕ V2d′j+1

).

Lemma 2.8.18 (Type D (p = 2)). Assume that p = 2. Let G = SO(V ) be aChevalley group of type Dl as de�ned above, where l ≥ 2. Let 1 ≤ c1 ≤ · · · ≤ ct,and 1 ≤ dt+1 ≤ d′t+1 ≤ · · · ≤ dt+s ≤ d′t+s, such that

∑ti=1 ci+

∑t+sj=t+1(dj+d′j) = l.

Set ci = di + d′i for t + 1 ≤ i ≤ t + s, set k1 = 1 and ki = 1 + c1 + · · · + ci−1 for1 1, set

vi =

ki+d′i−2∏

j=ki

xαj (1) · xεki+d′i−1−εki+ci−1(1) · xεki+d′i−1+εki+ci−1

(1).

For all t+ 1 ≤ i ≤ t+ s with d′i > 1, set

v′i =

ki+ci−2∏j=ki+d′i

xαj (1).

De�ne

ui =

1, if 1 ≤ i ≤ t and ci = 1.∏ki+ci−2j=ki

xαj (1), if 1 ≤ i ≤ t and ci > 1.

v′i, if t+ 1 ≤ i ≤ t+ s and di = 1.

viv′i, if t+ 1 ≤ i ≤ t+ s and di > 1.

Then u = u1 · · ·ut+s satis�es V ↓ K[u] =

t⊕i=1

W (ci)⊕t+s⊕j=t+1

(V (2dj) + V (2d′j)

).

What remains is to describe representatives in the cases where we have twounipotent conjugacy classes with the same decomposition V ↓ K[u]. For thesecases, representatives are given by the next two lemmas.

Lemma 2.8.19 (Type D, split classes). Let G = SO(V ) be a Chevalley group oftype Dl as de�ned above, where l ≥ 2. Let 1 < c1 ≤ · · · ≤ ct be such that ci is evenfor all i and

∑li=1 ci = l. Set k1 = 1 and ki = 1 + c1 + · · · + ci−1 for 1 2. Then:

(i) If p 6= 2, the unipotent elements u1 · · ·ut−1ut and u1 · · ·ut−1u′t are repre-

sentatives for the two classes of unipotent elements u with V ↓ K[u] =⊕ti=1 2 · Vci .

(ii) If p = 2, the unipotent elements u1 · · ·ut−1ut and u1 · · ·ut−1u′t are repre-

sentatives for the two classes of unipotent elements u with V ↓ K[u] =⊕ti=1W (ci).

Proof. For both claims, we see from Lemma 2.8.17 and Lemma 2.8.18 above thatfor u = u1 · · ·ut we have V ↓ K[u] =

⊕ti=1 2 · Vci if p 6= 2 and V ↓ K[u] =⊕t

i=1W (ci) if p = 2. Furthermore, for the graph automorphism τ of G inducedby the graph automorphism of ∆ swapping the two end nodes, we have τ(xαi(1)) =xαi(1) for 1 ≤ i ≤ l − 2, and τ(xαl−1

(1)) = xαl(1), τ(xαl(1)) = xαl−1(1). Thus

τ(u) = u1 · · ·ut−1u′t, so now the claim follows from Lemma 2.8.6.

2.8.4 Type Bl, l ≥ 2 (p = 2)

Suppose that p = 2. Let Φ be a simple root system of type Cl and let G be asimply connected Chevalley group over K with this root system. Then the dualroot system Φ∨ is a simple root system of type Bl. Denote by G∨ the simplyconnected Chevalley group over K with root system Φ∨, and denote by α∨ thedual root corresponding to a root α ∈ Φ. Since p = 2, it follows from [Ste68,Theorem 28] that there is an isogeny f : G→ G∨ of algebraic groups satisfying

f(xα(c)) =

{xα∨(c) if α is a long root,

xα∨(c2) if α is a short root.

for all α ∈ Φ and c ∈ K.Now by Lemma 2.1.2 (i), the map f de�nes a bijection between the unipotent

conjugacy classes of G and G∨. Furthermore, if ∆ is a base for Φ, then ∆∨ ={α∨ : α ∈ ∆} is a base for Φ∨, and (Φ∨)+ = {α∨ : α ∈ Φ+} is the set of positiveroots. Thus by using Lemma 2.8.12, we �nd representatives

∏α∈(Φ∨)+ xα∨(cα∨)

for unipotent conjugacy classes of G∨, as desired.

2.8.5 Type Bl, l ≥ 1 (p 6= 2)

α α =∑l

k=1 ckαi

εi − εj , 1 ≤ i < j ≤ l∑

i≤k≤j−1 αk

εi + εj , 1 ≤ i < j ≤ l∑

i≤k≤j−1 αk +∑

j≤k≤l 2αk

εi, 1 ≤ i ≤ l∑

i≤k≤l αk

Table 2.8: Type Bl, expressions for roots α ∈ Φ+ in terms of base ∆.

Assume that p 6= 2. Let e1, . . . , el, e0, e−l, . . . , e−1 be a basis for a complex vectorspace VC, and let VZ be the Z-lattice spanned by this basis. We have a non-degenerate symmetric bilinear form (−,−) on VC de�ned by (ei, e−i) = 1 =(e−i, ei), by (e0, e0) = 2, and (ei, ej) = 0 for i 6= −j. Let so(VC) be the Lie algebraformed by the linear endomorphisms X of VC satisfying (Xv,w) + (v,Xw) = 0for all v, w ∈ VC. Then so(VC) is a simple Lie algebra of type Bl. Let h bethe Cartan subalgebra formed by the diagonal matrices in so(VC). Then h ={diag(h1, . . . , hl, 0,−hl, . . . ,−h1) : hi ∈ C}. For 1 ≤ i ≤ l, de�ne maps εi : h→ Cby εi(h) = hi where h = diag(h1, . . . , hl, 0,−hl, . . . ,−h1). Now Φ = {±(εi ±εj) : 1 ≤ i < j ≤ l} ∪ {±εi : 1 ≤ i ≤ l} is the root system of so(VC), andΦ+ = {εi ± εj : 1 ≤ i < j ≤ l} ∪ {εi : 1 ≤ i ≤ l} is a system of positive roots. Thebase ∆ of Φ corresponding to Φ+ is ∆ = {α1, . . . , αl}, where αi = εi − εi+1 for

46 Chapter 2.

1 ≤ i < l and αl = εl. We give the expressions of roots α ∈ Φ+ in terms of ∆ inTable 2.8.

As before, for all i, j let Ei,j be the linear endomorphism on VC such thatEi,j(ej) = ei and Ei,j(ek) = 0 for k 6= j. Then a Chevalley basis for so(VC) withpositive structure constants for extraspecial pairs is given by


Xεi+εj = Ej,−i − Ei,−j for all i < j,

X−(εi+εj) = E−i,j − E−j,i for all i < j,

Xεi = 2Ei,0 − E0,−i for all i,

X−εi = E0,i − 2E−i,0 for all i.

Now VZ is a UZ-invariant lattice in VC. We de�ne V = VZ⊗ZK. Note that (−,−)also de�nes a non-degenerate symmetric form on V . Then we have the followingresult.

Lemma 2.8.20 ([Ree57, 7]). The Chevalley group of type Bl induced by V is equalto the group G = SO(V ) of invertible linear maps V → V with determinant 1 thatpreserve (−,−).

By abuse of notation we identify the basis (ei⊗1) of V with (ei). Let u ∈ G bea unipotent element. Since p 6= 2, by Proposition 2.3.1 and Proposition 2.3.2 theconjugacy class of u in G is uniquely determined by the Jordan block sizes of uon V . By Proposition 2.3.2 (ii), we know that each Jordan block of even size mustoccur with even multiplicity. Furthermore, since dimV is odd, the total number ofJordan blocks of odd size must be odd. Thus the conjugacy class of u is uniquelydetermined by the decomposition

V ↓ K[u] =t⊕i=1

2 · Vci ⊕s⊕j=1

(V2dj+1 ⊕ V2d′j+1

)⊕ V2ds+1+1,

where 1 ≤ c1 ≤ · · · ≤ ct are even, and 0 ≤ d1 ≤ d′1 ≤ · · · ≤ ds ≤ d′s ≤ ds+1, and∑ti=1 ci +

∑sj=1(dj + d′j + 1) + ds+1 = l.

Now letW1 be the subspace of V spanned by B1 = {e±j : 1 ≤ j ≤ l−ds+1} andlet W2 be the subspace of V spanned by B2 = {e±j : l− ds+1 + 1 ≤ j ≤ l} ∪ {e0}.We have an orthogonal direct sum V = W1⊕W2, so there is a naturally embeddedsubgroup SO(W1)×SO(W2) < SO(V ), where SO(Wi) is the subgroup of all g ∈ Gsuch that g(Wi) = Wi and gej = ej for all ej 6∈ Bi.

It follows then that a representative for the unipotent class determined bythe (possibly empty) sequences (ci), (dj), and (d′j) is given by u = u1u2, whereu1 ∈ SO(W1) and

W1 ↓ K[u1] =

t⊕i=1

2 · Vci ⊕s⊕j=1

(V2dj+1 ⊕ V2d′j+1

),

and where u2 ∈ SO(W2) and u2 acts on W2 with a single Jordan block of size2ds+1 + 1.

For �nding u1, note �rst that if dimW1 ≤ 2, then we can choose u1 = 1.Suppose then that dimW1 > 2 and write dimW1 = 2k. Then it is a consequenceof Lemma 2.8.13 that SO(W1) is a subsystem subgroup of type Dk with root

2.9. Computation of Jordan block sizes in irreducible representations 47

system {±(εi ± εj) : 1 ≤ i < j ≤ k}. The root system has a base {εi − εi+1 : 1 ≤i ≤ k− 1} ∪ {εk−1 + εk}, with set of positive roots {εi ± εj : 1 ≤ i < j ≤ k}. Nowone can apply Lemma 2.8.17 to �nd a suitable u1 ∈ SO(W1) as a product of rootelements.

We proceed to explain how to �nd u2. If ds+1 = 0, then it is clear that we canchoose u2 = 1. Suppose then that ds+1 > 0. Now it is a consequence of Lemma2.8.20 that SO(W2) is a subsystem subgroup of G with root system {±(εi ± εj) :l − ds+1 + 1 ≤ i < j ≤ l} ∪ {±εi : l − ds+1 + 1 ≤ i ≤ l} and a base {αi :l − ds+1 + 1 ≤ i ≤ l}. It follows from Proposition 2.3.3 and Lemma 2.8.1 that wecan choose u2 =

∏lj=l−ds+1+1 xαj (1).

2.9 Computation of Jordan block sizes in irreduciblerepresentations

For our solution of Problem 1.1.6, we have to answer the following question insome speci�c cases.

Problem 2.9.1. Let ϕ : G → GL(V ) be a non-trivial rational irreducible repre-sentation of G. For a unipotent element u ∈ G, what are the Jordan block sizes ofϕ(u)?

Since in characteristic p = 2 the Jordan block sizes do not determine when aunipotent element of a classical group (Sp(V ) or SO(V )) is distinguished (Section2.4), we will also have to consider the following question.

Problem 2.9.2. Assume that p = 2. Let ϕ : G → GL(V ) be a non-trivial ratio-nal irreducible representation of G with a non-degenerate G-invariant alternatingbilinear form. For a unipotent element u ∈ G, what is the conjugacy class of ϕ(u)in Sp(V )?

Recall that we are working over a �eld of positive characteristic. Thus in generalProblem 2.9.1 and Problem 2.9.2 are out of reach by current methods, as even�nding the dimensions of the rational irreducible representations is a major openproblem. However, for �xed G, a �xed unipotent conjugacy class C of G, and Vof small dimension, it is possible to calculate the Jordan block sizes of ϕ(u) foru ∈ C and the orthogonal decomposition of V ↓ K[u] with a computer. We willproceed to explain the methods we have used to do this.

First note that using the methods from 2.8, we can �nd a representative u =∏α∈Φ+ xα(cα) of C , where cα ∈ Z for all α ∈ Φ+. Let λ ∈ X(T )+ be non-zero

p-restricted dominant weight, and write λ =∑l

i=1 aiωi, where 0 ≤ ai ≤ p−1. Ourgoal is to determine the decomposition LG(λ) ↓ K[u].

Recall that the Weyl module VG(λ) has a unique maximal G-submodule M ,and VG(λ)/M ∼= LG(λ). Furthermore, M is also the unique maximal G(Fp)-module of VG(λ) by [Won72, Theorem 2D]. Since our representative u is in G(Fp),we do our computations with the �nite group G(Fp).

The G(Fp)-module VG(λ) can be constructed with the Chevalley construction.This can be done with MAGMA, which uses an implementation of the algorithmin [CMT04]. For example, the following MAGMA commands construct the repre-sentation r : G(Fp)→ GL(VG(ω2)) for G(Fp) of type A3:

48 Chapter 2.

G := GroupOfLieType ( "A3" , GF(p) : I sogeny := "SC" ) ;r := HighestWeightRepresentat ion (G, [ 0 , 1 , 0 ] ) ;

In particular, with respect to a �xed K-vector space basis of weight vectors(given by MAGMA), we can compute the matrix of xα(c) acting on VG(λ) for allα ∈ Φ and c ∈ Fp. In the example above, we would get the matrix of xα1+α2(1)acting on VG(λ) with the following commands. Note that here xα1+α2(1) = xα4(1)in the ordering of Φ+ used in MAGMA, which is the total order ≤ de�ned inSection 2.8.

g := e l t < G | <4,1>>;r ( g ) ;

Since VG(λ) has a unique maximal submodule, it follows that VG(λ)∗ has asimple socle, isomorphic to LG(λ)∗ and generated by a maximal vector f ∈ VG(λ)∗

of weight −w0(λ), which is unique up to a scalar. Then using the matrices givingthe action of xα(c) on VG(λ), we can compute a basis for Gf ∼= LG(λ)∗ and thusthe matrix of xα(c) acting on LG(λ)∗ for any α ∈ Φ and c ∈ Fp. By using theexpression of u as a product of the elements xα(c), we then get the matrix of uacting on LG(λ)∗. Now it is a simple matter of computing ranks of some matricesto �nd the Jordan block sizes of u acting on LG(λ)∗ (see Lemma 3.1.2). TheseJordan block sizes are the same as of u acting on LG(λ), giving us the Jordanblock sizes occurring LG(λ) ↓ K[u].

If p 6= 2 we now have all the information we need for our purposes. If p = 2and LG(λ) is self-dual, we also need a way to determine the conjugacy class ofthe image of u in Sp(LG(λ)). To this end, one can use the MAGMA commandInvar iantBi l inearForms to �nd a non-degenerate alternating bilinearform on LG(λ) invariant under the action of xα(c) for all α ∈ Φ and c ∈ K. It isthen straightforward to apply Lemma 2.4.5 to determine the conjugacy class of uin Sp(LG(λ)) (see Remark 2.4.9).

2.10 Action of triality on unipotent conjugacy classesof D4

In this section, we �nish the discussion on the action of graph automorphismsgiven in Section 2.8 and give the action of a graph automorphism of G = D4

induced by triality. We will do this by direct computation. For p = 3 the result isgiven in [LLS14, Proof of Lemma 3.2].

We begin by recalling the Chevalley construction from Section 2.8.3. Let theroot system Φ = {±(εi ± εj) : 1 ≤ i < j ≤ 4} of type D4 be as in 2.8.3, with base∆ = {α1, α2, α3, α4}, where αi = εi − εi+1 for 1 ≤ i ≤ 3, and α4 = ε3 + ε4. Wechoose a Chevalley basis {Xα : α ∈ Φ} ∪ {Hα : α ∈ ∆} as in Section 2.8.3.

Let G be a simply connected Chevalley group constructed with the Chevalleyconstruction, using this Chevalley basis. Let G′ = SO(V ) be the Chevalley groupconstructed using VC as in Section 2.8.3. Denote the root elements generating G′

by x′α(c), where α ∈ Φ and c ∈ K. By [Ste68, Corollary 5, pg.44], we have a

2.10. Action of triality on unipotent conjugacy classes of D4 49

surjective morphism ϕ : G → G′ such that ϕ(xα(c)) = x′α(c) for all α ∈ Φ andc ∈ K.

We will order the positive roots Φ+ = {α1, . . . , α12} with the total order ≤de�ned in Section 2.8. We give the order explicitly in Table 2.1. One �nds thatthe structure constants Nα,β for our Chevalley basis are as given in Table 2.2.

Now let σ′ : ∆ → ∆ be the triality graph automorphism of ∆, which acts onthe simple roots as follows.

α1 7→ α3

α2 7→ α2

α3 7→ α4

α4 7→ α1

Then by Lemma 2.8.4, there exists an isomorphism σ : G → G and signsεα = ±1 such that εα = 1 for all α ∈ ∆, and σ(xα(c)) = xσ′(α)(εαc) for all α ∈ Φand c ∈ K.

The signs εα for the isomorphism σ are uniquely determined, and can befound using the Chevalley commutator relations [Ste68, Corollary to Lemma 15]as follows. Since the root systemD4 is simply laced, the commutator relations havethe simple form [xα(c), xβ(d)] = xα+β(Nα,βcd) for all α, β ∈ Φ such that α+ β ∈Φ. Applying σ to both sides we get the relation εαεβNσ′(α),σ′(β) = εα+βNα,β .From this one can calculate that εα = 1 for all α, except α = ±(α1 + α2) andα = ±(α2 + α4), for which εα = −1.

Our goal next is to compute the action of σ on the unipotent conjugacy classesof G. The �rst step is to �nd representatives for all unipotent conjugacy classes.This can be done with the results of Section 2.8.3. We have listed representativesin Table 2.3 and Table 2.4. Note that there are some pairs of classes of unipotentelements u with same decomposition V ↓ K[u]. To distinguish between them, intables 2.3 and 2.4 we have labeled one of them with the symbol (V ↓ K[u])′ andthe other by (V ↓ K[u])′′.

Now using the signs εα determined above, we �nd σ(u) as a product of rootelements, for each unipotent class representative u ∈ G in tables 2.3 and 2.4. Usingthis and the map ϕ : G→ G′ de�ned above, we �nd explicit matrices for ϕ(σ(u)),which allows one to compute the decomposition V ↓ K[σ(u)] (see Remark 2.4.9).The end result of this computation is that the orthogonal decompositions are asgiven in Table 2.5 and Table 2.6.

What still remains is to check, for the split classes, which image of σ gives theclass (V ↓ K[u])′, and which one gives (V ↓ K[u])′′. Since σ is an automorphismof order 3, the orbits of σ acting on the unipotent classes have size 1 or 3. Fromthis fact and the decompositions V ↓ K[σ(u)] it is easy to deduce that the actionof σ on the unipotent classes is precisely as given in Table 2.5 and Table 2.6.

Remark 2.10.1. There are also many other ways to determine the conjugacyclass of σ(u) without relying on the computation of V ↓ K[σ(u)]. For example, itis immediate from Lemma 2.8.1 that the regular unipotent class is stabilized byσ.

We see in particular that in all cases all distinguished unipotent classes arestabilized by σ. Combining this observation with Lemma 2.8.5 and Lemma 2.8.6,we get the following corollary.

50 Chapter 2.

Proposition 2.10.2. Let G be a simple algebraic group and let σ be a graphautomorphism of G. Then

(i) σ �xes all distinguished unipotent classes of G.

(ii) Let V be a G-module and u ∈ G a distinguished unipotent element. Then V ↓K[u] ∼= V σ ↓ K[u]. In particular, the element u acts on V as a distinguishedunipotent element if and only if u acts on V σ as a distinguished unipotentelement.

α = αi α =∑4

i=1 kiαi α = εi ± εjα1 α1 ε1 − ε2

α2 α2 ε2 − ε3

α3 α3 ε3 − ε4

α4 α4 ε3 + ε4

α5 α1 + α2 ε1 − ε3

α6 α2 + α3 ε2 − ε4

α7 α2 + α4 ε2 + ε4

α8 α1 + α2 + α3 ε1 − ε4

α9 α1 + α2 + α4 ε1 + ε4

α10 α2 + α3 + α4 ε2 + ε3

α11 α1 + α2 + α3 + α4 ε1 + ε3

α12 α1 + 2α2 + α3 + α4 ε1 + ε2

Table 2.1: Positive roots of D4.

α1 α2 α3 α4 α5 α6 α7 α8 α9 α10 α11 α12

α1 1 1 1 1α2 -1 1 1 1α3 -1 -1 1 1α4 -1 -1 1 1α5 1 1 -1α6 -1 -1 1α7 -1 -1 1α8 -1 -1α9 -1 -1α10 -1 1α11 -1α12

Table 2.2: Structure constants Nαi,αj used for D4, with empty entries indicatingzero and roots ordered as in Table 2.1.

2.10. Action of triality on unipotent conjugacy classes of D4 51

Unipotent class RepresentativeV (2) + V (6) xα1(1)xα2(1)xα3(1)xα4(1)V (4)2 xα1(1)xε2−ε4(1)xε2+ε4(1)xα3(1)W (4)′ xα1(1)xα2(1)xα3(1)W (4)′′ xα1(1)xα2(1)xα4(1)W (1) + V (2) + V (4) xα1(1)xα2(1)xε2+ε3(1)W (1) +W (3) xα1(1)xα2(1)W (2) + V (2)2 xα1(1)xε1+ε2(1)xα3(1)(W (2)2)′ xα1(1)xα3(1)(W (2)2)′′ xα1(1)xα4(1)V (2)2 +W (1)2 xα1(1)xε1+ε2(1)W (2) +W (1)2 xα1(1)

Table 2.3: Representatives for unipotent classes of G = D4 when p = 2.

Unipotent class Representative[1, 7] xα1(1)xα2(1)xα3(1)xα4(1)[3, 5] xα1(1)xε2−ε4(1)xε2+ε4(1)xα3(1)xα4(−1)[13, 5] xα1(1)xε2−ε4(1)xε2+ε4(1)[42]′ xα1(1)xα2(1)xα3(1)[42]′′ xα1(1)xα2(1)xα4(1)[12, 32] xα1(1)xα2(1)[1, 22, 3] xα1(1)xε1+ε2(1)xα3(1)[15, 3] xα1(1)xε1+ε2(1)[24]′ xα1(1)xα3(1)[24]′′ xα1(1)xα4(1)[14, 22] xα1(1)

Table 2.4: Representatives for unipotent classes of G = D4 when p 6= 2.

Unipotent class of u Unipotent class of σ(u)

V (2) + V (6) V (2) + V (6)V (4)2 V (4)2

W (4)′ W (1) + V (2) + V (4)W (4)′′ W (4)′

W (1) + V (2) + V (4) W (4)′′

W (1) +W (3) W (1) +W (3)W (2) + V (2)2 W (2) + V (2)2

(W (2)2)′ W (1)2 + V (2)2

(W (2)2)′′ (W (2)2)′

W (1)2 + V (2)2 (W (2)2)′′

W (2) +W (1)2 W (2) +W (1)2

Table 2.5: Action of σ on unipotent classes of G = D4 when p = 2.

52 Chapter 2.

Unipotent class of u Unipotent class of σ(u)

[1, 7] [1, 7][3, 5] [3, 5][13, 5] [42]′′

[42]′ [13, 5][42]′′ [42]′

[12, 32] [12, 32][1, 22, 3] [1, 22, 3][15, 3] [24]′′

[24]′ [15, 3][24]′′ [24]′

[14, 22] [14, 22]

Table 2.6: Action of σ on unipotent classes of G = D4 when p 6= 2.

Chapter 3

Linear algebra

In this chapter, we give various lemmas about unipotent linear maps that will beneeded in the sequel.

Let f : V → V be a linear map. If W is a subspace of V invariant under f , wewill denote the restriction of f to W by fW and the linear map induced on V/Wby fV/W .

3.1 Basic lemmas about unipotent elements

De�nition 3.1.1. Let u ∈ GL(V ) be unipotent. For all m ≥ 1, we denote byrm(u) the number of Jordan blocks of size m in the Jordan decomposition of u.

Lemma 3.1.2. Let u ∈ GL(V ) be unipotent and denote X = u− 1. Then for allm ≥ 1, we have

rm(u) = rankXm+1 + rankXm−1 − 2 rankXm

andrm(u) = 2 dim kerXm − dim kerXm+1 − dim kerXm−1.

Proof. The formulas follow from the fact that dim kerXm =∑m

k=1 tk, where tk isthe number of Jordan blocks of size ≥ k [Jan04, 1.1].

Lemma 3.1.3. Let u be a unipotent matrix of order p. Then the number Jordanblocks of size p in the Jordan decomposition of u is equal to rank(u− 1)p−1.

Proof. By Lemma 3.1.2, we have that the number of Jordan blocks of size p isequal to rank(u−1)p+1+rank(u−1)p−1−2 rank(u−1)p. The claim follows because(u− 1)p = (u− 1)p+1 = 0.

The next lemma is observed in [Sup01, (1), pg. 2585]. We include the proof(which is easy) for completeness.

Lemma 3.1.4. Let u be a unipotent linear map on the vector space V and supposethat u has order p. Suppose that V has a �ltration V = W1 ⊇ W2 ⊇ · · · ⊇ Wt ⊇Wt+1 = 0 of K[u]-submodules. Then rp(u) ≥

∑ti=1 rp(uWi/Wi+1

).

Proof. Let X = u− 1. Writing X as a block triangular matrix, we have

53

54 Chapter 3.

Xp−1 =

(XWt)

p−1 ∗ ∗ · · · ∗0 (XWt−1/Wt

)p−1 ∗ · · · ∗

0 0. . . · · · ∗

......

.... . .

...0 0 0 · · · (XW1/W2

)p−1

so rankXp−1 ≥

∑ti=1 rank(XWi/Wi+1

)p−1 and the claim follows from Lemma3.1.3.

3.2 Jordan blocks in subspaces and quotients

Some of the irreducible representations LG(λ) of G that we have to consider inthe solution of Problem 1.1.6 are best understood as subquotients of certain inde-composable modules W . In many cases that are relevant to us, the module W willbe uniserial of the form LG(0)/LG(λ)/LG(0). For example, if G = SL(V ), thenW = V ⊗V ∗ = LG(0)/LG(ω1 +ωl)/LG(0) (uniserial) if p divides dimV . We havea decent understanding of what Jordan blocks sizes of u ∈ G acting on W looklike, and the idea is that we can use this information to study the Jordan blocksizes of u acting on LG(λ).

We begin with two lemmas which describe how the Jordan block sizes changewhen moving from the whole space to a subspace of codimension one, or to aquotient by a one-dimensional subspace.

Lemma 3.2.1. Let u ∈ GL(V ) be unipotent and denote X = u− 1. Suppose thatW ⊆ V is a subspace invariant under u such that dimV/W = 1. Let m ≥ 0 besuch that kerXm ⊆W and kerXm+1 6⊆W . Then

(a) if m = 0, we have

• r1(uW ) = r1(u)− 1,

• ri(uW ) = ri(u) for all i 6= 1.

(b) if m ≥ 1, we have

• rm+1(uW ) = rm+1(u)− 1,

• rm(uW ) = rm(u) + 1,

• ri(uW ) = ri(u) for all i 6= m,m+ 1.

Proof. Now kerXi ⊆ W for all 0 ≤ i ≤ m, which means that dim kerXiW =

dim kerXi for all 0 ≤ i ≤ m. By Lemma 3.1.2, this implies that for all 1 ≤ i ≤m− 1, we have ri(u) = ri(uW ).

Next note that kerXi 6⊆ W for all i ≥ m + 1, which means that V/W =(kerXi + W )/W ∼= W/ kerXi ∩W . Hence dim kerXi

W = dim kerXi − 1 for alli ≥ m+1. Thus by Lemma 3.1.2, we have rm(uW ) = rm(u)+1 if m ≥ 1. Similarlyrm+1(uW ) = rm+1(u)− 1 and ri(uW ) = ri(u) for all i ≥ m+ 2.

Lemma 3.2.2. Let u ∈ GL(V ) be unipotent and denote X = u− 1. Suppose thatW ⊆ V is a subspace invariant under u such that dimW = 1. Let m ≥ 0 be suchthat W ⊆ Xm(V ) but W 6⊆ Xm+1(V ). Then

3.2. Jordan blocks in subspaces and quotients 55

(a) if m = 0, we have

• r1(uV/W ) = r1(u)− 1,

• ri(uV/W ) = ri(u) for all i 6= 1.

(b) if m ≥ 1, we have

• rm+1(uV/W ) = rm+1(u)− 1,

• rm(uV/W ) = rm(u) + 1,

• ri(uV/W ) = ri(u) for all i 6= m,m+ 1.

Proof. Now XiV/W (V/W ) = (Xi(V ) + W )/W ∼= Xi(V )/(W ∩ Xi(V )). Thus for

0 ≤ i ≤ m we have rankXiV/W = rankXi − 1, because W ⊆ Xi(V ). Similarly

we have rankXiV/W = rankXi for i ≥ m + 1, because W 6⊆ Xi(V ) and thus

W ∩ Xi(V ) = 0. Therefore by Lemma 3.1.2, we have rm(uV/W ) = rm(u) + 1 ifm ≥ 1. Similarly one sees rm+1(uV/W ) = rm+1(u) − 1 and ri(uV/W ) = ri(u) forall i 6= m,m+ 1.

In general, if u ∈ GL(V ) is unipotent, then it is possible to �nd a �ltration0 = V0 ⊂ V1 ⊂ · · · ⊂ Vn = V of u-invariant subspaces such that dimVi/Vi−1 = 1for all 1 ≤ i ≤ n. For any such �ltration, we can use Lemma 3.2.1 and Lemma3.2.2 to describe, in terms of the Jordan block sizes of u on V , the possible Jordanblock sizes of u acting on any subquotient Vi/Vj . We record one such descriptionin the next lemma.

Lemma 3.2.3. Let u ∈ GL(V ) be unipotent and suppose that there exist u-invariant subspaces W ′ ⊆ W ⊆ V such that dimV/W = 1 and dimW ′ = 1.Denote the image of u in GL(W/W ′) by u′. Then one of the following holds:

(i) r1(u′) = r1(u)− 2, and rk(u′) = rk(u) for all k 6= 1.

(ii) There exists m ≥ 2 such that rm(u′) = rm(u)− 2, rm−1(u′) = rm−1(u′) + 2,and rk(u′) = rk(u) for all k 6∈ {m,m− 1}.

(iii) There exists m ≥ 3 such that rm(u′) = rm(u)− 1, rm−2(u′) = rm−2(u) + 1,and rk(u′) = rk(u) for all k 6∈ {m,m− 2}.

(iv) r2(u′) = r2(u)− 1, and rk(u′) = rk(u) for all k 6= 2.

(v) There exists m ≥ n + 2, n ≥ 2 such that rm(u′) = rm(u) − 1, rm−1(u′) =rm−1(u) + 1, rn(u′) = rn(u)− 1, rn−1(u′) = rn−1(u) + 1, and rk(u′) = rk(u)for all k 6∈ {m,m− 1, n, n− 1}.

(vi) There exists m ≥ 3 such that rm(u′) = rm(u)− 1, rm−1(u′) = rm−1(u) + 1,r1(u′) = r1(u)− 1, and rk(u′) = rk(u) for all k 6∈ {1,m,m− 1}.

Proof. The claim follows by �rst applying Lemma 3.2.1 to u and W , and thenLemma 3.2.2 on uW and W/W ′.

De�nition 3.2.4. Suppose that p 6= 2. We will say that the action of a unipotentelement u ∈ GL(V ) on V is inadmissible, if at least one of the following holds:

56 Chapter 3.

• Some Jordan block of u of size > 1 has multiplicity ≥ 3.

• There are ≥ 2 Jordan block sizes of u with multiplicity ≥ 2.

Otherwise we will say that the action of u on V is admissible.

De�nition 3.2.5. Suppose that p = 2. We will say that the action of a unipotentelement u ∈ GL(V ) on V is inadmissible, if at least one of the following holds:

• Some Jordan block size of u has multiplicity ≥ 4.

• There are ≥ 2 Jordan block sizes of u with multiplicity ≥ 3.

• There are ≥ 2 odd Jordan block sizes of u with multiplicity ≥ 2.

• Some Jordan block size of u has multiplicity ≥ 3, and u has a Jordan blockof odd size.

Otherwise we will say that the action of u on V is admissible.

The point of making these two de�nitions is in the two lemmas below, whichsay that in certain cases if the action of u is �inadmissible�, then u does not actas a distinguished unipotent element on a particular subquotient. These lemmaswill be applied in Chapter 5 when we consider irreducible representations such asLAl(ω1 + ωl) and LCl(ω2), which can be constructed as such subquotients.

Lemma 3.2.6. Suppose that p 6= 2. Let u ∈ GL(V ) and suppose that there existK[u]-submodules W ⊆ W ′ ⊆ V such that dimW = 1, dimV/W ′ = 1 and thatu leaves a non-degenerate symmetric form invariant on W ′/W . If the action ofu on V is inadmissible, then the image of u in SO(W ′/W ) is not a distinguishedunipotent element.

Proof. Denote the image of u in SO(W ′/W ) by u′. Suppose that the action of u onV is inadmissible. We will show that u′ has a Jordan block size of multiplicity ≥ 2,and thus cannot be a distinguished unipotent element in SO(W ′/W ) (Proposition2.3.4). To do this, we apply Lemma 3.2.3 and consider the di�erent possibilities(i) - (vi) to the Jordan block sizes of u′ given there. Below (i) - (vi) will alwaysrefer to the di�erent cases of Lemma 3.2.3, and m and n will be the integers incases (ii), (iii), (v), and (vi) of Lemma 3.2.3.

Suppose that the action of u on V is inadmissible. First, if u has some Jordanblock of size > 1 with multiplicity ≥ 3, then in all cases (i) - (vi) it is clear thatrk(u

′) ≥ 2 for some k.The other possibility is that two distinct block sizes of u have multiplicity ≥ 2,

say rk0(u), rk′0(u) ≥ 2 for some k0 > k′0 ≥ 1. In cases (i), (ii), (iii) and (iv) it isclear that we have rk0(u′) ≥ 2 or rk′0(u′) ≥ 2.

In case (v), if we had rk(u′) ≤ 1 for all k, then we must have m = k0 and

n = k′0. Furthermore, now rm(u) = 2, rn(u) = 2, so rm(u′) = rn(u′) = 1. Since weare assuming that u′ leaves a nondegenerate symmetric form invariant on W/W ′,the multiplicity of a block of even size must be even. Therefore both m and n areodd. But then m− 1 and n− 1 are even, so rm−1(u′) and rn−1(u′) are even. Sincerm−1(u′), rn−1(u′) are ≥ 1 in this case, we have in fact rm−1(u′), rn−1(u′) ≥ 2.

Finally consider the possibility of case (vi). Similarly to case (v), if we hadrk(u

′) ≤ 1 for all k, then we have m = k0 and 1 = k′0, and rm(u) = 2 and

3.2. Jordan blocks in subspaces and quotients 57

r1(u) = 2. Therefore rm(u′) = 1. Again since u′ leaves a nondegenerate symmetricform invariant on W/W ′, any block of even size has even multiplicity, so it followsthat m must be odd. Then m − 1 is even, so because rm−1(u′) ≥ 1 in this case,we have rm−1(u′) ≥ 2.

Lemma 3.2.7. Suppose that p = 2. Let u ∈ GL(V ) and suppose that there existK[u]-submodules W ⊆ W ′ ⊆ V such that dimW = 1, dimV/W ′ = 1 and that uleaves a non-degenerate alternating bilinear form invariant onW ′/W . If the actionof u on V is inadmissible, then the image of u in Sp(W ′/W ) is not a distinguishedunipotent element.

Proof. Denote the image of u in Sp(W ′/W ) by u′. Suppose that the action of u onV is inadmissible. We will show that u′ has either a Jordan block of odd size, or aJordan block of of multiplicity ≥ 3, and thus cannot be a distinguished unipotentelement in Sp(W ′/W ) (Proposition 2.4.4). To do this, we apply Lemma 3.2.3 andconsider the di�erent possibilities (i) - (vi) to the Jordan block sizes of u′ giventhere. Below (i) - (vi) will always refer to the di�erent cases of Lemma 3.2.3, andm and n will be the integers in cases (ii), (iii), (v), and (vi) of Lemma 3.2.3.

Suppose that the action of u on V is inadmissible. Consider �rst the case wheresome block size of u has multiplicity ≥ 4, say rk0(u) ≥ 4 for some k0 ≥ 1. In case(i) we have rk0(u′) ≥ 4 or r1(u′) ≥ 2, and so u′ is not distinguished in Sp(W ′/W ).Consider then case (ii). Ifm 6= k0, then rk0(u′) ≥ 4 and thus u′ is not distinguishedin Sp(W ′/W ). If m = k0, then rk0(u′) ≥ 2. So if k0 is odd, then u′ has odd blocksizes and cannot be distinguished in Sp(W ′/W ). On the other hand if k0 is even,then rk0−1(u′) ≥ 2 and again u′ has odd block sizes. In cases (iii), (iv), (v) and(vi) it is clear that rk0(u′) ≥ 3 and thus u′ is not distinguished in Sp(W ′/W ).

Suppose next that u has two distinct Jordan block sizes, each occurring withmultiplicity ≥ 3, say k′0 > k0 ≥ 1 with rk′0(u), rk0(u) ≥ 3. In all cases it is clearthat rk′0(u′), rk0(u) ≥ 1, so if k′0 or k0 is odd then u′ has odd block sizes and cannotbe distinguished in Sp(W ′/W ). Suppose then that k′0 and k0 are even. Then incases (i), (ii), (iii), (iv) and (vi) it is clear that rk′0(u′) ≥ 3 or rk0(u′) ≥ 3 so u′

cannot be distinguished in Sp(W ′/W ). In case (v) we also have rk′0(u′) ≥ 3 orrk0(u′) ≥ 3 unless k′0 = m and k0 = n. If k′0 = m and k0 = n, then rk0−1(u′) ≥ 1and so u′ has odd block sizes and cannot be distinguished in Sp(W ′/W ).

Consider next the case where u has ≥ 2 odd Jordan block sizes of u withmultiplicity ≥ 2. That is, suppose that there exist k′0 > k0 ≥ 1 odd such thatrk′0(u) ≥ 2 and rk0(u) ≥ 2. In all cases it is clear that rk′0(u′) ≥ 1 or rk0(u′) ≥ 1,so u′ has odd block sizes and cannot be distinguished in Sp(W ′/W ).

Finally, suppose that some Jordan block size of u has multiplicity ≥ 3, and thatu has a Jordan block of odd size. That is, assume that there exists k′0, k0 ≥ 1 suchthat k0 is odd, rk′0(u) ≥ 3 and rk0(u) ≥ 1. Now if k′0 is odd, then rk′0(u′) ≥ 1, and sou′ has an odd block size and cannot be distinguished in Sp(W ′/W ). Assume thenthat k′0 is even. In case (i) we have rk′0(u′) ≥ 3 and thus u′ is not distinguished inSp(W ′/W ). In cases (ii) and (vi) we have rk′0(u′) ≥ 3 if k′0 6= m, and rk′0−1(u′) ≥ 2if k′0 = m (giving odd block sizes), so u′ is not distinguished in Sp(W ′/W ).Similarly in case (v) rk′0(u′) ≥ 3 if k′0 6= m,n, and rk′0−1(u′) ≥ 2 if k′0 = mor k′0 = n (giving odd block sizes), so u′ is not distinguished in Sp(W ′/W ). Incase (iii), we have rk′0(u′) ≥ 3 if k′0 6= m and rk0(u′) ≥ 1 of k′0 = m, so u′ hasan odd block size and cannot be distinguished in Sp(W ′/W ). In case (iv) wehave rk0(u′) ≥ 1 and so u′ has odd block sizes and cannot be distinguished inSp(W ′/W ).

58 Chapter 3.

Lemma 3.2.8. Suppose that p = 2. Let u ∈ GL(V ) and suppose that there existK[u]-submodules W , W ′ of V such that dimW ′ = 1, V = W ′ ⊕W and that uleaves a non-degenerate alternating bilinear form invariant on W . If the actionof u on V is inadmissible, then the image of u in Sp(W ) is not a distinguishedunipotent element.

Proof. Now the Jordan blocks of the restriction uW of u to W are the same asthose of u acting on V , except that a Jordan block of size 1 is removed. If theaction of u on V is inadmissible, then clearly uW has a block of odd size or someblock of multiplicity ≥ 3.

3.3 Decomposition of tensor products

In this section, we give various results about the Jordan form of the tensor pro-duct of two unipotent matrices. To consider the Jordan decomposition of tensorproducts of unipotent matrices, it is convenient to express them in terms of therepresentation theory of a cyclic p-group. We will use the notation for indecom-posable K[u]-modules as described in Section 1.4, where u is a unipotent linearmap of order q = pα.

There is a large amount of literature concerning the decomposition of Vm ⊗Vn into a direct sum of indecomposables, for example [Sri64], [Ral66], [McF79],[Nor95], [Hou03], and [Bar11]. There is no explicit formula for the decomposition ofVm⊗Vn as there is in characteristic 0, but there are various recursive descriptionswhich su�ce for our purposes.

We will often study the decomposition of Vm ⊗ Vn, ∧2(Vn) and S2(Vn) usingcertain �nite sequences of integers, which are de�ned in terms of m, n, and p. Forthese �nite sequences of integers, the next de�nition gives various notation whichwill be convenient later.

De�nition 3.3.1 ([Bar11], [GPX16, De�nition 1]). Let s = (a1, . . . , an) and s′ =(b1, . . . , bm) be �nite sequences of integers. The following notation will be de�ned.

(i) For all k ∈ Z, set s+ k = (a1 + k, . . . , an + k).

(ii) De�ne s⊕ s′ = (a1, . . . , an, b1, . . . , bm).

(iii) De�ne the negative reverse of s by nr(s) = (−an, . . . ,−a1).

(iv) Denote by s> the subsequence of positive terms in s, and by s< the subse-quence of negative terms in s.

(v) The k-multiple of s is the sequence (ka1, . . . , ka1, . . . , kan, . . . kan) of lengthkn (each element and its multiplicity in the sequence is multiplied by k).

(vi) For m ∈ Z and k ∈ Z≥0, denote (m : k) = (m,m, . . . ,m), where m occurs ktimes. Here (m : 0) is the empty sequence ().

(vii) For 1 ≤ k ≤ n, the kth term of the sequence s will be denoted by s(k).

We will now describe one of the main results in [Bar11] which gives the de-composition of Vm⊗Vn into a direct sum of indecomposables. This information iscontained in a sequence sp(m,n) de�ned below.

3.3. Decomposition of tensor products 59

De�nition 3.3.2 ([Bar15, De�nition 1]). Let 0 ≤ m ≤ n be integers. The se-quence sp(m,n) of integers is de�ned recursively as follows. De�ne sp(0, n) =(0 : n). Assume now that 0 < m ≤ n and let k ≥ 0 be the integer such thatpk ≤ n < pk+1. Write n = bpk + d, where 0 < b pk+1. Then

s1 = (pk+1 : m+ n− pk+1)

s2 = sp(pk+1 − n, pk+1 −m)

Case (2): m+ n ≤ pk+1 and c+ d > pk. Then

s1 = ((a+ b+ 1)pk : c+ d− pk)s2 = sp((a+ b+ 1)pk − n, (a+ b+ 1)pk −m)

Case (3): m+ n ≤ pk+1, 1 ≤ c+ d ≤ pk, and a > 0. Then

s1 = sp(min(c, d),max(c, d)) + (a+ b)pk

s2 = sp((a+ b)pk − n, (a+ b)pk −m)

Case (4): m+ n ≤ pk+1, 1 ≤ c+ d ≤ pk, a = 0, and d > 0. Then

s1 = sp(m, bpk − d)< + 2bpk

s2 = (0 : n−m)

Case (5): m+ n ≤ pk+1, 1 ≤ c+ d ≤ pk, a = 0, and d = 0. Then

s1 = (bpk : m)

s2 = (0 : bpk −m)

Case (6): m+ n ≤ pk+1, c = 0, and d = 0. Then

s1 = ((a+ b− 1)pk : pk)

s2 = sp((a− 1)pk, (b− 1)pk)

For more detail and examples on how to apply De�nition 3.3.2, see [Bar11]and [Bar15].

Lemma 3.3.3 ([Bar11, Proposition 1]). Assume that 0 ≤ m ≤ n. Then

(i) sp(m,n) is a nonincreasing sequence of m+ n integers.

(ii) n−m+ 1 ≤ sp(m,n)(l) ≤ m+ n− 1 when 1 ≤ l ≤ m.

(iii) sp(m,n)(m+ n− l + 1) = −sp(m,n)(l) for 1 ≤ l ≤ m+ n.

60 Chapter 3.

(iv) sp(m,n)(l) > 0 for 1 ≤ l ≤ m, and sp(m,n)(l) < 0 for n + 1 ≤ l ≤ m + n,and sp(m,n)(l) = 0 otherwise.

Remark 3.3.4. For 0 < m ≤ n, in De�nition 3.3.2 we set sp(m,n) = s1⊕s2⊕s3,where s3 = nr(s1) and s1 and s2 are given depending on which of the cases (1)-(6)of De�nition 3.3.2 hold for m and n.

According to Lemma 3.3.3 (i) the sequence sp(m,n) is nonincreasing, so itfollows that s2(i) ≤ s1(j) for all i and j. In fact, something slightly stronger is true:we can show that s2(i) < s1(j) for all i and j. We omit the full details, but this iseasy to verify in each of the cases (1)-(6) of De�nition 3.3.2, using the fact (Lemma3.3.3) that for all 0 ≤ m′ ≤ n′, we have −m′−n′+ 1 ≤ sp(m′, n′)(k) ≤ m′+n′−1for all 1 ≤ k ≤ m′ + n′.

Now in terms of the sequence sp(m,n), the decomposition of Vm⊗ Vn is givenby the following result. Note that the statement makes sense in view of Lemma3.3.3 (iv).

Theorem 3.3.5 ([Bar11, Theorem 1]). Let 0 ≤ m ≤ n ≤ q. Then

Vm ⊗ Vn = ⊕mk=1Vsp(m,n)(k)

We will need the following description of Vm⊗Vn which holds in the case where1 ≤ m,n ≤ p. This is a consequence of [Fei82, ch. VIII, Theorem 2.7], as noted in[Sup09, Lemma 2.8]. It could also be deduced from Theorem 3.3.5.

Lemma 3.3.6. Let 1 ≤ m ≤ n ≤ p. Then

Vm ⊗ Vn ∼= ⊕h−1i=0 Vn−m+2i+1 ⊕N · Vp = ⊕hi=1Vn−m+2h−2i+1 ⊕N · Vp

where h = min{m, p− n} and N = max{0,m+ n− p}. In particular, Vm ⊗ Vp =m · Vp for all 1 ≤ m ≤ p.

The following corollary is an easy consequence of Lemma 3.3.6 and will beuseful later. It is also a special case of [Bar15, Theorem 2].

Corollary 3.3.7. Let 1 ≤ m ≤ n ≤ p. Then

(i) If p ≥ n+m− 1, then Vn ⊗ Vm ∼= ⊕m−1i=0 Vn−m+2i+1 = ⊕mi=1Vn+m−2i+1.

(ii) If p < n+m− 1, then Vn ⊗ Vm has ≥ 2 Jordan blocks of size p.

Proof. Let h and N be as in Lemma 3.3.6. If p > n + m − 1, then h = m andN = 0, so the claim follows immediately from Lemma 3.3.6. If p = n + m − 1,then h = m− 1 and N = 1, so by Lemma 3.3.6 we have

Vn ⊗ Vm ∼= ⊕m−2i=0 V2i+1 ⊕ Vp = ⊕m−1

i=0 Vn−m+2i+1

as claimed. Finally if p < n + m − 1, then N ≥ 2, so by Lemma 3.3.6 the tensorproduct Vn ⊗ Vm has ≥ 2 Jordan blocks of size p.

The following theorem is due to Glasby, Praeger and Xia. It reduces �ndingthe decomposition of Vm ⊗ Vn to the case where p does not divide gcd(m,n). Inthe case where m ≤ n ≤ p, this was proven by Renaud in [Ren79, Lemma 2.2].

Theorem 3.3.8 ([GPX16, Theorem 5]). Let 0 < m ≤ n ≤ q. Then for all k ≥ 0,the sequence sp(pkm, pkn) is the pk-multiple of sp(m,n).

The following results are well known and could be deduced using Theorem3.3.5. They also follow from [GPX16, Theorem 4, Table 1].

Lemma 3.3.9. Let n ≥ 2. Then

Vn ⊗ V2 =

{Vn ⊕ Vn if n ≡ 0 mod p

Vn−1 ⊕ Vn+1 if n 6≡ 0 mod p

Lemma 3.3.10. Assume that p 6= 2 and let n ≥ 3. Then

Vn ⊗ V3 =

Vn ⊕ Vn ⊕ Vn if n ≡ 0 mod p

Vn−1 ⊕ Vn−1 ⊕ Vn+2 if n ≡ 1 mod p

Vn−2 ⊕ Vn+1 ⊕ Vn+1 if n ≡ −1 mod p

Vn−2 ⊕ Vn ⊕ Vn+2 if n 6≡ 0, 1,−1 mod p

Lemma 3.3.11. Assume that p = 2 and let n ≥ 3. Then

Vn ⊗ V3 =

Vn ⊕ Vn ⊕ Vn if n ≡ 0 mod 4

Vn−1 ⊕ Vn−1 ⊕ Vn+2 if n ≡ 1 mod 4

Vn−2 ⊕ Vn ⊕ Vn+2 if n ≡ 2 mod 4

Vn−2 ⊕ Vn+1 ⊕ Vn+1 if n ≡ 3 mod 4

We present a result due to Barry which determines when Vm ⊗ Vn has norepeated blocks. For this we will need the following de�nition from [Bar15].

De�nition 3.3.12. Assume that p 6= 2. We de�ne the following sets of pairs ofintegers. Let

S = {(k, d) : 1 < k ≤ d ≤ p+ 1− k} ∪ {(k, p+ k − 1) : 1 < k ≤ p+ 1

2}.

For all integers t ≥ 2, we de�ne St = (T1 \ T2) ∪ T3, where

T1 = {(ipt−1 +pt−1 ± 1

2, jpt−1 +

pt−1 ± 1

2) : 1 ≤ i ≤ p− 1

2, i ≤ j ≤ p− i− 1}

T2 = {(ipt−1 +pt−1 + 1

2, ipt−1 +

pt−1 − 1

2) : 1 ≤ i ≤ p− 1

2}

T3 = {(ipt−1 +pt−1 + 1

2, ipt−1 +

pt−1 − 1

2+ pt) : 1 ≤ i ≤ p− 1

2}

We de�ne the set S to be the set of all pairs (m,n) and (n,m) of integerssuch that 1 ≤ m ≤ n and one of the following conditions hold:

(i) m = 1.

(ii) 1 < m ≤ p, and (m,n′) ∈ S, where n′ is the unique integer such thatm ≤ n′ ≤ p+m− 1 and n′ ≡ n mod p.

(iii) pt−1 < m ≤ pt (t ≥ 2), and (m,n′) ∈ St, where n′ is the unique integer suchthat m ≤ n′ ≤ pt +m− 1 and n′ ≡ n mod pt.

62 Chapter 3.

Using his results in [Bar11] (Theorem 3.3.5 above), Barry has shown the fol-lowing.

Theorem 3.3.13 ([Bar15, Theorem 2, Theorem 3, Corollary 1]). Assume thatp 6= 2. Let m,n ≥ 1. Then

(i) The tensor product Vm ⊗ Vn has no repeated blocks if and only if (m,n) iscontained in the set S of De�nition 3.3.12.

(ii) If Vm ⊗ Vn has no repeated blocks, then Vn ⊗ Vm ∼= ⊕min(m,n)i=1 Vn+m−2i+1.

Theorem 3.3.14 ([Bar15, Theorem 1]). Assume that p = 2. Let 1 ≤ m ≤ n.Then

(i) The tensor product Vm ⊗ Vn has no repeated blocks if and only if one of thefollowing conditions hold:

• m = 1,

• m = 2 and n ≡ 1 mod 2,

• m = 3 and n ≡ 2 mod 4.

(ii) If Vm ⊗ Vn has no repeated blocks, then Vn ⊗ Vm ∼= ⊕min(m,n)i=1 Vn+m−2i+1.

Later in this chapter we will give a result similar to Theorem 3.3.13 for ∧2(Vn)and S2(Vn) (Proposition 3.5.3).

With Theorem 3.3.13 and Theorem 3.3.14, one could actually give a descriptionof all K[u]-modules V and W such that V ⊗W has no repeated blocks. We �nishthis section by giving this description in a speci�c case which is relevant to Problem1.1.2, namely the case where p 6= 2 and all the Jordan block sizes in V and Ware of the same parity (see Lemma 3.3.17 below). At the end of this chapter, thisresult will be used in the proof of Theorem 1.1.4 in the tensor product subgroupcase.

Lemma 3.3.15. Assume that p 6= 2. Let m ≥ 1 and 1 ≤ n ≤ n′ be integers suchthat n ≡ n′ mod 2. Suppose that Vm ⊗ (Vn ⊕ Vn′) has no repeated blocks. Thenn′ − n ≥ 2m, and in particular n′ > m.

Proof. First note that Vm ⊗ (Vn ⊕ Vn′) = (Vm ⊗ Vn) ⊕ (Vm ⊗ Vn′), so it followsthat Vm ⊗ Vn and Vm ⊗ Vn′ have no repeated blocks.

Suppose that n′ ≤ m. In this case we have Vm⊗Vn = ⊕ni=1Vm+n−2i+1 and Vm⊗Vn′ = ⊕n′j=1Vm+n′−2j+1 by Theorem 3.3.13 (ii). The largest block size occurring inVm ⊗ Vn is m+ n− 1. We claim that Vm ⊗ Vn′ also has a block of size m+ n− 1.For this, note that the block sizes occurring in Vm⊗Vn′ are all the integers whichlie in the interval [m− n′ + 1,m+ n′ − 1], and which are congruent to m− n′ + 1mod 2. A straightforward veri�cation shows that m+ n− 1 is such an integer, soVm⊗ Vn′ also has a block of size m+n− 1. Thus Vm⊗ (Vn⊕ Vn′) has ≥ 2 Jordanblocks of size m+ n− 1, which is a contradiction.

Therefore we must have n′ > m, so now Vm ⊗ Vn ∼= ⊕min(m,n)i=1 Vm+n−2i+1 and

Vm ⊗ Vn′ ∼= ⊕mj=1Vn′+m−2j+1 by Theorem 3.3.13 (ii). The Jordan block sizes ofVm ⊗ Vn consist of integers in the interval [|m − n| + 1,m + n − 1] congruent ton + m − 1 mod 2, while the Jordan block sizes of Vm ⊗ Vn′ consist of integersin the interval [n′ − m + 1, n′ + m − 1] congruent to n′ + m − 1 mod 2. Now

n + m − 1 ≡ n′ + m − 1 mod 2, so in order for (Vm ⊗ Vn) ⊕ (Vm ⊗ Vn′) tohave no repeated blocks, the two intervals must be disjoint. This is equivalent tom+ n− 1 < n′ −m+ 1, which is equivalent to n′ − n ≥ 2m since n′ ≡ n mod 2.This completes the proof of the lemma.

Lemma 3.3.16. Assume that p 6= 2. Let m,m′, n, n′ ≥ 1 be integers such thatm ≡ m′ mod 2 and n ≡ n′ mod 2. Then (Vm ⊕ Vm′) ⊗ (Vn ⊕ Vn′) has repeatedblocks.

Proof. Without loss of generality, we can assume that m ≤ m′ and n ≤ n′. If(Vm⊕Vm′)⊗ (Vn⊕Vn′) has no repeated blocks, then the same is true for its directsummand Vm′ ⊗ (Vn ⊕ Vn′), so by Lemma 3.3.15 we have n′ > m′. On the otherhand, now the direct summand (Vm ⊕ Vm′)⊗ Vn′ also has no repeated blocks, soby Lemma 3.3.15 we have m′ > n′, which is a contradiction.

Lemma 3.3.17. Assume that p 6= 2. Let V1 and V2 be K[u]-modules such that1 < dimV1 ≤ dimV2, V1 ↓ K[u] =

⊕si=1 Vmi , V2 ↓ K[u] =

⊕tj=1 Vnj , where

1 ≤ m1 ≤ · · · ≤ ms and 1 ≤ n1 ≤ · · · ≤ nt are integers such that mi ≡ mi′ mod 2for all 1 ≤ i, i′ ≤ s and nj ≡ nj′ mod 2 for all 1 ≤ j, j′ ≤ t.

Suppose that s > 1 or t > 1. Then V1 ⊗ V2 has no repeated block sizes if andonly if the following conditions hold:

(i) s = 1, so V1 ↓ K[u] = Vm for m = m1;

(ii) ni − ni−1 ≥ 2m for all 2 ≤ i ≤ t;

(iii) (m,ni) is contained in the set S of De�nition 3.3.12, for all 1 ≤ i ≤ t.

Furthermore, when (i), (ii), and (iii) hold, we have

V1 ⊗ V2 =

min(m,n1)⊕j=1

Vn1+m−2j+1 ⊕t⊕i=2

m⊕j=1

Vni+m−2j+1.

Proof. Suppose that the conditions (i), (ii), and (iii) hold, so V1 = Vm and V2 =⊕tj=1Vnj , where 1 ≤ n1 < · · · < nt. Note that we have ni > m for all 2 ≤ i ≤ t by(ii). It follows then from condition (iii) and Theorem 3.3.13 (ii) that

V1 ⊗ V2 =

min(m,n1)⊕j=1

Vn1+m−2j+1 ⊕t⊕i=2

m⊕j=1

Vni+m−2j+1,

proving the last claim of the lemma. In order to show that V1⊗V2 has no repeatedblocks, note that the Jordan block sizes occurring in V1 ⊗ V2 are the integerscongruent to n1 +m+1 mod 2 which lie in one of the intervals [|n1−m|+1, n1 +m− 1], [ni −m+ 1, ni +m− 1], i ≥ 2. It follows from (ii) that these intervals arepairwise disjoint, so V1 ⊗ V2 has no repeated blocks.

For the �only if� direction of the claim, suppose that V1 ⊗ V2 has no repeatedblocks. First of all, we claim that t = 1 or s = 1. Indeed, if t, s > 1, then(Vm1⊕Vm2)⊗ (Vn1⊕Vn2) is a direct summand V1⊗V2, so we have a contradictionby Lemma 3.3.16. Next we show that s = 1. If s > 1, then we must have t = 1.Now note that (Vm1⊕Vm2)⊗Vn1 has no repeated blocks, being a direct summandof V1⊗V2. By Lemma 3.3.15 we havem2 > n1, but this contradicts the assumption

64 Chapter 3.

dimV1 ≤ dimV2 = n1. Therefore s = 1, in other words, condition (i) holds. Setm = m1, so V1 = Vm. Now

V1 ⊗ V2 =t⊕i=1

Vm ⊗ Vni ,

so it follows from Theorem 3.3.13 (i) that for all 1 ≤ i ≤ t, the pair (m,ni)is contained in the contained in the set S of De�nition 3.3.12. In other words,condition (iii) holds.

Since Vm ⊗ (Vn1 ⊕ Vn2) is a direct summand of V1 ⊗ V2, by Lemma 3.3.15 wehave n2 > m. Thus by Theorem 3.3.13 (ii), we have

V1 ⊗ V2 =

min(m,n1)⊕j=1

Vn1+m−2j+1 ⊕t⊕i=2

m⊕j=1

Vni+m−2j+1.

In order for V1⊗V2 to have no repeated blocks, it follows as in the �rst paragraphthat the intervals [|n1−m|+ 1, n1 +m−1], [ni−m+ 1, ni+m−1], i ≥ 2 must bepairwise disjoint. It is straightforward to verify that this is equivalent to condition(ii), which completes the proof of the lemma.

3.4 Jordan decompositions in tensor squares (p 6= 2)

In this section, we make the following assumption.

Assume that p 6= 2.

The purpose of this section is to describe and apply the results of Barry in[Bar11], which give the decomposition of the K[u]-modules ∧2(Vn) and S2(Vn)into a direct sum of indecomposables, where u is a unipotent linear map of orderq ≥ n.

Set sp(n) := sp(n, n) for all n ≥ 0. As a special case of Theorem 3.3.5, thepositive terms of the sequence sp(n) give the decomposition of Vn ⊗ Vn into adirect sum of indecomposables. That is,

Vn ⊗ Vn = ⊕nk=1Vsp(n)(k)

for all 0 < n ≤ q.It is easily checked from De�nition 3.3.2 that the following lemma holds.

Lemma 3.4.1. If n = 0, then sp(0) = (0 : 0) = (). Assume then that n > 0, andlet k ≥ 0 be such that pk ≤ n < pk+1, and write n = bpk + d, where 1 ≤ b pk+1.

sp(n) = (pk+1 : 2n− pk+1)⊕ sp(pk+1 − n)⊕ (−pk+1 : 2n− pk+1)

Case (2): 2n ≤ pk+1 and d ≥ pk+12 (in this case k > 0).

sp(n) = ((2b+1)pk : 2d−pk)⊕sp((2b+1)pk−n)⊕(−(2b+1)pk : 2d−pk)

3.4. Jordan decompositions in tensor squares (p 6= 2) 65

Case (3): 2n ≤ pk+1 and 1 ≤ d ≤ pk−12 (in this case k > 0).

sp(n) = (sp(d) + 2bpk)⊕ sp(2bpk − n)⊕ (sp(d)− 2bpk)

Case (4): 2n ≤ pk+1 and d = 0.

sp(n) = ((2b− 1)pk : pk)⊕ sp((b− 1)pk)⊕ (−(2b− 1)pk : pk)

The sequence sp(n) has the following properties, cf. Lemma 3.3.3.

Lemma 3.4.2. Assume that n > 0. Then

(i) sp(n) is a nonincreasing sequence of 2n integers.

(ii) 1 ≤ sp(n)(l) ≤ 2n− 1 when 1 ≤ l ≤ n.

(iii) sp(n)(2n+ 1− l) = −sp(n)(l) for 1 ≤ l ≤ 2n.

(iv) sp(n)(l) > 0 for 1 ≤ l ≤ n and sp(n)(l) < 0 for n+ 1 ≤ l ≤ 2n.

(v) sp(n)(j) is odd for all 1 ≤ j ≤ 2n.

Proof. Statements (i), (ii), (iii) and (iv) follow from Lemma 3.3.3. We proceed toprove (v) by induction on n. If n = 1, then sp(n) = (1,−1). Suppose then thatn > 1. Now pk ≤ n < pk+1 for some k ≥ 0, and n = bpk + d for 1 ≤ b < p and0 ≤ d < pk. If n is as in case (1), (2) or (4) of Lemma 3.4.1, then the claim followsby induction (note that pk+1, (2b+ 1)pk and (2b− 1)pk are odd). In case (3) theclaim also follows, since sp(d)+2bpk and sp(2bpk−n) are sequences of odd integersby induction.

Note that the largest entry in sp(n), i.e., the largest Jordan block occurringin Vn ⊗ Vn, is easily deduced from Lemma 3.4.1. We will also need to know thesmallest Jordan block sizes occurring in Vn ⊗ Vn. This information is given in thefollowing lemma.

Lemma 3.4.3. Let n > 1. Set α = νp(n) and β = max{νp(n − 1), νp(n + 1)}.Then the smallest entries of the sequence sp(n)> are given as follows.

(i) If α > 0, then sp(n)> = (. . . , pα, pα, . . . , pα) where pα occurs exactly pα

times in the sequence.

(ii) If α = 0 and β > 0, then sp(n)> = (. . . , pβ, pβ, . . . , pβ, 1) where pβ occursexactly pβ − 2 times in the sequence.

(iii) If α = 0 and β = 0, then sp(n)> = (. . . , 3, 1) where 3 occurs only once inthe sequence.

Proof. We proceed to prove the claim by induction on n. If n = 2, then it is easyto see from Lemma 3.4.1 that sp(n)> = (3, 1) for all p and that the claim holds.

Suppose that n > 2 and let k be such that pk ≤ n < pk+1. We �rst take careof two special cases, namely n = pk+1− 1 and n = pk. If n = pk+1− 1, then α = 0and β = k + 1. Furthermore,

sp(n) = (pk+1 : pk+1 − 2)⊕ (1,−1)⊕ (−pk+1 : pk+1 − 2)

66 Chapter 3.

by Lemma 3.4.1 (1), so the claim holds. For n = pk, we have sp(n) = (pk :pk)⊕ (−pk : pk) by Lemma 3.4.1 (4), so the claim holds.

Suppose then that pk < n < pk+1 − 1. By Lemma 3.4.1, we have

sp(n) = s1 ⊕ sp(n′)⊕ nr(s1),

where s1 is a sequence of positive integers and in cases (1), (2), (3) and (4) ofLemma 3.4.1 we have n′ = pk+1 − n, n′ = (2b + 1)pk − n, n′ = 2bpk − n andn′ = n − pk respectively. Note that every entry of the sequence sp(n′) is < thanany entry of the sequence s1 (Remark 3.3.4). Thus by applying induction on sp(n′),the claim for sp(n)> follows once we verify that all of the following hold: n′ > 1,νp(n

′) = νp(n) and {νp(n′ + 1), νp(n′ − 1)} = {νp(n+ 1), νp(n− 1)}.

To this end, using pk < n < pk+1 − 1, an easy check in cases (1)-(4) ofLemma 3.4.1 shows that n′ > 1. Next, note that we have n′ ≡ −n mod pk,so n′ ± 1 ≡ −(n ∓ 1) mod pk. Since 1 < n′ < n < pk+1 − 1, it follows thatνp(n

′) = νp(n) and νp(n± 1) = νp(n∓ 1).

In characteristic zero, the decomposition of the exterior and symmetric squareof a unipotent matrix is easy to describe. This is seen from the following result,which is presumably well known, but we give a proof for completeness.

Proposition 3.4.4. Let uC be a unipotent Jordan block acting on a C-vector spaceV of dimension n > 1.

(i) The Jordan block sizes of uC acting on S2(V ) are [2n−1, 2n−5, . . . , c], wherec = 1 if n is odd and c = 3 if n is even.

(ii) The Jordan block sizes of uC acting on ∧2(V ) are [2n−3, 2n−7, . . . , c], wherec = 3 if n is odd and c = 1 if n is even.

Proof. We apply the representation theory of SL2(C). In an irreducible represen-tation V of SL2(C), a nonidentity unipotent element of SL2(C) acts as a singleJordan block. Representations of SL2(C) over C are semisimple, so then the Jordanblock structure of uC acting on S2(V ) and ∧2(V ) is determined by the irreduciblesummands. These summands are determined by the weight multiplicities.

Fix the maximal torus formed by the diagonal matrices in SL2(C). We canidentify the weights with integers, and then in V we have a basis of weight vectorse1, e2, . . . , en where ei has weight n−2i+1. Now S2(V ) has a basis of weight vectorsgiven by eiej , with 1 ≤ i ≤ j ≤ n where eiej has weight 2n−2(i+j−1). It followsthat in S2(V ) the highest weight 2n− 2 has multiplicity 1, the weight 2n− 6 hasmultiplicity 2, and generally 2n− 4k+ 2 has multiplicity k for 1 ≤ k ≤ n+1

2 . Fromthis (i) follows.

Similarly in ∧2(V ) a basis of weight vectors is given by ei ∧ ej , with 1 ≤ i <j ≤ n, and where ei ∧ ej has weight 2n− 2(i+ j − 1). The highest weight 2n− 4has multiplicity 1, the weight 2n− 8 has multiplicity 2 and generally 2n− 4k hasmultiplicity k for 1 ≤ k ≤ n

2 . From this (ii) follows.

We continue with results in positive characteristic p 6= 2. The sequence sp(n)de�ned above can used to be describe the decomposition of S2(Vn) and ∧2(Vn),as seen in [Bar11]7.

7Note that Theorem 2 in [Bar11] has a typo: case (3) is missing the assumption 2n ≤(2b+ 1)pk ≤ pk+1, and in the decomposition of S2(Vn) the direct sum should run from j = 1 toj = d.

3.4. Jordan decompositions in tensor squares (p 6= 2) 67

Theorem 3.4.5 ([Bar11, Corollary 3]). Let 0 < n ≤ q. Then

S2(Vn) =

dn/2e⊕k=1

Vsp(n)(2k−1) and ∧2 (Vn) =

bn/2c⊕k=1

Vsp(n)(2k).

Theorem 3.4.6 ([Bar11, Theorem 2]). Let 0 < n ≤ q. Now pk ≤ n < pk+1 forsome k ≥ 0, and n = bpk + d for 1 ≤ b pk+1, then

∧2(Vn) = (n− pk+1 + 1

2)Vpk+1 ⊕ S2(Vpk+1−n)

and

S2(Vn) = (n− pk+1 − 1

2)Vpk+1 ⊕ ∧2(Vpk+1−n)

(2) If 2n ≤ pk+1 and d ≥ pk+12 , then

∧2(Vn) = (n− (2b+ 1)pk + 1

2)V(2b+1)pk ⊕ S2(V(2b+1)pk−n)

and

S2(Vn) = (n− (2b+ 1)pk − 1

2)V(2b+1)pk ⊕ ∧2(V(2b+1)pk−n)

(3) If 2n ≤ pk+1 and 1 ≤ d ≤ pk−12 , then

∧2(Vn) =

d⊕j=1

Vsp(d)(2j)+2bpk ⊕ ∧2(V2bpk−n)

and

S2(Vn) =d⊕j=1

Vsp(d)(2j−1)+2bpk ⊕ S2(V2bpk−n)

(4) If 2n ≤ pk+1 and d = 0, then

∧2(Vn) =pk − 1

2V(2b−1)pk ⊕ S2(V(b−1)pk)

and

S2(Vn) =pk + 1

2V(2b−1)pk ⊕ ∧2(V(b−1)pk)

Remark 3.4.7. In Theorem 3.4.6, we have formulated the cases di�erently thanBarry in [Bar11]. However, it is easy to see that (1), (2), (3) and (4) in Theorem3.4.6 are equivalent to

(1)' 2n > pk+1,

(2)' 1 ≤ b < p, 0 < d < pk and (2b+ 1)pk < 2n ≤ pk+1,

(3)' 1 ≤ b < p, 0 < d < pk and 2bpk < 2n ≤ (2b+ 1)pk ≤ pk+1

68 Chapter 3.

(4)' n = bpk and 2b < p

respectively. The cases (1)', (2)', (3)', and (4)' are the cases given in [Bar11,Theorem 2].

Proposition 3.4.8. Suppose that a block size t has multiplicity ≥ 2 in ∧2(Vn) orS2(Vn). Then t is a multiple of p.

Proof. It is straightforward to prove by induction on n, using Lemma 3.4.1, thatif some value t > 0 occurs twice in the sequence sp(n), then t is a multiple of p.This is also a consequence of [GPX15, Theorem 4]. In any case, from this fact andTheorem 3.4.5, the claim follows.

The following lemma is elementary, see for example [FH91, B.1, pg. 473].

Lemma 3.4.9. Let V and W be KG-modules for any group G. Then we have thefollowing isomorphisms of KG-modules.

(i) S2(V ⊕W ) ∼= S2(V )⊕ (V ⊗W )⊕ S2(W ),

(ii) ∧2(V ⊕W ) ∼= ∧2(V )⊕ (V ⊗W )⊕ ∧2(W )

Lemma 3.4.10. Let n > 1, let α = νp(n), and let β = max{νp(n−1), νp(n+ 1)}.Let m be the smallest block size occurring in ∧2(Vn). Then the following hold:

(i) If α = 0 and n is even, then m = 1. Furthermore, Vm has multiplicity 1 in∧2(Vn).

(ii) If α = 0, n is odd, and β = 0, then m = 3. Furthermore, Vm has multiplicity1 in ∧2(Vn).

(iii) If α = 0, n is odd, and β > 0, then m = pβ. Furthermore, Vm has multiplicitypβ−1

2 in ∧2(Vn).

(iv) If α > 0, then m = pα. Furthermore, Vm has multiplicity pα+12 in ∧2(Vn) if

n is even and multiplicity pα−12 if n is odd.

Proof. This is a straightforward consequence of Theorem 3.4.5 and Lemma 3.4.3.

Lemma 3.4.11. Let V = Vd1 ⊕ · · · ⊕ Vdt , where t ≥ 1 and di ≥ 1 for all i. Letα = νp(gcd(d1, . . . , dt)). If α > 0, then the smallest block size occurring in ∧2(V )is pα.

Proof. Suppose that α > 0, and for all i set αi = νp(di). By Lemma 3.4.9 we have

∧2(V ) ∼=t⊕i=1

∧2(Vdi)⊕⊕

1≤i<j≤tVdi ⊗ Vdj

as K[u]-modules. Now pα divides di and dj for all i ≤ j, so by Theorem 3.3.8 wehave the equality

min sp(di, dj)> = pα ·min sp(dipα,djpα

)>

3.5. Symmetric and exterior squares with no repeated blocks (p 6= 2) 69

which is always ≥ pα.Therefore by Theorem 3.3.5, in Vdi ⊗ Vdj the smallest Jordan block has size

≥ pα. By Lemma 3.4.10 the smallest block size occurring in ∧2(Vdi) is pαi ≥ pα.

Thus each block size occurring in the K[u]-module ∧2(V ) is ≥ pα. Furthermore,a block of size pα occurs in ∧2(V ) since α = αi for some i, which proves theclaim.

Lemma 3.4.12. Let n > 1, let α = νp(n), and let β = max{νp(n−1), νp(n+ 1)}.Let m be the smallest block size occurring in S2(Vn). Then the following hold:

(i) If α = 0 and n is odd, then m = 1. Furthermore, Vm has multiplicity 1 inS2(Vn).

(ii) If α = 0, n is even, and β = 0, then m = 3. Furthermore, Vm has multiplicity1 in S2(Vn).

(iii) If α = 0, n is even, and β > 0, then m = pβ. Furthermore, Vm has multipli-

city pβ−12 in S2(Vn).

(iv) If α > 0, then m = pα. Furthermore, Vm has multiplicity pα−12 in S2(Vn) if

n is even and multiplicity pα+12 if n is odd.

Proof. This is a straightforward consequence of Theorem 3.4.5 and Lemma 3.4.3.

Lemma 3.4.13. Let V = Vd1 ⊕ · · · ⊕ Vdt , where t ≥ 1 and di ≥ 1 for all i. Letα = νp(gcd(d1, . . . , dt)). If α > 0, then the smallest block size occurring in S2(V )is pα.

Proof. The same proof as in Lemma 3.4.11 works: just replace ∧2 with S2 andapply Lemma 3.4.12 instead of Lemma 3.4.10.

3.5 Symmetric and exterior squares with no repeatedblocks (p 6= 2)

As in 3.4, we retain the notation from Section 3.3 and make the following assump-tion.

Assume that p 6= 2.

In this section we determine all n > 1 such that S2(Vn) or ∧2(Vn) have norepeated blocks. First, in view of Proposition 3.4.4, we make the following de�ni-tions.

De�nition 3.5.1. We say that the decomposition of S2(Vn) is as in characteristic0 if

S2(Vn) = V2n−1 ⊕ V2n−5 ⊕ · · · ⊕ Vcwhere c = 1 if n is odd and c = 3 if n is even.

De�nition 3.5.2. We say that the decomposition of ∧2(Vn) is as in characteristic0 if

∧2(Vn) = V2n−3 ⊕ V2n−7 ⊕ · · · ⊕ Vc,where c = 3 if n is odd and c = 1 if n is even.

70 Chapter 3.

The main result of this section is the following proposition.

Proposition 3.5.3. Let n > 1, and V = ∧2(Vn) or V = S2(Vn). The decomposi-tion of V has no repeated blocks precisely in the following cases:

(i) p ≥ 2n− 1 for V = S2(Vn),

(ii) p ≥ 2n− 3 for V = ∧2(Vn),

(iii) n = p+ p−32 for V = ∧2(Vn),

(iv) n = bpk + pk±12 for some k ≥ 1, 0 ≤ b ≤ p−1

2 , for V = ∧2(Vn) and V =S2(Vn).

In all of the cases above, the decomposition of V is as in characteristic 0.Furthermore, if there are repeated blocks in V , then some block of size > 1 hasmultiplicity ≥ 2.

The proof is essentially an application of Theorem 3.4.6. The fact that someblock of size > 1 has multiplicity ≥ 2 when there are repeated blocks follows fromour proof, but also from Proposition 3.4.8. Similarly the fact that the decomposi-tion is as in characteristic 0 when there are no repeated blocks follows from theproof, but it is also possible to deduce a priori that this is the case; we omit thediscussion of how this could be done.

We begin by a series of lemmas which will be needed in the proof.

Lemma 3.5.4. Let p ≥ n.

(i) If p ≥ 2n− 1, then the decomposition of S2(Vn) is as in characteristic 0. Ifn ≤ p < 2n− 1, then S2(Vn) has ≥ 2 blocks of size p.

(ii) If p ≥ 2n − 3, then the decomposition of ∧2(Vn) is as in characteristic 0. Ifn ≤ p < 2n− 3, then ∧2(Vn) has ≥ 2 blocks of size p.

Proof. If p ≥ 2n−1, then it is immediate from Corollary 3.3.7 that sp(n)> = (2n−2j + 1)nj=1, so by Theorem 3.4.5 both S2(Vn) and ∧2(Vn) are as in characteristic0.

If p = 2n−3, then by Lemma 3.3.6 we have sp(n)> = (p, p, p)⊕(2n−2j+1)n−3j=3 .

Now it follows from Theorem 3.4.5 that ∧2(Vn) is as in characteristic 0 and S2(Vn)has 2 blocks of size p.

Suppose then that n ≤ p < 2n − 3. In this case it follows from Lemma 3.3.6that Vn ⊗ Vn has ≥ 5 blocks of size p. In particular, the �rst four terms of sp(n)>are equal to p, and thus by Theorem 3.4.5 both ∧2(Vn) and S2(Vn) have ≥ 2blocks of size p.

Lemma 3.5.5. Let n = bpk + pk±12 , where 0 ≤ b ≤ p−1

2 and k ≥ 1. Thensp(n) = (2n− 2j + 1)2n

j=1.

Proof. In the notation of De�nition 3.3.12, we have (n, n) ∈ S since (n, n) ∈T1 \T2, so it follows from Theorem 3.3.13 (ii) that sp(n)> = (2n−2j+1)nj=1. Nowthe claim follows from Lemma 3.4.2 (iii).

Lemma 3.5.6. Let n = p + p−32 . Then the decomposition of ∧2(Vn) is as in

characteristic 0, but S2(Vn) has ≥ 2 blocks of size p.

Proof. If p = 3, then the result follows from Lemma 3.5.4. Assume then thatp > 3. By Theorem 3.4.6 (3), we get (here d = p−3

2 )

∧2(Vn) = ⊕dj=1Vsp(d)(2j)+2p ⊕ ∧2(V p+32

)

= ⊕dj=1V2n−4j+1 ⊕ ∧2(V p+32

).

Here the last equality follows since p ≥ 2d−1 and thus sp(d) = (2d−2j+1)2dj=1

by Corollary 3.3.7. Now by Lemma 3.5.4 the decomposition of ∧2(V p+32

) is as in

characteristic 0, so the decomposition of ∧2(Vn) is also as in characteristic 0.For the symmetric square, applying Theorem 3.4.6 (2) gives

S2(Vn) = ⊕dj=1Vsp(d)(2j−1)+2p ⊕ S2(V p+32

).

Now S2(V p+32

) has ≥ 2 blocks of size p by Lemma 3.5.4, as claimed.

Lemma 3.5.7. Let n = bpk + pk±12 , where k ≥ 1 and 0 ≤ b ≤ p−1

2 . Then thedecompositions of ∧2(Vn) and S2(Vn) are as in characteristic 0.

Proof. According to Lemma 3.5.5, we have sp(n) = (2n − 2j + 1)2nj=1. Thus the

claim follows immediately from Theorem 3.4.5.

We are now ready to prove Proposition 3.5.3.

Proof of Proposition 3.5.3. The fact that in cases (i)-(iv) the decompositions areas in characteristic 0 follows from lemmas 3.5.4, 3.5.6 and 3.5.7. To prove that theyare the only cases where there are no repeated blocks, we proceed by inductionon n (case n = 1 is obvious).

If p ≥ n, then Proposition 3.5.3 follows from Lemma 3.5.4. Assume then thatp < n. Then there exists k ≥ 1 such that pk ≤ n < pk+1 and n = bpk + d, where1 ≤ b pk+1.For the symmetric square, by Theorem 3.4.6 we have

S2(Vn) = (n− pk+1 − 1

2)Vpk+1 ⊕ ∧2(Vpk+1−n),

which means that if S2(Vn) has no repeated blocks, then n − pk+1−12 ≤ 1. Since

2n > pk+1, it follows that 2n = pk+1 + 1, so n = pk+1+12 (case (iv) in 3.5.3).

For the exterior square, we see similarly with Theorem 3.4.6 that if ∧2(Vn)

has no repeated blocks, then n − pk+1+12 ≤ 1. Since 2n > pk+1, it folllows that

n = pk+1+12 (case (iv) in 3.5.3) or n = pk+1+3

2 . If n = pk+1+32 , then by Theorem

3.4.6 we have∧2(Vn) = Vpk+1 ⊕ S2(V pk+1−3

2

).

We are assuming that ∧2(Vn) has no repeated blocks, so the same must be

true for S2(V pk+1−32

) as well. Since pk+1−32 is not of the form given in 3.5.3 (iv),

72 Chapter 3.

by induction it must be of the form given in 3.5.3 (i). That is, we have p ≥2 · p

k+1−32 − 1 = pk+1 − 4. But this is not possible, since p ≥ 3 and k ≥ 1 gives

pk+1 − 4 ≥ p2 − 4 > p.

In the rest of the cases we have 2n ≤ pk+1, and we note that this implies b ≤ p−12 .

Case (2): 2n ≤ pk+1 and d ≥ pk+12 .

For the symmetric square, by Theorem 3.4.6

S2(Vn) = (n− (2b+ 1)pk − 1

2)V(2b+1)pk ⊕ ∧2(V(2b+1)pk−n),

which means that if S2(Vn) has no repeated blocks, then n− (2b+1)pk−12 ≤ 1. Since

n = bpk + d, this gives d ≤ pk+12 , so in fact d = pk+1

2 . We have b ≤ p−12 , so we are

in case (iv) of 3.5.3.For the exterior square, we see similarly with Theorem 3.4.6 that if ∧2(Vn) has

no repeated blocks, then n− (2b+1)pk+12 ≤ 1. Since n = bpk+d, this gives d ≤ pk+3

2 ,

and thus d = pk+12 or d = pk+3

2 . If d = pk+12 , then we are in case (iv) of 3.5.3 since

b ≤ p−12 . Suppose then that d = pk+3

2 . In this case (2b + 1)pk − n = n − 3, so

∧2(Vn) = V2n−3 ⊕ S2(Vn−3) by Theorem 3.4.6. Because n− 3 = bpk + pk−32 is not

of the form given in 3.5.3 (iv), by applying induction on S2(Vn−3) it follows thatp ≥ 2(n− 3)− 1 = 2n− 7 = (2b+ 1)pk − 4. But this is not possible, since p ≥ 3,k ≥ 1, and b ≥ 1 gives (2b+ 1)pk − 4 ≥ 3p− 4 > p.

Case (3): 2n ≤ pk+1 and 1 ≤ d ≤ pk−12 .

We consider �rst S2(Vn) and ∧2(Vn) in the case where b = 1.When b = 1, for the symmetric square we have

S2(Vn) = ⊕dj=1Vsp(d)(2j−1)+2pk ⊕ S2(Vpk−d)

by Theorem 3.4.6. Assuming that S2(Vn) has no repeated blocks, by the decom-position above the same must be true for S2(Vpk−d). Thus by applying inductionon pk − d, one of the following must hold:

• Case (i) of 3.5.3: p ≥ 2(pk − d)− 1.

• Case (iv) of 3.5.3: pk − d = cpk−1 + pk−1±12 , where 0 ≤ c ≤ p−1

2 and k > 1.

Suppose �rst that S2(Vpk−d) is as in case (i) of 3.5.3. If k > 1, we have

2(pk − d) − 1 ≥ pk > p. Thus k = 1 and p ≥ 2(p − d) − 1. This gives d ≥ p−12 ,

so d = p−12 and we are in case (iv) of 3.5.3. If S2(Vpk−d) is as in case (iv) of

3.5.3, then pk − d = cpk−1 + pk−1±12 , where 0 ≤ c ≤ p−1

2 and k > 1. In particular

pk− d ≤ pk+12 , so d ≥ pk−1

2 and therefore d = pk−12 . In other words, n is as in case

(iv) of 3.5.3.When b = 1, for the exterior square

∧2(Vn) = ⊕dj=1Vsp(d)(2j)+2pk ⊕ ∧2(Vpk−d)

by Theorem 3.4.6. Assuming that ∧2(Vn) has no repeated blocks, by the decom-position above the same must be true for ∧2(Vpk−d), so we can apply inductionon pk − d. Thus one of the following must hold:

• Case (ii) of 3.5.3: p ≥ 2(pk − d)− 3.

• Case (iii) of 3.5.3: pk − d = p+ p−32 .

• Case (iv) of 3.5.3: pk − d = cpk−1 + pk−1±12 , where 0 ≤ c ≤ p−1

2 and k > 1.

If we are in case (ii) of 3.5.3 for ∧2(Vpk−d), then p ≥ 2(pk − d) − 3 ≥ pk − 2,

so 2 ≥ pk − p. This forces k = 1, so p ≥ 2(p− d)− 3 and d ≥ p−32 . Hence d = p−1

2

or d = p−32 , so either n = p+ p−1

2 (case (iv) of 3.5.3) or n = p+ p−32 (case (iii) of

3.5.3). If we are in case (iii) of 3.5.3 for ∧2(Vpk−d), then pk−d = p+ p−3

2 . It is clear

in this situation that k > 1. Furthermore, we must have k = 2 since pk−d ≥ pk+12

and p+ p−32 < p2. Thus d = p2− 3(p−1)

2 ≥ p2

2 , contradicting the fact that d <p2

2 . Ifwe are in case (iv) of 3.5.3 for pk− d, then as in the case of the symmetric square,

we get pk − d ≤ pk+12 . This implies d ≤ pk−1

2 , so d = pk−12 and n is as in case (iv)

of 3.5.3.Next we consider the case where b > 1. For the symmetric square, we have

S2(Vn) = ⊕dj=1Vsp(d)(2j−1)+2bpk ⊕ S2(Vbpk−d)

by Theorem 3.4.6. Assuming that S2(Vn) has no repeated blocks, the same mustbe true for S2(Vbpk−d) as well. Now p < 2(bpk − d)− 1, so by induction it followsbpk − d must be as in case (iv) of 3.5.3. Since bpk − d = (b− 1)pk + (pk − d), this

means that pk − d = pk±12 , which gives d = pk∓1

2 . Thus n is also as in case (iv) of3.5.3.

For the exterior square, we have

∧2(Vn) = ⊕dj=1Vsp(d)(2j)+2bpk ⊕ ∧2(Vbpk−d)

by Theorem 3.4.6. Assuming that ∧2(Vn) has no repeated blocks, by inductionthe same is true for ∧2(Vbpk−d). Now p < 2(bpk− d)− 3, so by induction it followsbpk−d must be as in case (iii) or case (iv) of 3.5.3. If ∧2(Vbpk−d) is as in case (iv),we see as in the previous paragraph for the symmetric square that n is as in case(iv) of 3.5.3. If bpk − d = (b− 1)pk + (pk − d) is as in case (iii) of 3.5.3, it followsthat b = 2, k = 1 and p − d = p−3

2 . But then d = p+32 , which is in contradiction

with d ≤ p−12 .

Case (4): 2n ≤ pk+1 and d = 0.For the symmetric square, in this case

S2(Vn) =pk + 1

2V(2b−1)pk ⊕ ∧2(V(b−1)pk)

by Theorem 3.4.6. Now pk+12 ≥ 2, so S2(Vn) has ≥ 2 blocks of size (2b− 1)pk.

For the exterior square,

∧2(Vn) =pk − 1

2V(2b−1)pk ⊕ S2(V(b−1)pk)

by Theorem 3.4.6. Now if ∧2(Vn) has no repeated blocks, then pk−12 ≤ 1. Therefore

pk ≤ 3, which forces p = 3 and k = 1. Since b ≤ p−12 = 1, we have b = 1. Therefore

n = 3 and we are in case (ii) of 3.5.3.

74 Chapter 3.

3.6 Decomposing tensor products of form modules

Assume that p = 2.

Let u ∈ SL(V ) be a unipotent element. Suppose that V1 and V2 are K[u]-modules equipped with non-degenerate u-invariant alternating bilinear forms β1

and β2, respectively. Then there is a product form β = β1 ⊗ β2 on the tensorproduct V1⊗V2, de�ned by β(v⊗w, v′⊗w′) = β1(v, v′)β2(w,w′) for all v, v′ ∈ V1

and w,w′ ∈ V2. The purpose of this section is to give, in some small cases, thedecomposition of V (2m) ⊗ V (2l) with respect to the product form into orthogo-nally indecomposable (De�nition 2.4.3) summands. Although methods for �ndingthe Jordan block sizes of V (2m) ⊗ V (2l) can be found in the literature (see e.g.Theorem 3.3.5), it seems that in general the decomposition of this K[u]-moduleinto orthogonally indecomposable summands is not known. As the main resultof this section (Lemmas 3.6.2 - 3.6.4, Table 3.6) we will give the decompositionsfor m ∈ {1, 2, 3}. This will allow us to decide when tensor product subgroups ofclassical groups contain distinguished unipotent elements (Proposition 3.6.6).

To begin, we will give an explicit construction of V (2m) and V (2l). Lete1, . . . , em, e−m, . . . , e−1 be a basis for a vector space W over K, and de�ne analternating bilinear form on W by (ei, ej) = 1 if i = −j, and 0 otherwise. We willde�ne the action of u on W as follows:

ue1 = e1

uei = ei + ei−1 + · · ·+ e1 for all 2 ≤ i ≤ mue−m = e−m + em + em−1 + · · ·+ e1

ue−i = e−i + e−(i+1) for all 1 ≤ i ≤ m− 1

Then the form (−,−) is non-degenerate, u-invariant, andW ↓ K[u] = V (2m). ForV (2l), we use the same construction, but with di�erent notation for convenience.We denote a basis of V (2l) by f1, . . . , f2l, with alternating bilinear form de�nedby (fi, fj) = 1 if i+j = 2l+1 and 0 otherwise. The action of u on V (2l) is de�nedas above:

uf1 = f1

ufi = fi + fi−1 + · · ·+ f1 for all 2 ≤ i ≤ l + 1

ufi = fi + fi−1 for all l + 1 < i ≤ 2l

Denote by X the element u − 1 of K[u]. For determining the decompositionof a K[u]-module with a u-invariant form into orthogonally indecomposable sum-mands, we will need some knowledge about the action of the powers of X. For thesituations that we are about to consider, we will give the action of Xk explicitly.First, recall that the action of u on the tensor product V1 ⊗ V2 of K[u]-modulesVi is de�ned by u · (v1 ⊗ v2) = uv1 ⊗ uv2 for all vi ∈ Vi. Therefore if k = 2α andukv1 = v1, it follows from the identity Xk = uk−1 that Xk ·(v1⊗v2) = v1⊗Xkv2.In particular, this holds for all vi ∈ Vi if 2α ≥ dimV1. It follows then that it willbe enough to compute the action of Xk for 1 ≤ k < 2α, which will give formulasfor the action of Xk on V1 ⊗ V2 which depend on k mod 2α.

For example, consider the action of u on V (2)⊗ V (2l) as de�ned above. Firstof all, it is easy to see that for all v ∈ V (2l) we have X · (e1 ⊗ v) = e1 ⊗Xv andX ·(e−1⊗v) = (e−1+e1)⊗Xv+e1⊗v. Furthermore, now u2 acts trivially on V (2),

3.6. Decomposing tensor products of form modules 75

so X2 · (ei⊗ v) = ei⊗X2v for all v ∈ V (2l) and i = 1,−1. It follows then that weget the formulae of Table 3.1 which hold for all v ∈ V (2l). In the same way, onecomputes formulae for the action of Xk on V (4) ⊗ V (2l) and V (6) ⊗ V (2l). Wehave given these in Table 3.2 and Table 3.3 for later use.

It will also be useful for us to know the action of Xk on V (2l) explicitly. Thiscan be determined using the next lemma.

Lemma 3.6.1. Consider V (2l) with basis f1, . . . , f2l and the action of u on V (2l)de�ned as above. Then we have the following:

(i) Xkfj = 0 for all j ≤ k.

(ii) Xkfj ∈ 〈f1, . . . , fj−k〉 for all j > k.

(iii) Let 0 ≤ d ≤ l. Then Xdfj = 0 for all j ≤ d. For d < j ≤ l + 1, we have

Xdfj =∑j−d

t=1 µ(j)t ft, where µ

(j)t =

(j−1−tj−d−t

).

(iv) Let k ≤ l + 1. Then X2l−kfj = 0 for all j ≤ 2l − k. For 2l − k < j ≤ 2l, we

have X2l−kfj =∑k+j−2l

t=1 λ(j)t ft, where λ

(j)t =

(l−t

j+k−2l−t).

(v) Let k ≤ l+ 1 and let α ≥ 1 be such that 2α ≥ k. Then the coe�cients λ(j)t in

(iv) are determined by the value of l modulo 2α.

Proof. The claims (i) and (ii) are immediate from the fact that Xf1 = 0 andXfj ∈ 〈f1, . . . , fj−1〉 for all 1 < j ≤ 2l.

For (iii), we begin by considering the action of X on the X-invariant sub-space 〈f1, . . . , fl+1〉. The action of X on this subspace with respect to the basisf1, . . . , fl+1 is given by following matrix.

A =

0 1 · · · · · · 1

0 1 · · ·...

. . . . . ....

. . . 10

Now note that in (iii), we have µ(j)

t = µ(j−1)t−1 . Thus one observes that it is

enough to show that for all 1 ≤ d ≤ l, the matrix Ad has zero entries, except inthe (l + 1 − d) × (l + 1 − d) upper right corner which is upper triangular of theform

a1 a2 · · · · · · al+1−d

a1 a2 · · ·...

. . . . . ....

. . . a2

a1

where as =

(s+d−2s−1

)for all 1 ≤ s ≤ l + 1 − d. We proceed to prove this claim by

induction on d. For d = 1 this is clear. Suppose then that the claim holds for some1 ≤ d < l. By multiplying Ad with A, one calculates that Ad+1 = A ·Ad has zero

76 Chapter 3.

entries, except for the (l−d)× (l−d) upper right corner which is upper triangularof the form

a′1 a′2 · · · · · · a′l−d

a′1 a′2 · · ·...

. . . . . ....

. . . a′2a′1

where a′f = a1 + · · ·+ af for all 1 ≤ f ≤ l − d. Now

f∑s=1

(s+ d− 2

s− 1

)=

f−1∑s=0

(d− 1 + s

s

)=

(d+ f − 1

f − 1

)

by a standard combinatorial identity (�hockey-stick identity�)8, so it follows thata′f =

(f+d−1f−1

). Hence the claim holds for d+1 as well, so (iii) follows by induction.

For claim (iv), the fact that X2l−kfj = 0 for all j ≤ 2l − k follows from (i).Suppose then that 2l−k < j ≤ 2l. First, if j < l+1, then 2l−k ≤ l−1 which impliesthat k = l+1 and j = l. Therefore 2l−k = l−1, and the claim is easily veri�ed byapplying (iii) with d = l−1. Consider then j ≥ l+1. In this caseXj−(l+1)fj = fl+1,so X2l−kfj = X3l−k−j+1 ·Xj−(l+1)fj = X3l−k−j+1fl+1. Now it is immediate from

(iii) that X3l−k−j+1fl+1 =∑k+j−2t

t=1 λ(j)t ft with λ

(j)t =

(l−t

j+k−2l−t), proving (iv).

Finally for claim (v), note that for λ(j)t =

(l−t

j+k−2l−t)we have λ(j)

t = λ(j−1)t−1 , so it

follows from (iv) that the matrix of X2l−k with respect to the basis f1, . . . , f2l haszero entries, except in the k×k upper right corner which has the upper triangularform

bk bk−1 · · · · · · b1

bk bk−1 · · ·...

. . . . . ....

. . . bk−1

bk

where bt =

(l−tk−t)for all 1 ≤ t ≤ k. To prove (v), it su�ces to show for all

1 ≤ t ≤ k that bt mod 2 depends only on l mod 2α. In other words, we shouldshow that if l′ is such that l′ ≡ l mod 2α, then

(l′−tk−t)≡(l−tk−t)

mod 2 for all1 ≤ k ≤ l+ 1 and 1 ≤ t ≤ k. This congruence is a straightforward consequence ofLucas' theorem9.

In Table 3.4 and Table 3.5 we have given the action of X2l−k on V (2l) for1 ≤ k ≤ l+ 1 with 1 ≤ k ≤ 6. These tables were found by a computer calculation,using Lemma 3.6.1 (iv) and (v).

8The identity is a common exercise and follows easily from the identity(nk

)=(n−1k

)+(n−1k−1

),

see e.g. [Knu97, 1.2.6, (10)].9Lucas' theorem [Luc78, Section XXI] states that for two non-negative integers a and b

with expansions a =∑i≥0 aip

i and b =∑i≥0 bip

i in prime base p, we have the congruence(ab

)≡∏i≥0

(aibi

)mod p. Here we set

(mn

)= 0 if m < n. For a short proof, see for example

[Fin47].


ei ⊗ v Xk · (ei ⊗ v)

e1 ⊗ v e1 ⊗Xkv

e−1 ⊗ v e−1 ⊗Xkv, if k ≡ 0 mod 2

(e−1 + e1)⊗Xkv + e1 ⊗Xk−1v, if k ≡ 1 mod 2

Table 3.1: Action of Xk on V (2)⊗ V (2l).


e1 ⊗ v e1 ⊗Xkv

e2 ⊗ v e2 ⊗Xkv, if k ≡ 0 mod 2

(e2 + e1)⊗Xkv + e1 ⊗Xk−1v, if k ≡ 1 mod 2

e−2 ⊗ v e−2 ⊗Xkv, if k ≡ 0 mod 4

(e−2 + e2 + e1)⊗Xkv + (e2 + e1)⊗Xk−1v, if k ≡ 1 mod 4


(e−2 + e2)⊗Xkv + (e2 + e1)⊗Xk−1v + e1 ⊗Xk−2v, if k ≡ 3 mod 4

e−1 ⊗ v e−1 ⊗Xkv, if k ≡ 0 mod 4

(e−1 + e−2)⊗Xkv + e−2 ⊗Xk−1v, if k ≡ 1 mod 4

(e−1 + e2 + e1)⊗Xkv + (e2 + e1)⊗Xk−2v, if k ≡ 2 mod 4

(e−1 + e−2 + e2)⊗Xkv + (e−2 + e1)⊗Xk−1v + e2 ⊗Xk−2v +e1 ⊗Xk−3v,

if k ≡ 3 mod 4


V (2m)⊗ V (2l) V (2m)⊗ V (2l) ↓ K[u]

V (2)⊗ V (2l), l ≥ 1 W (2l), if l ≡ 0 mod 2V (2l)2, if l ≡ 1 mod 2

V (4)⊗ V (2l), l ≥ 2 W (2l)2, if l ≡ 0 mod 4W (2l − 2) +W (2l + 2), if l ≡ 1 mod 4W (2l) + V (2l)2, if l ≡ 2 mod 4W (2l − 2) +W (2l + 2), if l ≡ 3 mod 4

V (6)⊗ V (2l), l ≥ 3 W (2l)3, if l ≡ 0 mod 4W (2l − 2)2 + V (2l + 4)2, if l ≡ 1 mod 4W (2l − 4) +W (2l) +W (2l + 4), if l ≡ 2 mod 4V (2l − 4)2 +W (2l + 2)2, if l ≡ 3 mod 4

Table 3.6: Decomposition of V (2m)⊗ V (2l) for 1 ≤ m ≤ 3.

78 Chapter 3.


e1 ⊗ v e1 ⊗Xkv




(e3 + e2 + e1)⊗Xkv + (e2 + e1)⊗Xk−1v, if k ≡ 1 mod 4


(e3 + e2)⊗Xkv + (e2 + e1)⊗Xk−1v + e1 ⊗Xk−2v, if k ≡ 3 mod 4

e−3 ⊗ v e−3 ⊗Xkv, if k ≡ 0 mod 4

(e−3 + e3 + e2 + e1)⊗Xkv + (e3 + e2 + e1)⊗Xk−1v, if k ≡ 1 mod 4


(e−3 + e3)⊗Xkv + (e3 + e2)⊗Xk−1v + (e2 + e1)⊗Xk−2v +e1 ⊗Xk−3v

if k ≡ 3 mod 4

e−2 ⊗ v e−2 ⊗Xkv, if k ≡ 0 mod 8


(e−2 + e3 + e2 + e1)⊗Xkv + (e3 + e2 + e1)⊗Xk−2v, if k ≡ 2 mod 8

(e−2 + e−3 + e3 + e1)⊗Xkv+(e−3 + e2)⊗Xk−1v+(e3 + e1)⊗Xk−2v + e2 ⊗Xk−3v,

if k ≡ 3 mod 8


(e−2 + e−3 + e1)⊗Xkv + e−3 ⊗Xk−1v + e1 ⊗Xk−4v, if k ≡ 5 mod 8

(e−2 + e3 + e2)⊗Xkv+ (e3 + e2 + e1)⊗Xk−2v+ e1 ⊗Xk−4v, if k ≡ 6 mod 8

(e−2 + e−3 + e3) ⊗ Xkv + (e−3 + e2) ⊗ Xk−1v + (e3 + e1) ⊗Xk−2v + e2 ⊗Xk−3v + e1 ⊗Xk−4v,

if k ≡ 7 mod 8

e−1 ⊗ v e−1 ⊗Xkv, if k ≡ 0 mod 8



(e−1 + e−2 + e−3 + e3 + e2 + e1)⊗Xkv+(e−2 + e3 + e2 + e1)⊗Xk−1v+(e−3+e3+e2+e1)⊗Xk−2v+(e3+e2+e1)⊗Xk−3v,

if k ≡ 3 mod 8


(e−1 + e−2 + e2 + e1)⊗Xkv+(e−2 + e1)⊗Xk−1v+(e2 + e1)⊗Xk−4v + e1 ⊗Xk−5v,

if k ≡ 5 mod 8

(e−1 + e−3 + e2)⊗Xkv + e−3 ⊗Xk−2v + e2 ⊗Xk−4v, if k ≡ 6 mod 8

(e−1+e−2+e−3+e3)⊗Xkv+(e−2+e3+e2)⊗Xk−1v+(e−3+e3 + e2 + e1)⊗Xk−2v + (e3 + e2 + e1)⊗Xk−3v + (e2 + e1)⊗Xk−4v + e1 ⊗Xk−5v,

if k ≡ 7 mod 8



X2l−1(1)

(l ≥ 1)

X2l−2

(1 1

1

) (1 0

1

)(l ≥ 1) l ≡ 0 mod 2 l ≡ 1 mod 2

X2l−3

1 1 11 0

1

1 0 01 1

1

(l ≥ 2) l ≡ 0 mod 4 l ≡ 1 mod 41 1 0

1 01

1 0 11 1

1

l ≡ 2 mod 4 l ≡ 3 mod 4

X2l−4

1 1 1 1

1 0 11 1

1

1 0 0 0

1 1 11 0

1

(l ≥ 3) l ≡ 0 mod 4 l ≡ 1 mod 4

1 1 0 01 0 0

1 11

1 0 1 0

1 1 01 0

1

l ≡ 2 mod 4 l ≡ 3 mod 4

Table 3.4: Action of X2l−k on V (2l) for 1 ≤ k ≤ 4. Here the matrices representthe upper right corner of the matrix X2l−k with respect to the basis f1, . . . , f2l;the remaining entries in the matrix of X2l−k are zero.

Lemma 3.6.2. Let l ≥ 1. Then the decomposition of V (2)⊗V (2l) with respect tothe product form is as given in Table 3.6.

Proof. First of all, note that by Lemma 3.3.9 we have V2 ⊗ V2l = V2l ⊕ V2l, sothe Jordan block sizes in V (2) ⊗ V (2l) are as claimed. We next determine thedecomposition of V (2)⊗V (2l) into orthogonally indecomposable summands. Notefor the orthogonal decomposition, our claim is that ε(2l) = 0 if l ≡ 0 mod 2, andε(2l) = 1 if l ≡ 1 mod 2, see De�nition 2.4.6. We proceed to prove this usingLemma 2.4.8.

By Lemma 2.4.8, we have ε(2l) = 1 if and only if (X2l−1(ei ⊗ fj), ei ⊗ fj) 6=0 for some basis vector ei ⊗ fj of V (2) ⊗ V (2l). We thus proceed to compute(X2l−1(ei ⊗ fj), ei ⊗ fj) for all ei and fj .

Using Table 3.1, we see that, for all v ∈ V (2l),

(X2l−1 · (e1 ⊗ v), e1 ⊗ v) = 0, and

(X2l−1 · (e−1 ⊗ v), e−1 ⊗ v) = (X2l−1v, v) + (X2l−2v, v).

80 Chapter 3.

X2l−5

1 1 1 1 1

1 0 1 01 1 0

1 01

1 0 0 0 01 1 1 1

1 0 11 1

1

(l ≥ 4) l ≡ 0 mod 8 l ≡ 1 mod 8

1 1 0 0 01 0 0 0

1 1 11 0

1

1 0 1 0 01 1 0 0

1 0 01 1

1

l ≡ 2 mod 8 l ≡ 3 mod 8

1 1 1 1 01 0 1 0

1 1 01 0

1

1 0 0 0 11 1 1 1

1 0 11 1

1

l ≡ 4 mod 8 l ≡ 5 mod 8

1 1 0 0 11 0 0 0

1 1 11 0

1

1 0 1 0 11 1 0 0

1 0 01 1

1

l ≡ 6 mod 8 l ≡ 7 mod 8

X2l−6

1 1 1 1 1 1

1 0 1 0 11 1 0 0

1 0 01 1

1

1 0 0 0 0 0

1 1 1 1 11 0 1 0

1 1 01 0

1

(l ≥ 5) l ≡ 0 mod 8 l ≡ 1 mod 8

1 1 0 0 0 01 0 0 0 0

1 1 1 11 0 1

1 11

1 0 1 0 0 0

1 1 0 0 01 0 0 0

1 1 11 0

1

l ≡ 2 mod 8 l ≡ 3 mod 8

1 1 1 1 0 01 0 1 0 0

1 1 0 01 0 0

1 11

1 0 0 0 1 0

1 1 1 1 01 0 1 0

1 1 01 0

1

l ≡ 4 mod 8 l ≡ 5 mod 8

1 1 0 0 1 11 0 0 0 1

1 1 1 11 0 1

1 11

1 0 1 0 1 0

1 1 0 0 11 0 0 0

1 1 11 0

1

l ≡ 6 mod 8 l ≡ 7 mod 8

Table 3.5: Action of X2l−k on V (2l) for 5 ≤ k ≤ 6. Here the matrices representthe upper right corner of the matrix X2l−k with respect to the basis f1, . . . , f2l;the remaining entries in the matrix of X2l−k are zero.


Furthermore, with Table 3.4, one �nds that

(∗)

(X2l−1fj , fj) + (X2l−2fj , fj) = 0 for all j ≤ 2l − 1.

(X2l−1f2l, f2l) + (X2l−2f2l, f2l) =

{0, if l ≡ 0 mod 2.

1, if l ≡ 1 mod 2.

From (*) we conclude that ε(2l) = 0 if l ≡ 0 mod 2, and ε(2l) = 1 if l ≡ 1mod 2, �nishing the proof of the lemma.


Proof. As a consequence of Theorem 3.3.5, Theorem 3.3.8 and Lemma 3.3.9, wehave

V4 ⊗ V2l =

{4 · V2l if l ≡ 0 mod 2

2 · V2l−2 ⊕ 2 · V2l+2 if l ≡ 1 mod 2

so the Jordan block sizes in V (4)⊗ V (2l) are as claimed.We prove next that the decomposition of V (4) ⊗ V (2l) into orthogonally in-

decomposable summands is as claimed.We consider �rst l ≡ 0 mod 2. Note that here the claim is that ε(2l) = 0 if l ≡

0 mod 4 and ε(2l) = 1 if l ≡ 2 mod 4, see De�nition 2.4.6. Now V4⊗V2l = 4 ·V2l,so by Lemma 2.4.8, it will be enough to show that (X2l−1(ei ⊗ fj), ei ⊗ fj) = 0for all i and j if and only if l ≡ 0 mod 4. Since 2l− 1 ≡ 3 mod 4, with Table 3.2we see that

(X2l−1 · (e1 ⊗ v), e1 ⊗ v) = 0

(X2l−1 · (e2 ⊗ v), e2 ⊗ v) = 0

(X2l−1 · (e−2 ⊗ v), e−2 ⊗ v) = (X2l−1v, v) + (X2l−2v, v)

(X2l−1 · (e−1 ⊗ v), e−1 ⊗ v) = (X2l−2v, v) + (X2l−4v, v)

for all v ∈ V (2l). Now if l = 2, one computes with Table 3.4 thatX2l−2f2l = f1+f2

and X2l−4f2l = f2l, so (X2l−1 · (e−1 ⊗ f2l), e−1 ⊗ f2l) = 1 and thus ε(2l) = 1, asclaimed.

Suppose then that l > 2. Since l ≡ 0 mod 2, it follows from (*) in the proof ofLemma 3.6.2 that we have (X2l−1 ·(e−2⊗v), e−2⊗v) = (X2l−1v, v)+(X2l−2v, v) =0 for all v ∈ V (2l). Furthermore, with Table 3.4 one computes that (X2l−1 · (e−1⊗fj), e−1 ⊗ fj) = (X2l−2fj , fj) + (X2l−4fj , fj) = 0 for all j ≤ 2l − 1. Finally, withTable 3.4 we see that

(X2l−1 · (e−1 ⊗ f2l), e−1 ⊗ f2l) = (X2l−2f2l, f2l) + (X2l−4f2l, f2l)

=

{0, if l ≡ 0 mod 4.

1, if l ≡ 2 mod 4.

Hence ε(2l) = 0 if l ≡ 0 mod 4 and ε(2l) = 1 if l ≡ 2 mod 4. This completes theproof of the claim in the case where l ≡ 0 mod 2.

Next we consider l ≡ 1 mod 2. Here the claim is that ε(2l + 2) = 0 andε(2l − 2) = 0 (De�nition 2.4.6). To show that ε(2l + 2) = 0, by Lemma 2.4.8 it

82 Chapter 3.

will be enough to show that (X2l+1 · (ei ⊗ fj), ei ⊗ fj) = 0 for all i and j. Now2l + 1 ≡ 3 mod 4, so with Table 3.2 one gets

(X2l+1 · (e1 ⊗ v), e1 ⊗ v) = 0

(X2l+1 · (e2 ⊗ v), e2 ⊗ v) = 0

(X2l+1 · (e−2 ⊗ v), e−2 ⊗ v) = (X2l+1v, v) + (X2lv, v) = 0

(X2l+1 · (e−1 ⊗ v), e−1 ⊗ v) = (X2lv, v) + (X2l−2v, v) = (X2l−2v, v)

for all v ∈ V (2l). Since l ≡ 1 mod 2, we have (X2l−2fj , fj) = 0 for all j (Table3.4), so it follows that ε(2l + 2) = 0.

To show that ε(2l− 2) = 0, by Lemma 2.4.8 it will be enough to demonstratethat (X2l−3 · γ, γ) = 0 for all γ in a basis for the kernel of X2l−2 acting onV (4)⊗ V (2l). Now 2l − 2 ≡ 0 mod 4, so by Table 3.2 we have X2l−2 · (v ⊗ w) =v ⊗ X2l−2w for all v ∈ V (4), w ∈ V (2l). Hence a basis for the kernel of X2l−2

acting on V (4) ⊗ V (2l) is given by ei ⊗ fj , where j ≤ 2l − 2. Now 2l − 3 ≡ 3mod 4, so with Table 3.2 we �nd that

(X2l−3 · (e1 ⊗ v), e1 ⊗ v) = 0

(X2l−3 · (e2 ⊗ v), e2 ⊗ v) = 0

(X2l−3 · (e−2 ⊗ v), e−2 ⊗ v) = (X2l−3v, v) + (X2l−4v, v)

(X2l−3 · (e−1 ⊗ v), e−1 ⊗ v) = (X2l−4v, v) + (X2l−6v, v)

for all v ∈ V (2l).With Table 3.4, it is straightforward to verify that for l ≡ 1 mod 2, we have

(X2l−3 · (e−2⊗ fj), e−2⊗ fj) = (X2l−3fj , fj) + (X2l−4fj , fj) = 0 for all j ≤ 2l− 2.Similarly one veri�es for all l ≡ 1 mod 2 with l > 3 that (X2l−3 · (e−1⊗fj), e−1⊗fj) = (X2l−4fj , fj) + (X2l−6fj , fj) = 0 for all j ≤ 2l − 2. Finally, for l = 3a calculation shows that (X2l−4fj , fj) + (X2l−6fj , fj) = 0 for all j ≤ 2l − 2,completing the proof of the claim.


Proof. As a consequence of Theorem 3.3.8 and Lemma 3.3.11, we have

V6 ⊗ V2l =

6 · V2l if l ≡ 0 mod 4

4 · V2l−2 ⊕ 2 · V2l+4 if l ≡ 1 mod 4

2 · V2l−4 ⊕ 2 · V2l ⊕ 2 · V2l+4 if l ≡ 2 mod 4

2 · V2l−4 ⊕ 4 · V2l+4 if l ≡ 3 mod 4

so the Jordan block sizes in V (6)⊗ V (2l) are as claimed.We prove next that the decomposition of V (6) ⊗ V (2l) into orthogonally in-

decomposable summands is as claimed. We consider the di�erent possibilities forl mod 4.

l ≡ 0 mod 4:

Here the claim is that ε(2l) = 0 (De�nition 2.4.6). By Lemma 2.4.8, it will beenough to show that (X2l−1(ei⊗fj), ei⊗fj) = 0 for all ei and fj . Since 2l−1 ≡ 7

mod 8, with Table 3.3 one gets

(X2l−1 · (e1 ⊗ v), e1 ⊗ v) = 0

(X2l−1 · (e2 ⊗ v), e2 ⊗ v) = 0

(X2l−1 · (e3 ⊗ v), e3 ⊗ v) = 0

(X2l−1 · (e−3 ⊗ v), e−3 ⊗ v) = (X2l−1v, v) + (X2l−2v, v)

(X2l−1 · (e−2 ⊗ v), e−2 ⊗ v) = (X2l−2v, v) + (X2l−4v, v)

(X2l−1 · (e−1 ⊗ v), e−1 ⊗ v) = (X2l−3v, v) + (X2l−4v, v)

+ (X2l−5v, v) + (X2l−6v, v)

for all v ∈ V (2l). Since l ≡ 0 mod 2, it follows from (*) in the proof of Lemma3.6.2 that we have (X2l−1fj , fj) + (X2l−2fj , fj) = 0 for all j. Furthermore, onecomputes with Table 3.4 and Table 3.5 that for all j we have (X2l−2fj , fj) +(X2l−4fj , fj) = 0 and (X2l−3fj , fj)+(X2l−4fj , fj)+(X2l−5fj , fj)+(X2l−6fj , fj) =0. Hence ε(2l) = 0, as claimed.

l ≡ 1 mod 4:In this case, the claim is that ε(2l+4) = 1 and ε(2l−2) = 0 (De�nition 2.4.6). Toshow that ε(2l + 4) = 1, by Lemma 2.4.8 it will su�ce to show that (X2l+3(ei ⊗fj), ei ⊗ fj) 6= 0 for some ei ⊗ fj . Now 2l + 3 ≡ 5 mod 8, so with Table 3.3one �nds that (X2l+3(e−1 ⊗ v), e−1 ⊗ v) = (X2l−1v, v) + (X2l−2v, v). Now forv = f2l we have X2l−1v = f1 and X2l−2v = f2 (Table 3.4), so it follows that(X2l+3(e−1 ⊗ f2l), e−1 ⊗ f2l) = 1. Therefore ε(2l + 4) = 1.

To prove that ε(2l−2) = 0, by Lemma 2.4.8 we should show that (X2l−1·γ, γ) =0 for all γ ∈ V (6)⊗V (2l) in a basis for the kernel of X2l−2 acting on V (6)⊗V (2l).Now 2l−2 ≡ 0 mod 8, so by Table 3.3 we have X2l−2 ·(v⊗w) = v⊗X2l−2w for allv ∈ V (6), w ∈ V (2l). Hence a basis for the kernel of X2l−2 acting on V (6)⊗V (2l)is given by ei ⊗ fj , where j ≤ 2l− 2. Since 2l− 3 ≡ 7 mod 8, with Table 3.3 one�nds:

(X2l−3 · (e1 ⊗ v), e1 ⊗ v) = 0

(X2l−3 · (e2 ⊗ v), e2 ⊗ v) = 0

(X2l−3 · (e3 ⊗ v), e3 ⊗ v) = 0

(X2l−3 · (e−3 ⊗ v), e−3 ⊗ v) = (X2l−3v, v) + (X2l−4v, v)

(X2l−3 · (e−2 ⊗ v), e−2 ⊗ v) = (X2l−4v, v) + (X2l−6v, v)

(X2l−3 · (e−1 ⊗ v), e−1 ⊗ v) = (X2l−5v, v) + (X2l−6v, v)

+ (X2l−7v, v) + (X2l−8v, v)

Using Table 3.4 and Table 3.5, one veri�es for all j ≤ 2l − 2 that we have(X2l−3fj , fj) + (X2l−4fj , fj) = 0 and (X2l−4fj , fj) + (X2l−6fj , fj) = 0. We alsoclaim that for all j ≤ 2l−2 we have (X2l−5fj , fj)+(X2l−6fj , fj)+(X2l−7fj , fj)+(X2l−8fj , fj) = 0. If j ≤ l, then this is clear, since we have (Xkfj , fj) for anyk ≥ 0 by Lemma 3.6.1 (i) and (ii). If l < j ≤ 2l − 2, then we can writefj = X2fj+2 and verify using Table 3.4 and Table 3.5 that (X2l−5fj , fj) +(X2l−6fj , fj)+(X2l−7fj , fj)+(X2l−8fj , fj) = (X2l−3fj+2, fj)+(X2l−4fj+2, fj)+(X2l−5fj+2, fj) + (X2l−6fj+2, fj) = 0. It follows therefore that ε(2l − 2) = 0.

l ≡ 2 mod 4:

84 Chapter 3.

In this case, the claim is that ε(2l+4) = 0, ε(2l) = 0, and ε(2l−4) = 0 (De�nition2.4.6). To show that ε(2l+ 4) = 0, by Lemma 2.4.8 it will be enough to show that(X2l+3(ei ⊗ fj), ei ⊗ fj) = 0 for all ei and fj . Now 2l + 3 ≡ 7 mod 8, so withTable 3.3 we get

(X2l+3 · (e1 ⊗ v), e1 ⊗ v) = 0

(X2l+3 · (e2 ⊗ v), e2 ⊗ v) = 0

(X2l+3 · (e3 ⊗ v), e3 ⊗ v) = 0

(X2l+3 · (e−3 ⊗ v), e−3 ⊗ v) = (X2l+3v, v) + (X2l+2v, v) = 0

(X2l+3 · (e−2 ⊗ v), e−2 ⊗ v) = (X2l+2v, v) + (X2lv, v) = 0

(X2l+3 · (e−1 ⊗ v), e−1 ⊗ v) = (X2l+1v, v) + (X2lv, v) + (X2l−1v, v) + (X2l−2v, v)

= (X2l−1v, v) + (X2l−2v, v)

for all v ∈ V (2l). Since l ≡ 0 mod 2, it follows from (*) in the proof of Lemma3.6.2 that we have (X2l−1fj , fj) + (X2l−2fj , fj) = 0 for all j. Hence ε(2l− 4) = 0.

For showing that ε(2l) = 0, by Lemma 2.4.8 it will su�ce to show that(X2l−1v, v) = 0 for all v ∈ V (6) ⊗ V (2l) in a basis of the kernel of X2l actingon V (6)⊗ V (2l). Now 2l ≡ 4 mod 8, so with Table 3.3 one veri�es that

X2l · (ei ⊗ v) = 0 (i = 1, 2, 3,−3)

X2l · (e−2 ⊗ v) = e1 ⊗X2l−4v

X2l · (e−1 ⊗ v) = e2 ⊗X2l−4v

for all v ∈ V (2l). Hence a basis for the kernel of X2l acting on V (6) ⊗ V (2l) isgiven by the ei⊗ fj such that 1 ≤ j ≤ 2l− 4 if i = −2 or i = −1. Since 2l− 1 ≡ 3mod 8, with Table 3.3 we get

(X2l−1 · (e1 ⊗ v), e1 ⊗ v) = 0

(X2l−1 · (e2 ⊗ v), e2 ⊗ v) = 0

(X2l−1 · (e3 ⊗ v), e3 ⊗ v) = 0

(X2l−1 · (e−3 ⊗ v), e−3 ⊗ v) = (X2l−1v, v) + (X2l−2v, v)

(X2l−1 · (e−2 ⊗ v), e−2 ⊗ v) = (X2l−2v, v) + (X2l−4v, v)

(X2l−1 · (e−1 ⊗ v), e−1 ⊗ v) = (X2l−1v, v) + (X2l−2v, v)

+ (X2l−3v, v) + (X2l−4v, v)

for all v ∈ V (2l). Now l ≡ 0 mod 2, so it follows from (*) in the proof of Lemma3.6.2 that we have (X2l−1fj , fj) + (X2l−2fj , fj) = 0 for all fj . Furthermore,using Table 3.4 and Table 3.5, it is straightforward to verify that (X2l−2fj , fj) +(X2l−4fj , fj) = 0 and (X2l−3fj , fj)+(X2l−4fj , fj)+(X2l−5fj , fj)+(X2l−6fj , fj) =0 for all j ≤ 2l − 4. It follows then that ε(2l) = 0.

What remains in this case is to show that ε(2l − 4) = 0. By Lemma 2.4.8,it will su�ce to show that (X2l−5v, v) = 0 for all v in a basis of the kernel ofX2l−4 acting on V (6) ⊗ V (2l). Now 2l − 4 ≡ 0 mod 8, so by Table 3.3 we haveX2l−4 · (v ⊗ w) = v ⊗X2l−4w for all v ∈ V (6), w ∈ V (2l). Hence a basis for thekernel of X2l−4 acting on V (6)⊗ V (2l) is given by ei ⊗ fj , where j ≤ 2l− 4. Now

2l − 5 ≡ 7 mod 8, so with Table 3.3 we get

(X2l−5 · (e1 ⊗ v), e1 ⊗ v) = 0

(X2l−5 · (e2 ⊗ v), e2 ⊗ v) = 0

(X2l−5 · (e3 ⊗ v), e3 ⊗ v) = 0

(X2l−5 · (e−3 ⊗ v), e−3 ⊗ v) = (X2l−5v, v) + (X2l−6v, v)

(X2l−5 · (e−2 ⊗ v), e−2 ⊗ v) = (X2l−6v, v) + (X2l−8v, v)

(X2l−5 · (e−1 ⊗ v), e−1 ⊗ v) = (X2l−7v, v) + (X2l−8v, v)

+ (X2l−9v, v) + (X2l−10v, v)

With Table 3.5, it is easy to verify that (X2l−5fj , fj)+(X2l−6fj , fj) = 0 for allj ≤ 2l−4. We claim that (X2l−6fj , fj)+(X2l−8fj , fj) = 0 for all j ≤ 2l−4. If j ≤ l,this is clear, since we have (Xkfj , fj) for any k ≥ 0 by Lemma 3.6.1 (i) and (ii).If l < j ≤ 2l − 4, then we can write fj = X4fj+4 and use Table 3.4 to verify that(X2l−6fj , fj) + (X2l−8fj , fj) = (X2l−2fj+4, fj) + (X2l−4fj+4, fj) = 0. Similarly,one checks that (X2l−7fj , fj) + (X2l−8fj , fj) + (X2l−9fj , v) + (X2l−10fj , fj) = 0for all j ≤ 2l − 4. Therefore ε(2l − 4) = 0, as claimed.

l ≡ 3 mod 4:

In this case, the claim is that ε(2l+2) = 0 and ε(2l−4) = 1 (De�nition 2.4.6). Toshow that ε(2l + 2) = 0, by Lemma 2.4.8 it will su�ce to show that (X2l+1(ei ⊗fj), ei ⊗ fj)) = 0 for all ei and fj . Now 2l + 1 ≡ 7 mod 8, so with Table 3.3 onegets

(X2l+1 · (e1 ⊗ v), e1 ⊗ v) = 0

(X2l+1 · (e2 ⊗ v), e2 ⊗ v) = 0

(X2l+1 · (e3 ⊗ v), e3 ⊗ v) = 0

(X2l+1 · (e−3 ⊗ v), e−3 ⊗ v) = (X2l+1v, v) + (X2lv, v) = 0

(X2l+1 · (e−2 ⊗ v), e−2 ⊗ v) = (X2lv, v) + (X2l−2v, v) = (X2l−2v, v)

(X2l+1 · (e−1 ⊗ v), e−1 ⊗ v) = (X2l−1v, v) + (X2l−2v, v)

+ (X2l−3v, v) + (X2l−4v, v)

for all v ∈ V (2l). Now using Table 3.4, it is straightforward to verify for all j that(X2l−2fj , fj) = 0 and (X2l−1fj , fj)+(X2l−2fj , fj)+(X2l−3fj , fj)+(X2l−4fj , fj) =0. Thus ε(2l + 2) = 0.

Finally, we show that ε(2l− 4) = 1 by �nding a γ ∈ V (6)⊗ V (2l) such that γis in the kernel of X2l−4 acting on V (6) ⊗ V (2l), and such that (X2l−5γ, γ) = 1(Lemma 2.4.8). To this end, de�ne

γ = e2 ⊗ f2l + e−3 ⊗ (f2l−2 + f2l−4) + e−1 ⊗ f2l−4.

Now 2l − 4 ≡ 2 mod 8, so with Table 3.3 and Table 3.4 one computes

X2l−4 · (e2 ⊗ f2l) = e2 ⊗ f4

X2l−4 · (e−3 ⊗ (f2l−2 + f2l−4)) = e−3 ⊗ f2 + e2 ⊗ f4

X2l−4 · (e−1 ⊗ f2l−4) = e−3 ⊗ f2

86 Chapter 3.

and thusX2l−4 ·γ = 0. If l = 3, a calculation shows that (X2l−5 ·γ, γ) = (X ·γ, γ) =1. Suppose then that l > 3. With Table 3.3 and Table 3.5, one computes that

X2l−5 · (e2 ⊗ f2l) = (e2 + e1)⊗ (δf1 + f4 + f5)

+ e1 ⊗ (δf2 + f4 + f6)

X2l−5 · (e−3 ⊗ (f2l−2 + f2l−4)) = (e−3 + e3 + e2 + e1)⊗ (f2 + f3)

+ (e3 + e2 + e1)⊗ (f2 + f4)

X2l−5 · (e−1 ⊗ f2l−4) = (e−1 + e−2)⊗ f1 + e−2 ⊗ f2

where δ = 0 if l ≡ 3 mod 8 and δ = 1 if l ≡ 7 mod 8. Now it is straightforwardto verify that (X2l−5 · γ, γ) = 1.

We can now apply the decompositions given in 3.6 to determine when maximaltensor product subgroups contain distinguished unipotent elements.

Lemma 3.6.5. Let 1 < m ≤ n be even. Suppose that in the K[u]-module Vm⊗Vnall block sizes have multiplicity ≤ 2. Then m ≤ 6.

Proof. Write m = 2k and n = 2k′. By Theorem 3.3.5 and Theorem 3.3.8, we have

Vm ⊗ Vn = 2 · V2k1 ⊕ · · · ⊕ 2 · V2kt ,

where Vk⊗V ′k = Vk1⊕· · ·⊕Vkt . Thus if in Vm⊗Vn all block sizes have multiplicity≤ 2, it follows that in Vk ⊗ Vk′ all block sizes have multiplicity ≤ 1. By Theorem3.3.14, this implies that k ≤ 3, hence m ≤ 6, as claimed.

Proposition 3.6.6. Let V1 and V2 be vector spaces equipped with non-degeneratealternating bilinear forms β1 and β2, respectively. Assume that 1 < dimV1 ≤dimV2. Let ui ∈ Sp(Vi) be a unipotent element. Then with respect to the productform β1 ⊗ β2 on V1 ⊗ V2, the unipotent element u = u1 ⊗ u2 ∈ Sp(V1)⊗ Sp(V2) isa distinguished unipotent element of Sp(V1 ⊗ V2) if and only if the following hold:

• dimV1 = 2, and V1 ↓ K[u1] = V (2) (in other words, u1 ∈ Sp(V1) is a regularunipotent element).

• We have an orthogonal decomposition V2 ↓ K[u2] = V (2k1) + · · · + V (2kt),where t ≥ 1, 1 ≤ k1 < · · · < kt, and ki is odd for all 1 ≤ i ≤ t.

Furthermore, when these two conditions hold, we have V1⊗V2 ↓ K[u] = V (2k1)2 +· · ·+ V (2kt)

2.

Proof. Suppose that u = u1 ⊗ u2 is a distinguished unipotent element of Sp(V1 ⊗V2). First we claim that ui is a distinguished unipotent element in Sp(Vi) fori = 1, 2. Indeed, now u1⊗u2 = ϕ(u1, u2), where ϕ is the homomorphism Sp(V1)×Sp(V2)→ Sp(V1⊗ V2) of algebraic groups de�ned by (g1, g2) 7→ g1⊗ g2. It followsthen as in Lemma 2.1.2 (ii) that (u1, u2) cannot be centralized by a nontrivialtorus in Sp(V1) × Sp(V2), hence that ui is a distinguished unipotent element inSp(Vi) for i = 1, 2.

We �rst consider the case where Vi are orthogonally indecomposable as K[ui]-modules. Since ui is distinguished in Sp(Vi), by Proposition 2.4.4 we have V1 ↓K[u1] = V (2m) and V2 ↓ K[u2] = V (2n), where 1 ≤ m ≤ n.

3.7. Proof of Theorem 1.1.4 87

Since u1⊗u2 is a distinguished unipotent element of Sp(V1⊗V2), it follows fromProposition 2.4.4 that all Jordan block sizes in the K[u]-module V2m ⊗ V2n havemultiplicity ≤ 2. By Lemma 3.6.5, it follows that m ≤ 3. For m = 1, m = 2, andm = 3, one checks from Table 3.6 that u1⊗u2 is a distinguished unipotent element(as described in Proposition 2.4.4) if and only if m = 1 and n is odd. In this case,we have V1 ↓ K[u] = V (2) and V2 ↓ K[u] = V (2n), and V1 ⊗ V2 = V (2n)2 byTable 3.6. This proves the claim in the case where V1 and V2 are both orthogonallyindecomposable.

Consider then the case where we have orthogonal decompositions V1 = ⊕si=1Wi

and V2 = ⊕tj=1Zj whereWi and Zj are orthogonally indecomposableK[u]-modulesfor all i and j. This gives an orthogonal decomposition V1⊗V2 = ⊕i,j(Wi⊗Zj) ofK[u]-modules. Since u1⊗u2 is a distinguished unipotent element of Sp(V1⊗V2), itfollows that for all i and j, theK[u]-moduleWi⊗Zj corresponds to a distinguishedunipotent element of Sp(Wi⊗Zj). From the indecomposable case we have treatedbefore, it follows that for all i, j one of the following holds:

• Wi ↓ K[u] = V (2), and Zj ↓ K[u] = V (2k) for some k ≥ 1 odd;

• Wi ↓ K[u] = V (2k) for some k > 1 odd, and Zj ↓ K[u] = V (2).

Note that in both cases we have Wi ⊗ Zj ↓ K[u] = V (2k)2 by the indecompo-sable case. We show now that the second possibility cannot occur. Suppose thatfor some i, we have Wi ↓ K[u] = V (2k) for k > 1 odd. Then it follows that wehave Zj ↓ K[u] = V (2) for all j. Thus t = 1, as otherwise V1 ⊗ V2 ↓ K[u] wouldhave ≥ 4 Jordan blocks of size 2k. But then dimV1 ≥ dimWi > dimV2 = 2,contradicting our assumption dimV1 ≤ dimV2. Therefore Wi ↓ K[u] = V (2) forall i, and for all j we have Zj ↓ K[u] = V (2kj) where kj ≥ 1 is odd. Now we haves = 1, as otherwise (V1 ⊗ V2) ↓ K[u] would have ≥ 4 Jordan blocks of size 2kj forall j.

Thus V1 ↓ K[u] = V (2), and V1 ⊗ V2 = ⊕tj=1V (2kj)2. Again because each

Jordan block size has multiplicity ≤ 2 in V1 ⊗ V2, each kj must be distinct. Byordering the summands suitably, we have 1 ≤ k1 < · · · < kt, so V1 and V2 are asclaimed. This completes the proof of the proposition.

3.7 Proof of Theorem 1.1.4

In this section, we put results from previous sections together and prove Theorem1.1.4. For convenience, we recall the statement of Theorem 1.1.4:

Theorem 1.1.4. Let G = SL(V ) (dimV ≥ 2), G = Sp(V ) (dimV ≥ 2), orG = SO(V ) (dimV ≥ 5), where V is a �nite-dimensional vector space over K.Fix a distinguished unipotent element u ∈ G. Let X be one of the maximal closedconnected subgroup of G given in (a) - (f) of Theorem 1.1.3. Then the cases whereX contains a G-conjugate of u are precisely the following:

(a) G = SL(V ), X = Sp(V ) (dimV even) or X = SO(V ) (p 6= 2 and dimV odd),and u is a regular unipotent element.

(b) G = Sp(V ), p = 2, dimV > 2, X = SO(V ), and the number of Jordan blocksof u is even.

88 Chapter 3.

(c) G = SO(V ), p = 2, dimV is even, X is the stabilizer of a nonsingular 1-space,and u has a Jordan block of size 2.

(d) X is a maximal parabolic subgroup.

(e) G = Sp(V ) or G = SO(V ), V = W ⊕ W⊥ where W is a non-degeneratesubspace of V , X = stabG(W )◦, and V ↓ K[u] ∼=

⊕ti=1 Vdi ⊕

⊕sj=1 Vd′j for

integers di, d′j ≥ 1 such that the following conditions hold:

• dimW =∑t

i=1 di.

• If p = 2 and G = SO(V ), then t ≡ 0 mod 2.

(f) V = V1 ⊗ V2 with dimV1 ≤ dimV2 and one of the following holds:

(i) G = SO(V ), p = 2, dimV1 = 2, X = Sp(V1)⊗Sp(V2), and the orthogonaldecomposition V ↓ K[u] (Proposition 2.4.4) is equal to V (2d1)2 + · · · +V (2dt)

2 for some 1 ≤ d1 < · · · < dt such that di is odd for all i.

(ii) G = Sp(V ) or G = SO(V ), p 6= 2, X is as in Theorem 1.1.3 (ii), (iii) or(iv); and for m = dimV1 and n = dimV2 one of the following conditionshold:

• The pair (m,n) is contained in the set S of De�nition 3.3.12, andV ↓ K[u] ∼=

⊕mi=1 Vm+n−2i+1. (In this case V ↓ K[u] ∼= Vm ⊗ Vn).

• There exist integers 1 ≤ n1 < n2 < · · · < nt such that∑t

i=1 ni = n,ni ≡ ni′ mod 2 for all 1 ≤ i, i′ ≤ t, ni−ni−1 ≥ 2m for all 2 ≤ i ≤ t,the pair (m,ni) is contained in the set S of De�nition 3.3.12 forall 1 ≤ i ≤ t, and

V ↓ K[u] ∼=min(m,n1)⊕

j=1

Vn1+m−2j+1 ⊕t⊕i=2

m⊕j=1

Vni+m−2j+1.

(In this case V ↓ K[u] ∼= Vm ⊗ (Vn1 ⊕ Vn2 ⊕ · · · ⊕ Vnt)).

Proof. We begin with cases (a) - (d), and here we will see that claim followsfrom results which are more or less well known. For case (a), we know that theonly distinguished unipotent elements of SL(V ) are those with only one Jordanblock (Lemma 2.2.2), so the claim follows from Proposition 2.3.3 (ii)-(iii) andProposition 2.4.4 (v). Case (b) is a consequence of Proposition 2.4.4 (iii), andcase (c) follows from [LS12, 6.8]. For case (d), recall that every element of G iscontained in some Borel subgroup. Hence any parabolic subgroup of G intersectsevery conjugacy class of G. In particular, a maximal parabolic subgroup of Gintersects every unipotent conjugacy class of G.

For case (e), suppose that G = Sp(V ) or G = SO(V ). If V = W ⊕W⊥ andu ∈ stabG(W )◦, then it is very obvious that V ↓ K[u] ∼=

⊕ti=1 Vdi ⊕

⊕sj=1 Vd′j ,

where W ↓ K[u] ∼=⊕t

i=1 Vdi and W⊥ ↓ K[u] ∼=

⊕sj=1 Vd′j ; thus dimW =

∑ti=1 di

in this case. If p = 2 and G = SO(V ), then stabG(W )◦ = SO(W ) × SO(W⊥).Since the restriction of u to W is contained in SO(W ), it follows from Proposition2.4.4 (iii) that t ≡ 0 mod 2.

Conversely for (e), let V ↓ K[u] ∼=⊕t

i=1 Vdi ⊕⊕s

j=1 Vd′j , and assume that

t ≡ 0 mod 2 if p = 2 and G = SO(V ). By [LS12, Proposition 3.5] for p 6= 2 and

3.7. Proof of Theorem 1.1.4 89

Proposition 2.4.4 for p = 2, it is possible to write the decomposition of V ↓ K[u]as an orthogonal direct sum

V ↓ K[u] = V1 ⊕ · · · ⊕ Vt ⊕ V ′1 ⊕ · · · ⊕ V ′s ,

where Vi ∼= Vdi and V′j∼= Vd′j for all 1 ≤ i ≤ t and 1 ≤ j ≤ s. Then for Z =

⊕ti=1 Vi

we have Z⊥ =⊕s

j=1 V′j , so u ∈ stabG(Z) for V = Z ⊕Z⊥ with dimZ =

∑ti=1 di.

In fact, we have u ∈ stabG(Z)◦, when p 6= 2 or G = Sp(V ) this is true forany unipotent element of stabG(Z); when p = 2 and G = SO(V ), this followsfrom the assumption t ≡ 0 mod 2 and Proposition 2.4.4 (iii). Finally, any otherorthogonal decomposition V = W ⊕W⊥ with dimW = dimZ is conjugate to thedecomposition V = Z ⊕ Z⊥; i.e. there exists some g ∈ G such that g(Z) = W .Then gug−1 ∈ g stabG(Z)◦g−1 = stabG(W )◦, as desired.

What remains is to deal with the stabilizers of tensor decompositions in The-orem 1.1.3 (f). In this case, su�cency follows from Proposition 3.6.6 and Lemma3.3.17. We show that the conditions listed in our claim are necessary. For this,suppose that V = V1 ⊗ V2, where 1 < dimV1 ≤ dimV2. Let X be one of thesubgroups in Theorem 1.1.3 (f) and suppose that u ∈ X. Now X = X1 ⊗ X2,where for all i = 1, 2 we have Xi = SL(Vi), Xi = Sp(Vi), or Xi = SO(Vi). Since uis a distinguished unipotent element, it follows as in the beginning of the proof ofProposition 3.6.6 that u = u1 ⊗ u2, where ui is a distinguished unipotent elementof Xi.

We now consider di�erent possibilities for the subgroups X in Theorem 1.1.3(f). First suppose that G = SL(V ), so now Xi = SL(Vi) for i = 1, 2. Since ui isdistinguished in Xi, we have that ui acts on Vi with a single Jordan block (Lemma2.2.2). Similarly u acts on V with a single Jordan block. This is a contradiction,since it is well known that the tensor product of any two Jordan blocks of size > 1has ≥ 2 Jordan blocks (see e.g. Theorem 3.3.5 or [SS97, Lemma 1.5]).

Suppose then that G = Sp(V ) or G = SO(V ). If p = 2, then the only case inTheorem 1.1.3 (f) that applies is the one where G = SO(V ) and X = Sp(V1) ⊗Sp(V2). In this case the claim follows from Proposition 3.6.6. Consider then p 6= 2,so for all i = 1, 2 we have Xi = Sp(Vi) or Xi = SO(Vi). Since ui is distinguishedin Xi, it follows from Proposition 2.3.2 that V1 ↓ K[u1] =

⊕si=1 Vmi and V2 ↓

K[u2] =⊕t

j=1 Vnj , where 1 ≤ m1 < · · · < mt and 1 ≤ n1 < · · · < ns areintegers such that mi ≡ mi′ mod 2 and nj ≡ nj′ mod 2 for all 1 ≤ i, i′ ≤ sand 1 ≤ j, j′ ≤ t. Since u is distinguished in G, the tensor product V1 ⊗ V2 hasno repeated block sizes. Now the claim is immediate from Lemma 3.3.17. Thiscompletes the proof of the theorem.

Chapter 4

Some representation theory

The purpose of this chapter is to list some results on the representation theory ofsimple algebraic groups for later reference. Most important will be the results onrepresentations of SL2(K) (Section 4.2 and Section 4.6), which will be key in theproof of Theorem 1.1.10 for unipotent elements of order p.

4.1 Weyl modules and tilting modules

Recall the notation LG(λ) for the irreducible G-module with highest weight λ ∈X(T )+, and VG(λ) for the Weyl module of G with highest weight λ.

Proposition 4.1.1. Let λ, µ ∈ X(T )+ such that µ 6� λ. Then if 0 → LG(µ) →E → LG(λ) → 0 is a nonsplit extension of LG(λ) by LG(µ), we have E ∼=VG(λ)/W for some submodule W of VG(λ).

Proof. This follows from [Jan03, Lemma II.2.13, II.2.14].

De�nition 4.1.2. Let V be a G-module. A �ltration 0 = V0 ⊆ V1 ⊆ · · · ⊆ Vt = Vof submodules Vi of V is called a Weyl �ltration, if for all 1 ≤ i ≤ t we haveVi/Vi−1

∼= VG(λi) for some λi ∈ X(T )+. We say that V is a tilting module, if bothV and V ∗ have a Weyl �ltration.

The following facts about tilting modules will be useful in the sequel.

Theorem 4.1.3 ([Jan03, Corollary II.E.2], [Mat90]). Let V1 and V2 be G-modules.Then

(i) (Donkin) The direct sum V1 ⊕ V2 is a tilting module if and only if both V1

and V2 are tilting modules.

(ii) (Mathieu, Donkin) If V1 and V2 are tilting modules, then the tensor productV1 ⊗ V2 is tilting.

Theorem 4.1.4 ([Jan03, Lemma II.E.3, Lemma II.E.5, Proposition II.E.6]). Letλ ∈ X(T )+. Then:

(i) (Ringel) Up to isomorphism, there is a unique indecomposable tilting moduleTG(λ) with dimTG(λ)λ = 1 and such that TG(λ)µ 6= 0 implies µ � λ.

(ii) If Q is a tilting module with Qλ 6= 0 such that λ is maximal among theweights of Q, then TG(λ) is a direct summand of Q.

91

92 Chapter 4.

(iii) If Q is a tilting module, there exist unique integers n(ν) ≥ 0, almost all 0,such that

Q ∼=⊕

ν∈X(T )+

TG(ν)n(ν).

We call TG(λ) the indecomposable tilting module with highest weight λ. Thecomposition factors of TG(λ) are not known in general. For G of type A1, we havethe following result which gives a recursive description of the indecomposabletilting modules.

Theorem 4.1.5 ([Sei00, Lemma 2.3], [Don93, Example 2, pg. 47]). Assume thatG = SL2(K) and let c ∈ Z≥0 be a dominant weight. Then:

(i) If 0 ≤ c ≤ p − 1, the indecomposable tilting module TG(c) is irreducible, soTG(c) = LG(c).

(ii) If p ≤ c ≤ 2p − 2, the indecomposable tilting module TG(c) is uniserial ofdimension 2p, and TG(c) = LG(2p− 2− c)/LG(c)/LG(2p− 2− c).

(iii) (Donkin) If c > 2p− 2, then TG(c) ∼= TG(p− 1 + r)⊗ TG(s)[1], where s ≥ 1and 0 ≤ r ≤ p− 1 are such that c = sp+ (p− 1 + r).

4.2 Irreducible representations of SL2(K)

In this subsection, we give some basic results about irreducible representationsof SL2(K). Throughout we will identify the weights of SL2(K) with Z, and thedominant weights with Z≥0. We begin with the following lemma which is wellknown and follows from Steinberg's tensor product theorem.

Lemma 4.2.1. Let λ ∈ Z≥0 be a dominant weight of G = SL2(K). Write λ =∑i≥0 λip

i, where 0 ≤ λi ≤ p − 1. Then the weights occurring in the irreduciblemodule LG(λ) are precisely the ones of the form∑

i≥0

(λi − 2ki)pi

where 0 ≤ ki ≤ λi. Furthermore, each weight space of LG(λ) is one-dimensional.

For the rest of this section, we will give some technical lemmas about the actionof a unipotent element u of SL2(K) on irreducible representations of SL2(K).These will be used later in Section 5.13. The �rst lemma is well known.

Lemma 4.2.2. Let λ ∈ Z≥0 be a non-zero weight of G = SL2(K) such that0 ≤ λ ≤ p− 1. Then a non-identity unipotent element u ∈ G acts on LG(λ) witha single Jordan block of size λ+ 1.

Lemma 4.2.3. Let λ ∈ Z≥0 be a non-zero weight of G = SL2(K). Then a non-identity unipotent element u ∈ G acts on LG(λ) with Jordan blocks of distinct sizesif and only if λ = apk + bpl, where 0 ≤ k < l, 0 ≤ a, b ≤ p− 1 and a+ b ≤ p− 1.In this case u acts on LG(λ) with Jordan blocks

[a+ b+ 1, a+ b− 1, . . . , a+ b− 2d+ 1]

where d = min{a, b}.

4.2. Irreducible representations of SL2(K) 93

Proof. Write λ in base p, say λ =∑t

i=0 kipi, where 0 ≤ ki ≤ p − 1. Now by

Steinberg's tensor product theorem, we have

LG(λ) ∼=t⊗i=0

LG(ki)[i].

Since u acts on LG(ki) with a single Jordan block of size ki + 1 (Lemma 4.2.2), itfollows that

LG(λ) ↓ K[u] ∼=t⊗i=0

Vki+1.

It is now easy to see that to prove the claim, it is enough to show that the followingfacts hold:

(1) If 1 ≤ d1, d2 ≤ p and d1 + d2 > p+ 1, then Vd1 ⊗ Vd2 has repeated blocks.

(2) If 1 ≤ d1 ≤ d2 ≤ p and d1 +d2 ≤ p+1, we have Vd1⊗Vd2 ∼= ⊕d1−1i=0 Vd1+d2−2i−1.

(3) If s > 2 and 1 < d1 ≤ d2 ≤ · · · ≤ ds ≤ p, then the tensor product ⊗si=1Vdi hasrepeated blocks.

The claims (1) and (2) are Corollary 3.3.7 (i) and (ii), respectively. Whatremains is to show that ⊗si=1Vdi has repeated blocks if s ≥ 3 and 1 < d1 ≤ d2 ≤· · · ≤ ds ≤ p. Clearly it will be enough to do this for s = 3. For the sake ofcontradiction, suppose then that Vd1 ⊗ Vd2 ⊗ Vd3 has no repeated blocks for some1 < d1 ≤ d2 ≤ d3 ≤ p. Then Vd1 ⊗Vd2 has no repeated blocks, so applying (1) and(2) to Vd1 ⊗ Vd2 we get

Vd1 ⊗ Vd2 ⊗ Vd3 ∼= (Vd1+d2−1 ⊗ Vd3)⊕ (Vd1+d2−3 ⊗ Vd3)⊕ · · ·

Now Vd1+d2−1⊗Vd3 and Vd1+d2−3⊗Vd3 also have no repeated blocks, so againapplying (1) and (2) to these tensor products gives

Vd1+d2−1 ⊗ Vd3 ∼= Vd1+d2+d3−2 ⊕ Vd1+d2+d3−4 ⊕ · · ·

andVd1+d2−3 ⊗ Vd3 ∼= Vd1+d2+d3−4 ⊕ · · ·

Therefore the tensor product Vd1 ⊗ Vd2 ⊗ Vd3 has ≥ 2 Jordan blocks of size d1 +d2 + d3 − 4, contradiction.

Lemma 4.2.4. Assume that p ≥ 3. Let r ≥ p. Suppose that a nonidentity unipo-tent element u of SL2(K) acts on the module LG(r)⊕LG(r− 2) with no repeatedblocks. Then r = p+ 1, r = p, r = 2p or r = p+ pl with l ≥ 2.

Proof. By applying Lemma 4.2.3 to the module LG(r), it follows that r = apk+bpl,where 0 < a ≤ p− 1, 0 ≤ k < l, 0 ≤ b ≤ p− 1, and a+ b ≤ p− 1. If k ≥ 2, thenr − 2 is written in base p as

r − 2 = (p− 2) + (p− 1)p+ · · ·+ (p− 1)pk−1 + (a− 1)pk + bpl,

so u acts on LG(r− 2) with repeated blocks, by Lemma 4.2.3. Therefore we musthave k = 0 or k = 1.

94 Chapter 4.

Suppose that k = 0, so now r = a + bpl, where 0 < a ≤ p − 1, l ≥ 1, and0 < b ≤ p − 1. If a ≥ 2, then r − 2 = (a − 2) + bpl in base p, and so by Lemma4.2.3 we have

LG(r) ↓ K[u] = [a+ b+ 1, a+ b− 1, . . .]

LG(r − 2) ↓ K[u] = [a+ b− 1, . . .]

so u acts on LG(r)⊕LG(r−2) with two Jordan blocks of size a+b−1, contradiction.Therefore we must have a = 1. Now if l ≥ 2, then r − 2 is written in base p as

r − 2 = (p− 1) + (p− 1)p+ · · ·+ (p− 1)pl−1 + (b− 1)pl,

so u acts on LG(r − 2) with repeated blocks by Lemma 4.2.3. Therefore l = 1, sor = 1 + bp and r− 2 = (p− 1) + (b− 1)p in base p. Since u acts on LG(r− 2) withno repeated blocks, we have b = 1 by Lemma 4.2.3. That is, we have r = p+ 1, asclaimed.

Next we consider the case where k = 1. Now r = ap+bpl, where 0 < a ≤ p−1,l ≥ 2, and 0 ≤ b ≤ p − 1. Then r − 2 is written in base p as r − 2 = (p − 2) +(a− 1)p+ bpl. Since u acts on LG(r − 2) with no repeated blocks, it follows fromLemma 4.2.3 that either b = 0 and a ≤ 2, or a = 1 and b ≤ 1. In other words,we have r = p, r = 2p, or r = p + pl for l ≥ 2. This completes the proof of theclaim.

Lemma 4.2.5. Assume that p ≥ 5. Let r ≥ p. Suppose that a nonidentity unipo-tent element u of SL2(K) acts on the module LG(r)⊕LG(r− 4) with no repeatedblocks. Then one of the following holds:

(i) p | r.

(ii) r = p+ 1, r = p+ 2, r = p+ 3, r = 2p+ 1, r = 2p+ 2, or r = 1 + 3p.

(iii) r = a+ pl for some 4 ≤ a ≤ p− 2 and l ≥ 1.

Proof. By applying Lemma 4.2.3 to the module LG(r), it follows that r = apk+bpl,where 0 < a ≤ p − 1, 0 ≤ k < l, 0 ≤ b ≤ p − 1, and a + b ≤ p − 1. If k > 0,then we are in case (i) of the claim. Suppose then that k = 0, so now r = a+ bpl,where 0 < a ≤ p− 1, l > 0, and 0 0 since we areassuming that r ≥ p.

If a ≥ 4, then r− 4 is written in base p as r− 4 = (a− 4) + bpl. If b > 1, thenit follows from Lemma 4.2.3 that

LG(r) ↓ K[u] = [a+ b+ 1, a+ b− 1, a+ b− 3, . . .]

andLG(r − 4) ↓ K[u] = [a+ b− 3, . . .].

Thus LG(r)⊕LG(r−4) ↓ K[u] has two Jordan blocks of size a+b−3, contradiction.Therefore b = 1, so r = a + pl, where a ≥ 4. Since a + b ≤ p − 1, we have4 ≤ a ≤ p− 2, so r is as in case (iii) of the claim.

Consider then a < 4. If l > 1, then r − 4 is written in base p as

r − 4 = (p− 4 + a) + (p− 1)p+ · · ·+ (p− 1)pl−1 + (b− 1)pl,

4.3. Weyl group and weight orbits 95

so by Lemma 4.2.3 the element u acts on LG(r− 4) with repeated blocks, contra-diction. Therefore l = 1, so now r = a+ bpl, where 1 ≤ a ≤ 3. Furthermore, sincer− 4 is written in base p as r− 4 = (p− 4 + a) + (b− 1)p, it follows from Lemma4.2.3 that (p− 4 + a) + (b− 1) ≤ p− 1. This gives a+ b ≤ 4, and since 1 ≤ a ≤ 3and 1 ≤ b ≤ p − 1, the only possibilities for r are those given in case (ii) of theclaim.

Lemma 4.2.6. Assume that p ≥ 5. Let r ≥ p. Suppose that a nonidentity unipo-tent element u of SL2(k) acts on the module LG(r)⊕ LG(r − 2)⊕ LG(r − 4) withno repeated blocks. Then r = p or r = p+ 1.

Proof. According to Lemma 4.2.4, we must have r = p, r = p + 1, r = 2p, orr = p+ pl with l ≥ 2.

If r = 2p, then r − 2 = (p − 2) + p and r − 4 = (p − 4) + p, so by Lemma4.2.3 we have LG(r − 2) ↓ K[u] = [p− 2, p] and LG(r − 4) ↓ K[u] = [p− 4, p− 2].Therefore LG(r− 2)⊕LG(r− 4) ↓ K[u] has two blocks of size p− 2. If r = p+ pl

with l ≥ 2, the same argument shows that LG(r − 2)⊕ LG(r − 4) ↓ K[u] has twoblocks of size p − 2. Therefore the only possibilities are r = p and r = p + 1, asdesired.

4.3 Weyl group and weight orbits

Recall that W denotes the Weyl group of G. Fix a base ∆ = {α1, . . . , αl} for theroot system Φ of G, with respect to the maximal torus T . We will denote there�ection in W corresponding to a root α ∈ Φ by σα ∈W .

Let V be any G-module. Then we have a decomposition

V =⊕

µ∈X(T )

Vµ,

where Vµ denotes the T -weight space for weight µ ∈ X(T ). Now W acts on X(T ),and X(T )+ is a fundamental domain for this action [Hum72, Lemma 10.3B]. Thuswe have a decomposition

V =⊕

µ∈X(T )+

⊕w∈W

Vwµ.

This implies that dimV ≥ |Wµ| if µ ∈ X(T )+ and Vµ 6= 0, and this basicinequality will often be useful.

To describe the sizes of the orbits, we have the following lemma which followsfrom [Hum72, Lemma 10.3B].

Lemma 4.3.1. Let λ ∈ X(T )+ and write λ =∑l

i=1 aiωi, where ai ≥ 0. ThenStabW (λ) is generated by all σαi such that ai = 0.

Since the Weyl group is always generated by simple re�ections, from Lemma4.3.1 we see that the W -stabilizer of λ =

∑li=1 aiωi is the Weyl group of the

semisimple Lie algebra with Dynkin diagram ∆′ = {αi : ai = 0}. This makes thecomputation of |Wλ| = |W |/| StabW (λ)| easy, since the orders of Weyl groups ofindecomposable root systems are known (see for instance [Hum72, Table 12.1]).

96 Chapter 4.

4.4 Invariant forms on G-modules

In this section, we list some basic results on the existence of G-invariant bilinearforms on G-modules. We begin with the following de�nitions.

De�nition 4.4.1. Let V be a G-module. The following notation will be de�ned.

(i) Bil(V ) is the vector space of all bilinear forms on V .

(ii) For β ∈ Bil(V ) and g ∈ G, de�ne βg ∈ Bil(V ) by βg(v, w) = β(g−1v, g−1w).

(iii) Bil(V )G is the vector space of all G-invariant bilinear forms on V , in otherwords, the space of all β ∈ Bil(V ) such that βg = β.

De�nition 4.4.2. For λ ∈ X(T )+, we de�ne d(λ) =∑

α>0〈λ, α〉.

As seen in the next lemma and other lemmas below (Lemma 4.4.9, Lemma4.4.10), the quantity d(λ) in De�nition 4.4.2 will often be useful for describingwhen a G-module has a G-invariant alternating or symmetric bilinear form.

Lemma 4.4.3 ([Ste68, Lemma 78, Lemma 79]). Let λ ∈ X(T )+. Then:

(i) LG(λ)∗ ∼= LG(−w0λ), where w0 is the longest element in the Weyl group. Inparticular, LG(λ) is self-dual if and only if λ = −w0λ.

(ii) Assume that λ = −w0λ and p 6= 2. Then LG(λ) has a non-degenerate G-invariant bilinear form b, unique up to a scalar. Furthermore, the form b issymmetric if d(λ) ≡ 0 mod 2, and alternating if d(λ) ≡ 1 mod 2.

Corollary 4.4.4. Assume that p 6= 2. Let G = SL2(K) and �x a dominant weightλ ∈ Z≥0 of G. Then LG(λ) is self-dual, and LG(λ) is symplectic if λ is odd andorthogonal if λ is even.

By Lemma 4.4.3, in characteristic p 6= 2 deciding whether an irreducible mo-dule is symplectic or orthogonal is a straightforward computation with roots andweights. Let λ ∈ X(T )+ be a dominant weight and write λ =

∑li=1miωi, where

mi ∈ Z≥0. In Table 4.1, we give the value of d(λ) mod 2 (when λ = −w0(λ)) foreach simple type, in terms of the coe�cients mi.

In characteristic 2, it turns out that each nontrivial, irreducible self-dual mo-dule is symplectic, as shown by the following lemma.

Lemma 4.4.5 (Fong, [Fon74]). Assume that charK = 2. Let V be a nontrivial,irreducible self-dual representation of a group G. Then V has non-degenerate G-invariant bilinear form b, unique up to a scalar. Furthermore, b is alternating.

Lemma 4.4.5 shows that when p = 2, the image of any irreducible self-dualrepresentation of G lies in Sp(V ). Note that in the proof of Lemma 4.4.5 nothingabout algebraic groups is needed, so this holds in a much more general setting.Another lemma which is well known and also works in greater generality is thefollowing.

Lemma 4.4.6. Let V be a G-module. Then V ⊕ V ∗ has a non-degenerate G-invariant alternating bilinear form, and a non-degenerate G-invariant symmetricbilinear form.

4.4. Invariant forms on G-modules 97

Root system When is λ = −w0(λ)? d(λ) mod 2 when λ = −w0(λ)

Al (l ≥ 1) i� mi = ml−i+1 for all i0, when l is evenl+12 ·m l+1

2, when l is odd

Bl (l ≥ 2) always0, when l ≡ 0, 3 mod 4ml, when l ≡ 1, 2 mod 4.

Cl (l ≥ 2) always m1 +m3 +m5 + · · ·

Dl (l ≥ 4)l even: alwaysl odd: i� ml = ml−1

0, when l 6≡ 2 mod 4ml +ml−1, when l ≡ 2 mod 4.

G2 always 0F4 always 0E6 i� m1 = m6 and m3 = m5 0E7 always m2 +m5 +m7

E8 always 0

Table 4.1: Values of d(λ) modulo 2 for a weight λ =∑l

i=1miωi

Proof. De�ne a bilinear form on V ⊕V ∗ by 〈v+ f, v′+ f ′〉 = f ′(v) + cf(v′), wherec ∈ K is some �xed non-zero scalar. It is straightforward to verify that 〈−,−〉is a non-degenerate G-invariant bilinear form. Furthermore, the form 〈−,−〉 issymmetric if we choose c = 1 and alternating if we choose c = −1.

The following two lemmas are well known. For proofs, see for example [Bou59,�1, Dé�nition 12, pg. 30], [McG05], and [McG02].

Lemma 4.4.7. Suppose that a G-module V has a non-degenerate G-invariantalternating bilinear form (−,−). Then

(i) For all 1 ≤ k ≤ dimV , the exterior power ∧k(V ) admits a non-degenerateG-invariant bilinear form, which is alternating if k is odd and symmetric ifk is even.

(ii) If p = 0 or p > k, then the symmetric power Sk(V ) admits a non-degenerateG-invariant bilinear form, which is alternating if k is odd and symmetric ifk is even.

Lemma 4.4.8. Suppose that a G-module V has a non-degenerate G-invariantsymmetric form (−,−). Then

(i) For all 1 ≤ k ≤ dimV , the exterior power ∧k(V ) admits a non-degenerateG-invariant symmetric form.

(ii) If p = 0 or p > k, then the symmetric power Sk(V ) admits a non-degenerateG-invariant symmetric form.

The following lemma describes G-invariant bilinear forms on Weyl modules. Itis most likely well known, see for example [Jan73, Satz 8] for a similar result.

98 Chapter 4.

Lemma 4.4.9. Let λ ∈ X(T )+ and suppose that λ = −w0λ, where w0 is thelongest element in the Weyl group. Then:

(i) The Weyl module VG(λ) has a non-zero G-invariant bilinear form b, uniqueup to a scalar.

(ii) The radical rad b is the unique maximal submodule of VG(λ).

(iii) Assume that p 6= 2. Then b is symmetric if d(λ) ≡ 0 mod 2 and alternatingif d(λ) ≡ 1 mod 2.

(iv) Assume that p = 2. Then b is alternating.

Proof. We know that there exists a surjective G-morphism π : VG(λ) → LG(λ),which has the unique maximal submodule of VG(λ) as its kernel. It follows from[GN16, Lemma 4.3] that the map π induces an isomorphism Bil(LG(λ))G →Bil(VG(λ))G. Under this isomorphism, a bilinear form b ∈ Bil(LG(λ))G maps tobπ ∈ Bil(VG(λ))G, where bπ(v, v′) = b(π(v), π(v′)) for all v, v′ ∈ VG(λ).

Now Lemma 4.4.3 implies that Bil(LG(λ))G has dimension 1, so claim (i)follows. For claim (ii), let b ∈ Bil(LG(λ))G be non-zero, so that bπ ∈ Bil(VG(λ))G

is non-zero. It is clear that the radical rad bπ contains the kernel of π, which isthe unique maximal submodule of VG(λ). Since rad bπ is a G-submodule of VG(λ)and since bπ is non-zero, it follows that rad bπ = kerπ. This proves claim (ii).

Claim (iii) follows from the fact that bπ is symmetric if and only if b is symme-tric, and the fact that bπ is alternating if and only if b is alternating. Claim (iv)follows similarly, using Lemma 4.4.5.

We consider next G-invariant bilinear forms on tilting modules.

Lemma 4.4.10. Let λ ∈ X(T )+. Then:

(i) TG(λ)∗ ∼= TG(−w0λ), where w0 is the longest element in the Weyl group. Inparticular, TG(λ) is self-dual if and only if λ = −w0λ.

(ii) Assume that λ = −w0λ and p 6= 2. Then TG(λ) admits a non-degenerate G-invariant symmetric or alternating bilinear form b. Furthermore, any suchform b is symmetric if d(λ) = 0 mod 2 and alternating if d(λ) = 1 mod 2.

Proof. Claim (i) is [Jan03, Remark II.E.6].For claim (ii), assume that λ = −w0λ and p 6= 2. Now TG(λ) ∼= TG(λ)∗ by

(i). Furthermore, since TG(λ) is indecomposable, the ring EndG(TG(λ)) is a localring. It follows then from [QSSS76, Lemma 2.1] or [Wil76, Satz 2.11 (a)] thatthere exists a non-degenerate G-invariant bilinear form b on TG(λ) such that b issymmetric or alternating.

For the other part of claim (ii), let W ⊆ TG(λ) be a G-submodule such thatW ∼= VG(λ), see [Jan03, II.E.4]. We show �rst that W cannot be totally isotropicwith respect to b. For this, note thatWλ = 1. Thus ifW ⊆W⊥, then by V/W⊥ ∼=W ∗ we get dimTG(λ)λ ≥ 2, contradicting dimTG(λ)λ = 1 (Theorem 4.1.4 (i)).Therefore the restriction of bW of b to W is non-zero. This implies that b issymmetric if and only if bW is symmetric, and b is alternating if and only if bW isalternating. Now the claim follows from Lemma 4.4.9 (iii).

4.4. Invariant forms on G-modules 99

Corollary 4.4.11. Assume that p 6= 2. Let G = SL2(K) and n ∈ Z≥0. Then theindecomposable tilting module TG(n) has a non-degenerate G-invariant bilinearform, which is symmetric if n is even and alternating if n is odd.

Lemma 4.4.12. Assume that p = 2. Let G = SL2(K) and n ∈ Z>0. Then theindecomposable tilting module TG(n) has a non-degenerate G-invariant alternatingbilinear form.

Proof. If n = 1, then TG(n) = LG(1) (Theorem 4.1.5 (i)) and the claim followsfrom Lemma 4.4.5. For n = 2, we have TG(n) = LG(0)/LG(2)/LG(0) uniserialof dimension 4 by Theorem 4.1.5. The tensor product LG(1) ⊗ LG(1) is a tiltingmodule (Theorem 4.1.3) with highest weight 2, so it follows that LG(1)⊗LG(1) =TG(2) (Theorem 4.1.4). Since LG(1) has a non-degenerate G-invariant alternatingbilinear form, the same is true for the tensor product LG(1) ⊗ LG(1) (see forexample [KL90, 4.4, pg. 126-127]).

For n > 2, we will prove the claim by induction on n. Suppose that TG(n′) has anon-degenerate G-invariant alternating bilinear form for all 0 < n′ < n. If n = 2k,then TG(n) ∼= TG(2) ⊗ TG(k − 1)[1] by Theorem 4.1.5 (iii). Applying inductionto TG(2) and TG(k − 1), it follows that the tensor product TG(2) ⊗ TG(k − 1)[1]

has a non-degenerate G-invariant alternating bilinear form (again, for example by[KL90, 4.4, pg. 126-127]). If n = 2k+1, the claim follows with the same argument,since TG(n) ∼= TG(1)⊗ TG(k − 1)[1] by Theorem 4.1.5 (iii).

Lemma 4.4.13. Let λ ∈ X(T )+ such that λ = −w0λ. Let V be a uniserialG-module such that V = LG(0)/LG(λ)/LG(0), and suppose that V admits a non-degenerate G-invariant alternating bilinear form β. Then any non-degenerate G-invariant alternating bilinear form on V is a scalar multiple of β.

Proof. Let γ be some non-degenerateG-invariant alternating bilinear form on V . Itis straightforward to see (for example [Wil76, Satz 2.3]) that there exists a uniqueG-isomorphism ϕ ∈ EndG(V ) such that γ(v, v′) = β(ϕ(v), v′) for all v, v′ ∈ V .

By [GN16, Example 4.5], we have dim EndG(V ) = 2, and EndG(V ) has a basisconsisting of the identity map 1 : V → V and a map N : V → V such that Nis nilpotent and N(V ) is the unique 1-dimensional G-submodule of V . Then byreplacing γ with a scalar multiple if necessary, we can assume that ϕ = 1 + λ ·Nfor some λ ∈ K. Let z ∈ N(V ) be a non-zero vector, so N(V ) = 〈z〉. Since Vis indecomposable, it follows that 〈z〉 ⊆ 〈z〉⊥ with respect to the bilinear formβ. Furthermore, we have KerN = 〈z〉⊥ since V is uniserial. Thus we can �nd aw ∈ V such that w 6∈ 〈z〉⊥ and N(w) = z. Since β and γ are both alternating, wehave γ(w,w) = β(w,w) +λ ·β(Nw,w) = λ ·β(z, w) = 0. Since w 6∈ 〈z〉⊥, we haveβ(z, w) 6= 0 and so λ = 0. Therefore γ = β, and the lemma follows.

The following lemmas are useful in some cases for classifying subgroups ofclassical groups up to conjugacy. Below for a bilinear form β on a vector space V ,we denote by I (V, β) the group consisting of all g ∈ GL(V ) such that β(gv, gw) =β(v, w) for all v, w ∈ V (equivalently, βg = β, see De�nition 4.4.1).

Lemma 4.4.14. Let ρ : G→ GL(V ) and ψ : G→ GL(V ) be two representationsof G. Set ρ(G) = X and ψ(G) = Y . Suppose that there exists a non-degeneratebilinear form β on V such that X,Y < I (V, β). Assume that β is alternating orsymmetric, and assume that p 6= 2 if β is symmetric. Suppose that up to scalarmultiples, the form β is the unique non-degenerate X-invariant alternating or

100 Chapter 4.

symmetric bilinear form on V . Then if ρ and ψ are equivalent, the subgroups Xand Y are conjugate in I (V, β).

Proof. Suppose that ρ and ψ are equivalent. Then there exists a f ∈ GL(V ) suchthat fY f−1 = X. Since the form β is Y -invariant, the form βf is invariant underfY f−1 = X. By uniqueness of β, it follows that βf = λβ for some non-zeroscalar λ ∈ K. Now βg = β for g =

√λf . Hence g ∈ I (V, β), and it is clear that

gY g−1 = X.

Lemma 4.4.15. Assume that p 6= 2. Let ρ : G → GL(V ) and ψ : G → GL(V )be two equivalent representations of G such that the associated G-module V isindecomposable. Let X = ρ(G) and Y = ψ(G). Suppose that X,Y < I (V, β),where β is a non-degenerate alternating or symmetric bilinear form. Then X andY are conjugate in I (V, β).

Proof. Since the G-modules associated with ρ and ψ are isomorphic, this followsfrom the general result [Wil76, Satz 3.12] (alternatively, [QSSS76, Korollar 3.5]).

4.5 Weight multiplicities in VG(λ) and LG(λ)

The purpose of this subsection is to list some well known results about weightmultiplicities in Weyl modules and irreducible modules of G.

Let (−,−) be the usual inner product de�ned on X(T ) ⊗Z R, normalized sothat long roots in Φ have norm 1. For λ, µ ∈ X(T )+, we will de�ne

d(λ, µ) = (λ+ ρ, λ+ ρ)− (µ+ ρ, µ+ ρ)

where ρ is the half-sum of positive roots. The following useful result is a conse-quence of the strong linkage principle.

Proposition 4.5.1 (Linkage principle). Assume that p > 2, and that p > 3 ifG is of type G2. Let λ, µ ∈ X(T )+. If LG(µ) occurs as a composition factor ofVG(λ), then p divides the integer 2 · d(λ, µ).

Proof. See [Sei87, 6.2].

In Table 4.1 and Table 4.2, we have given the value of d(λ, µ) for some particu-lar λ � µ. In the tables we have also included the weight multiplicity mVG(λ)(µ) ofµ in VG(λ), which can be computed with Freudenthal's formula [Hum72, Theorem22.3]. This information will be used later in Section 5.13. The following two lem-mas will also be useful. In the �rst lemma, recall that we are using the Bourbakilabeling of the Dynkin diagrams, so in root systems of rank 2 the simple shortroot is α1.

Lemma 4.5.2. Suppose that G has rank 2. Let λ = c1ω1 + c2ω2 ∈ X(T )+, wherec1c2 6= 0. For µ = λ − α1 − α2, we have 0 < mLG(λ)(µ) ≤ 2. Furthermore,mLG(λ)(µ) = 1 if and only if one of the following holds:

(i) G has type A2 and p | 1 + c1 + c2,

(ii) G has type C2 and p | 2 + c1 + 2c2,

4.5. Weight multiplicities in VG(λ) and LG(λ) 101

(iii) G has type G2 and p | 3 + c1 + 3c2.

Proof. This is [Tes88, 1.35].

Lemma 4.5.3. Suppose that G is of type Al and λ = c1ω1 + clωl. Let µ =λ− α1 − α2 − · · · − αl. Then mVG(λ)(µ) = l, and

mLG(λ)(µ) =

{l − 1 if p | c1 + cl + l − 1,

l if p - c1 + cl + l − 1.

Proof. This a special case of [Sei87, 8.6].

The following basic lemma will be useful occasionally. We omit the proof whichis easy.

Lemma 4.5.4. Let λ, µ ∈ X(T )+. Suppose that for all λ � µ′ � µ, the Weylmodule VG(λ) does not have LG(µ′) as a composition factor. Then mLG(λ)(µ) =mVG(λ)(µ).

We know that if λ ∈ X(T )+, then the set of weights in VG(λ) is saturated.That is, if µ is a weight in VG(λ), then µ − iα is also a weight for all α ∈ Φ andi between 0 and 〈µ, α〉. With some mild assumptions on the characteristic, thesame is true for LG(λ) as well if λ is p-restricted.

Theorem 4.5.5 (Premet [Pre87]). Suppose that G is simple and (G, p) is not oneof (Bn, 2), (Cn, 2), (F4, 2) or (G2, 3). Then for any p-restricted λ ∈ X(T )+, theset of weights occurring in VG(λ) and the set of weights occurring in LG(λ) areequal. In particular, the set of weights in LG(λ) is saturated.

For the rest of this section, we will describe how some weight multiplicitiesmLG(λ)(µ) can be found by computing within an irreducible representation ofsome Levi factor of G. To explain this well known technique, we have to establishsome notation.

For α ∈ Φ, let Uα be the root subgroup associated with α. Fix some base ∆of the root system Φ of G. Now for any subset J ⊂ ∆, denote ΦJ = ZJ ∩Φ. Thenthe standard parabolic subgroup PJ associated with J is de�ned as PJ = LJQJ ,where

QJ = 〈Uα : α ∈ Φ+ \ ΦJ〉

andLJ = 〈T,Uα : α ∈ ΦJ〉.

Here QJ is the unipotent radical of PJ , and PJ is a semidirect product LJ nQJof algebraic groups (Levi decomposition).

Let L′J = [LJ , LJ ] be the commutator subgroup of LJ . Then L′J is a connectedsemisimple algebraic group with maximal torus TJ = T ∩ L′J . For a weight λ ∈X(T )+, we will denote the restriction of λ to TJ by λ′. Then Φ′J = {α′ : α ∈ ΦJ}is a root system of L′J with base ∆′ = {α′ : α ∈ J}. With this base ∆′, thefundamental dominant weights are ω′i for αi ∈ J .

Now the relation between weight multiplicities in LG(λ) and in LL′J (λ′) is given

by the corollary of the next proposition. Below we write LG(λ)QJ for the �xedpoint subspace of QJ on LG(λ).

102 Chapter 4.

Proposition 4.5.6 (Smith [Smi82]). Let J ⊆ ∆ and λ =∑l

i=1 ciωi ∈ X(T )+.Let PJ be the parabolic subgroup determined by J with Levi decomposition PJ =LJ nQJ . Then

LG(λ)QJ = 〈Vµ : µ = λ−∑αi∈J

kiαi, where ki ∈ Z≥0〉

andLG(λ)QJ ↓ L′J ∼= LL′J (λ′)

where λ′ =∑

αi∈J ciω′i.

Corollary 4.5.7. Let J ⊆ ∆ and λ =∑

αi∈J ciωi ∈ X(T )+. Let PJ be theparabolic subgroup determined by J with Levi decomposition PJ = LJ nQJ . ThenmLG(λ)(µ) = mLL′

J(λ′)(µ

′) for all µ = λ−∑

αi∈J kiαi, where ki ∈ Z≥0.

b µ ∈ X(T )+ d(λ, µ) mV(λ)(µ)

b ≥ 1 λ− α1 − α2 b+ 1 1b ≥ 3 λ− α1 − 2α2 2b 1b ≥ 2 λ− 2α1 − 2α2 2(3b+ 2) 2

Table 4.1: Type G2: some weight multiplicities mV(λ)(µ) for λ = bω2.

b µ ∈ X(T )+ d(λ, µ) mV(λ)(µ)

b ≥ 1 λ− α1 − α2 b+ 1 1b ≥ 3 λ− α1 − 2α2 2b 1b ≥ 5 λ− α1 − 3α2 3(b− 1) 1b ≥ 2 λ− 2α1 − 2α2 2(b+ 1) 2b ≥ 4 λ− 2α1 − 3α2 2(b− 4) 2b ≥ 7 λ− α1 − 4α2 6(b− 3) 2

Table 4.2: Type C2: some weight multiplicities mV(λ)(µ) for λ = bω2.

.

4.6 Representations of SL2(K)

In this section, let G be the algebraic group SL2(K) with natural module E. Fixalso a nonidentity unipotent element u ∈ G. Throughout we will identify theweights of a maximal torus of G with Z, and the dominant weights with Z≥0.

The main purpose of this section is to classify indecomposable G-modules Vwhere u acts on V with at most one Jordan block of size p. One consequence ofthis is a criterion for a representation of G to be semisimple. Speci�cally, we provethat a self-dual G-module V must be semisimple if u acts on V with at most oneJordan block of size p (Proposition 4.6.10). In Section 5.13, this result will be thestarting point for classifying irreducible representations of simple algebraic groupsin which a unipotent element of order p acts as a distinguished unipotent element.

We begin by considering Jordan block sizes in Weyl modules and tilting mo-dules of G.

4.6. Representations of SL2(K) 103

Lemma 4.6.1. Let m = qp+ r, where q ≥ 0 and 0 ≤ r < p. Then u acts on theWeyl module VG(m) with Jordan blocks [p, p, . . . , p, r+ 1], where p occurs q times.

Proof. The Weyl module VG(m) is isomorphic to Sm(E)∗, see for example [Jan03,II.2.16]. Therefore it su�ces to compute the Jordan blocks of u acting on thesymmetric power Sm(E).

Now there exists a basis x, y of E such that ux = x and uy = x + y. Letxm, xm−1y, . . . , ym be the basis induced on Sm(E). Then

u(xm−kyk) = xm−k(x+ y)k =k∑i=0

(k

i

)xm−iyi,

so with respect to this basis, the matrix of u acting on Sm(E) is the upper tri-angular Pascal matrix P = (

(i−1j−1

))0≤i,j≤m. Here we de�ne

(ij

)= 0 if i < j. The

Jordan form of the transpose of P is computed for example in [Cal02], and fromthis result we �nd that P has q Jordan blocks of size p and one Jordan block ofsize r + 1, as claimed.

Lemma 4.6.2. Let c ≥ p− 1. Then TG(c) ↓ K[u] = N · Vp, where N = dimTG(c)p .

In particular, TG(c) ↓ K[u] = Vp ⊕ Vp if p ≤ c ≤ 2p− 2.

Proof. If c = p − 1, then the claim follows from Theorem 4.1.5 (i) and Lemma4.2.2. If p ≤ c ≤ 2p − 2, the claim follows from [Sei00, Lemma 2.3], see [McN02,Proposition 5].

Consider then the case where c > 2p− 2. By Theorem 4.1.5 (iii), we have

TG(c) ∼= TG(p− 1 + r)⊗ TG(s)[1],

where s ≥ 1 and 0 ≤ r ≤ p − 1 are such that c = sp + (p − 1 + r). Sincep − 1 ≤ p − 1 + r ≤ 2p − 2, by the previous paragraph TG(s)[1] ↓ K[u] = N ′ · Vpfor N ′ = dimTG(s)

p . Therefore TG(c) ↓ K[u] ∼= N · Vp for N = dimTG(c)p by Lemma

3.3.6.

For studying representations of G which are not semisimple the starting pointis the following result, originally due to Cline [AJL83, Corollary 3.9].

Theorem 4.6.3 (Cline). Let λ =∑

i≥0 λipi and µ =

∑i≥0 µip

i be weights ofG, where 0 ≤ λi, µi < p. Then Ext1

G(LG(λ), LG(µ)) 6= 0 precisely when thereexists a k ≥ νp(λ + 1) such that µi = λi for all i 6= k, k + 1 and µk = p −2 − λk, µk+1 = λk+1 ± 1. In the case where Ext1

G(LG(λ), LG(µ)) 6= 0, we haveExt1

G(LG(λ), LG(µ)) ∼= K.

Note that if there exists a nonsplit extension of LG(λ) by LG(µ), it is uniqueup to isomorphism because Ext1

G(LG(λ), LG(µ)) ∼= K by Theorem 4.6.3.We will use the following result of Serre to construct some indecomposable

representations of G. The result is not speci�c to algebraic groups, and actuallyholds for any group H and �nite-dimensional representations V and W of H overK.

Theorem 4.6.4 ([Ser97]). Let V and W be G-modules. If V ⊗W is semisimple,then V is semisimple if p - dimW .

104 Chapter 4.

Proposition 4.6.5. Let V be a G-module which is a nonsplit extension of LG(λ)by LG(µ). Then u acts on V with at least one Jordan block of size p. If u acts onV with precisely one Jordan block of size p, then there exists p ≤ c ≤ 2p − 2 andl ≥ 0 such that one of the following holds:

(i) λ = cpl, µ = (2p− 2− c)pl and V ∼= VG(c)[l].

(ii) λ = (2p− 2− c)pl, µ = cpl and V ∼= (VG(c)∗)[l].

Moreover, if V , λ, µ, and c are as in case (i) or case (ii), then V is a nonsplitextension of LG(λ) by LG(µ), and u acts on V with Jordan blocks [p, c− p+ 1].

Proof. We begin by considering the claims about VG(c)[l] and (VG(c)∗)[l], wherep ≤ c ≤ 2p − 2 and l ≥ 0. It is well known (and easily seen by considering theweights in VG(c)) that VG(c) is a nonsplit extension

0→ LG(2p− 2− c)→ VG(c)→ LG(c)→ 0.

Since the irreducible representations of G are self-dual, we see that VG(c)∗ is anonsplit extension

0→ LG(c)→ VG(c)∗ → LG(2p− 2− c)→ 0.

Let λ = cpl and µ = (2p − 2 − c)pl, where l ≥ 0. By taking a Frobenius twist,we see that for all l ≥ 0 the module VG(c)[l] is a non-split extension of LG(λ)by LG(µ), and its dual (VG(c)∗)[l] is a non-split extension of LG(µ) by LG(λ).Furthermore, by Lemma 4.6.1 the unipotent element u ∈ G acts on VG(c) withJordan blocks [p, c − p + 1]. The Jordan block sizes are not changed by takinga dual or a Frobenius twist, so for all l ≥ 0 the element u acts on both VG(c)[l]

and (VG(c)∗)[l] with Jordan blocks [p, c−p+ 1]. This completes the proof that theproperties of VG(c)[l] and (VG(c)∗)[l] are as claimed.

Now let λ, µ ∈ Z≥0 be arbitrary weights, and let V be a nonsplit extension

0→ LG(µ)→ V → LG(λ)→ 0.

Assume that u acts on V with at most one Jordan block of size p. We note �rst thatto prove the proposition, it will be enough to show that there exist p ≤ c ≤ 2p− 2and l ≥ 0 such that λ and µ are as in case (i) or (ii) of the claim. Indeed, ifthis holds, then since Ext1

G(LG(λ), LG(µ)) ∼= K (Theorem 4.6.3), we must haveV ∼= VG(c)[l] or V ∼= (VG(c)∗)[l]. Furthermore, as seen in the �rst paragraph, thenu acts on V with Jordan blocks [p, c−p+1], in particular with exactly one Jordanblock of size p.

We proceed to show that λ and µ are as claimed. Write λ =∑

i≥0 λipi, where

0 ≤ λi ≤ p− 1 for all i. Recall that by the Steinberg tensor product theorem, wehave

LG(λ) ∼=⊗i≥0

LG(λi)[i]

Consider �rst the case where λl = p − 1 for some l. Then u acts on LG(λl)[l]

with a single Jordan block of size p. Now the tensor product of a Jordan blockof size p with any Jordan block of size c ≤ p consists of c Jordan blocks of size p(Lemma 3.3.6), so it follows that LG(λ) ↓ K[u] = N · Vp, where N = dimLG(λ)

p .

Since u acts on V with at most one Jordan block of size p, by Lemma 3.1.4 theaction of u on LG(λ) can have at most one Jordan block of size p. It follows thenthat N = 1, so λ = (p − 1)pl. Since V is a nonsplit extension, by Theorem 4.6.3the weight µ must be (p − 2)pl−1 + (p − 2)pl or (p − 1)pl + (p − 2)pk + pk+1

for some k 6= l, l − 1. By Lemma 3.1.4 the action of u on LG(µ) has no Jordanblocks of size p. With Lemma 3.3.6, one �nds that this happens only if p = 2 andµ = (p− 2)pl−1 + (p− 2)pl = 0. Then λ and µ are as in case (i) of the claim.

Thus we can assume that 0 ≤ λi ≤ p− 2 for all i. Write µ =∑

i≥0 µipi, where

0 ≤ µi ≤ p − 1 for all i. According to Theorem 4.6.3, there exists an l such thatwe have µi = λi for i 6= l, l + 1, and µl = p − 2 − λl, µl+1 = λl+1 ± 1. We canwrite λ = ζ+λlp

l +λl+1pl+1 and µ = ζ+µlp

l +µl+1pl+1. By the Steinberg tensor

product theorem, we have

LG(λ) ∼= LG(ζ)⊗ LG(λl + λl+1p)[l]

LG(µ) ∼= LG(ζ)⊗ LG(µl + µl+1p)[l]

Note that here p does not divide the dimension of LG(ζ), because we areassuming that λi < p − 1 for all i. Furthermore, by Theorem 4.6.3 there exists aG-module W which is a nonsplit extension

0→ LG(λl + λl+1p)→W → LG(µl + µl+1p)→ 0.

Therefore by Theorem 4.6.4, the module LG(ζ) ⊗W [l] is a nonsplit extension ofLG(λ) by LG(µ). Hence LG(ζ)⊗W [l] ∼= V , because a nonsplit extension of LG(λ)by LG(µ) is unique by Theorem 4.6.3.

Consider �rst the case where ζ = 0. Here V ∼= W [l], so without loss of generalitywe may assume that l = 0. Write λ = c+ dp and µ = (p− 2− c) + (d± 1)p, where0 ≤ c, d ≤ p− 2 and d > 0 if µ = (p− 2− c) + (d− 1)p.

We consider λ = c + dp and µ = (p − 2 − c) + (d − 1)p. Here dimV =dimLG(λ) + dimLG(µ) = λ + 1 = dimVG(λ). Now V is a nonsplit extension ofLG(λ) by LG(µ) and λ � µ, so by Proposition 4.1.1 we must have V ∼= VG(λ). ByLemma 4.6.1, the action of u on V has Jordan blocks [p, p, . . . , p, c + 1], where poccurs d times. Therefore u acts on V with at most one Jordan block of size p ifand only if d = 1, that is, when λ = c+ p and µ = (p− 2− c). Then λ and µ areas in case (i) of the claim.

Next consider the case where λ = c + dp and µ = (p − 2 − c) + (d + 1)p.In this case dimV = dimLG(λ) + dimLG(µ) = µ + 1 = dimVG(µ). BecauseV ∗ is a nonsplit extension of LG(µ) by LG(λ) and µ � λ, by Proposition 4.1.1we have V ∗ ∼= VG(µ). By Lemma 4.6.1, the action of u on V has Jordan blocks[p, p, . . . , p, p − (c + 1)], where p occurs d + 1 times. In this case u acts with atmost one Jordan block of size p if and only if d = 0, that is, when λ = c andµ = (p− 2− c) + p. Then λ and µ are as in case (ii) of the claim.

Finally we consider the possibility that ζ 6= 0. Since W is a nonsplit extensionof two irreducible modules, it follows from the ζ = 0 case that u acts on W withat least one Jordan block of size p. Now the tensor product of a Jordan block ofsize p with any Jordan block of size c ≤ p consists of c Jordan blocks of size p(Lemma 3.3.6). Therefore if ζ 6= 0, then u acts on V with more than one Jordanblock of size p, contradiction.

Lemma 4.6.6. Let p ≤ c ≤ 2p− 2. Then for all l ≥ 0:

106 Chapter 4.

(i) Ext1G(VG(c)[l], LG(c)[l]) = 0.

(ii) Ext1G(VG(c)[l], LG(2p− 2− c)[l]) = 0.

(iii) Ext1G(LG(c)[l], VG(c)[l]) = 0.

(iv) Ext1G(LG(2p− 2− c)[l], VG(c)[l]) ∼= K.

Furthermore, every nonsplit extension of LG(2p − 2 − c)[l] by VG(c)[l] is iso-morphic to TG(c)[l].

Proof. Set c′ = 2p−2−c. We note �rst that the last claim of the lemma follows from(iv), once we show that TG(c)[l] is a nonsplit extension of LG(c′)[l] by VG(c)[l]. Tothis end, by [Sei00, Lemma 2.3 (b)] the tilting module TG(c) is a nonsplit extensionof LG(c′) by VG(c). The extension stays nonsplit after taking a Frobenius twist,so TG(c)[l] is a nonsplit extension of LG(c′)[l] by VG(c)[l].

For claims (i) - (iv), we prove them �rst in the case where l = 0. In this case,claims (i) and (ii) follow from the fact Ext1

G(VG(λ), LG(µ)) = 0 for any µ ≤ λ[Jan03, II.2.14]. For (iii) and (iv), recall that there is an exact sequence

0→ LG(c′)→ VG(c)→ LG(c)→ 0.

Applying the functor HomG(LG(d),−) gives a long exact sequence

0→ HomG(LG(d), LG(c′))→ HomG(LG(d), VG(c))→ HomG(LG(d), LG(c))

→ Ext1G(LG(d), LG(c′))→ Ext1

G(LG(d), VG(c))→ Ext1G(LG(d), LG(c))→ · · ·

Considering this long exact sequence with d = c, we get an exact sequence

0→ K → K → Ext1G(LG(c), VG(c))→ 0

since Ext1G(LG(c), LG(c′)) ∼= K and Ext1

G(LG(c), LG(c)) = 0 by Theorem 4.6.3and [Jan03, II.2.12], respectively. This proves (iii).

With d = 2p− 2− c, we get an exact sequence

0→ Ext1G(LG(c′), VG(c))→ K.

Therefore dim Ext1G(LG(c′), VG(c)) ≤ 1. Thus to prove (iv), it will be enough to

show that there exists some nonsplit extension of LG(c′) by VG(c). For this, wehave already noted in the beginning of the proof that TG(c) is such an extension.

We consider then claims (i) - (iv) for l > 0. If p 6= 2, then the claims followfrom the case l = 0, since Ext1

G(X [l], Y [l]) ∼= Ext1G(X,Y ) for all G-modules X and

Y [Jan03, II.10.17]. Suppose then that p = 2. Note that then we must have c = 2,and c′ = 0. By [Jan03, Proposition II.10.16, Remark 12.2], for any G-modules Xand Y there exists a short exact sequence

0→ Ext1G(X,Y )→ Ext1

G(X [l], Y [l])

→ HomG(X,LG(1)⊗ Y )→ 0.

Thus claims (i) - (iv) will follow from the case l = 0 once we show thatHomG(X,LG(1)⊗ Y ) = 0 in the following cases:

(i)′ X = VG(2) and Y = LG(2),

(ii)′ X = VG(2) and Y = LG(0),

(iii)′ X = LG(2) and Y = VG(2),

(iv)′ X = LG(0) and Y = VG(2).

In all cases (i)′ - (iv)′, it is straightforward to see that X and LG(1)⊗Y have nocomposition factors in common. Thus HomG(X,LG(1)⊗ Y ) = 0, as claimed.

Lemma 4.6.7. Let W be a nonsplit extension of two irreducible G-modules, andlet Z be an irreducible G-module. If V is a nonsplit extension of Z by W , then Vis indecomposable.10

Proof. We can assume that W ⊆ V and V/W ∼= Z. Let W ′ ⊆ W be the uniqueirreducible submodule of W . If V is not indecomposable, then there exists anirreducible submodule Z ′ ⊆ V such that Z ′ 6= W ′. Then Z ′ ∩W = 0, so V =W ⊕Z ′. Therefore V ∼= W ⊕Z, which is impossible when V is a nonsplit extensionof Z by W .

Proposition 4.6.8. Let V be an indecomposable G-module. Then one of the fol-lowing holds:

(i) u acts on V with ≥ 2 Jordan blocks of size p.

(ii) V is irreducible.

(iii) V is isomorphic to a Frobenius twist of VG(c) or VG(c)∗, where p ≤ c ≤ 2p−2.Furthermore, u acts on V with Jordan blocks [p, c− p+ 1].

Proof. Let V be a counterexample of minimal dimension to the claim. Then V isnot irreducible and u acts on V with ≤ 1 Jordan block of size p. Note also that byLemma 3.1.4, the element u acts on any subquotient of V with ≤ 1 Jordan blockof size p. Therefore any proper subquotient of V must be as in (ii) or (iii) of theclaim.

Since V is not irreducible, there exists a subquotient Q of V which is a nonsplitextension of two irreducible G-modules. By Proposition 4.6.5, the subquotient Qis isomorphic to VG(c)[l] or (VG(c)∗)[l] for some l ≥ 0 and p ≤ c ≤ 2p− 2.

We are assuming that V is a counterexample, so there must be a subquotientQ′ of V which is a nonsplit extension of Q and some irreducible module Z. ByLemma 4.6.7, the subquotient Q′ is indecomposable, and so by the minimality ofV we actually have Q′ = V . By replacing V with V ∗ if necessary, this reduces usto the situation where V is a nonsplit extension of VG(c)[l] and some irreduciblemodule LG(d).

There must be a subquotient of V which is a nonsplit extension of LG(d)with LG(c)[l] or LG(2p − 2 − c)[l]. Therefore by Proposition 4.6.5, it follows thatLG(d) = LG(c)[l] or LG(d) = LG(2p − 2 − c)[l], respectively. Thus V ∼= TG(c)[l]

by Lemma 4.6.6. This gives us a contradiction, because u acts on TG(c) with twoJordan blocks of size p by Lemma 4.6.2.

Corollary 4.6.9. Let V be any representation of G. Suppose that u acts on Vwith all Jordan blocks of size < p. Then V is semisimple.

10Of course, the statement of the lemma and the proof given here work in a much more generalsetting.

108 Chapter 4.

Proof. By Proposition 4.6.8, the only indecomposable G-modules on which u actswith all Jordan blocks of size < p are the irreducible ones.

Proposition 4.6.10. Let V be a self-dual representation of G. Suppose that uacts on V with at most 1 Jordan block of size p. Then V is semisimple.

Proof. Write V = W1 ⊕ · · · ⊕Wt, where Wi are indecomposable G-modules. Sup-pose that Wi 6∼= W ∗i for some i. Now irreducible G-modules are self-dual, so Wi

is not irreducible and thus u acts on Wi with exactly 1 Jordan block of size p(propositions 4.6.5 and 4.6.8). On the other hand, V ∼= V ∗ ∼= W ∗1 ⊕ · · · ⊕W ∗t , sowe have Wj

∼= W ∗i for some i 6= j since the indecomposable summands are uniqueby the Krull-Schmidt theorem. But then Wi ⊕W ∗i is a summand of V , which is acontradiction since u acts on Wi ⊕W ∗i with two Jordan blocks of size p.

Thus Wi∼= W ∗i for all i, and so by Proposition 4.6.8 each Wi must be irredu-

cible, since VG(c) 6∼= VG(c)∗ for p ≤ c ≤ 2p− 2.

Chapter 5

Irreducible overgroups ofdistinguished unipotent elements

The purpose of this chapter is to prove Theorem 1.1.10 and Theorem 1.1.11. Thatis, we classify all pairs (G, u), where u ∈ G is a unipotent element and G is asimple subgroup of I (V ) = SL(V ), I (V ) = Sp(V ), or I (V ) = SO(V ) such that

• V is irreducible and tensor-indecomposable as a G-module,

• u is a distinguished unipotent element of I (V ).

When u has order > p, the basic idea of the proof is to �rst rule out mostof the irreducible representations of G with an elementary argument describedin Section 5.1-5.2, and then deal with the remaining representations case-by-case(Section 5.3-5.12). In the case of unipotent elements u order p (Section 5.13), we�rst reduce to the case where p is good for G. Then u is contained in a subgroupA < G of type A1 such that the labeled diagram of A is that of u (Theorem 2.6.8).With Proposition 4.6.10, one can then use the representation theory of SL2(K)and methods from [LST15] to complete the proof.

5.1 Reduction, unipotent elements of order > p (p 6= 2)

Assume that p 6= 2.

In this section, we reduce the proof of Theorem 1.1.10 in the case where uhas order > p to a small number of λ to consider. The reduction is based on thefollowing lemma, Premet's theorem (Theorem 4.5.5), and results of Lübeck givenin [Lüb01] and [Lüb17].

Lemma 5.1.1. Let ϕ : G → GL(V ) a representation of G. For a unipotentelement u ∈ G, let du be the size of the largest Jordan block of ϕ(u). Then

(i) If u ∈ G is unipotent with order pα, then du ≤ pα.

(ii) If u′ ∈ G is a regular unipotent element, then du ≤ du′ for all unipotentelements u ∈ G.

(iii) If V = L(λ) is irreducible and u ∈ G is a unipotent element, then du ≤mu(λ) + 1 (De�nition 2.7.2).

109

110 Chapter 5.

(iv) If a unipotent element u ∈ G acts on V as a distinguished unipotent element,

then dimV ≤ (du+1)2

4 if du is odd and dimV ≤ du(du+2)4 if du is even. In

any case if u acts on V as a distinguished unipotent element, then dimV ≤(du+1)2

4 .

Proof. Claims (i) and (ii) are given in Lemma 2.7.1, and claim (iii) is Theorem2.7.8.

For (iv), we know that if u acts on V as a distinguished unipotent element,then Jordan block sizes of ϕ(u) are distinct and all odd, or distinct and all even.Therefore for d = du, we have

dimV ≤ d+ (d− 2) + · · · =

{d(d+2)

4 if d is even,(d+1)2

4 if d is odd.

so the claim follows since d(d+2)4 < (d+1)2

4 .

Remark 5.1.2. Note that Lemma 5.1.1 (iv) holds under the slightly weaker as-sumption that all Jordan block sizes of u acting on V are distinct and either alleven or all odd.

Remark 5.1.3. This section is concerned with elements of order > p, but notethat Lemma 5.1.1 also makes sense when u has order p. We will occasionally applyit in this situation as well. However, when u has order > p and G is of classicaltype, we can get an upper bound on the order of u which only depends on rankGand not on p. This will be a key fact that we will use in subsections 5.1.1, 5.1.2and 5.1.3.

We give the reduction for each simple type in the subsections that follow.

5.1.1 Type Al

In this subsection, assume that G is simple of type Al. The only distinguishedunipotent class in G is the regular unipotent class, which acts with a single Jordanblock of size l+1 on the natural module of G (Proposition 2.3.4). Therefore it willbe enough to consider actions of some regular unipotent element u ∈ G (Lemma1.1.8). In this section we are only considering distinguished unipotent elements oforder > p. Hence for the rest of this subsection, we make the following assumption:

Assume that p ≤ l.

Note that our assumption implies that l ≥ 2. Our reduction for type Al isgiven by the results which follow.

Proposition 5.1.4. Assume that 2 ≤ l ≤ 14. Let u ∈ G be a regular unipotentelement and let λ ∈ X(T )+ be a nonzero and p-restricted. If u acts on L(λ) as adistinguished unipotent element, then one of the following holds.

(i) λ = ω1 or λ = ωl.

(ii) λ = ω1 + ωl.

(iii) l is odd and λ = ω l+12.

5.1. Reduction, unipotent elements of order > p (p 6= 2) 111

(iv) p = 3, l = 3 and λ = 2ω2.

Proof. If L(λ) is not self-dual, then we are done by Proposition 1.1.12. Assumethen that L(λ) is self-dual and suppose that u acts on L(λ) as a distinguishedunipotent element.

The regular unipotent element u acts on the natural module of G with a singleJordan block of size l+ 1 (Proposition 2.3.4). Therefore the order of u is equal toMl,p := ps+1, where s ≥ 1 is such that ps < l + 1 ≤ ps+1. Therefore

dimL(λ) ≤(Ml,p + 1)2

4

by Lemma 5.1.1. For all 2 ≤ l ≤ 14 and p ≤ l, checking self-dual irreducible

representations of dimension ≤ (Ml,p+1)2

4 in the tables given in [Lüb17], we end upwith representations given by (ii), (iii) and (iv) in our claim.

Lemma 5.1.5. If p = 3 and l = 3, then a regular unipotent element u ∈ G doesnot act as a distinguished unipotent element on L(2ω2).

Proof. A computer calculation with MAGMA (Section 2.9) shows that in this caseu acts on L(2ω2) with Jordan blocks [52, 9]; hence not as a distinguished unipotentelement11.

Lemma 5.1.6. Assume that l ≥ 15. Let µ =∑l

i=1 aiωi ∈ X(T )+. If one of the

following statements hold, then |Wµ| > (l2+1)2

4 .

(i) ai 6= 0, where l = 2i− 1.

(ii) aial+1−i 6= 0 for some 3 ≤ i ≤ b l2c.

(iii) a1a2al−1al 6= 0.

Proof. Write f(l) = (l2+1)2

4 . In what follows the W -orbit sizes are computed asdescribed in Section 4.3.

(i): Suppose that ai 6= 0 for l = 2i − 1. Now |Wµ| ≥ |W (ωi)| =(l+1i

)=(

2ii

),

so it will be enough to show that(

2ii

)> f(l) = f(2i − 1) for all l ≥ 15,

equivalently for all i ≥ 8.

Now for the central binomial coe�cient(

2ii

), we have the well known es-

timate(

2ii

)≥ 2i. This estimate can be seen by induction on i, using the

identity(

2ii

)= 2 · (2− 1

i )(

2(i−1)i−1

). One can verify that 2i > f(l) = f(2i− 1)

for all i ≥ 19, so(

2ii

)> f(l) = f(2i − 1) if i ≥ 19. For 8 ≤ i ≤ 18, the

inequality(

2ii

)> f(2i− 1) is seen by a calculation.

11For a computer-free proof, one can proceed as follows. It is possible to show that for thenatural module V of G, we have S2(∧2(V )) = L(0)/L(2ω2)/L(0) (uniserial). Furthermore, it iseasy to compute with Theorem 3.4.5 that for p = 3, we have S2(∧2(V4)) = [12, 52, 9]. Thus theaction of u on S2(∧2(V )) is inadmissible (De�nition 3.2.4), so it follows then from Lemma 3.2.6that u does not act as a distinguished unipotent element on L(2ω2).

112 Chapter 5.

(ii): Suppose that aial+1−i 6= 0, where 3 ≤ i ≤ b l2c. Now |Wµ| ≥ |W (ωi +

ωl+1−i)| =(l+12i

)(2ii

). For i = 3 and i = 4, one can verify that the inequality(

l+12i

)(2ii

)> f(l) holds for all l ≥ 9.

Consider then i ≥ 5. Note that |Wµ| ≥ |W (ωi)| =(l+1i

). Since 5 ≤ i ≤ b l2c,

we have(l+1i

)≥(l+15

). One can verify that the inequality

(l+15

)> f(l) holds

for all l ≥ 35, so it follows that |Wµ| > f(l) for all l ≥ 35. Finally for15 ≤ l ≤ 34, the inequality

(l+12i

)(2ii

)> f(l) can be veri�ed by calculation.

(iii): In this case |Wµ| ≥ |W (ω1 + ω2 + ωl−1 + ωl)| = (l+ 1)l(l− 1)(l− 2). Thusthe claim follows from the single variable polynomial inequality (l+ 1)l(l−1)(l − 2) > f(l), which holds for all l ≥ 4.

Proposition 5.1.7. Assume that l ≥ 15. Let u ∈ G be a regular unipotent elementand let λ ∈ X(T )+ be nonzero and p-restricted. If u acts on L(λ) as a distinguishedunipotent element, then one of the following holds.

(i) λ = ω1 or λ = ωl.

(ii) λ = ω1 + ωl.

Proof. If L(λ) is not self-dual, then we are done by Proposition 1.1.12. Assumethen that L(λ) is self-dual (that is, λ = −w0λ) and suppose that u acts on L(λ)as a distinguished unipotent element.

The regular unipotent element u acts on the natural module of G with a singleJordan block of size l+ 1 (Proposition 2.3.4). Therefore the order of u is equal tops+1, where s ≥ 1 is such that ps < l + 1 ≤ ps+1. Now ps+1 = p · ps ≤ l2 sinceps ≤ l, and thus

dimL(λ) ≤ (l2 + 1)2

4(*)

by Lemma 5.1.1. Furthermore, we also have

dimL(λ) ≤ (mu(λ) + 2)2

4(**)

by Lemma 5.1.1.Write λ =

∑li=1 aiωi, where 0 ≤ ai ≤ p−1 for all i. Since λ = −w0λ, by Table

4.1 we have ai = al+1−i for all i. We now consider various possibilities for thecoe�cients ai to rule out everything except (i) and (ii) in the claim. In all cases,this will be done by showing that there exists a weight µ =

∑li=1 a

′iωi ∈ X(T )+

such that µ � λ and |Wµ| > (l2+1)2

4 or |Wµ| > (mu(λ)+2)2

4 . Then by Premet'stheorem (Theorem 4.5.5) we have dimLG(λ) ≥ |Wµ|, so this will contradict (*)or (**). In what follows some speci�c W -orbit sizes of weights are computed using4.3.

Case (1): l odd and ai 6= 0 for i = l+12 :

In this case we can choose λ = µ, since then |Wµ| > (l2+1)2

4 by Lemma5.1.6 (i).


Case (2): ai 6= 0, where 3 ≤ i ≤ b l2c:Here we can also choose λ = µ, since ai = al+1−i and thus |Wµ| >(l2+1)2

4 by Lemma 5.1.6 (ii).

Case (3): a1a2 6= 0:

Now a1 = al and a2 = al−1, so similarly to cases (1) and (2), this followsfrom Lemma 5.1.6 (iii).

By the arguments given for cases (1)-(3), we are left to consider λ = bω1 + bωland λ = bω2 + bωl−1 for 1 ≤ b ≤ p − 1. We consider the various possibilities toconclude that λ = ω1 + ωl, which completes the proof.

Case (4): λ = bω2 + bωl−1, where b ≥ 2:

In this case, for µ = λ − α2 − αl−1 we have µ ∈ X(T )+ such that

a′3a′l−2 6= 0, so |Wµ| > (l2+1)2

4 by Lemma 5.1.6 (ii).

Case (5): λ = ω2 + ωl−1:

Here mu(λ) = 4l − 4 by Lemma 2.7.3. For µ = λ, one can verify that

the polynomial inequality |Wµ| = (l+1)!2!2!(l−3)! >

(mu(λ)+2)2

4 holds for alll ≥ 5.

Case (6): λ = bω1 + bωl, where b ≥ 3:

In this case, for µ = λ − α1 − αl we have µ ∈ X(T )+ such that

a′1a′2a′l−1a

′l 6= 0, so |Wµ| > (l2+1)2

4 by Lemma 5.1.6 (iii).

Case (7): λ = 2ω1 + 2ωl.

Here mu(λ) = 4l by Lemma 2.7.3. Now for µ = λ−α1−αl = ω2 +ωl−1

one can easily verify that the polynomial inequality |Wµ| = (l+1)!2!2!(l−3)! >

(mu(λ)+2)2

4 holds for all l ≥ 6.

We summarize our reduction for type A in the following proposition, whichis an immediate consequence of Proposition 5.1.4, Lemma 5.1.5 and Proposition5.1.7.

Proposition 5.1.8. Let u ∈ G be a regular unipotent element and let λ ∈ X(T )+

be nonzero and p-restricted. Assume that u has order > p. If u acts on LG(λ) asa distinguished unipotent element, then one of the following holds.

(i) λ = ω1 or λ = ωl.

(ii) λ = ω1 + ωl.

(iii) l is odd, 3 ≤ l ≤ 13, and λ = ω l+12.

Note that in case (i) of Proposition 5.1.8 a regular unipotent element acts asa distinguished unipotent element. Cases (ii) and (iii) of Proposition 5.1.8 will bedealt with in sections 5.3 and 5.12, respectively.

114 Chapter 5.

5.1.2 Types Bl and Cl

Let G be a simple algebraic group of type Bl or Cl. We know that G has unipotentelements of order > p if and only if a regular unipotent element u ∈ G has order> p. For type Bl and type Cl, a regular unipotent element has order > p if andonly if p < 2l + 1 and p < 2l respectively. Now p is odd, so this is equivalent top ≤ 2l − 1. Therefore we will make the following assumption for the rest of thissubsection.

Assume that p ≤ 2l − 1.

Note that a regular unipotent of G has order ps+1, where s ≥ 1 is such thatps < 2l + 1 ≤ ps+1. In particular, we have ps+1 = p · ps ≤ 2l · 2l = 4l2.

Thus if some unipotent element u ∈ G acts on a representation V as a distin-guished unipotent element, then

dimV ≤ 4l2(4l2 + 2)

4= 4l4 + 2l2 (*)

by Lemma 5.1.1. If V = LG(λ) and u ∈ G is a regular unipotent element, thenany unipotent element of G acts on V with largest block of size ≤ mu(λ) + 1 byLemma 5.1.1. Thus if some unipotent element of G acts on V as a distinguishedunipotent element, then

dimV ≤ (mu(λ) + 2)2

4(**)

by Lemma 5.1.1. Similarly to the proof of Proposition 5.1.7, our reduction is basedon applying the bound (*) on V = LG(λ) when λ =

∑li=1 aiωi with mu(λ) large,

and then applying (**) when mu(λ) is small.

Lemma 5.1.9. Assume that 3 ≤ l ≤ 9 and G = Bl. Let λ ∈ X(T )+ be nonzerop-restricted. If some unipotent element of G acts on LG(λ) as a distinguishedunipotent element, then one of the following holds.

(i) λ = ωi for some 1 ≤ i ≤ l.

(ii) λ = 2ω1.

(iii) λ = 3ω1.

(iv) p = 5, l = 3, and λ = 2ω3, λ = ω1 + ω3 or λ = ω2 + ω3.

Lemma 5.1.10. Assume that 2 ≤ l ≤ 9 and G = Cl. Let λ ∈ X(T )+ be nonzerop-restricted. If some unipotent element of G acts on LG(λ) as a distinguishedunipotent element, then one of the following holds.


(ii) λ = 2ω1.

(iii) λ = 3ω1.

(iv) p = 3, l = 2, and λ = 2ω2, λ = ω1 + ω2, or λ = 2ω1 + ω2.

(v) p = 5, l = 3, and λ = ω2 + ω3 or λ = 2ω3.


Proof of Lemma 5.1.9 and Lemma 5.1.10. Suppose that some unipotent elementof G acts on LG(λ) as a distinguished unipotent element. Let u ∈ G be a regularunipotent element. Then it follows from Lemma 5.1.1 that dimLG(λ) ≤M , where

M = (|u|+1)2

4 . For 2 ≤ l ≤ 9, we list the largest possible value of |u| and M inTable 5.1 (recall the assumption p ≤ 2l − 1).

Now the irreducible representations LG(λ) such that dimLG(λ) ≤ M arefound in [Lüb17]. In tables 5.2 - 5.5, we list all λ such that λ 6= 2ω1, 3ω1, ωi anddimLG(λ) ≤M . We have also included the minimal possible value of dimLG(λ).

One can verify that for most λ occurring in the tables, we have dimLG(λ) >(mu(λ)+2)2

4 , which contradicts (**). The cases in the tables for which dimLG(λ) ≤(mu(λ)+2)2

4 is a possibility are:

Case (1): G = C2, with λ ∈ {2ω2, ω1 + ω2, 2ω1 + ω2}.

Case (2): G = C3, with λ ∈ {ω1 + ω2, ω1 + ω3, ω2 + ω3, 2ω3}.

Case (3): G = B3, with λ ∈ {2ω3, ω1 + ω3, ω1 + ω2, ω2 + ω3}.

What remains is to verify the claim of the two lemmas for these cases. ForG = C2 (case (1)), the assumption p ≤ 2l − 1 implies p = 3, so case (1) is case(iv) of the claim.

ForG = C3 (case (2)), we have p = 3 or p = 5. If p = 3, then u has order 32, and

dimLG(λ) > 25 = (32+1)2

4 for all λ occurring in Table 5.2. Thus by Lemma 5.1.1,no unipotent element of G acts on LG(λ) as a distinguished unipotent elementwhen p = 3. To verify the claim of Lemma 5.1.10 for p = 5, we should show thatno unipotent element of G acts on LG(λ) as a distinguished unipotent elementwhen λ ∈ {ω1 + ω2, ω1 + ω3}. Using [Lüb01], one can verify for p = 5 that

dimLG(ω1 + ω2) = 64 and dimLG(ω1 + ω3) = 70, so dimLG(λ) > (mu(λ)+2)2

4 forλ ∈ {ω1 + ω2, ω1 + ω3}. Thus the claim follows from Lemma 5.1.1.

Finally we have to consider G = B3 (case (3)). Now p ≤ 2l − 1 implies that

p = 3 or p = 5. If p = 3, then u has order 32, and dimLG(λ) > 25 = (32+1)2

4for all λ occurring in Table 5.4. Thus by Lemma 5.1.1, no unipotent element ofG acts on LG(λ) as a distinguished unipotent element when p = 3, as desired.What remains is to show that for p = 5, no unipotent element of G acts as adistinguished unipotent element on LG(ω1 + ω2). For this the claim follows againfrom Lemma 5.1.1: we have dimLG(ω1 + ω2) = 105 when p > 3 by [Lüb01], so

dimLG(ω1 + ω2) > (mu(λ)+2)2

4 when p = 5.

Lemma 5.1.11. In the following cases no unipotent element u ∈ G of order > pacts on LG(λ) as a distinguished unipotent element:

(i) G = C2, p = 3, and λ = 2ω1 + ω2.

(ii) G = C3, p = 5, and λ ∈ {ω2 + ω3, 2ω3}.

(iii) G = B3, p = 5, and λ = ω2 + ω3.

116 Chapter 5.

G Largest possible value of |u| Largest possible value of MB2, C2 32 25B3, C3 52 169B4, C4 72 625B5, C5 72 625B6, C6 112 3721B7, C7 132 7225B8, C8 132 7225B9, C9 172 21025

Table 5.1: ForG = Bl andG = Cl (2 ≤ l ≤ 9), largest possible order |u| of a regularunipotent element u of order > p, and largest possible value of M = (|u|+1)2

4 .

Proof. In all of these cases, the only unipotent elements in G with order > p arethe regular ones. For a regular unipotent element u ∈ G, we have computed thedecompositions LG(λ) ↓ K[u] in cases (i) - (iii) with a computer program imple-mented in MAGMA (Section 2.9)12. The claim follows from these decompositions,which are given in Table 5.6.


i=1 aiωi ∈ X(T )+. If one of thefollowing statements hold, then |Wµ| > 4l4 + 2l2.

(i) ai 6= 0 for some i ≥ 19.

(ii) aiaj 6= 0 for some 1 ≤ i < j ≤ l − 1 such that j ≥ 5.

(iii) aial 6= 0 for some 2 ≤ i ≤ l − 2.

Proof. Write f(l) = 4l4 + 2l2. In what follows the W -orbit sizes are computedusing 4.3.

(i): Now |Wµ| ≥ |W (ωi)| = 2i(li

), so it will be enough to show that 2i

(li

)> f(l)

for all l ≥ i ≥ 19. We show �rst that this inequality holds for i = l,that is, we show that 2l > f(l) for all l ≥ 19. We do this by inductionon l. For l = 19 this is a calculation, and for l > 19 by induction we get2l = 2 · 2l−1 > 2 · f(l − 1) ≥ f(l), where the last inequality is an easilyveri�ed single variable polynomial inequality.

We now proceed to prove that 2i(li

)> f(l) for all l ≥ i ≥ 19 by induction

on i. For i = 19, one can verify the inequality by calculation. Consider theni > 19. The case i = l was dealt with in the previous paragraph, so supposethat i < l. In this case

2i(l

i

)= 2i

((l − 1

i

)+

(l − 1

i− 1

))> 2i

(l − 1

i− 1

)> 2 · f(l − 1)

by induction. Furthermore, 2 · f(l − 1) ≥ f(l) as mentioned in the previousparagraph, so 2i

(li

)> f(l), as claimed.

12One could also give a computer-free proof for many of the entries in Table 5.6, similarly toFootnote 11.

G λ Minimal possible value of dimLG(λ) mu(λ)

C2 2ω2 14 8ω1 + ω2 16 72ω1 + ω2 25 10

C3 ω1 + ω2 50 13ω1 + ω3 57 14ω2 + ω3 62 17

2ω3 63 182ω2 90 164ω1 126 20

C4 ω1 + ω2 112 19ω1 + ω4 240 23

2ω2 266 24ω1 + ω3 279 22ω3 + ω4 312 31

2ω4 313 324ω1 330 28

ω2 + ω3 504 27ω2 + ω4 513 282ω1 + ω2 558 26

C5 ω1 + ω2 210 252ω2 615 32

C6 ω1 + ω2 352 312ω2 1221 404ω1 1365 44

ω1 + ω3 1924 382ω1 + ω2 2847 42ω2 + ω3 3432 47ω1 + ω5 3638 46ω1 + ω6 3652 47

C7 ω1 + ω2 546 372ω2 2184 484ω1 2380 52

ω1 + ω3 3795 462ω1 + ω2 5355 50ω2 + ω3 7098 57

Table 5.2: For G of type Cl, 2 ≤ l ≤ 7 and 2 < p < 2l: irreducible p-restrictedrepresentations LG(λ) of dimension ≤ M , where M is as in Table 5.1 and λ 6=ωi, 2ω1, 3ω1.

118 Chapter 5.


C8 ω1 + ω2 800 432ω2 3620 564ω1 3876 60

ω1 + ω3 6749 54

C9 ω1 + ω2 1122 492ω2 5661 644ω1 5985 68

ω1 + ω3 11154 622ω1 + ω2 14193 66

Table 5.3: For G of type Cl, 8 ≤ l ≤ 9 and 2 < p < 2l: irreducible p-restrictedrepresentations LG(λ) of dimension ≤ M , where M is as in Table 5.1 and λ 6=ωi, 2ω1, 3ω1.


B3 2ω3 35 12ω1 + ω3 48 12ω1 + ω2 63 16ω2 + ω3 64 16

3ω3 104 182ω1 + ω3 120 18

2ω2 132 20ω1 + 2ω3 168 18

B4 ω1 + ω4 112 182ω4 126 20

ω1 + ω2 147 22ω2 + ω4 304 24ω3 + ω4 336 28

2ω2 369 284ω1 450 323ω4 544 30

ω1 + ω3 558 262ω1 + ω4 576 26ω2 + ω3 579 32

B5 ω1 + ω2 264 28ω1 + ω5 320 25

2ω5 462 30

Table 5.4: For G of type Bl, 3 ≤ l ≤ 5 and 2 < p < 2l + 1: irreducible p-restricted representations LG(λ) of dimension ≤ M , where M is as in Table 5.1and λ 6= ωi, 2ω1, 3ω1.

B6 ω1 + ω2 416 34ω1 + ω6 768 33

2ω2 1559 442ω6 1716 424ω1 1728 48

ω1 + ω3 2847 42ω2 + ω6 3392 43

B7 ω1 + ω2 650 40ω1 + ω7 1664 42

2ω2 2715 524ω1 2940 56

ω1 + ω3 5250 502ω7 6435 56

2ω1 + ω2 6798 54

B8 ω1 + ω2 935 46ω1 + ω8 4096 52

2ω2 4251 604ω1 4540 64

B9 ω1 + ω2 1273 522ω2 6763 684ω1 7124 72

ω1 + ω9 9216 63ω1 + ω3 14193 662ω1 + ω2 17424 70

Table 5.5: For G of type Bl, 6 ≤ l ≤ 9 and 2 < p < 2l + 1: irreducible p-restricted representations LG(λ) of dimension ≤ M , where M is as in Table 5.1and λ 6= ωi, 2ω1, 3ω1.

G λ LG(λ) ↓ K[u]

C2 (p = 3) 2ω2 [5, 9]ω1 + ω2 [2, 6, 8]2ω1 + ω2 [7, 92]

C3 (p = 5) ω2 + ω3 [102, 12, 152]2ω3 [102, 13, 152]

B3 (p = 5) 2ω3 [1, 5, 7, 9, 13]ω1 + ω3 [3, 5, 7, 9, 11, 13]ω2 + ω3 [5, 7, 102, 15, 17]

Table 5.6: For G = C2, G = C3, and G = B3; actions of a regular unipotentelement u ∈ G on some small irreducible representations LG(λ).

120 Chapter 5.

(ii): Now |Wµ| ≥ |W (ωj)|, so the claim follows from (i) if j ≥ 19. Consider then1 ≤ i < j ≤ l − 1 such that 5 ≤ j ≤ 18. Since |Wµ| ≥ |W (ωi + ωj)| =

2j(lj

)(ji

), it will be enough to verify that 2j

(lj

)(ji

)> f(l) for all l ≥ 10. For

all 1 ≤ i < j ≤ l − 1 with 5 ≤ j ≤ 18 (�nitely many cases to check), thispolynomial inequality is straightforward to verify by a calculation.

(iii): As in (ii), the claim follows from (i) if i ≥ 19. Consider then 2 ≤ i ≤ 18,where i ≤ l−2. Similarly to (ii), we have now have �nitely many i to check,so a calculation shows that |Wµ| ≥ |W (ωi +ωl)| = 2l

(li

)> f(l) holds for all

l ≥ 10.

Lemma 5.1.13. Assume that l ≥ 10. Let u ∈ G be a distinguished unipotentelement and let λ ∈ X(T )+ be nonzero and p-restricted. If u acts on L(λ) as adistinguished unipotent element, then λ = 2ω1 or λ = ωi for some 1 ≤ i ≤ l.


i=1 aiωi, where 0 ≤ ai ≤ p − 1. Set n = 2l + 1 if G has typeBl and n = 2l if G has type Cl. Suppose that some unipotent element u of G actson LG(λ) as a distinguished unipotent element.

Our strategy is to rule out the various possibilities for the coe�cients ai caseby case and eventually conclude that either λ = ωi or λ = 2ω1 (cf. Proposition5.1.7). Most cases will be ruled out by showing that there exists a weight µ =∑l

i=1 a′iωi ∈ X(T )+ such that µ � λ and |Wµ| > 4l4 + 2l2 or |Wµ| > (mu(λ)+2)2

4 .Then by Premet's theorem (Theorem 4.5.5) we have dimLG(λ) ≥ |Wµ|, so thiswill contradict (*) or (**). Throughout we apply 4.3 to compute the W -orbit sizes|Wµ|.

Case (1): aial 6= 0 for 1 ≤ i ≤ l − 1:

In this case for i = l−1, we set µ = λ−αl−1−αl so we have µ ∈ X(T )+

with a′l−2a′l 6= 0 if G has type Bl and a′l−2a

′l−1 6= 0 if G has type Cl. If

2 ≤ i ≤ l − 2, we set µ = λ. If i = 1 and a1 ≥ 2, we set µ = λ − α1

and then we have µ ∈ X(T )+ with a′2a′l 6= 0. If al ≥ 2, then we set

µ = λ−αl and we have µ ∈ X(T )+ with a′1a′l−1 6= 0. Now in these cases

we have |Wµ| > 4l4 + 2l2 by Lemma 5.1.12 (ii) or (iii).

What remains is the case where a1 = 1 and al = 1. By the previouscases, we can assume that λ = ω1 + ωl. If G is of type Cl, then λ � µfor µ = ω1 +ωl−2 ∈ X(T )+, and |Wµ| > 4l4 +2l2 by Lemma 5.1.12 (ii).If G is of type Bl, then mu(λ) = 2l+ l(l+1)

2 . It is easy to verify that for

µ = λ, we have |Wµ| = 2ll > (mu(λ)+2)2

4 for all l ≥ 6.

Case (2): aiaj 6= 0 for 1 ≤ i < j ≤ l − 1 with j ≥ 5:

Here |Wλ| > 4l4 + 2l2 by Lemma 5.1.12 (ii), so we can pick µ = λ.

Case (3): aia4 6= 0 for 1 ≤ i ≤ 3:

In this case if i = 3, we set µ = λ−α3−α4 and we have µ ∈ X(T )+ witha′2a′5 6= 0. If i = 2, we set µ = λ−α2−α3−α4 and we have µ ∈ X(T )+

with a′1a′5 6= 0. If a1 ≥ 2 or a4 ≥ 2, then we set µ = λ−α1−α2−α3−α4

and we have µ ∈ X(T )+ with a′1a′5 6= 0 or a′4a

′5 6= 0. Thus in these cases

|Wµ| > 4l4 + 2l2 by Lemma 5.1.12 (ii).


What remains is the case where a1 = 1, a4 = 1. By the previous casestreated, we can assume that λ = ω1 + ω4. In this case we �nd a weight

µ ∈ X(T )+ with λ � µ and |Wµ| > (mu(λ)+2)2

4 in Table 5.7.

Case (4): a2a3 6= 0:

In this case if a1 ≥ 1, a2 ≥ 2, or a3 ≥ 2, then for µ = λ − α1 − 2α2 −2α3 − α4 we have µ ∈ X(T )+ and a′1a

′5 6= 0, a′2a

′5 6= 0, or a′3a

′5 6= 0,

respectively. Thus if a1 ≥ 1, a2 ≥ 2, or a3 ≥ 2, then by Lemma 5.1.12(ii) we have |Wµ| > 4l4 + 2l2.

What remains is the possibility that a1 = 0, a2 = 1, and a3 = 1. By theprevious cases, we can assume that λ = ω2 + ω3. In this case we �nd a

weight µ ∈ X(T )+ with λ � µ and |Wµ| > (mu(λ)+2)2

4 in Table 5.7.

Case (5): a1a3 6= 0:

In this case if a1 ≥ 3, then for µ = λ − 2α1 − 2α2 − 2α3 − α4 we haveµ ∈ X(T )+ and a′1a

′5 6= 0. If a3 ≥ 2, then for µ = λ − α2 − 2α3 − α4

we have µ ∈ X(T )+ and a′1a′5 6= 0. Thus if a1 ≥ 3 or a3 ≥ 2, then by

Lemma 5.1.12 (ii) we have |Wµ| > 4l4 + 2l2.

Consider then a1 ≤ 2 and a3 = 1. By the previous cases, we can assumethat λ = a1ω1 +ω3. In this case we �nd a weight µ ∈ X(T )+ with λ � µand |Wµ| > (mu(λ)+2)2

4 in Table 5.7.

Case (6): a1a2 6= 0:

If a1 ≥ 4, then for µ = λ − 3α1 − 3α2 − 2α3 − α4 we have µ ∈ X(T )+

with a′1a′5 6= 0. If a2 ≥ 3, then for µ = λ− 2α1 − 4α2 − 3α3 − 2α4 − α5

we have µ ∈ X(T )+ with a′1a′6 6= 0. Thus if a1 ≥ 4 or a2 ≥ 3, then by

Lemma 5.1.12 (ii) we have |Wµ| > 4l4 + 2l2.

Consider then a1 ≤ 3 and a2 ≤ 2. By the previous cases, we can assumethat λ = a1ω1 + a2ω2. In this case we �nd a weight µ ∈ X(T )+ with

λ � µ and |Wµ| > (mu(λ)+2)2

4 in Table 5.7.

By the arguments given for cases (1)-(6), we are left to consider λ = bωi with1 ≤ b ≤ p− 1. We consider the various possibilities to conclude that if b > 1, thenb = 2 and i = 1, which completes the proof.

Case (7): λ = bωl, where b ≥ 2:

In this case if b ≥ 3, then for µ = λ − 2αl − αl−1 we have µ ∈ X(T )+.Furthermore, we have a′l−2a

′l 6= 0 if G has type Bl and a′l−2a

′l−1 6= 0 if

G has type Cl. Thus |Wµ| > 4l4 + 2l2 by Lemma 5.1.12 (ii) or (iii).

Consider then b = 2, so now λ = 2ωl. If G has type Cl, then λ � µ forµ = ωl−2 + ωl ∈ X(T )+, and |Wµ| > 4l4 + 2l2 by Lemma 5.1.12 (iii).Suppose then that G has type Bl. In this situation we need to considermore than one orbit of weights. First of all, we have mu(λ) = 2 l(l+1)

2 =l(l + 1). Furthermore, it is easy to see that λ � 0 and λ � ωi for all1 ≤ i ≤ l − 1. Now |Wωi| = 2i

(li

), so in this situation dimLG(2ωl) ≥∑l

i=0 2i(li

)= 3l. One can verify that dimLG(2ωl) ≥ 3l > (mu(λ)+2)2

4 forall l ≥ 6, which contradicts (**).

122 Chapter 5.

Case (8): λ = bωi, where 1 ≤ i ≤ l − 1 and b ≥ 2:

In this case if 4 ≤ i ≤ l − 1, we set µ = λ − αi and we have µ ∈X(T )+ with a′i−1a

′i+1 6= 0. If i = 3, we set µ = λ − α2 − 2α3 − α4

and we have µ ∈ X(T )+ with a′1a′5 6= 0. If i = 2 and b ≥ 3, we set

µ = λ − α1 − 3α2 − 2α3 − α4 and we have µ ∈ X(T )+ with a′1a′5 6= 0.

If i = 1 and b ≥ 6, then for µ = λ − 4α1 − 3α2 − 2α3 − α4 we haveµ ∈ X(T )+ with a′1a

′5 6= 0. In these cases we have |Wµ| > 4l4 + 2l2.by

Lemma 5.1.12 (ii) or (iii).

What remain are the cases λ = 3ω2 and λ = bω1 with 3 ≤ b ≤ 5. Forλ = 3ω2 and λ = bω1 with 4 ≤ b ≤ 5, we �nd a weight µ ∈ X(T )+ with

λ � µ and |Wµ| > (mu(λ)+2)2

4 in Table 5.7. Finally for λ = 3ω1, we note�rst that mu(λ) = 3n − 3. Furthermore, λ � µ for µ = ω3. Now one

veri�es that dimLG(3ω1) ≥ |Wλ|+ |Wµ| = 2l + 23(l3

)> (mu(λ)+2)2

4 forall l ≥ 10, which contradicts (**).

λ µ ≺ λ λ− µ |Wµ| mu(λ)

ω1 + ω4 ω5 α1 + α2 + α3 + α4 25(l5

)5n− 17

ω2 + ω3 ω5 α1 + 2α2 + 2α3 + α4 25(l5

)5n− 13

2ω1 + ω3 ω5 2α1 + 2α2 + 2α3 + α4 25(l5

)5n− 11

ω1 + ω3 ω4 α1 + α2 + α3 24(l4

)4n− 10

3ω1 + 2ω2 ω1 + ω6 3α1 + 4α2 + 3α3 + 2α4 + α5 26(l6

)· 6 7n− 11

2ω1 + 2ω2 ω1 + ω5 2α1 + 3α2 + 2α3 + α4 25(l5

)· 5 6n− 10

ω1 + 2ω2 ω5 2α1 + 3α2 + 2α3 + α4 25(l5

)5n− 9

3ω1 + ω2 ω5 3α1 + 3α2 + 2α3 + α4 25(l5

)5n− 7

2ω1 + ω2 ω4 2α1 + 2α2 + α3 24(l4

)4n− 6

ω1 + ω2 ω3 α1 + α2 23(l3

)3n− 5

2ω2 ω4 α1 + 2α2 + α3 24(l4

)4n− 8

5ω1 ω5 4α1 + 3α3 + 2α2 + α1 25(l5

)5n− 5

4ω1 ω4 3α1 + 2α2 + α3 24(l4

)4n− 4

Table 5.7: Type Bl (n = 2l+1) and type Cl (n = 2l): For some speci�c λ ∈ X(T )+,

weights µ ≺ λ such that µ ∈ X(T )+ and |Wµ| > (mu(λ)+2)2

4 for all l ≥ 10.

We summarize our reduction for type B and C in the following proposition,which is a corollary of Lemma 5.1.9, Lemma 5.1.10, Lemma 5.1.11, and Lemma5.1.13.

Proposition 5.1.14. Suppose that G is simple of type Bl (l ≥ 3) or type Cl(l ≥ 2). Let λ ∈ X(T )+ be nonzero and p-restricted. If some unipotent element

u ∈ G of order > p acts on LG(λ) as a distinguished unipotent element, then oneof the following holds.


(ii) λ = 2ω1.

(iii) λ = 3ω1.

(iv) G = C2, u is a regular unipotent element, p = 3, and λ ∈ {ω1 + ω2, 2ω2}.

(v) G = B3, u is a regular unipotent element, p = 5, and λ ∈ {2ω3, ω1 + ω3}.

Cases (i), (ii), and (iii) of Proposition 5.1.14 will be dealt with in sections 5.12,5.7, and 5.9 respectively. Note that by Table 5.6 in cases (iv) and (v) the elementu acts on LG(λ) as a distinguished unipotent element, and that we have recordedthese examples in Table 1.1.

5.1.3 Type Dl

Let G be a simple algebraic group of type Dl. In this subsection, we give thereduction for G using the same arguments as in section 5.1.2.

Now G has unipotent elements of order > p if and only if a regular unipotentelement of G has order > p. Furthermore, since G has type Dl, a regular unipotentelement has order > p if and only if p < 2l − 1. Thus we will make the followingassumption for the rest of this subsection.

Assume that p < 2l − 1.

Then a regular unipotent element of G has order ps+1, where s ≥ 1 is suchthat ps < 2l − 1 ≤ ps+1. As in 5.1.2, we see that ps+1 ≤ 4l2. Therefore if someunipotent element of G acts on a representation V as a distinguished unipotentelement, the inequalities (*) and (**) from subsection 5.1.2 must hold.

The following lemma gives our reduction for small l (cf. Lemma 5.1.4, Lemma5.1.9, Lemma 5.1.10).

Lemma 5.1.15. Assume that 4 ≤ l ≤ 9. Let λ ∈ X(T )+ be nonzero p-restricted. Ifsome unipotent element of G acts on LG(λ) as a distinguished unipotent element,then one of the following holds.


(ii) λ = 2ω1.

Proof. Suppose that some unipotent element of G acts on LG(λ) as a distinguishedunipotent element. Let u ∈ G be a regular unipotent element. Then it follows from

Lemma 5.1.1 that dimLG(λ) ≤M , where M = (|u|+1)2

4 . For 4 ≤ l ≤ 9, we list thelargest possible value of |u| andM in Table 5.8 (recall the assumption p < 2l−1).

Now the irreducible p-restricted representations LG(λ) such that dimLG(λ) ≤M are found in the tables of [Lüb01]. In Table 5.9, we list (based on [Lüb01]) allλ such that λ 6= 2ω1, ωi and dimLG(λ) ≤ M . We have also included the value ofmu(λ) in the tables, which is easily calculated with the table in Lemma 2.7.3.

In any case, one can verify that for all λ occurring in Table 5.9, we have

dimLG(λ) > (mu(λ)+2)2

4 , which would contradict (**). This implies the claim.

124 Chapter 5.

G Largest possible value of |u| Largest possible value of MD4 52 169D5 72 625D6 72 625D7 112 3721D8 132 7225D9 132 7225

Table 5.8: For G = Dl (4 ≤ l ≤ 9), largest possible order |u| of a regular unipotentelement u of order > p, and largest possible value of M = (|u|+1)2

4 .


i=1 aiωi ∈ X(T )+. If one of thefollowing statements hold, then |Wµ| > 4l4 + 2l2.

(i) ai 6= 0 for some i ≥ 20.

(ii) aiaj 6= 0 for some 1 ≤ i < j ≤ l − 2 such that j ≥ 5.

(iii) aial 6= 0 or aial−1 6= 0 for some 3 ≤ i ≤ l − 3.

(iv) a1alal−1 6= 0 or a2alal−1 6= 0.

(v) a1a2al−1 6= 0 or a1a2al 6= 0.

Proof. Write f(l) = 4l4 + 2l2. In what follows the W -orbit sizes are computedusing 4.3.

(i): If i ≤ l − 2, then the value of |Wωi| is the same as for type Bl and Cl,so the claim follows Lemma 5.1.12 (ii). For i = l − 1 and i = l, we have|Wµ| ≥ |Wωi| = 2l−1, so it will be enough to show that 2l−1 > f(l) forall l ≥ 20. This follows with the exact same proof as the one given for theinequality 2l > f(l) in Lemma 5.1.12 (ii).

(ii): Now the value of |W (ωi + ωj)| is the same as for type Bl and Cl, so theclaim follows from Lemma 5.1.12 (ii).

(iii): If l ≥ 20, the claim follows from (i). For 10 ≤ l ≤ 19, one can verify bya computer calculation that |Wµ| ≥ |W (ωi + ωl)| = 2l−1

(li

)> f(l) for all

3 ≤ i ≤ l − 3.

(iv), (v): As in (iii), the claim follows from (i) if l ≥ 20, and is easy to verify manuallyfor 10 ≤ l ≤ 19.

Lemma 5.1.17. Assume that l ≥ 10. Let u ∈ G be a distinguished unipotentelement and let λ ∈ X(T )+ be nonzero and p-restricted. If u acts on L(λ) as adistinguished unipotent element, then λ = 2ω1 or λ = ωi for some 1 ≤ i ≤ l.


i=1 aiωi, where 0 ≤ ai ≤ p− 1. Suppose that some unipotentelement u of G acts on LG(λ) as a distinguished unipotent element. We proceedto rule out various possibilities for ai as in the proofs of 5.1.7 and 5.1.13 to showthe claim. That is, in most cases we �nd a contradiction with (*) or (**) by

D4 ω1 + ω4 56 123ω1 104 18

ω1 + ω2 104 16ω1 + 2ω4 168 18

D5 2ω5 126 20ω1 + ω5 128 18ω1 + ω2 190 22

3ω1 210 24ω4 + ω5 210 20ω2 + ω5 544 24

3ω5 544 302ω2 559 28

2ω1 + ω5 576 264ω1 606 32

D6 ω1 + ω6 320 253ω1 340 30

ω1 + ω2 340 282ω6 462 30

D7 ω1 + ω2 532 343ω1 546 36

ω1 + ω7 768 332ω7 1716 422ω2 1975 444ω1 2275 48

ω6 + ω7 3003 42

D8 ω1 + ω2 768 403ω1 800 42

ω1 + ω8 1920 422ω2 3483 524ω1 3605 562ω8 6435 56

ω1 + ω3 6900 50

D9 3ω1 1104 48ω1 + ω2 1104 46ω1 + ω9 4096 52

2ω2 5490 604ω1 5644 64

Table 5.9: For G of type Dl, 4 ≤ l ≤ 9 and 2 < p < 2l − 1: irreducible p-restricted representations LG(λ) of dimension ≤ M , where M is as in Table 5.8and λ 6= 2ω1, ωi.

126 Chapter 5.

�nding a µ =∑l

i=1 a′iωi ∈ X(T )+ such that µ � λ, and |Wµ| > 4l4 + 2l2 or

|Wµ| > (mu(λ)+2)2

4 . Throughout we apply 4.3 to compute the W -orbit sizes |Wµ|.Note that if a′l = 0 and a′l−1 = 0, then the Weyl orbit size |Wµ| is the same

as for the corresponding weight (one with the same coe�cients a′i) for type Bl orCl. From this fact, we will see in several cases that the same arguments as thosegiven in the proof of Lemma 5.1.13 work.

Case (1): aial−1 6= 0 or aial 6= 0 for some 3 ≤ i ≤ l − 3:

Here |Wλ| > 4l4 + 2l2 by Lemma 5.1.16 (iii), so we can pick µ = λ.

Case (2): al−2al−1 6= 0 or al−2al 6= 0:

By applying a graph automorphism on λ, it is enough to consideral−2al−1 6= 0. In this case for µ = λ− αl−2 − αl−1 we have µ ∈ X(T )+

and a′l−3a′l 6= 0, so |Wµ| > 4l4 + 2l2 by Lemma 5.1.16 (iii).

Case (3): al−1al 6= 0:

If a1al−1al 6= 0 or a2al−1al 6= 0, then |Wλ| > 4l4 + 2l2 by Lemma5.1.16 (iv) and we can pick µ = λ. Otherwise by case (1) and (2) wecan assume that λ = al−1ωl−1 + alωl. If al−1 ≥ 2 or al ≥ 2, then forµ = λ − αl−2 − αl−1 − αl we have a′l−1a

′l−3 6= 0 or a′la

′l−3 6= 0, so

|Wµ| > 4l4 + 2l2 by Lemma 5.1.16 (iii).

Consider then al−1 = 1, al = 1, so now λ = ωl−1 + ωl. In this case

mu(λ) = l(l− 1) and one can verify that |Wλ| = 2l−1l > (mu(λ)+2)2

4 forall l ≥ 8.

Case (4): a2al−1 6= 0 or a2al 6= 0:

By applying a graph automorphism on λ, it is enough to considera2al−1 6= 0. If a1a2al−1 6= 0, then |Wλ| > 4l4 + 2l2 by Lemma 5.1.16(v), so we can pick µ = λ. Thus we can assume a1 = 0, and by cases(1), (2) and (3) we can assume λ = a2ω2 + al−1ωl−1.

If al−1 ≥ 2, then for µ = λ− αl−1 we have µ ∈ X(T )+ and a′2a′l−2 6= 0,

so |Wµ| > 4l4 +2l2 by Lemma 5.1.16 (ii). If a2 ≥ 2, then for µ = λ−α2

we have µ ∈ X(T )+ and a′3a′l−1 6= 0, so |Wµ| > 4l4 + 2l2 by Lemma

5.1.16 (iii).

Consider then a2 = 1, al−1 = 1, so now λ = ω2 + ωl−1. In this casemu(λ) = 2(2l−3)+ l(l−1)

2 and one can verify that |Wλ| = 2l−2l(l−1) >(mu(λ)+2)2

4 for all l ≥ 6.

Case (5): a1al−1 6= 0 or a1al 6= 0:

By applying a graph automorphism on λ, it is enough to considera1al−1 6= 0. By cases (1), (2), (3) and (4) we may assume that λ =a1ω1+al−1ωl−1. If a1 ≥ 3, then for µ = λ−2α1−α2 we have µ ∈ X(T )+

and a′3a′l−1 6= 0, so |Wµ| > 4l4 + 2l2 by Lemma 5.1.16 (iii). If al−1 ≥ 2,

then for µ = λ − αl−1 we have µ ∈ X(T )+ and a′1a′l−2 6= 0, so

|Wµ| > 4l4 + 2l2 by Lemma 5.1.16 (ii).

Consider then a1 ≤ 2 and al−1 = 1, so now λ = ω1 + ωl−1 or λ =

2ω1 +ωl−1. In this situation mu(λ) ≤ mu(2ω1 +ωl−1) = 2(2l−2)+ l(l−1)2

and one veri�es that |Wλ| = 2l−1l >(mu(2ω1+ωl−1)+2)2

4 ≥ (mu(λ)+2)2

4 forall l ≥ 8.

Cases (6)-(10): aiaj 6= 0 for some 1 ≤ i < j ≤ l − 2 with j ≥ 5; aia4 6= 0 for 1 ≤ i ≤ 3;a2a3 6= 0; a1a3 6= 0; or a1a2 6= 0:

In these cases, the exact same arguments as those given in cases (2)-(6)in the proof of Lemma 5.1.13 work.

By the arguments given for cases (1)-(10), we are left to consider λ = bωi with1 ≤ b ≤ p− 1. We consider the various possibilities to conclude that if b > 1, thenb = 2 and i = 1, which completes the proof.

Case (11): λ = bωl−1 or λ = bωl, where b ≥ 2:

By applying a graph automorphism on λ, it is enough to consider λ =bωl−1. If b ≥ 3, then for µ = λ− 2αl−1 − αl−2 we have µ ∈ X(T )+ anda′l−3a

′l 6= 0, so |Wµ| > 4l4 + 2l2 by Lemma 5.1.16 (iii).

Consider then b = 2, so λ = 2ωl−1. If l is even, then λ � µ for allµ = ωi ∈ X(T )+ with 1 ≤ i ≤ l − 2 even. Furthermore, a standardbinomial identity gives∑

1≤i≤l−2i even

|Wωi| =∑

1≤i≤l−2i even

2i(l

i

)=

3l + 1

2− 2l.

Similarly, if l is odd, then λ � µ for all µ = ωi ∈ X(T )+ with 1 ≤ i ≤l − 2 odd. Again a standard binomial identity gives∑

1≤i≤l−2i odd

|Wωi| =∑

1≤i≤l−2i odd

2i(l

i

)=

3l + 1

2− 2l.

Therefore it follows that dimLG(λ) ≥ 3l+12 − 2l. Now mu(λ) = l(l − 1)

and one veri�es that 3l+12 − 2l > (mu(λ)+2)2

4 for all l ≥ 6.

Case (12): λ = bωi, where 1 ≤ i ≤ l − 2 and b ≥ 2:

In this case, the exact same arguments as those given in case (8) in theproof of Lemma 5.1.13 work.

We summarize our reduction for type D in the following proposition, which isimmediate from Lemma 5.1.17 and Lemma 5.1.15.

Proposition 5.1.18. Let G = Dl, where l ≥ 4. Let λ ∈ X(T )+ be nonzero p-restricted. If some unipotent element u ∈ G of order > p acts on LG(λ) as adistinguished unipotent element, then one of the following holds.


(ii) λ = 2ω1.

Cases (i) and (ii) of Proposition 5.1.18 will be dealt with in sections 5.12 and5.7, respectively.

128 Chapter 5.

5.1.4 Type G2

We will prove the following proposition, which gives the claim of Theorem 1.1.10for G = G2 in the case where u has order > p (see Table 1.2).

Proposition 5.1.19. Let G = G2 and λ ∈ X(T )+ be nonzero p-restricted. Aunipotent element u ∈ G of order > p acts on LG(λ) as a distinguished unipotentelement if and only if u is regular and one of the following holds:

(i) λ = ω1 or λ = ω2.

(ii) p = 5 and λ = 2ω1.

(iii) p = 5 and λ = ω1 + 2ω2.

Proof. In G = G2, there are unipotent elements of order > p if and only if p ≤ 5(see Appendix A). Furthermore, when 2 p. Therefore it will be enough to only consider regular unipotentelements of G.

Let u ∈ G be a regular unipotent element and let λ ∈ X(T )+ be p-restricted.If p = 3, then u has order 32. Thus if u acts on LG(λ) as a distinguished unipotent

element, then dimLG(λ) ≤ (32+1)2

4 = 25 by Lemma 5.1.1. By [Lüb01], this impliesthat λ = ω1 or λ = ω2. In both cases u acts on LG(λ) with a single Jordan blockof size 7 (Proposition 1.1.12).

Suppose then that p = 5. Now u has order 52, so if u acts on LG(λ) as a

distinguished unipotent element, then dimLG(λ) ≤ (52+1)2

4 = 169 by Lemma5.1.1. It follows from the results in [Lüb01] that λ occurs in Table 5.10, where wehave also given the decomposition of LG(λ) ↓ K[u]. This data was obtained bya calculation with MAGMA (Section 2.9). We see that u acts as a distinguishedunipotent element precisely in the cases λ = ω1, λ = ω2, λ = 2ω1, and λ =ω1 + 2ω2.

λ dimLG(λ) LG(λ) ↓ K[u]

ω1 7 [7]ω2 14 [3,11]2ω1 27 [5,9,13]

ω1 + ω2 64 [5, 7, 102, 15, 17]3ω1 77 [52, 102, 13, 15, 19]2ω2 77 [1, 5, 102, 152, 21]

ω1 + 2ω2 97 [13,15,21,23,25]

Table 5.10: Action of a regular unipotent element u of G = G2 on some smallirreducible representations for p = 5.

5.1.5 Types F4, E6, E7, and E8

Suppose that G is simple of type F4, E6, E7, or E8. In this situation our reductionis essentially an application of Lemma 5.1.1 and results due to Lübeck, cf. Lemma5.1.9, Lemma 5.1.10, and Lemma 5.1.15.

Lemma 5.1.20. Suppose that G is simple of type F4, E6, E7, or E8. Let λ ∈X(T )+ be p-restricted. If some unipotent element of G with order > p acts onLG(λ) as a distinguished unipotent element, then λ = ωi.

Proof. Let u ∈ G be a regular unipotent element. If some unipotent element u ∈ Gacts on LG(λ) as a distinguished unipotent element, then by Lemma 5.1.1 we have

dimLG(λ) ≤ M , where M = (|u|+1)2

4 . We give the maximal possible value of |u|and M (when u has order > p) for G of type F4, E6, E7, and E8 in Table 5.11.

We have listed the λ 6= ωi such that dimLG(λ) ≤ M in tables 5.12 - 5.15,based on [Lüb17]. We have also included the value of mu(λ), which was computedusing Lemma 2.7.3. One can verify that

dimLG(λ) >(mu(λ) + 2)2

4

for all λ in tables 5.12 - 5.15, so by Lemma 5.1.1 for these λ no unipotent elementof G acts on LG(λ) as a distinguished unipotent element, as desired.

G Largest possible value of |u| Largest possible value of MF4 112 3721E6 112 3721E7 172 21025E8 292 177241

Table 5.11: ForG = F4 andG = El, largest possible order |u| of a regular unipotentelement u of order > p, and largest possible value of M = (|u|+1)2

4 .

λ Minimal possible value of dimLG(λ) mu(λ)

2ω4 298 322ω1 755 44

ω1 + ω4 1053 38ω3 + ω4 2404 46

3ω4 2651 48

Table 5.12: Type F4, irreducible representations LG(λ) of dimension ≤ 3721, withλ 6= ωi and 2 < p ≤ 11.


2ω1 324 32ω1 + ω6 572 32ω1 + ω2 1377 38ω1 + ω3 2404 46

2ω2 2430 443ω1 3002 48

Table 5.13: Type E6, irreducible representations LG(λ) of dimension ≤ 3721 upto graph automorphism, with λ 6= ωi and 2 < p ≤ 11.

130 Chapter 5.


2ω7 1330 54ω1 + ω7 5568 61

2ω1 5832 683ω7 18752 81

Table 5.14: Type E7, irreducible representations LG(λ) of dimension ≤ 21025,with λ 6= ωi and 2 < p ≤ 17.


2ω8 23125 116

Table 5.15: Type E8, irreducible representations LG(λ) of dimension ≤ 177241,with λ 6= ωi and 2 < p ≤ 29.

5.2 Reduction (p = 2)

Assume that p = 2.

In this section, we reduce the proof of Theorem 1.1.11 to a small number of λto consider. The reduction is based on the following elementary observation andthe results of Lübeck [Lüb01] on small-dimensional irreducible representations ofG.

Lemma 5.2.1. Let u ∈ G be a unipotent element and ϕ : G → GL(V ) a repre-sentation of G. Let d be the size of the largest Jordan block of ϕ(u). If u acts onV as a distinguished unipotent element, then dimV ≤ d(d+2)

2 .

Proof. We know that if u acts on V as a distinguished unipotent element, theneither ϕ(u) is a single Jordan block, or all Jordan block sizes of ϕ(u) are evenwith multiplicity ≤ 2 (Lemma 2.4.4 (ii)). If ϕ(u) is a single Jordan block, thendimV = d ≤ d(d+2)

2 . If all Jordan blocks of ϕ(u) are even with multiplicity ≤ 2,then

dimV ≤ d+ d+ (d− 2) + (d− 2) + · · ·+ 2 + 2 =d(d+ 2)

2

as desired.

We give the reduction for each simple type in the subsections that follow.

5.2.1 Type Al

In this subsection, assume that G is simple of type Al, l ≥ 1.

Proposition 5.2.2. Let u ∈ G be regular and let λ ∈ X(T )+ be nonzero 2-restricted. If u acts on L(λ) as a distinguished unipotent element, then one of thefollowing holds.

(i) λ = ω1 or λ = ωl.

(ii) l ≥ 2 and λ = ω1 + ωl.

(iii) l = 5 and λ = ω3.

5.2. Reduction (p = 2) 131

(iv) l = 3 and λ = ω2.

Proof. If l = 1, there is nothing to do, so we will assume that l ≥ 2. By Proposition1.1.12 we can assume that L(λ) is self-dual, that is, λ = −w0λ. Suppose that uacts on L(λ) as a distinguished unipotent element.

Now u acts on the natural module of G with a single Jordan block of sizel + 1. Therefore the order of u is equal to Ml := 2s+1, where 2s < l + 1 ≤ 2s+1.Then in any representation of G, the action of u has largest Jordan block of size≤ 2s+1 = 2 · 2s ≤ 2l. Thus by Lemma 5.2.1 we have dimL(λ) ≤ Ml(Ml+2)

2 ≤2l(2l+2)

2 = 2l2 + 2l.

Now 2l2 + 2l < l3

8 if l ≥ 17, so by [Lüb01, Theorem 5.1] we have λ = ω1 + ωlif l ≥ 17. In the cases where 2 ≤ l ≤ 16, checking irreducible representations ofdimension ≤ Ml(Ml+2)

2 in the tables given in [Lüb01], we �nd that the only self-dual ones besides L(ω1 + ωl) are L(ω3) in the case l = 5, and L(ω2) in the casel = 3.

Lemma 5.2.3. Let u ∈ G be regular. Then

(i) For l = 5, we have the orthogonal decomposition LG(ω3) ↓ K[u] = V (2)2 +W (8) (Proposition 2.4.4) with respect to any non-degenerate G-invariantalternating bilinear form on LG(ω3). In particular, the element u does notact as a distinguished unipotent element on LG(ω3).

(ii) For l = 3, we have the orthogonal decomposition LG(ω2) ↓ K[u] = V (2) +V (4) (Proposition 2.4.4) with respect to any non-degenerate G-invariant al-ternating bilinear form on LG(ω2). In particular, the element u acts as adistinguished unipotent element on LG(ω2).

Proof. Both (i) and (ii) can be veri�ed by a computation with MAGMA (Section2.9). Claim (ii) also follows from the fact that the image of the representation ρ :SL4(K)→ GL(LG(ω2)) is equal to SO4(K), and that a regular unipotent elementof SO4(K) has decomposition V (2) + V (4) on the natural module (Proposition2.4.4 (vi)).

Therefore for type Al, the claim of Theorem 1.1.11 is reduced to the caseλ = ω1 + ωl, which will be considered in Section 5.4.

5.2.2 Types Bl, Cl and Dl

In this subsection, assume that G is of type Bl (l ≥ 2), type Cl (l ≥ 2), or of typeDl (l ≥ 4).

Proposition 5.2.4. Let u ∈ G be distinguished and let λ ∈ X(T )+ be 2-restricted.If u acts on L(λ) as a distinguished unipotent element, then one of the followingholds.

(i) λ = ω1 or λ = ω2.

(ii) G = Bl or G = Cl, and λ = ωl where 3 ≤ l ≤ 7, or l = 9.

(iii) G = Dl and λ = ωl or λ = ωl−1, where l = 4, l = 6, l = 8, or l = 10.

(iv) l = 5 and λ = ω3.

132 Chapter 5.

Proof. Now u acts on the natural representation of G with largest Jordan block ofsize ≤ 2l. So if 2s < 2l ≤ 2s+1, the element u must have order ≤ 2s+1 = 2 ·2s ≤ 4l.

Therefore by Lemma 5.2.1, we have dimL(λ) ≤ 2s+1(2s+1+2)2 ≤ 4l(4l+2)

2 = 8l2 + 4l.Now 8l2 + 4l < l3 if l ≥ 9, so by [Lüb01, Theorem 5.1] we have λ = ω1 or

λ = ω2 if l ≥ 12. In the cases where 2 ≤ l ≤ 11, checking irreducible, self-dual

representations of dimension ≤ 2s+1(2s+1+2)2 given in [Lüb01], we �nd that the the

only ones besides ω1 and ω2 are those in (ii), (iii) and (iv) of the claim.

Lemma 5.2.5. Suppose that G is of type C5 and let u ∈ G be a distinguishedunipotent element. Then u does not act as a distinguished unipotent element onL(ω3).

Proof. We have dimL(ω3) = 100 for example by [Lüb01]. Thus if u has order≤ 23, then by Lemma 5.2.1 the action of u is not distinguished on L(ω3) (because23(23+2)

2 = 40 < 100). The only distinguished unipotent element of G with order> 23 is the regular unipotent element. A computation with MAGMA (Section 2.9)shows that the regular unipotent acts on L(ω3) with Jordan blocks [62, 10, 14, 164],and thus the action is not distinguished.

Therefore for types Bl, Cl and Dl, the claim of Theorem 1.1.11 is reduced tothe cases λ = ω2 and λ = ωl (l ≤ 10), which will be dealt with in Section 5.6 andSection 5.11, respectively.

5.2.3 Exceptional types

In this subsection we assume that G is of exceptional type.

Proposition 5.2.6. Let u ∈ G be distinguished and let λ ∈ X(T )+ be 2-restricted.If u acts on L(λ) as a distinguished unipotent element, then one of the followingholds.

(i) G = G2 and λ = ω1 or λ = ω2.

(ii) G = F4 and λ = ω1 or λ = ω4.

(iii) G = E6 and λ = ω2.

(iv) G = E7 and λ = ω1 or λ = ω7.

(v) G = E8 and λ = ω8.

Proof. As in the proof of Proposition 5.2.2, we can assume that L(λ) is self-dual.When G is of type G2, F4, E6, E7 or E8, then all distinguished unipotent

elements of G have order at most 23, 24, 24, 25, and 25 respectively (see tables inAppendix A). Then by Lemma 5.2.1, the dimension of L(λ) is at most 40, 144,144, 544, and 544 respectively. Going through the tables given in [Lüb01], one�nds that the irreducible self-dual 2-restricted modules of dimension at most thebound are those in (i), (ii), (iii), (iv), and (v) respectively.

For groups of exceptional type, Proposition 5.2.6 reduces the claim of Theorem1.1.11 to a small number of λ to consider. These representations are treated inAppendix B.

5.3. Representation LG(ω1 + ωl) for G of type Al (p 6= 2) 133

5.3 Representation LG(ω1 + ωl) for G of type Al (p 6= 2)

Assume that p 6= 2 and that G is of type Al (l ≥ 2).

Let V be the natural module of G and set n = l + 1 = dimV .

The purpose of this section is to determine when a unipotent element u ∈ G actsas a distinguished unipotent element on L(ω1 + ωl).

It is well known that we can �nd L(ω1 + ωl) as a subquotient of V ⊗ V ∗, asshown by the following lemma which also holds when p = 2. For a proof, one canapply [Sei87, Lemma 8.6] as in [McN98, Proposition 4.6.10 (a)].

Lemma 5.3.1. As G-modules, we have

V ⊗ V ∗ ∼=

{L(ω1 + ωl)⊕ L(0) if p | nL(0)/L(ω1 + ωl)/L(0) (uniserial) if p - n

We will now determine when a regular unipotent u ∈ G acts as a distinguishedunipotent element on L(ω1 + ωl).

Proposition 5.3.2. Let u ∈ G be regular. Then u acts on L(ω1 +ωl) as a distin-guished unipotent element if and only if one of the following holds:

(i) p ≥ 2n− 1,

(ii) n = bpk + pk±12 for some k ≥ 1 and 0 ≤ b ≤ p−1

2 .

In all of these cases, u acts on L(ω1 + ωl) with Jordan blocks [2n − 1, 2n −3, . . . , 3].

Proof. Recall the notation for K[u]-modules from Section 1.4. As a K[u]-module,we have V ∼= Vn and so V ⊗ V ∗ ∼= Vn ⊗ Vn.

Suppose �rst that p | n, so now V ⊗V ∗ ∼= L(0)/L(ω1+ωl)/L(0) as a G-module.By Theorem 3.3.5 and Lemma 3.4.3, the unipotent element u acts on V ⊗V ∗ withpα Jordan blocks of size pα, where α = νp(n). Thus the action of u on V ⊗ V ∗ isinadmissible (De�nition 3.2.4), so by Lemma 3.2.6 the unipotent element u doesnot act on L(ω1 + ωl) as a distinguished unipotent element.

Therefore it will be enough to consider the case where p - n. Now V ⊗ V ∗ ∼=L(ω1 + ωl) ⊕ L(0) as a G-module. Since in Vn ⊗ Vn block size 1 occurs withmultiplicity 1 (Lemma 3.4.3), it follows that u acts on L(ω1+ωl) as a distinguishedunipotent element if and only if Vn ⊗ Vn has no repeated blocks. Since p 6= 2, wehave Vn ⊗ Vn ∼= ∧2(Vn) ⊕ S2(Vn) as a K[u]-module, so the claim follows fromProposition 3.5.3 and Proposition 3.4.4.

5.4 Representation LG(ω1 + ωl) for G of type Al (p = 2)

Assume that p = 2 and that G is of type Al (l ≥ 2).

Let V be the natural module of G.

134 Chapter 5.

The purpose of this section is to determine when a unipotent element u ∈ Gacts as a distinguished unipotent element on L(ω1 + ωl). The answer is given byProposition 5.4.4; we have also recorded this in Table 1.3.

As a special case of Lemma 5.3.1, we have the following.


V ⊗ V ∗ ∼=

{L(ω1 + ωl)⊕ L(0) if dimV is odd

L(0)/L(ω1 + ωl)/L(0) (uniserial) if dimV is even

We will also be applying the following theorem, which tells how a tensor squareVn⊗ Vn splits into a sum of indecomposables. It could be easily deduced with themain result of [Bar11] (Theorem 3.3.5), but it is also given in [GL06, Corollary 3].

Theorem 5.4.2. Let n = 2k + s, where k ≥ 0 and 0 ≤ s < 2k. Then

(i) If s = 0, then Vn ⊗ Vn = n · Vn.

(ii) If s > 0, then Vn ⊗ Vn = (Vn−2s ⊗ Vn−2s)⊕ 2s · V2k+1.

Lemma 5.4.3. All Jordan blocks in Vn ⊗ Vn have multiplicity ≤ 3 if and only ifn = 1, 2, 3 or n = 5.

Proof. If n ≤ 5, we can easily decompose Vn ⊗ Vn using Theorem 5.4.2 and verifythe claim. We have given these decompositions in Table 5.1 for convenience.

Suppose then that n > 5 and write n = 2k + s, where k ≥ 2 and 0 ≤ s < 2k.If s = 0, then by Theorem 5.4.2 (i) the tensor square Vn ⊗ Vn has n ≥ 4 blocks ofsize n. Consider then 0 < s < 2k, so now

Vn ⊗ Vn = (Vn−2s ⊗ Vn−2s)⊕ 2s · V2k+1

by Theorem 5.4.2 (ii). Therefore if s ≥ 2, then Vn⊗Vn has ≥ 4 blocks of size 2k+1.What remains is the case where s = 1. In this case, we have n − 2s = 2k − 1 =2k−1 + (2k−1 − 1) and thus by theorem 5.4.2 (ii) the tensor square Vn−2s ⊗ Vn−2s

has 2(2k−1−1) = 2k−2 ≥ 4 blocks of size 2k (we have k ≥ 3 since we are assumingn > 5).

Proposition 5.4.4. Suppose that G = Al, l ≥ 2 and let u ∈ G be a regularunipotent element. Then u acts on L(ω1 +ωl) as a distinguished unipotent elementif and only if l = 2 or l = 4.

Proof. We know from lemma 5.4.1 that as a G-module, V ⊗V ∗ = L(ω1+ωl)⊕L(0)if 2 - l + 1 and V ⊗ V ∗ = L(0)/L(ω1 + ωl)/L(0) if 2 | l + 1.

Now u acts on V with a single Jordan block of size l + 1. Thus if l = 3 or ifl ≥ 5, then by Lemma 5.4.3 the action of u on V ⊗ V ∗ has some Jordan blockwith multiplicity ≥ 4. Hence the action of u on V ⊗V ∗ is inadmissible (De�nition3.2.5), so it follows from Lemma 3.2.7 and Lemma 3.2.8 that u does not act onL(ω1 + ωl) as a distinguished unipotent element in this case.

What remains is to consider l = 2 and l = 4. For these cases, a computationwith MAGMA (Section 2.9) shows the following.

• If l = 2, then we have the orthogonal decomposition (Proposition 2.4.4)L(ω1 + ω2) ↓ K[u] = V (4)2.

5.5. Representation LG(ω2) for G of classical type (p 6= 2) 135

• If l = 4, then we have the orthogonal decomposition (Proposition 2.4.4)L(ω1 + ω4) ↓ K[u] = V (4)2 + V (8)2.

Therefore in both of these cases u acts on L(ω1 + ωl) as a distinguished uni-potent element (Proposition 2.4.4), as claimed.

n Vn ⊗ Vn1 [1]2 [22]3 [1, 42]4 [44]5 [1, 42, 82]

Table 5.1: Decomposition of Vn ⊗ Vn for 1 ≤ n ≤ 5.

5.5 Representation LG(ω2) for G of classical type(p 6= 2)

Assume that p 6= 2.

Suppose that G is of type Bl, Cl or Dl with natural module V . In this section,we will determine when a unipotent element u ∈ G acts on L(ω2) as a distinguishedunipotent element. We will show that this can happen only when u is a regularunipotent element. Furthermore, with the unique exception of type C3 with p = 3,we will see that u acts on L(ω2) as a distinguished unipotent element only if itacts on ∧2(V ) with no repeated blocks.

5.5.1 Construction of LG(ω2)

In this subsection, we will describe the well known construction of LG(ω2) forG = Sp(V ) (type Cl) and G = SO(V ) (type Bl or Dl). For types Bl and Dl thisis easy, as seen in the following lemma.

Lemma 5.5.1. Let G = SO(V ). Then ∧2(V ) ∼= LG(ω2).

Proof. This is a consequence of [Sei87, 8.1 (a), 8.1 (b)], see for example [McN98,Proposition 4.2.2].

For the rest of this subsection, let G = Sp(V ), where dimV = 2l (l ≥ 2). Let(−,−) be a G-invariant alternating bilinear form on V . The following lemma showsthat the exterior square ∧2(V ) is not irreducible, but we can �nd the representationLG(ω2) as a subquotient of ∧2(V ).


∧2(V ) ∼=

{L(ω2)⊕ L(0) if p - l,L(0)/L(ω2)/L(0) (uniserial) if p | l.

Proof. See for example [McN98, Lemma 4.8.2].

136 Chapter 5.

We now proceed to describe the submodule structure of ∧2(V ) explicitly. Fixa symplectic basis e1, . . . , e2l of V such that

(ei, ej) =

{(−1)i if i+ j = 2l + 1,

0 otherwise.

De�ne γ ∈ ∧2(V ) by

γ =

l∑i=1

(−1)iei ∧ e2l+1−i.

Then γ spans the unique one-dimensional G-submodule of ∧2(V ), as seen by thefollowing lemma.

Lemma 5.5.3. The element γ is �xed by the action of G on ∧2(V ).

Proof. The form (−,−) induces a G-module isomorphism V → V ∗ de�ned byv 7→ (v,−). This in turn induces an isomorphism ψ : ∧2(V ) → ∧2(V ∗) of G-modules.

Let Alt(V ) be the space of alternating bilinear forms on V . Then Alt(V ) is a G-module with the action de�ned by (g ·β)(v, w) = β(g−1v, g−1w) for all β ∈ Alt(V ),g ∈ G, and v, w ∈ V . Now there is an isomorphism χ : ∧2(V ∗) → Alt(V ) of G-modules, de�ned by χ(f ∧f ′)(v, w) = f(v)f ′(w)−f(w)f ′(v) for all f, f ′ ∈ V ∗ andv, w ∈ V .

Therefore it is enough to show that the alternating bilinear form χψ(γ) = βis �xed by the action of G. That is, we should show that β is a G-invariantalternating bilinear form on V . To this end, a straightforward calculation on thebasis elements ei shows that β(v, w) = (v, w) for all v, w ∈ V .

We de�ne the linear map ϕ : ∧2(V )→ K by ϕ(v∧w) = (v, w) for all v, w ∈ V ;it is easily seen that ϕ is a surjective morphism of G-modules.

Now we can use the G-submodules kerϕ and 〈γ〉 to describe the submodulestructure of ∧2(V ). Note that we have ϕ(γ) = l. Therefore if p - l, then γ 6∈ kerϕand so ∧2(V ) = kerϕ⊕ 〈γ〉 as a G-module. In this case kerϕ ∼= L(ω2) by Lemma5.5.2. If p | l, then γ ∈ kerϕ, and thus kerϕ/〈γ〉 ∼= L(ω2) by Lemma 5.5.2.

5.5.2 Types Bl and Dl

Lemma 5.5.4. Let G = SO(V ) and let u ∈ G be a unipotent. Suppose that u isa distinguished unipotent element of G, so as a K[u]-module V = Vd1 ⊕ · · · ⊕ Vdt,where the di are distinct and odd (Proposition 2.3.4). Assume that t > 1. Then uacts on ∧2(V ) with no repeated blocks if and only if t = 2, and V = Vd⊕V1, where∧2(Vd) has no repeated blocks and d ≡ 1 mod 4.

Proof. Suppose that u acts on ∧2(V ) with no repeated blocks. Now as a K[u]-module, ∧2(V ) has ∧2(Vd1)⊕ · · · ⊕∧2(Vdt) as a direct summand by Lemma 3.4.9.Therefore u acts on ∧2(Vdi) with no repeated blocks for all i, and by Proposition3.5.3, each ∧2(Vdi) decomposes as in characteristic 0. In particular, since all di areodd, for all di > 1 we have a block of size 3 in ∧2(Vdi) (Proposition 3.4.4). Thusdi > 1 for at most one i, which implies t = 2 and V = Vd⊕V1 for some odd d > 1.By Lemma 3.4.9, we have

∧2(V ) = ∧2(Vd)⊕ Vd


as aK[u]-module. Since ∧2(Vd) has no repeated blocks, it decomposes as in charac-teristic 0 (Proposition 3.5.3). In particular, by Proposition 3.4.4, the K[u]-module∧2(Vd) has Vd as a summand if and only if 2d − 3 ≡ d mod 4, which happens ifand only if d ≡ 3 mod 4. Therefore u acts on ∧2(V ) with no repeated blocks ifand only if d ≡ 1 mod 4.

Proposition 5.5.5 (Type Bl). Let G be the orthogonal group SO(V ), wheredimV = 2l + 1 (l ≥ 2). Then

(a) A non-regular unipotent element of G does not act as a distinguished unipotentelement in the representation L(ω2).

(b) A regular unipotent element of G acts as a distinguished unipotent element inL(ω2) if and only if one of the following holds:

(i) p ≥ 4l − 1,

(ii) 2l + 1 = bpk + pk±12 , where k ≥ 1 and 0 ≤ b ≤ p−1

2 ,

(iii) 2l + 1 = p+ p−32 ,

Proof. Recall that ∧2(V ) ∼= L(ω2) (Lemma 5.5.1). Let u ∈ G be a distinguishedunipotent element.

If u is not regular, then the fact that dimV is odd implies that the actionof u on V has ≥ 3 Jordan blocks (Proposition 2.3.2). Then by Lemma 5.5.4 theaction of u on ∧2(V ) has repeated blocks, and thus the action is not distinguished,proving (a).

Suppose that u is regular, so now V ↓ K[u] = V2l+1. Then u acts on L(ω2) asa distinguished unipotent element if and only if u acts on ∧2(V ) with no repeatedblocks, so the claim follows from Proposition 3.5.3.

Proposition 5.5.6 (Type Dl). Let G be the orthogonal group SO(V ), wheredimV = 2l (l ≥ 4). Then


(b) A regular unipotent element of G acts as a distinguished unipotent element inL(ω2) if and only if l is odd and one of the following holds:

(i) p ≥ 4l − 5,

(ii) 2l − 1 = bpk + pk±12 , where k ≥ 1 and 0 ≤ b ≤ p−1

2 ,

(iii) 2l − 1 = p+ p−32 ,

Proof. Recall that ∧2(V ) ∼= L(ω2) (Lemma 5.5.1). Let u ∈ G be a distinguishedunipotent element. If u is not regular, then u acts on V with ≥ 2 Jordan blocksof size > 1 (Proposition 2.3.3). Then by Lemma 5.5.4, the action of u on ∧2(V )has repeated blocks and so the action is not distinguished, which proves (a).

Suppose next that u is regular, so V ↓ K[u] = V2l−1 ⊕ V1 (Proposition 2.3.3).In this case the claim follows from Lemma 5.5.4 and Proposition 3.5.3, since uacts on L(ω2) as a distinguished unipotent element if and only if u acts on ∧2(V )with no repeated blocks.

138 Chapter 5.

5.5.3 Type Cl

Lemma 5.5.7. Let G = Sp(V ) and let u ∈ G be a unipotent. Suppose that u isa distinguished unipotent element of G, so as a K[u]-module V = Vd1 ⊕ · · · ⊕ Vdt,where 0 < d1 < · · · < dt are even (Proposition 2.3.4). If t > 1, then u acts on∧2(V ) with some block of size > 1 having multiplicity ≥ 2.

Proof. It follows from Lemma 3.4.9 that ∧2(V ) ↓ K[u] has ∧2(Vd1 ⊕ Vd2) as adirect summand. Thus it will be enough to prove the claim in the case wheret = 2, say V = Vm ⊕ Vn with 0 < m < n even. Suppose that u acts on ∧2(V )such that all blocks of size > 1 have multiplicity ≤ 1. Now ∧2(V ) ↓ K[u] =∧2(Vm)⊕ (Vm ⊗ Vn)⊕ ∧2(Vn), so by Proposition 3.5.3 the action of u on ∧2(Vm)and ∧2(Vn) is as in characteristic 0.

Note that since m and n are both even, we have 2m− 3 ≡ 2n− 3 ≡ 1 mod 4,so every block in ∧2(Vm) must occur in ∧2(Vn) by Proposition 3.4.4. In particularif m ≥ 4, then both ∧2(Vm) and ∧2(Vn) have a block of size 5. Thus ∧2(V ) has≥ 2 blocks of size 5, contradiction.

If m = 2, then we have ∧2(V ) = ∧2(Vn) ⊕ (Vn ⊗ V2) ⊕ V1. Now by Lemma3.3.9 we have Vn ⊗ V2 = Vn−1 ⊕ Vn+1 if p - n and Vn ⊗ V2 = Vn ⊕ Vn if p | n.We have n ≥ 4, so n + 1 ≤ 2n − 3 and therefore either Vn−1 or Vn+1 occurs in∧2(Vn). Thus either a block of size n− 1, n or n+ 1 has multiplicity 2 in ∧2(V ),contradiction.

Lemma 5.5.8. Let G = Sp(V ) and let u ∈ G unipotent. Suppose that u is adistinguished unipotent element of G, so as a K[u]-module V = Vd1 ⊕ · · · ⊕ Vdt,where the di are distinct and even (Proposition 2.3.4). Assume that p | dimV , so∧2(V ) = L(0)/L(ω2)/L(0) as a G-module (Lemma 5.5.2). Let u0 be the image ofu in SL(∧2(V )) and u′′0 the image of u in SL(L(ω2)). Let α = νp(gcd(d1, . . . , dt)).Then for the Jordan block sizes rm(u′′0) (De�nition 3.1.1) the following hold:

(a) If α = 0, then r1(u′′0) = r1(u0)− 2 and rm(u′′0) = rm(u0) for all m > 1.

(b) If α > 0, then one of the following holds:

(i) rpα(u′′0) = rpα(u0) − 1, rpα−2(u′′0) = 1, and rm(u′′0) = rm(u0) for allm 6= pα, pα − 2.

(ii) rpα(u′′0) = rpα(u0) − 2, rpα−1(u′′0) = 2, and rm(u′′0) = rm(u0) for allm 6= pα, pα − 1.

Proof. We begin by constructing u as in Section 2.5. Let V = W1 ⊕ · · · ⊕Wt, anorthogonal direct sum, with dimWi = di. For each i, let e

(i)1 , . . . , e

(i)di

be a basisfor Wi such that

(e(i)x , e

(i)y ) =

{(−1)x if x+ y = di + 1,

0 otherwise.

Set ex = 0 for all x ≤ 0. Now de�ne a linear map u : V → V by (u − 1)e(i)x =

(u + 1)e(i)x−1 for all 1 ≤ i ≤ t and 1 ≤ x ≤ di. Then u ∈ Sp(V ) (see Section 2.5)

and V ↓ K[u] ∼= Vd1 ⊕ · · · ⊕ Vdt with Wi∼= Vdi .

Let ϕ : ∧2(V ) → K and γ ∈ ∧2(V ) be as in subsection 5.5.1, so nowkerϕ/〈γ〉 ∼= L(ω2) as G-modules.


Without loss of generality, assume that α = νp(d1). Set d = d1 and ex = e(1)x

for all 1 ≤ x ≤ d. Denote the restriction of u0 to kerϕ by u′0. We can and willconsider u′′0 to be the map induced by u0 on kerϕ/〈γ〉.

We consider �rst the case where α = 0. Now

γ1 =

d/2∑j=1

(−1)jej ∧ ed−j+1

is a �xed point for u0 by Lemma 5.5.3. Furthermore, we have ϕ(γ1) = d/2, soγ1 6∈ kerϕ since p does not divide d. Thus ker(u0− 1) 6⊆ kerϕ, so by Lemma 3.2.1we have r1(u′0) = r1(u0)− 1 and rm(u′0) = rm(u0) for all m > 1.

Since ∧2(V ) admits a non-degenerate G-invariant symmetric form (Lemma4.4.7), it follows from Proposition 2.3.2 (ii) that rm(u0) = rm(u′0) is even if m > 0is even. On the other hand, we also have a non-degenerate G-invariant symmetricform on kerϕ/〈γ〉 ∼= L(ω2) (see Table 4.1), so rm(u′′0) is even if m > 0 is even,as well. Thus when we look at the Jordan blocks of u′′0 in terms of those of u′0using Lemma 3.2.2, we see that the situation in Lemma 3.2.2 (b) cannot occur.Indeed, otherwise rm(u′′0) would be odd for some even m > 0. Thus we haver1(u′′0) = r1(u′0)− 1 = r1(u0)− 2 and rm(u′′0) = rm(u′0) = rm(u0) for all m > 1, asdesired.

Consider then the case where α > 0. Write d = pαk, where p does not dividek. Note that in this case the smallest Jordan block size of u acting on ∧2(V ) is pα

by Lemma 3.4.11.We will show next that ker(u0− 1)p

α 6⊆ kerϕ. Since (u0− 1)pα

= upα

0 − 1, thisis equivalent to �nding a �xed point for up

α

0 outside of kerϕ.Since (u− 1)ex = (u+ 1)ex−1, it follows that (u− 1)mex = (u+ 1)mex−m for

all m ≥ 1. In particular, (u− 1)pαex = (u+ 1)p

αex−pα , so

(upα − 1)ex = (up

α+ 1)ex−pα

for all x.Therefore, the subspace W of V with symplectic basis

e1, e1+pα , . . . , e(k−1)pα+1, epα , e2pα , . . . , ekpα

is a non-degenerate upα-invariant subspace. Then by Lemma 5.5.3, the element

γ′ ∈ ∧2(V ) de�ned by

γ′ =k−1∑j=0

(−1)j+1e1+jpα ∧ e(k−j)pα

is a �xed point for upα

0 . Furthermore, now ϕ(γ′) = k so γ′ 6∈ kerϕ.We have seen that ker(u0 − 1)p

α 6⊆ kerϕ and that the smallest Jordan blockof u0 has size pα, so it follows from Lemma 3.2.1 that rpα(u′0) = rpα(u0) − 1,rpα−1(u′0) = 1 and rm(u′0) = rm(u0) for all m 6= pα, pα − 1. Because kerϕ/〈γ〉 ∼=L(ω2) has a G-invariant symmetric form, it follows that rpα−1(u′′0) must be even.Thus when we look at the Jordan blocks of u′′0 in terms of those of u′0 usingLemma 3.2.2, we see that we must have m = pα − 1 or m = pα − 2 in Lemma3.2.2 for rpα−1(u′′0) to be even. This proves the claim, since these two possibilitiescorrespond to the situations (i) and (ii) given in case (b) of our claim.

140 Chapter 5.

Remark 5.5.9. In case (b) of Lemma 5.5.8, both cases (i) and (ii) can occur. Forexample, consider the case where p = 3 and G = Sp(V ) with dimV = 18 (typeC9). Here we have ∧2(V ) = L(0)/L(ω2)/L(0).

For u ∈ G with V ↓ K[u] = V18 (regular unipotent), one computes that

∧2(V ) ↓ K[u] = [95, 274]

L(ω2) ↓ K[u] = [7, 94, 274]

so the blocks are given as in Lemma 5.5.8 (b) (i). For u ∈ G with V ↓ K[u] =V6 ⊕ V12, one computes that

∧2(V ) ↓ K[u] = [34, 95, 155, 21]

L(ω2) ↓ K[u] = [22, 32, 95, 155, 21]

so here the blocks are given as in Lemma 5.5.8 (b) (ii).Although it is not necessary for the solution of our main problem, it would be

interesting to �nd a way to determine which of the cases (i) or (ii) occur in Lemma5.5.8 (b). One could also try to generalize Lemma 5.5.8 for non-distinguishedunipotent elements. For example, in our proof of Lemma 5.5.8 we never used thefact that the block sizes di are distinct, and indeed Lemma 5.5.8 holds for allu ∈ Sp(V ) which have all Jordan block sizes even.

Proposition 5.5.10 (Type Cl). Let G be the symplectic group Sp(V ), wheredimV = 2l (l ≥ 2). Then


(b) A regular unipotent element of G acts as a distinguished unipotent element inL(ω2) if and only if one of the following holds:

(i) p ≥ 4l − 3,

(ii) 2l = bpk + pk±12 , where k ≥ 1 and 0 ≤ b ≤ p−1

2 ,

(iii) 2l = p+ p−32 ,

(iv) p = 3, l = 3.

Proof. We consider the claims (a) and (b) �rst in the case where p - l. In thiscase ∧2(V ) = L(ω2) ⊕ L(0) (Lemma 5.5.2), so by Lemma 5.5.7 a non-regularunipotent u ∈ G acts on ∧2(V ), thus on L(ω2) with some block of size > 1 havingmultiplicity ≥ 2. This proves (a). If u ∈ G is regular, then V ↓ K[u] = V2l and itfollows from Proposition 3.5.3 that u acts on L(ω2) as a distinguished unipotentelement if and only if it acts on ∧2(V ) with no repeated blocks. Thus (b) followsfrom Proposition 3.5.3.

Consider then p | l, so now ∧2(V ) = L(0)/L(ω2)/L(0) as a G-module. Letu ∈ G be a distinguished unipotent element, say V ↓ K[u] = Vd1 ⊕· · ·⊕Vdt wheredi are distinct and even. Set α = νp(gcd(d1, . . . , dt)).

Suppose �rst that u is not regular, i.e. that t > 1. If α = 0, then it followsfrom Lemma 5.5.7 and Lemma 5.5.8 that u acts on L(ω2) with some Jordan blockof size > 1 having multiplicity ≥ 2. Consider then the case where α > 0. For all i,set αi = νp(di). It follows from Lemma 3.4.10 that u acts on ∧2(Vdi) with smallest

5.6. Representation LG(ω2) for G of type Cl (p = 2) 141

Jordan block size pαi , which occurs with multiplicity pαi+12 ≥ 2. Since ∧2(V ) has

∧2(Vd1)⊕ · · · ⊕ ∧2(Vdt) as a direct summand by Lemma 3.4.9, and since t > 1, itfollows that the action of u on ∧2(V ) is inadmissible (De�nition 3.2.4). By Lemma3.2.6, the action of u on L(ω2) is not distinguished. This proves (a).

For (b), suppose that u is regular, say V ↓ K[u] = Vd where d = dimV . Itfollows from Lemma 3.4.10 that u acts on ∧2(V ) with smallest Jordan block sizepα, which occurs with multiplicity pα+1

2 . Therefore if p > 3 or α > 1, the actionof u on ∧2(V ) is inadmissible and thus by Lemma 3.2.6, the action of u on L(ω2)is not distinguished.

Suppose then that p = 3 and α = 1, say d = 3k where 3 does not divide k.Now by Theorem 3.3.8, we have sp(d)> = (3λ1, 3λ1, 3λ1, . . . , 3λk, 3λk, 3λk) wheresp(k)> = (λ1, . . . , λk). Note that λk = 1 and λk′ > 1 for all k′ < k by Lemma3.4.3.

Since k is even, it follows from Theorem 3.4.5 that if k > 2, then u acts on∧2(V ) with ≥ 2 blocks of size 3λk and ≥ 2 blocks of size 3λk−2. Once again theaction of u on ∧2(V ) is inadmissible, so the action of u on L(ω2) is not distinguishedby Lemma 3.2.6.

Therefore the only possibility left is that k = 2, and this is precisely the case(b) (iv) in our claim. One can compute, either directly or e.g. with MAGMA(Section 2.9), that in this case we have

∧2(V ) ↓ K[u] = [3, 3, 9]

L(ω2) ↓ K[u] = [1, 3, 9]

so the action of u on L(ω2) is distinguished. This completes the proof of theproposition.

Remark 5.5.11. Let G = Sp(V ). In Proposition 5.5.10 (b) (i)-(iii), it is clearthat p - l, and thus V (ω2) = L(ω2) and ∧2(V ) ∼= L(ω2) ⊕ L(0) by Lemma 5.5.2.Therefore it follows from Proposition 5.5.10 that if some unipotent element u ∈ Gacts on L(ω2) as a distinguished unipotent element, then V (ω2) = L(ω2) unlessp = 3, l = 3, and u is a regular unipotent element of G. This re�ects a moregeneral phenomenon that is seen in Theorem 1.1.10. Looking at the p-restrictedhighest weights λ that occur in the statement of Theorem 1.1.10, one can observethat there are very few cases where V (λ) is not irreducible.

5.6 Representation LG(ω2) for G of type Cl (p = 2)

Assume that p = 2.

Suppose that G is of type Cl (l ≥ 2) and let V be the natural module of G. Inthis section, we determine when a unipotent element u ∈ G acts as a distinguishedunipotent element on L(ω2). The answer is given in Proposition 5.6.7 below andis also recorded in Table 1.3.

As in subsection 5.5.1, we can �nd the representation L(ω2) as a subquotientof the exterior square ∧2(V ). Lemma 5.5.2 also holds in characteristic two (e.g.by [McN98, Lemma 4.8.2]) and becomes the following.


∧2(V ) ∼=

{L(ω2)⊕ L(0) if l is odd

L(0)/L(ω2)/L(0) (uniserial) if l is even

142 Chapter 5.

Let u ∈ G be a distinguished unipotent element and suppose that u has order2α+1. We retain the notation for K[u]-modules from Section 5.3. We will applythe following result, which tells us how the exterior square of an indecomposableK[u]-module splits into a sum of indecomposables. It was proven by Gow andLa�ey in [GL06] and it is a special case of a more general result due to Himstedtand Symonds [HS14, Theorem 1.1].

Theorem 5.6.2. Let 0 < s ≤ 2n−1 and n ≥ 1. Then

∧2(V2n−1+s) = ∧2(V2n−1−s)⊕ V2n−s ⊕ (s− 1)V2n

Remark 5.6.3. In Theorem 5.6.2, we interpret V0 = 0, so s = 2n−1 gives∧2(V2n) = (2n−1 − 1)V2n ⊕ V2n−1 .

Lemma 5.6.4. Suppose that all blocks in ∧2(Vr) have multiplicity ≤ 3. Thenr ≤ 20.

Proof. Write r = 2n−1 + s, where n ≥ 1 and 0 < s ≤ 2n−1. Now

∧2(V2n−1+s) = ∧2(V2n−1−s)⊕ V2n−s ⊕ (s− 1)V2n

so s − 1 ≤ 3 and so s ≤ 4. On the other hand, 2n−1 − s = 2n−2 + 2n−2 − s soapplying 2n−2 − s − 1 ≤ 3 and so 2n−2 ≤ s + 4 ≤ 8. Therefore n ≤ 5 and thusr = 2n−1 + s ≤ 24 + 4 = 20.

Using Theorem 5.6.2, we can easily decompose ∧2(Vr) for 1 ≤ r ≤ 20 (seeTable 5.2) and get the following corollary.

Corollary 5.6.5. Suppose that all blocks in ∧2(Vr) have multiplicity ≤ 3, andthat at most one block has multiplicity 3. Then r ≤ 12.

In the proof of the next lemma, we will make use of Table 5.3, which gives thedecomposition of Vr ⊗ Vs for 2 ≤ r ≤ s ≤ 12 where r and s are even. The data intable 5.3 can be computed with a computer program (for example MAGMA), orby hand using recursive formulae as described for example in [Bar11, Theorem 1]or [GL06, Corollary 3].

Lemma 5.6.6. Let V = Vr1 ⊕ · · · ⊕ Vrt be a K[u]-module, where 2 ≤ r1 ≤ r2 ≤· · · ≤ rt are even. Then the action of u on ∧2(V ) is inadmissible (De�nition 3.2.5),unless one of the following holds.

(i) t = 1 and V = Vr, where r ≤ 12.

(ii) t = 2 and V = V2 ⊕ V2 or V = V2 ⊕ V10.

Proof. If V = Vr and the action of u on ∧2(V ) is admissible, then by Corollary5.6.5 we have r ≤ 12.

Consider then V = Vr1 ⊕ Vr2 , where 2 ≤ r1 ≤ r2 are even and supposethat the action of u on ∧2(V ) is admissible (De�nition 3.2.5). Now ∧2(V ) ∼=∧2(Vr1)⊕ (Vr1 ⊗ Vr2)⊕∧2(Vr2), so by (i) we must have r1, r2 ≤ 12. Furthermore,now the action of u on Vr1 ⊗ Vr2 is admissible, so by Table 5.3 the pair (r1, r2)is one of the following: (2, 2), (2, 4), (2, 6), (2, 8), (2, 10), (2, 12), (4, 6), (4, 10),(6, 12). Computing ∧2(Vr1 ⊕Vr2) in these cases using Table 5.2 and Table 5.3, onegets the following decompositions.

5.6. Representation LG(ω2) for G of type Cl (p = 2) 143

∧2(V2 ⊕ V2) = [12, 22]∧2(V2 ⊕ V4) = [1, 2, 43]∧2(V2 ⊕ V6) = [12, 63, 8]∧2(V2 ⊕ V8) = [1, 4, 85]∧2(V2 ⊕ V10) = [12, 6, 8, 102, 14, 16]∧2(V2 ⊕ V12) = [1, 2, 4, 123, 163]∧2(V4 ⊕ V6) = [1, 2, 43, 6, 83]∧2(V4 ⊕ V10) = [1, 2, 4, 6, 83, 122, 14, 16]∧2(V6 ⊕ V12) = [1, 2, 4, 6, 83, 123, 165]

Therefore (r1, r2) must be (2, 2) or (2, 10).Finally, suppose that V = Vr1 ⊕ Vr2 ⊕ · · · ⊕ Vrt where 2 ≤ r1 ≤ r2 ≤ · · · ≤ rt

are even and t ≥ 3. Note that ∧2(V ) has ∧2(Vr1 ⊕ Vr2) and ∧2(Vr2 ⊕ Vr3) as adirect summand. Thus if the action of u on ∧2(V ) is admissible, then the actionof u on each of ∧2(Vr1 ⊕ Vr2) and ∧2(Vr2 ⊕ Vr3) is also admissible. By (ii), thisimplies that Vr1 ⊕Vr2 ⊕Vr3 = V2⊕V2⊕V10 or V2⊕V2⊕V2. A computation showsthat

∧2(V2 ⊕ V2 ⊕ V2) = [13, 26]∧2(V2 ⊕ V2 ⊕ V10) = [13, 22, 6, 8, 104, 14, 16]

and then since ∧2(Vr1 ⊕ Vr2 ⊕ Vr3) is a direct summand of ∧2(V ), it follows thatthe action of u on ∧2(V ) is inadmissible.

Proposition 5.6.7. The action of u on L(ω2) is distinguished if and only if l andu occur in Table 5.1.

Proof. Suppose that the action of u on L(ω2) is distinguished. Now by lemmas5.6.1, 5.6.6, 3.2.7 and 3.2.8 one of the following holds:

• 2 ≤ l ≤ 6 and u is regular.

• l = 2 and u is in class (221).

• l = 6 and u is in class (21, 101).

If l = 4 and u is regular, then a computer calculation shows that the action ofu on L(ω2) has Jordan blocks [2, 83], so the action is not distinguished. If l = 6and u is regular, then the action of u on L(ω2) has Jordan blocks [4, 12, 163], sohere too the action is not distinguished.

The remaining cases are those in Table 5.1. In these cases, one can verify bya computation with MAGMA (Section 2.9) that the Jordan blocks are as in theclaim, and that u acts as a distinguished unipotent element on L(ω2).

144 Chapter 5.

l Class of u Jordan blocks of u acting on L(ω2)

2 regular [4]2 (22

1) [22]3 regular [6, 8]5 regular [6, 8, 14, 16]6 (21, 101) [6, 8, 102, 14, 16]

Table 5.1: Cases where a unipotent element u of G = Cl acts on L(ω2) as adistinguished unipotent element.

r ∧2(Vr)

1 02 [1]3 [3]4 [2, 4]5 [3, 7]6 [1, 6, 8]7 [5, 82]8 [4, 83]9 [5, 82, 15]10 [1, 6, 8, 14, 16]

r ∧2(Vr)

11 [3, 7, 13, 162]12 [2, 4, 12, 163]13 [3, 11, 164]14 [1, 10, 165]15 [9, 166]16 [8, 167]17 [9, 166, 31]18 [1, 10, 165, 30, 32]19 [3, 11, 164, 29, 322]20 [2, 4, 12, 163, 28, 323]

Table 5.2: Decomposition of ∧2(Vr) for 1 ≤ r ≤ 20

HHHH

HHrs

2 4 6 8 10 12

2 [22] [42] [62] [82] [102] [122]4 [44] [42, 82] [84] [82, 122] [124]6 [22, 84] [86] [84, 142] [82, 122, 162]8 [88] [86, 162] [84, 164]10 [22, 84, 164] [42, 82, 166]12 [44, 168]

Table 5.3: Decomposition of Vr ⊗ Vs for 2 ≤ r, s ≤ 12 even.

5.7 Representation LG(2ω1) for G of classical type(p 6= 2)

Assume that p 6= 2.

Suppose that G is of type Bl, Cl or Dl with natural module V . In this section,we will determine when a unipotent element u ∈ G acts on L(2ω1) as a distin-guished unipotent element. We will show that this can happen only when u is aregular unipotent element. Furthermore, we will also see that u acts on L(2ω1)as a distinguished unipotent element only if it acts on S2(V ) with no repeatedblocks. The results and proofs given are very similar to those found in Section 5.5.

5.7. Representation LG(2ω1) for G of classical type (p 6= 2) 145

5.7.1 Construction of LG(2ω1)

In this subsection, we will describe a construction of LG(2ω1) for G = Sp(V )(type Cl) and G = SO(V ) (type Bl or Dl). For type Cl this is easy, as seen in thefollowing lemma.

Lemma 5.7.1. Let G = Sp(V ). Then S2(V ) ∼= LG(2ω1).

Proof. See for example [McN98, Proposition 4.2.2].

For the rest of this subsection, let G = SO(V ) with dimV = n, and denote thesymmetric form on V by (−,−). In this situation the following lemma shows thatthe symmetric square S2(V ) is not irreducible, but we can �nd the representationLG(2ω1) as a subquotient of S2(V ) (cf. Lemma 5.5.2).


S2(V ) ∼=

{L(2ω1)⊕ L(0) if p - n,L(0)/L(2ω1)/L(0) (uniserial) if p | n.

Proof. See for example [McN98, Lemma 4.7.3].

We will now consider the submodule structure of S2(V ) more explicitly, simi-larly to the description of ∧2(V ) in 5.5.1.

Fix a basis e1, . . . , en of V such that

(ei, ej) =

{(−1)min{i,j} if i+ j = n+ 1,

0 otherwise.

Note that this form is exactly the same as the one given in Section 2.5 if n isodd.

If n = 2l + 1 (type Bl), we de�ne γ ∈ S2(V ) to be the vector(l∑

i=1

(−1)ieien+1−i

)+

1

2(−1)l+1e2

l+1.

Similarly if n = 2l (type Dl), we de�ne γ ∈ S2(V ) to be the vector

γ =

l∑i=1

(−1)ieien+1−i.

The following lemma shows that γ is �xed by the action of G.

Lemma 5.7.3. The element γ is �xed by the action of G on S2(V ).

Proof. (cf. Lemma 5.5.3) The form (−,−) on V induces a G-module isomorphismV → V ∗ de�ned by v 7→ (v,−). This in turn induces an isomorphism ψ : S2(V )→S2(V ∗) of G-modules.

Let Quad(V ) be the vector space of quadratic forms on V . Then Quad(V ) isa G-module with the action de�ned by (g ·Q)(v) = Q(g−1v) for all Q ∈ Quad(V ),g ∈ G and v ∈ V . Now there is an isomorphism χ : S2(V ∗) → Quad(V ) ofG-modules, de�ned by χ(ff ′)(v) = f(v)f ′(v) for all f, f ′ ∈ V ∗ and v ∈ V .

Therefore it is enough to show that χψ(γ) = Q is �xed by the action of G.That is, we should show that Q is a G-invariant quadratic form. To this end, astraightforward calculation on the basis elements ei shows that Q(v) = 1

2(v, v) forall v ∈ V .

146 Chapter 5.

De�ne a linear map ϕ : S2(V )→ K by ϕ(vw) = (v, w) for all v, w ∈ V . Thenϕ is a surjective morphism of G-modules. As in 5.5.1, we will use kerϕ and γ todescribe the submodule structure of S2(V ).

Note that we have ϕ(γ) = n/2. Therefore if p - n, then γ 6∈ kerϕ and soS2(V ) = kerϕ ⊕ 〈γ〉. In this case kerϕ ∼= L(2ω1) by Lemma 5.7.2. If p | n, thenγ ∈ kerϕ, and thus kerϕ/〈γ〉 ∼= L(2ω1) by Lemma 5.7.2.


Lemma 5.7.4. Let G = SO(V ) and let u ∈ G be a unipotent. Suppose that u isa distinguished unipotent element of G, so as a K[u]-module V = Vd1 ⊕ · · · ⊕ Vdt,where 0 < d1 < · · · < dt are odd (Proposition 2.3.4). If t > 1, then one of thefollowing holds:

(i) The element u acts on S2(V ) with some block of size > 1 having multiplicity≥ 2,

(ii) As a K[u]-module V = V1 ⊕ Vn, where n ≡ 3 mod 4 and S2(Vn) has norepeated blocks.

Proof. (cf. Lemma 5.5.7) It follows from Lemma 3.4.9 that S2(V ) ↓ K[u] hasS2(Vdi ⊕ Vdj ) as a direct summand for any i 6= j. Therefore it will be enough toprove the lemma in the case where t = 2, say V = Vm ⊕ Vn with 0 < m < n odd.

Suppose that u acts on S2(V ) such that all blocks of size > 1 have multiplicity≤ 1. Now S2(V ) ↓ K[u] = S2(Vm)⊕ (Vm ⊗ Vn)⊕ S2(Vn), so by Proposition 3.5.3the action of u on S2(Vm) and S2(Vn) is as in characteristic 0. Since m and n areboth odd, we have 2m− 1 ≡ 2n− 1 mod 4, so every block in S2(Vm) must occurin S2(Vn) by Proposition 3.4.4. Therefore m = 1, as otherwise both S2(Vm) andS2(Vn) would have a block of size 5.

Now S2(V ) ↓ K[u] = V1 ⊕ Vn ⊕ S2(Vn). Furthermore, a block of size n occursin S2(Vn) if and only if n ≡ 2n − 1 mod 4, which happens if and only if n ≡ 1mod 4. Hence we must have n ≡ 3 mod 4, and so (ii) of the lemma holds.

Lemma 5.7.5. Let G = SO(V ) and let u ∈ G unipotent. Suppose that u is adistinguished unipotent element of G, so as a K[u]-module V = Vd1 ⊕ · · · ⊕ Vdt,where 0 < d1 < · · · < dt are odd (Proposition 2.3.4). Assume that p | dimV , soS2(V ) = L(0)/L(2ω1)/L(0) as a G-module. Let u0 be the image of u in SL(S2(V ))and u′′0 the image of u in SL(L(2ω1)). Let α = νp(gcd(d1, . . . , dt)). Then for theJordan block sizes rm(u′′0) (De�nition 3.1.1) the following hold:

(a) If α = 0, then r1(u′′0) = r1(u0)− 2 and rm(u′′0) = rm(u0) for all m > 1.

(b) If α > 0, then one of the following holds:

(i) rpα(u′′0) = rpα(u0) − 1, rpα−2(u′′0) = 1, and rm(u′′0) = rm(u0) for allm 6= pα, pα − 2.

(ii) rpα(u′′0) = rpα(u0) − 2, rpα−1(u′′0) = 2, and rm(u′′0) = rm(u0) for allm 6= pα, pα − 1.


Proof. (cf. Lemma 5.5.8) We begin by constructing u as in Section 2.5. Let V =W1 ⊕ · · · ⊕ Wt as an orthogonal direct sum, with dimWi = di. For all i, lete

(i)1 , . . . , e

(i)di

be a basis for Wi such that

(e(i)x , e

(i)y ) =

{(−1)x if x+ y = di + 1,

0 otherwise.

Set ex = 0 for all x ≤ 0. Now de�ne a linear map u : V → V by (u − 1)e(i)x =

(u + 1)e(i)x−1. Then u ∈ SO(V ) (see Section 2.5) and V ↓ K[u] ∼= Vd1 ⊕ · · · ⊕ Vdt

with Wi∼= Vdi .

Let ϕ : S2(V ) → K be as in 5.5.1. Let γ ∈ S2(V ) be a generator for theone-dimensional G-submodule of S2(V ), which exists by Lemma 5.7.2. We havekerϕ/〈γ〉 ∼= L(2ω1).

Without loss of generality, assume that α = νp(d1). Set d = d1 and ex = e(1)x

for all 1 ≤ x ≤ d. Denote the restriction of u0 to kerϕ by u′0. We can and willconsider u′′0 to be the map induced by u0 on kerϕ/〈γ〉.

We consider �rst the case where α = 0. Write d = 2f + 1. Now

γ1 =

f∑j=1

(−1)jejed−j+1

+1

2(−1)f+1e2

f+1

is a �xed point for u0 by Lemma 5.7.3. Furthermore, we have ϕ(γ1) = d/2, soγ1 6∈ kerϕ since p does not divide d. Thus ker(u0− 1) 6⊆ kerϕ, so by Lemma 3.2.1we have r1(u′0) = r1(u0)− 1 and rm(u′0) = rm(u0) for all m > 1. Since we have aG-invariant symmetric form on S2(V ) (Lemma 4.4.8) and LG(2ω1) (Table 4.1), itfollows as in the proof of Lemma 5.5.8 (�fth paragraph) that r1(u′′0) = r1(u0)− 2and rm(u′′0) = rm(u0) for all m > 1, as desired.

Consider then the case where α > 0. Write d = pαk, where p does not dividek. Note that in this case the smallest Jordan block size of u acting on S2(V ) is pα

by Lemma 3.4.13.We will show next that ker(u0− 1)p

α 6⊆ kerϕ. Since (u0− 1)pα

= upα

0 − 1, thisis equivalent to �nding a �xed point for up

α

0 outside of kerϕ.Since (u − 1)ex = (u + 1)ex−1, it follows that (u − 1)mex = (u + 1)mex−k for

all m ≥ 1. In particular, (u− 1)pαex = (u+ 1)p

αex−pα , so

(upα − 1)ex = (up

α+ 1)ex−pα

for all x.Therefore the subspace W of V with basis

e1, e1+pα , . . . , e(k−1)pα+1, epα , e2pα , . . . , ekpα

is a non-degenerate upα-invariant subspace. By Lemma 5.7.3, the element γ′ ∈

S2(V ) de�ned by

γ′ =k−1∑j=0

(−1)j+1e1+jpαe(k−j)pα

is a �xed point for upα

0 . Now ϕ(γ′) = k, so γ′ 6∈ kerϕ since p does not divide k.We have shown that ker(u0 − 1)p

α 6⊆ kerϕ and that the smallest Jordanblock size of u0 is pα. Since we have a G-invariant symmetric form on S2(V )and LG(2ω1), the claim follows by arguing as in the end of the proof of Lemma5.5.8.

148 Chapter 5.

Remark 5.7.6. (cf. Remark 5.5.9) In case (b) of Lemma 5.7.5, both cases (i) and(ii) can occur. For example, consider the case where p = 5 and G = SO(V ) withdimV = 5 (type B2). Here we have S2(V ) = L(0)/L(2ω1)/L(0).

For u ∈ G with V ↓ K[u] = V5 (regular unipotent), one computes that

S2(V ) ↓ K[u] = [53]

L(2ω1) ↓ K[u] = [3, 52]

so the blocks are given as in Lemma 5.5.8 (b) (i).For an example of Lemma 5.5.8 (b) (ii), consider the case where p = 3 and

G = SO(V ) with dimV = 18 (type D9). Once again S2(V ) = L(0)/L(2ω1)/L(0).For a distinguished unipotent u ∈ G with V ↓ K[u] = V3 ⊕ V15, one computes

that

S2(V ) ↓ K[u] = [34, 9, 155, 21, 272]

L(2ω1) ↓ K[u] = [22, 32, 9, 155, 21, 272]

so the blocks are given as in Lemma 5.5.8 (b) (ii).

The following two propositions determine when for G = SO(V ) a unipotentelement u ∈ G acts on LG(2ω1) as a distinguished unipotent element. They willbe proven simultaneously.

Proposition 5.7.7 (Type Bl). Let G be the orthogonal group SO(V ), wheredimV = 2l + 1 (l ≥ 2). Then

(a) A non-regular unipotent element of G does not act as a distinguished elementin the representation L(2ω1).

(b) A regular unipotent element of G acts as a distinguished unipotent element inL(2ω1) if and only if one of the following holds:

(i) p ≥ 4l + 1,

(ii) 2l + 1 = bpk + pk±12 , where k ≥ 1 and 0 ≤ b ≤ p−1

2 .

Proposition 5.7.8 (Type Dl). Let G be the orthogonal group SO(V ), wheredimV = 2l (l ≥ 4). Then


(b) A regular unipotent element of G acts as a distinguished unipotent element inL(2ω1) if and only if l is even and one of the following holds:

(i) p ≥ 4l − 3,

(ii) 2l − 1 = bpk + pk±12 , where k ≥ 1 and 0 ≤ b ≤ p−1

2 .

Proof of Proposition 5.7.7 and Proposition 5.7.8. (cf. Proposition 5.5.10) Consi-der �rst the case where p - dimV , so S2(V ) = L(2ω1) ⊕ L(0) (Lemma 5.7.2).Note that a nonregular distinguished unipotent element of G acts on V with ≥ 2blocks of size > 1 (Proposition 2.3.4). Therefore by Lemma 5.7.4, a non-regular

unipotent u ∈ G acts on S2(V ), hence on L(2ω1) with some block of size > 1having multiplicity ≥ 2. This proves (a).

For (b), still assuming p - dimV , let u ∈ G be a regular unipotent element.Suppose �rst that dimV = 2l + 1, so V ↓ K[u] = V2l+1 (Proposition 2.3.3). Itfollows from Proposition 3.5.3 that u acts on L(2ω1) as a distinguished unipotentelement if and only if it acts on S2(V ) with no repeated blocks, and thus (b) followsfrom Proposition 3.5.3. Consider then dimV = 2l, so V ↓ K[u] = V2l−1 ⊕ V1

(Proposition 2.3.3). In this case, by Lemma 5.7.4 the action of u on L(2ω1) isdistinguished if and only if u acts on S2(V2l−1) with no repeated blocks and2l − 1 ≡ 3 mod 4. That is, if and only if u acts on S2(V2l−1) with no repeatedblocks and l is even. Once again (b) follows from Proposition 3.5.3.

Consider then the case where p | dimV , so S2(V ) = L(0)/L(2ω1)/L(0)(Lemma 5.7.2). Note that in the proposition being proven, we have p - dimVin (i)-(ii) of (b); therefore we prove next that no unipotent element of G acts as adistinguished unipotent element on L(2ω1). Let u ∈ G be a distinguished unipo-tent element, say V ↓ K[u] = Vd1 ⊕ · · · ⊕ Vdt where di are distinct and odd. Setα = νp(gcd(d1, . . . , dt)).

Suppose �rst that α = 0. Note that in this case t > 1. We will show thatu acts on S2(V ) with some Jordan block of size > 1 having multiplicity ≥ 2.Then it follows from Lemma 5.7.5 that u acts on L(2ω1) with some Jordan blockof size > 1 having multiplicity ≥ 2. Hence by Lemma 5.7.4 we can assume thatdimV = 2l, V ↓ K[u] = V2l−1 ⊕ V1, and S2(V2l−1) has no repeated blocks. Sincep divides dimV = 2l, it follows from Proposition 3.5.3 that

2l − 1 = bpk +pk ± 1

2,

for some k ≥ 1 and 0 ≤ b ≤ p−12 . If 2l − 1 = bpk + pk−1

2 , then 2l = bpk + pk+12 .

But then the fact that p | 2l implies that p | pk + 1, contradiction. Therefore

2l− 1 = bpk + pk+12 , and so 2l = bpk + pk+3

2 . Since p | 2l, it follows that p | pk + 3and hence p = 3. Because b = 0 or b = 1 in this case, we can assume that2l − 1 = 3k+1

2 for some k ≥ 1. Therefore 2l = 3k+32 , and so 3k + 3 ≡ 0 mod 4,

which means that k must be even. But then 3k + 3 ≡ 4 mod 8, so it follows thatl = 3k+3

4 is odd. Thus 2l− 1 ≡ 1 mod 4, and by Lemma 5.7.4 the element u actson S2(V ) with some Jordan block of size > 1 having multiplicity ≥ 2 (namely,block size 2l − 1 occurs with multiplicity two).

Finally consider the case where α > 0. Suppose �rst that t > 1. Set αi = νp(di).It follows from Lemma 3.4.12 that u acts on S2(Vdi) with smallest Jordan blocksize pαi , which occurs with multiplicity pαi+1

2 ≥ 2. Since S2(V ) has S2(Vd1) ⊕· · · ⊕ S2(Vdt) as a direct summand by Lemma 3.4.9, and since t > 1, it followsthat the action of u on S2(V ) is inadmissible (De�nition 3.2.4). By Lemma 3.2.6,the action of u on L(2ω1) is not distinguished.

What remains is the case where α > 0 and t = 1. In this case dimV = 2l+1 =d, where νp(d) = α, and V ↓ K[u] = Vd. It follows from Lemma 3.4.12 that u actson S2(V ) with smallest Jordan block size pα, which occurs with multiplicity pα+1

2 .Therefore if p > 3 or α > 1, the action of u on S2(V ) is inadmissible and thus byLemma 3.2.6, the action of u on L(2ω1) is not distinguished.

Suppose then that p = 3 and α = 1, say d = 3k where 3 does not divide k.Now by Theorem 3.3.8, we have sp(d)> = (3λ1, 3λ1, 3λ1, . . . , 3λk, 3λk, 3λk) wheresp(k)> = (λ1, . . . , λk). Note that λk = 1 and λk′ > 1 for all k′ < k by Lemma3.4.3.

150 Chapter 5.

Since d is odd, and l ≥ 2, we have k ≥ 3. It follows from Theorem 3.4.5 thatu acts on S2(V ) with ≥ 2 blocks of size 3λk and ≥ 2 blocks of size 3λk−2. Onceagain the action of u on S2(V ) is inadmissible, so the action of u on L(2ω1) is notdistinguished by Lemma 3.2.6. This completes the proof of the proposition.

Remark 5.7.9. (cf. Remark 5.5.11) Let G = SO(V ). It is clear in (b) (i) - (ii) ofProposition 5.7.7 and Proposition 5.7.8 that we have p - l. Thus it follows fromProposition 5.7.7 and Proposition 5.7.8 that if some unipotent element u ∈ Gacts on L(ω2) as a distinguished unipotent element, then V (ω2) = L(ω2) andS2(V ) ∼= L(ω2)⊕ L(0).

5.7.3 Type Cl

Lemma 5.7.10. Let G = Sp(V ) and let u ∈ G be unipotent. Suppose that u is adistinguished unipotent element of G, so as a K[u]-module V = Vd1 ⊕ · · · ⊕ Vdt,where 0 < d1 < · · · < dt are even (Proposition 2.3.4). If t > 1, then u acts onS2(V ) with repeated blocks.

Proof. Suppose that t > 1 and that u acts on S2(V ) with no repeated blocks.Now as a K[u]-module, S2(V ) has S2(Vd1)⊕· · ·⊕S2(Vdt) as a direct summand byLemma 3.4.9. Therefore u acts on S2(Vdi) with no repeated blocks for all i, andby Proposition 3.5.3, each S2(Vdi) decomposes as in characteristic 0. In particular,since all di are even, for all i we have a block of size 3 in S2(Vdi) (Proposition 3.4.4).Therefore u acts on S2(V ) with ≥ 2 Jordan blocks of size 3, contradiction.

Proposition 5.7.11 (Type Cl). Let G be the symplectic group Sp(V ), wheredimV = 2l (l ≥ 2). Then


(b) A regular unipotent element of G acts as a distinguished unipotent element inL(2ω1) if and only if one of the following holds:

(i) p ≥ 4l − 1,

(ii) 2l = bpk + pk±12 , where k ≥ 1 and 0 ≤ b ≤ p−1

2 .

Proof. Recall that S2(V ) ∼= L(2ω1) (Lemma 5.7.1). If u ∈ G is a distinguishedunipotent element that is not regular, then u acts on S2(V ) with repeated blocksby Lemma 5.7.10, and thus the action of u on L(2ω1) is not distinguished. Thisproves (a).

For (b), let u ∈ G be a regular unipotent element, so V ↓ K[u] = V2l (Propo-sition 2.3.3). Then u acts on L(2ω1) as a distinguished unipotent element if andonly if u acts on S2(V ) with no repeated blocks. Thus (b) follows from Proposition3.5.3.

5.8 Representation LG(ω3) for G of classical type(p 6= 2)

Assume that p 6= 2.

Suppose that G is of classical type with rankG ≥ 3. In this section, we willdetermine when a unipotent element u ∈ G acts on L(ω3) as a distinguished uni-potent element. The answer is given by the following proposition. In the statementwe have excluded G = D4, since in this case L(ω3) is a twist of L(ω1) by triality(Proposition 2.10.2 (ii)).

Proposition 5.8.1. Suppose that G is simple of type Al (l ≥ 3), Bl (l ≥ 3),Cl (l ≥ 3) or Dl (l ≥ 5). A unipotent element u ∈ G acts on the irreduciblerepresentation L(ω3) as a distinguished unipotent element if and only if u is aregular unipotent element and G, p are in Table 5.1.

G u of order > p u of order pA3 p = 3 p ≥ 5A5 p = 5 p ≥ 11C3 p = 3, 5 p ≥ 11C4 none p ≥ 17C5 none p ≥ 23B3 p = 5 p ≥ 13

Table 5.1: u ∈ G regular and acts on L(ω3) as a distinguished unipotent element

For the proof, we will use the construction of L(ω3) as a subquotient of ∧3(V ),where V is the natural module for G. Thus in some cases we will then have tocompute the decomposition of the K[u]-mdoule ∧3(Vd). These computations weredone with MAGMA and they are contained in tables at the end of this section.

5.8.1 Types Al, Bl, and Dl

For G of type Al, Bl and Dl, the claim in Proposition 5.8.1 is really a claim aboutthe exterior cube of a unipotent matrix, as seen from the next proposition.

Proposition 5.8.2. Let G = SL(V ) or G = SO(V ) with rankG ≥ 3. ThenL(ω3) ∼= ∧3(V ).

Proof. This is a consequence of [Sei87, 8.1], as noted in [McN98, Proposition 4.2.2].

Now let u ∈ GL(V ) be a unipotent element of order q and recall the notationV1, V2, . . ., Vq from Section 1.4 for indecomposable K[u]-modules.

Lemma 5.8.3. Suppose that u acts on ∧3(Vd) such that all Jordan block sizes aredistinct and all even or all odd. Then d ≤ 8.

Proof. We can consider u as a regular unipotent element of G = Ad−1, such thatV = Vd is the natural module of G. By Proposition 5.8.2 and Lemma 5.1.1 (iii), weknow that the largest Jordan block size of u acting on ∧3(Vd) is ≤ mu(ω3) + 1 =3d−8 (Lemma 2.7.3). One can verify that the polynomial inequality dim∧3(Vd) =(d3

)> (mu(ω3)+2)2

4 holds for all d ≥ 12. Thus by Lemma 5.1.1 (iv), if u acts on∧3(Vd) such that all Jordan block sizes are distinct and all even or all odd, thend < 12.

For 9 ≤ d ≤ 11, the claim is immediately seen from the decompositions givenin Table 5.2.

152 Chapter 5.

Note that when G is of type Al, the irreducible module L(ω3) is self-dual ifand only if l = 5 (Table 4.1). Therefore for type Al the claim of Proposition 5.8.1follows from Theorem 1.1.12, Lemma 5.8.3 and Table 5.2. In particular, we �ndthat a regular unipotent element of G = Al acts on L(ω3) as a distinguishedunipotent element if and only if l = 3, or l = 5 and p ≥ 5; these examples havealso been recorded in Table 1.1.

For G of type Bl or Dl, the claim of Proposition 5.8.1 follows from Proposition5.8.2 and the following lemma.

Lemma 5.8.4. Let G = SO(V ), where dimV ≥ 3. Let u ∈ G be a distinguishedunipotent element, so as a K[u]-module V = Vd1 ⊕ · · · ⊕ Vdt , where the di aredistinct and odd (Proposition 2.3.4). Then u acts on ∧3(V ) as a distinguishedunipotent element if and only if one of the following holds:

(i) V = V3, or V = V1 ⊕ V3, for any p;

(ii) V = V5, for p = 3 or p ≥ 7;

(iii) V = V7, for p = 5 or p ≥ 13.

Proof. If t = 1, then the claim follows immediately from Lemma 5.8.3 and thedata in Table 5.2. Suppose then that t > 1.

For any K[u]-modules W and W ′, we have an isomorphism

∧3(W ⊕W ′) ∼= ∧3(W )⊕ (∧2(W )⊗W ′)⊕ (W ⊗ ∧2(W ′))⊕ ∧3(W ′) (*)

of K[u]-modules (this is an elementary fact, see for example [FH91, B.1, pg. 473]).It follows from (*) that ∧3(Vdi ⊕ Vdj ) is a direct summand of ∧3(V ) for all i 6= j.Therefore it is enough to prove the lemma in the case where t = 2, say V = Vm⊕Vnwith 0 < m < n odd.

Suppose that u acts on ∧3(V ) as a distinguished unipotent element. We willshow that V = V1 ⊕ V3, which will prove the �only if� part of the lemma. Now∧3(V ) has no repeated blocks, so it follows from (*) that ∧3(Vs) and ∧2(Vs)have no repeated blocks for all s ∈ {m,n}. Therefore by Lemma 5.8.3, we havem,n ∈ {1, 3, 5, 7}. By Table 5.2 and Proposition 3.5.3, if ∧3(Vn) or ∧2(Vn) has norepeated blocks for n ∈ {1, 3, 5, 7}, then their decomposition is as in the followingtable.

n ∧2(Vn) ∧3(Vn)

1 0 03 [3] [1]5 [3, 7] [3, 7]7 [3, 7, 11] [1, 5, 7, 9, 13]

Consider �rst 1 < m < n. If (m,n) 6= (3, 5), we see from the table above that∧3(Vn) and ∧3(Vm) have block sizes in common, so it follows from (*) that u actson ∧3(V ) with repeated blocks, contradiction. For (m,n) = (3, 5), we have by (*)and the table above that

∧3(V3 ⊕ V5) ∼= V1 ⊕ (V3 ⊗ V5)⊕ (V3 ⊗ (V3 ⊕ V7))⊕ (V3 ⊕ V7).

However, then V3 ⊗ V5 has no repeated block sizes, so by Lemma 3.3.10 we haveV3 ⊗ V5

∼= V3 ⊕ V5 ⊕ V7. But then ∧3(V3 ⊕ V5) has at least two blocks of size 3,contradiction.

Thus we must have m = 1, and then ∧3(V ) ∼= ∧3(Vn) ⊕ ∧2(Vn) by (*). Ifn ∈ {5, 7}, then it follows from the table above that ∧3(Vn) and ∧2(Vn) haveblock sizes in common. Hence we must have n = 3, so V = V1 ⊕ V3, as claimed.

Finally for the other direction of the lemma, for V = V1⊕V3, we have ∧3(V ) ∼=∧3(V3) ⊕ ∧2(V3) by (*). It is clear that ∧3(V3) ∼= V1 for all p, and ∧2(V3) ∼= V3

for all p (e.g. by Proposition 3.5.3). Hence ∧3(V3 ⊕ V1) ∼= V3 ⊕ V1 for all p. Thiscompletes the proof of the lemma.

5.8.2 Type Cl

In this subsection, let G = Sp(V ), where dimV = 2l for some l ≥ 3. We can �ndL(ω3) as a subquotient of ∧3(V ), as seen by the following lemma.

Lemma 5.8.5. As a G-module, we have

∧3(V ) ∼=

{L(ω3)⊕ L(ω1) if l 6≡ 1 mod p,

L(ω1)/L(ω3)/L(ω1) (uniserial) if l ≡ 1 mod p.

In particular, we have dimL(ω3) ≥(

2l3

)− 4l.

Proof. This is well known, see for example [McN98, Lemma 4.8.2].

Lemma 5.8.6. Suppose that u ∈ G acts on L(ω3) as a distinguished unipotentelement. Then u is regular and 3 ≤ l ≤ 5.

Proof. Now for a regular unipotent element u ∈ G we have mu(ω3) = 3(2l− 3) =6l − 9 by Lemma 2.7.3, and thus by Lemma 5.8.5 we have

dimL(ω3) ≥(

2l

3

)− 4l >

(mu(ω3) + 2)2

4

if l ≥ 7. Therefore by Lemma 5.1.1, no unipotent element of G acts on L(ω3) as adistinguished unipotent element if l ≥ 7.

Consider then 3 ≤ l ≤ 6. In these cases for all non-regular distinguished u ∈ G,one can easily compute their labeled Dynkin diagram using Proposition 2.6.5, andthen compute mu(ω3) using the fact that

ω3 = α1 + 2α2 + 3α3 + · · ·+ 3αl−1 +3

2αl.

For convenience we have listed this information in Table 5.3. One veri�es that

dimL(ω3) <(mu(ω3) + 2)2

4

in all cases, so it follows from Lemma 5.1.1 that a non-regular u does not act onL(ω3) as a distinguished unipotent element.

Finally consider case where u is regular and l = 6, so now V ↓ K[u] = V12. ByLemma 5.8.5, in this case dimL(ω3) = 208 if p 6= 5, and dimL(ω3) = 196 if p = 5.

If p = 5, then u has order 52. But then dimL(ω3) = 196 and (52+1)2

4 = 169, soby Lemma 5.1.1 the element u does not act as a distinguished unipotent elementon L(ω3). Suppose then that p 6= 5. According to Lemma 5.8.5, we have ∧3(V ) ∼=L(ω3) ⊕ V . Now computing the action of u on L(ω3) is a matter of computing∧3(V12), and this is given in Table 5.2. One sees immediately from the table thatin all cases some block of size 6= 12 has multiplicity ≥ 2 in ∧3(V12), and thus udoes not act as a distinguished unipotent element on L(ω3).

154 Chapter 5.

We can now prove Proposition 5.8.1 for G of type Cl. Suppose that u ∈ Gis a distinguished unipotent element and that u acts on L(ω3) as a distinguishedunipotent element. By Lemma 5.8.6, the element u is a regular unipotent elementand 3 ≤ l ≤ 5, so V ↓ K[u] = V2l.

If l ≡ 1 mod p, then we have l = 4, p = 3, and dimL(ω3) = 40 by Lemma

5.8.5. In this case u has order 32 and (32+1)2

4 = 25 < 40, so by Lemma 5.1.1 theaction of u on L(ω3) is not distinguished.

If l 6≡ 1 mod p, then ∧3(V ) ∼= L(ω3) ⊕ V by Lemma 5.8.5. Now taking thedecomposition of ∧3(V2l) for 3 ≤ l ≤ 5 found in Table 5.2, removing a block ofsize 2l gives the decomposition of L(ω3) ↓ K[u]. The result of Proposition 5.8.1 isthen easily veri�ed.

5.8.3 Computations

Let G = SL(V ), where dimV = d ≥ 4. Fix a regular unipotent element u ∈ G,so now V ↓ K[u] = Vd. In Table 5.2, we have given the decomposition of theK[u]-module ∧3(Vd) into indecomposable summands. This table was generatedby a computer calculation, as follows.

We know that ∧3(V ) ∼= L(ω3), and by Lemma 2.7.3 we havemu(ω3) = 3d−9. Itfollows then from Lemma 2.7.9 that for all p > 3d−9, the decomposition of ∧3(Vd)is the same. Thus for any given d, we can �nd all the possible decompositions of∧3(Vd) with a �nite computation: �rst we compute ∧3(Vd) for all 2 3d − 9. For all of the entries in Table 5.2, thesecomputations can be quickly done with the aid of a computer program.

d ∧3(Vd) p4 [4] p ≥ 3

5 [3, 7] p = 3[52] p = 5[3, 7] p ≥ 7

6 [62, 8] p = 3[4, 6, 10] p = 5[6, 72] p = 7[4, 6, 10] p ≥ 11

7 [1, 7, 93] p = 3[1, 5, 7, 9, 13] p = 5[75] p = 7[1, 5, 7, 112] p = 11[1, 5, 7, 9, 13] p ≥ 13

8 [2, 96] p = 3[52, 8, 10, 12, 16] p = 5[6, 72, 8, 142] p = 7[4, 8, 114] p = 11[4, 6, 8, 12, 132] p = 13[4, 6, 8, 10, 12, 16] p ≥ 17

9 [3, 99] p = 3[52, 104, 15, 19] p = 5[3, 72, 9, 11, 142, 19] p = 7[7, 117] p = 11[3, 7, 9, 135] p = 13[3, 72, 9, 11, 13, 172] p = 17[3, 72, 9, 11, 13, 15, 19] p ≥ 19

d ∧3(Vd) p10 [2, 96, 10, 183] p = 3

[108, 202] p = 5[4, 72, 102, 12, 14, 16, 18, 22] p = 7[10, 1110] p = 11[6, 10, 138] p = 13[4, 6, 8, 102, 14, 174] p = 17[4, 6, 8, 102, 12, 14, 18, 192] p = 19[4, 6, 8, 102, 12, 14, 16, 18, 22] p ≥ 23

11 [1, 7, 94, 122, 184, 25] p = 3[1, 5, 9, 104, 153, 19, 21, 25] p = 5[1, 73, 9, 11, 13, 142, 17, 19, 21, 25] p = 7[1115] p = 11[9, 1312] p = 13[1, 5, 7, 9, 11, 13, 177] p = 17[1, 5, 7, 92, 11, 13, 15, 195] p = 19[1, 5, 7, 92, 11, 132, 15, 17, 19, 232] p = 23[1, 5, 7, 92, 11, 132, 15, 17, 19, 21, 25] p ≥ 29

12 [62, 8, 126, 183, 242, 26] p = 3[4, 6, 8, 102, 122, 152, 16, 18, 20, 24, 252] p = 5[74, 12, 145, 16, 212, 24, 28] p = 7[10, 1110, 12, 224] p = 11[12, 1316] p = 13[4, 8, 10, 12, 16, 1710] p = 17[4, 6, 8, 10, 122, 16, 198] p = 19[4, 6, 8, 102, 122, 14, 162, 20, 234] p = 23[4, 6, 8, 102, 122, 14, 162, 18, 20, 22, 24, 28] p ≥ 29

Table 5.2: Decomposition of ∧3(Vd) for 4 ≤ d ≤ 12.

5.9. Representation LG(3ω1) for G of classical type (p 6= 2, 3) 155

G u Dynkin diagram of u mu(ω3) dimL(ω3)

C6 [2, 10] 222202 21 208 (p 6= 5), 196 (p = 5)C6 [4, 8] 220202 15 208 (p 6= 5), 196 (p = 5)C6 [2, 4, 6] 202002 11 208 (p 6= 5), 196 (p = 5)C5 [2, 8] 22202 15 110C5 [4, 6] 20202 11 110C4 [2, 6] 2202 9 48 (p 6= 3), 40 (p = 3)C3 [2, 4] 202 5 14

Table 5.3: Values of mu(ω3) for non-regular distinguished unipotent elements u inCl, where 3 ≤ l ≤ 6.

5.9 Representation LG(3ω1) for G of classical type(p 6= 2, 3)

Assume that p 6= 2, 3.

Suppose that G is simple of classical type. In this section, we will determinewhen a unipotent element u ∈ G acts on L(3ω1) as a distinguished unipotentelement. The method used will be similar to that found in previous sections, inparticular Section 5.8.

The answer is given by the following proposition. Note that for type Al therepresentation L(3ω1) is not self-dual if l ≥ 2, so by Theorem 1.1.12 we only needto consider G of type Bl, Cl or Dl.

Proposition 5.9.1. Suppose that G is a simple algebraic group of type Bl (l ≥2), Cl (l ≥ 2) or Dl (l ≥ 4). A unipotent element u acts on the irreduciblerepresentation L(3ω1) as a distinguished unipotent element if and only if u is aregular unipotent element and G, p are as in Table 5.1.

G u of order > p u of order pC2 none p ≥ 11C3 none p ≥ 17B2 none p ≥ 13B3 none p ≥ 19

Table 5.1: u ∈ G regular and acts on L(3ω1) as a distinguished unipotent element

The proof of Proposition 5.9.1 will be given in the next two subsections.


In this subsection, let G = SO(V ), where dimV = 2l + 1 (l ≥ 2) or dimV =2l (l ≥ 4). We can �nd the irreducible module L(3ω1) as a subquotient of thesymmetric cube S3(V ), as seen in the following lemma which is well known.

Lemma 5.9.2. As a G-module, we have

S3(V ) ∼=

{L(3ω1)⊕ L(ω1) if p | dimV + 2,

L(ω1)/L(3ω1)/L(ω1) (uniserial) if p - dimV + 2.

156 Chapter 5.

In particular, we have dimL(3ω1) ≥(

dimV+23

)− 2 dimV .

Proof. See [McN98, Proposition 4.7.4].

Lemma 5.9.3. Let u ∈ G be a distinguished unipotent element. If u acts onL(3ω1) as a distinguished unipotent element, then u is regular and dimV = 5,dimV = 7 or dimV = 9.

Proof. Write n = dimV . Let u ∈ G be a regular unipotent element. By Lemma2.7.3 we have

mu(3ω1) = 3mu(ω1) =

{3(n− 1) if n is odd,

3(n− 2) if n is even.

Furthermore, by Lemma 5.9.2, we have dimL(3ω1) ≥(n+2

3

)− 2n. Now

dimL(3ω1) ≥(n+ 2

3

)− 2n >

(mu(3ω1) + 2)2

4(*)

if n = 8 or n ≥ 10, so it follows from Lemma 5.1.1 that no unipotent element ofG acts on L(3ω1) as a distinguished unipotent element if n = 8 or n ≥ 10.

If n ≤ 7, then all distinguished unipotent elements of G are regular by Proposi-tion 2.3.3. Note that we are assuming n ≥ 8 if n is odd, so we are done in this case.What remains is to consider the case where n = 9. If u ∈ G is a distinguished uni-potent element that is not regular, then by Proposition 2.3.3 the element u lies inthe unipotent class labeled by [1, 3, 5]. In this case mu(3ω1) = 3mu(ω1) = 12, anddimL(3ω1) ≥ 147 by Lemma 5.9.2, hence inequality (*) holds. By Lemma 5.1.1,the element u does not act on L(3ω1) as a distinguished unipotent element.

By Lemma 5.9.3, what remains is to verify Proposition 5.9.1 in the cases whereu ∈ G is a regular unipotent element and dimV = 5, dimV = 7 or dimV = 9.If p - dimV + 2, then S3(V ) ∼= L(3ω1) ⊕ V by Lemma 5.9.2 and then the claimfollows from the data in Table 5.2. Suppose then that p | dimV + 2. Then eitherdimV = 5 and p = 7, or dimV = 9 and p = 11. In both cases u has order p and

dimL(3ω1) > (p+1)2

4 , so u does not act as a distinguished unipotent element onL(3ω1) by Lemma 5.1.1. This completes the proof of Proposition 5.9.1 for G oftype Bl and Dl.

5.9.2 Type Cl

In this subsection, let G = Sp(V ), where dimV = 2l (l ≥ 2). In this case we haveL(3ω1) ∼= S3(V ) (see e.g. [Sei87, 1.14 and 8.1 (c)] or [McN98, Proposition 4.2.2(h)]), so dimL(3ω1) =

(2l+2

3

).

Lemma 5.9.4. Let u ∈ G be a unipotent. If u acts on L(3ω1) as a distinguishedunipotent element, then u is regular and l ≤ 4.

Proof. Let u ∈ G be a regular unipotent element. Then mu(3ω1) = 3mu(ω1) =3(2l − 1) = 6l − 3 by Lemma 2.7.3. Thus we have

dimL(3ω1) =

(2l + 2

3

)>

(mu(3ω1) + 2)2

4(*)

5.9. Representation LG(3ω1) for G of classical type (p 6= 2, 3) 157

for all l ≥ 5. It follows then from Lemma 5.1.1 that no unipotent element of Gacts on L(3ω1) as a distinguished unipotent element if l ≥ 5.

Suppose then that l ≤ 4 and let u ∈ G be a distinguished unipotent element.In this case, it is straightforward to verify that the inequality (*) holds if u ∈G is not a regular unipotent element. Indeed, if l = 2 then there are no non-regular distinguished unipotent elements. If l = 3, then u is in class labeled by[2, 4] and dimL(3ω1) = 56, mu(3ω1) = 9. If l = 4, then u is in class labeledby [2, 6] and dimL(3ω1) = 120, mu(3ω1) = 15. Therefore by Lemma 5.1.1, anon-regular unipotent element of G does not act on L(3ω1) as a distinguishedunipotent element.

The claim of Proposition 5.9.1 for G of type Cl follows now from Lemma 5.9.4and the information given in Table 5.2.

5.9.3 Computations

Let G = SL(V ), where dimV = d ≥ 4. Fix a regular unipotent element u ∈ G, sonow V ↓ K[u] = Vd. In Table 5.2, we have given the decomposition of the K[u]-module S3(Vd) into indecomposable summands. This table was generated by acomputer calculation, similarly to Table 5.2. Here we use the fact that S3(V ) ∼=L(3ω1) for p > 3, and mu(3ω1) = 3d − 3. It follows as in subsection 5.8.3 thatit will su�ce to compute S3(Vd) for all 2 3d− 3.

d S3(Vd) p4 [2, 3, 6, 9] p = 3

[54] p = 5[6, 72] p = 7[4, 6, 10] p ≥ 11

5 [3, 5, 93] p = 3[57] p = 5[75] p = 7[1, 5, 7, 112] p = 11[1, 5, 7, 9, 13] p ≥ 13

6 [3, 8, 95] p = 3[54, 102, 16] p = 5[78] p = 7[4, 8, 114] p = 11[4, 6, 8, 12, 132] p = 13[4, 6, 8, 10, 12, 16] p ≥ 17

7 [3, 99] p = 3[52, 7, 102, 13, 15, 19] p = 5[712] p = 7[7, 117] p = 11[3, 7, 9, 135] p = 13[3, 72, 9, 11, 13, 172] p = 17[3, 72, 9, 11, 13, 15, 19] p ≥ 19

d S3(Vd) p8 [3, 913] p = 3

[4, 6, 8, 102, 12, 152, 202] p = 5[78, 143, 22] p = 7[10, 1110] p = 11[6, 10, 138] p = 13[4, 6, 8, 102, 14, 174] p = 17[4, 6, 8, 102, 12, 14, 18, 192] p = 19[4, 6, 8, 102, 12, 14, 16, 18, 22] p ≥ 23

9 [3, 918] p = 3[1, 5, 9, 104, 152, 204] p = 5[75, 9, 144, 19, 21, 25] p = 7[1115] p = 11[9, 1312] p = 13[1, 5, 7, 9, 11, 13, 177] p = 17[1, 5, 7, 92, 11, 13, 15, 195] p = 19[1, 5, 7, 92, 11, 132, 15, 17, 19, 232] p = 23[1, 5, 7, 92, 11, 132, 15, 17, 19, 21, 25] p ≥ 29

10 [4, 913, 184, 27] p = 3[108, 207] p = 5[6, 72, 8, 10, 12, 143, 16, 18, 212, 24, 28] p = 7[1120] p = 11[12, 1316] p = 13[4, 8, 10, 12, 16, 1710] p = 17[4, 6, 8, 10, 122, 16, 198] p = 19[4, 6, 8, 102, 122, 14, 162, 20, 234] p = 23[4, 6, 8, 102, 122, 14, 162, 18, 20, 22, 24, 28] p ≥ 29

Table 5.2: Decomposition of S3(Vd) for 4 ≤ d ≤ 10.

158 Chapter 5.

5.10 Spin representations (p 6= 2)

Assume that p 6= 2.

Let G be simple of type Bl (l ≥ 3) or type Dl (l ≥ 4). In this section, wedetermine when a unipotent element u of G acts as a distinguished unipotentelement on the irreducible representation LG(ωl).

In fact, we do a bit more: we determine when a distinguished unipotent elementu of G acts on LG(ωl) such that all Jordan block sizes are distinct and all even orall odd (recall that LDl(ωl) is not self-dual if l is odd). The precise result is thefollowing proposition.

Proposition 5.10.1. Suppose that G is simple of type Bl (l ≥ 3) or Dl (l ≥ 4).A distinguished unipotent element u ∈ G acts on the irreducible representationLG(ωl) such that all Jordan block sizes are distinct and all even or all odd if andonly if G and p are as in Table 5.1.

G Class of u u of order > p u of order pD4 regular p = 3, 5 p ≥ 7D4 [5, 3] p = 3 p ≥ 5D5 regular p = 3, 5, 7 p ≥ 11D5 [7, 3] p = 3, 5 p ≥ 11D6 regular p = 3, 5, 7 p ≥ 17D6 [9, 3] p = 7 p ≥ 13D7 regular p = 3, 5 p ≥ 23D7 [11, 3] p = 7 p ≥ 17D8 regular p = 11 p ≥ 29D9 regular none p ≥ 37

B3 regular p = 3, 5 p ≥ 7B4 regular p = 3, 5, 7 p ≥ 11B5 regular p = 3, 5, 7 p ≥ 17B6 regular p = 3, 5 p ≥ 23B7 regular p = 11 p ≥ 29B8 regular none p ≥ 37

Table 5.1: For G = Bl and G = Dl: all cases where a distinguished unipotentelement u ∈ G acts on LG(ωl) such that all Jordan block sizes are distinct and alleven or all odd.

Lemma 5.10.2. Let G = Dl, l ≥ 5 and let u ∈ G be a distinguished unipotentelement. If u acts on LG(ωl) such that all Jordan block sizes are distinct and alleven or all odd, then one of the following holds.

(i) The element u is a regular unipotent element and l ≤ 10.

(ii) The element u lies in the class [2l − 3, 3] of G and l ≤ 8.

Proof. Let u ∈ G be a regular unipotent element. Thenmu(ωl) = l(l−1)2 by Lemma

2.7.3. From this one can verify that

dimLG(ωl) = 2l−1 >(mu(ωl) + 2)2

4(*)

5.10. Spin representations (p 6= 2) 159

for all l ≥ 11. Thus if some unipotent element of G acts on L(ωl) such that allJordan block sizes are distinct and all even or all odd, we have l ≤ 10 by Lemma5.1.1 (see Remark 5.1.2).

Suppose then that l ≤ 10 and let u ∈ G be a distinguished unipotent element.In this situation, it is straightforward to verify from Table 5.2 that the inequality(*) holds, except when we are in case (i) or (ii) of the claim. Thus the claim followsagain from Lemma 5.1.1.

With Lemma 5.10.2, the claim of Proposition 5.10.1 for G of type Dl, l ≥ 5,follows from the data given in Table 5.5 and Table 5.4. The Jordan block sizesgiven in these tables were computed with MAGMA (Section 2.9). In the case wherel = 4, Proposition 5.10.1 for G = Dl follows from Proposition 2.10.2 (i) sinceLG(ω4) is a twist of the natural representation by a triality graph automorphism.

For G of type Bl, we can reduce our computations to the case of type Dl withthe following lemma.

Lemma 5.10.3 ([Sei87, Table 1, IV1]). Let l ≥ 4 and consider Bl−1 < Dl natu-rally embedded. Then LDl(ωl) ↓ Bl−1

∼= LBl−1(ωl−1).

Note that Bl−1 naturally embedded in Dl contains a regular unipotent elementof Dl. Thus Proposition 5.10.1 for G of type Bl follows from the next lemma,combined with Lemma 5.10.3, and Proposition 5.10.1 for type Dl.

Lemma 5.10.4. Let G = Bl, l ≥ 3 and let u ∈ G be a distinguished unipotentelement. If u acts on LG(ωl) such that all Jordan block sizes are distinct and alleven or all odd, then u is a regular unipotent element and l ≤ 9.

Proof. We proceed as in Lemma 5.10.2. For a regular unipotent element u ∈ G,we have mu(ωl) = l(l+1)

2 by Lemma 2.7.3. From this we see that

dimLG(ωl) = 2l >(mu(ωl) + 2)2

4(*)

for all l ≥ 10. Thus if some unipotent element of G acts on L(ωl) such that allJordan block sizes are distinct and all even or all odd, we have l ≤ 9 by Lemma5.1.1.

Suppose then that l ≤ 9 and let u ∈ G be a distinguished unipotent element.As in Lemma 5.10.2, one veri�es from Table 5.3 that the inequality (*) holds forall non-regular distinguished unipotent elements of G. Thus the claim follows fromLemma 5.1.1.

160 Chapter 5.

G u Labeled diagram of u mu(ωl) dimLG(ωl)

D10 [19, 1] 2222222222 45 512[17, 3] 2222222022 37 512[15, 5] 2222202022 31 512[13, 7] 2220202022 27 512[11, 9] 2020202022 25 512[11, 5, 3, 1] 2220200200 19 512[9, 7, 3, 1] 2020200200 17 512

D9 [17, 1] 222222222 36 256[15, 3] 222222022 29 256[13, 5] 222202022 24 256[11, 7] 220202022 21 256[9, 5, 3, 1] 220200200 14 256

D8 [15, 1] 22222222 28 128[13, 3] 22222022 22 128[11, 5] 22202022 18 128[9, 7] 20202022 16 128[7, 5, 3, 1] 20200200 10 128

D7 [13, 1] 2222222 21 64[11, 3] 2222022 16 64[9, 5] 2202022 13 64

D6 [11, 1] 222222 15 32[9, 3] 222022 11 32[7, 5] 202022 9 32

D5 [9, 1] 22222 10 16[7, 3] 22022 7 16

Table 5.2: Values of mu(ωl) for distinguished unipotent elements u in Dl, where5 ≤ l ≤ 10.


G u Labeled diagram of u mu(ωl) dimLG(ωl)

B9 [19] 222222222 45 512[15, 3, 1] 222222020 29 512[13, 5, 1] 222202020 24 512[11, 7, 1] 220202020 21 512[11, 5, 3] 222020020 19 512[9, 7, 3] 202020020 17 512

B8 [17] 22222222 36 256[13, 3, 1] 22222020 22 256[11, 5, 1] 22202020 18 256[9, 7, 1] 20202020 16 256[9, 5, 3] 22020020 14 256

B7 [15] 2222222 28 128[11, 3, 1] 2222020 16 128[9, 5, 1] 2202020 13 128[7, 5, 3] 2020020 10 128

B6 [13] 222222 21 64[9, 3, 1] 222020 11 64[7, 5, 1] 202020 9 64

B5 [11] 22222 15 32[7, 3, 1] 22020 7 32

B4 [9] 2222 10 16[5, 3, 1] 2020 4 16

B3 [7] 222 6 8

Table 5.3: Values of mu(ωl) for distinguished unipotent elements u in Bl, where3 ≤ l ≤ 9.

162 Chapter 5.

G LG(ωl) ↓ K[u] pD10 [4, 94, 184, 22, 2714] p = 3

[102, 12, 152, 2518] p = 5[6, 72, 145, 16, 214, 22, 284, 32, 352, 40, 46] p = 7[10, 116, 14, 228, 28, 334, 40, 46] p = 11[4, 6, 10, 134, 16, 182, 20, 222, 263, 28, 30, 32, 34, 36, 40, 46] p = 13[10, 1714, 26, 347] p = 17[18, 1926] p = 19[4, 10, 16, 22, 2320] p = 23[4, 6, 102, 14, 162, 18, 20, 22, 28, 2912] p = 29[4, 6, 102, 12, 14, 16, 182, 20, 22, 24, 28, 3110] p = 31[4, 6, 102, 12, 14, 162, 182, 20, 222, 24, 26, 28, 30, 32, 36, 374] p = 37[4, 6, 102, 12, 14, 162, 182, 20, 222, 24, 26, 282, 30, 32, 34, 40, 412] p = 41[4, 6, 102, 12, 14, 162, 182, 20, 222, 24, 26, 282, 30, 32, 34, 36, 432] p = 43[4, 6, 102, 12, 14, 162, 182, 20, 222, 24, 26, 282, 30, 32, 34, 36, 40, 46] p ≥ 47

D9 [94, 13, 184, 275] p = 3[1, 7, 102, 15, 17, 21, 257] p = 5[1, 7, 9, 11, 142, 17, 19, 21, 23, 25, 27, 31, 37] p = 7[1, 7, 113, 15, 17, 19, 222, 25, 27, 31, 37] p = 11[1, 7, 9, 133, 17, 19, 21, 23, 262, 31, 37] p = 13[1, 1715] p = 17[9, 1913] p = 19[1, 7, 11, 13, 17, 239] p = 23[1, 7, 9, 11, 13, 15, 17, 19, 23, 25, 294] p = 29[1, 7, 9, 11, 13, 15, 17, 19, 21, 23, 27, 313] p = 31[1, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 31, 37] p ≥ 37

D8 [7, 9, 122, 182, 25, 27] p = 3[104, 202, 23, 25] p = 5[72, 142, 15, 212, 29] p = 7[5, 9, 11, 15, 17, 19, 23, 29] p = 11[5, 134, 19, 262] p = 13[9, 177] p = 17[5, 11, 17, 195] p = 19[5, 9, 11, 15, 19, 233] p = 23[5, 9, 11, 15, 17, 19, 23, 29] p ≥ 29

D7 [4, 10, 12, 16, 22] p = 3[4, 10, 12, 16, 22] p = 5[72, 142, 22] p = 7[4, 112, 16, 22] p = 11[12, 134] p = 13[4, 10, 16, 172] p = 17[4, 10, 12, 192] p = 19[4, 10, 12, 16, 22] p ≥ 23

D6 [6, 10, 16] p = 3[6, 10, 16] p = 5[6, 10, 16] p = 7[10, 112] p = 11[6, 132] p = 13[6, 10, 16] p ≥ 17

D5 [7, 9] p = 3[5, 11] p = 5[5, 11] p = 7[5, 11] p ≥ 11

Table 5.4: Action of a regular unipotent element u of G = Dl on LG(ωl), where5 ≤ l ≤ 10.


G LG(ωl) ↓ K[u] pD8 [3, 5, 9, 11, 122, 15, 17, 21, 23] p = 3

[3, 5, 102, 11, 13, 15, 17, 21, 23] p = 5[74, 144, 21, 23] p = 7[3, 5, 114, 15, 17, 222] p = 11[11, 139] p = 13[3, 5, 9, 11, 15, 175] p = 17[3, 5, 9, 112, 13, 194] p = 19[3, 5, 9, 112, 13, 15, 17, 21, 23] p ≥ 23

D7 [62, 9, 11, 15, 17] p = 3[5, 7, 102, 15, 17] p = 5[5, 7, 9, 11, 15, 17] p = 7[9, 115] p = 11[5, 7, 134] p = 13[5, 7, 9, 11, 15, 17] p ≥ 17

G K[u] ↓ LG(ωl) pD6 [6, 8, 92] p = 3

[52, 10, 12] p = 5[4, 6, 10, 12] p = 7[4, 6, 112] p = 11[4, 6, 10, 12] p ≥ 13

D5 [2, 6, 8] p = 3[2, 6, 8] p = 5[2, 72] p = 7[2, 6, 8] p ≥ 11

Table 5.5: Action of a unipotent element u in conjugacy class [2l− 3, 3] of G = Dl

on LG(ωl), where 5 ≤ l ≤ 8.

164 Chapter 5.

5.11 Spin representations (p = 2)

Assume that p = 2.

In this section we will determine for G of type Cl (l ≥ 3) and Dl (l ≥ 4)when a unipotent element u ∈ G acts as a distinguished unipotent element onthe irreducible representation LG(ωl) (and also LG(ωl−1) if G has type Dl). Theanswer is given by Proposition 5.11.3 and Proposition 5.11.4 below; it is alsorecorded in Table 1.3.

Since we are in characteristic two, we can and we will consider Dl < Cl as thesubsystem subgroup generated by the short root subgroups. The following lemmaallows us to reduce our computations to the case where G is of type Cl.

Lemma 5.11.1. LCl(ωl) ↓ Dl∼= LDl(ωl)⊕ LDl(ωl−1).

Proof. This is a consequence of [For96, Theorem 3.3, U6 in Table II].

Corollary 5.11.2. Let l ≥ 4 be even. Let u ∈ Dl < Cl be a distinguished unipotentelement. Suppose that the image of u lies in the conjugacy class labeled by

(d(1)n1

ε(d(1)), d(2)n2

ε(d(2)), . . . , d(t)ntε(d(t)))

of Sp(LCl(ωl)) (Corollary 2.4.7). Then the image of u in Sp(LDl(ωl)), and theimage of u in Sp(LDl(ωl−1)) lies in the conjugacy class labeled by

(d(1)n1/2ε(d(1)), d(2)

n2/2ε(d(2)), . . . , d(t)

nt/2ε(d(t))).

Proof. By Lemma 5.11.1, the module LCl(ωl) has a Dl-submodule W isomorphicto LDl(ωl). Note that W

∼= W ∗ because l is even (see Table 4.1). We claim thatW is a non-degenerate subspace with respect to any non-degenerate Cl-invariantform on LCl(ωl). If not, then W

⊥ ∩W 6= 0, so W ∩W⊥ = W and W ⊆W⊥ sinceW is irreducible as a Dl-module. But then V/W⊥ ∼= W ∗ ∼= W and so W occurstwice as a Dl-composition factor of LCl(ωl), contradicting Lemma 5.11.1.

Let τ : Dl → Dl be the usual automorphism of Dl induced by the graphautomorphism swapping the two end nodes of the Dynkin diagram of type Dl.Now LDl(ωl−1)τ ∼= LDl(ωl), so it follows from Proposition 2.10.2 (ii) that theconjugacy class of the image of u in Sp(LDl(ωl)) and Sp(LDl(ωl−1)) is the same.Thus by Lemma 5.11.1 each ni is even, and the image of u in Sp(LDl(ωl)) andSp(LDl(ωl−1)) lies in the conjugacy class labeled by

(d(1)n1/2ε1

, d(2)n2/2ε2

, . . . , d(t)nt/2εt)

for some εi ∈ {0, 1}.Finally, what we did in the �rst paragraph shows that LCl(ωl) ↓ Dl = LDl(ωl)⊕

LDl(ωl−1) as an orthogonal direct sum. Therefore we can conclude that as aK[u]-module, LCl(ωl) has V (d(i)) as an orthogonal direct summand if and only ifLDl(ωl) has V (d(i)) as an orthogonal direct summand. By Lemma 2.4.5, we haveεi = ε(d(i)) for all i.

Proposition 5.11.3. Suppose that G is of type Cl (l ≥ 3) and let u ∈ G be adistinguished unipotent element. Then u acts on L(ωl) as a distinguished unipotentelement if and only if u occurs as one of the boldface entries in Table 5.1.

5.11. Spin representations (p = 2) 165

Proof. Suppose that u acts on L(ωl) as a distinguished unipotent element. ByProposition 5.2.4, we have 3 ≤ l ≤ 7 or l = 9. For these cases, the orthogonaldecomposition of L(ωl) ↓ K[u] can be computed with MAGMA (Section 2.9), andthe decompositions are given in Table 5.1 and Table 5.2 (using the labeling givenin Corollary 2.4.7); one can easily verify the claim from the tables.

Proposition 5.11.4. Suppose that G is of type Dl (l ≥ 4) and let u ∈ G be adistinguished unipotent element. Then u acts on L(ωl) (or L(ωl−1)) as a distin-guished unipotent element if and only if l = 4 and u is in class (21, 61) or (42

1), orl = 6 and u is in class (21, 101).

Proof. Suppose that u acts on L(ωl) or L(ωl−1) as a distinguished unipotent ele-ment. Now l ∈ {4, 6, 8, 10} by Proposition 5.2.4. With Corollary 5.11.2 and Table5.1 - 5.3, one can verify that the proposition holds. We illustrate this with anexample and leave the rest to the reader. Consider a regular unipotent elementu of G when l = 10. Now u lies in the unipotent conjugacy class of C10 labeledby (21, 181), so we �nd in Table 5.3 that its image in Sp(LC10(ω10)) lies in theunipotent class labeled by

(221, 6

21, 102

1, 1421, 1624

0 , 1821, 222

1, 2621, 302

1, 32120 ).

Then by Corollary 5.11.2, the image of u in Sp(LG(ω10)) lies in the unipotent classlabeled by

(211, 61, 101, 141, 1612

0 , 181, 221, 261, 301, 3260).

In particular, we �nd that u does not act as a distinguished unipotent element onL(ω10).

Remark 5.11.5. In our proofs we have relied on tables 5.1 - 5.3, which werecomputed with MAGMA (Section 2.9). We note that most of the entries are notnecessary for our proof, but we have included them for completeness. For example,when G = C9, a non-regular distinguished unipotent element u ∈ G has order at

most 24; therefore (|u|+1)2

4 < 29 = dimLG(ω9) and so u does not act on LG(ω9) asa distinguished unipotent element by Lemma 5.2.1. Hence for G = C9, we actuallyonly have to compute the action of a regular unipotent element u ∈ G on LG(ω9).For G = Cl with 3 ≤ l ≤ 7, similar arguments can rule out most distinguishedunipotent classes without computing their action on LG(ωl). Similar remarks alsoapply in the case where G = Dl.

Furthermore, when G = Cl, the computations given in tables 5.1 - 5.3 can bereduced to the case where u ∈ G a regular unipotent element. For this, supposethat G = Sp(V ). If u ∈ G is a distinguished unipotent element which is notregular, then there is an orthogonal decomposition V = V1 ⊕ · · · ⊕ Vt such that uis contained in Sp(V1)×· · ·×Sp(Vt) and acts on each Vi with a single Jordan block.Set dimVi = 2di, so now Sp(Vi) is simple of type Cdi . We can write u = u1 · · ·ut,where ui ∈ Sp(Vi) is a regular unipotent element of Sp(Vi). It follows from [Sei87,Theorem 4.1] that the restriction of LG(ωl) to Sp(V1)×· · ·×Sp(Vt) is irreducible,and isomorphic to the outer tensor product

LCd1 (ωd1)⊗ · · · ⊗ LCdt (ωdt).

Thus if we know the orthogonal decompositions LCdi (ωdi) ↓ K[ui] for all 1 ≤ i ≤ t,by computing the orthogonal decomposition of their tensor product (see Section3.6) we can �nd the orthogonal decomposition of LG(ωl) ↓ K[u].

166 Chapter 5.

l Class of u Action of u on L(ωl)

3 regular 21,613 (21, 41) 42

0

4 regular 8214 (21,61) 221,6

21

4 (421) 44

1

4 (221, 41) 44

0

5 regular 21,61,101,1415 (21, 81) 84

0

5 (41, 61) 440, 8

20

5 (221, 61) 24

1, 641

5 (21, 421) 48

0

6 regular 421, 122

1, 1620

6 (21,101) 221,621,10

21,14

21

6 (41, 81) 880

6 (221, 81) 88

0

6 (621) 24

1, 641, 8

40

6 (21, 41, 61) 480, 8

40

6 (221, 4

21) 416

0

7 regular 21, 61, 101, 141, 1660

7 (21, 121) 440, 124

0, 1640

7 (41, 101) 440, 8

40, 124

0, 1620

7 (221, 101) 24

1, 641, 104

1, 1441

7 (61, 81) 8160

7 (21, 41, 81) 8160

7 (21, 621) 28

1, 681, 8

80

7 (421, 61) 416

0 , 880

7 (221, 41, 61) 416

0 , 880

Table 5.1: For G = Cl with 3 ≤ l ≤ 7, actions of distinguished unipotent elementsu ∈ G on LG(ωl).

5.11. Spin representations (p = 2) 167

l Conjugacy class Action of u on LG(ωl)

8 (161) 16161

8 (21, 141) 221, 6

21, 102

1, 1421, 1612

0

8 (41, 121) 481, 128

1, 1680

8 (221, 121) 48

0, 1280, 168

0

8 (61, 101) 241, 6

41, 8

80, 104

1, 1441, 164

0

8 (21, 41, 101) 480, 8

80, 128

0, 1640

8 (821) 832

1

8 (21, 61, 81) 8320

8 (421, 81) 832

0

8 (221, 41, 81) 832

0

8 (41, 621) 416

0 , 8240

8 (221, 6

21) 216

1 , 6161 , 8

160

8 (21, 421, 61) 432

0 , 8160

9 (181) 21, 61, 101, 141, 16120 , 181, 221, 261, 301, 326

0

9 (21, 161) 16320

9 (41, 141) 440, 8

40, 124

0, 16260

9 (221, 141) 24

1, 641, 104

1, 1441, 1624

0

9 (61, 121) 480, 8

80, 128

0, 16200

9 (21, 41, 121) 4160 , 1216

0 , 16160

9 (81, 101) 8320 , 1616

0

9 (21, 61, 101) 281, 6

81, 8

160 , 108

1, 1481, 168

0

9 (421, 101) 416

0 , 8160 , 1216

0 , 1680

9 (221, 41, 101) 416

0 , 8160 , 1216

0 , 1680

9 (21, 821) 864

0

9 (41, 61, 81) 8640

9 (221, 61, 81) 864

0

9 (21, 421, 81) 864

0

9 (21, 41, 621) 432

0 , 8480

9 (221, 4

21, 61) 464

0 , 8320

Table 5.2: For G = Cl with l ∈ {8, 9}, actions of distinguished unipotent elementsu ∈ G on LG(ωl).

168 Chapter 5.

Conjugacy class Action of u on LG(ω10)

(201) 441, 124

1, 1680, 204

1, 2841, 3220

0

(21, 181) 221, 6

21, 102

1, 1421, 1624

0 , 1821, 222

1, 2621, 302

1, 32120

(41, 161) 16640

(221, 161) 1664

0

(61, 141) 241, 6

41, 8

80, 104

1, 1441, 1652

0

(21, 41, 141) 480, 8

80, 128

0, 16520

(81, 121) 8320 , 1648

0

(21, 61, 121) 4160 , 8

160 , 1216

0 , 16400

(421, 121) 432

1 , 12321 , 1632

0

(221, 41, 121) 432

0 , 12320 , 1632

0

(1021) 28

1, 681, 8

160 , 108

1, 1481, 1640

0

(21, 81, 101) 8640 , 1632

0

(41, 61, 101) 4160 , 8

480 , 1216

0 , 16240

(221, 61, 101) 216

1 , 6161 , 8

320 , 1016

1 , 14161 , 1616

0

(21, 421, 101) 432

0 , 8320 , 1232

0 , 16160

(41, 821) 8128

0

(221, 8

21) 8128

0

(621, 81) 8128

0

(21, 41, 61, 81) 81280

(221, 4

21, 81) 8128

0

(421, 6

21) 464

0 , 8960

(221, 41, 6

21) 464

0 , 8960

Table 5.3: For G = C10, actions of distinguished unipotent elements u ∈ G onLG(ω10).

5.12. Fundamental irreducible representations (p 6= 2) 169

5.12 Fundamental irreducible representations (p 6= 2)

Assume that p 6= 2.

In this section we determine when a unipotent element u ∈ G of a simplealgebraic group G acts as a distinguished unipotent element on a irreducible re-presentation of fundamental highest weight ωi. The result is given by the followingproposition.

Proposition 5.12.1. Assume that p 6= 2. Let G be a simple algebraic group ofrank l ≥ 2 and let u ∈ G be a distinguished unipotent element. Then u acts onL(ωi) (1 ≤ i ≤ l) as a distinguished unipotent element if and only if G, u, p andωi are as in Table 5.1 or Table 5.2, where λ is given up to graph automorphism ofG.

Remark 5.12.2. Let u ∈ G be a distinguished unipotent element of order p andlet λ = ωi. Then Proposition 5.12.1 implies the following statement. The unipotentelement u acts on LG(λ) as a distinguished unipotent element if and only if G, u,and λ occur in Table 5.1 or Table 5.2 (up to a graph automorphism applied to λ),and p > mu(λ). In Section 5.13, we will generalize this fact (Proposition 5.13.1).

We will prove Proposition 5.12.1 for each simple type in the subsections thatfollow.

5.12.1 Type Al, Bl and Dl

Suppose that G is of type Al, Bl or Dl. For the representation L(ωl) of Bl andDl, the claim of Proposition 5.12.1 follows from Proposition 5.10.1. Then since thehalf-spin representation L(ωl−1) ofDl is a twist of L(ωl) by a graph automorphism,by Proposition 2.10.2 (ii) the claim follows also for the representation L(ωl−1) ofDl. Suppose then that LG(ωi) is not a spin or a half-spin representation. Thenit is well known (see for example [Sei87, 8.1] or [McN98, Proposition 4.2.2]) thatLG(ωi) ∼= VG(ωi) ∼= ∧i(V ), where V is the natural representation of G. For LG(ω1)the claim of Proposition 5.12.1 is clear. For L(ω2) and L(ω3), Proposition 5.12.1follows from 5.5.5, 5.5.6 and 5.8.1.

We proceed next to show that when i ≥ 4, no unipotent element of G actson ∧i(V ) as a distinguished unipotent element. Let n = dimV . Since we areinterested in self-dual representations, for type Al we can assume i = l+1

2 , so l isodd. Since we are assuming that LG(ωi) is not a spin representation, for type Blwe have i ≤ l − 1 = n−3

2 and for type Dl we have i ≤ l − 2 = n2 − 2. In any case,

we have n ≥ 2i.Let u′ ∈ G be a regular unipotent element. According to Lemma 5.1.1 and

Lemma 2.7.3, the unipotent element u acts on ∧i(V ) with largest Jordan block ofsize ≤ mu′(ωi) + 1 ≤ i(n− i) + 1. One can show that

dim∧i(V ) =

(n

i

)>

(i(n− i) + 2)2

4

if n ≥ 2i and i ≥ 5 or if i = 4 and n > 8, so by Lemma 5.1.1 we are done in thesecases.

What remains is to consider the case where i = 4 and n = 8. Since we areassuming that ωi is not a spin representation for Dl, we have that G = A7. In this

170 Chapter 5.

λ G u of order > p u of order p mu(λ)

ω1 Al, Bl, Cl, Dl any any

ω3 A5 p = 5 p ≥ 11 9

ω2 Bl, l ≥ 3 prop. 5.5.5 prop. 5.5.5 4l − 2

ωl Bl, 3 ≤ l ≤ 8 prop. 5.10.1 prop. 5.10.1 l(l+1)2

ω2 Cl prop. 5.5.10 prop. 5.5.10 4l − 4ω3 C3 p = 3, 5 p ≥ 11 9ω3 C4 none p ≥ 17 15ω3 C5 none p ≥ 23 21ω4 C4 none p ≥ 17 16ω5 C5 none p ≥ 29 25

ω2 Dl prop. 5.5.6 prop. 5.5.6 4l − 6

ωl D6,D8 prop. 5.10.1 prop. 5.10.1 l(l−1)2

ω1 G2 p = 3, 5 p ≥ 7 6ω2 G2 p = 3, 5 p ≥ 11 10

ω1 F4 p = 5, 7, 11 p ≥ 23 22ω4 F4 p = 3, 5, 7, 11 p ≥ 17 16

ω2 E6 p = 5, 7, 11 p ≥ 23 22ω1 E7 p = 11 p ≥ 37 34ω7 E7 p = 3, 5, 11, 13, 17 p ≥ 29 27ω8 E8 p = 11, 13, 17, 23 p ≥ 59 58

Table 5.1: p 6= 2: Cases where a regular unipotent element u ∈ G acts on LG(ωi)as a distinguished unipotent element.

λ G Conjugacy class of u u of order > p u of order p mu(λ)

ω1 Bl, Cl, Dl any distinguished any anyω6 D6 [9, 3] p = 7 p ≥ 13 11ω4 F4 F4(a1) p = 5, 7 p ≥ 11 10ω7 E7 E7(a1) p = 5, 13 p ≥ 23 21ω7 E7 E7(a2) p = 5, 7, 11 p ≥ 19 17ω8 E8 E8(a1) p = 11 p ≥ 47 46

Table 5.2: p 6= 2: Cases where a non-regular distinguished unipotent element u ∈ Gacts on LG(ωi) as a distinguished unipotent element.


case the only distinguished unipotent elements u ∈ G are the regular ones, whichact on V with a single Jordan block of size 8. A computation with MAGMA showsthat u does not act on ∧4(V ) as a distinguished unipotent element; for conveniencewe have listed the decomposition of ∧4(V ) ↓ K[u] for p ≥ 3 in Table 5.3.

p Decomposition of ∧4(V8)

p = 3 [7, 97]p = 5 [1, 52, 9, 102, 13, 17]p = 7 [1, 74, 13, 142]p = 11 [1, 5, 9, 115]p = 11 [1, 52, 9, 11, 133]p ≥ 17 [1, 52, 92, 11, 13, 17]

Table 5.3: Decomposition of ∧4(V8) as a K[u]-module.

5.12.2 Type Cl

Suppose that G is a simple algebraic group of type Cl. To prove Proposition 5.12.1,without loss of generality we can assume that G = Sp(V ), where dimV = 2l. HereV = L(ω1) = V (ω1), so for ω1 the claim is obvious. For L(ω2) and L(ω3), theclaim of Proposition 5.12.1 follows from Proposition 5.5.10 and Proposition 5.8.1,respectively.

For ωi, i ≥ 4 we will apply results due to Premet and Suprunenko, which give arecursive formula for the dimension of L(ωi). To state their results, we need to �rstmake some de�nitions. Let a, b ∈ Z≥0 and write a =

∑i≥0 aip

i and b =∑

i≥0 bipi

for the expansions of a and b in base p. We say that a contains b to base p if forall i ≥ 0 we have bi = ai or bi = 0. 13

For r ≥ 1, we de�ne Jp(r) to be the set of integers 0 ≤ j ≤ r such that j ≡ rmod 2 and l − j + 1 contains r−j

2 to base p. Furthermore, set

f0 = 1

andfk = dimL(ωk)

for all k ≥ 1. Now [PS83, Theorem 2] implies the following.

Theorem 5.12.3. Let 1 ≤ r ≤ l. Then in the Weyl module V (ωr), each composi-tion factor has multiplicity 1, and the set of composition factors is

{L(ωj) : j ∈ Jp(r)}.

Furthermore, the integers fr (r ≥ 2) satisfy the following recurrence relation.

fr =

(2l

r

)−(

2l

r − 2

)−

∑j∈Jp(r)−{r}

fj

Note that the recursive formula above makes computing the values of fr veryeasy. As a corollary (see Corollary 2 in [PS83]), we have the following which canalso be deduced from the results of Gow in [Gow98].

13Note that in [PS83] there is a typo, the de�nition on pg. 1313, line 9 should say �for everyi = 0, 1, . . . , n ..�

172 Chapter 5.

Corollary 5.12.4. If p > l, then fr =(

2lr

)−(

2lr−2

)for all 1 ≤ r ≤ l.

With the recursive formula we also get a lower bound for fr.

Corollary 5.12.5. Let r ≥ 4. Then fr ≥(

2lr

)− 2(

2lr−2

)+ 1.

Proof. Note that we have dimV (ωj) =(

2lj

)−(

2lj−2

)[Bou75, Ch. VIII, 13.3, pg.

203], so fj ≤(

2lj

)−(

2lj−2

). Since Jp(r) is a subset of the nonnegative integers

r, r − 2, r − 4, . . ., the recursive formula of Theorem 5.12.3 gives

fr ≥(

2l

r

)−(

2l

r − 2

)− fr−2 − fr−4 − · · ·

≥(

2l

r

)−(

2l

r − 2

)−((

2l

r − 2

)−(

2l

r − 4

))−((

2l

r − 4

)−(

2l

r − 6

))− · · ·

=

(2l

r

)− 2

(2l

r − 2

)+ C

where C = f0 = 1 if r is even and C = f1 = 2l if r is odd. In any case, we get thedesired inequality.

For i = 4 and i = 5, the claim of Proposition 5.12.1 for G will be proven inthe next two lemmas.

Lemma 5.12.6. Proposition 5.12.1 holds for G = Cl, l ≥ 4, and i = 4.

Proof. Let u ∈ G be a regular unipotent element, so now mu(ω4) = 4(2l − 4) =8l−16 by Lemma 2.7.3. By Corollary 5.12.5 we have dimLG(ω4) ≥

(2l4

)−2(

2l2

)+1.

One can verify that the polynomial inequality(

2l4

)− 2(

2l2

)+ 1 > (mu(ω4)+2)2

4 holdsor all l ≥ 6. Thus by Lemma 5.1.1, no unipotent element of G acts on LG(ω4) asa distinguished unipotent element if l ≥ 6.

For l = 4 and l = 5, one can compute mu′(ω4) for all non-regular distinguishedunipotent elements u′ ∈ G. This is a straightforward application of Proposition2.6.5, and the expression of ω4 as a sum of simple roots [Hum72, 13, Table 1]. Wehave listed this information, along with dimLG(ω4) (computed using Theorem5.12.3) in Table 5.4. One veri�es that

dimLG(ω4) >(mu′(ω4) + 2)2

4

in all cases, so it follows from Lemma 5.1.1 that a non-regular unipotent elementu′ ∈ G does not act on LG(ω4) as a distinguished unipotent element.

Now we still need to consider the action of a regular unipotent element onLG(ω4) in the cases where l = 4 and l = 5. Using MAGMA (Section 2.9), we havecomputed these actions for all p ≥ 3 and listed them in Table 5.6. One sees thatfor l = 4, a regular unipotent element acts on LG(ω4) as a distinguished unipotentelement if and only if p ≥ 17. Furthermore, for l = 5, the regular unipotent doesnot act on LG(ω4) as a distinguished unipotent element, as claimed.

Lemma 5.12.7. Proposition 5.12.1 holds for G = Cl and i = 5.

Proof. We proceed as in the proof of Lemma 5.12.6. Let u ∈ G be a regularunipotent element, so now mu(ω5) = 5(2l − 5) = 10l − 25 by Lemma 2.7.3. ByCorollary 5.12.5 we have dimLG(ω5) ≥

(2l5

)− 2

(2l3

)+ 1, and one veri�es that


the polynomial inequality(

2l5

)− 2(

2l3

)+ 1 > (mu(ω5)+2)2

4 holds or all l ≥ 6. Thusby Lemma 5.1.1, no unipotent element of G acts on LG(ω5) as a distinguishedunipotent element if l ≥ 6.

What remains is to consider the case where l = 5. As in the second paragraphof the proof of Lemma 5.12.6, one veri�es that

dimLG(ω5) >(mu′(ω5) + 2)2

4

for all non-regular distinguished unipotent elements u′ ∈ G; for convenience wehave listed the values of mu′(ω5) and LG(ω5) in Table 5.5. Thus no non-regularunipotent element of G acts on LG(ω5) as a distinguished unipotent element.Using MAGMA (Section 2.9) we have computed the action of a regular unipotenton LG(ω5) for G = C5. The Jordan block sizes are given in Table 5.7, and theclaim of Proposition 5.12.1 follows from this.

To �nish the proof of Proposition 5.12.1 for G, what remains is to consider thecase where i ≥ 6 and to show that in this case no unipotent element of G acts onLG(ωi) as a distinguished unipotent element. We proceed to do this.

Let u ∈ G be a regular unipotent element, so nowmu(ωi) = i(2l−i) by Lemma2.7.3.

We start by considering 6 ≤ i ≤ 18. First, if i ≤ l − 3, by calculation one can

verify the polynomial inequality(

2li

)− 2(

2li−2

)+ 1 > (mu(ωi)+2)2

4 for all 6 ≤ i ≤ 18.Therefore by Corollary 5.12.5 and Lemma 5.1.1, no unipotent element of G acts onLG(ωi) as a distinguished unipotent element. For 6 ≤ i ≤ 18 and i = l, i = l − 1,i = l − 2 with the formula of Theorem 5.12.3 one can compute fi = dimLG(ωi)

and check that fi >(mu(ωi)+2)2

4 , so we are again done by Lemma 5.1.1.What remains is to consider i ≥ 19. If p > l, then fi =

(2li

)−(

2li−2

)by Corollary

5.12.4. One can verify that(

2li

)−(

2li−2

)> (mu(ωi)+2)2

4 for l ≥ i ≥ 6, so by Lemma5.1.1 no unipotent element of G acts on L(ωi) as a distinguished unipotent elementin this case. Suppose then that p ≤ l. By the argument given in the beginning ofSection 5.1.2, we see that it will be enough to show that dimL(ωi) > 4l4 + 2l2.Now dimL(ωi) ≥ |Wωi| where W is the Weyl group, so dimL(ωi) > 4l4 + 2l2

follows from Lemma 5.1.12 (i). This completes the proof of Proposition 5.12.1 forG = Cl.


C5 [2, 8] 22202 16 121 (p = 3), 165 (p > 3)C5 [4, 6] 20202 12 121 (p = 3), 165 (p > 3)

C4 [2, 6] 2202 10 41 (p = 3), 42 (p > 3)

Table 5.4: Values of mu(ω4) for non-regular distinguished unipotent elements u inC4 and C5.

5.12.3 Type G2

In this subsection we prove Proposition 5.12.1 for G simple of type G2.Since p 6= 2, we have VG(ω1) = LG(ω1). Now the computations given in [Law95,

Table 1] prove the claim of Proposition 5.12.1 for LG(ω1).

174 Chapter 5.


C5 [2, 8] 22202 16 122 (p = 3), 132 (p > 3)C5 [4, 6] 20202 12 122 (p = 3), 132 (p > 3)

Table 5.5: Values of mu(ω5) for non-regular distinguished unipotent elements u inC5.

G p LG(ω4) ↓ K[u]

C4 p = 3 [5, 94]p = 5 [5, 102, 17]p = 7 [72, 142]p = 11 [9, 113]p = 13 [5, 11, 132]p ≥ 17 [5, 9, 11, 17]

C5 p = 3 [94, 13, 184]p = 5 [57, 157, 25]p = 7 [1, 73, 9, 11, 13, 142, 17, 19, 21, 25]p = 11 [1115]p = 13 [9, 1312]p = 17 [1, 5, 7, 9, 11, 13, 177]p = 19 [1, 5, 7, 92, 11, 13, 15, 195]p = 23 [1, 5, 7, 92, 11, 132, 15, 17, 19, 232]p ≥ 29 [1, 5, 7, 92, 11, 132, 15, 17, 19, 21, 25]

Table 5.6: Action of a regular unipotent u ∈ G on LG(ω4), for G = C4, G = C5.

p LG(ω5) ↓ K[u]

p = 3 [94, 14, 184]p = 5 [107, 202, 22]p = 7 [2, 72, 10, 143, 18, 20, 26]p = 11 [1112]p = 13 [2, 1310]p = 17 [2, 6, 10, 12, 176]p = 19 [2, 6, 8, 10, 14, 16, 194]p = 23 [2, 6, 8, 10, 12, 14, 16, 18, 232]p ≥ 29 [2, 6, 8, 10, 12, 14, 16, 18, 20, 26]

Table 5.7: Action of a regular unipotent u ∈ G on LG(ω5), for G = C5.


For LG(ω2) and p 6= 3, we have LG(ω2) ∼= L (G) and the claim follows from[Law95, Table 2]. If p = 3, then a computation with MAGMA (Section 2.9) provesthat the only distinguished action on LG(ω2) occurs for the regular unipotentelement of G, which acts on LG(ω2) with a single Jordan block of size 7. For adi�erent argument, one could also argue by using the exceptional isogeny G→ Gwhich exists when p = 3.

5.12.4 Type F4

Let G be simple of type F4. We proceed to prove Proposition 5.12.1 for G.Since p 6= 2, we have that VG(ω1) ∼= LG(ω1) ∼= L (G). Now the computations

given in [Law95, Table 4] prove the claim of Proposition 5.12.1 for LG(ω1).For the fundamental weight ω2, consider �rst p = 3. In this case the regular

unipotent element of G has order 33 (see Appendix A), and dimLG(ω2) = 1222

by [Lüb01]. Since 1222 > (33+1)2

4 , it follows from Lemma 5.1.1 that no unipotentelement of G acts on LG(ω2) as a distinguished unipotent element. Suppose thenthat p > 3. By Lemma 2.7.3, we havemu(ω2) = 42 for a regular unipotent element

u ∈ G. Now dimLG(ω2) = 1274 by [Lüb01] and 1222 > (42+2)2

4 , so by Lemma 5.1.1no unipotent element of G acts on LG(ω2) as a distinguished unipotent element.

Consider next the fundamental weight ω3. If p > 3, then dimLG(ω3) = 273by [Lüb01] and mu(ω3) = 30 for a regular unipotent u by Lemma 2.7.3. Since

273 > 272 = (30+2)2

4 , by Lemma 5.1.1 no unipotent element of G acts on LG(ω3)as a distinguished unipotent element. If p = 3, then dimL(ω3) = 196. The non-regular unipotent elements of G have order 32 in this case (Appendix A), and

196 > (32+1)2

4 , so by Lemma 5.1.1 it is enough to consider a regular unipotentelement u. A computation with MAGMA (Section 2.9) shows that when p = 3, aregular unipotent element acts on LG(ω3) with Jordan blocks [3, 96, 184, 19, 21, 27],hence not as a distinguished unipotent element.

For LG(ω4), the claim of Proposition 5.12.1 follows from [Law95, Table 3] ifp > 3 since VG(ω4) = LG(ω4) in this case. For p = 3, a computation with MAGMA(Section 2.9) shows that the distinguished unipotent classes F4, F4(a1), F4(a2),and F4(a3) act on LG(ω4) with Jordan blocks [1, 9, 15], [7, 92], [1, 3, 62, 9], and[12, 33, 53] respectively. Note that here the only distinguished action occurs for theregular unipotent class F4.

5.12.5 Type E6, E7 and E8

Let G be simple of type E6, E7 or E8 and let λ = ωi. Set m = mu(ωi) for a regularunipotent u ∈ G, and d = dimL(ωi). Then with Lemma 2.7.3 and the results ofLübeck in [Lüb01], we can compute values of m and a lower bound for d, whichwe list in Table 5.8.

By Lemma 5.1.1, we know that no unipotent element of G acts on L(ωi) as

a distinguished unipotent element if d > (m+2)2

4 . By calculating with the values

in Table 5.8, we �nd that d > (m+2)2

4 except when (G,ωi) is (E6, ω1), (E6, ω2),(E6, ω6), (E7, ω1), (E7, ω7) or (E8, ω8). Since it is enough to consider self-dualrepresentations by Theorem 1.1.12, it will be enough to consider (E6, ω2), (E7, ω1),(E7, ω7) and (E8, ω8). Now except for the case G = E6, p = 3, λ = ω2, in thesecases the claim of Proposition 5.12.1 follows from the computations of Lawther in[Law95].

176 Chapter 5.

λ E6 E7 E8

ω1 d = 27, m = 16 d = 133, m = 34 d = 3875, m = 92ω2 d ≥ 77, m = 22 d ≥ 856, m = 49 d > 100000, m = 136ω3 d = 352, m = 30 d ≥ 8512, m = 66 d > 100000, m = 182ω4 d ≥ 2771, m = 42 d > 100000, m = 96 d > 100000, m = 270ω5 d = 352, m = 30 d ≥ 25896, m = 75 d > 100000, m = 220ω6 d = 27, m = 16 d ≥ 1538, m = 52 d > 100000, m = 168ω7 d = 56, m = 27 d ≥ 30132, m = 114ω8 d = 248, m = 58

Table 5.8: Values of m = mu(ωi) and d = dimLG(ωi) for u a regular unipotentelement of G = En.

What remains is to consider the case where G = E6, p = 3, and λ = ω2. Inthis case, a computation with MAGMA (Section 2.9) shows that the distinguishedunipotent classes E6, E6(a1) and E6(a3) act on LG(ω2) with Jordan block sizes[1, 93, 152, 19], [5, 98], and [1, 33, 64, 7, 94] respectively. Therefore no unipotent classof E6 acts on LG(ω2) as a distinguished unipotent element when p = 3.

5.13 Unipotent elements of order p (p 6= 2)

Assume that p 6= 2.

In this section we prove Theorem 1.1.10 in the case where u has order p.Speci�cally, the result proven in this section is the following.

Theorem 5.13.1. Let u ∈ G be a distinguished unipotent element of order p, andlet λ ∈ X(T )+ be a nonzero, p-restricted highest weight. Then u acts on LG(λ) asa distinguished unipotent element if and only if p > mu(λ) and G, λ and u areas in the Table 5.1 or Table 5.2, with λ given up to graph automorphism of An orDn.

Remark 5.13.2. Note that the λ occurring in Table 5.1 and Table 5.2 are preciselythe ones that occur in the main result of [LST15]. Furthermore, the conditionp > mu(λ) implies that the action of u on VG(λ) is the same as the correspondingaction in characteristic 0 (Lemma 2.7.9). Therefore when u has order p, the resultof Theorem 1.1.10 is similar to the characteristic 0 result of [LST15].

As seen in the previous sections, the situation is less uniform when u has order> p. In this setting things are unpredictable, and it is more di�cult to determinewhen exactly the examples in [LST15] work in characteristic p. For unipotentelements of order > p, we also have a unique example of a λ that corresponds to adistinguished action of unipotent element, but does not occur in the main resultof [LST15] (see Theorem 1.1.10 (iii)).

Remark 5.13.3. In [Sup03, Predlo�enie 4] and [Sup05, Predlo�enie 2], Su-prunenko has announced results which have the consequence that if u ∈ G is aunipotent element of order p, λ ∈ X(T )+, and 2 < p ≤ mu(λ), then u acts onLG(λ) with ≥ 2 Jordan blocks of size p. Assuming this result, Theorem 5.13.1 forthe case where u is a regular unipotent element becomes an immediate consequenceof Corollary 2.7.6, Lemma 2.7.9, and the main result of [LST15].

5.13. Unipotent elements of order p (p 6= 2) 177

G λ Conjugacy class of u mu(λ)

An ω1 regular nAn, n ≥ 2 ω1 + ωn regular 2n

A1 any 0 ≤ c ≤ p− 1 regular cA5 ω3 regular 9

Bn ω1 anyB3 ω1 + ω3 regular 12B3 2ω3 regular 12B3 3ω1 regular 18

Bn, n ≥ 3 2ω1 regular 4nBn, n ≥ 3 ω2 regular 4n− 2

Bn, 3 ≤ n ≤ 8 ωn regular n(n+1)2

Cn ω1 anyC2 bω1 for 1 ≤ b ≤ 5 regular 3bC2 bω2 for 1 ≤ b ≤ 5 regular 4bC2 ω1 + ω2 regular 7C2 ω1 + 2ω2 regular 11C2 2ω1 + ω2 regular 10C3 3ω1 regular 15C3 ω3 regular 9C4 ω3 regular 15C5 ω3 regular 21C4 ω4 regular 16C5 ω5 regular 25

Cn, n ≥ 3 2ω1 regular 4n− 2Cn, n ≥ 3 ω2 regular 4n− 4

Dn ω1 anyD6 ω6 regular 15D6 ω6 [9, 3] 11D8 ω8 regular 28

Dn, n ≥ 4 even 2ω1 regular 4n− 4Dn, n ≥ 4 odd ω2 regular 4n− 6

Table 5.1: G of classical type

178 Chapter 5.

G λ Conjugacy class of u mu(λ)

G2 ω1 regular 6G2 ω2 regular 10G2 ω1 + ω2 regular 16G2 2ω1 regular 12G2 2ω2 regular 20G2 3ω1 regular 18

F4 ω1 regular 22F4 ω4 regular 16F4 ω4 F4(a1) 10

E6 ω2 regular 22

E7 ω1 regular 34E7 ω7 regular 27E7 ω7 E7(a1) 21E7 ω7 E7(a2) 17

E8 ω8 regular 58E8 ω8 E8(a1) 46

Table 5.2: G of exceptional type

However, since the full proof of Suprunenko's results remain unpublished afterher 2003 and 2005 announcements [Sup03] and [Sup05], we have decided to give aproof of Theorem 5.13.1 that does not rely on these results. Our proof of Theorem5.13.1 is based on applying Proposition 4.6.10, which allows us to use the methodsof Liebeck, Seitz and Testerman from [LST15]. In some cases we have to givedi�erent proofs, but the general steps of the proof follow [LST15].

The proof of Theorem 5.13.1 will be given in what follows. We begin by takingcare of some special cases.

Lemma 5.13.4. Theorem 5.13.1 holds if λ 6= −w0(λ).

Proof. If λ 6= −w0(λ), then LG(λ) is not self-dual, and so u acts on LG(λ) asa distinguished unipotent element if and only if u acts on LG(λ) with a singleJordan block. Thus the claim follows from Proposition 1.1.12.

Lemma 5.13.5. Theorem 5.13.1 holds if λ = ωi.

Proof. This follows from Proposition 5.12.1, see Remark 5.12.2.

Lemma 5.13.6. Theorem 5.13.1 holds if G is of type A1.

Proof. This follows from the fact that for all 0 ≤ c ≤ p−1, a nonidentity unipotentelement u ∈ G acts on LG(c) with a single Jordan block of size c + 1 (Lemma4.2.2).

Lemma 5.13.7. Theorem 5.13.1 holds if p = 3.

Proof. Let p = 3, let u ∈ G be a distinguished unipotent element of order 3 and letV = LG(λ) be a nontrivial, p-restricted irreducible representation of G. Lookingat the cases where 3 > mu(λ) in Table 5.1 and Table 5.2, we see that the claimof Theorem 5.13.1 is that u acts on V as a distinguished unipotent element if andonly if u is a regular unipotent element and (G,λ) is (A1, c), (A2, ω1), or (A2, ω2).

Su�ciency follows from Lemma 5.13.6 and Lemma 5.13.5. Suppose then thatu acts on V as a distinguished unipotent element. Since u has order 3, we havedimV ≤ 4 by Lemma 5.1.1. If G is of exceptional type, then by [Lüb01] anynontrivial irreducible representation has dimension ≥ 7. Thus G must be of clas-sical type. Because u is distinguished of order 3, the largest Jordan block of u inthe natural representation has size ≤ 3. Therefore G is either of type A1 or A2

and u is a regular unipotent element. For type A1 the claim follows from Lemma5.13.6. For type A2, by [Lüb01] the only nontrivial, p-restricted irreducible repre-sentations of dimension ≤ 4 are the natural representation LG(ω1) and its dualLG(ω2). This proves the claim.

Lemma 5.13.8. Theorem 5.13.1 holds if p = 5.

Proof. Let p = 5, let u ∈ G be a distinguished unipotent element of order 5 and letV = LG(λ) be a nontrivial, p-restricted irreducible representation of G. If λ = ωi,the result of Theorem 5.13.1 follows by Lemma 5.13.5. Assume then that λ 6= ωi.Now looking at the cases where 5 > mu(λ) and λ 6= ωi in Table 5.1 and Table 5.2,we see that the claim of Theorem 5.13.1 is that u acts on V as a distinguishedunipotent element if and only if (G,λ) is (A1, c) or (A2, ω1 + ω2).

Su�cency follows from Lemma 5.13.6 and Proposition 5.3.2. For the otherdirection, suppose that u acts on V as a distinguished unipotent element. Sinceu has order 5, we have dimV ≤ 9 by Lemma 5.1.1. If G is of exceptional type,then by [Lüb01] any nontrivial, p-restricted irreducible representation 6= LG(ωi)has dimension > 9. Therefore G must be of classical type. Because u has order 5,the largest Jordan block of u in the natural representation has size ≤ 5. Thereforeit follows that G is of type Al (l ≤ 4), type C2, type C3, type B4 or type D4. Sincewe are assuming λ 6= ωi, by [Lüb01] the only case where we have a nontrivial,p-restricted representation of dimension ≤ 9 is when G = A1 or G = A2 andλ = ω1 + ω2. This completes the proof of the claim.

It follows from the lemmas above that for the rest of this section, we can andwill make the following assumptions.

• p ≥ 7 (by Lemma 5.13.7 and Lemma 5.13.8)

• rankG ≥ 2 (by Lemma 5.13.6)

• LG(λ) is self-dual, ie. λ = −w0(λ) (by Lemma 5.13.4)

These assumptions will have some implications which we will apply throughoutthe proof. The �rst assumption implies that the characteristic p is good for G, sowe can apply facts from Section 2.6 here. Thus we can assume that u is containedin a subgroup A < G of type A1, such that a maximal torus TA < A is containedin the �xed maximal torus T of G. Furthermore, we can assume that TA gives usthe labeled Dynkin diagram associated with u (Theorem 2.6.8). Therefore we canuse the labeled diagram associated with u to determine how T -weights restrict to

180 Chapter 5.

TA-weights. Indeed, write λ ∈ X(T ) as λ =∑l

i=1 qiαi, where αi are the simpleroots and qi ∈ Q+ for all i. Identifying the TA-weights with integers in the usualway, we see that λ restricts to the TA-weight

∑2qi ∈ Z, where the sum runs over

the i such that αi has label 2 in the labeled Dynkin diagram associated with u.For the rest of this section we will denote V = LG(λ). According to Proposition

4.6.10, under our assumptions we also have the following.

• If u acts on V as a distinguished unipotent element, then V ↓ A is semisim-ple.

Throughout we shall also denote r = mu(λ), which is the TA-weight given bythe restriction of the T -weight λ (De�nition 2.7.2). Now r is the highest weightof V ↓ A, since all T -weights of V are of the form λ −

∑kiαi, where ki ∈ Z≥0.

Furthermore, the labeled Dynkin diagram of u has all labels equal to 0 or 2, so itfollows that all TA-weights of V ↓ A are of the form r − 2c for some c ∈ Z≥0.

In this section, we will often write T -weights of the form λ−k1α1−k2α2−· · · asλ−1k12k2 · · · . For example, we write λ−α1 = λ−1 and λ−2α2−α3 = λ−223. Fora T -weight µ, we will write µsi for its image under the re�ection si correspondingto the simple root αi.

Note also that since we are working in good characteristic, Premet's theorem(Theorem 4.5.5) applies and so the set of weights in LG(λ) is saturated. We willoften use this fact in this section without mentioning it explicitly.

Now under our assumptions we will be able to prove the following lemma from[LST15] in our setting (positive characteristic), and essentially the rest of proof ofTheorem 5.13.1 is based on applying it. The proof of the lemma is essentially thesame as the one given in [LST15].

Lemma 5.13.9 ([LST15, Lemma 2.2]). Write λ =∑ciωi. Suppose that u acts

on V as a distinguished unipotent element. Then

(i) If ci is nonzero, then αi has label 2.

(ii) At most two values of ci can be nonzero.

(iii) For c ≥ 0, the weight r − 2c occurs with multiplicity at most c+ 1.

(iv) If r−2 has multiplicity 1, then for all c ≥ 1 the weight r−2c has multiplicityat most c.

Proof. As noted before, under our assumptions the fact that u acts on V as adistinguished unipotent implies that V ↓ A is semisimple. Then it follows thateach composition factor of V ↓ A occurs with multiplicity one. Furthermore, thesimple summands that occur in V ↓ A are determined by the weights of A thatoccur in V ↓ A. Recall also (Lemma 4.2.1) that all weights of a simple moduleL(c) have multiplicity 1 and that they are contained in {c, c−2, · · · ,−(c−2),−c}.We now proceed to use these facts to prove the claims (i) - (iv).

(i) If ci 6= 0 and αi has label 0, then the weight r of V ↓ A is a�orded by weightsλ and λ− i of V . But since r is the highest weight in V ↓ A, it follows thatL(r) must occur at least twice as a composition factor of V ↓ A, which is acontradiction.


(ii) Suppose that i, j, k are distinct indices such that ci, cj , ck 6= 0. Now (i) impliesthat the nodes αi, αj and αk have label 2. Thus the weight r− 2 is a�ordedby λ − i, λ − j and λ − k. Since all the weights of V ↓ A are ≤ r, this canonly happen if either L(r) or L(r − 2) occurs twice as a composition factorof V ↓ A. As in (i), this is a contradiction.

(iii) Suppose that the weight r − 2c occurs with multiplicity ≥ c + 2 in V ↓ A.Since r−2c can only occur in L(s) for s ≥ r−2c, it follows that one of L(r),L(r − 2), . . ., L(r − 2c) must occur at least twice as a summand of V ↓ A,which is a contradiction.

(iv) We may assume that c ≥ 2. Suppose that the weight r − 2c occurs withmultiplicity ≥ c + 1 in V ↓ A. Since each simple summand of V ↓ A hasmultiplicity one, it follows that each of L(r), L(r − 2), . . ., L(r − 2c) mustoccur as a summand of V ↓ A. Furthermore, r − 2c occurs as a weight ineach of the modules L(r), L(r−2), . . ., L(r−2c). In particular, r−2c occursas a weight in L(r).

Now because we are assuming that r − 2 has multiplicity 1, it follows thatr − 2 does not occur as a weight of L(r), and therefore r ≥ p. Since L(r)⊕L(r−2)⊕L(r−4) occurs as a direct summand of V ↓ A, it follows by Lemma4.2.6 that r = p or r = p+ 1. But then r − 2c does not occur as a weight ofL(r), contradiction.

Remark 5.13.10. Lemma 5.13.9 (iv) corresponds to [LST15, Lemma 2.2 (iii)],but note that a somewhat di�erent proof is needed in positive characteristic. Thisis also true for many other lemmas in this section that are analogous to lemmasin [LST15]. For example, compare Lemma 5.13.23 and [LST15, Lemma 4.5 (i)].

Now according to Lemma 5.13.9 (ii) and Lemma 5.13.5, it will be enough toprove Theorem 5.13.1 in the the following cases.

Case (I): λ = ciωi + cjωj , where i 6= j and ci, cj ≥ 1.

Case (II): λ = bωi, where b ≥ 2.

When u is a regular unipotent (that is, when all nodes αi have label 2), we willprove Theorem 5.13.1 in case (I) in this subsection (Proposition 5.13.16). Case (II)for a regular unipotent element will be treated in subsection 5.13.1. The proofsfor a distinguished, non-regular unipotent element u will be given in subsection5.13.2. We will now proceed to prove some general facts about the two cases (I)and (II).

Lemma 5.13.11. Let λ = bωi, where b ≥ 1. Assume that u acts on V as adistinguished unipotent element. Then if all nodes adjacent to αi have label 2, theweight r − 2 of A occurs with multiplicity 1.

Proof. Note that αi has label 2 by Lemma 5.13.9 (i), so λ− i a�ords r−2. If r−2does not have multiplicity 1 in V ↓ A, then it would have to be a�orded by someλ− ij, where αj is a node adjacent to αi. In this situation the node αj would haveto have label 0.

182 Chapter 5.

Lemma 5.13.12 ([LST15, Lemma 2.3]). Let λ = bωi, where b ≥ 2. Suppose thatu acts on V as a distinguished unipotent element. Then

(i) The node αi is an end node with label 2.

(ii) The node adjacent to αi has label 2.

(iii) The weight r − 2 of V ↓ A has multiplicity 1.

Proof. (i) We know that αi has label 2 by Lemma 5.13.9 (i). The fact that αiis an end node follows with the same proof as in [LST15, Lemma 2.3 (i)].

(ii) If G has rank 2, then we know from the description of the labeled Dynkindiagrams (Section 2.6) that all nodes have label 2, unless G = G2 and u liesin class G2(a1). In this case α1 has label 0 and α2 has label 2, so we musthave λ = bω2. But then r − 2 is a�orded by λ − 2, λ − 12 and λ − 122,contradicting Lemma 5.13.9 (iii). If G has rank at least 3, then the claimfollows with the same proof as in [LST15, Lemma 2.3 (ii)].

(iii) This follows immediately from (i), (ii) and Lemma 5.13.11.

Lemma 5.13.13. Let λ = bωi, where b ≥ 2 and assume that u is a regularunipotent element. Suppose u acts on V as a distinguished unipotent element.Then r = 2 + p, r = 3 + p, r = 2 + 2p or r = a+ pk with 4 ≤ a ≤ p− 2 and k ≥ 1.

Proof. Now by Lemma 5.13.12, the node αi is an end node, the node αj adjacentto αi has label 2, and the weight r − 2 of V ↓ A occurs with multiplicity 1. Thusthe weight r−4 has multiplicity 2, being a�orded by the weights λ− i2 and λ− ij.

Since u is a regular unipotent element and λ is p-restricted, it follows fromLemma 2.7.4 that p - r. Therefore the weight r − 2 occurs in L(r). Thus L(r − 2)does not occur as a summand of V ↓ A, so L(r) ⊕ L(r − 4) must occur as asummand of V ↓ A. Furthermore, r − 4 must occur as a weight of L(r). Now theclaim follows from Lemma 4.2.5.

Lemma 5.13.14 ([LST15, Lemma 2.6]). Let λ = ciωi + cjωj, where i 6= j andci, cj ≥ 1. Suppose that u acts on V as a distinguished unipotent element. Then

(i) Nodes adjoining αi and αj have label 2.

(ii) If ci > 1 or cj > 1, then αi and αj are adjacent.

(iii) Either αi or αj is an end node.

(iv) If all nodes have label 2 and αi and αj are not adjacent, then αi and αj areboth end nodes.

Proof. (i) This follows as in [LST15, Lemma 2.6 (ii)].

(ii) This follows with the argument given in the beginning of the proof of [LST15,Lemma 2.6 (iii)].


(iii) We give a proof following [LST15, Lemma 2.6 (iv)]. Suppose that αi and αjare not end nodes, so now rankG ≥ 3. Then there exists a node αt 6= αjadjacent to αi and a node αs 6= αi adjacent to αj . By (i), these nodes musthave label 2, so λ − ij, λ − it and λ − js a�ord the weight r − 4. Nowci = cj = 1, as otherwise λ− i2 or λ− j2 would also a�ord the weight r− 4,contradicting Lemma 5.13.9 (iii). Similarly αi and αj must be adjacent, asotherwise there exists a node αu 6= αt adjacent to αi, and then λ− iu a�ordsthe weight r − 4.

We will show next that λ− ij has multiplicity 2, which contradicts Lemma5.13.9 (iii) and �nishes the proof. Now for a Levi factor L = A2, L = B2

and L = C2, in the representation LL(ω1 +ω2) the weight ω1 +ω2−α1−α2

has multiplicity 2 by Lemma 4.5.2, since we are assuming p ≥ 7. Thus byCorollary 4.5.7 the weight λ− ij has multiplicity 2 in V .

(iv) This follows with the same argument as [LST15, Lemma 2.6 (vi)].

Lemma 5.13.15. Let λ = ciωi + cjωj, where i 6= j and ci, cj ≥ 1. If u acts on Vas a distinguished unipotent element, then p > r.

Proof. Suppose that r ≥ p and that u acts on V as a distinguished unipotentelement. Now the weight r − 2 has multiplicity 2 in V ↓ A since it is a�orded byλ− i and λ− j. Therefore L(r − 2)⊕ L(r) occurs as a summand of V ↓ A, so byLemma 4.2.4 we have r = p, r = p+1, r = 2p, or r = p+pl for some l ≥ 2. On theother hand, now r− 2 must occur as a weight of L(r), so it follows that r = p+ 1.In this case r − 4 does not occur as a weight of L(r). Therefore the weight r − 4has multiplicity ≤ 2 in V ↓ A, as otherwise L(r−2) or L(r−4) would occur twiceas a summand. The rest of our proof is based on exploiting this fact.

Recall that by Lemma 5.13.14 (i) any node adjacent to αi and αj must havelabel 2. We claim that αi and αj are adjacent. If this is not the case, then thereexists a node αt 6= αj adjacent to αi and a node αs 6= αi adjacent to αj . Thenλ − it, λ − ij and λ − js a�ord r − 4, which is a contradiction since r − 4 hasmultiplicity ≤ 2 in V ↓ A.

Thus αi and αj must be adjacent, and by Lemma 5.13.14 (iii) either αi or αjis an end node. Suppose �rst that G has rank at least 3. Without loss of generalityassume that αi is an end node. Then there exists a node αk 6= αi adjacent to αj ,and then λ− ij and λ−jk a�ord the weight r−4. Thus we must have ci = cj = 1,as otherwise λ − i2 or λ − j2 would also a�ord the weight r − 4. As in the proofof Lemma 5.13.14 (iii), we can use Corollary 4.5.7 to conclude that λ − ij hasmultiplicity 2 in V , giving a contradiction.

Suppose then that G has rank 2, so λ = c1ω1 + c2ω2. We will show that λ− 12has multiplicity 2, which will show that c1 = c2 = 1 as otherwise λ− 12 or λ− 22

would also a�ord the weight r − 4. If G is of type G2, then r = 6c1 + 10c2 byLemma 2.7.3. Since r = p + 1, this implies p > c1 + 3c2 + 3, so by Lemma 4.5.2the weight λ − 12 has multiplicity 2. The same argument works also for G = A2

and G = C2 to show that λ− 12 has multiplicity 2.Finally, what remains is the case where G has rank 2 and λ = ω1 + ω2. If G

has type A2, then r = 4, so p = 3, but we are assuming that p ≥ 7. For G of typeC2 or G2, we have r = 7 and r = 16, respectively. In these cases r = p+ 1 is nota possibility.

184 Chapter 5.

Using Lemma 5.13.15 and the main result of [LST15], we can now prove The-orem 5.13.1 in the case where u is a regular unipotent element and λ = ciωi+cjωjwith i 6= j and ci, cj ≥ 1.

Proposition 5.13.16. Theorem 5.13.1 holds when u is a regular unipotent ele-ment and λ = ciωi + cjωj, where i 6= j and ci, cj ≥ 1.

Proof. Let u ∈ G be a regular unipotent element and let λ = ciωi+cjωj , where i 6=j and ci, cj ≥ 1. Suppose that u acts on LG(λ) as distinguished unipotent element.By Lemma 5.13.15, we have p > r, so by Corollary 2.7.6 we have VG(λ) = LG(λ).Furthermore, by Lemma 2.7.9 the action of u on VG(λ) has the same Jordan blocksizes as the corresponding action in characteristic 0. Thus the claim of Theorem5.13.1 follows from [LST15, Theorem 1].

Lemma 5.13.17 ([LST15, Lemma 2.6]). Let λ = ciωi + cjωj, where i 6= j andci, cj ≥ 1. Suppose that u acts on V as a distinguished unipotent element. Then

(i) If αi and αj are adjacent, then λ− ij has multiplicity 2.

(ii) Either ci = 1 or cj = 1.

(iii) If ci > 1 or cj > 1, then G has rank 2.

Proof. (i) According to Lemma 5.13.15, we have p > mu(λ) = 6c1 + 10c2. Thusif G has type G2, then it follows from Lemma 4.5.2 that the weight λ−ij hasmultiplicity 2. Suppose then that G does not have type G2. We always havemu(ωi) ≥ 2, so it follows that mu(λ) ≥ 1 + ci + cj , 1 + 2ci + cj , 1 + ci + 2cj .Then for L = A2, L = B2 and L = C2, in the representation LL(c1ω1+c2ω2),the weight ω1 + ω2 − α1 − α2 has multiplicity 2 by Lemma 4.5.2. Thus byCorollary 4.5.7, the weight λ− ij has multiplicity 2 in V .

(ii) Using (i), the claim follows with the argument given in the end of the proofof [LST15, Lemma 2.6 (iii)].

(iii) Again with (i), this follows with the same proof as [LST15, Lemma 2.6 (v)].

5.13.1 Case where u is regular and λ = bωi, b ≥ 2

In this section we will handle the case where λ = bωi, b ≥ 2. The result in thiscase is the following proposition (cf. [LST15, Proposition 4.1]), which we will proveusing lemmas given later in this section. This then �nishes the proof of Theorem5.13.1 in the case where u is a regular unipotent element.

Proposition 5.13.18. Suppose that λ = bωi, where b ≥ 2. Then a regular uni-potent element u ∈ G acts on V as a distinguished unipotent element if and onlyif p > mu(λ) and G and λ are as in the table below, with λ given up to graphautomorphism of D4.

For the proof, recall that we can assume that αi is an end node (Lemma 5.13.12(i)). Note also that now if u acts on V as a distinguished unipotent element, byLemma 5.13.12 (iii) the weight r− 2 of V ↓ A has multiplicity 1. Then by Lemma5.13.9 (iv), for all c ≥ 1 the weight r−2c of V ↓ Amust occur with multiplicity ≤ c.We will use these facts in the proofs of the lemmas that follow. In this subsectionu will always be a regular unipotent element of G.


λ G

2ω1 Bn, Cn, Dn (n even), G2

3ω1 C2, C3, B3, G2

bω1, b ≤ 5 C2

bω2, b ≤ 5 C2

2ω3 B3

2ω2 G2

Lemma 5.13.19. Suppose that G = Bn (n ≥ 3), G = Cn (n ≥ 3) or G = Dn

(n ≥ 4) and λ = bω1, where b ≥ 3. If u acts on V as a distinguished unipotentelement, then b = 3.

Proof. We proceed as in the end of the proof of [LST15, Lemma 4.4]. Suppose thatb ≥ 4. We show that the weight r − 8 of A occurs with multiplicity ≥ 5, whichwill give a contradiction by Lemma 5.13.9 (iv). If n ≥ 4, then r− 8 is a�orded byλ − 14, λ − 132, λ − 1222, λ − 1223 and λ − 1234. If n = 3, replacing the last ofthese weights by λ− 1232 (type B3) or λ− 1223 (type C3) gives us again a list of5 weights a�ording r − 8.

Lemma 5.13.20. Suppose that G = Bn with n ≥ 3 and λ = bωn with b ≥ 2. If uacts on V as a distinguished unipotent element, then n = 3 and b = 2.

Proof. We proceed similarly to [LST15, Lemma 4.3]. Now we have LB2(bω2) =VB2(bω2) = Sb(VB2(ω2)) (see e.g. [Sei87, 1.14 and 8.1 (c)] or [McN98, Proposition4.2.2.(h)]). Hence in LB2(bω2) the weight bω2 − α1 − 2α2 occurs with multiplicity2, so by using 4.5.7 we see that the weight λ − n2(n − 1) has multiplicity 2 inV ↓ A. So if b ≥ 3, the weight r − 6 has then multiplicity at least 4, since it isa�orded by λ− n2(n− 1), λ− n3 and λ− n(n− 1)(n− 2), contradicting Lemma5.13.9 (iv). Therefore b = 2.

If n ≥ 4, then r−8 has multiplicity at least 5, since it is a�orded by λ−n2(n−1)(n− 2), λ− n2(n− 1)2, λ− n(n− 1)(n− 2)(n− 3) and λ− n3(n− 1). This isagain a contradiction by Lemma 5.13.9 (iv). Therefore we must have b = 2 andn = 3.

Lemma 5.13.21. If G = Cn with n ≥ 3 and λ = bωn with b ≥ 2, then u does notact on V as a distinguished unipotent element.

Proof. We proceed again similarly to [LST15, Lemma 4.3]. Suppose that u actson V as a distinguished unipotent element. If b ≥ 3, then the weight r − 6 ofA is a�orded by λ − (n − 2)(n − 1)n, λ − (n − 1)n2, λ − n3 and λ − (n − 1)2n,contradicting Lemma 5.13.9 (iv).

Suppose then that b = 2. In this case if n ≥ 5, the weight r−12 has multiplicity

186 Chapter 5.

at least 7, since it is a�orded by the following weights:

λ− n2(n− 1)4

λ− n2(n− 1)3(n− 2)

λ− n2(n− 1)2(n− 2)2

λ− n2(n− 1)2(n− 2)(n− 3)

λ− n2(n− 1)(n− 2)(n− 3)(n− 4)

λ− n(n− 1)2(n− 2)2(n− 3)

λ− n(n− 1)2(n− 2)(n− 3)(n− 4)

This again contradicts 5.13.9 (iv).Consider n = 4 and λ = 2ω4, in which case mu(λ) = 32 by Lemma 2.7.3. Since

u is regular of order p we have p ≥ 2n = 8, so p > 7. Then according to [Lüb01],

we have dimV = 594 > 289 = (32+2)2

4 , so Lemma 5.1.1 gives a contradiction.In the case n = 3 a computation with MAGMA (Section 2.9), given in Table

5.3, shows that a regular unipotent element does not act on V = LG(2ω3) as adistinguished unipotent element.

p LG(2ω3) ↓ K[u]

p = 3 [3, 99]p = 5 [102, 13, 152]p = 7 [712]p = 11 [7, 117]p = 13 [3, 7, 9, 135]p = 17 [3, 72, 9, 11, 13, 172]p ≥ 19 [3, 72, 9, 11, 13, 15, 19]

Table 5.3: Action of a regular unipotent u ∈ G on LG(2ω3), for G = C3.

Lemma 5.13.22. If G = C2 and λ = bω1 with b ≥ 6, then u does not act on Vas a distinguished unipotent element.

Proof. In this case we have V = LG(λ) = VG(λ) = Sb(E) (see e.g. [Sei87, 1.14 and8.1 (c)] or [McN98, Proposition 4.2.2.(h)]), where E is the natural module for G.Suppose that u acts on V ↓ A as a distinguished unipotent element. The weightsthat occur in E are m, m − 2, m − 4 and m − 6, where m = 3. Now the weightr − 12 has multiplicity at least 7, because it is a�orded by symmetric powers ofvectors with weights as follows (each tuple below has length b):

(m, . . . ,m,m,m,m,m,m− 6,m− 6)

(m, . . . ,m,m,m,m,m− 2,m− 4,m− 6)

(m, . . . ,m,m,m,m− 2,m− 2,m− 2,m− 6)

(m, . . . ,m,m,m,m,m− 4,m− 4,m− 4)

(m, . . . ,m,m,m,m− 2,m− 2,m− 4,m− 4)

(m, . . . ,m,m,m− 2,m− 2,m− 2,m− 2,m− 4)

(m, . . . ,m,m− 2,m− 2,m− 2,m− 2,m− 2,m− 2)

This is a contradiction by Lemma 5.13.9 (iv).


Lemma 5.13.23. If G = C2 and λ = bω2 with b ≥ 6, then u does not act on Vas a distinguished unipotent element.

Proof. Suppose that u acts on V as a distinguished unipotent element. The resultsof Lübeck [Lüb01] imply that dimLG(bω2) ≥ 85 for b ≥ 6, so we can assume thatp > 17 by Lemma 5.1.1.

If b = 6, then a computation with MAGMA (Section 2.9), given in Table5.4, shows that u acts on V with repeated blocks. Assume then that b ≥ 7. Letr = mu(λ), so here r = 4b by Lemma 2.7.3.

Now the weight r−8 is a�orded by the weights λ−24, λ−123 and µ = λ−1222.One can verify that all weights λ � µ′ � µ have multiplicity 1 in VG(λ) (see Table4.2). Hence by Lemma 4.5.4, the weight µ occurs with multiplicity 2 in V unlessLG(µ) is a composition factor of VG(λ).

Consider �rst the case where LG(µ) is a composition factor of VG(λ). Now byProposition 4.5.1 and Table 4.2, we have that p divides b+1. Since λ is p-restricted,we have b = p− 1. Then r = 4b = 4p− 4 = 3p+ (p− 4), which is a contradictionby Lemma 5.13.13.

Thus LG(µ) does not occur as a composition factor of VG(λ), and so µ occurswith multiplicity 2 in LG(λ). We proceed to show next that r−10 has multiplicityat least 6 in V ↓ A, which will contradict Lemma 5.13.9 (iv). First of all, theweights ν1 = λ − 124 and ν2 = λ − 1223 a�ord the weight r − 10. Furthermore,by Table 4.2, the weights ν1 and ν2 have multiplicity 2 in VG(λ). One can verifythat all weights λ � µ′ � νi except for µ′ = µ have multiplicity 1 in VG(λ) (seeTable 4.2). Then because LG(µ) does not occur as a composition factor of VG(λ),it follows from Lemma 4.5.4 that νi has multiplicity 2 in LG(λ) if LG(νi) does notoccur as a composition factor of VG(λ). Now by Proposition 4.5.1 and Table 4.2,if LG(ν1) (respectively LG(ν2)) occurs as a composition factor of VG(λ), then pdivides b − 3 (respectively b − 4). Because b ≤ p − 1, it follows that p does notdivide b− 3 nor b− 4, so we can conclude that ν1 and ν2 both have multiplicity 2in LG(λ). Hence r−10 occurs in V ↓ A with multiplicity at least 6, being a�ordedby λ− 25, λ− 1322, ν1, and ν2.

p LG(6ω2) ↓ K[u]

p = 7 [720]p = 11 [1, 3, 5, 7, 1110]p = 13 [1, 3, 7, 9, 135]p = 17 [1, 7, 13, 177]p = 19 [1, 7, 9, 13, 15, 195]p = 23 [1, 7, 9, 132, 15, 17, 19, 232]p ≥ 29 [1, 7, 9, 132, 15, 17, 19, 21, 25]

Table 5.4: Action of a regular unipotent u ∈ G on LG(6ω2), for G = C2.

Lemma 5.13.24. Suppose that G = Dn and λ = bωn or λ = bωn−1, where b ≥ 2.If u acts on V as a distinguished unipotent element, then n = 4 and b = 2.

Proof. This follows exactly as in the end of the proof of [LST15, Lemma 4.3].

Lemma 5.13.25. If G = G2 and λ = bω1 with b ≥ 4, then u does not act on Vas a distinguished unipotent element.

188 Chapter 5.

Proof. ([LST15, Lemma 4.5 (ii)]) The representation LG(ω1) gives an embeddingG < B3, and now we have LB3(bω1) ↓ G ∼= V by [Sei87, Table 1, III1]. Since underthis embedding G2 contains a regular unipotent element of B3 (Theorem 1.1.12),the claim follows from Lemma 5.13.19.

Lemma 5.13.26. If G = G2 and λ = bω2 with b ≥ 3, then u does not act on Vas a distinguished unipotent element.

Proof. Here r = 10b by Lemma 2.7.3.Suppose that u acts on V as a distinguished unipotent element. Note �rst

that by looking at the results of Lübeck in [Lüb01], we have dimV ≥ 148 underour assumptions. It follows then from Lemma 5.1.1 that p > 23, as otherwise

dimV > (p+1)2

4 .Consider �rst b = 3. In this case p > 23 and [Lüb01] imply that dimV =

273 > 256 = (r+2)2

4 , which is a contradiction by Lemma 5.1.1. Therefore b ≥ 4.Now the weight r− 8 is a�orded by λ− 24, λ− 123, λ− 132 and µ = λ− 1222.

Note that µ occurs with multiplicity 2 in VG(λ) (Table 4.1). Furthermore, one canverify that all weights λ � µ′ � µ have multiplicity 1 in VG(λ) (see Table 4.1).Thus if LG(µ) does not occur as a composition factor of VG(λ), it follows fromLemma 4.5.4 that µ occurs with multiplicity 2 in V , contradicting Lemma 5.13.9(iv).

Hence LG(µ) must occur as a composition factor of VG(λ). By Proposition4.5.1 and Table 4.1, this implies that p divides 3b+2. Now b ≤ p−1, so 3b+2 = por 3b+ 2 = 2p. Thus r = 10b = 10p−20

3 or r = 20p−203 , which is a contradiction by

Lemma 5.13.13.

Lemma 5.13.27. If G = F4 and λ = bωi with b ≥ 2, then u does not act on Vas a distinguished unipotent element.

Proof. Suppose that u acts on V ↓ A as a distinguished unipotent element. As inthe results above, recall that by Lemma 5.13.12 the node αi is an end node, andthat the weight r − 2 of A has multiplicity 1. If b ≥ 3, then as in the proof of[LST15, Lemma 4.7], we see that the weight r−8 of V ↓ A occurs with multiplicityat least 5, contradicting Lemma 5.13.9 (iv).

Consider then the cases where b = 2. If λ = 2ω1, then r = 44 by Lemma 2.7.3.

By [Lüb01], we have dimV ≥ 755 > 529 = (r+2)2

4 , which is a contradiction byLemma 5.1.1. Finally if λ = 2ω4, then r = 32 by Lemma 2.7.3. Again by [Lüb01],

we have dimV ≥ 298 > 289 = (r+2)2

4 , which is again a contradiction by Lemma5.1.1.

Lemma 5.13.28. If G = E6, G = E7 or G = E8 and λ = bωi with b ≥ 2, then udoes not act on V as a distinguished unipotent element.

Proof. Suppose that u acts on V ↓ A as a distinguished unipotent element. Recallthat by Lemma 5.13.12 the node αi is an end node, and that the weight r − 2 ofA has multiplicity 1.

Consider �rst the case where i = 1. If b = 2, then by Lemma 2.7.3 we haver = 32, r = 68 and r = 184 when G = E6, G = E7, or G = E8 respecti-vely. Furthermore, it follows from [Lüb01] that dimV = 351, dimV ≥ 7370, and

dimV > 100000 respectively. In any case, we see that dimV > (r+2)2

4 , which is a


contradiction by Lemma 5.1.1. If b ≥ 3, then the weight r− 8 of V ↓ A is a�ordedby λ− 1232, λ− 1234, λ− 1234 and λ− 1345, contradicting Lemma 5.13.9 (iv).

Thus we must have i > 1. If i = 2, then the weight r−8 is a�orded by λ−2242,λ− 2243, λ− 2245, λ− 2431 and λ− 2435, which again contradicts Lemma 5.13.9(iv).

Therefore αi must be the last node in the Dynkin diagram of G, say λ = bωnwhere G = En. In this situation if b = 2, then by Lemma 2.7.3 we have r = 32,r = 54 and r = 118 when G = E6, G = E7, and G = E8 respectively. Furthermore,it follows from [Lüb01] that dimV = 351, dimV = 1463, and dimV > 100000

respectively. As earlier in the proof, we see that dimV > (r+2)2

4 , which is a con-tradiction by Lemma 5.1.1.

Hence b ≥ 3. In this case the weight r − 12 has multiplicity at least 7, beinga�orded by λ − n3(n − 1)3, λ − n3(n − 1)2(n − 2), λ − n3(n − 1)(n − 2)(n − 3),λ−n2(n−1)2(n−2)2, λ−n2(n−1)2(n−2)(n−3), λ−n2(n−1)(n−2)(n−3)(n−4),and λ− n(n− 1)(n− 2)(n− 3)(n− 4)(n− 5). This is a contradiction by Lemma5.13.9 (iv).

We now summarize the lemmas above and give the proof of Proposition 5.13.18.

Proof of Proposition 5.13.18. Suppose that λ = bωi, where b ≥ 2. Assume that aregular unipotent element u ∈ G acts on V = LG(λ) as a distinguished unipotentelement. It follows from Lemma 5.13.12 (i) that αi is an end node. Therefore ifG = An, then the claim of Proposition 5.13.18 follows from Lemma 5.13.4.

If G = Bn with n ≥ 3, then it follows from Lemma 5.13.19 and Lemma 5.13.20that λ = 2ω1, λ = 3ω1, or n = 3 and λ = 2ω3. Furthermore, for these cases theclaim of Proposition 5.13.18 follows from Proposition 5.7.7 (b) (ii), Proposition5.9.1, and Table 5.5, respectively. Here the decompositions given in Table 5.5 werecomputed with MAGMA (Section 2.9).

If G = B2 = C2, then it follows from Lemma 5.13.22 and Lemma 5.13.23that λ = bωi with b ≤ 5. For b ≤ 3 the claim of Proposition 5.13.18 follows fromProposition 5.7.11 (b) (ii) (for i = 1, b = 2), Proposition 5.7.7 (b) (ii) (for i = 2,b = 2), and Proposition 5.9.1 (for b = 3). For b = 4 and b = 5, the claim ofProposition 5.13.18 follows from Table 5.6, which was computed with MAGMA(Section 2.9).

If G = Cn with n ≥ 3, then it follows from Lemma 5.13.19 and Lemma 5.13.21that λ = 2ω1 or λ = 3ω1. For these cases the claim of Proposition 5.13.18 followsfrom Proposition 5.7.11 (b) (ii) and Proposition 5.9.1, respectively.

If G = Dn with n ≥ 4, then it follows from Lemma 5.13.19 and Lemma 5.13.24that λ = 2ω1, λ = 3ω1, or n = 4 and λ ∈ {2ω3, 2ω4}. For these cases the claim ofProposition 5.13.18 follows from Proposition 5.7.8 (b) (ii), Proposition 5.9.1, andProposition 2.10.2 (ii), respectively.

If G = G2, then it follows from Lemma 5.13.25 and Lemma 5.13.26 thatλ ∈ {2ω1, 3ω1, 2ω2}. In these cases the claim is a consequence of Table 5.7, whichwas computed with MAGMA (Section 2.9).

Finally if G = F4, G = E6, G = E7, and G = E8, we have a contradiction byLemma 5.13.27 or Lemma 5.13.28.

190 Chapter 5.

p LG(2ω3) ↓ K[u]

p = 3 [1, 7, 93]p = 5 [1, 5, 7, 9, 13]p = 7 [75]p = 11 [1, 5, 7, 112]p ≥ 13 [1, 5, 7, 9, 13]

Table 5.5: Action of a regular unipotent u ∈ G on LG(2ω3), for G = B3.

λ p LG(λ) ↓ K[u]

4ω1 p = 5 [57]p = 7 [75]p = 11 [1, 5, 7, 112]p ≥ 13 [1, 5, 7, 9, 13]

5ω1 p = 7 [78]p = 11 [4, 8, 114]p = 13 [4, 6, 8, 12, 132]p ≥ 17 [4, 6, 8, 10, 12, 16]

4ω2 p = 5 [511]p = 7 [5, 77]p = 11 [115]p = 13 [5, 11, 133]p ≥ 17 [5, 9, 11, 13, 17]

5ω2 p = 7 [713]p = 11 [3, 5, 9, 114]p = 13 [137]p = 17 [5, 9, 11, 15, 173]p = 19 [5, 9, 11, 13, 15, 192]p ≥ 23 [5, 9, 11, 13, 15, 17, 21]

Table 5.6: For G = C2 and u ∈ G a regular unipotent element, action of u onLG(λ) for λ = bωi with i ∈ {1, 2} and b ∈ {4, 5}.

5.13.2 Case where u is not regular

In this section, we keep our setup as before, but assume throughout that u is adistinguished unipotent element that is not regular. The result in this case is thefollowing proposition, which completes the proof of Theorem 5.13.1.

Proposition 5.13.29. A non-regular unipotent element u of order p acts as adistinguished unipotent element on V if and only if p > mu(λ) and G and λ areas in the table below, with λ given up to graph automorphism of Dn.


λ p LG(λ) ↓ K[u]

2ω1 p = 3 [93]p = 5 [5, 9, 13]p = 7 [5, 73]p = 11 [5, 112]p ≥ 13 [5, 9, 13]

3ω1 p = 5 [52, 102, 13, 15, 19]p = 7 [711]p = 11 [117]p = 13 [3, 9, 135]p = 17 [3, 7, 9, 11, 13, 172]p ≥ 19 [3, 7, 9, 11, 13, 15, 19]

2ω2 p = 3 [93]p = 5 [1, 5, 102, 152, 21]p = 7 [711]p = 11 [117]p = 13 [1, 11, 135]p = 17 [1, 5, 9, 11, 173]p = 19 [1, 5, 9, 11, 13, 192]p ≥ 23 [1, 5, 9, 11, 13, 17, 21]

Table 5.7: For G = G2 and u ∈ G a regular unipotent element, action of u onLG(λ) for λ = 2ω1, λ = 3ω1, and λ = 2ω2.

λ G class of uω1 Bn, Cn, Dn anyω6 D6 [3, 9]ω4 F4 F4(a1)ω7 E7 E7(a1), E7(a2)ω8 E8 E8(a1)

To prove Proposition 5.13.29, it will be enough to show in each case that udoes not act as a distinguished unipotent element on V when λ 6= ωi. After thatis done, the result follows from Lemma 5.13.5.

We begin by considering the case where G is of classical type. Because weare assuming that u is a non-regular distinguished unipotent element, we haverankG ≥ 3 and G is not of type An.

Lemma 5.13.30. Suppose that G = Bn, G = Cn or G = Dn and assume that uacts on V as a distinguished unipotent element. Then

(i) λ = bωi for some i.

(ii) If λ = bωi with b ≥ 2, then i = 1.

Proof. (i): We argue as in [LST15, Lemma 6.2 (iii)]. Suppose that λ 6= bωi. Thenby Lemma 5.13.9 (ii) we have λ = ciωi + cjωj where i 6= j and ci, cj ≥ 1.According to 5.13.14 (i), nodes adjoining αi and αj have label 2, so according

192 Chapter 5.

to Proposition 2.6.5 the nodes αi and αj occur in the initial string of 2's.Now since rankG ≥ 3, by Lemma 5.13.17 (iii) we have ci = cj = 1, andLemma 5.13.14 (iii) shows that either αi or αj is an end node. Because αiand αj occur in the initial string, without loss of generality we can assumei = 1 so λ = ω1 + ωj for some j > 1.

If j > 2, then there exists a node αk 6= αi adjacent to αj . Since αi and αjoccur in the initial string and not all labels are 2, the node αj is not an endnode. Therefore there exist distinct nodes αs, αt 6= αi which are adjacentto αj . Then the weight r − 4 of V ↓ A is a�orded by λ− 12, λ− 1j, λ− tjand λ− sj, which is a contradiction by Lemma 5.13.9 (iv).

Thus λ = ω1 +ω2, and so the labeled Dynkin diagram has at least three 2'sin the initial string. If the labeling starts with 2220 . . ., then r−4 is a�ordedby the weights λ − 12, λ − 23 and λ − 234. Here λ − 12 has multiplicity 2by Lemma 5.13.17 (i), so this contradicts Lemma 5.13.9 (iv). On the otherhand, if the labeling starts with 2222 . . ., then r − 6 is a�orded by λ− 123,λ− 234, λ− 122 and λ− 122. Here λ− 123 = (λ− 12)s3 has multiplicity 2by Lemma 5.13.17 (i), again a contradiction by Lemma 5.13.9 (iv).

(ii): This follows with the same argument as [LST15, Lemma 6.2 (iv)].

Lemma 5.13.31. Suppose that G = Bn, G = Cn or G = Dn. If λ = bω1 withb ≥ 2, then u does not act on V as a distinguished unipotent element.

Proof. If b = 2, the claim follows from Proposition 5.5.10 (a), Proposition 5.5.5(a) and Proposition 5.5.6 (a). If b = 3, then the claim follows from Proposition5.9.1.

Suppose then that b ≥ 4 and assume that u acts on V ↓ A as a distinguishedunipotent element. According to Lemma 5.13.12, the nodes α1 and α2 have label2, so n ≥ 3. Furthermore, by Lemma 5.13.12 (iii) the weight r − 2 of A hasmultiplicity 1. Suppose that α3 has label 0. In this case r−4 is a�orded by λ−12,λ−12 and λ−123, contradicting Lemma 5.13.9 (v). Therefore α3 has label 2, andso n ≥ 4. Suppose that α4 has label 0. Then r − 6 is a�orded by λ− 13, λ− 122,λ−123 and λ−1234, again a contradiction by Lemma 5.13.9 (v). Therefore α4 haslabel 2. Now r− 8 is a�orded by λ− 14, λ− 132, λ− 1222, λ− 1223 and λ− 1234,once again contradicting Lemma 5.13.9 (v) and completing our proof.

This completes the proof of Proposition 5.13.29 for G of classical type. If Gis of type Bn, Cn, or Dn and u acts on V = LG(λ) as a distinguished unipotentelement, then λ = ωi by Lemma 5.13.30 and Lemma 5.13.31.

We now move on to consider the exceptional groups.

Lemma 5.13.32. Let G = G2. If u acts on V as a distinguished unipotent ele-ment, then λ = ωi.

Proof. Suppose that λ 6= ωi. Since u is not regular, it cannot act on V as adistinguished unipotent element in view of Lemma 5.13.9 (i) and Lemma 5.13.12(ii).

Lemma 5.13.33. Let G = F4. Then if u acts on V as a distinguished unipotentelement, we have λ = ωi.

5.14. Proof of Theorem 1.1.10 and Theorem 1.1.11 193

Proof. We proceed similarly to [LST15, Lemma 6.7]. Suppose that u acts on V ↓ Aas a distinguished unipotent element. Consider �rst the possibility that λ = ciωi+cjωj , where i 6= j and ci, cj ≥ 1. Then by Lemma 5.13.14, the nodes αi and αjand nodes adjacent to them have label 2. But by looking at the possible labelingsgiven in Proposition 2.6.7, this means that all labels of the Dynkin diagram are2, contrary to our assumption that u is not regular.

Therefore λ = bωi for some i. If b ≥ 2, then by Lemma 5.13.12 the node αi is anend node. Furthermore, αi and the node adjacent to it have label 2. By Proposition2.6.7, this means that i = 1 and the labeled Dynkin diagram of u is 2202. Nowthe weight r − 4 is a�orded by λ− 12, λ− 12 and λ− 123, contradicting Lemma5.13.9 (v), since r − 2 has multiplicity 1 by Lemma 5.13.12. Thus λ = ωi.

Lemma 5.13.34. Let G = E6, G = E7 or G = E8. If u acts on V as a distin-guished unipotent element, then λ = ωi.

Proof. Assume that G = En and that u acts on V ↓ A as a distinguished unipotentelement.

Suppose �rst that ci, cj ≥ 1 for some i 6= j. Then λ = ciωi + cjωj by Lemma5.13.9 (ii). Now the arguments given in the beginning of the proof of [LST15,Lemma 6.8] show that we must have λ = ωn−1 +ωn. Since we are assuming p ≥ 7,by 4.5.7 and 4.5.3 the weight λ − n(n − 1) occurs with multiplicity 2. Then theargument given in [LST15, Proof of Lemma 6.8, paragraph 2] shows that we havea contradiction with Lemma 5.13.9 (iii).

Now consider λ = bωi. According to Lemma 5.13.12, the node αi is an end nodewith label 2, the node adjacent to αi has label 2, and the weight r − 2 of V ↓ Aoccurs with multiplicity 1. Now the arguments given in the proof of [LST15, Proofof Lemma 6.8, paragraphs 3-4] show that we have a contradiction with Lemma5.13.9 (iv) if b > 1. Therefore λ = ωi.

With Lemma 5.13.32, Lemma 5.13.33, and Lemma 5.13.34, the claim of Pro-position 5.13.29 for G of exceptional type follows. This also completes the proofof Theorem 5.13.1.

5.14 Proof of Theorem 1.1.10 and Theorem 1.1.11

In this section we will put results from previous sections together and prove The-orem 1.1.10 and Theorem 1.1.11.

Let u ∈ G be a unipotent element, and let λ ∈ X(T )+ be a non-zero p-restricted dominant weight.

We assume �rst that p 6= 2 and prove Theorem 1.1.10. Suppose that u actson LG(λ) as a distinguished unipotent element. It follows from Lemma 1.1.8 thatu is a distinguished unipotent element. If u has order p, then Theorem 1.1.10 isgiven by Theorem 5.13.1. Suppose then that u has order > p. In this case, for thedi�erent types, Theorem 1.1.10 is a consequence of the following results:

• G = Al (l ≥ 1): Here the only distinguished unipotent class is the regularone (Lemma 2.2.2), so u is a regular unipotent element. It follows fromProposition 5.1.4 and Lemma 5.1.5 that λ = ωi or λ = ω1 +ωl. In these casesthe claim of Theorem 1.1.10 follows from Proposition 5.12.1 and Proposition5.3.2, respectively.

194 Chapter 5.

• G = Bl or G = Cl (l ≥ 2): In this case, it follows from Proposition 5.1.14that either λ = ωi, λ = 2ω1, λ = 3ω1, Proposition 5.1.14 (iv) holds, orProposition 5.1.14 (v) holds. Now the claim of Theorem 1.1.10 follows fromProposition 5.12.1, Proposition 5.7.7, Proposition 5.7.11, Proposition 5.9.1,and Table 5.6.

• G = Dl (l ≥ 4): Here it follows from Proposition 5.1.18 that λ = ωi orλ = 2ω1. In these cases the claim of Theorem 1.1.10 follows from Proposition5.12.1 and Proposition 5.7.8, respectively.

• G = G2: In this case, the claim of Theorem 1.1.10 is given by Proposition5.1.19.

• G = F4, G = E6, G = E7, or G = E8: In this case, Theorem 1.1.10 followsfrom Lemma 5.1.20 and Proposition 5.12.1.

This completes the proof of Theorem 1.1.10.We assume next that p = 2 and proceed to prove Theorem 1.1.11. Suppose

that u acts on LG(λ) as a distinguished unipotent element. It follows from Lemma1.1.8 that u is a distinguished unipotent element. For each of the di�erent types,Theorem 1.1.11 is given by the following results:

• G = Al (l ≥ 1): Here u must be a regular unipotent element (Lemma 2.2.2).It follows from Proposition 5.2.2 and Lemma 5.2.2 (i) that λ = ω1, λ = ωl,λ = ω1 + ωl, or l = 3 and λ = ω2. In these cases, the claim follows fromTheorem 1.1.12, Proposition 5.4.4, and Lemma 5.2.2 (ii).

• G = Cl (l ≥ 2): It follows from Proposition 5.2.4 that λ = ω1, λ = ω2,λ = ωl, or l = 5 and λ = ω3. For λ = ω1 the claim is obvious since LG(ω1) isthe natural representation, and for rest of the cases the claim follows fromProposition 5.6.7, Proposition 5.11.3, and Lemma 5.2.5.

• G = Bl (l ≥ 2): It follows from Proposition 5.2.4 that λ = ωi. There existsan exceptional isogeny ϕ : Bl → Cl [Ste68, Theorem 28], and we haveLCl(ωi)

ϕ ∼= LBl(ωi) if 1 ≤ i ≤ l− 1, and LCl(ωi)ϕ ∼= LBl(ωl)

[1]. Thus in thiscase, the claim of Theorem 1.1.11 follows from the result for type Cl, whichwe have already proven.

• G = Dl (l ≥ 4): It follows from Proposition 5.2.4 that λ = ωi. In the casewhere λ = ωl−1 or λ = ωl, the result is given by Proposition 5.11.4. Supposethen that 1 ≤ i ≤ l− 2. Note that for a simple group H of type Cl, we haveG < H as the subsystem subgroup generated by short root subgroups, andLCl(ωi) ↓ G ∼= LG(ωi) for 1 ≤ i ≤ l−2 [Sei87, Theorem 4.1]. Thus the claimfollows from the result for type Cl.

• G simple of exceptional type: In this case, the claim is given by Proposition5.2.6 and the tables in Appendix B.

Chapter 6

Non-G-completely reducibleovergroups

Let G be a simple linear algebraic group and let u ∈ G be a distinguished uni-potent element. In this section, we study connected reductive subgroups X of Gcontaining u such that X is contained in some proper parabolic subgroup of G.Testerman and Zalesski [TZ13] have shown that no such X exist if u is a regularunipotent element. We will �nd that their result does not generalize to distinguis-hed unipotent elements and we will give several counterexamples in this section.However, we are able to give a complete list of X contained in a proper parabolicsubgroup in the case where u has order p (Theorem 6.2.12). In the case where uhas order > p, we have partial results. Furthermore, the general impression fromthe examples and results we have is that such subgroups X are quite rare and�nding a complete list of them should eventually be possible.

6.1 Preliminaries on G-complete reducibility

We begin by making a basic useful observation. Let X be a connected reductivesubgroup of G containing u, and assume that X is contained in a proper parabolicsubgroup P of G. Since every Levi factor of P is a centralizer of some non-trivialtorus, and since u is a distinguished unipotent element, it follows that X cannotbe contained in any Levi factor of P . In the terminology of the next de�nition dueto Serre, we say that X is a non-G-completely reducible subgroup of G.

De�nition 6.1.1 (Serre, [Ser03]). Let H be a closed subgroup of G. We say thatH is G-completely reducible (G-cr), if whenever H is contained in a parabolicsubgroup P of G, it is contained in a Levi factor of P . Otherwise we say thatH is non-G-completely reducible (non-G-cr). If H is not contained in any properparabolic subgroup of G, we say that H is G-irreducible (G-ir).

We record the observation made above in the next lemma.

Lemma 6.1.2. Let X < G be a reductive subgroup of G. Suppose that X containsa distinguished unipotent element of G. Then X is G-ir or X is non-G-cr.

Below we will list some well known properties of G-cr and G-ir subgroupsthat will be needed in the sequel. As seen from the next theorem, for classical

195

196 Chapter 6.

groups the concept of G-cr (G-ir) subgroups can be seen as a generalization ofsemisimplicity (irreducibility) in representation theory14.

Theorem 6.1.3 ([Ser05, 3.2.2]). Let G = SL(V ), G = Sp(V ), or G = SO(V ).Assume that p > 2 if G = Sp(V ) or G = SO(V ). Then a closed subgroup H < Gis G-cr if and only if V ↓ H is semisimple.

The following two results describe parabolic subgroups and G-ir subgroups insimple classical groups.

Theorem 6.1.4. Let G = SL(V ), G = Sp(V ), or G = SO(V ). Any parabolicsubgroup of G is of the form StabG(W1 ⊂ · · · ⊂ Wt), where W1 ⊂ · · · ⊂ Wt is a�ag of subspaces such that for all 1 ≤ i ≤ t, the subspace Wi is totally isotropic ifG = Sp(V ), and totally singular if G = SO(V ).

Proof. This is well known, for a proof, see for example [MT11, Proposition 12.13].

Theorem 6.1.5 ([LS96, pg. 32-33]). Let G = SL(V ), G = Sp(V ), or G = SO(V ).Let X < G be a connected reductive subgroup. Then X is G-ir if and only if oneof the following holds:

(i) G = SL(V ) and V ↓ X is irreducible;

(ii) G = Sp(V ) or G = SO(V ), and V ↓ X = V1 ⊥ · · · ⊥ Vt (orthogonal directsum), where:

• For all 1 ≤ i ≤ t, the Vi are non-degenerate subspaces and irreducibleX-modules,

• For all 1 ≤ i, j ≤ t with i 6= j, we have Vi 6∼= Vj as X-modules.

(iii) G = SO(V ), p = 2, dimV is even, and V ↓ X = V1 ⊥ V2 ⊥ · · · ⊥ Vt(orthgonal direct sum), where:

• V1 ↓ X is non-degenerate, the subgroup X stabilizes a nonsingular 1-space R of V1, and the image of the representation X → SO(R⊥) isSO(R⊥)-ir.

• For all 2 ≤ i ≤ t, the Vi are non-degenerate subspaces and irreducibleX-modules,

• For all 2 ≤ i, j ≤ t with i 6= j, we have Vi 6∼= Vj as X-modules.

6.2 Unipotent elements of order p

In this subsection, we consider unipotent elements of order p that are contained ina non-G-cr subgroup. As an application of results in Section 4.6 and recent workof Litterick and Thomas in [LT], we describe all unipotent elements that can becontained in some non-G-cr subgroup of type A1 when p is good for G. We will alsoshow that except for two known examples which occur in the case (G, p) = (C2, 2),

14For exceptional groups, reductive non-G-completely reducible subgroups only occur in smallcharacteristic [LS96, Theorem 1].

6.2. Unipotent elements of order p 197

any connected reductive subgroup containing a distinguished unipotent elementof order p must be G-ir (Theorem 6.2.12).

We begin by considering the situation where p is good forG. LetGC be a simplealgebraic group over C with the same root system Φ as G. In good characteristic,recall (De�nition 2.7.7) that for any unipotent element u ∈ G, we can de�ne aunipotent element uC ∈ GC which has the same labeled diagram as u.

De�nition 6.2.1. Assume that p is good for G. Let u ∈ G be a unipotent element.For uC ∈ GC, we de�ne N(uC) to be the largest Jordan block size of uC acting onthe adjoint representation of GC.

With the �order formula� proven by Testerman in [Tes95], Lawther has shownthe following result.

Theorem 6.2.2 ([Law95, Theorem 1]). Assume that p is good for G. Let u ∈ Gbe a unipotent element. Then u has order p if and only if N(uC) ≤ 2p− 1.

We now proceed to describe for each simple type when equality N(uC) = 2p−1holds. It turns out that with one exception (regular unipotent elements in G oftype Ap−1), equality holds if and only if the unipotent element is contained insome non-G-cr subgroup of type A1.

Lemma 6.2.3 (Type Al, l ≥ 2). Let G = SL(V ) and assume that dimV ≥ 3. Letu ∈ G be a unipotent element with V ↓ K[u] = Vd1 ⊕ · · · ⊕ Vdt, where d1 ≥ d2 ≥· · · ≥ dt > 0. Then

(i) N(uC) = 2d1 − 1.

(ii) N(uC) = 2p− 1 if and only if d1 = p.

(iii) u is contained in a non-G-cr subgroup X < G of type A1 if and only if d1 = pand t > 1.

Proof. Using the fact that the highest root in type Al is equal to ω1 +ωl, claim (i)can be seen from [Sup09, Proposition 1.5, Algorithm 1.6]. Claim (ii) is immediatefrom (i).

We consider claim (iii). For the �only if� part, suppose that u is contained ina non-G-cr subgroup X < G of type A1. Then u must have order p, so d1 ≤ p. Ifd1 < p, then by Corollary 4.6.9 the restriction V ↓ X is semisimple. By Theorem6.1.3 the subgroup X is G-cr, contradiction. Therefore d1 = p. Now if t = 1, thenV has dimension p and thus V ↓ X must be semisimple by [Jan97, (A)], again acontradiction by Theorem 6.1.3. Thus t > 1, giving the claim.

For the �if� part of claim (iii), suppose that d1 = p and t > 1. Let X = SL2(K)and �x a non-identity unipotent element v ∈ X. Consider �rst the case whered2 = p. By Lemma 4.6.2, for the indecomposable tilting X-module TX(p) we haveTX(p) ↓ K[v] = Vp ⊕ Vp. Furthermore, by Lemma 4.2.2 we have LX(di − 1) ↓K[u] = Vdi for all i. Thus we can identify V with the X-module

TX(p)⊕t⊕i=3

LX(di − 1),

which gives an embedding X < SL(V ) such that V ↓ K[v] = Vd1 ⊕ · · · ⊕ Vdt . Nowv and u are conjugate in G (Lemma 2.2.1), so by replacing X with a conjugate, we

198 Chapter 6.

can assume that u = v. Now V ↓ X is not semisimple since TX(p) is not semisimple(Theorem 4.1.5 (ii)), so by Theorem 6.1.3 the subgroup X is a non-G-cr subgroupof type A1 containing u.

Consider next the case where d2 < p. It follows from Lemma 4.6.1 that VX(p+d2−1) ↓ K[u] = Vp⊕Vd2 . As before, by Lemma 4.2.2 we have LX(di−1) ↓ K[u] =Vdi for all i. We can identify V with the X-module

VX(p+ d2 − 1)⊕t⊕i=3

LX(di − 1),

which gives an embedding X < SL(V ) such that V ↓ K[v] = Vd1 ⊕ · · · ⊕ Vdt . Asin the previous paragraph, we can assume that v = u. It follows from Theorem6.1.3 that X is a non-G-cr subgroup of type A1 containing u since VX(p+ d2− 1)is not semisimple (Proposition 4.6.5).

Lemma 6.2.4 (Type Cl, l ≥ 2). Let G = Sp(V ) and assume that p > 2 anddimV ≥ 4. Let u ∈ G be a unipotent element with V ↓ K[u] = Vd1 ⊕ · · · ⊕ Vdt,where d1 ≥ d2 ≥ · · · ≥ dt > 0. Then

(i) N(uC) = 2d1 − 1.

(ii) N(uC) = 2p− 1 if and only if d1 = p.

(iii) u is contained in a non-G-cr subgroup X < G of type A1 if and only ifd1 = p.

Proof. Claim (i), claim (ii), and the �only if� part of claim (iii) follow exactly asin the proof of Lemma 6.2.3.

For the �if� part of claim (iii), suppose that d1 = p. Let X = SL2(K) and �xa non-identity unipotent element v ∈ X. Now for unipotent elements in Sp(V ),odd Jordan block sizes must have even multiplicity (Proposition 2.3.2), so d2 = p.Furthermore, we can write

V ↓ K[u] = Vp ⊕ Vp ⊕r⊕i=1

2 · V2ai+1 ⊕s⊕j=1

V2bj

where ai ≥ 0 for all i, where bj ≥ 1 for all j, and 2 + 2r+ s = t. We proceed nextto realize each of these summands as X-modules with non-degenerate X-invariantalternating bilinear forms.

It follows from Corollary 4.4.11 that the tilting X-module TX(p) has a non-degenerate X-invariant alternating bilinear form, and from Lemma 4.6.2 thatTX(p) ↓ K[v] = Vp ⊕ Vp. For all 1 ≤ i ≤ r, it follows from Lemma 4.4.6 thatthe X-module LX(2ai)⊕ LX(2ai)

∗ has a non-degenerate X-invariant alternatingbilinear form, and from Lemma 4.2.2 that LX(2ai)⊕LX(2ai)

∗ ↓ K[v] = 2 ·V2ai+1.For all 1 ≤ j ≤ s, it follows from Corollary 4.4.4 that the X-module LX(2bj − 1)has a non-degenerateX-invariant alternating bilinear form, and from Lemma 4.2.2that LX(2bj − 1) ↓ K[v] = V2bj . Hence the X-module

W = TX(p)⊕r⊕i=1

(LX(2ai)⊕ LX(2ai)∗)⊕

s⊕j=1

LX(2bj − 1)

has a non-degenerate X-invariant alternating bilinear form andW ↓ K[v] = Vd1⊕· · · ⊕ Vdt . Thus by identifying V with W , we get an embedding X < Sp(V ) suchthat V ↓ K[v] = Vd1 ⊕ · · · ⊕ Vdt . By Proposition 2.3.1, the elements u and v areconjugate in G, so by replacing X with a conjugate we may assume that u = v.Now V ↓ X is not semisimple since TX(p) is not semisimple (Theorem 4.1.5 (ii)),so by Theorem 6.1.3 the subgroup X is a non-G-cr subgroup of type A1 containingu.

Lemma 6.2.5 (Type Bl (l ≥ 3) and type Dl (l ≥ 4)). Let G = SO(V ) andassume that p > 2 and dimV ≥ 7. Let u ∈ G be a unipotent element with V ↓K[u] = Vd1 ⊕ · · · ⊕ Vdt , where d1 ≥ d2 ≥ · · · ≥ dt > 0. Then

(i) N(uC) =

2d1 − 1, if d1 = d2.

2d1 − 2, if d2 = d1 − 1.

2d1 − 3, if d2 ≤ d1 − 2.

(ii) N(uC) = 2p− 1 if and only if d1 = d2 = p.

(iii) u is contained in a non-G-cr subgroup X < G of type A1 if and only ifd1 = d2 = p.

Proof. Using the fact that the highest root in type Bl (l ≥ 3) and type Dl (l ≥ 4)is equal to ω2, claim (i) follows with [Sup09, Proposition 1.5, Algorithm 1.6]. Forclaim (ii), it is immediate from (i) that d1 = d2 = p implies N(uC) = 2p− 1. Forthe other direction of (ii), suppose that N(uC) = 2p − 1. If d2 = d1 − 1, then by(i) we have N(uC) = 2d1 − 2, contradiction since 2p − 1 is odd. If d2 ≤ d1 − 2,then by (i) we have N(uC) = 2d1− 3 = 2p− 1, so d1 = p+ 1. But then d1 is even,which is a contradiction since even Jordan block sizes must have even multiplicityby Proposition 2.3.2. Therefore we must have d1 = d2. Now it is immediate from(i) and N(uC) = 2p− 1 that d1 = d2 = p.

For the �only if� part of claim (iii), suppose that u is contained in a non-G-crsubgroup X < G of type A1. If d1 < p or d2 < p, it follows that u acts on V with≤ 1 Jordan block of size p. Since X < SO(V ), the restriction V ↓ X is self-dual,and thus V ↓ X is semisimple by Proposition 4.6.10. Then by Theorem 6.1.3 thesubgroup X is G-cr, contradiction.

For the �if� part of claim (iii), suppose that d1 = d2 = p. Let X = SL2(K)and �x a non-identity unipotent element v ∈ X. By Proposition 2.3.2, the evenJordan block sizes of u have even multiplicity, so we can write

V ↓ K[u] = Vp ⊕ Vp ⊕r⊕i=1

2 · V2ai ⊕s⊕j=1

V2bj+1

where ai ≥ 1 for all i and bj ≥ 0 for all j, and 2+2r+s = t. Similarly to the proofof Lemma 6.2.4, we proceed to construct these summands as suitable X-moduleswith non-degenerate symmetric bilinear forms.

It follows from Corollary 4.4.11 that the tilting X-module TX(p + 1) has anon-degenerate X-invariant symmetric bilinear form, and from Lemma 4.6.2 thatTX(p+1) ↓ K[v] = Vp⊕Vp. For all 1 ≤ i ≤ r, it follows from Lemma 4.4.6 that theX-module LX(2ai−1)⊕LX(2ai−1)∗ has a non-degenerate X-invariant symmetricbilinear form, and from Lemma 4.2.2 that LX(2ai − 1) ⊕ LX(2ai − 1)∗ ↓ K[v] =2·V2ai . For all 1 ≤ j ≤ s, it follows from Corollary 4.4.4 that theX-module LX(2bj)

200 Chapter 6.

has a non-degenerate X-invariant symmetric bilinear form, and from Lemma 4.2.2that LX(2bj) ↓ K[v] = V2bj+1. Hence the X-module

W = TX(p+ 1)⊕r⊕i=1

(LX(2ai)⊕ LX(2ai)∗)⊕

s⊕j=1

LX(2bj − 1)

has a non-degenerate X-invariant symmetric bilinear form, and W ↓ K[v] =Vd1 ⊕ · · · ⊕ Vdt . By identifying V with W , and replacing X with PGL2(K) ifthe representation ρ : X → GL(W ) has nontrivial kernel, we get an embeddingX < SO(V ) such that V ↓ K[v] = Vd1 ⊕ · · · ⊕ Vdt . By Proposition 2.3.1, theunipotent elements u and v are conjugate in the full orthogonal group O(V ), soby replacing X with a conjugate we can assume that u = v. Now V ↓ X is notsemisimple since TX(p+ 1) is not semisimple (Theorem 4.1.5 (ii)), so by Theorem6.1.3 the subgroup X is a non-G-cr subgroup of type A1 containing u.

Lemma 6.2.6. Let G be a simple algebraic group of exceptional type and assumethat p is good for G. Let u ∈ G be a unipotent element of order p. Then u iscontained in a non-G-cr subgroup X < G of type A1 precisely in the followingcases:

(i) G = E6, p = 5, and u is in unipotent class A4 or A4A1.

(ii) G = E7, p = 5, and u is in unipotent class A4, A4A1, or A4A2.

(iii) G = E7, p = 7, and u is in unipotent class A6.

(iv) G = E8, p = 7, and u is in unipotent class A6 or A6A1.

Proof. Here we will rely heavily on the work of Litterick and Thomas in [LT]. Themain result of [LT] gives a complete list of non-G-cr subgroups X < G of typeA1, up to G-conjugacy. Thus to prove our claim, it will be enough to check foreach X which conjugacy class of unipotent elements of order p it intersects. Notethat there is only one such conjugacy class, since in X all non-identity unipotentelements are conjugate.

For each non-G-cr subgroupX, Litterick and Thomas give theX-module struc-ture of the restriction of the adjoint representation of G, see [LT, Table 11 - Table16]. Using results on the action of a non-identity unipotent element u ∈ X onX-modules (example given below), this allows us to compute the Jordan blocksizes of a non-identity unipotent element u ∈ X on the adjoint representation.Then by [Law95, Theorem 2], we can use the tables in [Law95] to identify the pre-cise conjugacy class of u in G. Doing this straightforward15 computation for eachnon-G-cr subgroup of type A1 given in [LT], one �nds that they can only containunipotent elements listed in (i) - (iv), and that all of the unipotent elements in (i)- (iv) are contained in some non-G-cr subgroup of type A1.

We give one example of how the computation is done, all the other compu-tations use similar methods. Let p = 5 and G = E6. Fix non-negative integers rand s such that rs = 0. We consider an A1 subgroup X of G, which is embeddedinto a maximal rank subgroup of type A1A5 via the representations LX(1)[r] (forA1 factor) and VX(5)[s] (for A5 factor). According to [LT], the subgroup X is

15But perhaps tedious, since there are 26 entries in the tables of [LT] to check.

non-G-cr, and by [LT, Table 11] restriction of the adjoint representation of G toX decomposes into a direct sum

LX(1)[r]⊗LX(9)[s]+LX(1)[r]⊗TX(5)[s]+LX(2)[r]+TX(10)[r]+TX(6)[s]+LX(4)[s].

Let u ∈ X be a non-identity unipotent element of X. For any X-module V ,the Jordan block sizes of u acting on V and any Frobenius twist of V are equal.Thus it will su�ce to compute the Jordan block sizes of u on acting the X-module

LX(1)⊗ LX(9) + LX(1)⊗ TX(5) + LX(2) + TX(10) + TX(6) + LX(4).

We proceed to �nd the K[u]-module decomposition for each of the summands.

• LX(1) ⊗ LX(9): By Steinberg's tensor product theorem, we have LX(9) ∼=LX(1)[1] ⊗ LX(4). Then by Lemma 4.2.2, we have LX(1)⊗ LX(9) ↓ K[u] ∼=V2 ⊗ V2 ⊗ V5, which decomposes to 4 · V5 by Lemma 3.3.6. Thus LX(1) ⊗LX(9) ↓ K[u] = 4 · V5.

• LX(1) ⊗ TX(5): Now LX(1) ↓ K[u] = V2 by Lemma 4.2.2, and TX(6) ↓K[u] = 2 · V5 by Lemma 4.6.2. Hence LX(1) ⊗ TX(5) ↓ K[u] = 4 · V5 byLemma 3.3.6.

• LX(2): Here LX(2) ↓ K[u] = V3 by Lemma 4.2.2.

• TX(10): With Theorem 4.1.5, one computes that TX(10) has dimension 20,so by Lemma 4.6.2 we have TX(10) ↓ K[u] = 4 · V5.

• TX(6): Here TX(6) ↓ K[u] = 2 · V5 by Lemma 4.6.2.

• LX(4): Here LX(4) ↓ K[u] = V5 by Lemma 4.2.2.

It follows then that L (G) ↓ K[u] = [3, 515]. By [Law95, Theorem 2, Table 6],the unipotent element u lies in the conjugacy class A4A1 of G.

If p is good forG, then using lemmas 6.2.3 - 6.2.6 we can now describe unipotentelements that are contained in a non-G-cr subgroup of order p in terms of N(uC).

Proposition 6.2.7. Let G be a simple algebraic group and assume that p is goodfor G. Let u ∈ G be a unipotent element. Then N(uC) = 2p− 1 if and only if oneof the following holds:

(i) G is of type Ap−1 and u ∈ G is a regular unipotent element,

(ii) u is contained in a non-G-cr subgroup X < G of type A1.

Proof. Suppose that N(uC) = 2p − 1. If G is simple of classical type, the claimfollows from (ii) and (iii) of lemmas 6.2.3 - 6.2.5. If G is simple of exceptional type,it follows from [Law95, pg. 4128] that the conjugacy class of u is given in one of(i) - (iv) of Lemma 6.2.6.

For the other direction, if (i) holds, then we have N(uC) = 2p− 1 by Lemma6.2.3 (i). Suppose then that (ii) holds. If G is simple of classical type, we haveN(uC) = 2p−1 by (ii) and (iii) of lemmas 6.2.3 - 6.2.5. If G is simple of exceptionaltype, we have N(uC) = 2p− 1 by Lemma 6.2.6 and [Law95, pg. 4128].

As an immediate corollary, we have the following.

202 Chapter 6.

Corollary 6.2.8. Let G be a simple algebraic group and assume that p is verygood for G. Let u ∈ G be a unipotent element. Then u is contained in a non-G-crsubgroup X < G of type A1 if and only if N(uC) = 2p− 1.

Our proof of Proposition 6.2.7 and Corollary 6.2.8 is heavily based on case-by-case checking. It would be interesting to see if there is a general way to prove theresult.

We now move on to consider non-G-cr subgroups containing distinguishedunipotent elements of order p. The following result shows that no such subgroupsexist if p is good for G.

Theorem 6.2.9. Let G be a simple algebraic group and assume that p is goodfor G. Let u ∈ G be a distinguished unipotent element of order p. If X < G is aconnected reductive subgroup of G containing u, then X is G-ir.

Proof. Let X be a connected reductive subgroup of G containing u. We consider�rst the case where X is simple of type A1. By Lemma 6.1.2, it will be enoughto show that X is G-cr. If G is simple of exceptional type, it is immediate fromLemma 6.2.6 that X is G-cr. If G is simple of classical type, one can easily see thatX is G-cr by using (iii) in lemmas 6.2.3 - 6.2.5, and the description of distinguishedunipotent classes of G in terms of Jordan block sizes (Proposition 2.3.4).

Consider then the general case where X is a connected reductive subgroup ofG containing u. Since u is not centralized by a nontrivial torus, the same mustbe true for X, so it follows that X is semisimple. Write X = X1 · · ·Xt, wherethe Xi are simple and commute pairwise. If p 6= 3 or if no Xi is of type G2, itfollows from [Tes95, Theorem 0.1] and [PST00, Theorem 5.1] that there exists aconnected simple subgroup X ′ < X of type A1 such that u ∈ X ′. It follows fromthe previous paragraph that X ′ is G-ir, so X must be G-ir as well.

Suppose then that p = 3 and that some Xi is of type G2. Since p is good forG, it follows that G is simple of classical type. Let V be the natural module for G.Now u is distinguished of order 3, so it follows from Lemma 5.1.1 that dimV ≤ 4.Since every non-trivial representation of a simple algebraic group of type G2 hasdimension > 4 (see e.g. [Lüb01]), it follows that no Xi can be simple of type G2,contradiction.

What remains then is to consider distinguished unipotent elements of order pin bad characteristic. There are only two cases where such elements exist (typeC2 for p = 2, and type G2 for p = 3, see proof of Theorem 6.2.12). The next twolemmas will deal with type C2 for p = 2.

Lemma 6.2.10. Assume that p = 2 and let G = SL2(K). Let V be a G-modulesuch that V ↓ K[u] = [2, 2] for a non-identity unipotent element u ∈ G. Then oneof the following holds:

(i) V is irreducible and isomorphic to LG(1)[n] ⊗ LG(1)[m], where 0 ≤ n < m.

(ii) V ∼= LG(1)[n] ⊕ LG(1)[m], where 0 ≤ n ≤ m.

(iii) V ∼= TG(2)[n] ∼= LG(1)[n] ⊗ LG(1)[n], where n ≥ 0.

Proof. If V is irreducible, then the fact that V has dimension 4 implies thatV ∼= LG(1)[n]⊗LG(1)[m] for some 0 ≤ n < m, so V is as in (i). Suppose then thatV is not irreducible. Then the possibilities for the composition factors of V are

(1) Two composition factors: LG(1)[n] and LG(1)[m] for some 0 ≤ n ≤ m.

(2) Three composition factors: LG(0) (twice), and LG(1)[n] for some n ≥ 0.

In case (1), we have V ∼= LG(1)[n]⊕LG(1)[m] since Ext1G(LG(1)[n], LG(1)[m]) =

0 = Ext1G(LG(1)[m], LG(1)[n]) by Theorem 4.6.3. Thus V is as in (ii).

Consider then case (2), where V has composition factors LG(0), LG(0), andLG(1)[n] for some n ≥ 0. If n = 0, then V ∼= LG(0) ⊕ LG(0) ⊕ LG(1), sinceExt1

G(LG(0), LG(0)) = 0 and Ext1G(LG(1), LG(0)) = 0 = Ext1

G(LG(0), LG(1)) byTheorem 4.6.3. In this case V ↓ K[u] = [1, 1, 2], contradicting the assumptionthat V ↓ K[u] = [2, 2]. Therefore V has composition factors LG(0), LG(0), andLG(1)[n+1] = LG(2)[n] for some n ≥ 0. We show next that V must be indecompo-sable. Suppose that V = W ⊕W ′ whereW,W ′ are proper non-zero G-submodulesof V . Since Ext1

G(LG(0), LG(0)) = 0, it follows that either W or W ′ has LG(0) asa direct summand. But then u acts on V with at least one Jordan block of size 1,contradicting the assumption that V ↓ K[u] = [2, 2]. Therefore V is indecompo-sable.

Since V is indecomposable and not irreducible, there must be a subquotient Qof V which is a non-split extension between two irreducible G-modules. Now Q is aproper subquotient of V , so u acts onQ with exactly one Jordan block of size 2, andthus the subquotient Q must be isomorphic to VG(2)[n] or (VG(2)∗)[n] (Proposition4.6.8). By replacing V with V ∗ if necessary, we may assume that Q is isomorphicto VG(2)[n]. Since V is indecomposable, it follows that V is a nonsplit extensionof VG(2)[n] and LG(0). By Lemma 4.6.6, it follows that V ∼= TG(2)[n]. Finally, asnoted in the proof of Lemma 4.4.12, we have LG(1) ⊗ LG(1) = TG(2). ThereforeLG(1)[n] ⊗ LG(1)[n] ∼= TG(2)[n]. This completes the proof of the lemma.

Lemma 6.2.11. Assume that p = 2. Let G = Sp(V ), where dimV = 4, so Gis simple of type C2. Fix a distinguished unipotent element u ∈ G of order p,so V ↓ K[u] = V (2)2 (Proposition 2.4.4). If X < G is connected reductive andu ∈ X, then X is G-ir unless one of the following holds:

(i) X is simple of type A1, embedded into G via LX(1) ⊥ LX(1) (orthogonaldirect sum).

(ii) X is simple of type A1, embedded into G via LX(1)⊗ LX(1) ∼= TX(2).

Furthermore, subgroups X < G in (i) and (ii) exist, contain a conjugate ofu, are contained in a proper parabolic subgroup, and the conditions (i) and (ii)determine X up to conjugacy in G.

Proof. Suppose that X < G is connected reductive and u ∈ X. We show thateither X is G-ir, or one of (i) or (ii) holds.

If X is normalized by a maximal torus of G, then X is G-cr by [BMR05,Proposition 3.20]. Since u is distinguished, it follows from Lemma 6.1.2 that X isG-ir.

Suppose then that X is not normalized by any maximal torus of G. Since Ghas rank 2, it follows that X is simple of type A1. Now there exists a rationalrepresentation ρ : SL2(K) → SL(V ) such that ρ(SL2(K)) = X. If V is an irre-ducible X-module, then by Theorem 6.1.5 the subgroup X is G-ir. Thus we mayassume that V is non-irreducible. Then by Lemma 6.2.10, as an SL2(K)-moduleV must be isomorphic to either

204 Chapter 6.

(1) LSL2(K)(1)[n] ⊕ LSL2(K)(1)[m], where 0 ≤ n ≤ m; or

(2) TSL2(K)(2)[n] ∼= LSL2(K)(1)[n] ⊗ LSL2(K)(1)[n], where n ≥ 0.

In case (1) we have ρ = ρ′ ◦Fn, where F is the usual Frobenius endomorphismand ρ′ : SL2(K)→ SL(V ) is a rational representation such that the correspondingSL2(K)-module V is isomorphic to LSL2(K)(1) ⊕ LSL2(K)(1)[m−n]. Similarly incase (2), we have ρ = ρ′ ◦ Fn, where F is the usual Frobenius endomorphism andρ′ : SL2(K) → SL(V ) is a rational representation such that the correspondingSL2(K)-module V is isomorphic to TSL2(K)(2). Since applying a Frobenius twistdoes not change the image of the representation ρ′, it follows that we may assumethat as an SL2(K)-module, V is isomorphic to either

(1)′ LSL2(K)(1)⊕ LSL2(K)(1)[n], where n ≥ 0; or

(2)′ TSL2(K)(2).

In case (2)′ we have X as in case (ii) of the claim. Consider then case (1)′.Here if n > 0, it follows from Theorem 6.1.5 (ii) that X is G-ir. Suppose thenthat n = 0, so V ∼= LSL2(K)(1) ⊕ LSL2(K)(1). Let W be a X-submodule of Vsuch that W ∼= LSL2(K)(1). If W is totally singular, then X is a Levi factor oftype A1 and a non-identity unipotent element u ∈ X satis�es V ↓ K[u] = W (2)[LS12, 6.1]. This contradicts the assumption V ↓ K[u] = V (2)2. Therefore W isnot totally singular, and thus it must be non-degenerate, since W is irreducibleas an X-module. Thus V = W ⊕W ′ as an orthogonal direct sum of X-modules,where W ∼= LSL2(K)(1) ∼= W ′. Thus X is as in case (i) of the claim.

We consider the existence and uniqueness claims for the subgroups X in (i)and (ii). We begin by considering subgroups X in (i). Now the SL2(K)-moduleLSL2(K)(1) has a non-degenerate SL2(K)-invariant alternating bilinear form byLemma 4.4.5. Therefore it is clear that we can �nd a representation ρ : SL2(K)→G with V = V1 ⊥ V2 (orthogonal direct sum) such that V1

∼= LSL2(K)(1) ∼= V2, sowe can choose X = ρ(SL2(K)).

We show next that such an X is the unique such subgroup up to G-conjugacy.For this, let Y = ρ′(SL2(K)) for some representation ρ′ : SL2(K) → G such thatV = W1 ⊥ W2, where W1 and W2 are Y -modules such that W1

∼= LSL2(K)(1) ∼=W2. It is well known that the orthogonal direct sum decompositions V1 ⊥ V2 andW1 ⊥W2 are conjugate under the action of G, in other words, there exists f ∈ Gsuch that f(W1) = V1 and f(W2) = V2. Then fY f−1 stabilizes V1 and V2, so byreplacing Y with fY f−1, we may assume that W1 = V1 and W2 = V2. Now fori = 1, 2, the restrictions of ρ and ρ′ to Vi are both isomorphic to LSL2(K)(1) asSL2(K)-modules, so there exists xi ∈ SL(Vi) such that xiρ′(g)x−1

i = ρ(g) for allg ∈ SL2(K). Since dimVi = 2, we have SL(Vi) = Sp(Vi), so for x = x1 ⊕ x2 wehave x ∈ G and xY x−1 = X. Thus X is unique up to G-conjugacy.

Next, note that since a non-identity unipotent element u ∈ X acts on themodule LSL2(K)(1) with a single Jordan block of size 2, it is clear that V ↓ K[u] =V (2)2. Then for subgroups X in (i), what still remains is to show that X iscontained in a proper parabolic subgroup of G. To this end, let W,W ′ be X-submodules of V such that V = W ⊥ W ′ with W ∼= L(1) ∼= W ′. Let (−,−) bethe non-degenerate alternating bilinear form used to de�ne G. There exists an X-equivariant isometry f : W →W ′, in other words, an isomorphism of X-modulessuch that (f(w1), f(w2)) = (w1, w2) for all wi ∈ W . Then it is straightforward to

6.3. Unipotent elements of order > p 205

check that Z = {w+ f(w) : w ∈W} is a totally singular X-submodule of V withZ ∼= LSL2(K)(1). By Theorem 6.1.4, the subgroup X lies in a proper parabolicsubgroup of G. This completes the proof of the claims for (i).

For the subgroup X in (ii), note that by Lemma 4.4.12 the tilting SL2(K)-module TSL2(K)(2) has a non-degenerate SL2(K)-invariant alternating bilinearform. Therefore there exists a representation ρ : SL2(K) → G, with V ↓ X ∼=TSL2(K)(2) for X = ρ(SL2(K)). Now by Lemma 4.4.13, a non-degenerate SL2(K)-invariant alternating bilinear form on TSL2(K)(2) is unique up to scalar multiples;thus it follows that a subgroup X < G of type A1 with V ↓ X ∼= TSL2(K)(2)is unique up to G-conjugacy (Lemma 4.4.14). Next note that LSL2(K)(1) has anon-degenerate SL2(K)-invariant alternating bilinear form by Lemma 4.4.5, soby uniqueness of the non-degenerate SL2(K)-invariant alternating bilinear formon TSL2(K)(2), we may choose the form on TSL2(K)(2) to be the product form onLSL2(K)(1)⊗LSL2(K)(1) = TSL2(K)(2). Now it follows from Table 3.6 that for a non-identity unipotent element u ∈ X, we have TSL2(K)(2) ↓ K[u] ∼= V (2) ⊗ V (2) ∼=V (2)2. Finally to show that X is contained in a parabolic subgroup, note thatV ∼= TSL2(K)(2) has a 1-dimensional X-submodule W (Theorem 4.1.5 (ii)). Thissubmodule is totally isotropic since any 1-dimensional subspace of V is, so X iscontained in a proper parabolic subgroup of G (Theorem 6.1.4). This completesthe proof of the claims for (ii).

We are now ready to prove the main result of this section.

Theorem 6.2.12. Let G be a simple algebraic group and let u ∈ G be a distinguis-hed unipotent element of order p. Let X < G be a connected reductive subgroup ofG containing u. If X is contained in a proper parabolic subgroup of G, then p = 2,G is simple of type C2, and X is as in Lemma 6.2.11.

Proof. Suppose that X is contained in a proper parabolic subgroup of G. Since uis a distinguished unipotent element, X must be non-G-cr (Lemma 6.1.2). Thenby Theorem 6.2.9 we have that p is bad for G.

Consider �rst the case where G is simple of exceptional type. Looking at theTables in Appendix A, the fact that p is bad for G and the fact that u is adistinguished unipotent element of order p implies that G = G2, p = 3, and u isin unipotent class G2(a1) or (A1)3. However, in this case it follows from [Ste10]that every connected reductive subgroup G is G-cr, contradicting the fact that Xis non-G-cr.

Suppose then that G is simple of classical type. Since p is bad for G, we havep = 2 and G is simple of type Bl (l ≥ 3), Cl (l ≥ 2), or Dl (l ≥ 4). Let V = L(ω1)be the natural irreducible representation of G. Now u is a distinguished unipotentelement of order p, so u acts on V with all Jordan block sizes even of size ≤ 2(Proposition 2.4.4). Therefore dimV ≤ 4, so l ≤ 2. Therefore G is simple of typeC2, and in this case the claim follows from Lemma 6.2.11.

6.3 Unipotent elements of order > p

In this subsection, we consider distinguished unipotent elements of order > p thatare contained in a non-G-cr subgroup. Unlike in the previous subsection where weconsidered unipotent elements of order p, we only have partial results.

206 Chapter 6.

Exceptional types

We begin by considering the case of simple groups of exceptional type. Relying onthe results of Litterick and Thomas in [LT], we get the following.

Theorem 6.3.1. Let G be a simple algebraic group of exceptional type and assumethat p is good for G. Let u ∈ G be a distinguished unipotent element. If X < G isa connected reductive subgroup of G containing u, then X is G-ir.

Proof. Let X be a connected reductive subgroup of G containing u. Suppose thatX is not G-ir. Since u is a distinguished unipotent element, it follows that X isnon-G-cr (Lemma 6.1.2). Now according to [LT, Theorem 1-4], one of the followingmust hold:

• p = 5, and X is semisimple of type A1, A1A1, or A1A2.

• p = 7, and X is semisimple of type A1, G2, A1A1, or A1G2.

We can see that in all of the cases above, unipotent elements of X have orderp. Indeed, this is true for simple algebraic groups of type A1 for all p, for type A2

for all p ≥ 3, and for type G2 for all p ≥ 7 (Appendix A). It follows then fromTheorem 6.2.9 that X is G-ir.

For type G2, we can give the following result as an easy corollary of [Ste10].

Theorem 6.3.2. Let G be a simple algebraic group of type G2. Let u ∈ G be adistinguished unipotent element. If X < G is a connected reductive subgroup of Gcontaining u, then X is G-ir.

Proof. Let X < G be a connected reductive subgroup of G containing u. If Xis not G-ir, then the fact that u is distinguished implies that X is non-G-cr. Itfollows from [Ste10] that p = 2 and X is simple of type A1. This is a contradiction,since now every distinguished unipotent element of G has order > 2 (AppendixA), and every non-identity unipotent element of X has order 2.

Let G be a simple algebraic group of exceptional type. By the two theoremsabove, for classifying connected reductive non-G-cr overgroups of distinguishedunipotent elements of G, what still remains is the case where G is simple of typeF4, E6, E7, or E8 in bad characteristic. In this case the connected reductive non-G-cr subgroups of G are not known in general, so more work is still needed.

Classical types

We �nish by discussing the situation for simple algebraic groups G of classicaltype. Here our work is still in progress, so we will mostly just give some examplesof simple non-G-cr subgroups containing a distinguished unipotent element.

For this problem, the basic reductions such as Lemma 5.2.1 and Lemma 5.1.1used in the irreducible case still work. With them, one can try to reduce to a smallnumber of possible composition factors that can occur, and then study extensionsbetween them to �nd a solution. In many cases we have a solution, and at least inthe case where X is simple, a complete solution should be doable in future work.When we have a connected semisimple algebraic group X = X1 · · ·Xt, one shouldtry to limit the number and the types of simple factors Xi that can occur. This isagain something which will be studied in future work.

Example 6.3.3. Assume that p = 3. We will show the following:

(i) For G = SO(V ) with dimV = 27, up to G-conjugacy there exists a uniqueX < G simple of type F4 such that V ↓ X ∼= TX(ω4);

(ii) Such an X is non-G-cr;

(iii) A regular unipotent element u ∈ X lies in the distinguished unipotent class[3, 9, 15] of G.

Let X be a simple algebraic group of type F4. We begin by showing thatTX(ω4) = LX(0)/LX(ω4)/LX(0) and that TX(ω4) is uniserial.

We use the usual embedding of X into a simply connected group Y of type E6.That is, we consider X as the centralizer of the involutory graph automorphismof Y induced by the nontrivial automorphism of the Dynkin diagram of Y .

Since in type E6 the fundamental highest weight ω1 is minuscule, we haveLY (ω1) = TY (ω1). According to [vdK01, Theorem 20], the restriction of everytilting module of Y to X is a tilting module for X. In particular, the restrictionLY (ω1) ↓ X is tilting for X. For the character of this restriction, we have

chLY (ω1) ↓ X = chVX(ω4) + chVX(0)

by [LS96, Table 8.7]. Thus LY (ω1) ↓ X has TX(ω4) as a direct summand (Theorem4.1.4 (ii)). Furthermore, since p = 3, we have VX(ω4) = LX(ω4)/LX(0) (see e.g.[Lüb17]). Now since TX(ω4) must have VX(ω4) as a submodule and VX(ω4)∗ as ahomomorphic image [Jan03, II.E.4], we conclude that

LY (ω1) ↓ X = TX(ω4) = LX(0)/LX(ω4)/LX(0)

and that TX(ω4) is uniserial.By Lemma 4.4.10, we have a non-degenerate X-invariant symmetric bilinear

form on TX(ω4). Now (i) follows from Lemma 4.4.15. For claim (ii), note thatTX(ω4) is an indecomposable and non-irreducible X-module, so it follows thatX is non-G-cr in G = SO(TX(ω1)). For (iii), let u ∈ X be a regular unipotentelement. From the way X is constructed as the stabilizer of a graph automorphismof Y , it is clear that the subgroup X contains a regular unipotent element of Y .Hence it follows that u is also a regular unipotent element of Y , see e.g. [TZ13,Lemma 2.1]. According to [Law95, Table 5], we have LY (ω1) ↓ K[u] = [3, 9, 15],hence TX(ω4) ↓ K[u] = [3, 9, 15].

We will omit the proof, but for simple algebraic groups of exceptional type, itturns out that Example 6.3.3 is the only example in odd characteristic16.

Theorem 6.3.4. Assume that p > 2. Let X be a simple algebraic group of excep-tional type. Suppose that X < G, where G is a simple algebraic group of classicaltype with natural module V . If X contains a distinguished unipotent element of G,then X is G-ir, unless p = 3, X = F4, and V ↓ X = TX(ω4).

We will now give some further examples in characteristic p = 2.

16For X simple of classical type, so far we have not found any examples in odd characteristic,and based on partial results and examples we conjecture that there are none. In other words, weconjecture that Theorem 6.3.4 holds for any simple algebraic group X, not just X of exceptionaltype.

208 Chapter 6.

Example 6.3.5. (cf. Example 6.3.3) Assume that p = 2. We will show that thefollowing hold:

(i) For G = Sp(V ) with dimV = 8, up to G-conjugacy we have a unique X < Gsimple of type G2 such that V ↓ X ∼= TX(ω1);

(ii) Such an X is non-G-cr.

(iii) A regular unipotent element u ∈ X lies in the distinguished unipotent classV (2) + V (6) of G.

Let X be a simple algebraic group of type G2. It is well known that we have anembedding of X into a simply connected group Y of type D4, by considering X asthe centralizer of a suitable triality automorphism. Let V be the natural module forY . We have that V ↓ X is tilting by [Bru98, Proposition 3.3 (vi)], so as in Example6.3.3, one �nds that V ↓ X = TX(ω1) and that TX(ω1) = LX(0)/LX(ω1)/LX(0)is uniserial.

Since V has a non-degenerate Y -invariant alternating bilinear form, we getan embedding X < G for G = Sp(V ). By Lemma 4.4.13 and Lemma 4.4.14, asubgroup X < G of type G2 with V ↓ X ∼= TX(ω1) is uniquely determined up toG-conjugacy, proving (i).

Claim (ii) follows from the fact that V is an indecomposable, non-irreducibleX-module. For (iii), let u ∈ X be a regular unipotent element. Now u is also aregular unipotent element of Y , so we have the orthogonal decomposition V ↓K[u] = V (2)+V (6) (Proposition 2.4.4 (vi)), and so u is a distinguished unipotentelement of G (Proposition 2.4.4 (ii)).

Example 6.3.6. We give two other examples similar to Example 6.3.3 and Ex-ample 6.3.5, without giving all the details. Let p = 2. Then one can show that thefollowing hold17:

(i) For G simply connected of type Cl (l ≥ 2) with natural module V , upto G-conjugacy we have a unique X < G simple of type Bl−1 such thatV ↓ X ∼= TX(ω1).

(ii) For G simply connected of type C67 with natural module V , up to G-conjugacy we have a unique X < G simple of type E7 such that V ↓ X ∼=TX(ω1).

Furthermore, in both (i) and (ii) one �nds that X is non-G-cr and a regularunipotent element of X is distinguished in G. In case (i) a regular unipotentelement of X is in the unipotent conjugacy class V (2) + V (2l − 2) of G, and incase (ii) in the unipotent conjugacy class V (2) +V (8) +V (10) +V (16) +V (18) +V (22) + V (26) + V (32) of G.

We �nish with the following more involved example for X = G2.

17We omit the full details, but to see this, it is enough to show that in both (i) and (ii), wehave that TX(ω1) = LX(0)/LX(ω1)/LX(0) is uniserial and has a non-degenerate X-invariantalternating bilinear form. Then (i) and (ii) follow from Lemma 4.4.13 and Lemma 4.4.14.


Example 6.3.7. Assume that p = 2. For G = Sp14(K), we will describe a familyof non-G-cr simple subgroups of type G2, parametrized by Z≥0 and K \ {0, 1},such that the subgroups are pairwise non-conjugate and each of them contain adistinguished unipotent element of G.

Let X be a simple algebraic group of type G2 and let u ∈ X be a regularunipotent element. Let n ≥ 1 be an integer.

It is well known (and straightforward to compute) that the Weyl module V (ω1)for X is a nonsplit extension

0→ K → V (ω1)→ L(ω1)→ 0. (*)

Applying the functor HomX(L(ω1)[n],−) to (*) gives an exact sequence

0→ Ext1X(L(ω1)[n],K)→ Ext1

X(L(ω1)[n], V (ω1))→ Ext1X(L(ω1)[n], L(ω1)).

It follows from the main result of [Sin92] that H1(X,L(ω1) ⊗ L(ω1)[n]) = 0, soExt1

X(L(ω1)[n], L(ω1)) = 0. Similarly Ext1X(L(ω1)[n],K) ∼= H1(X,L(ω1)[n]) ∼= K

by [Sin92]. Plugging this into the exact sequence above, we get the isomorphismExt1

X(L(ω1)[n], V (ω1)) ∼= K. Thus up to isomorphism, there is a unique X-moduleW which is a nonsplit extension

0→ V (ω1)→W → L(ω1)[n] → 0.

This short exact sequence gives the long exact sequence in cohomology

0→ H1(X,V (ω1))→ H1(X,W )→ H1(X,L(ω1)[n])

→ H2(X,V (ω1))→ · · · (**)

We will show next using (**) that H1(X,W ) ∼= K2. For this, �rst note thaton the short exact sequence (*) the long exact sequence in cohomology givesH i(X,L(ω1)) ∼= H i(X,V (ω1)) for all i ≥ 1, since H i(X,K) = 0 for all i ≥ 1 by[Jan03, II.4.13]. Therefore we have H1(X,V (ω1)) ∼= K, since H1(X,L(ω1)) ∼= Kby the main result of [Sin92]. Furthermore, we have H1(X,L(ω1)[n]) ∼= K asnoted in the previous paragraph. Plugging this into (**), we see that to showH1(X,W ) ∼= K2, it will be enough to show that H2(X,V (ω1)) = 0. To this end,for the short exact sequence

0→ L(ω1)→ V (ω1)∗ → K → 0,

the long exact sequence in cohomology gives H i(X,L(ω1)) = 0 for all i ≥ 2, sinceH i(X,K) = 0 and H i(X,V (ω1)∗) = 0 for all i ≥ 1 by [Jan03, II.4.13]. Thus infact H i(X,V (ω1)) = 0 for all i ≥ 2.

For c ∈ H1(X,W ) ∼= Ext1X(K,W ), denote by Vc a representative

0→W → Vc → K → 0

of the equivalence class of extensions corresponding to c. We omit the proof,but with explicit computations one can show that as a K-vector space, the �rstcohomology group H1(X,W ) has a basis {a, b} with the following properties:

(i) Each nonsplit extension 0→W → V → K → 0 is isomorphic to exactly oneof the modules in the set {Va} ∪ {Va+λ·b : λ ∈ K}.

210 Chapter 6.

(ii) Va and Vb are not self-dual.

(iii) For all λ ∈ K \{0}, the module Va+λ·b is self-dual, and has a non-degenerateX-invariant alternating bilinear form which is unique up to a scalar.

(iv) Va ↓ K[u] = [2, 62] and Vb ↓ K[u] = [2, 62].

(v) For all λ ∈ K \ {0}, with respect to any non-degenerate X-invariant al-ternating bilinear form on Va+λ·b, we have the orthogonal decomposition(Proposition 2.4.4)

Va+λ·b ↓ K[u] =

{V (2) + V (6)2, if λ ∈ K \ {0, 1}.W (1) + V (6)2, if λ = 1.

Now let G = Sp(V ) be a simple algebraic group of type C7, so dimV = 14.By (iii) above, for all λ ∈ K \ {0} there exists a subgroup Xn,λ < G such thatXn,λ is simple of type G2 and V ↓ Xn,λ

∼= Va+λ·b. Note that by (iii) and Lemma4.4.14, such an Xn,λ is uniquely determined up to G-conjugacy. Since Va+λ·b isindecomposable and not irreducible as an Xn,λ-module, it follows that Xn,λ isnon-G-cr. If λ ∈ K \ {0, 1}, it follows from (v) that the regular unipotent elementof Xn,λ is contained in the distinguished unipotent class V (2) + V (6)2 of G.

In conclusion, what we have is that

{Xn,λ : n ∈ Z≥1, λ ∈ K \ {0, 1}}

is an in�nite family of pairwise non-conjugate, non-G-cr subgroups of G, each ofwhich is simple of typeG2, and each of which intersects the distinguished unipotentconjugacy class V (2) + V (6)2 of G.

Appendix A

Orders of distinguished unipotentelements

Here we list the orders of distinguished unipotent elements in exceptional groupsin characteristic p > 0. All of them have order p, except the ones listed in thetables below. In the following, an empty entry means that the element has orderp. The orders can be found by using the results of Lawther in [Law95] and [Law98].

Unipotent class 2 3 5G2 23 32 52

G2(a1) 22

(A1)3 - 3 -

Table A.1: Type G2

Unipotent class 2 3 5 7 11F4 24 33 52 72 112

F4(a1) 23 32 52 72

F4(a2) 23 32 52

F4(a3) 22 32

(C3(a1))2 22 - - - -(A2A1)2 22 - - - -

Table A.2: Type F4

Unipotent class 2 3 5 7 11E6 24 33 52 72 112

E6(a1) 24 32 52 72

E6(a3) 23 32 52

Table A.3: Type E6

211

212 Appendix A.

Unipotent class 2 3 5 7 11 13 17E7 25 33 52 72 112 132 172

E7(a1) 24 33 52 72 112 132

E7(a2) 24 33 52 72 112

E7(a3) 24 33 52 72

E7(a4) 23 32 52 72

E7(a5) 23 32 52

Table A.4: Type E7

Unipotent class 2 3 5 7 11 13 17 19 23 29E8 25 34 53 72 112 132 172 192 232 292

E8(a1) 25 33 52 72 112 132 172 192 232

E8(a2) 25 33 52 72 112 132 172 192

E8(a3) 25 33 52 72 112 132 172

E8(a4) 24 33 52 72 112 132

E8(b4) 24 33 52 72 112 132

E8(a5) 24 33 52 72 112

E8(b5) 24 33 52 72 112

E8(a6) 24 33 52 72

E8(b6) 24 32 52 72

E8(a7) 23 32 52

(A7)3 - 32 - - - - - - - -(D7(a1))2 24 - - - - - - - - -(D5A2)2 23 - - - - - - - - -

Table A.5: Type E8

Appendix B

Actions in some irreduciblerepresentations (p = 2)

Assume that p = 2.

Let G be a simple algebraic group of exceptional type and let ϕ : G→ GL(V )be a self-dual irreducible representation of G. Then V has a non-degenerate, G-invariant symplectic form (Lemma 4.4.5), and thus ϕ(G) lies in Sp(V ). In thetables below, we give for each unipotent u ∈ G the conjugacy class of ϕ(u) inSp(V ) for some small self-dual irreducible representations of G. Speci�cally, wegive the conjugacy classes in the cases where

• G = G2 and V = L(ω1), V = L(ω2).

• G = F4 and V = L(ω1), V = L(ω4).

• G = E6 and V = L(ω2).

• G = E7 and V = L(ω1), V = L(ω7).

• G = E8 and V = L(ω8).

For these representations, the Jordan blocks of ϕ(u) were already given inarbitrary characteristic by Lawther in [Law95] [Law98].

Note that when G = F4, there exists an exceptional isogeny τ : G→ G [Ste68,Theorem 28]. Then any irreducible rational G-module induces another irreducibleG-module V τ by twisting with τ , and in particular LG(ω1)τ ∼= LG(ω4). Thus theentries in tables B.3 and B.4 are almost the same, except some of the classes areswapped by τ .

In the tables below, we label the conjugacy classes in Sp(V ) as in corollary2.4.7 (i). The computations were done with MAGMA (Section 2.9). In the tableswe have bolded the cases where the action is distinguished. Note that there are afew cases where all Jordan block sizes are even and occur with multiplicity ≤ 2,but where the action is not distinguished (classes E7(a1), E7(a2) and E7(a3) forG = E7, V = L(ω7))

213

214 Appendix B.

Class of u action on LG2(ω1)

G2 61G2(a1) 32

0

A1 231

A1 120, 2

20

Table B.1: G = G2 and V = L(ω1)

Class of u action on LG2(ω2)

G2 61,81G2(a1) 32

1, 421

A1 120, 2

61

A1 120, 2

61

Table B.2: G = G2 and V = L(ω2)

Class of u action on LF4(ω4)

F4 101,161F4(a1) 21, 8

31

F4(a2) 520, 8

21

F4(a3) 120, 4

61

C3 631, 81

B3 120, 2

31, 6

31

(C3(a1))2 231, 4

51

C3(a1) 231, 4

51

(A2A1)2 220, 3

20, 4

41

A2A1 220, 3

20, 4

41


A2A1 261, 3

20, 4

20

(B2)2 140, 21, 4

51

B2 140, 21, 4

51

A2 360, 4

21

A2 180, 3

60

A1A1 120, 2

121

(A1)2 160, 2

101

A1 160, 2

101

A1 1140 , 2

60

Table B.3: G = F4 and V = L(ω4)


F4 101,161F4(a1) 21, 8

31

F4(a2) 520, 8

21

F4(a3) 120, 4

61

C3 120, 2

31, 6

31

B3 631, 81

(C3(a1))2 231, 4

51

C3(a1) 140, 21, 4

51

(A2A1)2 220, 3

20, 4

41

A2A1 261, 3

20, 4

20


A2A1 220, 3

20, 4

41

(B2)2 231, 4

51

B2 140, 21, 4

51

A2 180, 3

60

A2 360, 4

21

A1A1 120, 2

121

(A1)2 160, 2

101

A1 1140 , 2

60

A1 160, 2

101

Table B.4: G = F4 and V = L(ω1)

215

Class of u action on LE6(ω2)

E6 61, 81, 1641

E6(a1) 421, 8

21, 112

0, 1621

E6(a3) 320, 4

21, 8

81

D5 61, 891

D5(a1) 481, 61, 8

51

A5 120, 2

61, 8

81

A4A1 241, 4

41, 5

20, 6

20, 8

41

A22A1 12

0, 261, 4

161

D4 280, 6

91, 81

D4(a1) 320, 4

181


A4 140, 3

40, 4

21, 5

20, 7

40, 8

21

A3A1 120, 2

61, 4

161

A2A21 216

1 , 320, 4

101

A22 114

0 , 4161

A3 160, 2

40, 4

161

A2A1 180, 2

101 , 3

60, 4

81

A31 12

0, 2381

A2 1160 , 3

180 , 4

21

A21 114

0 , 2321

A1 1340 , 2

221

Table B.5: G = E6 and V = L(ω2)


E7 21,101,181,261E7(a1) 101, 141, 162

0

E7(a2) 221, 102

1, 1620

E7(a3) 621, 8

20, 142

1

E7(a4) 420, 8

60

E7(a5) 221, 4

20, 6

20, 8

40

E6 140, 102

0, 1620

E6(a1) 120, 5

20, 9

20, 132

0

E6(a3) 140, 5

40, 8

40

D6 231, 61, 103

1, 141

D6(a1) 420, 8

60

D6(a2) 221, 6

61, 8

20

A6 520, 7

20, 8

40

D5A1 241, 8

60

D5(a1)A1 220, 4

60, 6

21, 8

20

A5A1 221, 6

60, 8

20

A4A2 340, 4

40, 7

40

A3A2A1 241, 4

120

D5 140, 2

20, 8

60

D5(a1) 140, 4

60, 6

20, 8

20

(A5)′ 140, 6

60, 8

20

(A5)′′ 221, 6

60, 8

20

D4(a1)A1 240, 4

120


D4A1 2101 , 6

61

A4A1 120, 2

20, 3

20, 4

20, 5

20, 6

20, 7

20

A3A2 220, 3

40, 4

100

(A3A2)2 240, 4

120

A3A21 28

1, 4100

A22A1 14

0, 240, 3

40, 4

80

A2A31 216

1 , 460

D4 180, 2

60, 6

60

D4(a1) 180, 4

120

A4 160, 3

20, 5

60, 7

20

(A3A1)′ 140, 2

60, 4

100

(A3A1)′′ 281, 4

100

A22 14

0, 3120 , 4

40

A2A21 14

0, 2120 , 3

40, 4

40

A41 228

1

A3 1120 , 2

20, 4

100

A2A1 180, 2

80, 3

80, 4

20

(A31)′ 18

0, 2240

(A31)′′ 228

1

A2 1200 , 3

120

A21 116

0 , 2200

A1 1320 , 2

120


216 Appendix B.


E7 81,101,161,181,221,261,321

E7(a1) 61, 141, 1671

E7(a2) 21, 61, 81, 1021, 16

61

E7(a3) 21, 861, 10

21, 141, 16

31

E7(a4) 21, 41, 61, 8151

E7(a5) 21, 320, 4

41, 6

20, 8

121

E6 120, 61, 81, 1020, 16

61

E6(a1) 421, 520, 8

21, 9

20, 11

20, 13

20, 16

21

E6(a3) 120, 320, 4

21, 5

40, 8

121

D6 120, 231, 6

31, 81, 10

41, 14

31, 161

D6(a1) 120, 41, 61, 8151

D6(a2) 120, 271, 6

61, 8

101

A6 320, 720, 8

141

D5A1 231, 61, 8151

D5(a1)A1 21, 4141 , 631, 8

71

A5A1 120, 271, 6

60, 8

101

A4A2 120, 4121 , 540, 7

20, 8

61

A3A2A1 231, 320, 4

301

D5 120, 220, 61, 8

151

D5(a1) 120, 4141 , 631, 8

71

(A5)′ 140, 2

61, 6

60, 8

101

(A5)′′ 1140 , 21, 6

60, 8

101

D4(a1)A1 140, 21, 320, 4

301


D4A1 140, 2151 , 6151 , 81

A4A1 261, 320, 4

61, 5

40, 6

40, 7

20, 8

41

A3A2 120, 220, 3

20, 4

301

(A3A2)2 120, 221, 3

20, 4

301

A3A21 120, 2

131 , 4261

A22A1 140, 2

101 , 340, 4

241

A2A31 2291 , 360, 4

141

D4 160, 2140 , 6151 , 81

D4(a1) 160, 320, 4

301

A4 180, 360, 4

21, 5

80, 7

60, 8

21

(A3A1)′ 140, 2

121 , 4261

(A3A1)′′ 1140 , 271, 4

261

A22 1160 , 3120 , 4201

A2A21 120, 2

281 , 360, 4

141

A41 160, 2

631

A3 1160 , 260, 4261

A2A1 1140 , 2181 , 3140 , 4101(A3

1)′ 180, 2

621

(A31)′′ 1260 , 2531

A2 1340 , 3300 , 421A2

1 1280 , 2521A1 1640 , 2341


217

Class of u action on LE8(ω8)E8 241, 3271E8(a1) 121, 1631, 281, 3251E8(a2) 141, 1671, 261, 3231E8(a3) 841, 141, 1631, 2221, 2420, 301, 321E8(a4) 1221, 16141E8(a5) 621, 8

61, 1421, 16101

E8(a6) 720, 8141 , 1320, 1661

E8(a7) 340, 4121 , 7

40, 8

201

E8(b4) 120, 21, 61, 1021, 1431, 16111E8(b5) 320, 4

21, 61, 8

31, 1020, 1220, 16101

E8(b6) 481, 720, 8

61, 1140, 1240, 1520, 1621

E7 120, 231, 81, 1031, 161, 1831, 221, 2631, 321

E7(a1) 120, 21, 61, 1020, 1431, 16111E7(a2) 120, 2

61, 61, 81, 1061, 16101

E7(a3) 120, 221, 6

40, 8

101 , 1021, 1451, 1631

E7(a4) 140, 21, 451, 61, 8

271

E7(a5) 120, 261, 3

20, 4

81, 6

60, 8

201

D7 220, 461, 1291, 1671

(D7(a1))2 120, 221, 6

41, 8

101 , 1021, 1451, 1631

D7(a1) 481, 61, 871, 1020, 1260, 141, 1631

D7(a2) 641, 8281

A7 420, 8301

E6A1 120, 261, 61, 81, 1060, 16101

E6(a1)A1 241, 441, 5

20, 6

20, 8

41, 9

20, 1020, 1120, 1220, 1320, 1420, 1621

E6(a3)A1 120, 261, 3

20, 4

61, 5

40, 6

40, 8

201

A6A1 221, 320, 4

20, 6

40, 7

20, 8

241

(D5A2)2 231, 451, 61, 8

271

D5A2 220, 320, 4

41, 61, 8

271

D5(a1)A2 320, 4301 , 6

31, 8

131

A4A3 320, 4301 , 6

20, 7

20, 8

121

A4A2A1 120, 261, 4

241 , 5

40, 6

40, 7

20, 8

101

E6 1140 , 61, 81, 1060, 16101E6(a1) 180, 4

21, 5

60, 8

21, 9

60, 1120, 1360, 1621


218 Appendix B.


E6(a3) 1140 , 320, 4

21, 5

120 , 8

201

D6 140, 2101 , 6

51, 81, 10101 , 1451, 161

D6(a1) 140, 21, 451, 61, 8

271

D6(a2) 140, 2121 , 6

181 , 8

141

A6 140, 320, 5

40, 7

60, 8

221

D5A1 120, 2121 , 61, 8

271

D5(a1)A1 120, 261, 4

261 , 6

71, 8

111

A5A1 140, 2121 , 6

180 , 8

141

(D4A2)2 120, 261, 4

261 , 6

71, 8

111

D4A2 2140 , 360, 4

141 , 6

151 , 8

71

D4(a1)A2 380, 4561

A4A2 160, 380, 4

201 , 5

40, 7

100 , 8

61

A4A21 2161 , 3

20, 4

141 , 5

40, 6

120 , 7

20, 8

61

A23 240, 4

601

A3A2A1 120, 2121 , 3

20, 4

541

A22A

21 140, 2

281 , 3

40, 4

441

D5 1140 , 260, 61, 8

271

D5(a1) 1140 , 4261 , 6

71, 8

111

A5 1160 , 261, 6

180 , 8

141

D4A1 160, 2361 , 6

271 , 81

D4(a1)A1 160, 2101 , 3

20, 4

541

A4A1 180, 2101 , 3

60, 4

101 , 5

80, 6

80, 7

60, 8

41

(A3A2)2 160, 2101 , 3

20, 4

541

A3A2 160, 260, 3

100 , 4

501

A3A21 140, 2

301 , 4

461

A22A1 1160 , 2

181 , 3

120 , 4

401

A2A31 120, 2

621 , 3

60, 4

261

D4 1260 , 2260 , 6

271 , 81

D4(a1) 1260 , 320, 4

541

A4 1240 , 3100 , 4

21, 5

200 , 7

100 , 8

21

A3A1 1160 , 2241 , 4

461

A22 1280 , 3

360 , 4

281

A2A21 1140 , 2

521 , 3

140 , 4

221

A41 180, 2

1201

A3 1440 , 2100 , 4

461

A2A1 1340 , 2341 , 3

300 , 4

141

A31 1280 , 2

1101

A2 1780 , 3540 , 4

21

A21 1640 , 2

921

A1 11320 , 2581

Table B.9: G = E8 and V = L(ω8) (continued)

Appendix C

Ordering of the positive roots

Let Φ be an indecomposable root system with base ∆ = {α1, . . . , αl} and set ofpositive roots Φ+. We assume that the simple roots are ordered as in the standardBourbaki labeling of the Dynkin diagram, see for example [Hum72, 11.4, pg. 58].

In this appendix we give, for Φ of exceptional type, the total ordering ≤ onΦ+ that was de�ned in Section 2.8 (following [Car72, 2.1]). We recall that forα, β ∈ Φ+, we set α ≤ β if and only if one of the following holds:

• α = β,

• There exists an integer 1 ≤ k ≤ l such that β−α =∑k

i=1 ciαi, where ci ∈ Zfor all 1 ≤ i ≤ k and ck > 0.

Set N = |Φ+|. Writing Φ+ = {β1, . . . , βN}, where β1 < β2 < · · · < βN withrespect to the total order ≤, we give the expression of βi for all 1 ≤ i ≤ N asa sum of the simple roots αi in the tables that follow. In the tables we use thenotation βi = (k1, k2, . . . , kl), for βi =

∑lj=1 kjαj .

β1 = (1, 0)β2 = (0, 1)β3 = (1, 1)β4 = (2, 1)β5 = (3, 1)β6 = (3, 2)

Table C.1: Ordering of the positive roots of the root system G2.

219

220 Appendix C.

β1 = (1, 0, 0, 0)β2 = (0, 1, 0, 0)β3 = (0, 0, 1, 0)β4 = (0, 0, 0, 1)β5 = (1, 1, 0, 0)β6 = (0, 1, 1, 0)β7 = (0, 0, 1, 1)β8 = (1, 1, 1, 0)β9 = (0, 1, 2, 0)β10 = (0, 1, 1, 1)β11 = (1, 1, 2, 0)β12 = (1, 1, 1, 1)

β13 = (0, 1, 2, 1)β14 = (1, 2, 2, 0)β15 = (1, 1, 2, 1)β16 = (0, 1, 2, 2)β17 = (1, 2, 2, 1)β18 = (1, 1, 2, 2)β19 = (1, 2, 3, 1)β20 = (1, 2, 2, 2)β21 = (1, 2, 3, 2)β22 = (1, 2, 4, 2)β23 = (1, 3, 4, 2)β24 = (2, 3, 4, 2)

Table C.2: Ordering of the positive roots of the root system F4.

β1 = (1, 0, 0, 0, 0, 0)β2 = (0, 1, 0, 0, 0, 0)β3 = (0, 0, 1, 0, 0, 0)β4 = (0, 0, 0, 1, 0, 0)β5 = (0, 0, 0, 0, 1, 0)β6 = (0, 0, 0, 0, 0, 1)β7 = (1, 0, 1, 0, 0, 0)β8 = (0, 1, 0, 1, 0, 0)β9 = (0, 0, 1, 1, 0, 0)β10 = (0, 0, 0, 1, 1, 0)β11 = (0, 0, 0, 0, 1, 1)β12 = (1, 0, 1, 1, 0, 0)β13 = (0, 1, 1, 1, 0, 0)β14 = (0, 1, 0, 1, 1, 0)β15 = (0, 0, 1, 1, 1, 0)β16 = (0, 0, 0, 1, 1, 1)β17 = (1, 1, 1, 1, 0, 0)β18 = (1, 0, 1, 1, 1, 0)

β19 = (0, 1, 1, 1, 1, 0)β20 = (0, 1, 0, 1, 1, 1)β21 = (0, 0, 1, 1, 1, 1)β22 = (1, 1, 1, 1, 1, 0)β23 = (1, 0, 1, 1, 1, 1)β24 = (0, 1, 1, 2, 1, 0)β25 = (0, 1, 1, 1, 1, 1)β26 = (1, 1, 1, 2, 1, 0)β27 = (1, 1, 1, 1, 1, 1)β28 = (0, 1, 1, 2, 1, 1)β29 = (1, 1, 2, 2, 1, 0)β30 = (1, 1, 1, 2, 1, 1)β31 = (0, 1, 1, 2, 2, 1)β32 = (1, 1, 2, 2, 1, 1)β33 = (1, 1, 1, 2, 2, 1)β34 = (1, 1, 2, 2, 2, 1)β35 = (1, 1, 2, 3, 2, 1)β36 = (1, 2, 2, 3, 2, 1)

Table C.3: Ordering of the positive roots of the root system E6.

221

β1 = (1, 0, 0, 0, 0, 0, 0)β2 = (0, 1, 0, 0, 0, 0, 0)β3 = (0, 0, 1, 0, 0, 0, 0)β4 = (0, 0, 0, 1, 0, 0, 0)β5 = (0, 0, 0, 0, 1, 0, 0)β6 = (0, 0, 0, 0, 0, 1, 0)β7 = (0, 0, 0, 0, 0, 0, 1)β8 = (1, 0, 1, 0, 0, 0, 0)β9 = (0, 1, 0, 1, 0, 0, 0)β10 = (0, 0, 1, 1, 0, 0, 0)β11 = (0, 0, 0, 1, 1, 0, 0)β12 = (0, 0, 0, 0, 1, 1, 0)β13 = (0, 0, 0, 0, 0, 1, 1)β14 = (1, 0, 1, 1, 0, 0, 0)β15 = (0, 1, 1, 1, 0, 0, 0)β16 = (0, 1, 0, 1, 1, 0, 0)β17 = (0, 0, 1, 1, 1, 0, 0)β18 = (0, 0, 0, 1, 1, 1, 0)β19 = (0, 0, 0, 0, 1, 1, 1)β20 = (1, 1, 1, 1, 0, 0, 0)β21 = (1, 0, 1, 1, 1, 0, 0)β22 = (0, 1, 1, 1, 1, 0, 0)β23 = (0, 1, 0, 1, 1, 1, 0)β24 = (0, 0, 1, 1, 1, 1, 0)β25 = (0, 0, 0, 1, 1, 1, 1)β26 = (1, 1, 1, 1, 1, 0, 0)β27 = (1, 0, 1, 1, 1, 1, 0)β28 = (0, 1, 1, 2, 1, 0, 0)β29 = (0, 1, 1, 1, 1, 1, 0)β30 = (0, 1, 0, 1, 1, 1, 1)β31 = (0, 0, 1, 1, 1, 1, 1)β32 = (1, 1, 1, 2, 1, 0, 0)

β33 = (1, 1, 1, 1, 1, 1, 0)β34 = (1, 0, 1, 1, 1, 1, 1)β35 = (0, 1, 1, 2, 1, 1, 0)β36 = (0, 1, 1, 1, 1, 1, 1)β37 = (1, 1, 2, 2, 1, 0, 0)β38 = (1, 1, 1, 2, 1, 1, 0)β39 = (1, 1, 1, 1, 1, 1, 1)β40 = (0, 1, 1, 2, 2, 1, 0)β41 = (0, 1, 1, 2, 1, 1, 1)β42 = (1, 1, 2, 2, 1, 1, 0)β43 = (1, 1, 1, 2, 2, 1, 0)β44 = (1, 1, 1, 2, 1, 1, 1)β45 = (0, 1, 1, 2, 2, 1, 1)β46 = (1, 1, 2, 2, 2, 1, 0)β47 = (1, 1, 2, 2, 1, 1, 1)β48 = (1, 1, 1, 2, 2, 1, 1)β49 = (0, 1, 1, 2, 2, 2, 1)β50 = (1, 1, 2, 3, 2, 1, 0)β51 = (1, 1, 2, 2, 2, 1, 1)β52 = (1, 1, 1, 2, 2, 2, 1)β53 = (1, 2, 2, 3, 2, 1, 0)β54 = (1, 1, 2, 3, 2, 1, 1)β55 = (1, 1, 2, 2, 2, 2, 1)β56 = (1, 2, 2, 3, 2, 1, 1)β57 = (1, 1, 2, 3, 2, 2, 1)β58 = (1, 2, 2, 3, 2, 2, 1)β59 = (1, 1, 2, 3, 3, 2, 1)β60 = (1, 2, 2, 3, 3, 2, 1)β61 = (1, 2, 2, 4, 3, 2, 1)β62 = (1, 2, 3, 4, 3, 2, 1)β63 = (2, 2, 3, 4, 3, 2, 1)


222 Appendix C.

β1 = (1, 0, 0, 0, 0, 0, 0, 0)β2 = (0, 1, 0, 0, 0, 0, 0, 0)β3 = (0, 0, 1, 0, 0, 0, 0, 0)β4 = (0, 0, 0, 1, 0, 0, 0, 0)β5 = (0, 0, 0, 0, 1, 0, 0, 0)β6 = (0, 0, 0, 0, 0, 1, 0, 0)β7 = (0, 0, 0, 0, 0, 0, 1, 0)β8 = (0, 0, 0, 0, 0, 0, 0, 1)β9 = (1, 0, 1, 0, 0, 0, 0, 0)β10 = (0, 1, 0, 1, 0, 0, 0, 0)β11 = (0, 0, 1, 1, 0, 0, 0, 0)β12 = (0, 0, 0, 1, 1, 0, 0, 0)β13 = (0, 0, 0, 0, 1, 1, 0, 0)β14 = (0, 0, 0, 0, 0, 1, 1, 0)β15 = (0, 0, 0, 0, 0, 0, 1, 1)β16 = (1, 0, 1, 1, 0, 0, 0, 0)β17 = (0, 1, 1, 1, 0, 0, 0, 0)β18 = (0, 1, 0, 1, 1, 0, 0, 0)β19 = (0, 0, 1, 1, 1, 0, 0, 0)β20 = (0, 0, 0, 1, 1, 1, 0, 0)β21 = (0, 0, 0, 0, 1, 1, 1, 0)β22 = (0, 0, 0, 0, 0, 1, 1, 1)β23 = (1, 1, 1, 1, 0, 0, 0, 0)β24 = (1, 0, 1, 1, 1, 0, 0, 0)β25 = (0, 1, 1, 1, 1, 0, 0, 0)β26 = (0, 1, 0, 1, 1, 1, 0, 0)β27 = (0, 0, 1, 1, 1, 1, 0, 0)β28 = (0, 0, 0, 1, 1, 1, 1, 0)β29 = (0, 0, 0, 0, 1, 1, 1, 1)β30 = (1, 1, 1, 1, 1, 0, 0, 0)



223



Table C.6: Ordering of the positive roots of the root system E8 (continued).

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Curriculum vitae

Name Mikko Korhonen

Nationality Finland

Contact details mikko.korhonen@ep�[email protected]+41 21 693 03 98 (EPFL)

Education Ph.D. in Mathematics, 2014 - 2018École Polytechnique Fédérale de LausanneThesis supervisor: Donna Testerman

M.Sc. in Mathematics, 2012 - 2014University of OuluThesis supervisor: Markku Niemenmaa

B.Sc. in Mathematics, 2008 - 2012University of OuluMinor in Information Processing Science

Research interests Linear algebraic groups: representation theory, sub-group structure, properties of unipotent elements.

Publications Invariant forms on irreducible modules of simple alge-braic groups. J. Algebra, 480:385-422, 2017.

Unipotent elements forcing irreducibility in linear al-gebraic groups. J. Group Theory, to appear.

Language skills Finnish (native), English (�uent), French (intermedi-ate)

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