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3 inIllimmlummiil

3 1293 01427 9172

This is to certify that the

dissertation entitled

Quadratic Representations for Groups of Lie Type
Over Fields of Characteristic Two

presented by
Timothy F. Englund

has been accepted towards fulfillment
of the requirements for

ph.D. degree in Mathematics

 

Date 5]. /g/q7

 

MS U is an Affirmative Action/Equal Opportunity Institution 0-12771

 

LIBRARY

Michigan State
Unlvorslty

 

 

 

QUADRATIC REPRESENTATIONS FOR GROUPS OF

LIE TYPE OVER FIELDS OF CHARACTERISTIC TWO

By

Timothy F. Englund

A Dissertation

Submitted to
Michigan State University
in partial fulfillment of the requirements

for the degree of

DOCTOR OF PHILOSOPHY

Department of Mathematics

1997

ABSTRACT

QUADRATIC REPRESENTATIONS FOR GROUPS OF
LIE TYPE OVER FIELDS OF CHARACTERISTIC TWO

By

Timothy F. Englund

Let K be a field of characteristic two, G be a group of Lie type defined over K,
and let V be an irreducible K G-module. By a theorem due to Steinberg we know
that V E“ @361 Vi, where I is an arbitrary index set and each V,- is an algebraic
conjugate of a restricted K G-module. Now suppose that G contains a fours-group
which acts quadratically on V. We determine then that III 3 2. Moreover, by using
the weight structure of the modules and information about the parabolic subgroups
of G we determine which restricted modules are possible when |I| = 2 and, with some
restrictions on A, when II | = 1. In all cases the restricted modules are fundamental

modules, and in many cases the majority of these are ruled out.

Dedicated to the memory of my grandfather, Frederick A. Dostal

iii

ACKNOWLEDGMENTS

I wish to express my heartfelt appreciation to my advisor, Professor Ulrich Meier-
frankenfeld, for his invaluable guidance, advice, cajoling, and corrections at all stages
of this project.

I am also very grateful to Professors J. Hall, R. Phillips, S. Schuur, and C. Wei]
for serving on my committee.

A very special word of thanks goes to my friends and my family (especially Arnold)
for their continuous encouragement, understanding, constant moral support, and
timely distractions.

iv

TABLE OF CONTENTS

LIST OF FIGURES

LIST OF TABLES

1

2

8

9

Introduction

Setup, Notation, etc.

Root Systems

General Lemmas

Linear Dependence

Linearly Dependently Acting Fours-groups

A Result Concerning Root Systems and Weights
Linear Independence

The Classical Groups

10 En(K)

11 F4(K)

BIBLIOGRAPHY

vi

vii

11

19

28

38

45

52

57

64

73

77

79

LIST OF FIGURES

3.1 Labeling of the Dynkin Diagrams .................... 13
3.2 The Rank 2 Indecomposable Root Systems ............... 14

vi

3.1
3.2
3.3
3.4
3.5

LIST OF TABLES

The roots of E6 .............................. 15
The roots of E7 .............................. 16
The roots of E8 .............................. 17
The roots of F4 .............................. 18
The highest long and short roots in (I) .................. 18

vii

CHAPTER 1

Introduction

Let V be a vector space over a field K. Then a subgroup A g G S GLK(V) is said
to act quadratically on V if [V, A, A] = 0. V is called a quadratic representation for
G.

In [1], Michael Aschbacher mentions the following question about finite groups G
with 02(G) = 1 and faithful GF(2)G—modules V: “ Do there exist 4-subgroups A
of G acting quadratically on V; that is, with [V, A, A] = 0? Determine the triples
(G, V, A) with this property.”

In this paper we attempt to answer this question when G is a group of Lie type
defined over a field with even characteristic.

Considerations of quadratic action where first made by John Thompson in [15].
There he classified finite irreducible subgroups of GLK(V) generated by quadratically
acting elements for fields K with char(K) = p Z 5. He determined that for p 2 5, the
groups satisfying the above conditions are groups of Lie type defined over a field of
characteristic p. Ho Chat—Yin solved a similar problem with a few restrictions added
for the case of p = 3 in [7]. Completing the picture when p is odd, A.A. Premet and
ID. Suprunenko classified the irreducible quadratic representations of groups of Lie
type over fields of odd characteristics in [10].

Quadratic GF(2)-representations are addressed in [1], [8], and [9] by Michael

Aschbacher, Ulrich Meierfrankenfeld, and Gernot Stroth. Of course, a different
set of criteria is needed here since every involution acts quadratically on a GF(2)-
representation. Consequently, quadratically acting fours-groups were considered in-
stead. The alternating groups, the sporadic groups, and groups of Lie type over fields
with odd characteristic containing quadratically acting fours-groups were considered
by Ulrich Meierfrankenfeld and Gernot Stroth in [8] and [9]. There it was determined
which of the above groups admitted quadratic representations and they indicated
these representations.

As was mentioned, in this paper we address the question for groups of Lie type
defined over fields of even characteristic. This situation was examined by Gernot
Stroth in [13] under assumptions which were essentially equivalent to the assumption
that if A is a quadratically acting fours group, then A intersects nontrivially, but is
not contained in, a root subgroup. This restriction is indicative of the fact that some
restraints on the types of fours-groups that should be considered are necessary. For
example, a fours-group contained in a root subgroup would tend to act quadratically
on too many representations to make classifying them worthwhile. Towards that end

we make the following definition:

Definition 1 Let V be a vector space over a field K with characteristic two and
suppose that a and b are commuting involution; in GLK(V). We say that A —- (a, b)
acts linearly dependently on V if there exists 7 E K such that [2), a] = ’y[v, b] for all

’UEV.

Clearly this is a strong restriction on V and A. In fact, when K is a field of even
characteristic and G is a Chevalley group over K, we are able classify the irreducible
K G-modules for which G contains a linearly dependently acting fours-group A. We
record these in the following theorem. (The ordering used for the root systems is

given in Figure 3.1.)

Theorem 1 If K is a field of even characteristic, G is a Chevalley group over K,
V is an irreducible K G-module, and A is fours-subgroup of G which acts linearly
dependently on V, then [K | 2 [A] Z 4 and up to algebraic conjugates of V one of the

following holds:
1. G E A1(K) and V is a fundamental module.
2. G 2’ B,(K) and one of the following is true:

(a) r,- is a long root and V is the natural or spin module, or

(b) r,- is a short root and V is a fundamental module.
3. G ’5 D,(K) and V is the natural or a half-spin module.
4. G E” E5(K) and V = V()\1) or V(/\6).
5. G ’.-‘_-’ E7(K) and V = V()\7).
6. G ":14 F4(K) and one of the following is true:

(a) r,- is a long root and V = V()\4), or

(b) r,- is a short root and V = V01).

7. G g G2(K), r,- is short, and V = V()\2), the natural module.

We accomplish the proof of this theorem by finding a conjugate of A which is
contained in a minimal parabolic subgroup of G and which acts nontrivially and
linearly dependently on all the chief factors of V for this parabolic. We note then
that the constant 7 E K associated with this linearly dependent action on V must
remain constant for all these chief factors. Using this we are able to eliminate all but
a very few possible values for the weight of V since the weight of V determines the

weight of the various chief factors for the parabolic subgroups.

The list of modules obtained in the theorem proves useful because of the following
lemma which is central in this paper. To see its significance, recall that by The
Steinberg Tensor Product Theorem if V is any irreducible K G-module (K and G
as above), then V ’=-‘1 @361 V,, where I is some indexing set and V, is an algebraic

conjugate of a restricted module for G.

Lemma 1 Let G be a Chevalley group over K, char(K) = 2. Let V 2 ® V,- be

tel
a K G-module with each V,- nontrivial and irreducible. Also let A S G be an abelian
two-subgroup, IA] 2 4, which acts quadratically on V. Then [I I S 2. If [I I = 2, then

A acts linearly dependently on each V}. Moreover, if A acts linearly dependently on

V, then [I] = 1.

The proof of the lemma follows from a few very straight forward calculations.

Once we have the above list of potential modules, we are then able to determine
the structure of fours-groups which may act linearly dependently. To do this, in most
of the cases, we find subgroups X E’ Sl2(K) and Y S CG(X) such that A S X x Y.
A theorem due to Steinberg in [12] states that every chief factor of V for X x Y is
isomorphic to one of the form V1 (8') V2 where V1 is an irreducible K X -module and V2
is an irreducible K Y-module. However, the lemma above implies then that A must
act trivially on one of these two and so often we are able to conclude that A S X and

then in a root subgroup of G. We summarize this result in the following corollary.

Corollary 1 If K is a field of even characteristic, G is a Chevalley group over K,
V is an irreducible K G-module, and A is fours-subgroup of G which acts linearly

dependently on V, then one of the following holds:

1. If G g A1(K) and if V is not an algebraic conjugate of the natural module for

G, then A is contained in a root subgroup.

2. If G g BI(K) and if V is not an algebraic conjugate of the natural module for

G, then one of the following is true:

( a ) V is not an algebraic conjugate of the spin module for G and A is contained

in a short root subgroup of G.
(b) l 2 3, V is an algebraic conjugate of the spin module for G and A is

contained in a root subgroup of G.

3. If G 2-“ D1(K) and V is not an algebraic conjugate of the natural module for G,

then A is contained in the product of three commuting root subgroups of G.
4. IfG ET E6(K) or E7(K), then A is contained in a root subgroup of G.
5. IfG E“ F4(K), then one of the following is true:

(a) V is an algebraic conjugate of V()\1) and A is contained in a long root
subgroup of G.
(b) V is an algebraic conjugate of V(A4) and A is contained in a short root

subgroup of G.

6. IfG '5 G2(K), then A is contained in a short root subgroup of G.

After the following definition, we are almost in a position to state the main result

of the paper.

Definition 2 Let G be a group of Lie-type and A S G a fours-group.

1. We say that A is a linearly dependent fours-group in G if one of the following
hold:

( a j G E’ A1(K) or DI(K) and A acts linearly dependently on the natural module
for G.

(b) G E“ 82(K) and A act linearly dependently on either the natural or the

spin module for G.

(c) G E BI(K), l 2 3 and either A acts linearly dependently on the natural
module for G or there crisis a rank two connected parabolic subgroup, say
P, of G such that A is contained in the Levi complement of P and is

linearly dependent there.

(d) G 91 E6(K), E7(K), or E8(K) and there exists a proper connected parabolic
subgroup M1 such that A9 S L J and is linearly dependent there for some

9 E G.
(e) G 1%“ F4(K) or G2(K) and A is contained in a root subgroup of G.

( f ) G E“ G‘(K) is a twisted Chevalley group and A is linearly dependent when

considered as a subgroup of G(K).

2. We say that A is a linearly independent fours-group if A is not a linearly de-

pendent fours— group.

It follows from the corollary above that except for E8(K), a fours—group is linearly
dependent if and only if it acts linearly dependently on some irreducible K G—module.

Moreover, it follows from the lemma above that if A is linearly independent and
if it acts quadratically on V, then V is an algebraic conjugate of a restricted module
for G.

To prove the main result of the paper, first we show that if G E’ A2(K) or 32(K)
and if A is linearly independent and acts quadratically on V, then V must be an
algebraic conjugate of a fundamental module for G. Then, in an inductive manner,
we show that if G is an arbitrary Chevalley group and A S G is linearly indepen-
dent, then there exists a rank two connected parabolic subgroup, say P, such that
A02(P) / 02(1)) remains a linearly independent subgroup of 02'(P)/02(P). By the
above, it follows that all the nontrivial chief factors of V for the Levi complement of
P must be fundamental modules. This fact is enough to eliminate the majority of

the possible values of the weight of V, leaving us with the desired list.

We now state the main result.

Theorem 2 Let G = G(K) be a Chevalley group defined over K, char(K) = 2
and let V a nontrivial, irreducible K G-module. If there exists a linearly independent
fours-group A which act quadratically on V, then up to algebraic conjugacy of V, one

of the following is true:
1. G 91 A¢(K) and V is a fundamental module.
2. G 1—‘1 BI(K) and V is a fundamental module.
3. G E’ DI(K) and V is the natural or a half-spin module.
4. G g E6(K) and V = V()\1) or V0.6).
5. G E’ E7(K) and V = V(/\7).
6. G .1": F4(K) and V = V()\1) or V()\4).

7. G E’ G2(K) and V = V()\2).

Corollary 2 Let k be the algebraic closure ofK and let G0 = Gi(K) S G(k) be a
twisted Chevalley group. Let V be a nontrivial, irreducible K Go—module and suppose
that Go contains a linearly independent fours-group which acts quadratically on V.
Then V is obtained from the restriction to Go of a rational representation V’ of G (k),

where V’ is one of the representation from the conclusion of Theorem 2.

The corollary follows easily from the main theorem and a result of Steinberg.
In the next two chapters we record the notation used throughout the paper. The
results due to Steinberg mentioned above are recorded in Chapter 3 as Theorems 4.13

and 4.14, and can be found in [12].

CHAPTER 2

Setup, Notation, etc.

Throughout this paper we are concerned with determining the structure of certain
modules, involutions, and subgroups of groups of Lie type. For the Chevalley groups
which correspond to classical groups we use the structure of the natural module for the
group to obtain this information. For the exceptional Chevalley groups, on the other
hand, we utilize the (B, N)-structure of the group. Consequently, in this chapter we
record notation and a few basic facts and important theorems concerning the weight
structure of certain modules.

Except where noted otherwise we use the following set of abbreviations which was,

for the most part, adapted from [4]. Let
K be a field with char(K) = 2,
G be a group of Lie type defined over K,
V be an non-trivial, irreducible GK -module,
A be a fours-group, A S G, which acts quadratically on V.

Suppose now that G is a Chevalley group. Within the semi-simple Lie algebra L

corresponding to G, we let

(I) be a root system of L,

H be a fundamental root system, II g (I),

(PT be the set of positive roots in (I),

(DJ be the root subsystem of (I) generated by J C_: H,

{Xm H, | a E (I),r E 11} be a Chevalley basis for L,

W be the Weyl group of (I),

w, be the reflection in the hyperplane orthogonal to the root r,
WJ be the subgroup of W generated by {wr | r E J C_: H},

{/\,-} be a set of fundamental weights corresponding to II.

Thus, (/\,-,rj) = 6,];
Within G itself we let
X, be the root subgroup of G corresponding to r E (I),
H be the diagonal subgroup of G,
U be the unipotent subgroup of G generated by the positive root subgroups,
B be the Borel subgroup of G with B = UH,
N be the monomial subgroup of G with N / H E W,
NJ be the inverse image in N of WJ in W and J C_: II.

We define our notation for the parabolic subgroups of G as follows: Let J Q 11. Then
define

PJ = (B,X_,. | 7‘ e J),

MJ = (B,X_,|r¢J),

10

LJ = (Xi, | r Q“ J), the Levi complement of MJ,
UJ=<XTITEJH®+),

Q1 = 02(MJ)-
Thus, My = QJLJH.
Note that My 2 PH“.

Lastly, if r, s E Q with r 7S is, then we denote the root subsystem of Q generated by
r and s by (ir, is). That is, (ir, is) = (Zr + Z8) 0 Q.

Furthermore, we say that (ir, is) has type

A2(long) if both r and s are long roots and (ir, is) is a root system of type A2.
A2(short) if both 7‘ and s are short roots and (ir, is) is a root system of type A2.
B2 if (ir, is) is a root system of type 82.

L _L L if both roots are long roots and are perpendicular to each other.

L _L S if one root is a long root and the other is a short root and they are perpen-

dicular.

S _L S if both roots are short and perpendicular.

Notice that if w E W, then (ir, is)” has the same type as (ir, is).

One last word about some of the terminology used in this paper. When discussing
the classical groups we will often mention “the natural module.” By this we are
referring to the vector space relative to which G is most often defined as a subgroup
of non-singular linear transformations with determinant equal to one, preserving some
particular form and which, relative to the ordering of the root systems given in the
next section, is denoted by V(/\1). Moreover, if G E’ B((K), then V0.1) is identified
with “the spin module for G” while if G E’ D1(K), then V()\,_1) and V0.1) are

identified with the “half-spin modules for G.”

CHAPTER 3

Root Systems

Basic facts involving root systems, parabolic subgroups, and the Levi decomposition
will be used repeatedly throughout this paper, especially when discussing the ex-
ceptional groups. Consequently, we need a list of some of the roots in Q. We have
based the the labeling of the roots in each root system on the labeling of the Dynkin
Diagrams given in Figure 3.1 below.

For the classical groups we have given an explicit description of the root systems
of type Al, 8;, D1 and G2 below.

Figure 3.2 shows the roots in system of rank two expressed as integral combinations
of fundamental roots.

For the systems of type E5, E7, E8, and F4, on the other hand, we have explicitly
listed a subset of the root system. These are given in Tables 3.1, 3.2, 3.3, and 3.4.

In Figure 3.5 we have given a list of the highest long and short roots in each root

system.

11

12

The Classical Root Systems

Type Al

Let e0, e1, . . . , e; be an orthonormal basis of a Euclidean space with dimension l + 1.
For 1 S i S l, let r, = e,-_1 — e,-. Then the set {r,- | 1 S i S l} is a fundamental system

of type A, and the set {ei — e, | i # j, 0 S i,j S l} is the full set of roots.

Type Bl

Let e1, 62, . . . , e, be an orthonormal basis of a Euclidean space with dimension 1. For
1SiSl—1,1etr,-=ez—el+1andletrzzez. Thentheset{r,-|1SiSl}isa
fundamental system of type B; and the set {ie,- i ej, ie, | i -+‘ j, 0 S i,j S l} is the

full set of roots.

Type Dz
Let e1,e2,...,e1 be as above. Let r,- = e,- —e,+1 for 1 S i S l—l and let r1: e,_1+e1.
Then the set {r,- | 1 S i S l} is a fundamental system of type D, and the set

{ie,- i e, | i 74 j,0 S i,j S l} is the full set of roots.

(:22

13

Figure 3.1. Labeling of the Dynkin Diagrams

 

 

 

 

 

7‘1 7‘2 7‘3 Tn—i Tn
c c c c o
7"1 7"2 7" 3 Tn—i Tn
C C 3> -——4t::$:::=.
Tn—l
T1 7‘2 T3
Tn
7‘1 7‘3 7‘ 4 7‘ 5 Tn
t r c —o
7‘2
7‘ 1 7‘2 7‘ 3 7" 4
O——-O=:=D——o
7‘1 7‘ 2

14

terns
t Sys
Roo

able
Indecompos

k 2

Ran

The

3.2.

e

Figur

 

 

 

 

 

 

 

 

 

2T2
7‘1 +
+ T2
7‘1
T1
7‘1

7‘2

‘7‘1 + 2 T1 + T2
7‘1
7‘2
27'1 + 3T2

7‘1

 

— 3T2 7‘1 — 2T2

 

 

15

Table 3.1. The roots of E6

Roots in Q+ for Q of type E6, where if r E Q+ and if r = 2:1 n,r, then we represent
r in the table as r = n1n3n4n5n6. We have ranked them approximately according

n2
to their height.
,3, = 10000 r2 = 00000 r3 = 01000 n = 00100
0 1 0 0
r5 = 00010 r6 = 00001 r7 = 01100 r8 = 00110
0 0 0 0
79 = 01110 r10 = 00100 7‘11 = 01100 7'12 = 00110
0 1 1 1
r = 01110 r = 01210 T = 11000 T =11100
13 1 14 1 15 0 16 0
r = 11100 r = 11110 r = 11110 r = 11210
17 1 18 0 19 1 20 1
r 212210 7m 2 00011 r = 00111 7‘ = 00111
21 1 22 0 23 0 24 1
7—25 2 01111 T26 = 01111 T27 = 01211 T28 = 01221
0 1 1 1
r = 11111 r = 11111 r = 11211 r = 12211
29 0 30 1 31 1 32 1
7‘33 2 11%21 7‘34 2 12%21 T35 = 12:]??ZI T36 = 12:32].

16

Table 3.2. The roots of E7

Roots in Q+ for Q of type E7 which are either fundamental roots or for which the
coefficient of r7 is nonzero. Again, if r E Q+ and if r 2 23:1 n,r,~, then we represent r
in the table as r = n1n3n4n5n6n7. We have ranked them approximately according

n2
to their height.

r1 = 100000 r2 = 000000 r3 = 010000 r4 = 001000

0 1 0 0
r5 = 000100 r5 = 000010 r7 = 000001 r8 = 000011

0 0 0 0
r =000111 r =001111 r =001111 r =011111

9 0 10 0 11 1 12 0

r13 = 011111
1

r = 111111
17 0

r21 = 122111
1

7'25 =123211
1

r = 123221
29 2

r33 = 2331321

r14 = 012111
1

T18 = 111111
1

7'22 = 122211
1

7'25 = 12?221

T30 = 12332 1

r15 = 012211
1
r19 = 112111
1
7‘23 2 112221
1
7‘27 = 12?321

r31 = 12121321

T16 2' 012221
1

r20 = 113211

7'24 = 122221
1

r = 123211
28 2

T32 = 1342132 1

17

Table 3.3. The roots of E8

Roots in Q+ of type E8 which are either fundamental roots or for which the coefiicient
of r8 is nonzero. Once more, if r E Q+ and if r = 2le niri, then we denote r in the

table as r = n1n3n4n5n6n7ng.
n2

r1 = 1080000 r2 = 0030000 r3 = 0180000 r4 = 0060000

r5 = 0081000 r5 = 0080100 r7 = 0080010 r8 = 0080001

Roots for which n1 = 0 and n8 # 0.

r9 = 0080011
r13 = 0031111
r17 = 0131111

r10 = 0080111

7‘ = 0111111
14 0

r18 = 0132111

Roots for which n1 2 1 and n8 75 0.

r21 = 1161111
r25 = 1132111
r29 = 1232111
r33 = 1232211
r37 = 1132221
r41 = 1233221
r45 = 121213221
r49 = 1234321

r22 = 1131111
r25 = 1232111
r30 = 1232211
r34 = 1233211
r38 = 1232221
r42 = 1233221
r45 = 1333221
r50 = 131214321

Roots for which n1 2 2 and n8 74 0.

r53 = 234213211
r57 = 2334321
T61 "—" 2434321

7'55 2 2435432

r54 = 231213221
r58 = 245214321
r52 = 2435321

7‘11 = 0081111
7‘15 2 00111111
r19 = 0132211

r23 = 1131111
r27 = 1132211
r31 = 1232111
r35 = 121212111
r39 = 1232221
r43 = 1233321
r47 = 1233321
r51 = 1334321

7'55 = 231213321
7'59 = 2334321
r53 = 2435421

r12 = 0061111
r16 = 0131111
7'20 = 0132221

r24 = 1231111
r23 = 1232211
r32 = 1233211
r36 = 1333211
r40 = 1232221
r44 = 1233321
r48 = 1333321
r52 = 1334321

T55 = 231214321
T50 = 2434321
r54 = 2435431

18

Table 3.4. The roots of F4

Roots in <I>+ for (I) of type F4:

T1 = 1000
T5 = 1100
T9 = 1120
7‘13 = 1220
T16 = 1122
T18 = 1222
r20 = 1242
T22 = 1342
7‘24 = 2342

T2 = 0100
T6 = 0120
7‘10 2 0122

T3 = 0010
T7 = 0110
r11 = 1110
r14 = 1111
7‘17 = 1121
T19 = 1221
T21 = 1231
T23 = 1232

74 = 0001
7‘8 = 0011
7‘12 = 0111
T15 = 0121

Notice that the first two columns contain the long roots while the second two contain

the shorts roots.

Table 3.5. The highest long and short roots in (I)

If we follow the preceding convention for describing roots as the sum of fundamental
roots, then the following is a list of the highest short and long roots in (I).

(1)

Highest Short Root

Highest Long Root

 

Al
Bl
Dz
E6
E7
E8
F 4
G 2

 

111...
111...

1

1
11

1232
12

 

111...1
122...2
122.. 1

. 21
12321
1
234321
2
2435432

2342
23

CHAPTER 4

General Lemmas

We now prove a number of general lemmas which will be useful later.

Lemma 4.1 Let r E <I>+.

1. If r is either a short root or all the roots in (I) have the same length, then r can
be written as the sum of fundamental roots r = ril + riz + + rik in such a

way that for each I g k, r’ = ril + riz + . . . + Ti, 6 (I), and |r| = |r’|.

2. If r is a long root in (I), then r can be written as the integral linear combination
of fundamental roots
I?!2 |’I"l2 M2

= r- + - +...+
ITi1|2 n ITi2|2 12 |"‘ikl2

 

 

 

Tik

in such a way that for each l g k,

,_ |r|2 ITI2 ITI2
_ —|Ti1|27”i1 + ——|Ti2|27‘i2 + . . . + I’m-1'27.“ E (p,

and |r| = |r’|.

Proof: Suppose r = eren njrj. We induct on Brien nj (the height of r). If

eren nj 2 1, then we are clearly done, so assume that 25-611 n,- > 1. Since (r, r) > 0,

19

20

n,- > 0 for all n,, and (r,r) = Zia-en nj(r,rj), it follows that for some r,- E H,

(r,r,—) > 0. As (ir, in) is a root subsystem of type A2, 82, or G2, one can easily

check that if |r| _<_ |le for all r, 6 ll, then wr,(r) = r — r,. Similarly, if |r| > |rj| for
IT M2

2
some rj E H, then one sees that TrTIl-Z E Z, and wn(r) = r — mm. The lemma follows

by induction applied to w,,(r).

Lemma 4.2 There is a unique conjugacy class ( possibly empty) of roots subsystems
of type A2{long), A2 (short), 82, L _L S, and S _L S in Q under W. Moreover, if
Q 7é B; or D1, then there is also a unique conjugacy class of type L _L L. If Q = B,
with l _>_ 2 or D; with l 2 5, then there are two conjugacy classes of type L J. L. If

Q = D4, there are three conjugacy classes.

Proof: Let r,s,a, and fl 6 Q and suppose that (:tr,:l:s) and (imifl) have the
same type. We assume that |r| = |al. Then there is a w E W such that aw = r, and
so (:ta, imw = (ir, :tfiw). In particular, it suffices to show that Stabw(r) has the

indicated number of orbits on the set
{7 E Q I (ir, in) has the same type as (ir, is”.

If r is a long root and Q 75 D4, then the result follows from [5], Lemma 4.2 and
Propositions 4.2, 6.5, and 6.16 (possibly from [3] also).

Note that the two conjugacy classes in B; of type L .L L are a result of the fact
that sometimes the root subsystem is contained in a larger subsystem of type 82, and
sometimes it is not. For example, (:i:(el — 62), i(el + 62)) (_I (:l:(el — e2), :l:e2), which
has type .32. On the other hand (:l:(el — 62), i(€3 — 64)) is not contained in any such
subsystem. We will call the first conjugacy class (L _L L)1 and the second (L _L L)2.

Also, if Q = D, with l 2 5, then the two conjugacy classes have representatives
{:t(e,- — ej),:t(e,- + e,)} and {:l:(e,- + ej),:l:(e;c + em)} where {i,j} H {m, k} = (0.

Similarly, we will call the first conjugacy class (L _L L)1 and the second (L J. L)2.

21

Now suppose that Q = D4 and (ir, is) has type L J_ L. Then one can easily

check that the conjugacy classes are the following:

{i(e,- — ej), i(e, + e,)}

{3503135 81),i(€k i 81) I {i,j} n {k, l} I 0}
{i(ei i ej)1i(ek 2F 81) l {2,3} m {k’ l} = Q}

We will call these conjugacy classes (L _L L)1, (L _L L)2, and (L J. L);;, respectively.

Lastly, suppose that r is a short root. If Q = 3;, then as each two short roots
are perpendicular and are contained in a root subsystem of type B2, and as all the
root systems of type 82 are conjugate, the result follows. If Q = F4, then the result
follows because of the graph automorphism that switches long and short roots while

preserving angles between them.

Corollary 4.3 Let r, 3, oz, and [3 E Q and suppose that (ir, is) has type 82, L _L L,
L _L S, or S _L S. Moreover, suppose that (ir, is) and (ia,i[3) are in the same

conjugacy class of Q under W and that (r, s) = (0:, 6). Then there exists w E W such
that {r, 3}“ = {0, 6}.

Proof: As this is true for each set of roots {7,6} 3 (id, id) and as there exists

w’ E W such that {r, s}“" C; (id, id), the claim is clear.

Lemma 4.4 Let G be a Chevalley group and suppose I 7é M S U E Syl2(G). Let P1
and P2 be parabolic subgroups of G containing U such that G = (P1, P2). Then there
exists g E G andi E {1,2} such that M9 g P,- and M9 Z 02(P,).

Proof: Let X = {R S U | for all g E G, R9 g P,- implies that R9 Q 02(1),» and
let Y = (X) _<_ G. Choose R E X. Since R S U 3 P1, it follows that R S 02(P1).
Thus 02(P1) <1P1 implies that (RP 1) g 02(P1) s U. But note that (RP!) S U implies

22

that (RPI) E X, by definition of X. Hence P1 3 NG(Y). Similarly, P2 3 NG(Y). So
G = (P1, P2) implies that Y <1 G. Thus, Y S U implies that Y S 02(G) = 1, proving

the claim.

At various points in the paper we will have occasion to explicitly write an involu-
tion as the product of elements of root subgroups. Towards that end we include the

following lemmas.

Lemma 4.5 Let r E Q, Q sré G2, and g E G \ CG(X,.). Then (X,,X,€’) _<_ G is
either a 2—group or is conjugate in G to (X,, X _r) 9’. SL2(K). More specifically, it is
isomorphic to either S12(K), a 2-Sylow subgroup of Sl3(K), or K x K ( considered as

an additive group).

Proof: Since all roots of the same length are conjugate under W, we can assume
without loss that if r is a long root, then it is the highest long root or that if it is a short
root, then it is the highest short root. In either case, we have that B S Ng(X,). So
write g = blnbg with b1,b2 E B, n E N. Then (X,.,X,?) = (Xan‘nb?) = (X,,X,I‘)b2.

An inspection of the various root systems now yields the result.

Lemma 4.6 Let G be a Chevalley group, G $5 G2(K), and let (a, b) = A be a fours-
group in G. If there exists a root subgroup, say X, such that a E CG(X) but such that
b E CG(X), then (ACCW) either contains a nontrivial element of a root subgroup or

an element which is conjugate to an element of the form xa(1)x5(1) where a, 6 E Q,

lal = Ifll, and a _L 6.

Proof: Let W = (X, X“). By Lemma 4.5, W is either isomorphic to Sl2(K), a Sylow
2-subgroup of L3(K), or K x K. We consider each, case by case.

Case 1: W E“ 312(K).

As Y = ([X, a]) is normal in (X, a), Y is normal in W. Moreover, because Y contains

at least one involution, Y is not contained in the center of W. Hence Y = W. Thus

23

because Y S (ACCW), we can choose an element x in a root subgroup of W which

satisfies the claim of the lemma.

Case 2: W is isomorphic to a Sylow 2—subgroup of L3(K).
Again, Y = ([X, a]) is normal in W. Moreover, because Z (W) = [X a, X] we see that
Y j<_ Z(W). Hence, Z(W) g Y and so as Z(W) is a root subgroup the claim is also

satisfied in this case.

Case 3: W 9.” K x K, viewed as an additive group.
Without loss we can assume that X = X,, where r is the highest long or short root
in Q so that B s NG(X). If we write a = bl'flbg, with b1,b2 E B and n E N, then
X “X = X "b2X = (X "X )b2 which is clearly conjugate to X "X . Thus, we can choose
an element 1 75 x E [X , a] S (ACG(b)) which is conjugate to an element in Xan for
some roots 01,5 E Q, lal 2 IS] and such that a + B is not a root.

Now suppose that a and ,8 are not perpendicular. Then as a + 3 is not a root,

(1 — fl is a root, and as Xng g X:””, the claim follows.

Lemma 4.7 Let X, be a root subgroup in G, X = (X,,X_r) and suppose that
NG(X,) = M,. Then 02'(NG(X)) = L, x X and 02'(GG(X)) = L,.

Proof: Let g E NG(X). Since X is doubly transitive on its Sylow 2-subgroups, there
exists x E X such that X5.” --= X, and X3: = X_,.. That is, gx E NG(X,)DNG(X_,.) =

LiH, and so g E L,H X , proving the lemma.

Lemma 4.8 Let a E G be an involution, G $4 G2(K) and let r denote the highest
long root in Q. If applicable, let 3 denote the highest short root in Q. Also let
N(X,) = M,- and NG(X5) = Mj. Then there exists a conjugate a’ of a such that
either (XE;,X_T) = (Xi,) and a’ E XrL, or (XE’S,X_3) = (Xis) and a’ E XsLj.

Proof: By Lemma 4.4 applied several times to (a), there is a minimal parabolic Pk

and a conjugate, say a”, of a such that a” E U\02(Pk). As (X_rk, Xa” ) is clearly not

24

a 2-group, it follows from Lemma 4.5 that it is conjugate to (er,X_rk) % SL2(q).
Moreover, since by Lemma 4.2 (er,X_,.k) is conjugate to (X,,X_,.) if rk is a long
root, or (XS,X_S) if rk is a short root, it follows that for some conjugate, a’, of a,
we have either (XE',,X_,.) = (X,,X_,.) or (XE’3,X_8) = (X3,X_3). The lemma now

follows by Lemma 4.7.

Suppose that all the roots in Q have the same length. We choose roots
81,82, . . .,s,, E Q as follows: Let 31 be the highest weight root in Q and suppose
that NG(X,,) = M. Let L be the Levi subgroup of M and let Q1 9 Q be the root
subsystem corresponding to L. If G is not an orthogonal group, then Q1 is a con-
nected root subsystem. In this case, choose 82 to be the highest root in Q1. On the
other hand, if G is an orthogonal group, then Q1 = Q2 UQ3 where both Q2 and Q3 are
connected root subsystems and Q3 has type A2. In this case choose 32 to be the high-
est root in Q2 and 33 to be the unique positive root in Q3. We continue the selection
of the roots by considering 82 as we considered 31 above, until the Levi complement

of normalizer of X," is trivial.

Lemma 4.9 If all the roots in Q have the same length, then every involution in G
is conjugate to one of the form x3,1(1)x3,2(1)---xs (1) for some {i1,i2,---,i,-} g

‘1‘
{1,2,~-,n}.
Proof: This follows from Lemma 4.8 and our choice of 31,32, . . . sn.

Lemma 4.10 Suppose V is a vector space over some field K. Moreover, suppose

that V = U?=1Vi; where each V,- is a proper subspace of V. Then |K| S n.

Proof: Without loss, we assume that n above is minimal. Thus V,- ,Q_ U?=1 V,- for
#1
each 1 S j S n. Choose x E V1 \ (U?=2 V,) and y E V2 \ (V1 U UL; V,) and choose A1

and A2 E K, with A1 aé A2. If there exist an i such that Alx + y and Agx + y E Vi,

25

then we see that both x and y are in V,, contradicting our choice of x and y. Hence,

n 2 |{Ax+y I A E K}| = |K| , proving the lemma.

We will now record a few results involving the weight structure of a K G-module.
All of the theorems cited below may be found in [12].

Let G be a universal Chevalley group defined over a field K (arbitrary), L be
the associated Lie Algebra, and let A be a weight such that (A, r) E Z+ U {0} for all
r E Q+. By [12], Theorem 39(e) (page 209), there exists a unique irreducible rational
K G-module, say V, for which A is the highest weight. We will now present a brief
description of the construction of V given there.

Let A also represent the corresponding weight on L, so A(H,_) = (A, r) E Z+ U {0},
for all r E Q+. By [12], Theorem 3(e) (page 14), there exists an irreducible L—module,
say (p, V’), with A as the highest weight. Let v+ denote a nonzero highest weight
vector in V’. Let U be the universal enveloping algebra of L and then let Liz be the
subalgebra of U generated by {X,'."/m! | r E Q}. By [12], Theorem 2, Corollary 1
(page 17), there exists a lattice M contained in V’ which is invariant under LIZ. In
fact, M = Uzv+.

Now define VK = M <82 K and let G’ be the Chevalley group constructed as an
automorphism group inside of LIZ. Then by [12], Theorem 7, Corollary 1, there is a
rational homomorphism (b : G —> G’ such that ¢(x,.(t)) = x’,(t) for all r E Q and
t E K. This give us a representation of G on V’. This representation, however, need
not be irreducible, but it does contain v+ which has weight A. Let V” be the smallest
submodule of V’ containing 21+, and let V’” be the maximal proper submodule of V”.
Then V” / V’” is the required K G-module.

Now suppose that A S G is a fours-group and V(A) is an irreducible K G-module
with highest weight A upon which A acts quadratically. Suppose that k is the algebraic

closure of K. Let G k denote the Chevalley group derived from the same representation

26

as G, only over It instead of K. Similarly, denote Vk’, Vk", and Vk’” in the construction

given above. Then V’ S Vk’, V” S Vé’, and V’” S V,,’”. Therefore, if we identify G as

III

a subgroup of G1,, then it follows that A acts quadratically on Vk”/Vk as well.

Thus we have proven the following proposition.

Proposition 4.11 Let k be the algebraic closure of K and let V be an irreducible
quadratic KG-module. Then the kG-module gotten by extending V remains a

quadratic module for G with the same weight structure as V.

Lemma 4.12 Let Q be a connected root system containing only one root length, r
be the highest root in Q and let G be the associated Chevalley group defined over K.
Also suppose that M,- = NG(X,.) and that P is a connected parabolic containing both
U and (Xiril- Then X = (Xf) is the unique irreducible K P-module in Q,- = 02(P).

Moreover, X has weight A,-.

Proof: Let Y be a P-module in Q. Because U is a 2—group acting on Y, Gy(U) 75 1.
Thus, because Z (U) = X, and because H acts irreducibly on X,., it follows that
XTSYandszSY.

Also, M,- = NG(X,) implies that if r,- is a fundamental root and if (Xirj) S
GG(Xr), then j = i. And as x,.(t)"'i(’\) = x,.(At), it follows that X is isomorphic, as a
KP-module, to V(A,).

Lastly, the following two theorems due to Robert Steinberg are used throughout
this paper. They are stated for an arbitrary field with nonzero characteristic equal

to p.

Theorem 4.13 (The Steinberg Tensor Product Theorem) Assume that G is
a universal Chevalley group with |II| = l. Let R be the set of p’ irreducible rational
representations of G for which the highest weight A satisfies 0 S (A, n) S p — 1 for all

r,- E H. Then every irreducible rational representation of G can be written uniquely

27

®§io p, o F rj where p, E 7?. and Fr denotes the Frobenius map which replaces the

matrix entries of the elements of G with their pth power.
This is Theorem 41 (page 217) of [12].

Theorem 4.14 Let G be a finite universal Chevalley group or one of its twisted
analogues constructed as the set of fixed points of an automorphism of the form
x,(t) -—> xp,(it‘7(’)). Then the HTS“ q(r) irreducible rational representations of the
including algebraic group (got by extending the base field to its algebraic closure) for
which the highest weights satisfy 0 S (A, r) S q(r)—1, for all r E II, remain irreducible

and distinct on restriction to G and form a complete set.

This is Theorem 43 (page 230) of [12].

CHAPTER 5

Linear Dependence

Definition 5.1 (Linearly Dependent Action Of A Group On A Module)
Let G be a group, K an arbitrary field, and V a K G-module. We say that G acts

linearly dependently on V if for each a, b EG there exists A E K (possibly zero) such
that either [v, a] = A[v, b] for all v E V, or [v, b] = A[v, a] for all v E V.

Proposition 5.2 Suppose V is a vector space over K, K a field of characteristic
two, and a, b E GLK(V) are involutions. Then (a, b) acts linearly dependently on V

if and only if a and b normalize the same subspaces of V.

Proof: Suppose that there exists A E K such that [v, b] = A[v,a] for all v E V. As
1 7A b E GLK(V), A aé 0. Now let W S V be a subspace and suppose that W“ = W.
Choose x E W. Then [x, b] = A[x, a] implies that x” = Ax“ + (A + 1)x E W and so b
normalizes W as well.

So suppose now that a and b normalize the same subspaces of V. Choose x E
V \ Cv(a). Then as a normalizes the 1-dimensional subspace K (x“ + x), b does as
well. Thus, because b is an involution, (x“ + x)” = x“ + 1:. Hence v“” + v“ +v” + v = 0
for all v E V. Similarly, vb“ + v“ + v” + v = 0 for all v E V, and so v“” = vb“, for
all v E V. Now choose y E V. Because a normalizes K (y, ya), there exists scalars

A and u E K such that y” = Ay“ + uy. Hence ya” 2 pg“ + Ay. Thus because

28

29

9“ + 31b + y = Slab = #31“ + Ay implies that y” = (u + 1)ya + (A + 1)y, it follows that

u = A + 1, and so y” = Ay“ + (A +1)y. Therefore, [v, b] = A[v, a] for all v E V.

Remark: Suppose that dimK(V) = k < 00 and that a and b E GLK(V) are
involutions such (a, b) acts linearly dependently on V. Then there exists a scalar

0 7E A E K and a basis for V relative to which a and b have the following matrix

form:
I; 0 0 It 0 O
a = 0 Ik—Ql O a b = 0 Ik—2l 0 a
I, 0 1, AI, 0 I;

where l = dimK([V, a]).

Lemma 5.3 Let A be a 2-group, ]A| _>_ 4; K afield with char(K) = 2; V = ®i€l V,,
with each V, a KA-module, such that [V, A, A] = 0; also let J = {i E I | [V,, A] 74$ 0}.

Then the following are true:
1. If |J| _>_ 2, then A acts linearly dependently on each V,.

2. If a,b E A and if there is anj E J and A E K such that [v, a] = A[v,b], for all

v E V], then for alli E I and v E V}, [v,a] = A[v,b].

3. If IJ] Z 3, or ifA acts linearly dependently on V and |J| 2 2, then A = 0 or 1.
That is, there exists a subgroup H < A with [A : H] = 2 such that [V,, H] = 0,
for all i E I .

Proof: Assume that |J| Z 2, and choose a,b E A:1 with [V,,b] 95 0 for some j E J.
Note that if i E I \ J, then by definition of J, [v,c] = 0 for all v E V,- and for all
c E A, and so of course [v,a] = A[v,b] for all A E K,v E V,. Now choose i E J \ {j}.
Without loss, assume that i = 1 and j = 2. We wish to consider the action of A
on V’ = V1® V2, so set I’ = I\{1,2} and let V” = ®i€pV§ so that V = V’ <8) V”.
Choose 0 aé w E C'v~(A). Then [V, A, A] = 0 implies that [V’ (X) w,A, A] = O, and

30

so [V’, A,A] <8) w = 0. Thus w aé 0 implies that [V’, A, A] = O. In particular, for all

v E V1,w E V2, we have [(v,w),a, b] = 0. That is,

(2)“, wa”) + (1)“, w“) + (vb, w”) + (v, w) = 0

for all v E V1 and w E V23.
Similarly we can show that [V,”, A, A] = 0, n = 1 or 2; so that v“” = v“ + v” + v

for all v E V1 or V2. Substituting this into the equation above, we see that

(1)“ +v” + v, w“ + w” + w) + (va,w°) + (vb,wb) + (v,w) = 0

which is equivalent to

(v“, w” + w) + (vb, w“ + w) + (v, w“ + w”) = 0

for all 22 E V1 and w E V2. In particular, as w“ + w” = (w“ + w) + (wb + w), we get
that

(v“+v,wb+w) = (vb+v,wa+w)

forallvEVlandwEVg.

Because of our choice of V2, we may choose w E V2 such that w” + w aé 0.
Case 1: [V1,a] 75 0.

Choose v E V1 such that v“ + v 5A 0. As (v“ + v,wb + w) 75 0 implies that
(vb + v, w“ + w) # 0 as well, we see that there exists A E K“ such that [v,a] = A[v, b]
and [w, a] = A[w, b]. Since our choice of v E V1 \ CV, (a) was arbitrary, it follows that
[v, a] = A[v, b], for all v E V1\C'v1 (a). Now choose v E Cv,(a). Then (v“+v, w”+w) = 0
implies that (vb+v, w“ +w) = 0; so wa+w # 0 yields v”+v = 0. Thus [v, a] = A[v, b]

for all v E V1. Similarly, [w, a] = A[w, b] for all w E V2.

31

Case 2: [V1,a] = 0.

First suppose that [V2,a] aé 0. Then choose c E A with [V1,c] 96 0. By Case 1
applied to {a,c}, there exists u E K * such that [v,c] = u[v,a] for all v E V1 or V2,
contrary to [V1,a] = 0 and [V1,c] # 0. Thus [V2,a] = 0, and so [v, a] = O[v,b] for all
v E V1 or V2, proving the first part of the lemma.

Note that we have just proven that for all a E A, [V,, a] 7Q 0 for some i E J if and
only if [V,-,a] 75 0 for everyj E J. That is, CA(V,-) = CA(V,-) for all i,j E J. Let
H = GA(V,-) for some j E J. Then A/H acts faithfully on each V},i E J. To finish
the lemma, we may assume that ]A] Z 2 and that A is faithful on each V,-, i E J.

Suppose now that |J| 2 3 and choose a,b E A”. Without loss assume that
{1,2,3} 9 J. Then as A acts quadratically on (V1 (8) V2) <8) V3, and as [V3,a] 74 0,
it follows from the above that there exists A E K * such that [v, b] = A[v, a] for all
v E V1 (8) V2 or V3.

Thus by considering V1 (8) V2 in place of V, to prove part (3) of the lemma we
may assume that A acts linearly dependently on V and that {1,2} Q J. Moreover,
we assume that ]A] > 2. Choose a,b E All with a 74 b. As before, let V’ = V1 (8 V2
and V” = ®k6,\{1,2} Vk so that V = V’ (8) V”. Note that A acts linearly dependently
on V’. Thus, there exist A and u E K * such that [v, b] = A[v,a] for all v E V’, and
[2), b] = n[v, a] for all v E V1 or V2.

Because A is faithful on both V1 and V2, for n = 1 and 2 we choose on E Vn such
that v: aé on. Let zn = [vma], so of, = 2n + U". Then, [(v1,v2),b] = A[(v1,v2),a] =
A(v‘1‘,v§‘) + A(v1,v2) = A(zl,v2) + A(v1,22) + A(z1,zg). However, we also have that
[v,,, b] = u(vf,+vn) = uzn, and so of, = uzn+vn. In particular, we see that [(v1, v2), b] =

(’U’ffvg) + (v13v2) = ”(zl) U2) + ”(211)22) + ”2(219 Z2)‘ Hence:

/\(Z1a ”2) + /\(01,Z2) + M21, 22) = ”(21.02) + #001,132) + [12(21, 22),

32

and so

(213 (A + I‘ll)? + (A + IRAQI?) = (v13 (A + “)22)'

Moreover, as [vma] = zn, and la] = 2’" for some m E Z, on and 2,, are linearly
independent. Thus it follows that A + u = 0 = A + u2, and so u 75 0 implies that
u = A = 1. But then [v,b] = u[v, a] for all v E V1 and V2 implies that v“ = v” for
all v E V1 and V2. Therefore, since A acts faithfully, we get that a z b, contradicting

our choice of a and b and completing the proof of the lemma.

For the remainder of this section, we adopt the following hypothesis: Let G be a

Chevalley group over K, as before we assume that char(K) = 2. Let V = ® V,- be

iel
a K G—module with each V,- nontrivial and irreducible, I some index set; A S G an
elementary abelian 2—subgroup with IA] _>_ 4 such that A acts linearly dependently on

V.
Lemma 5.4 A acts faithfully on each V, and II ] = 1

Proof: Suppose there is an i E I and 1 sé a E A such that [V,-,a] = 0. Then
(a0) S CG(V,-). Let M = G/C’G(V,-). Since the char(K) = 2, Z(G) has odd order
and so CCU/i) S Z (G) In particular, we see that M is a 2-group. By definition,
CM(V,-) = 1. However, as ]M| = 2", for some k E Z, CM(V,-) 7E 1 unless M = 1.
Consequently, G must centralize V,, a contradiction. Therefore, A acts faithfully on

each V,. It follows from the second part of lemma 5.3 that ]I] = 1.

Remark: Because A acts faithfully on V,, it follow that if a,b E An and A E K
such that [v, a] = A[v, b] for all v E V,, then A # 0. Moreover, if c E A such that

[v, c] = u]v, b] for all v E V,, then A = u if and only if a = 0.
Lemma 5.5 Let 1 3A a,b E A. Then CG(a) = CG(b).

Proof: Let g E Gg(a). Since A acts linearly dependently on V, there exist A E K ’1

such that [v,a] = A[v,b], for all v E V. Hence A(v”9 + v” + v9 + v) = A[v, b, g] =

33

[v,a,g] = [v,g,a] = A[v,g,b] = A(v9” + v” + v9 + v). Thus, v”9 = vg” for all v E V.

Therefore 9 E CG(b) too, since G acts faithfully on V.

Lemma 5.6 Let G and V be as above. Ifr 2 eren njrj E Q, then h,(A) and
<r-.r~>

Hrjen h,j (Anj (or) ) act identically on V.

Proof: By [12], Lemma 19 (page 27), h,(A) acts as multiplication by A0”) on

the weight space V” of V, where, (u,r) = u(H,) and H, 2 (72:7). Thus since

7" = 2.3a; "in,

 

H_ an_ 272173,”) Hr.-

rJ-EII rj EH

So because u is a linear functional, it follows that h,(A) and

Hh ("”14“)

73‘ EH

act identically on V“, and hence on all of V since V is the sum of its weight spaces.

Lemma 5.7 Let S E Syl2(G) with A S S. Let P1 and P2 be parabolic subgroups of
G containing S such that G = (P1, P2). Then there exists g E G and i E {1, 2} such

that A9 S P,- with A9 D 02(P,) = 1.

Proof: By Lemma 4.4 there exists g E G and i E {1,2} such that A9 S P,, but
A9 S 02(1),). Suppose I S a E A9 H 02(1),). Let Y be a chief factor in V for P,-.
Then a E 02(P,-) implies that [Y, a] = 0. Let 1 75 b E A9. Then by the above remark,
there exists A E K n such that [v,a] = A[v, b] for all v E Y. Thus [Y, b] = 0 as well
and so we see that [Y, A9] = O for every chief factor Y of V for P,-. However, this is

equivalent to A9 S 02(1),), contrary to the above and proving the lemma.

Remark: It follows from Lemma 5.7 and induction that there exists an r,- E II and

g E G such that A9 S P,, and A9 n 02(P,,) = 1. Without loss we can assume that

34

A S P“. and A D 02(P,,) = 1 since if A acts linearly dependently on V, then so do

all of its conjugates.

Theorem 5.8 If G, V, A, and r,- E H are as above, then |K] 2 IA] 2 4 and up to

algebraic conjugates of V one of the following hold:
1. G 2’ A1(K) and V is a fundamental module.
2. G 9: B1(K) and one of the following is true:

(a) r,- is a long root and V is the natural or spin module, or

(b) r,- is a short root and V is a fundamental module.
3. G E D1(K) and V is the natural or a half-spin module.
4. G E“ E6(K) and V = V(Al) or V(A6).
5. G E“ E7(K) and V = V(A7).
6. G E F4(K) and one of the following is true:

(a) r, is a long root and V = V(A4), or

(b) r,- is a short root and V = V(Al).
7. G E’ G2(K), r,- is short, and V = V(Ag), the natural module.

Proof: It follows from 4.13 and Lemma 5.4 that V 2’ V(A)C where V(A) is a restricted
module for G and c E Aut(K). We can assume without loss that V is a restricted
module.

Let L = (X in) 91 S l2(K ) and let Y be a non-trivial chief factor in V for P“. Also
let A g L be such that A02(P,,) = A02(P,,). Since [Y, 02(P,,)] = 0, A acts linearly

dependently on Y. Let V, denote the natural module for SL2(K). So as Y is also

35

a chief factor for L and since V,, is the unique non-trivial restricted module for L, it
follows as above from Lemma 5.4 that Y E Vn", for some a E Aut(K).

Let T = {a E Aut(K) | V: ’:‘=’ Y where Y is a chief factor of V for L}. Choose
a E T, a E A”, and Y a chief factor of V for L such that Y 2 Vn". Then for all
b E A, there exists Ab E K such that [v,b] = Ab]v, a] for all u E Y. Note that
by the remark following Lemma 5.4, Ab = AC if and only if b = c. Thus if we let
K(A),, 2 {Ab ] b E A}, then ]K(A),,] = ]A]. Thus as K(A),, S K, we see that
|K I _>_ IAI 2 4.

Now choose 7 E T and Y’ E“ V: a chief factor of V for L. Then for each v E Y’,

[v, b] = Ag“17]v, a]. In particular, Ab = Alf-17, and so we see that

A” = A7 for all A E K(A),,, ”y E T (5.1)

Now let 0 7e v E C'V(U), the highest weight space of V. Suppose r E Q with
]r] = |r,-| such that h,(A)v 75 v for some A E K. Let L’ = (Xh). Since char(K) = 2
and since |r] = |r,-], there exists w E W such that x,,(t)"’ = x,(t) for all t E K,
and Lw = L’. Since h,(A)v 79 v, there is a non-trivial chief factor Y/ X of V for Pfif
such that v E Y \ X. But then (Y/ X )w-1 is a non-trivial chief factor for L and so,
as above, (Y/X)“’_l E’ Vn" as KL—modules for some a E T. Let '27 = v + X. Then
[W"1,X,,] = ["17,X,]w_1 = 0 since X, S U. Hence W4 E C(Y/X)w-1(Xri) implies that

1 1

UV] is in the highest weight space of (Y/X)w-’. Thus, h,,(A)u‘“_ = You" . In

particular, since h,(A) = h,,(A)‘” we see that h,(A)? = A06. Therefore,

h,(A)v = A"v for all A E K. (5.2)

36

Let J = {r,- E II ] h,j(A)v # v for some A E K}. Since V is a restricted module, it

follows from Lemma 5.6 that

A)v= Uh, (Am (..., #))v=AI”v, (5.3)

where a: z, 6, 72,7214

If ]r] = ]r,] and if h,(A )v 7Q v, we define o,(A) = X” for all A E K. It then follows
from equations (5.2) and (5.3) above that o, E Aut(K). In particular, if we choose
another 3 E Q with ]s] = ]r,-] such that h,(A)v S v for some A and s = ererl sjrj,

then it follows from equations (5.1) and (5.2) that for all A E K (A),,,

(r r) (r- r)
_J’_.L_ _J’_.l_
AZT‘I en8j(, r) __/\erenn1(rir:) (5.4)

For convenience of notation, let 191.2%: Ff]:— for all r,- E II.
Now, choose r E Q+ with |r] = ]r,] and such that if r = eren nJ-rj, then ZUEJ n]

is maximal among all such roots.
Case 1: Suppose ]r,] S ]r,] for all r,» E II.

We claim that either 23.6, n,- = 1 or (7.) from the statement of the Theorem is
true (i.e. G ”=1 G2(K), etc.). So, suppose not. Then by Lemma 4.1 applied to r, we
see that there are roots s,s’ E Q+ with IS] = ]s’] = ]r,-| such that ifs = eren sjrj
and s’ = Err 6Hs’ r,, then 23'er =2 and ZjEJs’ ]- =1, and such that if k E J
with 3’], S 0, then 3;, # 0 either. Hence, it follows from equation 5.3 that there exists
j,k E J (not necessarily distinct) such that h,(A)v = Apk+piv and h3:(A)v = Apkv.
Thus, by equation 5.4, Apk+Pj = A” for all A E K (A), and so A” = 1 for all

A E K(A),,a. However, as ]r,] 2 ]r,-] we see that p,- - I I *1, 2, or 3. But if p, =1 or

In]2 _

2, we get a contradiction to ]K(A),,s] = IA] 2 4.

37

To finish this case, we need only show that G L—‘L’ G2(q) and 1 E J is not a
possibility, so suppose it is. First, assume that {1, 2} = J. Then if we let s = r2 and
s’ = r1 + 2r2, we see that h,(A)v = Av while hs:(A)v = A521, and so A5 = A for all
A E K (A),,,, yielding a contradiction similar to the one above. Thus we can assume
that J = {1}. So, if r = r1 +r2, then a, E Aut(K), but o,(A) = A3, for all A, a
contradiction since by Proposition 4.11 we can assume that K is algebraically closed.

Hence if G = G2(K), then J = {2}.

Case 2: Suppose ]r,-] > ]r,] for some r,- E II.

Then we claim that 2351777 IIrZIl: = 1. So, suppose not. By Lemma 4.1 there

 

are roots 3 and s' E Q with ]s] = ]s’] = ]r,|, such that ifs = 2136113173 and if

s’ = Z1367, s’jrj, then ZJEJ stT’EIITZ = 2 and ZjEJ s’jITrgl—I; = 1. It follows from equation
5.3 that h,(A)v = A21) and hsr(A)v = Av. And so, as above, we get that A2 = A for all
A E K(A),,_,, contrary to ]A] _>_ 4.

An inspection of the various roots of maximal height in each root system, as

recorded in Table 3.5 now yields the result, proving the theorem.

CHAPTER 6

Linearly Dependently Acting

Fours- groups

In this chapter we prove the following corollary to Theorem 5.8:

Corollary 6.1 Suppose G, A, and V are as in Theorem 5.8. Then the following are

true:

1. If G 2’ A1(K ) and if V is not an algebraic conjugate of the natural module for

G, then A is contained in a root subgroup.

2. If G ’=’ B1(K ) and if V is not an algebraic conjugate of the natural module for

G, then one of the following is true:

( a ) V is not an algebraic conjugate of the spin module for G and A is contained
in a short root subgroup of G.
(b) l 2 3, V is an algebraic conjugate of the spin module for G and A is

contained in a root subgroup of G.

3. If G ’=’ Dz(K) and V is not an algebraic conjugate of the natural module for G,

then A is contained in the product of two commuting root subgroups of G.

4. IfG ’=’ E6(K) or E7(K), then A is contained in a root subgroup of G.

38

39

5. IfG ’=’ F4(K), then one of the following is true:

(a) V is an algebraic conjugate of V(Al) and A is contained in a long root

subgroup of G.

(b) V is an algebraic conjugate of V(A4) and A is contained in a short root

subgroup of G.
6. IfG ’=’ G2(K), then A is contained in a short root subgroup of G.

The proof of this corollary will follow from the several lemmas to follow. Conse-
quently, throughout the chapter we will assume that G, A, and V are as in Theorem
5.8. The information on certain parabolic subgroups used below can be found in [5]
or can be easily verified computationally.

Assume first that G $4 G2(K). We will handle G E“ G2(K) separately at the end
of the chapter.

Choose 1 75 a E A. By Lemma 4.8 we may assume there exists a root, say t E Q,

such that X = (XE,,X-,) = (Xfl) ’=’ Sl2(K) and [X,,a] = 1.

Lemma 6.2 IfG E’ A1(K), Dz(K), En(K), or F4(K) or ifG '=’ B1(K) andt is a
short root, then A S X X C’G(X).

Proof:

Case 1: G ”=’ A](K).

Without loss we may assume that t = r1. Thus NG(X) = (X x GG(X))H = L2H.
By [11], Y = CV(Q2) is a nontrivial irreducible K Mg-mOdLIIG. Moreover, because M2
is a maximal parabolic subgroup of G and because Y is a proper subspace of V, it
follows that M2 = NG(Y). Thus A S M2 since a E NG(Y) implies that A S NG(Y),

by Proposition 5.2.

40

Now let Q; = (X _, ] X, E Q2). Then M2— = LgHQ; is also a maximal parabolic
subgroup of G containing a. By an argument identical to one used above, A S M;
as well. Hence, A S 02’(M2 0 M2“) = L2.

Case 2: G g A,(K) or F4(K).

Then as a E CG(Xt), A S GG(X,) as well by Lemma 5.5. Now let Y = CV(X).
Because a E NG(Y), A S NG(Y) too, by Proposition 5.2. Thus if A S X x GG(X),
then 02(NG(Xt))fl(NG(Y)\Xt) S 1. Hence because 02(NG(X,))/Xt is an irreducible
module for the Levi compliment of NG(Xt), it follow that G = (NG(Xt),X_t) S
NG(Y), a contradiction.

Case 3: G E” F4(K).

Because of the graph automorphism of F4, we may assume without loss that t is a
short root. Moreover, we may assume that t is the highest short root in Q, namely
t = r23. As in Case 2, we have that A S M4 0 NG(Y), where M4 = NG(X,)
and Y = GV(X). However, although Q4/Xt is an indecomposable module for M4,
it is not irreducible. Rather, M4 acts irreducibly on X,, S/Xt, and 624/ S where
S = Z(Q4) = (X,,0,X,,6,X,,8,X,20,X,22,X,23,X,24). Thus, by an argument similar
to the one used in Case 2, we may assume that A S (L4, S).

Now let 31 = —r24, 82 = r1, 33 = r2, and s4 = r3. Then one can easily check that
Q’ = (i81,i82, isg, is4) is a root subsystem of type B4 and that A S (L4,S) S
(X, ] s E Q’) C‘—_‘ B4(K). Thus, by Case 2, we get that A S X X Ca(X), proving the

lemma.

Lemma 6.3 If G is as in Lemma 6.2, then Corollary 6.1 holds.

Proof: We assume that if t is a long root, then it is the highest long root and if t is
a short root, then it is the highest short root. Let J Q H such that NG(Xt) = M J.

Thus we have A S X x LJ by Lemmas 4.7 and 6.2.

41

Now let A be the weight of V and assume that A Q ZAJ- for some r,- E J. (Recall
that by Theorem 5.8, A is a fundamental weight.) By Lemma 5.6 neither X nor L J
centralize C'v(U); so choose a non-trivial chief factor Y/ W of V for X x L J such that
Cv(U) (_Z Y \ W. Then by the corollary to Lemma 68 in [12], Y/W “=’ V1 <83 V2 as
a K (X x L J)-module, where V1 is a nontrivial irreducible K X -module and V2 is a
nontrivial irreducible K L J-module. Suppose that [V2, A] 74 0. Then by Lemma 5.3
there exists 1 75 c E A such that [V1,c] = [V2,c] = 0. That is, c normalizes every
subspace of V1. However, this contradicts Proposition 5.2 since the rest of A clearly
does not. Hence, [V2, A] = 0.

Now, if M J is a connected parabolic subgroup, then [V2, A] = 0 implies that A S X
and so, in particular, A S X,.

Thus we may assume that G ’=’ D1(K). Let A S LJ be such that AX/X = AX/X
in NG(X)/X. Then [1] = Sl2(K) X Dz_2(K) ifl 2 5 and L] E Sl2(K) x Sl2(K) x
Sl2(K) if l = 4. Then as A E ZA1_1 or ZA, by assumption, A must be contained in
the Sl2(K) factor in the first case or in the product of at most two Sl2(K) factors in
the second case.

Now assume that G ’=’ D4(K) and suppose that A E ZA4. Then by the above,
A S (X,,) x (X,3) x (X,). However, by Lemma 4.2, <iT1,:lZT3> is conjugate to
(i(e1+ e4), i(e2 + e3)) = (i(r1+ r2 + r4), i(r2 + r3 + r4)) and so we see that either
_A S (X,,) x (X,), or A S (X,,) x (X,). In either case, however, we get that A is
contained in the product of two root subgroups. Similarly if A E ZA3 or, in fact,

A E ZAI.

Lemma 6.4 IfG ’5 B((K), l 2 3, and ifA E ZAI, then A is contained in a root

subgroup of G.

Proof: Without loss we may assume that t is the highest long root in Q. Then

NG(X) S M2. As in the lemma above, we have A S NG(CV(X)) 0 M2.

42

Let s be the highest short root in Q. Also let Q0 = {r E Q ] if r = Enennfl‘i,
then n1 = 0} and let Q0 = (X, ] r E Q3) 0 Q2. Note that Q0 is a root subsystem of
type B;_1 and that t — s is the longest short root there. Thus we get that L12 acts
irreducibly on Q0 / X ,_S.

Similarly, L12 acts irreducibly on (Q1 (1 Q2)/(Xt,X8). Thus as conjugation by
elements in (Xin) interchanges elements in Q0 with elements in Q1 (‘1 Q2. It follows
that L2 acts irreducibly on Q2/Z where Z = (Xt,X3, Xt_,).

In particular, as in Lemma 6.2, we get that A S ZL2. However, note that Q’ =
{it, is, i(t——s), irl} is a root subsystem of type B2. So, if we let M = (X, ] r E Q’),
then Z L2 = M X L12 2 B2(K) x B1_2(K). Therefore, since both of the above factors
act nontrivially on C'v(U), it follows as above that A S M.

Lastly, since B2( K) has three conjugacy classes of involutions with representatives
x,(1), x3(1), and xt(1)x3(1), we may assume that A contains one of these. However,
note that if we let c denote this element and if c is one of the two latter types of
involutions, then (X is, X -3) is certainly not a 2—group and so we may apply Lemma
6.2. Thus we have c = x,(1), but then as Z(CM(c)) = X,, we get that A S X,, by

Lemma 5.5.

Corollary 6.5 Let G ’5 82(K) and A be as in Theorem 5.8. Also let V = V(Al)
denote the natural module for G. If A acts linearly dependently on both the natural
and the spin module for G, then A is contained in a root subgroup of G. Moreover,

if A acts linearly dependently on either the natural or the spin module for G, then

Proof: Let (1) be the graph automorphism of G. Then as

43

(b : x,,(t) +—> x,2(t)
$n+r2(t) I" x,,+2,2(t)
x,.,(t) r—+ 32,, (t2)
~Tn+2r2(t) ’—’ $r1+r2ft2),

it follows that V‘f’" = V(Ag).

Now assume that A acts linearly dependently on both V(Al) and V(Ag). Then
it follows from the above that both A and A4’ must act linearly dependently on V.
However, because G contains three conjugacy classes of involutions with representa-

tives

23,,(1), $T2(1)i and $r1+r2(1)$r1+2r2(1)a

we may assume that A contains one of these. Thus because A acts linearly depen—
dently on V we see that A must be contained in one of the following sets: X,,, X,.,,
or {x,,+,2(t)x,,+2,,(t) | t E K}. However, the last of the above sets is not normalized
by (I), and so can not contain A. Therefore A must be contained in a root subgroup,
proving the first part of the corollary.

The second part of the corollary now follows since, by the above, either A or A4’
acts linearly dependently on V. Thus, A must be contained in either one of the three
aforementioned sets or in the set {x,,+,2 (t2)x,,+2,,(t) | A E K} and so, in particular,
the elements of A are conjugate under H. Hence if 1 S a, b E A with b’I = a for some
h E H, then [V, a] = [V, b"] = [V,b]" = [V, b] since H acts as the group of diagonal
matrices on V. Therefore, [V, a] = [V, A] for all 1 S a E A.

Lemma 6.6 Let G ’=’ G2(q), A, and V be as in Theorem 5.8. Then A is contained

in a short root subgroup.

Proof: By [14], (8.1) G has two conjugacy classes of involutions with representatives

t = x2a+b(1) and z = x3a+2b(1). Moreover, it follows from [14], (3.3), (8.5), and

 

44

the commutator relations developed in [14] that Z (CG(t)) = X2a+b and Z (00(2)) =
X3a+2b-

Without loss we can assume that either t or z is an element of A. If t E A, then
by 5.5 and the above we see that A is contained in a short root subgroup and we are
done. So assume now that z E A. Then again by 5.5 and the above we see that A
is contained in a long root subgroup. However, this implies that a conjugate of A is
contained in P,1 \ 02(1)”), contradicting Theorem 5.8 and proving this lemma and

Theorem 6.1.

CHAPTER 7

A Result Concerning Root

Systems and Weights

Throughout this chapter, assume that G is a Chevalley group, G S G2(K), and Q is
the root system associated with G. As usual, let H be a fundamental root system in

Q.

Definition 7.1 Let A be a weight of an irreducible module for G. Define
Al = {rE Q ] (A,r) =0}.

Let J be a rank two root subsystem of Q. Recall that in Lemma 4.2 we determined
the orbit of J in Q under W. We will now prove the following lemma in which we

determine all weights A such that Al F] Jw S (l for all w E W.

Lemma 7.2 Let J g Q be a rank 2 root subsystem. If A is a weight such that
AJ' 0 J‘” S (b, for all w E W, then A is an integral multiple of one of the fundamental

weights listed in the table below.

45

 

46

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Q J Possible Weights
A2(long) A,- where 1 S i S n
A‘ L 1 L A1 or A,
A2(long) A1 or A;
B2 A,-where1SiSl
(L _L L)1 A1
8’ (L 1 L)2 A,
L _L S A1
S _L S A1
A2(long) A1, A3, or A4
(L i L)1 A3 07' /\4
D4 (L 1 L)2 A1 or A3
(L _L L);, A1 or A4
A2(long) A1, A1_1, or A;
Dl,l25 (L_LL)1 A1
(L J_ L)2 /\l—l 07‘ AI
A2(long) A1 or A5
E6 L _L L None are possible
14200719) A7
E7 L _L L None are possible
A2(long) None are possible
E8 L _L L None are possible

 

 

 

47

 

 

 

 

 

 

 

 

Q J Possible Weights
A2(long) A4
A2(short) A1
B2 A1 or A4
F4 L _L L None are possible
L .L S None are possible
S _L S None are possible

 

This lemma will follow from the next several lemmas.

Lemma 7.3 Let J Q Q, with J ’2‘: A2 or B2. Choose r]- E II. Also,
1. if J "=1 A2 (long) or B2, let r = ngEH n,r,~ be the highest long root in Q, or
2. if J 2 A2 (short), let r = Znen nir, be the highest short root in Q.

Ifnj=1 or ifJE’Bg andnj S2, then K(Aj)Jw Sillfor alle W.

Proof: Let w E W. Choose s,t E J‘” to be positive roots such that ]s] = ]t] and
s — t E J'”. Note that if J 2 B2, then s and t must both be short roots. Suppose
that s = Znen sir, and t = Znen t,r,-. If either 3, or t,- is zero, then K(Aj)Jw S 0,
so assume that sj,tj S 0.

Suppose first that n, = 1. Then as r was chosen to have maximal height, 0 <
sj,tj S n, and so s, = t,- =1. Hence (Aj,s — t) = 0, and so K(Aj)Jw S (0.

Now suppose that J ’=’ 82 and n,- S 2. Then because s+t is a long root in J'” and
because r was chosen to have maximal height, we see that s, + t,- S 2. Consequently,

as above, 3,- = t,- = 1 and so K(Aj)Jw S (0.

Hence we have found a sufficient condition, when J is connected, for a fundamental
weight to be included in the table above. We will show now that these are in fact the

only weights that should be included in the table.

48

Definition 7.4 For a weight A and r = Enen n,r, E Q, define

3A0”) = Z nz-lr‘z-IQ/IV‘IQ-

TiEIl\AJ'

Lemma 7.5 Suppose that J is a connected rank 2 root subsystem in Q and that A is
a weight such that Al 0 J‘” S (ll, for all w E W. If J has type A2 (long), let r be the
highest long root in Q. Also, if J has type A2 (short) or B2, let r be the highest short
root in Q. IfJ g B2, then S,\(r) S 2. Otherwise, S,\(r) = 1.

Proof: Suppose not.

Case 1: Assume that all the roots in Q have the same length and so J E A2.

By Lemma 4.1, there exists r’ E Q and r, E H with r’ - r, E Q, SA(r’) = 2, and
S ,\(r’ — r,) = 1. Thus, if J’ = (r’,r,-), then J’ ’-‘=’ A2 implies that there exists w E W
such that J‘” = J’. However, AL 0 J’ = (b, contrary to the hypothesis.

Case 2: Assume that Q = B; and J ’=’ A2.

Since S,\(r) > 1, A a! ZAI. Then, as in Case 1, there exists long roots r’ E Q and
r,- E II such that r’ — r,- E Q and such that S A(r’ ) = 2 and SA(r,-) = 1 which brings us

to the same contradiction as above.

Case 3: Assume that Q = B, and J E“ B2.

As above, A ¢ ZA; and so by Lemma 4.1 there exists a short root r’ E Q and a
long root r,- E II such that S,\(r’) > S),(r’ -- r,-) > 0. Now, since J’ = (ir’,ir,-) =
i{r’, r,-, r’ — r,, r’ + r,}, we see that Al F) J’ = 0, contrary to the hypothesis.

Case 4: Assume that Q = F4.

Here it seems easiest to simply give explicit sets of roots which generate connected

rank 2 root subsystems and which eliminate all but the desired values of A.

1. Suppose that J has type A2(long)

Let J’ = i{r13,r16,r24} = i{1220,1122,2342}. Thus, if Al 0 J’ S (ll, then

49

A E ZA4 and so S),(r) = 1.

2. Suppose that J has type A2(short)
LEL J, = i{T12,T17,T23} = i{Olll, 1121,1232}. Thus, as above, A E ZAl.

3. Suppose that J has type B2
Let J’ = i{r10,r11,r23,r24} = i{0122, 1110, 1232,2342}. Thus, as in the two

cases above, A E ZA1 or ZA4.

Remark: It follows from Lemma 7.3 and 7.5 that if J is a connected rank 2 root
subsystem and A is a weight such that Al 0 J‘” S (b, for all w E W, then A is an

integral multiple of one of the weights in Lemma 7.2.

Lemma 7.6 Suppose that Q S B, or D; and let J be a root subsystem of type L _L L,
L _L S, or S J. S. Then A is a weight such that Al F) JI" S 0 for all w E W if and
only ifQ = A) and A E ZAI or ZAI.

Proof: Because of the graph automorphism of F4, we may assume that J contains a
long root. Thus without loss we may assume that J contains the highest long root,
say 7'. As (A, r) S 0 it follows that every root with the appropriate length in ri is in
Al.

Now choose r, E H\Al. Then r, 9! rJ” and so ]r,] = ]r]. As above we see that every
root with the appropriate length in rgL is then also in Al. However, unless Q = A),

ri is a maximal root subsystem in Q, in which case Q = (rgL,ri), a contradiction.

Hence Q = A, and A E Z(Al +A1). If (A,rl) S 0, then (r2,r3, . . . ,r1) = (rli,ri) S Al
and so A E ZAI. Similarly, if (A,r)) S 0, then A E ZA).

Moreover, we notice that if r E Q+ such that (r, A1) S 0, then r = r1+r’ = e0 — e,-
for some root r’ and 1 S j S l. Thus ifr and s are roots such that (r, A1) S O S (3, A1),
then r ,1 s. In particular, we see that Al can not be eliminated from the list of possible

weights. Similarly, neither can A].

50

As in Lemma 7.5 it again seems easiest for Q = B, or D) to explicitly list a set of

roots which will eliminate all but the desired weights.

Assume that Q = B).

1. Suppose first that J has type (L i L)1.
Let r = r1+r2+- - -+r,_2+2r1_1+2r, = el+e,._1 and s = r2+r3+---+r1_1+2r1=

e2 + e). J’ = (r, s) has type (L _L L)1 and so if K(A)JI S (b, then A E ZAI.

Similarly as in Lemma 7.6, we note that (r, A1) S 0 if and only if r = iel i e,
for some j. Thus, as above, we see that Al can not be eliminated from the list

of possible weights.

2. Suppose now that J has type (L _L L)2.
Let r = r1+r2+- - '+Tl—2+T1_1 = el—el and s = r1+r2+- - -+rz—1+2r, = e1+el.

J’ = (r, s) has type (L _L L)2 and so if Al F] J’ S 0, then A E ZAl.
Moreover, we note that as one of the roots, say r must be of the form 6,- — ej, it

follows that if r = 22:, nkrk, then n] = 0. Hence A; may not be deleted from

the list of possible roots.

3. Suppose that J has type (L J. S).
Letr=r2+r3+~~rl=e2 and lets=r1+r2+~--rl_1+2r¢=e1+e[. Thus,

A E ZA1 and, by the above, A1 can not be eliminated.

4. Suppose that J has type (S _L S).
Let r =r1+r2+~~r1= el and let s=r2+r3+~~rl = e+2. As above, we
see that only A1 has the desired property. Also because the only short root r

such that (Al, r) S 0 is el, it follows that A1 may not be eliminated.

Now assume that Q = D], l 2 5.

51

1. Suppose that J has type (L J. L)1.
Let r =r1+r2+---rz =e1+e,_1 and let 8 =r2+---rz_2+2r]_2+r1_1+rl =

e2 + e1_2. Then A E ZAI which, as above, can not be eliminated.

2. Suppose that J has type (L J. L)2.
Let r = r1+r2+---+r1_1= e1 -e, and let 3 = r1+r2+---+rz_2+r) = e1+el.

Thus A E ZA]_1 or ZA].

Now, as one of the roots, say r, must be of the form e,—ej, then if r = 21:, nkrk,
n) = 0. Thus A, can not be eliminated from our list. Similarly if r is of
the form 6,- + ej, r = 22:171ka and if nz_1 S 0, then max{i,j} S l. Thus
8 = e,—ej = Zl=1mkrk with m]_1 = 0, and so Az_1 can not be eliminated. Note
that the preceding statement also follows because of the graph automorphism

of Q.

Lastly assume that Q E“ D4.
Here the result follows from inspection of the various conjugacy classes which are
given explicitly in Lemma 4.2.

Lemma 7.2 now follows from the preceding arguments.

CHAPTER 8

Linear Independence

Definition 8.1 Let G be a group of Lie-type and A S G a fours-group.

1. We say that A is a linearly dependent fours-group in G if one of the following
holds:

(a) G ’5 A)(K) or D)(K) and A acts linearly dependently on the natural module
for G.

(b) G ’=’ B2(K) and A act linearly dependently on either the natural or the

spin module for G.

(c) G E“ B1(K), l 2 3 and either A acts linearly dependently on the natural
module for G or there exists a rank two connected parabolic subgroup, say
P, of G such that A is contained in the Levi complement of P and is

linearly dependent there.

(d) G ’=’ E6(K), E7(K), or E8(K) and there exists a proper connected parabolic
subgroup M J such that A9 S L J and is linearly dependent there for some

9 E G.
(e) G ’=’ F4(K) or G2(K) and A is contained in a root subgroup of G.

(f) G ’=’ G’(K) is a twisted Chevalley group and A is linearly dependent when

considered as a subgroup of G(K).

52

53

2. We say that A is a linearly independent fours-group if A is not a linearly de—

pendent fours- group.

Proposition 8.2 Let G, A, and V be as in Theorem 5.8. Then A is a linearly

dependent fours- group.

Proof: This follows from Corollary 6.1 and the definition above.

We now state the main theorem of the paper.

Theorem 8.3 Let G = G(K) be a Chevalley group defined over K, char(K) = 2
and let V a nontrivial, irreducible K G-module. If there exists a linearly independent
fours-group A which act quadratically on V, then up to algebraic conjugacy of V, one

of the following is true:
1. G ’=’ A1(K) and V is a fundamental module.
2. G 9.“ B1(K) and V is a fundamental module.
3. G ’=’ D)(K) and V is the natural or a half-spin module.
4. G ’=’ E6(K) and V = V(Al) or V(Ae).
5. G ’5 E7(K) and V = V(A7).
6. G ’—_‘-’ F4(K) and V = V(Al) or V(A4).

7. G '=’ G2(K) and V = V(Ag).

Corollary 8.4 Let k be the algebraic closure of K and let G0 = G’(K) S G(k) be a
twisted Chevalley group. Let V be a nontrivial, irreducible K Go-module and suppose
that Go contains a linearly independent fours-group which acts quadratically on V.
Then V is obtained from the restriction to Go of a rational representation V’ of G (k),

where V’ is an arbitrary representation from the conclusion of Theorem 8. 3.

54

Proof: The corollary follows directly from Theorem 4.14 and Theorem 8.3.

The proof of Theorem 8.3 will follow from the next several lemmas and sections.
Lemma 8.5 V is an algebraic conjugate of a restricted module for G.

Proof: Suppose not. Then, by Theorems 4.13 and 4.14, V is the tensor product
of two or more algebraic conjugates of restricted modules. However, A acts linearly
dependently on each those restricted modules by Lemma 5.3, and so by Proposition

8.2, A is a linearly dependent fours—group, contrary to the hypothesis of the Theorem.

Remark: It follows from Lemma 8.5 that we may assume that V is an restricted

module.

Lemma 8.6 Suppose G = A2(K). Then V is isomorphic to the natural module for
G.

Proof: Let VN denote the natural module for G. Because a is an involution and
VN is 3—dimensional, we may assume that a is in a root subgroup, say X,,+,2. Also,
as CG(X,,+,2) S B we have that A S U. In particular, U = (X,,,X,2,X,,+,2) and
]X,,,X,,] = X,,+,2 implies that either A S 02(P,,) or 02(P,,). Without loss, we
assume that A S 02(P,,), and b E 02(P,,) \ X,,+,2. Then for all g E N,, we have
a9 e U\02(P,,) and b9 e P,2 \ U.

Let X = GV(02(P,,)) and note that because X is U-invariant and because
CV(U) is 1-dimensional it follows that C'v(U) S X. If C'U(X) S 02(P,,), then
all of (C'U(X)P'1) = O2’(P,,) acts trivially on GV(U) in which case A = An, and
we are done. So assume that CU(X) < 02(P,,), but GV(U) S X. Then, for each
9 E N,2, [X,ag] S 0. Moreover, [X,ag] = [X,02(P,,)a9] is invariant under U and
so CV(U) S [X , a9]. In particular, we see that CV(U) is centralized by (U, b9) for all

g E N,,. Hence, A = A,1 and we are done.

55

Lemma 8.7 Let G = B2(K). Then V is an algebraic conjugate of either the natural

or the spin module.

Proof: Suppose not. Then it follows from the definition of linearly independent,
Theorem 5.8 and from Theorem 4.13 that V is an algebraic conjugate of V(A1 + A2).
Recall that r1 is a long root and r2 is a short root. Let qfi be the graph automorphism

of G induced by the graph automorphism of B2. Then as

9’5 : mm“) H 35m“)

33m (t) H @1032)
we see that V(A1 + A2)“1 is isomorphic to V(Al) <8) V(Ag)2 as a KG-module. Thus
V is isomorphic to VA, <8) VAT However by Lemma 5.3, this implies that A must act
linearly dependently on both the natural and the spin module for G, contrary to A

being a linearly independent fours-group, proving the lemma.

Definition 8.8 Suppose that A is a weight for G and w E W. Then we define A” to
be the weight defined by Aw(H,) = A(Hw—1(,)).

Lemma 8.9 Let G be a Chevalley group and V = V(A) be a nontrivial irreducible
K G-module with highest weight A. Also let PJ be a parabolic subgroup of G. Then for
all w E W, Aw is conjugate under W] to the highest weight of some chief factor of V

for PJ. That is, each orbit of WJ on {Am ] w E W} corresponds to a chief factor.

Proof: Let M K = NG(V,\). For convenience of notation, in this lemma only we let
LJ = (Xi, ] r E J) and Q1 = 02(PJ) instead of the usual definition of LJ and QJ.
Note that because QJw-l flQ+ is a positive root system for J‘”_1, ((pr—l flQ+)I" =
Q J D (Q+)“’ is a positive root system in Q J. Thus we get that L J (7 MX is a parabolic
subgroup of L J. Hence there exists w’ E WJ such that P = Q JH (L J H ME)“ is a

parabolic subgroup of G with B S P. So now let X / Y be a chief factor of V for PJ

56

such that (VAWI + Y/Y) fl (X/Y) S 0. Then
(VAww’ + Y)8 S (VAww’ + Y)QJH(LJD’WK’)w’ = (VAurw’ + Y)LJn’wl?)w, = V ww’ + Y,

since L J normalizes Y and My“ normalizes VAWI. Thus, it follows that Aww’ is the

highest weight of X / Y.

Proposition 8.10 If G, V, and A are as in Theorem 8.3 and if there exists a con-
nected rank two parabolic subgroup, say PJ, of G such that A Q O2I(PJ) \02(PJ) and
such that A02(PJ)/02(PJ) is a linearly independent fours-group in 02’(PJ)/02(PJ),
then Al 0 J‘“ S (l), for all w E W, where A is the highest weight in V

Proof: It follows from Lemmas 8.6 and 8.7 that if 'y is the highest weight of any chief
factor of V for P], then 7i 0 J S (ll. Thus, by Lemma 8.9, (Aw)i F) J S (l, for all
w E W. However, this is clearly equivalent to the condition that Al 0 J w S (l, for all

wEW.

Remark: Therefore we see that to prove Theorem 8.3 for G ’S G2(K), it is sufficient
to show that there exists a connected rank two parabolic subgroup which contains a

linearly independent fours-group. We will do exactly that in the next sections.

CHAPTER 9

The Classical Groups

Lemma 9.1 Let G = An(K) and let A be a fours-group in G. Let V be the natu-
ral module for G and suppose that A acts linearly independently on V. Then there
is a 3-dimensional subspace X S V normalized by A upon which A acts linearly

independently.

Proof: Suppose not.

Case 1: There exists 1) E V such that (22") is 4-dimensional.

Then v, v“, v”, and v“” are linearly independent vectors in V. Let X = (v“ + v, v” +
v, W + v). Then X is a 3—dimensional subspace normalized by A, and as v“ + v E
Nv(a) \ Nv(b), we see that N X(a) S N x(b), and so A act linearly independently on
X by Lemma 5.2.

Case 2: There exists v E V such that (vA) is 3—dimensional.

Then X = (0") satisfies the claim of the lemma since if [v, b] E K [v, a], then v” E
(v, v“). However, v” E (v, v“) implies that (vA) is only 2-dimensional, contrary to the
assumption.

Case 3: dimK((vA)) S 2 for all v E V.

Choose v 9! C'V(a). Then v and v“ are linearly independent and so span (0"). Thus,

there exists A and 'y E K such that v” = Av + 7v“. Then vaI’ = Av“ + 7v and

57

58

v = Av” +7va”. Substituting, we get v = A(Av +7v“) +7(Av“ +7v) = (A2 +72)v. S0,
A2 +72 = 1 and so A = 7 +1. Thus v” = (7+ 1)v +7v“, and so [v,b] = 7[v,a].
Suppose now that Gv(a) S Cv(b) and choose w E Cv(a) \ Cv(b). Then v + w E’

Gv(a) implies, by the above, that there exists a E K such that

7[v,a] + [w, b] = [v, b] + [w,b] = [v + w,b] = a[v + w,a] = a[v,a],

since w E Cv(a). Hence, 0 S [w,b] E K[v,a]. In particular, X = (v,w,v“ + v) is an
A-invariant subspace of V. Moreover, w E Cv(a)\Cv(b) implies that GX(a) S C’X(b),
and so A acts linearly independently on X, again contrary to our assumptions.

Thus we may now assume that dimK((vA)) S 2 for all u E V and that Cv(a) =
Cv(b).

Choose v,w ¢ Cv(a). Then there exists 7,u E K such that [v, b] = 7[v,a] and
[w, b] = u[w, a]. We claim that 7 = )1. Let W = K (v, w), the subspace generated by
v and w.

Suppose that W H Cv(a) S 0. Then there exists 01,6 E K n such that av +
Bw E Gv(a). Since W is only 2-dimensional and as W Z Gv(a), it follows that
Cw(a) = K(av + 6w). Thus 0 = [av + 6w, b] = a7[v, a] + 6u[w, a] = [07v + Buw, b]
implies that 7 = u.

Thus we may assume that WflC’v(a) = 0. In particular, there exists 7] E K I‘ such

that [v + w, b] = 77[v + w, a]. But then

n[v, a] + n[w, a] = 77[v + w, a] = [v + w, b] = [v, b] + [w, b] = 7[v, a] + u[w, a]

from which it follows that [(r) + 7)v + (r) + u)w, a] = 0. Hence, 7 = n = u. Therefore

A acts linearly dependently on V contrary to our initial hypothesis.

59

Lemma 9.2 Let G '=’ B](K) and suppose that A contains a transvection, but that A is
not contained in a short root subgroup. Then there exists a rank 2 connected parabolic
subgroup of type B2 which contains a linearly independent fours-group which also acts

quadratically on V

Proof: Let r and s be the highest long and short roots in Q, respectively. We may,
without loss, assume that l S a E A 0 X,, but that A S X,. Because char(K) = 2
we have that NG(X,) = M1. Moreover, because 621/ X, is the unique minimal normal
subgroup in M1 /X,, we get that there exist 1 S x E X, H (ACCIal). Therefore, as
the subgroup A’ = (a, x) acts quadratically on V and as (ir, is) has type Bg, the

lemma is proven.

Lemma 9.3 Let G = B,(K) withl Z 3 and ]K] > 6. Let A S G be afours-group and
suppose that A contains no transvections and acts linearly independently on (V, (-, )),
the natural module for G. Then A acts linearly independently on either some singular

3-space in V or on xi/Kx for some x E CV(A).

Proof: Suppose not. First we note that if CV(A) is nonsingular we may choose
x,y E GV(A) with (x,y) S 0 and let W = (x,y). Then as V = W 69 WJ- and
Wi ”-3 xi / K x, we get a contradiction. Hence GV(A) is singular.

The rest of the proof of the lemma will follow from several claims. Choose a, b E A
such that A = (a, b).
Claim 1: [V, A, A] = 0.
Proof: Suppose not. Then choose v E V with [v, a, b] S 0. If c is any involution, then
because dim([W, Cl) S 1/2dim(W) for any c—invariant subspaces W of V, it follows
that X = (vA) is 4-dimensional. In particular, [X , A] is a 3—dimensional A-invariant
subspace upon which A acts linearly independently. Now if there is a z E CV(A) \
[X, A], then [X, A] S [V, A] = CR/(A)i S zi implies that [X, a] + Kz/Kz S zi/Kz,

contrary to the above.

 

60

On the other hand, if GV(A) S [X, A] S X, then C'v(A) S CX(A). But then
dim(Gv(a)) 2 l and dim(Cv(A)) 2 U2 imply that l/2 S 1 and so l S 2, a contradic-
tion. Therefore [V, A, A] = 0.

Let Q = U{C’v(c) |1S c E A}.

Claim 2: If x E V \ Q, then there exists 0 S A, E K such that [x, a] = A,[x, b].
Proof: Let U = [K x, A] = [K x, a] + [K x, b] and suppose that U is 2-dimensional.
Note that U S CV (A) implies that U is singular. Moreover, note that if x E U ‘I', then
K x+U is a 3-dimensional singular subspace upon which A acts linearly independently,
contrary to our assumption. Thus we may assume that x ¢ U i. Then xi 0 U is 1-
dimensional. Moreover, since CV(A) = [V, A]i is at least 3-dimensional, we get that
dim(xi fl C’V(A)) 2 2. In particular, xi 0 GV(A)) Q U. Let 2 E (xi (1 GV(A)) \ U.
z E CV(A) implies that GV(A)L S 2*; so we have U S [V, A] S GV(A)i S zi. Hence
K x + U + K 2 / K z is a 3-space in zl / K z upon which A acts linearly independently,
contrary to our assumption. Thus, U is 1-dimensional and so for each x E V\Q there
exists 0 S A, E K such that [x, a] = A,[x, b].

Claim 3: If x, y E V such that (Kx + Ky) 0 Q = 0, then A, = Ay.

Proof: As [x, a] + [y, b] = A,+y([x, b] + [y, b]) and as [x,a] + [y, a] = A,[x, b] + Ay[y, b],
we see that [(A, + A,+y)x + (A, + A,+y)y, b] = 0. Thus (Kx + Ky) fl 9 = 0 implies
that A, = A,+y = Ay.

Claim 4: If x, y E V \ Q, then there exists 2 E V such that (Kx + K2) 0 Q = 0 and
(Kz+Ky)flQ=O.

Proof: Suppose not. Thus for all z E V, either Kz S Kx + Q or Kz S Ky + 0.
That is, V S {Kx + Q} U {Ky + 9}. Hence,

v g {Kx + Cv(a)} u {Kx + Cv(a)} u {Kx + Cv(b)} u {Kx + ovum}

U{Ky + Gv(a)} u {Ky + Cv(b)} u {Ky + CV(ab)}

 

61

Because A does not contain any transvections, we get that dim(Cv (c)) < 2l — 1
for all 1 S c E A. Hence V is the union of 6 proper subspaces, contrary to ]K | > 6, by
Lemma 4.10. Therefore, A, = A,, for all x, y E V\f2 and so also for all x,y E (V\f2).
Claim 5: V = (V \ 9).

Proof: If not, then let X = (V \ (2). Then V is the union of 4 proper subspaces,
namely, X, Gv(a) Gv(b), and Gv(ab) again a contradiction. Therefore, A acts linearly

dependently on V, contrary to our initial hypothesis and proving the lemma.

Lemma 9.4 Let G = 02,,(K), n 2 4, K an algebraically closed field with char(K) =
2, and let A be a fours-group in G. Let (V, q, (-, )) be the natural module for G and
suppose that A acts linearly independently on V. Then A acts linearly independently

on either some singular 3—dimensional subspace of V or on xi / K x for some x E

CV(A).

Proof: Suppose not. The proof of the lemma will follow from several claims.

Claim 1: Suppose that v E V such that [v, A, A] S 0 and let X = (vA). Then there
exists 2 E GV(A) \ [X, A] with q(z) = 0.
Proof: Note that X is 4—dimensional and [X , A] is a 3—dimensional A-invariant sub—
space upon which A acts linearly independently. Now we note that if U is any
singular subspace of V, then U contains an isotropic hyperplane. This is because if
we choose u E U such that q(u) S 0, then for all w E U there exists 7 E K such
that q(w + 7u) = q(w) + 72q(u) = 0, since K is algebraically closed. Thus if the
dimension of GV(A) is greater than or equal to 3, then the claim is clearly true as
O[X.A1(A) = (v + vCIL + vb + v“”) is 1-dimensional. Similarly the claim is true if there
exists w E V with (w + w“ + wI’ + waI’) S (v + v‘1 + v” + vab) since u + u“ + uI’ + u“”
is a singular vector for all u E V.

Thus, we may assume that dim(Gv(A)) S 2 and dim([V, A, A]) = 1. Now note

that if 1 S c E A, then ]c] = 2 implies that dim(Cw(c)) Z %dim(W) for all c—invariant

 

62

subspaces W of V. Hence 2 Z dim(Cv(A)) = dim(CCV(a)(b)) Z édim(CV(a)) 2 %n
implies that n = 4, and dim(CV(a)) = 4 too. Thus dim([V,a]) = 4. But [V, a, b]
1-dimensional implies that O[v,a](b) is 3-dimensional. However, O[V,a](b) S CV(A),
contrary to the above.

Claim 2: [V, A, A] = 0.

Proof: Suppose not. Choose [v, A, A] S 0 and let (X A). Choose 2 E Cv(A) \ [X , A]
with q(z) = 0. Then [X, A] S [V, A] = CR/(A)i S zi implies that [X, A] + Kz/Kz is
a 3—space in zi / K 2 upon which A acts linearly independently, contrary to the above.

Let Q = U{Cv(C) ] 1 S c E A}.

Claim 3: Let x E V \ 52. Then there exists 0 S A, E K such that [x, a] = A,[x, b].
Proof: Let U = [K x,A] = [K x,a] + [K x, b] and suppose U is 2-dimensional. First
suppose that x J. U. We note that U is singular because q(x + x“) = q(x) + q(x“) +
(x, x“) = (x, x“) = (x, x+x“) = 0 since x J. U. Now suppose that for each y E CV(A)
and for all A E K, q(x+ Ay) S 0. Then K algebraically closed and q(x+Ay) = q(x) +
A2q(y) + A(x, y) implies that x E Cv(A)J- and that q(y) = 0 for all y E CV(A). Thus
because [V, A, A] = 0 implies that dim(CV(A)) 2 n we get that CV(A) is a maximal
singular subspace of V. But then CIR/(A)i = GV(A), contrary to x E CV (A)i\GV(A).
Hence we choose 2 E x+GV(A) with q(z) = 0. However, C’V(A) J. [V, A] then implies
that K 2 + U is a singular 3-space upon which A acts linearly independently, contrary
to our assumption.

Hence we can assume that x )1 U. Then xi (‘1 U is 1-dimensional. More-
over, because [V, A, A] = 0 and n 2 4 implies that CV(A) = [V, A]i is at least
4—dimensional, we get that xi 0 CV(A) is at least 3-dimensional. Hence there exists
2 E (xi 0 CV(A)) \ U with q(z) = 0. But then Kx + U + Kz/Kz is a 3-dimensional
space in zi / K 2 upon which A acts linearly independently, contrary to the above.

Therefore, U is 1-dimensional and so there exists A, E K such that [x, a] = A,[x, b].

 

 

63

The remainder of the proof follows exactly as in the proof of Lemma 9.3.

 

 

CHAPTER 10

En(K)

Choose a, b E A such that (a, b) = A.

Lemma 10.1 Let G = En(K), n = 6, 7, or 8, and suppose X, is a root subgroup of
G with [A O X,] = 2. Then there exists a connected rank two parabolic subgroup of G

which contains a linearly independent quadratically acting fours-group.

Proof: Suppose that b E X, and let X = (ACGIbl). As 02(NG(X,))/X, is the unique
minimal subgroup of CG(X,)/X,, it follows that 02(NG(X,)) S X. Thus we can
choose a root subgroup X, S 02(NG(X,)) such that r and s are contained in a root
subsystem of type A2 and choose an involution x E X,. A’ = (x, b) is then the

required fours— group.

Proposition 10.2 If there exists a root subgroup, say X, such that a 9! CG(X) but
such that b E Ga(X), then either there exists a connected rank two parabolic subgroup
of G which contains a linearly independent quadratically acting fours-group or else

we may assume that there exists roots (1,6 E Q, with ]a] = ]6] and a .L 6 such that
a = $a(1)$g(1).
Proof: This follows directly from Lemmas 4.6 and 10.1.

To complete the proof of Theorem 8.3 for the case when G g En(K), it suffices

to assume that a = xa(1)x5(1) E A, where a and 6 are as in Proposition 10.2.

64

 

 

65

We now investigate the structure of (ACG(“)). In particular, we shall prove that
if a and b centralize different root subgroups, then (ACCIal) contains a fours-group
which intersects nontrivially, but is not contained in, a root subgroup and which also
acts quadratically on V.

Before we can do this, however, we first determine CG(a). Towards this end we
make a helpful, albeit nonobvious, choice for a and 6 as follows:
for G = E6(K) we choose or = r32 and 6 = r33,
for G = E7(K) we choose a = r27 and 6 = r29,
for G = E8(K) we choose a = r53 and 6 = r59.

Recall that by Lemma 4.2 all sets of roots {a, 6 E Q | or J. 6} are conjugate under
W.

The following lemmas, Lemmas 10.3, 10.4, and 10.5, were adapted from [5] where

they appeared in a more general context.

Lemma 10.3 Let G = E6(K). Then GG(a) S M”, and 02’(Cg(a)) = UOLO where
U0 = 02(M1,5) = 62le and L0 ’=’ Sp6(K). Moreover, U6 = Z(Uo) = Q1 0 Q6
is isomorphic to the natural module for L15 ’=’ D4(K) 91 S0; (K) with root ele—
ments corresponding to isotropic vectors and a = x,32(1)x,33(1) corresponding to an

anisotropic vector.

Proof: Since Q1 and Q5 are abelian and since X,32, X,33 S Q1 0 Q6, it is clear that

U0 = Q1626 S Ca(a). Now let

L0 = (X,2,X,4,x,3(A)x,5(A),w2,w4,w3w5 ] A E K).

Then L0 S L1,4 and, using Table 3.1 and the Chevalley Commutator Formula, one
sees that L0 S Cc(a) and that L1,.) S Ca(a). Moreover, L1,4 "S D4(K) and L0 can be

obtained as the set of fixed points of the graph automorphism of order 2 of L. Hence,

 

 

 

66

Lo g 3300 g 5176le-

Let X = UOLO. Then U0 = 02(X). We claim that M1 and M6 are the only
parabolic subgroups of G which contain X. So suppose that X S M = Mf’ , for some
9 E G. As X involves an Sp6(K), we see that i = 1 or 6. Thus because 02(M) is
abelian and 02(X) = U0 is not, it follows that X is contained in a proper parabolic
subgroup of M. Hence X S Mf’ fl Mf’c,

is, X S (M10 M6)"9 for some h E M,-. In particular, 02(X) S 02((M1 fl M6)"9).

where {i,j} = {1,6} and k E Mf’. That

But 02(X) = 02(M1 0 M5) and so hg E NG(02(M1 0 M6)) = M1 0 M6. Thus
M6” = M? = M,. Therefore, the only parabolic subgroups which can contain X
are M1 and M6. On the other hand its easy to check that neither X_,1 nor X _,6
centralize a. Thus, Cg(a) S M1 0 M5 = Mm.

Now again using Table 3.1 and the Chevalley Commutator Formula, one can check
that U’ = Q1 DQG = Z(Uo) = (X," | 29 S i S 36). In particular, X,” E U6 where r36
is the highest root in Q. Let Y = (X6325) By Lemma 4.12, Y is the unique irreducible
L15 module in U0 and has weight A2. Thus Y must be the natural module for L”, 2’
SO; (K) and hence is eight dimensional. On the other hand, X,36 S Z (U0) implies
that Y S Z (U0) which is also eight dimensional. Thus U6 = Z (U0) is isomorphic to
the natural module for L15.

Moreover, if S E Syl2(SO,'3+ (K )) and VN is the natural module, then CVN(S) is
an isotropic 1-dimensional subspace. Hence X,36 corresponds to an isotropic vector
and therefore because all the root elements in U6 are conjugate they all correspond
to isotropic vectors too. In particular, since a is not conjugate to a root element by
assumption, a corresponds to an anisotropic vector. Hence, because the centralizer of
an anisotropic vector in SO; (K) is isomorphic to Sp6(K), L0 = GL,’6(a). Therefore,

02’(Cg(a)) = UoLo-

Lemma 10.4 Let G = E7(K). Then GG(a) S M6 and 02’(Ca(a)) = UoLo where
U0 = Q6 = 02(M6) and L0 '5 Sp8(K)xSL2(K). Moreover, U6 = Z(Uo) is isomorphic

67

to the natural module for L63 "S D5(K) ’=" S066(K) with root elements corresponding

to isotropic vectors and a = x,.,,(1)x,.,9(1) corresponding to an anisotropic vector.

Proof: Recall that L5 ’S D5(K) x A1(K) 91 SOfo(K) x SL2(K). Let L0 be the
direct product of the SL2(K) factor of L5 with the fixed point group of the graph

automorphism of the D5(K) factor. Thus,

L0 : <Xr1,X1-3,Xr4,Ir2(A)$,-5()\),wl,W3,’IU4,’IU2U}5 ] A E K) X <Xi‘r7)

= Spst) X 51/2le-

Let U0 = Q6. Using Table 3.2 and the Chevalley Commutator Formula, one can
easily check that UOLO S GG(a), but that L6 S CG(a). Let X = UoLo. We claim
that M6 is the unique maximal parabolic subgroup containing X. Suppose that
X S M = Mf’ for some maximal parabolic subgroup M and g E G. Because X
involves an Sp3(K) x SL2(K), it follows that i = 6 or 7.

Suppose i = 7. Then as 02(M7) is abelian while 02(X) = Q6 is not. It follows that
X must be contained in a proper parabolic subgroup of M of the form M7gflM6’h, where
h E M—fl. However, by comparing orders, we see that Sp8(K) x SL2(K) S M79 0 Mg".

Thus i = 6. Now if 02(X) = U0 S 02(M6’), then X must be contained in a proper
parabolic subgroup of M5. Thus X S Mg 0 M6”, for some h E M6. But this yields
the same contradiction as above. Hence, 02(X) = U0 S 02(M6I). That is, since
U0 = 02(M6), we have that 02(M6) S 02(M6I) and hence g E M6. Therefore, M6 is
the unique maximal parabolic of G containing X, and thus CG(a) S M6 as well.

Now using Table 3.2 and the Chevalley Commutator Formula, one can easily check
that

U6 = 2(6),) = (X,, | i = 23, 24, 26, 27, 29, 30, 31, 32, 33).

68

In particular, because X,33 S U6, where r33 is the highest root in Q, it fol-
lows from Lemma 4.12 that Y = (X327) is isomorphic to the natural module for
L6;- '=’ SOfLo(K). Thus as U6 is ten dimensional, we see that Y = U6. As in the
case of E5(K), because the centralizer of a Sylow subgroup of S 0:6(K ) in the natural
module is an isotropic vector, we get that all the root elements of U6 correspond to
isotropic vectors and so a must correspond to an anisotropic vector.

Moreover, because the centralizer of an anisotropic vector in SOibe) is isomor-

phic to Sp8(K), we see that L0 = CL6 (a). Therefore 02’(Cg(a)) = UOLO.

Lemma 10.5 Let G = E8(K). Then CG(a) S M1 and 02’(GG(a)) = UOLO where
U0 = 02(M1) = Q1 and L0 ’=’ Sp12(K). Moreover, U6 = Z(Uo) is isomorphic to
the natural module for L1 E“ D7(K) ’=’ SOf;(K) with root elements corresponding to

isotropic vectors and a = x,58(1)x,59(1) corresponding to an anisotropic vector.

Proof: Recall that L1 g D7(K). Let L0 be the set of fixed points of the graph

automorphism of L1. Thus

L0 = (x’r2(A)$T3 (A), X”: era XTG‘) X‘r'n Xr‘ga
w2w31w41w53w63w7aw8 ] A E K)

Sp12(K).

Also let U0 = Q1. Using Table 3.3 and the Chevalley Commutator Formula, one can
easily check that X = UOLO S CG(a) but L1 S CG(a).

We claim that M1 is the unique maximal parabolic subgroup which contains X.
So suppose that X S M = Mf’ for some maximal parabolic subgroup M and g E G.
Because X / U0 9:“ Sp12(K), it follows by comparing the orders that i = 1 or 8. Now if
U0 S 02(M), then we get that X S M," 0 M39” with {i,j} = {1,8} for some h E Mf.

However this is clearly impossible. Thus we can assume that U0 S 02(M).

 

 

69

Hence because [626] < [Qi] = |U6], U0 S Q8. Therefore i = 1, and as in the
previous cases, we can conclude that M1 is the unique maximal parabolic subgroup
of G which contains X.

Now using Table 3.3 and the Chevalley Commutator Formula, one can easily check

that

U6 = Z(Qi)
8

: (Xr, ] if Ti = anrj, then 711 = 2).

i=1

In particular, because X,65 S U6, where r65 is the highest root in Q, it follows from
Lemma 4.12 that Y = (Xr’gg) is isomorphic to the natural module for L1 g SOII;(K).
Thus as U6 has dimension 14, Y = U6. As in the case of E6(K), because the centralizer
of a Sylow subgroup of SOit,(K) in the natural module is an isotropic vector, we
get that all the root elements of U6 correspond to isotropic vectors and so a must
correspond to an anisotropic vector. Moreover, as the centralize of an anisotropic
vector in SOfC,(K) is isomorphic to Sp10(K), we see that L0 = CL, (a). Therefore

02’(CG(G)) = UoLo-

Lemma 10.6 If a and b centralize different subgroups, then G contains a quadrati-
cally action fours-group A’ and a root subgroup X such that ]A’flX | = 2. In particular,

V is one of the modules in Lemma 7.2.

Proof: Suppose that a = $a(1)$g(1). Let Z = {xa(A)xg(A) ] A E K}. One can see
from the description of CG(a) that Z S Z (GG(a)) and that all the elements of Z are

conjugate in H.

Case 1: bEZ.

70

Then LOUD S CG(b) implies that 02’(Cg(a)) S 02’(Cg(b)). But then because a
and b are conjugate to each other, 02’(C'G(a)) = 02’(Gg(b)), contrary to the hypoth-

esis.

Case 2: bE U6\Z.

Recall that U6 has the structure of a natural module for S 0;",(K ), for some m and
that L0 is the centralizer of an anisotropic vector in U6. For n = 6 or 8, let M = L0.
For n = 7, recall that L0 E Sp8(K) x SL2(K) and let M be the Sp8(K) factor.
In all cases then, M E Sp2m_2(K). Now, when U6 is viewed as an M K-module,
it contains M K —submodules X and Y with X S Y such that X is 1-dimensional
and is spanned by an isotropic vector, Y/ X is (2m — 2)-dimensional, and U6 / Y is
1-dimensional. it follows that (bM + Z /Z ) is isomorphic to the natural module for
M. In particular, (bLO) must contain an isotropic vector. Therefore, because isotropic

vectors correspond to root elements in U6, (ACCIal) must contain a root element.

Case 3: b E U0 \ U6.

Because U6 = Z (U0) and because U0 is generated by root subgroups, it follows that
there is a root subgroup, X S U0 such that [X , b] S 1. We claim that [X , b] D Z = 1.

Consider the set of all triples (R, S,x) where R and S are commuting root sub-
groups in G and x is an involution in R‘S" which is not contained in any root sub-
group. By Lemmas 4.5 and 4.6 there are perpendicular roots r,s E Q+ such that
RS is conjugate to X,X,. Thus, it follows from Lemma 4.2 that all such triples are
conjugate in G. In particular, for a fixed involution x, we get that CG(x) is conjugate
on the set of all pairs of root subgroups R and S such that x E R'I'S“. Therefore,
since U6 is invariant under CG(a), it follows that if R and S are root subgroups such
that a E R*S*, then R, S S U6. That is, [R, b] D Z S 1 only if R S U6. Consequently,
[X, b] D Z = 1. We choose x E [X, b] with ]x] = 2 and let A’ = (a, x) By Case 2,

(A’CG(“)) contains a root element.

 

71

Case 4: G = E6(K) or E8(K) and b g? U0, or G = E7(K) and b E MW \ U0.
Suppose that [b, U0] S U’ = Z (U0). Then by the 3—Subgroup Lemma, [b, U6] = 1,
contrary to U6 being a faithful module for L0 if G = E6(K) or E8(K) and for L53 if
G = E7(K). Hence we can choose a root subgroup X S U0 \ U6 and x E [b, X] such
that x E U0 \ U6 and ]x] = 2. We then apply Case 3 to A’ = (a, x).
Case 5: G = E7(K) and b ¢ MW.
We can assume that b E U and so we see that X = (X _,7, X f ,7) is not a 2-group.

Thus, by Lemma 4.5 X E SL2(K) and so we can choose a root element x in (ACCIal).

Then A’ = (x, a) is a quadratically acting fours-group containing a root element.

Lemma 10.7 Let G = En(K), n = 6, 7, or 8, and suppose that a and b centralize

the same root subgroups in G. Then A is linearly dependent.

Proof: Suppose there exists a root subgroup X, such that A centralizes (X,, X _,).
If NG(X,) = M,, then by Lemma 4.7, A S L,. The result then follows by induction
on the root systems since the statement is clearly true for A1(K) and Dl(K).

First we consider G = E6(K). By Lemma 4.9 we may assume that

a : $T36(1)x1'29(1)xr9(1)$r4(1)'

Now let Q’ = (irg, ir4, irg, :lZT'Qg). Then one can easily check that {r4, r9, r29, r35} Q
Q’ and that Q’ has type D4. Thus, without loss, we may assume that a E D4(K). We
claim that every involution in D4(K) centralizes a root subgroup and its corresponding
negative root subgroup.

Let {01,02,03,a4} be fundamental root system for D4 with highest root r =
a1 + 2012 + a3 + a4 and assume that a’ is an involution in D4(K). By Lemma 4.9 we

may assume that a’ = x,(1)xa,(1)x,,3(1)xa,(1). However, if we let

9 = $02+03(1)$01+02 (1)13a2+a4 (1)33014-024-03 (1),

72

then a9 = xa,(1)x03(1)x05(1) which centralizes (X,,X_,) and so the claim follows,
proving the lemma for E6(K).

Next we consider G = E7(K). As above, by Lemma 4.9, we may assume that a =
x,33(1)x,,6(1)x,6(1)a’ where a’ is an involution in L1,6,7 E D4(K). However, as we saw
above, every involution in D4(K) centralizes a root subgroup and its corresponding
negative root subgroup, so the lemma is also proven for E7(K).

Lastly, if G = E8(K), then because 02’(CG((X,65,X_,65))) = L8 E E7(K), the

lemma follows by the above.

CHAPTER 11
F4(K)

Choose a, b E A such that (a, b) = A. Also let Q be a root system of type F4. Recall
from Table 3.4 that r23 and r24 are the highest short and long roots in Q, respectively.

For ease of notation, let 3 = r23 and r = r24.

Lemma 11.1 Let G = F4(K) and suppose there exists a root subgroup X of G such
that ]AflX] = 2. Then there exists a connected rank 2 parabolic subgroup of G of type

B2 which contains a linearly independent fours-group which also acts quadratically on

V.

Proof: Suppose that 1 S a E A 0 X. Because of the graph automorphism of G,
we may assume that X = X,. Then because Z (02(NG(X,))) is the unique minimal
normal subgroup of NG(X,) and because X, S Z (02(NG(X,))), we have that X, S
(ACCIal). Hence if we chose an involution x E X,, then A’ = (x, b) is the fours-group

claimed above.

Lemma 11.2 Let G = F4(K) and suppose that a and b centralize different root
subgroups in G. Then there exists a connected rank 2 parabolic subgroup of G of type

B2 which contains a linearly independent fours—group which also acts quadratically on

V.

73

74

Proof: First suppose that A contains a nontrivial element of a root subgroup. As
above, we can assume that 1 S a E A n X,. Thus, because CG(a) = O2’(M1) and
because Z (02’(M 1)) = X,, it follows that ]A n X,] = 2, and so the lemma follows
from Lemma 11.1.

Hence, by Lemma 4.6 we may assume that a = $a(1)$3(1) where (1,6 E Q,
]a] = ]6] and a is perpendicular to 6. Because there exists unique conjugacy classes
of type L .L L and S _L S in Q, there exists a root subsystem, say J Q Q, of type
B2 such that (1,6 E J. Moreover, because there exists a unique conjugacy class
with type B2, we may assume that J = (ir, is). Lastly, because B2(K) has three
conjugacy classes of involutions with representatives x,(1), x,(1), and x,(1)x,(1), we
may assume that a = x,(1)x,(1).

Note that because a E Z (U ), GG(a) S PJ for some J Q Q. One can easily check
then that CG(a) = 02’(M1,4).

Suppose that b E X;X;. Then as H is transitive on XfiXg, there exists h E H
such that a" = b. But then CG(b) = CG(a)’I = CG(a), contrary to the assumptions of
the lemma since CG(a) = 02’(M1,4) is generated by root subgroups.

Thus we may assume that b E’ X,X,. Because Z(Ql) fl Z(Q4) = X,X,, it follows
that either b E’ Z (Q1) or Z (Q4). Also because of the graph automorphism of G
which interchanges Q1 and Q4 we can assume without loss that b ¢ Z (Q4). Then
(bQ4) 0 Q, S Z(Q4). Thus as Qf, = X,, we can choose 1 S x E [(bQ4),Q4] n X S

(ACGI‘Il). A’ = (a, x) is then the required fours-group.

Lemma 11.3 Let G = F4(K) and suppose that a and b centralize the same root
subgroups. Then either V is as claimed in Theorem 8.3, or A is contained in a root

subgroup of G and hence is linearly dependent.

Proof: Suppose that V has weight A. Let ct denote the graph automorphism of G.

75

By [6] G has four conjugacy classes of involutions with representatives t = x,(1),
u = x,(1), tu = x,(1)x,(1), and v = x,,9(1)x,20(1).

Now, if t E A, then because GG(t) = 02’(M 1) is generated by root subgroups and
because Z(O2’(M1)) = X,, it follows that A S X,.

Similarly if u E A, then A S X 8.

If tu E A, then because Z(CG(tu)) = X,X,, by the above, we get that A S X,X,.
Let J = (ir,is). Then J has type 82. Because A is not contained in a root
subgroup, it follows from Corollary 6.5 and Lemma 8.7 that every nontrivial chief
factor of V for (X in X is) must either be the tensor product of algebraic conjugates
of the natural module or the tensor product of algebraic conjugates of the spin module.
In particular, Al (1 JI" S (b for all w E W . Thus, by Lemma 7.2, A in an integral
multiple of either A1 or A4. Moreover, because A is not contained in a root subgroup
of G, it follows from Theorem 6.1 that V must be a restricted module. Therefore V
is an algebraic conjugate of either V(Al) or V(A4).

Thus we may assume that v E A. By [6], Cg(v) is generated by root subgroups
and Z(C'G(v)) = X,,,Xm. Thus A S X,,ng. Suppose that b = x,,9(71)x,20(72).
Because a4’ = a and 6” = x,,9(72)x,,0(7f) we can assume that 71 S 72. Now as r19 is
a short root and r20 is a long root and because r19 J. r20, it follows from Lemma 4.2
that there exists w E W such that {r19,r20}“’ = {r1,r3}. Hence Aw S L4 E Sp6(K),
the Levi compliment of M4. Note that because 71 S 72 we can assume that A” does
not act linearly dependently on the natural module for Sp6(K). Moreover, if there
exists a rank two parabolic subgroup of type B2 such that A is contained in the Levi
complement and is linearly dependent there, then the result follows from the above
cases. Thus we may assume that A‘” is linearly independent in L4. Therefore, it
follows from Lemma 9.3 that there exist a rank 2 connected root subsystem, say J,
such that Al n J‘”’ S (l) for all w’ E W. By Lemma 7.2, A is an integral multiple of

either A1 or A4. Moreover, because A is not contained in a root subgroup, it follows

76

as above that V is an algebraic conjugate of either V(Al) or V(A4), as claimed.

CHAPTER 12

G2(K)

Lastly we consider G2(K). Because of the unique structure of its associated root
system it is easiest to do a small amount of computation rather than develop an

approach similar to that used in the previous cases.

Lemma 12.1 Let G = G2(q). IfA is not contained in a root subgroup of G, then V

is an algebraic conjugate of the natural module for G.

Proof: Assume that A is not contained in a root subgroup of G. We claim that,
without loss, we may assume that Q1 = 02(M1) S (ACCIal) for some a E A. By

[14] G has two conjugacy classes of involutions with representatives x2,,+3,2(1) and
$r1+2r2(1)-

Case 1: a = x2,,+3,,(1) E A.

Since a E Z (U), CG(a) S P] for some J Q Q. One can easily check then that
CG(a) = O2'(M1) = QlLl, where L1 = (X,,,X_,,) E Sl2(K). It follows from the
commutator relations that 61 = C21/X2,,+3,2 is abelian, and so may be regarded
as a 4—dimensional K Ll-module with multiplication by a scalar, say 7, defined as
conjugation by h2,,+3,2 (7) Moreover, because X,,+3,2 S 061992), X,2 E Syl2(L1)

_L .
and because x,,+3,,(7)h'2(*) = x,,+3,,(A37), we see that (X,1’+3,.2) contains a K L1-

submodule isomorphic to Vn (8) V3, where V,, is the natural module for S L2(K ) Since

77

 

78

]K] > 2, this submodule is irreducible and because it is four dimensional, it must be

all of GI. In particular, since A S X2,,+3,2, Q1 S (ACCIal).

Case 2: a = x,,+2,.,(1) E A

Then, again by [14], GG(a) = U2L2, where L2 = (Xin), U2 = X,,.rg,2 X X,,+3,2 x
X2,,+3,2, and L2 acts irreducibly on Q = X,,+3,2 >< X2,,+3,2. So, A S X,,”,2 implies
that Q S (ACCI‘Il). In particular, we can choose an element, say x, of a long root

subgroup in Q and apply Case 1 to the fours-group A’ = (a, x), proving our claim.

Thus we have [V, X2,,+3,2, Q1] = 0. Let g = w,,+3,2. Then Xé’rflm = X,,, implies
that [V,X,,,Q51I] = 0 too.

Let V0 = Cv(02(P,,)). By [11], V0 is an irreducible module for (X,,,X_,,). So,
W = [V0,X,,] is centralized by both Q5] and 02(P,,). However, ( 517,02(P,,)) = G.
Thus W = 0, and so also [V0,02’(P,,)] = 0. Therefore, as V was assumed to be a

restricted module, we see that A = A2. That is, V is the natural module for G.

 

BIBLIOGRAPHY

 

BIBLIOGRAPHY

[1] M. Aschbacher, GF(2)-representations of finite groups. Amer. J. Math. 104
(1982), 683-771.

[2] M. Aschbacher, G. Seitz, Involutions in Chevalley Groups over Fields of Even
Order. Nagoya Math J., 63 (1976), 1-91.

[3] N. Bourbaki, Elements de Mathe'matique. Fasc. XXXI V, Groupes et Alge’bras de
Lie, Chaps. 4,5,6. Actualités Sci. Indust., no 1337, Hermann, Paris, 1968 (MR
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