Classical groups

Citation for published version (APA):

Higman, D. G., & Taylor, D. E. (1978). Classical groups. (EUT report. WSK, Dept. of Mathematics and Computing Science; Vol. 78-WSK-04). Eindhoven University of Technology.

Document status and date: Published: 01/01/1978

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ONDERAFDELING DER WISKUNDE Classical Groups by D.G. Higman with an appendix by D.E. Taylor DEPARTMENT OF MATHEMATICS

r---"·-·--·---,

I

~

~),

'.~ ~,

T

K'

\

r-'-"-'~~"

61)

6 41

f.'

.. ,,/

f _ _ _ .0 . . . . ~.,' ,~ . . . ,.. . ( " _ _ . . . . w'~ _ _ •

I

T'"'

v'"

i

,!

q . .:

:~

..

1',

1

I

III" } ( " _"', I ~_ 'i ~ .," . . . ' , _ , , _ • ~"_U'-" ' ... _ ... _ ,. _ . , _ , _ . _ T.H.-Report 78-WSK-04 August 1978

These are notes taken by W. Haemers and H. Wilbrink of an introductory course on classical groups (over commutative fields) given in the

spring semester 1978 at the Department of Mathematics of the Technological University Eindhoven.

The main goal was the determination of the normal structure (assuming positive index in the unitary and orthogonal cases) by the method intro-duced by Iwasawa for the linear case and applied by Tamagawa to orthogonal groups.

Because of time cons.iderations orthogonal groups over fields of characteristic 2 were omitted.

Some discussion of the sporadic isomorphisms is included.

An appendix by D.E. Taylor contains a uniform treatment of generic isomorphisms

and a construction of the Suzuki groups.

1. Gro~ actions and Iwasawa' s lemma

Let G be a group with identity 1, say. An action of G on a set X

#

0

is a map: G x X ~ X, (g,x) I~ gx, such that

1) (gh)x

=

g(hx)

(g,h E: G, x € X) •

2) 1x

=

x

One can easily verify that an action of G on X is equivalent to a homomorphism: G ~

Ex'

where

Lx

denotes the symmetric group on

X.

The kernel of an action is the kernel of the corresponding homomorphism. An action is faithful if the ker-nel is trivial (= {i}). If the action is faithful then G is isomorphic to a subgroup of

Ex'

i.e. a permutation group. If the action has kernel

K

then G induces a faithful action of G/K on X.

A G-set (G-space) is a set X

#

~ with a given action of

G

on X. Two G-sets X

and Yare isomorphic iff there is a bijection Ip: X ~ Y such that <p (gx)

=

glp (x) (g E G, x E X). Two actions of G are equivalent if the corresponding G-spaces are isomorphic.

A subset Y

S

X is stable or a G-subspace (if Y #~) if gy E Y for all y E Y, g E G. If Y S X is stable and Y

#

~ then G acts on Y.

Example. JH = {z t. (J:

I

1m (z) > O}, the upper half plane of ct.

SL 2 (JR) is the group of all matrices

[~ ~J

with a,b,c,d E lR and ad - bc

=

1.

Let this group act on

<t

:= (J: U {oo} in the following way; if g ::

[~ ~J

E SL

2 (JR2)

~ az + b . [

I

and z E ct then gz := cz + d • It follows eas1ly that Im(gz)

=

Im(z)/ cz + d . which implies thatJH is stable under SL

2(JR). The kernel of this action is {±I}, so PSL

2(JR) := SL2(JR)/{±I} acts faithfully on

i.

SL2(Z) is the subgroup of SL2OR) with coefficients in Z. G := SL

2(Z)/{±I} is the modular group. One can identify SL

2 (JR) with SL (JR2) I the group of all linear transformations of lR

2

with determi-nant 1. SL(JR2) acts on the points of the projective line based onlR2, Le. the

2 1 - dim subspaces of lR •

Let X be a G-space, x,y E X. Define

x ~ y :~ (gx

=

y for some g E G) •

'I'hen ~ is an equivalence relation; the equivalence classes are the orbits. The action is transitive if there is only 1 orbit (= X). Each orbit is stable and

transitive, each G-space is uniquely partitioned into a disjoint union of tran-sitive G-spaces« Let X be a G-space, H S G, Y S X, then HY := {hy

I

h E H, Y E Y},

is a transitive G-space according to (g,hH) ~ (gh)H, the natural action. Its kernel is the core of H in G, i.e. the join of all normal subgroups of G con-tained in H. A G-space is homogeneous if it is isomorphic with one of the form

G/H for some H ~ G.

Let X be a G-space, x € X, Y £ X, Y

'#

~.

G := {g € G

I

gx = x} is the stabilizer of x in G, x

Gy := NG<Y) := {g E G , gY - y} is the (set-wise) stabilizer or

normalizer of Y, := {g € G

I

gy ;:: g,

Vy

E y};:: n y€Y G is the pOintwise y stabilizer or centralizer of Y • 0... -1 For H £ G, g € G, ~1:1 :;:: gHg is a conjugate of H in G.

The following properties are obvious:

1.1. a) G

x' Gy ' GCYJ are subgroups of G. b) Gx = G{x}

=

G[{x}] (x E X). c) G

CXJ is the kernel of the action of G on X.

d) Gy acts on Y with kernel G[yJ. The corresponding permutation group will be denoted by GY so we have an exact sequence

with GY a permutation group on Y, GY

~

Gy/G[y]. In particular we have an exact sequence

(An exact sequence is a sequence of homomorphisms ••• -+ G

~

H

t

K -+ ••• kernel of 1./J.

such that image of cp

For example 1 -+ G[y] tive etc.)

-+ Gy means that the homomorphism G[yJ -+ _{Gy is}

injec-e) g(G )

=

G g(G)

=

x gx ' Y

1.2. Let X be a G-space, x c X then Gx ::::: GIG as G-spaces. x

Proof. Take the map gx -+ gG

x for the isomorphism.

As a corollary we have

1.3. Every transitive G-space is homogeneous.

1.4. If H,K ~ G then G/H and G/K are isomorphic G-spaces iff Hand K are conjugate. Proof. If H = glK let the isomorphism

~

be defined by

9 E G .

gl Conversely if <p is an isomorphism and <p(H) == glK i t follows that H == K.

0

Assume G tra

x

(i.e. G acts transitively on X). An (imprimitive) block is a subset B of X, such that g8 n 8 ~ ~ implies gB

=

8 for all 9 E G. The blocks ~, {x}, X are trivial blocks. The action is imRrimitive if there exists a non-trivial block, primitive otherwise. If 8 ~ ~ is a block, then {gB

I

9 E G} is

a partition of X into blocks and G acts transitively on this set of blocks ac-cording to (g,hB) H' ghB.

1.5. Suppose G tra X and let x € X. The map 81~ G

B is an isomorphism of the lattice of blocks containing x onto the lattice of subgroups of G containing G

x (the inverse map is HI~ H for all G ~ H ~ G).

x x

As a corollary .to 1.5 we have

1.6. G pri X (i.e. G acts primitively on X) iff G is a maximal subgroup for some

x

1. 7.

(hence for all) x IE X.

If G pri X and N ~ G then N ~ _{G[x] or N tra X.}

Proof. Take x E X and suppose N _{-$, G[x] then N} -$, G (since N ~ G implies

x x

N

₌

gN ::; gG == G for all 9 IE G) . Hence, by 1.6, G

=

NG

.

If g € G then

x gx x

9 :: _{nh for some n IE N, h IE G so gx = nhx} ::: nx.

0

x

k

Let X be a set and keN, k ~ 1. We denote by X the k-fold Cartesian product of X with itself,

[~]

the set of all

(xl/"'~)

€ xk with Xi

~

Xj

(l~i<j~k),

X

(k) the set of all k-subsets of X.

An action of G on X induces

Remark. Take

(x11""~)

E

[~].

Theset{(Yl'···'Yk)

I

{Y1""'Yk} ==

:::

{xl""/~}}

C

[~]

is an imprimitive block for

Ex'

The action on this set

X

Let X be a G-space and k E:N, k ?: 1. The action is regular if G tra X and

X

G == 1, for all x € X (note that if G is faithful and regular on X, then for

x

any x € X the map g~ gx is a bijection of G onto X). The action is k-fold

transitive or k-transitive (notation: G k-tra X) if G tra

[~].

The action is sharply k-(fold) transitive if G acts regularly on

[~J.

The action is

k-homo-X

qeneous if G tra (k)' In particular G I-tra X means G tra X. Clearly G k-tra X implies G (k - 1) -tra X.

1.8. G 2-tra X implies G pri X.

Proof. Let B be a block, IBI ~ 2. Take x,y E Bt x ~ y and let z ~ X\{x}. There

exists agE G such that gx == x and gy

=

z.

From x

=

gx E B

n

gB i t follows that B == gB and so

z

=

gy € gB

=

B i.e. B

=

X.

o

Let G be a group. The derived or commutator subgroup G' of G is the intersec-tion of all N ~ G such that G/N is Abelian. It follows that

G'

=

<[g,h] := ghg-1h-1 I g,h € G> ,

the group generated by the commutators of G. Of course G/G' is Abelian, and G is Abelian iff G' == 1. We say that G is simple if the only normal subgroups of G are 1 and G itself.

1.9. (Iwasawa's lemma). Let G pri X, x E X. Assume there exists A(X) ~ G , such

x

that A(x) is Abelian and G

=

<gA(x) 9 € G>. Then

a) N ~ G implies N ~ G[XJ or N ~ _{G' •}

b) If G

=

G' then G/G[X] is simple. Proof.

NA(x). Indeed, let 9 E G, a) If N ~ G[X] then N ~ G

x so G

=

NGx' We claim: G

=

since 9 == nh for some n E N, h E G we have gA(X)

=

x

nh n

A(x)

=

A(x) S NA(x) and so G == <gA(X) I 9 € G> ~ NA(x) S G i.e. NA(x) = G. Now G/N == NA(x) /N Q,I

A(x)/N n A(x), which is Abelian, so N ~ G'.

b) Suppose

N

~ G/G[X] then

N

= N/G[X] with G[XJ ~ N ~ G. I f

N :!

1 then N ~ G[XJ hence by a) N ~ G'. From G = G' i t now follows that N

=

G, Le.

N

== G/G[XJ'O

2. The general linear ~roup

Let V be a vectorspace over a fieldF, dim V

=

n , 2 $ n < ~. GL(V) := the group of all non-singular linear transformations of V. This section is devot-ed to finding the normal subgroups of GL(V) •

n

Let v

1' .. .,vn be a basis of V, T c GL{V), 'r{vi ) :::

I

ajivj , with aij f F. j=1

The map Tt+ A

=

(a, ,) is an isomorphism of GL(V) onto GL{n~) := GL (F) :=

~J n

the general linear grou~ (of degree n over F) := the group of all non-singu-lar n x n-matrices. LetF* be the multiplicative group of the non-zero ele-ments of F. The determinant map det: GL(V) +F* is a group homomorphism and is onto. The kernel of det is SL(V)

=

{T E GL(V) I det T

=

1} so we have an exact sequence

det *

1 + SL (V) + GL (V) + F .". 1

and GL(V)/SL(V) ~F* is Abelian (hence SL(V) ~ GL(V) '). SL(V) ~ SL(n~) := SLn (IF) == the special linear group (of degree n over F) := the group of all n x n-matrices with coefficients in F and determinant 1. GL (V) acts fai thful-lyon V#:= V\{O}, GL(V)

~

E

II

V

2.1. GL(V) acts faithfully and regularly on the set of all ordered bases of V. Thus there is a 1-1 correspondence between GL(V) and the set of ordered bases of V.

I f

/:IF

I

=

q < 00 then I GL(V)

I

== # ordered bases of V = (qn _ 1) (qn _ q) (qn _ q2) .••

n n-l ••• (q - q ), so IGL(V)

I

n II i=1

(qi _ 1), ISL(V)

1=

IGL{V)

I

iJF*1

n IT i=2 i (q - 1)

2.2. If x € V# then {ax

I

a EF*} is an imprimitive block. The corresponding

ac-tion of GL (V) on blocks is equivalent to the action of GL (V) on the pOints of the projective space based on V i.e. on the 1-dimensional subspaces of V. Note that if

iFl

= 2, then the action of GL(V) on

Y#

is equivalent to the action on the points of the projective space.

The projective space PV based on V is the lattice of subspaces of V. If V has dimension n then PV has dimension n - 1. The subspaces of V are the linear va-rieties or flats of PV. The codimension of a k-flat (k+1 dimensional subspace of V) := codimension of the corresponding subspace (= n - k - 1). Dictionary:

point line plane

PV

k-dim lin variety} k-flat hyperplane

v

l-dim subspace 2-dim subspace 3-dim subspace (k + 1) -dim subspace hyperplane (through 0)

A line of PV contains

PFI

+ 1 points. GL(V) acts on the k-flats of PV for all k. Look at the action on the points (O-flats) of PV. We have an exact sequence

1 + Z(V) + GL{V) + BGL(V) + 1 ,

where Z(V) is the kernel of this action, and PGL(V) := GL(V)points

=

GL(V)!Z(V) PGL(V) is the projective general linear group (of degree n ~JF).

2.3.

Z(V)

=

all nonzero scalar transformations {aI

I

a E

F*}.

Proof. Clearly {aI

I

a €JF*} ~ Z(V). Suppose v

1, ••• ,vn is a basis of V. Let

*

T be an element of Z(V). Then T(V

i )

=

aivi for some ai € JF , and T(V1 + ••• +Vn) := ::: a(v

1 + ••• +vn) for some a € JF. Hence a

=

a1

=

a2 = •••

=

an and T

=

aI.

0

PGL(V)

=

PGL(n;IF) :== GL(n;IF) !{aI I a € JF*}.

2.4. If

iFl

=

q then

I

PGL (V)

I

""-r'::::-7:-:7+

=

I

GL (V)

I

= (q - 1) ISL(V) I n II i=2 i (q - 1)

2.5. Z(V)

=

center of GL(V)

=

centralizer of SL(V) in GL(V). (If G is a group,

H ~ G then CG(H) :== {g E G

I

gh

=

hg, Vb E H} is the centralizer of H in G;

CG(G) is the center of G.)

~. Clearly Z(V) ~ CGL(V) (GL(V» ~ CGL(V) (SL(V». Suppose A € GL(n,F)

cen-tralizes SL(n,F) then A(I + E

ij)

=

(I + Eij)A for all i ~ j. (Eij is the ma-trix with a 1 in the (i,j)-position and ots elsewhere.) Hence AEij

=

EijA for all i ~ j and so A == aI, a E JF.

o

SL(V) acts on the points of PV. We have an exact sequence 1 + ZO(V) + SL(V) + PSL(V) + 1

where Zo(V) is the kernel of the action. PSL(V) == SL(V)points

=

SL(V)!ZO(V) and PSL(V) ~ PGL(V) ~ Lpts'

{ I

*

n }

2.6 ,ZO (V) = z (V) n SL (V)

=

center of SL (V) aI a ( IF , a

=

1 ::::: the group of the n-th roots of unity in IF. Define Z (n,1F) := {aI

I

a (IF*} = the group of all non-singular n x n scalar matrices, Zo (n,1F) := Z (n,1F) n SL (n,1F) then

PSL(V) ::::: PSL(n,lF) := SL(n,1F)/ZO{n,lF) •

PSL(n,lF) is the projective special linear group of degree n overlF.

2.7. If IlFl q then

I

PSL (V)

I

n IT i=2 i (q - 1) where d (n,q-l).

With ()F*) n := {an

I

a E IF*} we have the following commutative diagram in which

the rows and columns are all exact. (Notice that Z(V)

::::: p* .)

1 1 1 t t t 1 -+ (F*) n -+ IF* -+- IF*/(IF*)n -+ 1 t t t 1 -+-

z

(V) -+ GL(V) -+ PGL(V) -+ 1 t t t 1 -+-

_Zo

_(V) -+ SL (V) -+- P$L (V) -+- ₁ t t t 1 1 1

q < 00 we sometimes write GL(n,q) instead of GL(n,lF) etc. We have

seen:

I

GL (n,q)

I

I

SL(n,q) I n IT i=1 i (q - 1) (n) IPGL(n,q)

I

q 2 (n) _n n i IT (q - 1) i=2 1 i IpSL(n,q)I _=d"q 2 IT (q -1), d = (n,q - 1)

.

i=2

Remark. Notice the order coincidence ISL(n,q)/

=

IpGL(n,q)/. In general (i.e. iff d

=

(n,q - 1) 1= 1) PGL(n,q)

i

SL (n,q) since the center of PGL(n,q) is 1.

An n-simplex in PV is a set of n + 1 points no n of which are in a hyperplane. For any n-simplex {P

1, .•• ,Pn+1} we can P.

=

<v.>, 1 ~ i $ n, and P 1

=

~L~ 1

~ ~ n+ ~=

larly on the set of ordered simplices.

choose a basis v

1, ••• ,vn of V such that v.>. This implies that GL(V) acts

regu-~

2.8. GL(V) 2-tra pts of PV.

If n ~ 3 then GL(V) and SL(V) not 3-tra on pts of PV (there are collinear and non-collinear triples of pts).

If n

=

2 then GL(V) sharply 3-tra pts (in this case we usually view PV as the setlF U {oo} in such a way that the point < (x

1,x2» corresponds to x1/X2ElF if x

2 # 0 and < (1,0»++- 00. Thus

(~ ~)

€ Gt.(V) induces the MObius transformation

(x 1+ ax +

~) ~:

PGL (V) • Notice that (x t+- ax +

~)

€ PSL(V) iff ad - bc "" 0 (= a

ex + ex +

square in IF) ) •

If q is even then PSL(2,q)

=

PGL(2,q) acts sharply 3-tra onpts.

If q is odd then IpSL(2,q)I

=

~IPGL(2/q)1 and PSL(2,q) is not 3-tra on pts. If q == -1 (4) then PSL(2,q) is 3-hom0<;Jeneous.(Let x

1,x2,x3 be threedistinctpts of IF u {oo}. Define 9

1,92 € PGL(2,q) by.

then gl ({x

1,x2,x3})

=

92({x1,x2,x3})

=

{O,l,oo}. Since -1 ,

0

either

(Xi - x

2) (X2 - X3) (x3 - xl) = 0 or -1 (xl - )(2) (x 2 - x3) (xj - xl) = 0 i.e. either

9

1 or 92 is in PSL(2,q).).

SL(V) acts 2-tra on the pts of PV.

The normal structure of GL(V)

*

*

Let V denote the space of all linear functionals of V. Let e € V, ~ € V and

define the map L "" L V ~ V by x ~ x

+

~(x)c. Clearly L is a linear trans-(j),c

formation of V, and

to

_,e

=

1.

2.9. I f ep(c) -1 then the kernel of t - <C>.

2.10. If ep(c) , -1 then the kernel of T =

{oJ

hence T € GL(V).

Proof. Take x € ker T, then x

= -ql

(x) c hence x € <c>; kernel of T -:; {O}.

ker T = <c> • e -ep(e)c ~ (j)(c) = -1.

The linear transformation T is called a transvection if q>(c)

=

O. If q> ~ 0 then the kernel of q> is a hyperplane. This hyperplane contains e iff T is

(j),e a transvection. A transveetion T has matrix

if we choose a basis v

1, ... ,vn of V such that v1, .•• ,vn_1 E ker ~ and v1

=

c. The determinant of a transvection equals 1, hence SL(V) contains all transvec-tions. On the other hand all matrices

1

"

... ...

"

..

and representent transveetions. 1

'.

*

*

"" *

_'1

* •

.

...

*

"1

*

In particular I + aE . ., a ElF, i -F j, represent transvections. If !.p(e)

I

{O,-i} 1J

T

!.p,C is called a dilatation. If v1; ••• ,vn is a basis of V such that, viEker ~

(15 i s n-l) and v

n

"

1

= c then the matrix of the dilatation T

!.p,C

1 +!.p (c) .

On the other hand any matrix

1

"1

*

(

... 1

is

with

*

i:

0,1 represents a dilatation. The determinant of T equals 1 +qJ(c)

I

{O,l}

so T GL(V)\SL(V). The following properties are easily verified.

1 -. T fixes

IP, c eV(~ry vector in the kernel

! - .[ _{fixos overy s\.lbspace containing}

qJ,c

*

2.12. a) T

=

T (a c·lF). Ql,ac a;:p,c b) If ;:P1 (c)

=

0 then T T !.p1 'c Cjl2'C c) For any T E GL(V): T(T )

=

!.p,c T • qJl+;:P2'c T -1 ~T ,Tc of !.p. c.

Let H be a hyperplane of PV, let P be a point in H. Define the set of trans-vections X P := {T

I

qJ(H}

=

0, <c>

=

pl.

2.13. XH,p ~ SL(V)H,P and ~,P g GL(VlH,p for T (~.p) XT(H) ,T(P) for all T € GL(V) •

2.14. ~,P ::: (JF,+) by result 2.12.

2 .• 15. Let L be a line such that L n H = P then ~,P acts faithfull,y and regularly on the points of L\{P}.

Proof. Suppose P,

Q

and R are distinct points of L, P

=

<p>,

Q

=

<q>,

R = <ap + q>. Choose

(jl

E V

*

such that H

=

ker cp and ql(q) = a then 't

ql,p E ~,P

and T (q)

=

q + ep (q) p

=

q + ap :. T (Q)

=

R. I f T E _~,p fixes any point

ep, p tp ,p

not on H then T

=

1. Suppose T fixes

Q "

H, Q = <q>, T(q)

₌

aq for some a ElF

then T(q) = aq = q + (jl(q)p i.e. (jl(q) =: 0 hence (jl

=

0 and so T

=

1.

Let P be a point. Define Xp := {T

I

(jl E

v*,

ep{c) = 0, P

=

<c>}. <P,C

2.17. ~ =: u

H

P~H

XT(P) for all T € GL(V).

xp,H,a partition (i.e. ~ H n ~ H

, 1 ' 2

1, H 1

#:

H 2) •

2.18. ~ acts regularly on the points different from P of any line L through P.

o

Proof. Let H be a hyperplane such that H n L

=

P, then ~,P tra L\{P}, and ~,P ~ ~, hence ~ tra L\{P}. Suppose T E ~ fixes Q

=

L, Q

#:

P. From 2.17 we see that T E ~,P for some hyperplane H containing P. If L C H then T acts trivially on H and so T acts trivially on L. If L c H then T = 1 by

2.15. []

*

2.19. Let c E V, P

=

<C>. Define the homomorphism ~: V +IF by ~«(jl) (jl (c), then

~ ::: kernel ~.

Proof. The isomorphism is given by T 1+ (jl.

(jl,c

o

*

2.20. GL (V) is genera ted by the transformations T , ep E V , C E V\ {O}. SL (V) is

ep,c

generated by the transvections T (jl E

v*,

C E V\{O},

(jl(c)

=

O.

(jl,c

f

0]

Any A E

GL(n~)

can be reduced to the form

l~

*

(where

*

= 1 iff A E SL(n~», by elementary row operations of the form: add a multiple of

one row to a different row. Each such operation can be obtained by multipli-cation with a matrix of the form I + aE .. , i

#:

j .

0

1.J

2.21. As a corollary we have SL{V) =

<T~

I

T E SL(V», indeed

~~

=

~(p)1

and SL(V) is transitive on the points.

We have the following structure GL(V Z(V) n SL(V)

=

Zo(V 1

*

~ lF , Abelian PSL(V) central

We shall obtain the simplicity of PSL(V) from Iwasawa's lemma applied to the action of SL(V) on the points. We have SL(V)

=

<T~

I

T e SL(V» with

~ 9 SL(V)p and (from 2.19) Xp is Abelian. So we still have to show: 1) SL(V) is primitive on the points, and

2) SL(V)

=

SL(V) '.

2.22. SL(V) acts 2-transitively on the points.

Proof. We show that SL(V)p is transitive on the points different from P. Take distinct points P, Q and R. Suppose P, Q and R are on one line L. Take a hy-perplane H such that H n L

=

P, then ~,P takes Q to R, and ~,P S SL(V)p' Suppose P, Q and R are not collinear. Let L be the line through Q and R. Take a point S

=

L, S ~ Q,R and a hyperplane H containing P such that H n L

=

S,

then ~,S fixes P and moves Q to R.

o

2.23. In case n ~ 3 then SL(V)

=

SL(V)'.

*

Proof. [I + aE_ij, I + bEjkJ

=

I + abEik for all a,b elF , i,j and k distinct (note that E, ,Ekfl

=

0, and E. ,E'k ::: E'k for all distinct i,j,k and L In

par-~) N -1 ~J ) 2

ticular (I + aE, .) = I - aE,. for all i ~ j). With respect to a sui table

ba-2J 2)

sis, every transvection can be written as I + abE. ,_

0

1.)

2.24. If n = 2 and

IFI

~ 4 then SL(V)

=

SL(V) '.

~<?!.

Let i' = ;

~)

be a transvection. Take a E IF*

b =

T-

then T

=

[(~ ~_1L [~ ~)J.

a - 1

~

J

such that a2

~

1. Put

o

2.25. I f (n,lF) ~ (2,GF{2», (2,GF(3» then PSL(n,lF) is si:mple.

~. Apply Iwasawa's lemma.

o

2.26, If (n,lF) ~ (2,GF(2», (2,GF(3» and N $ GL(V) then N g GL(V) iff N $ Z(V) or

Proof.

a) GL(V) , SL(V). Indeed GL(V)/SL(V) :::: IF

*

is Abelian, hence GL(V) , S SL(V)

=

= SL(V) , S GL(V) '.

b) If N S Z(V) then obviously N 9 GL(V). If N ~ GL(V) , then N/GL(V) , 9 GL(V)/ GL(V)t, which is Abelian, so N 9 GL(V}.

c) Let N 9 GL(V), N

1

Z(V). Define

N

:= NZ(V)/Z(V) S PGL(V). Suppose N n SL{V) S

Zo(V) then

N

n PSL(V)

=

1 and so Nand PSL(V) commute elementwise. Moreover N is transitive on points. Fix a point P, take G € PSL(V)p' n E

N

then Gn(p}

=

nG(p) n(p), so PSL(V)p acts trivially on the points, a

contradic-tion. Hence N n SL(V} ~ Zo(V) and so N ~ SL(V) by the first part of Iwasa-wa's lemma.

Order cOincidences and sporadic isomorphisms involving PSL(n,g) and A

_--

_m

1) _{PSL (2,2)} ~ ~ - 3 order: 6 2) PSL (2,3) "'A _{- 4} 12 3) PSL (2,4) PSL(2,S) ~A - 5 60 4) PSL (2,7) :::: PSL(3,2} 168 5) PSL(2,9) ~A - 6 360 6) PSL (3,4)

t

PSL(4,2) "'A - 8 20160

Result 1, 2 and PSL(2,4) ,..., A are

- 5 straightforward. Using Sylow's theorem

o

one can prove that there is only one simple group of order 60 and 168 {cf. [7J, p. 183-1S5}, this proves 3) and 4}. It is easy to prove that the centers of the sylow-2 subgroups of PSL(3,4) and PSL(4,2) have order 4 and 2 resp., this proves the first part of 64 For the details, and for result 5 we refer to [5J or [7J.We will now sketch a proof of PSL(4,2) :::: AS due to A. Wagner

(on collineation groups of Projective Spaces I, MATH. Z.

?!i,

411-42h (1961»: the projective plane of order 2 ( Fano plane ) can be represented by the array

1 234 567 2 3 4 5 6 7 1 456 7 123

Let A7 act on this array to planes of order 2. Define a

1A71

.

rod ce - 15 different proJ"ective p u IpSL(3,2} 1

-new incidence structure

P,

whose "points" are the 15 planes, and whose "lines" are the 3S triples out of {1, •.• , 7}. A "line" is incident with a "point" if the corresponding triple represents a line of the corresponding Fano plane. By verification i t follows that P is a projective space, hence P

=

PG(3,2), whose automorphism group is PSL(4,2). Thus we have A7 S PSL(4,2) with index 8, hence PSL(4,2) S ES' so 1'SL(4,2) ::::

Aa'

Alternative proofs of all sporadic isomorphisms involving alternating and classical groups will be given at a later date.

We remark on some natural questions arising from our geometrical discussion of GL(V). (Details can be found in

[lJ, [3J

and [6}.)We assume n ~

3.

For the case n = 2 we refer to

[3J.

A collineation of PV is a permutation of the points which induces a permuta-tion of the lines. The group of all collineapermuta-tions of PV is denoted by ColI PV. Of course PSL(V) S PGL(V) ~ Coll PV.

Q~ti~ns

1) What is the analytic description of ColI PV?

2) What is the synthetic description of PGL(V) and PSL(V)?

About 1). Let T E Aut IF. A T-semilinear transformation of V is a map T: V -+ V

*

such that T(x+y) = T(x) + T(y), T(ax) T(a)T{x) for all x,y E V, a ElF •

If S o-semilinear and T T-semilinear then ST(ax) = S(T(a)T(x»

=

o~(a)ST(x),

hence ST is oT-semilinear.

We define fL(V) := the group of all non-singular semilinear transformations of V. rL(V) acts on the pointsj Z(V) is the kernel. We have:

1 t AutlF t 1 -+ Z(V) -+ rL(V) -+ prL(V) -+ 1 (exact) t GL(V) t 1 1) PfL(V)

=

ColI PV.

About 2). A collineation a of PV is central if 0 fixes all points of some hyperplane H and all lines through some point P. If G

:f.

1 then Hand Pare uniquely determined and P together with the points of H is the complete set of fixed points of G. H is called the axis and P the center. A central colli-neation is called an elation in case P c H. A projectivity is the product of central collineations, a perspectivity is the product of elations.

2) A central collineation is induced by exactly one linear transformation of the form T

<P,C

3) PGL(V) is the group of all projectivities. PSL(V) is the group of all pers-pecti vi ties.

3. Pairings and bilinear forms a) Dual spaces

LetF be a field and let V and W be vector spaces overF. Define Hom(V,W) :=

HO~;(V,W) := the vector space of all linear transformations from V to W

(ad-dition and scalar multiplication defined pointwise). Suppose the dimensions of V and Ware finite, let v

1, ••• ,vm and w1, ••• ,wn be bases for V and W res-pectively. Then Tij(v

k)

=

0kjWi defines Tij ( Hom(V,W) and {Tij 1 ~ i 5 n,

1 $ j ~ m} is a basis of Hom(V,W). So the dimension of Hom(V,W) equals mn. In case W F

we

write V* := Hom(V~) and V* is called the dual space of V.

Ifv₁,···,v_m is a basis of V then the dual basis v

*

*

*

1, ••• ,vm of V is defined

*

by vi (V

j )

: =

_*

0.. ( _1.J i , j

=

1, .... , m) • The map v. 1.

»

v. determines an isomorphism 1.

*

of V with V • If the dimension is infinite then V and V* are not isomorphic. **

There is a natural isomorphism cr of V onto a subspace of V , namely cr(x){f} :=f(x), X E V, f E V

*

If the dimension of V is finite then 0: V ~ V** (i.e. 0 is an isomorphism of **

V and V ).

b) Pairings

Let V and W be finite dimensional vectorspaces over the fieldF. Bil(V,W)

de-*

notes the space of all bilinear maps f: V x W -+ F (Example: W

=

V ; (x, cp) i-+

cp (x) ) •

Let v

1, •.. ,vm and w1, ••• ,wn be bases of V and W resp., let f E Bil(V,W) and A the m x n matrix (f(v.,w.». The map f~ A is an isomorphism of Bil(V,W)

1. J

and IF • Define new bases for V and W by v~ := l:~ 1 p .. v. and w! := L~ 1 q, ,w, •

mxn 1. t J= 1.) J 1. )= 1.J J

Put P == (p .. ), Q == (q .. ) then (f(v~ ,w~» == PAQ • We speak of f E Bil(V,W) as

1.J 1.) 1. J

a pairing of V and W. Fix a pairing f E Bil(V,W). We define f(x,y) 0, Vy E W} and L f := {x <: V R f := {y E W f(x,y)

=

0, Vx E V}

the left and right kernel of f.

We now have the linear maps

*

<Pf: V -l- W

,

_{<Pf (x) (y)} := f(x,y)

*

1jJf! W +V, 1jJf(Y) (x) := f(x,y)

x <:: V, YEW.

Note that the kernel of <Pf L

f and the kernel of ~f == Rf • We have the

3.1.

f ++ £P

f

Again fix f € Bil(V,W). Let a 5 V, K 5 Wand define

and al :;:; lK ₅ vl

₌

1 (Hl) Wi a 1 Vi _Kl al := {y E W lK := {x € V ::> _{H2 implies} :0; _{K2 implies}

Rf'

lW

=

Lf' :?: H, ViI :s; Vi (lK}l f(x,y)

=

0, Vx E a} f(x,y)

=

0, Vy E K} Hl 1 VH,H 1,H2 :0; V. 1 ~ H2, lK ~ 1 VK,K 1,K2 $ W. 1 K2, :?: K, VK ::> w.

3.2. If H $ V then there is a linear injection W/Hl -+ H*, so codim Hl :0; dim a.

Proof. The map f

1: H x W/H

l

-+lFt (x,al + y) 1-+ f(x,y)

ing of a and w/Hl. Suppose Hl + Y € R

f then f(x,y)

=

is a well defined

pair-o

for all x E H. Hence

y ( Hl. so al + y

=

Hl 0 in W/Hl.

Thi~

implies R f

=

a hence o/f : W/H l -+ H* 1 1 is injective.

o

3.3. If L

f :: 0 and K <; W t then we have an injection lK -+ (W/K) * so dim lK < codim K.

Proof. The map f

2: lK x

w/K

-+IF, (x,K+y) 1-+ f(x,y) is a well defined pairing.

Suppose implies

then f(x,y)

=

0 for all YEW. Hence x E L

f so x

=

O. This hence £P

f : K -+ (W/K)* is injective. 0

2

3.4. If L

f

=

a and H :s; V then codim Hl

=

dim a and l(Hl) := a.

Proof. dim J'(Hl.) (3s:3) codim Hl. (3~2) dim H (3~1) dim lCHl). Hence

codim Hl

=

dim H := dim l(Hl.).

o

3.5. If L

f "" a then codim Rf := dim V.

I

~. R

f

=

V~, so take H := V in 3.4.

3.6, codim L

f := codim Rf•

Proof. The map fa: V/Lf x W, (L

f + x,y) 1-+ f(x,y) is a well defined pairing.

L_f

=

a and R

f Rf , hence dim V/Lf

=

co dim Rf by 3.5.

0

a

0

Call f nonde2enerate if L

f := Rf

= a

(this is equivalent to dim V

and det A # 0 for any matrix A of f) .

3.7. I f f is a nondegenerate pairing of V and W, then i) dim V

=

dim W.

ii) _{Q/f: V} ::;:W

*

and lj! f: W+V

*

.

iii) dim H codim H.1, for all H :s: V.

dim K :::: _codim .1KI _{for all K}

_{:s: W.}

iv) .1 (H.1) H, for all H :s: V.

( lK).1 = K, f

or

a 11 K ~

w.

v) H » H.1 is a 1-1 inclusion reversing map from the subspaces of V to the subspaces of W. The inverse is KI~ .1K •

*

Example. The pairing of V and V defined by (x,A) i-+ A(X) is nondegeneratej

**

*

*

~: V ~ V is a (the natural isomorphism) j

W:

V ~ V is the identity.

c) Bilinear forms

Let V denote a finite dimensional vectorspace over the fieldF. Let f be a bilinear form on V (i.e. a pairing of V with V) •

be two bases of V such that vi' ;: I~ 1 p .. v., i =

1.= l.) J

Let v1, ••• ,vn and vi""'v~ I, •••

,n.

Put P = (P .. ),

1.)

A = (f(v.,v.» then (f(v~ ,v~» PAP t • Det A,

1. J 1. ) determined up to a nonzero

square inF, is called the discriminant of f.

3.8. The following are equivalent: f is nondegeneratei L

_O.

f R f = O. ~

*

Ip f: V ->- V ; ~

*

l/If: V ->- V ; The discriminant of f f

o.

3.9. If f is a nondegenerate bilinear form on V and H :s: V, then dim H ;: codim s.1 =

codim .1S f .1 (H.1)

=

(.1H).1 H and the map H t-+ H.1 is an inclusion reversing

per-mutation of the subspaces of V with inverse H » .1H (in projective terminology: the mapH» H.1 is a correlation of PV (n ? 3».

n

t

In

Example. V m, f(x,y) ;: xy

= .

1 x.y., f(x,y)

=

0 iff f(y,x)

=

0 iff x

- 1.= 1. 1.

and yare orthogonal. Take H smn then H.1 = .1S is the orthogonal complement of H. dim H + dim H.1

=

n, H n H1 = 0, mn = H $ H1.

In general we say x is orthogonal to y, notation x 1 y, iff f(x,y)

=

O. It can happen that x 1 y whilst Y

i

x. We call f reflexive if f(x,y)

=

0 iff f(y,x)

=

0 for all x,y E V (so 1 is a symmetric relation). If f is reflexive then IH

=

Hl for all H ~ V. It is possible that HI n H

F

0 therefore we pre-fer to call Hl the perp(endicular) of H (rather than the orthogonal comple-ment).

If a nondegenerate bilinear form f on V is reflexive, then the correlation H ~ Hl, H s V has period two, i.e. Hll

=

H. A correlation of period two is a polarity.

Let f E Bil(V,V) then

i) f is symmetric if f(x,y) ::: f(y,x) for all x,y E

v.

ii) f is skew-s~etric if f(x,y)

=

-f(y,x) for all x,y E

v.

iii) f is alternate (symplectic) if f(x,x)

=

0 for all x E V.

3.10. If f is alternate then f is skew-symmetric.

Proof. 0

=

f(x+y,x+y)

=

f(x/y) + f{y,x) , for all x,y E V.

3.11. If char. IF ::: 2: f is symmetric iff f is skew-symmetric.

3.12. If char. IF

'f:

2: f is alternate iff f is skew-symmetric. Proof. "<:=": f(x,x) = -f(x,x), hence 2f(x,x) = 0, thus f(x,x)

3.13. f E Bil(V,V) is reflexive iff f is symmetric or alternate.

o.

Proof. ... ': It is clear that symmetriC and alternate forms are reflexive.

":?": Assume f is reflexive. Then for all a,b,c E: V:

f(a,f(a,c)b - f(a,b)c)

=

f(a,c)f(a,b) - f(a,b)f(a,c) 0, hence f(f(a,c)b - f(a,b)c,a)

=

0, i.e.

f(a,c)f(b,a) - f(a,b)f(c,a)

=

0, for all a,b,c E V • Take a c in ("'): f(a,a) (f(b,a) - f(a,b»

=

O. Thus

f(b,a)

F

f(a,b) implies f(a,a) f(b,b) :: 0, for all a,b E V .

o

o

Assume f is not symmetric, i.e. f(a,b)

#-

f(b,a) for some a,b E V. Then f(a,a)

=

f(b,b) = O. We wish to prove that f(c/c)

=

0, for all C E V. Assume

f(e,c) ~ 0 for some c V. From (**) i t follows f(a/c) ::: f(c/a) and f(b/c) ::

== f(etb). Then by (*) f(a,c) (f(b,a) - f(a,b» ::: 0, hence f(a,c) ::: 0 ::: f(c,a) and similarly f(b/c) :: 0 :: f(c,b). Now we have

f(a + c,b) f(a,b) + f(c,b)

=

f(a,b) f(b,a + c)

=

f(b,a) + f(b,c) = f(b,a)

By f(a,b) f; f(b,a) and (**) we have f(a+c,a+c) 0, but f(a+e,a+c) ""

=

f(a,a) + f(a,c) + f(c,a)

+

f(e,c)

=

f(e,e) = 0

# .

0

d) Quadratic forms

A quadratic form is a map Q: V +~, such that

i) Q(ax)

=

a Q(x), for all a 2 E~, X E V.

ii) f(x,y) := Q(x+y) - Q(x) - Q(Y), x,y E V defines a bilinear form f on V.

3.14. f is symmetric.

3.15. f(x,x)

=

2Q(x) for all x € V.

3.16. If charF ~ 2: Q(x) ~f(x,x), Q is uniquely determined by f. Moreover if f is any symmetric bilinear form on V then Q(x)

=

~f(x,x) is a quadratic form having f as its associated bilinear form.

3.17. If char IF 2: f(x,x) = 0 for all x E V, i.e. f is alternate.

e) Reflexive bilinear form spaces

A pair (V,f) , with V a finite dimensional vectorspace over the fieldF, and f a reflexive bilinear form on V is called a reflexive bilinear form space. We say that (V,f) is symplectic if f is alternate, orthogonal if f is symme-tric and char IF f; 2. We assume char IF f; 2 if f is symmetric: symmetric non-alternate bilinear forms in case aharlF

=

2 are explicitly excluded.

An isometry of (V,f) into (V',f') is an injective linear map T: V + V' such that f'(T(x),T(y»

=

f(x,y), for all x,y t V. The radical of (V,f) is

rad (V,f) := VI; (V,f) is nondegenerate if rad(V,f)

=

0, i.e. if f is nonde-generate. We take the following point of view: f is fixed; speak of the space V, meaning the reflexive bilinear form space (V,f) and say that V is

symplec-tic, alternate, nondegenerate etc. according as (V,f) has the corresponding property_

3.18. If U S V then (U,f U x U) is a reflexive bilinear form space of the same type as V, and rad U

=

U nul.

I f V = VI @ • • • @ V and the V. are pairwise orthogonal then V is the

orthogo-r ~

nal direct sum of Vl""'V

r and we write V = V1 L ••• L Vr ' Given reflexive bilinear form spaces (Vi,f

i ), i l, .•• ,r we can define a bilinear form f on the direct sum V

=

V

1 @ • • • $ V by f(x,y) r

=

L7

~= 1f.(x. ,y.), ~ ~ ~

x = xl + •.• + x

r' y

=

Yl +.,.+ Yr' xi'Yi € Vi' which is reflexive if the (Vi,f

i) are all of the same type. Identifying Vi with a subspace of V as usu-al we have V

=

V_l L ••• L V

r '

3.19. Suppose V i) rad V

VI + •.. + Vr with Vi orthogonal to Vj for all i ~ j.

rad VI + •• ,+ rad V_r '

ii) If V. is nondegenerate for i = 1, .•• ,r then V is nondegenerate and ~

V = V

l 1. ••• L Vr '

3.20. a) The map V/rad V x V/rad V +F defined by (rad V + x,rad V + Y) ~+ f(x,y) is a well defined nondegenerate bilinear form on V/rad V.

b) If V

=

rad V @ U then U is nondegenerate and V

=

rad V L U and the natural isomorphism U -+ V/rad V, u 1+ rad V + u is an isometry.

3.21. Suppose V = V

1 L .. L Vr' U

=

U1 L •• 1. Ur' U and V spaces over the same field F. Let S.: V. + U. be an isometry 1 ~ i ~ r. We can define an isometry S; V+U

~ ~ ~

by Sex)

=

Sl (Xl) + •.. + 5_r (x_r) for x

=

Xl + ••• + xr E V, Xi E Vi' S is called the orthogonal direct sum of the 5

i and we write S

=

S1 L ••• L Sr' 3.22. If V

=

V

l L ••• L V and S. is an isometry of V. r ~ ~ + V., 1 ~ ~ i ~ r then S

=

5

1 1. ••• 1. Sr is an isometry of V onto V and det S

=

det S1 .•. det Sr' If

T _Tl .L ••• 1. Tr , where Ti is an isometry of Vi + Vi then ST

=

S1 Tl i . • • 1. SrTr

3.23. If V is nondegenerate and U ~ V then a) U1.1. U and dim U + dim U1. dim V. b) rad U rad u1.

=

U nuL,

c) U is nondegenerate iff u1. _{is nondegenerate.}

d) U is nondegenerate iff V :: U 1 u1,

3,24. If V := U 1. W with u,W nondegenerate then W u1.,

(Note t.hat we did not use char :IF '" 2 so far).

x E V is isot.ropic if (x,x) = 0 (notation: (x,y) := f(x,y». u ~ V is isotro-pic if U

=

0 or if there exists a nonzero vector x E U which is isotropic.

U ~ V is totally isotropic if (x/y)

=

0 for all x,y £ U, i.e. if rad U

=

U.

Note: A point P (of PV), i.e. a 1-dim subspace of V,is isotropic iff it is degenerate iff P is spanned by an isotropic vector.

3.25. If V is orthogonal and every vector of V is isotropic then V is totally iso-tropic.

Proof. f is symmetric. Every vector of V is isotropic means f is alternate, hence skew-symmetric. Therefore f = 0 (charlF:f. 2!). II

Let P be a point. P is isotropic iff P £ pI (pI is the polar hyperplane of

p).

V

is symplectic iff every pOint is isotropic. A line (2-dim subspace) is hyperbolic if i t is nondegenerate and isotropic.

3.26. a) The hyperbolic lines are those of the form P + Q, where P and Q are nonor-thogonal isotropic points.

b) The totally isotropic lines are those of the form P + Q, where P and Q are orthogonal isotropic points.

Proof.

a) Suppose L is hyperbolic, then there exists an isotropic point p

=

<p>s L. Let R <r> be a second point on L. RIP would imply P S rad L

=

0 so Rip i.e. (p/r) :f.

O.

If V is symplectic we have nothing to prove. Assume V is orthogonal. Let q := ap + r, a ElF, then (q,q) = 2a(p,r) + (r/r) so

k (r/r) th

ta e a = - en (q,q)

2 (p, r)

o

and Q := <q> is isotropic and L

=

P + Q.

IfL P + Q with P and Q isotropic (p,q)

=

a :f. 0 then L has discriminant det[+O oa

J

±a2 :f. 0, so L is hyperbolic.

_a

b) Trivial. D

An ordered pair P,Q of points is hyperbolic if P and Q are isotropic and not orthogonal. An ordered pair of vectors U,v is hyperbolic if (u,u)

=

(v,v) =0 and (u,v) = 1. A line is hyperbolic if it passes through a hyperbolic pair of points, i.e. iff i t is spanned by a hyperbolic pair of vectors.

Structure of reflexive bilinear form spaces 3.27. Let V be a symplectic space. Then

a) V is an orthogonal direct sum

where P l ' ... , are isotropic points and L

b) If V is decomposed as in a) then

PrQof.

a) call a subspace u S V indecomposable if i t is not an orthogonal direct sum of proper subspaces. Certainly V is an orthogonal direct sum of indecompo-sable subspaces. By 3.20 b) rad U

=

U or rad U

=

0 i.e. U is totally iso-tropic or nondegenerate. If U is totally isotropic then U is a point. If U is nondegenerate then dim U ~ 2. Let P be a point of U then there exists a point Q C U with

Q

t

P. Now L := P + Q 1s a hyperbolic line, L is nonde-generate, so U

=

L ~ (L~ n U) i.e. U

=

L. This proves a).

b) According to 3.19

o

The codimension of rad V (= 2r) is the rank of V.

3.28. TWo symplectic spaces overF are isometric iff they have the same dimension and rank. A nondegenerate symplectic space V has even dimension.

3.29. An orthogonal space is an orthogonal direct sum

with P. isotropic 1 ~ i S s , Q. not isotropic 1 S i ~ r.

L L

Proof. We only need to determine the indecomposable subspaces. If U is total-ly isotropic then U is a point. If U is nondegenerate then there exists a non-isotropic point P in U bij 3.25. So U

=

P ~ (P~ n U) hence U is a point.

0

3.30. Let V be a nondegenerate space, U S V. Choose a complement W for rad U,

U

=

rad U l'W and a basis u

1' •.• ,u_r of rad U. Put Pi: <u

i >, 1 S; i Sr. Then

1) there exists pairwise orthogonal hy-perbolic lines L1, •.. ,L

r all orthogo-nal to W such that P. c L" 1 $ i ~ r.

L - L

Thus

U

Ll ~ L2 ~ ••• ~ Lr ~ W is a non-degenerate subspace containing U.

rad U

~.P~4-'-_-_---.p

1 2 r

L L •••

-I 2 L r W

2) If a: U ~ VI is an isometry of U onto some nondegenerate space V' then a can be extended to an isometry 0: U ~ V' •

Proof.

1) We use induction on r. There is nothing to prove if r = 0 (U

=

U). Assume r > 0 and put U

o

.1 l. W, so rad U

_{o =}

<u 1, ••• ,ur_1> == rad UO' Since P

¢

U we r -

a

= <u₁'···,u r_1>

have UOl. c pol so

- r

Now L := P + X is a

r r hyperbolic

there exists a pOint X ~ U5 with X

i

Pre line and L c U

o

.1. Since Ll. is

nondegene-r - r

rate we may apply the induction hypothesis to U

_o

=

rad U

_o

.1 W ~ Lr to get .1 pairwise orthogonal lines L

1, ••• ,Lr_1 all orthogonal to W such that Pi:: Li

1 ~ i ~ r-l. This completes the induction. 2) Let L1, ..• ,L

r be the hyperbolic lines constructed in 1). Then Li = <ui,vi>, u.,v. a hyperbolic pair, 1 ~ i ~ r. Let U· := a(U), u~ :== a(u]..}, 1 5 i s r,

]. ]. J.

then ui"."u~ is a basis of rad U', U' = rad U' .1 W, with W· := a(W). Put

U'

= Li .1 ••• .1 L~ .1 Wi where Ll

=

<ui,vi>' uitVi a hyperbolic pair, applying

1) _{to U' • Then o(u.)}

=

u~, a(v.) = VJ.~' 15 i ~ r, o/w == a/w is an isometry

]. 1. 1.

extending a.

o

3.31. If V is a nondegenerate symmetric space and x and yare nonisotropic vectors such that (x,x) = (y,y) then there exists an isometry T of V such ;that T (x) == y. Proof. Since V is symmetric we have x+y l. x-Yo Since not both (x+y,x+y) =

"" 2«x,x) + (x,y» and (x-y,x-y)

=

2«x,x) - (x,y» can be 0 one of x+y and x - y is nonisotropic. Let z == x + E Y withe = ±1 such that z is nonisotropic. Then V <z> .1 H, H

=

<z> l. and x - ey E H. Let ~ = T~,Z' H

=

ker ~, ~(z)

=

-2.

Then ~ 1H l. - 1 _<z> so ~ is an isometry, and ~(z)

=

~(x + ey) -x - €y,

~ (x - EY) == x - ey hence ~(x)

=

-EY. So if 8 = -1 we can take T

=

~, if

E = +1 we can take T

=

-1 ~.

V

o

3.32. (Witt's theorem). Let V and V' be nondegenerate spaces and let p: V + V· be an isometry of V onto V' and a: U + Vi an isometry of a subspace U of V into V', then

°

can be extended to an isometry

0:

V + V'.

By 3.30 we may assume that U is nondegenerate •

.1 .1 1.

Case V is symplectic: V == U 1. U , V' == u' .1 (U') where U' := a(U). U and

(u·)J. are nondegenerate symplectic spaces of the same dimension. Hence by

.1 .1

-3.28 there exists an isometry T of U onto (U') • Then a := 0 .1 T is an iso-metry of V extending o'

Case V is orthogonal: Induction on dim U. If dim U == 1 an obvious application of 3.31. Assume dim U > 1, then U P.l W,

W = pJ. n U, P nonisotropic point. Let U' := o(U), P' := o(P), W' := o(W),

01 :=

alp,

02 :=

olw.

p V ---_-_..:.._-_-_-_-_-_ ... ~ V I

I

a

,

o

u

---~)

u'

/ \"1

~J\

u <12 w---~

____

_+.

W'

f"V IV I"oJ .1

By induction we have an extension

°

1: V ... V' of 01 and 01 (P) = p' so 01 (P ) =

l l l

=

(pI) , As p is nondegenerate and W C P we may apply induction hypothesis

to W : pl, 02: W ... (p,)l to get an isometry 0"2: pl ... (p,)l extending

°

2, Now V = P l pl and 8" := 01 .l 0'2 is an isometry of V onto V' extending (J,

0

Let V be nondegenerate. We

may

define: ~ V := max dim of a totally isotro-pic subspace of V. By 3.23 a) it follows that index V ~ ~ dim V, with

equali-ty if V is symplectic because of 3.27.

3.33. All maximal totally isotropic subspaces of V have the same dimension, so index V = the dimension of any such subspace.

Proof, Apply Witt's theorem.

3.34. Let V be a nondegenerate space of index r. Then

o

1) V

=

H2r 1 W, where H

2r is an orthogonal direct sum of r hyperbolic lines and W is nonisotropic.

2) The geometry of W is independent of the choice of H 2r, Proof.

1) Let U be a totally isotropic subspace of dim r. By 3.30 there exis ts H 2r .::. (.1. An H

2s has a totally isotropic subspace of dim s, so 2r is the max dimen-sion of such a subspace. Moreover I V = H 2r .L W, W : =

H~r

and if 0

F

x <: W

is isotropic then <H

2r,x> contains a totally isotropic subspace of dimr+1# •

2) Follows from Witt's theorem: If Hir is a second such sum of hyperbolic lines, then certainly there is an isometry a of H

2r onto H2r• The (J extends to an isometry

a

of V and

cr

(H~r)

(H

4. The symplectic group

Let (V,f) be a nondegenerate reflexive bilinear form space. The group of all isometries of (V,f) is denoted by Sp(V) if (V,f) is symplectic and by O(V,f) if (V,f) is orthogonal. Sp(V) is called the sysplectic group, O(V,f) the or-thogonal group. Let v

1, .•• ,vn be a basis of V, and let E

=

(f(vi,vj » be the

corresponding matrix of f. Let T E GL(V) and A ( GL(n~) the matrix of T with respect to this basis. Then T is an isometry iff AEAt = E.

We define

Sp(n~)

:= {A E

GL(n~)

I AEAt

=

E} if (V,f) is of symplectic type, O(nJF,f) := {A E GL(nJF) 1 ABAt

=

E} if (V,f) is of orthogonal type. Clearly Sp(V) ~ Sp(n~) and O(V,f) ~ O(n~,f) •

4.1. Isometries of (V,f) have determinant ±1.

Assume (V,f) is symplectic

By 3.27 V has even dimension n = 2r and a Symplectic basis u

1,u_1,u2

,u_

2"" ••• ,ur,u_

r' such that (u1,u_1) - •.• - (ur,u_r) = 1, (ui,Uj) =

a

if i + j ~ O.

4.2. Sp(V) acts faithfully and regularly on the ordered symplectic bases.

Assume IF = IF , We can determine 1 Sp (V) 1 by counting the ordered symplectic

q

bases. Define L := <u

1,u_1>, then L is an hyperbolic line and L.l.=:<u2,u_2, •• ••• ,ur,u_

r> is a nondegenerate symplectic space of dimension 2 (r - 1). Let

~(r) denote the number of ordered symplectic bases, then ~(r)

=

(#

hyperbo-lic pairs of vectors). <p(r-1). Suppose u,v is a hyperbohyperbo-lic pair of vectors, tllen U,W is a hyperbolic

tile number of hyperbolic

• <p (r - 1), and 4.3. ISp(n,GF(q» I with <p(1) (E.) 2 n/2 2 q •

n

i=1

pair iff (u,v-w) = 0, Le. iff v-w (; <u>.l.. Hence

2r 2r-l 2r 2r-l

pairs equals (q - l)q , so <per) = (q - l)q

2 2 r 2i

:::: (q -1)q we find <per)

=

qr

n

(q -1).

i=1 (q2i _ 1) •

If n = 2 then ISp(2,GF(q» I q(q2 _ 1)

=

ISL(2,GF(q» I. In fact: 4.4. Sp(2~) ~ SL(2~) for any fieldlF.

Proof. Choose a hyperbolic pair of vectors as

d f . (a b) t

an or any matr1x A:::: c d we have ABA

=

Le. i f f A E SL(2,lF).

a basis of V. Then E =

(_~ ~)

( a

ad-bc) = E iff ad - be = 1

bc-ad

a

We consider tile action of Sp(v) on the points of PV. We have an exact sequence 1 -+ Zp (V) -+ Sp (V) -+ PSp (V) -r 1 ,

where PSp(V) := Sp(V)Points, Zp(V) = kernel of this action = Z(V) n Sp(V). A scalar transformation AI is in Sp(V) iff AI E AI

=

E, i.e. iff A = ±1. Hence Zp (V) = {±1}. I f IF = GF (q) then (!!.) 2 n/2 IpSp(n,GF(q»

I

=

-;iISP(n,GF(q» I

=

~

q 2 IT 1=1 2i (q - 1), d

=

(2,q-l).

We know already that isometries have determinant +1 or -1. We shall show that symplectic isometries have determinant +1. First of all we determine the trans-formations T which are in Sp(V) • Assume ~

#

0, so T := T

#

1 and let

CP,P Ijl,P

H ;= ker cP, P ;= <p> then

T (Q) = Q, for all Q

.s.

H, hence 1" (Sl) = Q 1, I for all Q

.s.

H, hence

P

_.s.

Q 1, I for all Q

.s.

H, Q

#

H.L I hence Q

.s.

P 1, I for all Q

.s.

Hf Q

#

H.L , hence

H\H.L c p.L so H ::: p.L

.

This shows that T E ~ ,pl' Conversely if P

=

<p> is any point and 1 # T E Xp ,p.L then T(X) = x + a(p,x)p for Some a ElF*, and T is an isometry for (T(X),T(Y» =

2

= (x,y) + a(p,x) (P,y) + a(x,P) (P,Y) + a (P,X) (P,y) (P,P) ::: (x,y).

4.5. Xp,p.L ~ Sp(V) for all points P.

If T E Sp(V) then T(P.L)

The elements of Xp,p.L are called symplectic transvections. We know XPIP.L~ OF,+)

is Abelian.

4.7. Sp(V) is generated by the symplectic transvections so Sp(V)

=

<T(~,P.L)

I

T E Sp(V» since Sp(V)tls transitive on the points of PV •

_.

~. Let G(~) be the subgroup of Sp(V) generated by symplectic transvections. We show that G(V) is transitive on the ordered symplectic bases •

•

1) G(V) is transitive on the vectors ~ O. Let u,v E V\{O}.

Case u

t

v: P := <u + v> is a point on L = <u,v> and P~

n

L

=

P.

We

know that ~,pJ.. moves u to v.

Case u J.. v: There exists a vector w such that w

t

u and w

t

v, because <u> .Lu <v>~ ~ V. Move u to wand w to v.

2) G(V) is transitive on the hyperbolic pairs of vectors. Let u,v; u'v' be hyperbolic pairs of vectors. By 1) we may assume u

=

u'. We must show the existance of T E G{V) such that T{U)

=

u, T(v)

=

v'. Let P := <U>,

Q: <v> 1 Q' : = <v' > •

Case P,Q,Q' collinear: L := P + Q + Q' is a line. Then P~ n L P and

~,p.L moves v to v· and fixes u.

Case P/Q,g' not collinear: Let R := (Q + Q') n p.L. Suppose Q

t

R~ then

XR1R~

moves v to v' and fixes u. If Q

=

R~

we take a point Q" on P + Q', Q" ~ P,Q'.

~ J..

Let R' := (Q + Q")

n

P . From Q

=

R it follows that R c Q~. Then (P + R) n Q~

=

R

as p

i

Ql. Hence R'

i

QJ.. as R' S P + R,

R' :; R. Thus X

R" (R') ~ moves Q to Q" and

Qrt can be moved to Q' •

Q

3) G(V) is transitive on ordered symplectic bases. Let u

1,u l " " ' u , u - r - r and vl,v_1, ••• ,vr,v_

r be two symplectic bases. We can assume u1 = v1' u_1 =v_l

.1

by 2). Then, if L = <u

1,u_1>, L = <u2,u 2' ••• ,u,u - r -r > = <v2,v 2' .•. ,v,v - r - r > By induction there exists a 0 IE G(L~) such that o(ui)

=

v., i = ±2, •.• ,±r.

1.

Put T = lL ~ 0 then T maps the one basis to the other. Clearly T E G(V) for

if A is a symplectic transvection of L.l then lL .L A is a symplectic

trans-vection of V.

4.8. T <:: Sp(V) implies det T == 1.

4.9. Center of Sp(V)

=

{±l}.

T

Proof. Let T be in the center of Sp(V) then (~,pl)

Therefore T(P)

=

P for all P, i.e. T E Zp(V) = {±l}. 4.10. Sp(V)

1:±1l

center Sp (V) } PSp(V), simple?

[]

~,P~ for all points P.

In order to prove simplicity of PSp(V) we want to apply Iwasawa's lemma to the action of Sp(V) on the points of PV. So we still have to prove

a) Sp(V) acts primitively on the points. b) Sp{V) is perfect.

Because Sp(2~) ~ SL(2~} there will be exceptions to b) and we can restrict our investigations to n ~ 4. Now n

=

4,

PFI

=

2 is the first case and

Isp(4,2)1

=

24(22_1)(24_1) =

6~

= IE61. In fact 4.11. Sp(4,2) ~ PSp(4,2} ~ [6 so PSp(4,2) is not simple.

This is an immediate corollary of:

4.12. I:

2n+2 ;S; Sp(2n,2) .

Proof. Let X be a set of 2n + 2 points. The partitions of X into two subsets with an even number of points form a vectorspace of dimension 2n over GF(2) if we define addition by

{A,X\A} + {B,X\B} := {A , B, X\(A T B)}, A,B

=.

X, IAI = IBI - 0(2).

(A f B == symmetric difference of A and B = (A u B) \ (A n B» •

We can define a nondegenerate symplectic form on this vectorspace by ({A, X\A}, {B,X\B} ) := _{IA n BI mod 2, A,B}

:=

_{X, IAI} = IBI :.: 0 (2) •

I t is clear that 1:

2n+2 leaves this form invariant, hence E2n+2 S Sp(2n,2).

There is also a nice proof of 4.11 using the isomorphism PSL(4,2)

=

AS' Construct a polarity of PG(3, 2) using A

6• A6 commutes with this polarity:.

A6 s PSp(4,2) so E6 ~ PSp(4,2) .

2

Let X denote ~le set of pOints of PV, 1

=

1X the diagonal subset of X • a

1 := {(P,Q)

i l l

P l- Q}, Ci.2 := {(P,Q)

I

P

i

Q} • Clearly x2 := 1 u a 1 U C( 2 moreover by 4.2 we have

1-kJ

4.13. 1, (1

1 and (1.2 are the orbits for the componentwise action of Sp(V) on x2 Le.

2

X /Sp(V) = {1,a

1,a2}.

Note. 4.13 means that Sp(V) has rank 3 in its action on X i.e. PSp(V) is a rank 3 permutation group on X (if G tra X than G is said to be of rank r if

Let P € X, define for i E {1,2} pa

i := {Q I (P,Q) E ail so Pai is the set of vertices in the graph (X,a.) adjacent to P.

~

Note. (x,a

1), (X,U2) is a pair of complementary strongly regular graphs or equivalently (X,{l,a

1,a2}) is an association scheme with 2 treatments.

4.15. Sp(V) pri X.

Proof. Let B be an imprimitive block, IBI > 1. We have to prove that B X. Let P E B, if B n Pa

i

#

~ then Pai

S

B. Moreover Qai

S

B for every Q E B since

SP(V)B tra B. Case B n pa

l

1J!:

Let R be any point not in {p} U Pal = p.1, take Q € (R + P).1 then R E Qa

1 and Q E Pal S B, hence R E Qa1 S B i.e. B

=

X. Case B n Pa~: Let R be any point not in {p} U Pa

2• Take Q € X\(p.1 U R.1) then Q E Pa

2 and R E Qa2 S B, :. B X.

o

Note. The essential thing in the above proof is that (X,U

t) and (x,a2) are

shown to be connected. The general statement is: Suppose G tra X then G pri X iff all graphs (X,a) are connected, a E

x

2/G, a

#

1.

4.16. Sp(n,JF), n = 2r is perfect unless (n,

/IFI)

= (2,2), (2,3), (4,2).

Proof. Suppose Sp(n,JF) is perfect for some n ~ 2 and let T € ~,p.1 be a sym-plectic transvection in Sp(n+ 2,JF). Let L be a hyperbolic line such that P S L..l then V

=

L .1 L..l and T = lL .1 a where 0 = T I L..l is a symplectic trans-vection in Sp (L.1). Then 0 is a product of commutators in Sp (L.1). If A, j.1 E Sp (L.1)

then lL .1 (A,j.1)

=

(lL .1 A,lL .1 j.1), hence T is a product of commutators in

Sp(n + 2,JF) :. Sp(n

+

2,JF) is perfect. So

(*) IfSP(n,JF) is perfect so is Sp(m,JF) for all m ~ n • By 4.4 and 2.24 we have:

(** ) I f

/IF I

> 3 then Sp (n"F) is perfect for all n ~ 2 •

It remains to show that Sp(4,3) and Sp(6,2} are perfect. In each case it suf-fices to show the existence of a single transvections

#

1 in the commutator subgroup (if 1 # T E Sp(V)' is a symplectic transvection then T

some P. Since Xp ,p.1 ~ OF,+) has order 2 or 3, T generates ~,p.1

Therefore XT(P) ,T(P)'

T~,p..l

s

T(Sp(V)')

=

Sp(V) , for all T E

E ~,p..l for

so X .1 -;; Sp (V) 1

--P,P