12 I. QUADRATIC FORMS AND CLIFFORD ALGEBRAS

subspaces X, Y and elements g, L By Theorem 1.2 we can find an element 7 of

O^ such that X 7 = Y, /17 = /c, and gj = £. It X ^ {0}, then we can take 7 from

5 0 ^ by modifying it by a suitable element of 0(p(X).

1.6. Let us now express various things by matrices. Given (V, p) with degener-

ate or nondegenerate /?, take an F-basis {ei}f=1 of V. For x = ^™=1 &e* ^ ^ w ^ n

£i G F, let Xo be the element of F^ whose components are £1, ... , £n. Let /?o

be the n x n-matrix [^(e^, e^)]. . Then tpQ = y?o and ip(x, y) = #0^0 * t?/o- We

call /? o ^

e

matrix that represents p with respect to {e^}^=1, or simply a matrix

representing p when the basis is not specified. Clearly p is nondegenerate if and

only if det(/?o) 7^ 0. Hereafter, we shall often use the same letter for a quadratic

form and the matrix representing it when the basis is fixed. If we change the basis,

then ifo is changed into apQ-la with a G GLn(F). We can define an isomorphism

a ^ a o of GL(V) onto GLn(F) by (xa)o =

XQCXO.

If p is nondegenerate, the map

a ^ a

0

gives an isomorphism of O^ onto the group

(1-3) O(p0) = {Pe GLn(F) I f3p0 •tP = Vo }•

If (W, I/J) = (V, (p) 0 {V', (/?'), and p resp. /?' is represented by po resp. ^Q, then

ij) is represented by diag[(/?0, Po]-

It is an easy exercise to show that every degenerate or nondegenerate p can be

represented by a diagonal matrix. In particular, we have a diagonal representation

of a nondegenerate p with respect to a basis {hi}f=1 if and only if ip(hi, hj) = Q5^

for every i and j with C{ G F x . In such a case we call {hi}^=1 an orthogonal

basis of V with respect to ip.

If V = Yl^iiFfi + ^ e 0 ^s a S P ^ Witt decomposition, we shall often use the

expression (iJm, 2~1r]m) instead of (V, ip), where

' 0 l

m

lm 0

which is twice the matrix representing the quadratic form with respect to the ba-

sis {/1, ... , /

m

, ei, ... , e

m

} . Also we shall often put V =

YHLI

Ffi a n d

^

=

X ^ i ^ei s o t n a t ^m = / ' + /.

Suppose now / ? is nondegenerate; put F

x 2

= { a

2

| a G F

x

} . Take po as above.

Since det(a^

0

•

tot)

G

det((/?0)Fx2,

the coset

de\J{po)Fx2

in

Fx/Fx2

is completely

determined by (p. We call the coset

det((/?o)Fx2

the discriminant of /?, or of

(V, p), and denote it by d(ip) or d(V, p). If n is odd, then for c e

Fx

we have

d(c(/?) =

cnd(p)

= cd(/). Therefore, given ip with odd n, we can always take

c G

Fx

so that d(cy) is represented by any given element of F

x

. If n is even, by

the discriminant field of p, or of (V, (/?), we understand the extension

(1.5a) K0 =

F({(-l)n'2 det(^o)}1/2)

of F This is either F itself or a quadratic extension of F We then define the

discriminant algebra K of /?, or of (V, ip), by

f if0 if K0 ^ F,

If C is the restriction of £ to a core subspace Z of V, then the discriminant field

and algebra of £ coincide with those of p.

(1-4) »*,