with an F-linear bijection / of V onto V'. We also denote by (V, p) ® (V, (pf) the
structure (W, '*/) given by W = V ® V and ij.i(x + x\ y + y') = p(x, y) + (pf(xf, y')
for x, y £ V and x', y' G V. Clearly (p 0 p7 is nondegenerate if and only if both
(p and (// are nondegenerate. In such a case, we can view O^xO^ as a subgroup
of O*®*'. The element (a, /J) of O^ x O1^ viewed as an element of O*®*' will be
denoted by a x /? or by (a, /?).
Theorem 1.2. Let (V, y) = (S, a) © (T, r) = (S', a') 0
(T7, T7)
with iioncte-
generate p. If(S, a) is isomorphic to
(57, a7),
then (T, r) is isomorphic to
(T7, r7).
PROOF. If there is an element / of O^ such that Sf = S", then clearly / gives
an isomorphism of (T, r) to
(T7, r7).
We prove the existence of / by induction on
dim(S). Suppose dim(5) = 1; then we can put S = Fx and S' = Fy with elements
x and y such that ip[x] = (p[y] ^ 0. Since /?[# -f y] + p[# y] = 4p[#] ^ 0, we have
p[x + ?/] 7^ 0 or y?[x 2/] 7^ 0. Changing y for —y if necessary, we may assume that
(p[x + y] ^ 0. Let [7 = ((Ffc + y ) ^ . Define / G GLfV) by u / =-u ioi u e U and
(x+y)f = x + y. Clearly / G O^. Since x y G C/, we have (x y)f = y x, so that
xf y, which proves the case where dim(S) = 1. Next suppose dim(5) = m 1.
Let h be an isomorphism of (S, a) to (S",
Take any x G 5 such that p[x] ^ 0;
put y = xh, R = SO (Fxy, and
= Rh. Then 5 = Fx 0 i? and
= Fy 0
By the above result we can find an element g of O^ such that Fyg = Fx. Now
both i? and R'g are contained in
and hg gives a bijection of R onto
that leaves ip invariant. Since dim(i?) = m 1, by induction
has an
element k that maps it! onto R'g. Extend fe to an element £ of GL(V) by putting
= x. Then £ G O^ and we see that
= S\ which completes the proof.
This is called Witt's theorem. It can be generalized to the case of a hermitian
form over a division algebra with an involution; see [S97, Theorem 1.2 and Lemma
1.3], for example.
Lemma 1.3. Given (V, p) with a nondegenerate p as above, let I be a totally
isotropic subspace of V of dimension ra, and let {ei}^L1 be an F-basis of L Then
we can find a subspace Z of V and a set of elements { / i } ^ such that
(1.2a) V^Z + Y^iiFei + Ffi),
(1.2b) (p(eu ej) = p{fi, fj) = 0, 2ip(eu fj) = Sij for every i and j ,
(1.2c) Z = { v G V | ip(eu v) = (p(fa v) = 0 for every i } .
Conversely suppose that V has 2m elements e$, fi satisfying (1.2b); put I =
YlT=\Fei and
I' = Y^iL\Ffi'i define Z by (1.2c); let £ denote the restriction
of (p to Z. Then I and V are totally isotropic subspaces of V of dimension m,
V = Z 0 V 0 /, and £ is nondegenerate.
PROOF. The converse part can easily be verified. We prove the first part by
induction on m. Let be a nonzero element of V such that p[ei] = 0. Since
(p is nondegenerate, we can find an element x such that p(ei, x) = 1. Put /i =
Then ip[fi] = 0 and /?(ei, f\) = 1. Given such {e-i, / i } , define Z
by (1.2c) with m = 1. Then the converse part of our lemma gives (1.2a) in the case
m = 1. If we have / = YLlLi
^ei w
wi 1* then the for i 1 belong to the
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