1 Quadratic Spaces 295(i) dimU+dimU⊥=dimV;
(ii)U⊥⊥=U;
(iii)U⊥is non-singular if and only if U is non-singular.
Proof There exist vectorsvm+ 1 ,...,vn∈Vsuch thatv 1 ,...,vnform a basis forV.
If we putαjk=(vj,vk)then, sinceVis non-singular, then×nsymmetric matrix
A = (αjk)is non-singular. Hence, for anyη 1 ,...,ηn ∈ F,there existunique
ξ 1 ,...,ξn∈Fsuch thatv=ξ 1 v 1 +···+ξnvnsatisfies
(v 1 ,v)=η 1 ,...,(vn,v)=ηn.This proves the first part of the proposition.
By takingU=〈v 1 ,...,vm〉andη 1 =···=ηm=0, we see that dimU⊥=n−m.
ReplacingUbyU⊥, we obtain dimU⊥⊥=dimU. Since it is obvious thatU⊆U⊥⊥,
this impliesU=U⊥⊥.SinceUnon-singular meansU∩U⊥={ 0 }, (iii) follows at
once from (ii).
We now introduce some further definitions. A vectoruis said to beisotropicif
u=0and(u,u)=0. A subspaceUofV is said to beisotropicif it contains an
isotropic vector andanisotropicotherwise. A subspaceUofVis said to betotally
isotropicif every nonzero vector inUis isotropic, i.e. ifU⊆U⊥. According to these
definitions, the trivial subspace{ 0 }is both anisotropic and totally isotropic.
A quadratic spaceV over a fieldFis said to beuniversalif it represents every
γ∈F×, i.e. if for eachγ∈F×there is a vectorv∈Vsuch that(v,v)=γ.
Proposition 5If a non-singular quadratic space V is isotropic, then it is universal.
Proof SinceVis isotropic, it contains a vectoru=0 such that(u,u)=0. Since
Vis non-singular, it contains a vectorwsuch that(u,w)=0. Thenwis linearly
independent ofuand by replacingwby a scalar multiple we may assume(u,w)=1.
Ifv=αu+w,then(v,v)=γforα={γ−(w,w)}/2.
On the other hand, a non-singular universal quadratic space need not be isotropic.
As an example, take F to be the finite field with three elements and V the
2-dimensional quadratic space corresponding to the quadratic formξ 12 +ξ 22.
Proposition 6A non-singular quadratic form f(ξ 1 ,...,ξn)with coefficients from a
field F representsγ∈F×if and only if the quadratic form
g(ξ 0 ,ξ 1 ,...,ξn)=−γξ 02 +f(ξ 1 ,...,ξn)is isotropic.
Proof Obviously if f(x 1 ,...,xn)=γandx 0 =1, theng(x 0 ,x 1 ,...,xn)=0.
Suppose on the other hand thatg(x 0 ,x 1 ,...,xn)=0forsomexj∈F, not all zero.
Ifx 0 =0, thenfcertainly representsγ.Ifx 0 =0, thenfis isotropic and hence, by
Proposition 5, it still representsγ.
Proposition 7Let V be a non-singular isotropic quadratic space. If V=U⊥W, then
there existsγ∈F×such that, for some u∈U andw∈W,
(u,u)=γ, (w,w)=−γ.