Number Theory: An Introduction to Mathematics

1 Quadratic Spaces 299

Proposition 14Let V be a non-singular quadratic space. If U is a totally isotropic
subspace with basis u 1 ,...,um, then there exists a totally isotropic subspace U′with
basis u′ 1 ,...,u′msuch that

(uj,u′k)= 1 or 0 according as j=kor j=k.

Hence U∩U′={ 0 }and

U+U′=H 1 ⊥···⊥Hm,

where Hjis the hyperbolic plane with basis uj,u′j( 1 ≤j≤m).

Proof Suppose first thatm=1. SinceVis non-singular, there exists a vectorv∈V
such that(u 1 ,v)=0. The subspaceH 1 spanned byu 1 ,vis a hyperbolic plane and
hence, by Proposition 13, it contains a vectoru′ 1 such that(u′ 1 ,u′ 1 )=0,(u 1 ,u′ 1 )=1.
This proves the proposition form=1.
Suppose now thatm>1 and the result holds for all smaller values ofm.LetW
be the totally isotropic subspace with basisu 2 ,...,um. By Proposition 4, there exists
a vectorv ∈ W⊥such that (u 1 ,v)= 0. The subspaceH 1 spanned byu 1 ,vis a
hyperbolic plane and hence it contains a vectoru′ 1 such that(u′ 1 ,u′ 1 )=0,(u 1 ,u′ 1 )=1.
SinceH 1 is non-singular,H 1 ⊥is also non-singular andV=H 1 ⊥H 1 ⊥.SinceW⊆H 1 ⊥,
the result now follows by applying the induction hypothesis to the subspaceWof the
quadratic spaceH 1 ⊥.

Proposition 15Any quadratic space V can be represented as an orthogonal sum

V=V⊥⊥H 1 ⊥···⊥Hm⊥V 0 ,

where H 1 ,...,Hmare hyperbolic planes and the subspace V 0 is anisotropic.

Proof LetV 1 be any subspace supplementary toV⊥.ThenV 1 is non-singular, by the
definition ofV⊥.IfV 1 is anisotropic, we can takem=0andV 0 =V 1 .OtherwiseV 1
contains an isotropic vector and hence also a hyperbolic planeH 1 , by Proposition 14.
By Proposition 3,

V 1 =H 1 ⊥V 2 ,

whereV 2 =H 1 ⊥∩V 1 is non-singular. IfV 2 is anisotropic, we can takeV 0 =V 2 .Other-
wise we repeat the process. After finitely many steps we must obtain a representation
of the required form, possibly withV 0 ={ 0 }.

LetVandV′be quadratic spaces over the same fieldF. The quadratic spaces
V,V′are said to beisometricif there exists a linear mapφ:V →V′which is an
isometry,i.e.itisbijectiveand

(φv,φv)=(v,v) for allv∈V.

By (1), this implies

(φu,φv)=(u,v) for allu,v∈V.