302 VII The Arithmetic of Quadratic Forms
Proof This follows at once from Proposition 18, sinceU′is also non-singular and
V=U⊥U⊥=U′⊥U′⊥. The first statement of Proposition 18 is known asWitt’s extension theoremand
the second statement asWitt’s cancellation theorem. It was Corollary 19 which was
actually proved by Witt (1937).
There is also another version of the extension theorem, stating that ifφ:U→U′
is an isometry between two subspacesU,U′of anon-singularquadratic spaceV,
then there exists an isometryψ:V →Vsuch thatψu=φufor allu∈U.For
non-singularUthis has just been proved, and the singular case can be reduced to the
non-singular by applying (several times, if necessary) the following lemma.
Lemma 20Let V be a non-singular quadratic space. If U,U′are singular subspaces
of V and if there exists an isometryφ:U→U′, then there exist subspacesU ̄,U ̄′,
properly containing U,U′respectively and an isometryφ ̄ : U ̄ → U ̄′such that
φ ̄u=φu for all u∈U.
Proof By hypothesis there exists a nonzero vectoru 1 ∈U∩U⊥.ThenUhas a basis
u 1 ,...,umwithu 1 as first vector. By Proposition 4, there exists a vectorw∈Vsuch
that
(u 1 ,w)= 1 ,(uj,w)=0for1<j≤m.Moreover we may assume that(w,w) = 0, by replacingwbyw−αu 1 , with
α=(w,w)/2. IfWis the 1-dimensional subspace spanned byw,thenU∩W={ 0 }
andU ̄=U+WcontainsUproperly.
The same construction can be applied toU′, with the basisφu 1 ,...,φum,to
obtain an isotropic vectorw′and a subspaceU ̄′ = U′+W′. The linear map
φ ̄:U ̄→U ̄′defined by
φ ̄uj=φuj( 1 ≤j≤m), φw ̄ =w′,is easily seen to have the required properties.
As an application of Proposition 18, we willconsider the uniqueness of the repre-
sentation obtained in Proposition 15.
Proposition 21Suppose the quadratic space V can be represented as an orthogonal
sum
V=U⊥H⊥V 0 ,where U is totally isotropic, H is the orthogonal sum of m hyperbolic planes, and the
subspace V 0 is anisotropic.
Then U =V⊥,m=indV−dimV⊥, and V 0 is uniquely determined up to an
isometry.
Proof SinceHandV 0 are non-singular, so also isW=H⊥V 0. Hence, by the remark
after the proof of Proposition 3,U=W⊥.SinceU⊆U⊥, it follows thatU⊆V⊥.In
factU=V⊥,sinceW∩V⊥={ 0 }.