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Let $(V,Q)$ be a nondegenerate quadratic space and let $W$ be a subspace of $V$. We then have $V = W \oplus W^\perp$. If an isometry of $W$ - for the restriction of $Q$ to $W$ - and an isometry of $W^\perp$ - for the restriction of $Q$ to $W^\perp$ - are given, under which conditions can we conclude that these two isometries are in fact the restrictions of one and the same isometry of $(V,Q)$?

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I'm stupid... We don't need to have $V = W \oplus W^\perp$! –  Evariste Jun 8 '11 at 9:41

1 Answer 1

up vote 2 down vote accepted

It seems as if isometries always lift.

Let $T:W \rightarrow W$ and $T':W^{\perp} \rightarrow W^{\perp}$ be isometries of $(W,Q|_{W}), (W^{\perp},Q|_{W^{\perp}}),$ respectively, and consider $T\oplus T':V \rightarrow V$. We claim $T\oplus T'$ is an isometry of $(V,Q).$

Let $v\in V.$ Then $v = w + w_0$ for unique vectors $w\in W$ and $w_0 \in W^{\perp}.$ Let $B_Q$ be the bilinear form associated to $Q.$ Then

\begin{align*} Q(T\oplus T'v) &= Q(Tw + T'w_0) \\ & = B_Q(Tw + T'w_0, Tw + T'w_0) \\ &= B_Q(Tw,Tw) + B_Q(Tw,T'w_0) + B_Q(T'w_0,Tw) + B_Q(T'w_0,T'w_0) \\ &= B_Q(Tw,Tw) + B_Q(T'w_0,T'w_0) \\ & = B_Q(w,w) + B_Q(w_0,w_0) \\ & = B_Q(w,w) + B_Q(w,w_0) + B_Q(w_0,w) + B_Q(w_0,w_0) \\ & = B_Q(w + w_0, w + w_0) \\ & = Q(v).
\end{align*}

Hence, $T\oplus T'$ is an isometry of $(V,Q).$

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