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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 $V = W \oplus W^\perp$!2011-06-08

1 Answers 1

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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 $(V,Q).$