Let $U,W$ be maximal completely isotropic subspaces of a finitely dimensional quadratic space $(V, q)$ over a field $\mathbb{K}$, $\operatorname{char} \mathbb{K} \neq 2$. Prove that any $q$-isometry $g\colon U \rightarrow W$ can be extended to a $q$-isometry $h \colon V \rightarrow V$
In fact, any isomorphism $g \colon U \rightarrow W$ is an isometry since the subspaces are completely isotropic, so the form restricted to $U$ or $W$ is a zero form. Besides, I guess it would be a good start to prove it if $\operatorname{codim} U = 1 $, in this way maybe we could even do this inductively.
Still, I don't know how to do this even in $\operatorname{codim} U = 1 $. Can you give me a hint?
In the case that $(V,q)$ is regular (or non-degenerate or nonsingular), this follows easily from Witt's cancellation theorem, which says the following:
Thus, since $V$ is regular, we know that $U \perp U^\perp = V = W\perp W^\perp$. Since $U\cong W$, the theorem tells us $U^\perp \cong W^\perp$. Taking the direct sum of these two isometries (i.e. the one between $U$ and $W$ and the one between $U^\perp$ and $W^\perp$) gives us the desired extension.
I haven't looked much at solving the problem for forms which are not necessarily regular, but I think if I were you, I'd look at the proof of Witt's Cancellation Theorem to get some ideas. Hope this helps!