Singular quadratic space

Let $(V,b)$ a symmetric bilinear space. An old theorem of Witt says that if $(V,b)$ is regular, then given a subspace $W$ of $V$ and an isometry $\sigma: W \to V$, there exists an isometry $\Sigma: V \to V$ which extends $\sigma$.

Does this necessarily fail if $(V,b)$ is singular - e.g. if the radical $V^\perp$ is one-dimensional?

flag	asked Jul 2 at 14:33 Wanderer 2,969●12●28

I'm not sure I've understood you correctly, but it seems to me that over $\mathbb{R}$ the result fails if and only if the signature of the bilinear form on $V/V^{\perp}$ is indefinite. Take an element of $V^{\perp}$ and send it to a null element of $V$ not in $V^{\perp}$ - this of course cannot extend. Over $\mathbb{C}$ there are always null vectors... this may not be what you mean. – Paul Reynolds Jul 2 at 15:43

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