Witt extension Theorem Stronger Version

Question

This is Theorem 1.5.5 from Scharlau's book Quadratic and Hermitian Forms (Witt's theorem).

Let $(V, b)$ be a regular symmetric bilinear space.

Let $W$ be a subspace of $V$ and $\sigma \colon W \to V$ an isometry. Then there exists an isometry $\tau \colon V \to V$ which extends $\sigma $ i.e $\tau$ restricted to $W$ is $\sigma $. $\tau$ is product of reflections which extends to $\sigma$.

How to prove above result?

I don't understand what book suggest in second paragraph when $W$ is not totally isotropic they prove above result by induction. Can someone please explain the case for which $W$ is not totally isotropic. Also I want some explanation of totally isotropic subspace. What is step where we use induction in totally isotropic subspace and how we use it? What are reflection in second case? Please someone help me out.

Answer 1

DISCLAIMER: Take this answer with a grain of salt. This topic is new to me and basically I mostly rely only on the stuff I've learned here on the site when answering a few other questions from the same book.

5.3 Theorem. (Witt) Let $(V,b)$ be a regular symmetric bilinear space. Let $W$ be a subspace of $V$ and $\sigma \colon W\to V$ an isometry. Then there exists and isometry $\Sigma\colon V\to V$ which extends $\sigma$, that is $\Sigma|_W=\sigma$.

5.5 Theorem. Under the hypotheses of 5.3 there exists a product of reflections $\Sigma$ which extends $\sigma$.

You have asked about one part of this proof (and you have also mentioned in chat that basically you're after a more detailed breakdown of this proof):

I don't understand what book suggest in second paragraph when $W$ is not totally isotropic they prove above result by induction. Can someone please explain the case for which $W$ is not totally isotropic.

The part of proof you asked above:

We now proceed by induction on $m=m(W):=\dim W+2\dim(W\cap W^\bot)$. (In this way we can avoid some technical steps in the classical proofs.)

Since I do not know the classical proofs, I cannot comment much on this. Probably somebody who is more familiar with theory of quadratic forms can say more, but it seems that goal of the question is mainly to understand this proof.

For $m=0$ we have $W=0$ and can take $\Sigma=id$.

I guess that the case $W=0$ can be considered trivial.

We assume now $m>0$ and consider first the case that $W$ is not totally isotropic. By 5.6 there is an anisotropic vector $x\in W$.

This is just an application of Lemma 5.6, not much to say here.

Let $W_1=x^\bot\cap W$ so that $W=xK\oplus W_1$.

From Lemma 3.4 we know $V=xK \oplus x^\bot$. By taking intersection with $W$ we get $W=xK \oplus (x^\bot\cap W)$. In fact we know more than that - we know that $V= xK \perp x^\bot$ and $W= xK\perp W_1$.

Since $m(W_1)=m(W)-1$ we can apply the induction hypothesis to $\sigma_1=\sigma|_W$.

We have $\dim(W)=1+\dim(W_1)$, hence $\dim(W_1)=\dim(W)-1$.

At the same time, we have $W_1\cap W_1^\bot=W\cap W^\bot$. Indeed \begin{align*} W\cap W^\bot &= (xK \oplus W_1) \cap (xK \oplus W_1)^\bot \\ &= (xK \oplus W_1) \cap x^\bot \cap W_1^\bot \\ &= (xK \cap x^\bot \cap W_1^\bot)\oplus (W_1 \cap x^\bot \cap W_1^\bot) \\ &= W_1 \cap x^\bot \cap W_1^\bot \\ &= W_1 \cap W_1^\bot \end{align*}

There is a product of reflections $\Sigma_1$ extending $\sigma_1$.

This is just application of the induction hypothesis. (Notice that a restriction of an isometry is again an isometry. So the induction hypothesis can be applied.)

The vectors $x$ and $y:=(\Sigma_1^{-1}\sigma)x$ lie in $W_1^\bot$.

The fact that $x\in W_1^\bot$ follows simply from $W_1\subseteq x^\bot$.

Since $\Sigma_1$ extends $\sigma_1$, any element of the space $W_1$ is invariant under $\Sigma_1^{-1}\sigma$. I.e., $\Sigma_1^{-1}\sigma(w)=\Sigma_1^{-1}\sigma_1(w)=w$ for any $w\in W_1$.

Now using the fact that $\Sigma_1^{-1}\sigma$ is an isometry and any isometry maps orthogonal element to an orthogonal element, we get from $x\in W_1^\bot$ that $y\in W_1^\bot$.

By the first step of the proof there is a product of (at most 2) reflections of the form $\tau_z$, $z\in W_1^\bot$ which maps $x$ to $y$.

This is application of the fact proved in the first paragraph of the proof (the case $W=xK$, $x$ anisotropic).

These reflections act as an identity on $W_1$.

The first part of Lemma 5.2 says that $\tau_z$ acts as identity on $z^\bot$.

Then $\Sigma:=\Sigma_1\Sigma_2$ is the map we want to find: $$\Sigma_1\Sigma_2(x\alpha+x_1) =\Sigma_1(y\alpha+W_1) =\sigma(y\alpha)+\sigma x_1 =\sigma(x\alpha+x_1).$$

This is a direct computation using $\Sigma_2x=y$ and $\Sigma_1|_{W_1}=\Sigma_2|_{W_1}=id_{W_1}$.