Understanding Witt's Theorem I just began to learn something about classical polar spaces and now, I'm trying to understand three implications of Witt's theorem.
Let $V$ be an $m$-dimensional vector space over a field $K$ equipped with a non-degenerate quadratic, symmetric, alternating or Hermitian form. Then, Witt's theorem says that every isometry between two subspaces $U_1$ and $U_2$ of $V$ extends to an isometry of $V$.
I read that from Witt's theorem we can deduce some facts about the maximal isotropic subspaces of $V$ and the isometry group of $V$. But I don't see why. So, here are my three question:


*

*Why does Witt's theorem imply that every maximal totally isotropic subspace of $V$ has the same dimension, which defines the rank of a polar space?

*Why does Witt's theorem imply that every totally isotropic subspace lies in a maximal totally isotropic subspace?

*Why does Witt's theorem imply that the isometry group of $V$ acts transitively on every set consisting of totally isotropic subspaces of the same dimension?

 A: The main thing to note in order to prove all of the statements you mentioned is the following:
Lemma. Let $U_1,U_2\subseteq V$ be totally isotropic subspaces, and $T:U_1\to U_2$ a bijective linear map. Then $T$ is an isometry. 
The proof of the lemma is easy, since for any $u,w\in U_1$ we have that 
$$0=(u,w)=(Tu,Tw)=0,$$
where the first equality holds since $U_1$ is totally isotropic, and the last one since  $U_2$ is.
Now, (3) follows, since any two subspaces of the same dimension are isomorphic, as vector spaces, which, by the lemma, is an isometry of these two space, and extends to an isometry of $V$. Thus, given two totally isometric subspaces of the same dimension, we have an isometry of $V$ mapping one to onto the other.
(1) follows from the lemma as well. For, if we assume towards contradiction that $U_1,U_2\subseteq V$ are maximally isotropic subspaces and $\dim U_1<\dim U_2$. Then, by linear algebra, we have an injective map $T:U_1\to U_2$ which, by the Lemma and Witt's theorem, extends to an isometry of $U_1$ onto a proper subspaces of $U_2$, both of which are maximally isotropic. A contradiction.
Finally, to prove (2), by the same argument as (1), it suffices to prove that some maximally isotropic subspace exists. For, given such a subspace $U$, and any totally isotropic subspace $W\subseteq V$, we may find an isometry $S:V\to V$ such that $SW\subseteq V$, and hence $W$ is included in the maximally isotropic subspace $S^{-1}V$. The existence of such a subspace follows, for example, from the existence of Darboux bases.
