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Let $(V,q)$ be a non-degenerate quadratic space. Then we know that for any $d$ with $0 \leq 2d \leq \dim V$, the group $O(q)$ of isometries of $(V,q)$ acts transitively on the set of totally isotropic subspaces of $V$ of dimension equal to $d$. (This is a consequence of a theorem of Witt.)

Is the same statement always true (maybe under some additional conditions) for $SO(q)$?

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    $\begingroup$ This is true if $\dim V$ is odd: take your map from $O(q)$ and compose with minus identity if necessary. $\endgroup$
    – algori
    Commented May 31, 2011 at 1:20
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    $\begingroup$ Jose -- at first I agreed with this, but then I've realized that this map does not preserve the quadratic form $xy$. I've also just deleted an answer which purported to prove the statement by induction on the dimension of $V$. The "proof" started by saying that the statement in the case of dimension $\leq 2$ is clear, while in fact it is false in general: take the form $xy$. It has two isotropic lines. A linear map that swaps the lines has the form $(x,y)\mapsto (ay,bx)$. In order for this to preserve the form we must have $ab=1$. However, the determinant is $-ab$. $\endgroup$
    – algori
    Commented May 31, 2011 at 4:46
  • $\begingroup$ algori: you're right, of course. I've deleted my comment. $\endgroup$
    – José Figueroa-O'Farrill
    Commented May 31, 2011 at 11:05

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If the dimension of $V$ is even, $2n$, then there are two families of totally isotropic subspaces of dimension $n$ (if the form is split otherwise there may be no such subspace at all). Two such subspaces $W$ and $W'$ belong to the same family precisely when the parity of $\dim W\cap W'$ is the same as that of $n$. The two families are stable under $\mathrm{SO}(q)$ (and permuted by the rest of $\mathrm O(q)$). There certainly is an elementary proof of this fact yet I think that the "real" reason comes from algebraic geometry. The two families are disjoin algebraic subvarieties of the Grassmannian of $n$-spaces in $V$ and $\mathrm{SO}(q)$ is connected and hence must preserve them.

In any case when $n=1$ everything is very elementary; we can choose an isotropic basis $e_1,e_2$ with $e_1\cdot e_2=1$ and then $\mathrm{SO}(q)$ consists of the diagonal matrices of determinant $1$ which hence fixes the two isotropic subspaces (spanned by $e_1$ and $e_2$ respectively) while the orthogonal matrices of determinant $-1$ permute them.

In all other case $\mathrm{SO}(q)$ acts transitively on the totally isotropic spaces of fixed dimension (which is easily proven).

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