Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $SO(q)$?

share|improve this question
1  
This is true if $\dim V$ is odd: take your map from $O(q)$ and compose with minus identity if necessary. –  algori May 31 '11 at 1:20
2  
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 $-ab$. –  algori May 31 '11 at 4:46
    
algori: you're right, of course. I've deleted my comment. –  José Figueroa-O'Farrill May 31 '11 at 11:05

1 Answer 1

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).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.