2
$\begingroup$

I am trying to figure out the number of orbits of a two-point stabilizer of the action of $Sp(2m,2)$ on its two orbits $\Omega^+$ and $\Omega^-$ as detailed in Dixon and Mortimer's "Permutation Groups", the paper https://doi.org/10.1016/S0021-8693(02)00551-3, as well as this question https://mathoverflow.net/a/85094/172098. In the last one, Derek Holt says that the two-point stabilizers will have four orbits, with the two nontrivial ones given by the vectors orthogonal and not orthogonal to the fixed isotropic point. However, I cannot seem to figure out how to show this.

I am new to the area, but here is my (possibly mistaken) understanding so far. Let us fix a symplectic form $b$ on $V:=(GF(2))^{2m}$. The group $Sp(2m,2)\leq GL(2m,2)$ which preserves $b$ acts on the set $\Omega$ of quadratic forms on $V$ and has two orbits: $\Omega^+$ and $\Omega^-$. These are respectively the orbits of the forms $q^+=\sum_i^m x_i y_i$ and $q^-=x_1^2 + y_1 ^2 + \sum_i^m x_i y_i$. Take any quadratic form $q$ and for now assume that $q\in \Omega^+$. Then the subgroup $O(V,q)$ of $Sp(2m,2)$ which fixes $q$ can be shown (by Witt's Lemma) to act transitively on $\Omega^+\setminus\lbrace{q\rbrace}$. This can be done because there is a natural bijection from $V$ to $\Omega$ such that the actions of $O(V,q)$ on each will be isomorphic. In particular, $\Omega^+$ will correspond to the isotropic vectors in $V$ with respect to $q$ and $\Omega^-$ will correspond to the anisotropic vectors.

Thus, to show the desired goal that a two-point stabilizer of the action of $Sp(2m,2)$ on $\Omega^+$ will have four orbits, it suffices to pick a nonzero isotropic vector $v\in V$ with respect to a bilinear form $q\in \Omega^+$ and show that its stabilizer $O(V,q)_v$ on $O(V,q)$ will have two orbits on $V\setminus \lbrace 0,v\rbrace$. Derek Holt's claim was that these two orbits are precisely those that are orthogonal and not orthogonal to $v$ (with respect to the bilinear form $b$), and this is what I cannot show. Of course, it is clear that if $w$ is orthogonal to $v$, then so is any member of its orbit in $O(V,q)_v$; an analogous statement holds for when $w$ is not orthogonal to $v$. However, I am unable to show that all isotropic vectors (distinct from 0 and $v$) that are orthogonal (resp. not orthogonal) to $v$ will necessarily be in the same orbit of $O(V,q)_v$. A completion of this idea, an alternative approach, a reference confirming this result, and any corrections to the above will be greatly appreciated.

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy