# The number of orbits of a two-point stabilizer of the symplectic group $Sp(2m,2)$

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.