What is the number of all skew symmetric bilinear forms on $m$ dimensional space $V$ over $\mathbb{F}_q$ with rank equal to $2r$?

Question

Let $V$ be a vector space of dimension $m$ over the finite field $\mathbb{F}_q.$ Then I want to find the number of all skew symmetric bilinear forms on $V$ with rank equal to $2r$ ($0 \leq 2r \leq m$).

I need some help. Thanks in advance.

Answer 1

I'll assume that $q \neq 2$ so that skew-symmetric forms are alternating. Let

$$ S := \{ \beta \colon V \times V \rightarrow \mathbb{F}_q \, | \, \beta \text{ is alternating and has rank } 2r \}. $$

We want to find $|S|$. Note that $G = \operatorname{GL}(V)$ acts on $S$ from the right by

$$ (\beta \cdot A)(u,v) := \beta(Au, Av) $$

and this action is transitive because any two alternating forms with the same rank are congruent. Choose some $\beta_0 \in S$ and let $G_{\beta_0} = \{ g \in G \, | \, \beta_0 \cdot g = \beta_0 \}$ be the stabilizer of $\beta_0$. Since the action is transitive, we have

$$ |S| = \frac{|G|}{|G_{\beta_0}|} $$

so we are left with calculating the size of the stabilizer $G_{\beta_0}$. Since it doesn't really matter with which $\beta_0 \in S$ we work, assume that $V = \mathbb{F}_q^m$ and that $\beta_0$ is represented with respect to the standard basis by the matrix

$$ \Omega_{r,m} := \begin{pmatrix} 0_{r \times r} & I_{r \times r} & 0_{r \times (m - 2r)} \\ -I_{r \times r} & 0_{r \times r} & 0_{r \times (m - 2r)} \\ 0_{(m - 2r) \times r} & 0_{(m - 2r) \times r} & 0_{(m - 2r) \times (m - 2r)} \end{pmatrix} = \begin{pmatrix} \Omega_{r} & 0_{2r \times (m - r)} \\ 0_{(m - r) \times 2r} & 0_{(m - r) \times (m - r)} \end{pmatrix} $$

where $$\Omega_r = \begin{pmatrix} 0 & I_{r \times r} \\ -I_{r \times r} & 0 \end{pmatrix} $$ is the matrix representing the standard non-degenerate symplectic form on a $2r$-dimensional space. Let $X \in \operatorname{GL}(m,q)$ and decompose it as a block matrix

$$ X = \begin{pmatrix} A_{2r \times 2r} & B_{2r \times (m - r)} \\ C_{(m-2r) \times 2r} & D_{(m - 2r) \times (m - 2r)} \end{pmatrix}. $$

Then $X \in G_{\beta_0}$ iff $X \Omega_{r,m} X^T = \Omega_{r,m}$. Expanding everything in blocks, we get

$$ X \Omega_{r,m} X^T = \begin{pmatrix} A \Omega_r A^T & A \Omega_r C^T \\ C \Omega_r A^T & C \Omega_r C^T \end{pmatrix} = \begin{pmatrix} \Omega_{r} & 0 \\ 0 & 0 \end{pmatrix}. $$

The equation $A \Omega_r A^T = \Omega_r$ implies that $A$ is symplectic and in particular is of full rank. Then the equation $A \Omega_r C^T = 0$ implies that $C = 0$ (because $A \Omega_r$ has full rank). Hence,

$$ G_{\beta_0} = \left \{ \begin{pmatrix} A & B \\ 0 & D \end{pmatrix} \, \big| \, A \in \operatorname{Sp}(2r,q), \, B \in M_{(2r) \times (m - 2r)}(\mathbb{F}_q), \,D \in \operatorname{GL}(2m - r, q) \right \} $$

and so

$$ |S| = \frac{|\operatorname{GL}(m,q)|}{|\operatorname{Sp}(2r,q)| |M_{(2r) \times (m - 2r)}(\mathbb{F}_q)||\operatorname{GL}(m - 2r, q)|}= \frac{(q^m - 1) \cdots (q^{m - 2r + 1} - 1)}{(q^2 - 1) \cdots (q^{2r} - 1) q^{r - r^2}}.$$

For a sanity check, when $m = 3$ and $r = 1$ this formula gives us $q^3 - 1$ which is indeed the number of antisymmetric matrices

$$ \begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix} $$

of rank two.