# Orbits of a symplectic group on its Lie algebra in the finite field case

The classical problem regarding the action of symplectic group on its Lie algebra gives rise to the following question in the finite field case.

Let $\mathbb F_p$ be a finite field. Then the symplectic group over $\mathbb F_p$ acts by conjugation on the set of matrices over $\mathbb F_p$ that satisfy $\Omega A + A^t \Omega = 0$, $\Omega$ is the skew symmetric matrix

$$\begin{pmatrix} 0 & I \\\\ -I & 0 \end{pmatrix}$$

where $I$ is identity matrix. What are the orbits of this action?

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This problem is answered in a paper by Burgoyne and Cushman. I don't have the reference to hand.

This also came up in Classification of adjoint orbits for orthogonal and symplectic Lie algebras?

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Thanks for the reference. – Pooja Jul 11 '10 at 13:00

It won't tell you everything about the orbits, but the decomposition of $\mathfrak{sp}_{2n}\textbf{F}_p$ as an $\text{Sp}_{2n}\textbf{F}_p$-representation is known. This can be found in Hogeweij, "Almost-classical Lie algebras. I." Nederl. Akad. Wetensch. Indag. Math. 44 (1982), no. 4, 441-452, but it is hard to extract the answer from that paper, so I'll briefly give the argument.

If $p$ is odd, then $\mathfrak{sp}_{2n}\textbf{F}_p$ is irreducible with highest weight $2\omega_1$, where $\omega_1,\ldots,\omega_n$ are the fundamental weights for $\text{Sp}_{2n}\textbf{F}_p$.

If $p=2$, we proceed as follows. Let $H\approx \textbf{F}_p^{2n}$ be the standard representation of $\text{Sp}_{2n}\textbf{F}_p$. Note that as a subspace of $\mathfrak{gl}_{2n}\textbf{F}_p\approx H^*\otimes H$, the condition defining $\mathfrak{sp}_{2n}\textbf{F}_p$ describes exactly the subspace $\text{Sym}^2 H$ inside $H\otimes H\approx H^*\otimes H$, so we are looking at orbits of $\text{Sp}_{2n}\textbf{F}_p$ on $\text{Sym}^2 H$.

In characteristic 2 we have an embedding of $H$ into $\text{Sym}^2 H$ by $x\mapsto x\cdot x$, which is linear since $(x+y)^2 = x^2+y^2$ (in general this twists by Frobenius but we are over $\textbf{F}_2$). Since $x\cdot y=y\cdot x=-y\cdot x$, the quotient $\text{Sym}^2 H/H$ is isomorphic to $\bigwedge^2 H$. Now $\bigwedge^2 H$ has two invariant subrepresentations. One is trivial, spanned by the vector $\omega=a_1\wedge b_1+\cdots+a_n\wedge b_n$. The other is the kernel $K$ of the contraction $\bigwedge^2 H\to \textbf{F}_2$, defined by $a_i\wedge b_i\mapsto 1$, $a_i\wedge a_j\mapsto 0$, $b_i\wedge b_j\mapsto 0$, and $a_i\wedge b_j\mapsto 0$. Note that under this contraction $\omega$ is taken to $n\in \textbf{F}_2$; thus $\omega$ is contained in $K$ iff $n$ is even. Finally, $K$ is irreducible when $n$ is odd, and $K/\langle\omega\rangle$ is irreducible when $n$ is even.

If I'm not mistaken, this means the invariant subrepresentations are thus just $H$, $\langle\omega\rangle$, $H\oplus \langle\omega\rangle$, and $H+K$ (the kernel of the contraction $\text{Sym}^2 H\to \textbf{F}_2$).

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