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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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2 Answers 2

up vote 3 down vote accepted

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
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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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