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?


2 Answers 2


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?


The subspace in question is $\mathfrak{sp}_{2n}\textbf{F}_p$; explicitly, it consists of matrices of the form $\begin{pmatrix}A&B\\C&-A^\top\end{pmatrix}$ where $B=B^{\top}$ and $C=C^{\top}$.

Although it won't tell you everything about the orbits, the description 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. [Edit: I am no longer sure how to extract the argument below from that paper, but I believe everything below is self-contained and can be directly verified except the assertions that various representations are irreducible.]

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 $H\approx \textbf{F}_2^{2n}$ be the standard representation of $\text{Sp}_{2n}\textbf{F}_2$, so the explicit description of $\mathfrak{sp}_{2n}\textbf{F}_2$ above identifies it with a subspace of $\mathfrak{gl}_{2n}\textbf{F}_2\approx H^*\otimes H$. The symplectic form gives an isomorphism $H^*\cong H$ so we can identify $\mathfrak{sp}_{2n}\textbf{F}_2$ with a subspace of $H\otimes H$. The corresponding subspace is precisely the invariant subspace $\Gamma^2 H=(H\otimes H)^{\mathbb{Z}/2}$; so in particular the original question is asking about the orbits of $\text{Sp}_{2n}\textbf{F}_2$ on $\mathfrak{sp}_{2n}\textbf{F}_2\cong\Gamma^2 H$. [I am grateful to Andy Putman for pointing out a mistake in the earlier version of this answer, where I mistakenly had $\text{Sym}^2 H$ instead of $\Gamma^2 H$.]

First, let us discuss the structure of $\Gamma^2 V$ as a $\text{GL}(V)$ representation in characteristic 2, before specializing to $\text{Sp}_{2n}\textbf{F}_2$. In characteristic 2 we have short exact sequences of $\text{GL}(V)$-representations: $$0\to \textstyle \bigwedge^2 V\to \bigotimes^2 V\to \text{Sym}^2 V\to 0$$ $$\textstyle 0\to \Gamma^2 V\to \bigotimes^2 V\to \bigwedge^2 V\to 0$$ and also $$\textstyle0\to V(1)\to \text{Sym}^2 V\to \bigwedge^2 V\to 0$$ $$\textstyle0\to \bigwedge^2 V\to \Gamma^2 V\to V(1)\to 0$$ (Our interest will be in the last one.) Note that here $V(1)$ denotes the "Frobenius twist" of $V$, which is the same vector space as $V$ but with the $\text{GL}(V)$ action twisted by Frobenius (so if $V$ is a vector space over $\mathbf{F}_2$ itself, there is no difference). The embedding $V(1)\to \text{Sym}^2 V$ sends $x\mapsto x\cdot x$, which is linear since $(x+y)^2 = x^2+y^2$ (but note that it does not commute with the action of e.g. scalar matrices in $\text{GL}(V)$, which is why the domain must be $V(1)$ instead of just $V$).

Returning to $\text{Sp}_{2n}\textbf{F}_2$, we have an exact sequence $0\to \bigwedge^2 H\to \Gamma^2 H\to H\to 0$. Note that in terms of our description $\mathfrak{sp}_{2n}\textbf{F}_2=\begin{pmatrix}A&B=B^{\top}\\C=C^{\top}&-A^\top\end{pmatrix}$, the subspace $\bigwedge^2 H$ corresponds to those matrices for which both $B$ and $C$ have all diagonal entries equal to 0. So we need to consider $\bigwedge^2 H$ as an $\mathfrak{sp}_{2n}\textbf{F}_2$-representation. This has two invariant subrepresentations.

First, there is a trivial representation spanned by the vector $\omega=a_1\wedge b_1+\cdots+a_n\wedge b_n$ (in our matrix representation, this is the matrix $\begin{pmatrix}I&0\\0&I\end{pmatrix}$).

Second, there is the kernel $K$ of the contraction $c\colon \bigwedge^2 H\to \textbf{F}_2$, defined by $a_i\wedge b_i\mapsto 1$, $a_i\wedge a_j\mapsto 0$, $b_i\wedge b_j\mapsto 0$, and $a_i\wedge b_j\mapsto 0$. (In terms of matrices, this contraction sends $\begin{pmatrix}A&B\\C&-A^\top\end{pmatrix}\mapsto \text{tr} A$, so $K$ is the subspace where $A$ has even trace and $B$ and $C$ have no diagonal entries.)

Note that under the contraction $c$ the invariant element $\omega$ is taken to $n\in \textbf{F}_2$. Therefore when $n$ is odd the element $\omega$ provides a section of $c$, giving a direct sum decomposition $\bigwedge^2 H\cong \textbf{F}_2\oplus K$. In this case $K$ is irreducible.

When $n$ is even, on the other hand, we have $\langle\omega\rangle\subset K\subset \bigwedge^2 H$; in this case $K/\langle\omega\rangle$ is irreducible.

If I'm not mistaken, the invariant subrepresentations of $\mathfrak{sp}_{2n}\textbf{F}_2$ are (regardless of the parity of $n$) $\langle\omega\rangle$, $K$, and $\bigwedge^2 H$. (There could possibly be some that project onto $H$, but I believe/suspect any such subrepresentation must be everything.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.