From symmetric groups to symplectic groups?

Question

The question is self-contained finite group theory but the motivation requires more background.

The finite groups I am interested in are the groups $Sp(n,F_3)$. For $n$ even these are the usual symplectic groups over the field with three elements. For $n$ odd these are the odd symplectic groups. These are a semi-direct product of a symplectic group with a Heisenberg group and we have a sequence of subgroups $Sp(n,F_3)\rightarrow Sp(n+1,F_3)$.

Consider one of these inclusions and look at induction/restriction of irreducible complex representations. My question is: take an irreducible representation of the subgroup and induce. Is this representation a direct sum of pair-wise non-isomorphic irreducible representations (sometimes this is called multiplicity free)?

I can prove this for $n$ even because the irreducible representations of $Sp(2m+1,F_3)$ can be constructed using Clifford theory (known to physicists as Mackey theory). I can use the computer for small $n$. So my question is really for $n$ odd.

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I can give more information which hints at my interest. Define a sequence of groups $G(n)$ by a sequence of finite presentations so that we have surjective homomorphisms $B(n)\rightarrow G(n)$ where $B(n)$ is the usual braid group. Take generators $\sigma_1,\ldots ,\sigma_{n-1}$ and the Artin relations. In addition take $\sigma_i^3$ and for $n\ge 5$ take $(\sigma_1\sigma_2\sigma_3\sigma_4)^{10}=1$. Then we have $G(n)=Sp(n-1,F_3)$.

Then we also have $G(n)\times G(m)\rightarrow G(n+m)$ compatible with $B(n)\times B(m)\rightarrow B(n+m)$. This is similar to the symmetric groups.

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I am aware that this question can be generalised. I have deliberately restricted to a simple example as I would like to have one case fully worked out before generalising. If you have a proof of the above and your proof generalises, that's different!

Answer 1

Without working it out completely, I can't give a yes or no answer.

But here's a start: Let $G = Sp(2n)$ (a group of $2n$ by $2n$ matrices, in the notation I use) and let $K = Sp(2n-2) \ltimes H$, where $H$ is the appropriate Heisenberg group. Let's work over any finite field of odd characteristic -- I'd guess that if what you want is true, then it's true in this generality.

You wish to show (by Frobenius reciprocity) that restriction from $G$ to $K$ is multiplicity-free. A standard way to accomplish this would be to prove the following:

Claim: The convolution ring $A = C[K \backslash G / K]$ of $K$-bi-invariant functions on $G$ is commutative.

In other words, try to prove that $(G,K)$ is a Gelfand pair. The standard method involves accomplishing the following:

Task: Find an anti-involution $\sigma$ of $G$ (meaning $\sigma^2 = Id$ and $\sigma(gh) = \sigma(h) \sigma(g)$, such that every $K$-double-coset $K g K$ in $G$ is stable under $\sigma$, i.e., if $\sigma(g) \in K g K$ for all $g \in G$.

Such an involution yields an anti-automorphism of $A$, making it commutative -- this is the "Gelfand-Kazhdan method".

My advice: try something like conjugation by a matrix (the matrix $J$ defining the symplectic form, perhaps) followed by the transpose, for the anti-involution. It's up to you to analyze the double cosets, but I bet it's been done, at least in low rank (2n = 4, perhaps). The double cosets can get unwieldy, but in your case it almost suffices to analyze the double cosets for the "Heisenerg parabolic" $P$ containing $K$. The $P$ double cosets in $G$ can be analyzed via the Weyl group.

Final advice: the "odd symplectic groups" that you refer to are often called Jacobi groups in the literature due to their relevance to Jacobi forms. Someone may have worked out some of this already!