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!
No comments:
Post a Comment