# Finite subgroups of the symplectic group that don't act on simple principally polarized abelian varieties

I'm sorry if this question is awkwardly phrased -- I'm very much an amateur at algebraic geometry, but this question came up in my research.

Here goes. Let $\mathbb{H}^g$ be the genus $g$ Siegel upper half plane. Thus the symplectic group $Sp_{2g}(\mathbb{Z})$ acts on $\mathbb{H}^g$ with quotient the moduli space of principally polarized abelian varieties (PPAV's). Let $\Gamma < Sp_{2g}(\mathbb{Z})$ be a finite subgroup. Define $X(\Gamma) \subset \mathbb{H}^g$ to be the set of all $p$ such that $\gamma(p)=p$ for all $\gamma \in \Gamma$. The set $X(\Gamma)$ is then an analytic subvariety and $\Gamma$ acts as a group of automorphisms of the PPAV's corresponding to the points of $X(\Gamma)$.

Recall that a PPAV is simple if it doesn't split as a nontrivial direct sum of PPAV's. I'm interested in finite subgroups $\Gamma < Sp_{2g}(\mathbb{Z})$ such that none of the PPAV's corresponding to points of $X(\Gamma)$ are simple. A dream would be to have a classification of such subgroups, but even interesting examples would be very helpful.

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If the subalgebra of $M_{2g \times 2g}(\mathbb{Q})$ generated by $\Gamma$ is not a division ring then $\Gamma$ cannot act on a simple abelian variety. A wild conjecture might be that this is a necessary and sufficient condition. –  ulrich Apr 28 '10 at 6:20
Notational comment: I would always use "simple" to mean "not isogenous to a nontrivial direct sum of PPAVs," which is a stronger condition than "doesn't split as a nontrivial direct sum of PPAVs." –  JSE Apr 28 '10 at 16:42
As for the question, I think unknown google's suggestion is a good one. I would go look at Mumford's book on abelian varieties where he classifies the possible algebras that can arise as End(A) tensor Q for A a simple PPAV. (Don't have the book in front of me, so don't know the page.) Then one can start thinking about what finite subgroups these algebras have...? –  JSE Apr 28 '10 at 16:43
Do you want the splitting to be equivariant? uknown's criterion is heading in that direction. The condition for an action on an equivariantly simple (assuming JSE's notion of simple) is that only one or two types of irrep appear, a self-dual one or a dual pair. But if you as $S_3$ to act on g=2 with 2 copies of the 2d irrep, then I think it has to be (isogenous to) a product of isomorphic elliptic curves. –  Ben Wieland Apr 28 '10 at 20:17
@JSE : That notion of simplicity is fine by me. @Ben : I don't require the splitting to be equivariant. In fact, I find nonequivariant splittings to be even more interesting. –  Andy Putman Apr 28 '10 at 23:23