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Let $G$ be the reductive group $\operatorname{GSp}_{4}$. Let $\pi$ be a smooth admissible cuspidal representation of $\operatorname{GSp}_{4}(\mathbb{A}^{(\infty)})$ of dominant weight. Assume, for caution, that $\pi$ satisfies a multiplicity one hypothesis.

Fix $p$ an odd prime. To $\pi$ is attached a $p$-adic representation $\rho$ of the absolute Galois group of $\mathbb{Q}$ unramified outside a finite set of finite places and such that the characteristic polynomial of the Frobenius morphisms $Fr\ell$ for $\ell$ outside this set coincides with the Euler factor at $\ell$ of the degree 4 $L$-function of $\pi$. This Galois representation occurs in the degree 3 cohomology of the étale cohomology of a Siegel-Shimura variety.

The image of complex conjugation under $\rho$ is semi-simple so can be chosen to be diagonal with eigenvalues 1 and -1. How many $-1$ are there?

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I once went to a talk by Tilouine where he discussed some similar questions. You might want to look up some of his work. –  ulrich Apr 15 '10 at 12:38
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up vote 4 down vote accepted

The preprint "Conjecture de type de Serre et formes compagnons pour $GSp_4$" by Florian Herzig and Tilouine (available here) indicates that there should be two 1's and two -1's.

EDIT: Having come in to work today, I could look up the more precise statement given in chapter 9 of Tilouine's book "Deformations of Galois representations and Hecke algebras" (incorrectly referenced in the aformentioned article of Herzig–Tilouine). The statement is the following: Let $G=GSp(4)$, let $\pi$ be a regular algebraic cuspidal automorphic representation of $G(\mathbf{A}_F)$, where $\mathbf{A}_F$ is the adele ring of a number field $F$. Let $\bar{\rho}$ be the associated mod $p$ Galois representation. Let $v$ be a real place of $F$. Then the $G$-conjugacy class of ${\bar\rho}(c_v)$ (where $c_v$ is a complex conjugation at $v$) contains the matrix $\text{diag}(1,1,-1,-1)$.

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Thanks a lot! I will wait a bit more see if someone knows something unconditional (and do some research myself). If nothing new comes out of it, I'll accept your answer. –  Olivier Apr 16 '10 at 8:48
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