# Is the Galois x Hecke action on cohomology of Shimura varieties semi-simple?

Given a reductive group $G/\mathbf Q$ (+ additional data), and a compact open subgroup $K\subset G(\mathbf A^\infty)$, there is a standard construction that produces a Shimura variety $S$ and if we choose a rational representation $\xi : G\to Aut(V)$, we obtain a certain $l$-adic local system $\mathcal L$ on $S$. Moreover, all of this can be done over some fixed number field $E$.

Question 1: If $S$ is proper, is it true that the $Gal(\overline E/E)\times H(G,K)$ action on $H^*(S\otimes\overline E, L)$ is semisimple? (where $H(G,K)$ is the corresponding Hecke algebra).

Question 2: What happens if $S$ is not proper? Is the action semi-simple at least if we restrict to parabolic cohomology (the image of compactly supported cohomology inside usual cohomology)?

References for places where this is discussed would also be great.

Thanks

• I don't know the answer, but for the non-proper case, it is the intersection cohomology (of min compactification) that people are interested in: this is the one that is related to §L^2§-cohom and hence Lie algebra cohom, and therefore automorphic rep. via trace formula. Jun 8, 2011 at 12:18

To get a Shimura variety, the reductive group $G$ should satisfy some axioms. In fact, you should begin not just with $G$, but with a Shimura datum for $G$.
Leaving that aside, the Hecke action will be semi-simple (if we omit Hecke operators at primes dividing the level); more generally, one could take the limit over all levels, and then get a $G(\mathbb A^{\infty})$-action, which will be semi-simple (e.g. by comparison with automorphic forms; and here I am supposing $S$ compact for the moment).
Each $G(\mathbb A^{\infty})$-rep. will appear with some multiplicity, and this multiplicity space carries the $Gal(\overline{E}/E)$-action. If this action is irreducible, which is the case e.g. for modular curves (admittedly non-compact, but ignore that for the moment!) or Shimura curves, then it is certainly semi-simple. But in more general situations it need not be irreducible, and then its semi-simplicity is a certain case of the Tate conjecture, and I'm pretty sure that nothing will be known about it.