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It is conjectured that the action of the Frobenius acting on the etale cohomology of an algebraic variety over a finite field is semisimple. A first approximation of my question is :

What is the importance of this statement ?

It is only an approximation because this question is rather unprecise. More precisely, I would like to know two different things :

1) Is there some conjectural statements which are implied by the semisimplicity conjecture ?

More generally :

2)What the semisimplicity property means ?

I ask for an arithmetic or geometric interpretation of the semisimplicity of the Frobenius. For example, the eigenvalues of the Frobenius are important because they control the number of points of the variety over finite fields. I have the impress that often people only care of eigenvalues of the Frobenius (often there are remarks of the form : "if the representation is not semisimple, take the semisimplification"). Is there some situation where the conjectural semisimplicity would be an important thing ?

Remark : I don't want an answer like "it is important because it is implied by the standard conjectures and so if it would be true it would be a confirmation of a part of a general philosophy ..." I would like some "concrete" example of the use of the semisimplicity property (maybe in the cases where it is actually known : abelian varieties ...)

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    $\begingroup$ The question applies more generally: after knowing the characteristic polynomial of a matrix, why it is important to know its Jordan form? I guess knowing how the space decomposes into generalized eigenspaces will usually be helpful, in various situations. In the case of etale cohomology, as Yuri said, if you know there are many eigenvectors for an eigenvalue, then (modulo Tate conjecture) it will lead to many algebraic cycles, which is usually a difficult thing in alg. geom. This also applies to K3 (a recent result?). $\endgroup$
    – shenghao
    Sep 2, 2012 at 13:53

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Here is a conjectural statement that follows from the combination of the semisimplicity conjecture and the Tate conjecture on algebraic cycles.

Let $X$ and $Y$ be geometrically irreducible smooth projective varieties over a finite field $F$. Suppose that $X$ and $Y$ have the same number of $F^{\prime}$-points for all finite overfields $F^{\prime}$ of $F$. Then there exist a finite overfield $F_0$ of $F$ and a geometrically irreducible closed $F_0$-subvariety $Z\subset X \times Y$ such that $\dim(X)=\dim(Y)=\dim(Z)$ and both projections map $Z \to X$ and $Z \to Y$ are surjective.

The semisimplicity is known to be true for abelian varieties (Weil). Combining the semisimplicity with Tate's theorem on homomorphisms, one may deduce that if $X$ and $Y$ are abelian varieties with the same number of $F^{\prime}$-points (for all $F^{\prime}$) then they are isogenous over $F$.

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  • $\begingroup$ Do you mean $Z\to X$ and $Z\to Y$ are also finite or something else ? Otherwise $Z=X\times Y$ is always surjective on $X$ and $Y$. $\endgroup$
    – Qing Liu
    Sep 1, 2012 at 21:11
  • $\begingroup$ Thanks for pointing out: I also claim (please see the corrected version) that $\dim(Z)=\dim(X)=\dim(Y)$. $\endgroup$
    – Yuri Zarhin
    Sep 1, 2012 at 21:20

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