Let $X_0$ be a variety defined over a finite field of characteristic $p \neq l$. Is it true, that the action of the frobenius on the ladic cohomology $H_l^*(X)$ is semisimple (say for smooth $X_0$)? If not, what would be a counterexample?

Let $X$ be a smooth projective variety over a finite field $\mathbb{F}_{q}$ of caracteristic $p$ and let $l$ be a prime number different from p. We consider the following statement : (A) The action of the Frobenius on the etale cohomology $H^{i}_{et}( X_{\overline{\mathbb{F}}_{q}}, \mathbb{Q}_{l})$ is semisimple. How to suppress the projective hypothesis is the subject of the mathoverflow question link text (A) is true in the following cases : 1) $X=A$ an abelian variety (and so for $X$ a curve via the jacobian). As mentionned in comment by Emerton, it is a consequence of the Weil's work on the Riemann hypothesis in this case. Fix a polarization on $A$. For $x$ an endomorphism of $X$ which gives an endomorphim on $H^{1}_{et}$, we can define an endomorphism $x'$ (' : "Rosati involution") by $x' = *x^{T}* $ where in the middle we have the transposition with respect to the intersection product and * comes from the duality theory of abelian varieties ( the polarisation gives a identification between $H^{1}_{et}(A)$ and $H^{1}_{et}(\check{A})$). Weil proved that $Tr(xx')>0$ if $x$ is nonzero. Let $F$ be the (geometric) Frobenius. For $x = q^{1/2}F$, we have x'=$x^{1}$. So $Tr(aa')$ is a definite positive bilinear form on the $\mathbb{Q}$ algebra generated by x and is preserved by multiplication by $x$ : so multiplication by $x$ is unitary which shows that $x$ is semisimple (and eigenvalues of modulus one gives the Riemann hypothesis). 2) $X$ a K3 surface. As mentionned in comment by shenghao, it is a consequence of the work of Deligne : link text The result is deduced from the case of abelian varieties via the KugaSatake construction (of course there is a nontrivial thing to do because KugaSatake construction is a priori of transcendental nature but Deligne did it). For $X$ general, (A) is conjectured. It is a consequence of standard conjectures. More precisely, things should work as in the case of abelian varieties. We can still define $x \mapsto x'$ at the cohomological level but $Tr(xx')>0$ is conjectural : a standard conjecture of Lefschetz type implies $x'$ algebraic if $x$ is, which permits to use a trace formula expressing $Tr(xx')$ as an intersection product. The positivity should then be a consequence of a standard conjecture of Hodge type. For more details, as mentionned in comment by Damian Rössler, see Kleiman "The standard conjectures" (whose some details depend on Kleiman, "Algebraic cycles and the Weil conjectures"). 

