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Let $X$ be a smooth projective variety over a field $k$. Then if $\ell\neq \text{char} k$, $k$ is finite, and $X$ is an abelian variety it was shown by Weil that the $\ell$-adic cohomology of $X_{k^{sep}}$ is a semisimple representation of $G_k$. The conjectural reason for this, if I've understood this correctly, is that the $\ell$-adic representation gives an equivalence of the category of "geometric" Galois representations to the category of pure motives (see for instance), and we furthermore conjecture that the later category is semisimple.

What I would be curious about is the "bad coefficient" case, that is to say do we expect $H^i(X_{k^{sep}},\mathbb{F}_p)$ to be semisimple as $G_k$ representation? By this argument, we know that the $p$-adic Tate module gives a semisimple representation, but, unless I'm mistaken, we would need an additional argument, as the minimal polynomial might become divisible by a square mod $p$. I suspect this might be a way of providing a counter-example for the semisimplicity of $H^1(A_{k^{sep}},\mathbb{F}_p)$, but I'm stuck at that point.

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    $\begingroup$ Even if $\ell\neq \mathrm{char} k$, there is no reason that $H^i(X_{k^{\mathrm{sep}}},\mathbb F_\ell)$ is semisimple, because of the reduction modulo $\ell$ issue. In fact, I think this representation can be pretty arbitrary. The same ought to happen mod $p$. $\endgroup$ Mar 14 at 21:13
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Take an elliptic curve $E$ with a (single) rational $2$-torsion point, say $y^2 = x(x^2+1)$ over $\mathbb{Q}$ (or even $\mathbb{R}$). Then the Galois action on $E[2]$ factors through a group of order $2$ and so is not semisimple.

An easier example is the variety $x^2+1=0$ and $H^0(X,\mathbb{F}_2)$.

If you insist that the characteristic of the field equals $p$ you could take $x^2+x+1=0$ over $\mathbb{F}_2$ and look at $H^0(X,\mathbb{F}_2)$.

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