This question is inspired by the question: Example of non-projective variety with non-semisimple Frobenius action on etale cohomology?

Let $K$ be a number field (or finitely generated field of characteristic $0$). If I am not mistaken, it is not expected for general varieties to have semisimple Galois representations on their $\ell$-adic cohomology.

Is there a known example of a variety $X/K$ such that the Galois representation on the $\ell$-adic cohomology is not semisimple?

If $X$ is not proper, I guess we should work with cohomology with compact support.

Two subquestions:

  • Are there examples where $X$ is smooth?
  • Are there examples where $X$ is proper?

1 Answer 1


Here's an example, if I'm not mistaken. Let $E / K$ be an elliptic curve and $x \in E$ a non-torsion $K$-point. Then the image of the divisor $\{x\} - \{\infty\}$ under the etale cycle class map is a nontrivial class in $H^1(K, H^1(E_{\bar K}, \mathbf{Q}_\ell)(1))$ and thus corresponds to a non-split extension of $H^1(E_{\bar K}, \mathbf{Q}_\ell)(1)$ by $\mathbf{Q}_\ell$; and this non-semisimple Galois representation is realized in the etale cohomology of the (smooth, non-proper) variety $E - \{x, \infty\}$.

EDIT: There are two ways of seeing that we obtain a class in $H^1(K, H^1(E_{\bar K}, \mathbf{Q}_\ell)(1))$. On the one hand, identifying $H^1(E_{\bar K}, \mathbf{Z}_\ell)(1)$ with the $\ell$-adic Tate module $T_\ell(E)$, this is simply the Kummer map (the inverse limit of the boundary maps associated to the sequence $0 \to E[\ell^n] \to E(\overline{K}) \to^{\times \ell^n} E(\overline{K}) \to 0$). The more high-powered approach is that we have an etale cycle class $$ CH^1(E) \to H^2_{et}(E, \mathbf{Q}_\ell(1)) $$ and the composite with the edge-map of the Hochschild--Serre exact sequence, going into $H^2_{et}(E_{\overline{K}}, \mathbf{Q}_\ell)(1))^{G_K} = \mathbf{Q}_\ell$, is just the degree of the divisor, which is 0; so one lands in the next step of the filtration of $H^2_{et}(E, \mathbf{Q}_\ell(1))$ induced by the spectral sequence, which is $ H^1(K, H^1(E_{\bar K}, \mathbf{Q}_\ell(1)))$.

To see how this class is manifested in the cohomology of the open variety $E \setminus Z$, where $Z = \{x, \infty\}$, consider the excision exact sequence $$ 0 \to H^1(E_{\bar K}) \to H^1(E_{\bar K} - Z_{\bar K}) \to H^2_Z(E_{\bar K}) \to H^2(E_{\bar K}) \to \dots $$

The cycle class of the divisor $\{x\} - \{\infty\}$ gives an element of $H^2_Z(E_{\bar K})(1)$ whose image in $H^2(E_{\bar K})(1)$ is trivial, so by pullback we get a short exact sequence $$ 0 \to H^1(E_{\bar K})(1) \to W \to \mathbf{Q}_\ell \to 0 $$ where $W$ is a subspace of $H^1(E_{\bar K} - Z_{\bar K})(1)$. This gives a geometric realization of the extension of Galois representations associated to the class in $ H^1(K, H^1(E_{\bar K}, \mathbf{Q}_\ell(1)))$ coming from the spectral sequence.

I've worked with non-compact supports here, but everything is smooth of dimension 1, so the compactly-supported cohomology $H^i_c$ is just the dual of the non-compact $H^{2-i}$, by Poincare duality.

  • $\begingroup$ Would you mind to expand a bit, or maybe give a reference? The divisor gives a class in the Galois invariants, why does it also give a class in the $H^{1}(K, \_)$? Furthermore, is this non-split extension realized in the $\ell$-adic cohomology with compact support ($f_{!}$) or without ($f_{*}$)? Is there an easy way to see how it is realized in the cohomology? Anyway, thanks for the answer! $\endgroup$
    – jmc
    May 20, 2014 at 9:52
  • $\begingroup$ Thank you very much for your edit! Explaining how the cycle class arises was really helpful. I will study the excision sequence a bit (because I do not know what $H_{Z}^{2}(E_{\bar{K}})$ means), but it all seems very plausible from the analogue perspective of singular cohomology. I wish I could give you more than one upvote now. $\endgroup$
    – jmc
    May 20, 2014 at 11:05
  • 6
    $\begingroup$ A variation of this works if you want an example which is proper and not smooth - take an elliptic curve and glue together the origin with a non-torsion point, then $H^1$ is not semisimple. Also I recommend Carlson, "Extensions of mixed Hodge structures" for a concrete explanation of how the "same" extensions look on the Hodge theory side. $\endgroup$ May 20, 2014 at 11:08
  • $\begingroup$ @DanPetersen – Thanks for your comment! I already knew about the Hodge side (although I did not have examples ready at hand, so I'll have a look at Carlson). The answer for the Galois side over finite fields took me a bit by surprise, which prompted this question. (In retrospect, the situation is entirely as one expects, I just had to think about it a bit longer.) $\endgroup$
    – jmc
    May 20, 2014 at 17:08
  • 2
    $\begingroup$ @jmc Conjecturally the following should be true, at least over a number field: a short exact sequence of mixed motives splits if and only if any one of its $\ell$-adic or mixed Hodge realizations splits. So you should expect any construction of something non-semisimple on the Hodge side to also work on the $\ell$-adic side. $\endgroup$ May 21, 2014 at 7:24

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