$\DeclareMathOperator{\char}{char}\DeclareMathOperator{\gal}{Gal}$ Let $P$ be a smooth projective variety over a field $K$ (one may certainly assume that $K$ is perfect; the case $K=\mathbb{Q}$ already seems to be interesting enough). For some $\ell\neq \char K$, $n>0$, should the $n$-th $\mathbb Q_\ell$-adic Galois cohomology of $X_{K^{sep}}$ be semi-simple as a $\gal(K)$-representation? Certainly; no proof of this fact is known, so I would rather like to know whether it is related with some 'motivic' conjectures.

Some remarks:

  1. For a finite $K$ one can consider the 'motivic' Frobenius; thus the conjecture follows from standard (motivic) ones. Yet this argument does not seem to work already for $K=\mathbb Q$.

  2. It is certainly tempting to apply some polarizability argument. Yet my impression is that polarizability can only be applied to Hodge structures (in general); is this true?

Upd. It seems (see the comment of Ulrich) that 'my conjecture' is wrong for $K= \mathbb Q_\ell$; this settles my question. Yet I wonder where I can find the details for this example (when is the representation corresponding to an elliptic curve with multiplicative reduction indecomposable).

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    $\begingroup$ Mikhail, if you assume Tate, as Will suggested, then the subrepresentations would be motivic. Also, regarding 2, there is a conjecture that the image of $Gal(\matbb{Q})$ in $GL(H^i(X_{et}, \mathbb{Q}_\ell)$ is the $\mathbb{Q}_\ell$ points of the Mumford-Tate group of the Hodge structure on $H^i(X)$. If this were true then semisimplicity would follow from Hodge theory, but establishing this would perhaps be harder than establishing semisimplicity directly. $\endgroup$ – Donu Arapura Feb 1 '13 at 23:01
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    $\begingroup$ @Donu: This is not exactly true. If you consider the motive $M$ of rank 2 attached to an elliptic curve $E$ over a quadratic imaginary field $K$ with complex multiplication by $K$ defined over $K$, then $M_{\mathbb Q_l}$ will have sub-representations of dimension that are not motivic for half of the $l$'s, namely those which are split in $K$. Now your argument will work I think after extending the field of coefficients. Since semi-simplicity is invariant by extension of fields of the coefficients (which is not the case of course of irreducibility), your and Will's argument is saved. $\endgroup$ – Joël Feb 2 '13 at 0:39
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    $\begingroup$ Semi-simplicity certainly does not hold for arbitrary $K$: for example, it does not hold for elliptic curves with multiplicative reduction over a $p$-adic field $K$ $\endgroup$ – naf Feb 2 '13 at 6:41
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    $\begingroup$ Mikhail: The Galois representation for elliptic curves with multiplicative reduction is easy to understand using the theory of Tate curves. One gets that the Tate module (which is equal to $H^1$ upto twist) in the the case of split multiplicative reduction is a (non-trivial) extension of $\mathbb{Z}_p$ by $\mathbb{Z}_p(1)$. (See, for example, Silverman's book "Advanced Topics in the Arithemetic of Elliptic Curves" p. 452) $\endgroup$ – naf Feb 2 '13 at 12:35
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    $\begingroup$ Minor correction: In my previous comment $p$ should be $l$ (an arbitrary prime) and the extension is non-split -- the image of the Galois group is non-abelian -- even with $\mathbb{Z}$ replaced by $\mathbb{Q}$. $\endgroup$ – naf Feb 2 '13 at 13:07

For a base field finitely generated over the prime field, the Tate conjecture implies that the $l$-adic realization gives an equivalence from the category of pure motives tensor $\mathbb{Q}_l$ to the category of $l$-adic Galois representations generated by the cohomology of smooth projective algebraic varieties over the field. The standard conjectures imply that the category of pure motives is semisimple, and hence also the category of Galois representations. (Actually, all you need for this is the Tate conjecture plus the conjecture that numerical equivalence and l-adic homological equivalence coincide, which are both in Tate's original Woods Hole article.)

For other base fields, the Galois representations need not be semisimple -- as ulrich explains in his comments.

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