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Suppose $X$ is a smooth projective variety defined over $\mathbb{Q}$, and the pure Hodge structure on $H^{2n}(X)$ is of a very simple form, \begin{equation} \mathbb{Q}(-n)^{b^{2n}} \end{equation} where $b^{2n}$ is just $\text{dim}\,H^{2n}(X)$. Under what conditions is the etale cohomology $H^{2n}_{et}(X,\mathbb{Q}_{\ell})$ also isomorphic to \begin{equation} \mathbb{Q}_{\ell}(-n)^{b^{2n}} \end{equation} as representations of $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$?

This is however not always true, consider a smooth quadratic surface $S$ of $\mathbb{P}^3_{\mathbb{Q}}$ defined by a quadratic polynomials with determinant say 7. Then the Hodge structure on $H^2(S)\simeq H^2 (\mathbb{P}^1 \times \mathbb{P}^1)$ is $\mathbb{Q}(-1)^{2}$, while $H^{2}_{et}(S,\mathbb{Q}_{\ell})$ is \begin{equation} \mathbb{Q}_{\ell}(-1)^{2} \otimes \chi \end{equation} where $\chi$ is a Dirichlet character.

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    $\begingroup$ This is expected to always be true after passing to a finite extension of $\mathbb{Q}$ (for example, in the case of your quadric, the extension of $\mathbb{Q}$ defined by the kernel of $\chi$); it would follow from the conjecture that Hodge cycles are absolute Hodge, for example (which is weaker than the Hodge conjecture mentioned by anon below). $\endgroup$ – Daniel Litt Feb 28 '18 at 0:57
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Assuming the Hodge conjecture, it's true if and only if the relevant algebraic cycles are defined over $\mathbb{Q}$. Conjecturally, the cohomologies are different realizations of the same motive. The condition on the Hodge cohomology tells you the motive is a Tate twist of an Artin motive, and you want the Artin motive to be trivial.

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