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Suppose $X$ is a variety over $\mathbb{Q}$ and it has a Galois twist $X'$, i.e. $X$ is isomorphic to $X'$ over $\overline{\mathbb{Q}}$. The set of isomorphism classes of twists of $X$ is classified by $H^1(\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}), \text{Aut}(X_{\overline{\mathbb{Q}}}))$.

If $X'$ corresponds to a class $c \in H^1(\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}), \text{Aut}(X_{\overline{\mathbb{Q}}}))$, what is the relation between the Galois representations $H^q_{et}(X_{\overline{\mathbb{Q}}},\mathbb{Q}_{\ell})$ and $H^q_{et}(X'_{\overline{\mathbb{Q}}},\mathbb{Q}_{\ell})$?

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    $\begingroup$ That should probably be $H^q_{\operatorname{\acute et}}(X_{\bar{\mathbb Q}}, \mathbb Q_{\ell})$ (base changed to the algebraic closure). $\endgroup$
    – R. van Dobben de Bruyn
    Commented Nov 4, 2017 at 0:39
  • $\begingroup$ Thank you! The Galois action of $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ on the etale cohomology should be different! $\endgroup$
    – Wenzhe
    Commented Nov 4, 2017 at 11:30
  • $\begingroup$ I don't know what can be expected between these representations. Already in dimension 0, the regular representation of any finite quotient of the absolute galois group can appear as $H^0(\mathrm{Spec} (L\otimes_{\mathbb{Q}} \bar{\mathbb{Q}}), \mathbb{Q}_\ell)$, where $L/\mathbb{Q}$ is a finite extension. $\endgroup$
    – Wille Liu
    Commented Nov 5, 2017 at 18:33
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    $\begingroup$ Have you worked a simple case, for example $X$ is an abelian variety and the cocycle $c:\text{Gal}(\overline{\mathbb Q}/\mathbb Q)\to\text{Aut}(X_{\overline{\mathbb Q}})$ takes values in automorphisms that fix $0$, so $X'$ is also an abelian variety. Then $H^1_{et}(X_{\overline{\mathbb Q}},\mathbb Q_\ell)$ is dual to the Tate module $T_\ell(X)$. Fixing a $\overline{\mathbb Q}$-isomorphism $f:X\to X'$, it's easy enough to untangle how the Galois representations on $T_\ell(X)$ and $T_\ell(X')$ are related. $\endgroup$
    – Joe Silverman
    Commented Nov 5, 2017 at 18:42
  • $\begingroup$ @JoeSilverman Thank you. I have not worked it out, but someone has told me the result, it is twisted by a Dirichlet character, and I will have a try! $\endgroup$
    – Wenzhe
    Commented Nov 6, 2017 at 16:23

1 Answer 1

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Just follow your nose. Functoriality of cohomology gives a map $\operatorname{Aut}(X_\bar{\mathbb{Q}})\to \operatorname{Aut}\big(H^q_{\text{ét}}(X_{\bar{\mathbb{Q}}}, \mathbb{Q}_\ell)\big)$ (actually, the functoriality is contravariant, so gives a map to the opposite group -- but for groups, $G \cong G^{op}$ via $g\mapsto g^{-1}$). This map is compatible with Galois action, hence you get (again by functoriality) a map $$H^1\big(\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q}),\operatorname{Aut}(X_{\bar{\mathbb{Q}}})\big)\to H^1\big(\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q}),\operatorname{Aut}(H^q_{\text{ét}}(X_{\bar{\mathbb{Q}}}))\,\big).$$ The RHS once again classifies Galois twists (now of a vector space).

There is a topological way of seeing why this functoriality should hold. Very informally, you should think of an equivariant object (e.g. $X_{\bar{\mathbb{Q}}}$) of any category with $\Gamma$ action (for $\Gamma$ the Galois group) as a local system of objects in a local system of categories over a topological space $S$ with $\pi_1(S) = \Gamma.$ The object over the basepoint $s_0\in S$ is $X_{\bar{\mathbb{Q}}},$ and the Galois action $\Gamma\to \operatorname{Aut}(X_{\bar{\mathbb{Q}}})$ is encoded by the monodromy; a twist in $H^1(S, \operatorname{Aut})$ is some class over $S$ which modifies the local glueing data between the copies of $X_{\bar{\mathbb{Q}}}$ on intersections locally by some elements of the relevant automorphism group. Now any functor of $\Gamma$-equivariant categories gives maps of all the relevant local data over $S$, and in particular takes a twist to a twist.

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