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 $H^q_{et}(X'_{\overline{\mathbb{Q}}},\mathbb{Q}_{\ell})$?

  • 1
    $\begingroup$ 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 Nov 4 '17 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 Nov 4 '17 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 Liou Nov 5 '17 at 18:33
  • 2
    $\begingroup$ 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 Nov 5 '17 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 Nov 6 '17 at 16:23

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.