Timeline for Galois twist of a variety
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 6, 2017 at 16:23 | comment | added | Wenzhe | @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! | |
| Nov 5, 2017 at 19:21 | answer | added | Dmitry Vaintrob | timeline score: 5 | |
| Nov 5, 2017 at 18:42 | comment | added | Joe Silverman | 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. | |
| Nov 5, 2017 at 18:33 | comment | added | Wille Liu | 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. | |
| Nov 5, 2017 at 18:21 | history | edited | David Loeffler | CC BY-SA 3.0 |
added 56 characters in body
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| Nov 4, 2017 at 11:30 | comment | added | Wenzhe | Thank you! The Galois action of $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ on the etale cohomology should be different! | |
| Nov 4, 2017 at 0:39 | comment | added | R. van Dobben de Bruyn | That should probably be $H^q_{\operatorname{\acute et}}(X_{\bar{\mathbb Q}}, \mathbb Q_{\ell})$ (base changed to the algebraic closure). | |
| Nov 3, 2017 at 21:39 | history | asked | Wenzhe | CC BY-SA 3.0 |