Timeline for Are the Galois actions on automorphisms of twists isomorphic?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 14, 2020 at 18:39 | comment | added | Qiaochu Yuan | @Asvin: you may be interested in reading a sequence of blog posts I wrote about how to describe Galois descent in terms of homotopy fixed points. The punchline is here but you may have to go back a few posts for context: qchu.wordpress.com/2015/11/17/forms-and-galois-cohomology | |
| Oct 14, 2020 at 18:27 | comment | added | Asvin | Right, I understand from the twist point of view why the forms are isomorphic (and one can define an explicit isomorphism on $H^1$) but I was wondering if this was the reflection of some more general construction in group cohomology. | |
| Oct 14, 2020 at 18:25 | comment | added | Qiaochu Yuan | @Asvin: when you stipulate that $X$ and $Y$ are isomorphic over $\bar{k}$ you're imposing exactly that they have the same forms but that's all you're imposing. If you work through the proof that $H^1$ classifies forms what you'll see is that different $1$-cocycles are used to twist the Galois action (starting from a base; in other words the bijection requires picking a $k$-form) and taking fixed points of that action is what produces the different forms. | |
| Oct 14, 2020 at 18:15 | comment | added | Asvin | That's great, thanks! I thought of the twisted $\mathbb P^1$ example too but I too did not know how to compute the automorphism group of the conic over the reals. It's a little strange to me to be having two different groups with $\Gamma$ actions that nevertheless give the same $H^1$. Is there a purely group theoretic explanation of what's going on here? For instance, is there some classification of what sort of different $\Gamma$ actions can you get by twisting? | |
| Oct 14, 2020 at 18:13 | vote | accept | Asvin | ||
| Oct 14, 2020 at 10:39 | comment | added | R.P. | @AchimKrause I think that is correct. Another way to see it is to write out the automorphisms of $X$ over $\mathbb{C}$ in terms of their corresponding ring isomorphisms: they are given by $x \mapsto x$, $x \mapsto -x$, $x \mapsto (2x+i)(x-i)$, $x \mapsto (2x-i)(x+i)$, $x \mapsto (-2x-i)(x-i)$, and $x \mapsto (2x+i)(x+i)$ if I haven't made any mistakes. Now it is easy to see that only the first two are stable under complex conjugation. For me, the Galois action is always easier to understand on the ring level, since the action there is just given by the action on the coefficients. | |
| Oct 14, 2020 at 8:14 | comment | added | Achim Krause | I think you can even get a hands-on $0$-dimensional counterexample: Let $X = \operatorname{Spec}\mathbb{R}[x]/(x^2 + 1)x$ and $Y=\operatorname{Spec}\mathbb{R}[x]/(x^2-1)x$. Then I'd think that over $\mathbb{C}$ both have automorphism group $\Sigma_3$, but the action is trivial for $Y$ and conjugation with a transposition for $X$. (Am I messing anything up here?) | |
| Oct 14, 2020 at 8:07 | comment | added | R.P. | You can see (at least intuitively) that twisting can kill (or introduce) automorphisms over the base field: by twisting the Fermat cubic $x^3+y^3+z^3=0$ you obtain e.g. $ax^3+by^3+cz^3=0$, where all of the symmetry is broken. | |
| Oct 14, 2020 at 7:42 | history | undeleted | Qiaochu Yuan | ||
| Oct 14, 2020 at 7:42 | history | edited | Qiaochu Yuan | CC BY-SA 4.0 |
added 111 characters in body
|
| Oct 14, 2020 at 7:37 | history | deleted | Qiaochu Yuan | via Vote | |
| Oct 14, 2020 at 7:09 | history | answered | Qiaochu Yuan | CC BY-SA 4.0 |