Timeline for Crystalline exact sequence in Galois cohomology
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 5, 2023 at 16:41 | comment | added | kindasorta | Very very curious to see an argument, whether from a reference or an answer. | |
| May 5, 2023 at 8:08 | comment | added | Satan's Minion | If $A, B, C$ are crystalline representations, your statement is true for highly nontrivial reasons. I'm quite certain it fails in general. (Sorry, I'm on my phone in the subway, can't look up references.) | |
| May 4, 2023 at 12:44 | comment | added | kindasorta | I could, but if I'm honest I don't know how to prove this statement even in the case when $A,B,C$ are replaced with arbitrary (or crystalline) representations. Should that be obvious? | |
| May 4, 2023 at 12:19 | comment | added | Satan's Minion | Hmm, ok. You said they were group schemes. Then I'm confused about the meaning of $H^1_f(G,A)$ in this particular setup. Can you define it precisely for us? | |
| May 4, 2023 at 12:11 | comment | added | kindasorta | Why should it not make sense? I said that these groups come equipped with a (non-trivial) $G$-action. The case I care about is when this is a conjugation action, so, for example, $1\in A$ is always $G$-fixed. | |
| May 4, 2023 at 12:09 | comment | added | Satan's Minion | Uhhh $A^G$ doesn't make sense; do you mean $A(\overline{\mathbf{Q}_p})^G$? | |
| May 4, 2023 at 9:33 | history | edited | kindasorta | CC BY-SA 4.0 |
added 2 characters in body
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| May 4, 2023 at 8:09 | comment | added | Chris Wuthrich | I guess $G$ should be the Galois group of $\mathbb{Q}_p$. It would be good to add (a reference to) the definition of $H^1_f$ in the non-abelian case. | |
| May 4, 2023 at 6:54 | history | asked | kindasorta | CC BY-SA 4.0 |