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Let $G$ be the absolute Galois group of $\mathbb{Q}_p$, and let $1\longrightarrow A\longrightarrow B\longrightarrow C\longrightarrow 1$ be a short exact sequence of (non-abelian) algebraic group schemes over $\mathbb{Q}_p$ equipped with a $G$ action (the maps are $G$-equivariant).

It is well known that there exists a long exact sequence in Galois cohomology:

$0\longrightarrow A^G\longrightarrow B^G\longrightarrow C^G\longrightarrow H^1(G,A)\longrightarrow H^1(G,B)\longrightarrow H^1(G,C).$

I am interested in knowing whether there is a similar exact sequence in Crystalline Galois cohomology, i.e. if I add the crystalline restriction, I still have an exact sequence:

$0\longrightarrow A^G\longrightarrow B^G\longrightarrow C^G\longrightarrow H^1_f(G,A)\longrightarrow H^1_f(G,B)\longrightarrow H^1_f(G,C).$

Many thanks in advance.

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  • $\begingroup$ 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. $\endgroup$ Commented May 4, 2023 at 8:09
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    $\begingroup$ Uhhh $A^G$ doesn't make sense; do you mean $A(\overline{\mathbf{Q}_p})^G$? $\endgroup$ Commented May 4, 2023 at 12:09
  • $\begingroup$ 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. $\endgroup$
    – kindasorta
    Commented May 4, 2023 at 12:11
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    $\begingroup$ 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? $\endgroup$ Commented May 4, 2023 at 12:19
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    $\begingroup$ 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.) $\endgroup$ Commented May 5, 2023 at 8:08

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