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Let $K = \mathbb{Q}_p$, $G = G_K$ its absolute Galois group. Let $$1\longrightarrow A\longrightarrow B\longrightarrow C\longrightarrow 1$$ be a split exact sequence of (not necessarily abelian) group schemes over $K$, equipped with a $G$ action. The exact sequence is $G$-equivariant.

Let $B_{cr}$ denote the crystalline period ring, and consider the commutative diagram of Galois cohomology sets: $$H^0(G,C)\longrightarrow H^1(G,A)\longrightarrow H^1(G,B)\longrightarrow H^1(G,C)\\ \downarrow\quad\quad\quad\quad\quad\quad\downarrow\quad\quad\quad\quad\quad\quad\downarrow\quad\quad\quad\quad\quad\quad\downarrow\\ H^0(G,C(B_{cr}))\longrightarrow H^1(G,A(B_{cr}))\longrightarrow H^1(G,B(B_{cr}))\longrightarrow H^1(G,C(B_{cr})). $$ Denote by $H^1_f(G, B)$ the kernel of the vertical arrow $H^1(G,B)\longrightarrow H^1(G,B(B_{cr}))$, and we define $H^1_f(G,S), H^1_f(G,A)$ in a similar way (as the kernel of the corresponding vertical arrow).

The commutativity of the diagram, together with the exactness of the rows, implies that the restriction of the horizontal morphism $\pi: H^1(G,B)\longrightarrow H^1(G,C)$ to $H^1_f(G,B)$ lands in $H^1_f(G,C)$. Now, let $[c]\in H^1_f(G,B)$ be a cocycle class. I want to try an express the crystalline fibre of the map $\pi$ over $\pi([c])$, i.e., I want to express $\pi^{-1}(\pi([c]))\cap H^1_f(G,B)$ in terms of cohomology sets of certain Serre twists.

In Serre's "Galois Cohomology", he says that the fibre $\pi^{-1}(\pi([c]))$, is in bijection with the set $H^1(G,{}_cA)/H^0(G,{}_cC)$, where ${}_cA$ is the Serre twist of $A$ by $c$, which, as an algebraic group scheme is isomorphic to $A$, but the Galois action is twisted by $c$ as follows. If $\widetilde{a}$ is the element of ${}_cA$ corresponding to $a\in A$, then ${}^g\widetilde{a} = c(g)^{-1}\widetilde{{}^ga}c(g)$, where $\widetilde{{}^ga}$ is the element of ${}_cA$ that corresponds to the element ${}^ga\in A$, and ${}_cC$ is defined similarly.

Denote the restriction of $\pi$ to $H^1_f(G,B)$ by $\pi_f$, then I imagine that $\pi^{-1}_f(\pi_f([c]))$ is bijective with the fibred product: $$H^1(G,{}_cA)/H^0(G,{}_cC)\times_{H^1(G, B(B_{cr}))} *,$$

where $*$ is trivial cocycle of $H^1(G,B)$.

My question is if, similar to the non-crystalline characterization of Serre, there is a bijection between the set $\pi^{-1}_f(\pi_f([c]))$ and $H^1_f(G,{}_cA)/H^0_f(G,{}_cC)$.

Thanks in advance!

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    $\begingroup$ I'm pretty confused by your vertical maps, because $B_{crys}$ is not a $\overline{K}$-algebra... $\endgroup$ Commented Aug 30, 2023 at 7:53
  • $\begingroup$ $B_{cr}$ is a $K$ algebra, so a $K$-point of $A,B,C$ is naturally a $B_{cr}$-point as well, and that's the meaning of the vertical maps. $\endgroup$
    – kindasorta
    Commented Aug 30, 2023 at 8:18
  • $\begingroup$ In that case, please spell out your exact meaning of $H^1(G,A)$ for me. $\endgroup$ Commented Aug 30, 2023 at 9:16
  • $\begingroup$ A cocycle $c\in Z^1(G, A)$ is a continuous map $c: G\longrightarrow A(K)$, satisfying the cocycle condition, i.e. $A$ comes equipped with a $G$-action, and we need for every $g,h\in G$ that $c(gh) = c(g){}^gc(h)$. The set/groupoid $H^1(G,A)$ is obtained when we mod out by the action of the coboundaries, so two cocycles $c_1, c_2\in Z^1(G, A)$ are equivalent if there exists an $a\in A(K)$ such that for every $g\in G$, $c_2(g) = a^{-1}c_1(g){}^ga$. $\endgroup$
    – kindasorta
    Commented Aug 30, 2023 at 10:12
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    $\begingroup$ "i.e. $A$ comes equipped with a $G$-action" Is this extra data? If so it doesn't seem to be mentioned in your question. $\endgroup$ Commented Aug 30, 2023 at 12:23

1 Answer 1

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So it seems I managed to find an elementary proof of this fact.

The fibre $\pi_f^{-1}(\pi_f([c]))$ is in bijection with the fibred product $H^1(G,{}_cA)/H^0(G,{}_cC)\otimes_{H^1(G,B(B_{cr}))}*$.

Now, since $c\in Z^1(G, B)$, and we have a natural injection, $B\longrightarrow B(B_{cr})$, we get a map $Z^1(G,B)\longrightarrow Z^1(G, B(B_{cr}))$, and by abuse of notation, we denote the image of $c$ under the map of cocycle sets by $c$ as well.

The fibre of the morphism $H^1(G, B(B_{cr}))\longrightarrow H^1(G, C(B_{cr}))$ above $c$, by Serre's lemma, is in bijection with $H^1(G, {}_cA(B_{cr}))/H^0(G, {}_cC(B_{cr}))$. In particular, there is an injection:

$$H^1(G, {}_cA(B_{cr}))/H^0(G, {}_cC(B_{cr}))\longrightarrow H^1(G, B(B_{cr}),$$

and the morphism $H^1(G, {}_cA)/H^0(G, {}_cC)\longrightarrow H^1(G, B(B_{cr}))$ factors through: $$H^1(G, {}_cA)/H^0(G, {}_cC)\longrightarrow H^1(G, {}_cA(B_{cr}))/H^0(G, {}_cC(B_{cr}))\longrightarrow H^1(G, B(B_{cr}).$$ Since the latter map is injective, the kernel of the composition is in bijection with the kernel of the first map: $H^1(G, {}_cA)/H^0(G, {}_cC)\longrightarrow H^1(G, {}_cA(B_{cr}))/H^0(G, {}_cC(B_{cr}))$.

Now, we obtain a commutative diagram: $$H^1(G, {}_cA)\longrightarrow H^1(G, {}_cA(B_{cr}))\\ \downarrow\quad\quad\quad\quad\quad\quad\quad\quad \downarrow\\ H^1(G, {}_cA)/H^0(G, {}_cC)\longrightarrow H^1(G, {}_cA(B_{cr}))/H^0(G, {}_cC(B_{cr})). $$ Now, we are trying to find the kernel of the bottom map. If $c'\in H^1(G, {}_cA)$ is in the kernel of the top arrow, then by the commutativity of the diagram, its projection onto $H^1(G, {}_cA)/H^0(G, {}_cC)$ is in the kernel of the bottom arrow as well, which means that $\pi_f^{-1}(\pi_f([c]))\supseteq H^1_f(G, {}_cA)/H^0(G, {}_cC)$.

Alternatively, start with a cocycle $\tilde{c}$ in the kernel of the bottom arrow, since both vertical arrows are surjective, there exists a cocycle $c'$ projecting onto it. Its image under the top arrow must be in the $H^0(G, {}_cC(B_{cr}))$-orbit of the trivial cocycle of $H^1(G, {}_cA(B_{cr}))$.

I claim that this orbit is trivial: let $\gamma\in ({}_cC(B_{cr}))^G$, and denote the trivial cocycle of $H^1(G, {}_cA(B_{cr}))$ by $t$. $\gamma$ acts on $t$ via: $(\gamma\circ t)(g) = \gamma^{-1}t(g){}^g\gamma$, but $\gamma$ being a $G$-invariant means that ${}^g\gamma = \gamma$. Since $t = *$, $t(g) = e$, which implies $(\gamma\circ t)(g) = e$, and therefore $\gamma\circ t = t$.

We conclude that all the cocycles in the preimage of an element of the kernel of the bottom row are crystalline cocycles, which gives set containment in the other direction, and completes the proof.

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