Denote by $G_p$ a choice of an absolute Galois group of $Q_p$, the field of $p$-adic numbers. Consider a continuous representations of $G_p$ on a $3$-dimensional $Q_p$ vector space that is a successive extension of the trivial character, the cyclotomic character ($\chi_p$) and the square of the cyclotomic character, i.e. there are exact sequences of $G_p$-modules :

$$0 \to Q_p(2) \to V \to W \to 0 $$ and $$0 \to U \to V \to Q_p \to 0$$

where $W$ is an extension $0 \to Q_p(1) \to W \to Q_p \to 0$ and $U$ is an extension $0 \to Q_p(2) \to U \to Q_p(1) \to 0$. Here $Q_p(n)$ denote a 1 dimensional $Q_p$ vector space endowed with an action of $G_p$ via the $n$-th power of the cyclotomic character.

Assume $U$ and $W$ to be crystalline, that is the $Q_p$-vector spaces $D_{crys}(U) = (B_{crys} \otimes_{Q_p} U)^{G_p}$ and $D_{crys}(W) = (B_{crys} \otimes_{Q_p} W)^{G_p}$ are both of dimension 2 ($B_{crys}$ is one of the ring of $p$-adic periods introduced by Fontaine).

In this case, is $V$ crystalline ?

It is the case if and only if the cocycle $[V] \in H^1(G_p, U)$ determined by (the extension determined by) $V$ is in the kernel of the map $H^1(G_p, U) \to H^1(G_p, B_{crys}\otimes_{Q_p}U)$, which is usually denoted by $H^1_{crys}(G_p, U)$.

One can show that in our case $H^{1}_{crys}(G_p,U)$ is 2 dimensional (there is a formula involving the dimension of $U$, the dimension of its invariants under $G_p$ and the Hodge-Tate weights). Now $H^{1}_{crys}(G_p,Q_p(2))$ and $H^{1}_{crys}(G_p,Q_p)$ are both 1 dimensional over $Q_p$ so we have an exact sequence of $Q_p$ vector spaces $$0 \to H^{1}_{crys}(G_p,Q_p(2)) \to H^{1}_{crys}(G_p,U) \to H^{1}_{crys}(G_p,Q_p) \to 0.$$

From there I don't know if there is more one can say...

  • $\begingroup$ You don't say anything about having tried to just do the calculation on the semi-linear algebra side. (As you know, part of the power of these notions is to turn Galois-theoretic calculations into (semi-)linear algebra ones that can be easier or more tractable, all the more so over a ground field like $\mathbf{Q}_p$ for which the Frobenius operator is the identity and all semi-linearity is just usual linearity.) Have you tried to do the work on the linear algebra side rather than by looking at Galois cohomology? $\endgroup$
    – user74230
    Feb 8, 2015 at 19:06
  • $\begingroup$ This is a good remark. Actually I didn't think of trying to do the calculations because I had in mind the case where the base field is not necessarily $Q_p$... $\endgroup$
    – user65490
    Feb 8, 2015 at 19:34

1 Answer 1


One has a natural exact sequence: $$0 \rightarrow H^1(G_p, \mathbb{Q}_p(2)) \rightarrow H^1(G_p,U) \xrightarrow{s} H^1(G_p,\mathbb{Q}_p(1)) \rightarrow 0,$$ where $ H^1(G_p, \mathbb{Q}_p(2)) \cong H^1_{crys}(G_p,\mathbb{Q}_p(2))$, $\dim H^1(G_p,U)=\dim H^1_{crys}(G_p,U)+1$, $\dim H^1(G_p, \mathbb{Q}_p(1))=\dim H^1_{crys}(G_p, \mathbb{Q}_p(1))+1$. So $s^{-1}\big(H^1_{crys}(G_p, \mathbb{Q}_p(1))\big)=H^1_{crys}(G_p,U)$. In other words, if $U$ is crystalline, then $V$ is crystalline $\Leftrightarrow$ $W$ is crystalline.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.