4
$\begingroup$

It is my understanding that that every $p$-adic representation of the absolute Galois group of a finite extension $K$ of $\mathbb{Q}_p$ can be described in term of its associated $(\varphi,\Gamma)$-module over the Robba ring $\mathcal{R}_K$.

This is due to Cherbonnier-Colmez and Kedlaya.

Do we have a similar picture for integral representations ? (i.e. representations which coefficients lie in $\mathbb{Z}_p$). If not, is there any reason that such a thing does not exist ?

$\endgroup$

1 Answer 1

5
$\begingroup$

The theory of $(\varphi, \Gamma)$-modules works for $\mathbf{Z}_p$-linear representations, but one has to use a slightly different coefficient ring $\mathbf{A}_K$. This is explained very clearly in section 1.4 of the first Cherbonnier--Colmez paper ("Representations $p$-adiques surconvergentes") which you cite in your question. This goes all the way back to Fontaine's work in the late 80's, and thus pre-dates the Robba ring theory by about 10 years.

The difficulty is that $\mathbf{A}_K$ does not embed in the Robba ring. There is a subring $\mathbf{A}_K^\dagger \subset \mathbf{A}_K$ which does embed in $\mathcal{R}_K$, and the main result of Cherbonnier--Colmez is that every etale $(\varphi, \Gamma)$-module over $\mathbf{A}_K$ can be descended to $\mathbf{A}_K^\dagger$. But $\mathbf{A}_K^\dagger$ and its cousin $\mathbf{B}_K^\dagger = \mathbf{A}_K^\dagger[1/p]$ are both much smaller than the Robba ring, and the Robba ring does not have any natural integral subring --there is no nice ring $A \subset \mathcal{R}_K$ such that $\mathcal{R}_K = A[1/p]$ and $A$ is $p$-adically separated.

$\endgroup$
2
  • $\begingroup$ Thanks for your answer ! So there is no hope for a modulo $p$ version of trianguline representations for example ? $\endgroup$
    – user33624
    Jan 18, 2015 at 11:30
  • $\begingroup$ Not as far as I'm aware. $\endgroup$ Jan 18, 2015 at 12:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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