12
$\begingroup$

The following question came up in a discussion with a colleague about local Galois representations:

To what extent is the classification of continuous $p$-adic representations of $G_{\mathbf{Q}_{\ell}}$ for $\ell\neq p$ similar to the classification of tamely ramified $p$-adic representations for $\ell=p$?

More precisely, let $\rho: G_{\mathbf{Q}_{\ell}}\rightarrow \mathrm{GL}_n(\mathbf{Q}_p)$ be a (continuous) $p$-adic representation. If $\ell\neq p$, then Grothendieck proved (using the observation that such a representation kills an open subgroup of wild inertia) that $\rho$ is determined by the associated Weil-Deligne representation (see, for example, the notes of Brinon and Conrad, pg. 111, or Taylor's 2002 ICM article).

When $\ell=p$ and $\rho$ is trivial on the wild inertia subgroup, is it the case that $\rho$ is necessarily de Rham?

What seems clear to me is that if one assumes that $\rho$ is Hodge-Tate, then the only Hodge-Tate weight is zero. If indeed $\rho$ were de Rham = pst, then the associated filtered $(\phi,N)$-module would have trivial filtration, and so one ``ought" to be able to recover it from the attached Weil-Deligne representation. In other words, the classification of $p$-adic representations of $G_{\mathbf{Q}_{\ell}}$ for $\ell\neq p$ is literally the same as the case $\ell=p$, provided one throws in the (rather drastic) condition that wild inertia is killed (or at least some open subgroup of it is killed).

Does this sound correct?

$\endgroup$
5
  • $\begingroup$ Isn't Hodge-Tate automatic? At least I think Sen proved that $\mathbb C_p$-admissibility is the same inertia having finite image. And a tamely ramified $p$-adic Galois rep. has finite image. (The codomain is locally pro-$p$ and tame inertia is prime to $p$.) $\endgroup$
    – fherzig
    Jan 20, 2011 at 15:29
  • $\begingroup$ I meant finite image of inertia. $\endgroup$
    – fherzig
    Jan 20, 2011 at 15:30
  • 1
    $\begingroup$ But if inertia has finite image isn't there then a finite extension $K/\mathbb Q_p$ such that the restriction of $\rho$ to $I_K$ is trivial, hence $\rho|_{G_K}$ is crystalline (you can check that on inertia)? So $\rho$ is potentially crystalline. I may be wrong. $\endgroup$
    – fherzig
    Jan 20, 2011 at 15:38
  • $\begingroup$ Dear Florian, I think that you're completely correct. As you write, if wild inertia is trivial, then the image of inertia is finite, hence $\rho$ is potentially crystalline with HT weights equal to $0$, and so is determined by the associated Weil--Deligne representation. $\endgroup$
    – Emerton
    Jan 20, 2011 at 15:56
  • $\begingroup$ P.S. You should post this as an answer! Best wishes, Matt $\endgroup$
    – Emerton
    Jan 20, 2011 at 15:57

1 Answer 1

16
$\begingroup$

It's true that if $\rho$ is tamely ramified, then $\rho$ is de Rham. In fact, it's even potentially crystalline with all Hodge-Tate weights equal to 0.

First, note that $\rho(I_{\mathbb Q_p})$ is finite. The reason is that the image of $\rho$ lands in $GL_n(\mathbb Z_p)$, which has a pro-$p$ subgroup of finite index, namely the principal congruence subgroup $1+pM_n(\mathbb Z_p)$. Since $I_{\mathbb Q_p} \to GL_n(\mathbb Z_p)$ factors through a prime-to-$p$ group (tame inertia) by assumption, $\rho(I_{\mathbb Q_p})$ injects into $GL_n(\mathbb F_p)$, hence it is finite.

It follows that there is a finite extension $K/\mathbb Q_p$ such that $\rho|_{I_K}$ is trivial. (The kernel of ($\rho$ restricted to $I_{{\mathbb Q}_p}$) corresponds to a finite (tame) extension of $\mathbb Q_p^{nr}$ and we can choose $K$ such that that extension is contained in $\mathbb Q_p^{nr} \cdot K$.)

It's a general fact that $\rho|_{I_K}$ crystalline implies $\rho|_{G_K}$ crystalline, hence $\rho$ is potentially crystalline. (Added: this follows from Hilbert's theorem 90. If $L/K$ is a Galois extension and $Gal(L/K)$ acts semilinearly and continuously on a finite-dimensional $L$-vector space $V$, then $V$ has a basis that is $Gal(L/K)$-invariant.)

In particular, you only get WD representations with $N = 0$.

$\endgroup$
1
  • 1
    $\begingroup$ Dear Florian, Thank you for the very nice answer! $\endgroup$
    – B. Cais
    Jan 20, 2011 at 20:15

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.