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?

  • $\begingroup$ 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


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 $N = 0$.

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

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