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?



Tamely ramified p-adic Galois representations
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 $N = 0$.


