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Let $K$ be a local field and $\rho: G_K \to \operatorname{GL}_n(\mathbb{C})$ be a Galois representation where $G_K$ denotes the absolute Galois group of $K$.

We call a Galois representation $\rho$ unramified if $\rho(I_K)=1$ where $I_K$ denotes the inertia subgroup of $G_K$.

Question: Why does there exist a finite extension $K'/K$, so that the restriction of $\rho$ on $G_{K'}$ is unramified?

If $\rho$ would be an Artin representation, then, by definition, there exists a finite Galois extension $F/K$ such that $\operatorname{Gal}(\bar{K}/F) \subset \ker{\rho}$, i.e. $\rho$ comes from a representation $\operatorname{Gal}(F/K) \to \operatorname{GL}_n(\mathbb{C})$. Then $\rho|_F : \operatorname{Gal}(F/F) = \{1\} \to \operatorname{GL}_n(\mathbb{C})$ is obviously unramified.

But how can we approach the case of a general Weil representation? I heard that one can do this by noticing that $I_K$ is profinite and $W_K$ is the semidirect product of $I_K$ and $\mathbb{Z}$ where $\mathbb{Z}$ is the cyclic group generated by the Frobenius automorphism $x \mapsto x^{|k|}$ on $\bar{k}$, the algebraic closure of the residue field $k$ of $K$.

Could you please help me to explain this problem to me? Thank you in advance!

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  • $\begingroup$ Let $L = K(\zeta_{p^\infty-1})$ so that $I_K = Gal(\overline{K}/L)$. Are you saying that if the image of $\rho|_{I_K}$ is infinite then it is dense in some topological subgroup of $GL_n(\mathbb{C})$ having not so many normal subgroups of finite index ? $\endgroup$ – reuns Oct 19 '18 at 1:47
  • $\begingroup$ For me, an Artin representation is, by definition, a continuous representation $G_K \to \mathrm{GL}_n(\Bbb C)$. So what do you mean by "general case"? Are you interested in $l$-adic representations? $\endgroup$ – Watson Oct 21 '18 at 9:04
  • $\begingroup$ Thank you for your edit. So by "general Galois representation", you mean a continuous morphism $G_K \to GL_n(F)$ for some field $F$, not necessarily $F \subset \Bbb C$ (for instance $F = \Bbb Q_p$), right? $\endgroup$ – Watson Oct 21 '18 at 11:48
  • $\begingroup$ I think this will be wrong in general : the $p$-adic cyclotomic character $\chi_p : G_{\Bbb Q_p} \to \Bbb Q_p^{\times}$ is not potentially unramified. The best I know is that a representation $G_{\Bbb Q_p} \to GL_n(\Bbb Q_l)$ is potentially semi-stable, when $l \neq p$. $\endgroup$ – Watson Oct 21 '18 at 12:08
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$\DeclareMathOperator{\ur}{ur}$ $\DeclareMathOperator{\Gal}{Gal}$Let $K^{\ur}$ be the maximal unramified extension of $K$, so the inertia group $I_K$ is equal to $\Gal(\overline{K}/K^{\ur})$. The restriction of $\rho$ to the compact group $I_K$ has open kernel, so there exists a finite Galois extension $E$ of $K^{\ur}$ such that $\rho|_{I_K}$ is well defined on $\Gal(E/K^{\ur})$.

Claim 1: We can take $E$ to be the compositum of $K^{\ur}$ and a finite Galois extension $L$ of $K$.

Proof: If $E_1$ is another finite Galois extension of $K^{\ur}$ containing $E$, then we have $\Gal(\overline{K}/E_1) \subset \Gal(\overline{K}/E)$, so $\rho|_{I_K}$ is also well defined on $\Gal(\overline{K}/E_1)$. Write $E = K^{\ur}(\beta)$, and take $L$ to be the Galois closure of $K(\beta)/K$. Then $LK^{\ur}$ contains $E$ and does the trick. $\blacksquare$

Claim 2: $L^{\ur}$, the maximal unramified extension of $L$ inside $\overline{L} = \overline{K}$, is equal to $E = LK^{\ur}$.

Proof: If $M$ is a finite unramified extension of $K$, then $LM$ is a finite unramified extension of $L$. This shows $LK^{\ur} \subseteq L^{\ur}$. The residue field of $LK^{\ur}$ is separably closed, giving us equality. $\blacksquare$

Now consider the restriction of $\rho$ to the local Weil group $W_L$, which sits inside $\Gal(\overline{K}/L)$. The local Weil group $W_L$ contains its inertia group $I_L = \Gal(\overline{K}/L^{\ur}) = \Gal(\overline{K}/LK^{\ur}) = \Gal(\overline{K}/E)$, on which $\rho$ is trivial.

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  • $\begingroup$ why is $\ker(\rho|_{I_L})$ open in $I_K$ ? (open subgroups are of finite index) $\endgroup$ – reuns Oct 19 '18 at 3:40
  • $\begingroup$ I assume OP wants $\rho$ to be continuous, and a continuous homomorphism of a profinite group like $I_K$ into $\operatorname{GL}_n(\mathbb C)$ must have open kernel by a no small subgroup argument. $\endgroup$ – D_S Oct 19 '18 at 3:51
  • $\begingroup$ Sure $\rho$ is meant to be continuous. Do you look at the sub-Lie algebras (of the connected component of the identity in $\rho(I_K)$ which, if infinite, is a compact Lie group) to show there aren't many Lie subgroups of finite index ? $\endgroup$ – reuns Oct 19 '18 at 8:32
  • $\begingroup$ You don't know yet that $\rho(I_K)$ is a Lie subgroup of $\operatorname{GL}_n(\mathbb C)$. But you can use the Lie algebra and exponential map to show that if $G$ is any smooth Lie group, there exists an open neighborhood $U$ of $1_G$ which contains no subgroup except the trivial one. See Jason Devito's answer here mathoverflow.net/questions/103783/no-small-subgroups-argument $\endgroup$ – D_S Oct 19 '18 at 14:16
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    $\begingroup$ Before the question was edited, I assumed the "general case" OP asked about was a continuous representation $\rho$ of the Weil group $W_K$, whose image is not necessarily finite dimensional. $\endgroup$ – D_S Oct 21 '18 at 13:56

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