# Why can we always restrict an Galois representation so that it becomes unramified?

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 $$K$$.

• 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 ? – reuns Oct 19 '18 at 1:47
• 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? – Watson Oct 21 '18 at 9:04
• 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? – Watson Oct 21 '18 at 11:48
• 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 $l \neq p$. – Watson Oct 21 '18 at 12:08

$$\DeclareMathOperator{\ur}{ur}$$ $$\DeclareMathOperator{\Gal}{Gal}$$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.

• why is $\ker(\rho|_{I_L})$ open in $I_K$ ? (open subgroups are of finite index) – reuns Oct 19 '18 at 3:40
• 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. – D_S Oct 19 '18 at 3:51
• 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 ? – reuns Oct 19 '18 at 8:32
• 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 – D_S Oct 19 '18 at 14:16
• 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. – D_S Oct 21 '18 at 13:56