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Let $K$ be a non-Archimedean local field and $W_K$ be the Weil group of $K$. We consider a representation $\rho: W_K \to \operatorname{GL}_n(\mathbb{C})$ between two topological groups. Here, $\operatorname{GL}_n(\mathbb{C})$ is endowed with the discrete topology and $W_K$ has this weird topology such that $$ 1 \to I_K \to W_K \to \mathbb{Z} \to 1 $$ becomes a short exact sequence of topological spaces where $I_K$ is the inertia subgroup of the absolute Galois group $G_K$ and $\mathbb{Z}$ denotes the subgroup generated by the Frobenius element $x \mapsto x^{|k|}$ of the absolute Galois group $G_k$ of the residue field $k$ of $K$.

I would like to show the equivalence of the following two statements:

  • $\rho$ is continuous,
  • $\rho(I_K)$ is finite.

If one of these two criteria is satisfied, we will call $\rho$ a Weil representation.

Attempts and Ideas:

  • If $\rho$ is continuous, then $\rho$ maps compact subsets of $W_K$ to compact subsets of $\operatorname{GL}_n(\mathbb{C})$. If I manage to show that $I_K \subseteq W_K$ is compact, then $\rho(I_K)$ is finite since $\operatorname{GL}_n(\mathbb{C})$ has the discrete topology.
  • I have no clue for the other direction. It seems like we need an argument from the theory of topological spaces. However, I have little to no knowledge about it.

Could you please help me with this problem? Thank you!

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Let's take a step back for a second and ask: if $H$ is a discrete topological group and $G$ is an arbitrary topological group, what does it mean for a homomorphism $\rho:G\to H$ to be continuous?

Since $\{e\}$ is open in $H$, we certainly need $\ker \rho$ to be open in $G$. And this condition is sufficient too: if $H_0\subset H$ is a subset (and hence an open subset) of $H$, then $$\rho^{-1}(H) = \bigcup_{h\in H}\bigcup_{g\in \rho^{-1}(h)}g\ker(\rho)$$ is a union of open sets, so is open.

Now, everything is simpler. Here are some hints:

  • If $\rho$ is continuous, then $\rho|_{I_K}$ is continuous, so $\rho|_{I_K}:I_K\to \mathrm{GL}_n(\mathbb C)$ has open kernel. But $I_K$ is profinite, so the kernel must be a finite index normal subgroup.

  • If $\rho(I_K)$ is finite, then $\ker(\rho)\cap I_K$ is a finite index normal subgroup, and hence is open. The remainder of the Weil group is determined by the Frobenius element.

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  • $\begingroup$ Thank you for your response! I understood the first part completely. Could you please explain why $I_K$ being profinite implies that the kernel is a finite index normal subgroup? I still have trouble to understand how the topology of inverse limits work. $\endgroup$ – Diglett Jan 4 at 12:30
  • $\begingroup$ I would like to gather everything I know: I only know that in a compact topological group (so esp. profinite groups) a subgroup is open if and only if it is closed of finite index. I understand that $\operatorname{ker}(\rho|_{I_K}) = I_K \cap \operatorname{ker}(\rho)$ has finite index if $\rho(I_K)$ is finite, so the kernel must be closed too if I want to convince myself that it is open. In the books I looked up it was never mentioned that a finite index normal subgroup of a profinite group is open. Could you please recommend me a reference where your claim is mentioned? $\endgroup$ – Diglett Jan 6 at 9:38
  • $\begingroup$ What I said isn't true in general. But it is true for the absolute Galois group, and in particular for the inertia group. Galois theory ensures a bijective correspondence between Galois extensions of $K$ and closed normal subgroups of $G_K$. If $H\subset I_K$ is finite, then by Galois theory + ramification theory, $H$ corresponds to a finite extension of $K$, and hence $H$ is closed and of finite index, so is open. $\endgroup$ – Mathmo123 Jan 7 at 10:01
  • $\begingroup$ Okay thanks, I think I understand now why $I_K \cap \operatorname{ker}(\rho)$ is open in $I_K$. Because $I_K$ is open in $W_K$, the intersection of $I_K$ and the kernel must be open in $W_K$ too. But I am not able to proceed to show that the kernel is open in $W_K$ which would give me the continuity of $\rho$. All I know is that the quotient of $W_K/I_K$ is the cyclic subgroup generated by a Frobenius element. Could you please explain how this is useful here? $\endgroup$ – Diglett Jan 7 at 15:17
  • $\begingroup$ You can use the fact that $W_K =\bigcup _{i\in\mathbb Z}\phi^i I_K$ where $\phi$ is the Frobenius. $\endgroup$ – Mathmo123 Jan 7 at 18:44

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