# Characterization for the continuity of Weil representations

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.

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.

• 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. – Diglett Jan 4 at 12:30
• 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? – Diglett Jan 6 at 9:38
• 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. – Mathmo123 Jan 7 at 10:01
• 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? – Diglett Jan 7 at 15:17
• You can use the fact that $W_K =\bigcup _{i\in\mathbb Z}\phi^i I_K$ where $\phi$ is the Frobenius. – Mathmo123 Jan 7 at 18:44