Characterization for the continuity of Weil representations

Question

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!

Answer 1

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.