The image of an Artin representation is finite

Question

First let us define the necessary terms for the problem:

Definition

• Let $G$ be a profinite group and let $k$ be a topological field. An $n$-dimensional representation of $G$ over $k$ is a continuous homomorphism of groups $\rho: G \to \operatorname{GL}_n(k)$.

• An $n$-dimensional representation of $G$ over $k$ is called an Artin representation if $k \subset \mathbb{C}$, i.e. $k$ is a topological subfield of $\mathbb{C}$.

Problem: Now I would like to show that the image of an Artin representation is finite.

My own progress so far:

• Since $\require{enclose} \enclose{horizontalstrike}{G}$ is a profinite group we know that $\require{enclose} \enclose{horizontalstrike}{G}$ is isomorphic to an inverse limit of a projective system of finitely many groups $\require{enclose} \enclose{horizontalstrike}{\{G_1,\dots,G_m\}}$. In other words,

$$ \require{enclose} \enclose{horizontalstrike}{ G = \varprojlim_{i} G_i = \left\{ (\gamma_1, \dots, \gamma_m) \in \prod_{i=1}^m G_i \: \middle| \: f^j_i(\gamma_j)=\gamma_i \: \forall j \geq i \right\}.}$$

That means that we can insert $m$-tuples $\require{enclose} \enclose{horizontalstrike}{ (\gamma_1,\dots,\gamma_m) }$ in our presentation $\rho$ which gives us elements $\require{enclose} \enclose{horizontalstrike}{ \rho(\gamma_1,\dots,\gamma_m) \in \operatorname{GL}_n(k) }$.

Edit: I made a huge mistake! In a projective limit, the groups are finite, not the system of groups! Now I have no idea at all again.

But now I do not know how to continue so solve this problem anymore.

My specific questions (besides "How can I show this statement?"):

• Why exactly do I need that $k$ is a topological subfield in $\mathbb{C}$?
• How do I handle the group homomorphisms $f^j_i : G_j \to G_i$ in our projective system? Do they play an important role for our problem?

Any help is really appreciated.

Answer 1

This comes down to the "no small subgroup argument."

Lemma: Let $H = \operatorname{GL}_n(k)$, where $k$ is a topological subfield of $\mathbb C$. There exists an open neighborhood $U$ of the identity matrix $I$ which contains no subgroups of $H$ besides the trivial subgroup.

A consequence of the lemma is that if $\rho: G \rightarrow \operatorname{GL}_n(k)$ is a continuous homomorphism, then the kernel of $\rho$ is an open subgroup of $G$. Since $G$ is profinite, it is compact, so the kernel of $\rho$ is of finite index in $G$. This immediately implies that the image of $\rho$ is finite.

To see that the kernel of $\rho$ is open, take an open neighborhood $U$ as in the lemma, and consider the preimage $\rho^{-1}(U)$, which is an open neighborhood of $1_G$. It's a general fact about profinite groups that inside any open neighborhood of the identity, you can find an open subgroup. So let $N$ be an open subgroup of $G$ contained in $\rho^{-1}(U)$. Then $\rho(N)$ is a subgroup of $\operatorname{GL}_n(k)$ which is contained in $U$. By the hypothesis on $U$, $\rho(N)$ must be the trivial group, i.e. $N$ must be contained in $\operatorname{Ker}\rho$. Thus $\operatorname{Ker} \rho$ contains an open subgroup, so it is itself open.

Proof of the lemma: It suffices to prove the lemma in the case $k = \mathbb C$, since if $U$ is an open neighborhood of $I$ which works for $\operatorname{GL}_n(\mathbb C)$, then $U \cap \operatorname{GL}_n(k)$ is one which works for $\operatorname{GL}_n(k)$.

The case $n =1$ (where you are working with $\operatorname{GL}_1(\mathbb C) = \mathbb C^{\ast}$) is obvious.

For arbitrary $n$, I have seen several proofs of this. My favorite one comes from differential geometry. The same link provides alternative proofs.