Complex Galois Representations are Finite

Solution 1:

I didn't read through the details of your argument, but the structure is correct, and is the standard one. The point is that, among locally compact groups, there are two extremes: Lie groups, which have no small subgroups, i.e. sufficiently small neighbourhoods of $1$ contain no non-trivial subgroup; and profinite groups, in which every neighbourhood of the identity contains an open subgroup. As you observed in your argument, any continuous homomorphism between a profinite group and a Lie group then necessarily has finite image, because of their incompatible nature: "no small subgroups" vs. "arbitrarily small subgroups".

This type of consideration (whether or not a given group admits arbitrarily small non-trivial subgroups or not) is a standard one in the theory of locally compact groups.

Solution 2:

Here is an alternative proof (here $G$ is an arbitrary profinite group):

Since $\rho: G\rightarrow GL_n(\mathbb{C})$ is continuous, its kernel is closed, so by factoring through the quotient we may assume $\rho$ is injective. Now any continuous injective map from a compact space to a Hausdorff space is a homeomorphism onto its image, and the image is closed in the ambient space. Thus, $G$ is topologically isomorphic to $\rho(G)$, which is a closed subgroup of a Lie group, hence is a Lie subgroup. Then we are done since any profinite Lie group must be finite.

Solution 3:

I don't think there is an easier proof. It is probably stated as obvious because it is very well-known, and thus considered standard.

The proof of Lemma 1 is not complete: why does $m^n$ escape the neighbourhood? (the exponential map is not globally injective). You can show that if $m^n, n \in \mathbb{Z}$ is bounded, then $m$ is diagonalisable, and its eigenvalues have modulus one. From there you just have to treat the case $d=1$.

Concerning your second question, you just showed that if $\rho$ is continuous for the usual topology, it is continuous for the discrete topology, so $\rho^{-1}\left( \{1\} \right)$ is open. Note that it is harder to directly prove that $\rho^{-1}\left( \{1\} \right)$ is open than to prove the continuity for the standard continuity, so "suffices" is a bit of a lie here.