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In A First Course in Modular Forms, Diamond and Shurman leave as an exercise ($9.3.3$) that every complex Galois representation is finite. While I think I have worked through this exercise here, this solutions seems sort of strange to me, because most other sources just state this fact as if it is obvious, giving no indication as to why it is true. However, this solution does not seem very obvious to me. (Of course, that could just be because I haven't spent enough time thinking about these things.)

Is there a better way to understand why complex Galois representations are finite? I've heard it informally explained that this happens because the topologies on $\mathbf{G}_{\bar{\mathbf{Q}}}$ and $GL_{d}(\mathbf{C})$ are incompatible, with too many open sets on the latter. While the proof I posted above does illustrate this, I'm wondering if there's another proof that may be more to my taste.

Bonus/Related Question: it has been my assumption all along that the topology on $GL_{d}(\mathbf{C})$ is the standard one. However, another comment D&S make leads me to wonder if I'm wrong. In discussing the relationship between Dirichlet characters and $1$-dimensional representations $\rho : \mathbf{G}_{\bar{\mathbf{Q}}} \to \mathbf{C}^*$, they say that to check continuity of $\rho$, it suffices to check that $\rho^{-1}(1)$ is open. This seems to imply that $C^{*}$ is given the discrete topology. Can anyone clear this up for me?

Thank you in advance!

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Just a comment: the solutions imply that the result holds if you put either the discrete topology on $\operatorname{GL}_n(\mathbb{C})$ or its usual Euclidean topology: in fact you get a real Lie group either way. –  Pete L. Clark May 13 '11 at 1:41
Yes, you're absolutely right. The issue about which topology is used is actually relevant to a slightly different context, so perhaps I shouldn't have lumped it in with this one, but thanks for pointing that out. –  Jeff May 13 '11 at 1:52
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3 Answers

up vote 10 down vote accepted

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.

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Thank you, I found this very helpful. I'm starting to understand the general philosophy at work here. –  Jeff May 13 '11 at 2:21
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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.

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Thanks for this, especially your comments on finishing off that lemma. –  Jeff May 13 '11 at 2:22
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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.

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I think a better proof is the following. Pick an open set U containing the identity in GL_n(C) and also containing no nontrivial subgroup of GL_n(C). The inverse image of U is an open subset of G containing the identity in G, so that inverse image contains an open subgroup H (as G is profinite). Then rho(H) is a subgroup of GL_n(C) inside U, hence rho(H) is trivial. That shows rho kills a finite index subgroup of G (open subgroups of compact groups have finite index). Therefore the image of rho is the image of the map G/H ---> GL_n(C) induced by rho, which has finite image since G/H is finite. –  KCd May 13 '11 at 2:58
@Kcd: very cute :) –  Mariano Suárez-Alvarez May 13 '11 at 3:07
@KCd: Dear Keith, This is the solution that the OP gave in the linked answer. I agree that it is the natural proof (as you can tell from my answer!). Best wishes, –  Matt E May 13 '11 at 3:25
Thank you for this different perspective! It's also nice to see the proof generalized to arbitrary profinite groups. –  Jeff May 13 '11 at 4:00
@Kcd: Your proof is much cleaner (and more general) than mine. Thanks for posting that. You and Matt have successfully convinced me that this proof is the natural proof. (But I was still glad to see Kevin's!) –  Jeff May 13 '11 at 4:13
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