Last year I wrote up how to prove that the fundamental group of a (connected) topological group was abelian. Since Lie groups are topological groups, they also have abelian fundamental groups, but I think there is a much neater way to prove this fact using smooth things. Here it is:

Lemma 1: A connected Lie group that acts smoothly on a discrete space is must be a trivial action.

Proof: Suppose our action is by . Consider a point with non-trivial orbit. Then where is non-empty. Thus , a disconnection of . Thus all orbits are trivial.

Lemma 2: A discrete normal subgroup of a Lie group (from now on always assumed connected) is central.

Proof: Denote the subgroup . Then is a smooth action on (a discrete set). Thus by Lemma 1, for all . Thus is central, and in particular abelian.

Now for the main theorem. Let be a connected Lie group. Let be the universal cover of . Then the covering map is a group homomorphism. Since is simply connected, the covering is normal and hence . By virtue of being normal, we also get that acts transitively on the fibers of . In particular, on the set , which is discrete being the fiber of a discrete bundle. But this set is , which is a normal subgroup. I.e. a discrete normal subgroup, which by Lemma 2 is abelian.

Fix . Then we get an ismorphism by where is the unique covering automorphism that takes to . Thus is abelian which means is abelian.

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I am a mathematics graduate student fascinated in how all my interests fit together.

September 8, 2010 at 3:38 pm

This is a nice proof, but I feel like the Eckmann-Hilton argument is somehow a more fundamental fact about mathematics inasmuch as it applies to rather general objects.

July 13, 2011 at 8:57 am

This is very nice! I think in Lemma 1 you mean to say that the conclusion is that the action is trivial, and not the group.

July 13, 2011 at 9:30 am

Thanks. That is what I meant.

March 22, 2013 at 4:58 am

This is probably trivial, but how do you see that $x \mapsto \phi_x$ is group homomorphism? Or equivalently $\phi_x(y) = xy$ for all $x,y \in Ker p$?

March 22, 2013 at 10:02 am

It’s OK, I can see it now, it’s because of the uniqueness of deck transformation sending $q$ to $x$.