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 $x$.