Lie groups have abelian fundamental group

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 ${\theta: G\times M \rightarrow M}$ by ${\theta(g,x)=g\cdot x}$. Consider a point ${q}$ with non-trivial orbit. Then ${\text{im}\theta_{q}=\{q\}\cup A}$ where ${A}$ is non-empty. Thus ${G=\theta_{q}^{-1}(q)\cup \theta_q^{-1}(A)}$, a disconnection of ${G}$. 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 ${H}$. Then ${\theta(g,h)=ghg^{-1}}$ is a smooth action on ${H}$ (a discrete set). Thus by Lemma 1, ${ghg^{-1}=h}$ for all ${g\in G}$. Thus ${H}$ is central, and in particular abelian.

Now for the main theorem. Let ${G}$ be a connected Lie group. Let ${U}$ be the universal cover of ${G}$. Then the covering map ${p: U\rightarrow G}$ is a group homomorphism. Since ${U}$ is simply connected, the covering is normal and hence ${Aut_p(U)\simeq \pi_1(G, e)}$. By virtue of being normal, we also get that ${Aut_p(U)}$ acts transitively on the fibers of ${p}$. In particular, on the set ${p^{-1}(e)}$, which is discrete being the fiber of a discrete bundle. But this set is ${\ker p}$, which is a normal subgroup. I.e. a discrete normal subgroup, which by Lemma 2 is abelian.

Fix ${q\in \ker p}$. Then we get an ismorphism ${\ker p \simeq Aut_p(U)}$ by ${x\mapsto \phi_x}$ where ${\phi_x}$ is the unique covering automorphism that takes ${q}$ to ${x}$. Thus ${Aut_p(U)}$ is abelian which means ${\pi_1(G,e)}$ is abelian.

5 thoughts on “Lie groups have abelian fundamental group”

1. 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.

2. 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.

3. hilbertthm90 says:

Thanks. That is what I meant.

4. rafaelm says:

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 $x,y \in Ker p$?

5. rafaelm says:

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