algebraic topology, manifolds, topology

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

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

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