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Specifically, Warner says to prove that if a Lie hom $\phi:G\to H$ has a discrete kernel then the kernel is actually contained in the center. (I can do this part.) The exercise then says to use this fact to prove that the fundamental group of a Lie group is abelian. (I'm familiar with other proofs of this, e.g. Hilton-Eckmann.) I suppose the idea here is to exhibit the fundamental group as the kernel of some hom, but I just don't have any ideas on how to do that.

  • What about the universal cover? – Steve D Jul 20 '18 at 04:51
  • Well, the kernel of $\pi : \tilde{G} \to G$ is discrete. I also know the deck transformation group of $\tilde{G}$ is isomorphic to $\pi_1(G)$. I also know the deck transformation group acts freely and transitively on the fibres (such as the kernel). I know this gives a bijection between $ker(\phi) \times \pi_1(G)$ and $ker(\phi) \times ker(\phi)$, but I don't see how this plays nice with the group structure on $ker(\phi)$. –  Jul 20 '18 at 05:30
  • Let $k$ be in the kernel. Then the map $g\mapsto gk$ is a deck transformation, and this idea gives a bijection from the kernel to the fundamental group. Show this is a homomorphism. – Steve D Jul 20 '18 at 13:05
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    I believe this just boils down to the standard concatenation of paths is homotopic to group multiplication of paths. Thanks! –  Jul 20 '18 at 15:33

2 Answers2

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Let $\rho:\widetilde{G}\rightarrow G$ be the universal covering homomorphism. You already know the kernel $K$ is a discrete central subgroup of $\widetilde{G}$. We have the following

Lemma: For every $k\in K$, the map $\phi_k:\widetilde{G}\rightarrow\widetilde{G}$ given by $g\mapsto gk$ is a deck transformation. The map $K\rightarrow\operatorname{Aut}(\rho)$ is a group isomorphism.

Your question is solved by this, because $\operatorname{Aut}(\rho)$ is isomorphic to $\pi_1(G)$.

Proof of Lemma: $\phi_k$ is clearly a homeomorphism, and it is a deck transformation because $\rho(gk)=\rho(g)$. The map $k\mapsto\phi_k$ is injective, because $\phi_k(1)=\phi_h(1)$ implies $k=h$. It is surjective because if $\phi$ is any deck transformation, then $\phi(1)=k\in K$ since $\rho\phi(1)=1$. But then $\phi\equiv\phi_k$, because both are deck transformations, agreeing at a point.

Finally, the map is a homomorphism because \begin{align*} \phi_{kh}(g) &= gkh\\ &= ghk\\ &= \phi_k\phi_h(g) \end{align*} where the second equality follows because $K$ is central.

Steve D
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I assume that $f:G\rightarrow H$ is a morphism of Lie groups and $G, H$ are connected. Suppose that $Kerf$ is discrete and let $g\in Kerf$. Consider $f_g(x) =xgx^{-1}$ it is continous this implies that the image of $f_g$ is connected since $G$ is connected. Since $Kerf$ is discrete and the image of $f_g$ is contained in $Kerf$ since $Kerf$ is normal, we deduce that the image of $Kerf$ contains only one element $1g1^{-1}=g$. This implies that $Kerf$ is central.

If $H$ is the universal cover of $G$, and $f$ the covering map, the kernel of $f$ is discrete since it is a local homeomorphism. The first part implies that $Kerf=\pi_1(H)$ is abelian.

Tsemo Aristide
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  • You've skipped over what is probably the important part: showing the kernel and the fundamental group are isomorphic. – Steve D Jul 20 '18 at 13:23