In a paper I'm reading, we have a compact Lie group $G$ and he says "We can identify $\pi_1(G)$ with the kernel of $\widetilde{G} \rightarrow G$.
I can't seem to find anything that says that the covering map for the universal cover of a Lie group is a homomorphism. The statement doesn't make sense to me if it's not a homomorphism, however. If it's not a homomorphism, what does this sentence mean?
Best Answer
If $G$ is a connected Lie group, then $\tilde G$ (the universal covering of $G$) is a simply connected manifild. But it is not a Lie group. One must define there a groups structure. But, if $p\colon\tilde G\longrightarrow G$ is the covering map and if you fix some element $\tilde e$ of $p^{-1}(e)$, then there is one and only one group structure on $\tilde G$ such that $\tilde e$ is the identity element of $\tilde G$ and that $p$ is a group homomorphism. With respect with this group structure, yes, $\tilde G$ is a Lie group. Furthermore, is $\tilde e'$ is another element of $p^{-1}(e)$, then $\tilde G$ endowed with these group structures (the one induced by $\tilde e$ and the one induced by $\tilde e'$) is twice the same group (in the sense that there is a group isomorphism between $\tilde G$ endowed with the first group structure and $\tilde G$ endowed with the second one).