# Prove that the universal covering space of a Lie group is unique

This is a problem in Lee's book, introduction to smooth manifold. I am trying to show that the universal covering space of a connected Lie group is unique.

Suppose $G$ is a Lie group, and $\tilde{G}$ and $\hat{G}$ are universal covering groups of $G$. Therefore, the smooth covering maps $\pi:\tilde{G}\to G$ and $\hat{\pi}:\hat{G}\to G$ are Lie group homomorphisms.

I am trying to show there exists a Lie group isomorphism $\Psi:\tilde{G}\to\hat{G}$ such that $\hat{\pi}\circ\Psi=\pi$. The following is my attempt.

I know a universal covering of a manifold is unique. It means that there exists a diffeomorphism of $\Psi:\tilde{G}\to\hat{G}$ such that $\hat{\pi}\circ\Psi=\pi$. What is left is to show $\Psi$ is actually a homomorphism, which exactly means that $$\Psi\circ\tilde{m}=\hat{m}\circ(\Psi\times\Psi)$$ where $\tilde{m}$ and $\hat{m}$ are the multiplication maps in $\tilde{G}$ and $\hat{G}$, respectively.

I can show that both $\Psi\circ\tilde{m}$ and $\hat{m}\circ(\Psi\times\Psi)$ are lifts of the same map $m\circ(\pi\times\pi)$, where $m$ is the multiplication map of $G$. Hence, it suffices to show that $\Psi\circ\tilde{m}$ and $\hat{m}\circ(\Psi\times\Psi)$ agree at some point. I want to show they agree on $(\tilde{e},\tilde{e})$, where $\tilde{e}$ is the unity in $\tilde{G}$.

To prove that, it suffices to show $\Psi(\tilde{e})=\hat{e}$. Now, the proof is almost done. However, I cannot show it is true.

I made two futile following attempt.

First, I notice that $\pi(\tilde{e})=\hat{\pi}(\hat{e})$. Therefore, $\pi$ has a unique lifting map $F:\tilde{G}\to\hat{G}$ such that $F(\tilde{e})=\hat{e}$. Also, $\Psi$ is a lift of $\pi$. However, I do not know how to show they are the same.

Second, I notice that $\hat{\pi}\circ\Psi(\tilde{e})=e$. Therefore, I choose a neighborhood $U$ of $e$ evenly covered by $\hat{\pi}$. Clearly, $\hat{e},\Psi(\tilde{e})\in\hat{\pi}^{-1}(U)$. However, I do not know how to prove they belong to the same component of $\hat{\pi}^{-1}(U)$.

I know I must be missing something obvious, since I think the proof is almost done. Could anyone point it out? Thank you in advance.

• (sorry if not relevant, I did not read more than the title) Just to give a reference: in Daniel Bump's "Lie groups" , Proposition 13.1 p.72 – Noix07 Apr 23 '15 at 21:03
• How about the example of $Pin^{\pm}(n)$ to $O(n)$? Is that unique? – annie marie heart May 3 '17 at 16:23

The problem is that the map $\Psi$ is not unique, but you can force it to have $\Psi(\tilde e)=\hat e$ in a unique way. The universal covering space of a space is indeed unique up to isomorphism, but not up to a unique isomorphism. For instance, the covering map $\mathbf R \to S^1$ has many automorphisms (which are in bijection with $\pi_1(S^1) = \mathbf Z$). You obtain unicity of the isomorphism only in the category of pointed spaces. Thus, in fact, given the pointed space $(G,e)$ and your two covering spaces $(\tilde G, \tilde e)$ and $(\hat G, \hat e)$, the universal property of the covering space implies that there is a unique map of pointed spaces $\Psi: (\tilde G, \tilde e) \to (\hat G, \hat e)$.
• I am still confused. How could I just force $\Psi$ to satisfy $\Psi(\tilde{e})=\hat{e}$? The theorem I used just ensure the existence of $\Psi$. So, how to correct my proof? Thanks! – YYF May 28 '13 at 13:35
• Frankily speaking, I don't fully understand. What I understand is that the isomorphism is not unique. Therefore, I have some freedom to manipulate it such that $\Psi(\tilde{e})=\hat{e}$. Actually, based on this idea, I think I figured out a correct proof. However, I don't understand about the situation about two pointed spaces. – YYF May 28 '13 at 20:28
• @Y.Fan It's just a way to express the universal property of the universal covering space. $\Psi$ becomes unique only if you specify that $\Psi(\tilde e) = \hat e$, and you have the freedom to do this. That's what the universal property says. I hope that helps! – Bruno Joyal May 28 '13 at 21:01