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.

  • $\begingroup$ (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 $\endgroup$
    – Noix07
    Apr 23 '15 at 21:03
  • $\begingroup$ How about the example of $Pin^{\pm}(n)$ to $O(n)$? Is that unique? $\endgroup$ 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 $\Psi: (\tilde G, \tilde e) \to (\hat G, \hat e)$.

  • $\begingroup$ 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! $\endgroup$
    – YYF
    May 28 '13 at 13:35
  • $\begingroup$ I've already figured out a correct proof! Thank you! $\endgroup$
    – YYF
    May 28 '13 at 16:45
  • $\begingroup$ @Y.Fan: You are welcome. However did you understand what I was trying to say? $\endgroup$ May 28 '13 at 18:47
  • $\begingroup$ 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. $\endgroup$
    – YYF
    May 28 '13 at 20:28
  • $\begingroup$ @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! $\endgroup$ May 28 '13 at 21:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.