1
$\begingroup$

Two Lie groups $G_1, G_2$ have homeomorphic universal covers $\tilde{G_1}, \tilde{G_2}$ respectively if and only if the corresponding Lie algebras $\frak{g_1}, \frak{g_2}$ are isomorphic as Lie groups.

This is based off of the Lie algebra-Lie group correspondence but I am not sure how to really show this is true. A proof of one direction would be very helpful.

$\endgroup$
  • 1
    $\begingroup$ What do you mean by the "corresponding Lie algebras are isomorphic as Lie groups"? $\endgroup$ – Bombyx mori May 22 '16 at 22:47
  • 1
    $\begingroup$ @grayQuant you're confusing the words "homeomorphic" and "isomorphic" $\endgroup$ – YCor May 23 '16 at 16:16
3
$\begingroup$

This is not true, any simply-connected $n$-dimensional nilpotent Lie group is homeomorphic to $R^n$, but its Lie algebra is not always commutative. Thus is not isomorphic to the Lie algebra of the $n$-dimensional simply connected commutative Lie group $R^n$.

$\endgroup$
  • $\begingroup$ I'm confused, the lie algebras are supoised to be isomorphic not commutative. Don't understand your answer… $\endgroup$ – grayQuant May 22 '16 at 22:37
  • 3
    $\begingroup$ The Lie algebra of $R^n$ is commutative, if a nilpotent Lie group $N$ has dimension $n$, it is homeomorphic to $R^n$, but $R^n$ and $N$ does not have always isomorphic Lie algebras $\endgroup$ – Tsemo Aristide May 22 '16 at 22:39

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.