# Prove that two Lie groups have homeomorphic universal covers if and only if their corresponding Lie algebra are isomorphic

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.

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

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 $R^n$.

• I'm confused, the lie algebras are supoised to be isomorphic not commutative. Don't understand your answer… – grayQuant May 22 '16 at 22:37
• 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 – Tsemo Aristide May 22 '16 at 22:39