# Lie Group and Universal Covering Groups

Is every Lie group realized as the quotient of its universal covering group by a discrete group of isometries? Basically, a Lie group analog for the uniformization theorem. It seems reasonable but I'm rather ignorant to the theory of Lie groups and can't seem to find a reference.

Edit: I should have mentioned when the Lie group has a universal cover (i.e. path connected, locally path-connected, and semi-locally path connected)

The word "isometries" isn't really appropriate here, because there's no canonical metric on an arbitrary Lie group. But something very close to what you suggested is true: Every Lie group $G$ has a universal covering group that is also a Lie group, unique up to isomorphism; and $G$ is isomorphic to the quotient of its universal covering group by a discrete central subgroup. (One place to find a proof of this is my Introduction to Smooth Manifolds (2nd ed.), Theorem 21.32 and Problem 21.18.)
• Thank you very much for the help! Also, I came across the following fact which seems related. There is a short exact sequence $1\rightarrow \pi_1(X)\rightarrow \tilde{X}\rightarrow X\rightarrow 1$ where $\tilde{X}$ is the universal covering group of $X$. – Joseph Zambrano Mar 15 '14 at 16:24
• @Jyrki: "Isometry" means distance-preserving, so I don't think you'd want to define isometries that way for Lie groups. For example, considering $\mathbb R^n$ as a Lie group under addition, Haar measure is just ordinary Euclidean volume. Every linear map with determinant $1$ preserves Haar measure, but most of them do not preserve distances. – Jack Lee Mar 15 '14 at 22:16
• @Joseph: Yes, this follows from the result I stated above, together with the fact that if a discrete group $\Gamma$ acts continuously and properly on a simply connected space $X$, then the fundamental group of the quotient space $X/\Gamma$ is isomorphic to $\Gamma$. – Jack Lee Mar 15 '14 at 22:20
• Thank you so much for the help. One more related question if its not too much trouble. Is there a notion of quasi-isometry for Lie groups in the following sense (I understand that there is not canonical metric): if $X\equiv \tilde{X}/\pi_1(X)$ where $X$ is a topological group, then is $\tilde{X}$ "quasi-isometric" to $\pi_1(X)$. – Joseph Zambrano Mar 16 '14 at 16:26