# Covering spaces of Lie groups

In a paper of Tom Bridgeland's, he describes an action by the universal over $G:=\tilde{GL^+}(2,\mathbb{R})$ using a description of $G$ I find unintuitive. Namely, he indexes write the fiber over $T\in GL^+(2,\mathbb{R})$ as the increasing maps $f:\mathbb{R}\to\mathbb{R}$ such that $f(r+1)=f(r)+1$ and $T$ and $f$ induce the same maps on $S^1$ when seen as a quotient of $\mathbb{R}$ or a subspace of $\mathbb{R}^2$ as appropriate.

Now, I can see that this does give a description of the appropriate covering space, since for every $T$ the fiber's really just $\mathbb{Z}$ ($f$ is determined by $T$ and by which element of $\mathbb{Z}+\theta$ it takes $0$ to, where $\theta$ is the argument of $T((1,0))$.) But I don't see how this is a particularly natural way to think about $G$. Is it an instance of a more general construction of covering groups, for instance? Or is Bridgeland just being very clever in writing $\mathbb{Z}$ so as to make his action easy to define?

• I think, he exploits two things here: 1. One of the most natural ways to describe groups is to identify them as certain transformation groups. 2. The Lie group in question retracts to its maximal compact subgroup which is the circle, hence describing its universal cover mostly reduces to describing the universal cover of this circle subgroup. Thus, there are two circles appearing here. In higher dimensions thus trick does not work since special orthogonal group can no longer be identified with the sphere on which it acts. – Moishe Kohan Mar 10 '14 at 18:21