TITLE: How to "lift" a transitive group action on a manifold?
QUESTION [8 upvotes]: Let $M=G/H$ be a homogeneous manifold, with $G$ connected Lie group. Suppose that  $\widetilde{M}$ is a covering of $M$.

QUESTION: is there a general prescription to obtain a Lie group  $\widetilde{G}$, starting from $G$, in such a way that $\widetilde{M}=\widetilde{G}/\widetilde{H}$?

Using the case of $M=G=S^1$ and $H=1$ as a toy model (and, for instance, $\widetilde{M}=\widetilde{G}=\mathbb{{R}}$) we see that $\widetilde{G}$ has two remarkable properties:
1) it contains the group $\Gamma=\mathbb{{Z}}$ of "gauge symmetries" of $\widetilde{M}\to M$;
2) its factor by $\Gamma$ returns the original group $G$.
So, I guess that the  two properties above are enough to characterise $\widetilde{G}$ but I'm not able to prove it. I'm sure it's a well-known result, but I can't find any reference (I could not get which book is this "Bredon" mentioned here: lifting group action). In the case that my guess is correct, I'd like to understand if there is a constructive way to obtain $\widetilde{G}$, e.g., by realising the Lie algebra $\frak{g}$ as vector fields on $M$, lifting them to $\widetilde{M}$, and then take the group generated by their flows.

REPLY [3 votes]: The Palais theorem  assumes  that  the manifold $\tilde M$ is  compact.
The positive  answer   gives proposition 6   of  the  Onishchik  book "Topology of   transitive transformation  groups".
It  states :
 For  any  action  of  a Lie  group $G$  on a manifold and  any  covering
$\pi : N \to M$  there is  an action  of  the  universal  cover $\tilde{G}$ on
$N$   which  cover  the  action  of $G$ on $M$, i.e.  such  that  the  projection $\pi : N \to M$ is $\tilde{G}$-equivariant.