Let $M=G/H$ be a homogeneous manifold, with $G$ connected Lie group. Suppose that $\widetilde{M}$ is a covering of $M$.



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> QUESTION: is there a general prescription to obtain a Lie group $\widetilde{G}$, starting from $G$, in such a way that $\widetilde{M}=\widetilde{G}/\widetilde{H}$?
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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](https://mathoverflow.net/questions/158341/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.



How to "lift" a transitive group action on a manifold?
If $G$ is connected, take its Lie algebra, acting as vector fields on $M$. Lift the vector fields by the covering map. There is a unique connected Lie group $\tilde{G}$ acting on $\tilde{M}$ whose Lie algebra has this action, by a theorem of Dick Palais: <http://en.wikipedia.org/wiki/Lie%E2%80%93Palais_theorem>


