# How to "lift" a transitive group action on a manifold?

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.

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
• Indeed. This is more or less what I suggested in my last lines. I'm aware of Lie-Palais theorem, though you need to require some extra topological conditions from $\widetilde{M}$. Still, I'd like to see some universal property characterising $\widetilde{G}$, like, e.g., "it is the unique group admitting $G$ as a factor by covering transformations", or something like that, and/or some algebraic way to construct it out of the available data! Sep 29, 2014 at 15:28
• The only extra condition you need on $\tilde{M}$ is that it is connected. Sep 29, 2014 at 15:39
• The fundamental group of $G$ lies inside the universal covering group of $G$. The group $\Gamma$ you want is the subgroup of $\pi_1(G)$ which maps trivially to $\pi_1(M)$ when a loop in $G$ carries a base point of $M$ around. Sep 29, 2014 at 15:43
• Don't you need $\widetilde{M}$ to be compact? Sep 29, 2014 at 20:35
• @G_infinity: you cannot always embed every Lie algebra into the Lie algebra of vector fields on every manifold. For example, you cannot embed the Lie algebra of $SU(2)$ into the vector fields on the circle, or else $SU(2)$ would have a nontrivial action on the circle, and by compactness would have a compact orbit, so the circle itself, with stabilizer of a point a closed subgroup of codimension 1, i.e. dimension 2. But the Lie algebra of $SU(2)$ is cross product, so you can picture the cross product on a plane in 3-dimensions: no invariant plane. So no subgroup of dimension 2. Sep 30, 2014 at 9:57
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.