# lifting group action

Let a Lie group $G$ act on a manifold $M$. Under which condition(s) we have a lifting action of $G$ on the universal covering of $M$?

By Bredon, I already know that there is a group $\widetilde{G}$, which is a covering of $G$, acts on $\widetilde{M}$, the universal covering of $M$.

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If you know $\tilde G$, you should just check that $\ker[\tilde G\to G]$ acts on $\tilde M$ trivially. – Alex Degtyarev Feb 22 '14 at 11:52

Here is an answer not referring to $\tilde G$ explicitly. I assume $G$ connected.
The given action $G\times M\to M$ defines a map $\pi_1(G)\times\pi_1(M)\to\pi_1(M)$, which can be restricted to a homomorphism $\rho\colon\pi_1(G)\to\pi_1(M)$. Geometrically, realize a class in $\pi_1(G,1)$ by a loop in $G$ and act by this loop on a point $*\in M$: you get a loop in $M$, which represents a class in $\pi_1(M,*)$. Well, it is more or less obvious (representing the points of $\tilde M$ as homotopy classes of paths in $M$) that the action lifts iff $\rho$ is trivial, $\operatorname{Im}\rho=\{1\}$.