Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $M$.

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

1 Answer 1

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, $\operatorname{Im}\rho=\{1\}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.