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Let $\tilde{G}$ be universal cover of a lie group $G$. Then we can easily lift any action of $G$ on a connected manifold $X$ to an action of $\tilde{G}$ on the universal cover $\tilde X$ of $X$. (cf. Notes on lifting group actions James Montaldi and Juan-Pablo Ortega October 2008) If we are given that the action of $G$ on $X$ is free then under what condition/s the lifted action would be free.?

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migrated from mathoverflow.net Oct 31 '13 at 3:04

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If $G$ acts freely on $X$ its orbits are copies of $G$, so you can ask if the fundamental group of $G$ injects in the fundamental group of $X$. This doesn't seem like a research problem to me. –  Ryan Budney Oct 29 '13 at 17:29

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