Let $G$ be a Lie group which acts smoothly, freely and properly in a smooth manifold $M$. If the group $G$ is compact and the quotient smooth manifold $M/G$ is compact, is $M$ compact?
• I proved that using only the hypothesis which the Lie group in the conditions Of the problem assures that $M$ is compact.
• I found three examples where the conjecture is valid, but I did not find counterexamples.
Yes. Under these hypotheses, the quotient map $M\to M/G$ is a principal $G$-bundle, i.e. it is locally trivial (see Local triviality of principal bundles, for instance). Since $M/G$ is compact, it can be covered by finitely many open subsets $U_1,\dots,U_n$ for which $M\to M/G$ is trivial over the closures $\overline{U_1},\dots,\overline{U_n}$. Then $M$ is covered by sets which are homeomorphic to $\overline{U_1}\times G,\dots \overline{U_n}\times G$. Each of these sets is compact, and so $M$ is compact.
