Differential Topology – Does Quotient Manifold Theorem Provide a Fibration?

complex-geometrydifferential-topologyfiber-bundleslie-groupsmanifolds

It's known that if $G$ is a Lie group and $H\subseteq G$ is a closed subgroup, then the quotient map $p\colon G\to G/H$ is in fact a principal $H$-bundle, which follows from the existence of local sections.

However, in our course on complex manifolds, the professor claims that in general, if $M$ is a smooth (or complex analytic) manifold and the Lie group $G$ acts smoothly (or holomorphically), freely and properly on $M$, then $M/G$ is a smooth (or complex analytic) manifold and the quotient map $p\colon M\to M/G$ is a principal $G$-bundle.

I doubt this result since I didn't find any reference for this via google. I wonder whether there's a counterexample for this, and a corrected version of this theorem, or some references on this theorem?

Any help? Thanks!

Best Answer

I don't know much about complex manifolds, but this is certainly true for smooth manifolds. From John Lee's Smooth Manifolds:

Theorem 21.10. Suppose $G$ is a Lie group acting smoothly, freely, and properly on a smooth manifold $M$. Then the orbit space $M / G$ is a topological manifold of dimension equal to $\text{dim} M - \text{dim} G$, and has a unique smooth structure with the property that the quotient map $\pi: M \to M / G$ is a smooth submersion.

See also Natural Operations in Differential Geometry by Kolar, Michor, Slovak:

Lemma 10.3. Let $p: P \to M$ be a surjective submersion (a fibred manifold) and let $G$ be a Lie group which acts freely on $P$ from the right such that the orbits of the action are exactly the fibres $p^{-1}(x)$ of $p$. Then $p: P \to M$ is a principal $G$-bundle.

Together these results give you the desired conclusion.

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