Vtyjkyuk

The accepted answer in

Projection map between the Stiefel manifold and the Grassmanian

says

If $G$ is a Lie group and $X$ is a manifold on which $G$ acts freely and properly, then $X/G$ has a natural manifold structure and the projection $Xrightarrow X/G$ is a principal $G$-bundle. In particular, $Xrightarrow X/G$ is a fibre bundle with fibre $G$.

without a reference. I wonder where I can find such a result? Does this hold in the infinite dimensional setting, where $X$ is locally Banach spaces, as well?

(I am already struggling with seemingly different definitions of principal fibre bundle etc. Questions asked/answered by others have been helpful but a source for the result above will be very helpful, too.)

Yes, the result is true and proved in many differential geometry books. This is definitely in John Lee's Smooth Manifolds book, for instance.

The Banach manifold case is harder to track down, but is in Bourbaki, Lie Groups and Lie Algebras as Proposition 3.1.5.10. The following is not precisely a translation (I have changed it slightly to make the assumptions more clear), but is close.

Suppose $G$ is a Banach Lie group and $X$ a Banach manifold on which $G$ acts. Suppose that $G$ acts freely and properly on $X$, and such that if $rho(x): T_eG to T_xX$ is the differential of the action at the point $x$, the image of $rho(x)$ is closed and has some closed complement for all $x$. Then the quotient topology on $X/G$ is Hausdorff, and there is a unique smooth structure on $X/G$ such that $pi: X to X/G$ is a submersion. Furthermore, $X to X/G$ is a principal $G$-bundle.

So the main difference in the Banach case is that the orbits should have complemented tangent space. in the Hilbert case, this is just saying the orbits have closed tangent space.