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Let $M$ be a smooth manifold and let $G$ be a compact Lie group acting smoothly and freely over $M$. Let $\pi:M\rightarrow M/G$ be the canonical projection, and endow $M/G$ with the unique differentiable structure such that it is a smooth manifold and $\pi$ is a surjective submersion.

I would like to prove that, in the previous situation, the map $\pi$ satisfies the path lifting property, in the sense that for each smooth path $\gamma:I\rightarrow M/G$ defined in a compact interval $I$, there exists a smooth lift $\tilde\gamma:I\rightarrow M$ with $\pi\circ\tilde\gamma=\gamma$.

This is an exercise from the book Mathematical Gauge Theory from Mark J. D. Hamilton. It appears in a chapter on group actions, preceding those about fiber bundles, connections, curvature, etc. So apparently this can be proved attending only to the basic topological properties of the objects at hand.

In this SE post the same question is asked; however, the answer given is not complete. At first, this is what I thought the answer might look like; since surjective submersions admit local sections, we can first take an open cover of the trace of $\gamma$ with open sets for which we can find sections of $\pi$, then we take a finite subcover due to the compactness of the trace, and then we patch toghether the images of the pieces of the curve via the aforamentiones sections, using the fact that, by construction, $G$ acts transitively on the fibers of $\pi$. However, this approach is not quite perfect, for there might be problems with differentiability at the points where we patch the pieces toghether. Unfortunately, it doesn't look like this problem can be remedied.

I am out of ideas, so I would really appreciate any kind of help at this point.

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    $\begingroup$ It suffices to trivialize the restriction of the bundle to the curve, i.e the pullback bundle $\gamma^* \pi$ over $I$. $\endgroup$
    – Tobias Diez
    Commented Mar 30, 2022 at 21:26
  • $\begingroup$ $M$ is a principal $G$-bundle over $M/G$; choose a connection and consider a horizontal lift. $\endgroup$
    – Konrad Waldorf
    Commented Mar 31, 2022 at 9:23
  • $\begingroup$ @KonradWaldorf Yes, that is exactly what I first thought about. However I think there should be a simpler way to solve this problem; the author of the book where I found this problem included it on a chapter previous to the introduction of bundles and connections. I know that with the method you mentioned one can prove what I asked, but supposedly there is a more elementary approach to it; that is what I am really looking for. $\endgroup$
    – Akerbeltz
    Commented Mar 31, 2022 at 11:45
  • $\begingroup$ You can just reuse parts of the proof that horizontal lifts exist (once you have a connection). Usually, you first trivialize the pullback bundle over the curve to write the horizontal path equation as an ODE on the structure group. Instead of using the solution to this ODE, you can also simply take the curve that for all times sits at the identity element (or some other arbitrary smooth curve on G) to obtain some smooth lift. $\endgroup$
    – Tobias Diez
    Commented Mar 31, 2022 at 15:08
  • $\begingroup$ Choose local trivializations of $M \to M/G$ along $\gamma$, as $I$ is compact, finitely many suffice. Under each l.t., there is a smooth lift $(\gamma(t),1)\in U \times G$. On the overlaps, the lifts differ by a constant group element. Perform finitely many shifts by these groups elements, and you'll have a smooth lift. $\endgroup$
    – Konrad Waldorf
    Commented Apr 1, 2022 at 5:48

1 Answer 1

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Bredon [Bre72,Theorem II.6.2] shows that for any Hausdorff space $X$ with the action of a compact Lie group $K$, the natural map $X \to X/K$ has the path-lifting property.

This follows since there is a slice at each point. Bredon attributes this to Montgomery and Yang (The existence of a slice. Ann. of Math. 65 (1957), 108-116).

The general philosophy is that whenever there are slices, there is path-lifting.

Addendum (after it was pointed out I misread the question):

So the above result applies to the situation when $X$ is a manifold and $K$ is acting smoothly and freely to produce a lifted path. But that path may not be smooth a priori.

However, unless I am missing something (I often am, and so please forgive me for that), the proof (I just re-read) in Bredon can be adapted line-by-line to show the resulting path can be taken to be smooth if $K$ acts smoothly and freely on a smooth manifold $X$.

Let $f:I\to X/K$ be a smooth path. $X/K$ is smooth since $K$ is acting freely and properly (since $K$ is compact). Let $\pi:X\to X/K$ be the projection (which is smooth). There are smooth slices (Koszul, 1953) and so there exists a smooth slice $\sigma:I\to f^*X$, which gives your desired smooth path through the pull-back diagram.

[Bre72] Bredon, Glen E. Introduction to compact transformation groups. Pure and Applied Mathematics, 46. Academic Press, New York, 1972.

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    $\begingroup$ The question is about smooth path lifting. $\endgroup$
    – abx
    Commented Mar 31, 2022 at 4:19
  • $\begingroup$ Yes, I am aware of the result from Bredon, but it is proved for continuous actions not necessarily acting on manifolds. I don't know how this may apply to my case, producing smooth lifts. $\endgroup$
    – Akerbeltz
    Commented Mar 31, 2022 at 8:04
  • $\begingroup$ Well, the result applies to smooth manifolds (so there is a lifted path in your setting). But sure, it doesn't imply the lifted path is necessarily smooth. I didn't notice you wanted the path to be smooth when I read your question. Sorry. $\endgroup$
    – Sean Lawton
    Commented Mar 31, 2022 at 12:41
  • $\begingroup$ I just read the proof in Bredon, and it seems to me that you can adapt it step by step to show the resulting path is smooth if you assume $G$ is acting smoothly and freely. You pull back the path and take a smooth slice. To show there is a smooth slice you need smooth local slices, which you have since $M\to M/G$ is principal bundle (or you can use the smooth slice theorem, Koszul 1953). $\endgroup$
    – Sean Lawton
    Commented Mar 31, 2022 at 13:01
  • $\begingroup$ I don't have the book at hand, but seemingly you don't assume that the action is free in the first paragraph? $\endgroup$
    – Z. M
    Commented Apr 3, 2022 at 12:31

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