1
$\begingroup$

Let $\mathbb{R}$ act freely and smoothly on a manifold $M$, such that the space of orbits $M/\mathbb{R}$ is a smooth manifold and the projection $M \to M/\mathbb{R}$ is a smooth submersion.

Is it then true that the $\mathbb{R}$-action is proper? I suspect that it is, but have trouble explicitly showing it. Any advice?

I see that this is some kind of converse statement to the Quotient Manifold Theorem, however that didn't help me so far.

$\endgroup$
5
  • $\begingroup$ Does this help? math.stackexchange.com/questions/3233/… $\endgroup$
    – AlexD
    Nov 7, 2021 at 12:23
  • $\begingroup$ Thank you, that would be a way to do it. I am just wondering if for this simple example $G=\mathbb{R}$ there is a direct way, without first proving that $M$ is a principal bundle. $\endgroup$
    – Oliver
    Nov 7, 2021 at 14:13
  • $\begingroup$ I am not at all convinced that the argument in the link is a valid proof. Details are missing. $\endgroup$
    – Moishe Kohan
    Nov 8, 2021 at 4:07
  • 1
    $\begingroup$ I wrote details for the linked answer. You can avoid using the terminology "principal fiber bundle" but, even in the case when $G={\mathbb R}$, the proof will go through the same steps. $\endgroup$
    – Moishe Kohan
    Nov 8, 2021 at 19:22
  • $\begingroup$ Thank you very much! $\endgroup$
    – Oliver
    Nov 11, 2021 at 15:07

0

Reset to default

You must log in to answer this question.