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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.

Oliver
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  • Does this help? https://math.stackexchange.com/questions/3233/converse-to-quotient-manifold-theorem-exercise-in-lee-smooth-manifolds – AlexD Nov 07 '21 at 12:23
  • 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. – Oliver Nov 07 '21 at 14:13
  • I am not at all convinced that the argument in the link is a valid proof. Details are missing. – Moishe Kohan Nov 08 '21 at 04:07
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    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. – Moishe Kohan Nov 08 '21 at 19:22
  • Thank you very much! – Oliver Nov 11 '21 at 15:07

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