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Jun 15, 2020 at 7:27 history edited CommunityBot
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Jul 14, 2019 at 12:22 history edited Peter Michor CC BY-SA 4.0
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Jul 14, 2019 at 11:03 history edited Peter Michor CC BY-SA 4.0
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Jul 14, 2019 at 10:51 comment added Peter Michor Okay, this convinces me. I shall adapt my answer.
Jul 14, 2019 at 10:35 comment added Misha Proper actions have Hausdorff quotient spaces. What you explained nicely in your linked note is that given an incomplete vector field on a manifold, you can extend it to a complete vector field on a (potentially) non-Hausdorff manifold. What happens in the 3-holed sphere example is that the quotient is obviously non-Hausdorff and, hence, you cannot get a proper $R$-action (even though, individual trajectories are proper, of course). What you have here is a variation on the standard example of a non-proper $R$-action on the punctured affine plane, given by $(t, (x,y))\mapsto (e^tx, e^{-t}y)$.
Jul 14, 2019 at 9:36 comment added Peter Michor But the $\mathbb R$-bundle is over a non-Hausdorff space, in general. In particular, in the 3 punctured sphere: Where 1 orbit becomes two, the two cannot be separated. See www.mat.univie.ac.at/~michor/vect-mf.pdf
Jul 14, 2019 at 9:05 comment added Misha If you have a proper $R$-action then $M$ is diffeomorphic to the total space of an $R$-bundle; of course, many noncompact manifolds do not admit such.
Jul 14, 2019 at 8:43 history edited Peter Michor CC BY-SA 4.0
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Jul 14, 2019 at 8:11 comment added Peter Michor True, one needs a vectorfield without periodic orbit.
Jul 14, 2019 at 7:53 comment added o r If I look at the vector field $y\partial_x - x\partial_y$ then it should be complete on $\Bbb R^2-\{0\}$, where it has no zeros. But I think the flow is not proper, as for any point $x$ we have $\varphi_X^{2\pi n}(x)=x$. Is that correct?
Jul 14, 2019 at 7:38 history edited Peter Michor CC BY-SA 4.0
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Jul 14, 2019 at 7:29 history answered Peter Michor CC BY-SA 4.0