Timeline for Possible isometry groups of open manifolds
Current License: CC BY-SA 4.0
12 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
|
|
Jul 14, 2019 at 12:22 | history | edited | Peter Michor | CC BY-SA 4.0 |
added 199 characters in body
|
Jul 14, 2019 at 11:03 | history | edited | Peter Michor | CC BY-SA 4.0 |
deleted 692 characters in body
|
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 |
added 441 characters in body
|
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 |
added 142 characters in body
|
Jul 14, 2019 at 7:29 | history | answered | Peter Michor | CC BY-SA 4.0 |