Timeline for Path lifting property for $\pi:M\rightarrow M/G$ for $G$ compact Lie acting smoothly and freely
Current License: CC BY-SA 4.0
9 events
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| Apr 3, 2022 at 17:03 | comment | added | Sean Lawton | @Z.M That's right, for continuous path-lifting there are no assumptions on the action except that it is continuous. | |
| Apr 3, 2022 at 12:31 | comment | added | Z. M | I don't have the book at hand, but seemingly you don't assume that the action is free in the first paragraph? | |
| Mar 31, 2022 at 13:33 | history | edited | Sean Lawton | CC BY-SA 4.0 |
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| Mar 31, 2022 at 13:01 | comment | added | Sean Lawton | 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). | |
| Mar 31, 2022 at 12:41 | comment | added | Sean Lawton | 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. | |
| Mar 31, 2022 at 8:04 | comment | added | Akerbeltz | 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. | |
| Mar 31, 2022 at 4:19 | comment | added | abx | The question is about smooth path lifting. | |
| Mar 30, 2022 at 23:33 | history | edited | Sean Lawton | CC BY-SA 4.0 |
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| Mar 30, 2022 at 23:20 | history | answered | Sean Lawton | CC BY-SA 4.0 |