Timeline for The group of isometries of a manifold is a Lie group, isn't it?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Nov 4, 2020 at 9:34 | comment | added | abx | @YCor: Very good, thanks! | |
| Nov 4, 2020 at 8:48 | comment | added | YCor | @abx It's indeed known that the isometry group of every connected locally compact metric space is locally compact. It's due to van Dantzig and van der Waerden (1928) and can be found in the Kobayashi–Nomizu book. | |
| Nov 4, 2020 at 7:56 | comment | added | abx | @YCor: I am a little bit worried about the other direction: assuming the Hilbert-Smith conjecture, if you want to prove that the isometry group is a Lie group, don't you need to prove first that it is locally compact? How does one do that if $M$ is not compact? | |
| Nov 3, 2020 at 9:55 | history | edited | YCor | CC BY-SA 4.0 |
added remarks on HS conjecture
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| Nov 3, 2020 at 0:33 | vote | accept | aglearner | ||
| Nov 3, 2020 at 0:30 | comment | added | YCor | The best I'm aware of is at Tao's blog | |
| Nov 3, 2020 at 0:23 | comment | added | aglearner | Thanks! Maybe a last question, do you know some good and readable reference on Hilbert-Smith conjecture? | |
| Nov 3, 2020 at 0:21 | comment | added | YCor | @aglearner Yes. Because every continuous homomorphism of a profinite group into any Lie group has an open kernel. | |
| Nov 3, 2020 at 0:10 | comment | added | aglearner | So, when you write "If OP's assertion holds, then the isometry group being a Lie group, the map has an open kernel", you mean that this Lie group is a finite group? I don't quite understand why this sentence holds... | |
| Nov 3, 2020 at 0:05 | comment | added | YCor | Yes open kernel = kernel is open (= an open subgroup, which in a profinite group implies finite index). | |
| Nov 3, 2020 at 0:00 | comment | added | YCor | Actually for the HS conjecture it's enough to check it for $\mathbf{Z}_p$ for all $p$, for which the Haar measure is explicit (the various equivalences between versions of the HS conjecture are far harder than the existence of Haar measure on a compact group, anyway). | |
| Nov 2, 2020 at 23:57 | comment | added | aglearner | Thanks, that's very interesting. I guess, this is a classical fact that a profinite group has a Haar measure, is it easy to prove? Also, when you say "open kernel" you mean that the kernel of the action is an open subset of the group $G$? | |
| Nov 2, 2020 at 23:44 | history | answered | YCor | CC BY-SA 4.0 |