Timeline for Spherical space form conjecture
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10 events
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| Sep 17, 2020 at 16:24 | comment | added | George K | @Moishe Kohan I do. I hoped that the fact that the topological and the smooth category coincide in 3 dimensions might yield something good, so I thought to ask. Meanwhile, I found a few papers on the matter and I see that this is not the case. You only have that smooth actions are smoothly conjugate to subgroups of O(4) and that locally linear actions are topologically conjugate to subgroups of O(4) (and O(4) of course further reduces to SO(4) for actions which preserve orientation). At least these are the only results that I found. Thank you all again. | |
| Sep 17, 2020 at 16:03 | comment | added | Moishe Kohan | @GeorgeK: Do you understand that your question is based on a false premise? | |
| Sep 17, 2020 at 14:57 | comment | added | Moishe Kohan | @IgorBelegradek Yes, one can prove that topological 3d orbifolds are smoothable. I am not sure if this is in the literature though. But it should be a good separate question. | |
| Sep 17, 2020 at 12:53 | comment | added | Igor Belegradek | I wonder if every locally linear (not necessarily free) action of a finite group on $S^3$ is conjugate to a linear action. The Ricci flow proof seems to fail in locally linear setting. | |
| Sep 17, 2020 at 11:11 | comment | added | mme | No. To rephrase Moishe: If your action is free, then it is topologically conjugate to a linear action (smoothly conjugate if the action was smooth to begin with). Non-free topological actions need not be smoothable. Eg, the double of the (non-manifold) Alexander horned ball is homeomorphic to $S^3$, and hence $S^3$ carries an involution with non-locally-flat fixed set, so this action is not even topologically conjugate to a smooth action. | |
| Sep 17, 2020 at 11:04 | comment | added | George K | @MoisheKohan Thank you. Does that mean that the spherical space form conjecture holds for topological, non-smooth actions over S^3 as well? | |
| Sep 17, 2020 at 9:46 | comment | added | Moishe Kohan | The SFS conjecture is about free finite group actions. If you drop the freeness assumption then in the smooth category every finite group action is still conjugate to an orthogonal action, but in the topological category this is no longer true (Bing's examples).It is a rather old theorem (Moise+Munkres) that in dimension 3 TOP=DIFF: every topological manifold admits a smooth structure and this structure is unique up to diffeomorphism. See the references I gave here. | |
| Sep 17, 2020 at 8:50 | comment | added | YCor | I think it's been known for a while (long before Hamilton–Perelman and even Thurston) that in dimension 3 the smooth and topological Poincaré conjectures are equivalent. Unfortunately I don't even seen this mentioned in Wikipedia's Poincaré conjecture page. Hopefully somebody will provide a reference and more precise statement. | |
| Sep 17, 2020 at 8:47 | history | edited | YCor | CC BY-SA 4.0 |
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| Sep 17, 2020 at 8:24 | history | asked | George K | CC BY-SA 4.0 |