Timeline for Faithful locally free circle actions on a torus must be free?
Current License: CC BY-SA 4.0
11 events
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Oct 13, 2023 at 11:02 | history | edited | YCor | CC BY-SA 4.0 |
to conform with title and answer, reformuated question to the positive
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Jan 9, 2023 at 4:36 | vote | accept | Chan Ki Fung | ||
Jan 8, 2023 at 20:45 | answer | added | Nick L | timeline score: 10 | |
Jan 8, 2023 at 10:21 | history | edited | Chan Ki Fung | CC BY-SA 4.0 |
deleted 1 character in body; edited tags
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Jan 7, 2023 at 23:44 | comment | added | Moishe Kohan | In a little while: It comes from the essential uniqueness of Seifert fibrations on 3-manifolds. | |
Jan 7, 2023 at 23:39 | comment | added | Chan Ki Fung | @MoisheKohan Could you please briefly explain the argument for dim 3? | |
Jan 7, 2023 at 14:42 | comment | added | Moishe Kohan | This is true when $n\le 3$, I do not know about dimension 4 and higher. | |
Jan 7, 2023 at 11:03 | history | edited | Chan Ki Fung | CC BY-SA 4.0 |
added 6 characters in body
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Jan 7, 2023 at 9:17 | comment | added | Chan Ki Fung | Yes I assume the action is smooth, but I am not sure whether there exists a transversal $T^{n-1}$. In fact, I am not sure $T^n/S^1$ is smooth. | |
Jan 7, 2023 at 8:34 | comment | added | Sam Nead | I assume that your action is smooth? If so, it seems reasonable to guess that a smooth, faithful, locally free action has a "global section" -- a torus $T^{n-1}$ which meets all orbits, exactly once, transversely. If there is a section, then the action should be conjugate to the action which rotates the first coordinate and fixes the others. | |
Jan 7, 2023 at 3:25 | history | asked | Chan Ki Fung | CC BY-SA 4.0 |