Timeline for Fundamental group of the homeomorphism group of a compact manifold
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Sep 23 at 18:55 | comment | added | William of Baskerville | It turned out that what YCor suggested in his first comment is true. Every connected separable locally contractible topological group has a countable fundamental group. The proof goes by looking at the universal cover, first noticing that it is separable, being generated by a separable identity neighbourhood, and then recalling that the fundamental group of the original group is in bijective correspondence with the fibres. These are countable by separability of the universal cover. | |
| Jul 10 at 13:54 | vote | accept | William of Baskerville | ||
| Jul 10 at 12:21 | answer | added | Tyrone | timeline score: 7 | |
| Jul 10 at 4:13 | comment | added | skupers | By Lemma 8.6 of arxiv.org/abs/2205.01755 it is countable. It need not be finitely-generated, e.g. by Example 3 of msp.org/pjm/1976/67-2/p09.xhtml. | |
| Jul 9 at 21:30 | comment | added | YCor | @KevinCasto yes, and this is an abuse of language to ignore basepoints. But topological groups have a canonical basepoint, so things are clear-cut. | |
| Jul 9 at 21:01 | comment | added | Kevin Casto | @YCor Sure, but extremely often people speak of "the fundamental group of a space" (for example, in this question) because we are used to connected spaces where we can ignore basepoint issues. | |
| Jul 9 at 20:56 | comment | added | YCor | @KevinCasto $\pi_1$ is defined for any based topological space, no need of connectedness. Of course this "ignores" all other path components. | |
| Jul 9 at 20:55 | comment | added | YCor | So, in the space of based loops, the (based) homotopy relation seems to be open, so that there are countably many equivalence classes. | |
| Jul 9 at 19:30 | comment | added | William of Baskerville | Sure, it is all about the identity component. | |
| Jul 9 at 19:28 | comment | added | Kevin Casto | Normally $\mathcal H(X)$ is disconnected, for example if $X$ is a surface than the connected components comprise the mapping class group. So normally you consider $\pi_1$ of the identity component (ofc since $\mathcal H$ is a group, all components have the same fund gp). | |
| Jul 9 at 18:36 | comment | added | William of Baskerville | It is locally contractible - this result is due to A. V. Černavskij (1969). | |
| Jul 9 at 18:31 | comment | added | YCor | Maybe it is locally contractible and every locally contractible Polish group has a countable fundamental group? | |
| Jul 9 at 18:22 | history | asked | William of Baskerville | CC BY-SA 4.0 |