Timeline for Homotopy properties of Lie groups
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 10, 2022 at 12:40 | answer | added | Vitali Kapovitch | timeline score: 8 | |
| Mar 10, 2022 at 12:00 | vote | accept | Arshak Aivazian | ||
| Mar 10, 2022 at 11:13 | answer | added | Jesper Grodal | timeline score: 14 | |
| Mar 2, 2022 at 11:57 | comment | added | Arshak Aivazian | @YCor For every $n > 1$, the covering $p \colon X \to B$ establishes an isomorphism $\pi_n(p) \colon \pi_n(X) \to \pi_n(B)$ | |
| Mar 2, 2022 at 11:56 | comment | added | Tyrone | There are manifolds whose homotopy groups are those of a compact Lie groups, but which are not homotopy equivalent to any Lie group. (Neither does homology, cohomology, nor even unstable $\mathcal{A}_p$-algebra structure distinguish these objects.) | |
| Mar 2, 2022 at 10:13 | comment | added | YCor | @AivazianArshak since these are only products up to covering, I'm not sure exactly how one deduces the homotopy groups from those of factors. But I guess it's known. | |
| Mar 2, 2022 at 10:08 | comment | added | Arshak Aivazian | Hmm, indeed, in particular, the nth homotopy groups of any Lie group are simply some products of the nth homotopy groups of compact, simply connected, simple Lie groups. | |
| Mar 2, 2022 at 10:02 | comment | added | Mark Grant | Just riffing off @YCor's first comment: Looking at en.wikipedia.org/wiki/Compact_group#Classification, it seems that your manifold would have to be finitely covered by something of the homotopy type of a product of tori, $Sp(n)$'s, $SU(n)$'s, $\operatorname{Spin}(n)$'s and exceptional Lie groups. In particular it's universal cover is homotopy equivalent to a product of $Sp(n)$'s, $SU(n)$'s, $\operatorname{Spin}(n)$'s and exceptional Lie groups. | |
| Mar 2, 2022 at 9:12 | comment | added | Arshak Aivazian | I know about it from Hatcher, thanks | |
| Mar 2, 2022 at 9:11 | comment | added | YCor | This doesn't answer your question but you might be interested in the notion of H-space. | |
| Mar 2, 2022 at 9:09 | comment | added | Arshak Aivazian | @YCor Yes, I assume $G$ is connected, added to question text, thanks. | |
| Mar 2, 2022 at 9:06 | history | edited | Arshak Aivazian | CC BY-SA 4.0 |
added 55 characters in body
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| Mar 2, 2022 at 8:18 | comment | added | YCor | You seem to assume $G$ connected. Indeed $G$ is homotopy equivalent to a connected compact Lie group (which are far from arbitrary among compact smooth manifolds). These are classified and, for instance, their rational homotopy groups are known. | |
| Mar 2, 2022 at 7:33 | history | asked | Arshak Aivazian | CC BY-SA 4.0 |