Timeline for Homotopy classes of maps to Lie groups
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 20, 2017 at 20:15 | comment | added | Michael Albanese | The following fact is being used implicitly, but I think it is worth mentioning. If $G$ is a path-connected topological group, then for any pointed space $X$, we have $[X, G] = [X, G]_{\bullet}$ where the former denotes the set of free homotopy classes of maps $X \to G$, and the latter denotes the set of basepoint-preserving homotopy classes of maps. In general, $[X, Y]$ is not the same as $[X, Y]_{\bullet}$: the set $[X, Y]$ is a quotient of $[X, Y]_{\bullet}$ by a $\pi_1(Y)$-action. | |
| Apr 17, 2013 at 17:19 | vote | accept | Peter Kravchuk | ||
| Apr 16, 2013 at 21:57 | answer | added | Anton Fetisov | timeline score: 3 | |
| Apr 16, 2013 at 7:24 | answer | added | Neil Strickland | timeline score: 16 | |
| Apr 16, 2013 at 6:16 | comment | added | Ben McKay | Up to finite cover, every compact Lie group splits into a product of circles and compact simple groups, and we just get products of homotopy classes for each, so we can reduce to the case of $G$ a compact simple Lie group. | |
| Apr 16, 2013 at 0:44 | answer | added | Allen Knutson | timeline score: 5 | |
| Apr 15, 2013 at 21:32 | comment | added | Ben McKay | Since every connected Lie group $G$ retracts to a maximal compact subgroup $K$, if $M$ is connected then $[M,G]=[M,K]$, so we can assume that $M$ is compact. | |
| Apr 15, 2013 at 21:19 | history | asked | Peter Kravchuk | CC BY-SA 3.0 |