Timeline for What does $\mathrm{Conf}_n(M)^{h S_n}$ look like?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 25 at 17:02 | comment | added | Mark Grant | @ConnorMalin: Thanks. I was able to work that out from the context. But it obviously is not the first category that comes to mind when thinking about equivariant homotopy theory. | |
| Nov 23 at 15:28 | comment | added | Connor Malin | @MarkGrant If it wasn't clear, I was writing in the homotopy category of Borel G-spaces, i.e. G-spaces where we invert G-equivariant maps which are underlying weak equivalences. | |
| Nov 23 at 15:08 | comment | added | Mark Grant | This is a nice answer, but I found it a bit confusing since the point $*$ is not $G$-homotopy equivalent to $EG$. Would it not be better to say "provides an equivariant splitting of the classifying map $X\to EG$"? | |
| Nov 23 at 14:13 | vote | accept | Nicolas Guès | ||
| Nov 23 at 14:13 | comment | added | Nicolas Guès | Great answer, thank you! | |
| Nov 23 at 8:44 | comment | added | Connor Malin | @R.vanDobbendeBruyn Functors preserve retractions, and so if a complex $X$ retracts onto a point in the homotopy category of $G$-spaces, it implies that $H_\ast (X_{hG})$ retracts onto $H_\ast(\ast_{hG})=H_\ast(BG)$. That implies that $H_\ast (X_{hG})$ is nonzero in all the degrees $H_\ast(BG)$ is. If the action of $G$ on $X$ is free and nice enough at the pointset level, then $X_{hG}\simeq X_G$. If $X$ is finite dimensional, this implies $H_\ast(X_{hG})$ is bounded, so if $H_\ast(BG)$ is unbounded such a map cannot exist. | |
| Nov 22 at 22:27 | comment | added | R. van Dobben de Bruyn | Could you maybe include a little more detail about what the actual obstruction is? It's not clear to me why unboundedness prevents this map from having a section. | |
| Nov 22 at 19:58 | comment | added | Chris Schommer-Pries | that is a nice argument! | |
| Nov 22 at 18:13 | history | answered | Connor Malin | CC BY-SA 4.0 |