Timeline for Bredon cohomology of a permutation action on $S^3$
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Aug 17, 2020 at 23:58 | comment | added | Steve Costenoble | I'm not sure I understand exactly what you're suggesting, but watch out for the fact that the quotient of $(D(V),S(V))$ by $G$ may have cohomology in several degrees, both even and odd, so I don't see that the spectral sequence necessarily collapses. | |
| Aug 17, 2020 at 17:37 | comment | added | Grisha Taroyan | Thank you for great reference! I had another question about possibly non-trivial representation on the cells. If I restrict to the case of the coefficient system $\underline{\mathbb{Z}}$ I'm just computing the cohomology of the quotient space. Assume that cellular structure is preserved in a sense that the image of a cell under a group action is once again a cell. Then the spectral sequence of the filtration on the quotient space collapses at $E_2$ for dimensional reasons, just like in the case of cellular cochains. Am I getting it wrong? | |
| Aug 17, 2020 at 17:21 | comment | added | Steve Costenoble | One other situation where you can use arbitrary cells: At least for $G = \mathbb{Z}/p$, if you have a space built out of cells of the form $G/H\times D(V)$ where $V$ is even-dimensional, but can vary, then, with some additional assumptions, you can conclude that the $RO(G)$-graded cohomology is a free module over the cohomology of a point, though the generators may not be where you expect them to be. I think this was first proved by Ferland and Lewis. | |
| Aug 17, 2020 at 14:47 | comment | added | Steve Costenoble | Not easily, at least not if you want to calculate the integer-graded part of the cohomology. Any filtration gives rise to a spectral sequence that could, theoretically, be used for computation, but that's not going to be straightforward. There is a notion of $G$-CW($V$) complexes, using cells of the form $G/H\times D(V)$, but that calculates the cohomology in grading $V$. I think Stefan Waner first noticed this, Gaunce Lewis published an exposition, and then Stefan and I published a book using a very generalized version. | |
| Aug 17, 2020 at 12:28 | comment | added | Grisha Taroyan | Is there a way to work with "cells" with nontrivial action of the stabilizer? I'm now aware that any cell with nontrivial action can be subdivided into "good" G-cells, but is there a way to avoid this step? | |
| Aug 17, 2020 at 10:24 | vote | accept | Grisha Taroyan | ||
| Aug 17, 2020 at 10:23 | comment | added | Steve Costenoble | All the cells in a $G$-CW complex have to have the form $G/H\times D^n$ where $G$ acts trivially on $D^n$. $T\times T$ doesn't have that form, it looks like the disc of a nontrivial representation of $G$. | |
| Aug 17, 2020 at 8:15 | comment | added | Grisha Taroyan | It is not a G-CW structure because fixed points are not a G-CW subspace? | |
| Aug 17, 2020 at 7:22 | vote | accept | Grisha Taroyan | ||
| Aug 17, 2020 at 8:14 | |||||
| Aug 16, 2020 at 23:04 | history | answered | Steve Costenoble | CC BY-SA 4.0 |