Timeline for Gel'fand's and Fuks' calculation of cohomology of formal vector fields - isomorphic spectral sequences yield isomorphic cohomology?
Current License: CC BY-SA 4.0
11 events
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| Aug 3, 2018 at 16:31 | comment | added | Pedro | Ideally, one wants a map $F$ of filtered complexes from the complex that computes $H^*(W_n,\mathbb R)$ to the one that computes $H^*(X_n,\mathbb R)$. In the situation of the authors, where the $E_2$-pages are isomorphic (by this I mean the map induced by $F$ on that page is an isomorphism), you get that the map $F_\infty$ on the $E_\infty$ page is an isomorphism. Since this is just the map $\mathsf{gr}\,H^*(F)$, weak conditions on the filtrations will guarantee that $F$ is a quasi-isomorphism. Else, you can only say dimensions match. The first scenario is of course much more desirable. | |
| Aug 3, 2018 at 14:27 | comment | added | user126256 | Oh, that's way more simple than I thought! I wasn't aware the extension problem didn't occur for field coefficients. Should've done the calculation myself. Thank you, that was all I needed! | |
| Aug 3, 2018 at 14:09 | comment | added | mme | The $E^\infty$ page is the associated graded of cohomology (there don't seem to be any convergence issues). Then if you're looking to calculate rank of cohomology in each degree you are finished, because when you work over a field there are no extension problems; a filtered vector space is determined up to isomorphism by its associated graded. So you just need to read rank off of the diagonals. The ring structure is not completely determined but it seems this is taken care of by another step. | |
| Aug 3, 2018 at 11:32 | history | edited | user126256 | CC BY-SA 4.0 |
Added a few details.
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| Aug 3, 2018 at 11:21 | comment | added | user126256 | How do you mean? Would there be a way to derive from this fact that there's an isomorphism-inducing map on the chain-level or so? I must admit, I don't have a large body of experience with spectral sequence shenanigans. | |
| Aug 3, 2018 at 11:11 | history | edited | user126256 | CC BY-SA 4.0 |
One index was in the wrong position
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| Aug 3, 2018 at 11:10 | comment | added | Jeff Strom | Could it just be because $E_1=E_2$? | |
| Aug 3, 2018 at 11:06 | history | edited | user126256 | CC BY-SA 4.0 |
It's "Fuks'" and not "Fuks's". I fuksed up the title.
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| Aug 3, 2018 at 10:00 | history | edited | user126256 | CC BY-SA 4.0 |
Bolded up things to make it clear what the question is
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| Aug 3, 2018 at 9:15 | review | First posts | |||
| Aug 3, 2018 at 9:35 | |||||
| Aug 3, 2018 at 9:14 | history | asked | user126256 | CC BY-SA 4.0 |