Timeline for Stable cohomology of mapping class group with coefficients in $H^{\otimes n}$
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 21, 2022 at 14:31 | vote | accept | Fred | ||
| Sep 21, 2022 at 12:09 | answer | added | archipelago | timeline score: 4 | |
| Sep 20, 2022 at 21:34 | comment | added | Fred | @archipelago: Ah, that's exactly what I was looking for. If you post it as an answer, I'll accept it. Thanks! | |
| Sep 20, 2022 at 20:37 | comment | added | archipelago | I recommend to have a look at Appendix B of Randal-Williams' "Cohomology of automorphism groups of free groups with twisted coefficients". The stable description of the graded $\Sigma_q$-module $H^*(\Gamma_g;H^{\otimes q})$ appears in the middle of page 1471. | |
| Sep 20, 2022 at 19:49 | history | edited | Fred | CC BY-SA 4.0 |
deleted 14 characters in body
|
| Sep 20, 2022 at 19:44 | history | edited | Fred | CC BY-SA 4.0 |
deleted 14 characters in body
|
| Sep 20, 2022 at 19:39 | comment | added | Fred | @DanPetersen: Whoops, you're absolutely right! I'll work out the correct answer at least for $H$ (maybe it will be a pain now to get $H^{\otimes 2}$) and update the question. | |
| Sep 20, 2022 at 19:32 | comment | added | Dan Petersen | Stably, the cohomology ring of $M_g$ is a polynomial algebra in the $\kappa$ classes whereas the cohomology ring of $M_{g,n}$ is a polynomial algebra in the $\kappa$ classes as well as the $n$ classes $\psi_1,\ldots,\psi_n$. In your calculation you haven't taken into account the higher powers of $\psi$. | |
| Sep 20, 2022 at 19:15 | comment | added | Fred | More generally, I would be interested in any nice description. If a generating function like you describe is the best possible answer, then that's fine with me. | |
| Sep 20, 2022 at 19:13 | comment | added | Fred | point just introduces a new stable generator in $H^2$ corresponding to the first Chern class of the vertical tangent bundle. | |
| Sep 20, 2022 at 19:13 | comment | added | Fred | @DanPetersen: Are you sure? Remember that we're just working rationally. The proof for $H$ uses the fact that the universal curve $\mathcal{M}_{g,1} \rightarrow \mathcal{M}_g$ is an algebraic fiber bundle with projective fiber, so the spectral sequence for it degenerates. I think this means that $H^k(\mathcal{M}_{g,1})$ is the direct sum of $H^k(\mathcal{M}_g)$ and $H^{k-1}(\mathcal{M}_g;H)$ and $H^{k-2}(\mathcal{M}_g)$. But it's easy to see that we also have $H^k(\mathcal{M}_{g,1})$ equal to the direct sum of $H^k(\mathcal{M}_g)$ and $H^{k-2}(\mathcal{M}_g)$ -- the marked (continued) | |
| Sep 20, 2022 at 19:09 | comment | added | Dan Petersen | One can give a kind of generating series for these cohomology groups in terms of plethysm of symmetric functions, would you be interested in this? | |
| Sep 20, 2022 at 19:08 | comment | added | Dan Petersen | Your calculations by hand are definitely wrong. The cohomology with coefficients in $H$ does not vanish stably. Neither does the cohomology with coefficients in the two nontrivial representations of weight two. | |
| S Sep 20, 2022 at 17:39 | review | First questions | |||
| Sep 20, 2022 at 17:52 | |||||
| S Sep 20, 2022 at 17:39 | history | asked | Fred | CC BY-SA 4.0 |