Timeline for Computing the cardinality of cohomology groups
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 17, 2013 at 21:32 | answer | added | Derek Holt | timeline score: 6 | |
| Feb 17, 2013 at 8:34 | comment | added | Gregor Samsa | @Martin Brandenburg: I only claim that $|H^{2n}(G,\mathbb{Z})| = |\mathbb{Z}/k\mathbb{Z}| = k$ and did not intend to suggest that $|A/kA| = k$ or $\leq k$ holds in general. | |
| Feb 17, 2013 at 6:20 | answer | added | Mariano Suárez-Álvarez | timeline score: 2 | |
| Feb 17, 2013 at 3:05 | comment | added | Mariano Suárez-Álvarez | (And that is enough to show that the Poincare-Hilbert series of cohomology coverges, say) | |
| Feb 17, 2013 at 2:55 | comment | added | Mariano Suárez-Álvarez | There are silly bounds, assuming some information on the coefficients. For example, taking $A=\mathbb Z$ one has that the complex which computes cohomology in terms of the bar resolution is made up of free abelian groups of a rank one can make precise, so each cohomology group is generated by at most that number of elements; since we know multiplication by $|G|$ is zero on cohomology, this gives a bound. | |
| Feb 17, 2013 at 1:45 | history | edited | user9072 |
tags
|
|
| Feb 16, 2013 at 23:28 | answer | added | Chris Gerig | timeline score: 3 | |
| Feb 16, 2013 at 21:07 | history | asked | Gregor Samsa | CC BY-SA 3.0 |