Timeline for Another group cohomology cup product question
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 4, 2010 at 1:08 | answer | added | Ralph | timeline score: 3 | |
| Nov 2, 2010 at 2:40 | answer | added | grok | timeline score: 2 | |
| Nov 1, 2010 at 22:39 | comment | added | Mariano Suárez-Álvarez | The usual description of $H^1$ can be obtained at once from computing it using the Gruenberg resolution. To multiply, you probably want to compare that resolution with the bar resolution (this is one way to do this...) and induce corresponding isomorphisms on cohomology. Everything can be made explicit. | |
| Nov 1, 2010 at 20:12 | answer | added | John Palmieri | timeline score: 0 | |
| Nov 1, 2010 at 19:30 | comment | added | Josh | @Mariano: But the problem is, I don't know how to multiply elements of H^1 or H^2 without knowing what they are as cocycles. | |
| Nov 1, 2010 at 19:01 | comment | added | Torsten Ekedahl | A maybe not so wild guess; the coproduct map $H_2(G)\to \Lambda^2 H_1(G)$ is given by $R\cap[F,F]/[F,R] \to [F,F]/([F,R],[F,[F,F]]) = \Lambda^2(F/R)$. | |
| Nov 1, 2010 at 18:46 | comment | added | Mariano Suárez-Álvarez | (The ismorphism of $H_1$ with $H^1$ is quite unnatural...) What you want to do is doable: write down explicitely all the isomorphisms involed, and compose them. | |
| Nov 1, 2010 at 18:31 | history | asked | Josh | CC BY-SA 2.5 |