Timeline for Group Extensions and Line Bundles on $BG$
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 7, 2010 at 14:12 | vote | accept | Steve | ||
| Dec 6, 2010 at 22:13 | answer | added | Konrad Waldorf | timeline score: 5 | |
| Dec 6, 2010 at 22:05 | comment | added | Konrad Waldorf | ... and "line bundle" should probably be "hermitian line bundle". | |
| Dec 6, 2010 at 22:01 | comment | added | Konrad Waldorf | I think the statements in the question are correct given that $H^2(G,\mathbb{Z})$ is supposed to denoted group cohomology, while $H^2(X,\mathbb{Z})$ denotes ordinary cohomology. I'd better replace $H^2(G,\mathbb{Z})$ right away by $H^2(BG,\mathbb{Z})$. | |
| Dec 6, 2010 at 21:59 | answer | added | Mariano Suárez-Álvarez | timeline score: 7 | |
| Dec 6, 2010 at 21:58 | answer | added | Oscar Randal-Williams | timeline score: 15 | |
| Dec 6, 2010 at 21:46 | comment | added | Somnath Basu | I guess you meant line bundles on $G$ and not $BG$. And assuming that the action of $G$ on $\mathbb{Z}$ is trivial, $H^2(G;\mathbb{Z})$ classifies central extensions of $G$ by $\mathbb{Z}$. And finally, if your $G$ is a simply connected Lie group then $H^2(G;\mathbb{Z})=0$. | |
| Dec 6, 2010 at 21:24 | history | asked | Steve | CC BY-SA 2.5 |