Timeline for $H^4(BG,\mathbb Z)$ torsion free for $G$ a connected Lie group
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 2, 2014 at 12:33 | vote | accept | André Henriques | ||
| Oct 2, 2014 at 4:17 | comment | added | Matthias Wendt | @AntonFetisov: not quite. For homology, $H_3(K(\pi,2),\mathbb{Z})=0$ by the Hurewicz theorem. André's calculation is about cohomology $H^3(BG,\mathbb{Z})$ whose torsion, by the universal coefficient formula, comes from the torsion in $H_2(BG,\mathbb{Z})\cong\pi_1(G)$. I think my argument applies not just to simply connected Lie groups. | |
| Oct 2, 2014 at 4:15 | comment | added | Matthias Wendt | @DanPetersen: I don't think so. The group $G$ is connected, so $BG$ is simply connected. In this situation, the Hurewicz theorem applied to $BG$ gives a bijection $\pi_2(BG)\to H_2(BG)$ and a surjection $\pi_3(BG)\to H_3(BG)$. The latter is the thing I use to represent any homology class by some bundle. | |
| Oct 2, 2014 at 1:00 | comment | added | Anton Fetisov | Since $\pi_3(BG) = \pi_1(BG) = 0$, $H_3(BG, \mathbb Z) = H_3( K(\pi, 2), \mathbb Z) = \pi_{tor}$, where $\pi \equiv \pi_1(G)$ and $\pi_{tor}$ is its torsion part (see André's calculation). Thus this reasoning covers only the simply connected case, which is obvious. | |
| Oct 1, 2014 at 21:24 | comment | added | Dan Petersen | Don't you need $G$ to be simply connected to apply the Hurewicz theorem? | |
| Oct 1, 2014 at 19:48 | comment | added | André Henriques | Note that this argument proves the slightly stronger statement $H_3(BG,\mathbb Z)=0$. | |
| Oct 1, 2014 at 18:52 | comment | added | Ulrich Pennig | I like this argument. | |
| Oct 1, 2014 at 18:09 | history | answered | Matthias Wendt | CC BY-SA 3.0 |