Timeline for $H^4(BG,\mathbb Z)$ torsion free for $G$ a connected Lie group
Current License: CC BY-SA 3.0
23 events
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| Feb 17, 2018 at 0:08 | history | edited | André Henriques | CC BY-SA 3.0 |
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| Feb 19, 2016 at 11:33 | history | edited | André Henriques | CC BY-SA 3.0 |
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| Dec 20, 2014 at 23:57 | history | edited | André Henriques | CC BY-SA 3.0 |
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| Oct 2, 2014 at 12:33 | vote | accept | André Henriques | ||
| Oct 1, 2014 at 18:09 | answer | added | Matthias Wendt | timeline score: 37 | |
| Sep 9, 2014 at 2:12 | answer | added | Ben Wieland | timeline score: 12 | |
| Sep 8, 2014 at 22:46 | comment | added | André Henriques | @Ben: this is a great observation! I think that you should post it as an answer as it offers a very different and possibly quite useful perspective on my problem. I would still count it as "computational" though... in particular, it doesn't provide a very good understanding of what $H^4(G)$ really is, apart from the fact that it's torsion-free. (I would be delighted if you could contradict that last statement) | |
| Sep 8, 2014 at 18:25 | comment | added | Ben Wieland | @guest those pages only explicitly deal with the simply connected case, but I think that analysis of the spectral sequence it uses $H^*(BK;H^*(K/T))\Rightarrow H^*(BT)$ is easier than the one André gave. | |
| Sep 8, 2014 at 18:08 | comment | added | André Henriques | @Ben: $G=SO(3)$ is an easier example. | |
| Sep 8, 2014 at 17:51 | comment | added | Ben Wieland | @Chris, no, the statement is not for all even degrees. $SO(3)\times SO(3)$ has torsion in degree 6. | |
| Sep 8, 2014 at 17:26 | comment | added | Chris Gerig | My flaw: The 2nd "=" is not true, because I cannot divide by $\chi(G/T)$ with $\mathbb{R}/\mathbb{Z}$-coefficients (whereas it is true with $\mathbb{Q}$-coefficients). | |
| Sep 8, 2014 at 17:07 | comment | added | guest | Pages 45-46 of Deligne's paper are of relevance here numdam.org/numdam-bin/fitem?id=PMIHES_1996__84__35_0 | |
| Sep 8, 2014 at 13:12 | history | edited | André Henriques | CC BY-SA 3.0 |
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| Sep 8, 2014 at 12:15 | history | edited | André Henriques | CC BY-SA 3.0 |
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| Sep 8, 2014 at 9:15 | comment | added | Chris Gerig | Quick thought, probably flawed: $Tors(H^4(BG;Z)) = H^3(BG;\mathbb{R}/Z) = H^3(BT;\mathbb{R}/Z)^W$ where $T\subset G$ is maximal torus and $W$ is Weyl group, and I think $H^*(BT;\mathbb{R}/Z)$ has only even-dimensional generators? The first "=" is taken from the Dijkgraaf-Witten paper, I don't immediately see why it's true, but I do see that $H^k(BG;\mathbb{R}/Z)\hookrightarrow H^{k+1}(BG;Z)$ when $k$ is odd, with image given by kernel of $H^{k+1}(BG;Z)\to H^{k+1}(BG;\mathbb{R})$ which at least contains the torsion elements. Perhaps then $Tors(H^\text{even}(BG;\mathbb{Z})) = 0$? Doubtful. | |
| Sep 7, 2014 at 21:49 | history | edited | André Henriques | CC BY-SA 3.0 |
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| Sep 7, 2014 at 21:44 | history | edited | André Henriques | CC BY-SA 3.0 |
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| Sep 7, 2014 at 21:39 | history | edited | André Henriques | CC BY-SA 3.0 |
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| Sep 5, 2014 at 21:51 | comment | added | Dylan Wilson | but that's perhaps just as computational, and only covers the simply-connected case. | |
| Sep 5, 2014 at 21:50 | comment | added | Dylan Wilson | If $G$ is simply connected then $\pi_i(G)$ is torsion-free for $i<4$ since $\pi_2$ of a Lie group is zero and $\pi_3$ is always torsion-free. Thus $\pi_i(BG)$ is torsion-free for $i\le 4$ and the result follows from mod C Hurewicz. | |
| Sep 5, 2014 at 18:03 | comment | added | Matthias Wendt | @DanielBarter: in some sense yes. Cohomology of a finite groups is always annihilated by the order of the group, by a transfer argument. An easy explicit example of torsion in $H^4$ for finite groups appears e.g. for cyclic and dihedral groups. So the connectedness assumption is relevant. | |
| Sep 5, 2014 at 16:33 | comment | added | Daniel Barter | I am well out of my depth here, but is there a stupid reason why this is not true for finite groups? | |
| Sep 5, 2014 at 15:42 | history | asked | André Henriques | CC BY-SA 3.0 |