Timeline for $H^4(BG,\mathbb Z)$ torsion free for $G$ a connected Lie group
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 21, 2014 at 22:08 | history | edited | André Henriques | CC BY-SA 3.0 |
deleted 3 characters in body
|
| Sep 19, 2014 at 15:19 | comment | added | Ben Wieland | I went back to Deligne and he gives an easy proof using just the one spectral sequence. Since everything on the 4 line is torsion-free, $H^4(BT)$ is the sum of those pieces, including $H^4(BG)$. So the image must saturate over $\mathbb Z$. In particular, if you know the rational result, the integral result follows. But identifying the image doesn't seem very closely related to knowing that there is no kernel. | |
| Sep 15, 2014 at 17:16 | history | edited | André Henriques | CC BY-SA 3.0 |
added 85 characters in body
|
| Sep 10, 2014 at 11:56 | comment | added | André Henriques | I don't know that it exactly answers the question, as the argument is still quite computational, but it certainly provides a very valuable piece of information. | |
| Sep 10, 2014 at 5:51 | comment | added | Ben Wieland | That makes sense, though I am still very confused. That answers your question, right? It identifies the group with a group that is obviously torsion-free. | |
| Sep 9, 2014 at 22:15 | comment | added | André Henriques | I think that even in the non-simply connected case $H^4(BK)=H^4(BT)^W$. The argument goes roughly as follows. Consider a finite cover $\tilde K$ of $K$ which is a product of some simply connected group and some torus. Let $A$ be the kernel of the projection map from $\tilde K$ to $K$. Then compare the Serre spectral sequences for $B\tilde K\to BK\to K(A,2)$ and $B\tilde T\to BT\to K(A,2)$. Key fact: the map of SS is injective in all the relevant bi-degrees, namely (0,2) (0,4) (3,0) (3,2) (5,0). The $E_\infty$ term for $H^4(BK)$ is therefore given by $H^4(BT)\cap H^4(B\tilde K)$. | |
| Sep 9, 2014 at 20:07 | comment | added | Ben Wieland | I was confused. Now I am so confused that not only do I not have an example that doesn't fill up the invariants, but my analyses reach contradictions. | |
| Sep 9, 2014 at 14:13 | comment | added | André Henriques | Ben: I'm not sure what you mean by "cf the characteristic class $\frac{p_1}2$". Could you please elaborate? | |
| Sep 9, 2014 at 2:12 | history | answered | Ben Wieland | CC BY-SA 3.0 |