Timeline for Cohomology ring of BG
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Mar 1, 2016 at 1:31 | comment | added | jdc | Because $T$ is path-connected, left multiplication by $t$ has Lefschetz number equal to the Euler characteristic. Since $t \neq 1 \in T$ acts without fixed points on the sphere and $C$, these have Euler characteristic 0. In particular, the sphere is odd-dimensional. By excision, the relative Euler characteristic $\chi(G/N,C) = 1$. The long exact sequence of a pair then gives $\chi(G/N) = 1$. Since $G/T \to G/N$ is a finite covering with fiber $W$ and $H^{\mathrm{odd}}(G/T) = 0$, it follows $H^{\mathrm{odd}}(G/N) = 0$ and so $\dim_{\mathbb Q} H^*(G/N) = 1$. | |
| Mar 1, 2016 at 1:29 | comment | added | jdc | For posterity, here is a proof borrowed from Mimura and Toda. It relies on already knowing $H^*(G/T)$ is concentrated in even degree, something I believe was initially due to Borel. First, show the only fixed point of left multiplication by any $t \in T \setminus \{1\}$ on $G/N$ is the identity coset $1N$. Now $T$ acts by isometries on the normal $\mathfrak n^\perp \cong \mathfrak g /\mathfrak n$, preserving a small sphere. The exponential of this sphere is another sphere, $T$-invariantly dividing $G/N$ into a disk and a complement $C$. | |
| Dec 20, 2012 at 20:27 | comment | added | Craig Westerland | You can avoid the first spectral sequence if you have another way of talking yourself into believing that the cohomology of $BN$ is the same as the $W$-invariants of $H^*(BT)$, e.g., using the transfer. For the latter spectral sequence, you can use the fact that the Euler class of $G/T$ is nonzero, as Chris Gerig indicates in his answer. | |
| Dec 20, 2012 at 20:21 | comment | added | Paul Siegel | This looks very nice, thanks! My only apprehension - apart from the fact that I'm not sure either how to calculate the rational homology of $G/N$ in general - is that I would like to avoid spectral sequences when I type this up. That might be awkward in this case... | |
| Dec 20, 2012 at 20:03 | history | answered | Craig Westerland | CC BY-SA 3.0 |