Timeline for Homology of classifying space of spin group BSpin(n)
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Feb 23, 2021 at 13:10 | comment | added | Christoph Weis | This is an old thread, but it's worth pointing out that the statement needs amendment for n=4. As $\mathrm{Spin}(4)$ is isomorphic to $\mathrm{Spin}(3) \times \mathrm{Spin}(3)$, $\pi_3(\mathrm{Spin}(4))= \mathbb{Z}^2$. Following the argument in this answer one now concludes that indeed, $H^4(\mathrm{BSpin}(4);\mathbb{Z})$ is free on two generators. (See e.g. the nLab entry on Spin(4).) | |
| Feb 19, 2013 at 7:21 | comment | added | Somnath Basu | @Xiao-Gang : Write the spectral sequence (in homology) for the fibration $G\to EG\to BG$. It will follow from the $E^5$ and $E^6$ page that $H_5(BSpin(n);\mathbb{Z})=H_4(Spin(n);\mathbb{Z})$ and $H_6(BSpin(n);\mathbb{Z})=H_5(Spin(n);\mathbb{Z})$. It follows from Milnor-Moore (with your input on the homotopy groups of $Spin(n)$) that these are torsion groups. Now you just have to compute these! | |
| Feb 18, 2013 at 14:06 | comment | added | Xiao-Gang Wen | Do we have any result on $H_5(BSpin(n);\mathbb{Z})$ and $H_6(BSpin(n);\mathbb{Z})$ (say for n=10)? We know that $\pi_d(Spin(10))$ are $0,0,\mathbb{Z},0,0,0$ for $d=1,2,3,4,5,6$, which are simple. | |
| Nov 30, 2012 at 4:59 | comment | added | Peter May | It may be worth commenting that a beautiful paper of Quillen ``The mod 2 cohomology rings of extra-special 2-groups and the spinor groups'' computes the mod 2 cohomology of BSpin(n). Unusually, this is done without knowing the mod 2 cohomology of Spin(n) as a Hopf algebra. That is computed by Zabrodsky and myself (math.uchicago.edu/~may/PAPERS/19.pdf). | |
| Nov 30, 2012 at 3:33 | vote | accept | HBS | ||
| Nov 30, 2012 at 3:33 | comment | added | HBS | I didn't thought about Hurewicz theorem and I guess that answers my question. Thanks a lot. | |
| Nov 30, 2012 at 3:02 | history | answered | Somnath Basu | CC BY-SA 3.0 |