5
$\begingroup$

The Theorem 1.5 and 1.6 of

Brown, Edgar H., Jr. The cohomology of BSOn and BOn with integer coefficients. Proc. Amer. Math. Soc. 85 (1982), no. 2, 283–288.

give a general answer for $H^d(BSO_n,Z)$ and $H^d(BO_n,Z)$, which are very complicated. I wonder what are $H^d(BSO_\infty,Z)$ and $H^d(BO_\infty,Z)$ for $d=1,2,3,4,5$?

Improve this question
$\endgroup$
1
  • 4
    $\begingroup$ The inclusions $BO(n)\to BO$ and $BSO(n)\to BSO$ are $n$-connected. So the $d$-dimensional cohomology in the infinite case is the same as for $n$ very large compared to $d$. There's no getting round the fact that these groups are complicated! But I suppose the description of the cohomology ring is simpler in the infinite case. $\endgroup$
    – Mark Grant
    May 5, 2014 at 11:54

1 Answer 1

Reset to default
6
$\begingroup$

The Theorem 1.5 and 1.6 you quote give the answer.

More precisely, for $SO$, in the range $d<6$, the only polynomial generators are $p_1$ which has degree 4, $\delta(w_2)$ with degree 3 and $\delta(w_4)$ with degree 5. The only relations are $2\delta(w_{2i})=0$, which gives $$H^d(BSO_{\infty};\mathbb{Z})\cong 0,\quad (d=1,2),$$ $$H^d(BSO_{\infty};\mathbb{Z})\cong \mathbb{Z}/2 ,\quad (d=3,5), $$ $$H^4(BSO_{\infty};\mathbb{Z})\cong \mathbb{Z}.$$

In the case of $BO$, there are more generators $\delta(w_1)$ and $\delta(w_1w_2)$ in degrees 2 and 4. Thus in degrees 4 and 5 we also have products $\delta(w_1)^2$ and $\delta(w_1)\delta(w_2)$.

All of these lead to $$H^1(BO_{\infty};\mathbb{Z})\cong 0,$$ $$H^i(BO_{\infty};\mathbb{Z})\cong \mathbb{Z}/2, \quad (i=2,3)$$ $$H^4(BO_{\infty};\mathbb{Z})\cong \mathbb{Z}\oplus (\mathbb{Z}/2)^2,$$ $$H^5(BO_{\infty};\mathbb{Z})\cong \mathbb{Z}/2\oplus \mathbb{Z}/2.$$

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thank you for the answer. I am not familiar with those polynomial rings. But I wonder do you also have $\delta(w_4)$ term for the $BSO$ and $BO$ cases, and the $\delta(w_1)\delta(w_1)$ for the $BO$ case? May be there are few more $Z/2$'s. $\endgroup$
    – Xiao-Gang Wen
    May 8, 2014 at 1:16
  • $\begingroup$ You are right, we have $\delta (w_4)$ in $H^5(BSO)$ and $H^5(BO)$, as well as $\delta (w_1)^2$ in $H^4(BO)$. I will edit the answer $\endgroup$
    – user43326
    May 8, 2014 at 9:29
  • 1
    $\begingroup$ I do wonder if that $H^5$ in the first block, which originally came before the $H^4$ (I re-ordered them in ascending dimension) was meant to be an $H^3$...? $\endgroup$
    – David Roberts ♦
    Apr 1, 2019 at 13:15
  • 1
    $\begingroup$ @DavidRoberts thank you for the comment and edit. Actually it was both $H^3$ and $H^5$, I corrected. $\endgroup$
    – user43326
    Apr 5, 2019 at 7:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.