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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 $d=1,2,3,4,5$?

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    $\begingroup$ 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 '14 at 11:54
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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 $$H^1(BO_{\infty};\mathbb{Z})\cong 0,$$ $$H^i(BO_{\infty};\mathbb{Z})\cong \mathbb{Z}/2, \quad (i=2,3)$$ $$H^4(BO_{\infty};\mathbb{Z})\cong \mathbb{Z}\oplus (\mathbb{Z}/2)^2,$$ $$H^5(BO_{\infty};\mathbb{Z})\cong \mathbb{Z}/2\oplus \mathbb{Z}/2.$$

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  • $\begingroup$ 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 '14 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 '14 at 9:29
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    $\begingroup$ 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 at 13:15
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    $\begingroup$ @DavidRoberts thank you for the comment and edit. Actually it was both $H^3$ and $H^5$, I corrected. $\endgroup$ – user43326 Apr 5 at 7:17

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