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While dealing with $BO(n)$, $BSO(n)$ and $BSpin(n)$ with the universal coefficient theorem and Künneth formula, I came to have the following question:

The universal coefficient says $H^n(X;M)\cong \hom(H_{n}(X;\mathbb{Z}),M)\oplus {\rm Ext}^{1} (H_{n-1}(X;\mathbb{Z},M))$ for a $\mathbb{Z}$-module $M$.

When $X=BSpin(n)$, we know that $H^4(BSpin(n);\mathbb{Z})\cong \mathbb{Z}$ and it seems likely that once we know what $H_p(BSpin(n);\mathbb{Z})$ would be for $p=3,4$ we might be able to retrieve this isomorphism with the aid of universal coefficient theorem.

So what would be $H_p(BSpin(n);\mathbb{Z})$ at least for $p=0,1,2,3,4$?

(I have to say that the question is not about how to prove the isomorphism $H^4(BSpin(4);\mathbb{Z})\cong \mathbb{Z}$.)

Thanks!

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  • $\begingroup$ The isomorphism you say this question is not about is not an isomorphism - $H^4(\mathrm{BSpin}(4);\mathbb{Z})\cong \mathbb{Z}^2$. See also my comment on the accepted answer. $\endgroup$
    – Christoph Weis
    Commented Feb 23, 2021 at 13:13

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Consider the classifying space $EG$ of a given group $G$. In your case, $G={Spin}(n)$. We have a fibration $G\to EG\to BG$ where $EG$ is contractible and $G$ acts freely on it. Therefore, the long exact sequence in homotopy groups tells you that $\pi_j(BG)\cong \pi_{j-1}(G)$. But $G={Spin}(n)$ is a Lie group which is simply connected. It is also classically known that $\pi_2(G)=0$ for finite dimensional Lie groups $G$. And, we also know that $\pi_3({Spin}(n))=\mathbb{Z}$. This implies that $\pi_j(B {Spin}(n))=0$ for $j=0,1,2,3$. Moreover, $\pi_4(B {Spin}(n))\cong H_4(B{Spin}(n);\mathbb{Z})$ by Hurewicz theorem. This is also isomorphic to $\pi_3({Spin}(n))$. The universal coefficient theorem now tells you that $H^4(B{Spin}(n);\mathbb{Z})=\mathbb{Z}$.

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  • $\begingroup$ I didn't thought about Hurewicz theorem and I guess that answers my question. Thanks a lot. $\endgroup$
    – HBS
    Commented Nov 30, 2012 at 3:33
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    $\begingroup$ 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). $\endgroup$
    – Peter May
    Commented Nov 30, 2012 at 4:59
  • $\begingroup$ 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. $\endgroup$
    – Xiao-Gang Wen
    Commented Feb 18, 2013 at 14:06
  • $\begingroup$ @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! $\endgroup$
    – Somnath Basu
    Commented Feb 19, 2013 at 7:21
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    $\begingroup$ 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).) $\endgroup$
    – Christoph Weis
    Commented Feb 23, 2021 at 13:10

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