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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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up vote 6 down vote accepted

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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I didn't thought about Hurewicz theorem and I guess that answers my question. Thanks a lot. –  hbs Nov 30 '12 at 3:33
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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). –  Peter May Nov 30 '12 at 4:59
    
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. –  Xiao-Gang Wen Feb 18 '13 at 14:06
    
@Xiao-Gang : Write the spectral sequence (in homology) for the fibration $G\to EG\to BG$. It will follow from the $E^5$ and $E^$ page that $H_5(BSpin(n);\mathbb{Z})=H_4(Spin(n);\Z)$ and $H_6(BSpin(n);\mathbb{Z})=H_5(Spin(n);\Z)$. 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! –  Somnath Basu Feb 19 '13 at 7:19
    
@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! –  Somnath Basu Feb 19 '13 at 7:21
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