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It is known that group cohomology class $H^d[U(1),Z]$ is Z for even d and 0 for odd d. Do we know $H^d[G,Z]$ for $G=SO(3)$, $SU(2)$ and other compact Lie group?

Also is the Borel-group-cohomology class $H^d[G,R]$ alway trivial for compact Lie group $G$?

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Your first example suggests you are talking about the cohomology of the classifying space $BG$. If so, why not ask this? Also, your first example gives a counter-example for your second question (unless I'm misinterpreting what the "Borel-group-cohomology class" is). Please clarify. –  Mark Grant Sep 14 '11 at 13:55
    
Dear Mark, Thanks for the question. I just know the algebraic definition of group cohomology and I am not familiar with classifying space (I am a physicist). By Borel-group-cohomology of $H^d[G,R]$ I mean that we take cochains as measurable function over $G$. The issue of "continuity" comes up since both the group $G$ and the module $R$ are continuous. If $H^d[G,R]$ is trivial, I am hoping to get $H^{d+1}[G,Z] = H^d[G,U(1)]$, again $H^d[G,U(1)]$ is the Borel-group-cohomology described above. I learned those from some math papers that I only half understand. I hope they are right. –  Xiao-Gang Wen Sep 14 '11 at 15:07
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3 Answers

up vote 7 down vote accepted

For the group $SU(2)=S^3$ we just have $H^*(BSU(2);\mathbb{Z})=\mathbb{Z}[c_2]$ (where $c_2\in H^4$). More generally, for all $n$ we have \begin{align*} H^*(BU(n);\mathbb{Z}) &= \mathbb{Z}[c_1,\dotsc,c_n] \\\\ H^*(BSU(n);\mathbb{Z}) &= \mathbb{Z}[c_2,\dotsc,c_n] \\\\ H^*(BSp(n);\mathbb{Z}) &= \mathbb{Z}[p_1,\dotsc,p_n] \end{align*} with $c_i\in H^{2i}$ and $p_i\in H^{4i}$.

Now let $V$ be the tautological $3$-plane bundle over the space $X=BSO(3)$. This has Stiefel-Whitney classes $w_2\in H^2(X;\mathbb{Z}/2)$ and $w_3\in H^3(X;\mathbb{Z}/2)$. There is also a Bockstein element $v=\beta(w_2)\in H^3(X;\mathbb{Z})$ (which satisfies $2v=0$) and a Chern class $c=c_2(\mathbb{C}\otimes V)\in H^4(X;\mathbb{Z})$. The mod two reduction map $\rho$ satisfies $\rho(v)=Sq^1(w_2)=w_3$ and $\rho(c)=w_2^2$. If I've got everything straight, one can check using the Bockstein spectral sequence that $$ H^*(BSO(3);\mathbb{Z}) = \mathbb{Z}[v,c]/(2v). $$

It is not possible to be similarly explicit about $H^*(BSO(n);\mathbb{Z})$ for general $n$ (although $H^*(BSO(n);\mathbb{Z}/2)$ and $H^*(BSO(n);\mathbb{Q})$ are fairly straightforward).

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A collection of relevant references (general, standard as well as specific ones) is here:

http://ncatlab.org/nlab/show/group+cohomology#OnTopologicalGroups

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Check out

http://www.math.cornell.edu/~hatcher/SO/comments.pdf

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Thank you very much, Igor. But I am a physicist. May I ask how to obtain $H^d[G,Z]$ from its torsion free part and $H^d[G,Z_2]$. In particular, what are $H^4[SO(3),Z]$ and $H^5[SO(3),Z]$? I guess $H^3[SO(3),Z]=Z_2$ corresponding to the projective representations of $SO(3)$ and $H^2[SO(3),Z]=Z_1$ corresponding the absence of non-trivial 1D representation of $SO(3)$. ( I am using $H^{d+1}[SO(3),Z] = H^d[SO(3),U(1)]$ here, if I use $H^d[SO(3),U(1)]$ to mean Borel group cohomology where the conchains are measurable functions over $SO(3)$.) –  Xiao-Gang Wen Sep 14 '11 at 14:54
    
In planetmath.org/encyclopedia/GroupCohomology3.html , it is stated that $H^d[SU(2),Z]=Z$ if $d=0$ mod 4 and 0 other wise. This seems contradict with the result in your note: The torsion free part of $H^d[SO(3),Z]$ is Z for $d=0$ mod 3 is zero other wise. (I am guessing that the torsion free part of $H^d[SU(2),Z]$ and $H^d[SO(3),Z]$ are the same, and I thought $H^4[G,Z]$ contains a $Z$ for any Lie group.) Please correct me if I misunderstand your note –  Xiao-Gang Wen Sep 14 '11 at 23:48
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After thinking more, now I feel that math.cornell.edu/~hatcher/SO/comments.pdf may not be the answer. It calculates cohomology of the topological space of $SO(n)$. What I am seeking is the group cohomology of $SO(3)$ which is the cohomology of the topological space $BSO(3)$ -- the classifying space of $SO(3)$. –  Xiao-Gang Wen Sep 15 '11 at 0:14
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