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Let $G$ be a topological group. In some cases, e.g. when $G$ is discrete or when the spaces $G^n$ are locally contractible and the coefficients are discrete, the cohomology of the classifying space $BG$ is the group cohomology of $G$. So, for simplicity, let us assume that $G$ is discrete.

My question: is there a nice explicit space $B_{\leq k}G$ that is functorial in $G$, such that $H^n(B_{\leq k} G, M) = H^n (BG, M)$ for $n \leq k$ and $0$ for $n>k$? Here $M$ is a $G$-module.

For example, for $G = S^1$, $BG = CP^\infty$ and a possible choice for $B_{\leq 2} S^1$ is $CP^1$. (If it simplifies the question, feel free to assume $M = \mathbf Z$.)

Thank you.

(Edit: In response to Ralph's comment, let us assume $G$ is discrete to simplify, but an answer for non-discrete groups would be interesting, too.)

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    $\begingroup$ In general the cohomology of the topological space $BG$ doesn't equal the group cohomology of $G$. However, they agree if $G$ is discrete. $\endgroup$
    – Ralph
    Commented Apr 25, 2013 at 21:58
  • $\begingroup$ Thanks. I have updated the question accordingly, even though the agreement for $H^*(BG,-)$ and $H^*(G, -)$ also holds in other cases that I am (actually) interested in. $\endgroup$
    – Jakob
    Commented Apr 25, 2013 at 22:40
  • $\begingroup$ What are these cases where $H^\ast(BG,-)=H^\ast(G,-)$ you have in mind ? $\endgroup$
    – Ralph
    Commented Apr 25, 2013 at 23:02
  • $\begingroup$ I was mostly thinking of $S^1$. $\endgroup$
    – Jakob
    Commented Apr 26, 2013 at 2:22
  • $\begingroup$ Note that $H^\ast(BS^1,-)\neq H^\ast(S^1,-)$ (where the latter is group cohomology). For example, $H^2(BS^1,\mathbb{Q})=\mathbb{Q}$, while $H^2(S^1,\mathbb{Q})=Hom_\mathbb{Q}(\mathbb{R},\mathbb{Q})$ has as $\mathbb{Q}$-dimension the cardinality of the power set of the reals. An interesting paper that compares the homology of $BG$ and the group homology of $G$ for Lie groups $G$ is Milnor: On the homology of Lie groups made discrete. Comment. Math. Helv. 58(1983), 72-85. $\endgroup$
    – Ralph
    Commented Apr 26, 2013 at 6:02

1 Answer 1

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There are several functorial models for $BG$, see for example [Adem, Milgram: Cohomology of Finite Groups, Chapter II] where $$BG = \coprod_{i=0}^\infty \sigma^i \times G^i/(\text{relations})$$ with $\sigma^i=\lbrace (x_1,...,x_i)\mid 0\le x_1 \le \cdots \le x_i \le 1\rbrace$ the standard $i$-simplex.

Now assume $G$ is discrete. Then $$B_nG := \coprod_{i=0}^n \sigma^i \times G^i/(\text{relations})$$ is the n-skeleton of $BG$ and depends functorially on $G$. Since the cohomology in degree $\lt n$ of a CW complex is determined by the $n$-skeleton, we obtain $H^k(B_nG;M)=H^k(BG;M)$ for $0 \le k < n$ and $H^k(B_nG;M)=0$ for $k> n$ and all (local) coefficients $M$.

In general $H^n(B_nG;M)\neq H^n(BG;M)$, but there is an exact sequence: Write $BG=EG/G$ $(EG$ is described explicitely in Adem-Milgram) and let $E_nG$ be the $n$-skeleton of $EG$. Then the following sequence is exact (Cartan, Eilenberg: Homologica Algebra, XVI, §9, Appl. 1): $$0 \to H^n(BG;M)\to H^n(B_nG;M)\to H^n(E_nG;M)^G \to H^{n+1}(BG;M)\to 0$$

Remark: In case that $G$ is not discrete, the above has to be adjusted accordingly. For example, if $G$ is (topologically) a $m$-dimensional CW complex then $B_nG$ is the $n(m+1)$-skeleton of $BG$.

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  • $\begingroup$ I took the liberty to editing it: I changed a "<" to \lt. It seems to have done the trick. $\endgroup$
    – José Figueroa-O'Farrill
    Commented Apr 25, 2013 at 23:13
  • $\begingroup$ Thank you, Ralph! This might be as close as one can get. A question concerning your last remark: for $G=S^1$, you say that $B_1 S^1 = S^1$ is the 2-skeleton of $BS^1 = CP^\infty$. Is there some typo here--I don't see how that can be true. I expect $sk_2 CP^\infty = CP^1$, instead. $\endgroup$
    – Jakob
    Commented Apr 26, 2013 at 2:26
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    $\begingroup$ I get $B_1S^1=\lbrace \ast\rbrace \; \coprod\; [0,1]\times S^1/\sim$ with the relations $$[0,1] \times 1 \sim \lbrace \ast\rbrace,\;0 \times S^1 \sim \lbrace \ast\rbrace,\;1\times S^1\sim \lbrace\ast\rbrace$$ This looks to me like $B_1S^1=S^2\cong CP^1$. $\endgroup$
    – Ralph
    Commented Apr 26, 2013 at 7:20

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