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Recall that an Eilenberg-Maclane space $K(G, n)$ is characterized by $\pi_i(K(G,n)) = G$ if $i=n$ and is trivial otherwise. (Of course $G$ should be abelian if $n>1$.)

I'm aware that computing $H^j(K(G,n), \mathbb Z)$ for general $j$ and $n$ is not so easy (see, e.g., here), but I'm hoping that for certain small values of $j$ and $n$ it's easier.

My question: Is there a good reference for $H^j(K(G,2), \mathbb Z)$, where $j \le 4$ and $G$ is finite abelian (or just cyclic)?

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up vote 9 down vote accepted

For a finite cyclic group G, in the range you ask for you get cohomology groups $$\mathbb{Z}, 0, 0, G \cong Ext(G, \mathbb{Z}), 0.$$ One sees this by for example computing the Leray--Serre spectral sequence for $$K(G, 1) \to * \to K(G,2).$$

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Are you sure about that last zero? – Tyler Lawson Apr 13 '11 at 15:52
In cohomology? I think so: there is nothing that could kill it in the SS. – Oscar Randal-Williams Apr 13 '11 at 16:39
That group is zero when $G$ is cyclic. – Oscar Randal-Williams Apr 13 '11 at 18:01
For a low-dimensional calculation like this one can give a direct argument without using the spectral sequence. When $G$ is cyclic, one can build a $K(G,2)$ with 3-skeleton consisting of a 3-cell attached to $S^2$ by a map of nonzero degree. This forces $H_3$ to be zero. Since one knows the homology groups are finite in positive dimensions, the universal coefficient theorem then gives $H^4=0$, and of course $H^3=H_2=G$. (The usual proof that the homology groups are finite uses spectral sequences, but tom Dieck's recent algebraic topology textbook gives a proof avoiding this.) – Allen Hatcher Apr 14 '11 at 14:50
More generally for arbitrary abelian G and A we have $H^j(K(G,2);A) =$ $A$, 0, $Hom(G,A)$, $Ext(G;A)$, for $j = 0,1,2,3$ and $H^4(K(G,2);A) =$ the quadratic maps from G to A. This very classical and goes back to calculations of Whitehead and Mac Lane on classifying simply connected 2-types. You should especially look up Mac Lane's "Abelian Cohomology" which gives an explicit cocycle way of computing this. – Chris Schommer-Pries Apr 14 '11 at 15:14

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