Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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)?

share|improve this question
    
I deleted my answer as I misread the question. –  Jim Conant Apr 13 '11 at 14:44
add comment

1 Answer

up vote 8 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 $$K(G, 1) \to * \to K(G,2).$$

share|improve this answer
    
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
1  
That group is zero when $G$ is cyclic. –  Oscar Randal-Williams Apr 13 '11 at 18:01
1  
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
1  
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
show 3 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.