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Can I find somewhere a table of the (first few) cohomology groups of $K(\mathbb{Z},n)$ with integer coefficients?

It seems like a natural counterpart to the table of the homotopy groups of spheres, but I couldn’t find anything. I’m aware of exposé 11, année 7 in the Cartan seminar where the homology of Eilenberg-MacLane spaces is computed, and I guess I could adapt it to the case where the group is $\mathbb{Z}$ and use the universal coefficient theorem to get the cohomology, but it’s not completely trivial, and I would be surprised that nobody thought about doing it before me.

Guillaume Brunerie
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    I would guess that books about cohomology operations would be a good place to look – Daniel Barter Apr 27 '16 at 16:20
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    The answer to the stable question (i.e. the integral cohomology of $H\mathbb{Z}$) is here: https://mathoverflow.net/questions/50519/integral-cohomology-stable-operations. Maybe there's something of interest for you there. – Denis Nardin Apr 27 '16 at 16:20
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    I'll add that the lack of interest compared to the case of $F_p$ is perhaps due to the fact that $H\mathbb{Z}$ is not a flat cohomology theory, and so it has no nice Adams spectral sequence that could serve as an application of the structure of the integral Steenrod algebra. – Denis Nardin Apr 27 '16 at 16:26
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    There is a paper "Integral cohomology operations" by Stan Kochman that treats the stable case, and this agrees with the unstable case through a range. However, the answer is unpleasant. I think that all possible applications can be done more cleanly using a combination of $H\mathbb{Z}/p$ and $H\mathbb{Q}$, or sometimes by using $K$-theory or complex cobordism instead. – Neil Strickland Apr 27 '16 at 16:44

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As far as I know, there is no complete source besides the Cartan seminar. Of course, the homology computation gives the cohomology computation immediately. However, if you are only interested in the first few groups, then you might be able to get away with just iteratively using the Serre spectral sequences for the fibrations $$K(\mathbb{Z},n-1)\rightarrow\star\rightarrow K(\mathbb{Z},n).$$ Since you know the integral cohomology of $K(\mathbb{Z},2)$ with its ring structure, and since the spectral sequences play well with the ring structure, it is not a difficult exercise to compute $H^*(K(\mathbb{Z},n),\mathbb{Z})$ for $*\leq 3n$ or so for any given $n\geq 3$.

Benjamin Antieau
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There are several computations carried out explicitly in the paper

Samuel Eilenberg and Saunders Mac Lane, MR 65162 On the groups $H(\Pi,n)$. II. Methods of computation, Ann. of Math. (2) 60 (1954), 49--139.

For instance, I once wanted to know the groups $H_{4+i}(K(\mathbb{Z},4);\mathbb{Z})$ for $i\le 3$, and was able to extract the answer from Section 24.

Mark Grant
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