# Integral cohomology ring of K(Z,3)

Computing the cohomology of Eilenberg Maclane spaces is a feasible but difficult problem in algebraic topology. The general answer is quite complicated (see the MO answer and the reference therein https://mathoverflow.net/a/24759/184 )

However the cohomology of K(Z,2) is quite simple, it is just a polynomial algebra on a generator in degree two. I am wondering about the next easiest case.

What is the integral cohomology ring of K(Z,3)?

You can get quite far with spectral sequence calculations, but if this is worked out in detail somewhere, why reinvent the wheel?

• I don't know about the integral cohomology, but the mod $p$ cohomology of $K(Z,n)$ in terms of Milnor basis (instead of Cartan-Adem basis) has been worked out by Tamanoi (JPAA 137, pp.153-199). This might make the computation of Bockstein spectral sequence easier. – user43326 Aug 31 '15 at 9:28
• Actually in Cartan seminar the homology of $K(Z,n)$ is computed, so by using universal coefficient theorem, you get at least additive structure of the cohomology. However, I have no idea on the ring structure. – user43326 Aug 31 '15 at 9:49
• Let me reproduce here a recent reference in a comment to an answer to the linked question: "Derived functors of the divided power functors" by Lawrence Breen, Roman Mikhailov and Antoine Touzé (Geometry & Topology 20 (2016) 257–352, arXiv preprint 1312.5676) contains in particular a description of $H_*(K(A,3);\mathbb Z)$ – მამუკა ჯიბლაძე Nov 30 '17 at 14:48

Given a complex oriented cohomology theory $E$, one can define a formal scheme of Weil pairings on the associated formal group, as explained in the paper "Weil pairings and Morava $K$-theory" by Matthew Ando and me. If we let $R_E$ denote the ring of functions on this scheme, then there is a natural map $R_E\to E^*K(\mathbb{Z},3)$. This is an isomorphism if $E$ is Morava $K$-theory or Morava $E$-theory. I think that it is also an isomorphism for $E=MU$ or $E=kU$ but not for $E=H$. However, there is a natural short exact sequence $$kU^*K(\mathbb{Z},3)/v \to H^*K(\mathbb{Z},3) \to \text{ann}(v,kU^*K(\mathbb{Z},3))$$ (where $v$ is the standard generator of $\pi_2kU=kU^{-2}(\text{point})$). I think that this is probably an effective way to understand $H^*K(\mathbb{Z},3)$.
• There is a fibration $K(\mathbb{Q}/\mathbb{Z},2)\to K(\mathbb{Z},3)\to K(\mathbb{Q},3)$. Here $K(\mathbb{Q},3)$ is the rationalisation of $S^3$ and is not so hard to understand.
• $K(\mathbb{Q}/\mathbb{Z},2)$ is the colimit of the spaces $K(\mathbb{Z}/n,2)$.
• One can understand $K(\mathbb{Z}/n,2)$ using the multiplication map $K(\mathbb{Z}/n,1)\times K(\mathbb{Z}/n,1)\to K(\mathbb{Z}/n,2)$.