Integral cohomology ring of K(Z,3)

Question

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?

Answer 1

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 $H^*K(\mathbb{Z},3)$.

Some other things that are going on in the background here:

• 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 $K(\mathbb{Z}/n,1)\times K(\mathbb{Z}/n,1)\to K(\mathbb{Z}/n,2)$.