# Cohomology of the Eilenberg-Maclane space $K(\mathbb{R},1)$

Let $$\mathbb{R}$$ be the reals as an abelian group. A connected topological space $$X$$ is called an Eilenberg–MacLane space of homotopy type $$K(\mathbb{R},1)$$, if it has fundamental group isomorphic to $$\mathbb{R}$$ and all other homotopy groups trivial. Such a space exists, is a CW-complex, and is unique up to a weak homotopy equivalence.

1)Where can I find references to the cohomology of $$K(\mathbb{R},1)$$?

2)Could we calculate it from basic algebraic topology tools?

My interest on this space relies in the fact that $$K(\mathbb{R},1)=B\mathbb{R}^\delta$$, i.e., it is also the classifying space for the group $$\mathbb{R}^\delta$$, which is the real numbers with the discrete topology.

Also, it has relations with:

Friedlander-Milnor's Conjecture: Let $$G$$ be a Lie group, and let denote $$G^\delta$$ the same group with the discrete topology. Then the map $$H^*(BG,\mathbb{Z}_p)\to H^*(BG^\delta,\mathbb{Z}_p)$$ is an isomorphism for any $$p$$.

The map $$\mathbb{R}^{\delta}\to \mathbb{R}$$ is continuous, then we have also a continuous map at classifying space level $$B\mathbb{R}^{\delta}\to B\mathbb{R}$$. It is known that the map $$H^*(B\mathbb{R},\mathbb{Z})\to H^{*}(B\mathbb{R}^{\delta},\mathbb{Z})$$ is injective, but the proof comes from the solution of the Friedlander-Milnor's Conjecture for nilpotent Lie groups. I would like to see how this woks in this easiest example of $$\mathbb{R}$$.

Let's start with of $$K(\mathbb{Q},1)$$. We can explicitly construct a $$K(\mathbb{Q},1)$$ as the mapping telescope of a sequence of maps $$K(\mathbb{Z},1)\to K(\mathbb{Z},1)\to\dots$$ which parallel writing $$\mathbb{Q}$$ as the union of a sequence of cylic subgroups. From this construction we can compute that $$H_1(K(\mathbb{Q},1);\mathbb{Z})\cong\mathbb{Q}$$ and $$H_n(K(\mathbb{Q},1);\mathbb{Z})=0$$ for $$n>1$$, since $$K(\mathbb{Z},1)$$ is just $$S^1$$. See Group homology of the rationals for more details.
Now, $$\mathbb{R}$$ is just a direct sum of uncountably many copies of $$\mathbb{Q}$$. So, we can write it as a filtered colimit of copies of $$\mathbb{Q}^n$$ for finite $$n$$, and thus write $$K(\mathbb{R},1)$$ as a filtered homotopy colimit of $$K(\mathbb{Q}^n,1)$$'s. By the Künneth formula, we find that $$H_i(K(\mathbb{Q}^n,1);\mathbb{Z})$$ is a vector space over $$\mathbb{Q}$$ for all $$i>0$$. It follows that the same is true of $$K(\mathbb{R},1)$$, since any filtered colimit of abelian groups which are $$\mathbb{Q}$$-vector spaces is a $$\mathbb{Q}$$-vector space. It then follows immediately that $$H^i(K(\mathbb{R},1);\mathbb{Z}_p)$$ is trivial for $$i>0$$, since if $$V$$ is a $$\mathbb{Q}$$-vector space then $$\operatorname{Hom}(V,\mathbb{Z}_p)$$ and $$\operatorname{Ext}(V,\mathbb{Z}_p)$$ are trivial since they are both $$p$$-divisible and $$p$$-torsion.
With torsion-free coefficients, describing the cohomology is more complicated. Keeping track of the computations with $$H_i(K(\mathbb{Q}^n,1);\mathbb{Z})$$ above more carefully, we can see that for $$i>0$$, $$H_i(K(\mathbb{R},1);\mathbb{Z})$$ has as a basis over $$\mathbb{Q}$$ corresponding to subsets of size $$i$$ from a fixed basis for $$\mathbb{R}$$ over $$\mathbb{Q}$$. You can then use this to show that $$H^*(K(\mathbb{R},1);\mathbb{Q})$$ is a completed exterior algebra on generators corresponding to a a basis for $$\mathbb{R}$$ over $$\mathbb{Q}$$ (i.e., the inverse limit of the exterior algebras on finitely many of the generators at a time). Or, without choosing a basis, $$H^n(K(\mathbb{R},1);\mathbb{Q})$$ can be described as the space of alternating $$n$$-linear forms on $$\mathbb{R}$$ as a vector space over $$\mathbb{Q}$$, with the cup product corresponding to the natural product on such forms.
With coefficients in $$\mathbb{Z}$$ things are more complicated and I don't know a simple description of $$H^*(K(\mathbb{R},1);\mathbb{Z})$$. Note though that the fact that $$H^*(B\mathbb{R},\mathbb{Z})\to H^{*}(B\mathbb{R}^{\delta},\mathbb{Z})$$ is injective is totally trivial, since $$B\mathbb{R}$$ is contractible so its cohomology is trivial.
• Sorry, I am quite confused with fact that $Ext(V,\mathbb{Z})$ is trivial, because this paper shows that $Ext(\mathbb{Q},\mathbb{Z})$ is isomorphic to $\mathbb{R}$. – melomm Dec 5 '19 at 18:13
• Thanks @Eric Wofsey, you are right the injectivity is trivial since $B\mathbb{R}$ is contractible. I will add a comment in the question, because my interest goes deeper, I am looking for certain characteristic classes of flat $\mathbb{R}$-bundles. Then I will appreciate a more concrete exposition about the ring cohomology $H^*(K(\mathbb{R},1),\mathbb{Z})$. – melomm Dec 5 '19 at 18:33