5

To elaborate on the question from the title, $\mathbb{R}_\delta$ is the additive group of real numbers (without any topology) and $K(\mathbb{R}_\delta,n)$ is an Eilenberg-MacLane space. I would like to know what its integral (co-)homology is, but I'm also happy about any results in this direction.

Bill Thurston has stated in a comment (Nontrivial finite group with trivial group homologies?) that the homology of $B\mathbb{R}_\delta = K(\mathbb{R}_\delta,1)$ has rank $2^\omega$ in each degree, but I don't know why this holds and I would also like to have results about $n>1$.

The background behind this question is that I'd like to show that $K(\mathbb{R}_\delta,n)$ cannot be a retract of a (smooth) manifold, and I'm hoping that (co-)homology will help me to figure this out.

Community
  • 1
Alexander Körschgen
  • 943
  • 2
    Doesn't homology commute with filtered colimits? Because $\mathbb{R}$ is surely a filtered colimit over its finite dimensional $\mathbb{Q}$ vector subspaces, and the cohomology of $K(\mathbb{Q},n)$ is easy :) – Achim Krause Jul 16 '15 at 19:28
  • 8
    $K(\mathbb R_\delta,n)$ cannot be a retract of a manifolds since a manifold is a countable CW complex and hence has countable homotopy groups. – Marc Hoyois Jul 16 '15 at 19:40
  • Achim, I think that homology commutes with filtered colimits if all your maps are cofibrations (so its a filtered _ho_colim). In this case, it should work by using appropriate models for the EM-spaces. So one should be able to reproduce André's computation after using a spectral sequence argument to compute the (co-)homology of the $K(\mathbb{Q},n)$. – Alexander Körschgen Jul 20 '15 at 18:40
  • Marc, that is also a nice way to solve my original problem, thanks. – Alexander Körschgen Jul 20 '15 at 18:42

1 Answers1

10

Given any $\mathbb Q$-vector space $A$ (in your case, $A=\mathbb R_\delta$), the integral homology of $K(A,n)$, which is the same as its rational homology, is given by the cofree graded-cocommutative coalgebra cogenerated by $A$ in degree $n$. In particular, for $n$ an even number, we have $$ H_m(K(A,n),\mathbb Z)=H_m(K(A,n),\mathbb Q)= \begin{cases}Sym^k A &\text{if $m=kn$} \\0 &\text{otherwise}\end {cases} $$

The rational cohomology of $K(A,n)$ is obtained by the universal coefficient theorem (UCT): $$ H^m(K(A,n),\mathbb Q)=\hom(H_m(K(A,n),\mathbb Q),\mathbb Q). $$ The integral cohomology is more messy to describe: it involves a non-trivial $ext$-term in the UCT short exact sequence (see the wikipedia entry on the UCT for more details).

André Henriques
  • 42,480
  • 2
    The rational cohomology can be thought of as a "completed" version of the free graded-commutative algebra on the $\mathbb{Q}$-dual $A^\vee$. Concretely, if ${b_i}$ is a basis for $A$ and ${c_i}$ are the dual functionals in $A^\vee$, $H^*(K(A,n),\mathbb{Q})$ is the ring of formal power series in the $c_i$ (commuting if $n$ is even, anticommuting if $n$ is odd). – Eric Wofsey Jul 16 '15 at 20:06
  • Ok, thanks guys! André, did you do this computation in the way that Achim outlined above? – Alexander Körschgen Jul 20 '15 at 18:45
  • Yes: Achim outlined the proof. – André Henriques Jul 21 '15 at 12:44