# Classify $K(\pi,n)$ that are manifolds

Inspired by `Naturally occuring' $K(\pi, n)$ spaces, for $n \geq 2$. and When is a classifying space a topological manifold?, I'd like to formulate a precise question:

For which $n \in \mathbb{Z}_{\ge 2}$ and (isomorphism classes of) groups $\pi$ can the homotopy type $K(\pi,n)$ be represented by a topological manifold?

By a smooth manifold?

In case it's not clear, I'm looking to see if one can prove that one has a complete list.

Loosely, there's a kind of uncertainty principle relating homotopy and cohomology: it's hard for a space to simultaneously have few homotopy groups and few cohomology groups. So on the one hand it's hard for classifying spaces $B^n A$ to have bounded cohomology (I think they never have bounded cohomology if $n$ is even?), and on the other hand it's hard for finite CW complexes to have bounded homotopy.
The cohomology of classifying spaces $B^n A$ is extensively studied because they describe cohomology operations, so presumably someone who's more familiar with these can tell you more about no-go results from this direction. It's not hard to show that $B^n \mathbb{Z}$ has nonzero cohomology in arbitrarily high degrees when $n$ is even; you can take the cohomology operations to be cup powers. Similarly it's not hard to show that $B^n \mathbb{Z}_2$ has nonzero cohomology in arbitrarily high degrees for all $n$; here you can also take cup powers, but Steenrod operations are also available.
• The case $K(\mathbb Z, 2)$ motivates the possibly more interesting question: when is a $K(G,n)$ a Hilbert manifold? – Kevin Casto Mar 11 '16 at 8:31
• Every countable, locally finite simplicial complex is homotopy equivalent to an open subset of the standard Hilbert space. Every $K(G,n)$ for $G$ finitely generated is of this form. (If you only care about "closed" Hilbert manifolds, I must disappoint you: every (infinite-dimensional) Hilbert manifold is diffeomorphic to an open subset of the standard Hilbert space.) – Lennart Meier Mar 11 '16 at 9:22