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.


The answer is that this never happens for manifolds which are of finite type in the sense that they are homotopy equivalent to finite CW complexes. Serre showed that a simply connected finite CW complex has infinitely many nonzero homotopy groups.

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.

  • $\begingroup$ The case $K(\mathbb Z, 2)$ motivates the possibly more interesting question: when is a $K(G,n)$ a Hilbert manifold? $\endgroup$ – Kevin Casto Mar 11 '16 at 8:31
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    $\begingroup$ 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.) $\endgroup$ – Lennart Meier Mar 11 '16 at 9:22

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