Eilenberg-Maclane space, when can $K(\pi, 1)$ be constructed as a compact manifold? See here.


  Let $\pi$ be any group. Construct a connected CW complex $K(\pi, 1)$ such that $\pi_1(K(\pi, 1)) = \pi$ and $\pi_q(K(\pi, 1)) = 0$ for $q \neq 1$.


It is rarely the case that $K(\pi, 1)$ can be constructed as a compact manifold. What is a necessary condition on $\pi$ for this to happen?
$\pi$ has to be torsions free:

Let $M$ be a compact aspherical manifold and $C$ a finite cyclic subgroup of its fundamental group. As $M$ is aspherical, its universal covering $\tilde{M}$ is contractible, so the quotient $\tilde{M}/C$ provides a finite dimensional model for the classifying space $BC$ of $C$. In particular, $C$ has nonzero homology groups only in finitely many degrees. This implies that $C$ is trivial.
If $\pi$ acts freely and properly on $R^n$ and the quotient is compact. Example $\pi=Z^n$ $K(Z^n,1)$ is the n-dimensional torus.