Given an integer $n$ and a group $G$ (abelian if $n \geq 2$), it's always possible to construct a $K(G,n)$ as a cell complex. The standard procedure is to choose a presentation $\langle S | R \rangle$ of $G$, construct a wedge sum of $n$-spheres (one for each generator in $S$), then attach an $(n+1)$-cell for each relation in $R$ via an attaching map determined by that relation. This gives you an $(n-1)$-connected space with $\pi_n = G$, and you continue attaching higher dimensional cells to kill off the higher homotopy groups.
This approach works, at least in theory, but relies on knowing what the next highest homotopy group at each stage is. The details get cumbersome as you attach more cells, and it's not clear to me that you can hope to get an explicit cell structure in this way. However, for some cases like $K(\mathbb{Z},2)=\mathbb{C} P^\infty$, the cell structure is rather simple to describe: $\mathbb{C} P^\infty$ has exactly one cell in every even dimension, and the attaching map for the $2k$-cell corresponds to the fiber bundle $S^{2k-1} \rightarrow \mathbb{C} P^{k-1}$.
My question is this: are there any other examples of $K(G,n)$'s that turn out to be familiar spaces like $S^1$, $\mathbb{C} P^\infty$, and $\mathbb{R} P^\infty$? Or just ones with easy to understand cell structures at all?
Best Answer
For $K(G,1)$ spaces there are some geometric methods. The key thing about the $K(G,1)$ property is that a connected CW complex possesses that property if and only if the universal covering space of that complex is contractible. And there are several examples of theorems of the form "such-and-such hypotheses imply that the universal covering space is contractible".
Here's one such class of examples, coming from Riemannian geometry. Let $M$ be a smooth connected $m$-manifold, and let $g$ be a complete Riemannian metric on $M$ (for what it's worth, all smooth compact manifolds have finite CW-complex structures). If all sectional curvatures of $g$ are non-positive, then $M$ is a $K(G,1)$ space. The reason this is true is because the universal cover $\widetilde M$ is simply connected $m$-manifold, the lift $\tilde g$ is a complete Riemannian metric with nonpositive sectional curvatures, and now one applies the Cartan-Hadamard theorem to conclude that $\widetilde M$ is diffeomorphic to $\mathbb R^m$ and is therefore contractible, and so $M$ is a $K(G,1)$ space.
For some very specific examples, if $m=2$ and if $M$ is any connected 2-manifold that is not homeomorphic to $\mathbb S^2$ or $\mathbb R P^2$ then $M$ is a $K(G,1)$ space (because all such surfaces have a complete Riemannian metric of constant curvature $0$ or $-1$, by an application of the Riemann mapping theorem). So, for example, for each $g \ge 2$ the closed, oriented surface of genus $g$ is a $K(G,1)$ space.