Given an Abelian group $G$ and a positive integer $n$, the Eilenberg-MacLane space $K(G,n)$ is a topological space such that $\pi_n(K(G,n))=G$, while $\pi_m(K(G,n))=0$ if $m\neq n$. For $n=1$ this coincides with the classifying space $BG$, and for many $G$ (for instance $U(1), \mathbb{Z}$ and cyclic groups) this has simple geometric realizations (for instance $K(U(1),1)=\mathbb{CP}^{\infty}$ while for $\mathbb{Z}_p$ is an infinite dimensional Lens space).
Is there some case in which $K(G,n)$, $n>1$ has a similarly simple geometric realization?
Side question: it is often written $K(G,n)=B^nG$, where $B^nG=B(B^{n-1}G)$ but I never understood this notation, since I know what is $B(\text{something})$ is $\text{something}$ is a group, but not in other cases. Could someone explain this to me?
Best Answer
Provided $G$ is an $E_n$ space, then $BG$, $B^2G$, $\dots$, $B^{n-1}G$ are topological groups, so $B^nG = B(B^{n-1}G)$ is defined, see this question. Moreover, if $G$ is a discrete abelian group, then $B^nG$ is defined for all $n$ and it is a $K(G, n)$, see this question. Note, if $G$ is not equipped with the discrete topology, then $B^nG$ need not be a $K(G, n)$. For example, a model for $BU(1)$ is $\mathbb{CP}^{\infty}$ which is not a $K(U(1), 1)$ as you claimed. Instead, we have a fibration $U(1) \to EU(1) \to BU(1)$ from which it follows that $\pi_k(BU(1)) \cong \pi_{k-1}(U(1))$, so $BU(1)$ is in fact a $K(\mathbb{Z}, 2)$. Another way to see this is that $U(1)$ itself is a model for $B\mathbb{Z}$, so $BU(1) = B^2\mathbb{Z}$.
As for explicit models of $K(G, n)$ for $n > 1$, aside from $\mathbb{CP}^{\infty}$, these are hard to come by. One surprising one is $Top/PL$ which is a $K(\mathbb{Z}_2, 3)$ which gives rise to the Kirby-Siebenmann invariant. Also see this question.