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I am interested in knowing the explicit answers of Eilenberg–MacLane space $K(G,n)$ for explicit group examples ($G=$ the entries given at the top row) given below. Can someone fill in the Table?

$$\begin{array}{c|c|c|c|c|c|c|} G & \mathbb{Z}^n & \mathbb{Z} & \mathbb{R} &\mathbb{R}/\mathbb{Z}=U(1) & \mathbb{Z}_2 & \mathbb{Z}_n \\ \hline K(G,1) & T^n & S^1 & & CP^{\infty} & RP^{\infty} & S^{\infty}/\mathbb{Z}_n \\ \hline K(G,2)& & CP^{\infty} & & & & \\ \hline K(G,3)& & & & & & \\ \hline \end{array}$$

wonderich
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    $\mathbb{CP}^{\infty}$ is not a $K(U(1), 1)$. It is $BU(1)$, but the construction of classifying spaces takes the topology into account while Eilenberg-MacLane spaces are about discrete groups. Anyway, in general these things don't have very explicit descriptions. – Qiaochu Yuan May 21 '18 at 03:59
  • @ Qiaochu Yuan, What is the difference between $BG$ and $K(G,1)$ in general, and in this case? – wonderich May 21 '18 at 04:05
  • They're the same if $G$ is discrete, but $BG$ continues to make sense if $G$ a topological group (and $BU(1) \cong \mathbb{CP}^{\infty}$) whereas for $K(G, 1)$ to make sense $G$ must be a discrete group. You can consider $U(1)$ with the discrete topology but then $K(U(1)_d, 1)$ is not $\mathbb{CP}^{\infty}$ anymore. – Qiaochu Yuan May 21 '18 at 05:06
  • $K(\mathbb{Z}^n, 2) = K(\mathbb{Z}, 2)^n = (\mathbb{CP}^{\infty})^n$. – Michael Albanese May 21 '18 at 11:07
  • One model for $K(\mathbb{Z}_2, 3)$ is $TOP/PL$. – Michael Albanese May 21 '18 at 11:14
  • @ Michael Albanese and Qiaochu Yuan, is that true that $K(G,n)=B^nG$ only when $G$ is discrete? This equality does not hold when $G$ is continuous, also for other $n=1,2,3,...$? – wonderich May 21 '18 at 14:46
  • If you put spaces in the after the 'at' symbol, the user doesn't get notified. – Michael Albanese May 22 '18 at 17:26
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    As for your question, if $B^nG$ is defined, then $\pi_{i+n}(B^nG) = \pi_i(G)$ for every $i$, see this answer. So if $G$ is a group for which $B^nG$ is defined (that is, it is an $E_n$ space) and $\pi_i(G) = 0$ for $i > 0$, then $B^nG = K(\pi_0(G), 1)$. If $B^nG = K(G, 1)$, then $K(\pi_0(G), 1) = K(G, 1)$ so $\pi_0(G) \cong G$ as topological groups. As $\pi_0(G)$ is discrete, so is $G$. That is, $B^nG = K(G, 1)$ if and only if $G$ is discrete. – Michael Albanese May 22 '18 at 17:41

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