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Question: How should we interpret and understand the classifying space $B^nG$? Is that Eilenberg-MacLane space $K(G,n)$?

What one can learn about $BG$ follows the basic: A classifying space $BG$ of a topological group $G$ is the quotient of a weakly contractible space $EG$ (i.e. a topological space for which all its homotopy groups are trivial) by a free action of $G$. It has the property that any $G$ principal bundle over a paracompact manifold is isomorphic to a pullback of the principal bundle $EG \to BG$. For a discrete group $G$, $BG$ is, roughly speaking, a path-connected topological space $X$ such that the fundamental group of $X$ is isomorphic to $G$ and the higher homotopy groups of $X$ are trivial, that is, $BG$ is an Eilenberg-MacLane space, or a $K(G,1)$.

So if $G$ is a finite discrete group, what are the conditions of the higher homotopy groups of $B^nG$?

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    $\begingroup$ Yes, when $G$ is a discrete abelian group, $B^n G = K(G,n)$. $\endgroup$
    – user98602
    Jan 29, 2017 at 23:42
  • $\begingroup$ Thanks so much, Mike Miller, how about in general, do we have some relations between the twos, and what are additional terms? $\endgroup$
    – wonderich
    Jan 30, 2017 at 0:51
  • $\begingroup$ @Mike Miller, how about in general? $\endgroup$
    – wonderich
    Oct 6, 2018 at 2:07
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    $\begingroup$ $K(G,n)$ does not mean anything when $G$ is not abelian and $n > 1$. When $G$ is a topological abelian group, $B^n G$ is a topological abelian group, with $\Omega^n B^n G \simeq G$. You can say things about the specific homotopy groups. When $G$ is not a discrete group, $K(G,n)$ does not mean anything. $\endgroup$
    – user98602
    Oct 6, 2018 at 3:51

1 Answer 1

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If $G$ is a topological group, then there is a universal principal $G$-bundle $EG \to BG$ where $EG$ is weakly contractible. Using the long exact sequence in homotopy, we see that

$$\dots \to \pi_{i+1}(EG) \to \pi_{i+1}(BG) \to \pi_i(G) \to \pi_i(EG) \to \dots$$

As $EG$ is weakly contractible, $\pi_{i+1}(EG) = 0$ and $\pi_i(EG) = 0$, so $\pi_{i+1}(BG) = \pi_i(G)$.

Now, if $BG$ is a topological group (which happens if and only if $G$ is an $E_2$ space), then the above argument shows that $\pi_{i+2}(B^2G) = \pi_{i+2}(B(BG)) = \pi_{i+1}(BG) = \pi_i(G)$.

In general, if $G$ is an $E_k$ space, then $B^kG$ is defined and satisfies $\pi_{i+k}(B^kG) = \pi_i(G)$.

If $G$ is a discrete group, then $\pi_0(G) = G$ and $\pi_i(G) = 0$ for $i > 0$, so $\pi_k(B^kG) = G$ and $\pi_i(B^kG) = 0$ for $i \neq k$. Therefore $B^kG$ is an Eilenberg-MacLane space $K(G, n)$.

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