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How do we put a group structure on $K(G,n)$ that makes it a topological group?

I know that $\Omega K(G,n+1)=K(G,n)$ and since we have a product of loops this makes $K(G,n)$ into a H-space. But what about being a topological group?

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See algori's answer to this thread: mathoverflow.net/questions/12469/group-structure-on-cpinfinty basically, your question was answered by John Milnor in his paper "universal bundles I" –  Ryan Budney Jun 15 '11 at 18:20
    
@Grigory: well, I have to admit that I don't understand the issues well enough to write a comprehensive answer, so I would be much more comfortable if you supplied the correct details instead. –  Qiaochu Yuan Jun 15 '11 at 19:31

3 Answers 3

up vote 5 down vote accepted

// This is essentially the same answer that Qiaochu Yuan deleted for some reason. Hence CW.

By the Dold-Thom theorem $\pi_\bullet(\mathbb Z[X])=H_\bullet(X)$, where $\mathbb Z[X]$ is the free abelian group generated by $X$. Note that $\mathbb Z[X]$ is never connected ($\pi_0(\mathbb Z[X])=H_0(X)\supset\mathbb Z$) but it has reduced version: (fix a point $x_0\in X$ and define) $\widetilde{\mathbb Z[X]}:=\mathbb Z[X]/\mathbb Z[x_0]$ — and $\pi_\bullet\widetilde{\mathbb Z[X]}=\tilde H_\bullet(X)$.

// Remark 1. Actually (for connected $X$, at least), $\widetilde{\mathbb Z[X]}$ is homotopy equivalent to the free abelian monoid generated by $X$ with $x_0$ as the unit (aka $Sym^\infty(X):=\operatorname{colim}_{+x_0}Sym^n(X)$) that is used in the formulation of the Dold-Thom theorem in, say, nLab.

Now, it's easy to construct a Moore space $M(G,n)$ s.t. $\tilde H_n(M(G,n))=G$ and all other (reduced) homology groups of $M(G,n)$ are zero. And by Dold-Thom theorem topological abelian group $\widetilde{\mathbb Z[M(G,n)]}$ has homotopy type $K(G,n)$.

// Remark 2. In particular, $\widetilde{\mathbb Z[S^2]}$ is a topological abelian group with homotopy type of $\mathbb CP^\infty$. (Looks more explicit to me than most answers on MO.)

Upd. It's even easier to use $\widetilde{G[S^n]}\cong K(G,n)$.


Reference: M. C. McCord, Classifying spaces and infinite symmetric products, Transactions of the American Mathematical Society 146 (1969), 273–298.

See def. in the strart of Sec. 5 ($B(G,X)$ is my $\widetilde {G[X]}$), Prop. 6.6 (if $G$ is abelian group, so is $B(G,X)$), Cor. 10.6 ($\widetilde{G[S^n]}$ is an abelian group of homotopy type $K(G,n)$), Thm 11.4 (Dold-Thom type theorem I use; stated for any based $X$ having homotopy type of CW complex).


Another way to find homotopy type of $\mathbb Z[X]$ is to observe that if $X$ is cofibrant (e.g is a CW complex), $|X^\Delta|\to X$ (where $X^\Delta$ is the simplicial set of singular simplices in $X$) is a homotopy equivalence, so $\mathbb Z[X]\cong |\mathbb Z[X^\Delta]|$; now, by Dold-Kan correspondence homotopy groups of $\mathbb Z[X^\Delta]$ coincide with homology group of the corresponding complex — which is exactly singular complex of $X$.

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I'm still a bit doubtful as to the precise hypotheses on $X$ that are involved here. I agree that morally this should work, but as I already pointed out in a comment to Qiaochu's answer Dold and Thom didn't prove that result in this generality. I must say that I don't see the nlab as a very reliable source and while they're looking nice, the notes Qiaochu directed me to still haven't dispelled my doubts. Can you provide me with a reputable source in which the Dold-Thom theorem is stated and proved in the form that you use it? –  t.b. Jun 15 '11 at 23:50
    
@Theo I hope, Transactions of AMS is reputable enough source. –  Grigory M Jun 16 '11 at 7:15
    
Yes, definitely :) Thank you very much. –  t.b. Jun 16 '11 at 8:14
    
@Grigory M: what is $\mathbb Z[x_0]$ isn't this just $x_0$, recall that in other notation $\mathbb Z[X]$ is just the infinite symmetric product $SP^{\infty}(X)$ the direct limit over $n$ of $SP^{n}(X)=X^n/S_n$ where $S_n$ is the symmetric group. –  palio Jun 16 '11 at 16:20
    
@palio No-no, $\mathbb Z[x_0]=\mathbb Z$ (free abelian group generated on 1 generator, naturally); and $SP^\infty(X)$ is not $\mathbb Z[X]$ but $\widetilde{\mathbb Z[X]}$. –  Grigory M Jun 16 '11 at 17:39

The standard classifying space functor $B$ from topological groups to topological spaces is product preserving, so it takes abelian topological groups to abelian topological groups. Start with an abelian group $G$ as a discrete topological group, so a $K(G,0)$. Apply the functor $B$ iteratively $n$ times to reach $B^nG$, which is an abelian topological group and a $K(G,n)$.

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Well, an interval is a $K(\{e\},1)$, where $\{e\}$ is the trivial group, and an interval has no group structure that can make it a topological group (every continuous map has a fixed point). So there certainly are some $K(G,n)$'s out there that cannot be made into topological groups. Maybe if you refine the question there could be some sort of answer.

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I think the question should be made "how do we put a group structure on a space homotopy equivalent to $K(G, n)$?" (which is itself, of course, only well-defined up to homotopy). –  Qiaochu Yuan Jun 15 '11 at 16:54

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