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Suppose I have a grouplike $A_{\infty}$-space $G$ that carries an additional structure as a topological group (which does not coincide with the $H$-space structure of the $A_{\infty}$-part). Denote the $H$-space multiplication by $*$ and the group multiplication by $\cdot$ and suppose that $$ (a*b) \cdot (c *d) = (a \cdot c) * (b \cdot d) $$ also known as the Eckmann-Hilton condition holds (equality, no "up to homotopy" here). Moreover, the identity of the group is a homotopy unit for the $H$-space structure.

Is this enough to deduce that $BG$ coincides with the space $B(G_{*})$, i.e. that the classifying space of the topological group is (up to homotopy of course) the delooping of the $A_{\infty}$-space? Do I get homotopy commutativity as well?

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up vote 15 down vote accepted

EDIT: Here is a counterexample to the stated question.

I'm going to start with a topological group $G$ which is a product of Eilenberg-Mac Lane spaces. Specifically, I'm going to choose $G \simeq K(\mathbb{Z},2) \times K(\mathbb{Z},7)$. The Dold-Thom theorem allows me to actually construct this as an abelian topological group: I take a based CW-complex with $H_2 = \mathbb{Z}$, $H_7 = \mathbb{Z}$, and other (reduced) homology groups zero (e.g. $S^2 \vee S^7$). Then I take the free topological abelian group on it, and this has the homotopy groups I want.

  • The classifying space of $G$ using this multiplication is a $K(\mathbb{Z},3) \times K(\mathbb{Z},8)$.

  • $G$ has a unique H-space structure. Because Eilenberg-Mac Lane spaces represent cohomology, $[G \times G, G]$ is $H^2(G \times G) \times H^7(G \times G)$. A calculation with the Kunneth formula tells you that the restrictions to each factor makes the map $$[G \times G, G] \to [G,G] \times [G,G]$$ into an isomorphism. Thus, up to homotopy there is a unique binary multiplication that's homotopy unital; the group multiplication on $G$ is a representative.

  • As a result, if I construct any other $A_\infty$-structure on $G$, the binary multiplication is homotopic to the multiplication of $G$. Since $A_\infty$-structures are insensitive to such a change by homotopy (yes, this is handwaving) I could equally well assume that the $A_\infty$-structure starts with the multiplication of $G$. Because $G$ is abelian, this satisfies the interchange law.

  • Since $A_\infty$-structures pass along homotopy equivalences, it suffices for me to find a space $Y$ and a homotopy equivalence $G \to \Omega Y$ where $Y$ is not $K(\mathbb{Z},3) \times K(\mathbb{Z},8)$.

  • There is a map $f: K(\mathbb{Z},3) \to K(\mathbb{Z},9)$ classifying the cohomology operation $x \mapsto x^3$ (which is nontrivial, and 2-torsion). I'm going to let $Y$ be the homotopy fiber of $f$. Then $\Omega Y$ is the homotopy fiber of a map $\Omega f: K(\mathbb{Z},2) \to K(\mathbb{Z},8)$ which is 2-torsion, and hence $\Omega f$ is trivial. Therefore $\Omega Y$ is homotopy equivalent to the homotopy fiber of the trivial map, and is a product homotopy equivalent to $G$.

  • The map $f$ is the first k-invariant in the Postnikov tower of $Y$, and is nontrivial, so $Y$ is not a product of Eilenberg-Mac Lane spaces.

An exercise for the reader who is not in bed right now, for whatever reason, is to trace through why several simpler choices of cohomology operation (e.g. $Sq^2: H^2 \to H^4$) don't make this argument work.


Below follows my original post. I assumed that the interchange law held for the entire $A_\infty$-structure, meaning that for any of the $n$-ary operations $m_n$ we would have $$ m_n(x_1 \ast y_1, \ldots, x_n \ast y_n) = m_n(x_1, \ldots, x_n) \ast m_n(y_1, \ldots, y_n). $$ However, the question only says that the H-space structure itself satisfies the interchange law, and not the higher coherences.

The Eckmann-Hilton condition implies that you can take classifying spaces twice, and the order of taking classifying spaces doesn't matter. You can write $BBG$ for this iterated classifying space. Both $B(G)$ and $B(G_*)$ are grouplike, and so the group completion theorem implies that the natural maps $B(G) \to \Omega BBG$ and $B(G_*) \to \Omega BBG$ are weak equivalences. Thus, under the standard types of point-set assumptions on $G$, both classifying spaces are weakly equivalent.

In addition, this implies that, because $G$ is grouplike, it is weakly equivalent to the 2-fold loop space $\Omega^2 BBG$ with either H-space structure, and hence both multiplications are homotopy commutative.

This is related to work of Brun, Fiedorowicz, and Vogt in "On the multiplicative structure of topological Hochschild homology." (preprint version here) They show that the "tensor product" of the associative operad with the little $n$-cubes operad is weakly equivalent to the little $(n+1)$-cubes operad. This means that having an associative multiplication and an $A_\infty$-multiplication is, up to homotopy equivalence, basically the same as having an $E_2$-multiplication.

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Thanks Tyler! I think the higher compatibility conditions also hold in the case I am considering. That counterexample is very enlightening. –  Ulrich Pennig Oct 31 '11 at 16:37
    
Shouldn't it be possible to deduce the existence of homotopy inverses from the Eckmann-Hilton condition as well? We know that G is a group: Can we run through the Eckmann-Hilton clock (up to homotopy) to deduce that homotopy inverses exist for the H-space structure? –  Ulrich Pennig Nov 8 '11 at 18:52
    
@Ulrich: Yes. An H-space structure on G is the same as a lift of the functor $X \mapsto [X,G]$ to monoids; it factors through a group-valued functor if and only if a homotopy inverse exists. Since you're assuming the H-space structure is the same on both ends, you do have that homotopy commutativity and a homotopy inverse exist for your $A_\infty$-structure. –  Tyler Lawson Nov 8 '11 at 18:57
    
Thanks again. Eckmann-Hilton really seems to be a "buy one, get one free"-condition :-)! –  Ulrich Pennig Nov 8 '11 at 23:05
    
One last question about your second argument: You say, that both $BG$ and $BG_*$ are grouplike. But I only know this for G itself. Why is this true for $BG$ and $BG_*$? –  Ulrich Pennig Nov 12 '11 at 12:46

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