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Let $[X,S^3]$ be the set of homotopy classes of maps from the pointed CW complex X to the pointed 3-sphere. The group structure on S^3 endows this set with a group structure. Is this group ever noncommutative?

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1 Answer 1

up vote 8 down vote accepted

Yes, by Yoneda's lemma. To generalize a bit, you have a group object $G$ in some category with finite products, and you ask whether the functor it represents is pointwise abelian. This is the same as asking whether two natural transformations $[-,G\times G]=[-,G]\times[-,G]\to[-,G]$ are equal (one being $(f,g)\mapsto f\cdot g$ and the other being $(f,g)\mapsto g\cdot f$). By Yoneda, this is this case iff the representing maps $G\times G\to G$ are equal. That is, the group $[X,G]$ is always abelian iff the multiplication map on $G$ is itself commutative. In the case of $G=S^3$ in the homotopy category of spaces, the group operation is not (homotopy-)commutative, though I don't know an easy proof of that off the top of my head.

If you want an explicit example of an $X$ that makes $[X,G]$ nonabelian, you just have to unravel the proof of Yoneda's lemma. That means you set $X=G\times G$ and consider the identity map $G\times G\to G\times G$, which corresponds the two projection maps $p_0,p_1\in[G\times G,G]$ when you identify $[G\times G,G\times G]=[G\times G,G]\times [G\times G,G]$. So in your case, the group $[S^3\times S^3, S^3]$ is nonabelian, and in particular the two projection maps do not commute.

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Eric, if $[S^3 \times S^3,S^3]$ is a noncommutative group, which noncommutative group is it? (It looks like an extension of a subgroup of $[S^3$ wedge $S^3,S^3] = \mathbf{Z}+\mathbf{Z}$ by a quotient of $\pi_1([S^5,S^3]) = \mathbf{Z}/12$ to me.) Which element is $gh$ and which element is $hg$? –  ya-tayr Mar 14 at 2:56
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That's a great question that I hope someone else can answer! –  Eric Wofsey Mar 14 at 2:57
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That $S^3$ is not homotopy commutative was apparently proved in [Samelson,H.,"Groups and wpaces of loops, Comm.Math.Helv.,28,278-87 (1954)] –  Mariano Suárez-Alvarez Mar 14 at 5:22
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@SpecialUnit2, why a subgroup of $Z\oplus Z$ and not the whole thing, and why a quotient of $Z/12$ and not the group itself? –  Mariano Suárez-Alvarez Mar 14 at 5:52
    
@MarianoSuárez-Alvarez, he was considering a certain long exact sequence from which it was not a priori clear that the corresponding maps are surjective and injective. They are, however, consider my answer to mathoverflow.net/questions/160297/… –  Achim Krause Mar 14 at 6:57

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