10
$\begingroup$

There are twelve continuous maps $S^3\times S^3\to S^3$ up to homotopy that make the three-sphere $S^3$ into an H-space. This follows from a result of James [1], which says that if there exists one such multiplication on $S^n$ then the homotopy classes of multiplications that make $S^n$ into an H-space are in bijection with $\pi_{2n}(S^n)$. For $n=3$, we have $\pi_6(S^3) \cong \mathbb{Z}/12$. Not all of these other multiplications are homotopy associative, but eight of them are, and hence necessarily also have homotopy inverses.

Question. Is there an example in the literature of a space $X$, and two homotopy associative multiplications $m,m':S^3\times S^3\to S^3$, such that the groups $[X,(S^3,m)]$ and $[X,(S^3,m')]$ have been calculated explicitly and are not isomorphic? Necessarily, $X$ cannot be a suspension.

I am also interested in the same question but with $S^3$ replaced with any other H-space.

[1] James, I. M. "Multiplication on spheres (II)." Transactions of the American Mathematical Society (1957): 545-558.

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ If $X$ is a closed oriented $3$-manifold, then maps $X\to S^3$ are classified by their degree. Since the degree can be detected homologically, and since any multiplication $m:S^3\times S^3\to S^3$ must do the obvious thing on third homology, I think all the groups are isomorphic when $X$ is an oriented $3$-manifold. (I leave this comment in case anyone else was thinking of using Hopf's theorem.) $\endgroup$
    – Mark Grant
    Mar 17, 2015 at 10:19

1 Answer 1

Reset to default
5
$\begingroup$

This is semiexplicit. For any H-space $G$ with multiplication $\mu$, the projection maps $p_1, p_2: G\times G\to G$ have the property that $$ [p_1] \cdot [p_2] = [\mu] \in [G\times G, G]. $$ So if you have two different multiplications $\mu_1$ and $\mu_2$ on $S^3$ with induced multiplications $*_1$ and $*_2$ on $[-,S^3]$, you'll have $$ [p_1]*_1 [p_2] = [\mu_1] \neq [\mu_2] = [p_1]*_2 [p_2] $$ in $[S^3\times S^3, S^3]$.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Nice! And you might hope to compute the groups $[S^3\times S^3,S^3]$, using the cofibration sequence $S^3\vee S^3\to S^3\times S^3 \to S^3\wedge S^3 = S^6$. $\endgroup$
    – Mark Grant
    Mar 17, 2015 at 13:54
  • 2
    $\begingroup$ This group is computed for the standard $H$-space structure here. $\endgroup$
    – Eric Wofsey
    Mar 17, 2015 at 18:55
  • $\begingroup$ Great! This is explicit enough for me. However, does this actually imply that the two resulting groups are not isomorphic? Perhaps the isomorphisms to the same abelian group are different on the underlying sets? $\endgroup$
    – user1437
    Mar 18, 2015 at 19:12
  • $\begingroup$ Good point, and not one I have resolved yet. I sort of answered the question I thought you asked, not the one you did ask. Still thinking. $\endgroup$
    – Jeff Strom
    Mar 18, 2015 at 20:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.