# Two H-space structures on S^3 and [X,S^3] different as groups for each: Explicit Example?

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.

• 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.) – Mark Grant Mar 17 '15 at 10:19

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 $[S^3\times S^3, S^3]$.
• Nice! And you might hope to compute the groups $[S^3\times S^3,S^3]$, using the cofibration sequence $S^3\vee S^3\to S^3\times S^3 \to S^3\wedge S^3 = S^6$. – Mark Grant Mar 17 '15 at 13:54
• This group is computed for the standard $H$-space structure here. – Eric Wofsey Mar 17 '15 at 18:55