Fix a prime $p\geq 5$ and an integer $n>0$. All spaces in this question are implicitly $p$-localized. Consider the spaces $X=J_{p^n-1}S^2$ (the $p^n-1$'th stage in the James construction $JS^2\simeq\Omega S^3$) and $Y=\Omega X$. These appear naturally in a number of applications. The loop sum operation makes $Y$ into an $H$-space. Is it known whether this is commutative? (Here and elsewhere, "commutative" means "commutative up to homotopy".)

One basic idea is that the loop space of any $H$-space is a commutative $H$-space. However, it is standard that $H^*(X;\mathbb{Q})=\mathbb{Q}[x]/(x^{p^n})$, and it is easy to see that this does not admit any Hopf algebra structure, so $X$ is not an $H$-space.

On the other hand, there is a James-Hopf map $h\colon JS^2\to JS^{2p^n}$. A well-known calculation in cohomology shows that $X$ is the fibre of $h$, so $Y$ is the fibre of $\Omega h$. The domain of $\Omega h$ is $\Omega JS^2\simeq\Omega^2S^3\simeq\Omega^3\mathbb{H}P^\infty$. The codomain is $\Omega JS^{2p^n}\simeq\Omega^2S^{2p^n+1}$. Here $S^{2p^n+1}$ is not a loop space, but it is an old theorem that it admits a commutative product (as does any odd-dimensional $p$-local sphere). Thus, the domain and codomain of $\Omega h$ have some extra commutativity to spare. On the other hand, $h$ is not a loop map, so $\Omega h$ is not obviously a double loop map, so it may be that the extra structure on the (co)domain cannot be brought into play.

One can check that the map $H_*(Y;\mathbb{Z}/p)\to H_*(\Omega^2S^3;\mathbb{Z}/p)$ is injective, and $H_*(\Omega^2S^3;\mathbb{Z}/p)$ is commutative, so there is no obvious primary homological obstruction to commutativity of $Y$.

There is a canonical map $JS^2\to\mathbb{C}P^\infty$ which is a rational equivalence. This restricts to give a rational equivalence $X\to\mathbb{C}P^{p^n-1}$, which in turn gives a rational equivalence $Y\to\Omega\mathbb{C}P^{p^n-1}$. Using the fibration $S^{2p^n-1}\to\mathbb{C}P^{p^n-1}\to\mathbb{C}P^\infty$ one can check that $Y$ is rationally equivalent to $\Omega S^{2p^n-1}\times S^1$ and thus to $K(\mathbb{Q},2p^n-2)\times K(\mathbb{Q},1)$. This has an obvious commutative product, and I think it works out that this is the only possible product up to homotopy. We therefore deduce that $Y_{\mathbb{Q}}$ is commutative, but I do not think that this approach gives useful information integrally.


1 Answer 1


This was answered in the affirmative by Brayton Gray in his paper Homotopy Commutativity and the EHP Sequence. Specifically he shows that for all $n$ the space $\Omega J_{p^s-1} S^{2n}$ is homotopy commutative for $s\geq 1$ when localised at any prime $p\geq 3$. Moreover he claims to be able to show that $\Omega J_{jp^s-1}S^{2n}$ is homotopy commmutative for $s\geq 1$ and $j\leq p$ odd, although he does not give a full proof.

In the same paper he also obtains results on the homotopy commutativitivy of the classifying space $B_{2n-1,r}$ of the iterated suspension.

  • $\begingroup$ Gray's argument is to prove that the inversion map $x \mapsto x^{-1}$ is an H-space map (using a general lemma about involutions of fiber sequences of H-spaces applied to the EHP fiber sequence). This raises the question of finding the best $k$ such that $x \mapsto x^{-1}$ is an $A_k$-space map. I'd guess $k=p-1$. $\endgroup$ Sep 23, 2018 at 14:39

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