# Is $\Omega J_{p^n-1}S^2$ commutative up to homotopy?

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.

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.
• 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 $k=p-1$. Sep 23, 2018 at 14:39