# Is $\mathbb{H}P^\infty_{(p)}$ an H-space?

Put $X=\mathbb{H}P^\infty$ (so $X$ classifies quaternionic line bundles, and $\Omega X=S^3$). There is no obvious reason for $X$ to be an H-space, because the tensor product of quaternionic vector spaces is not naturally a quaternionic vector space. Below I will prove that there is no nonobvious H-space structure. However, the obstruction that I use has order $12$ and so vanishes if we localise at a prime $p>3$. My guess is that $X_{(p)}$ is not an H-space for any prime $p$; does anyone know a proof of that?

Note that $H^*(X)=\mathbb{Z}[y]$ with $|y|=4$, and this has a Hopf algebra structure given by $\psi(y)=y\otimes 1+1\otimes y$, which is compatible with all Steenrod operations. Thus, there do not seem to be any primary obstructions.

However, if $X$ were an H-space then $S^3=\Omega X$ would have two commuting binary operations with the same identity and so (by a standard argument) they would be the same and would be commutative. However, it is known that $S^3$ is not homotopy commutative: the commutator map $S^6=S^3\wedge S^3\to S^3$ is the standard generator $\nu'$ of $\pi_6(S^3)\simeq\mathbb{Z}/12$.

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Isn't it conceivable that $S^3$ has several H-structures, one of which is homotopy commutative? Then you might be detecting the wrong structure. –  Jeff Strom Oct 24 '12 at 16:18
Actually, $\pi_6(S^3)$ acts on $[S^3 \times S^3, S^3]$ and the orbit of one H-structure is all the H-structures. –  Jeff Strom Oct 24 '12 at 16:20
No. If it were an $H$-space, there would be self maps of $\mathbb{H}P^\infty_{(p)}$ inducing multiplication by $k$ in degree $4$ homology for all integers $k$. But this is not the case by a Theorem of S. Feder and S. Gitler in "Mappings of quaternionic projective spaces", Bol. Soc. Mat. Mex. 34 (1975) 12-18. Using Adams operations in complex $K$-theory they show that such a $k$ must be a $p$-adic square.