most recent 30 from http://mathoverflow.net 2013-05-21T18:42:22Z http://mathoverflow.net/feeds/question/109444 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109444/is-mathbbhp-infty-p-an-h-space Neil Strickland 2012-10-12T09:29:08Z 2013-04-15T22:08:49Z

<p>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?</p> <p>Note that <code>$H^*(X)=\mathbb{Z}[y]$</code> 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.</p> <p>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 $\pi_6(S^3)\simeq\mathbb{Z}/12$. </p>

http://mathoverflow.net/questions/109444/is-mathbbhp-infty-p-an-h-space/127659#127659 Gustavo Granja 2013-04-15T22:08:49Z 2013-04-15T22:08:49Z

<p>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.</p>