# Mapping a loop space to quaternionic projective space

Let $$\mathbf{H}P^\infty$$ denote the infinite-dimensional quaternionic projective space. The inclusion of its bottom cell defines a map $$S^4 \to \mathbf{H}P^\infty$$. Does this extend to a map $$\Omega S^5 \to \mathbf{H}P^\infty = BSU(2)$$?

Since $$\Omega S^5$$ is the James construction on $$S^4$$, this question would be very easy to answer (in the positive) if $$\mathbf{H}P^\infty$$ was a homotopy associative H-space --- but it's known that this is not true. (If $$Y$$ is a homotopy associative H-space, then any map $$X\to Y$$ from a path-connected space $$X$$ admits a unique extension to a H-space map $$\Omega \Sigma X \to Y$$.) However, the composite $$S^4\to BSU(2) \to BSU$$ does extend to a map $$f_\xi:\Omega S^5\to BSU$$ classifying a bundle $$\xi$$ over $$\Omega S^5$$; it is easy to see that the Chern classes $$c_i(\xi)$$ vanish for $$i\geq 3$$, so the map $$f_\xi$$ factors, at least on cohomology, through $$BSU(2)$$.

One natural expectation for the desired map is that it gives a map of fiber sequences from the EHP fiber sequence $$S^2 \to \Omega S^3 \to \Omega S^5$$ to the Hopf invariant fiber sequence $$S^2 \to \mathbf{C}P^\infty = BS^1 \to \mathbf{H}P^\infty = BS^3$$ via the map $$\Omega S^3 \to \mathbf{C}P^\infty$$ extending the inclusion of the bottom cell of the target. In fact, thinking along these lines shows that we'd get the desired map if $$S^2$$ was a loop space, which it isn't.

An approach to constructing the desired map comes from equivariant considerations. Namely, the bottom $$C_2$$-equivariant cell of $$\mathbf{C}P^\infty$$ under the complex conjugation action is the one-point compactification $$S^\rho$$ of the regular representation $$\rho$$ of $$C_2$$. This gives a map $$\Omega S^{\rho+1} \to \mathbf{C}P^\infty$$, and hence a map $$(\Omega S^{\rho+1})_{hC_2} \to (\mathbf{C}P^\infty)_{hC_2} = \mathbf{H}P^\infty$$. To get the desired map, it therefore suffices to construct a nonequivariant map $$\Omega S^5 \to (\Omega S^{\rho+1})_{hC_2}$$, but it's not clear to me how/whether such a map exists.

I'd like to remark that looping the map $$\Omega S^5\to \mathbf{H}P^\infty$$ defines a map $$\Omega^2 S^5\to S^3$$. This is already known to exist: it is the map appearing in work of Cohen-Moore-Neisendorfer.

• This isn’t relevant to your question, but you seem to claim that \Omega\Sigma X is the “free homotopy associative H-space” on X, which certainly isn’t true, right? Am I misunderstanding something? – Dylan Wilson May 5 at 20:21
• @DylanWilson This is in fact true (at least if X is path-connected), and was the original statement proved by James (Theorem 1.11 in his "Reduced product spaces"). – skd May 5 at 20:38
• Theorem 1.11 in that paper says that the James construction is the free topological monoid with the basepoint of X acting as the identity, which is not the same... did you mean to cite a different theorem? I’m really having trouble believing the statement is true... – Dylan Wilson May 5 at 23:43
• Maybe you mean 1.8? That says H-spaces X are a retract of \Loops\Sigma X, but surely the splitting is not unique. You can use that to build a map like the one you mention though. Anyway- I’m happy to forget about it, just wanted to make sure there wasn’t some contradiction lurking in my brain. – Dylan Wilson May 5 at 23:49
• The inclusion of the bottom cell does not even extend to the 8-skeleton. The attaching map of the 8-cell in the James construction is the Whitehead product [i_4,i_4] which is twice the Hopf map plus the suspension of the Blakers-Massey element (the generator of \pi_6(S^3)). You would need the Whitehead product to be twice the Hopf map in order for the extension to the 8-skeleton to exist. – Gustavo Granja May 6 at 14:06