# 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 $$\Omega^2 S^5\to S^3$$. Such a map is known to exist if we require that it be degree $$2$$ on the bottom cell of $$\Omega^2 S^5$$ (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? May 5 '19 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 '19 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... May 5 '19 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. May 5 '19 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. May 6 '19 at 14:06

[I should've posted this answer a long time ago, given that it was answered in the comments.] In order to extend the map $$S^4\to \mathbf{H}P^\infty$$ to a map from $$J_2(S^4)$$, the composite of $$[\iota_4,\iota_4]:S^7\to S^4$$ with the inclusion of $$S^4$$ into $$\mathbf{H}P^\infty$$ must be null. This is not true: the Whitehead product $$[\iota_4, \iota_4]\in \pi_7(S^4)$$ would have to be $$2\nu$$ to get the desired extension, but Equation 5.8 of Toda's composition methods book says that $$[\iota_4, \iota_4] = \pm (2\nu - \Sigma \nu')$$, where $$\nu'\in \pi_6(S^3)$$ is the Blakers-Massey element.
As Gustavo Granja points out in the comments, there is a $$p$$-local analogue of this (with $$p>2$$): the map $$S^4\to \mathbf{H}P^\infty$$ extends to a map $$J_{(p-1)/2}(S^4)\to \mathbf{H}P^\infty$$, and the composite with the $$(p+1)/2$$-fold Whitehead product $$[\iota_4, \cdots, \iota_4]: S^{2p+1}\to J_{(p-1)/2}(S^4)$$ produces $$\alpha_1\in \pi_{2p+1}(\mathbf{H}P^\infty)$$. (See also this answer: Is $\mathbb{H}P^\infty_{(p)}$ an H-space?.) This implies that there is no extension to a map $$J_{(p+1)/2}(S^4)\to \mathbf{H}P^\infty$$.
• More generally, localized at an odd prime $p$, $\mathbb{H}P^{\frac{p-1}{2}}$ is equivalent to the $(p-1)/2$-th stage of the James construction $J_{\frac{p-1}{2}} S^5$ as there are no obstructions to extending the identity on the bottom cell. This map can't be extended to $J^{\frac{p+1}{2}} S^5$ because $P^1$ acts non-trivially on the generator of $H^4(\mathbb{H}P^\infty;\mathbb Z/p)$. May 12 '20 at 12:58
• Yes, should be $S^4$. May 12 '20 at 18:52