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).

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    $\begingroup$ 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? $\endgroup$ May 5, 2019 at 20:21
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    $\begingroup$ @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"). $\endgroup$
    – skd
    May 5, 2019 at 20:38
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    $\begingroup$ 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... $\endgroup$ May 5, 2019 at 23:43
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    $\begingroup$ 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. $\endgroup$ May 5, 2019 at 23:49
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    $\begingroup$ 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. $\endgroup$ May 6, 2019 at 14:06

1 Answer 1


[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 $J_{(p+1)/2}(S^4)\to \mathbf{H}P^\infty$.

  • $\begingroup$ 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 $H^4(\mathbb{H}P^\infty;\mathbb Z/p)$. $\endgroup$ May 12, 2020 at 12:58
  • $\begingroup$ Yep! Edited my answer. (Also, I think you meant to write S^4 where you wrote S^5.) $\endgroup$
    – skd
    May 12, 2020 at 15:42
  • $\begingroup$ Yes, should be $S^4$. $\endgroup$ May 12, 2020 at 18:52

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