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Let $G_n$ be the space of homotopy-equivalences of $S^{n-1}$. Evaluation produces a map $G_{n} \to S^{n-1}$. For $n = 2m+1$, I would like to understand the induced map on $\pi_{4m-1}$. More precisely, what I would really like to know are the Hopf invariants of elements in the image of $\pi_{4m-1} G_{2m+1} \to \pi_{4m-1} S^{2m}$.

For $m = 1$, the classical Hopf fibration gives rise to a generator of $\pi_3 S^2 = \bf Z$, and this element even lies in the image of $\pi_3 SO(3) \to \pi_3S^2$ which factors through $\pi_3 G_3 \to \pi_3 S^2$.

For $m = 2$, I do not know the answer, but I can prove that every map lying in the image of $\pi_7 SO(5) \to \pi_7S^4 \cong {\bf Z} \oplus {\bf Z}/12\bf Z$ has Hopf invariant divisible by $12$. Moreover, the question whether $1$ is attained might be equivalent to asking whether ${\bf H}P^3$ is homotopy equivalent to the total space of a fibration with fiber $S^4$ and base $S^8$.

I do not know what happens for $m \geq 3$. The case $m = 4$, where $\pi_{15} S^8$ contains an element of Hopf invariant 1, would be especially interesting. Maybe there is also a relation to the non-existence of ${\bf O}P^3$?

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    $\begingroup$ The fibre of the evaluation map $e:G_n\rightarrow S^n$ is $\Omega^nS^n$. The fibration connecting map $\partial:\Omega S^n\rightarrow\Omega^nS^n$ acts on homotopy groups to send $\alpha\in\pi_k\Omega S^n\cong\pi_{k+1}S^n$ to the Whitehead product $[\alpha,\iota_n]\in\pi_{n+k}S^n\cong\pi_{k}\Omega S^n$. This was worked out by G. Lang in The evaluation map and EHP sequences, Pacific J. Math. Volume 44, Number 1 (1973), 201-210. See the last section, where the interactions between the evaluation sequence and James's EHP sequence are discussed. $\endgroup$
    – Tyrone
    Commented Aug 17, 2020 at 14:29
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    $\begingroup$ This is very interesting, thanks! So to fully answer my question one needs to determine the order of $[e,\iota] \in \pi_{6m-2} S^{2m}$, where $e \in \pi_{2m}S^{2m}$ is a generator and $\iota \in \pi_{4m-1} S^{2m}$ generates $\pi_{4m-1} S^{2m} / \text{torsion} \cong \bf Z$, right? $\endgroup$
    – Jens Reinhold
    Commented Aug 17, 2020 at 15:15
  • $\begingroup$ Tyrone, would you be interested in converting (and possibly expanding) your comment into an answer? $\endgroup$
    – Jens Reinhold
    Commented Aug 18, 2020 at 13:29

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