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This is more or less the same question as [ What is the generator of $\pi_9(S^3)$? ], except what I would like to know is if it is possible to describe this map in a way not only topologists can make sense of.

[EDIT] (Following the advice of Ryan Budney.) A purely geometric construction like the famous Hopf fibration for $\pi_3(S^2)$ would be perfect. (Something like a map $S^9\to S^2$ or $S^9\to S^3$ which may be written down in equations.) But I understand that there is little hope for that.

A less explicit but probably more reasonable approach is to try and represent this homotopy class by a framed 7-manifold in ${\mathbb R}^9$ following Pontryagin. In fact, any information about such a manifold may be of help. Is it really complicated?

I am not really familiar with the work of Jie Wu, but what I have read this far makes sense to me. So, the answer can also be along this lines, but if so I would like to see more then hints. (This computation looks horrendous, and I probably cannot handle it by myself.)

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    $\begingroup$ You might find some algebraic descriptions in the work of Jie Wu. For example, facts listed on his home page imply that $\pi_9(S^2)$ is the center of the group $G(n)$ with generators $x_1$, ..., $x_9$ and relations $x_1\cdots x_9=1$ and moreover all commutators involving every generator trivial. Or also, $\pi_9(S^2)$ is the $9$th homology group of the group of Brunnian braids. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented May 27, 2019 at 15:38
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    $\begingroup$ As far as I know the isomorphism Jie Wu is using is rather indirect, making it difficult to see what precisely the map $S^9 \to S^2$ is. $\endgroup$
    – Ryan Budney
    Commented May 27, 2019 at 17:53
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    $\begingroup$ It's pretty complicated in terms of Wu's isomorphism as well; for example, there's a number of easy-to-obtain elements corresponding to iteration of nontrivial stable element in $\pi^s_1$, which can be obtained as $\alpha_1 = [x_1, x_2]$ = Hopf map, $\alpha_2 = [[x_1, x_2], [x_1, x_3]] = \pi_4(S^2)$ and so on; as someone may know, fifth iterate $S^7 \to S^2$ becomes trivial, and you have to find less obvious elements in the centre of $G(n)$ to represent generator of $\pi_9$. $\endgroup$
    – Denis T
    Commented May 27, 2019 at 18:01
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    $\begingroup$ The Hopf fibration implies that $\pi_9(S^2) = \pi_9(S^3)$, so the map is just the composite $S^9 \to S^3 \xrightarrow{\eta} S^2$, where the first map is the generator of $\pi_9(S^3) \cong \mathbf{Z}/3$. $\endgroup$
    – skd
    Commented May 28, 2019 at 1:38
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    $\begingroup$ I think people are mis-understanding the question. You might need to add some details of what an answer should look like, Alex. $\endgroup$
    – Ryan Budney
    Commented May 28, 2019 at 2:31

2 Answers 2

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I posted an answer to the previous question describing one way to view a generator of $\pi_9 S^3$. From the Hopf fibration $$S^1\to S^3 \overset{\eta}{\to} S^2$$ and fibration long exact sequence, we know that $$\pi_9 S^1 = 0 \to \pi_9 S^3 \overset{\eta}{\to} \pi_9 S^2\to \pi_8 S^1=0 $$ is exact, so $\pi_9 S^3 \cong \pi_9 S^2$ via the Hopf map $\eta$ (comment made by @skd above).

I'm not sure if these answers are only understandable by topologists, but all the information is available in Hatcher's book Algebraic Topology (however, some of the information is not proved there, e.g. Bott periodicity).

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  • $\begingroup$ Thank you. What I want is to have a picture of the geometry. (Even if only a bit. Then, at least, I would not have to completely trust a topologist that all of this works.) To do this your way I would need to make sense of (1) the map $S^8\to S^5$ in $\pi_8 S^5$ and of (2) the geometry of $\pi_8 S^5\to \pi_9S^3$. Both things look rather obscure to me (even if I more or less understand the argument on the formal level). $\endgroup$
    – Alex Gavrilov
    Commented May 29, 2019 at 10:49
  • $\begingroup$ @AlexGavrilov The element $nu$ is very explicit, and hence so is its 3-fold suspension (so the generator of $\pi_8(S^5)$). The other map is harder to describe, but in principle could be computed with a connection, which is essentially canonical from the homogeneity of the situation. I’ll try to add some details tomorrow. $\endgroup$
    – Ian Agol
    Commented May 29, 2019 at 11:19
  • $\begingroup$ @AlexGavrilov: a new answer to the other question gives a fairly explicit realization of the generator for $\pi_9 S^3$. mathoverflow.net/a/332750/1345 $\endgroup$
    – Ian Agol
    Commented May 29, 2019 at 16:28
  • $\begingroup$ I think this description is good enough for me. (And now I do not mind moderators closing my question as a duplicate.) $\endgroup$
    – Alex Gavrilov
    Commented May 30, 2019 at 10:52
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Since my old answer is referenced here What is the generator of $\pi_9(S^3)$?, I spent a little time trying to figure out what it says about this. "Toda's sequence" is a $p$-local fiber sequence ($p$ any prime) of the form $$ S^{2n-1} \to \Omega \widehat{S}^{2n} \to \Omega S^{2pn-1}, $$ where $\widehat{S}^{2n}$ is a certain space I don't need to care about yet; see https://en.wikipedia.org/wiki/EHP_spectral_sequence. This backs up one step to a $p$-local fiber sequence $$ \Omega^2 S^{2pn-1} \xrightarrow{f} S^{2n-1} \to \Omega \widehat{S}^{2n}. $$ The bottom non-trivial homotopy group of the fiber is $\mathbb{Z}$, and so we get a map $$ \mathbb{Z}=\pi_{2pn-1}S^{2pn-1}=\pi_{2pn-3} \Omega^2 S^{2pn-1} \xrightarrow{f_*} \pi_{2pn-3}S^{2n-1}. $$ When $p=3$ and $n=2$ this gives $\mathbb{Z}=\pi_{9}\Omega^2S^{11}\to \pi_9S^3$, which by the argument I gave in my other answer surjects with image $\mathbb{Z}/3$. So we want to compute the image of $f_*$ in this dimension, i.e., the effect on $f$ on the bottom cell of $\Omega^2 S^{2pn-1}$.

Now, when $p=2$, "Toda's sequence" is the James's EHP sequence (with $\widehat{S}^{2n}=S^{2n}$). In this case $f_*$ is $\mathbb{Z}\to \pi_{4n-3}S^{2n-1}$. We know what the image of the generator is in this case: it is the "Whitehead square" $[\iota_{2n-1}, \iota_{2n-1}]$, which can be described geometrically as the attaching map of the $4n+2$-cell of $S^{2n-1}\times S^{2n-1}$.

I then tried to figure out what Toda's sequence actually is, in hopes of finding a geometric description of the map, and I failed. Everyone seems to refer to Toda's Composition Methods book for this sequence, but the statement given on Wikipedia is not actually there. However, the idea of the sequence (together with the other Toda sequence described on wikipedia) seems to be essentially 13.1 in Toda's book. In turn, 13.1 is not proved in Toda's book, but rather in the paper

Toda, Hirosi: On the double suspension $E^2$, J. Inst. Polytech. Osaka City Univ. Ser. A. 7 (1956), 103–145.

The fiber sequence I am interested in is essentially Theorem 7.6 of Toda's paper.

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  • $\begingroup$ Thanks to the person who provide the reference. (Our library doesn't have that journal, and therefore didn't tell me it is freely available.) $\endgroup$
    – Charles Rezk
    Commented May 29, 2019 at 15:49

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