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What is $\pi_{31}(S^2)$, the 31st homotopy group of the 2-sphere ?


This question has a physics motivation:

  1. There are relations between (2nd and 3rd) Hopf fibrations and (2- and 3-) qubits (quantum bits) entanglement; see Pinilla and Luthra - Hopf Fibration and Quantum Entanglement in Qubit Systems.

  2. Maybe there are relations between classification of qubits entanglements and sphere homotopy groups, and we are interested in the classification of 4-qubits entanglements.


I tried fo find the solution on the net, with help of math fans, but without success.

Wikipedia gives only to the 22nd group homotopy of the 2-sphere.

This article of John Baez gives interesting references, like Allen Hatcher, Stable homotopy groups of spheres or a link with braids (Berrick, Cohen, Wong, and Wu - Configurations, braids, and homotopy groups). One speaks of a book of Kochman Stanley O.: Stable Homotopy Groups of Spheres: A Computer-Assisted Approach.


But I am totally unable to find the answer.

A subsidiary question would be: Until what rank do we know these high homotopy group of the 2-sphere?

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    $\begingroup$ In what way the two things you list as motivation are motivation for asking what the 31st homotopy group is? $\endgroup$ May 17, 2013 at 7:34
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    $\begingroup$ But there is no fourth Hopf fibration! $\endgroup$ May 17, 2013 at 7:45
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    $\begingroup$ @David: $\pi_{31}(S^2)$ is well out of the stable range. There is a homomorphism to the 29th stable group, but it is unlikely to be injective or surjective. $\endgroup$ May 17, 2013 at 7:54
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    $\begingroup$ You could deduce some information on $\pi_{32}(S^3)$ from Bob Bruners unstable chart on the bottom of math.wayne.edu/~rrb/cohom/index.html (if that's of any help) $\endgroup$ May 17, 2013 at 8:08
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    $\begingroup$ «For separable states, the original Hilbert space $S^7$ simplifies to $S^2\times S^2$» Reading physics papers requires a lot of restraint :-) $\endgroup$ May 17, 2013 at 8:31

2 Answers 2

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My apologies for updating this very old question.

As already mentioned, the 31st homotopy group of $S^2$ is the same as the 31st homotopy group of $S^3$. Serre's mod-C theory shows that this is a finite abelian group, and moreover that for any prime $p$ the first $p$-primary torsion occurs in $\pi_{2p} S^3$. This means that we don't have to check any primes $p$ for which $2p > 31$, which leaves us with the primes $p=3,5,7,11,13$.

The main tool for calculating these is the EHP spectral sequence. For $S^3$ the EHP sequence is pretty simple: there is a long exact sequence $$ \dots \to \pi_{n-1}(S^{2p-1}) \to \pi_{n}(S^3) \to \pi_{n}(S^{2p+1}) \to \pi_{n-2}(S^{2p-1}) \to \dots $$ that goes through $n > 3$.

The next point is the stable range. At odd primes and for odd spheres, the stable range is larger than that given by the Freudenthal theorem: if $n$ is odd, the stabilization map $\pi_{n+k}(S^n) \to \pi_k^S(S^0)$ is an epimorphism if $k=(n+1)(p-1) - 2$ and an isomorphism if $k < (n+1)(p-1)-2$. This means that all the groups relevant to computing $\pi_{31}(S^3)$ appear in the stable ranges of $\pi_*(S^{2p \pm 1})$ for $p \geq 5$, and those stable ranges are moreover very sparse (they only include the "image of J", which is well-known). If I have calculated correctly, $\pi_{31}(S^3)$ includes no $p$-torsion for $p \geq 5$.

That leaves only $p=3$, where we have to switch to a different algorithm ("I cannot do this, so I better look up what Toda calculated"). Toda showed that the 3-primary part of $\pi_{31}(S^3)$ is $\Bbb Z/3$. I found this in his 2003 text, "Unstable 3-primary Homotopy Groups of Spheres", but that was simply because I had it handier.

Combining this with the information from Mark Grant's answer gives $\Bbb Z/3 \oplus \Bbb Z/2 \oplus \Bbb Z/2 \oplus \Bbb Z/2 \oplus \Bbb Z/2$.

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    $\begingroup$ Wow, that's ironic that Toda's text was handy, given how hard it is to find! (I note this answer from 2016: mathoverflow.net/a/258475/4177) Here's the link from a comment there in case people are trying to look: ci.nii.ac.jp/naid/110009673506/en $\endgroup$
    – David Roberts
    Aug 25, 2021 at 3:13
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    $\begingroup$ @DavidRoberts That question from 2016 is more-or-less why I had access to the book in the first place (Professor Minato Yasuo very kindly contacted me and helped me find access to it). These degrees are low enough, however, that I believe this appears in more immediately accessible work of Toda's (possibly "Composition methods...", or several of his papers from the 1960s). $\endgroup$ Aug 25, 2021 at 15:10
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One simple observation is that $\pi_{31}(S^2)\cong\pi_{31}(S^3)$, by the long exact sequence of the Hopf fibration.

The homotopy groups $\pi_i(S^3)$ for $i\le 64$ are apparently computed in:

Curtis, Edward B.,Mahowald, Mark, The unstable Adams spectral sequence for $S^3$, Algebraic topology (Evanston, IL, 1988), 125–162, Contemp. Math., 96, Amer. Math. Soc., Providence, RI, 1989.

Unfortunately I wasn't able to get a hold of that reference to check for an explicit answer. Maybe someone has it on their shelf and can check.

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    $\begingroup$ The beginning of the article can apparently be read on Google Books $\endgroup$
    – Trimok
    May 17, 2013 at 9:34
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    $\begingroup$ I don't know if it is the correct result, but it seems that the answer should be $Z_2 + Z_2 + Z_2 + Z_2$ (Found page 204, in 1st paper reference below) Nobuyuki Oda, On the 2.Components of the Unstable Homotopy Groups of Spheres I. Source: Proc. Japan Acad. Ser. A Math. Sci. Volume 53, Number 7 (1977), 215-218. 1st paper $\endgroup$
    – Trimok
    May 17, 2013 at 10:29
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    $\begingroup$ @Trimok: Good find! Note that this is only supposed to be the $2$_primary part; there may be other summands of the form $\mathbb{Z}_{p^n}$ with $p$ an odd prime. $\endgroup$
    – Mark Grant
    May 17, 2013 at 11:01
  • $\begingroup$ @Mark : Yes, I understand this is not the final answer... $\endgroup$
    – Trimok
    May 17, 2013 at 11:04
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    $\begingroup$ I checked, and unfortunately the reference you've listed only seems to compute the 2-primary part as well. $\endgroup$
    – mme
    Sep 4, 2014 at 1:09

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