# Homotopy groups of $S^2$

in the paper

Foundations of the theory of bounded cohomology,

by N.V. Ivanov, the author considers the complex of bounded singular cochains on a simply connected CW-complex $X$, and constructs a chain homotopy between the identity and the null map. The construction of this homotopy involves the description of a Postnikov system for the space considered. In some sense, $S^2$ represents the easiest nontrivial case of interest for this construction, and I was just trying to figure out what is happening in this case. Since the existence of a contracting homotopy obviously implies the vanishing of bounded cohomology, this is somewaht related to understanding why the bounded cohomology of $S^2$ vanishes.

A first step in constructing the needed Postnikiv system is the computation of the homotopy groups of $X$, so the following question came into my mind:

Do there exists integers $n\neq 0,1$ such that $\pi_n(S^2)=0$?

I gave a look around, and I did not find the answer to this question, but I am not an expert of the subject, so I don't even know if this is an open problem.

In

Berrick, A. J., Cohen, F. R., Wong, Y. L., Wu, J., Configurations, braids, and homotopy groups, J. Amer. Math. Soc. 19 (2006), no. 2, 265–326

it is stated that $\pi_n(S^2)$ is known for every $n\leq 64$, and Wikipedia's table http://en.wikipedia.org/wiki/Homotopy_groups_of_spheres#Table_of_homotopy_groups shows that $\pi_n (S^2)$ is non-trivial for $n\leq 21$.

• Probably an open problem... – André Henriques May 6 '11 at 16:13
• Do you have any particular reason to think all the homotopy groups are nontrivial, except for the low-dimensional evidence? – Tilman May 6 '11 at 16:21
• Why are you interested? – Todd Trimble May 6 '11 at 16:54
• @Todd: beyond finiteness of homotopy groups of spheres this seems to be one of the simplest questions you could ask about these groups. So I find it a pretty natural and elementary question. If I was to guess, because of the Berrick-Cohen-Wong-Wu theorem, perhaps Roberto is interested in properties of Brunnian braids. – Ryan Budney May 6 '11 at 19:13
• @Tilman: Are you saying that the set of $n$ such that $\pi_n(S^0)_{(p)}^{stable}\not = 0$ is infinite? That wold be pretty surprising to me... – André Henriques May 6 '11 at 23:38

## 2 Answers

SERGEI O. IVANOV, ROMAN MIKHAILOV, AND JIE WU have recently(2nd June 2015) published a paper in arxive giving a proof that for $n\geq2$, $\pi_n(S^2)$ is non-zero. You can look at it in the following link.

Sergei O. Ivanov, Roman Mikhailov, Jie Wu, On nontriviality of homotopy groups of spheres, arXiv:1506.00952

• I'm pretty sure you meant $S^2$ rather than $S^n$ and have edited accordingly. Change it back if I've misunderstood. – Jeremy Rickard Aug 8 '15 at 15:57
• @JeremyRickard yes,that was a typo...thank you for the correction... – Ripan Saha Aug 8 '15 at 17:41
• I edited to give a) a human readable identification of the paper and b) a link to the abstract page instead of the pdf. – David Roberts Aug 19 '15 at 0:56
• In Griffiths-Morgan "Rational homotopy theory and differential forms" 1981, it is written in Ch.7 that "it is an amazing result of E. Curtis that $\pi_i(S^2) \neq 0$ for all $i\geq 2$." I guess this is a misattribution, since Curtis in his 1969 paper "Some nonzero homotopy groups of spheres" proved the statement for $i \neq 1$ mod 8 (as mentioned in the Ivanov--Mikhailov--Wu paper). – Aleksandar Milivojevic Mar 15 '19 at 17:35

I don't believe the answer to this question is known. There are various things one can say that are related. For example, there are known non-zero elements of known order from the image of the J homomorphism in all dimensions congruent to 3 mod 4 (by which I mean $\pi_{2+n}(S^2)$ with n congruent to 3 mod 4).

So none of those groups is zero, and if you like, you can then say that there can't be more than three consecutive zero groups.

There are other conclusions like this that one can draw, but I don't know how to show that all dimensions congruent to k mod 4 are non-zero for any k other than 3.