Homotopy groups of $S^2$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T09:29:02Z http://mathoverflow.net/feeds/question/64131 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/64131/homotopy-groups-of-s2 Homotopy groups of $S^2$ Roberto Frigerio 2011-05-06T15:34:45Z 2011-05-09T09:09:13Z <p>Dear all, </p> <p>in the paper </p> <p>Foundations of the theory of bounded cohomology,</p> <p>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.</p> <p>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:</p> <p>Do there exists integers $n\neq 0,1$ such that $\pi_n(S^2)=0$? </p> <p>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.</p> <p>In </p> <p>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</p> <p>it is stated that $\pi_n(S^2)$ is known for every $n\leq 64$, and Wikipedia's table <a href="http://en.wikipedia.org/wiki/Homotopy_groups_of_spheres#Table_of_homotopy_groups" rel="nofollow">http://en.wikipedia.org/wiki/Homotopy_groups_of_spheres#Table_of_homotopy_groups</a> shows that $\pi_n (S^2)$ is non-trivial for $n\leq 21$. </p> http://mathoverflow.net/questions/64131/homotopy-groups-of-s2/64152#64152 Answer by Hal Sadofsky for Homotopy groups of $S^2$ Hal Sadofsky 2011-05-06T20:20:48Z 2011-05-06T21:11:57Z <p>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).</p> <p>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. </p> <p>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.</p>