Finiteness of stable homotopy groups of spheres - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T14:59:45Z http://mathoverflow.net/feeds/question/105575 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105575/finiteness-of-stable-homotopy-groups-of-spheres Finiteness of stable homotopy groups of spheres Andreas Holmstrom 2012-08-26T21:33:13Z 2012-08-26T22:53:49Z <p>Since the work of Serre in the early 50's on homotopy groups of spheres, it is known that the homotopy group $\pi_k(S^n)$ is finite, except when $k=n$ (in which case the group is $\mathbb{Z}$), or when $n$ is even and $k=2n-1$ (in which case the group is the direct sum of $\mathbb{Z}$ and a finite group). As a consequence, the stable homotopy groups $\pi_k^s$ are finite groups for $k>0$, and $\pi_0^s \cong \mathbb{Z}$. </p> <p>The work of Serre was done before anyone knew about stable homotopy theory and chromatic methods, and this makes me wonder about the following questions. </p> <p>Question 1: Is it possible to use methods from stable/chromatic homotopy theory to prove finiteness of stable homotopy groups of spheres directly, without having to compute any unstable homotopy groups of spheres? </p> <p>Question 2: Is there any philosophical or conceptual reason for why these groups should be finite?</p> http://mathoverflow.net/questions/105575/finiteness-of-stable-homotopy-groups-of-spheres/105583#105583 Answer by Peter May for Finiteness of stable homotopy groups of spheres Peter May 2012-08-26T22:53:49Z 2012-08-26T22:53:49Z <p>I agree with Ryan that Serre's proof can be viewed as perfectly conceptual, but here is a modern version. Accept from Serre that the homotopy groups of spheres are finitely generated. Let <code>$k\colon S^n \longrightarrow K(\mathbf{Z},n)$</code> be the canonical map. We know how to rationalize spaces and maps. The rationalization of $k$ is a map $k_{0}\colon S^n_{0}\longrightarrow K(\mathbf{Q},n)$. If $n$ is odd, $k_0$ is an isomorphism on rational cohomology and therefore an equivalence. If $n$ is even, a very little cohomological calculation shows that the fiber of $k_{0}$ is $K(\mathbf{Q},2n-1)$. Since the homotopy groups of spheres are finitely generated, the kernel of the map on homotopy groups induced by the rationalization $S^n\longrightarrow S^n_{0}$ is finite in each degree. The rank of the free part is immediate from what I've said about $S^n_{0}$. Serre's theorem follows. This uses no calculation of unstable homotopy groups except maybe deep down that $\pi_n(S^n) = \mathbf{Z}$. </p>