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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 $\pi_0^s \cong \mathbb{Z}$.

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.

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?

Question 2: Is there any philosophical or conceptual reason for why these groups should be finite?

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Regarding Q2: Is there a reason why you think the existing proofs of finiteness should not be considered as "conceptual"? It's unclear to me how you can discount them from being conceptual, as they're proofs. –  Ryan Budney Aug 26 '12 at 22:16
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... moreover the techniques were Serre's dissertation, and were a key part of why he was awarded the Fields Medal. –  Ryan Budney Aug 26 '12 at 22:39
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A reason for the appearances of copies of Z may be seen from the perspective of rational homotopy theory. Namely the (reduced) cohomology of the sphere is an exterior algebra on a single generator x say and so the minimal model is given by a DGA with one generator in the case n is odd and 2 generators when n is even to force x^2 to be 0 in cohomology (the DGA is graded commutative). One then needs to know a bound on the p-component that can appear (I believe the first appreance of an element of order p is at 2p-3) and I am sure someone has a good conceptual explanation of this. –  Callan McGill Aug 26 '12 at 23:10
    
I have to agree with Callan's comment, and the attitude is really implicit in Peter's answer: Use rational homotopy theory. One might think of this as being a little bit like chromatic homotopy theory, it is the "0th layer" of chromatic homotopy theory. –  Sean Tilson Aug 27 '12 at 2:07
    
Thanks for all the comments and thanks to Peter May for an excellent answer! @Ryan: Sure, a proof is a proof, and Serre's work is amazing, but I still think it is meaningful to ask for conceptual reasons for believing something. Part of my motivation is that there are many unproven conjectures suggesting that this or that cohomological invariant should be finite (or finitely generated) and it would be interesting to hear the reasons (if any!) that experts believe such conjectures. –  Andreas Holmstrom Aug 29 '12 at 17:56
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2 Answers 2

up vote 22 down vote accepted

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 $k\colon S^n \longrightarrow K(\mathbf{Z},n)$ 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 $\pi_n(S^n) = \mathbf{Z}$.

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Do you even need to do the calculation in the even case to get the stable result? You already know that there's no rational homotopy above degree $n$ when $n$ is odd, and when a given stable homotopy group stabilizes it has to have the same stable value whether $n$ is even or odd. –  Qiaochu Yuan Jun 29 at 2:21
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There is a surprisingly different proof using absolutely nothing by Podkorytov.

Which one is more conseptual I can not decide. (I like Serre's proof more.)

Few years ago Gromov suggested me to find a "geometric" proof (perhaps using cobordisms of framed immersions.)

You can download Podkorytov's paper here: http://link.springer.com/journal/10958/110/4/page/1

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Does anyone know of a free version of this paper? I can only find such a thing in russian... –  Dylan Wilson Jul 19 '13 at 17:39
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