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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    $\begingroup$ 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. $\endgroup$ Aug 26, 2012 at 22:16
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    $\begingroup$ ... moreover the techniques were Serre's dissertation, and were a key part of why he was awarded the Fields Medal. $\endgroup$ Aug 26, 2012 at 22:39
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    $\begingroup$ 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. $\endgroup$ Aug 26, 2012 at 23:10
  • $\begingroup$ 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. $\endgroup$ Aug 27, 2012 at 2:07
  • $\begingroup$ 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. $\endgroup$ Aug 29, 2012 at 17:56

3 Answers 3


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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    $\begingroup$ 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. $\endgroup$ Jun 29, 2014 at 2:21
  • $\begingroup$ This leaves aside one half of the proof, namely finite generation of these homotopy groups. Can this part be considered as conceptual too? $\endgroup$
    – YCor
    Oct 31, 2019 at 16:37

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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    $\begingroup$ Does anyone know of a free version of this paper? I can only find such a thing in russian... $\endgroup$ Jul 19, 2013 at 17:39

Let me phrase a proof of Serre's computation of the rational stable homotopy groups of spheres as stably as I can:

For every spectrum $X$, we can define its rationalization $X_{\mathbb{Q}}$ as the homotopy colimit $$\mathrm{hocolim}\, X \xrightarrow{2} X \xrightarrow{3} X \xrightarrow{4} \cdots .$$ As such a directed homotopy colimit translates into a usual colimit after applying homotopy groups, we obtain $\pi_* X_{\mathbb{Q}} \cong (\pi_* X) \otimes \mathbb{Q}$.

By a form of Hurewicz, we know that the negative homotopy groups of the sphere spectrum $\mathbb{S}$ are $0$ and its zeroth homotopy group is $\mathbb{Z}$. Thus, we obtain a Postnikov truncation map $\mathbb{S} \to H\mathbb{Z}$. After rationalization, this induces a map $\mathbb{S}_{\mathbb{Q}} \to H\mathbb{Q}$. We want to show that this map is an equivalence.

As these are bounded below spectra, Hurewicz/Whitehead II imply that we just need to check on homology. As homology also translates directed homotopy colimits into colimits, we obtain that $H_*(S_{\mathbb{Q}}; \mathbb{Z}) \cong \mathbb{Q}$ (concentrated in degree $0$). It suffices thus to show that the integral homology of $H\mathbb{Q}$ is the same (as the map is then automatically an isomorphism as it is one on $\pi_0$).

As the homology of $H\mathbb{Q}$ coincides with the homotopy of $H\mathbb{Q} \otimes H\mathbb{Z}$, we see that we can equally well compute the rational homology of $H\mathbb{Z}$. The $i$-th such homology is isomorphic to the colimit over $\widetilde{H}_{i+n}(K(\mathbb{Z}, n); \mathbb{Q})$. For $n$ odd, these groups are concentrated in degree $n$ and are $\mathbb{Q}$ there, implying the result. (This is the only place where I find it to be necessary to apply an unstable result.)

For the finite generation, I likewise need one input result, namely that the homology of $H\mathbb{Z}$ is degreewise finitely generated. Using unstable methods, this is again easy to prove as we obtain $H_i(H\mathbb{Z}; \mathbb{Z})$ as the colimit of $\widetilde{H}_{i+n}(K(\mathbb{Z}, n); \mathbb{Z})$. Each of the groups is finitely generated by an inductive argument using the Serre spectral sequence and the sequence stabilizes.

The argument is now the following: We prove simultaneously by induction that the $n$-th homotopy group of $\mathbb{S}$ is finitely generated and that the connective cover $\tau_{\geq n+1}\mathbb{S}$ has degreewise finitely generated homology. Hurewicz implies that $\pi_n \mathbb{S} \cong \pi_n\tau_{\geq n}\mathbb{S} \cong H_n(\tau_{\geq n}\mathbb{S}; \mathbb{Z})$. The latter group is finitely generated by assumption and hence also $\pi_n\mathbb{S}$. Thus $H\pi_n\mathbb{S}$ has degreewise finitely generated homology. Thus, the same is true for the fiber of $\tau_{\geq n}\mathbb{S} \to H\pi_n\mathbb{S}$, namely $\tau_{\geq n+1}\mathbb{S}$, which finishes the induction step.

  • $\begingroup$ This is a very nice way of phrasing the arguments! $\endgroup$ Oct 31, 2019 at 23:12

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