Finiteness of stable homotopy groups of spheres

Question

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?

Answer 1

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 $\pi_n(S^n) = \mathbf{Z}$.

Answer 2

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

Answer 3

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.