Growth of stable homotopy groups of spheres

Question

Let ${}_2\pi_n^S$ denote the $2$-power torsion subgroup of $n$th stable homotopy group of the sphere spectrum. Its order is a power of $2$: $$|{}_2\pi_n^S|=2^{k_n}.$$

Question: What is known about growth of $k_n$? Is it polynomial? What is the best estimation?

Remark: Of course, any estimation on ${\rm Ext}^{s,t}_{\mathcal A_2}(\mathbb F_2,\mathbb F_2)$ implies an estimation on $k_n.$ The Lambda algebra gives an exponential estimation, it is not interesting. I'm not very familiar with the May spectral sequence. Does it give a better estimation?

Answer 1

There is work by Boedigheimer and Henn that bounds the size of unstable homotopy groups of spheres or rather of the number of $p$-local summands (i.e. the dimension after tensoring with $\mathbb{F}_p$). The bound is again exponential, namely $3^{q-n/2}$ for $\mathrm{dim}_{\mathbb{F}_p}\pi_q(S^n)\otimes \mathbb{F}_p$. There is a slight improvement in later work by Henn, but the bound is still exponential as I understand it.

Looking at the data, the growth of the stable homotopy groups seems to be less than exponential though. According to Isaksen's charts (with possible miscounts by myself) the sequence of the first few $k_n$ is:

1 1 3 0 0 1 4 2 3 1 3 0 0 2 6 2 4 4 4 3 2 2 8 2 2 2 3 1 0 1 8 4 5 5 5 1 2 3 9 7 5 5 3 3 7 4 10

Particularly big ones are $k_{15} = 6$, $k_{23} = 8$ and $k_{47} = 10$. The contribution of the image of $J$ is $5$, $4$ and $5$ respectively in these degrees. While the image of $J$ should dominate in low degrees, elements of higher Adams-Novikov filtration should become more and more dominant. All in all, the data does not really look like an exponentially growing sequence, but who knows with our limited knowledge?

Edit: I incorporated Allen Hatcher's corrections to my sequence of $k_n$.

Answer 2

Since posting the preprint Tim mentions, I found Iriye's 1987 paper `On the ranks of homotopy groups of a space'. Iriye says (Theorem 1 and Remark 2) that it is not hard to replace the $2^\frac{k}{p-1}$ in Tim's answer with a $3^\frac{k}{2p-3}$. This is slightly better than the bound I proved, and I suspect that the proof is cleaner (though Iriye doesn't give a proof, and I haven't checked it). I have updated the Arxiv submission to acknowledge Iriye. Of course, this is still an exponential bound.

Answer 3

A recent preprint of Boyde improves this bound, showing that

$$\log_p(\#\pi_{n+k}(S^n)_{(p)}) \leq c_p 2^{k/(p-1)}$$

where $c_p = \frac{1}{4}2^{1/(p-1)}$

Note that this bound depends on $k$ and not $n$, so it stabilizes to show that

$$\log_p(\#\pi^s_{k}(\mathbb S)_{(p)}) \leq c_p 2^{k/(p-1)}$$

In his introduction, Boyde mentions some earlier results of Henn which are better than the ones that Lennart mentions, in that they depend only on $k$ and so stabilize. The citation is to the same single-author paper of Henn that Lennart links to.

Of course, this is still an exponential bound.

Answer 4

A subexponential bound is available (using only things known 40 years ago).

Thanks to John Palmieri over here for pointing out that the $E_1$ term of the May spectral sequence is a commutative polynomial algebra and so ought to have graded dimension which counts some sort of partition, and for subsequently pointing out that on account of $h_{1,0}$, this observation must be supplemented with some information about vanishing lines in the Adams spectral sequence.

Indeed, the May $E_1$ term $V^{\ast\ast\ast}$ is a polynomial algebra in $h_{ij}$ for $i\geq 1, j \geq 0$, with tridegree $|h_{ij}| =(s,t,u) = (1,2^j(2^i-1),i)$, i.e. bidegree $(s,t) = (1,2^j(2^i-1))$ in the Adams $E_2$, i.e. degree $t-s = 2^j(2^i-1)-1$ in the stable stems. Note that $h_{1,0}$ has bidegree $(s,t) = (1,1)$; all other $h_{i,j}$'s have $t-s > 0$.

Let $W^{\ast\ast\ast} \subseteq V^{\ast\ast\ast}$ be the subalgebra generated by the $h_{i,j}$'s other than $h_{1,0}$. Keeping just the last grading $k = t-s$, we see that $\dim W^k$ counts the number of ways of partitioning $k$ using positive integers of the form $2^j(2^i-1)-1$, i.e. positive integers whose binary expression contains exactly one zero (since the numbers $i$ and $j$ are uniquely determined by the quantity $2^j(2^i-1)-1$). This is less than the total number of partitions, and hence subexponential. The estimate via total partitions tells us that $\dim W^k \leq \exp(c\sqrt{k})$ for some $c>0$, but since the allowed parts for partitioning are exponentially sparse like in the Steenrod algebra, I'd guess that an upper bound of the form $\dim W^k \leq \exp(c(\log k)^2)$ actually follows from this if one works through the combinatorics, just as it does with the Steenrod algebra.

Adding back in the generator $h_{1,0}$, we see that $V^k$ is infinite-dimensional for all $k$. But because the $s$-grading of $h_{1,0}$ is still positive, we can use the fact that the $E_2$ page of the Adams spectral sequence (which I'm just calling $Ext^{\ast,\ast}$) has a vanishing line, in the sense that $Ext^{s,t} = 0$ for $0 < t-s < 2s + d$ for some constant $d$. I believe this implies that every element of $Ext^{s,t}$ with $t-s = k$ can be written as a sum of monomials $h_{1,0}^a\prod h_{i,j}^{b_{i,j}}$ where $a < k/2 - d/2$, so that we essentially have $\dim (\oplus_{t-s = k} Ext^{s,t}) \leq \sum_{a=0}^{k/2} \exp(c\sqrt{k-a})$. Because subexponential functions are closed under integration, it follows that the Adams $E_2$ page already has subexponential growth, and hence that $\log_2 |\pi_k\mathbb S_{(2)}|$ likewise has subexponential growth.

There is an odd primary analog of this too. See Ravenel's green book for a version of the May spectral sequence at odd primes where the $E_1$ term is a commutative polynomial algebra.

The dimensions of the graded parts of $W^k$ start off, if I coded things correctly, as:

degree $k$	bound on $\dim W^k$
0	1
1	1
2	2
3	3
4	5
5	7
6	11
7	15
8	21
9	28
10	38
11	49
12	65
13	83
14	107
15	136
16	172
17	215
18	269
19	332

This sequence must be well-known, as I think it is the basis for Bruner's Ext software. So I'm surprised that I can't find it in OEIS. Perhaps I have made a mistake.

---

Here is an earlier version of this answer:

Thanks to Nicholas Kuhn over here for pointing out that the dimension of the Lambda algebra can be used to bound the size of the $E_2$ term of the Adams spectral sequence.

The dimensions of the graded pieces of the Lambda algebra appear in the OEIS: https://oeis.org/A049285 and a bit of searching reveals that Tangora computed the asymptotics in Level number sequences of trees and the Lambda algebra, where I think he also considers the odd-primary case. Unfortunately, the results are stated in terms of generating functions, and a quick look has not allowed me to find a place where he actually states the asymptotic upper bound he gets on the dimension of the Lambda algebra. But his work is based on Flajolet and Prodinger's Level number sequences for trees, and if I'm deciphering things correctly, it looks like the bound is still exponential.