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?

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    $\begingroup$ The full group or mod the image of j? For the former it's going to be dominated by the image of j, which is a question about growth of Bernoulli numbers. $\endgroup$ – Noah Snyder Jan 26 '18 at 20:06
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    $\begingroup$ @NoahSnyder Is that true? The order of the image of $j$ is the denominator of $B_{2n}/4n$. By the Von Stuadt-Clausen theorem, the denominator of $B_{2n}$ is a multiple of $2$ but not $4$, so the $2$-adic valuation of the order of the image of $j$ is $v_2(8n)=3+\log_2(n)$ and grows only logarithmically with $n$. $\endgroup$ – Will Sawin Jan 26 '18 at 20:23
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    $\begingroup$ (Edit: whoops, this is a bound for the torsion exponent, not the size of the whole group- sorry!) I think the best that's known is in this paper by Akhil Mathew: msp.org/agt/2016/16-2/agt-v16-n2-p12-s.pdf. It provides an essentially sharp, universal bound for the torsion in the kernel of the Hurewicz homomorphism. At 2, the kernel is annhilated by 2^{(n/2)+3}. For the sphere itself perhaps one could improve on that, but I don't think anything qualitatively better than `k_n is roughly linear in n' is known. $\endgroup$ – Dylan Wilson Jan 26 '18 at 22:13
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    $\begingroup$ @WillSawin: Looks like you're right, the small numbers were misleading about growth. I should have actually looked at the growth of Bernoulli numbers before commenting $\endgroup$ – Noah Snyder Jan 27 '18 at 21:23
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    $\begingroup$ I want to mention a recent result of Burklund: The exponents of the stable homotopy groups grow at most linearly. See Theorem 3.8 here: arxiv.org/pdf/1910.14116.pdf $\endgroup$ – Lennart Meier May 25 at 7:01

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 $k_n$.

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  • $\begingroup$ A couple small corrections to the original list of $k_n$'s in this answer: (1) Delete the third zero in the list to make the initial terms 1 1 3 0 0 1 4. (2) Change the pair 8 6 near the end of the list to 9 7. $\endgroup$ – Allen Hatcher Jan 30 '18 at 20:50
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    $\begingroup$ Since this post has been bumped: one should also add the recent paper arxiv.org/pdf/2001.04247.pdf. Conjecture 2.2 there says that k_1 + ... + k_n = O(n^2), so k_n = O(n). $\endgroup$ – skd May 23 at 20:03
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    $\begingroup$ @skd I'm glad to hear there's an explicit conjecture along these lines! One quibble -- although $k_1 + \dots + k_n = O(n^2)$ implies that $k_n \sim n$ "on average", Isaksen-Wang-Xu don't actually conjecture that $k_n = O(n)$. I get the sense that they expect $k_n$ is "random" enough that $\limsup k_n / n$ might be $\infty$ (and $\liminf k_n / n$ might be 0), even though the sum $k_1 + \dots + k_n$ smooths this out and is $O(n^2)$. $\endgroup$ – Tim Campion May 24 at 17:16

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.

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