How nilpotent is the ring of stable homotopy groups of spheres?

Question

Are there any known or conjectured bounds on the exponent $d(r)$ such that $x^{d(r)} = 0$ for all $x \in \pi_r^S(S^0)$?

Answer 1

Let $\alpha \in \pi_s(S^0)$ for $s > 0$ be an element of positive degree in the stable stems. Then $\alpha$ has positive-dimensional filtration in the Adams-Novikov spectral sequence: in other words, it is annihilated by complex bordism.

The $E_\infty$-page of the Adams-Novikov spectral sequence for the sphere is known to have a vanishing "curve" of slope tending to zero as $t -s \to \infty $ (this is essentially the Devinatz-Hopkins-Smith nilpotence theorem). The exact behavior of this curve is not known, but there is a conjecture of Hopkins (from this talk) that it looks like $s \simeq \sqrt{t-s}$. In this case, any element in $\pi_t$ would have nilpotence order $O(t)$, by considering the spectral sequence. This conjecture is (as far as I know) unproven.

It might be worth noting that explicit nilpotence-type results have been proven using power operations (this is the sort of technique that led to Nishida's proof that order $p$ elements are nilpotent), for instance the Toda relation in this paper.

Answer 2

If one uses the Adams-Spectral sequence based on cohmology theories other than $BP$ it is possible to say a little more. In "A vanishing line in the $BP \langle 1 \rangle$-Adams spectral sequence" Jesús González shows that when $p$ is odd the $E_2$-page of the $BP\langle 1 \rangle$-Adams spectral sequence for the sphere spectrum has a vanishing line of slope $(p^2-p-1)^{-1}$. This is used to show that the $E_\infty$-term of the classic ASS has a vanishing line of slope $(2p-1)/[(2p-2)(p^2-p-1)]$. The result you are looking for is then Corollary 4.4:

In the $n$-th stable stem, any $p$-local homotopy class not detected by the $J$-spectrum is annihilated by $p^{\text{min}\{4,p\}+ \epsilon_n+S(n)}$, where $q=2p-2$, $S(n)$ is the first integer larger than or equal to $(q+1)n/[q(p^2-p-1)]$ and $\epsilon_n = 0$ unless $n+2$ is divisible by $q$, in which case $\epsilon_n$ is the highest power of $p$ dividing $n+2$.