# stable homotopy groups and zeta function

I have heard during a discussion that there is a well known relation between the stable homotopy groups of a sphere (more precisely the order of stable homotopy groups of localized sphere spectrum with respect to some homology theory $E$) and the values of the zeta function at some integers.

$$|\pi_{i}^{s}L_{E}\mathbb{S}| =^{?} \zeta(-n)$$ I'm not sure that I understood well, I will be glad if someone can explain this relation.

• Perhaps this is what you mean. Take $E = K(1)$, the first Morava $K$-theory. You can find the homotopy groups $\pi_*L_{K(1)}S^0$, at least when $p$ is odd, in, for example, Lurie's course notes (math.harvard.edu/~lurie/252xnotes/Lecture35.pdf). The order of the cyclic summand that appears can be expressed as the denominator of a certain expression involving Bernoulli numbers. This is related to the image of $J$, see en.wikipedia.org/wiki/J-homomorphism – Drew Heard Sep 13 '15 at 9:58
• @DrewHeard: as this question has been highly upvoted, you should probably promote your comment to an answer. – Neil Strickland Sep 13 '15 at 12:29

Here is a slightly more fleshed out version of my comment. Let $K(1)$ be the first Morava $K$-theory. When $p$ is odd one can calculate the homotopy groups of the $K(1)$-localised sphere spectrum to be $$\pi_nL_{K(1)}\mathbb{S} = \begin{cases} \mathbb{Z}_p, &n=0,1\\ \mathbb{Z}/p^{\nu_p(t')+1} & n=2(p-1)t'-1, t' \in \mathbb{Z}. \end{cases}$$ Here $\nu_p(x)$ is the $p$-adic valuation of $x$.
Following Adams define a function $m(l)$ by $$\nu_p(m(l)) = \begin{cases} 0 & l \not \equiv 0 \mod (2(p-1)) \\ 1+ \nu_p(l) & l \equiv 0 \mod (2(p-1)). \end{cases}$$ Adams shows (following Milnor and Kervaire) that $m(2s)$ is the denominator of $\beta_{2s}/4s$, where $\beta_s$ is the $s$-th Bernoulli number, and the fraction is expressed in the lowest possible form.
Using standard properties of $\nu_p(x)$ there is an equivalence $\nu_p(t')+1 = \nu_p((n+1)/2)+1$. Since $(n+1)/2 \equiv 0 \mod (2(p-1))$ we see $$\nu_p(m\left(2\cdot \frac{n+1}{4}\right)) = \nu_p((n+1)/2)+1$$ and so the order of $\pi_nL_{K(1)}S^0$ is the denominator of $\beta_{(n+1)/2}/(n+1)$.
Edit: Let me try and say something about the image of $J$ then. This is a homomoprhism $J:\pi_nSO \to \pi_n\mathbb{S}$. When $n=4k-1$ the order of the image of $J$ is cyclic of order the denominator of $\beta_{2k}/4k$. Let $\text{Im}(J_n)_p$ denote the image of the composite $\pi_nSO \to \pi_n\mathbb{S} \to \pi_n \mathbb{S}_{(p)}$. I believe this is meant to be isomorphic to $\pi_nL_{K(1)}\mathbb{S}$ for $n>1$.