18
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ 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 $\endgroup$ – Drew Heard Sep 13 '15 at 9:58
  • 2
    $\begingroup$ @DrewHeard: as this question has been highly upvoted, you should probably promote your comment to an answer. $\endgroup$ – Neil Strickland Sep 13 '15 at 12:29
12
$\begingroup$

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 $n>1$.

$\endgroup$
  • $\begingroup$ Would you mind enhancing this nice answer with some elaborations on your comment regarding the image of J? $\endgroup$ – Konrad Voelkel Sep 14 '15 at 18:00
  • 2
    $\begingroup$ @KonradVoelkel - I've added what I believe to be the connection, hopefully it is correct! $\endgroup$ – Drew Heard Sep 14 '15 at 18:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.