The explicit formula for the zeta function, e.g.

$$\psi_0(x) = \dfrac{1}{2\pi i}\int_{\sigma-i \infty}^{\sigma-i \infty}\left(-\dfrac{\zeta'(s)}{\zeta(s)}\right)\dfrac{x^s}{s}ds=x-\sum_\rho\frac{x^\rho}{\rho} - \log(2\pi) -\dfrac{1}{2}\log(1-x^{-2})$$

(where $\psi_0$ is the normalized Chebyshev function, $\sigma > 1$) relating zeroes of the $\zeta$ function and primes is certainly of much importance.

How does one understand this beyond what's normally given by a standard textbook on analytic number theory?

According to Connes

The noncommutative space of adele classes of a global field provides a framework to interpret the explicit formulas of Riemann-Weil in number theory as a trace formula.

Can someone elaborate on this? How do we interpret this as a trace formula?

  • 3
    $\begingroup$ It is $\psi_0(x) =\displaystyle \frac{1}{2i \pi} \int_{\sigma-i \infty}^{\sigma+i\infty} (-\frac{\zeta'(s)}{\zeta(s)})\frac{x^s}{s}ds$ whenever $\sigma > 1$ (aka the inverse Mellin/Laplace transform). And how do you understand the explicit formula ? Because what Connes wrote is one of the most abstract (and complicated) way to understand it. $\endgroup$
    – reuns
    Dec 4, 2016 at 11:12
  • $\begingroup$ @user1952009 Oops. I copied the formula from Wikipedia page of explicit formula en.wikipedia.org/wiki/… and didn't notice that. My personal understanding does not go beyond contour integrals, which is why I'm kinda unhappy because it feels like coincidence (but I guess too important to be just a coincidence). $\endgroup$
    – h__
    Dec 4, 2016 at 19:54
  • 2
    $\begingroup$ About "How do we interpret this as a trace formula?": well we don't... otherwise, the Riemann Hypothesis would follow immediately. $\endgroup$ Dec 4, 2016 at 21:29

1 Answer 1


The analogy between zeta functions and trace formulas goes at least to Selberg, when he proved his famous trace formula for hyperbolic surfaces and the result turned out to resemble Weil's generalization of Riemann's explicit formula quite a bit.

Riemann-Weil explicit formula:

\begin{equation*} \begin{split} \sum_\gamma h(\gamma)=\frac{1}{2\pi}\int_{-\infty}^{+\infty} h(r) \frac{\Gamma'}{\Gamma}\left(\frac{1}{4}+\frac{1}{2}ir \right)dr &+h\left(\frac{i}{2}\right)+h\left(-\frac{i}{2}\right)\\ &-g(0)\ln\pi-2\sum_{n=1}^\infty\frac{\Lambda(n)}{\sqrt{n}}g(\ln n) \end{split} \end{equation*} Selberg trace formula:

\begin{equation*} \begin{split} \sum_{n=0}^\infty h(r_n)=\frac{\mu(F)}{4\pi}\int_{-\infty}^{+\infty} rh(r) \tanh (\pi r)dr &+\sum_n \Lambda(n) g(\ln N(n)) \end{split} \end{equation*}

There's a big literature on this type of question, and the interplay of number theory, spectral analysis, mathematical physics... I recomend section 3 of Lagarias' survey "The Riemann Hypothesis: Arithmetic and Geometry" for references.

Connes approach started with

  • "Formule de trace en géométrie non-commutative et hypothèse de Riemann" (1996)

And was completed (as far as I know) in

The main result says that given a global field $K$ and a character $\alpha=\prod_v \alpha$ of the space of adele classes $A/K$, and any adecuate test function $h$, we have:

$$\underbrace{\widehat{h}(0)+\widehat{h}(1)-\sum \widehat{h}(\mathcal{X},\rho)}_{\text{spectral side}}=\underbrace{\sum_v \int_{K_v^*}' \frac{h(u^{-1})}{|1-u|}d^*u}_{\text{arithmetic side}}$$

  • $\begingroup$ what is $\sum \hat{h}(\mathcal{X},\rho)$, a sum over the non-trivial zeros of a Hecke L-function ? $\endgroup$
    – reuns
    Dec 4, 2016 at 20:09
  • $\begingroup$ Yes, $\widehat{h}(\mathcal{X},\rho)=\int h(u) \mathcal{X} |u|^zd^*u$, and sum is over the nontrivial zeros of $L(\mathcal{X},s)$, $\mathcal{X}$ a Hecke character. $\endgroup$
    – Myshkin
    Dec 4, 2016 at 20:18
  • $\begingroup$ what role does $\rho$ play in $\widehat{h}(\mathcal{X},\rho)=\int h(u) \mathcal{X} |u|^zd^*u$ ? $\endgroup$
    – h__
    Dec 4, 2016 at 21:43
  • $\begingroup$ @h__ You're right, there was a typo on my comment. It is $\widehat{h}(\mathcal{X},\rho)=\int h(u)\mathcal{X}^\rho d^* u$. $\endgroup$
    – Myshkin
    Dec 4, 2016 at 21:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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