# Geometric / physical / probabilistic interpretations of Riemann zeta(n>1)?

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What are some physical, geometric, or probabilistic interpretations of the values of the Riemann zeta function at the positive integers greater than one?

I've found some examples:

1) In MO-Q111339 on a Tamagawa number, GH states

$$\mathrm{vol}(\mathrm{SL}_2(\mathbb{R})/\mathrm{SL}_2(\mathbb{Z}))=\zeta(2).$$

2) In "Quantum Gauge Theories in Two Dimensions," Edward Witten derives

$$\mathrm{vol}(\mathcal M)=\frac{2}{(\sqrt{2}\:\pi)^{2g-2}}\zeta(2g-2)$$

from a volume form for the moduli space $\mathcal M$ of flat connections on a gauge group ($G=SU(2)$) bundle over a compact two-dimensional manifold, a Riemann surface of genus $g$, and, for a connected sum of an orientable surface of genus $g$ with $k$ Klein bottles and $r$ copies of the projective plane $RP^2$, he derives

$$\mathrm{vol}(\mathcal M)=\frac{2(1-2^{1-(2g-2+2k+r)})}{(\sqrt{2}\:\pi)^{2g-2+2k+r}} \zeta(2g-2+2k+r).$$

3) In Wikipedia on the Stefan-Boltzmann law, the black body irradiance (total energy radiated per unit surface area of a black body per unit time) is given as

$$j^{*}=2\pi\:3!\zeta(4)\:\frac{(kT)^{4}}{c^{2}h^{3}}.$$

(In n-dimensional space, it's proportional to $n!\zeta(n+1)$, and Planck's law for the electromagnetic energy density inside the 3-D black body has an extra factor of $4/c$.)

4) In "Feynman's Sunshine Numbers," David Broadhurst gives the rate per unit surface area at which a black body at temperature $T$ emits photons as

$$2\pi\:2!\zeta(3)\:\frac{(kT)^{3}}{c^{2}h^{3}}.$$

(And the density of photons inside the body has an extra factor of $4/c$.)

Motivation: I'm motivated not only by general interest, but also by MO-Q111165 and MO-Q111770. Determinants (volumes) of adjacency matrices and, therefore, the cycle index polynomials (CIPs) for the symmetric group pop up in statistical physics, e.g., in Potts q-color field theory and scaling random cluster model, and the CIPS can be "rescaled" to obtain the complete Bell polynomials (OEIS-A036040) which are related to the cumulant expansion polynomials (OEIS-A127671), both of which are related to statistical correlations and their diagrammatics (see references in OEIS-A036040).

5) The $p_n(z)$ of MO-Q111165 seem formally related to the Chern classes $c_{k}(V)$ of a direct (infinite) sum of line bundles $\:\:\:\: V=L_{1}\oplus L_2\oplus ...\:.$ :

With $x_{i}=c_{1}(L_i)$, the first Chern classes,

$$p_k(z)=k!\:c_{k}(V)=k!\:e_{k}(x_{1},x_{2}, ...),$$

where $e_{k}$ are elementary symmetric polynomials. The $\zeta(n)$ can be identified as the power sums of the first Chern classes, and then, for example,

$$3!\:c_{3}(V)=p_3(z)=(z+\gamma)^3-3\zeta(2)(z+\gamma)+2\zeta(3)$$ $$4!\:c_{4}(V)=p_4(z)=(z+\gamma)^4-6\zeta(2)(z+\gamma)^2+8\zeta(3)(z+\gamma)+3[\zeta^2(2)-2\zeta(4)].$$

Update (Nov. 16, 2012): Just found the sequence in a thesis by R. Lu, "Regularized Equivariant Euler Classes and Gamma Functions," which discusses the relationship to Chern and Pontrjagin classes.

See also "An integral lift of the Gamma-genus" and "The motivic Thom isomorphism" by Jack Morava and "Hodge theoretic aspects of mirror symmetry" by L. Katzarkov, M. Kontsevich, and T. Pantev.

This post imported from StackExchange at 2014-04-07 13:24 (UCT), posted by SE-user Tom Copeland
retagged Apr 19, 2014
Email me at jsinick@gmail.com and I'll send you a relevant (unpublished) document.

This post imported from StackExchange at 2014-04-07 13:24 (UCT), posted by SE-user Jonah Sinick
This isn't exactly physical or geometric but I've always liked the interpretation of $1/\zeta(2)$ as the (limiting) probability that two independent, uniform random natural numbers are coprime. And indeed likewise for $k > 2$ different natural numbers.

This post imported from StackExchange at 2014-04-07 13:24 (UCT), posted by SE-user Oliver Nash
Well, geometrically, that can be described as the (limiting) probability that a (uniformly) randomly chosen lattice point is visible from the origin.

This post imported from StackExchange at 2014-04-07 13:24 (UCT), posted by SE-user DavidLHarden

This post imported from StackExchange at 2014-04-07 13:24 (UCT), posted by SE-user Tom Copeland
One can generalize GH's claim, in general there is a computation due to Langlands which calculate the volume of fundamental domains via (poles of constant terms of-) Eisenstein series. In the arithmetic situation, those constant terms are related to zeta functions, see for example Lapid's notes here - math.huji.ac.il/~erezla/papers/Utah.pdf

This post imported from StackExchange at 2014-04-07 13:24 (UCT), posted by SE-user Asaf
Probably a good introduction for a mathematician to Witten's work is "Flat Connections On Oriented 2-Manifolds" by Lisa Jeffrey.

This post imported from StackExchange at 2014-04-07 13:24 (UCT), posted by SE-user Tom Copeland
I'm surprised no one has already mentioned the deep and mysterious fact that $\zeta(s)$ coincides, for all $\Re s>1$, with $\sum_n n^{-s}$

This post imported from StackExchange at 2014-04-07 13:24 (UCT), posted by SE-user Piero D'Ancona

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Bourgade, Fujita & Yor shows to get Zeta functions from Cauchy Random Variables for even values and the $\chi_4$ L-functions for odd values. For some reason they always come in this pair.

This proof is simplified by Luigi Pace for $\zeta(2)$. The Cauchy Random variable is $$p_X (x) = \frac{2}{1+x^2}$$

when we look at the ration of two such random variables $Y = X/X'$. $$p_Y(y) = \frac{4}{\pi^2} \frac{\log y}{y^2-1}$$ Then observe $\mathbb{P}(Y \geq 1) = \mathbb{P}(X < X') = \frac{1}{2}$. So they compute $$\sum_{k=0}^\infty \frac{1}{(2k+1)^2}= \int_0^1 \frac{-\log y}{1 - y^2} = \mathbb{P}(Y \geq 1)= \frac{\pi^2}{8}$$

I learned through a blog a proof using 2D Brownian motion at least for the case $\zeta(2)$.

Suppose that $f: \mathbb{C} \to \mathbb{C}$ is an analytic function on the neighbourhood of the unit disk. This
function maps the unit disk to with boundary where . A two dimensional brownian motion started at $f(0)$ takes on average time $$\mathbb{E}[\tau] = \sum_{k \geq 1} |a_k|^2$$ to exit domain $f(\mathbb{D})$ where $f(z) = \sum_{k \geq 0} a_k z^k$ and $\tau = \inf \{ t > 0: B_t \in \partial f(\mathbb{D}) \}$ is the hitting time of the boundary .

You can get $\zeta(2)$ by considering Brownian motion on the strip $\{ x+iy: |x| < \pi/2 \}$ and evaluating the left and right sides. The Brownian motion exit time is $\tau = \pi^2/4$ and $$f(z) = \log\left(\frac{1-z}{1+z}\right) = -2\left(z + \frac{z^3}{3} + \frac{z^5}{5} + \dots \right)$$ maps the strip to the unit disk.

This style is traced to the arXiv article by Greg Markowsky.

Also check out this paper by Noam Elkies who relates them to Alternating permutations. One can show:

\begin{eqnarray*} \sum_{k=0}^\infty \frac{1}{(2k+1)^2} &=& \sum_{k= 0}^\infty \int_0^1 \int_0^1 (xy)^{2k}dx\, dy \\\\ &=& \int_0^1 \int_0^1 \left( \sum_{k= 0}^\infty(xy)^{2k} \right)dx \, dy = \int_0^1 \int_0^1 \frac{ dx \, dy}{1 - (xy)^2} \end{eqnarray*} Then he does the strange Calabi substitution: $x = \frac{\sin u}{\cos v} ,y = \frac{\sin v }{\cos u}$

and recovers a calculus identity: $\int_0^1 \int_0^1 \frac{ dx \, dy}{1 - (xy)^2} = \int_{u+v < \pi/2} 1 \, du \, dv = \frac{\pi^2}{8}$

This proof is extended to higher dimensions in Elkies' paper.

You can then study the transform $T: L^2[0,\pi/2] \to L^2[0,\pi/2]$, the characteristic function of a triangle.

$(Tf)(x)=\int_0^{\pi/2 -x} f(t) \, dt$

and ask when does $Tf = \lambda f$. The spectrum of this operator is

$\lambda = \frac{1}{4k+1} , f_\lambda(x) = cos (4k+1)u$

Then one can take the trace of $T^n$ and compare to the volume of a polytope:

\begin{eqnarray} \sum_{k=-\infty}^\infty \frac{1}{(4k+1)^k}&=& \sum_\lambda \langle f |T^n | f \rangle \\\\ & =& \mathrm{Vol}\bigg(\{0 < x_1 > x_2 < x_3 > \dots < x_{n-1} > x_n > \frac{\pi}{2}\}\bigg) \end{eqnarray} The volume of this polytope can be expressed in terms of alternating permutations.

I first learned of this iterated integral idea in Stanley's survey on Alternating Permutations, but also in some papers by Chebikin on Parking Functions, this seems to be an example of a chain polytope.

What other L-functions can take neat values like $L(k) \in \mathbb{Q}\pi^k$ where $k \in \mathbb{Z}$ ? Possibly need an algebraic extension $K / \mathbb{Q}$. This post imported from StackExchange at 2014-04-07 13:24 (UCT), posted by SE-user john mangual
answered Nov 12, 2012 by (310 points)
Nice. I've known about the relations among the Bernoulli numbers, $\zeta(2n)$, $\zeta(1−2n)$, polylogarithms, Eulerian numbers, Euler numbers, Genocchi numbers, Zag, and other special number arrays (OEIS-A131758 oeis.org/A131758) for quite a while, but I didn't know about the random variable connections.

This post imported from StackExchange at 2014-04-07 13:24 (UCT), posted by SE-user Tom Copeland
The explicit expression for the kernel of $T^n$ is given in arxiv.org/abs/1207.2055

This post imported from StackExchange at 2014-04-07 13:24 (UCT), posted by SE-user Zurab Silagadze
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I wrote an article about this very subject titled Zeta Values in Geometry and Topology three years ago. My thinking on the points in the article has evolved, in particular, I'm fairly convinced that Questions 0.1-0.4 aren't fruitful lines of inquiry. Still, the material therein is fascinating to me.

This post imported from StackExchange at 2014-04-07 13:24 (UCT), posted by SE-user Jonah Sinick
answered Nov 11, 2012 by (40 points)
But a sort of limited reflection formula applies: $$\frac{2}{(2\pi)^{2n}}\:(2n-1)!\:\zeta(2n)=(-1)^{n+1}\frac{B_{2n}}{2n}=(-1)^{n}‌​\zeta(1-2n)$$.

This post imported from StackExchange at 2014-04-07 13:24 (UCT), posted by SE-user Tom Copeland
@ Tom - you can edit my answer and add them if you'd like. I'm pressed for time right now.

This post imported from StackExchange at 2014-04-07 13:24 (UCT), posted by SE-user Jonah Sinick
See also Itzykson and Zuber's "Matrix Integration and Combinatorics of the Modular Group" lpthe.jussieu.fr/~zuber/MesPapiers/iz_CMP90.pdf

This post imported from StackExchange at 2014-04-07 13:24 (UCT), posted by SE-user Tom Copeland

The refection formula in my other comment was in response to another user's comment (since deleted) that Jonah addressed only values of zeta at the negative integers.

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Here is a generalization of GH's answer concerning Tamagawa numbers:

Reference Paul Garett's notes: http://www.math.umn.edu/~garrett/m/v/volumes.pdf

$$vol(SL(n,\mathbb{Z}) \backslash SL(n, \mathbb{R})) = \zeta(2) \zeta(3) \zeta(4) \zeta(5) \cdots \zeta(n).$$

$$vol(Sp(n,\mathbb{Z})\backslash Sp(n, \mathbb{R})) = \zeta(2) \zeta(4) \zeta(6) \zeta(8) \cdots \zeta(2n)$$

This works for number field and global function fields analogous. There you have to replace Riemann zeta by the Dedekind zeta or Hasse-Weil zeta function and additionally root numbers show up.

This post imported from StackExchange at 2014-04-07 13:24 (UCT), posted by SE-user plusepsilon.de
answered Sep 26, 2013 by (20 points)
Nice reference for Paul Garrett's comment.

This post imported from StackExchange at 2014-04-07 13:24 (UCT), posted by SE-user Tom Copeland
I see Paul Garrett's comment just now:/

This post imported from StackExchange at 2014-04-07 13:24 (UCT), posted by SE-user plusepsilon.de
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There is a relation to the famous Veneziano amplitudes of nascent string theory (cf. Kholodenko, New Strings for old Veneziano amplitudes) through the factorials via the Euler beta function integral

$$B(s,\alpha) = \frac{(s-1)!(\alpha-1)!}{(s+\alpha-1)!} x^{s+\alpha-1} |_{x=1} = \int_0^x t^{s-1} \; (x-t)^{\alpha-1} \; dt |_{x=1}.$$

This can be morphed into the core Riemann-Liouville fractional integroderivative of fractional calculus (analytically continued)

$$D_x^{-s} x^{\alpha-1} |_{x=1} = \int_0^x \frac{t^{s-1}}{(s-1)!} \; (x-t)^{\alpha-1} \; dt |_{x=1}= \frac{(\alpha-1)!}{(s+\alpha-1)!} x^{s+\alpha-1} |_{x=1}\; ,$$

which can be expressed in terms of the infinitesimal generator

$$R_x = -\log(x) + \psi(1 + xD_x) = -(\log(x) + \gamma) - \sum_{n \ge 1} (-1)^n \zeta(n+1) \; (xD_x)^n \; ,$$

incorporating $\zeta(n>1)$ using well-known formulas for the digamma, or Psi, function $\psi(\beta)$ with $\gamma =- \frac{d\beta!}{d\beta} |_{\beta=0}= \psi(1)$, the Euler-Mascheroni constant, as

$$D_x^{-s} x^{\alpha-1} = e^{-sR_x} x^{\alpha-1} \; .$$

answered Sep 22, 2015 by (290 points)
edited Sep 22, 2015
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Elaborating on Nash's comment:

Oliver, special case of Zipf's law, right? That leads to the Zipf–Mandelbrot law that has a probability mass function of $$f(k;N,1,s)=\displaystyle\frac{\frac{1}{(k+1)^s}}{\sum_{i=1}^{N}\frac{1}{(i+1)^s}}$$ and then back to $\mathrm{vol}(\mathcal M)$ for the Klein bottles and particle statistics through $$(1-2^{1-s})\zeta(s)=\sum_{n=0}^{\infty } \frac{1}{2^{n+1}} \sum_{k=0}^{n}(-1)^k \binom{n}{k}\frac{1}{(k+1)^s}$$ $$=\eta(s)=\int_{0}^{\infty }\frac{1}{\exp(x)+1}\frac{x^{s-1}}{(s-1)!}dx$$

where $\eta(s)$ is the Dirichlet eta function, and so the Klein bottle manifolds seem connected to fermions and Fermi-Dirac statistics (as apropos Möbius twists), whereas the orientable Riemann manifolds seem related to bosons and Bose-Einstein statistics.

And, Alan Gut in "Some remarks on the zeta distribution" defines the random variable $U$ with probability mass function (choose your favorite $\sigma= 2, 3, ...$)

$$P(U_\sigma)=\frac{1}{\zeta(\sigma)n^\sigma}$$

and says, "The main point is that, for $\sigma>1$, one can view the normalized zeta function $\varphi_{\sigma}(t)=\frac{\zeta(\sigma\:+\:i\:t)) }{\zeta(\sigma)}$ as the characteristic function of, as it turns out, a compound Poisson distribution. "

He shows how the moments and cumulants of the distribution (related to OEIS A036040 and A127671) given as functions of $\zeta(\sigma)$ and its derivatives are related to the von Mangoldt and Moebius functions and re-derives (and extends) an identity due to Selberg.

On a tangent, the zeta values can be used to translate the Gamma-genus:

With $$R_z = z+\gamma + \sum_{n=1}^{\infty } (-1)^n\zeta (n+1)(d/dz)^n,$$

then $$\displaystyle \exp(\omega\:R_z)\frac{e^{(t\:z)}}{t!}=\exp{(\omega\:d/dt)}\frac{e^{(t\:z)}}{t!}=\frac{e^{((t+\omega)\:z)}}{(t+\omega)!}$$

This post imported from StackExchange at 2014-04-07 13:24 (UCT), posted by SE-user Tom Copeland
answered Nov 11, 2012 by (290 points)

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