MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


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


(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


(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.

share|cite|improve this question
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. – Oliver Nash Nov 11 '12 at 10:18
Well, geometrically, that can be described as the (limiting) probability that a (uniformly) randomly chosen lattice point is visible from the origin. – DavidLHarden Nov 11 '12 at 11:06
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 - – Asaf Nov 11 '12 at 14:46
We should note that Siegel by the 1940s proved that a (natural-normalization) of the volume of $SL(n,\mathbb Z)\backslash SL(n,\mathbb R)$ is $\zeta(2)\zeta(3)\zeta(4)\zeta(5)...\zeta(n)$, and a natural volume of $Sp(2n,\mathbb Z)\backslash Sp(2n,\mathbb R)$ is $\zeta(2)\zeta(4)\zeta(6)... \zeta(2n)$. These are predecessors of the Langlands result mentioned. – paul garrett Nov 11 '12 at 17:34

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 $K / \mathbb{Q}$.

share|cite|improve this answer
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 for quite a while, but I didn't know about the random variable connections. – Tom Copeland Nov 12 '12 at 12:05
The explicit expression for the kernel of $T^n$ is given in – Zurab Silagadze Sep 26 '13 at 13:45
@ZurabSilagadze, have you noted the connection between the formula for your kernel and those in earlier papers by Hodges and Sukumar I reference in…? – Tom Copeland Sep 28 '15 at 19:37
@TomCopeland No, I was not aware of this paper. Thanks for indicating this connection. – Zurab Silagadze Sep 29 '15 at 7:35
See also – Tom Copeland Oct 3 '15 at 5:31

Here's a physical interpretation of ζ(2): Suppose you're stuck in stopped traffic on a long road, and that you're standing just in front of your car. Suppose further that all the cars behind you are blowing their horns at you. Then the noise you hear is ζ(2) times louder than the sound of the first car's horn. Of course, there's a similar visual interpretation, maybe more realistic, using the illumination given by a long string of Christmas lights.

share|cite|improve this answer

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

Reference Paul Garett's notes:

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

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

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.

share|cite|improve this answer
Nice reference for Paul Garrett's comment. – Tom Copeland Sep 26 '13 at 9:18
I see Paul Garrett's comment just now:/ – Marc Palm Sep 26 '13 at 9:22

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.

share|cite|improve this answer
But a sort of limited reflection formula applies: $$\frac{2}{(2\pi)^{2n}}\:(2n-1)!\:\zeta(2n)=(-1)^{n+1}\frac{B_{2n}}{2n}=(-1)^{n}‌​\zeta(1-2n)$$. – Tom Copeland Nov 12 '12 at 11:42
@ Tom - you can edit my answer and add them if you'd like. I'm pressed for time right now. – Jonah Sinick Nov 12 '12 at 18:44
See also Itzykson and Zuber's "Matrix Integration and Combinatorics of the Modular Group" – Tom Copeland Nov 29 '12 at 3:32
My comment above was in response to a comment (since deleted) by another user that Jonah's work applied only to zeta at negative n. – Tom Copeland Apr 7 '14 at 23:33

This isn't about the Riemann zeta function, but instead has to do with the zeta function of a number field $K$. But I found it unexpected. Let $\zeta_K(s)=\sum_{\mathfrak a\ne0} N\mathfrak a^{-s}$ be the usual zeta function, where we sum over all non-zero integral ideals, and let $\zeta_K^{\text{prin}}(s) = \sum_{(\alpha)\ne0} N\alpha^{-s}$ be the same, except now we sum only over principal ideals. Then the probability that an elliptic curve over $K$ has a global minimal Weierstrass equation is more-or-less equal to $$\frac{\zeta_K^{\text{prin}}(10)}{\zeta_K(10)},$$ where the probability is counted by taking elliptic curve $y^2=x^3+Ax+B$ with $A$ and $B$ in appropriate boxes. See:

  • The density of elliptic curves having a global minimal Weierstrass equation, Ebru Bekyel, Journal of Number Theory, Volume 109, Issue 1, November 2004, Pages 41–58 doi:10.1016/j.jnt.2004.06.003
share|cite|improve this answer
Wow, where does the $10$ come from? Product of small primes? – David Roberts Sep 22 '15 at 3:50
@DavidRoberts It seems to come from 10=4+6, since $A$ has weight 4 and $B$ has weight 6. But I must admit that I expected to see $12=\text{LCM}(4,6)$, as happens more commonly with modular form. – Joe Silverman Sep 22 '15 at 11:06

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 $\sigma= 2, 3, ...$)


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

share|cite|improve this answer for Nash's comment. – Tom Copeland Jun 1 '15 at 15:04… for Harden's comment. – Tom Copeland Jun 1 '15 at 17:44
See also… – Tom Copeland Sep 28 '15 at 4:47

You asked for a geometric interpretation. Let $\gamma_\pm(a)$ be the two geometric shapes described

by the transcendental equation $e^{-y}\pm~e^{-\large x^a}=1$. Then their areas for $~a=\dfrac1n~$ are

$$ \begin{align} A_+&=-\int_0^\infty\ln\Big(1-e^{-\large x^a}\Big)dx=\Gamma(1+n)\cdot\zeta(1+n),\\ A_-&=-\int_0^\infty\ln\Big(1+e^{-\large x^a}\Big)dx=\Gamma(1+n)\cdot\zeta(1+n)\cdot(1-2^{-n}). \end{align} $$

Since the various polynomial curves, such as circles $(x^2+y^2=r^2)$, ellipses & hyperbolae

$\bigg(\dfrac xa\bigg)^2\pm\bigg(\dfrac yb\bigg)^2=1$, superellipses $\bigg(\dfrac xa\bigg)^n+\bigg(\dfrac yb\bigg)^m=1$, etc., have been studied for

centuries, it seemed only natural to expand this line of thought by inquiring about the properties

of exponential ones, such as $a^{\large x^n}\pm b^{\large y^m}=1$.

share|cite|improve this answer

The probability that a ``random" number is $n$-free (AKA does not have any integer $b$ so that $b^n$ divides it) is given by $1/\zeta(n)$

share|cite|improve this answer
Reference? Similar result for the Dedekind zeta function in ref in my comment to my elaboration on Nash's answer below. – Tom Copeland Sep 24 '15 at 14:07
I don't know. Its kinda common knowledge so I guess wikipeida. – Daniel Parry Sep 24 '15 at 16:47

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

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

$$$$ $$$$ For easy reference, from the paper, Veneziano's 4-particle scattering amplitude is proportional to

$$A(s,t,u) = V(s,t) + V(s,u) + V(t,u) \; ,$$

where $V(s,t) = B(-\alpha(s),-\alpha(t))$, in which the arguments of beta are Regge trajectories.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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