## Axioms for Riemann $\zeta$ function

Are there any set of axioms that completely characterize the Riemann zeta function? i.e. like Ressayre axioms for the exponential function in an exponential field or functional equations.

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 Do you allow axioms like: "analytic except for a pole at 1" ?? – Gerald Edgar Jul 18 at 13:35 Yes of course. Analitycity is first-order. – Math-player Jul 18 at 13:44 ...I mean after invoking Cauchy-Riemann conditions. – Math-player Jul 19 at 9:55

Perhaps you are looking for something like Hamburger's Theorem?

It states, essentially, that the only Dirichlet series with a finite number of singularities satisfying the same functional equation as the zeta-function is the zeta-function. You can find the details in Titchmarsh's book.

Googling I found the following link: http://www.mat.univie.ac.at/~esiprpr/Zetaproc/patterson.pdf

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There are in fact at least two axiomatic characterizations of $\zeta(s).$ One of them is given by Hecke and one of them was given by Hamburger.

Hamburger's Theorem states: Suppose $D(s)$ is a dirichlet series, convergent in some half plane and whose coefficients have polynomial growth. If  1) There exists a polynomial $P(s)$ so that $P(s)D(s)$ is entire and of finite order.

2) $D(s/2)$ is also dirichlet series. That is, the coefficients of $D(s)$ are supported on squares.

3) If $\xi(s)= \pi^{-s}\Gamma(s)D(s)$ then $\xi(1/2-s)=\xi(s).$

Then $D(s)=C\zeta(2s).$

Hecke's version states: Suppose $D(s)$ is a dirichlet series, convergent in some half plane and whose coefficients have polynomial growth. If  1) $(s-1/2)D(s)$ is entire and of finite order.

2) The coefficients of $D(s)$ have arbetrary support.

3) If $\xi(s)= \pi^{-s}\Gamma(s)D(s)$ then $\xi(1/2-s)=\xi(s).$

Then $D(s)=C\zeta(2s).$

A little extra information: These theorems cannot be combined. The so-called "Big Mac Theorem" where 1) There exists a polynomial $P(s)$ so that $P(s)D(s)$ is entire and of finite order.

2) The coefficients of $D(s)$ have arbetrary support.

3) If $\xi(s)= \pi^{-s}\Gamma(s)D(s)$ then $\xi(1/2-s)=\xi(s).$

produces infinitely many linearly independent dirichlet series!

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Hamburger's Theorem (see Titchmarsh 'Theory of the Riemann Zeta Function' $\S$ 2.13) is in some sense an axiomatic characterization of $\zeta(s)$ among all Dirichlet series by its functional equation. It says:

Let $f(s)=\sum_n a_n n^{-s}$ a Dirichlet series absolutely convergent for $\sigma>1$ such that for some polynomial $P(s)$, $G(s)=P(s)f(s)$ is an integral function of finite order. Suppose $$f(s)\Gamma(s/2)\pi^{-s/2}=g(1-s)\Gamma((1-s)/2)\pi^{-(1-s)/2}$$ where $g(1-s)=\sum_n b_n n^{1-s}$ is absolutely convergent for $\sigma<-\alpha<0$. Then $$f(s)=C\zeta(s)$$ for some constant $C$.

Hamburger's theorem was a motivation for Hecke's study of Dirichlet series with functional equations generally, leading to his work on automorphic forms.

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oops, too late. – Stopple Jul 18 at 15:54

I don't know if this is in the spirit you're looking for, but there is the Selberg class -- an attempt at axiomatizing $L$-functions, requiring a Dirichlet series, functional equation of a certain type, analyticity, and an Euler product (typically) -- and it would be possible to impose extra conditions to isolate the zeta function. In particular, all degree 1 elements are known to come from Dirichlet L-functions (this was proved by Kaczorowski and Perelli, and then reproved by Soundararajan). Thus, requiring the degree and the conductor to both be 1 should isolate the zeta function.

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If it is possible, I would like to remove any conditions related to the infinite sum. Because giving an infinite sum completely characterizes the zeta function, but we are unable to express it in first order as I suppose. – Math-player Jul 18 at 13:37
Fair enough. It's also worth noting that Hamburger's theorem, as mentioned by Micah and Stopple, is the better way of stating my answer, since the conditions that the degree and conductor are 1 restrict the functional equation to being exactly the one satisfied by zeta, and is thus subject to Hamburger's result. – rlo Jul 18 at 18:18
Robert: Do any (or all) of the proofs which classify degree one elements in the Selberg class in some way assume the existence of an Euler product? – Micah Milinovich Jul 18 at 23:29
As I stated it, yes, that is necessary. However, Kaczorowski and Perelli define what they call the extended Selberg class, where there is no Euler product, and they provide a classification of the elements of degree up to 1. Degree 0 elements are Dirichlet polynomials satisfying a certain symmetry condition, and degree 1 elements are linear combinations of the product of a degree 0 element with a (potentially shifted) Dirichlet L-function. Sound's proof also works for this class, and shows that the Dirichlet coefficients are periodic, so that multiplicativity implies Dirichlet character. – rlo Jul 19 at 2:29
Thanks for the clarification... – Micah Milinovich Jul 19 at 19:20