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I'll admit from the outset that this question is slightly vague. The actual question appears at the end of the post.

The explicit formula of Guinand and Weil can be written in the following way:

For 'nice' g (i.e. in $C_c^\infty(\mathbb{R})$)

$$ \sum_\gamma \hat{g}(\gamma/2\pi) - \int_\mathbb{R} \frac{\Omega(\xi)}{2\pi}\hat{g}(\xi/2\pi) d\xi = \int_\mathbb{R} [g(x)+g(-x)] e^{-x/2}d(e^x-\psi(e^x)), $$ (1)

where the sum is over those $\gamma$ such that $1/2+i\gamma$ a non-trivial zero of the Riemann Zeta function, $\psi(x) = \sum_{n\leq x} \Lambda(n)$ is the Chebyshev prime counting function, and $$\Omega(\xi) = \tfrac{1}{2}\tfrac{\Gamma'}{\Gamma}(1/4+i\xi/2) + \tfrac{1}{2}\tfrac{\Gamma'}{\Gamma}(1/4-i\xi/2) - \log \pi. $$

Here $\gamma$ can possibly be complex.

It is usually proven using a contour integral to capture the zeroes of the Zeta function, then evaluating the integral a different way, making use of the reflection formula along with the arithmetical meaning of $\zeta(s)$ for $\Re s > 1$. (See for instance Montgomery and Vaughan, Multiplicative Number Theory.)

$\Omega(\xi)/2\pi \sim \log \xi /2\pi$, and is the mean density for the number of zeroes to occur in the critical strip with real part $\xi$. On the assumption of the Riemann hypothesis, the left hand side takes the nice form: $$ \int_\mathbb{R}\hat{g}(\xi/2\pi)\bigg(\sum_\gamma \delta(\xi-\gamma) - \frac{\Omega(\xi)}{2\pi}\bigg) d\xi. $$

The explicit formula therefore expresses a Fourier duality between the error term in the Chebyshev prime counting function and the error term in the zero counting function. The structural reason why this duality arises is not really apparent to me from the contour integral proof above, and is what I'm really getting at with this question.

That said, left at this the question is a little imprecise, and there is something of a lie here because the form of the explicit formula where this becomes apparent involves assuming the Riemann hypothesis. Therefore:

Question: Is there a way not making use of entire function theory proper to show that there exist numbers $\gamma$ with $|\Im \gamma| \leq 1/2$, so that (1) is true?

A proof using harmonic analysis over the adeles would get bonus points.

One reason to be interested in a question like this beyond what I've elaborated above is to ask to what extent explicit formulas like (1) can be replicated for the 'Beurling primes.'

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I refer to en.wikipedia.org/wiki/Beurling_zeta_function. It seems that a similar formula for the Beurling zeta function is not possible, i.e. you really need a functional equation to get an explicit description somewhere for real part small. It is very unlikely that there exists a functional equation for the spectral zeta function of some operator in general. Here, the Beurling zeta function is given by choosing the Beurling primes being the eigenvalues of some operator. –  Marc Palm Apr 24 '11 at 15:05
There are functional equations of a rather limited sort, see section 3 of arxiv.org/PS_cache/math/pdf/0410/0410270v1.pdf. One has to define the notion of dual primes, and a lot of the elegance seems to be irretrievably lost. Could you elaborate on the point that we're choosing the Beurling primes to be the eigenvalues of some operator? This sounds interesting, but is mostly lost on me... Thanks again. –  Brad Rodgers Apr 25 '11 at 6:09
This is nothing deep with the eigenvalues, you just define $\sum\limits_{j} \lambda_j^{-s}$ for the set of eigenvalues of some operator. Of course, there should not be to many eigenvalues. There are also some work on explicit formulas by Lang-Jorgenson in some LNM. –  Marc Palm Apr 25 '11 at 7:18

1 Answer 1

up vote 9 down vote accepted

The Riemann zeta function is given for $Re s>1$

$$\zeta(s) = \prod\limits_{p \; prime} ( 1-p^{-s}) $$

This product converges absolutely in $Re s >1$, hence it does not vanish in $Re s>1$. Actually the product also converges locally uniformly, which implies that $\zeta(s)$ is holomorphic for $Re s>1$.

The functional equation follows from the Poisson summation formula, which is a Fourier theoretic argument. The functional equation is given by $$ \Lambda(s) = \pi^{-s/2} \Gamma(s/2) \zeta(s) = \Lambda(1-s).$$ This implies that all zeros lie inside $0 \leq Re s \leq 1$. We will need to evaluate a certain integral $$ * = \int\limits_{Re\; s = 1 + \epsilon/2 \cup Re \; s = -\epsilon/2} G(s) \frac{\Lambda'}{\Lambda} (s) d s$$

We have now for a meromorphic function $F$ and a holomorphic function $G$ in some region $G$ containing the closure of a simply connected region $O$, that the contour integral $$\frac{1}{2 \pi i} \int\limits_{\partial O} G(z) \frac{F'}{F}(z) d z = \sum\limits_{\rho \; zero \; of \; F \; in \; O} G( \rho) - \sum\limits_{\nu \; pol \; of \; F \; in \; O} G( \nu),$$ where $\partial O$ denotes the boundary of $O$ and is a Jordan curve by assumption. For this identity to be valid $F$ must have no zero and no pol on the boundary of $O$.

This is what I call the weighted argument principle and can be derived in the same lines as the argument principle. It follows for entire functions easily from the Hadamard factorization theorem, but is a purely local property.

Apply this to the function $F = \Lambda$ and $G$ being holomorphic in $ - \epsilon \leq Re\; s \leq 1$ with $\epsilon>0$ and certain restrictions of the growth. We choose the contour $C=C(T)$ being the boundary of $ - \epsilon/2 < Re\;s < 1 + \epsilon/2$ and $| Im \;s | \leq T$, where \zeta does not vanish on $Im \; s = \pm T$. This give an expression for $*$ for $T \rightarrow \infty$ involving the nontrivial zeros of $\zeta$.

Using the Euler product, we can also derive a nice explicit expression $$ \frac{\Lambda'}{\Lambda} (s) = (log \Lambda(s))'= -1/2 \log \pi +\frac{1}{2} \frac{\Gamma'}{\Gamma}(s/2) - \sum\limits_{p \; prime} \frac{p^{-s} \log p }{1-p^{-s}}.$$ This gives an expression for $*$ for $T \rightarrow \infty$ involving the primes.

Assuming certain boundedness conditions on $\Lambda(s)$ and $G(s)$ in $-\epsilon \leq Re \; s \leq 1+\epsilon$, we are actually allowed to choose $C(T)$ with $T \rightarrow \infty$ and derive the formula as the limit. The boundedness conditions for $\Lambda$ follow from the Hadamard three lines principle or the Phragmen Lindeloeff principle (this is not entire function theory, but this only a complex analysis argument), then the explicit formula follows by choosing $G$ appropiately.

Remark: Inserting a Gaussian function into the explicit formula allows to derive the functional equation for $\zeta$, hence the functional equation is equivalent to Weil's explicit formula. Actually Samuel Patterson states in his famous book on $\zeta$ that they are actually both equivalent to the Poisson summation formula, but I do not know how? Of course the Poisson summation implies Functional equation of $\zeta$... How to go back?

So the philosophy is: Functional equation + Euler product = explicit formula. Another example for this is the relation of the Selberg Zeta function and Selberg trace formula.

You are right that the entire function theory implies that $\Lambda$ has necessary many zeros, but not too many, since it is of exponential type $1$ because of the factor $\Gamma$. If you want to derive this without using the merophorphicity of $\Lambda$, you might want to try to deduce this without knowledge over the primes and by inserting an approppiate chosen test function in the explicit formula. I have never seen this been worked out, but choosing an appropiate function $g$ being supported in $ - \log 2 < t < \log 2$ (so no contribution by finite primes) etc. should lead to a rough asymptotic of the zeros without any information used about the primes, but possibly a weaker error term than in the classical van Mangoldt estimate, which I expect to be a square root of the actual main term. Look at similiar techniques used in Werner Mueller and Erez Lapid's article Chapter 2 of http://www.math.uni-bonn.de/people/mueller/papers/orbint09.11.pdf for the Weyl law. The Selberg trace formula has many analogies with the explicit formula. In Iwaniec - Spectral methods in automorphic forms, you can find an argument using Tauberian theorems, which is weaker, but you get only the main term and no error term.

One interesting, but technical derivation of the explicit formula using only the languages of the adeles, harmonic analysis and no entire function theory at all was given by Ralf Meyer: http://arxiv.org/pdf/math/0311468v3

However, the Fourier transform of a function with certain growth properties has some holomorphicity conditions, so you basically will just hide the complex analysis arguments, but be able to deduce the results from Fourier analysis only. If you want to derive the prime number theorem from Fourier analysis, you might want to consider the treatement in Rudin - Functional analysis, which is based on real analysis only. Also there are elementary proofs of the prime number theorem, which I have no idea if they apply to Beurling primes.

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Very nice answer! thanks for the insightful explanations! –  SGP Apr 24 '11 at 14:11
Thanks! The article by Meyer at the end is I think exactly what I'm looking for; basically I'm looking to translate the complex analysis to harmonic analysis in as 'canonical' a way as possible. It will take me a little while to digest it though... –  Brad Rodgers Apr 24 '11 at 20:44
Perhaps this article of Meyer arxiv.org/abs/math/0412277 gives a better start, since it only deals with the Riemann Zeta function without introducing nuclear bornological vector spaces and such... The adeles allow in his other article to threat all Hecke L-functions simultaneously, which makes everything only more conceptual so far. I am not sure how you want to construct your adeles or local fields for generalized primes anyway, if they are not associated to norms of prime ideals in a number field. Are you willing to share some rough ideas about your plans here? –  Marc Palm Apr 25 '11 at 11:27
and here the equivalence of poisson summation formula and functional equation: mathoverflow.net/questions/62969/… –  Marc Palm Apr 26 '11 at 12:40
Well, there's not a deep plan at this point; I'd just like to understand the explicit formula slightly better. (I have more analytic facility with harmonic analysis than complex analysis, for instance.) One reason I am interested in these things is to study statistical properties of the Riemann Zeta function; one can generally convert such statements into arithmetical statements, and it is elucidating to see if these statements are still true for the Beurling primes (often times they are!). (cont.) –  Brad Rodgers May 1 '11 at 22:42

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