Background: Beurling's systems of numbers

Beurling considered a sequence of reals $1<x_1<x_2<\cdots <$ as "primes" and then the ordered sequence of all products of these "primes" as "integers". (We can even allow repetitions for the "primes".) In this setting Beurling and subsequently others considered analogs of the prime number theorem and of the Riemann hypothesis.

For a square free Beurling integer $x$ we can consider its "Möbius function", $\mu (x)$ to be $-1$ if $x$ is the product of odd number of "primes" and $1$ if $x$ is the product of even number of primes.

The questions

Question 1: Consider the assertion of the prime number theorem $$\sum \{\mu (y): y \le x\}=o(x).\label{1}\tag{1}$$ What are general conditions that are known to guarantee the assertion of the PNT? What are some useful examples that violates the PNT?

Question 2: Consider the assertion of RH, for every $\delta >0$ $$\sum \{\mu (y): y \le x\}=o(x^{1/2+\delta}).\label{2}\tag{2}$$ What is known (I suppose that mainly negative results) about conditions that guarantee or do not guarantee RH?

Question 3: To what generality does the formulations of the PNT and RH in terms of zeta functions extend and is equivalent to the Mobius function formulation?

Question 4: If you take every (usual) prime with probability 1/2. What is the status of \eqref{1} and \eqref{2}?

Question 5: If you take $x_n=n \log n$ what is the status of \eqref{1} and \eqref{2}?

Question 6: If you take $x_n$ to be a tiny tiny perturbation of $p_n$ (so the Mobius function is never 0) what is the status of \eqref{1} and \eqref{2}? (Maybe this is obvious.)

Question 7: If you take $x_n$ to be a tiny tiny perturbation of $n$ (so the Mobius function is never 0) what is the status of \eqref{1} and \eqref{2}?


2 Answers 2


Here's a partial answer and some information about your Question 1.

Let $N(x)$ denote the number of 'generalised integers' corresponding to the generalised primes (i.e. products of the shape $x_1^{a_1}\cdots x_k^{a_k}$ for some non-negative integers $a_i$) at most $x$.

Zhang (A generalization of Halász's theorem to Beurling's generalized integers and its application, Illinois J. Math. 31(4) 1987, pp. 645-664) showed that if there are some constants $A>0$ and $\gamma>1$ such that

$$ N(x) = Ax + O\left( \frac{x}{(\log x)^\gamma}\right)$$

then $\sum_{n\leq x}\mu(n)= o(x)$ (where $\mu$ is defined as in your question).

This is a curious result, since Beurling showed that $N(x) = Ax +O(x/(\log x)^\gamma)$ with $\gamma>3/2$ was sufficient to obtain the more usual form of the PNT, $\pi(x) \sim x/\log x$, and Diamond (A set of generalized numbers showing Beurling’s theorem to be sharp, Illinois J. Math., vol. 14 (1970), pp. 29-34.) gave an example to show that this fails with $\gamma=3/2$.

That is, there is a choice of $x_i$ such that $N(x) = Ax +O(x/(\log x)^{3/2})$ for some constant $A>0$, but the asymptotic $\pi(x)\sim x/\log x$ fails to hold. Zhang's result shows that we still have $\sum_{n\leq x}\mu(n)=o(x)$ for this example, however - so there is some regime where these different forms of the PNT are no longer equivalent!

DeBruyne, Diamond, and Vindas (https://arxiv.org/abs/1609.03504) have recently shown that least one side of this equivalence still holds, showing that if $\pi(x)\sim x/\log x$ and $N(x)\ll x$ then $\sum_{n\leq x}\mu(n)=o(x)$. Section 4 of this paper contains some relevant examples and constructions you might find interesting.

  • $\begingroup$ Great answer! Many thanks, Thomas. (BTW, shouldn't we think about $N(x)$ as the number of 'generalised integers' below $x$?) $\endgroup$
    – Gil Kalai
    Nov 5, 2021 at 20:41
  • $\begingroup$ @GilKalai Yes, definitely, generalised integers was what I meant to write, thanks! Corrected. $\endgroup$ Nov 5, 2021 at 21:04
  • 1
    $\begingroup$ @GilKalai, you might find arxiv.org/abs/1902.03870 interesting. Corollary 1.2 shows that positive density and a Chebyshev estimate imply the result (1) of your question. The paper is an improvement of Zhang's result mentioned by Thomas Bloom. $\endgroup$
    – F_M_
    Apr 1, 2022 at 7:49

We can express the inverse zeta function $$ \prod_{i} (1 - x_i^{-s}) = \sum_ x \mu(x) x^{-s} = \int_1^{\infty} \left(\sum_{y \leq x } \mu(y) \right) s x^{-s-1} dx $$

From this integral representation, we immediately get the implication that the RH in the form $\sum_{y \leq x } \mu(y) = O( x^{ 1/2+ \delta})$ for all $\delta>0$ implies RH in the form that the inverse zeta function is holomorphic for $\operatorname{Re}(s) > 1/2$, as then the integral representation is convergent in that range. I believe the reverse implication can be proven by the usual contour integral argument, maybe with some assumption that the $x_n$ do not grow too much more slowly than the primes, so that the zeta function can be controlled to the right of the critical strip.

Similary, the prime number theorem in the form $\sum_{y \leq x}\mu(y) = o(x)$ implies the inverse zeta function is $ \int_{1}^{\infty} s o (x^{-s} ) dx = o ( |s|/ (\operatorname{Re} s -1))$ where the $o$ goes to $0$ as $\operatorname{Re}(s)$ goes to $1$. This is because $ \int_{1}^{\infty} s x^{-s} dx = s / (s-1)$ and as $\operatorname{Re} s$ goes to $1$, most of the mass is supported on large values of $x$ where $o(x)$ is much smaller than $x$. I'm not sure if this necessary condition for PNT is also sufficient or if more is needed.

For questions 4, 5, 6, and 7, the game is going to be to work with the dlog of the zeta function

$$ \sum_i \frac{ d \log (1- x_i^{-s})^{-1} }{ds} = \sum_i \sum_{j=1}^{\infty} x_i^{-js} \log x_i = \sum_{j=1}^{\infty} \int_1^{\infty} \left( \sum_{x_i \leq x^{1/j}} \log x_i \right) s x^{-s-1} dx $$

and plug in estimates for the sum to evaluate the integral.

For question (4), the sum $ \sum_{y \leq x^{1/j}} \log x_i $ will of course be close to half the number of primes up to $x^{1/j}$, with square-root error. The square-root error will integrate to a holomorphic function in the $\operatorname{Re }s >1/2$ region, so the Beurling zeta function will be equal to the square root of the usual zeta function up to a nonvanishing holomorphic factor. In particular, its inverse will not be holomorphic, as it has nontrivial monodromy around $s=1$. If we instead take two copies of $p_i$ with probability $1/2$ and no copies with probability zero, then the Riemann hypothesis for the modified zeta function follows from Riemann for the original zeta function.

For question (5) and (7), I think something similar happens. We have a precise asymptotic for the number of Beurling primes up to $x$, with small error term, but the main term looks different from the main term in the prime number theorem, and so we won't get a Riemann hypothesis the way you've formulated it.

For example, for question (7), if we take $x_n$ to be a small perturbation of $n$, then the log of the inverse zeta function is close to $\sum_n \log (1- n^{-s}) = -\sum_n \sum_{j=1}^{\infty} n^{-js} / j = -\sum_{j=1}^{\infty} \zeta(js) / j$. For $s$ near $1$, the dominant term is $-\zeta(s)$, which is close to $-1/(s-1)$, so the inverse zeta function itself is equal to $e^{-1/(s-1)}$ up to a nonvanishing holomorphic factor. When the real part of $s$ is positive, the real part of $-1/(s-1)$ is always negative, so the growth rate of this function is not so large and I'm not sure whether (1) holds, but as there is an essential singularity at (2),

However, it's possible to give a sequence with a non-arithmetic definition that unconditionally satisfies the Riemann hypothesis.

If we define $x_n$ inductively by $x_1 =2$, $x_{n+1} = x_n + \log x_n$ for $n>1$, then $\sum_{i=1}^n \log x_i = x_{n+1} - 2$ so $\sum_{x_i \leq x} \log x_i = x + O (\log x)$ which means the logarithmic derivative of the zeta function is close to $s/(1-s)$. Exponentiating, the inverse zeta function is going to be $(s-1)$ times something holomorphic on $ \operatorname {Re} s>1/2$, which by a contour integral should give the Riemann hypothesis.

If we take half the primes at random, the zeta function we get will be, up to a nonvanishing holomorphic factor, the square root of the usual zeta function. So zeta inverse will, near $s=1$, look like $(s-1)^{1/2}$. This has a singularity at $s=1$, but a very mild one - in particular the function grows slower than $1/(s-1)$. I suspect when you do the contour integral to estimate the Mobius sum here, you get the prime number theorem, i.e. it satisfies (1).

  • $\begingroup$ Great Answer! Many thanks, Will. $\endgroup$
    – Gil Kalai
    Nov 7, 2021 at 0:17

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.