Heuristic argument for the Riemann Hypothesis

Question

Is there a heuristic argument that supports the validity of the Riemann hypothesis or are we just relying on numerical evidence? Moreover, what is the strongest theorem that supports the validity of RH?

I have heard that a certain strain of number theorist believes the evidence for RH is indeed quite thin. But maybe someone who knows more about that could post an answer to that effect. — Sam Hopkins, Sep 03 '19 at 13:49
@SamHopkins : In Derbyshire's book Prime Obsession, he says that Martin Huxley is a skeptic, citing Littlewood's heuristic that a longstanding conjecture in analysis generally turns out to be false. He also says that Andrew Odlyzko is completely agnostic about RH. He quotes Odlyzko as saying that when a certain function called the S function gets large then it might threaten the truth of RH. Selberg proved that S is unbounded, but it grows very slowly, and the regime where it might threaten RH is far beyond computational reach. — Timothy Chow, Sep 07 '19 at 03:05

score 41 · Answer 1 · answered Aug 31 '19 at 07:39

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The Riemann hypothesis is true, if primes are random in certain ways.

answered Aug 31 '19 at 07:39

Pace Nielsen

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If one had all but finitely many zeros on the critical line would this still hold? – Mark Schultz-Wu Aug 31 '19 at 07:50
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@Mark: A single zeta zero with real part greater than $1/2$ would already cause big irregularities in the distribution of prime numbers. – GH from MO Aug 31 '19 at 07:55
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"Cause", eh? I have to warn students: "If P then Q" does not mean "P causes Q". – Gerald Edgar Aug 31 '19 at 13:02
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@Gerald Edgar, is there a principled definition of "P causes Q" when P and Q are logical propositions? – Solveit Sep 01 '19 at 05:56
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To elaborate a bit on the comment by GH from MO, each zero off the line yields an irregularity that makes itself felt asymptotically. It is not the case that asymptotically, we can ignore a finite number of zeros. Of course if the zero is very close to the line then it might not be detectable numerically until you've gone a long way out. – Timothy Chow Sep 01 '19 at 15:35
This borders on being a link-only answer, no? – R.. GitHub STOP HELPING ICE Sep 01 '19 at 16:19
@ChanBae: When P and Q are logical propositions, it may be a poor choice of wording to talk about "causation", but I don't think its use qualifies as a correlation/causation fallacy like Gerald seems to have implied. – R.. GitHub STOP HELPING ICE Sep 01 '19 at 16:21
@Pace Nielsen, thanks for the nice link. Could you elaborate whether the argument in the link "If Riemann was right, there will be no zeros for Re(σ)>1/2, and hence we could choose σ to be as little as 1/2+ε" applies to zeros with multiplicity? If zeta function has zeros off the 1/2 line such that the residue of 1/zeta is zero, would that have an impact on prime distribution? – Michael Sep 03 '19 at 20:33
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@Michael Yes to both questions. It's not the residue of $\zeta$ which matters, but the residue of $x^s/\zeta(s)$, which can only be zero for all $x$ if $\zeta$ is nonvanishing there. – Will Sawin Sep 04 '19 at 03:47
@WillSawin do you mean the residue of $-(\zeta'(s)/\zeta(s))(x^s/s)$? Logarithmic derivatives of $\zeta(s)$ (or $L$-functions) are what naturally occur in standard explicit formulas, and the function $\zeta'(s)/\zeta(s)$ (like any logarithmic derivative) has at worst simple poles, with the residue of $\zeta'(s)/\zeta(s)$ at each number in $\mathbf C$ being the order of vanishing (positive, negative, or zero) of $\zeta(s)$ at that number, so logarithmic derivatives of meromorphic functions never have residue $0$ at a pole. – user1728 Sep 04 '19 at 13:39
@user1728 If you follow the link, he's talking about the explicit formula for the sum of Mobius, not the explicit formula for prime counting. So I tried to answer for the Mobius function. I agree with everything you said about the primes/von Mangoldt and logarithmic derivatives. – Will Sawin Sep 04 '19 at 13:42

score 32 · Answer 2 · answered Aug 31 '19 at 07:44

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There are many theoretical results that support the (generalized) Riemann hypothesis:

zero density estimates for the zeros of certain $L$-functions
infinitely many zeros on the critical line of certain $L$-functions
nonnegativity of the central value of certain $L$-functions
subconvex bounds for certain $L$-functions
strong lower and upper bounds for certain moments of certain $L$-functions
the (proven) Riemann hypothesis for the $L$-functions of algebraic varieties over finite fields

and so on. Each of these items is a research topic on its own, with hundreds of articles. Use Google, Wikipedia, MathSciNet etc. to learn more about them.

answered Aug 31 '19 at 07:44

GH from MO

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@MustafaSaid: I guess my point is that many powerful consequences of the Riemann hypothesis have been verified independently. Some of these address the distribution of the zeros directly. In my view, these results are much stronger evidence than the numeric verification of the Riemann hypothesis up to a certain height. Add to these that the algebraic geometric analogue of the Riemann hypothesis is a theorem thanks to Weil, Deligne, etc. I am talking about infinitely many $L$-functions in each of the bullet points in my anwer. – GH from MO Aug 31 '19 at 07:58
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And not only are there infinitely many zeta zeros on the critical line, but the proportion thereof has been shown to exceed 41%. – Sylvain JULIEN Aug 31 '19 at 07:58
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But where is the heuristic argument? The GUE connection is more along the lines of what I’m looking . – Mustafa Said Aug 31 '19 at 08:02
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@MustafaSaid: I reacted to "are we just relying on numerical evidence". The answer is a clear "no". – GH from MO Aug 31 '19 at 08:03
Why do you consider nonnegativity of a central $L$-value evidence towards GRH? – user1728 Sep 01 '19 at 01:44
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@user1728: In this case, one considers an $L$-function $L(s)$ which is holomorphic on $\mathbb{C}$ and real-valued on $\mathbb{R}$. The Riemann Hypothesis for $L(s)$ implies that $L(s)>0$ for $s>1/2$, and hence $L(1/2)\geq 0$. So if we can prove the nonnegativity of $L(1/2)$ unconditionally, then we could confirm an important prediction of GRH. I say "important", because nonnegativity usually has deep applications, while proving it requires a similarly deep understanding of the arithmetic structure behind $L(s)$. For infinitely many $L$-functions, $L(1/2)\geq 0$ is a theorem! – GH from MO Sep 01 '19 at 02:13

score 19 · Answer 3 · answered Aug 31 '19 at 22:19

The function field model came up in GH from MO's answer, and then also in user54038's. I just want to add some detail to explain how good of an analogy the function field model is.

The Riemann zeta function does not stand alone, but instead is the simplest member of a series of wider classes of zeta and $L$-functions of deep relevance in number theory - among them Dedekind zeta funcitons, Dirichlet $L$-functions, Hecke $L$-functions, Artin $L$-functions, elliptic curve $L$-functions, automorphic $L$-functions. (More precisely each of these fall into two wide classes, those being the motivic $L$-functions and the automorphic $L$-functions, which are conjectured to more-or-less agree, so you could also say there is a single class of $L$-functions with many interesting sub-classes). For all of these, the Riemann hypothesis has been conjectured.

There are direct analogues of all these in the function field setting, with the interrelationships among them basically the same. The Riemann hypothesis was proven for all of them - the first couple by Weil, the later ones by Deligne, and the automorphic ones by Drinfeld, L. Lafforgue, and V. Lafforgue.

Furthermore properties known and conjectured of these zeta and $L$-functions, in particular those to do with the distribution of the zeroes, match up in straightforward ways. As far as I know, all known results about the distribution of zeroes of $\zeta$ and $L$-functions can be proven also over function fields by essentially the same arguments, and much stronger extensions of these results, fitting what we conjecture over $\mathbb Q$, are known in the "large $q$ aspect".

So the function field check on the Riemann hypothesis is not just a single analogy between two functions but a web of related analogies that all seem remarkably consistent with each other.

(Just a tiny quibble, that probably Dwork and Grothendieck deserve considerable credit for the run-up to Deligne's proof. Probably others whom I don't know...) — paul garrett, Aug 31 '19 at 22:51
@paulgarrett Sure. The proof of the Riemann hypothesis is due to Deligne, but of course this was just the final piece of the puzzle. For the foundations of etale cohomology he used, one should probably credit the SGA school, in particular the authors of the relevant texts - at least Grothendieck, Raynaud, Artin, Verdier, Deligne again, Illusie, and Katz. While Dwork proved one of the Weil conjectures, it wasn't Riemann, and his methods weren't directly used in Deligne's work, though his ideas. We could also mention Rankin, Lefschetz, and some others. — Will Sawin, Aug 31 '19 at 23:30
are these other R.H. in some way equivalent to the asymptotic behaviour of some sort of generalized Merten functions, as mentioned in Pace Nielsen's and JoshuaZ's answers? — user347489, Sep 01 '19 at 03:08
@user347489 Sure, just sum the coefficients of the inverse of the $L$-function. — Will Sawin, Sep 01 '19 at 03:09

polfosol · Answer 4 · 2019-09-04T08:04:02.460

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I am not sure that this is a proper answer for your question. Sorry in advance if it is not.

Other answers provided some good examples of heuristic arguments supporting the RH. But I want to point at a heuristic result that puts some serious doubts on it. Since I am not so familiar with the technical details, it is better to quote directly from this answer:

The De Bruijn-Newman constant $\Lambda$ was defined and upper bounded by $\Lambda\le\frac{1}2$ in 1950. After 58 years of work, in 2008 this upper bound was finally improved to... $\Lambda<\frac{1}2$ (a 0% improvement) in a 26-page paper. The best known upper bound is currently $\Lambda<0.22$. The Riemann hypothesis was known to be equivalent to $\Lambda\le 0$, so if it's true then we've got quite a ways to go.

But Terrence Tao and Brad Rogers recently proved that $\Lambda\ge 0$. So in Tao's words (actually, Newman's words quoted by Tao):

If the Riemann hypothesis is true, then it's 'just barely' true.

edited Sep 04 '19 at 08:04

answered Sep 02 '19 at 09:44

polfosol

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Or, if the Riemann hypothesis is true (so $\Lambda = 0$), then we've got quite a ways to go. – Jesse Elliott Sep 02 '19 at 09:59
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I'm not familiar with the De Bruijn-Newman constant, but naïvely I don't see why a proof of $\Lambda\geq0$ would make $\Lambda\leq0$ any more dubious. In fact, for something as fundamental as the Riemann Hypothesis, it would seem quite fitting that it be “exactly spot on just true” rather than merely “true on the boundary of something more mundane”. – leftaroundabout Sep 03 '19 at 09:27
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@leftaroundabout : The intuition is that Lehmer pairs (unusually close consecutive zeros of $Z(t)$) look like they're "nearly" counterexamples to the Riemann hypothesis. Roughly speaking, $\Lambda\ge 0$ means that these near-counterexamples come very close to being true counterexamples. So if one's intuition is that the behavior of $Z(t)$ is modeled well by some "random process" then one might worry that some "random fluctuation" might cause one of the extrema of $Z(t)$ to fall just short of crossing the $t$-axis. (BTW the phrase "just barely" goes back to Newman and isn't Tao's coinage.) – Timothy Chow Sep 03 '19 at 21:21
@TimothyChow thanks for the tip about the last quote. I edited the answer. – polfosol Sep 04 '19 at 08:04

score 12 · Answer 5 · answered Aug 31 '19 at 17:36

Besides all the theorems already mentioned I would like to add the Bombieri-Vinogradov theorem. This can be roughly thought of as an averaged version of the Generalised Riemann Hypothesis and thus gives strong support for GRH. Even more, in many proves it can be used to replace the Riemann Hypothesis and so make theorems unconditional.

score 11 · Answer 6 · edited Sep 01 '19 at 00:30

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There have been some good answers already given but I want to note another aspect, namely a heuristic involving the Möbius function. Let $\mu(n)$ be the Möbius function. The Riemann Hypothesis is equivalent to the claim that for any $\epsilon >0$ one has that $$\sum_{1 \leq n \leq x}\mu(n) = O(x^{1/2+\epsilon}).$$ This equivalence stems from making an explicit integral for $1/\zeta(s)$ in terms of the Möbius function which converges up the the 1/2 line if one has the above bound on the sum.

Now, it is reasonable to guess/hope/assume that $\mu(n)$ is essentially random in the sense that the non-zero values behave essentially like a fair coin with heads corresponding to 1 and tails corresponding to –1. It turns out that if one has a fair coin, and one keeps flipping it, then the difference between the number of heads and tails after $x$ flips will with probability 1 be $O(x^{1/2+\epsilon})$. So if the Möbius function behaves like a random coin flip, or even close to a random coin flip we should expect RH to hold with probability 1.

Note that this isn't the only thing satisfying about this framing of the Riemann Hypothesis. This sort of shows one major reason that RH is important: a lot of sieves and other ways to get a handle on the primes involve inclusion/exclusion arguments, which is essentially what $\mu(n)$ is for. So in a moral sense RH says that if one is doing inclusion/exclusion with primes, one cannot get very large deviations in how much at any given stage one needs to include or exclude.

(Edit to add: actually this is essentially the same sort of approach as noted in the link by Pace above.)

edited Sep 01 '19 at 00:30

Community

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answered Aug 31 '19 at 16:21

JoshuaZ

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For iid coin flips the law of the iterated logarithm gives that the sum is $O(\sqrt{n \log \log(n)})$ with probability $1$. Is it known that the partial sums of the Mobius function are not $O(\sqrt{n \log\log(n)})$? – burtonpeterj Aug 31 '19 at 17:42
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@burtonpeterj This sum is sometimes called the Mertens function. According to Wikipedia, there is a conjecture by Steve Gonek that this sum should by $O(\sqrt{n} (\log \log \log n)^{5/4})$, so it would grow even a little bit slower than the coin flips heuristic would suggest. – Marc Aug 31 '19 at 18:22
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Let's write $M$ for the Mertens function. Apparently the best lower bounds for $\limsup_{n \to \infty} |M(n)|n^{-\frac{1}{2}}$ are less than $2$. Moreover, recent progress on this seems incremental. Given that, why is it considered so implausible that $|M(n)| = O(\sqrt{n})$? – burtonpeterj Aug 31 '19 at 18:45
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@burtonpeterj Probably based on heuristics on the linear independence / random distribution of the zeroes. – Will Sawin Aug 31 '19 at 20:26
What does $|M(n)| = O(\sqrt{n})$ say about structure in the zeroes? – burtonpeterj Aug 31 '19 at 20:27
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Gonek and Ng have conjectured that $\limsup_{x \to \infty} \frac{M(x)}{\sqrt{x}(\log \log \log x)^{5/4}}$ (resp., $\liminf_{x \to \infty} \frac{M(x)}{\sqrt{x}(\log \log \log x)^{5/4}}$) is finite and positive (resp., finite and negative), which would imply that $M(x) = \Omega_{\pm} (\sqrt{x})$. Good and Churchhouse and L'evy have made (I think less plausible) conjectures that imply that $\limsup_{x \to \infty} \frac{|M(x)|}{(x \log \log x)^{1/2}}$ is finite and positive, which also implies $M(x) = \Omega_{\pm} (\sqrt{x})$. The law of iterated logarithm is at play here I think. – Jesse Elliott Aug 31 '19 at 20:58
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$\dfrac{1}{\zeta(s)} = s \displaystyle \int_{1}^{+ \infty} \dfrac{M(x)}{x^{s+1}} dx , \quad \Re(s) > 1$ .. the equality is true also for $s=1$ (Von Mongoldt 1897). – Lagrida Yassine Aug 31 '19 at 21:51

score 11 · Answer 7 · answered Aug 31 '19 at 20:11

Bombieri reviews the evidence for the Riemann hypothesis in his problem description for the Millenium prize. In section III, he gives the numerical evidence as one of three pieces of evidence, the second being results on the density in the complex plane of hypothetical counterexamples, and the third that "it is known that more than 40% of nontrivial zeros of $\zeta(s)$ are simple and satisfy the Riemann hypothesis". However, in section IV, he says that "It may be said that the best evidence in favor of the Riemann hypothesis derives from the corresponding theory which has been developed in the context of algebraic varieties over finite fields." He gives special attention to Deligne's theorem. So, the strongest evidence is not the numerical evidence or even a heuristic argument, but rather an analogous theorem about a different system. At least, that's Bombieri's opinion.

Heuristic argument for the Riemann Hypothesis

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