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Attempts to prove the Riemann Hypothesis

So I'm compiling a list of all the attacks and current approaches to Riemann Hypothesis. Can anyone provide me sources (or give their thoughts on possible proofs of it) on promising attacks on Riemann Hypothesis?

My current understanding is that the field of one element is the most popular approach to RH.

It would be good if someone started a Polymath project with the aim of proving RH. Surely, if everyone discussing possible ways to prove RH, it would be proven in about a year or so or at least people would have made a bit more progress towards the proof or disproof.

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    $\begingroup$ "Surely, ..., , it would be proven in about a year". That's a lot of wishful thinking. $\endgroup$
    – lhf
    Oct 29, 2011 at 17:34
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    $\begingroup$ Consider the whole literature on the RH as a huge polymath project! $\endgroup$
    – lhf
    Oct 29, 2011 at 17:50
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    $\begingroup$ It ain't gonna happen. I don't think you understand either the difficulties inherent in the RH or the nature of mathematical research. There are only a small number of mathematicians with the skill-set to approach the RH, and they already know each other. They don't need the internet to generate their collaborations. It isn't like the other polymath projects, which focused on problems that are accessible with minimal background and which everyone knew were "just out of reach". There's a reason none of them resulted in bombshell results (or even in papers in top journals). $\endgroup$
    – Adam Smith
    Oct 29, 2011 at 19:02
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    $\begingroup$ As for unconventional approaches: I've found these three papers to be terribly interesting... $\endgroup$ Oct 29, 2011 at 20:04
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    $\begingroup$ @simplicity: Keep in mind that exactly the same argument can be made about the continuum hypothesis: if it is independent of the axioms, then you cannot find an uncountable subset of $\mathbb{R}$ which is not $c$, thus you prove it....The problem of independence is not that simple. If it is independent, it means that we can have two different mathematical models, so that RH is true in one and not true in the other.... The independence doesn't mean that you cannot find a zero, it only means that you don't have enough info to decide if there is a zero off the line... $\endgroup$
    – N. S.
    Oct 29, 2011 at 20:40

6 Answers 6

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"My current understanding is that the field of one element is the most popular approach to RH."

Analytic number theory, with ideas from algebraic geometry, random matrix theory, and any other areas that might be relevant, is the only approach known to have produced any concrete results toward RH. The random matrix theory in particular has produced a lot of new constraints and specific, provable ideas about the distribution of zeros on the critical line.

The field of one element is, for now, a speculative area of algebraic geometry whose foundations are not set. It is more an inspiration for research on more definite mathematical objects (e.g., is there a tensor product of zeta functions) than a well-defined topic of research in itself.

(I'll add here some response to the comments. Research on $F_1$ is, as Matt E writes, "serious" and conducted with various sophisticated intentions in mind, such as perfecting the analogies between number theory and geometry, proving the Riemann hypothesis, understanding quantum groups, or realizing parts of combinatorics as geometry over $F_1$. This was all proposed in Manin's lectures at Columbia 20 years ago which were instrumental in bringing the idea into fashion in recent years. However, as serious and sophisticated as this research is, the idea that a suitable notion of $\Bbb{F_1}$ exists as a deeper base for algebraic geometry, or that this line of research can be developed to cover new varieties beyond the original example of Weyl groups of reductive groups (or flag varieties and other examples with simple $q$-enumerations) --- or the hope that all this can help prove the Riemann hypothesis --- is a speculative enterprise and one whose foundations have not been established.)

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    $\begingroup$ Apparently there are strange things going on with the field of one element. A friend of mine told me of a talk by a graduate student working on $\mathbb F_1$ that started with the words "Contrary to popular belief, the field of one element contains two elements..." $\endgroup$ Oct 29, 2011 at 20:51
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    $\begingroup$ Dear zyx, I largely agree with your answer, but there are some people working seriously on $\mathbb F_1$; e.g. Jim Borger at ANU, and Alain Connes (who I just saw give a talk yesterday at IHP with concrete ideas related to $\mathbb F_1$). Regards, $\endgroup$
    – Matt E
    Oct 29, 2011 at 20:52
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    $\begingroup$ @Matt E: I agree, but most work on F_1 is without any overt connection to the Riemann hypothesis (vs Weyl groups or stable homotopy). This is true of papers by Borger, Durov, Dietmar, Toen-Vaquie and most work not coauthored by Connes. One has to admire Connes' ability to say something new about seemingly any area of mathematics, but I don't think any analytic facts or conjectures about $\zeta(s)$ have yet emerged from the $F_1$ approach. Noncommutative geometry's connection to arithmetic geometry seems promising (esp. the relation to Consani's thesis) but does not yet involve F_1 per se. $\endgroup$
    – zyx
    Oct 30, 2011 at 0:28
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    $\begingroup$ @Matt: further comments added in answer. While I don't want to throw cold water on the OP's mention of F_1, it would be remiss to avoid mention of the fact that as promising as the subject may be, it is quite speculative (e.g., there isn't a clear way of determining what the right foundations are, unless of course one approach suddenly achieves a breakthrough such as a new foundation for Arakelov theory, a proof of RH, a quantitative elucidation of the analogies between primes and knots in a 3-manifold, etc). $\endgroup$
    – zyx
    Oct 30, 2011 at 1:00
  • $\begingroup$ Dear zyx, I agree that the current state of the theory of $\mathbb F_1$ seems to be very far from RH. I think you've summarized the situation well in your answer and comments. Regards, $\endgroup$
    – Matt E
    Oct 30, 2011 at 2:16
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It should be worth pointing out that, Alain Connes attacked the problem from a very different plane(http://arxiv.org/abs/math/9811068), following Weil and Haran's path, he tried to construct an "index theory" in Arithmetical context linking Arithmetical data with Spectral properties of a certain operator which is closed and unbounded and whose spectrum consists of imaginary parts of the zeroes of Hecke's L function with Grossen-character. Essentially he reconstructed a theory similar to Selberg's, he found a trace formula equivalent to the RH using Weil's explicit formulae.

Actually Shai Haran also stated a "similar" trace formula in his AMS paper "On Riemann's zeta function", One can also find a derivation of his trace formula in his book The Mysteries of the Real Prime.

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Also check out the following paper, on the Baez-Duarte Criterion :

http://www.man.poznan.pl/cmst/2008/v_14_1/cmst_47-54a.pdf

In particular, look at that graph of Fig 2 on Page 5 of 8.

If someone can prove that it stays an obvious cosine function, as the associated formulae suggest, with non-increasing amplitude and no new trends creeping for larger n to throw it out, then perhaps that would be enough to prove RH !

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An attack by physics based on an inverse spectral problem yields that the inverse of the potential is $V^{-1}(x) = 2\sqrt \pi \frac{d^{1/2}}{dx^{1/2}} \operatorname{Arg} \xi (1/2+i \sqrt x)$.

In this case also the Theta functions classical and semiclassical are almost equal $$ \Theta (t)= \sum_n \exp(-t E_n) = \iint_C \,dp \, dx \, \exp(-tp^2 - tV(x)) .$$ This is one of the best approximation to RH.

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I am hesitating to write this answer since I know too little about the whole subject. However it is already a couple of years ago that I stumbled on a interesting paper on the arXiv https://arxiv.org/pdf/1703.03827.pdf by Vladimir Blinovsky .This paper seems to me like a simple and appealing idea to tackle the problem. It just seems strange for me that this idea has not attracted much publicity since it has been lying around on the arXiv for a quite long time. As a matter of fact this idea may be even described in plain words without resorting to mathematics. Let me do this now.

Let us fix the notation: \begin{equation} \zeta(s) := \sum\limits_{n=0}^\infty \frac{1}{n^s} \quad (A) \end{equation} where $s\in {\mathbb C}$ and $Re[s] > 1$. By analytic continuation (as originally carried out by Riemann https://en.wikipedia.org/wiki/File:Ueber_die_Anzahl_der_Primzahlen_unter_einer_gegebenen_Gr%C3%B6sse.pdf by deforming the integration contour in the complex plane and using Cauchy theorem) we obtain the functional equation: \begin{equation} \zeta(s) = 2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \Gamma(1-s) \zeta(1-s) \quad (B) \end{equation} for $s\in {\mathbb C}$ and $s\neq1$.

Let us define the squared module of the zeta function as follows: \begin{equation} K(\sigma,T) := \left| \zeta(s)\right|^2 \end{equation} where $s=\sigma + \imath T$. Clearly since $\zeta(s)$ is continuous and smooth the same holds for the function $K$, i.e. it too is continuous and smooth.

Now comes the important part. What Blinovsky claims in his paper is that the function $K(\sigma,T)$ above is convex in the variable $\sigma$. In other words he claims that \begin{equation} \frac{\partial^2 K(\sigma,T)}{\partial \sigma^2} \ge 0 \quad (C) \end{equation} for $\sigma \in (0,1)$ and $T$ being big enough, i.e. being bigger than some number which is independent of $\sigma$. Note that the property (C) along with functional equation (B) immediately implies the Riemann hypothesis being true. Indeed let us assume that there is some stray zero off the critical line at some $(\sigma = \sigma_0,T)$ where $\sigma_0 \in (1/2,1)$. Then from the functional equation there must be another zero at $(1-\sigma,T)$ where $1-\sigma_0 \in (0,1/2)$. But this would actually mean that the set $\left\{\xi \in (1-\sigma_0,\sigma_0) | K(\xi,T)\right\}$ is either strictly negative which it cannot be since per definition the function $K$ is non-negative or that the set in question is identically equal to zero which again is impossible because the function is smooth per definition.

Now the question appears how does Blinovsky prove the convexity? He starts from a certain integral representation of the zeta function and then by changing variables appropriately and then differentiating with respect to $\sigma$ twice he ends up with a following neat expression . We have: \begin{equation} \frac{\partial^2 K(\sigma,T)}{\partial \sigma^2} = \frac{8 }{\pi T} \left(\int\limits_0^1 f(h,T) \cos\left( \frac{\pi h}{2}\right)^2 dh - \frac{1}{2} \int\limits_0^1 f(h,T) dh \right) \end{equation} where $f(h,T)$ is some complicated function expressed through an improper integral and an infinite sum, a function to long to be written down in here without the benefit of any insight. Now since $\int\limits_0^1 \cos(\pi h/2)^2 dh = 1/2$ proving the convexity property is equivalent to the Fortuin–Kasteleyn–Ginibre (FKG) inequality https://en.wikipedia.org/wiki/FKG_inequality . Since the squared cosine is monotonically decreasing in $h\in(0,1)$ the only thing one needs to do is to prove that $f(h,T)$ too is monotonically decreasing in the domain in question. Blinovsky proceeds to show that by looking at the derivative $\partial_h f(h,T)$ and proving that it is negative for all $h \in(0,1)$ and for $T$ being big enough.

Having said all this my question would be is this approach to this problem sound or instead is there anything internally flawed in here that won't make this thread worth pursuing at all.

I would really appreciate some feedback. Thank you in advance for it.

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  • $\begingroup$ the result that $\begin{equation}K(\sigma,T) := \left| \zeta(s)\right|^2 \end{equation}$ is convex in $0<\sigma<1$ for fixed large $T$ contradicts the universality of zeta so is false $\endgroup$
    – Conrad
    Mar 31, 2022 at 14:54
  • $\begingroup$ @Conrad: Interesting. So let me rephrase what you are saying. There is a theorem that the zeta function can approximate any holomorphic function arbitrarily well and this theorem contradicts the statement that the function in question be convex(How?). On the other hand I remember Blinowsky did some long winded calculations with Mathematica in his paper in order to "prove" convexity. I guess it means that he must have made an error somewhere in those calculations. $\endgroup$
    – Przemo
    Apr 1, 2022 at 9:08
  • $\begingroup$ Universality means that if you take any holomorphic function that doesn't vanish on a disc in the right part of the strip (usually one uses a disc centered at $3/4$ of radius $1/4$ but that is not essential - what is essential is no zeroes there and right part) then infinitely many vertical translates of zeta approximate it uniformly as well as we want in any smaller disc; now take any highly non convex square modulus such (eg $f(3/4)=1, f(5/8)=f(7/8)=1/2$ and approximate it well enough by infinitely many vertical translates of zeta and of course for each of those $T$ convexity will fail $\endgroup$
    – Conrad
    Apr 1, 2022 at 13:01
  • $\begingroup$ Universality precludes any regularity for $\zeta, |\zeta|$ etc in the right part of the critical strip and it holds for the $L$ functions based on characters, while it also holds strongly for nontrivial linear combinations (same $q$) and derivatives of such (so they approximate any function whether it has zeroes or not - see Bagchi's paper link.springer.com/article/10.1007/BF01161980 ;this being said, $|\xi|$ has regularity properties in both the right and left hand strip and there are well knwown equivalences to RH of the behaviour of its modulus on horizontal rays $\endgroup$
    – Conrad
    Apr 1, 2022 at 13:17
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    $\begingroup$ Bagchi's generalization of Voronin's theorem implies lots of reults including the fact that the Davenport Heilbronn function (that one that precludes proofs of RH based on the functional equation only) or the derivative of zeta have tons of zeroes in the right-hand strip so the supremum of the real parts of those zeroes is $1$, so coming back to the post here, obviously Blinovsky's claim is wrong so nobody is going to bother and check where it fails as it has no chance of being corrected $\endgroup$
    – Conrad
    Apr 1, 2022 at 13:27
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Now that this thread has been resurrected a few days ago, one could add Deninger's approach using foliated spaces. Some references are

Deninger, Christopher. On the nature of the "explicit formulas'' in analytic number theory—a simple example. Number theoretic methods (Iizuka, 2001), 97–118, Dev. Math., 8, Kluwer Acad. Publ., Dordrecht, 2002.

Leichtnam, Eric. On the analogy between arithmetic geometry and foliated spaces. Rend. Mat. Appl. (7) 28 (2008), no. 2, 163–188.

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