19

A few years ago, I found on arXiv an article in which the authors (I think they were at least two to write it) claimed to have proven that the non trivial zeros of the Riemann zeta function were all simple using the concept of Riemann surfaces. But unfortunately, I just can't find it back. Does someone know if such a result has been published and widely accepted by the mathematical community? Thank you in advance.

GH from MO
  • 98,751
Sylvain JULIEN
  • 6,960
  • 4
    Not that this answers your question, but a multiple zero is in some sense even less likely than one would expect if the zeros were "randomly distributed" (given their average density): they seem to repel each other, so if you plot the zero spacings scaled to average 1 then their density approaches $0$ as the distance approaches $0$. [I hedge with "in some sense" because even random spacing would make the probability of a coincidence $0$. NB I'm trying to minimize confusion by using "zero" for a root of $\zeta(s)$ and "$0$" for the smallest nonnegative real number.] – Noam D. Elkies Apr 29 '16 at 03:03

3 Answers3

36

This is widely open. Moreover, I think we will prove the Riemann Hypothesis much earlier than the simplicity of the zeros (if true). The latter is somehow much more accidental, the only reasonable argument I know in favor of it is "why would two zeros ever coincide"? Note, however, that some automorphic $L$-functions do have multiple zeros. If I recall correctly, even a Dedekind $L$-function can have a multiple zero at the center.

GH from MO
  • 98,751
  • 5
    though there are results like a certain percentage of the zeros are simple... – shenghao Mar 28 '11 at 00:03
  • 4
    I am no expert, but can a Dedekind zeta function vanish multiply if there is no Armitage/Serre phenomenon, with a root number of -1 for an Artin representation in the decomposition? The MAGMA L-functions handbook code has an example. http://magma.maths.usyd.edu.au/magma/handbook/text/1385#15208 – Junkie Mar 28 '11 at 00:31
  • 35
    @Junkie: "I am no expert, but can a Dedekind zeta function vanish multiply if there is no Armitage/Serre phenomenon"...That sounds suspiciously like a question an expert would ask! – Pete L. Clark Mar 28 '11 at 02:50
  • 1
    Junkie: would you be willing to believe "even motivic weight implies nonvanishing at the central point, unless $\varepsilon=-1$, in which case the zero has order one"? I would. – David Hansen Mar 28 '11 at 03:10
  • (Sorry, I forgot to restrict to irreducible stuff.) – David Hansen Mar 28 '11 at 03:12
  • 4
    I let you guys sort this out. – GH from MO Mar 28 '11 at 03:21
  • 10
    For a Galois extension, the Dedekind zeta function factors as a product over all the Artin L-functions of the irreducible representations, with multiplicity equal to the dimension. Thus for dim>1 (ie non-Abelian Galois groups) the Dedekind zeta function is forced to have multiple zeros, not just at 1/2 but all the way up the critical line. All this goes back to Artin, much before Armitage/Serre. – Stopple Mar 28 '11 at 15:26
  • 1
    The whole point about Serre and then Armitage is they give and produce a specific condition for the root number, with it $-1$ depending on whether some ideal class is square if I can remember it.

    It "looks" as the $\zeta$-function vanishes at $s=1/2$ in this case and that's not expected from general notions of motives (again I am no expert, but $L$-functions should only vanish at integers when normalized right), though the phenomenon occurs because the Artin $L$-function has a shift I think.

    The sign is also tied up with a normal basis. http://www.springerlink.com/content/qq85212001764553/

     – Junkie Apr 02 '11 at 05:14
  • @David Hansen. If I understand, I think Stein disproved what you say back in 2000. There is a modular form on $\Gamma_1(122)$ with quadratic character and the sign is unitary (random root of unity) but the $L$-function vanishes. My bad, now I realize this is ODD motivic weight, but I will leave Stein's example. $$chi:=DirichletGroup(122).1;$$ $$N:=Newforms(ModularForms(chi));$$ $$f:=ComplexEmbeddings(N[1][1])[1][1];$$ $$L:=LSeries(f);$$ $$CheckFunctionalEquation(L);$$ $$Sign(L);$$ $$Evaluate(L,1);$$ – Junkie Apr 02 '11 at 05:32
  • @DavidHansen what about elliptic curves of rank $3$ (or for that matter even $2$) and higher? – Noam D. Elkies Apr 29 '16 at 02:59
  • in Friedlander, lectures 2006, dicussing the Siegel zero at p.103, he says we can get a zero of order $n$ at $s=1/2$ from the L-function of an elliptic curve, and also gives some references @NoamD.Elkies – reuns Aug 16 '16 at 22:55
11

To the best of my knowledge, it is still an open question as to whether all the zeros are simple. If you could find that article....

For what it's worth, any number of "proofs" of the Riemann Hypothesis have appeared on the ArXiv. Here are a few (I've not included three more that were withdrawn by the authors).

1006.0381 The Riemann Hypothesis, Ilgar Sh. Jabbarov (Dzhabbarov)

0906.4604 A Proof for the Riemann Hypothesis, Ruiming Zhang

0903.3973 Concerning Riemann Hypothesis, Raghunath Acharya

0802.1764 Riemann Hypothesis may be proved by induction, R. M. Abrarov, S. M. Abrarov [EDIT: It appears that this paper does not actually claim a proof of RH - see Gregory's answer to the question (and my comment on Gregory's answer).]

0801.4072 The Riemann Hypothesis and the Nontrivial Zeros of the General L-Functions, Fayang Qiu

0801.0633 From Bombieri's Mean Value Theorem to the Riemann Hypothesis, Fu-Gao Song

0709.1389 One page proof of the Riemann hypothesis, Andrzej Madrecki

math/0308001 A Geometric Proof of Generalized Riemann Hypothesis, Kaida Shi

math/9909153 Riemann Hypothesis, Chengyan Liu

Gerry Myerson
  • 39,024
  • 8
    Nice collection! I particularly liked "One page proof of the Riemann hypothesis" and "Riemann Hypothesis may be proved by induction". Here is a generalization: "One page induction proof of the Riemann Hypothesis AND the Twin Prime Conjecture". Can you beat that? Perhaps "Three-line proof that Peano Arithmetic is inconsistent"? – GH from MO Mar 28 '11 at 01:13
  • 1
    @GH Maybe it would be beaten by something like Zagier's title. Say something like "A One Sentence Proof Of The Riemann Hypothesis" =) – Adrian Barquero-Sanchez Mar 28 '11 at 01:56
  • 5
    @GH, did you notice that "One page proof of the Riemann hypothesis" is 17 pages long? – Gerry Myerson Mar 28 '11 at 03:19
  • 6
    Nope :-) I am sure 16 pages are for non-experts and then there is 1 page of beef. – GH from MO Mar 28 '11 at 03:23
  • 3
    Please note in article - 0802.1764, Riemann Hypothesis may be proved by induction, by R. M. Abrarov and S. M. Abrarov - authors did not claim the proof of RH. They only suggested that induction procedure may be used for RH. This is a nice paper. It contains useful equations that were not known in number theory. – Gregory Apr 02 '11 at 05:06
  • 2
    Here is the arxiv page: http://arxiv.org/abs/0802.1764 – B R Apr 02 '11 at 17:28
  • 2
    My apologies for including this article in my list. I was misled by the title, by the sentence "At least one of these identities may be applied to prove the Riemann Hypothesis by induction" in the abstract, and by the statement, "The Induction Procedure can be applied over and over again for further validation of (19). Hence the Riemann Hypothesis is justified" toward the end of the paper. But the last sentence of the article seems to say they have only found a condition which, if true, implies RH. So far as I can tell, the authors publish only in the arXiv. – Gerry Myerson Apr 04 '11 at 00:04
4

I finally managed to find back the article I was talking about. Just click on the green link in the first message of the following link: link text

Sylvain JULIEN
  • 6,960
  • 1
    For convenience, I'm adding a link to the original Arxiv page. The has been updated twice with respect to the version you link (no, apparently the main result wasn't withdrawn or amended): http://arxiv.org/abs/0911.5138 – Federico Poloni Mar 28 '11 at 11:45
  • 1
    It appears that this paper has been very recently accepted by the International Journal of Mathematics and Mathematical Sciences, as per http://www.hindawi.com/journals/ijmms/aip/985323/ – Gerry Myerson Mar 28 '11 at 23:27
  • 3
    I don't claim to have read it carefully, but a quick glance through makes me very suspicious: the arguments seem to abstract somehow, and there are quite a lot of computer-generated pictures that look as though they may be intended as substitutes for rigorous proofs. Basically, I can't find the beef anywhere -- for instance, I don't see any sign of hard estimates. Coupling that with GH's comment, I am left thinking I can safely ignore this one unless the experts suddenly get excited about it. – gowers Apr 02 '11 at 09:42
  • 1
    Already the very beginning makes me very suspicious. They make a claim about Riemann surfaces and give two references. One is a page in Ahlfors book, where we find no theorem, but an informal discussion of Riemann surfaces. Ahlfors says on the previous page "This idea leads to the notion of a Riemann surface. It is not our intention to give, in this connection, a rigorous definition of this notion. For our purposes it is sufficient to introduce Riemann surfaces in a purely descriptive manner. (cont. in next comment) – GH from MO Apr 02 '11 at 14:12
  • (cont. from previous comment) We are free to do so as long as we use them merely for purposes of illustration, and never in logical proofs." The second one is a conference proceeding publication by the authors which does not even show up in MathSciNet. Also, it does not look good on the second author that his 33 publications published over 40 years only have 4 citations in MathSciNet. The first author is 80 years old which raises further doubt about the credibility of the paper.

    In short: This paper is simply wrong by all likelihood.

     – GH from MO Apr 02 '11 at 14:17
  • 1
    @gowers, @GH, I wonder whether it would be appropriate for you to convey your misgivings to the editors of the journal in question. – Gerry Myerson Apr 04 '11 at 00:11
  • 1
    The ArXIV abstract says: "For the Riemann Zeta function the outstanding question of the multiplicity of its zeros, as well as of the zeros of its derivative is answered." This statement doesn't appear in the published version. $ $ Conjecture 11 published was Theorem 5 in ARXIV: ("All the non real zeros of $\zeta'$ are situated in the right half plane"). $ $ The Theorem 6 of the ARXIV version "Theorem 6. All the zeros of $\zeta$ are simple zeros" is also missing in the published version. So I think the published version has been tamed from the ARXIV hype. – Junkie Apr 04 '11 at 07:39
  • @Junkie, thanks, that's good news. I don't think I have access to the published version. – Gerry Myerson Apr 04 '11 at 14:00