Two equivalent versions of the Riemann Hypothesis
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00:00
MathematikKategorie <Mathematik>Vorlesung/Konferenz
01:33
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08:09
Numerische MathematikOrdnung <Mathematik>PrimzahlMengenlehreAsymptotikKategorie <Mathematik>Ganze ZahlAuswahlaxiomBeweistheorieFunktionalGeradeGrenzwertberechnungLeistung <Physik>ResultanteStichprobenfehlerTermTheoremWurzel <Mathematik>GammafunktionDistributionenraumPunktNegative ZahlSchätzfunktionVorzeichen <Mathematik>PrimzahlsatzRiemannsche VermutungPolstelleEinsVorlesung/Konferenz
17:20
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26:31
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28:59
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43:42
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51:04
Folge <Mathematik>Ordnung <Mathematik>VektorraumBerechenbare FunktionNebenbedingungÄquivalenzklasseArithmetisches MittelBetafunktionBeweistheorieGeradeLeistung <Physik>MultiplikationTermParametersystemGammafunktionDistributionenraumGrößenordnungKomplexe EbeneRiemannsche VermutungPolstelleMultiplikationsoperatorVorlesung/Konferenz
57:39
Numerische MathematikMengenlehreWahrscheinlichkeitsverteilungProdukt <Mathematik>VariableGanze ZahlUnendlichkeitAuswahlaxiomBetafunktionErwartungswertFunktionalInverser LimesKette <Mathematik>LogarithmusStellenringWärmeausdehnungStochastische AbhängigkeitGewicht <Ausgleichsrechnung>GammafunktionSummierbarkeitPunktDichte <Physik>SchätzfunktionMultiplikationsoperatorVorlesung/Konferenz
01:06:43
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Transkript: Englisch(automatisch erzeugt)
00:15
Thank you Jean-François, and thank you to the organizers for organizing this incredibly rich summer school
00:26
and for giving me the opportunity to speak. But before I start the talk I would like to use my modest voice to say how sad it is for mathematics, for science and for mankind in general
00:41
to lose a woman like Maryam Mirzakhani. Besides her genius for mathematics that we cannot copy, her courage and dedication to science are an example for all of us.
01:03
So this talk will not treat any spectral property
01:34
but if we admit that the Riemann hypothesis is connected to almost everything in mathematics
01:42
so maybe it's in the scope. So I'd like to introduce this famous problem.
02:03
So the story starts in 1859 when Bernard Riemann is elected at the German Academy of Science and he writes on that occasion an eight-page paper which will have a tremendous impact over all the generations.
02:36
This paper, the title is about the distribution of prime numbers
02:43
and what he discovers is that this distribution of prime numbers is connected with what will be called the Riemann zeta function. This is it.
03:10
Before Riemann this function was considered as a function defined on R, the real numbers.
03:25
It's a novelty to consider it as a function starting from complex numbers. And in this eight-pages paper, in this writing you see that this has sense only
03:47
when the real part of S is strictly bigger than one. And what he is able to do is to extend this definition to the all complex plane minus one.
04:10
So maybe since it is a summer school I could try to be as didactical as possible and show you the way it is possible to extend this function.
04:30
First of all you can make some Abell transformation and say that this sum is equal to the sum of all n greater than one of n times n to the minus S minus n plus one to the minus S
05:06
and transform it like this is an integral from n to n plus one of n. You write it as the integer part of x on this interval
05:27
and then put the right function with this one.
05:43
So you can change it as this.
06:06
You can write this integer part as x minus plus x.
06:33
So what you get is that this is equal to S divided by S minus one
06:43
and this is just the integration of bounded function divided by x to the S plus one. So it is defined for complex numbers whose real part now is strictly positive.
07:07
So we gained some extra domain
07:21
and so the price to pay is that we have here a pole. So Riemann with this and the extra functional relation
07:42
he is also proving in this 8 pages paper which connects the value of data at S with the value of data at one minus S
08:08
is able to define completely data on the all set C minus one.
08:21
Another remark here that you can see that when S belongs to the set of even negative numbers
08:41
then the value of data vanishes because of the presence of sinus pi S divided by two. But for positive numbers you see that it's not the case
09:02
but this comes from the presence of pole for gamma for negative numbers. So you have this picture.
09:34
So data is not defined as a pole at one and has roots for even negative numbers.
09:46
And what the conjecture says is that the other roots of data are concentrated on this real line. The first one is one half plus i times 14, something like that.
10:27
And also in this 8 pages paper he is able to connect the properties of data with not exactly the distribution of the prime numbers
10:40
but the distribution of the powers of prime numbers. And in order to get really information on the distribution of prime numbers we have to wait almost 40 years
11:01
with the results by Adamar and Lavallee-Poussin who are able to prove that if you denote by pi of x
11:26
the number of prime numbers which are smaller than x pi of x is equivalent to x divided by log of x plus infinity.
11:51
But still this doesn't really connect. The proof connects using a big achievement
12:01
using complex analyses, the Riemann zeta function. But the Riemann hypothesis is really linked
12:22
with the distribution of prime numbers caused by the theorem of von Koch in 1901 which states that the Riemann hypothesis is equivalent to some final asymptotics of pi of x
12:50
which is following actually this function is equivalent to that one
13:06
but this is not that one plus an error term which is small o of x to the one half plus epsilon for every epsilon strictly positive.
13:28
So it says that you can really have a fine estimation of the probability for an integer to be prime.
13:47
So today we are not going to ask this question about what is the probability for one integer to be prime. We are going to ask another question which is if you take two random integers
14:05
what is the probability that they are co-prime?
14:21
So like that this is the probability that two random integers are co-prime.
14:48
Of course if you don't define the way you choose them it doesn't make sense. So maybe you can start with the most natural choice one would like to take.
15:05
Consider x and y distributed like in the von Kortelram uniformly on one n.
15:29
And one can be said in this case. So you have one, let's say one.
16:07
So you have all these points which are... So this is zero. All these points are okay. All these ones also are okay. And you see for four you have this one, one and three.
16:27
Only these ones. Etc. So if we denote by p of n this probability you have n square points
16:44
and for each k between one and n
17:04
you have the number of coordinates which might be less than k which are co-prime with k. It's phi of k.
17:22
And you have also the same on the other side here. So the number for each k you have two times phi of k points which are okay. And you have only one here
17:52
so you have to subtract one if you want to be completely exact.
18:01
And this we know actually the limit of p of n when n goes to infinity is known. And we know this limit is equal to 6 divided by p to the square. Maybe I can give you an intuition of
18:22
why the limit is equal to 6 divided by p to the square. It's rather simple. Admit that this set of points let's denote it by q
18:42
which are points of n to the square with co-prime coordinates Admit that q has a density. Then you can write n square
19:17
as the disjoint union of q
19:24
2q, 3q, etc. But if q has an asymptotic density then the asymptotic density of q is d plus this asymptotic density will be d divided by 4
19:48
and the sum of all of them will be 1. So that d is equal to 6 divided by p square. It's not a proof since I didn't prove that
20:01
it had a density actually. But if we want but this is true, you can find a proof of it in the Adrienne Wright book Introduction to the Theory of Numbers.
20:23
And if we want to have a result in the spirit of Fancourt then you see that it's not you have some fluctuations which are quite strong.
20:40
Imagine that you have some p of n minus 1 and then the new n is a prime number. Then all at once you will get an extra phi of n would be equal to n
21:02
and you will have a fluctuation of 1 divided by n which is, we feel these are quite huge fluctuations of this quantity. So the fluctuations of this quantity are quite tough to study
21:22
and they were studied by Mertens in 1874 I think and who proved that the
21:44
the oscillations were not bigger than log n divided by n and this was ameliorated by Walfish in 1963 and the log n was log n to the 2 third and the fluctuations were also bounded from below
22:10
by Conway and it's a very but no result like that like as clear as Fancourt in the case
22:22
when the distribution you take is uniform on one n. So we change the probability distribution and take another one which is for probabilistic rather natural since you take x and y
22:43
geometrically distributed. So that's what we do and I can state the result I obtained
23:12
with Julien myself.
23:30
Julien is now teaching mathematics but I hopefully will go on doing research. He's a very talented guy.
23:43
So we take x and y two independent random variables geometrically distributed
24:01
with some parameter 1 minus e to the minus beta I could have taken beta but it's more convenient like that.
24:25
And so the Riemann hypothesis is equivalent to the fact that the probability that these two random variables are co-prime
24:43
is when beta goes to zero because if it goes to zero it makes you choose large integers which is interesting. So you recover this 6 divided by pi to the square
25:03
plus
25:23
so this takes the place of this integral and if you have a narrow term which is beta to the three-halves minus epsilon for all epsilon
25:42
then the Riemann hypothesis is true. So I can show you the proof of it which is surprisingly rather not so complicated
26:03
and at least much simpler than than van Gogh's result. So I guess that you would prefer me to show you this way
26:27
because if you are able to prove this then you will be able to finish the job and for me it's more convenient.
27:10
You mean this should be the parameter? When I will be not teaching I will change mine.
27:40
So you can write explicitly this probability and find that it's equal to that.
28:16
So if we denote this sum by f of beta
28:30
this expansion is equivalent to that one
28:51
so that what we have to prove is
29:20
that if we know this then the Riemann hypothesis is true. So what we are going to take is the milling transform of f of beta.
29:55
So it is equal to the sum of when you take the milling transform of e to the minus beta times x plus y
30:07
what you get is 1 divided by x plus 1 x plus y to the s and some gamma term in factor.
30:27
So this sum is not so nice because of the indices which are not a nice subset of n to the square
30:41
but the trick, the only trick here is here and it consists in multiplying and dividing by theta of s
31:06
and when you do that
31:25
this sum is much simpler since you get the sum over all pairs of x and y
31:41
and this is simple to compute
32:04
using the data function since you can sum on pairs of x and y having a given sum and then what you have got is
32:29
for a given k the number of pairs whose sum is k is equal to k minus 1
32:45
so what you get is beta s minus 1 minus tau s
33:04
So this is interesting since you see you have here the denominator so that's the reason why we are going to see the Riemann conjecture appear
33:22
and this is really due to the fact that the indices were taken only on these co-prime pairs and this trick.
33:48
Now it's already finished since we go back to the left hand side
34:02
and write that this integral from 0 to infinity we split it into the integral from 0 to 1 of f of beta minus 6
34:34
plus this integral of that from 0 to 1
34:43
which is equal to don't tell me I have to compute it
35:06
plus the integral from 1 to infinity of f of beta times beta but this is, since you see what is inside f of beta
35:25
you have some of exponentials so the convergence doesn't make any problem and this is holomorphic on C and with this assumption on f of beta
35:44
you can see that you have a power of beta to the s minus 1
36:02
minus one half plus epsilon so that you see that it's holomorphic on complex numbers whose real part is larger than one half plus epsilon so it means that if this is defined
36:29
on all complex numbers whose real part is larger than one half plus epsilon then data doesn't vanish on that region and by symmetry it does not vanish on the symmetric region
36:45
with respect to one half either The converse is a little bit harder
37:18
but not much more but it relies on
37:29
an inverse mailing transform formula that writes f of beta in terms of a contour integral of this function
37:48
and then you apply the residue theorem and get what you want maybe I can write you the inverse
38:28
for all strictly positive C you have that and with an assumption on data
38:45
and applying the residue theorem then you can find the expansion on f of beta
39:12
you see yes and you apply it for at least
39:27
for all C strictly bigger than you have to avoid strictly bigger than two I would say in order to avoid and for S equal to two
39:47
you have here a pole which is one it was the computation in the introduction and the residue you see here is gamma of two which is one
40:01
divided by zeta of two which is pi squared divided by six so you get back the and since you don't have any if you don't have any zero
40:22
in some strip then you will see that the term after will be O of one divided by
40:40
so in the title I said two equivalent versions of the Riemann hypothesis so I gave only one so I have to give you the second one which actually was the first one because this version is the result of
41:11
trying to simplify as much as possible the one I'm going to present now
41:22
so it deals with a problem in combinatorics that has nothing to do with this which is the following
42:13
I want to count
42:24
the number of convex polygonal lines and increasing ones which join the origin zero to the point of coordinates n,n
43:02
I take another one ah, no
43:23
I take the first one if not, it won't be interesting to inclination one, two
43:43
so you see I does not distinguish this this line with that line
44:00
for me these are the same only the shape matters and so it's a problem which was considered in the mid of the 90s by Baranyi
44:21
Vershik and Sinai separately more or less but with interactions and what they proved was that
44:42
the number of such chains was equal to the exponential of three times
45:00
theta of three divided by theta of two to the one third times n to the two thirds plus some small o of n to the two thirds so they proved this
45:30
and mainly
45:51
it seems it has nothing to do with the previous problem but if you think the way one can code
46:02
lines like that you can say the following I enumerate all the vectors I have to use the first one is the vector three one
46:21
the second one is the vector two two but I don't say it's the vector two two because if I had the vector one one twice I wouldn't be able to distinguish it so I say it's not the vector two two
46:41
but it's the vector one one that I have used twice this one I have used only once this I have used twice and the last one cannot be divided
47:03
actually I use it only once so the coding of complex chains is given by a function from this
47:22
starting from the set Q of co-prime pairs to zero, one, two, three, etc.
47:40
and actually maybe I state just before giving an idea of the proof I state what our result is
48:01
with Julien our result is twofold the first result we have is a real asymptotic equality for capital N of n
48:28
and the second one is an equivalent with Riemann conjecture
48:42
maybe I state it first and it says the following if you are able to prove that
49:24
then you get the Riemann hypothesis but if you suppose the Riemann hypothesis to be true
49:43
then you have a precise equivalent for N of n which is some explicit constant C divided by N to the
50:00
sum power plus the sum over all the roots
50:26
non-trivial roots of zeta of, well, I write it because I have some it has some interest
50:44
times N divided by this I call kappa
51:00
so you see with rho having real part one-half you see that the order of magnitude of this term is N to the one-sixth and if you look more carefully at what is in front
51:22
of these terms you have this gamma of rho I don't know if you remember where is the first zero the first non-trivial zero was fourteen but gamma of rho
51:42
decreases like e to the minus the second coordinate of the imaginary part of rho so the first the largest term in front of these
52:02
more or less oscillating powers is of the order e to the minus fourteen so it's something like ten to the minus ten something like that so if you see
52:21
in practice you cannot see this, you have to wait for ten to the N equal to ten to the sixty if you want this term to compete with this one
52:43
a nice guy put on this after our papers was posted he made some computation put the sequence on the
53:02
OEIS page and we can see that this is a good equivalent but we never see that so you cannot even
53:24
have an idea of the truth or not of the Riemann hypothesis by this means and simulations so maybe I can say one word
53:44
about the proof so it's a real equivalent
54:03
it comes from the computation of the pole at S equals zero and then
54:21
it's it's strange because you assume that this is this line and if this line is true then in fact you get something much later yes because you can read
54:41
the spirit you can read the empty zero region on the power that you have so the assemblage is true even if Riemann hypothesis fails?
55:01
no no no it relies it's true only if so what we use is but we we use the method of Sinai but we use in
55:25
a more extensive way and which is the following it's quite a nice idea that actually what Sinai says is that
55:42
forget about this constraint of finishing at n and we'll leave the chains free but we will pick them at random according to some distribution
56:03
which is the following for x and y being co-prime we so we have this omega of x
56:21
y which is a multiplicity of this vector xy we use in the chain and so we decide that all these all these multiplicities will be random will be also independent
56:41
and will be independent and geometrically distributed with some parameters I cannot say
57:01
but I can say hello and it starts also at zero so it's even worse and the parameter is e to the minus beta
57:24
times xy so
57:44
that with some beta you fix some beta strictly positive and under this random choice
58:00
you see that the probability for some chain to appear is equal to
58:21
the product of all these but k is nothing but omega of xy so we have here the sum of x plus y
58:41
times omega of x plus y times but this
59:01
for all the chains which are ending at n this is nothing but the first coordinate of the final point and this is the second coordinate of the final point so this is equal to 2n
59:22
and all the under this probability distribution all the chains which are ending at the same point have the same weight so that the probability
59:43
that the final point is equal to nn
01:00:00
n, I'm sorry, would be equal to this. But since all the chains have the same power,
01:00:20
I will have the number of them times this function, which is a function of beta.
01:00:47
So n of n would be equal to, well, you see this e to the 2 beta n times z of beta times this
01:01:04
probability. But this probability, actually, the final point is nothing but the sum of independent integer variables. So it's with local limit theorem, you can have a precise
01:01:24
estimation of this. You can also calibrate beta such that the expectation of the final point will be exactly equal to nn. So what we meant to do is have a very accurate way of dealing,
01:01:53
of estimating z of beta, when beta goes to zero, because beta will have to go to zero if I want
01:02:00
the mean final point to be equal to nn. So what was Sinai doing? He was estimating z of beta with only the use that the set of co-prime pairs has an asymptotic density. So it was a very weak
01:02:32
material he used. So with this only thing, you can only have that.
01:02:42
But what we did was actually that we could write an explicit integral writing for z of beta, which is more or less similar as this one. I don't know if I can show you, it's some
01:03:12
few or four lines. So we are concerned by this. So it's a little bit harder than the first
01:04:18
because you use the expansion of the logarithm and then you use the inverse milling transform
01:04:48
here, which says that e to the minus z is equal to the limit when t goes to infinity
01:05:03
for all c strictly positive z. So you put it in there. And so what we get
01:05:44
is gamma of s, s is s, but z is this, x plus y, s. And you have also the
01:06:35
sum of all the co-prime pairs, but now we are not afraid of it anymore because we know the
01:06:45
trick. So this sum will make some data of s plus one factor appear, gamma of s.
01:07:15
This is like before with the trick, beta of s. And you have some data plus data s minus one. So you see, when you want to estimate
01:07:47
log of z when beta goes to zero, then you see that you are looking at the residue. And you see the first one comes from s equal to two. From this, two minus one is one.
01:08:08
And so you see residue which is theta of three divided by theta of two. It reminds you of theta divided by b to the square. And gamma of two is one. But you have much more than that
01:08:33
because you can use contour and examine all the residues.
01:09:07
Are there any comments or questions? When you apply the residue, do you need some control on the growth of the
01:09:31
gamma? The gamma makes all the things very secure. What is technical is when you
01:09:43
want to have your contour and you have to cross the critical line, then we need some a fancy result of Valieron in the 1920s.
01:10:08
So you had this nice picture of the hand-biting grid with the points in Q color, points that I call prime color. So my question is, does this have a value ministram limit?
01:10:24
Ah, I just learned what it was. No, no, no, no, no, no. How do you define the connections? So you look at it from a random point and see if that planned random color graph is a
01:10:45
distribution of it. So you know that the probability that the point is converging to something, which is this 6 over pi squared, but you can ask where is the probability
01:11:00
of the point in its neighbor or something like that. Ah, yes, yes, yes, yes, I see.
01:11:28
Question about your fast theorem. Can you replace geometry by some other distribution? Ah, it's a very good question because you can do it, but not
01:11:45
with some ad hoc distribution. I think with a zeta distribution. It's not long and it's funny.
01:12:05
Ah, so the data distribution is, actually the parameter is simple. You take it and, well, I can try. It has some interest at the end.
01:12:38
What you get is for x, y, co-prime.
01:12:56
And then again, you use the trick. You multiply and divide by 1 over k to 2s
01:13:16
in order to, and what you get is nothing but this is equal to that.
01:13:35
So then you have, it's completely explicit in that case. And you see that when s goes to 1,
01:13:42
which is the case when you have large, the probability goes to 6. Ah, but for other, it's good for the fluctuations to be not to
01:14:10
large that the distribution rapidly decrease.
01:14:23
If you take a triple of an integer... Yes, it was the same way. But what I did not try was a number of integers which is growing to infinity in the small o of the parameter. I didn't try this.
01:14:45
But for finite, it goes the same way. So let's just thank you.