 Thank you, Jean-François, and thank you to the organizers for organizing this incredibly rich summer school, and for giving me the opportunity to speak. But before I start the talk, I like to use my modest voice to say how sad it is for mathematics, for science, and for mankind in general to lose a woman like Marie Mirzacani. Besides her genius for mathematics that we cannot copy, her courage and dedication to science are an example for all of us. So this talk will not treat any spectral property, but if we admit that the Riemann hypothesis is connected to almost everything in mathematics, so maybe it's in the scope. So I'd like to introduce this famous problem. 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. And this paper is, the title is about the distribution of prime numbers, and what he discovers is that this distribution of prime numbers is connected with the, what would be called the Riemann zeta function. But this is it. But before Riemann, this function was considered as a function defined on R, on the real numbers, and he, it's a very, it's a novelty that to consider it as a function starting from complex numbers. And in this eight-page paper he, well you see in this writing you see that this has since only when the real part of S is strictly bigger than one. And what he's able to do is to extend this definition to the whole complex plane minus one. So maybe since it is a summer school, I could try to be as didactical as possible and show you how the way it is possible to extend this function. First of all, you can make some Abel transformation and say that this sum is equal to the sum over all n greater than one of n times n to the minus s, minus n plus one to the minus s. And transform it like this is an integral from n to n plus one of n is you write it as the integer part of x on this interval. And then put the right function which is this one. So you can change it as this, you can write, you can write this integer part as x minus plus. So what you get is that this is equal to s divided by s minus one. And this is just the integration of a bounded function divided by divided by x to the s plus one. So it is defined for complex numbers whose real part now is strictly positive. So we gained some extra domain and so the price to pay is that we have here a pole. And so remain with this and the extra functional relation is also proving in this eight pages paper which connects the value of data at s with the value of data at one minus s is able to define completely data on the whole set c minus one. Another remark here that you can see that when s belongs to the set of even negative numbers then the value of data vanishes because of the presence of a sinus by s divided by two. But for positive numbers you see that it's not the case but this comes from the presence of pole for gamma for negative numbers. So you have this picture so data is not defined has a pole at one and has roots for negative from even negative numbers. And what the conjectures says that the other roots of data are concentrated on this real line. The first one known is around 14 second 48. One half plus i times 14 something like that. And in this also in this eight pages paper he is able to connect the properties of data with not exactly the distribution of the prime numbers 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 with the results by Adamard and Lavali Poussin who are able to prove that if you denote by pi of x the number of prime numbers which are smaller than x pi of x is equivalent to x divided by log of x plus infinity. But still this doesn't really connect if yeah the proof connects use is a big achievement using complex analysis the Riemann zeta function and but the Riemann hypothesis is really linked with the distribution of prime numbers by the the theorem of von Koch in 1901 which states that the Riemann hypothesis is equivalent to some some final asymptotics of pi of x which is following so actually this function is equivalent to that one but this is not that one plus an error term which is a small o of x to the one half plus epsilon for every epsilon strictly positive. So it says that you can really have have a fine estimation of the probability for an integer to be prime. 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 what if you take two random integers what is the probability that they are co-prime okay so this like that this what is the probability integers are co-prime. Of course if you don't define the 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 consider x and y distributed like in the von Koch theorem uniformly on one n okay and one can one can be said in this case so you have one let's say one so you have all that these points which which are so this is zero all these points are okay all these ones also are okay and you see for four you have all this one one and three only that that ones etc so so if we denote by p of n this probability well you have n n square point and you just for each k between one and n you have the the number of of coordinates be less than k which are co-prime with k it's phi of k so and you have also the same on the other side here so the the the number for each k you have two times phi of k points which are okay and you you have only one here so you have to to subtract one if you want to be completely exact and and this we know actually the the limit of p of n when n goes to infinity is known and we know it's this limit is equal to six divided by p to the square maybe i can give you an intuition of why is it is equal to to the limit is equal to six divided by p to the square it's rather simple admit that this this this set of points the set of points let's denoted by q which are points of n to the square with co-prime coordinates admit that q has a density then you can you can write you can write n square as the union the disjoint union of q two q three q etc but if q has an asymptotic density then the asymptotic density of q is is d well let's say that's a d plus this asymptotic density will be d divided by four etc and the sum of all of them will be one so that d is equal to six divided by p square i it's not approved since we i didn't prove that that it had a density actually but if we want but this is true you can find a proof of it in the ardy ardy and white book introduction to the theory of numbers and if if we if we want to have a result in the spirit of phone call then you see that it's not you have some fluctuations which are quite strong imagine that the new f you have some p of n minus one and then the new n is a prime number then at once you all at once you you you will get an extra five of n will be equal to n and you you will have a fluctuation of one divided by n which is we we feel these these are quite huge oscillation fluctuations of this of this quantity so the the fluctuations of this quantity were are quite tough to study and they were studied by by uh martins in uh 1874 i think and uh who's who proved that the the the the oscillations were uh not bigger than log n divided by n and uh this this was uh ameliorated by uh valfish in 1963 and the log n was log n to the 2 3rd and the the fluctuations were also bounded from from below by conway and uh it's a very but not no result like that like uh like as clear as phone call in the case when the distribution you take is uniform on uh on one n so we change the probability distribution and take another one which is uh for probability whether natural since uh you take x and y geometrically distributed so that's what we do and i can state the result i obtained with junior is now teaching mathematics but uh i hopefully will go on uh doing research in a very talented guy um so we take x and y two independent random variables geometrically distributed with some parameter one minus e to the minus beta well i could have taken beta but it's uh more convenient like that and so the Riemann hypothesis is equivalent to the fact that the probability that these two random variables are co-prime is when beta goes to zero because beta goes to zero it makes you choose a large large integer which is the interest so you recover this six divided by pi to the square plus so this takes a place of this this integral and if you have a neural term which is uh beta to the 3 3 3 3 3 halves minus epsilon for all epsilon uh then uh uh the Riemann hypothesis is true so i can show you the proof of it which uh is uh surprisingly rather uh not so complicated and at least uh much much much simpler than than uh then funkoff or funkoff result so i guess that uh you will prefer uh me to to show you the this this way because if you are able to prove this and you will finish you'll be able to finish the job so and for me it's more convenient so uh uh i i be you mean uh this this should be the parameter maybe okay when i when i will be not teaching i will change my you you can write explicitly this probability and find that uh it's um and and find uh that is equal to that so if we denote this sum by f of beta this expansion is uh equivalent to uh that one so that so that uh what we have to prove is that if we know this then uh the Riemann hypothesis is true so what we are going to to take is the median transform of f of beta so it is equal to the sum of when you take the median transform of of e to the minus beta times x plus y what you get is one divided by x plus uh one x plus y to the s and some uh gamma term in factor okay so this sum is not so nice because of the of the indices which are not uh a nice subset of uh n to the square but the the trick uh the only trick here is here and is consists in multiplying and dividing by data of s and when when you do that it's uh this sum is much simpler since you get the sum over all pairs of x and y and this and this is simple to to to compute uh using the data function since uh you can you can uh sum uh on pairs of x and y having uh a given sum and then what you have got is uh uh for for a given uh k the number of uh of pairs with sum is k is equal to k minus one but that what you get is s minus one so uh so essentially this is interesting since you you you see you you have here the the data the denominator so that's the reason why we are going to to to see the Riemann conjecture appear and this is really due to the fact that the indices were taken only on on on this uh coprime pairs so uh and and and and this trick now it's already most finished since we go back to the left hand side and right that this uh integral from zero to infinity we split it into the integral from zero to one of f of beta minus plus plus uh this in integral from the integral of that from zero to one which is equal to don't tell me i i have to compute it it's uh plus the integral from uh one to infinity f of beta times beta but this is since you see what is inside f of beta you have some of exponentials so it's the it doesn't make the convergence doesn't make any problem and this is oromorphic on on c and with this assumption on on f of beta you can see that you have a power of beta to the to the s minus one minus one half plus epsilon so that you see that it's olomorphic on complex numbers whose real part is larger than one half plus epsilon so it means that if this is defined on all complex numbers whose real part is larger than one half plus epsilon then data doesn't vanishes on that region and by symmetry that it does not vanishes on on the symmetric region with respect to one half either the other the converse is um is a concept is is a little bit harder but not much much more but it's it relies on um it it relies on a inverse mailing transform formula that that um writes f of beta in terms of of a contour integral of this of this function and then if you and then you apply residue theorem and get the what you what you want well maybe just I can write you the inverse this is for all strictly positive c you have that and with assumption of on on on data and applying the residue theorem then you you will uh you you you can find the the the the expansion on f the this expansion on f of beta uh you see uh yes and you apply it for for some at least for all c strictly bigger than you have to avoid strictly bigger than than two I would say in order to to to avoid and for s equal to two you have you have here a pole which is one it's it was the the the computation in the introduction and the residue you see here is gamma of two which is one divided by zeta of two which is uh pi squared divided by six so you you you get back the and since you don't have any if you don't have any zero in in some strip then you you will see that the the term after will be uh will be uh of one divided by so in the title I said two equivalent versions of the Riemann hypothesis so I gave only one so I have to to give you the the second one which actually was the first one because um this version is the result of of trying to simplify as as much as as possible the the the one I'm going to to present now so uh so it it it deals with um um a problem in combinatorics that has nothing to do with uh with with this which is the the following um I I want to count the number of convex polygonal lines which and um increasing ones which uh join uh the origin zero to the point of coordinates n n no no I I take another one no uh I take the first one if I if not it won't be interesting from the inclination and so uh you you see I I I does not distinguish this this line with that line okay for me this these are the same only the shape matters and so uh it's a problem which was considered in the mid of the 90s by um uh Barani uh Versich and uh Sinai separately more or less but with interactions and um and what they proved was that the number of of of of third chains was uh equal to the exponential of three times data or three divided by data of two to the one third times n to the two-third plus some o of small o of n to the two-third so they proved this and um um mainly uh what it seems it has nothing to do with the previous problem but if you think the way one can code uh lines like that you you can say the following well I will I enumerate all the all the vectors I I have to use the first one is the vector three one the second one is the vector two two but I I don't say it's the vector two two because it would because if I had the the vector one one twice I wouldn't be able to distinguish it so I say it's it's not the vector two two but it's a vector one one that I have used twice this one I have used only one only once and this I have used twice and the last one cannot be divided actually I use it I use I use it only once so so the the coding of context chains is is given by a function from this starting from the the set q of prime pairs to uh zero one two three etc and um actually um so we see the maybe I I I state just before um giving an idea of the proof I I I state what our result is uh with julien the our result is is is twofold the uh the the first result we have is a real um uh uh asymptotic equality for uh n of uh for n of capital n of n and uh and the second one is the um the the uh an equivalence with Riemann uh the Riemann conjecture for some uh maybe I I set it first and says the third ring you are able to to to prove that then and you you get the Riemann hypothesis but if you suppose the Riemann hypothesis to be true then you have uh precise equivalent of for n of n which is some explicit constant c divided by n to the some power plus the sum over all the roots not trivial roots of of zeta of well I write it because I have some it has some interest despite times n divided by uh this I call kappa so you see with rho having uh real part one half you'll see that uh the the order of magnitude of of this term is n to the one sixth and if you look more carefully at was what is in front of of this of of these terms you have this gamma of rho but I don't know if you remember where is the first zero the first coordinate for the first non trivial zero was 14 and and 14 but but gamma of of rho decreases like e to the e to the minus the the second coordinate of the imaginary part of rho so the the first the the large the largest term in front of this uh more or less oscillating powers is of of the order e to the minus 14 so it's something like 10 to the minus 10 something like that so uh if you see in practice uh you cannot see this this you have to wait for 10 to the n quad to 10 to the 60 if you you want this term to compete with this this one so uh uh a nice guy uh put on this uh after our uh papers uh was posted he made some computation put the put the put the the sequence on the o o e i s uh page and uh we can we can see that uh uh this is a good equivalent uh for but uh this never we never see that so you cannot even uh uh have an idea of the the truth of not or not of the Riemann hypothesis by by this mean and and simulations so um maybe i can say one word about about uh the the proof so um we it's a real equivalent yes yes it comes from uh uh um the the computation of the pole at s e of poles at s uh equals zero and uh well it's uh it's it's it's because you assumed that okay you said Riemann hypothesis is this line and if this line is true then in fact you'll get something much better yes yes yes because well you you can read uh you can read uh or the the the spirit is you can read the the the the the empty uh zero region on the on the on the on the power that you you have so the assumption is true also even if Riemann hypothesis is true? no no no it it it relies yes if it's it's true uh only if uh hypothesis is true so um what we use is uh but we we put it the we we use the method of sin i but we we use in uh um more extensive way and which is the following and uh it's a quite a nice idea that um actually what sin i say is that forget about this constraint of finishing at nn and uh we'll uh leave the chain uh the chains free but uh we we we will pick them at random according to to some to some distribution which um is the following for uh x and y uh being a co-prime we so we have this omega of x y which is a multiplicity of this vector x y we use in the chain and so we decide that uh all this uh all these multiplicities will be random will be also independent and um will be independent and geometrically distributed with some parameter i i cannot say but i can say the low and it starts also at zero so it's even worse than and the the parameter is e to the minus beta times x y so that with some beta we you fix some beta strictly positive and under this uh random choice you see that the probability for some chain to to appear is equal to well uh the product of all this but what k is nothing but omega of x y so we we have here the sum of x plus y times omega of x plus y okay times but this for all the chains which are ending at nn 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 and all the under this uh probability distribution all uh all the chains who which are ending at the same point uh have the same weight okay so that the probability that the final point is equal to nn uh it's uh it's a small n i'm sorry trouble will be equal to this but since all the chains have the same power i will have the number of them times this this this function which is a function of beta so now the the the so n of n would be equal to uh well you see this e to the 2 beta n times z of beta times this probability but this probability actually you you can it's it's it's the final point is nothing but the the the the sum of uh independent integer variable so it's um with a local limit theorem you have you can have a precise estimation of this uh you can also calibrate uh beta so such that the expectation of the final point will be exactly equal to to nn so um what remains to to to do is have a very accurate uh way of dealing of estimating z of beta when beta goes to zero because beta will have to go to zero if i want uh the the mean final point to be equal to nn and this so what was sin i doing he was estimating z of beta with only the use that the the set of coprime pairs has an asymptotic density so it was a very weak uh material uh he used so uh with this only uh thing you can only have that but what we did was actually that we we we could write an explicit uh integral uh writing for z of beta and uh which is which is uh more or less similar at this this uh similar as this one i don't know if i can show you it's some three or four lines but uh sorry we are concerned i'm concerned by by this so uh it's a little bit harder than the the first because you uh you use the expansion of the logarithm and then you use the uh inverse milling transform here which says that e to the minus z is equal to the limit when t goes to infinity for all uh c strictly positive so you you you put in it in there and uh so what we get is gamma of s s is uh s is s but z z is this that's why and you have also the the sum over all the coprime pairs but now we are not afraid of it anymore because we know the trick so this sum will um make some data of s plus one uh in factor appear gamma of s this uh is uh like before uh with the trick uh what data of s and you have uh some data plus data s minus one so you see when you want to estimate log of z when beta is goes to zero then you see you see that you you are you are looking at the the residue and you see the first one comes from the from for uh s equal to two from this two minus one is one and so you see residue which is data of of three divided by data of two it reminds you divided by b to the to the to the square and gamma of two is one uh but you you have much more than that because you can uh you can use contour and uh uh examine all the residues for for uh uh okay so it's uh lecture are there any comments or questions when you apply the residue you need some control on the growth of the function of the imaginary large imaginary um um no um as as long as you are not on the the gamma the the gamma is uh is very made or all the things very secure what what is um technical is when you when you when you want to you have your contour and you you have to cross the critical uh line then uh we need some uh fancy result of of uh valier on in 1920s so you had this nice picture of the n by n grid with the points in q color color like points that are copper and color and so so my question is does this have a very mini shamblin i just i just learned what what what what it was no no no no no um what how do you define the connections so you look at look at it from a random point and see if that's a random color graph has a distribution so you know that the probability that the point is going to is this convergent to something which is this 6 over by square but that you can ask where is the probability of the point and in its neighbor ah yes yes yes yes i see ah yes oh congress uh yes uh right now question about the other fast about the other phase theorem. So can you replace geometry by some other distribution, like Poisson? It's a very good question, because you can do it. And you can do it, but not with some ad hoc distribution. I think with a zeta distribution, it's not long, and it's funny. So the zeta distribution is, actually, the parameter is simple. You take it. And well, I can try. It has some interest at the end. What you get is for x, y, co-prime, OK? And then, again, you use the trick. You multiply and divide by 1 over k to 2s in order to. And what you get is nothing, but this is equal to that. Then you have, it's completely explicit in that case. And you see that when s goes to 1, which is a case when you have large, the probability goes to 6, 9, 10. But for other, it's good for the fluctuations to be not too large that the distribution rapidly decreases. But. Yes, it was the same way, yes. But what I did not try was a number of integers which is going to infinity in the small o of the parameter. I didn't try this. But for finite, it goes the same way. But it's thank to nothing.