 ᴋᴾᴜᴡ ᴅᴓᴓᴢᴁᴇᴄᴀ ᴀᴏᴀᴀᴄ ᴒᴛ ᴄ ᴀᴜᴇ ᴅᴏ ᴀᴇᴄ ᴇ�ᴀᴈᴛᴁᴇ ᴜᴀᴤᴈ ᴇᴀᴀᴅ ᴀᴀᴇ ᴄᴏᴏ ᴜᴀᴂᴀᴅ ᴍᴀᴜᴅᴋivo, Apprentice, O Each of, 5.090 HeavyOkay. Let me first tell you the statement. And what is O zero So there is a celebrated result If we consider the Riemann Zeta function, we should define by a conversion series for real part of S, bigger than one. What you see here is that when real part of S is one, I mean as you know Zeta of S can be an analytic continuum and on the real line real part of S equals one, Zeta is not zero. In the usual picture, we write S equals sigma plus IT, we have the point one here. On the right of this line real part of S equals one, Zeta is defined by a conversion product, so it's not zero. And the statement is that it's also not zero on this line. Well, in fact, after this work of Lavalipoussin and Dahmar, there was a lot more work in analytic number theory in the 30s. In this book of Davenport, which is called multiplicative number theory, you will find some result about region where Zeta of S is not zero. So maybe the best result, at least according to this book at that time, is the result by Vinogradov and Korkov, which has a following. Assume alpha is bigger than two-thirds, then there exists a constant such that one minus C log T to the alpha sigma is bigger than this, then Zeta of S is not zero. The region they discussed, if I'm not mistaken, is a region like this. So in this region, there is no zero. And now your symmetries, there is a symmetry by functional equation and symmetry by complex conjugation. So you can extend this result by symmetry and you get that there is no zero in this region, as we just said, the symmetry of this region and the symmetry here. So finally, there is no zero here. And of course, we have those first zero, which is around 14, in real part of S equals 12. One half. So in particular, if we look at the real line, there is nowhere a zero. Now we can consider Dirichlet characters. That is morphisms from Z mod Q star to C star. And the L function attached to chi. Ls chi is the sum at least one of chi of n over n to the s. And we have a similar problem. Namely, the series converges for real part of S because of one. It has a magnetic continuation. And we can look for a region in the critical strip where it does not vanish. I mean, there is Adama La Valet-Poussin is valid for L, but we are interested in thinking of zero inside the critical strip. In this book of Dalian Porte, he gives the following theorem. Consider the region for something positive. Consider the region where region R, where sigma, is at least 1 minus c over log dt, d being the conductor. If t is at least one, 1 minus c over log t, not take absolute value because t is positive. If t is positive and less than one. Then we have the picture of this region. So there is this region here, which depends only on t. Excuse me, this is log d here. So here we have a fixed lower one for the abscissa. And above it you have the kind of figure we had in the z function case. So we have the region R. And then when we have to distinguish two cases. First case is that if k is complex, I will define complex by saying that it's not real and I will tell you what real means. But that is a general case. If k is complex and l is as k as no zero in R. And second, if k is real, which means that the image of k, the character from zero, k is a character from z, from q star to c star. So I confused q and d, I'm sorry. So if I have a character from z, from d star to c star. Assume the image is real then. Well it's not the case that there is no zero in this region. There could be one zero. So a less guy can have at most one. And it will be a simple and real. So s equals sigma zero in R. So when the character is real, there could be a zero here. So nobody has proved till now that there is no zero in this region. These zeros are what's called zero zeros. I mean, one expects they are not there because one expects a generalized hypothesis. But it's not known for this special zero which might exist in this region. Okay. Sigma zero is called a z-goal zero. So in fact there are two cases. In case R2, I don't know. K of n could be either the residue quadratic residue of d and n or minus d and n. In which case we say that's a real quadratic and here is a complete quadratic. So there are two types of quadratic characters which are involved in the hypothesis ii. And here is a theorem of Stark and Grovill. C.R.M. by Grovill and Stark with some contribution of Elkis. Grovill and Stark proved the result of this kind and Elkis gave a new idea to my progress. So the theorem of the following assumes either congesture a or congesture b which I explained in the first talk. A true is true, one of them. Then one k. One k is complex quadratic. There is no z-goal zero. If you remember in the first talk I gave two congestures a which I attributed to Parshin and Moribay and then b which was more precise in the sense that for ABC it gives one because of epsilon. So one of them is enough to avoid z-goal zero. But on the other hand there is a result by a claim by Moshizuki and a congesture by Vojta. In which yes the ABC congesture for number field was something like hABC less than one plus epsilon nABC plus log delta of the number field plus some function which I call A of epsilon and the degree of k of q. So this statement is stronger than or slightly stronger than congesture b say but there is a difference. Is that in congesture b here we have linear terms. So Moshizuki claims Vojta result and he doesn't know whether the correction by the constant here is linear or the degree. And what happens is that without this linearity we don't have congesture result. This has no implication about z-goal zero. We shall see the proof that what is needed to go primary to those equal zeros is precisely a term here of linear type. Otherwise we don't have a conclusion. Okay. The proof of Granville and Sack and Elkies requires several sort of results. Some of them being pretty old that is dating from the 30s and I will not explain all of it. I will just take the results I need from the 30s. So I will talk about binary quadratic forms. This is an old topic starting with Gauss maybe. Namely you consider a quadratic form a squared plus b xy plus c y squared. I mean abc here are not the abc of the abc congesture. I hope it will not create any confusion. Assume the discriminant is minus d we are in the imaginary case and abc are integers. This is what we call binary quadratic form. So there is a notion due to Gauss which says that q is reduced when well first ab minus a less than b less a less than c or 0 less than b less than a equals c. I mean the question that Gauss was thinking of is when is the integer representable by binary quadratic form and the main result to start with is that you need to check this question only for reduced quadratic form because sl2z acts on binary quadratic form by change of variable, linear change of variable. If I have a matrix with integral coefficient in two variables I apply it to s y get new number x prime y prime and clearly if q represents an integer the transform under sl2z of the quadratic form will also represent this integer. So we go from one quadratic form to another one and they have the same set same type of representation of integers. So the theorem of Gauss is as follows. So that's called redactron theory proposition. There are two parts. Any binary quadratic form is equivalent under the action of sl2z to a reduced one. Second, there is finitely many reduced quadratic forms. So if you are interested in the question when binary quadratic form will represent some given integer you have to check only finitely quadratic form those obtained from this proposition. D is fixed for a given value of D there exists finitely many reduced quadratic form. So we write h of minus D for the number of reduced quadratic form of this component minus D. While this number is known better with the following definition let k be q where you add the square root of minus D then h of minus D is a class number and this question of studying quadratic form is closely related to study ideals in the integers of k. I will not do that for instance the number of classes is the same. So we have the following fact if we call z k the z function of k so we consider ideals in O different from 0 we consider the cardinality of O by A where O are the integers in k so that's ok so we get the z function of the number field k while it's related to L s k first so z times L s k so L s k is the quotient of the z function of k but the z function of Riemann and we have the following connection with quadratic form namely that z k of s and the sum for all reduced forms of z cubed of s where z cubed is defined as follows the sum for x, y and z x and y not both 0 q of x, y to the power minus s so we have the obvious z function of the quadratic form and it relates to the z function of the number field by these formulas ok so you see that if you study 0s of L you could instead study 0s of z function of quadratic form that the only thing I will say about the result of Malheur which is from the certificate theorem assume exists a constant a such that h of minus z bigger equal to a square root d divided by log d sum over q because I will introduce annotations maybe so the form q equals i x square plus b x square plus c square I will call a b c again so assume I have a what 1 over a so q is reduced and I take the sum 1 over a sum of the inverse of the coefficient of s square then assume we have such inequality h minus d at least a constant times this then ls chi has no equal to 0 where chi is a imaginary character of discriminant minus d so all we need to show is the following is that the hypothesis of Malheur's theorem whole so the theorem of Mark-Villain's theorem and Elkis is the following exists assume congesche a or congesche b whole we have some form of a b c which is valid whole then hypothesis in Malheur's theorem apply namely there exists a positive such that h minus d is at least a square d divided by log d sum over q reduced with coefficient a for x square 1 over a so I think I will stop because then I will go with the proof of that are there some questions say it again when you say reduced it means in the complex case oh yes this condition here I mean you can go from q to some complex number by saying that this is x minus tau y x minus tau bar y and tau is in the upper plane and reduced here is the same as tau belongs to the region here which is a fundamental group of sl2z but as a connection between reduction of quadratic form is the style of Gauss and the reduction of quadratic form in the tradition of Ziegel and others so the assertion is that since we have a quadratic form we can learn here and then there are finally many of these which can be represented by quadratic form q with integral coefficient so there are finally many points here which we will use a finally many class of reduced quadratic form so in the paper in my talk this theorem of Mallory's black box when one can read it but this does not explain so much I got an interesting email by Dimitrov Vasili who told me that instead of Mallory one could use a result by Colmez about the falting height of electric curve with cm there is a paper by Colmez with this title and it seems that in this paper it does the analytic number theory that we are using here in other words this papers are recycled into new in the new result but the authors of that paper they use Mallory's theorem there is a little problem they claim that the statement in the Mallory's is not exactly that it's a little weaker and then they say but it can be shown that it's a thing to give that I have not checked myself how easy it is to improve on Mallory's result to get this theorem so I start the proof and the first thing I want to discuss is complex multiplication so as usual we are at q equals exponential 2 pi i tau and we can consider the j function j of tau which is 1 plus 2 4 0 sum n at this one sum d device n of d3 not the same d qn and take a cube of the numerator divided by the delta function namely q times product n at this one 1 minus qn 2024 so if we develop in powers of q we get 1 over q plus 7 for 4 plus 1 over 6 887 plus a series in powers of q with integral coefficients 1q equals a squared plus bxy plus c y squared is reduced we attach to it so number 2q which I mentioned just before because question of Cartier so 2q is minus b plus c root minus d over 2a so now we can evaluate j on tauq and we have the classical result j of tauq is an algebraic integer which spans the Hilbert the Hilbert class field the Hilbert class field edge of k the field generated by j of tauq is the maximum unverified abelian extension of k that's the first one and the second is when q runs over reduced forms so we will change q then j after q runs over the conjugate of one of them it runs in a in a Galois orbit so let's call tau0 one choice of a reduced form and there are h minus d a reduced quadratic form and h minus d conjugate of j tauq so h minus d is indeed the degree of h over k now there is a result due to Gauss again is that if q is reduced qa is less than square root d over 3 so we have q we call the exponential of 2 pi i tauq which is e minus pi as the absolute value e minus pi square root d over q qi a so j is basically like 1 over q and q is essentially the exponential of square root d over so we get an estimate about the height of j tau by definition this is a product where q reduced of max g after q1 and we normalize by dividing by the degree of the extension so 1 over h minus d because we have an estimate on j so this is the same this is equivalent to exponential 1 over h minus d sum over q reduced with first coefficient a pi square root d over a what? you say tau0 is tau of q0 it's a choice it's a arbitrary choice ok so what we shall do is we shall prove the following the circle we shall prove the following inequality h of g tauq or tau0 for any choice is less than a constant d to some power so we know that the height is essentially the sum and we shall prove that the height is bounded by a power of d so I claim this is all we need because if you look at if we look at the estimate for h we get that a prime log d so the logarithm of the right hand side of this inequality here a prime of d is bigger log h and log h is given by the previous estimation 1 over h minus d pi square root d sigma 1 over a which is so log d but I forgot h minus d actually so h minus d is over there so there is h minus d I put it on the left h minus d is at least some constant times square root d over log d sum 1 over a so this inequality over there that the height is bounded by a power of d is what we need to check the hypothesis of malar that is with this inequality we deduce the theorem over there what? h is a malar the measure of the number h is the height it's the maximum of the supremum of absolute value of coordinate and we normalize it by dividing by the degree ok so the next ingredient will be to discuss the height of cm Lp curve namely assume e the Lp curve over q with complex multiplication by k so o of k is a sum of the under morphism of e we call k maybe I call it k prime at this point or k e mod 2 the field obtained by taking coordinates of points of order 2 so we have k which is a we add to k some coordinates namely the coordinates of points of order 2 in the Lp curve k is a quadratic field what? k is a quadratic field k is h so h I saw the mistake so we consider h which is a Hilbert field and k is a quadratic field and h is a Hilbert field and now we add to it the coordinates of points of order 2 when the coordinates when the points of order 2 are rational we have an equation of le genre namely y squared equals 4x x minus 1 x minus lambda it's easy to see that we have now 4 points of order 2 that's the standard drawing we have 0 and 1 which are of order 2 there is another component where there is lambda and the first point of order 2 is infinity so once we have added coordinates of points of order 2 we have le genre equation and in the invariant lambda lambda depends on the electric curve so it can be deduced from the J invariant of the electric curve so we have the following equation so the J of E with the J invariant of E then f of lambda equals 0 where f of x 2 5 6 x squared minus x plus 1 to the cube minus J invariant of the electric curve x squared minus x over 2 so this equation of degree 6 in x expresses lambda as a root depending on in J in the opposite way J is a rational function of lambda what? in the opposite way J of E is a rational function it's more usually people write it this way as I say J of E equals 2 5 6 so from J to lambda I start 1 to 1 because there are 6 solutions to this equation so the solution are the following lambda 1 over lambda 1 minus lambda 1 over 1 lambda minus 1 over lambda and lambda divided by lambda minus 1 I mean this is a usual function and the fact is that lambda is really the ratio of the 4 points of order 2 so if you want abstract definition of lambda we say it's the cross ratio of the 4 points of order 2 and p1 and then if you choose some order on the variable you get the different expression here ok so we want to burn from above the height of J to but of course J to is the same as J of E chosen to and we are the following step 1 which is as the maximum of J and 1 is equivalent is up to consent the product over all roots of F alpha equals 0 maximum of alpha and 1 so to prove this you prove it by discussing similar cases so you assume that lambda is chosen so that lambda is maximum what above the 6 values in the first case lambda is really big that lambda is small excuse me lambda less than 2 so then it's not hard to see that if alpha is 0 alpha will be between 1 over 2 and 2 1 minus lambda 1 over lambda is bigger than 1 over 2 1 minus lambda is less than 1 over 2 etc so because of the explicit formula when you have a 0 it is it is lying in the interval 1 half to 2 well this means that the right hand side product F alpha equals 0 max alpha 1 is basically a constant is over 1 and now if I look at the left hand side that is max J and 1 I also get something bounded I know that J can be expressed this way and lambda is between 1 half and 2 so J is bounded so in that case it is true that the maximum of J and 1 is the same as the product of the maximum of alpha and 1 second case lambda which is maximum among the solution of 0 is at least 2 then we have the following well first the maximum of 1 and lambda is lambda similarly the maximum of 1 and 1 minus lambda is lambda well it is equivalent to lambda and now if alpha equals 1 over lambda 1 over 1 minus lambda lambda minus 1 over lambda or lambda over lambda minus 1 in all these cases the maximum of alpha which is 1 is bounded because 1 over lambda is less and 1, 2 and etc so what we get is that the product of all rules of f of maximum of alpha and 1 is equivalent to lambda square there are 2 alpha which contribute to this product namely lambda and 1 minus lambda so this tells us about this right hand side which is lambda square and the left hand side but the maximum of j and 1 is also of the order of lambda square because of this equation for j j is the polynomial of degree 6 divided by polynomial of degree 4 so it's like lambda to the 2 the name to this equation so we have shown that the left hand side of alpha and 1 is the same as right hand side product of all alpha of maximum of alpha and 1 this is the first step we wanted to perform this quantity is equivalent to this quantity so we have not yet used abc construction so it's the time of doing it and it's a very strange choice of ab and c in the abc equation so the choice was proposed by Elkis namely we write lambda plus 1 minus lambda equals 1 this is certainly an equation a plus b equals c well we know by condition a or condition b we know that h so maybe I will use a exponential height I forgot was h logarithmic or exponential in the first part I give it logarithmic well it doesn't matter but it's logarithmic yeah so let's recall to some constant times n plus b times the determinant of the field where we are working which is h of lambda some linear terms is a degree okay what is the meaning of n lambda etc so that's like n the conductor so for q this is a product of all prime dividing abc and in general it's a product of prime ideals which make a difference between abc we had a definition last week of the conductor of abc okay so on the right hand side of this abc statement we have several terms and usually we are interested in the conductor but in fact in this case the conductor will be very simple it will be a constant why is it it's because this conductor lambda n1 minus lambda are in fact two are almost units there are two units up to a power of two there are units in the field therefore the prime which make a difference I mean the prime coming in the discriminant are just two yes so let's assume to see that let's assume that x is equal to 3y is the equation satisfied by lambda we consider the equation satisfied by lambda and we consider a solution assume it has an attribute of a relation at 3 then you can say the following x square minus x plus 1 to the cube is like 1 and you get a congruence that this is congruent to g of a g of v x minus 1 squared but x is 9 x is 3y so it's 9 y squared so we get the contradiction because here everything is integral and we have a 9 and there is no line on the left hand side so it's impossible it's impossible that 3 divides x and the same argument works for any prime different from 2 for any prime different from 2 x is a unit and if x is a unit the contribution to discriminant is 1 so where in the case where this expression here I should say 0 hello guys me so this time 0 does not contribute what contributes on the other hand will be the degree so we are left with the following h of lambda 1 minus lambda is less or equal let's put a put log here so as in equality times the constant so plus c times the degree so the abc statement is very different from usual namely the first time does not contribute and only the second time contributes okay so you need the logarithm of h less than okay third step we shall use the change of variable in classical height already there was already theory of height and this procedure follows you first define the height of a point in the projective space and when you have a lambda on a variety you embed the variety by power of L in the projective space therefore you induce the height from the projective space to the sub-variety you have to show that it does not depend on choices up to bounded constant but that's the direct way to define the height of a point on a variety so the first step really is the height of a point in projective space and this is done as follows assume I have a point at 0xn in pn of k where k is a number field then the height so now I will not take the logarithmic because I will take the exponential height and this is the product of the all places of the sub-variety of the coordinate by definition the height of a point in projective space is obtained from its coordinate this way and of course if you change the coordinate by a factor hrx will not change because of the product form ok now we can change the variable in this expression name is what sub hr goes from 1 to 1 or 0 to 1 0 to 1 thank you so let's see 0 fr hr r homogenous polynomial of degree and coefficient in k so let's see we have a point in pn of k and we can look at its image so this is the polynomial in k x0 xn homogenous of degree m so we can apply the polynomial change of variables consider phi of x which is the family phi 0 phi r of x and we shall take the height of this point so we assume that phi 0 x et cetera phi rx are not all 0 so let's define a point in pr of k then we have the following inequality the height of phi of x so this point of r plus 1 homogenous coordinates is as I call to a constant which can be explicit times the height of x to the power m so the map is essentially of degree m is of degree m and the height is raised to the power m ok so we find this independent of x independent of x ok so you find this as this is stated in the book of serobot model way so let's apply this change of variable to the following situation I take phi 0 lambda 0 lambda 1 lambda 0 lambda 0 lambda 0 square minus lambda 1 lambda 2 to the square to the cube phi 1 is by definition lambda 0 lambda 1 lambda 2 to the power 2 I will consider and now we suppose that m equals 6 I mean here we have 3 times 2 here 2 times 3 so the degree of phi 0 and phi 1 is 6 and we apply this formula so we find that h of phi of lambda on the point lambda 0 lambda 1 and lambda 2 so phi is about from p2 to p1 it's necessary to recall constant times the height lambda 0 lambda 1 lambda 2 to the 6 by this inequality and now we specialize that is we take some value of lambda 0 lambda 1 and lambda 2 in this inequality so take lambda 0 equals lambda lambda 1 equals 1 minus lambda and lambda 2 equals 1 so phi 0 we become lambda square minus lambda minus 1 to the cube so it will give us numerator and this one we will give you lambda 1 minus lambda to the 2 so it's not hard to see just the homogenization of the previous one what? you are from lambda to j and you homogenize it I homogenize on purpose to apply the change available so this is basically what happens so phi so the left hand side h of phi of lambda 0 lambda 1 and lambda 2 becomes h j of 1 by the equation star and it's bounded by the inequality by h of lambda 1 minus lambda and 1 lambda 0, lambda 1, lambda 2 to the power 6 thank you ok so we have h of j 1 less than h lambda 1 minus lambda 1 and at some point that is step 2 we get the discriminant so this is less than the discriminant to some power called maybe 6B the discriminant of h lambda over q times some constant so we apply abc from this step 2 so we are almost done we want it to that h j 1 is bounded by a power of d we get it's bounded by the discriminant of h lambda over q so we have to prove and I will not do it the following proposition which is proved in the paper of gonville and stack after the value of discriminant of h lambda over q is less recall to a power of d times constant so now we have that h of j 1 h of j 1 is less than discriminant so is less than a power of d this is what we did the proof of this proposition is little difficult for me to read because of a ray class field and the like so well it can be proved but the right in order of interesting and the constant depends on what the constant is inequality depends on what it's over x this is valid for what kind of x and what which kind of constant in fact they consider a problem of this kind and they get 6 square root of d but we have to shift from their choice of an equation to the a key's choice and I don't know which constant I think what comes out is the natural constant this is valid for which kind of x x so you have the input class h yes and you add John to it what kind of thing you add h of x it's input class this is h of points of order 2 so nothing happens between k and h by definition of h and then you have to control what happens by adding groups of 0 and so on the point of order 2 yes are there some questions so thank you for your attention