 thrust bl , hal twaw and r r r r r ffen ți 叛 ți                                                                                                                                                                                                                       Ḣ w Ḣ rᴓḤᴇ æ. Ḣ wḢ xἔ wᴍ ᴠᴏḣwn dᵜˤ k Ḥ m᱄ oḠ Ụ wộwᵇcᶦ wᵃwa wḎ wᵍ yᴏchᧈ gᵬ 깔� bunun ḏ wᵇ lᵜ reductions ll phy an, feat. K cheese appeara meteor liᵓ n ᵸw ᵘw HIV j ᵸwᵤ iz m-j, හයක්ක්කතා ක්ඳළමටක්ගාසු නවීමටය ඔබක්කතාක්ත කරම්නීක්රක්ගීස්ක්රක කර්ක්රවක්කත කරලකතාක්කදු ලකතාක්කතයෙකතාගගකතාර peripheral main注意 – dies Associative of X by S. So this is a line model on X which on the smooth axis are relative differential. And arcela defining metric on omega and I gave definition of this metric, so, let's call omega two the self interface over the megabar. This is a real number, an invariant of the asthmatic surface x. We said that it's a theorem of Wilmore that omega 2 bar is positive. And we made a conjecture about an upper bound for omega 2. So assume g equals 6, and we shall see today why it's g equals 6. Then for every positive real number epsilon, there exists a constant, beta of epsilon, such that omega 2 is at most 3 over 2 plus epsilon times dk plus beta epsilon times qq. The constant beta epsilon is assumed to varies in a bounded way when taking x in a projective family, and we shall see a more specific application of this condition today. So we announced the following theorem that if we have the conjecture b, if conjecture b holds, then the abc conjecture is true, where abc means precise one plus epsilon exponent in the bound on the, well, if we have a plus b equals c, then c is less than the constant times n one plus epsilon. The point was to have the correct exponent in the abc conjecture. So this is essentially what we did last time. And we started to state and prove a lemma, which goes as follows. So again k is the number field, s is the spectrum of k. And now we consider y, a smooth projective curve over k. So we have a curve on the field k. We can define the pika group, which is isomorphism classes of line models equipped with the operation of tensor product and taken modulo isomorphism of line models. The pika of y is a set of isomorphism classes of line models on y with the tensor product as an operation. So we assume that there is a model of this picture. So ys is a proper curve over s, which is a model. So ys tensor k os is y. Then we have an arithmetic pika group, which is the set of emission line models on ys. The operation is the tensor product. Of course, we can take the tensor product of two metrics to get the metric and the tensor product of line models. And this is modulo isomorphism, where isomorphism between L bar and M bar means that L and M are isomorphic and the isomorphism preserves the metric. So we do the following construction. We assume, well I should have, yeah, assume we have a finite extension of k. We write t for the spectrum of O f. And we suppose that we have a point in yf, a point on y defined over f. Then by the evaluative criterion of compactness, of properness, then we have a morphism from t to is, which generically is given by the point. On pika of y, I have a map which is called the degree. Let me write a degree in the xk. If I have a line model on a smooth projective curve, I have this degree. I mean, if you have a complex number, you can just consider the integral of c1 of the line model. But it takes this algebraically. It is given by the number of zeros minus the number of poles. So we have the degree from the pika group of y to z. And the lemma says the following. Let xs, a class in the pika group of ys, turns the q, assume that the restriction of xs to the generic fiber of ys over s, which is y by assumption, assumes x satisfies the conditions that the degree is negative. Then for any point, any number field exception of k, for any point p in yk, in yf, we have an inequality, namely the degree, the arithmetic degree of the restriction of xs to t. We have u, morphogen from t to ys. And here we have xs, a class in the arithmetic group of ys. We put it back and we get a real number, which is the arithmetic degree of the pullback. We assume that, generically, the degree is negative. And you conclude that this is less or equal to o of r, where o of r means constant time r, where the constant does not depend on the choice of f or p. So that's the lemma. And let's see how it goes. We are taking excess in the pika group of ys, turns the q, but this means that the multiple of xs is a line model, Hermitian line model, and we can certainly take a multiple in order to prove the lemma. So we can assume xs is a class of a line model, but I will be interested in the dual line model. So the line model l of os, ls to os bar. So this is the dual of an Hermitian line model, ls. So if we consider x, which is ls over y, the degree k of x, so we consider the dual. So it's the same as the dual of the restriction of l to s. And the degree of x is negative. So the degree of x is minus the degree of the line model l. So l is such that this degree is negative. The degree of l, excuse me, is positive. We are on a curve, so to have a positive degree is that l, example. On the other hand, we are considering, we can consider the inverse image of l bar s, and this degree, we write it as h ls of p, and this is kind of height for the point p. p is the rational point, an algebraic point on the curve, and the height of it is obtained by pulling back the Hermitian line model ls and taking the degree on t. So this is what is called the falting site. This is the definition. So what we are saying is that if l is an pole, the falting site is bounded from below because we shall take minus xs. So a few l examples, then the falting site is bounded from below. So we wrote a paper with Bost and Gile where we prove these conditions that if the line model is an pole, then the falting site is bounded from below. So let me give a hint for the proof of this session. So l is an pole, so there exists some big power of l, such that l to the n is o of 1 for some projective embedding of y and pk. So here we have o of 1 and o of 1 has a metric, which is the fubinistic metric on the line model. So that's the metric on o of 1. What's the remark is that the height with respect to o of 1 with the fubinistic metric is positive. This is given by a explicit formula, log of sum of square divided by 1 square, so it's positive. Now we can compare the fubinistic metric with the metric which is given on l. So assume that l bar is o of 1 times h where h is a fubinistic metric multiplied by a function exponential of phi where phi is a smooth function of the complex function of y. If I have two metric, I can look at the quotient of this two metric and I will get a still infinity function. But now yc is compact, so phi is bounded. So the two metric h and fubinistic are equal up to bounded amount. And this is all we need to get a lower bound for the height of p with respect to ls. We still have to do something at finite places and at finite places all I will say is that ls differs from omega 1 by bounded amount. I will not use the full proof of that. So we have that the fubinistic height is positive and the height and model we consider differs from fubinistic study by bounded amount. So we know that the height of p is not positive but at least bounded from below by some expressions like ar. So this is what we prove in detail in this paper with Boston Gillet. But this is nothing else than the degree ua per star of xs minus ar. So we get that the degree of ua per star xs is at most a times r. And this is the lemma. So the theory of heights and the fact that they vary by bounded amount when you change the metric or you change the model of the line model, this fact implies the lemma. That is before our Kepler theory the theory of height was modulo bounded amount. One defines the height which is called the logarithmic height which I will come to in a minute which is well defined up to o of 1 because when you change the metric of the model you don't want the height to change too much it changes by bounded amount. Now we'll do some construction which will relate the height of a point to omega2 or some curve. And this is the construction due to Kodair. I will talk about Kodair vibration. So this is the following. Assume I have a number field which I call k0. Assume I have a curve, a smooth projective. Geometrically connected. Curve over K0. Assume that the genus of C is 2. Then we have the following theorem of Kodair which is explained in an article of Mireille Martin Deschamps which is the following. There exists a finite extension of K0 alpha, an etal morphism finite etal map over K and on Y2 we have a family of curves. So C is a curve, Y2 is a curve and Y2 will be the basis of a family of curves. So we get a surface with basis Y2. So there exists a finite etal map of this kind and a morphism, which I will write by 2 from X2 to Y2, a smooth morphism with fibers, curves of genus 6. So by 2 is supposed to be non-animal trivial. That is you cannot trivialize by 2 by some finite extension of Y2. Fibers of Y2 are smooth, proper, geometrically connected of genus 6. So the genus are we called gamma. So this is why C's come in. I mean we start with the curve of genus 2. We have this family of the fibers of genus 6. So let me explain this confession of Kodair. So K has to be large enough for the existence of by 2. So we consider a K and we call Y0 the curve obtained by a standard scale from K0 to K. Now we consider Y1 to Y0, a degree 2 etal cover. So this might not exist on K0 but when K is large enough, there exists non-trivial degree 2 etal cover of the curve Y0. If I consider the genus of Y1, well since I have a etal cover I have 2G1-2 equals 2, 2G-2 where G is the genus of Y0. Since G equals 2, we get G1 equals 3. So we have a curve of degree 3 which is the etal cover of degree 2 of this curve of genus 2. We call J1 the Jacobian of Y1. And we do a second construction which is the following. We consider the multiplication by 2 in the Jacobian. We assume that K is large enough for X1 to be contained in J1 by choosing a point. So we have X1 go to J1. We consider a cover by the multiplication by 2 in the Jacobian. And we consider a connected component of the fibre product. So we call, I mean Y1 here and Y1 here. So we consider Y2 to be a connected component of the fibre product. So of Y1 cos G1 over G1 where here I have multiplication by 2. So we can consider the product of Y1 with itself. So it's a surface of course. And in this surface we have a divisor which I call gamma which is the product of Y1 with itself over Y0. So 2 points in Y1 with the same image in Y0. So for instance we can look at the diagonal where we have twice the same point. We can also look at the graph of the multiplication by 2. So we have in fact two components in gamma. One is the diagonal, the other one is the graph. But in any case, gamma is the divisor in Y1 cross Y1. Now we look at the map from Y2 cross Y1 in Y1 cross Y1 and we call gamma tilde the investment of gamma in this surface. What is the arrow? Y2 cross Y1 to Y1 cross Y1. Y2 is the last curve we considered. It maps to Y1 by definition. So Y2 cross Y1 maps to Y1 cross Y1. This is your question. Maybe I should write things upside down. Y2 cross Y1 is larger than Y1 cross Y1. Okay? And then there is the following fact which is proved in the references I am telling you. If K, the number field, is large enough, the line model L over Y2 cross Y1 tells that the square of L is the line model attached to this divisor gamma tilde. So it's O of gamma tilde. The line model, so we have divisor attached to the divisor we have a line model and the fact is that this line model has a square root. So this is following the construction that we get this property. So we choose according to this equation and then there exists a surface which maps to Y2 in fact it maps to Y2 cross Y1. The morphism is degree 2 and it's quantified gamma tilde. In order to define the map from Y1 to J1 do you need the original point on Y1? We shall map X2 to Y2 by taking the first question. Is it your question? No, I mean the map from Y1 to J1 do you need the original point? It's given by the stress of the point. This map here inclusion of a curve in the check command. For this you have to choose a point on Y1. So we make some choices in the construction for instance at this step we choose a point and now we choose L. So the construction will not be canonical. It will depend on some choices. So on Y2 cross Y1 I have this divisor gamma tilde. I choose a square root of this divisor namely L and then I get a surface which is the morphism of degree 2 ramified along gamma tilde. So this is the standard construction when you study cyclic covers of curves so the construction of X2 is as follows we call L the total space of the line model L so L maps to Y2 cross Y1 biomorphism called rho and we can consider two sections namely we have the section on L of the pullback of L curve by totology if I am in L I have chosen a section of the line model L so that the totological section consider T which is a section on Y2 cross Y1 of O of gamma tilde so the canonical section I have a divider on this surface I have a line model and this line model is given by a section with a section here is the definition of X2 it's a divisor in L so L was a relative dimension 1 on Y2 cross Y1 so Y2 cross Y1 is dimension 2 and L is dimension 3 we have a divisor in L which is a divisor of equation S squared equals phi of a star of T rho of a star of T so we look in the space attached to the line model L on Y2 cross Y1 it's a dimension 3 manifold and on this manifold we define a divisor by this equation S squared equals rho star of T so the general theory of cyclic covers says that to disconstruction attaches to a square root the two-cover ramified along gamma tilde and this is a suggestion to give the line the square root or to give the cover is equivalent so now we have X2 it is marked by the projection of L2Y2 cross Y1 to this product here I have the first projection 2Y2 and by definition the composite is what I call pi2 so this is the map in the statement of in the statement of the theorem so we can show that this is not isotree rule I mean we are not taking only products at some point we are actually doing something so we don't have a product family it's not isotree rule and we can consider the fibers of pi2 so we consider a point let's say in Y2 K bar we can consider the inverse image and the situation is this I have taken a fiber on the first component therefore it maps to the second component so X2Y maps to Y1 by the first projection or the second projection excuse me so it's a map of degree 2 and since X2 to Y2 is ramified the same is true for the fibers the fibers X2Y to Y1 is ramified at 2 points so it's ramified along Y cross Y1 intercession diameter and this consists of 2 points the diameter has 2 components the diagonal and the graph of the evolution if I take the intersection with Y cross Y1 I will get 2 points so called gamma the genus of the fiber there is a so called uvis formula I don't know where is the t which tells us how to find the genus of the ramified cover so I get the following 2 gamma minus 2 to twice 2G1 minus 2 plus the ramification so the ramification has index 2 2 minus 1 twice so I get plus 2 by the ramification of the morphism from X to Y to Y1 so G1 is 3 and gamma is 6 here is number 6 so let me remind you the notation we have a curve C over K0 we consider Y0 which is C times K over K0 we have a cover etal cover of degree 2 by Y1 and then we have an etal cover of Y1 such that there exist a family of curves on Y2 with curves of genus 6 so what we have to do now is to choose models for these varieties of the number field because we want to do height and so forth therefore we have to work on doing our integers so let S be the spectrum of K and we want to extend the calvary construction to S over S so the first thing we can do is the following so it's a statement which I learned from Abramovich it's the following if alpha from Y2 to Y1 is large enough that is to say I can replace Y2 to Y1 by Y2 to Y1 where Y2 to Y2 is an etal cover I have the freedom of making Y2 little bigger so by doing so we have the following property for any finite extension F of K for any point P in Y2F whatever Y2 minus 1 of P has a stable model on T equals PECOF so this is a general statement of the existence of stable models I mean this is well known when the base is dimension 1 by a finite extension I can always assume some instability I forgot the name of that so here we do it in the family by an etal cover of Y1 I mean we make an etal cover of Y2 and then we have the existence of models for every point we have a model but we want something more namely we want a model for the family so to do this a second fact which is due to Mireille Deschamps in the paper Mireille Martin Deschamps in the paper I quoted which is that at this we can extend the picture to an open neighborhood that is exist you in an open subset such that the map from X2 to Y2 extends over U this means that we have a map X2U goes to Y2U over the open subset U I mean this is usually the case if you make a construction genetically at least on some open the construction will be the same but of course bad thing can happen outside U so now we would like to compactify the picture so X2U over S is some compactification of the map Y2U goes to U I mean there are securities in Y2U outside the inverse image of U and now there is another theorem which is due to above average is that we can extend the model by some alteration so exist an alteration I will define this in a minute so it's a morphism FES from YS to IU bar it's a morphism of S and a stable curve XS over YS so after some alteration we are able to extend the family to a stable family so YS is dimension 2 now of the scheme and XS is dimension 3 so the result of above average an art FES an alteration means the following when FES is proper search active and generically finite there is some famous work of De Jong who replaces resolution of singularities by alteration of singularities in all characteristics so the same type of construction is done by Abramovich and Hort to extend the stable family from the generic fiber to a model so if I think of this picture so I call FES Pi as the morphism here and generically I have Pi from X to Y generically this family of stable curve over the curve Y this marks 2I2 and here I2 which I constructed in a previous hour and this diagram is commutative is Cartesian this X is the Cartesian product of X2 and Y over Y2 what happens is that stable models are unique so whatever construction you give of stable models there is only one and here you can either construct it by Hort Abramovich construction or by pulling back the stable curve X2 over Y2 ok so this is the end of the Kodara vibration construction and now I want to prove the main step in the proof of serial B that is MBC the step we take is called effective model that findings prove that a curve of genus at least 2 has finitely many points but there is no bound known on the size of this point an effective model would give a bound on the size of the solutions the height of the solution the height the number of solutions there is a bound very good but there is a bound what is not known is to bound the coordinate of the solution so what when you say the size is not clear yes yes so I will say that I will show that angle congestion B then we have a bound on the size of the solution ok so let's do the following K0 is another field C over K0 is a smooth projective geometric connected curve of genus 2 over the field K0 of genus 2 let's consider M an upper line bundle and C we can look at the set of algebraic point of C C of Q bar and there is a map from the set of algebraic point to real numbers which is called the logarithmic height so HM is called the logarithmic of the curve attached to M so I said what it was in the last hour it depends only on M does not depend on the choice of the metric or model but it is well defined up to a bounded amount so that is the whole theory of height where you define the logarithmic height as a quantity defined modulo bounded to a sum K0 is another field so the theorem is that the contrast should be all then for any epsilon positive exist beta epsilon such that for any field containing K0 for any point P in C of F the logarithmic height of P is bounded as follows it is less or equal to 1 4 plus epsilon times the degree of M d of K maybe I mean d of K0 d of K0 d of F over the degree of F of Q plus O of 1 so the theorem says that if we have a bounded omega2 we have a bounded height of points on curve of the two in that case at least d of F this minute d of F was defined as log of this this minute ok so to prove the theorem is to consider the Kodaila construction that is we introduce this curve y0, y1, y2 and the model xs over ys so we have L0 which is M on y0 plus a metric plus any metric on y1 we have the pullback the same way we have the line model on the generic fiber of ys so call it y and you assume that you have extended the metric and the line model to a model to the model ys ok now we shall use the lemma which I first move in the preceding hour so define xs as follows so I will use additive notation for the pika group so this is the line model L minus degree of L0 1 plus epsilon over 6 and here I have a line model namely here I have stable curve here I have a relative direction shift omega s I take the Akelov metric on this relative direction shift so I have omega bar s on xs xs is a family of curves on ys so we have the construction by the line model omega bar s, omega bar s on the base which is an arithmetic surface so this is an element in the arithmetic pika group of ys times q I assume that epsilon is in q ok so we shall apply the lemma to xs for this we consider the restriction of xs to the generic fiber we have the degree of x which is the degree of L minus degree of L0 1 plus epsilon over 6 and here we have omega gamma omega we are now looking at this construction but generically we have x to y we have omega on x and we have a line model omega omega on y which is the algebraic counterpart of what I was saying in the arithmetic case geometrically y is a curve xs surface so here I have dimension 1 and dimension 2 here I have dimension 2 ys is an arithmetic surface and xs is dimension 3 but don't give us despairing from xs to ys because the collective dimension is 1 ok so now look at omega 2 which is by definition degree k of omega omega is the line model on y which is the smooth projective curve therefore there is a degree called omega 2 and what you can show is that it's the same as omega 2 for x2 here I have y and here I have x and now here I have morphism of degrees of degrees of y over y2 by definition so omega omega is here I claim this is the same as omega omega on y2 multiplied by the degree this is because some diagram is commutative and Cartesian actually so here we have a Cartesian diagram here we have omega here we have omega 2 therefore the inverse image of omega 2 square is omega 2 and the degree is multiplied by the degree of the ascension ok and this is the same so this can be computed I mean omega 2 square which is really the intersection of c1 over omega 2 is itself on the surface y2 can be computed and 1 times 6 1 times 12 actually so we get here 6 degrees of y over y0 so now if I look at the expression degree k over x I get that this is equal to degree k over l minus degree k over 0 multiplied by the degree of y over y0 times 6 times 1 plus epsilon times 1 plus epsilon and it's not hard to see that the degree of l is the degree of l0 multiplied by the degree of the ascension just because l is the pullback of l0 so here I get minus epsilon degree k of l0 of y over y0 and this is negative so we have checked the condition of the lemma that xs define this way such that its degree of the generic fiber is strictly negative how did you get the 6 this is the computation which I am not explaining so is it not the same as the degree l minus 6 it's related what we have to do with the geometry of this ramified cover and you want to compute the ascension of omega in the cover it's not simply the fact that omega g does 6 it's very close it comes from the fact that omega is given by the reviews from the so and then you compute the ascension ok so we can conclude so by the lemma we know that the degree hat of u' of xs is at most o of r because we know that the degree of xs of the generic fiber is strictly negative ok so what does this inequality mean by the definition of xs this is the degree of u' of l bar s less or equal to degree of 0 1 plus epsilon over 6 degree of omega bar s omega bar s and the pullback ok so now we will go down in the tower which I raised the tower here I mean we are computing here at this point and then we shall go down to zero so we do it as follows by assumption we have a field f which is the final ascension of k0 and we have a point p in c of f so the first fact which I will not prove which is proved in the paper I referred to last time that is we can assume that k equals k0 that is the first step where we extend k0 to k is the identity so now we have we don't have to worry about c anymore just y0 and we can say the following we have a point in cf which is now the same as y0 of f and we want to lift y0 all the way to xs in order to compare the height over there well what happens is that the map from y2 to y0 is surjective or the map of scheme so this means that there exists an extension f2 and a point which I write p2 on y2 of f2 such that well the two points p and p2 are essentially the same namely I have y2 f2 these maps into y2 no no I have I have y0 of f0 of f where there is p I can look at this image in y0 of f2 I look at y2 of f2 and p2 here so it's not exactly the case that the image of p2 is p but they become equal when they extend the scalars so we get the same point whether we go this way or this way so we say that we have extended p 2y2 f0 what is f0? an extension of f0 what is f0? f0 what are you doing here? maybe it will be cool no no so I choose a point downstairs and I lift it progressively 2y2 and then 2y oh you start at this and then you go to y2 this is a statement so we are on f0 and we go up 2y2 and next we have to go from y2 to y so we do the same construction I'm sorry for the technicality of so again the map from y2 to y2 is subjective so it's essentially subjective on rational point so there exists an extension f of f2 such that there exists a point p yf such that if I consider yf I can go to yf2 I can also go from y2f2 to yf2 now I have p2 here I think that's okay so the point p and p2 correspond by this diagram it's not exactly the case that the image of p2 but up to finalization that's the case okay now let t be the spectrum of f call xt the stable model of xp the fibroid p of x we have a diagram like this and this diagram is cartesian by the unicity of stable model on t we have also another cartesian product namely xt goes to xt2 so I extend scalars from t2 to t I get the same stable model okay so we have this inequality degree of restriction of Ls is less or equal to degree at zero one perception of Lv6 times this degree here so really degree of u over star omega bar s omega bar s is the same other degree of omega t omega t because this diagram is competitive and cartesian omega s I get omega t and the same for this assertion so we have degree hat u star L bar s is at most degree at zero one plus epsilon of s6 degree hat over omega bar t which is nothing else than the arithmetic insertion product of omega bar t with itself now let's use the other cartesian product here to prove that omega 2 omega bar 2 omega bar t omega bar t is the same as omega t2 omega bar t2 times the degree f over f2 so this is by this product here now we apply congestion now we apply congestion b that is we have omega omega 2 which is an upper bound for the height and by congestion b omega bar square is bounded above from by log of this discriminant congestion b implies that for any epsilon there is a constant beta epsilon such that omega bar 2 omega bar t2 square is at most 3 over 2 plus epsilon d of f2 plus beta epsilon degree of f2 over q this is what congestion b says that this right hand side here can be bounded above by log of this discriminant so we are almost done because if we put this inequality together with the one we have already for degree of u up a star of l bar s we get the following up a star ls is at most f over f2 degree l0 and now we have 3 over 2 times 1 over 6 so this is 1 over 4 1 over 4 plus epsilon df2 over the degree of f2 over q plus some constant over 1 and beta epsilon so we have almost proved that the height of a point is bounded above by the discriminant but we need, it's not exactly the statement we want because we want a statement on f0 we want a statement downstairs so here we use the fact that y2 to y1 is etal to get that df2 over f2 over q it's less than df0 over f0 over q plus over 1 so this is a statement in the paper by Morabayi we know that y2 to y1 is etal therefore at the level of fields the only ramified primes are those already ramified in f0 and those ramified those sitting outside no, outside the open set where the map is etal so we have a bounded set which is independent of the choice of points which is the open set where this map is etal etal finite finite is the complement of 0 yes yes ok so we can replace the right hand side here by df0 over f0 and the last fact is that the height of p is the same as the height for 0 of p0 what do I mean by the height I mean precisely this number the degree of the pullback of L bar S on t and as I said in the proof of the lemma this degree does not depend on the metric up to bounded constant and does not depend on the model up to bounded amount so the logarithmic height is really the Akeler height up to constant constant and this formula is true by adjunction because L is the inverse image of L0 and p0 is the image of p so finally we get the following H L0 p0 less or equal to degree L0 1 plus epsilon over 4 dF0 over 0 over q plus over 1 the height of the point which started within the theorem is bounded above by the discriminant so this is what we wanted to show that the height is bounded by the discriminant times the degree so I hope the rest of the course will be less painful than today are there questions what the bigger of one all the constant depends on the compactness argument I mean when we compare to height we say that they are equal up to constant by taking the supremum of phi so this is very badly controlled the constant is obtained by sup on the compact set it would require a lot of work to give an explicit constant the family I was talking about in the conjecture is precisely the Kodara family we apply to the Kodara family the statement of conjecture B so it remains bounded in this family which is more precise than saying it is bounded in some project family so next time we shall see that effective model implies ABC thank you very much