 Last time, I do a reduction on how to reduce the average convex conjecture to the highs for CM points and Schmura curve. So on Schmura curve, this is very similar to modular curve. So in this case, we can do it in a similar way as a modular curve. So this is a formula, so something called the Charler-Selberg formula on Schmura curve. And the first approved by Leach, Charler-Selberg, in the last century or whatever, depends what do you mean. So this new formula was approved in 2008 as a byproduct of a Charler-Selberg formula. Xinyi didn't publish it because we really don't know why this is the important formula. And so in today's lecture, I will first put a statement on the theorem. It's already very complicated. Then I move back to modular curve. So just show you in the cluster case. How can you prove it directly using the modular forms? Then I will move to Schmura curve and then give you an idea why this does not work. So we will talk about the Gorozagir formula. So somehow the interesting thing for the Schmura curve is that to prove Charler-Selberg formula and the Gorozagir formula is the same difficulty. So there's nothing, say, one of them the simpler form or other. So actually, the most important difficulty when you prove Gorozagir formula is the self-intersection. So when you have a self-intersection, if you don't have modular form, then you have to figure out what you're going to do. So I talk about the statement of Charler-Selberg formula on Schmura curve. So the definition is very long. The reason is I needed to give a, so there's f to the total real field, number field. And to define a Schmura curve, you need to have a set. So this is a set of places of f. And then there's the finite set. And actually, the first condition you want to be with odd cardinality. Second is to include all accumulative places. So this is the odd set, include accumulative places. Then this data will give you a Schmura curve defined over f. So I will give a really formal definition, but it's not really important. So the idea is the following. So I can just write one formal definition. Is that it started with b is incoherent, quaternion algebra over a dels, a f. So incoherent with ramification set, sigma. So this is sigma is odd. Incoherent means that this will never be a best change. Come from a quaternion algebra of f. So really mean that we can write this one as formally a plus a i plus a j plus a k. As usual, so i squared, g squared, k squared is in a cross. And ij equals k equals negative ji. So that's a typical quaternion algebra. And so that's something quaternion algebra used when we formulated the Grosage formula. And this somehow is a very convenient thing. The reason is when we formulate the quaternion algebra, we consider it as an orthogonal Schmura variety. So this itself is viewed as a space. So if u, the b infinity means you take all the infinity out, maybe bf there. So just the finite part of b. And for any u inside of bf cross the open compact. Then this one gives us some kind of Schmura curve, x of u. So it's a bit strange. The thing that, so okay, so this curve is described by the following property. Such as that if for any f inside of c, in fact for any place. Let me try to say where the eraser, which has a uniformization at OVA inside of sigma. So usually for example, the modular curve has a uniformization at a academic place. But that's not a uniformization at any other periodic place. So there's a sigma that plays a role that the Schmura curve sometimes have the uniformization, the sigma, by either a concrete of a plane or a dream field. So in some sense that, so it's a c minus r or cv minus what? It's fv, right? If it is cv, it's fv, it's algebraic closure that take a completion. Of course this is just set, you have to put some analytical structure in this set. More concretely, for example, so if v is in sigma, then you have a b of v defined, b of v defined to be the quaternion algebra of f with the ramification set. Sigma take v out, right? When the take a v out, it becomes the even set, right? So you do have this thing, then, well, so the bv turns out with a, it's the same. You can fix isomorphism, you can fix isomorphism between this thing. For example, if v is infinity, well, let me try to write down one simple thing. Suppose it's infinity here, this f is isomorphic to bf, and then, so this x at v as an analytical variety, so this is the Riemann surface, right? It's equal to b over v cross, you take the c, take this out. Then you cross bf cross, take u out. So this is, if we really write down, so this is the sum, it's the finite sum of a plane, mod of some gamma i, right? So this is the finite, just join union. The reason I do not write this thing is each of individual, this one, each of them of the connected component is defined, not defined of f, but defined of a billion extension of f, right? So, but if they combine together, we'll be defined of f. In fact, it's connected of f, but not geometrically connected. So this is, the good is pretty convenient to write all of them together, right? And we even know which one is. So in fact, so it's not important, but just to give you an idea of that. So if you really want to study there, the x u, there's all the shift, right? So this actually is a billion extension. So what's the action of bv on v on vf? Could be if you take all the parameters out. What? What's the action of bv on vf? No, bv is a quaternion algebra, who is ramifying set as a second mark. But bf is only finite part, right? So you only take the finite part on bv to act on vf. Yeah, so this one is equal to, I needed to define the action, this one over that. So this one will be equal to bv, right? So this one will act on that. So I fix isomorphism. That's the only reason I want to do that. It seems that this is the definition of some more called different from your book. My book? Oh, OK. No, I just give a description. So what is the uniformization that our commutant place is? And the interesting thing that the way I'm writing down is v at our commutant place. You're also going to do it when v is not our commutant, right? OK, so this is a billion extension of f. If you write it as fu, for example, the Galois group of fu of f is by class field theory will be corresponding to this af cross. You take what have f cross, take infinity cross, take the norm over u, right? So this is reduced norm, maybe the reduced norm. So you have a quaternion b of v, right? You have a quaternion v to f cross. You have the reduced norm, right? So this actually is very explicit. So you really know which algebra is. So this is a very nice thing. And the important part is the following. So by the theory of Karajou, Greenfield Karajou, so that's actually this xu has a canonical integral model. A canonical model is also easy to describe. The reason is when u is small, your curve will have a genius greater than 2. So you will have an integral model. So their theory is that these minimum integral models actually are compatible when you make this system, right? So this is actually a minimal model given by the integral models. So this is actually equal to the minimal integral model when the genus of the kinetic component, genus of geometric kinetic component, is greater than 2. So in some sense, we can think about this to give you an idea that, OK, so x is integral models. So this gives you probably the simple way. You have two opposite ways to describe the question. So you can either talk about stack when the minimal level, well, you can take the limit, all xu there, this is the way I usually like. Or if u is really maximal, then you get x of u as a stack. Because when the genus of u is not small, the minimal smooth model is not a part of it. So in that case, so integral model, let me say xu has cannot integral model xu when u is small. So if u is not small, you just consider as a stack. Stack means that it has a covering by some groups. OK, so this is a, so anyway, this gives you a well-defined differential. So this gives you l of u is inside the picot group of x of u turns out to be q. So this is basically the stack differential. You can have a stack differential. So I mean that if you have xu, you're covered by, over the base, covered by piece of a piece. Then you move the level structure. When you have level structure, you can define the omega of u somewhere, because the eta from there. But you cannot find the eta cover directly from whole scheme. But you can put a piece by piece. So you can put two, for example, two prime numbers, one prime to p, one prime to q. Then in that term, you can put a p-level stretch and a q-level structure separately of the base scheme. Then you rest the level high enough, some power of the p. Then you get atomomorphism. There, the genius corrects them, too. Then you have a differential. You push forward, divide by the degree you have. So you will get this thing. So why do you need two primes? What? Why do you need two primes? Or because you need the atom map. But I mean the power of p and q are large enough. So you want to define, there is a Hodger bond so that it's compatible with the pullback, right? So you mean this atom will be now only when the power of p and q are both large? No, to construct the area u, you need to take a coverant. Let me just give you one example. Even the upper plane, even x of 1, how do you define the Hodger bond there? This one is covered by, say, xp, xq, right? So yeah, this is covered by, but this is over the spec of z, right? So you cover that by, this is, OK, that gives you some idea to do that. So this is a stack over this end, right? This stack is, so this is covered by spec of z1 over p, right? Cross union spec 1 over q. So let's assume p and q both are greater than 3, a co-prime to each other, right? Over here, you have x of p, right? OK, I have x of q, right? So these are two are really a scheme there. So this is the one you covered here, right? So you take an integral model, for example, you take an integral model, you take an integral model, then you will have this x1 is covered by x of p, right? x of q. OK, so the common part. The common part is covered by x of pq, right? The common part has x of pq. So you don't worry that much. So you can define the bundle by, well, omega x of p. You push forward, OK? So there's f there. The f equals fp, fq. fp, you push forward, it's a bundle. But you have to divide by degree, right? Degree of fp, right? Then you also can define fq omega x of q, 1 divided by degree of fq, right? You get two bundles downstairs, the q coefficient bundle. But these two bundles, you can also get different this one, right? So you get three bundles, fpq xpq divided by degree of fpq. But these bundles only are leaves over here. These bundle leaves are there. But the common part is isomorphized to this part. So this is one isomorphized to this one. So this is over, well, over z join the one over pq, right? z join one over pq. Yeah, then you glue two shifts together, get a shift, and you x1, right? I mean, the answer is actually very simple. The answer is just this error of one, just the usual one. If you take there, if one is isomorphized to p1, I mean, the single this one is not an x, it's a p1. The plus, I don't know, plus the cusp, right? I know. 1 minus half of i plus 1 minus 1 third, I forget. It's probably 1 zeta 3, right? It's just the two things added together. And if you take a classification, you need to add a cusp to infinity. So this is basically your hajj bundle, right? But a hajj bundle is constructed this way. OK, so let's just give you the idea. For Shimura curve, you construct the integral models and also construct the hajj bundle. Then, of course, you can define the fault and the height. So if p is any point, if x, x of u, then you can define the degree of p. Let me see. You can define the degree of the hajj bundle, ru, right? So pu here, right? Well, this will give you a point. But if you have a compatible family, say u1 equals xu2, your hajj bundle, ru2, if a pi here, pullback is ru1. So you computed this as a possible pu1 here, pu2 there. So you computed the height here and a computed height there is the same, right? There's no difference. So this will give you. So this is. The point is that by the Infant Carion, all these minimal integral models are compatible. In some sense, the natural map extends to integral models, minimal models, when the genus is bigger than 2. Yes. This is the point. This is the point. I mean, the point of that. Is that the statement of Dreamfield Carion? This is a statement in my paper. But essential ingredients are work of Carion and Dreamfield. They never talk about that in this experience. I mean, Carion works on locally when pi is not in sigma. Dreamfield working on pi is in sigma. The Carion basically is a generalization of work of a mezzocats, Deline, Rappaport. This kind of work is a four-piece split. Four-piece non-splitter that's a Dreamfield. Your Dreamfield will find out, actually, the strict-speaking Cerenik Dreamfield. Cerenik, he just finds purely by geometric method. And Dreamfield figured out how to define a modular problem. But the point is that only the integral or minimal model are compatible. Right, right, right, right. This is the proof in my paper, in this average code method. It took a while to prove that. OK, so we can normalize this thing by, so we define this thing. We can normalize, if we normalize this thing by, so let's define the definition. So this h, l, u, p, u, I don't know why I'm in there. Take a half. I want to, half probably means, oh yeah, this is a weight two forms. You take a half, I mean that you only consider weight one form. F of p, u over q of degree, l, u, p, u. OK, I write a chi less or less notation for the head there. Oh yeah, I needed to, OK, I need to put a metric there. So l, u has a metric at a Hamilton place, but v divides infinity. You see, l, u, when I have c there, is essentially extension contains omega x, u. Now, I mean, it's essentially, when you use small, it's essentially x, u of c at a v. So you just define d z is normally twice of y for z in the upper plane. So that's the way you define. And later on, we'll figure the reason why I put a 2y there. So the theorem that, so this is what I call that, the Chola-Sebel formula in 2008 is that, so it has some assumptions. This assumption probably not that important, but it proves that they assume that the sigma, the kinetic sigma is greater than 3, I guess. And there's no, and they're, OK, I forget about that stuff. So I need to define the same points first. I have not yet defined the same points. So this is the height for arbitrary points. So the same points is the following. So let E over F is the same quadratic extension. So you assume this also, assume that such as that, the E when you take a dose can be embedded to be, OK. So in this case, this is a very interesting thing because this x here I defined as all the project limit of this thing has action by, this has action by of B F cross, right? Because I took all the limits there. So, but this thing, well, of course, they have action by E cross. So, so then if you have this thing, then we can define this x of E cross. So this one is the same point by E. So this is a pro-scheme of dimension zero, OK. And with the action, see, this thing is fixed by E cross. But, primarily, this thing has action by even this part, right? So this actually has action by, with action by E whatever, turns the F cross because this one commutes with that. So, so this is very interesting. I mean, this is the following way. It can be a really beautiful way to describe there. So there, what is the reciprocity law? Reciprocity law is that this x E cross, of course, is a scheme, is a scheme defined as F, right? Because I'm just a fixed point. Pro-scheme, yeah, the pro-scheme defined as F. Pro-scheme is also a scheme. In this one, a scheme. Because every is finite, or more than finite, right? So this is actually, so this has action by, if you, so x E cross, take F bar, has action by Galois group, F bar over F, right? Also has action by, as I just said, E, let's write E hat cross. E hat cross, it's a simple way. Then write this one equal to E hat cross. So you have two actions. These two actions commute to each other. And then, of course, then you have reciprocity law. The reason is this action, this action is to make this one to be homogeneous space. So this is actually, I somewhat forget you. So there is a homogeneous space of E hat cross, modular. Let me try to see. Module F cross, module F hat cross. F cross hat is what? It's just a bar. Bar means the closure of F cross and AF, I mean, F hat cross. That's a complicated thing. The reason is that over Q, this thing is the same. Here is a different. So this thing, it's somehow isomorphic too. So these two actions, action with this one, this one, this is a pretty homogeneous space. So this is induced on homomorphism of Galois F bar of F to this E of this thing. So in fact, this is factor 0, Galois of E AB of F. So this is the reciprocity law you get. So this is the standard. Well, sorry, there is a something because my E does not exist there. It's fixed point by E. I'm a bit confused about what I'm doing here. Say it again. You already fixed by E. Yeah, so this is fixed by E. What? We don't see the reflex field. Reflex field? Of the CM point. So we should have an action of Galois of the reflex field of the CM point. OK. So maybe what I'm trying to say is I probably get something wrong. So here has this one. Let me write it more precisely. We can go back to correct it. So this X E cross has two orbit and Galois E bar of E. So they just have two orbit there. So each orbit is a homogeneous space, principle homogeneous space, and a Galois group of E AB of E. So that's what I'm trying to say. I'm a bit confused what I'm writing here. Because this is not a standard class field theory, right? Because this is not a bilion, right? So what I'm trying to say is the principle homogeneous space and the action is given by the reciprocity law, the standard reciprocity law, the Galois group, E AB of E isomorphic to, well, E hat cross, right? Modular E cross, I usually take a bar, right? Because I want to make sure that, yeah. So that's what I really mean. But this one is equal to E cross F cross and a bar. It's the same. OK, so now I write a theorem. So this is the UN 2008. So basically say if that P be any point of X E cross, then, well, first we assume a bunch of things, right? But assume that, as I said, the cardinality is equal to 3 and no place. Let me try to say that the discriminant, so there's no place of F is ramified. No finite place ramified in both B and E. OK? So this one, just a basic assumption. If P is the same points, then there's a height of P is equal to negative 1 half L prime of eta 0, eta 0 minus 1 quarter log discriminant, that have E, that have F, plus log data of B. So there's a data, E data F, data of B, absolute discriminant. So we're not going to describe. So you compare with the, you see, this is the term. It depends on quaternion algebra. Every other term seems completely free there. OK, so this is pretty much a very simple formula there. And as Emmanuel just asked, the risk procedure law, for this kind of Schmuhrer curve is simple. It's very simple. The way we describe it is very simple. So if you want to form a hard theory point of view, for example, if I take it at B of V, so that even started with this group, if you transfer real numbers, this isomorphic to M2 of real cross this Hamiltonian G minus 1. So the hard theory, in this case, is given by the simple way. If C cross goes to this group, cross will be GL2 of real number cross H, G minus 1, is given by much a simple thing. If we write Z equals X plus YI, this goes to, OK, I want this thing to be have weight minus 1. So we get of X, Y, next Y, X, right? That's the way you get. Then you have 1 there, minus 1. So that's a hard theory. So this simple hard, so this has a weight, negative 1 here, I mean 0 here. So this is guaranteed this is not a PR type, because you have it with 0 somewhere. But it's only have one component here. So these make a reciprocative law very simple. OK, so this is a very complicated sense. The reason I spent much of time to talk about that is that if you just for some, if I graduate students, probably never see this shmuraka, so I spend some time. So this is the simplest curve, simplest generalization of modular curve, where the reciprocative law is just as simple as possible. And the typical situation people work on is the PR type. PR type, the advantage you have a billion varieties there, the disadvantage that the reciprocative law usually is more complicated. Another thing is the important, this shmuraka does not parametrize anything, not even hard structure. The reason is this stupid F cross bar there. But if you replace group by smaller group, you say I don't want to make such a bigger group, well you have bigger problem. So if I change the structure, let me see how can I do. If I, it's harder for me, if I can cut this group by say determinants is in queue, right, then parametrize, but I have to change the structure as well. Can there is a construction by duty where you can change the center of... Change the center, change something you can make so that it parametrizes some hard structure. But it's a parametric hard structure in infinite level. Yeah, so you can put infinite level so that there's no bar there. Then at certain infinite level, then you have hard structure. You can parametrize many things. So this is a very interesting phenomenon. So you get a shmuraka does not parametrize anything in the final level. You have moved to infinite level. Okay, so that's been so much time here. So now I want to, so the classical, I talk about classical, try to save a formula. So I'm just to give you one idea. So that's at least something you can do. So the F equals the Q, sigma is just one place. That's the case we're doing. So this one, the B of infinity is just the matrix, right? So we get, we are also the Q cross of a bar is the Q cross. Everything's very nice. So that's probably only a case it happened, right? So we will get, we will get G2 of Q bar, Q hat will be the standard decomposition, G2 Q, G2 of Z hat. So in this way you get X infinity, this project limit. So this is the limit X of N of C. If you take all the project limit, you get this thing. G2 of Z join with up or lower half plan. So this is the C minus R cross G2 of Z hat. Okay? So this has many components, right? And so this is actually, this one, this is a modular X infinity of C is a modular of elliptic curves of A phi. So A over C is the elliptic curve. A phi is just a trivialization of a tether module, right? It just trivializes your tether module. But if I don't put a hat here, if I no hat here, that's upper plan, right? So you say, if I don't have this thing, I just want trivialize this, then I get a upper plan. But if I put a hat there, then I get a G2 G hat, right? So if I know, so this C take R out, this is a modular without a hat there. Okay, so now I'm thinking about the proof, and the original proof. So let's see. So now I try the original proof of Charles Ebert. So I had an original, I mean, it's just the classical proof. I don't know how original it is. So the idea is the following. Idea is we can describe the shift very well. Idea is that this X1, this is actually the bundle of degree, is one over 12. This is the bundle of C, it's a fractional bundle. So 12 of the power, of course, isomorphic to O1, if you think X1 isomorphic to P1, for example. So at least we have one section, right? So then I can write this section precisely as a following. Okay, so this is the basic idea. So to take some more precise, let's do the following. So there's this X1, it's a parametrized modular stack of univtacus, so we have a universal family. I write A here, so A of X1, so this is the universal family. And the first one, there's a Kodara Spencer map, omega A of X1, right? So this is, define this one to be, you have a section here, E, you pull back omega A of X1, then you have omega A of X1, you have to transfer two, this is isomorphic to the Hajj bundle at infinity. All right, so this is the stack you want. And the interesting thing you have to describe this one in very precise way, you can normalize this isomorphic in the tether curve, at a tether curve, the tether curve is the GM, right? Module Q of Z, so you find A of Q, for example, so this is the curve over Z in the power series, Lorentz series of Q, right? At tether curve, you can just describe this morphism precisely, so then write this morphism called the Kodara Spencer map. Should that be a push forward, push forward by the universal map? It's the same. Well, you want it to be a sheep on the set first. I don't want it to be a sheep on X1. Yes. It's M is X1. Then it's X1. Say again, I don't understand. Then it's M down, on the one hand. Oh, yeah, X1. Yeah, it's just, it's the same thing. The other one is, but if you finish it, oh, it's going to go back, but it's the same thing. It's the same, because the Ivarian... No, this is a shift on here, right? I pull back, get a shift X1, but you need a section. No, if you have a scheme X or Y, omega X or Y is not a shift on Y. It's a shift on X. Yes, but you want a shift on Y. Yeah, so I have a user section to pull it back, right? It basically is the view of the algebra, right? OK. Coolly algebra would have somebody call that. But you're right. I mean, right this way, for elliptic curve, there's no difference. For narrow model, there is a difference. But narrow model has many connected components. So you have to determine, you know, you take a push forward, your shift may have more sections than the section come from Orinian, right? So I just need to describe the sense, the Cordera-Spencer map. I want a Cordera-Spencer map of du over u square equals dq over q, OK? So that's the normalization you supposed to get, right? So u is the coordinate of GM. du over u is the invariant differential of aq. Then dq over q is the differential here. So it's of a complex number. For example, if you take z over q map to complex number, right? For example, the q goes to become eta 2 pi i of tau. Tau's alpha plan, if we do this thing, then you will get the aq, maybe c, z plus z tau, right? z tau plus z, right? If we use this coordinate, then the Cordera-Spencer map of 2 pi i d z square, maybe 2 pi i d tau. So that's an interesting thing. So this is why many 2 pi i here be really careful. How many 2 pi i are we going to do? And so, OK, so now from this one you can calculate. So we know this one has a metric. This bundle has a metric using, so let's say omega a has a metric, plus a metric, right? v plus a metric, alpha equals 2 pi plus i here, I guess. d, I mean, alpha wedge, alpha bond of a tau, a of c, right? Suppose this alpha is a section of a elliptic curve, just take the wedge product, right? So this side has a metric. Induce this side as a metric. So this metric, you can show that ks of, I mean, ks is isometric when the d tau has a norm defined by d tau equals twice of imaginary of tau. You can do the calculation figure out this will be the truth. So this will give you a well-defined thing here. So I just want to give you a concrete calculation in a modular curve case. The next thing I want to do, OK, so I need to calculate the fault in the height. The fault in the height is so, OK, so the omega a over 12, as I say, the degree 1 has a section, non-zero section. The section is given by the following way, is essentially, and is given by L equals to data of q du over u the 12 of the power, OK? So you can write down this way. So that's usually, I mean, so section, whose expansion, yeah, whose restriction on tether curve is given by this way. So this is the q product 1 minus q to the n, m from 1 to infinity du over u, the 12 of the power. The Coderre Spencer map of L will be more interesting, right? I call it this, what are we doing? It becomes q 1 minus q to the n 24, yeah, sorry, 24 there, dq over q to the sixth power, right? So that's the one you want to do. OK, so this is interesting. So we have a concrete section, our loss to compute a fault in the height. So now, so if a, for example, if a over L is on elliptic curve with semi-stable reduction, then, of course, this defined a morphism, right? This is defined a morphism spec of over L to x of 1, right? So you can pull back to the shift, right? So what do you can pull back to the following thing? Well, so this L, so you pull back, say, this is whatever, it's i here, i pull back of L, the device of this thing is the usual, it's exactly the, the neuron differential. This is discriminant, data of a, so this is the discriminant device of a. So because of this reason, you can compute a fault in the height in a rather explicit way. So this gives you explicit computation. So the 12 L over q fault in the height of a will be the degree of omega a, the 12 of the power, right? So that's just doing it. And this will, if you write everything down, be the log of the discriminant error of q, the discriminant of a, the minus summation of sigma L to c, a log L to sigma here. So it's just to compute the height. Well, I mean the error of sigma is nothing, it's just a log of delta q sigma and the 4 pi imaginary of tau sigma, whatever, 6 power, right? So you get a very explicit formula for fault in the height. So if your elliptic curve has cm, so if e, if a has cm, then of course, a good reduction of delta a is the trivial. So we don't have any sense of the fault in the height of a just 1 minus, so a error can be right at h, the behavioral class field. Then you have very much a simple formula. You have 12 there, h of q summation of sigma h of c there, log of delta q to sigma imaginary of tau sigma 4, whatever, 6 of this sense. So this one is a modular function. It's just a well-defined function of our plan, right? So the question will become, compute fault in the height become, how can I compute this function? And you can do naively. The reason is this is an infinite product, right? You can go there, just take a formal expansion, take a log, compute it in. Then you immediately realize that this is a kind of constant series. So in fact, so this is a modular function by Conex Limiter Formula. You can prove Conex Limiter Formula pretty easily. If you write it down, you feel, let me write it down there. But Conex Limiter Formula you will have, if we define a standard e tau of s to be summation mn not equal to 0, 0 of 1 to the s m tau plus n to the 2s. OK, so this is for s, a real part of s bigger than 1. It's convergent. It has a pole at s equals 1. It has a pole at s equals 1. So if you take the pole out, this is the formula, so e tau s minus pi of s minus 1. If you take expansion of that, you will get the formula imaginary, I guess that's exactly what we want of tau of 6 power plus o s minus 1. So this gives you an idea. So something you calculated is essentially ascension series. OK, so the proof of Charles' several formula is behind. Evaluate ascension series and cn points. So then it goes by the heck. So you need to calculate summation of e tau of s. For example, if you tau the cn point of discriminant of delta, for example, you calculate this thing. And this is pretty easy to calculate. This is the heck of sin. This is essentially the zeta function of the k of s. So if a has cn by k, so this is just a heck of work. So then you find out, just bring everything inside pretty much immediately you get there. So from this one, you will get the Charles' several formula. So that's the classical proof. Well, the most lucky thing, you have a dedicated eta function. Can be used to calculate for the height. And once you get something so concrete, then another lucky thing, this modular function has something connection to ascension series. For ascension series, you can evaluate the cn points. So there's a sequence of very good stuff happened here. OK, so this method does not work. So this method doesn't work on any other Schmrodinger curve, even for Schmrodinger curve of q. The reason is there's no cusp. You don't have modular form. So that's all this thing happened. But one interesting part of the thing is that if you, for hyperliptic curve, you do have a formula. But I do not see anybody spend time to study. For hyperliptic curve, you can have a formula similar to the formula right there. But I do not see people spend time to study the modular function in the second variable. So you still can get some modular functions. And that's not for genus 2, though, right? For any hyperliptic curve. In fact, you see my paper goes Schmrodinger I have a fault in the high Schmrodinger as a high Gaussian cycle plus some concrete terms. The Gaussian cycle vanishes for hyperliptic curves. So that thing, yeah, so they have everything very concrete. But we do not know how to, I mean, I didn't see people spend time to evaluate this modular function. What is the modular function? What? What is the modular function? You read my paper. So I say there's a case to do that. But there's a good reason you're not supposed to know how to compute a falling height precisely. If you note morally, it means you know how to prove ABC conjecture. Because there's a local term, somehow, unless it's a hyperliptic curve, they always have a secret bundle in the middle. This is the global invariance. There's no way to write a global invariance as some local invariance. So some global invariance has no way to write down the local invariance. If you know how to evaluate that, you will know how to compute the global invariance. The global invariance could be omega square, could be something else. So this is just to give you an idea. So don't try to show that every falling height can be computed. This is impossible. So now I talk about the break a little bit. Maybe take a break for five minutes. So somebody asked me a question. Why any other? I mean, it only works for modular curve, not for Shimura curve, but you see. But somehow, people work in a Shimura curve anyway using something called the Jack-O'-Lanterns correspondence. The Jack-O'-Lanterns correspondence, through speaking, is about talking about their relation between Hecker algebras. So in other words, even there is no explicit construction module form, but they do have explicit construction of Hecker algebra. So the second method I'm going to talk about, we don't use the module form anymore. We will use Hecker algebra to do the calculation. So this is somehow it's implicitly started in there. So let me write down the rough idea. The rough idea, another idea is the Korozake formula. The Korozake formula is about calculating the isogenes between elliptic curves. Instead of, so if you roughly speaking, what are you going to say? Roughly speaking, if you think about it, the falling height of an elliptic curve, as I said, that is a degree of this omega e. But you can think about it if you, this is a one-half of a degree where is omega, so roughly speaking, x1 at e. So the bi-Azengi formula, so this is somehow you can think about it, next to one-half is the p of e dot p of e, the Azengi formula at a junction. So you want to compute the safety section of a boundary there. So the falling height, you can view it as the computation of certain section at a resimated surface, safety section. We don't know how to compute it because, as I said that, because you need a module form to compute it. You need to have a form to push the section away. So this dedicated other function is you use it to push it away. Now you don't know how to do that. Well, but there are ways to calculate, say, the p of e dot tn p of e if, what I'm trying to say, this is a heck of it. If the p of e has no self-n isogenic, because these two are tn of p of e, the summation of t of e prime. So e has an isogenic e prime. Let me try to write it down. Let me write it down. e mod of c, c is a subgroup of e, the cognitive of c is n. You can calculate this thing. So at least you can calculate this thing there. That's the one thing you don't know how to compute. But you can compute the intersection of one lifted curve with this thing, because this is calculated isogenic. So the idea would be that maybe something would know how to compute. It's enough to recover something we don't know how to compute. The reason is a following. The reason is that all of the tn here for n inside n is not a London cycle. They have a relation between them. So this actually, this is a Hecker-Oprah apparatus. And a form, a module form, or way two. So this implies that t1 is a linear combination of ti's for many tms. So this is the secret idea you want to use there. So I want to compute 7. I don't know how to compute it. But I can compute isogenic between two of them. I mean, if you give me two lifted curve, I can calculate how many isogenes they can have. This is not hard, because this is just a module of problem. It's just a content by hand. So this is interesting thing. Well, so the key, well, so this is done. But the calculation, this P e t m P e is the main gradient of the proof, in the proof of Golov-Zagyov formula. So that's the, in some sense, that the Golov-Zagyov formula doing exactly this job, trying to compute their intersection between one elliptic curve and the Hecker-Oprah's and elliptic curve, right? And so secretly, Golov-Zagyov formula must give another proof of Chala-Sever formula. That actually is true. If you go back to read the Golov-Zagyov of the paper, in the middle of somewhere, they already computed it, right? But they also use the dedicated function to compute it. And then when we reread the paper, I find that that part is not necessary. So now, my lecture is, we are going to review the Golov-Zagyov formula that I tell you why it do not need to compute the self-intersection. So let's write down a little bit of Golov-Zagyov formula. OK, so the fixed n be a positive integer. So x0 of n, well, this is just a moduli of elliptic curves. So this is just a moduli of isogeny of, say, a1 to a2 over phi. So this is elliptic curves. So the kernel of phi, you want to basically the g0 of n, the Jacobian of x0 of n. So the tm, the Hecker operators of a1 to a2 is just an older image like that. So c is subgroup of a1, subgroup c joined with kernel of phi is trivial. The cardinality of c is m. So you write down a1 of c goes to phi a2 of c. So it's just Hecker operator defined by this way. So tm acts on x0 Jacobian, g0 of n, and also And so on the tangent model, so omega1 g0 of n over q. So that's exactly the cusp form, g0 of n. So this is the cusp form of a2. So gm, this is the geometric way, and you can x everything. The beautiful thing of this action is the following. The one really key observation that you Hecker operates acts on Jacobian. Its action is follow the action in the cusp forms. So for any 2.x0 of n, you can formulate a generating function to be the summation x take a 0, y tm, y take infinity. I just use the original formulation, 0, y minus infinity. Take the Neuron type of hard pairing, take a qm there, m from 1 to infinity, you can form these theories. The key observation is that zxy is a cusp form of a2. So this is a very interesting thing. And the proof is not very hard. It's just formally followed from this property. It's not difficult thing. So the question is, so the whole work, that's exactly what we want. We want to calculate the Hecker operators of a cm points and a tm cm points. And the first term is something we want to put is a fault in the height. But first term, we don't know how to compute it. But the whole thing together, we know what's going on. So this is a cusp form of a2. OK, so that's great. So this will tell you that the coefficients has a lot of relations among them. So there's many relations. So the major work of Gross-Dagier is to identify this form when both x and y are cm points. So there's no hope. For example, if I opt to give you x and y, I don't believe there's any clue how to write down the cusp form. But x and y are cm points. This cusp form is actually a cm point. And they do this thing as a guideline. So they're motivated by BSD construction. So otherwise, there's no way you can check that. So essential result is the following. So we're not going to write them very precisely. So they found out, I will not write it down. So roughly speaking, they found that z, x, y, the bosom cm, these are absolutely linear combinations of two forms. I mean, OK, so linear combination projection, polymorphic projection of two forms, one is asymptomatic series e tau s times theta of tau. So this is asymptomatic series, a derivative, for example, at 1. So this is e tau s is asymptomatic series of weight equals 1. Theta is a series of weight of 1. So they found out this thing. They found out this thing because they believe BSD conjecture is true. Once they believe it's true, you can write the other side. Then, OK, the whole paper, 150 pages, but no, just calculate. 10 by 10, those two are completely equal. OK, so they identify this thing. So let me write down, OK, so let me write down these parts give a name called phi of x, y. So there is a, so there is a, so anyway, so they are concrete. So the z, x, and y equal phi, x, and y. So this is the main formula. So this is explicit cuss form, where two. And this one I usually give a, called a name called analytical kernel. So this is called a geometric kernel. And this kernel, if your integration of phi against f, usually give you the derivative error function. So the Gonzaga formula, basically, you construct a geometric kernel by learning how to pair in a very cheap way to get a cuss form. Then you're trying to say this cuss form actually will connect to the error function because you believe BSD conjecture. But this side is very explicit. So you see, anyway, before you prove anything, you already know both sides are cuss forms, or where two. So I'm not trying to show that equal. The key observation for us is that to prove two cuss form equal, you don't really need to compute every term. You just compute the difference. It's sufficiently good. So that's exactly what I want to take. Anyway, I'm going to write down the Gonzaga formula. Let me write it down before I do everything. So the consequence of the calculation is so-called the Gonzaga formula. So roughly speaking, if f of f is a morphism, x0 of n to e, then if k is a cm, k over q is a cm, satisfies so-called the Hengelen condition. This Hengelen condition is a name given by a birch. It's not given by Hengelen. Hengelen never used this condition. Actually, when yi, t, and i work on a congruent number problem, we actually go back to original Hengelen paper. We have to drop the Hengelen condition. Hengelen condition basically means the following. Means that there is an idea such that o k of n is a morphism mod nz. So of course, from here, you can find out what's going on. So then, if you have this thing, then you can formulate a c mod n to c mod o of k. You get a point. If you get a this way, you get a point. This will be in such x0 of n. So the formula basically says that this actually is defined over here by the class field H over k. So if you write it on, let's give a name, write a y to be f of x. You take all the conjugates sigma inside a Galois of hk of k. So this will be inside e of k. So you can compute the narrated-heighted pair in there. The Grozegorff formula says that the derivative of e base change to k at 1 is essentially equal to the narrated-heighted pair in y and y, the right k here. So this is the formula almost conjectured by Birch. So according to the Birch's account that Grozegorff proved the formula before they made the conjecture. So in the process of making the conjecture, it's about to send the conjecture to Groz, but they already proved the formula. So this is the same here. Now I want to explain that if the Grozegorff are not working hard enough, they're going to prove the Cholotsebo formula easily. So what is the idea to prove Cholotsebo formula? So one claim I want to say is the following. So one claim is that, well, I want to compute x tm of x, so x is the same point. I want to compute this same. As I say, this one, you have two parts. One is compute the computer, you can write tm of x. You call some parts of x. Let's call the multiplicity, rm of x plus some other tm, say, 0 of x. You can write this way. So this one is just joined with x. You say you're on a generative fiber, generative fiber. So that's something I want to do. So if you compute this thing, you put in this one inside here, you will get rm x x plus tm 0 of x times x. So this one is computable, right? And this one is not computable directly. But you need a differential formula to compute it. You're setting a section, you need to push a section away. So this needs a module forms, differential forms, right? Differential forms to push, needed to push x away. I mean, you need to move your section, seven seconds moved away. This is a beautiful process of following. So you will get this one. So you suppose you will get x tm of x qm. So this is a module form, over 2. You call to summation xxrm of qm plus x tm of 0x qm. So you get some formula like that. So this is computable. OK, so assume I want to prove this thing equal to some phi. I assume this is supposed to prove this one equal to phi. This is some energy kernel, right? I mean, suppose this is known. And I want to compute this thing. The very interesting part is this one is a vector 1 form. So the end of the story is that, well, something I do not know how to compute is actually the vector 1 form. But my equality is a vector 2 form equal to vector 2. Suppose I did this computation. This one, suppose this one equal to phi 0, for example. I compute something equal to phi 0. Then I know this one equal to some phi 1 plus phi 0. Suppose this is a vector 1, right? Suppose this is a vector 1. The difference will be, you know, the difference will be funny. The difference will be this is a vector 1, this is a vector 1. Before I do any calculation, the difference will be a vector 1, right? So there, OK, so let me try to write down here. Write down there the argument. I will go back to give you more precise information. The precise information will be that, so precise information is the following. So this energy kernel gm of the qm, so this is the formula I want to prove because of phi. So before I prove everything, I know both sides are where 2 forms, right? And I know this one equals phi 1 plus phi 0. Phi 1 is a vector 1 form. Phi 0 is nothing. I really don't know what the phi 0 is. Phi 0 is some concrete power series. I also know this side equal to summation x times summation rm qm, right? I know this is a vector 1 plus the summation of x tm 0 of x qm, right? Assume summation x tm 0 x qm equals phi 0, then this way we'll conclude that these two are equal and these two are equal, right? The reason is if this is the true, what will happen? If this is the true, the difference of these two things is that where 2 form, a simultaneous of where 1 form. So where 2 form cannot be equal to where 1 form, unless both of them are vanished, right? Understand? So that's the point. The point is that, so this one, tm x qm equal phi 0 implies the sigma x tm x qm minus phi equals, right? You take a difference of them, is x rm qm minus phi 1. So this side is the where 2. This side is the where 1, right? So that's a very nice thing happen. So the only thing, where 2 form can equal to where 1 form, both of them vanish. So both must vanish. So you get a Kurozake formula and a Chalabal formula simultaneously. But that's actually the proof for Schmura curve. That's the only way for Schmura curve to prove Kurozake formula. You have to prove Chalabal formula simultaneously. You cannot have separate one way by other way. So now I go back to say why this one is where 1 form, right? This is not difficult to calculate. Let's do the calculation. The calculation is rather easy. So you have this fact that they both vanish. What do you conclude from that? Oh, this vanish means the Kurozake formula. Because I prove this one equal this one. This vanish, it means Chalabal formula. So I know how to compute a set up in the section of x. So this will give you Kurozake. This will give you Chalabal. So you're saying you know in advance what should be phi. Right. You know in advance what should be formula. Because this one, I know how to calculate. So some say I do not know how to calculate. I know it's a wet one form. So something I do not know how to calculate, I know what the shape of this thing is. You get rid of 50 pages and what are you? Something like that. So anything you use, data adder function can be erased. You do not really need it for that part. But anyway, when you do Schumerer curve, then you must do it this way. So I mean, it's not because we found this thing. When we work on Schumerer curve, this is the first thing we have to think about it, self-intersection. So self-intersection is always the most difficult thing to handle. And when Brilinsky even worked on high ways, he could not even define the self-intersection. So this is why RLKL theory is very important for this Kurozake type thing. So I need to calculate. I will finish my lecture by calculating the RM. Because this is an interesting thing. It is a completely elementary. So you want to compute that RM is a multiplicity of x in Tm of x. So this is what is called the cardinality of isogenicity between phi e equals to e. Suppose x represents e. The e, you say, is the c modulo of k. Everything is very concrete. So the kernel of phi is an esomorphic to, no, cardinality equals m. Degree of phi equals m. It's a very easy thing. So OK, if the phi is equal to e, so there was nothing, it just phi inside o of k because n of e is k. So there's a norm of phi is m. So that's exactly, well, that's the way one forms. This is just a part of it. So this is RM. It's just a part of element in o of k whose norm is m. So this is, so we know there's a summation q to the norm of x, x inside o of k plus maybe plus something, I don't know, w over 2, whatever. So this is the Wetter 1. It's a set of series of Wetter 1. So this is trivially true. So the sum of things we do not know how to calculate is given a Wetter 1 form inside. But our calculation about a Wetter 2 forms. OK, so I didn't talk about it. The proof, of course, for Shimura curve now is very clear. You put it in the Shimura curve, you compute it the same. So in general case, yeah, so this method for Shimura curve follows, no, for Shimura curve. So this method, yeah, this method works for Shimura curve. So because we do not need differential form to calculate selfless section. So now we ask, what's the secret that we use here? The secret we use is the following. Secretly use that we do not have the section, obvious section for differential forms. But we have, we can move HECO operators. We secretly use that a Tn, the T1 can be right on the summation of Tmi. I mean, basically, what do we use? We move a cycle by this end, r for i. So somehow, you are working on the x tm, xx, you can write down sigma x tm ix. Somehow, this side, you know how to use it. So secretly, we move HECO operators instead of doing differential forms. And this second philosophy also agree with the other group of the proof, the proof by Andrietta Bernhoud, and Goren, and Mata Pussy, Perra. What do they do? They make it even more explicit. They're using Boucher's method. The Boucher's method is the construction sections of differential forms. Section differential forms with precise zeros and poles. But one disadvantage of the method is that only it can work for Schmrock variety of orthogonal type over Q. So you cannot really use that method to do anything more than that. But this method is using modularity of HECO operators. But that's like a much more common thing happened there. And recently, I'm trying to do some story for U21. And so maybe in November, if I finish successfully, I can talk about the proof, maybe the focal length, or some hentamodular, some rational numbers. Looks like it's doable. OK, so I should stop here. In this case, somehow we have to know with advance what is with fire and where you'll get to know what you should find. OK, so that's a very interesting thing. You asked the right question. So when Grocer's idea that they are proof, that they're completely motivated by BSD. When we do it, we're not modified BSD. We're modified by Vaspirier formula. So the Vaspirier formula give us exactly the same thing. So in 1982, maybe 86. So independently, there's a Grocer's idea of work and the Vaspirier work. Vaspirier's work is a completely general for arbitrary Schmura, arbitrary Quaternion algebra would have. And for a long time, we do not know that's the relation. So in fact, the Grocer's even wrote their own formula against a similar time, completely trying to figure out what the Grocer's idea is. And probably after 10 years, how many years? Probably after 10 years, we know the Grocer's formula subset of the Vaspirier formula. So the Schmura curve in our book is just to use this formulation. So this is over x0 of n. So then we get the diagram. So this is the local intersection. Then there's a fiber product, where is the Grocer's idea of a Schmura curve? So this is done in the 2000, 13. It's very long late. So it's exactly formulation there. So what's crocheted is in the compact case, no? Vaspirier is no assumption. No assumption. No assumption on anything. But Vaspirier is not working on any Schmura curve. We're going to Quaternion algebra, the modular form of Quaternion algebra. So there is a recipe. So the Vaspirier's way is the following. It started with any e of f, any quadratic extension of number fields. It starts with any representation of pi for gl2 of f, and a chi gl1 of e, you start two representations. And you make only such as that the pi times chi is a kind of self-dual. That's the only condition you assume, because you're going to work on the center of your error function. Once you have this thing, this will give you a sigma. Sigma is the final set of places in the beginning of my lecture. In Vaspirier's case, the sigma is always even, because he started with the Quaternion of a field. In my case, sigma is always odd. So you can say the formula we proved is called arithmetical Vaspirier formula, for example. So that's the same thing the other side completely. So this is not really modified by BSD. For example, actually, this point of view is important, because the next lecture I will talk about Yun and Zhang's work, Zhu Wei Yun and Wei Zhang's work. They figured out how to do arbitrary derivative. But that's nothing. BSD doesn't cover arbitrary derivative. BSD supports the cover of the leading term. So the Gorosage formula is one very special formula in a very big spectrum there. And there is the same thing. You need to calculate instead of computing the isogenic between elliptic curve and you just computing the isogenic between stuccoes. And again, we don't know how to compute the self-intersection of stuccoes. So you need a high cooperative to move them away. Yeah, so you ask a really nice question. So the new one is not modified by BSD at all.