 We are happy to welcome you to one more session of the Math Associates, and not from the ITP. Our speaker today is Ayel Pachukti from the Universidad Nacional de Córdoba, and he's speaking about modularity of some geometric objects, transmission, reasons, and the state of the art. Please go ahead. Okay, so first of all I want to thank Alejandra for inviting me to give this talk. I was told that this was supposed to be a talk mostly aimed for PhD students and postdocs, so I'm going to try to give some ideas and examples, and not so many proofs. Actually, I don't think I have any proof in my talk, so I apologize for that. So let me start with some simple example. So let's consider the projective line. So I'm going to denote it by P1. And I'm going to take P and E prime and N a positive number. And we know that if we want to count the number of points in P1 of F, P to the N, then such number is P to the N plus 1. I think everyone is more or less happy with this fact. I mean, we know that points are always of the form. We can choose the points of the form A1 and the point 0, 1, sorry, 1, 0. And then you have P to the N possibilities for the first case. And you have an extra one for the second one. And then there has to be a set of function. What you have, what you want to do is when you have a geometric object, you want to count the number of points over the different extensions, all extensions starting from the finite field where your variety is defined. And then you want to put all this information in some analytic function. Well, this is just, at first, it's just a formal power series. And then you want to see if you can extract information from there. And in this particular case, I mean, using this formula over here, we know that we can split the sum. We're going to have a term with the shape P times T to the N over N plus one term, which is just T to the N over N. And then if we use the formula for the logarithm, I'm just talking about formal power series. Then we can see that when you compute this theta function and it has only two times, one term, one minus T, which comes from this contribution, one term, one minus P times T coming from this contribution over here. And this is something that is related to base conjecture. I'm going to say a few words later. But this works for a particular prime. So this is like the first part of the story. And now what we would like to do is to consider all different primes at the same time. So this is the second baby example. What we do is we gather the information from all the local contributions together. And now we don't want this just to be anything, something formal, but we want this to be a natural function. So we have to evaluate and try to put some constraints so that this keeps a complex function. And so what we do is we consider the local pieces we defined before, and we evaluate the variable T in P to the minus S. And this is for conventions. And then we take the product of all primes. And I'm cheating here a little bit when I'm going to talk about the general picture. We have to be a little more careful. I mean, what happens here is that the projective line has good reduction at all primes. So this is why we can just do these without any extra hypotheses. And then remember that we had one minus T and one minus P times T. So when we replace T by P to the minus S, we get one minus P to the minus S times one over one minus P to the one minus S. And this is something that is well known. The first product corresponds to the zeta function at S and the second product corresponds to the zeta function at S minus one. And now if you study the properties of the zeta function, you know that this is a function which converges in some half complex plane. So with a half complex plane, I mean all points whose real part is bigger than something. And in which half plane it converges, it depends on which of the two parts you are looking at. So for example, the first one converges for real part of S greater than one. But what is really interesting of this zeta function, which is something that Riemann proved, is that it can be extended to the whole complex plane. And it satisfies some functional equation. And actually it has a pole at S equals one. That's why you can, the series doesn't converge to the left of S equals one. And it has preceded one. And the other part has a pole in shifting. And we're going to talk a little bit about these shifts and how they appear in different situations. But so what is important here is that you can extend it and you have a functional equation. This is the two important properties of this example. What happens if we want to go to a more general situation? So as I said, I started with the projective line, not the affine line. So we're going to look at smooth projective varieties. So these are given by equations in some projective space of dimension D. And I'm going to work over any finite field. And we can do something similar to what we did before. We start counting the number of points over FQ. Now Q can be a power of P. It depends where you're working. And then you look at all its extensions. So it's given by powers of Q. Then you count the number of points and you put this into this power series. And you look at the exponential. And then you ask what properties does this zeta function satisfy? Remember that in the previous case, it had a really simple expression. It was just a portion of polynomials. It was given by 1 over 1 minus T and 1 minus P times T. So you might wonder what happens in general. And so this is part of waste-based conjecture. So it has more or less three different parts. The first part of the conjecture is that it's given by a rational function. So by a rational function, it means that it's a quotient of polynomials. This part was proved by Borg around 1960. And note that it has some indices, the polynomials, P0, P2D, P1, P2D minus 1. So the even ones goes in the denominator. And the odd ones goes in the numerator. And each one of these polynomials have some particular properties. They're integral polynomials, so they have integer coefficients. And the zeta function satisfies a functional equation, something similar to what it happened before. And what's most important is that in each one of these polynomials, the roots have absolute value q to the i over 2. And here, let me explain just a little bit what I mean with this part over here. So this is something that is relating to different objects. So suppose you start with some variety that it's defined over some number field, for example, over the rational or something like this. Then you can look at its reduction and then you're working over a finite field over fq. But also you can look at it, for example, as a complex variety, and you can just compute some other comologies, for example, like the veticomology. And what it's telling you here is that if you compute the veticomology of your global object, then its dimension coincides with the degree of your polynomial. So recall that we started with P1, right? So it's the projected plane. So P1 has dimension 1, and we can compute the veticomology. H0 is 1, H1 is 0, and H2 is 1, right? So what it's going to say, this conjecture is that when you want to compute the theta function, it has to be a quotient, well, d is equal to 1, so you have P0, P1, and P2. Since H1 is 0, it means that P1 has degree 0, so there is nothing in the numerator, which is exactly what happened in a previous situation. Since the H0 is 1, it means that you need to have a degree 1 polynomial, and if we look at the third condition here, it says that the root has to have absolute value q to the 0. So it has to be the polynomial 1 minus t. And the last fact over here means that the P2 is also degree 1, and its roots needs to have absolute value q to the 1 to the first power, so then it's given by 1 minus p times t, or q times t, right? So the root is 1 over q, which has absolute value q. So what this base conjecture says is that you can read all the information, and it has this prescribed properties. Of course, knowing just the absolute value of the roots of a polynomial doesn't give you the polynomial in general. So the projective line is a very particular, easy example. But also, for example, you can construct this data function just by counting points, and in particular, you can recover the beta numbers from this formula. So it combines a combinatorial part with a geometric part, so this is really interesting. Okay, and then we can do something. Okay, so let me show you an example, and not so easy example. And as I said, most of the talk will be about examples. So let's start with an elliptic curve. An elliptic curve is just an equation of this form. It's y squared equals a cubic polynomial, and I'm going to assume the characteristic of our base field is not two or three, so I can write it in an easy way. It really doesn't matter. Now, if we go back to base conjecture, then we can compute the beta numbers of our elliptic curve. The h0 is equals to the h2, which is one, but the h1 is two. So what this means is that if I want to compute the theta function, then I'm going to have, well, two parts in the denominator, one corresponding to the h0 and one corresponding to the h2, which are already given by this information and the information on the absolute value of the roots. And then in the numerator, I'm going to have a degree two polynomial. Okay, and I know that the roots have absolute value q to the one-half, so the product has half absolute value q. And in particular, for the theta function to be given by this particular portion of polynomials, it implies that if I want to compute the number of points, then it has to be given by this particular formula, where alpha and beta are the roots of the polynomial appearing in the numerator. So this somehow, you get this for free from base conjectures, but you don't know which are the numbers alpha and beta. And so what you can do, know that, okay, know that if you call a q of e, then the sum of the two roots. Now, if you compute the number of points, you look at this formula here and it tells you that the number of points is going to be one plus q minus the aq of e, the sum of the two roots. And in particular, if you just count the number of roots, so the number of points of your elliptic curve mod q, then you know the number of roots over any extension of fq. So you can recover, it's not that you can only recover your theta function, but you can also give a formula for the number of points over any finite extension of fq, okay? And of course, the way to prove it is going all the way around, right? You start counting points and you prove that the formula holds, and you can see from how I'm doing like reverse engineering. So once I know what the expected results should be, I can show you how you get this information. And I want to stress something important here is that for computing the whole theta function, you only need to count the number of points of your elliptic curve over fq, okay? In general, if your polynomial has degree, has higher degree, then this will not be enough, right? Because in this case, we want to compute two numbers, alpha and beta. And the point is that we know that the product has to be q, and we're computing in sum. So since I only have two numbers, if I know the sum and the product, I can know the two of them precisely. But if I have, I don't know, ten different roots, then I need to start counting over many extensions, and then these computations becomes extremely challenging from a computational point of view. Okay. I don't know if there are questions if I'm supposed to ask for questions, or I should go ahead. Alejandra? Go ahead. Okay. Okay. If there is any questions, just post it, and Alejandra will let me know. Yes, anything. Okay. So what do we do now? So we just do the same thing. We define the global has a bail, a theta function. And what we do now is we just take the product of all the local contributions and we do exactly the same thing as we did for the projected plane. We substitute variables at variable t in p to the minus s. And here we have to be a little careful with the primes where our smooth projective variety has bad reduction. Okay. So this is something that I'm not going to talk about it, but you have to think that if you, if you start with a variety, then there are many different instances that may occur. For example, it might be that you start with an equation which gives you something which is a singular when you reduce mod p, but it is something, but the issue is that the question is not good enough. Enough. You might take a change of variables and get a better equation where now your curve or your variety has good reduction. So this is something that might occur. And so you have to be really careful with this. But at least for all the primes of good reduction where, where you take your equations and you reduce mod p and you get the non singular variety, then you do precisely the same thing. Okay. And now what happens is that now the theta function, it converges for some half plane. Okay. So the situation is pretty much the same as it happened with the primes theta function. And it comes from the fact that we know base conjecture. And so, so we know on these parts here that they appear different polynomials and we know the absolute value of its roots. And this will give you, give us enough information to be sure that this converges in some half plane. And so the conjecture is that what happens with this theta function and we expect all of them to extend to a metamorphic function on the complex numbers. And we also expect it to satisfy some functional equation. Okay. So remember that here we have to be a little careful. I'm being a little vague here. For example, when we computed the theta function of p1, remember that it was given as the second function of s times the theta function of s minus one. Okay. So when we saying a functional equation, so, I mean, I cannot put the function equation for the two pieces together, right? I mean, each one satisfies a different functional equation, but still they do satisfy functional equations and the functional equation should be something that has to do with the different in the comology degrees that appear. So I will come back to this later. Okay. So let's go back to elliptic curves. If I start with my elliptic curve, then I can compute the theta function and remember that when I computed the theta function of the elliptic curve in terms of t, then it had a part which was like one minus t and one minus p times t. And in the numerator I had a degree two polynomial. Okay. Do you want to add something Alejandra? No. Okay. So in the numerator I had a degree two polynomial. So these two parts here will give me again the theta function at s and the theta function at s minus one. And the part coming from the degree two polynomial is what I'm going to call the L series of the my elliptic curve. So by definition, it is given by the product over all primes, just the good ones of one divided by one minus a p of e times p to the minus s. Remember that I had a t here and then p to the one minus two s because I have p times t square. Okay. And this converges for the real part of s greater than three halves. And these three halves here comes from the estimate that the absolute value of a p of e is less than a two square root of p. So this is has a spell. And now the natural question is that so I know that the two terms in the numerator, which comes from the theta function, I know that they do extend to the whole complex numbers and they satisfy the functional equations and this is due to Riemann. And what happens with the denominator? Can I say or prove that this is a series extends to the whole complex numbers? And furthermore, I mean, is it automorphic or it has poles because for theta function, we expect poles to appear. But what happens with this series? So the story goes somehow this way. If I start with a curve which has complex multiplication. So complex multiplication means that the automorphism rings. So all the maps from the elliptic curve to itself is bigger than just multiplying by integers. Then what happens is that the series in this case is related somehow to something similar to the theta function, which is you have to add a character, a character. And it factors as the product of the series of two characters, one character needs conjugate. And what happens in this case is that you get the extension from the extension of the series of the character. So I think this is due to Heke mostly, but I don't know the precise reference. So using a similar proof than that given by Riemann, you can prove that if your curve has complex multiplication, then the series satisfies expected properties, but it is automorphic. This is something that happens with any character. So if you start also with a Dirichlet character, then the series extends automorphically. It doesn't have a pole. The post comes from the theta function, not from the series given by Dirichlet characters. Okay. And this will happen for high dimensional things, but if the curve doesn't have complex multiplications, so these ideas of Riemann are not enough. We really don't know how to extend this series or how to prove anything out of them. So I want to change subjects a little bit completely, and I want to mention what we know how to do. Okay. So I'm sure most of you already read something on this, but let me just recall what is a modular form. I'm being baked all around the talk, so forgive me for that. So I'm going to take K to be a positive integer and a weight K modular form. It's just a form from the upper half plane. Okay. So the complex numbers with positive imaginary part to the complex numbers would satisfy some functional equation, which is given by if you take a matrix A, B, C, D and S2 of C, we call that S2 of C at two by two matrices with integral coefficients and determinant one. Then this group adds by a omotetes, I think it's called. So sending C to AC plus B over CC plus D. And what you want to do is when you evaluate the function F at an image of an omotete, you want it to be related to the function to the value of F at C. Okay. And it changes by CC plus D to the K. So the story of these functions come from then to understand all of the differentials on quotient of the upper half plane. And for, I just wrote it for S2 of C, but we'll also be interested in subgroups of this, like gamma naught of N or gamma one of N, where gamma naught of N, you ask, you impose the extra conditions that the place two one is divisible by N. See, it's a little smaller subgroup. So it doesn't matter. And also you have to impose some olomorphic condition at CASPs. Okay. So I will say just one word about this. So if you look with this definition, if you look at the matrix 1101, it acts sending C to C plus one. And this part here is trivial. So it means that your function is invariant and the translation. And in particular, it has a fully expansion, which is given by some sum of powers of e to the two by I N. If you take another subgroup like gamma one of N, you have to be a little careful and then something may appear in the denominator, but it's not important. And the condition of binolomorphic implies that I want the sum to start from zero. This is the definition. Okay. And for most of the things I'm going to say, I will only consider forms which are called CAS forms, in which case you also impose the condition that the a zero is equals to zero. Okay. This has to do with the olomorphic differential being olomorphic at the CASP, but I don't want to get too technical. And then to a modular form, you can attach what is called the melin transform. It's given by a transform, but it's not important how it's defined. It's an integral that you have to take a convolution with some exponential function. But the result is what's called the N series of the N function, which is given by the same sum. Note that the coefficients are the same ones as I had before. I start from one because I only consider CASP at once, but then you have N to the S in the denominator instead of an exponential. And again, you can prove that if you, your form is a modular form of weight K, then this sum converges for some half plane. Okay. Real part of S bigger than one plus K over two. Okay. I don't want to, to give all the definitions, but there's something important that you, you have an involution in this space. It's roughly speaking given by the matrix zero minus one and zero. If you look at the level gamma zero of N and it's, since it's an involution, so it squares the identity, then you can split the space between the forms which has eigenvalue one and eigenvalue minus one, or you have to add something if you look gamma one. And then looking at, if you start with a form which is an eigenform for this involution, then you can prove that it's a series extends to allomorphic function on the whole complex plane. Okay. And of course, since if you want only extension, then you don't need to be an involution because any form can be written as sum of eigenforms. So if you can extend the eigenforce, you can extend all of them. But the important part here is that it also satisfies a functional equation, which is what we are expecting coming from the Riemann theta function. Okay. So extension holds for all of them. But if you want some functional equation, you need to stick to some particular ones. Okay. So this is something which is somehow related to Riemann's theta function because you have something which is like a Dirichlet sum. And then you know that you have functional equation and you have that extension. It can be extended. Okay. And so what is the goal? The goal is to try to relate these analytic functions with geometry. So in particular, if I can prove that if I start with, I don't know, something like an elliptic curve and the series matches the series of one of these forms over here, then I get extension and functional equation for free. So somehow the analytic side propagates its nice properties to the geometric side. And this is somehow the goal. So let me go back to elliptic curves. So I didn't write it here. K is going to be equals to two. So suppose we start with an eigenform for the headquarters. It doesn't matter too much what it is. I would say a few words. And let's suppose that it has rational free expansions. Okay. So remember that when we started with a modular form, you can write it as a free expansion. A n times e to the two pi i n. So I'm asking for these numbers here to be rational. So you can always normalize your form so that all these coefficients lie in a finite extension in a number field, the finite extensions of q. And I'm asking them for precisely be rational. And in this situation, then there is a construction due to Eichler and Shimura. So it comes from an old construction of Abel and Chakobi that to a modular form, you can attach an elliptic curve. Okay. And this is pretty explicit. It comes from how the complex points of an elliptic curve are at Taurus. So they have genus one. And so what you do is you just compute this by integrating some periods against the differential form, the olomorphic differential coming from your modular form. And when you compute this integral, it gives you a period lattice. And then the elliptic curve is just the quotient of the complex numbers by this period lattice. And an important property that Eichler and Shimura proved is that the L series matched. Okay. So they prove that you have to be very careful with the price of bad reduction. But that's still okay. And what these Hiccup readers have to do. So how what happens is that you have this nice modular curves, which are the quotients XC of N, which are just the quotients of the upper half plane by gamma naught of N. And this is a curve. So you can look at this Chakobi and when you look at the Chakobi and it's a variety of really high dimension. But what happens is that this Chakobi has a lot of endomorphisms appearing on them. And so here you have something acting, which is what is called the Hiccup readers. And so since you have all these endomorphisms in your variety, then you can try to split it into different pieces according to these endomorphisms. And what happens is that this part here is a search in us to products of different billion varieties, which are most less useful. And your elliptic curve appears somehow in this way. So this is where the Hiccup operators appear. And as I said, I don't want to get too technical. So the point that here is that you have a rule that if you start with an eigenform with rational Fourier coefficients, you can attach to it an elliptic curve. And then the Schumann-Tanijama conjecture or questions, but it was a conjecture, is whether all curves or rational curves appear in this way. So is it true that you can construct any elliptic curve from modular phones? And so this has a lot of advantages. Of course, one of them is that we can translate since DL series matched and we know that the left-hand side can be extended, it has an elliptic continuations. We get the same for the right-hand side for free, but also it has other implications that maybe they're not so trivial. What happens, for example, if you want to make tables of elliptic curves? So if you want to make tables, you want to order them in some reasonable way. There's something called a conductor. And then you want to be sure whether your tables are complete or not. And somehow the point is that computing modular phones is easy. You just have to do some computation. I mean, you have to triangulate some variety, some curve. And so in particular, if this is true, you can always also prove that you have complete tables of elliptic curves, which is what happens, for example, in the LMFDB where we're looking at all Cormona's tables. And the answer to this question is yes. And this is pretty much the theorem of Weiss and Taylor Weiss. And what they proved is that this is true for semi-stable elliptic curves, but this will later generalize to all elliptic guys by a real Cormon and Taylor. And in particular, the Hasselbeckon texture is true. The series of an elliptic curve does extend and it satisfies a functional equation. So what happens if we want to generalize this strategy? So we started with objects, with geometric objects. We only consider genus 0, like P1, genus 1, like elliptic curves. And we didn't do anything else. So if we want to generalize these results, we can go in two different directions. So somehow the point here is we have an analytic part and we have a geometric part. And the question is when we want to generalize some, in one side some computations are easier than on the other one. For example, we can instead of working over Q, we can say what happens if I take an extension of Q. So on the geometric side, then it's clear what we have to do because we start with a schema of a variety of a K, which has some points. So it's clear what we're doing. But what happens here? What should we put instead of modular forms? So if we want to change the base field, the analytic side is a little harder to describe. So it's convenient to talk about the modular forms in the automorphic language. And what you have to do is you have to look at representations of some groups, of some redacted groups. And in the case of elliptic curves, they correspond to automorphic forms on the other groups of Q. And then once you understand this, which I think it was a huge step in history, once you understand what you're trying to do, then if you want to increase your base field, then what you have to do is you look at automorphic forms or automorphic representations and you just change the Q by K. But it's not completely clear from the picture of modular forms. But I mean, it comes from, you can look at the upper half plane as a quotient of GL2 of R, and then this is more or less where this GL2 group appears. And so what do we know in this particular situation? Well, there are some interesting results during the last couple of years if we have a field K, which is totally real. So totally real means that in turn, you can look at any field, you can write it as adding some element to Q number fields. And then what you're asking, this is going to be root of a polynomial and you ask all the roots of this polynomial to be real. This is what a totally real field means. And if it is quadratic or of degree three, then we do know modularity of ellipticals. So for real quadratic fields, this is a theorem of Freitas, Leung and Sechsek. And for cubic fields, it is a result of Derrick, Nachman and Sechsek. And this is from 2018, I think, or something like this. So has a conjecture R true, but if we take a totally real field or a CM extension, a CM extension is a totally imaginary extension of a totally real field, then what we do know is that any elliptic curve over K is what's called potentially modular. So the notion of potentially modular, I think that came, I think this was an idea of Taylor. And it means, I don't want to give the precise definition, but it means that it is attached to a modular form, not over K, but over some field extension and discussed to do with how you prove modularity. You don't get risk of modularity otherwise. But there is a really nice result and all results of Braga that implies that if you prove that something is potentially modular, then you can prove extension of the Hasse-Bael theta function. But the part that you do not get, you do not get allomorphicity. You only get that you can extend it in a meromorphic way. You do get functional equation, but you might have poles. So the idea is that you write your Hasse-Bael function as a portion of different Hasse-Bael functions. Each one of them are allomorphic, but the denominator might have zeros. So we do expect things to be allomorphic, but in this case, we only get meromorphic continuation. Okay? And this is... So the CM case, it's really hard one. It's also from the 2018, and it has many authors, so I'm going to mention all of them. Okay. And if you want to look at imaginary quadratic fields, then you can somehow try to prove modularity for each given case. So this is something that Cremona has been doing, computing tables of Bianchi modular forms. And what we did with Luis Dulefe and Lucio Geberov was to prove that most of them, many of them are modular, giving an algorithm to check modularity, and then you can replace meromorphic by allomorphic. Okay. So let me talk a little bit about what happens when we try to compute the data function. So we're going to increase the genus. So if we start with a genus 2 curve, and we try to compute the data function, then the formula is going to be similar. We're going to get two data functions in the numerator, but the denominator is harder. Now it's going to be a degree 4 polynomial. So it's even challenging to try to compute it. Okay? It's not so easy, and it takes more time. The way to think about it is that when you start with a curve and you look at its Jacobian, it's a surface of dimension, the genus. So it's going to be an ability of surface. And the modularity of the Jacobian was conjectured by Yoshida, and Brumer and Kramer somehow gave a refined conjecture, where the main contribution was to give a specific description of the level of the modular form you are searching for. In this case, you're searching for signal modular forms. So you can think of this as automotive representations of GSB4, or you can think of it as functions in some generalized upper half plane. I don't want to get too much into details, because I want to say a few more things. And so they gave a recipe for the precise level where this should appear, and they also made many computations on surfaces giving a list of the ones that appear for the first values in the semi-stable case. And so more or less the idea is, the idea is the same as I mentioned before, you have an analytic object where you know that the series can be extended and it's a very functional equation, and you have a geometric object, and somehow you want to match these two ones. And the way everything starts is that if you know where to search, you can just start playing with examples. So you compute some of the surfaces, and then you go and try to search for the automotive forms and see whether they match or not. So there is a big issue here, which is that all these forms are going to be of weight two, and automotive forms for GSB4 of weight two are non-comological. So this is like trying to compute weight one forms for the ones who know this. And the problem is that they are extremely hard to compute. I mean, you cannot just desolate some variety or something and try to compute in this way. You need extra tricks, and it's really time consuming. So computing the first 100 free coefficients might take a couple of weeks. And what we do now nowadays is that all ability surfaces are potentially modular. This is a result of a boxer, a Gallegari, Guillain Piloni from, I think, 2018 also at the end of 18. And what we did in this joint article with Bruma, Myself, Puer, Tornaria, Boyd and Yuan is to give a recipe to adapt an algorithm, which is faulting search method, to prove modularity of ability surfaces. So if you give me an ability surface with some, not so restricted hypotheses, you need the image of the residual representation at two to be not too small. And you give me a candidate for the automorphi form to prove that the two series actually do match. Okay. And for cups of cheese greater than three, then not much is not. Okay, so this is really a huge jump and there are some reasons. There is a post by Frank Gallegari explaining that nowadays techniques are not enough to do this. Unless the Jacobi has many automorphisms. Remember that if we started with an elliptical with complex multiplication, then it comes from some Hecke character. So in this case you get modularity, but because it's not something genuine, so somehow it's something weird. Okay. And the extra five minutes I have or so, I want to talk about some other examples that appear and I think it's interesting and maybe interesting for the audience particularly. So as I said, I don't want to be too technical in anything, so I'm not going to give the precise definition, but I'm going to say what a Gallegari variety satisfies. So a Gallegari variety of dimension D, what it satisfies is that when you want to compute the dimension of the homology groups, then you have a Hodge decomposition. And what it satisfies is that, so when you do this Hodge diagram, then all the elements in the orders of your diamond, the dimensions are zero. Okay. And the middle one is going to be one. So I'm going to talk about examples mostly. So for example, if you start with a one-dimensional Gallegari variety, then it's going to be something like this. I mean, this condition here imposes no condition at all. D is equal to one, so there is no integer between zero and one, so it doesn't impose any condition. And this condition here imposes that the H10 has to be one. So the Hodge diagram is precisely that of an elliptic curve. So these are precisely the elliptic curves. Well, if you have a two-dimensional Gallegari variety and you look at the Hodge diagram, then you have to put here zeroes. This is the first condition. And the second condition is that you have to put here ones. And, well, the 20 comes from some criteria. It doesn't matter. And note that. So I would like to compute the theta function of this Gallegari variety and see whether it extends or it doesn't extend. And so I'm playing here with something related to the proof of the base conjecture, which is there is a relation between counting points and computing comology. So we already mentioned the beta comology. Now this has to do with the tau comology, where representations appear. And so your theta function is going to, these two parts here are going to give you the terms set of s and set of one minus s in our previous example. And then this part here is somehow like the new interesting and hard part to compute and where we don't know whether they have a better function extends or not. And what happens that the non-severic group, so this comes from cycles inside your variety modulo equivalence. This gives you a contribution to this H2. So this part here is the H2. This is H0, H1, H2, H3 and H4. And you have a duality. So that's why it's symmetric. And these contributions, you know that all these parts there contribute something which is like a theta function. It might not be divine over Q, but it is like some extra theta function factors. This is called the algebraic part. And then you have something new, which is called the transcendental part. Okay. So if you never saw this non-severic group, it really doesn't matter. But if this non-severic group has maximum rank, the maximum number it can get is 20, then know that the remainder part, which is something that we do not understand, it is two-dimensional, right? I mean, the total number is 22. 1 plus 20 plus 1 is 22. I have 20, then I have two remaining. And then when I look at this color representation, then I'm going to have a 20-dimensional part coming from the Neuronsseveric group, which I know it extends and satisfies some functional equation. And then I have this extra two-dimensional part, which is like the weird one. And if I can prove that this extends to the whole complex plane and satisfies the functional equations, I get the whole Hasebel conjecture for free. And in this case, so this is something proved by Leibniz, this two-dimensional part is actually a modular. It comes from a modular form of weight three, and some Neven typos. But nowadays we have something even better. We have something which is called the SES conjecture, which says somehow that any two-dimensional representation over the rationals, with some minor hypotheses, comes from a modular form. You need it to be... even the determinant... the determinant at complex conjugation has to be minus one. You need it to be odd, sorry. And you see... so for this particular... these ones, which are called extremal Calabiyao two-dimensional, when the rank is as big as it can be, you also get modularity. Okay? So again, you have something which comes from an algebraic part, which is some sort of a theta function, and then you have a remaining part which comes from a modular form. So you see this is really interesting because the whole picture here, this is what I was saying before, when you want to count the number of points, you have some parts coming from theta functions and then the middle ones, I mean, it depends on the dimension, then it might be harder to understand, but sometimes this... it splits as different parts and some parts are easier to understand. Okay? And also... there are... there are some results of Lybne, Matthias Schultz and Yui, which prove that for some other smaller ranks, the modularity is also known, and what happens is that... I mean, if you have... as I said before, if you have a four-dimensional representation, it's not clear whether it should be modular or not, but if you have something which has many endomorphisms, then it should come somehow from a Hecke character. So what they prove is that in many instance, these L-series come from Hecke characters, so we know the whole picture. And let me talk... I'm almost done. Let me talk about three-folds. So before I was talking about Calabria's surfaces, so dimension two, and I want to talk about dimension three. So a Calabria three-fold, it's called rigid, if the H2-1, so this number here is zero. Okay? Note that if this number is here by symmetry, this other number here is zero as well. So it tells you that the middle part, which is in general the hardest one, it's going to be two-dimensional as well. This H1-1 somehow is algebraic, it comes from cycle, so you do know that it has some analytic continuation or functional equation, but so what happens is that when you have a rigid Calabria three-fold, then the H3 is going to be two again. And so, again, we get that it is modular, and it comes from a wait for modular four. And this result was mainly proven by Julefe and Majoharmayoum, but again it follows from seskin checkers nowadays, so this result is older than seskin checkers. Okay? And let me go back just one second. Note that when you had the surface, the weight was three and now that we have something of dimension three, the weight is four. So this is one plus D in general. Okay? And for some non-reached Calabria three-fold, what happens is that sometimes so non-reached means that these numbers are non-zero and then this part here might be the dimension of this might be quite large, but there are some instances or many examples in the literature where the H3 representation is reducible. So somehow it decomposes at smaller dimensional parts and then so you might expect to prove modularity as well. So there are some really interesting examples of Bernet and Bangui examples on this sort of phenomenon. Let me finish with another example, which is the Kontani-Scholten-Kintiq. So you look at this polynomial, which is HHF polynomial into variables. And then you look the Kintiq given by p of x1, x2 equals p of x3, x4. The question is not so important. You have to desingularize this, like I think it has 121 singular points or 120, I don't remember exactly. So you have to desingularize this and then you compute the Hodge diagram and know that this is huge. Okay. So this has to do with what I said before, this number 141 here is telling you that some polynomial, which we have degree 141. So if you want to compute the theta function, it's extremely expensive. Because even if you know that the dimension of each different piece, you need to count the number of fp fp square over many different finite fields to be able to deduce precisely what are the polynomials. So this is also a computational challenge to try to compute the theta function of these guys over here. And the interesting part, the H3, so somehow this guy here is algebraic. So this guy here, it is H3. This is four-dimensional. So you have a four-dimensional representation. And what Konsenial Schulten realized is that when you look at this four-dimensional representation and then you restrict this to q square root of 5, the other group of q bar by q square root of 5, it splits as a direct sum of two two-dimensional ones. So now you're in a nice situation because it is something of dimension two, but over a quadratic extension. And what we proved that it actually comes from a modular form, but it is a Hebrew modular form and the way it is 2 and 4 and the level is 30. So again, since it's related to some object, so this is again a case where you have a four-dimensional representation but it's not as big as it can be because when you go to a quadratic extension, it splits as two two-dimensional ones. So we proved that it comes from a Hebrew modular form and so again, you get that this function satisfies the conjecture. Okay, that's everything. Thank you. Thank you so much. So are there any questions? You can either unmute yourself or write it on the chat. And in the meanwhile, I'm going to also send a poll so that you could answer before what you think about your questions. So I ask a question, but so for example, in the case of the K3 surfaces can you say something about the ophthalmorphisms when you have the modularity or when you expect the modularity, can you say something about the ophthalmorphisms of the K3 because it depends on the lattice, so I don't know if that's the thing that comes to my mind. So that would be my question. You mean the K3 surface? Yes. So when you say this path here, so I think so I didn't look at the proof of these ue and chute these results where they get the CM forms, but what happens in many instances that your surface, you look at the endomorphisms, I mean you see for example in this equation here you see it's pretty symmetric, right? So you know that you have a lot of endomorphisms already appearing there and so sometimes what you try to do is you try to see whether you have enough endomorphisms to ensure that your variety should come from something which has complex multiplication. So I mean the way to prove this is that so they give some conditions for the ranks of these endomorphisms, there's a very group to be exceptional and so on and in one of these cases they somehow present families and they prove that any such variety belongs to one of the families and then they study the families on themselves. So it has to do, I think this is what you were asking, it has to do with equations. You have to know the equations to be able to say something. So I don't know if there is any theoretical information that from extraordinary endomorphisms, but I think what they do is they just compute the family and they do it on the family itself. Okay. Thank you so much. Are there any other questions? So if not, we thank Arielle again.