 Okay, thank you very much for the opportunity to speak. So I'm going to start by giving some notations. So F is going to be a casp eigen form, quite K for gamma naught n. P is going to be a prime in odd prime. And so I'm going to denote KF, the field of Fourier coefficient of F. And so I see it inside Q bar, and I fix in meetings of Q bar into Q p bar. And I choose a isomorphism with C. So I can see element in Q bar and Q p bar in C in this way. So I'm going to fix K, a periodic field containing KF. And so O is going to be the ring of integers. And I'm going to call OFP the intersection, like that. So I'm going to denote by rho F as a Galois or obsession attached to F by the ring. So I'm going to call VF the space of the representation. So VF is a morphic to K squared. And so I'm going to- Over C or is it- Here? Yeah, sorry. F is of weight equal to 2. Okay, so there is no confidence. So, okay. And so LF is going to be inside. VF is going to be a stable lattice. And I'm going to call rho F bar the residual position. And so we are going to assume that the image of rho F bar is sufficiently large. And for simplicity, I'm just going to assume that it contains SL2FP. And maybe there are weaker hypotheses, but just to make life simpler, we'll just assume this. And I will make another assumption. So this is one assumption. Another assumption is that rho F, so F is we call N primitive. And basically that means that the ramification of rho F bar is the same as the ramification of rho F. And our next assumption is that K is less than P, so this is a Fontaine Lafay type condition. Okay, so under this condition, there is a serum, which is essentially due to Wiles, Taylor Wiles, and Ida, which say that the adjointal function of rho F at 1 is basically equal to the size of the block at the thermo group of the residual representation. And so this is a special case of a general conjecture of Locato. And so the way that this is proved is by there are two important ingredients. So one is to relate H1F of Q to add rho F to an automorphic class group, which is what Ida called the congruence module. And this step is done by the work of Wiles and Taylor Wiles. And then there is an argument which is done by Hida, which proved that this automorphic class group, I mean the cardinality of this automorphic class group is equal to this adjointal value. So we can try to generalize this kind of formula Now if you take alpha, a Jewish character, the question that was raised by Ida is, so do we have a similar formula for the adjoint of rho F twisted by this character at 1? And so the question that he raised was not a formula that relates this adjointal value to a similar group, but to an automorphic class group. So this is equal to some automorphic. It's the cardinality of some automorphic class group. So I'm going to explain how we can answer this question for certain character alpha, and especially for quadratic character. So if some more notations, so I'm going to use this one now. So now F of our queue is going to be an abelian extension. And so we are going to consider two cases. So F is totally real, or F is imaginary quadratic. And so X is going to be the modular curve, X not of n. And I'm going to call X F. So the locally symmetric space for gl2 over F of level, so my gamma not of n, but of course for the field F here. So let n is equal to k k minus 2. And I'm going to call Ln of A. It's going to be the symmetric, sorry, the tensor product over all the invadings of F into q bar of sim n of a squared. And so we have an action of gamma not for F on this space. And we can consider the comology of X F on the local system attached to. So we can also consider a space sk F for gamma not n. It's going to be a space of cuss form for gl2. F of level gamma not n for F and parallel weights k. So if F is totally real, so this is a space of Hilbert modular forms. And if F is imaginary quadratic, this is the so-called space of Junki modular forms. Yeah, yes, it is. So we can define space, we can define echo operators acting on those spaces. So at HF, n, so the echo algebra are acting on those spaces. And so here we have only echo operators away from n. So it's just a spherical echo algebra, it's not a product of a spherical actual algebra at places not dividing n. And so we have a base change map that goes from the echo algebra for F to the echo algebra for Q. So this is just come from the arter-closal existence of the base change for gl2. And so here we have a map going to call lambda goes to KF, which has come from the original modular form F. And so we consider M the maximum. So I'm going to call this map lambda F. And I'm going to call M the maximal ideal corresponding to lambda mod P, or lambda F mod P. And consider T, TF, the localization of the echo algebra here over F at this MF. And T is the corresponding localization at M. And then we have a map here to O. So I'm just going to call this again lambda. This is going to be lambda F. So now we can define the congruence modules. So this is something that we call C1 lambda F, C1 lambda F. So you take the, yeah. So here I consider that this is the echo algebra over OFP. Well, this is the area I have to extend the scalar here. So here this is omega TF over O. Don't say O, it's lambda F. And you can define similarly C1 lambda as C omega T over O. Don't say O by lambda. There is another way to, so this is the same as PF, modulo PF squared, and this is P modulo P squared, where P is the kernel of lambda and PF is the kernel of lambda. So this map, lambda F, corresponds to the action of the operator for F on the base change of F to gl to F. So this here, so if you call F hat, the base change of F to gl to F. So lambda F corresponds to the action of the echo algebra on F hat. And so because this is a base change, so you have an action, so it's an action of the Galois group of F over Q on this module of C1 lambda F. So if you take alpha, a character of this Galois group, then you can consider C1 alpha of lambda F, which is the alpha part of this module. So what is that conjecture? So the conjecture, my header, is the following. So if I assume that alpha is quadratic, then set by alpha at 1 times L, so divided by omega F plus omega F minus is the same. You can't get it with the L value, not just the Galois. Yes, thank you. Of lambda F. So this is when alpha is quadratic real, and if alpha is, so this is alpha real, quadratic, and if alpha is imaginary quadratic. So here is a similar conjecture where this product is replaced by something else that I'm going to define. But it turns out that actually the conjecture, in that case, is false. Yes, because this is not the correct number here. Replaced by another congruent number. Of course, you can make this conjecture for any, not necessarily for, imagine not necessarily for quadratic characters. So now I'm going to define the periods. So you have, so I'm going to define epsilon. Epsilon is going to be a collection of sine index on the embeddings of F into Q bar. And so if you consider J, the subset such that epsilon sigma equals plus 1, you can define a map omega J, omega epsilon of F. So F here is going to be SKF inside HD. So here I'm going to assume the case F is real. Yes, totally real. And actually, so you have an action here of this group here on this homology for each real place. You have a complex conjugation that acts here. And so you can decompose the beta-comology. Now we take coefficient in OFP. And you can take the epsilon part here. So here, actually, this guy belongs to the epsilon part here. And this epsilon part is actually rank 1. And so we can define the delta epsilon F basis. And so we can define the period omega F epsilon. So by the formula, omega epsilon of F is equal to omega F epsilon times delta epsilon F. So when F is equal to Q, so when F is equal to Q, then we have just 2 sine plus 1 or minus 1. And it's how we can define omega F plus and omega F minus. Excuse me? I would say the ethical Q. Yes. So in that case, epsilon is just an element in plus or minus 1. Oh, no, no, this is F. This is a modular formula. So this F is over Q. The boldface F is over F. And then there is the case of F, boldF equals F at, which is the base change. So the conjecture is the following. And so I'm just stating the conjecture with the hypothesis that I have stated in the beginning. So in particular, in general, those periods should depend of the choice of the latest inside the Betty chronology or the Betty realization of the motif. But here we assume that OF bar is absolutely irreducible. So therefore, there is no choice of lattices. So in that case, we can state the following conjecture. So if F is totally real, so omega F hat epsilon should be equal up to an amount in O cross to omega F plus to the power D plus epsilon, omega F minus epsilon. So we are D plus the mod. So we are D plus epsilon is a quantity of sigma such that epsilon sigma is plus 1 and D minus. And so basically this conjecture is just a special case of general conjecture of periodulations of motifs, of base change of motifs. So here is a theorem that we can prove. So assume that the Galois group of F over Q is more thick to Z mod 2, some power. Then the conjecture for epsilon such that D plus epsilon equals D minus epsilon. So the proof of this conjecture, the main argument is when m is equal to 1. So we just have a quadratic extension. And then there is an induction argument over the dimension. But the essential case is when the extension is quadratic. So a corollary of this, and actually this is a corollary, but this is also proved at the same time as the case where this extension is quadratic. So it's a following theorem. So assume F real quadratic, then it has conjecture for alpha as a corresponding quadratic character. In fact, it follows from i equals t theorem. In that case, in fact, we have the formula that gamma of f is the size of the block Hato-Selma group for Hato-F, crystal by alpha. What time did I start? So let me explain now what kind of result we can get when F is imaginary quadratic. So now I'm going to assume F imaginary quadratic. So when F is totally real. Are you going to explain the proof of all of this? Of this theorem? Yes, yes, I'm going to explain this. So when F is imaginary quadratic, then in that case we, so in the total real case, the cuspital converges essentially concentrated in middle degree, which is the HD. But in the imaginary quadratic case, then the cuspital converges is essentially supported on two degrees, which are 1 and 2. So in that case, xf is an hyperbolic space of dimension 3 for r. And so we have two Eichler-Chimura maps for i equals 1 and 2. And so if we look at the lambda part here, this is going to be, so for i equals 1 and 2, this is rank 1 over 4fp. And so we can define two periods, omega 1, F, omega 2, F, exactly using the same kind of formula. Let me call, yes, modulo torsion. Thank you. So exactly the same formula that omega F, or maybe a 2. So in that case, so the theorem that we get is the following. So it's the same as hypothesis before. Yes, so there is an hypothesis I forgot to mention is that p has to be prime to the discriminant of the field. So the term is that omega 1 of F hat is gone. So now, so contrary to the case where F is total real, the L function, and now this add f, add row f at 1, is not a critical value. And so in that case, the period is not going to be this critical period here for this adjoint L function. So in fact, so something that, let's say this first part, the second part is that so omega 2 of F hat. So this is equal to something that I'm going to call eta FF-sharp, which is a congruence norm. This is the cardinality of a congruence module, not the one that I have defined before, measuring the congruences between F hat and the Enquimodular form G, which are not a base change. So in general, this number divides the cardinality of c1 alpha of lambda f. But there is no equality if tf is not complete intersection. And so in that case, actually, there is a work of Caligari-Garati, which I said that, so this tf is never complete intersection, except if we have only characteristic p-forms. So in fact, that means that we have never this equality here when we take a base change. What happened to the congruences with other forms that overcube? So they are controlled by the same number, which is the one that I defined at the beginning, which is the adjoint L value at 1 divided by omega f plus omega f minus. So this is because the alpha, the trist pi alpha, they don't appear? Yeah, they don't appear. 20 minutes. So I'm going to explain a bit of formalism between congruence modules. So in general, if you take r, a finite, or algebra, and so we assume that it is reduced. And so you, if you don't serve with the field of fraction, rk, so this is just rk, this is going to be semi-simple. So you can't find something made in terms of product of two semi-simple k-algebra. And so you can write a is the image of r into ak. So in that case, I'm just fixing on the morphism. So you can have an invading of this. And so it's the image in ak. So there is no torsion. There is no torsion. So in that case, so a cross b divided by r, you can see that this isomorphic to a divided by r intersection a cross 0. And this is the same as b divided by r intersection 0 cross b. Yeah, r is not available, yes. R is finite or algebra reduced. And yes, no torsion. Thank you. So you see that this is an isomorphism if and only if this module is trivial. And when it's not, you see that that means that you have congruences between characters that occur on the ak factor and characters that occur on the b factor. So this is a congruence module between a and b. Depending on r. Yes, depending on a, yes. So if m is r module, you have a similarity composition according to a and b. And you can denote by a. So let's call, if I call lambda a the map from r to a. And lambda b is a similar map. So you can define m lambda a is going to be the image of m. The map mk to r to ak. And you can denote by m lambda a as the intersection of this guy with m. And I'm just going to write this m. So if there is no confusion, I just write m a and m subscript or subscript a like this. And so in that case, you can see that the equivalent of this isomorphism here is the morphism m a m b. Yes, yes. So yes, those modules now are finite. Yeah, yeah, yeah. Yeah, yeah, yeah. All right, so for, so now if we put our stuff in the case where now we take r equals tf. So we know that we have a map from tf to t and to o. So in particular, you can define an isomorphism is tk cos something that I'm going to call t sharp. And that corresponds to forms which are not base change. All the forms here are forms which come from base change. And our souls do not come in from base change. And t sharp is just going to be the image of tf into t sharp k. So you can define, so I'm going to denote by c naught. So this module is going to denote this module by c naught a. And of course, this is the same as c So you can define, for instance, c naught lambda of tf. So this is just, now the continuity of this is just something that I'm going to be lambda f hat. This is the congruent number, f hat. And we can also consider the same for just t. So c naught, so this is lambda f here, lambda of t. This is just lambda f. And this is just the congruent number for f. So the original theorem of FIDA, it proved that actually the cardinality of c naught, lambda, and for the module, it takes the common g of x with coefficient l, l, o. And so we show that this is the same as this adjunct L value. And so from the result of Taylor-Wiles, this module is, sorry, I forgot to localize that m. So from the work of Taylor-Wiles, this module is 3 of, well, let me take a plus. This module is 3 of Frank 1 over the HECA algebra. So this is the usual f, is the congruent number. And because it's result from the work of Wiles and Taylor-Wiles that the HECA algebra in that case is complete intersection, then this is equal to the same thing to the cardinality of now the c1. And this can be essentially, this c1 of lambda is essentially the dual, the Portuguese dual of the Selma group. So it's how the bokeh to conjecture is proved. So you can do something similar for a totally real field. And under some assumption, you can prove also r equals 3 of theorem and that r is and t are complete intersection. And you can also generalize the work of Hedard. It was done by Dimitrov to show a formula of this type. But in that case, so in the case of t total real, the formula is going to be c0 over f of hd. And then you have to take the epsilon part is here, you divide by omega f hat epsilon omega f hat minus epsilon. And so you expect, so if you have this parallel relation, you see that now you can decompose this product as a product of those integral values twisted by all the character of the totally real fields. And after you have to separate everything. And so you have to, so in the case, in the real quality case, what I can define actually a congruence number that measures the congruences between base change and non-base change. So you can define, so if you have this module m, this tf module, so you can define, you look at m lambda and then you question by, so mt and then you take lambda. So mt is going to be, so this lambda f, sorry. So this mt now is a t module. So therefore you can consider it's up lambda part. And this, so let's call this eta lambda sharp of m, so measure congruences forms which are non-base change for the given module m. And so using this formalism, you can show that, so if you look at eta lambda for the module m, then this divides eta, sorry, lambda f, lambda for the mt times, sorry I should put the lambda f here. So now this, so when m is equal to hd epsilon by this formula, so I'm going to assume now that f is quadratic, you can show therefore that l of adjoint of f, 1 f of adjoint of f twisted by alpha at 1, divided by omega f epsilon, omega f hat minus epsilon. So divides, so this is just adjoint of divided by omega f plus omega f minus. And so here you have this number. You have this divisibility that comes from this formula of Dimitrov, and this formula which is very formal from this, from those definitions. And so there is another ingredient, so there is one one one more ingredient which is the following. So if you take g, so nilbert modular form for f quadratic, and so you look at omega of g, omega epsilon of g, so this is in that case, this is a two form, and so you have the modular curve that you can embed in the nilbert modular surface. So this gives you a two cycle inside xf, and you can integrate this two form over this surface here. And so in that case you get the two, you get the following. So you get zero if g is not a base change, and otherwise you get the adjointal value of f twisted by alpha at one if g is f hat. So using this you can show that, so you see that if you have a congruences between a form g which is not a base change on f hat modulo p, that tells you that this value is zero modulo p. So therefore that tells you that this eta is of lambda of of of h2, okay, so here epsilon has to be equal, has to be plus minus or minus plus epsilon. So divides the adjointal value of f so by alpha at one divided by omega f hat epsilon. So using this visibility on this one plus some arguments using gornivacal work you can deduce that omega f hat epsilon when epsilon is you know of this case is is the same as omega f plus omega. And so that gives you the period relation exactly in the case in the quadratic case. And by some induction argument you can do the general case when f is is just of type 222. Here this is modulo o fp and so you can you you have a similar form exactly with the same argument you can show in the case in the imaginary quadratic case that the omega two sorry the omega one f hat is is is is equivalent to omega f plus omega f minus. So I wanted to speak about the Baylinson conjecture in the case of adjoint of f twisted by alpha but I don't think I have time so to stop here. So in the imaginary quadratic case do you need the relative trace formula or do you? No no this is the same you have the same the same statement. Yeah I mean it's no it's not using the relative trace formula to show this. Basically what you do is that you integrate over over x g times some Eisenstein series and in that case this is related to the assail function and so when there is a base change the assail function as a pole and and and the Eisenstein series also as also a pole. So when you take the residue you get exactly this formula because the assail function is going to be in the case of a base change is going to be this adjoint function times etc function. So the pole is so the residue gives you this when this is a base change and the residue is zero when it's not. For what are the meanings of the gamma factor? So the gamma factor is so each time that so you have a motif and you look at the odd number you can define gamma factors. So in that case the gamma factor of ad f twisted by alpha s is so gamma s plus k minus 1 gamma r of s and then you you plus 1 in the if alpha is Israel and plus 0 if alpha is what yes yes and the gamma c is just 2 pi minus s gamma s and gamma r. So basically you have some power of pi and so if you look at so in the case of alpha imaginary quadratic you you can so in that case if you look at the the motif f m f is a motif attached to f when you look at this adjoint and adjoint motif twisted by alpha then in that case you have you have the motif ecomology which is a wrong one and so you can define a regulator let's say rf alpha and then in that case the bellinson conjecture blockato conjecture tells you that the adjoint l value f twisted by alpha at 1 times the gamma factor divided by omega f plus omega f minus times this regulator alpha is is the size of some blockato semi group and so here what yes this is a special case of bellinson here and basically you you expect that this is the same as omega 2 f half f hat and so basically that tells you that in that case bellinson conjecture of prasadavan cottage conjecture are equivalent but this is more of this is an integral I mean bellinson and vancatech prasadavan are saying a rational result this is an inequality result which means this is not a result means this is an inequality conjecture blockato it turns out that this this period is not equal to this period up to an integer up to a unit again because because in that case it does conjecture is not true what if you use first your cell door back on the cell tube well this is because I want to look at when I look at the galore optician here I want that so the galore optician here is going to be o f tensor o f so I want that this is sufficiently big but as I said I mean you you don't really need to assume that you contain s l 2 f what you want is that this is something dimension three over the c bar plus plus a character so this you need to have some this just to have also some r equal c theorem that you need to have visibility but as I said s l 2 f p is is probably not required