 Donc, encore une fois, comme toutes les lectures précédentes, je serai seulement descriptif. Donc, cette lecture est plus comme un thésor pour les gens qui travaillent dans l'analythique de la théorie pour investir dans l'adélique, parce que je crois qu'il y a un bon délai de idées de l'analythique de la théorie, qui peut être transposée avec un grand profit dans l'analythique de l'analythique. Donc, racontez-vous ce que nous voulons faire. Ok, donc, nous avons, donc, dirigé par Duke-Ciorem, nous avons un hortogonal groupe, et ce que je dirai peut être plus général. Et donc, nous avons ensuite un espace d'analythique. Et donc, ce que j'ai expliqué la dernière fois, c'était que cet espace d'analythique est un hortogonal, avec, donc, m'a-t-il appelé le G entre les brackets. Et il admite une mesure, qui est donnée comme question de la mesure de l'armée sur ce groupe divisé par la mesure discrète sur ce groupe discrète, donc appelé R. Et en ce cas, cette mesure de l'armée, donc, cette mesure de la mesure de l'armée, c'est correctement invariant. Et donc, je n'ai pas dit ça. Et finit. Et donc, nous pouvons normaliser ça pour être une probabilité. Et donc, ce que j'ai expliqué la dernière fois, c'était que, quand la forme quadratique est anisotropique, cette question est, en fait, compacte. Donc, il n'y a pas de surprise que la mesure est finite. Mais en fait, c'est vrai en plus de la généralité. Et c'est le... C'est-à-dire Borel, Dari, Chandra, finit. Ok, non. Il ne peut pas être toujours finit. En fait, la finite, en plus de la mesure, n'est 2 et q est isotropique. q est, je dirais, isotropique sur les rationaux. Mais, mais en ce cas, la groupe orthogonale est isomorphique. Le groupe orthogonal est isomorphique à l'équipe multiplicative et l'équipe adélique de l'équipe multiplicative est juste la unité de l'adélse de l'adélse, et vous voyez que cette question est isomorphique à R star modulo gl1 of z, qui est z star, qui est le groupe de groupes multiplicatifs de positive réel nombre. Donc, ici, la mesure de l'armée n'est, bien sûr, n'est pas finie, mais c'est la seule case. Donc, ici, on travaille principalement quand q est de franc, plus grand que 2, et je vais expliquer si il y a un petit problème. Donc, nous avons cette chose, et ensuite, nous avons un groupe sub, donc le stabilisateur de un point rationnel, et nous voulons, donc ensuite, nous avons aussi le quotient. Donc, ensuite, cet h est un groupe orthogonal dans une variable n-1. Donc, peut-être que la mesure n'est finie, mais on s'assume que ce que c'est h, que c'est q restricté à l'autocomplément de x est anisotropique, ou que n est plus grand que 3, et qu'on n'a pas de problème. Donc, ensuite, nous avons, encore une mesure finie. Et donc, nous voulons prendre une fonction, ça continue de supporter la fonction compacte sur g. Pour notre problème, nous demandons que cette fonction soit invariée par le stabilisateur de la lattice que nous sommes intéressés dans. Et nous voulons montrer que l'intégral de phi au long h, au long de cet h, donc en fait, nous devons ajouter un petit élément addélic qui dépend de notre choix de ce point réel x-naut et de la lattice l'alprime dans lequel x-q s'éteint. Et donc, intégrer ça. Et nous voulons montrer que c'est convergé à l'intégral de la fonction, mais au long de l'autocomplément groupe. Et même, donc ici, on va dire, donc même, nous pouvons assumer que l'intégral de phi est 0, parce que quand phi est une fonction constante, c'est obus, parce que les mesures sont normales pour être la mesure de probabilité. Donc, nous pouvons même rester dans ce cas. Donc, nous voulons, sous cette condition, nous voulons montrer que cet intégral se convertit à 0, comme les grosses et les conditions additionnelles que nous avons mentionnées dans la première lecture. Ok, donc, je veux maintenant décrire cet espace. Et donc, peut-être, donc juste récollection sur l'espace quadratique de l'eau. Donc, c'est juste, c'est utile, parce que, en fait, le cas le plus intéressant pour ce problème est l'eau. Donc, les cas de l'eau sont les cas de l'eau de l'eau. Donc, je ne sais pas si, donc, quand n est 2, et si vous avez une espace quadratique de l'eau de l'eau 2, alors une espace quadratique de l'eau de l'eau 2 est toujours similaire. Donc, ce qui signifie isométrique de l'eau, c'est toujours similaire à l'espace quadratique de la forme K times la norme de K pour K times l'espace quadratique et tel algebra. Donc, ce qui signifie simplement que K times Q times Q, ou un espace quadratique, et n est juste la norme de cet étal algebra, qui, donc, est la norme sur le espace quadratique, ou le produit ou les coordonnées, si vous êtes dans cette situation. Donc, et puis, vous pouvez aussi compter ce qu'est le groupe orthogonal. Alors, le groupe orthogonal sera isomorphique à l'algebraic groupe, donc c'est l'algebraic de l'élément de K de norme 1. Donc, et simplement, l'action est, vous avez lambda en K, donc non-zero élément, et vous avez X en K, et puis, si vous considérez la map X, vous vous avez lambda X, alors, c'est l'autogonal similité des facteurs similités, la norme de lambda. Donc, c'est le groupe 2 K. Donc, si vous êtes dans le groupe 3, l'espace quadratique est toujours, donc, similaire à l'espace alternatif de la forme B0 de NrB, where B is Aver, so is a degree for central simple algebra over Q, meaning that B is Aver's algebra of 2 by 2 matrices or quaternion algebra over Q. So, Nr is the norm, so in that case, the norm is just determinant. Ok, so B0, this is the subspace of trace 0 element, and the quadratic form Q is just either the determinant if you are in the matrix case or the norm of the quaternion algebra, let's say Z times the involution of the quaternion algebra. So, you have this description and you have also a nice description of the orthogonal group. So, SOQ, it's isomorphic to the, so to this is algebraic group of associated to the invertible element, the unit of B, and here this is the center, so you take the quotient of this group by the scalar quaternion, so I will write it P B star. So, it's an algebraic group whose points over Q are just the element of the Q, so the element of the algebra, the invertible element of the algebraic set divided by the scalar element. And so, maybe, yeah, and the action is given by this, so if you have, so, W, yeah, let's say, W in B star and if you take an element of trace 0, then you have, you associate to this element, it's conjugate W, Z, W minus 1, and then you get a new trace 0 quaternion whose norm is the norm of Z. And, okay, so maybe I should say, so for N, N is 4, there is analogous description, so in terms of, so ever you have the quaternary space given by a quaternion algebra equipped with its norm, or, so this is if the discriminant of Q is a square, and if it's not a square, then you have a description, which I will write BK plus times the norm, where K is the quadrat, okay, it's slightly false, but let me skew of, in fact of the square root of a multiple of the discriminant, so if, and so BK is, and BK plus is a certain four dimensional subspace inside BK, and we equip this space with the norm, and then any quaternary quadratic space is similar to one of these two spaces. In fact, okay, if you want to take K to be the split quadratic algebra, so, which is what you would get, you have, this description is valid uniformly whether the discriminant is a square or not, and you have, so, and then, so the orthogonal group SOQ can be described in terms of BK star of the action of this algebra of quaternions over K. So, this is standard, away, so, I think you can find this in Kitahoka's book on quadratic form, so, and the proof goes by considering the Clifford algebra associated to any of these quadratic space, and so it's not, it's a classical, but, so it's just to mention that for low rank quadratic space, so what will come up, a central object will be quaternion algebra and groups associated to quaternion algebra, so maybe, so now, so, automorphic forms, so, I go back now that one has an idea of what the structure of G is, automorphic forms on G, on GA are nice functions on the quotient GA modulo GQ, and just I want to give examples, maybe, of automorphic forms for various kind of groups, so, just to have an idea, examples, so, let's take, so, G, okay, let's take the case of A, so, the additive group modulo Q, so, it can be realized as a, as a, as a, as a, okay, can identify it with NA over NQ, where N is the group of unipotent matrices, so, you have, so, it, yeah, okay, and I recall you, you have A mod Q mod Z hat, which is identified with R mod Z as an additive group, and so, then, if you take a, a character of this circle group, then, because you have this identification, this psi of N correspond to a character, additive character on A mod Q, which is a trivial on Z hat, and maybe, we just work out what, what happens exactly, so, if you let us compute, so, consider XR in R inside A, so, then, you have, by definition, psi N of XR is E of N XR, but, what does it give if you compute, again, sauvage elements, so, now, if you take XP in QP, so, then, your XP, you can decompose it, okay, so, you take the, the pay-addict expansion, okay, so, here, this is an element of ZP, and this, this is an element of Q, because it's a fine, so, okay, AK, this is an element of Q, and, in that case, let's take phi N of XP, then, it's defined by, exactly, as E of minus N of the so-called, the fractional part of, of this pay-addict element, and, of course, ZP being contained into ZP hat, if you evaluate psi N again, this element, you just get one, okay, so, the way, the reason for this computation is, because, you know that your character is trivial on Z hat, and also, it has to be trivial on Q, on the Q embedded diagonally, maybe, I want also to do another example, so, which is now, let's me do G is GL1, so, G of A is A star, and so, we have that, this identification, G of A modulo G of Q modulo G of Z hat, it's identified with, so, in other term, if you give yourself a character of the multiplicative group of the positive real number, automatically, you will get a character on V squashant, so, on A star modulo Q star, which will be trivial over Z hat star, and so, the characters of, so, the characters of A star will be, so, they are all of the shape, they are just the absolute value, the real absolute value to some power, and so, once you give one of these, you obtain a character on V squashant, and so, what is this character precisely, this character, so, on A star is, the follow is of the shape, is given by the following formula, so, you take now an idyllic element, invertible, and you map it to the following function, sorry, to the power S, which is, by definition, the product over all, so, times the product over all prime of the payadic absolute value of the P component to the power S, and so, you verify that, in that way, you obtain a character, which is, which will be trivial on Q star, and trivial on Z hat star, and of the right, of the right shape, so, this character, so, this character here, so, the product of all the absolute values over all the component is called the modulus, or the idyllic modulus, so, this is another example of a automorphic form on, so, and here, you are on V squashant, so, other examples are given by diriclet characters, and, okay, so, what, so, suppose you have a diriclet character, so, then you have the following isomorphism, so, you can show that it's just the same kind of reasoning, but it's isomorphic to A star modulo Q star modulo, okay, KFQ, with KFQ is what we called previously the principal congruent subgroups, so, it's 1 plus Q Z hat star, or, if you prefer, it's the product over all primes not dividing Q of the ZP star times the product over all primes dividing Q of 1 plus Q ZP, so, and this is a multiplicative subgroup of Z star, and its index is precisely the cardinality of it, and so, in that way, from any diriclet character, you obtain, so, key, multiplicative diriclet character gives nadelic character of A star mod Q star, which will be trivial on this open compact subgroup, and, in fact, you have a objective map between diriclet characters are in bjection with finite order characters of A star modulo Q star, so, okay, yeah, thank you. Plus, yeah, okay, and now, if you want to obtain all the characters of this quotient of A star modulo Q star, what you have to do is to multiply a finite order adelic character with a character of that shape, and then you get all description. Okay, and so, okay, I will finish the example on that board, and so, okay. Now, if you go further, if you take finite extension of Q, a field, then this field defines, okay, an algebraic group, an algebraic torus, which I know tk, which is a sub-torus of the, okay, of the linear transformation of k, so, views as a Q vector field, and simply the action is given by x in k, and lambda in k star, you have this linear map, give lambda x, so, k star acts linearly on k by multiplication, and the k star is a group of rational point of some torus inside this GLK, so, which is, which you may seen as belonging to GLNQ by choosing a Q basis of k, so, fancy terms tk is the restriction from k to Q of scalar from k to Q of the multiplicative group, and what happens is that a-Q characters, so, classical a-Q characters of k correspond bijectively to, so, continuous characters of tk a modulo tk Q. And, so, when you look at, say, look in, say, don't know, Ziegler lecture or Ake's lecture on the definition of Ake character, usually it's a big mess, because your character has to satisfy some condition for, you have a lot of condition of your character, and these conditions are made to ensure that your corresponding a-delic character is trivial on the rational element. Ok, and finally, last example, which will be, ok, f in sk of sl2z a-Q eigenform. Ok, so, then, for you, f of gamma z equals g of gamma z to the k f of z. And, so, how does one associate to this a-Q eigenform, an automorphic form? So, we observe that the quotient h mod sl2z is identified with, so, this is what Valentin explains in a much more general context. So, and, simply, this map is given by g goes to, let's say, g i gamma. So, you have this identification and, ok, and let me write this here as pgl2r modulo pgl2z modulo pso2r et, here, I can identify this with an a-delic quotient pgl2a modulo pgl2q et, ici, modulo pso2r pgl2z. Ok, so, it's getting very, very small. So, ok, you have this identification of the modular surface with a-delic quotient so, of pgl2 by the rational points, and on the right, you have the product of the compact group, compact real group by this compact finite-delic subgroup. And, now, and so, how do I associate a function on that quotient? Then, it is, because of all this identification, it suffice to define a function on, yeah, not exactly on that quotient. But, so, to my function modular form f of z, I associate f tilde of g defined by g of, so it's a function on the group and, so, what you can check is that you can check that for all gamma in, say, in gl2z, f tilde of gamma g is f tilde of g let's say positive determinant, just to, for this to be well defined. So, then, it means that f tilde is a function on, and, ok, and it's not a function on the, on exactly this double quotient but, if you look at what happens, so, let's take kappa matrix, so, where the angle theta is only defined modulo p and, then, if you look at f tilde of g kappa, what you obtain is that it will be e minus k of kappa of f tilde of g where e minus k of kappa is e of minus k of theta. So, k is even and theta is defined modulo p, so, this is well defined. And, so, this, so, e minus, this, e minus k of kappa is a character of pso2r. So, what you get from a holomorphic weight k modular form is a function on this quotient which transform in a special way under the right action of this compact real subgroup. And, so, in that way, you see that you obtain from f tilde gives you, so, f tilde defines a function on modulo pgl2zat transforming which is an eigen function for the right action of pso2r. And, the eigen value for this action in this is this character which is determined by the weight. And, ok, and if your modular form is an eigen form, then the f tilde will be a genuine automorphic form. Ok, so, I return now to g is so2q and so we want to understand the space of continuous function on this quotient. And, we would like to understand the function which are moreover invariant by some open compact subgroup. And, for this, it's, in fact, the starting point is to understand first the L2 space. And, this L2 space is, so, unitary representation of g of a for, by, for the right action, multiplication. So, simply, so, if you have a function and you have a g, g phi is a new function which to g prime associate a phi of g prime g. So, you see, if your function is a function on ga, which is gq invariant, this new function will still remain gq invariant. And, the representation is unitary, so, simply because, so, unitary, simply because the measure that is taken here is R measure which is right invariant. And, so, the main thing to do, so, and, so, ok, one thing which is good is that, so, which is why, so, and, ok, in fact, you can upgrade this from representation of ga, and this is a standard things to representation of the, of, of an algebra. Say, for instance, the algebra of continuous compactly supported function of ga. So, with the product replaced by the convolution. So, simply, if you take a function continuous compactly supported, ok, the f is not the same f as a modular form, but if you take such a function, you can define the operator. And, so, just by this convolution trick, so, if, if suppose that your continuous function, which is a function on this product is written as a product of a function on ga times a function over the, on g, on the finite addels. Yeah, and suppose that this function is a smooth and this function here, for instance, suppose you take that it is a characteristic function of principal congruence subgroups, this small neighborhood of the identity, which is an open compact subgroup, then Rf phi will be smooth in the so-called jR variable, and kf right invariant in the, in the jf variable. So, by this convolution trick, your L2 space of function, which are just measurable, is generated by very nice smooth functions. Ok, so, basically it's, so, then if you understand the L2 space, you will also understand very well the smooth functions, which is really what we need in this, in this and the continuous functions. And, ok, also, I should say that the HEC operators have a Nadelic description, so, which is the following, so, for, ok, for p sufficiently large, then you can define the, and gp in g of qp, then you can define the HEC operator Tp, T of gp, to be the operator, the convolution operator associated to the following, to the characteristic function of the double class. And so, for, again, if p sufficiently large, the algebra, this algebra of operator will be commutative and made of normal, of normal, even in that case, if g is an orthogonal group, self-adjoint operators. And, so, HEC, because these are HEC operators, HEC eigenforms, so, for, say, pg for classical HEC eigenforms, will be, the corresponding function will be a eigenform for the family of all such HEC operators. Ok, so, I don't have much, much time, but I will try to explain what is, and, so, now I can write, ok, suppose that q is anisotropic and N is greater than 3, which is our case of interest. Then, so, this representation decomposes into an orthogonal sum of irreducible representations, and, so, this orthogonal sum is following, so, this part is very simple to describe. It's a direct sum of one-dimensional vector spaces. It's a chi compose with the norm. So, let's say q is anisotropic, which means that, I recall you, the ternary space corresponds to a quaternion algebra, plus condition. So, this space is this, and chi, this function is, so, you have a character, so, chi, it's a character on the idles, a finite order, and, so, this function is a function with two g in, let's say, b star of a goes to chi. So, g correspond to a quaternion, so, you take the norm of this quaternion, a deli quaternion, you compose with the character, and you obtain a function. Ok, so, and this, here, you, this is an infinite sum over all the echo characters, which, which, ok, which can live there, or which are, say, defined non-trivial thing here. And, ok, L2 cuspidal is a northogonal sum where p has, so, infinite dimensional representations, infinite dimensionals, so, unitary representation of gl of a. And, but, these representations are of, of quite a special type, because they are made, so, they admit a realization made of vectors, which are, which are functions, which are invariant of, of g of a. And, so, these representations are the, so, infinite dimensional cuspidal, cuspidal representations of ga. So, these are all, not all the unitary representation of that group, would be too big, but, so, if the, if q is isotropic, you have a further component coming from Eisenstein series. And, so, then, ok, and, so, so, then, we are interested in the v pi, which admits in the pi, or the v pi, which admits g of L hat, right invariant vectors. Ok, not, not all of, of them, and, and, and there there are only finitely many, no, no, sorry, not finitely many. And, so, for such, ok, either the finite dimensional ones or the infinite ones. So, now, to, so, to study, so, this integral, that we want to study this convergence, we may assume that phi belongs to one of the, that phi is either of the shape, chi compose with the norm or phi belongs to one of these v pi. And, so, then, you have a different situation, so, this case is relatively easy. So, I will do the first case. So, this case, so, we are looking at the integral along h of chi of, say, norm of h, and then there is a, there is a g x naught L prime dh. So, this, you can factor out this term. And, so, we are left just with this integral. And, ok, and, so, what we would, but then this is a, and in a sense we would like to show that this thing goes to zero when d is sufficiently large. So, what happens, and, and, so, for chi, of course for chi, not trivial, not the trivial character. So, then, what can happen is that, so, this, ok, chi, so, just to exist, this chi has to be a further two. So, then, is associated to some quadratic field, because it's a character of order two, so it corresponds to a quadratic Dirichlet character, and, so, corresponds to a quadratic field. So, and, what happens is that if h, which I, I have recalled in the beginning, is associated to a quadratic field to the same quadratic field, so, then, it will imply that chi of the norm on h will be constant equal to one, and so therefore this integral won't be zero. So, then, this explains the exclusion from the theorem of exceptional square classes for, for d, because d determines the quadratic field and, ok, so, which is, this is bad, because we would like the integral to converge to zero. So, and just in a milli, so, I'm very late, so, I finish with just the remaining case. So, now, if phi is in v pi is contained in L2 cusp, so, then, there is a formula of vats purgé, and, ok, and phi plus phi is nice, is nice, but we, we may assume that, then, there is a formula vats purgé, which tell you the following, so, the integral we are interested in, a square, and, so, can be returned up to a constant as an L function of a modular form plus a sort of finite product, ok, we will not write what this is. For, for f, aqueur eigenform, holomorphic, no, not necessarily holomorphic, aqueur eigenform, with the same eigenvalues aqueur p for almost every p. And, this is just the aqueur L function of f, and the aqueur L function of f time key, and, this is the value of the quadratic, of a quadratic character attached to d times d to the 1f. And, so, so, the fact that this integral over h goes to 0 is follows among other things to the following bound, so, which is called, of course, a subconvex bound plus bounds for these things that I have not written, and not, no time to describe here, but which are much easier to get. So, of course, I could have formulated everything in a delict terms, no time anymore. So, so, this is a so-called subconvex bound. And, so, such a bound was obtained for the first time by, by Ivanieck for holomorphic forms, from which Duke could trade a great deal of quadrat, ternary quadratic forms, and then Duke extended Ivanieck's bound to over kind of standard form, mass forms, and so on. And, so, then, together, we shoot the pillow, then they could get more general result, in particular, as a principle. So, when the d was square free. So, to remove the square free assumption, then you have to deal with these local things that I have not described. And here, in that case, maybe this, okay, I should say, but now there is a purely a delict treatment of these health functions, which is due to Venkatesh, which works over general number fields. So, in that way, you get the, the, the theorem that I have stated about ternary forms for over general number fields. Okay, it's over, over time. So, thank you. Any questions or comments? Yes? I have a question about what's on the board behind that board about the appearance of the continuous spectrum or not. I'm probably just confused about translating the classical picture. Okay, so, so, if, if Q is anisotropic, I thought, I thought whether or not the, the continuous spectrum shows up or not depends on whether G is compact or not. Yes, exactly. And, but if Q is anisotropic, it could fail to be isotropic just at a finite place, but the group could still be non-compact. Non, non, non. So, if Q is, so, it, anisotropic is anisotropic, Q anisotropic over the rational numbers. So, you could be isotropic at a real place or at some other finite place, but the global quotient, gm-odulogy Q, will be compact, and then its spectrum will be discrete. So, because all these operators, these RF will be compact operators. So, so, the condition is that, to have Eisenstein series is that you are, that you have, that you are isotropic, and if you consider more general group, it means that your group contains a rational unipotent element. And, and then the Eisenstein series comes from these unipotent elements. Another question. Can you give a few references for today? Okay. Yeah. Okay. So, maybe references, so for, okay. So, for analytic number theorists, maybe if you want to get used to Adels, maybe just to start, I would suggest there is a recent paper of Browning and Pancache Vichet, who was there last week, who's a do Adélique Circle Method. So, there is no automorphic form. It's, it's, say, or the only automorphic form that could appear are the additive characters over the Adels that I have described. So, that's an intro. So, because they prove an analog of a skinner on finding points on cubic varieties in, don't know, maybe nine or ten variables over number fields. So, then, to have a presentation of automorphic forms over the Adels, I would start with the book of Gelbart at Princeton University Press. So, which is a really a gentle book with, at the end, a very nice application. And then, I, maybe I would follow this up with the book of Bump at Cambridge. And maybe there are also the book of Goldfeld, which could be, okay. Now, if you want to enter into the analytic number theory that enters into this proof, now there is no book yet. But, so then, but I would suggest Akshai paper. I think it's something like 205 in Annals of Math. Oh, oh, I forgot. There are nice books by Knightly Ely. For instance, they have a manuscript who give Nadelic treatment of the Kuznetsov trace formula. So, if you do a classical modular form, this book is very interesting. What? Very gentle. There are several books. And okay, so then you have Venkatesh. And if you want to look deeper into the subconvexity problem, then this is my paper and with Akshai. So, probably you can get from the staff at IHES. And okay, and then, I should say, maybe some small advertisement, book in preparation. So, the list of authors is not completely determined. But, okay, so there is at least one author. And I hope more book in preparation on Duke's CRM. So, I would say in general. And so, you can find the first free chapter of that book on my web page. And these free chapters describe more or less all what I have explained in this lecture. So, the four chapters about the adults will come maybe soon, maybe in one or two weeks. And then there will be further chapters. Okay.