 Je vous remercie, donc ma lecture sera de vous partager, particulièrement avec les jeunes étudiants. Tout le plaisir que j'avais à étudier cette question sur le quartier, de ne mettre rien dans la question concernant le groupe de classe, pour exemple. Quand j'ai préparé cette lecture, je pensais que quand je suis étudiant, j'avais deux superviseurs, Désouyé et Yvagnet, et Yvagnet m'a dit que quand vous ne savez pas ce qu'il faut faire, c'est un numéro de classe. Et maintenant, je me dis que quand vous ne savez pas ce qu'il faut faire, c'est un groupe de classe, et c'est fantastique. Nous ne connaissons rien, tout est nouveau, et le challenge est de mettre, je dirais pas étrangement, une mathématique. Vous avez le plaisir, et vous commencez bien sûr, et c'est très excitant. Pour le moment, je suis beaucoup plus excité par cela que par la théorie primaire. Donc, maintenant, on commence. Donc, quelques basiques, des faxes. Donc, je commence par une extension finite de Q. C'est le ring de l'intagère. Et nous considérons que c'est idéal. Et donc, je considère que c'est un set non-zero idéal de OK. Et puis, vous dites que I et J sont équivalents, si et non seulement si. Il existe alpha et alpha prime, qui sont belonging à OK, comme quand vous multipliez, quand vous avez cette équité. Donc, c'est une relation équivalente. Et ce qui est très surprisant, c'est que vous considérez un groupe classique, et c'est le theorem zero, ce qui est cet état, avec cette relation, c'est une groupe finite de Abelian pour la loi induite par la multiplication, la composition, la multiplication des idéaux, très classiques. Et ce que nous avons, c'est CLD, ce que je n'appelle pas CLD. Donc, vous ne devez pas faire des confusions. C'est ce que nous appelons un groupe classique ordinaire. Ah oui, vous avez bien compris, CLK. Donc, cette CLK, à moins de 3 questions. Je voudrais avoir 3 types de mathématiques qui sont intéressées par ça. Tout d'abord, c'est l'algebra, des gens de la théorie algebraique. Et maintenant, on peut dire des gens de l'analytique et des gens d'un point de vue computationnel. Donc, c'est très intéressant, il y a 3 points de vue. Et je serai plus concerné pas avec le cas d'un fil quadratique. Qu'est-ce que c'est? Alors, vous impose K over Q equals 2. Et ce qui est important est le discriminant. Le discriminant fondamental. Donc, je vais le dire D. Donc, comment vous buildez le discriminant fondamental? Vous commencez par D star square 3. Et vous impose E. Donc, si c'est square 3, 1, 2, or 3, mode 4. Et c'est positif ou négatif. Et D est equal le fondamental... Vous obtenez 4 D star. Et vous multipliez. Donc, D est equal à D star or 4 D star. Et ici, quand D est congruent à 1, mode 4, et c'est 1, 2, or 3, mode 4. Donc, nous savons exactement ce qu'il y a dans le cas de quadratique. Nous pouvons enumerer tous les fields qu'il y a sans problème. Donc, ça signifie que vous considérez toute la square root d. Et il y a un CRM, qui est très facile. Et... pour ce set de discriminants, c'est le CRM. C'est... 0 de 1. C'est 3 par pi square x plus capital O de la square root de x. Ce n'est pas très bas. Et c'est le même type de base à la square root de x. Et c'est important. Vous pouvez... Cette formulae, donc, c'est une petite modification modifiée. Nous allons l'utiliser. Vous pouvez le faire pour D négative. Vous pouvez imposer des congruences. D congruent à 1, mode 4, ou D congruent à 4, mode 8, ou D congruent à 0, mode 8. Vous avez exactement le même type de formulae. C'est important parce que c'est surprise pour moi. Oui. Les gens doivent être conscients de cette formulae. C'est très rare. Mon question est que je ne peux pas nkx être la cardinale de l'optimité. Vous vous donne un sens à cette formulae comme le discriminant de k est moins que capital X. Vous comptez combien d'extensions de k que vous fixez sont les discriminants moins que x. Et la question, pour 2, c'est con. Et bien sûr, vous pouvez mettre un minus. Et il y a très peu de résultats. Je vérifie avec Bella Bass. Donc, pour k, c'est due à Davenport-Iron. Donc, vous voyez au début de les années 70. Pour k, c'est plus recent. C'est due à plusieurs gens. Olivier, Diazidiaz, Cohen et Bargava. Et K plus 5. C'est un couple de Bargava. Et c'est tout. Donc, j'insiste sur ce problème. Les gens sont capables de compter des familles. Mais ce type, comme je le sais, de la généralité, nous ne savons pas comment compter. Donc, je continue avec le 2 quand k est q root d. Donc, je vous rappelle des factures fondamentales que vous avez un formulae easy avec z omega. Avec omega c'est d plus root d divided by 2. Bien sûr, vous avez un problème x plus y. Et donc, ici, nous devons mettre x et y en q. Et ça already devient une différence. Donc, bien sûr, nous parlons de ça. Nous avons un set de unités. C'est complètement différent quand d est d positive. Ici, vous savez qu'il y a un numéro w qui est equal à 2, 4 ou 6, qui compte le nombre de unités. Pour d positive, vous avez quelque chose comme plus ou minus 1 multiplié par epsilon d n belonging to z. Et cette epsilon d est ce que nous appelons les unités fondamentales qui sont liées à l'équation qui est le maire de Real Quadratik Field. Il y a un autre groupe de classe qui arrive. C ld sera le groupe de classe ordinaire de q de root d. Vous avez le groupe de classe narrow qui est vous considérez que vous avez une nouvelle relation avec un n et vous demandez que alpha e est equal à alpha prime g avec norme alpha prime positive. Donc, c'est un groupe de classe narrow et parfois, vous êtes très puzzés d'imaginer ce que les gens ont dans la tête de la classe du groupe de classe. Donc, comme vous pouvez l'imaginer donc, quand d est négatif il n'y a pas de problème. Et ce qui s'occupe c ld non c ld est un groupe de c ld avec index 1 ou 2 et c'est une question très difficile de voir quand on a une équalité. Donc, oui une proposition facile donc, c'est que ces deux groupes de classe coïncident. Donc, la première possibilité est que d est négatif ou d positive et la norme de epsilon d est equal à minus 1. Donc, donc, on peut demander ce qu'est-ce est-ce que cela se préconise? D'abord, pour voir si on fait confusion. Donc, la question est quand est-ce solvable? Ce n'est pas si facile c'est très difficile c'est que n epsilon d c'est solvable est équivalent à l'équation de paramètres d star c'est solvable donc, vous devez prendre une queue parce que quand vous pensez de l'intagère vous avez numéro 2 ici donc vous devez avoir l'équation de plus ou minus 4 et maintenant oui? Donc, cela implique tout le facteur primaire p divise d star implique p est equal à 2 ou p correspond à 1 mod 4 et toute la difficulté est que la converse est forte ce n'est pas un local global c'est sur z donc j'ai écrit ces choses donc c'est la cardinale de d entre 0 et x comme que cld excuse-moi cld est appelé cd c'est très rare c'est un procès c'est un procès généralement on a c'est un gros groupe euh la cardinale c'est equal à 2x cld cardinale donc, qu'est ce que nous savons nous savons le numéro class numéro class donc quand d est négatif pas de confusion donc généralement vous trouverez la notation h et c'est w qui est equal à 2 4 ou 6 l1d euh square root of d divided by 2pi et pour d positive la formule est seulement pour ce que l'on appelle h d qui est le groupe ordinaire class et ici vous avez square root d l1 ok log of epsilon of d qui est notre nightmare et euh euh, oui, un autre un autre ce qui est intéressant c'est que bien sûr vous parlez d'une pensée de Gauss Gauss c'est une formule quadratique bien sûr cd peut être interprété euh binary quadratique form over z via the action of sl2 of z ok euh so I would like to now to give some conjecture so I would like to say yes why is it the nightmare because the nightmare is here because so many formulas you have always formula when h of d log epsilon d are married this is the case it's very when we can separate them we are doing I would not say silly thing but we are not approaching the difficulty the real difficulty of saying same thing h of d alone or log of epsilon d alone now I write some some conjecture to make the state of art conjecture that I which I would like to to stress and they will concern mainly euh the deep positive so I suppose deep positive and certainly the most famous one is due to Gauss question infinitely many p subjet with p congruent to 1 mode 4 subjet h of p equal 1 so that means there are infinitely many real quadratic field with which are principle c is positive it's a usual prime ok the second one is strange I remember it was quoted to me by Cohen I discuss about that we do not know this thing so you can I would like to write the name of Cohen there exist infinitely many subjet h of p different from 1 we do not know that it's strange at least that's strange we have this one always prime you see we will meet some day the parity the parity question the third one so of course this one is was implicitly in the certainly in the brain of a lot of people so I would like to so just at the same moment three people logical coincidences I say Sarnac, Houlet et Cohen wrote these two conjectures so there exist c1 strictly positive subjet when you sum the class number between 0 and x it is asymptotic to some c0x log2x so you see I was speaking to you of marriage and now I speak of divorce so I make comments there exist c2 c1 on the right ah yes that's false I stop my talk so h of p so I write p congruent to 1.4 is equivalent to I think the constant would be 1 over 8 something like that c2x so and now the fifth one so I I write it in a very so here also the convergence of several mathematicians the three I quoted so prove prove prove that epsilon d that most epsilon d concentrate exponential of square root of d so I do not define what means concentrate and you can guess what is a nightmare for computational point of view you take a d reasonable suppose with 3 digits and what is epsilon d we see the difficulty so in fact this conjecture ah say they are equivalent but they have the same feeling as soon as you have that you may think that in the class number formula where is it ah yes the log epsilon d is almost square root of d so you think that's a huge consolation so in fact they are almost equivalent to none of this one so to say that the number of classes which odd exponent is very vaguely around log of d ok and I shall explain that after so by prime number theorem you guess that here there is something else appears the two part the even part which is something that a 2 power omega of d something like log of d ok and so this one so they are very challenging question and I would say to what is known I say what I like about 1 and 2 so nothing precise is known about 1 and 2 but I have a strange result with Karim Behlabath which says that there exist so it's a theorem not a feeling these are conjectures there exist a positive proportion of p congruence to 1 mod 4 such that 3 does not divide h of p so it's a very few result with h of p infinitely and a positive proportion of prime and Rick will ask me what about parity you see but this sequence is very typical we can't escape and you see you may ask so with this thing he said nothing here nothing here of course and so it was prove around this here and people know this result and it's always known a question to prove that if you take p congruence to 1 mod 4 then such that 3 does not divide h of minus p so it's so we have the feeling that real quadratic field are much more complicated and here an example of something we know about real and it is not known about complex so the idea is about the number of difference we will see with about the number of element in the 3 part so now about this one yes so I have a result about this one it's a yes so it's a it is I would say it's something it's due to myself and improved by Ross so which prove that for almost all d epsilon d is greater than d power 3 minus delta it's it's very I would say it's ridiculous because when you compare this number with this one it's like the height of my tree in my garden it's the distance to the sun nevertheless if someone prove with 4 it will be accepted I'm sure because you see the question is suppose you improve that which type of mathematics you improve Ross use a determinant method ok you see it's anything can arrive in this question ok yes so now we little by little converging with Cohen and Lenstra so what about d negative d negative I think about this question we know much more and our philosophy is that the odd part is something like square root of d so it's completely different the class number and this one for almost all d for almost d so excuse me yes yes yes yes but yes but but the number 3 is not so easy you have to push to get to 3 your technique yes it's not for all d yes it's for almost all d ok for almost all d and Enrique to come to my mind the question suppose you know that infinitely many prime of the form n squared plus 1 many questions here are solved I don't know if you will be talking later on about the so maybe I would like to say something the separation of h of d from the particular unit is very well done in the particular place for it and because you have asymptotic for the I mean a lot of the length of your desert and so you can average class number according to the length of your desert then you should separate etc and other observation is never recorded in liberty is that if you look at that contribution in the place for the physical format of those the main it's a very special experience so that the ones to comment what you are saying that I insist about this type of summation it's what you have in my people some people may result about the log of so now we go to so always a quadratic field what do we know about the prong so first of all I must define what is so an abelian group and what is the prong of this it is a dimension of hp of a divided by p a so I use also sometimes the notation the cyclic group of order p by nu and if you write so it's maybe it's to have a feeling so suppose that you write as a product of all the prime nu to 1 to infinity of a cyclic group and which one appearing with a frequency alpha p nu then the prong is a sum of alpha p nu so what is well known is the theorem of Gauss always M which says that the 2 wrong of cd is equal to the number of prime divisor of d minus 1 so in some sense we know any question about the 2 wrong is a question about omega of d so it's rather simple so it's solid it and now what about the literature what about cl of d what is the 2 wrong of cl of d so here we have problem in the literature because they are false proof but they are correct one so I know that cl of d is a factor group of cd of index 1 and 2 so it may affect the 2 parts but it may affect the 2 wrong or maybe a factor c8 may become a factor c4 ok this is not sufficient to guess what is the 2 wrong so I recall that if you have that so we have always the 2 class group coincide ok and is this your theorem now I think you too I think we found a proof correct proof in Arthur and so suppose that d positive and epsilon d is equal to 1 so the 2 class group are not the same then the 2 statements are equivalent so the first one is c of d is equal to the cyclic group multiplied by cl of d so that means that here the 2 wrong is lowered by 1 and this is equivalent to there exist a p which divides d such that c if ok, this is equivalent in other words, if xy minus d star is locally solvable but not no, it's it's useless what I say yes so we have to guess so we know how behave the 2 wrong of the ordinary class group now we arrive at nstra heuristique so it's number 2 so it's so it's published in literature notes in mathematics in some journal arithmetic in this year and a lot of people have thought about these heuristics they have been generalized some of them have been proven extending the situation so I shall not enter in the algebraic interpretation very superficial thing so they use a computer first of all and they count the proportion of d negative up to the capacity of the computer the d negative such that the 3 part of cd is c9 is something like n now is another group of cardinality 9 and here they find that it is only 1 so how can you explain that so first of all you but you must be careful in this approach with computational because the convergence is very very slow, I will give you examples so the ratio of these two numbers is 8 and 8 is correct and this is 9 yes I like it and what is this this is the cardinality of the automorphism of c3 square divided by the cardinality of automorphism of c and the 3 the exponent is here ok so what I find to explain you what is behind so they are the the brilliant intuition to affect to some group some question where you want to meet some group not the cardinality because here the cardinality but this weight the inverse of the cardinality so you see here this group has much much more automorphism than this one so it appears much oftenly less oftenly and so some easy so some cardinality I want to to you must be very careful this cardinality is p r minus 1 p r minus p r minus 1 this is easy and it's natural this formula is classical it will be natural we will always find formula with the eta function so this function appear in theory of partition and so the first one is you take eta k of t is the product this is a real function ok t power minus i and you suppose that of course t is less than 1 and eta infinity of t is equal to the product i greater than 1 1 minus t minus i so this product is very quickly convergent, we will benefit from that and for instance with this notation this quantity is p power r square eta r of p so you see when I write coenlens traconjecture on a envie de voir cette fonction on le met fréquent qu'est ce qu'il y a j'ai écrit coenlens traconjecture dans la façon dont j'aime et je j'ai j'ai ok donc laissez p prime à peu près 3 ils étaient fréquentés par la prime p equals 2 parce que Gauss était ici et donc d est négatif donc j'utilise leur notation c5 parce qu'il y a d'autres conjectures la probabilité donc vous pouvez imaginer ce que je veux dire par la probabilité que le p rank de cd est equal à r cette probabilité est equal à p p product 1 minus p minus k minus 2 de k c6 il dit que l'avantage pour alpha integer r est equal à 0 l'avantage d'une expectation d'une expectation d'un produit fou pas fou mais je dis le p rank de cd minus donc quand vous summez de cette probabilité est equal à 1 donc quand vous voyez ça vous pouvez vous rappeler c'est l'origine de ça et pour le field ils s'appellent c9 c10 donc la probabilité est un peu modifiée c'est p minus r r plus 1 c'est presque utilisé c'est la probabilité non c'est p minus alpha donc ce sont les de quen l'instra dans le papier original maintenant on doit faire des recommandations ce qui est connu ce qui n'est pas connu donc ce qu'ils ont à soutenir leur conjecture c'est le time c'est la data et cette conjecture nous nous offrons seulement un cas donc nous nous offrons seulement un cas c'est à dire nous nous Bloody Several dans le port de Davenport qui est C'est ce que j'ai dit. Elles ne leur donnent que l'expression ici, donc elles ne l'ont qu'à prouver. Je ne sais plus. C'est ce que j'ai dit. C'est ce que j'ai dit. Je pense que c'est alpha equal à 1 ou alpha equal à 0? Alpha equal à 1. Oui. Ok. C'était seulement prouvé pour ça. Il n'y a pas beaucoup de change, je dirais. Mais nous avons une extension qui confirme l'intuition de Cohen et Lenstra. Donc, j'aimerais faire quelques communs. Le premier, si vous le croyez, vous devez voir, vous devez avoir en mind que cette coefficient va à 0 très rapidement, très rapidement. Donc, pour exemple, je doute que quelqu'un, un computer, c'est-à-dire un D avec les trois rounds de CD, c'est-à-dire 10. C'est-à-dire que vous avez la première coefficient. La deuxième est boundée. La première option, p power, c'est-à-dire 3 power minus 100, c'est très très petit. Donc, je pense que c'est vrai pour des spécialistes de compétition. Je pense que ce n'est pas connu. Et ce n'est pas surprenant. C'est très difficile. Et, premièrement, avec mes collègues clonaires, je dirais que vous devez... Je n'aurais pas critisé, bien sûr, mais commandez c6 et c10. Donc, on fait des observations. Je n'appelle pas nkp, c'est la cardinale des... Je ne sais pas ce que c'est en anglais. Linear vector spaces, sub spaces, of any dimension of hp power k. Yes. Of any dimension. So, the computation is almost done. So, you have a nkp is equal to the sum from l equals to 0 to k. And you have a ratio pk minus 1, pk minus p minus l minus 1 divided by pl minus pl minus 1. So, the eta function is here, the eta k function. So, c6, so it's only a playing with, I would say, a formula. So, c6 is equivalent to, for all alpha, for all alpha integer, the average value, that means here, what I wrote the expectation of p power alpha cd is equal to this number of vector subspaces. So, we have the moment for someone from analytic number theory prefer computing moments than this expression which is full of algebraic meaning in terms of automorphism. And now, c10 is equivalent for all alpha integer. So, it is slightly modified, n alpha plus 1p minus n alpha over p, over p power alpha. And in some lecture, I will explain in our proof about the forum with Clunus, we see this vector space. We do not see the automorphism, but we see this linear vector space. Okay, so, now, we made some remark with Clunus, because, and I think that Cohen and Lestra are not very precise. Maybe it was clear in the, I don't know. So, it's not, I would say, a proposition. Proposition is that c6 implies c5 and c10 implies c9. So, and I have this version of c6. So, as soon as you are able to compute all the moment, you know the probability. Here, also. It's not surprising, but it's not evident, I would say. It's not evident. And the converse, we cannot prove, of course, we do not know what is the rate of convergence of this probability. From probability, if you write that, you cannot deduce the expectation. So, nevertheless, I like this proof. And because, I must admit that sometimes we meet the paper of East Brown concerning the two Selmer group. They have the same type of solution. And we meet an infinite linear system. I went, so, I write what is it. So, I take only the negative. Oh, and so, when using Lebesgue-dominated convergence theorem and so on and so on. So, I call dr, the probability, the proportion, that the p rank of cd is equal to r. So, we use, so, I can't, it's not evident, but it is not deep. So, you meet this infinite linear system, which is d0 plus d1 plus d2 is equal to n0p, which is 1, of course. So, the sum of probability is equal to 1. So, and after it's more d0 plus pd1 plus pd2 plus p2d2 plus pq, and so on. And d0 plus p4. So, it's like, I would say, an infinite van der Mond determinant. Is Brown consider that like a van der Mond, with screeners, we find another technique. So, I would give some, to prove that it has, to check that it has a solution. And in fact, that the solution is unique. So, first of all, check that dr is equal to the probability I wrote, which is here, is a solution. So, I was very happy. We had to open, first of all, RDN write book in the partition function. And so, we find this, of course, it's classical. So, you know, if you see the partition and not the n' prime, you know that this is 1 over, and it is exactly the function eta infinity, 1 over x. And now, there is a classical extension, always in RDN write, which is a lemma. So, certainly, they say it was known to Euler. So, this is equal to 1 plus x divided by 1 minus x square plus x4, such a formula. And in fact, we required another one, which is, we found it in, excuse me. We can write this formula as eta infinity of 1 over x is equal to the sum xk square divided by eta. So, we are in the, I would say, in the kingdom of partition function. So, to check, this is not sufficient, this function. We have to generalize this equality. So, we have to generalize it. And it was proved already. And the generalization is this one. So, you take, so suppose d positive integer. So, we will, we will slightly, here you have the, the sequence of square here. We will slightly modify this, it will not be r square, but r, r minus t. But it gives something different, which is that always the partition function p of n xn. So, it is the sum from r equals t to infinity xr r minus t. And then, I write it because, and so you finish 1 minus x, xr minus t plus n. So, it is not in rd little wood, not in rd n right. So, this formula, you can use it to check that it satisfies this infinite linear equation system. And I like the proof of that, because you are exhausted. So, we make drawing. Ok, to explain with what we call derpy rectangle. So, we found it in a book of Contest. So, I take 1929 and I decompose it like 8 plus 7 plus 4 plus 4 plus 3 plus 2 plus 1. And I make the following drawing corresponding to 1, 2, 3, 4, 5, 6, 7, 8. Après, après je... 2, 3. So, what is the, so this is a usual description of a partition. So, we go maybe in the 19th century, so nothing deep. So, what is a derpy rectangle with defect t equals 1? So, you take the largest rectangle you can put inside this drawing with size, size r minus 1. And the drawing is the following, it's 4, 3. You see, the largest rectangle is here. 4, 3. Ok, you cannot... You see, there is one difference between the... Longueur et largeur, I don't know English well. So, when you are... What happens is that you are going to play to each partition of that tab. You will associate, first of all, a derpy rectangle and partition. Here, what happens? Here, you have 3. Ok? So, you have a certain number of... Here, there are r minus t points. And then, suppose that here you have l integer, l points, and here m points. What is here? You see the drawing here, you have a partition of this number m with some amount less than this number, less than 3. You see? So, you analyze this drawing like that. You see, you have the partition of the number l with some amount less than... Here, it is 4. Here, it is less than 3, even if it is 2. So, you have a bi-direction, and this gives this very integrated formula, which is the key of the fact that the dr... r solution, the dr, I propose. And now, about the unicity. So, the unicity for me is strange, because you are... So, I claim that this system I call S1. So, it's the first part. The second one, S1, S minus has, at most, one solution in positive di. And the key, what is important, that nkp is something like is less, less, capital O of pk2 over 2, something like that, yes. It's only n. What happens is that the second, what we call the second part of here, which is n1p, et so on, goes to infinity, but not too quickly. And here, we will use... So, I shall not explain Jensen formula to create... We built a complex variable function. And there are systems of that type which have not a unique solution, for instance. So, suppose here I replace... Suppose, here, I replace, here, but some ck. So, it will be c0, c1. So, it may have more than one solution, may have more than a unique positive solution. So, I give you... So, I shall take ck equal the sign, hyperbolic sign of pi p power k. And, of course, you will check... Excuse me. So, you will find... Excuse-moi. So, to find solution, you use Taylor expansion. You use Taylor expansion of the function of sign, hyperbolic sign of pa k at the... xk. And you will find another solution if you consider the Taylor expansion of sign, hyperbolic sign of pi xk plus a usual, the usual sign function. So, you use Taylor expansion of these both functions. You will see that the ck, what I call the ck, yes, it has this value. And the Taylor coefficients are not the same and positive. And you see, this function goes to infinity. It's absolutely not the same sign. It's something that exponential pi p power k. So, it goes to infinity much, much quicker. So, it was a pleasure with Qunos to make a Jensen formula. Okay. But he's brown, I don't know the technique. So, thank you. Allé, c'est fini. Avez-vous questions or comments? If you know the probability, the same type of vector, you can't get the second. No, you do not know the rate of convergence of the probability. It's a theory of momentum. And the edge, you see, it does a theory of momentum implies that we know the probability. It's work here. You know. If I have a rate of convergence, if I say more, I do not say probability, but the number is equal to a very good error term. Yes, I can do. But, you see what I mean?