 Oui, je continue ma investigation sur Coenlens-Strahe heuristique. Je vous ai dit que l'avarice de classe de l'équipe de quatathique. Alors, ce que je sais de la distribution des armes de carenques, P rank of the ordinary class group of the field Q of square root of P. This is a new class group in the narrow meaning of Q of square root of D, where D is a fundamental discriminant. So what is known? So only for P equals 4 is completely known, the asymptotic behavior. And only, so as you know 4 is not a prime. So this extension is due to Gert and his Fouvery-Cleuner's theorem. And what about P equals 3? Yes, what is not about P equals 3? It is not known. We will only, so we will quote, I shall explain to you the work of Davenport-Ilebron around this year. Maybe someone told me it was the very last paper of Davenport. And what they prove, they are unable to prove the distribution of the three ranks. They are only able to prove the first moment, which is that as usual D is a fundamental discriminant. Ok, so the theorem is the following one. First of all, if you are assuming about discriminant, which are negative, they prove this thing. So I write it on the form, ok. And for positive D less than x, they prove the asymptotic for third. And what is 2? 2 is equal, for me, is equal to N13, which is the cardinality of all sub-vector spaces of F3 power 1. And this fourth third is equal to N23 minus N13, divided by, I put 3 power 1, that's 3, ok. So what is this number? It's the cardinality of sub-vector spaces of F3 power 2. So it's this year. So I must stress the fact that nothing precisely in the context is known at the moment. For instance, no one knows the asymptotics of 3, the power, ok. So we see that your people know very well this conjecture. And for, so there is a case of Coenlandstra, which is known, it corresponds to the two ranks of cubic fil. But it is not in the purpose of my talk, and this is due to, at least, Bargava and someone else, I think. So Bargava, of course, as you know, brought something very new in all this context of understanding, forms, a link between forms and algebraic numbers theory. But apparently, it gives nothing new for this context. So for instance, this is too weak. We do not know the distribution law of the function of this function. But for the four ranks, we know it because we have all the numbers. So we are very, very few things. So something which is remarkable since this year is a result of several people in that direction. So I write in a piece and maybe some. We prove that concerning the three ranks, that the three ranks, if you want the three parts for some delta positive. So there exists a delta positive such that, you see D, you may have in mind that D power 1F is something like the class number, it's something like the odd part of the class group. So they prove that the class group is not eaten by this number. And for instance, when I think about pierce proof, pierce proof requires very sophisticated upper bounds for exponential terms, deline theory if you want. So this is maybe ten years ago. So I remember what I told you at the beginning of my talk when you do not know what to do, do class number and so on. And it is, you can do, many things can be information. So I told you, so we do not know, probability that the three ranks of CD is equal to R is not known, is not proved. Because we have only one moment, we would have all the moment to deduce. You see the theory, when you have the moment, you have the probability, provided that the moment do not go too quickly to infinity. I explained that to you. So for P equals 4, everything is known, ok? And yes, so the proof. Let us speak of Davenport-Hyburn. So I recall you something about, certainly you learned, and what is known about quadratic field versus quadratic form. So as you know, quadratic form was the kingdom of Gauss, if you want, and quadratic field after some years after. And there is a link, as you know, between the class number of quadratic form and the class number of ideals. And the link is very, so you start from, and to make it simpler as D-negative, for instance. And you write I am ideal as I alpha is equal to Z A1, direct sum of Z A2, so ideal of the ring of integer. And you suppose that you call alpha equals A1, A2. And you can, by reordering, you can suppose that you choose the way of writing that. And so what happens is that to this alpha, which is a pair of generators of this ideal, you associate the quadratic form, which is N, the norm of A1x plus A2y, divided by the norm of I alpha. Ok, so it's a classical result, which means that the two class numbers coincide, quadratic field and quadratic form. So what do we know is that this thing f alpha is a quadratic form with discriminant D, and is definitely positive, I do not write, but what is important is that if you pass from alpha, A1, A2, 2, f alpha prime equals biomatrice of SL2 of Z, then the quadratic form, then f alpha, if I write, isomorphic equivalent to f alpha prime by the action of SL2 of Z. Ok, so this is classical and the discriminant is the same and so on and so on. Ok, so let's create a bijection. So the theory almost exists for cubic extension, so I shall make a drawing and first write the name of Delon Fadiath and then Davenport Ibron. So when Davenport Ibron wrote their paper, several papers, they were not aware of the work of Delon Fadiath before the Second World War so now we systematically quote them. I did not know, I would say that Delon Fadiath did not prove the theorem which is above. They were studying the link between cubic fields, I would say cubic orders and cubic form. So I shall make a drawing here. I start from cubic fields. The first circle will be cubic fields via the Galois conjugation and here I put binary cubic form via the action of GL2 of Z. So what I have in mind here. So how do you create from a cubic field, a cubic form, a binary cubic form. It looks like this. So you suppose that Q is equal to 3. Ok, it's a cubic field. C'est peut-être qu'il y a peut-être une conjugation différente. Peut-être qu'il y a un Galois. Donc je considère, j'ai écrit le ring de l'intagère sur les 4. Donc c'est le ring de l'intagère. Ok. Et maintenant j'ai créé une formule cubique dans cette façon suivante de la formule cubique qui est d-1 ½. Donc d est la discrédite. Ok. Et ici c'est la norme. Donc je l'ai écrit plutôt que omega-omega prime x plus theta minus theta prime y. Donc je ferme ce truc. Et vous... Ok. Et ainsi. Donc vous pouvez dire oui, c'est une norme. Vous pouvez l'écrire. Ces sont les conjugates de omega. Les autres sont les conjugates de theta. Donc, bien sûr, ne vous inquiétez pas sur le signage de ces... si c'est négatif, c'est tout. Donc je dis que c'est une formule cubique. Ok. Fk est une formule cubique de z. Donc cette coefficient est de z. C'est réduisible. Et c'est très important. Donc avec la discrédite c'est equal à k. Ok. Fk x, y est primitaire. C'est à dire que vous n'avez pas de facteur coefficient. C'est belong to z. If k1 not conjugate to k2, then fk1 and... yes. And fk2 are not gl2 of z equivalent. And the third n. If k1 is conjugate to k2, then fk1 and fk2 are so are equivalent via the action of gl2 of z. So you see the picture looks like this one. That is quite different. So in some sense you see you may have omega, some other omega and theta. If you take another one you will have, you will recover a cubic form but which are gl2 of z equivalent. So you have to be careful. You see gl2 for quadratic form is sl2. So sometimes there will be a factor 2 as a denominator. So now we have we create a bijection and what I forgot to say it's a discriminant. Yes. And discriminant this is important and discriminant of the field k is equal to the discriminant of this cubic form. So what is the discriminant of a cubic form? You have a definition but if I call it axq plus b what is the discriminant? It's a delta of f so is equal to a quartic polynomial homogenous which is highly degenerate and it preserve the discriminant. So now we are we can pass here and here we have something which preserve the discriminant. So what about so what I prove is it that it is injective, what about surjectivity. The surjectivity it was quite delicate to prove is that right? Irreducible. Irreducible. That means we do not factorize. So it was quite difficult to create when this function I call it phi. When is phi surjective? So it makes several years to Davenport and Heilbronn to find the condition of surjectivity and they are only local local condition and so in the case so it will in the case where so so you have to restrict to some some subclasses so if I suppose I impose that discriminant of the field is a fundamental discriminant discriminant fundamental discriminant of quadratic field it is a surjective. So yes, you see you have an application here it's a discriminant preserving ok and the situation is more easier when we work over fundamental discriminant and it is what I want to use. So yes, so you see the landscape begin and now how can we count the number of cubic form via the action of gl2ofz so if gl2ofz this is not sl2ofz so there may be a factor 2 downstairs so to simplify I shall suppose that d positive correct, yes so to this quadratic cubic form you see with a b cd belonging to z4 you associate its action so yes you define h of s to be oh, I made no error so this is the action so the action of the cubic form and you see that it is a quadratic form that we write on the form x2 plus bxy and yes, so this action as a following property that discriminant of h of s is equal to minus 3 discriminant of f what's the definition of this one? you are so the action of f is that yes applied to ok ok yes there is an error, I see there are too many squares here I think it's partial square of f over partial x2 times partial square of f over you multiply two ringers and subtract in your chart so so is it correct? no and here only one no should be two something like that something like that yes or no? no we think the f should be inside yes here you want something like that excuse me I'm not in bad shape I'm not in good shape so it's the action I think the first one was the right one except for the missing f this one d2z sur dx2 d2z sur dx2 d2z sur dx2 d2z sur dx2 ah d'accord ok moins d2z moins d2z sur dx c'est quoi? donc vous vous mettez le f dans la parenthèse carré voilà le tout au carré ok et puis c'est ok it's not easy to be an old man believe me it's a mess but what happen but it's seriously serious be careful so you arrive at at the quadratic form ok you agree Milovich? ok thank you I have a fun club here that's nice so so what happens here is that this situation you see I suppose that the discriminant d is positive so I arrive here and the discriminant is negative and this is a quadratic form quadratic form you know so what we have to write is that classify this f is the same thing as classifying the h of f and these are quadratic form and we know how to deal so to be more precise yes so if you write h of s I wrote it here then the value of a equals to b square minus 3ac b equal bc minus minus nc 3ad so what is important is this last step so in fact classifying this binary cubic form of gl2 of z so it was due I don't know which people it's the same thing as classifying the h via the action of gl2 of z and I can give you the value of a b square ok and c is equal to c square minus ok c'est similary and so you see the application and now you see in fact we arrive at the the problem is the same as classifying quadratic form via the action of gl2 of z we know well about sl2 gl2 of z is not more complicated so in fact you know the story you have even it is not exactly nice you know that the classification you will meet certainness some inequality of that type you see these are quadratic form with negative discriminant which are the easiest one so so what we arrive so here we arrive at quadratic form with discriminant negative and via the action of gl2 of z which is I would say the half of the action of sl2 of z so we arrive at this lemma cubic form cubic form of gl2 of z we arrive at counting points abcd such that minus a ok or a positive delta of abcd is equal to d discriminant from which I was starting so you have a some problem you know it is not exactly but it is really what is important so what is you see cubic field by conjugation so you make this with discriminant negative and here the trick was to use the action ok and for instance maybe we said 20 years ago created new tables of cubic field using this bijection you see it was rather working with this expression so now we are almost finished we must end for the moment we do not see the 3 wrongs the 3 wrongs will arrive here here so what is it let this lemma proposition which starts from a suppose that d is a fundamental discriminant then 3 power the 3 wrongs of the class group of q square root of d minus 1 of the 3 minus 1 is equal to the isomorphice classes of cubic field is equal to the of cubic field with discriminant ok so we count subgroup of index 3 and we use class field theory so here we arrive at the quantity the 3 wrongs of cd minus 1 over 2 and this bijection is very special for 5 it does not exist ok so we arrive at this number ok so you make this drawing and here you are happy because here you are see this formula I wrote yes here we arrive at counting points in a volume of air power 4 ok so you see it's really and I'm using you do that if you are here you have this problem you go to cubic field cubic forms and then you taking the action you arrive at quadratic form via the action and we know what is the fundamental domain for quadratic form ok so so now we continue we arrive little by little at analytic number theory ok so this class field theory you see and so let v of x so for d negative there is another way of counting which is not so easy to explain but it works so you see this whole diagram takes I would say I would say 20 at least 20 years for Davenport and Dilebron they were first dealing with cubic form and then after coming to cubic fields so if I denote by v of x to be the cardinality of a, b, c, d point in z power 4 a positive minus a less than c yes or ok so I add a positive delta of a, b, c, d between 0 and x and delta of a, b, c, d is a device of I wrote it somewhere ah yes delta of I define above so you have to count points in that region so when you know v of x you know so it's almost with very tiny cardinality of v of x is the sum over c, d minus 1 over 2 ok so rather speaking we have a number of points in this crazy volume in f4 you can't integer points with the number a, b, c this one and so with all this way ok and you say that this thing is exactly so you recognize you see the fundamental domain for negative for quadratic form with negative discriminant this is very deep result quantity about the 3 rank of a quadratic field so now we are happy because analytic number theory shows it's nose so we are we arrive at counting ah yes we arrive at the problem of counting points in a volume ok so crazy volume in r4 the second condition or at the end it should be equal to c or a equal to c yes you are right exactly you have in mind the fundamental domain when you consider quadratic form you see all this one ok so it's not exactly there is some tiny difference but it is useless to speak that time so here we see that v of x is a volume in r4 so we are counting integer point in a volume so so and so we have the philosophy that the number of points is equal to the volume of v of x but as you know it's not exactly the case it's more difficult to do that so ah so what happened about this volume this volume first of all there is a problem for this volume in r4 it has a cusp a cusp that means that you may find very we cannot you see we have a may be very large ok so you have to you have to cut it so it's Davenport and Albron did that so ok so now what I want to say so we have the feeling that the number of points is equal to the volume ok c'est pas this is not clear at all see we have these a b c are given by that the discriminant is given by that so you have to use some lemma so you see to generalize this problem in dimension 1 ok in dimension 1 so suppose I ask you how many integer points there are in this segment ok so you have to count so you know it's almost the length of the the sum of the length of the interval but you have an error term which corresponds to the number of connected component and here it is bounded so in fact we use so what we call leap sheets Davenport use principle ok which says that I make you a drawing so suppose you have an area like that and I ask you how many integer points in v so leap sheets principle it says that it is the area of v plus capital O of the sum of length of projection of this of this of this area on the axis you see so you bound this ok ok yes so this is really so what we call leap sheet principle it is more sophisticated than that you have to count to take into account this phenomenon also and what they did Davenport and Heilbronn so in fact there will be a bad error term yes there is a plus 1 because you count the value of dimension 0 ok you know that ok so this yes leap sheets principle so when you have that you are able to count and in fact there is a bad error term if you follow that there is a bad error term when you count this problem because you have a condition to be square free see d is square free almost square free ok so we arrive at this yes at the sum of excuse me yes so we arrive at this formula formula minus 1 is equivalent to 1 over 6 4 if you count negative d so that means x tends to infinity and when you have that you have something like when you count cubic form no cubic fields up to isomorphism and you ask the discriminant to be so you find that it is asymptotic to x so you have to be more careful about local condition yes so for negative d so I wrote it yes this is positive and for negative d you have 3 times more so in fact this was quite interesting question and people like Bargava several people Shankar, Bellabas we are more precise in this formula and in fact they were able to find for instance for this one they have a first term which is c0x plus c1x plus capital O of x power using the fresh ideas that Bargava introduced in this counting process in algebraic number theory ok, so this is the first yes to give you to check something so now I would like to give you some hint about the proof I don't know if you have time about the theorem which is due to Bellabas and Fouri, such that there exist c0 strictly positive such that the cardinality of p less than x p congruent to 1 mod 4 3 does not divide h of p is greater than c0x divided by log x ok so so if you remember during my first lecture excuse me I forgot to tell you we have the same thing now for negative discriminant but when you are doing negative you pass to positive discriminant we do not like that so here so how do we do that so it was and we have not the same statement so excuse me I forgot to tell you that h of p is a class number and such assessment does not exist we are not able to prove it for for negative discriminant so we are mixing sieve and algebraic number theory so for instance Bellabas proof it will be nice to call h of d to be 3 powers or 3 ranks of cd minus 1 divided by 3 minus 1 so for instance we will use sieve and analytic number theory so for instance we have this first formula so in arithmetic progression so we are counting such that q divided d this is asymptotic to 1 nu q over q and what is this and nu of q is a so I must impose q square 3 and nu of q is a multiplicative function which is product of pi divided by q divided by p plus 1 so first of all we have to use this thing and this is true uniformly for q less than x power and so when we want to use sieve we will use average distribution and to use so h of d is this number ok so I shall erase the blackboard so we will use we have to push this argument here since time is going I shall only write so equation is the following one before using sieve ok so what we will use is find the number of solution to the equation so did I define it delta I can't remember yes delta of a, b, c, d congruent to 0 mod p where a, b, c, d belong to some box b is a box in air power 4 so it's a very difficult problem you see but we have a property because delta of a, b, c, d it's a discriminant so that means that if you have this is equal to 0 you know that it has a double factor you have x cube ok suppose it has a yes that means if the discriminant is equal to 0 that means it has a double factor that means that you can write it something like i and so you have a parametrization and this is a crucial part of the proof and now I would like to say some words we use the parametrization and now so a key point is this CRM so we use a notion of quasi discriminant so what is the of order p that means that if that means it is delta is a quasi discriminant of order p see if it is congruent to 1 mode 4 and p2 divides delta implies p greater than capital p so you see we have to meet the difficulty of to be square free it's very boring when you have in mind that we know that a cubic polynomial we know in one variable infinitely many often with the right frequency and it's a deep theorem not trivial we do not know the same thing for quartic and so we are here yes and we prove yes and we prove for this sequence a level of distribution of order so a level of distribution for quasi discriminant order p it's 3 8 ok so I can so p will be fixed p fix but ending to infinity as you show so I think so actually I did I write point yes so you use sieve and the fact differential and we did not know so you see this theorem is almost that I would say 20 years old and there is for the moment no extension to suppose here we are working with negative positive discriminant that's surprising because we are feeling that positive discriminant real quadratic fields are more difficult than imaginary and this theorem no one knows how to extend it to negative p if you want you see you impose p congruent to 3 mod 4 and you consider h of minus p we are not able to prove that and in fact here we pay so I have no time we benefit from negative discriminant ok so and here we benefit from the fact that on average the 3 rank of this part has a tendency to be smaller when you consider positive d rather than negative d so so yes exactly yes yes but you see in some sense what is very fascinating I come back to my first lecture so let's go back to ok gauss so conjecture does exist infinitely many p such that h of p is equal to 1 ok and you have different from 1 and not to be silly I impose p congruent to 1 mod 4 you see this one is gauss this one was cohen and suggested to me and no one knows how to prove that this one is what we prove belabar sandai sign of sin in this direction and we have not the dual for with a negative p if you want to see because it has a tendency the 3 part has a tendency to be larger for negative d and so it's more difficult to to make it equal to 0 in some sense and it's quite important did I write some 3 over 8 no no no no ok yes ok and what is important is that 3 over 8 so it's in some sense or it's a pti ok so you see it's a pleasure to explain that because we are picking mathematics everywhere that end with the exponential some class field and so on so on and so many things are to be done around this question you say you can say well yes what about the 5 round yes just do it or you will see there is a problem in my diagram somewhere no problem you see we are mixing so it's a ok I hope that my lecture were clear enough but we see I had great pleasure to do that because we are mixing so many types of mathematics and for it's not to see my theorem of Bella Bath is almost 20 years old and people know the problem of x and y with a negative p but it's a block somewhere ok so thank you questions I have a process conjecture concerning some of the individual components so you want to say what about the class what do you want so the p around and ok so you can what you can say you may guess that independence the 3 parts, the 5th part the 7th part for example no one has something against that you see class number is terrible when you go it's a great pleasure to do that because you see the history you see first of all Gauss and then boom and not a conjecture dating from 2 weeks for instance you meet not that no no I remember so Yvanierck wrote a report about Bella Bath Thesis and he said Gauss would have listened for me it's a compliment for Bella Bath Thesis ok class number I would say you see you know this terrible H of p and this one you see very fascinating and Bargavam is making a lot of work but in the context of what I explained there are very very few you can say me what about the 5th trunk where does the arrow where they are broken here so just to remark so Bargavam had counted Quintic Fields but if you're able to restrict to Quintic Fields whose galop closure is to the hydro group of order 10 then you could get some maybe you could get some results of the 5th ring yes exactly but it is a sub family yes exactly yes you see and for instance I have a paper also with a sub family we are counting asymptotically but as far as I know the big question the big set we have not it asymptotically and I would like to say it's a part of number 3 where you can put many many things very fascinating and for instance a real quadratic field ok, c'est always very normal