So P is a prime greater than 3, and we would like to prove that, what I call C6, to refer to the original work of Cohen-Lenzstra, C10, here I'm summing, so these are conjecture,

which would be true for all primes greater than 3, for all k equals 1, 2, ok?

And yes, D is a fundamental discriminant of a quadratic field, C D is a narrow class group.

If you confuse it with an ordinary class group, it is not drama, because I explained that they are frequently the same, and so we hope that this is asymptotic to n k p,

ok, 1, and here n k plus 1 p minus n k p divided by p power k, ok, so these are conjecture,

and so what is n k p? This is not the original version, it's only equivalent form, n k p is the cardinality of the set of the linear vector spaces, all linear vector spaces of f p

power k of any dimension. So this is when you have the moment, I'll explain to you

how we can get the probability that the k rank, the p rank to be equal to r, using I would say what we call the theory of format, ok? So that's not all, so what

happened in the history? So there is something very, very important, I think, in the influence of Gert, around, so Gert, around this year, said, you see Quenlenstra with p greater

than 3, because it's not a problem to deal with the two rank of the class group of q of square root of D, because by Gauss theory, we know that the number of prime

divisor minus 1. So the idea of Gert, say, in fact, we can extend the Quenlenstra heuristic to p equals 2, but instead of considering C D, you consider C D square, ok? So what

does that mean? That means that now, I shall explain some comments, so you put here p greater than 2, C D the square, the square, and the same, the same asymptotics. Why?

Because it's easy to see that if p is greater or equal to 3, the p rank of C D square is a

p rank of C D, and so what is the two rank of C D, C D square? It is what we call the 4 rank, the 4 rank, and if you write, so the 4 rank, if you have a finite

abelian group, the 4 rank is a dimension of f2 of, a dimension of f2 of what I wrote if it is additive 2a over 4a, or in other way, if you write this abelian group as a product

over all prime, product from mu equals 1 to infinity, a product of cyclic group with the frequency alpha p mu, then the 4 rank is equal to the exponent alpha 2 mu from

mu equals 2 to infinity. So it was, I would say, an audacious extension, I would say,

so we say, of course we do not know the 2 rank is blocked by Gauss, but we can say something about the 4 rank, and this is, and he was right. What I explained, so I guess,

so of course you replace it, it's what I did here, you take now 4a p greater than 2, and you have in mind that p equals 2 is equal to the 4 rank. Okay, so what was remarkable for

in Gert's work is that he proved something which was really in the, so, D positive, so you see C6 corresponds to D negative, the other one is for D positive, and I recall

a formula that you already met, is that the probability, for instance, for D negative, the probability for a negative D such as the 4 rank is equal to R is 2 power minus R square,

eta infinity of 2, the product of 1 minus 2 minus k power minus 2 from k equals 1 to R, so this is, so when you have a C6, you have that, so I write it only for negative D,

and the function eta infinity is the product 1 minus t minus j, or t power j, and you, from g equals 1 to infinity, and you suppose that t is less than 1. This is this function

that you met in Venkatesh lecture, which is, which has a strong link with partition function, and so on and so on. So, in fact, with Kuehnert and Mies some time ago, we saw the Cohen-Lenzstra conjecture, in the case p equals 2, means the 4 rank,

the description of the 4 rank now is known, and exactly fits to the prediction of Cohen-Lenzstra. And what, yes, so what did Gert prove? So, not only he had the intuition of extending,

but he proved something which is very attractive, so it's around, I think,

maybe around this year, so he proved, so you take, I would say, a strange way of counting. You count D, for instance, I write for negative D, for the mental negative D, such that the 4 rank of C D is equal to R, and such that the number of prime divisor

is equal to K. I count in set, you see, I block the number of prime divisor, I can count about prime, so now it divides by the cardinality of D,

okay, such that omega of D is equal to K, okay, so this is, and now what, it takes a double limit, I think it takes, and it makes the number of prime divisor go

to infinity, so you see, he counts, first of all, he measures the size of the integer, and then he blocks the number of prime divisor, and then the number of prime divisor goes to

infinity, and what is very encouraging, that he found that this limit is exactly 2 minus R square, eta infinity of 2, the product of 1 minus 2 minus K power square, so the first time,

I think about Cohen length heuristics, that we see this, I would say, not crazy, this very difficult function, and in fact, he proves that, but he has not moments,

he has not moments, and what he uses is something with the theory of Rede matrices, so these are, we say what is Rede matrix, so it's something which is around 1930, I think, so it's rather old, and what is a Rede matrix, so you, very vaguely,

so it's round, so you consider, so I simplify, so I forgot to tell that he has the same thing for positive D, so you write D equals p1 pt, rather speaking, and you consider the matrix M,

which has element MEG, and MEG is equal to the Legendre symbol, p e over p j, so I don't go into details about what happens about 2, so to give you,

very vaguely speaking, what happens, so Rede by the theory of Rede matrices, we know that the four rank of C D is, I don't know if it is t plus 1, t minus the rank of

the Rede matrices, okay, so you have a huge matrix, which is not exactly symmetric, because of, we can say that, reciprocity law for Jacobi symbol, no, for Legendre symbol,

so yes, so it was very positive, you see, we see this formula, but we can say that it is a strange way of counting, you see, it's like counting, remember Ivan Jekes talk about H of D,

you have a natural way, which is according to the size of D, and then you can count according to the length of geodesic, here, it counts about something, you see, the size of the matrix are blocked, and then they go to infinity, so yes,

so what I want to tell you is that, about this, is a theorem of myself and Kuehnertz, which says that, in fact, that C6, C10 are true, and the probability that the four

prong of C of D is equal to R is coincide, exactly, you're absolutely right, yes,

and it coincides with Kuehn-Lenztra prediction, so we can say that, in my mind,

it's impossible to go from Gers theorem to the other way of counting, so I give you some hints of the proof, yes, so in fact, I recall it, I shall concentrate of the proof

of these things, that means another theorem, which is the same I wrote it, so I write, so that means, I shall more concentrate of negative D, is equivalent, depends on K,

and K2, and so you go X to infinity, and I make some comments, first of all,

so in fact, we will prove that this thing minus this thing, so this thing, you see, I told you, yes, this thing is around something like some constant C naught X, and if you go that in that direction, we prove that the error term is capital O of X,

very, X, log X, so tiny power, minus 2, minus K, plus epsilon, so you see,

you take K equals 10, it's very, so the error term is very small, and something which is very surprising, if I write it here, you will not be surprised, but in the proof,

we are surprised that this is true, this is asymptotic, if you add on both sides, so, excuse me, the same thing for D positive, but of course, you replace this by N K plus 1,

2 minus N K, 2 divided by 2 power K, and so what is very surprising, not here, but in the current of the proof that I can, so it's D, yes, so you see, for me,

there are six type of fundamental discriminants, the positive, the negative, and then the power of 2 which appears, no power of 2, 2 to the power 2, or 2 to the power 3, so you can, 1, 2, 0, mod 8, so I say it's not surprising, but in the proof,

it's surprising, for me, it's a miracle, I don't understand what happens when you are doing combinatorics, yes, so okay, so now, what is the approach,

so you see, I must say that we invented the world because implicitly, all that I'm going to say during 10 minutes is implicitly in many papers,

like Redé, of course, was our guy, and we were, I would say we had chance to put that in the context of method that analytic number theory could say something, so I'm going to make some algebraic number theory, which is absolutely not a revolution,

so I write like, first of all, I write d on the form p1, pt, and maybe here,

if it is 2, you may have the exponent, you may have the exponent 1, 2, or 3, okay, and what I consider now is, so I consider pi, the prime ideal, which are, so that means p1,

pi, yes, like that, okay, pi, okay, and now, I consider the set of curly b, which is a set of p1, e1, pt, et, with e1 and all the ei between 0 and 1, okay,

so first of all, something which is, goes, that means that if you take c, d, and you divide it by cd square, so you consider this group, quotient group,

each class, so I write each class of, as exactly two representative in b, okay,

and so you see how many, you have 2 power t, divided by 2, and you arrive, what I wrote somewhere, about goes somewhere, yes, you see, it's another way to state that.

So now, I shall, so something which are now classical, so lemma, so always the fundamental, so the first one is the cardinality is that 2 power r of c of d is, no,

is a cardinality of the square, square of classes such that before his principle,

I am counting elements, and now something which is also classical is that,

is equal to one half the cardinality of b in curly b such that you can write this

as a product of an ideal, and alpha belongs to the ring of integer, and is an ideal of, okay, so here it's only to discuss, to translate this phenomenon, and to state,

so this is absolutely not new, so I shall introduce a symbol which is something like Hilbert symbol, so you take a and b, not integer, rational number, and you say that it is equal to

1 or 0 if, so it is a, if the quadratic form x square minus has non-trivial solution, and 0 otherwise, so what is, what is our detecting, so you see when we are here,

we are happy, we are certainly quote Legendre theorem in some, in some minutes,

that will be the case, and so for our characterization is the following one, so a proposition, so I say it's, what we did is to put in an acceptable way,

so it says that, so 2 to the power is equal to one half the cardinality of b, b positive, b square free, b divides d, b, this symbol b minus b prime is equal to 1,

where b prime is square free, square free, and such that b d is equal to b prime c square,

for some, for some c greater or equal to 1, so it's, it's very difficult to understand, not very difficult, yes, so what about all this stuff, all this stuff is always due to,

this problem here, is always due to the prime 2, I remember Cohen told me, but in that context, you need that context, remember 2 is not a prime,

but 4 is a prime, it's exactly what I have in mind, so you may, so you don't understand what that means, no, yes you understand of course, but you must have in mind, if d is congruent to 1 mod 4, the b prime, and I take d negative,

so I shall give the proof for d negative, and d congruent to 1 mod 4, in fact, the segment is equivalent to the following thing, is equal to one half, the cardinality of b, such that, b positive, b divides d, and the symbol b,

minus d over p, is equal to 1, yes, okay, so you take only the division,

this stuff of c square is only for the power of 2, so yes, now how to prove that, so,

yes, so you must think that, first you have an idea, that all the norm, all the ideas which have norm b, are in the set curly b, okay, the one half come from,

I say, this one half which is here, and now you play, so using that, we want, so the proof reduces to a square is equal to alpha b, it's what I wrote here,

if and only if, b minus d over b, is equal to 1, so two direction can be done, I take only one, so I shall do in the direction for instance, so you start from,

you have a lot of easy relation, because you are solving, you see about the definition, where I put the symbol, yes, you see, you can multiply by your square, it has a lot of evident, obvious property, and here, when you do that, in that sense,

so you know, so you start from this, is equal to 1, so by easy transformation, you see that it is, d b, and you know that the equation minus x, minus d y square,

minus b z square, is solvable in q, that means that, in fact, solvable, so that means that b is a norm in q, square root of d, so by clearing denominators,

we know that there is this alpha, such that, alpha in the amount of ok, such that norm of alpha is equal to this b, w square, so the norm, and w belongs to z,

so after, without, yes, and then, you can, by, you can suppose for n a p, alpha p,

does not belong to ok, you see, by clearing denominators, and then, it's rather boring to explain, and after you finish by using that all the, yes, you use the fact that ideals with norm p square are either generated by p,

or the square of an ideal of norm p, so it's rather, I would say, not boring,

but you have it rather, and you arrive, so I suppose that, yes, you arrive at this equality here, and you see, the fact is that b, when with this b, you have, all the representative are in curly b. In the opposite direction, so, another one,

if you want to do the other direction, so you suppose that a square is equal to alpha b, so with the same, with the property I said, so that means that norm of alpha is equal to

b norm of, this is either divided by b square, so that means that, so the norm, we know how is norm of an element, so give x square minus d y square minus 1 over b z square

is solvable, and so you have 1 equals db equals b minus b prime, so it's rather classical algebraic

number theory, so I insist about that fact, we did not invent the world, but we were lucky to push in the direction of quadratic form, and now, a very easy consequence of that,

so, boom, so it's, I like this statement, because, so, you see the fact of this b prime,

I wrote somewhere, spoils the exposition, you see the square, it's always to do with power of 2, so let us concentrate for d negative, d congruent to 1 mod 4, so you see what happened, I say, I am counting the b square 3, which are d by d, and b prime here is inverse,

so what does that mean, that means that, that means that the equation minus b plus d over b,

solvable, so here we know a criterion, we don't speak of as a principle, local global is okay, we refer to Legendre, and Legendre says that this thing is solvable if and only if,

so you check the sign, and you see the local position, so that means here, this thing has to be a square modulo this thing, and this thing has to be a square, so with this criteria, so yes, so d fundamental discriminant, so what we have is 2 c of d is

equal to 1 half, cardinality, so of, you write, okay, and you, excuse me, I can't remember if it is,

so you write, I guess, okay, this is correct, because d is negative, yes, you write it,

and r is a square modulo b, and b a square modulo r, so you take, so I use Legendre,

okay, to pass this, and you have a very, you see, and we are lucky to find this, which was, which is not written explicitly, and so there are, I told you, I told you, there are six types, for me, there are six types of fundamental discriminant,

and this one, this expression, is the most simplest, most simple one, for instance, suppose that d is positive, and d congruent to 0 mod 8, so we have, so we have this

formula, so we will have problem with the number 8, so cardinality, so you write d on the form, d equals 8 a b, so you write, you, you ask to minus 2 a to be a square modulo b,

and b a square modulo r, plus one half cardinality, so you write the same thing,

minus a square mod b, and to be, and to be a square modulo r, okay, so you see,

it's what I said to you, for me, it's not a miracle, but I'm very surprised that our proof will, we have the same formula when you insert the other condition,

when did I write then, the condition, and you see, with such a mess, all the coefficient will match together the symbol 2 over p at the end, it exactly matches to coin and let's try a guess, so now we go more to analysis, so yes, so now,

so I will, I will concentrate for d negative, d congruent to 1 mod 4, and so it's very easy,

now we see, you are happy with this thing, because we see that Le Jean symbol Jacobi will arrive, we guess that we will use consolation about a character sum and so on and so, what we call in France, in French, le grand jeu, so what did I write, yes, so okay, yes, okay,

so we write it, so I use this easy criteria, r is a square modulo b, and I suppose that a is

co-priming this, you detect this by this formula, which is product, so you p divides b, yes, 1 plus a over p, so I suppose to be correct that b is positive, b odd, to be more square

free to be, this thing is equal to 1, otherwise it is equal to 0, okay, so yes,

and always this thing, if a is a square modulo p, so did I write, I must to be sure not to

have problem, a b is equal to 1, so you see when you factorize fundamental discriminant, particularly if it is odd, you are sure to have a co-prime component, because it is square free, so yes, okay, so now I continue the analysis and I concentrate always for this case, okay,

and what I want to say is that with this, okay, we will have this exact formula,

we have the one half, 2 power of omega of d, and then we have the product minus d equals a b,

and here you, pardon, the sum p divides b, 1 plus a over p, and here I, the product p divides a, so you see you have this decomposition, now minus d is positive, you have no problem,

and so it is a square free, so you develop, and you use a Jacobi symbol, and it's rather simple, that you are, it is 2 omega of d, so I should put absolute value, yes, and you write multiply

by, so you see, you expand this product, and you glue, you will meet at the denominator, and the divisor of b, and the divisor of a, you put that together, and you obtain that it is

something like d equals a 1, a 2, b 1, b 2, so you will have a 1, a 2, divided by b 1,

b 1, b 2, divided by a 1, so you see this a 1 will be a divisor of a, and so on, the complementary is a 2, something like that, and so of course, you write it like,

sum over d, so I should have rather put minus minus d, yes, and so you factorize it at a 2, b 1, b 2, a 1, a 1 over b 1, b 1 over a 1, so you may think, yes, you know that it is

reciprocity law, but we don't use it, for the moment we don't use it, and I write only the fundamental first equality, which is that, after some work, which is that, so the moment of order

1, moment of order 1, which is the sum of 2 power rkc of d, so I take minus d less than x,

and I add d congruent to 1, mod 4 is equal to 1, 1 half, so we will have a triple sum,

so such that a 1, a 2, b 1, b 2 is less than x, so a fundamental discriminant,

okay, and here you put a 2 over b 1, so b 2 over a 1, a 1 over b 1, b 1 over a 1, and here I must

not forget the coefficient, which is 1 over 2 omega of a 1, a 2, b 1, b 2, okay, so you see, suddenly split minus d in four components, and then follow the equality, there's nothing deep,

what is crucial is corollary, how to detect the form, so now we have the philosophy, the following philosophy, Jacobi symbol m over n always oscillates

unless m equals 1, or n equals 1, so you may be shocked, but when it is a square,

it does not oscillate, yes, I know, but square, it's not frequent in the topic I am doing, so what happens here, so we see that, so you see, I want to guess what is the main term,

what is the main term, the main term I say is you are detecting when this product of forcing does not oscillate, so you take a 1 equals 1, a 2 equals 1, so it gives the first one, a 1

equals 1, a 2 is equal to 1, I found the second one, which is a 1 is equal to 1, b 1 is equal to 1, b 1 equals to 1, so these one are very easy to guess, so three main terms, another one is more delicate to see, it correspond to

a 2 equals 1 and b 2 equals 1, so in that case, so a 2 equals 1, b 2 equals 1, so you recognize you have only this symbol, but this symbol, it's the numerator and numerator

from 1, but however, it does not oscillate, because what do we know, we know that we have minus d, so that means minus a 1, a 2, b 1, b 2 is congruent to 1 mod 4, so a 2 equals 1,

b 2 equals 1, so that means that a 1, b 1 is congruent to 3 mod 4, and this symbol is equal to

1, because the product is congruent to 3 mod 4, and this thing, yes, in that case, this symbol is equal to 1, so we have four, you see, we see four main terms only with my philosophy,

okay, and now what to do with this main term, this main term, we do like that, so for instance, you see a 1, a 2, so you block this thing, does not oscillate, it one half the sum of 1, 2 omega of a b, some congruencies, and you sum, excuse me, and here

you have 2 power omega a b, okay, so you do that, you have four main terms, and it's an exercise to see you have, now you check what happens, you obtain the main term is asymptotic,

4 over 2, asymptotic to 4 over 2, the sum over minus d less than x, d congruent to 1 mod 4 of 1, and what is 4 over 2, it is 2, okay, and I think it is

the number of all the, yes, all the vector spaces in f2, in f2, so the 0 and f2 alone, so you see, it's very simple, and we think that, no, I, so I am cheating a little because

what about 0 term, so there are two difficulties now I want to present, is how to raise to any power, so we cross difficulties that is Brown had already,

when he computed the size of the two-selmer group of an elliptic curve which has a congruent problem, and if you see the, you see his papers, there are two papers of him, the first one in where he makes the first moment several time pass, and then he make all the,

all the, all the exponent, so you see, so we have an exact formula, and yes, and now what about, so I have to raise this formula which is here,

raise this formula to power k, so it's a nightmare, you can, you see, you have 4, you take the square, you have 16, okay, and then what we use, and we were,

we follow this idea, is to use f2, f2 square to code the decomposition, that means that instead of a1, so you write d, so minus d as a d00, d01, d01,

and you consider this, this thing belonging to f2 square, this indices, and now you are ready to

go to the kth power, however it's not easy, absolutely not, so you will see that the sum c of d will be something like, I'm very vague voluntarily, so you take indices in f2

power 2k, and then v2k, here you put du over dv, which is a symbol of

Jacobi symbol, here you put vq of uv, and then so d less than x, blah blah blah, here you put the product of du, the product of dv less equal to x, and what is, you see,

you don't know what happened when you raised to the power k, which you see, for instance, you see that, for instance, a1 over a2 does not appear, see, and so what is phi k of uv,

it's phi k of uv is a quadratic form, no, not u, it's u1 plus v1, comma u1 plus v2 plus

u2k plus minus 1 plus v2k minus 1, it's, yes, it's something like that, plus v2k, so

what is important, it's a quadratic, so you see, either it's in modulo with value in f2, so if the exponent is equal to zero, you say this Jacobi symbol does not appear,

if the exponent is equal to one, you say it appears, and this way of coding, you see, you are sure that if you, let me write that, yes, so, so you can say that

if you consider that is equal to one, if and only if, only one of the symbols du over dv

or dv over du appears, okay, so we have an excellent way of coding, and via f2, and what happens, I spoke to you of vector spaces, so I would say how do they appear, the number,

the total number, so when you have this, so it's, I write it, the vector subspaces appear

when studying orthogonality, orthogonality in, for this, for vk, so I was quite happy, it's a nightmare for me to study quadratic form over f2, you know why, and so we have

a quadratic form, there is some orthogonality, and then these vector spaces appear, so now, so yes, so you do, you do the job, and then you see, it's for me, it's a maze, you see, you compute, boom, it's a maze, and then at the end, just appear what a quaternary

guessed, and particularly, when you enter, you see, you have six types of discriminants, suppose I put here a minus a somewhere, what would be the effect at the end, it has no effect,

everything is okay, so now, what about, yes, so I wrote to see combinatorics, I would say, f2 power 2k, so I would like to make two remarks, you see, now when you are doing

computational number theory, people want to check by using a computer, suppose x equals 10 power 10, you compute what the heuristics are, what you obtain, here it's really a catastrophe,

why? Because let's think that you want to compute the second moment, you see, so for instance, something like that, okay, so you go to my interpretation, you see,

when I take the first moment, I have already four variables, so here there will be, when you expand, 16 variables, which are of the type, see, d0, 0, 0, 1, I don't know,

so you have a product of these Jacobi symbols, and so when you want to have oscillation, you want to be sure that the d has a non-trivial

decomposition, so that means that d at least 16 prime divisor, okay, so you see what is an integer which has at least 16 prime divisor, it's rather large, and if you want a lot,

that means that on average that you know that omega of n is something like log log n, so for instance, if you want to see oscillation of this sum, you have to work with x which are of the type e to the power 16, so we see nothing, nothing, and this phenomenon

is one quoted also in his Selmer paper, okay, so the combinatorics can be done, so now what about analytic tools for themselves, they are rather, I would say, classical, not classical,

but I put them here, so two main tools, I say, oh, so the first one is, I would say, coming from a l function theory, so that means around, I would say, siegel-valvish,

type, you see what we call siegel-valvish is something, some oscillation theorem, for instance, about n over d, d is fixed, and then here what we have, it's not like that, it's something like

sum over m, m over d, and what is d in our context, d will be something which will be, so I count n, m less than x, d will be something like log x to power a,

and you see this function 2 of 1 over 2 omega of m is as difficult, I was in some point of view, that the divisor function, so to do that, you have several technique combinatorics, but you have, for instance, take the square root of this function, and taking the square root, that means

that you have to control the zero-free region, okay, so it's not, this thing is not trivial at all, first ingredients, so that means that our result is ineffective, but, and the second one

is what I call double oscillation, double oscillation theorem, so this double, so I give you

an example, I think it's well known, so I think the oldest one is due to Heil-Bron, and

famous paper at Actaric Metica that he calls that large sheave inequality for quadratic character. So it's very simple. So you take this double sum. So you take alpha m, beta n, and to be sure I write this to be sure that they are

odd, square free, and I sum m around capital M and around capital N. Yes, and I multiply by the Jacobi symbol m over n. So what do we know is that we have the feeling that m over

n does not factorize as a product of two functions. So it's proved in this way. So if I suppose that alpha m is less than 1, beta n is equal to 1. This thing is less than mn, which is the

trivial bound, and then you can gain that provided that m and n are greater than log of mn power a.

So this result is rather easy to prove. It's only using Cauchy-Schwarz, of course, as usual. You interpret summation, and either you use Polyovinogradov, or you use the fact

that a character, when you sum it over all a period, is 0. So this theorem has, so this is the first one, the second one, and the second one has absolutely not the strength of this one. So thank you.

Are there questions? What is the source of the small error term that you get? Excuse me, what is it? The source of the small error term.

Yeah, why you... You say the small power of the log? Yes. How can I say that? So the variables are so, it's come from here. Yes, it's come from here. The logarithm, the tiny power has come, and there's so...

How can I say that? The room of variation of variable is very small. But it's very... So the tools are very classical, simple, but to arrive at that, to say, you see, we take a scissor, and you cut some spaces,

and then it oscillates, and the... So you mean the decomposition of how it's... So when you decompose the sum into... Yes, you are square, you have to take care of many things.

I can't explain in one second why the error term is so small, I would say. It's not a multiple representation of question cards? No, no, no, only one. No, no, no, no. Yes, the 2 power k is like, you see, it's like the f2 power 2k.

This is the same, it has to do with this 2k. I can't take me a quarter of an hour to explain. So you see, we have to deal when... So you see, the m and the n will be some part of the d, of this du, dv, such that,

and either you take that or this that, without, of course, you must keep the main term.