 Alors, je me souviens de ce que j'ai étudiant sur l'uristique de Cohen-Lenzra. Donc, P est le prime-getter des trois, et on aimerait prouver ce que j'ai appelé C6 pour référer au travail original de Cohen-Lenzra, C10. Donc, ce sont toutes ces conjectures, ce qui devrait être vrai pour tout le prime-getter des trois, pour tout le K equals 1, 2, OK. Et, oui, D est un discriminant fondamental de l'équatétique. Cd est un groupe narroclas. Si vous confusez avec un groupe ordinaire, ce n'est pas un drame, parce que j'explique qu'ils sont souvent les mêmes. Et donc, nous espérons que c'est asymptotique pour NkP, OK, 1. Et ici, Nk plus 1 P minus NkP divided by P power K, OK. Donc, ce sont les conjectures. Et donc, ce n'est NkP. Ce n'est pas la version originale. C'est seulement une forme équivalente. NkP est la cardinale de la ligne de vector spécifique, la ligne de vector spécifique de l'Esprit Power K, de la dimension née. Donc, c'est quand vous avez le moment, j'explique à vous, comment nous pouvons obtenir la probabilité que l'Esprit Power K, l'Esprit Power P, soit equal à R, avec ce que nous appelons la théorie de la forme. OK. Donc, ce n'est pas tout. Donc, ce qui s'est passé dans l'histoire, c'est quelque chose de très, très important. Je pense que c'est l'influence de Gert. Donc, Gert a dit que quand l'Esprit Power K est plus grand que 3, ce n'est pas un problème d'essayer avec les deux rangs de la classe de groupes de Q de l'équivalent de D. Parce que par la théorie de Gauss, nous savons que c'est le nombre de prime divisor minus 1. Donc, l'idée de Gert dit, en fait, nous pouvons externer la théorie de Gauss-le-Saint-Huristique à P est equal à 2, mais au lieu de considérer que c'est D, vous considérez que c'est D squared. OK. Donc, qu'est-ce que cela veut dire? Cela veut dire que maintenant, je vais expliquer quelques commentaires. Donc, vous mettez ici P greater than 2, c'est D squared, et le même asymptotex. Pourquoi? Parce que c'est facile de voir que si P est plus ou moins que 2, 3, le P rank de CD squared est le P rank de CD. Et donc, qu'est-ce que les deux rangs de CD squared sont ce que l'on appelle le four rank. Le four rank. Et si vous write, donc, le four rank, si vous avez un groupe finitif, le four rank est la dimension de la dimension de la dimension de ce que je dis, si c'est additive 2a ou d'autre façon, si vous write ce groupe habélien comme le produit du tout primaire, produit de nu equals 1 à l'infinité, un produit de groupes de groupes avec la fréquence alpha p nu, puis le four rank est le exponent alpha 2 nu de nu equals 2 à l'infinité. Donc, je dirais une extension audacieuse, je dirais, donc on dirait, bien sûr, les deux rangs sont bloqués par un groupe mais on peut dire quelque chose sur le four rank et c'est, et il était correct. Ce que j'explique, donc, bien sûr, vous le réplacez, c'est ce que j'ai fait ici, vous l'avez maintenant pour A, P greater than 2 et vous avez en mind que P equals 2 est equal au four rank. Ok. Donc, ce qui était remarquable dans le travail de Gert c'est qu'il prouve quelque chose qui était réellement dans le POPOM. Donc, c'est positive. Donc, vous voyez, c6 correspond à d négative. Ici, l'autre est pour d positive et je vous rappelle une formule que vous avez déjà met c'est que la probabilité, pour exemple, pour d négative, la probabilité pour négative d subjet le four rank est equal à R c'est 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 a 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. You see this function that you met in Venkatesh lecture which has a strong link with partition function and so on and so on. So, in fact with Cluner's enemies some time ago we saw the Cohen-Lenzra conjecture in the case p equals 2 which means that the 4 wrongs, the disruption or the 4 wrongs now is known and exactly fits to the prediction of Cohen-Lenzra and what yes so what did Gerth prove? So, not only he had intuition of extended but he proved something which is very attractive so it's around I think maybe around this year so he proved Je dirais que c'est une étrange façon de combattre, vous comptez d, pour l'instant je l'écoute, pour négative d, pour la montagne négative d, comme le 4 rang de cd est equal à r et comme le nombre de prime divisor est equal à k, je compte dans le set, vous voyez, je bloque le nombre de prime divisor, pour l'instant je ne peux pas raconter le nombre de prime, donc maintenant il divise par la cardinale de d, ok ? c'est le nombre de prime divisor, et puis il bloque le nombre de prime divisor et puis le nombre de prime divisor je pense que c'est, et il fait le nombre de prime divisor, il y a de l'infinité, vous voyez, il ne compte, avant tout il mesure le nombre de prime divisor et puis il bloque le nombre de prime divisor et puis le nombre de prime divisor va à l'infinité, et ce qui est très encourageant, c'est qu'il a trouvé que ce limiter est exactement 2-r² à l'infinité de 2, le produit de 1-2-k². C'est la première fois que je pense qu'il y a de l'infinité de l'infinité de l'infinité de l'infinité, donc c'est plutôt de l'olde, et ce sont les matrices de redame, donc vous êtes très végélés, donc c'est le ronde. Donc vous considérez, je l'ai simplifié, donc je n'ai oublié de dire que c'est la même chose pour positive d. Vous avez écrit d equals p1 pt, et vous considérez la matrice m, qui a l'élément m e g, et m e g est equal à le le gendre symbole p e j. Alors, je ne vais pas dans les détails sur ce qui s'occupe de 2-2, donc pour vous donner très végélés, en parlant de ce qui s'occupe. Donc redame, par la théorie de les matrices redames, nous savons que le fort ronde de cd est, je ne sais pas si c'est t plus 1, t minus le ronde de les matrices redames. Ok, donc vous avez une grande matrice, qui n'est pas exactement symétrique, parce que, on peut dire que c'est un symbole de jacobi, non, pour le gendre symbole. Donc, oui, c'est très positif, vous voyez, on voit cette formule, mais on peut dire que c'est une étude étrange. C'est comme si, vous vous souvenez, Yvaniex, on parle d'H of D, vous avez une manière naturelle, qui est, selon le size de D, et puis vous pouvez compter, selon le length de la géodésie. Ici, il compte sur quelque chose, vous voyez, le size des matrices sont bloquées, et puis elles vont à l'infinité. Donc, oui, donc, ce que je veux vous dire, c'est que ceci est un théorème de moi-même et de Cluner, qui dit, en fait, que c6, c10 sont vrais, et la probabilité que le 4 rang de C of D est equal à R est coincé, exactement, vous êtes absolument right, oui. C'est coincé avec la prediction de Cohen-Lenz-Stra. Donc, vous pouvez dire que, dans mon cerveau, c'est impossible d'aller d'une manière naturelle, sur le théorème de Geur, pour une autre manière de compter. Donc, je vous donne un peu d'intes de la preuve, oui. Donc, en fait, je le rappelle, je dois concentrer sur la preuve de ces choses, c'est-à-dire un autre théorème, qui est la même chose que j'ai écrit. Donc, c'est-à-dire que je dois plus concentrer sur le négatif D, c'est équivalent. Il dépend de K et de K2. Et donc, vous allez X à l'infinité. Et je fais quelques commentaires. Donc, en fait, nous allons prouver que ces choses, minus ces choses, donc ces choses, vous voyez, j'ai dit oui, oui, oui, oui, ces choses sont autour d'une chose, comme quelque chose de constant C0X. Et si vous allez dans cette direction, nous prouvez que le terme d'arrêt est capital O of X, X log X, donc, tiny power minus 2 minus K plus epsilon. Donc, vous voyez, vous avez K equals 10, c'est très bien. Donc, votre terme est très petit. Et quelque chose qui est très surprise, si je l'ai écrit ici, vous ne serez pas surpris, mais dans la preuve, nous sommes surpris, que c'est vrai, c'est asymptotique, si vous ajoutez sur les deux côtés, excusez-moi, c'est la même chose pour le positif, mais bien sûr, vous réplacez ça par NK plus 1, 2 minus NK, 2 divided by 2 power K. Et donc, ce qui est très surprise, c'est pas ici, mais dans la preuve, que je peux... Donc, c'est D, oui. Donc, vous voyez, pour moi, il y a six types de discriminations fondamentales, le positif, le négatif, et puis, le power of 2, qui arrive. No power of 2, 2 to the power 2, or 2 to the power 3. You can go on to 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 word because implicitly, all that I'm going to say during 10 minutes is implicitly in many papers like Reday, of course, all that 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 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... it goes, it goes. That means that if you take CD, and you divide it by CD squared, so you consider this group, cushion group, each class, so I write each class of... as exactly two representatives in B, okay? And so, you see, how many... you have two power T divided by two, and you arrive at what I wrote somewhere about Gauss, 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 D, fundamental. So, the first one is... the cardinality is that two power R of C of D is... no, is a cardinality of the square... square of classes, such that B4 is principal. I am counting elements. And now, something which is also classical is that... is equal to 1,5. The cardinality of B in curly B such that you can write this ideal 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. It's... So, I shall introduce a symbol which is something like a Hilbert symbol. So, you take A and B integer... not integer, rational number and you say that it is equal to 1 or 0 if... if the quadratic form x square minus a 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 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 exploitable way. So, you see that... So, 2 to the power is equal to 1 half the cardinality of B B positive B square free B device 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... 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. It's 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 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 1 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. You take only the division. This stuff of C squared is only for the power of 2. So, yes. Now, how to prove that? So, yes. So, you must think that all first to have an idea that all the all the norm all the ideal which have norm B are in the set curly B. Ok? The 1 half come from I say this 1 half which is here and now you play so I... So, using that ok? So, the proof reduces to S squared is equal to alpha B what I wrote here if and only if B minus D over B is equal to 1. So, 2 directions can be done. I take only 1. So, I shall do in that direction you start from... you see you have a lot of easy relation because you are solving the definition where I put the symbol yes. You see you can multiply by your square interverge 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 D B hence you know that the equation 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 a square root of 2. So, by clearing denominators we know that there is this alpha such that alpha of okay such that norm of alpha is equal to this B W square so the norm and W belongs to Z okay after without yes and then you can you can suppose for NAP alpha P does not belong to okay you see by clearing denominators and then it's rather boring to explain and after you finish are using that all the yes 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 all the representatives 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 property I said so that means that norm of alpha is equal to B norm of this is ideal divided by B square so that means that so the norm we know how norm of an element so x square minus dy square minus 1 over B z square is solvable and so you have 1 equals dB dB prime so so it's rather classical algebraic number theory so I insist about that fact we did not invented we did not invent the word but we were lucky to push in the direction of quadratic form and now a very easy consequence of that so boom so I like this statement because so you see the fact of this B prime I wrote somewhere spoil 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 device 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 do not speak of as a principle local global is ok we refer to Lejean and Lejean 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 cool so with this criteria so yes so d fundamental discriminant so what we have is a 2 c of d is equal to 1 half cardinality so of you write ok and you excuse me I can't remember if it is so you write ah yes ok yes it's correct because d is negative yes you write it and r is a square modulo B and B a square modulo r so you say so I use le genre ok to pass this and you have a very you see and we are lucky to find this which was which is not written explicitly et so I told you types type of fundamental discriminant and this one this expression is the most simplest 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 8ab so you write you ask to minus to a to be a square modulo B and B a square modulo r plus 1 half cardinality so you write the same thing minus a a square mod B and to be and to end and to be a square modulo r ok so you see it's what I said to you for me it's a not a miracle but a very surprise our proof we have the same formula when you insert other condition when you write them 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 draw guest so now we go more to what is this yes so now so 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 the legend symbol jacobi will arrive we guess that we will use consolation about character sum and so on what we call in France le grand jeu so what did I write yes so yes ok yes ok so we write it so I use this az criteria r is a square modulo b and I suppose that a is b 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 for instance to be this thing is equal to 1 otherwise it is equal to 0 ok so yes ok so I always see 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 yes ok so ok so so now I continue the analysis and I concentrate always for this case ok and what I want to say is that with this we will have this exact formula so we have the one half 2 power omega of d and then we have the product minus d equals a b and here you 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 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's 2 omega of d so I should put absolute value yes and you write multiplied by so you see you you expand this product and you glue you will meet at the denominator any divisor of b any divisor of a you put that together and you obtain that it is something like d equals a1 a2 b1 b2 so you will have a1 a2 divided by b1 b1b2 divided by a1 so you see this a1 will be a divisor of a and so on the complementary a2 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 a2b1 b2 a1 a1 over b1 b1 over a1 so you may see yes you know that it is a 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 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 a1, a2, b1 b2 is less than x so a fundamental discriminant ok and here you put a2 over b1 so b2 over a1 a1 over b1 b1 over a1 and here I must not forget the coefficient which is 1 over 2 omega of a1, a2, b1, b2 ok so you see it's only split minus d in 4 components and then follow the equality the equality it's nothing deep what is crucial is corollary we have the philosophy the following philosophy jacobi symbol m over n always oscillates unless 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 I want to guess what is the main term what is the main term the main term I say you are detecting when this product of forcing does not oscillate so you take a1 equals 1 a2 equals 1 so it gives the first one a1 equals 1 a2 is equal to 1 I found the second one which is a1 is equal to 1 b1 is equal to 1 b1 equals to 1 so these one are very easy to guess so 3 main terms another one is more delicate to see it corresponds to a2 equals 1 and b2 equals 1 so in that case a2 equals 1 b2 equals 1 so you recognize you have only this symbol but this symbol it's the numerator and numerator different from 1 but however it is it does not oscillate because what do we know we know that we have minus d a1 a2 b1 b2 is congruent to 1 mod 4 so a2 equals 1 b2 equals 1 so that means that a1 b1 is congruent to 3 mod 4 and this and this symbol is equal to 1 because the product is congruent to 3 mod 4 so in that case this symbol is equal to 1 so we have 4 we see 4 main terms only with my philosophy ok and now what to do with this main term this main term we do like that so for instance you see a1 a2 so you block this thing does not oscillate so sum of 1 2 omega of a b some congruences and here you have 2 power omega a b ok so you do that you have 4 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 asymptotique to 4 over 2 the sum over minus d less than x d congruence to 1 mod 4 of 1 and what is 4 over 2 it is 2 ok and I think it is the number of all the yes all the vector spaces in f2 so they are 0 and f2 alone so you see it's very simple and we think that no I am cheating a little because what about the other term so there are 2 difficulties now I want to present is how to raise to any power so we cross difficulties that is Brown had already when he when he computed the size of the 2-cell mer group of an elliptic curve which has a congruent problem and if you see you see his papers there are 2 papers of him the first one in invention where he makes the first moment several time pass and then he makes 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 2 power k so it's a nightmare you can you see you have 4, you take the square you have 16 ok and then what we use and 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 d00 d01 d01 and you consider this this thing belonging to f2 square et now you are ready to go to the case power however it's not easy absolutely not so you will see that the sum c of d will be something like I am very vague voluntarily so you take indices f2 power 2k and then v 2k 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 du du du du product dv less than x et là vous voyez ce qui ne arrive pas lorsque vous터 place du course k pour ce moment à1 à2 neoon à1 et puis la série ckv ckv c'est une forme quadratique, non, c'est u1 plus v1, u1 plus v2 plus u2k plus minus 1 plus v2k minus 1. Yes, it's something like that, plus plus v2k. So, what is important, it's a quadratic. You see, either it's in modular, with value in F2. So, if the exponent is equal to 0, you say, this rich Jacobi symbol does not appear. If the exponent is equal to 1, you say it appears. And this way of coding, you see, you are sure that if you, when I write that, yes. So, you can say that if you consider that is equal to 1, if and only if only one of the symbols du over dv or dv over du appears. Ok, so we have an excellent way of coding via F2. And what happens, I spoke to you of vector spaces. So, I would say how do they appear, the total number. So, when you have this, I write it. The vector subspaces appear when studying orthogonality. Orthogonality for this 4 phi k. 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, yes. So, you do, you do the job. And then, you see, it's for me, it's a mess. You see, you compute, it's a mess. And then, at the end, it just appears. What Quentin and Stray 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 ok. So, now, what about, yes, so I wrote to see combinatorics. I would say F2. 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 to x equals 10 power 10. You compute what the restics are, what you obtain. Here, it's really a catastrophe. Because let's, let's think that you want to compute the second moment. You see, so, for instance, something like that, ok? 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 types, see? D00001. 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 non-trivial decomposition. So, that means that D, at least, 16 prime divisor. Ok, 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 types, e to power e to the power 16. So, you see, nothing, nothing. And this phenomenon is round-quoted also in his Selmer paper. Ok? So, the combinatorics can be done. So, now, what about analytic tools for themselves? They are rather, I would say, classical. But I put them here. So, two main tools. I say how. So, the first one is, I would say, coming from a L function theory. So, that means around, I would say, Ziegel-Valfisch type. You see what we call Ziegel-Valfisch is 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, I count n, m less than x, d will be something like log x to the power a. And you see this function, 2 over 1 over 2 Mfm, is as difficult, in some point of view, that the divisor function. So, to do that, you have several techniques 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. So, it's not, this thing is not trivial at all, first ingredients, so that means that our result is ineffective. And the second one is what I call double oscillation, double oscillation theorem. So, this double oscillation theorem, so I give you an example, I think it's well known. I think the oldest one is due to Isle-Bron, and, for instance, Utila proved that the stronger one is due to Isle-Bron, which famous paper, Attac-Taritmetica, that he calls that launch of inequality for a 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, you use Cauchy-Schwarz, of course, as usual. You interpret summation, and either you use Polyavinogradoff or you use the fact that a character, when you sum it over, all the period is zero. So, this theorem as... So, this is the first one, the second one and the second one as absolutely not the strength of this one. Ok, c'est fini. Yes. Bon. So, thank you. I have a question. What is the source of the small error term? Excuse me, what is it? The source of the small error term. Yeah. Why you... You say the small power of log? Yes. How can I say that? So, the variables are so... It's come from here. Yes, it's come from here. The logarit, the tiny power, it's come from... And they're so... How can I say that? The room of variation of variable is very small. But it's very... So, genetic tools are very classical simple, but to arrive at that, to say, you see we take a season of... and you cut some spaces and then it oscillates and the... you know, it's... So, you mean the decomposition of... So, when you decompose the sum... Yes, you are... 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. But it's not a multiple application of question cards. No, no, no, no. Only one, 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... of this... DU, DV, such that. And either you take that or this that without, of course, you must keep the main term.