 Merci beaucoup à l'organisateur, et surtout à Gérard pour cette belle invitation. Oh, je vois que c'est la dernière page, donc j'ai besoin d'aller au début, je suis désolé pour ça. Ok, on y va. Je vais aussi parler de multiples états. Donc, grâce à l'organisateur, je vais aller plus vite au début. Donc, comme vous le savez, ici sont les multiples états. Vous avez vu ces étérétiques, et si vous parlez de multiples états, les NJs sont des intégres positifs. Et même si vous voulez une série de converges, vous avez besoin d'un N1 plus que deux. Ok, vous pouvez aller plus vite. Vous pouvez aller aux NJs positifs ou négatifs, mais vous devez avoir un N1 plus, etc. à l'aide d'un NJ, plus fort que un J pour un NJ, de 1 à K. Ok, donc, comme vous le savez, l'intégre K est le détail, et le summe des NJs est le weight. Ok, donc, je vais juste faire une petite surveille historique, mais pour un ancien. Je commence avec Euler, en 18e siècle. Il a vu multiples états dans les dents 1 et 2, et il a donné beaucoup de relations entre eux. Ce n'est pas tout, nous savons des relations de l'authentif maintenant, mais jusqu'à, je pense, 15 ans auparavant, la seule chose que nous savions était déjà de l'Euler. Et puis, il y a un grand détail jusqu'à 1980, mais ce détail n'est pas complètement élevé, donc, pour exemple, Min indiquait à moi que vous pouvez trouver des exemples de plus en plus dans le travail de Nielsen, au début de l'année dernière. Et puis, jusqu'à ma connaissance, les multiples états apparaissent, et, au début de l'année, les fonctions résurgentes, le volume 2. Ok. Alors, qu'est-ce que la fonction des multiples états ? Maintenant, les arguments peuvent être complexes. C'est le même somme, d'excepter que les déchets sont complexes. Et puis, il se trouve que c'est une fonction méromorphique dans K-Variable. Et c'est le travail de ces trois recherches japonaises, Akiyama, Egami et Tanigawa. Ces séries converges ont donné la partie réelle de ces déchets qui sont plus grandes que J pour Nij. Et vous pouvez externer la méromorphique pour la whole C-K avec la subvariative de la singularité, qui est décrivée comme suivi. Dans K-Depth, le locus singular de K-Depth est défini comme suivi. C'est celle des J, comme ça, si Z1 est equal à 1 ou Z1 plus Z2 s'élange à cet état, vous venez de 2x2, d'excepter 2, 1, 0. Ou, il existe un J, de 3 à K, comme ça, la partie réelle est plus petite ou est equal à un J. Vous pouvez voir la différence entre K-Depth 1, 2 et 3. C'est lié à la facture que la fonction single zeta peut continuer analytiquement tout le monde, d'excepter 1. Et pour Zeta double, ces hauts viennent de la facture que le nombre de Bernoulli, le nombre de Haute-Bernoulli, accepte le premier. Je vais décrire les relations parce qu'on va voir des généralisations de ces plus tard. Bien sûr, vous savez que le produit de deux single zetas est le nombre de trois termes, deux double zetas et le termes diagonaux, qui est encore un single zeta. Et bien sûr, la preuve est ici. Vous coupez le domaine dans trois parties, M1 plus grand que M2, M1 plus petit que M2 et M1 est equal à M2. Les termes diagonaux peuvent s'appliquer. Ok. Je peux décrire la relation des généralisations de quasi-choues donc vous multipliez deux multiples zetas, une de dépesse P, une de dépesse Q et puis vous obtenez des termes de dépesse P plus Q minus quelque chose. Ce small r est le nombre de contractions que vous avez dans votre terme et le nombre est indexé sur tous les choues de quasi-choues les choues de quasi-choues de type r. Ok. Et ces sont des projections sigma, de 1 à 2 P plus Q en 2, 1 à 2 P plus Q minus le type subject à la même relation que les choues. Mais ici, un sigma dans le premier rau correspond à un sigma dans le second rau et cela vous donne vos termes de contractions. Et ici, les arguments dans ces multiples zetas sont définis par le nombre de ce nombre ici. Et ce nombre, si vous pensez à ça, il contient seulement un, peut-être deux termes mais pas plus. Ok. Donc, vous probablement votre représentation intégrale des valeurs multiples zetas par rapport à Maxime Konsevich et qui est décédée par Donzaguet dans son papier 1994. Ok. Donc, la valeur multiple zeta ici, maintenant, les integers sont positifs. Il est définit par cette intégrale intégrale. Donc, vous intégrer autant que la valeur et vous intégrer les formes différentes d'ti par t ou d'ti par 1 minus t dans cet ordre. Donc, ici, vous avez des termes N1 tous ensemble. Après ça, vous avez des termes N2 tous ensemble et ici, des termes Nk tous ensemble. Et vous avez des formes différentes d'ti par t parce que il n'y a pas de分享 mondial avec des charges de cartes, mais on n'a pas de communication d' Chevr respect rapporte designer shoot pour le producteur de deux simplisées, et ce producteur de deux simplisées décompose dans 6 simplisées 4 dimensionnelles et ensuite vous avez 6 termes, 4 x zeta de 3 1 plus 2 x zeta de 2 2. Ok, les relations régularisées sont un peu plus drôle. Le plus simple est déjà connu par Euler, zeta de 2 1 est equal à zeta de 3. Et comment obtenez-vous cela ? Vous utilisez zeta de 1. Ok, vous vous dites que zeta de 1 n'existe pas. Ok, c'est juste une indétermination. Alors, bien sûr, dans le stock de Min, vous pouvez décider que zeta de 1 est equal à 0. Mais ici, c'est juste une indétermination. Et puis, si vous write down zeta of 1, 2, or zeta of anything starting with a 1, c'est aussi divergence. Donc maintenant, je write down this product in two different ways. So, the upper line is just by formally using the quasi shuffle relation. And the bottom line is just formally by using the shuffle relation. And then, if you equate the two members, the red terms disappear, like in a chemical reaction. And you get, of course, this one disappears with that one. And you get Euler's relation zeta of 3 is equal to zeta of 2, 1. Ok, so here I briefly explained the double shuffle relations, or double shuffle plus regularisation. And it is conjectured that no other relations occur. For example, you have duality relations coming from the change of variable Tj goes to 1 minus Tj in the integral presentation. And these duality relations should be deducible from the double shuffle. For example, this is the first example of duality. And as we have seen, it is also double shuffle. But you have all other duality relations like this. In principle, it should be expressible. But as far as I know, it is still an open question. Do these duality relations come from the double shuffle? Ok. Yeah. Ok, so multiple polylogarithms just in one variable. So Min already explained us, so I go very quickly here. The multiple polylogarithms add some T between 0 and 1. It's defined by this sum or it's defined by this integral. So it's the same for multiple zetas except that instead of going from 0 to 1, you stop at T. And so even if the multiple zeta value would be divergent, if you stop at T, strictly smaller than 1, it still converges here. Ok, so now I want to re-express this multiple polylogarithms in an operator way. So this is the formula you can see in the bottom of the slide. So you start with the constant function equal to 1. You apply it the operator big y, which is multiplying by this function, 1 over 1 minus T. Then you apply the operator r, which is the primitive operator. r of f is equal to the primitive of f vanishing at 0. This is the weight 0 rotabaxter operator. Ok, so and then you continue like this. You apply now the operator big x, which is multiplying by small x, this function here, 1 over T. And then operator r, etc. So it's another way to rewrite the conceviche integral representation of multiple polylogs. Ok, so of course you know when T is equal to 1, you recover the multiple zeta. Ok, so we have this representation by words. You introduce the infinite alphabet y. y star is a set of words with letters in y. You look at the linear span of the words. And you look at the quasi shuffle product. So my point of view is somehow dual to the one explained by Min. Instead of using concatenation product and D shuffle co-product, I will use the quasi shuffle product and deconcatenation co-product if you like. Ok, so here is the quasi shuffle product. So it's like the quasi shuffle relations. You sum up over the same quasi shuffles and the recipe looks very similar. You have contraction terms because of this R here. Ok, so for later use, once I have defined this quasi shuffle product, I can stick to G9 shuffles, I mean I can stick to R equal to 0. So I take a smaller sum and this also defines an associative product which is an ordinary shuffle product. But still on this infinite alphabet. So here is an example of quasi shuffle. Here I have five terms including these two contraction terms. But if I discard them, I have the shuffle product. Ok, so this is traditional notation. Y star convergent is equal to the set of words which don't start with Y1. And for any word, for any convergent word, I can define zeta quasi shuffle of the word is equal to zeta of the corresponding row of indices. But what I have gained here is that I can extend this linearly. And the quasi shuffle relations are equivalent to the fact that zeta quasi shuffle is a character for the quasi shuffle product. It's morphism. Of course, you would like to go from convergent words to any word. That's what Min explained, how to regularize the words which are not convergent. Ok, so this is just an example of quasi shuffle relation rewritten in terms of this quasi shuffle product. Ok, I think I can go quite quickly on this. Ok, so now I'll quote the first result I obtained with Sylvie Pecha some ten years ago about extensions to arguments of any sign. Ok, because the quasi shuffle product you can extend it to the words with letters in z. Ok, so counting for arguments of positive or negative sign. Ok, so what we proved is the following. There's a character from this extended algebra with the quasi shuffle product into the complex numbers but first of all, phi of v is equal to zeta quasi shuffle of v for any convergent word. Ok, secondly, for any words with letters positive or negative such that the multiple zeta can be defined by analytic continuation then the character coincide with this analytically extended value. So in particular, phi of minus n is equal to zeta of minus n. So minus Bernoulli n plus 1 over n plus 1. So adepts to phi of minus n minus n prime is equal to zeta minus n minus n prime. So there's a value here. If n plus n prime is odd and it's given by this expression here in terms of Bernoulli numbers. So we got a table of values in depth 2 and you have two kinds of values. You have these diagonals here. You have all the same values except the very last one in the northeast of the table. For example this one. Ok, these values are obtained by analytic continuation. The other diagonals are not. And for example, you have also a paper by Ligu and Bin Sang a little before us. They also have a table of value. It is different. But of course the good diagonals coincide because it's analytic continuation. Ok. Let me give a sketch of proof. So it's a concrimer regularisation and renormalisation basically. Ok, so we have the quasi shuffle product. Together with deconcatenation you get a connected filter of algebra. The filtration by the depth. And then I extend this hopfalgebra to letters in the complex numbers. So I extend it much more. And I look at the hopfalgebra, the same hopfalgebra with the quasi shuffle product replaced by the shuffle product. Ok. And as a result by Michael Hoffman, these two hopfalgebras are isomorphic and they are some explicit isomorphism. So now this big arm map is a regularisation map. So we define it on the shuffle version of the hopfalgebra. So it's very simple. Big R of a word is the same word but with the indices shifted by Z. Like this. Ok. And then we will use this Hoffman exponential with this which is the hopfalse isomorphism. And then from the regularisation map, I get a regularisation R tilde by twisting by the Hoffman exponential. So now this twisted regularisation respect the quasi shuffle product because the first regularisation would respect the shuffle product. So that's it. No surprise. And now we define the character big phi from this algebra with the quasi shuffle product defined by this. So again, zeta quasi shuffle. Well, it is the same. So by defining this, it's not, it's a non trivial result, but we can prove that it takes values into the meromorphic functions in one complex variable. And then, we define zeta quasi shuffle. So we define zeta quasi shuffle. So we define zeta quasi shuffle. So we define zeta quasi shuffle. So we define zeta quasi shuffle. On Allies values into the meromorphic functions in one complex variable. And then we have the Birkoff sometime or decomposition or big, big fi character and bid composバ兒 Major like this. So this is the convolution product and it pieces are still characters so we're shot relations are still verify their. et 5 plus de v est homomorphique à z equal à 0 pour une word v et c'est ce que l'on veut parce que maintenant ok, ce sont les formuleurs pour obtenir les deux pcs, 5 minus et 5 plus, et peut-être que j'ai réglé rapidement comment ça fonctionne votre méromorphique fonction algebra est split into two parts, the pole parts a minus and the formal series part a plus ok, and this is so called minimal subtraction scheme in physics and then you can define recursively 5 minus and 5 plus with these formulas here, the red term is called Bogolubov preparation map and then if you take the opposite of projecting on the pole part you get through the counter terms and projecting on the formal series part gives you the renormalised character and then the character we look for, you just evaluate the plus part at z equal to 0, this you can do and it still verifies quasi shuffle relations ok, so we were quite happy with this result and then a few years after that we wanted with Kurushe Brahimifard and Johannes Dinger we wanted to describe all solution of the problem describe the set of characters which extend multiple zeta functions in that sense, I mean by respecting the quasi shuffle relations and there's a renormalisation group behind this ok, and let me start with a very general result, I start with a commutative connected filtered hopf algebra and a will be any commutative unitary k algebra and ga will be the group of characters with values in a given the product is given by the convolution and we have this conil-potence property which is crucial to make some recursive arguments ok, so the proposition goes as follows, you suppose that big n is the right co-ideal with respect to the reduced co-product the reduced co-product is this delta tilde when you start from delta and extract it to extreme terms ok, and then if you look at this set ta you look at those characters alpha which vanish on the right co-ideal and this is a subgroup of ga and the proof is quite simple, it's a five line proof essentially you want, if you take two elements alpha and beta in the ta you want to prove that alpha beta minus one is still in ta you prove it is a subgroup and essentially you say that the inverse of a character is given by composition by the antipode you replace the antipode by its recursive expression and then you end up with alpha minus beta of w prime ok, here and now you use the right co-ideal property to say that this vanishes ok, so we call it the renormalisation group associated to the co-ideal n and then comes this trick you look at a partially defined character, that means it's defined only on the co-ideal n and you suppose that it respect the unit and it's multiplicative whenever the three terms are defined I mean whenever the three terms belong to n and now I look at x zeta a is equal to those phi's in the group ga so that phi restricted to n is equal to zeta ok, and it turns out that this is a principal homogeneous space so this we proved by with Kurushi Brahimifat, Johan Zinger and Jan Chang Sao this is a principal homogeneous space that means that this left action here is free and transitive and the proof is quite similar to the proof that ta was a group ok, I'll skip this ok, now we apply this general framework to our multiple zetas so the base field is the rational numbers so hopf algebra is this one with the arguments of any sign I remember you and the target algebra will be just the complex numbers the right co-ideal is the linear span of non-singular words that means that the words which write down like this and which fulfill these three conditions which are the contrary of the singularity conditions of the beginning of the talk right, so n1 has to be different from 1 n1 plus n2 has to be outside this set and this partial sum has to be outside this set ok, and this right ok, this big n is obviously a right co-ideal for the concatenation if you look at these three conditions here and there's a supplementary property it's also stable by contractions like this ok, you gather some consecutive letters and you sum them up like this and n is stable by these properties ok, so sigma will be the complementary of n ok, and so sigma k will be those words so that the corresponding row belongs to the singular locus at depth k and the partially defined character zeta will be zeta ok, will be our multiple zeta values or either by convergence of the iterated sum or by analytic continuation it can also happen ok, so now the set of all solutions to our initial problem is given by the renormalization group acting on one solution fortunately we had proved with Sylvie Pecha that one solution exists so we investigated a little bit this group just to say that it's very big it's infinite dimensional if you look already at depth 3 you see there's a lot of room inside it ok, so now I jump to Q analogs ok, so let me remind what is Jackson integral it's given by this ugly formula but maybe it's more easy to grasp it I like this so here is an example with Q equal to one half and here Q equal to three quarters roughly ok and it's quite obvious that when Q goes to one the classical limit gives you the ordinary Riemann integral ok but here our Q will be considered as an indeterminate now I can define a weight minus one rotorback store operator by summing up like this PQ of f of variable t will be f of t plus f of Qt plus f of Q square t etc operator Qt is invertible because this is due to our algebra A A is this algebra of a formal series in two variables but starting with t ok, and PQ is invertible and the inverse is given by the Q difference operator f of t minus f of Qt you have a modified lab needs rule for the Q so this is lab needs identity but you have a diagonal term here and we end up with three equivalent identities one for the piece, one for the decent, one mixed ok, so now I can define multiple Q polylogs the same way you replace the Riemann integral operator by the Jackson integral operator ok but this is exactly the same recipe so for later use I will need the operator y bar of multiplication by this function t over 1 minus t but for the moment I just mimic the formula for the polylogarithms replacing the Riemann integral operator by the Jackson integral operator ok, so now I will speak about the Ono Okuda Zudilin Q multiple zeta values model so we call that multiple zeta is given by evaluating the polylog at 1 here we evaluate the Q polylog at Q and in terms of nested sum you have this formula so in the numerator you have Q to the m1 and in the denominator you have powers of Q integers ok, so for convergent words when Q goes to 1 you recover the ordinary multiple zeta when Q goes to 1 in a reasonable way ok ok, we have an alternative description in terms of the operator PQ ah ok, so ZQ bar is the modified multiple zeta this is ZQ multiplied by 1 minus Q to the minus the weight ok, so this is this iterated sum and it is given by this flow of operators but here I use the operator big y bar, which is the operator of multiplication by this function we have other models for QMZVs I just mentioned a few the Schlesinger model, it's the same but you evaluate the Q polylog at 1 not at Q you have the Jaub Bradley model the numerator is more complicated Q to this power here and you have the Barman-Kuhn multiple divisor functions model maybe a few other ones ok, but we like this model by Onno, Okuda and Zudelin because it's defined for arguments of any sign if you take nJs in Z it still makes sense that you even have this estimate when Q is a complex of modules strictly smaller than 1 all the sums converge and you have this and you still have this formula this operator formula for computing the Z bar Q ok of course with negative powers of P are positive powers of D ok, here are some examples zeta Q bar of 0 bar without bar is the same it's Q over 1 minus Q a row of 0 it's just elevated to the power K zeta bar Q of minus 1 is just this so it's Q over 1 minus Q minus Q square over 1 minus Q square and interestingly enough Z bar Q of 1 is this Lambert series sum of Q to the m over 1 minus Q to the m ok, and here not only we have meaning for any signs but we have a complete double shuffle picture actually Takeyama had also a double shuffle picture but in the Bradley model so the picture is more complicated and it doesn't extend to arguments of any sign ok, so first of all I describe the Q shuffle relations ok maybe I have to hurry a bit so you look at the alphabet here with 3 letters but you suppose that dp is equal to pd is equal to 1 ok and we have a third letter y so any non empty word writes uniquely like this p to the n1y up to p to the nky with the nj's positive or negative now define the bar Q shuffle of a word by the Q shuffle of the corresponding row of nj's and extend linearly ok and you have a Q shuffle product on this space of words and it's defined recursively by these 4 relations and these relations are just abstractions of these relations we have with respect to these 3 operators ok and we were able to prove that the product is commutative and associative and the Q shuffle relations are just written like this the bar Q shuffle is a character for this shuffle product so we have also Q quasai shuffle so now y tilde is the infinite alphabet but in both directions with nz the internal product is the same zi contracted with zj will give you zi plus j y tilde starts the set of words with letters in y tilde star will be the ordinary quasai shuffle and our quasai shuffle product is the ordinary quasai shuffle moved by this operation here just this shift on the first letter of the world ok and I define the quasai shuffle z bar function value like this and I extend linearly and the Q quasai shuffle relations write like this it's a character for this quasai shuffle product so I give you an example of Q quasai shuffle relations for two letters the product of single Q zeta functions is given by the ordinary quasai shuffle term plus the weight drop term ok so the weight is not conserved and if you if you go to the ordinary well non modified Q zetas you have this 1 minus Q in front appearing so you can it's coherent with the fact that when the limit Q goes to 1 the weight drop term disappears and you get the ordinary quasai shuffle relation ok but here you have no regularization relations you have this swap or this bridge I don't know how to well this change of coding between the two the two alphabets big R the Gothic R and here is the recipe and the change of coding writes like this you jump from quasai shuffle zetas to shuffle zetas with this but that's all this is defined for arguments of any signs so no problem so we have quasai shuffle relations shuffle relations, change of coding and that's all so I give you an example of computation using double Q shuffle I look at this product ok and there's one appearing here so when Q goes to 1 maybe something strange will happen so let's see so first of all I use the quasai shuffle relation so I have not 3 terms but 6 terms the ordinary quasai shuffle term and the weight drop term so using Q shuffle so I translate into the shuffle words this product ok and so I have to shuffle the word py with the word ppy and then I use the recipe given by the Q shuffle formulas here no sorry here we go and ok in a few lines at the end I have the Q shuffle bar of this row this combination of words and I translate back to the bars ok so now this term above is equal to this term below there are quite a lot of nice constellations and at the end you end up with this ok if you remove the bars you have this formula here and when Q goes to 1 you go back to the Euler relation which was obtained by regularisation but here our regulator is the parameter Q so to speak ok and interestingly enough this identity can be found in work by Bailey in 1936 so this was pointed out to us by Vadim Zudelin and Bailey attributes this identity to Bell and it was learnt by Hardy to him so it's quite interesting ok so it's time to conclude maybe there are other relations than the double shuffles it seems that there are also duality relations but I have no time to develop this we would like a more combinatorial description of the Q-shuffle product and we would like to find a compatible co-product ok this is related to this naive question I spoke about how to to renormalise quasi shuffle relations we succeeded in that in some way what would mean extend the shuffle relations to the negative side that means that you will have to shuffle 2 packets of cards containing a negative number of cards so it doesn't make sense naively but with this Q parameter it does make sense in some way so we would like to find a hopf algebra structure here this is work in not progress for the moment but we we have still the project to go back to this and yeah ok and so I said you to use the parameter Q is a regular later what about renormalising when Q goes to 1 and ok interestingly enough when I when I gave this talk 2 weeks ago David Broadhurst asked me what about the author roots of units well good question but there's a very nice preprint by Dorigoni and Klein Schmidt about the behaviour of the Lambert series for Q goes to 1 or Q goes to the roots of units so at least for zeta Q of 1 we have a complete picture here and maybe the task would be to do the same for the author Q multiple zeta values and this could be a good research programme for the future thank you very much thank you Dominique very talk so yeah yeah yeah maybe but you have to say if you want to add something no no no I have finished thank you very much just wanted to have the gallery back ok questions I have a question yeah ok let's go for the negative with the LSE zeta at negative LSE you say that you obtain by electric prolongation so do you compare the result obtained by law in this PhD thesis values obtained by finite path of asymptotic expansion mm-hmm yeah they are quite different of your rational number but in a way it is coherent with the process obtained by finite path of asymptotic expansion ok but you say you obtain by this you say analytic continuation is coherent with this process of asymptotic expansion because the value gamma come from yeah ok asymptotic expansion is not come from analytic expansion except for another kind of analytic extension ok continuation but outside this point where analytic extension is possible you have a kind of freedom of choice which is encoded in this renormalization group clearly what that mean exactly but if we we don't have a coherent between shuffle and shuffle you already say and so that with analytic continuation I don't see why it's mm-hmm ok and before that you say me that the rational number you obtain you just satisfy the shuffle product and not shuffle product yeah yeah do you confirm that yes I do confirm more than that the shuffle product at least naively has no meaning in the negative side but no how already say the meaning of shuffle product give two process, analytic process to obtain rational number ok ok always the process of cellular expansion texture finite part analytic extension part and we get again aven-line theorem ok but we have also arguments on a no rational numbers anymore ok ok ok mm-hmm mm-hmm mm-hmm mm-hmm mm-hmm mm-hmm mm-hmm ok ok ok ok ok ok ok je ne sais pas je ne follow le lien mais peut-être il y a un certain nous étudie la sorollie j'ai juste une question mm-hmm mm-hmm tu as parlé d'un Q-staffel product je suis désolé je pensais d'un product Q-staffel qu'on a dans le non commutatif symétrique mais ce n'est pas le même au moins présenté comme ça il y a un Q-staffel Q-staffel ça génère à la route de l'unité ma question est tu as dit que c'était commutatif et associatif oui ce n'est pas trivial quand tu vois les relations tu as est-ce que ton argument est direct ou tu as des questions pour Q-staffel pour Q-staffel c'est assez facile c'est un transport de l'ordinaire Q-staffel tu peux voir cette expression ici ce n'est pas trop difficile pour l'opération associative tu as raison pour Q-staffel ce n'est pas Q-staffel sorry Q-staffel pour Q-staffel pour Q-staffel avec toutes ces relations ok ok ok pour Q-staffel ah oui c'était avant oui c'est prouvé par l'induction sur le nombre de lettres dans l'expression j'ai compris c'est un genre de parole direct par direct computation pour commutativité c'est facile pour associativité c'est dur mais c'est un peu long mais ok Chapoton nous a dit mais pourquoi n'est-ce pas que la représentation que tu as est faite et puis ce sera mais nous n'avons pas réussi à croiser que la représentation est faite donc nous devons rédouer l'objet pour qu'on aille de ce genre