 Merci beaucoup à Eric et Karin pour l'invitation et à Happy Birthday Dork. Je suis très heureux d'en parler. Bonne occasion. Je vais vous partager un travail joint avec Sylvie Pecha, Kourouche, Johannes Zinger, Yan Cheng Cao et Jaime Castillo Medina. C'est disparu dans plusieurs papiers durant les dernières 10 ans. Ah, ok, je vais scroller comme ça, oui. Ok, donc je pense que la plupart de vous connaissent plusieurs valeurs de data, donc elles sont données par cette liste d'intérêts là-bas. Et les NJs sont des integers positifs et la première doit être plus grande que 2, pour que la série se convertisse. Et ça fait un peu de sens sur ça. Si vous avez un N1+, etc., pour un NJ, c'est strictement plus grand que un J pour un NJ, vous pouvez faire le sens de cette série. Donc, comme vous le savez, l'intagère K est equal au détail et le summe de l'argument est appelé le weight. Ok. Peut-être un peu moins bien connu, c'est la fonction multiple. Vous pouvez laisser l'argument aller au nombre complexe et ensuite vous faites le sens de la suivante série. Bien sûr, ça ne convertit pas à chaque fois, mais le domaine de convergence a été étendu par ces 3 valeurs 20 ans plus tard. La série convertisse a donné la vraie partie du summe du J, c'est strictement plus grand que un J. Et plus que ça, ces autorités ont prouvé que vous pouvez externer ça dans une fonction homomorphique. Vous pouvez externer ça dans une fonction homomorphique de valeurs K complexes dans la whole C to the K. Et c'est possible de décrire précisément où la singularité est. Et elles sont décriées ici. Donc, la singularité de l'homogénial est parmi les parties homogénieuses, en parlant d'un J, et ce sont ceux d'A to the K, dans la C to the K, avec E to Z1 equal to 1, or Z1 plus Z2 is in this set, or there's a J in 3 up to K, so that Z1 up to plus plus ZJ is the sum is smaller than J, smaller equal than J. Ok, here you recognize the fact that the simple zeta function is homomorphic everywhere except at Z equal to 1. Here you jump from 2 to 2, it's related to the fact that the Bernoulli numbers with odd rank all vanish except the first one. And from dimension 3, you don't have any gaps like this. Ok, so... Of course, you probably know quasi-shuffle relations. Here is an example. If you multiply 2 simple zeta values, you get a linear combination of double and simple zeta values. And you get into 3 parts because you sum over M1 and M2 so you decide whether M1 is strictly bigger than M2 or converse, or whether you are on the diagonal M1 equal to M2. Ok, this gives you these 3 terms. Ok. That's possible to write the most general quasi-shuffle relations, so to speak. So you multiply together 2 multiple zeta values of depth P and Q respectively. And you sum over all the quasi-shuffles which are surjective maps from 1 up to P plus Q on to 1 up to P plus Q minus the number of contractions. And these surjections are subject to shuffle-like, like, like, like, like, shuffle-like constraint. Sigma 1, strictly smaller than etc. Sigma P, the same for the second packet. But you can have contraction terms. In that case, you sum the arguments. Dominique, Frédéric Chappelton asked in the Q&A, he suggested if you full screened your slides, Ah, ok, ok, ok. Let me see, yeah, yeah, yeah. Ok. Let me try just a second. Play d'écran. Is it better? Ah, yeah. Ok. Merci Frédéric. Ok, so let's continue. So much for these quasi-shuffle relations. And of course, you may have to sum maybe two terms but not more, ok? Because it can happen that some sigma in the first row is equal to some sigma in the second row. Then you have to sum. Ok. So now there are shuffle relations because of the integral representation. So, most of you also know that you can have multiple zeta value by means of an iterative integral and you iterate the same the same number of integrals as the weight and you integrate dt1 over t1 and then up to n1 minus 1 and then dtn1 up to over 1 minus tn1 and then you start with n2, etc. Ok. And then, sorry, from this representation you have a second way to express the product of two multiple zetas as a linear combination of multiple zetas. These are the shuffle relations. So let me give the canonical example. You multiply zeta of 2 by itself and then you have to integrate fourfold integral of this form and you integrate over a product of simplices. And this product of simplices you can rearrange it into six different simplices and it amounts to shuffle two words with two letters and you get 4 times zeta of 3,1 plus 2 times zeta of 2,2. And then you have also regularization relations. The first of one was already known by Euler zeta of 2,1 is equal to zeta of 3 and you obtain it as follows. Of course, zeta of 1 doesn't exist. But ok, you say it's something. It's something determinate. Zeta of 1,2 it's also divergent. So it's marked in red. And you apply quasi shuffle relation formally and also shuffle relations formally. Ok, and then you can equate the two right hand side terms. And of course, these two infinite terms cancel and you are left with zeta of 3 is equal to zeta of 2,1. Ok, so of course there is a big conjecture related to the period conjecture of Konsevich and Nagy saying that no other relations occur among multiple zeta values. There has been some steps in that direction. They might look tiny but they are really crucial. It was done about 20 years ago by Rivoyle and Zudeline after the first pioneering work d'Apéry who proved that zeta of 3 is irrational. Ok, so it's a good point to make some historical remarks. Double zeta values were already known by Leonardo Euler as well as a lot of relations relating double and simple ones. I say almost all because you have some, for example at weight 12 you have some relations coming from modular forms by Gengel Kaneko and Zagier, for example. There was not known in the 18th century but apart from these very difficult relations a lot of them were already known. And then there's a gap which is not completely empty I could cite, for example works by Nielsen but then I directly jump to the 80s and I should mention Jean Écal because up to my knowledge it's the first place when you can see the multiple zetas in a whole in a full generality in les fonctions résurgentes in the second volume. And then of course came these works by Don Zagier, Michael Hoffman and the integral representation is attributed to Maxim Konsevich et Don Zagier in his 1994 paper in ICM and this was the starting point to the modern approach with periods for mixed state motifs. And of course I must mention the breakthrough by Francis Brown 8 years ago. Now we know that any multiple zeta value is a linear combination with rational coefficient of multiple zeta values with arguments equal to 2 or 3. This was the conjecture by Hoffman. Ok, so very quickly I mentioned multiple polylogarithms in one variable so you start with integral representation of multiple zetas but instead of integrating 0 to 1, you stop at some t between 0 and 1 and quite easy computation can show you that you have also a nested some representation of these multiple polylogarithms. Ok but I want to express these polylogarithms in another way so I introduce these two functions of t 1 over 1 minus t and I introduce the operators multiplication by these functions big x is multiplication by small x this map big y is multiplication by small y and the operator r is the integral the primitive operator this is well known it is a rotabaxter operator 0 et avec these three operators you can define you can find a formula for your polylogarithms you start with a constant function equal to 1 and you reply to it this row of operators ok and this gives you the polylogarithms and of course when your row of integers is a converging row when you set variable t to 1 you recover the multiple zeta ok there's the word description ok maybe you know all this I'll be quite quick you look at an alphabet with an infinite number of letters y star is the set of words with letters in y you look at the linear span and you define the quasi shuffle product which is just an abstract way of rewriting the quasi shuffle relations the sum is looking the same because on the span of y itself you have a kind of internal product given by y i multiplied by y j is equal to y i plus j and then you can define the quasi shuffle product like this and for later use you can also use the shuffle product this is the same sum but you stick to r equal to 0 I mean you stick to ordinary shuffles and you want to discard all the contraction terms so here is an example of quasi shuffle product and an example of shuffle product because here you have these two contraction terms when you multiply y2 by y3 or y2 by y1 you get these two contracted terms in the shuffle product you discard them oh I keep trying to write my talk ok things that I probably should try to understand but I don't want ok so shall I continue ok so I I call convergent words the words which doesn't which don't start with y1 and for me zeta quasi shuffle of words is just zeta of the corresponding arguments but then you can extend zeta quasi shuffle linearly and applying to linear combination of words and the quasi shuffle relations are just summed up by this relation ok so here is an example so this is the ordinary way of writing the quasi shuffle relation and this is the word way you can see the correspondence I go to to the work to give a meaning to this sum and we succeeded but just looking at the quasi shuffle relations so what is about exactly we can find the character of this algebra q of z equipped with the quasi shuffle product with values in the complex number and q on z, q of z is the the span of words with letters in z ok reflecting that I also want non positive arguments in my extending multiple zeta values ok so there exists a character so that phi of v is equal to zeta quasi shuffle of v for any convergent word converging word and also sorry if you take a word so that zeta of n1 up to nk can be defined by analytic continuation then I want that the value of the character coincide with this analytically continued value so in particular phi of minus 1 minus n is equal to the non value for zeta of minus n and for in depth 2 you have also some values which can be reached by analytic continuation and our character coincide with this so here is the table we obtained in depth 2 so you can see for example some diagonals some diagonals they have the same argument everywhere except at the upper upper right corner these diagonals are values obtained by analytic continuation this one also etc the other diagonals they are not obtained that way and for example there is a work by Liguon Bin Chang which was published a little bit before us and the table of values is different but of course the values obtained by analytic continuation they are the same fortunately but I'll go back to the fact that there's no uniqueness for this problem ok so maybe I'll skip the sketch of proof because time is running but just a few words we first go to regularization that means that we were able to obtain a character with value with values in the miromorphic functions this was not so trivial to prove that this intermediate character this regularized character takes values in the miromorphic functions it has to respect quasi shuffle and not shuffle that's why we have to twist with the so called non-exponential but all this works and then we used Birkhoff Conkramer decomposition of our character with values in the miromorphic functions we use minimal subtraction scheme and and then we evaluate the positive part 0 so this is a quick reminder of the Birkhoff Conkramer decomposition you can write your character like this as a convolution product of two characters 5 plus takes values in the functions which are holomorphic at 0 and this one basically takes values in the pole part and there are recursive formulas for these two characters 5 minus and 5 plus for 5 minus you project you will project this red expression on the pole part and this is the Birkhoff's preparation map and the same expression if you project on the on the other part on the positive part you get the positive character 5 plus and now here is the definition of our normalised multiple zeta values you evaluate 5 plus at z equal to 0 ok so now we have one solution we even have one parameter family of solution because you can you can decide what zeta of one is for example but now we want to describe all solution to the problem the problem was to find multiple zeta values at any argument of any sign so that the quasi shuffle relations are still verified ok so this is our work with Kurush and Johannes Zinger and there's a renormalisation group behind all this ok so you can start with any commutative connected filter of algebra and you look at commutative unital K algebra and you look at the group of characters of H with values in A the product is given by convolution and here is our first result you look at the right co-ideal with respect to the reduced ok so you want the delta delta tilde of N included in N tensor H and the co-unit applied to N is equal to 0 then this set is a subgroup subgroup of the group of character so this set is the set of the characters which vanish on this right co-ideal and this is a subgroup of the group of character so here is the proof it's not very complicated you you just take alpha and beta in TA you want that alpha multiplied by beta inverse is still in TA so you compute this on an element of the co-ideal and you want to find 0 so you you say that ok beta minus 1 is given by composing with antipode so now you display what is convolution ok this is swiddler's style notation and then you replace the antipode by its recursive expression here ok and then you can you can compare this term with that one here and you recover alpha of w minus beta of w plus a sum of alpha minus beta of w prime and because n is the right co-ideal these terms here vanish and the whole sum above vanish vanishes and you are done ok so we call it the renormalization group associated to the co-ideal n and now we look at zeta for us for the moment it's a partially defined character that means that it sends one and it's multiplicative whenever each term is defined ok and you look at xi zeta a is equal to those phi's in the group of characters so that phi restricted to the co-ideal is equal to zeta and we showed that it's a principal homogenous space on the renormalization group ta so we have a free transitive action and the proof is ok I'll skip it but it's quite the same than the proof that ta was a group ok so now we apply that to our special case the base field will be the field of regionals our hopf algebra is q of z star where z is the alphabet on relative integers and the target algebra will be just the complex numbers and the right co-ideal is the liner span of non singular words that means words y is equal to sorry w yn1 up to ynk is in big n if and only if n1 is different from 1 n1 plus m2 is not in this set and you have this third condition in depth bigger than 3 so there are the conditions to the definition of the singular locus so these are called non singular words and if you think a little bit about it big n is the right co-ideal for the deconcatenation co-product what matters is the left part of the decomposition if you want and moreover it is stable by contractions so sigma will be the complementary the set of singular words and it's related to the singular locus of the beginning ok and now our partially different character zeta will be zeta will be the multiple zeta value whenever it exists either by a direct convergent or by analytic continuation and now we can say that the set of all solution to our initial problem is given by x zeta c is equal to the renormalization group acting on a particular solution unfortunately we know that a particular solution exists ok and ok what else not so much a part that we are convinced that this group is very big at least it's infinite dimensional ok now so to speak we have renormalize the quasi shuffle relation what about the shuffle relations of course it's a more delicate question because shuffle relations if you go to the negative side that means that you have to shuffle 2 packets of cards with a negative number of cards which doesn't make so much sense so fortunately by going to q analogs we can have a reasonable glimpse on this so I start with a Jackson integral everybody knows ok here I have illustrated it with 2 values of q q equal to 1 half q equal to 3 quarters and you can see that if q goes to 1 your Jackson integral goes to the ordinary Riemann integral ok but the parameter q you can consider it as an indeterminate and it becomes the Jackson integral operator becomes a non-demorphism of this algebra tq of tq the I will use 2 indeterminates ok I will need a weight minus 1 rotabax operator given by pq so it is f of t minus plus f of qt plus f of q square t etc ok and here in our algebra a it's invertible and the inverse is given by dq f of t equal to f of t minus f of qt so we have modified Leibniz rule which it's just comes from the minus 1 weight minus 1 rotabax operator rotabax identity sorry and we have 3 identities which are equivalent because pq is the inverse of dq ok so now I define multiple polylogarithmes with parameter q so I recall the definition of the ordinary multiple polylogarithmes if you replace the Riemann integral operator by the Jackson integral operator you get your q multiple polylogarithmes ok so for later use I also need y bar of t which is equal to 1 minus t and big y bar will be the operator of multiplication by this function and here comes the Onno-Okuda-Zudelin model for q multiple z values so here is the definition you evaluate the q multiple polylogarithmes at q and explicitly this gives you this q of n1 up to nk this is also nested sum but you use q to the m1 on the numerator and you use quantum integers at the denominators ok of course when q goes to 1 in a reasonable way you recover your ordinary multiple z values when the world converges you can also express them in terms of the operator y bar sorry maybe I skipped something ah yes ok you see we have zqs and we have zq bars ok this is just a normalization of my multiple zetas by the 1-q up to minus the weight ok and this modified q multiple zetas can be expressed in terms of the operators pq and operator y bar ok other models for q multiple z values exist you evaluate your q multiple polylogarithmes at 1, not at q the Chao Bradley model like this and also the Barman-Kuhn model with multiple divisor functions and now Henrik Barman and Wolf-Kuhn have developed a model free approach of all these q multiple zeta values which is very nice but I have no time to develop this I'll stick to the Ono Okudazu-Dilin model ok because this model is interesting because you can extend to arguments of any sign it still makes sense that you have an estimate here I mean for us now q will be a complex number of modules strictly smaller than 1 and for any argument of any sign this formula for your modified q multiple zeta still makes sense provided you stick to the fact that the inverse of pq is equal to dq here ok so here are some examples with argument 0 or minus 1 for example and with corrosion we were able to describe double q shuffle relations and just before us Yoshihiro Takeyama did this in the Bradley model but only in the positive side and the expression is more complicated in this model ok so first of all the q shuffle relations we need an alphabet with 3 letters but with d equal to the inverse of p so a non-empty word with these letters writes uniquely like this with positive or negative powers of p and this is the definition of the shuffle zeta shuffle q zeta word description ok so now by abstracting the relation we had here I define a shuffle product on this word and we prove that it's commutative and associative and the q shuffle relations can be written like this the product of 2 q multiple zetas is the q multiple zeta of the shuffle in that sense of 2 words ok we have also q quasi shuffle relations so this quasi shuffle product here can be deduced from your ordinary quasi shuffle product by a kind of translation operator like this and at the end of the day you have q quasi shuffle relations like this and for example if you have 2 letters you have the 3 ordinary quasi shuffle terms plus 3 other terms with a weight drop ok, the weight is not conserved but if you look at the non modified q multiple zetas you have these 6 terms but the weight drop terms have 1 minus q in front so when q goes to 1 you end up with the ordinary quasi shuffle relations which is not a surprise of course ok there are no regularization relations because all this is defined for arguments of any sign so that's ok and you just have a change of coding from the quasi shuffle picture to the shuffle picture and repeat the summary of the situation you have quasi shuffle relations shuffle relations and that's it ok let me finish by an example of commutation I want to multiply zeta q of 1 with zeta q of 2 so with q quasi shuffle I have my 6 terms the 3 ordinary terms and the 3 weight drop terms with q shuffle I translated into words and then I use the shuffle relations in that sense so you have to play with this these rules here and at the end of the day you have this linear combination of words so you get this sum of the 4 terms ok so now I have to compare this result with the result above and you will have quite a lot of cancellations and you end up with this so it looks like very much the Euler's relation plus weight drop term this is more apparent in the non modified setting and interestingly enough of course you recover Euler's relation when q goes to 1 and this was already settled in the 30s by Bailey we thank Vadim Zoudilin for having pointed this to us and Bailey in this article attributes this identity to Bell and it was learned to him by Hardy it's quite interesting but now we have an algebraic framework for this ok so some open problems it would be interesting to find out other relations maybe there are some other relations among q multiple zetas than double shuffle we don't know and most importantly we would like to find a compatible co-product because ok this parameter q re yields a natural regularization of your multiple zeta values and you have a double shuffle picture which is very nice but when q goes to 1 in general all this explodes unless your world is convergent so we would like to have a renormalization and why not apply the Birkhoff concrime renormalization but for the moment we didn't succeed because we lack a reasonable co-product compatible with all this structure ok here are some references and thank you very much for your attention alright thank you Dominique are there questions a question in the chat actually in the chat it says what are the geometric or quantum or arithmetic applications of q MZT models ah this we don't know what are the geometric applications applications ok is it possible to precise a little bit the question here I will let ask or unmute and they can ask themselves potentially anyway I think in the chat it was more general geometric or quantum or arithmetic applications so any applications ok so our motivation was of course first of all to understand this model which was rather new at that point and once we notice that this model makes sense for arguments of any sign of course we wanted to understand that more closely but for other applications maybe it would be interesting to go to q analog of kz equations but we have to do that but we didn't do that yet alright I see David has a question yeah Dominique do you have anything to say about the limit q goes to a root of unity on the unit circle ah interesting no no but ah non that's a very good question yeah I think that was a question by Goulsh in the chat could you say a few more words on Henrik and Ulth model free project oh as far as you remember as far as you remember I would have to to dive into their work again but ah yes I mentioned this bracket notation related to this multiple derivative functions and they have a more general model with double brackets and they have a conjecture up to my knowledge it's not solved yet their conjecture is that double brackets coincide with simple brackets you don't create new objects by going to this more general setting and ok just a comment I saw Henrik Bachmann is among the viewers so anybody has questions to Henrik maybe they can catch him later in the break also I was just going to ask quickly about the stuff that not even from the main part of the talk but that nice table of rational numbers I can't remember the nice table of rational numbers what can you say about interpretations or patterns in those I mean clearly I see some factorials in the denominators and things like that but what can you say about that table ok it's it's mostly based on Bernoulli numbers so for example in the previous page you can see the values at these famous nice diagonals so you can see Bernoulli numbers in the other diagonals the formula is more complicated but you also have some of products of Bernoulli numbers so that's why you can see 691 691 here all these kinds of things Henrik can say a word if he wants I've set him so he can unmute yes hi there Dominic how are you thank you for your talk and the question of course was what about this generalized model so the idea is just that we take arbitrary polynomials in the numerator and then consider the space spent by these models but it turns out that I mean you mentioned for example this Schlesinger model so it turns out this space is also just spent by the Schlesinger model if you allow for indices if you allow them to be 0 and then this conjecture which Dominic mentioned this whole space of the Schlesinger model where you allow 0 as an index is the same as the space where all entries need to be greater equal to 1 so this is equivalent to the conjecture Dominic mentioned but yeah so this generalized idea is just we don't fix a model and just consider the space spent by these types of q-series and all these models which were mentioned and today are in there yeah but in fact among these models some do make sense at arguments of any sign and some do not so that's why we liked very much this model by Zoudeline Ono and Okuda I mean you consider and somehow in bigger space when you allow these negative indices we just consider this positive or positive part alright well in the interest of time why don't we all thank Dominic again