 Alors, bonjour, à l'abord, j'aimerais remercier les organisateurs pour m'inviter. Yann, en fait, m'a dit qu'il a dit que c'est un workshop informel et que je devrais présenter des works qui ne sont pas finis, et c'est plus ou moins ce que je vais parler de, des choses qui ne sont pas finis. Donc, il y aura des nouveaux résultats, mais assez techniques, donc je ne les insiste pas. Mais je vais essayer de vous montrer une picture de où les choses vont. Donc, tout d'abord, topologie correction et résurgence. La résurgence est l'hôpital de cette conférence. J'espère qu'à présent, vous tous connaissez plus ou moins ce que c'est. Topologie correction, alors que certains d'entre vous sont experts dans la salle, mais j'espère que certains d'entre vous ne connaissent pas tout ce que c'est. Donc, let's me start by, basically, what is this about, topologie correction? Well, first, in topologie correction, there is the word recursion. Recursion means that you have some initial data and the recursion produces a sequence. In fact, for me, it will be a double, so instead of being indexed by one index, it will be a double index that I will call g that runs from 0 to infinity and n that runs from 0 to infinity and it will be a recursion on both integers. So, in topology correction, what is the initial data? The initial data is called a spectral curve. A spectral curve, you can think of it as a complex curve or plane curve together with some additional structure that I will explain very shortly. And the topology correction, so let's call it s, a spectral curve. The topology correction will produce a sequence that I will call omega g n of s so with g on n running from 0 to infinity. Ok, so it will produce this double sequence that we call the invariance or the topology correction invariance of s. And it is so, so it is, well, the reason why it's called topological, it's basically because omega g n will be a certain universal function, let me call it t r, of all the omega g prime n prime such that g prime 2g prime plus n prime minus 2 strictly less than 2g minus 2 plus n. So, in fact, it is a recursion on 2g minus 2 plus n and this is one of the reason why it's called topological. But in fact, in this talk, I will just talk about what the topological recursion is computing, what it is useful for, why it is, it can have some relationship to resultant questions, but I will not explain how it is computed. That's not the purpose of this talk. So, how to compute? So, I'm not going to write the recursion. So, first, what is a spectral curve? What is a spectral curve? Well, the easiest thing to think about when you think of a spectral curve is the locus of zero of some polynomial equation. So, take the locus of zero of a polynomial equation. So, it's a subset of c cross c, let's say. So, in c cross c, in the x, y, so you have a curve. In fact, x and y are in c. So, it's in fact a two-dimensional. And in fact, c cross c is a four-dimensional space. And you should think of it as a kind of surface. So, in fact, it's a surface. Well, okay, it's just a picture. So, but in fact, this restriction by the locus of zero of a polynomial equation is a little bit too restrictive for my purposes. And I will define something slightly more general. So, the spectral curve, s, will be the data of four things. A remand surface, sigma. And let me write for the moment just things. And I will explain what they are. So, sigma is a remand surface. It does not need to be compact, neither connected. It can have boundaries. It can be disconnected. It can be just a union of disks. X is a map from sigma to another remand surface that I will call sigma zero. And in fact, I will more or less always take it to be Cp1, the remand sphere, which is called the base. And so, x is analytic, holomorphic. So, it's an holomorphic projection from sigma to sigma zero, which makes sigma a rivemified cover of, let's say, the remand sphere. So, sigma is a ramified cover of a remand sphere with this x. And basically, there are ramification points. Y. Well, in this picture here in the set, zero locus of equations. So, basically, the projection is just, when you have a point xy here, you just project it to the plane. Y would be projected to the other direction. But in fact, the other direction, y is sort of here as an element of C, of a complex plane. But it's better, in fact, to look not at y, but at the one form, y dx. Y dx is a one form. So, let's me write it y dx. It's a one form on sigma. And that will assume neuromorphic. In fact, we can do even more general than neuromorphic. And in fact, it's due to the fact that C is also isomorphic to the cotangent space of C. T star C is isomorphic to C just by multiplying by dx. So, in fact, y dx belongs to T star C. And we have a fourth thing. So, let me... Sigma is a curve in a cotangent bundle plus sigma zero. Well, this defines... So, initially, sigma was just defined as a curve. And we require that it has a map to the cotangent bundle. But will we need a curve in C star plus C star as well? Sometimes, yes. In fact... In fact, as I said, sigma is not necessarily compact. So, you can just remove pieces from sigma where you can have a central singularities. And in particular, you can remove the places where you would go to zero or infinity in C star. So, you just put boundaries instead. So, it's the cotangent bundle, but the curve is not compact. So, yes. So, the idea is that you have your remain surface sigma. And it has... So, indeed, there is a map from that to the total cotangent bundle. And so, to each point P of sigma, you associate a point on that curve, which has both an X projection. This is X of P, and this is Y dx at P. The fourth data of a spectral curve, this B, is in some sense, it can be thought of as a choice of cycles here. As a marking of cycles. Well, in fact, again, it's a little bit too restrictive to think of a marking of cycles. So, what we introduced is something slightly more general. For compact curves, it is nearly equivalent to a marking of cycles, but for non-compact curves, I need something more general. And B will be one transform one form on sigma cross sigma. It is symmetric. And it has a double pole on diagonal. So, this means that basically, if you take two points, P and Q, B of PQ, in any local coordinate. So, let me say that P is a local coordinate. So, you want something like dP times dQ. It's a one transform one form. There is a double pole at P equals Q. And plus, something analytic. So, you want basically that this has no pole at P equals Q. And the wall has no other pole than the diagonal. And I will insist that I will put a coefficient one in front here. So, in fact, you can say that B belongs to H0 of sigma cross sigma to the canonical bundle to the square, the symmetric tensor products, twisted by a pole, a double pole on the diagonal. Ok. Ok. Ok. Which means that yes, the first cake corresponds to the first projection. The first second cake corresponds to the second projection. Ok. So here again we could write this. Ok. But I didn't want to insist too much on that. So... So this is my initial data, it's a spectral curve. Un exemple, quand la curve sigma est torreuse, il y a un candidat naturel pour le b, b de pq. Donc, p et q, p et q, p, c, c, cochanté par z plus tau z. C'est le complexe plan cochanté par la variation z equals z plus 1 et z equals z plus 2 plus tau. Et puis, il y a la function de vayage p minus q, donc c'est quelque chose qui a une double pole, que p s'appelle q, x dp dq et vous pouvez ajouter une constance habituelle ici. C'est tout ce que vous pouvez faire sur la surface de la surface compacture, et le choix de la séquence est quasiment équivalent à la séquence de la marque de cycles. Donc, c'est la notion de la curve spectra. Donc maintenant, la correction topologique, la correction topologique, comme je l'ai dit, associée à une curve spectra s, elle associée à la séquence omega gn de s, une séquence avec gn. Encore une fois, je ne vais pas écrire la définition, mais juste me dire que ce sont les premiers deux membres de cette séquence. Donc la première séquence omega 0, c'est juste la forme 1 de y dx, omega 0, c'est la forme 1 de b. Donc, basically, y dx et b servent comme les termes initiales dans cette séquence. Et puis, il y a un procédure récursif pour trouver tous les autres gn omega. Je ne vais pas l'écrire, mais je vais juste mentionner que l'omega gn de s est un produit de formes dans le sigmar de la force n, donc c'est une forme, c'est une forme, c'est sur la forme n, c'est une forme en forme de la première, une forme en forme de la seconde, une forme en forme de la seconde, une forme en forme de la troisième et ceci. Et c'est des formes simétriques. C'est-à-dire, sigma n k sigma tensor n en symétrique et il s'expose à des points de ramification de l'ordre. Donc ça veut dire que, localement, omega gn de s évalué à ce point p1 pn sera quelque chose comme dp1 tensor dpn x, ici, la coefficient est une fonction symétrique de tous les pIs et avec des points de ramification, seulement à des points de ramification. Pardon ? Non, ici, ce sont les points de ramification, les points sont... Ok, ici, on va continuer. Donc c'est pour n, c'est positif. Et pour n, c'est 0. La forme 0 est juste un scalar, c'est juste un nombre complexe. La forme 0, en fait, donc, omega g0 de s est juste un scalar. Et en fait, c'est dénoté fg of s en vérité. Donc c'est juste un nombre complexe. Pour une courbe spectra, c'est juste un nombre complexe. Et donc, on a une séquence de nombre complexe. Maintenant, ok, c'est, je vais juste expliquer ce qu'est la compétition de la compétition, c'est la compétition de la séquence des formes et des numéros. Et pourquoi est-ce que c'est utile ? C'est utile parce que ces séquences apparaissent dans beaucoup de applications dans la géométrie animerique, dans les matrices rondes, dans les combinaultes. Et je vais donner quelques exemples très sûrement. Je vais donner quelques exemples très sûrement, et quelques seraient liées aux notes. Mais avant ça, je veux donner une autre définition. Donc, c'est le lien avec la notion de résurgence. Au-delà de cette collection de numéros, nous allons faire une série, une série formale. Donc, nous allons introduire une autre définition. Juste avant d'aller au suivant. Donc, je vais définir ce que c'est la résumé de la séquence de la séquence. Donc, si vous avez une séquence de la séquence S, vous définissez la séquence de lambda, par définition, c'est que vous gardez le même sigma, le même x, vous multipliez le y du x par lambda, et vous gardez le même b. Donc, c'est juste que le spectro-curve de la séquence de lambda signifie que vous risquez juste la forme 1. Et là, il y a un théorème, qui est que l'Omega gn de lambda S est... donc l'Omega gn est homogéniou de 2-2g-n, l'Omega gn de S. Si vous connaissez la définition de la séquence, il y a un peu de trigel, juste parce que la forme 1 est seulement dans le dénominateur. Donc, maintenant, je vais définir quelques autres objets. Donc, je vais coller la fonction tau de la séquence spectro-curve. Et je vais donner cette définition Z. Je ne vais pas coller tau, parce que ce n'est pas encore la fonction tau. Mais ce n'est pas exactement celle-là. C'est une série formale. Donc, ça dépend d'un paramètre A. Et d'autres petits paramètres, que je vais appeler h bar. Et très sûrement, je vais changer la notation. Mais en définition, c'est l'exponential de l'Omega gn de 0 à l'infinité de h bar à la 2g-2fg de S. Donc, en fait, je vais coller z à l'infinité de h bar à l'infinité de 1s. Donc, c'est une série formale. C'est une série formale. Et donc, j'aimerais coller la fonction tau. Mais, d'abord, si cette série formale est divergent, ce n'est pas encore une série formale. Et donc, je vais coller la fonction tau. Et donc, je vais coller la fonction tau. Et donc, si cette série formale est divergent, ce n'est pas encore une fonction. Donc, en fait, c'est là où nous rencontrons les questions de résurgence. Comment pouvons-nous donner un sens à cette série si c'est typiquement factorialement divergent? Et puis, en fait, ce n'est pas la fin de la définition. Et nous savons que si c'est une série divergent, il y a une faible partie, plus, je vais coller la série formale, avec typiquement des termes expérimentaux, en h bar. Donc, typiquement, des termes expérimentaux minus 1 par h bar a. Ce qui n'est pas correct, mais en fait, j'ai une plus complète définition, qui est celle qu'on a formée avec Marcos, où ici, il y a des termes additionnels qui sont expérimentés en termes de l'invariance de l'invariance de la récursion topologique, mais je ne veux pas les écrire, mais je sais juste qu'il y a d'autres termes. Et je voudrais aussi définir une fonction d'alimentation, ou en fait, aussi parfois, je n'ai pas appelé une fonction d'alimentation, une fonction d'alimentation, j'ai appelé le psi, et ça dépend de mon h bar minus 1 s, et ça dépend de la pointe p sur la curve. Donc, vous choisissez p, une pointe sur la curve, p s'agit de sigma, ici, et c'est défini comme expérimentation des termes double, des termes de g et de n, h bar à 2g minus 2 plus n par n factorial. Et donc, il y aura une omega gn. Et puisque la omega gn est une n-forme sur sigma n, donc c'est une forme dans la première variable ou une forme dans la seconde variable, vous pouvez intégrer la première variable sur un certain path, vous pouvez intégrer la deuxième variable sur un certain path, et je intégrerai la première variable d'infinité à p, la deuxième variable d'infinité à p, tout le variable sur le même path d'infinité à p, où l'infinité est un point spécial sur la curve, typiquement l'une qui a une image x à l'infinité. Mais ce n'est pas trop important pour le moment, en fait, dans mes papiers, ce n'est pas exactement la bonne définition, ceci est un peu l'approximation de la définition actuelle et encore plus les termes de transseries. Alors, pourquoi est-ce que c'est intéressant ? Et donc, le premier remarque, sorry, nous avons défini des objets, donc j'ai brûlé le langage des systèmes intégrés, la fonction des caractéristiques, parce que, en fait, ce que c'est pensé, mais je ne pense pas qu'on peut dire que c'est totalement prouvé, mais c'est prouvé case-by-case, dans beaucoup de cas, dans les systèmes intégrés. Et, en fait, la caractéristique topologique serait de toute façon la façon dont l'expansion de les systèmes intégrés classiques. Mais, comme je l'ai dit, la première difficulté ici est que, typiquement, ceci est le résultat récent que je parle de, c'est un théorème, fg d'O de let me call it beta prime g factorial, times r to the minus g where r positive on beta prime the jevres index is bounded by 5. In fact, in many applications, we know that it should be beta prime equals 2 in practice. So, they grow like 2g factorial in practice. And they bound what I was able to prove for the moment is 5. But I know I did some overestimation but it's enough already to say that a Borel transform does exist and is convergent in a certain disc. Now, it doesn't say if it is endlessly continuable to infinity. And for that, we have to re-understand the geometry. And the fact that we have a recursion for the coefficient means that we have some information for the geometry. So, first, let me now go to some examples. And also the omega g and grow factorially. So, the Baker-Aqueuse function also has the same issue of some ability. Excuse me, you have defined omega 0, 1, and not fg. So, it's hard to argue. They are defined in my previous papers et c'est important to say what the definition is. I just believe it, it's a black box. You have a spectral curve, you put it on your computer, you press the button, you get fg. I'm going to give some examples which would be more concrete. So, let me give some on these facts, that's what I was going to do now. Example. So, example 1 spectral curve. So, y square minus x equals 0. So, plotted in the xy plane just this square root curve. There is obviously only one ramification point here. And let's me translate it into my language. So, it corresponds to the case where c Michaels. Well, let's say c. X is the map c1 which maps point z to z square. And y of z is z. So, you can check that indeed y square minus x equals 0. And the b of z1 z2 is dz1 dz2 over z1 minus z2 to the square. Ok. So, this is that very simple curve. Now, I can tell you what the fgs are for this curve. In fact, the fgs are not so interesting. So, let me write the omega gn's. What is the omega gn for that curve? So, that was proved that omega gn for that curve. So, it's a function of n points of the curve. So, n points in the complex plane. And so, it's product i equals 1 to n of dz i. So, it's a one form in each variable. And there are poles. We know that the poles are only at the ramification point. And the ramification point corresponds to z equals 0. So, basically, we expect that there are poles of this form z i to the 2d i plus 2. In fact, you can check that this is always an odd form. Sorry. 2d i plus 2. Let me normalise by the coefficient 2d i plus 1 double factorial. And so, we have a sum over d1 dn. And here, you have a coefficient which depends on those numbers, d1 d2 dn, the degrees. And it depends also on g and n. And this coefficient is tau d1 tau dn dg. And it is some very important number in algebraic geometry, in enumerative geometry. These are called the written Konsevich intersection numbers. Well, let me call bit numbers. They are typically irrational numbers. They belong to q. And these are just the coefficients. And in fact, they are non vanishing only if d1 plus dn equals 3g minus 3 plus n. So, basically, what you get is a polynomial in the 1 over z i squares. The sum here is finite. So, this is an example of what you would get. In fact, in these examples, the fg's are very trivial. Basically, fg equals 0 except f0. I think f0 is 2 thirds. So, the would-be tau function would be exponential 2 thirds of h bar to the minus 2. So, it's not a very interesting quantity. But the Baker-Achiseur function is more interesting. Let me get the other board. So, if you compute carefully the Baker-Achiseur function, so psi of z equals exponential sum of g and n h bar to the 2g minus 2 plus n over n factorial. You integrate from infinity to z of omega gn. So, you just integrate those forms. You get some formal power series of h bar. In fact, it's possible to recognize the coefficient of that formal power series of h bar. They are exactly the coefficient of the asymptotic expansion of the every function with a power minus 2 thirds z square. In fact, remember that z square was what I called x x equals z square. So, basically, it's the every function. Just this formal series with those coefficients are just the coefficients of the every function. So, in particular, it's satisfied h bar d over dx to the square minus x psi equals 0. Formally, if you look at this curve here, formally, it's like l'entre-operator est plus ou moins de la même chose que l'arrière de l'arrière d'h bar d dx. Donc, parfois, c'est le curve classique et c'est la quantisation de ce curve. Donc, c'est le curve quantum. Il y a un système intégrable associé à ça et c'est le RQKDV. Il y a un système intégrable. Donc, en fait, let me give another example. So, well, the resultant properties of the every function or the Borel transform of the every function is kind of trivial. This is just an exponential. So, I'm not going to spend time on explaining what is the resultant properties of the every function. It's very, very trivial. It's not worth. So, let me go to the last board. So, this example was more, so the example with this every curve was more to show you that indeed, those definitions of the objects defined by topological recursion are relevant for algebraic geometry for enumerative geometry for integral system. So, even with the simplest spectral curve, you already get very interesting numbers. So, now let me go to another example which is the example of knots. So, here I draw the figure of eight knot. It vaguely looks like an eight. That's why it called the figure of eight knot. It is the knot used to rope when you do rock climbing. It's a practical knot. And let me have it here. I wanted to introduce the notion of John's polynomial. Okay. I don't have the expression of the John's polynomial of the figure of eight knot. Probably Stavros would know it by heart, but okay. Associated to a knot, there is a notion of John's polynomial. It's a polynomial of Q. In fact of Q on Q minus 1. It's a Laurent polynomial. Well, the exact expression of the polynomial of the J of that knot is something like Q to the 2. Okay. Don't remember. Well, okay. And there is plus, I think, Q to minus 2 or something. I mean, goes, the degree grows between 2 and minus 2 or something like that. But there is a notion of colored. There is another notion, which is the notion of colored John's polynomial on JN of Q is also a polynomial. It's a polynomial of Q on Q minus 1 with degree, whose degree. Yes, they have integer coefficients. Sorry. Yes, so they belong to Z of Q on Q minus 1. Whose degree grows with N. In fact, here, the John's polynomial is the case N equals 2. N is related to a representation to a young diagram with N minus 1 boxes. Okay. And it's because it's related to the representation theory of SU2 or SL2C. Okay. I don't want to explain what is the John's polynomials, but it's a very important object in the theory of knots in no-dimensional topology. But a big question, a big open question is the asymptotic expansion. Expansion in a regime where you want to send N to infinity Q to 1 and such that N log Q remains finite. So let me call it U and which is O of 1. On Q, you write it exponential H bar and basically this is a limit. So you are looking at a limit where H bar goes to 0 and where N is U H bar. U over H bar. Yes. So, what is the asymptotic of the John's polynomials in this regime? And in 95 CASHIF made a conjecture about the leading order, the leading asymptotic order, so JN of Q behaves like exponential H bar to the minus 1. And some quantity, some certain function, let me call it S minus 1 of U. Certain function of U. And this function of U is such that the S minus 1 of U over D U, let me call it Y of U. Sorry. Let me call it Y of U is solution of a certain algebraic equation. Solution of some equation of an algebraic equation. So there is a certain polynomial that's called the A polynomial. I should have called it X1. OK. U on... Well let me call it X. So let me call it that Y of X is the solution of some algebraic equation. So leave some space here. Equal 0 where A is a certain polynomial but in fact it's not a polynomial equation rating X on Y but E to the X on E to the Y. So it's indeed the VCC star cosy star. So for the figure of 8 0, the A of XY is just Y plus Y minus 1 minus X2 minus X minus 2 plus X plus X minus 1 plus 2. This one I know by heart. So this can be used by so this is so this will define a spectral curve. So this defines a curve a compact Riemann surface. This defines a projection to the base which is X. This defines a one form which is Y dx and there is a natural choice of marking of cycles which defines a B. So this gives our spectral curve. So the A polynomial gives a spectral curve. And we see that basically this Cache-IF asymptotique says that this is 8 bar to the minus 1 integral to U of omega 0 1 of that spectral curve. Ok, I didn't say what is the other point I'm cheating you a little bit in this formula. But so then the next step was how to guess what is the rest of the asymptotique expansion and so I have nothing to erase here nearly and the guess made by so this so then digraph Foujian Malabé says that Jn of Q has this asymptotique expansion while here just let's say it's U and with that spectral curve. So the omega Jn being computed by the topology correction from that spectral curve this would give this expression would give the asymptotique expansion of the Jones polynomial. So we have Foujian-Malabé in 2010 well they made the correction not only for the figure of 8 knots but basically for all knots all hyperbolic knots so in fact when digraph Foujian-Malabé computed that so they computed the first U omega Jn from the topology correction includes the data of this symmetric form B so what in this example where do you this curve here you can check that this is a torus and there is a natural choice of an A cycle on the torus while here associated to this projection there is a kind of natural choice of A and B cycle associated to that representation it's not only a curve it's a plane curve so you have a x-coordinate, a y-coordinate so this makes a kind of kind of natural choice and then this gives a B so you take the B which is normalized on that cycle so basically the B is the Weierstrass function plus a constant and the constant should be chosen such that this is normalized on the A cycle so basically this is Weierstrass function plus the G2 Eisenstein series ok ok so so you from topology correction you can compute the first two omega Gn's you can integrate them that gives some very non trivial functions of you and these are exactly the functions that appear in the asymptotic expansion of the knot but they have a problem the functions are the good ones but the coefficients in front of them are not exactly good so the graph Fougian Manabé invented a kind of what they called renormalization procedure to adjust the coefficient at each step but in fact in fact it's because they use that expression and they forgot the trans series terms so then this is times so sum over n of exponential let's say n times something and with exponential 2 pi i tau n square and 1 plus so h bar minus 1 n times something 1 plus on in fact here it was the b cycle in fact b minus tau a no sorry b cycles 1 plus h bar so this is very the full series of corrections so somehow they took only the term n equals 0 in their asymptotic expansion so without all the trans series corrections and it was wrong it was nearly good but not exactly and in fact for knots varies so if you compute this b cycle integral and also for knots in fact people were interested only in the case where h bar is 2 pi i over an integer so basically that was the relevant thing that people want to study for knots and it turns out that this b cycle integral is precisely 4 pi square times an integer so in fact this exponential here is exactly a multiplier well the phase here is exactly a multiplier of 2 pi i so in fact for knots these trans series corrections are in fact of order 1 and that's why they change the coefficients in the expansion and in fact if you take that into account you get exactly the good asymptotic expansion of the Jones polynomial without having to do any renormalization so in fact this trans series part is absolutely needed and we've borrowed on myself in 2012 we checked that the conjecture is ok for ok up to o o of h bar 4 which is a very non trivial test so we are totally confident that this conjecture is correct but proving it is really a very very big challenge in fact I think at the moment no one has we say the beginning of an idea to prove it I'm not sure maybe I'm wrong but so this should say Q difference equation well one way would be to prove that this would satisfy your Q difference equation but this is hard to prove ok this is one possible way but I don't know how to do it order by order in powers of h bar you can check that it satisfies the Q difference equation so this is another application of topological regression and an example where you see that the trans series part is absolutely essential and it would be very good to totally understand it I want to go to another Q difference equation is linear however you only know it at the constant and the constant is the volume conjecture yes ok so indeed that would not be enough ok so let me go to another example I wanted to talk so in fact yesterday Ricardo talked about the matrix model integral I'm not going to talk about that and let me talk about the DOZZ formula in CFT in conformal field theory so in fact this has to do with SL 2C space of connection the super what's the name so dawn auto zamolochikov and it goes back to maybe 70s or 80s or in natural old basis this formula but it computes something which is related to the space of connection of a principle of 2C principle bundle bundle on the sphere so CP1 minus 3 points so we have your sphere 3 points Z1 Z2 Z3 it's always possible to choose them to be 0, 1, infinity for example and you are concerned with a connection so you want to look at a connection on this SL2C bundle so a connection is something like d minus phi of x dx where phi of x is a matrix in the algebra SL2C so basically it means it's a 2x2 matrix whose trace is 0 and it will be a function connection meaning that it has poles simple poles at those 3 points so it will be sum over i equals 1 2 3 of phi i over x minus zi and where phi i belongs to SL2C so meaning that trace of phi i equals 0 so it's a 2x2 matrix whose trace is 0 and I also require that there is no pole at infinity meaning that sum of phi i must be 0 so this is a connection in fact it's interesting to look at the eigenvalues of phi i so the eigenvalues of phi i are just alpha i minus alpha i ok so they are just so since the trace is 0 there are 2 eigenvalues and they are just opposite to each other you can think of it in fact think of the diagonal matrix alpha i minus alpha i which belongs to the carton algebra ok and there are some numbers that play an important role alpha i square is one half of trace of phi i square it's called the Casimir's or in fact in conformal theory alpha i are called the charges or the impulsions or it depends so in fact if you want to fix the eigenvalues of the 3 phi i up to a gauge transformation basically that determines totally the phi i so in fact you have almost once you have fixed the eigenvalues basically you have no freedom in choosing the phi i up to gauge transformations so now there is a quantity in conformal theory that people are interested in and it's called well it could be the tau function is something which in conformal theories denoted v alpha 1 of z1 v alpha 2 of z2 v alpha 3 of z3 it's called the product of 3 vertex operators located at z1 z2 z3 with charges alpha 1 alpha 2 alpha 3 it's a certain quantity and basically you can be so first it was it was computed exactly as a function of alpha i and zi by Don Otto Zamolchikov Zamolchikov and I will give you the exact expression in a minute but what is the relationship to sorry do you consider these arbitrary because they are the z you state usually in particular sorry do you consider these arbitrary yes distinct yes distinct on the charges also distinct and not related by a multiplier of by an integer shift ok ok and so in fact so in fact very soon I'm going to I want to do an asymptotic expansion with a small parameter so where can we put a small parameter a way to put a small parameter is to multiply things by h bar here and it's called an h bar connection and you see it's more or less equivalent to putting the one average bar in front of the phi and so basically introducing the h bar is exactly equivalent to changing the alpha i to h bar minus one alpha i so it's called the heavy limit the charges become large so it's the limit of large charges so if you put large charges complex here well for the moment the charges just Eigen values of complex matrix they can be complex I'm in SL2C not in SL2R so you have this quantity and it was completed by Don Otto so I hope I will get the formula correct so first there is a trivial z i dependence so there is 1 over z1 minus z2 to the power alpha 1 square plus alpha 2 square minus alpha 3 square times h bar minus 2 h bar minus 2 ok you just rescale all the h bar ok let me not put h bar here ok well I mean the final formula will just be I mean the asymptotic will just be in the limit where you rescale every alpha by h bar so times well there is 2 equivalent terms with 2, 3 and 3, 1 ok you just do permutations and times the coefficient and the coefficient is non trivial and the coefficient it turns out is a product of bounds g functions 2 alpha 1 I think plus 1 g of 2 alpha 2 I get it slightly wrong maybe it was in fact minus 1 and g of alpha 1 plus alpha 2 plus alpha 3 minus 1 g of alpha 1 minus alpha 2 plus alpha 3 well let's say plus alpha 3 minus 1 times g so again in the denominator g of alpha 1 minus alpha 2 plus alpha 3 minus 1 and g of alpha 1 plus alpha 2 minus alpha 3 minus 1 I think this is it this minus 1 I think is a very unfortunate notation of the bounds function and it would be much more convenient if bounds would have defined the function shifted by 1 all formulas would be a little bit simpler I think so the claim is that so first the claim is that when you do the asymptotic expansion of this first this coincides with the topology correction tau function would be tau function that I defined before so basically this tau should be equal to exponential sum of h bar to the 2g minus 2fg of some spectral curve and what is the spectral curve is the one you would get by just solving the equation determinant of y minus phi of x equals 0 and you see that because so basically which is equivalent to saying that y square equals 1 half of trace of phi of x to a square and you see that is this is something which has a double pole i equals 1 to 3 y square over x minus zi to a square and there is also some coefficient in front of with simple poles and the beta i's are some functions of the alpha i's I'm not writing them but basically this is the spectral curve and the claim is that if you take that spectral curve it has genu 0 compute the topology correction invariance fg then you correctly get the asymptotic expansion of the down auto asymptotic function so this is the claim and just to have an idea of what what is the topology of that curve one way to see that curve is ok you had the 3 point function of an SL2 theory so since it's SL2 you just do a doubling of this graph ok and now you've thickened the graph into a surface so it looks like something like that it's a genu 0 surface in fact all what I'm saying here for the 3 point function of SL2C would work for SLN on for n point function typically if you have 4 point function with SL2 you draw a graph with 4 points you double it and when you thicken it you see that what you get would be a genus 1 curve when you thicken it you would get something like sorry I need to do that to my conclusion and if you would do a SL3 for instance or SL3 you would do this and it would again be a genus 1 curve for example ok and so I'm just going to say a few words about resurgence for that I need one more so I'm going to conclude very shortly about resurgence so where are the poles where are the poles of borale transform sorry what are the poles of borale transform and basically the poles of the borale transform are going to be periods of those curves so in fact the borale transform of the Bard's function is explicitly known log of g of x plus 1 is integral over e to the let's call it h bar minus 1x easy to be h bar s ds 0 to infinity 1 over s d over ds of 1 over e to the s minus 1 minus 1 over s minus 1 over 12 I think minus s over 12, sorry I think this is correct, the last two terms are just there to kill the pole at s equals 0 and you see that all the poles are at s equals 2 pi i n 1 without 0 so it's a little bit like André was presenting so all the poles on the imaginary axis but except 0 so when you apply this you see that the poles of the borale transform for the DOZZ formula has only poles the borale transform has only poles and you see that the arguments of bounds functions contain 2 alpha 1 2 alpha 2 or alpha 1 plus alpha 2 plus alpha 3 and so on and in that curve they are exactly the integrals over some cycles so typically for instance take this cycle so here let me write the period so for each gamma let me write 1 over 2 pi i integral over gamma of y dx so here we get alpha 1 minus alpha 1 if you take that one or that one and here if you take that one you will get 2 alpha 1 so this is one of the similarities one of the poles we have here on alpha 1 plus alpha 2 alpha 1 plus alpha 2 minus alpha 3 also corresponds to a certain cycle and basically they all correspond to some cycles you can now play the same game with the 4 point function or with an SL3 or theory or anything again you would have some typically some singularities of the borale transform corresponding to such cycles but you have also those non trivial cycles those non contractible cycles will also play a role in the borale transform there is a big difference here between these kind of cycles that you have in the 3 point function where you see the borale transform is very trivial and resurgence properties are very trivial and in SL4 sorry in 4 point function you have something much more non trivial that happens but my time is over and so what I just so it was just to give you a small glimpse of why topological recursion is useful it is useful for many problems for non generative algebraic geometry or random matrices or conformal field theory or integrable systems it's very useful it appears everywhere and also it has very deep questions related to resultant questions and just first to make sense of those theories and also but because topological recursion is very closely related to geometry it can give a geometric understanding of those borale transforms and in fact what I'm working on at the moment is a method to extract just from the spectral curve all the information about the borale transform so all the singularities and basically all the properties of the borale transform this is underway if this is not finished I hope to finish soon but at least in this kind of formula DOZZ and so on it gives exactly what we want thank you for your attention we have a resurgence property for logarithms we go to one function but formally making exponent which is kind of too rough because already for gamma function we get a logarithms for gamma function gamma function itself and it's not just really much smaller but in fact so I didn't have time it would require much more than 1 hour to explain but in the topology correction framework there is a natural way of saying which question to put in front of each cycle exactly and when you take the log or when you don't take the log it's very in the framework so indeed sometimes it's easier to take the log and sometimes it's easier not to take the log for instance with every function it's easier not to take the log yeah in the gamma function if I remember correctly when you take the log you have a sequence of poles and if you take the gamma itself then it becomes one branch cut is it something like that yes something like that appears yeah also for the bounce function basically there is just one cut does the side function always satisfy a differential equation at least formally always satisfy a differential equation ok it always satisfies an equation well first of all topology correction itself so since it's a recursion among the coefficient can be rephrased in a linear equation satisfied by the tau function or by the wipe by the psi function but it's more like an ODE an infinite number of parameters so it can be rephrased in that way but still it's enough to get the recursion the question by your recursion and in some cases indeed it reduces to a nice ODE which is called the quantum curve this is well ok the answer is yes and no it depends what we are talking about maybe I should say that this is no when the curve is hydrogen it's definitely no stop telling people that in hydrogen this is true because it's not ok basically it really depends ok there are extra data that you can put to make sure that the answer would be yes but in many examples in fact the answer is no there is no differential equation in the case of notes what is expected is a difference equation but this is not proved it is an over conjecture this trans series correction how do you get that long spectrum ok well with Marcos we gave a formula which still has some three parameters the characteristic of the theta function basically so the trans series terms look like the expansion of a theta function on theta derivatives I can tell you in matrix models they are really natural that's why we introduced them because in matrix models we saw them so we kind of conjectured that they should be there all the time and they are precisely what is needed to make the tau function first to make it really a tau function to have good modular properties ok I can give you the definition if you want it's not so complicated there needs to be mentioned that these corrections are well defined for the A polynomial but in general it's not so easy to define they are only valid for good two curves I mean a special point in the model space at a generic point in the model space these corrections are not really a trans series I mean are not really so clearly well defined ok maybe we can discuss that later time but I have made some progress for I think I know better how to define them in general we also know how to define in better but this is a generic point in the model space but it's different I think the way we define it is really working for this particular curve in general it's one has to be curve thanks speaker