 Merci Stavros, c'est mon plaisir de vous remercier les organisateurs pour cette conférence. Je suis très heureux d'être invité à participer. Je veux rappeler sur le travail joint avec Frédéric Fauvé et Ricardo Skiappa. C'est un travail en progrès sur l'exact WKB Method et la résurgence parallèle. C'est un sujet qui a été initié par André Boros et Jean-Hicard dans les années 70. Je l'ai mentionné juste en 1983, un paper très célèbre par André. En 1984, il y avait ce paper « Sainte application des fonctions résurgentes » et l'une de l'application était dédicée au problème de WKB. En 1994, il y avait aussi une très intéressante preuve sur les produits weightés. Il y avait aussi le « Nice school de la barre dilangée femme » et il y avait un paper dédié de DDP, qui a été créé en 1993. Il y avait aussi la « Kyoto school de Kawa et Takei » Je l'ai mentionné ici, en 2005, une monographie par eux. Mais il y avait beaucoup de collaborateurs et il y avait des collègues. C'est Koyke, qui s'est passé malheureusement l'année dernière. Plus recentement, il y avait beaucoup d'activités et de renouvelations d'interesse. C'est une sélection, une bibliographie très partielle. L'article par Gaiotto Moore « Nights in 2008 » était certainement très intéressant pour WKB et la formulae d'allumage du monde et le spectrum de BPS et le conservateur de Bellemande pour la formulae d'allumage du monde. Je l'ai mentionné ici « Pasquetti's kiappa » parce que c'est une théorie topologique et « Garufalidis » c'est « Kappayef Marignot » parce que la connexion avec Pelleuville et Stanton J'aime particulièrement « Iwaki Nakanishi 2014 » un bon service sur WKB analysie et la connexion avec les algebras de cluster et très récentement, cette année, Gaiotto Marignot-Choux sur les questions de TBA et les problèmes de Riemann-Hilbert Juste pour dire qu'il y a beaucoup d'activités, de réneaux d'intérêt sur ce sujet. Donc notre rêve est d'expliquer si possible où la résurgence vient d'utiliser un outil inventé par Jean Écale une construction inductive d'élémentaire et non-trivé résurgence pour laquelle le molde de formalisme est particulièrement efficace donc je vais essayer d'expliquer cela à la fin du talk Le message c'est que ces moldes ont aidé à élucider la structure résurgence ils ont aidé à établir les équations brides qui sont derrière les automorphismes ou la formule DDP et ils ont aussi induit les constructions récursives des fonctionnalités en point de vue dans l'esprit d'autres récursions que nous avons écoutées donc ce serait l'end de tout le talk Alors, nous allons commencer avec l'équation de l'équation de l'onimation de l'onimation de l'onimation dans cette forme donc la forme de l'onimation de l'onimation est la variable l'Eta est l'un des barres de l'âge donc c'est un paramètre large et ici nous prenons pour Q un polynomial polynomial plus polynomial en 1 à l'Eta c'est plus général donc nous avons dans un peu plus de termes géométriques des termes méromorphiques des différences quadratiques Qdx² sur la surface compacte c'est-à-dire sur la sphère qui est régulièrement à l'extérieur des sets finitifs donc bien sûr, il y a le pool à l'infinité et les 0s de la function Q comptent comme cégurités pour les différences quadratiques donc elles sont appelées les points de tournage donc nous avons une courbe complexe qui est dénotée par c.2 donc c. est juste le set régulier et c.2 c'est la double couvercle de c. sur laquelle nous avons une courbe square de notre différente quadratique c'est-à-dire Pdx parce que donc je vais me faire Write c'est-à-dire que c'est-à-dire que c'est-à-dire que c'est-à-dire que c'est-à-dire que c'est-à-dire que c'est-à-dire que c'est-à-dire que cette extension deides d'une forme classique dans une forme formal sur la courbe complexe l'AMDA LORD dans laquelle voir la courbique en 1 prophese et vous avez un série formale dans l'une d'un eta une série de l'9 et je la n'ai option j'ai pour ne voulue car ce sera évidemmentちは entrenté le changement co precisely et les coefficients seront engagementants et il sera homomorphique sur la courbe complexe. La définition sera récordée au moment. Nous voulons étudier cette série formale en eta avec des fonctions homomorphiques en X. Nous voulons étudier la transforme boréale avec respect aux paramètres eta. C'est la résurgence paramétrique. Donc, ici est la définition. La transforme boréale est réplacée en eta par xi. Donc, on obtient, en tout cas, une fonction de xi et xi homomorphiques pour xi de modulé small enough. Donc, la rule est qu'on divise par les facturaux comme ceci. Et cela devrait être sans doute continuable avec respect aux xi, dans le sens que la continuation analytique existe en large avec seulement la singularité isolée. Nous répétons l'équation de Schrödinger. Ce n'est pas seulement que nous produirons ce lambda. Je ne vous dis pas encore ce que c'est. Mais nous produirons des solutions formales pour l'équation de Schrödinger dans la forme de WKB ansatz. Il y a une exponentielle i eta a où a est la fonction classique de la fonction actuelle, l'intervalle de q à 1,5 qui est bien définie sur notre curve complexe. Mais bien sûr, la primitive est multivalue sur la curve complexe. Nous avons cette q à minus 1,4 et ensuite nous avons 5 plus ou 5 minus si nous prenons l'opposite branche de q à 1,5. Et 5 plus et 5 minus sont destinés à commencer avec l'une constante. Donc, ces coefficients phi 0, phi 1, comme a seront multivalues et la série 5 plus et 5 minus sont déterminées seulement à un factor multipliqué de x indépendant, parce que c'est une équation linéaire. Nous serons intéressés à la transformation boréale avec respect au paramètre eta pour certaines choisis de normalisation puisqu'il y a cette question de déterminer une solution que nous serons de la transformation boréale. Peut-être que certaines choisis sont mieux que d'autres. Donc, je répète la résurgence consiste d'être endlessly continuable dans la variable xi. Dépendant les x, peut-être cette propriété s'occupera pour une générique x, mais pas tous les x. Par exemple, x ne devrait pas s'occuper de trois points. Simple exemple. Je répète la formule de WK par contre, ici. Donc, nous avons plus ou minus i eta a of x q to the minus 1, 4, c'est très classique, phi plus or minus of x eta et bien sûr, vous voyez que cette a of x c'est juste l'intégrale de lambda 0 de quelques points de référence x0 que nous fixons pour tous. Par exemple, il n'y a que un point de tournage et rien d'autre. Point de tournage de multiplicité m, à l'origine, nous appelons xj les points de tournage et en ce cas, vous computez la fonction action a of x vous computez q to the minus 1, 4 et il s'agit qu'il y a une ferme ferme pour la transformation de ces séries, phi plus et phi minus. Vous voyez que j'ai installé ce facteur, ce monomial eta to the minus mu minus 1 c'est le nombre selon la multiplicité m donc c'est juste convenant, juste pour... de cette façon, nous avons une fonction algebraique de xi. Vous voyez que si vous replacez phi plus par 1, la transformation borale de ce monomial est juste xi à la mu divided par gamma of plus 1. Ici, quand vous avez la transformation borale et vous observez qu'indeed la chose est convergente dans la plane xi et en ce cas, il y a une bonne continuation analytique c'est algebraique, phi plus la transformation borale phi plus a seulement une singularité à 2i a of x et phi minus a seulement une singularité à minus 2i a of x maintenant, quelle est la singularité de phi plus let me write this it should be clear that it will be proportional to the borale of phi minus indeed let's have a look borale of eta minus mu minus 1 phi plus you can factor out these 2i a of x let's put it in front 2i a of x power minus mu divided by gamma of mu plus 1 and we have this xi of mu xi to the mu and 2i a of x minus xi power mu now what does it mean to consider the singularity at that point it means that we replace xi by 2i a plus zeta and we want that xi approach approach 2i a we want zeta to approach 0 so you see that this will become exponential i pi mu zeta to the mu and this will become 2i a plus zeta power mu so in fact exactly you find the same formula as here so in fact if you it's an exercise in alien calculus if you manipulate this alien operator defined by ecal which is just the the smart way of measuring singularities in the borale plane you find this formula so while it's an exercise you have this exponential i pi mu but then you have to study the dictionary between the singularity and the function in the xi I mean you have the notion of major the singularity of is defined by a major and then you take the minor so when you compute you find this sign function incidentally that formula was written by Jean Rical in 81 in the second book of the series of on resurgence function in quite slightly different context it was in the context of equational resurgence dealing with Riccati equation we'll see the Riccati equation in a moment so particular case m is 1 this is the quantum area equation parametric area equation you find the stokes constant this number here this proportionality factor will be minus i in that case because mu is minus 1 over 6 and for the quantum harmonic oscillator you find different number ok that is really the simplest example you can dream of so now to the associated Riccati equation we consider the logarithmic derivative of the wave function psi and it's an exercise it's very clear that you agree that if i eta p is exactly the logarithmic derivative then its derivative is written like this and so the Schrodinger equation is equivalent to that Riccati equation p squared equal to q plus i eta inverse derivative of p and now there are only two formal solutions to that equation indeed it's clear that you can choose to start with plus or minus q to the one half and then once you have chosen that the branch the rest of the service is determined here is the next term it so happens that it is a logarithmic derivative and then plus or minus this etc so you find two well defined formal series by the way they are related because it's Galoisian I mean if you go you can go from one branch to the other and so you exchange the two solutions by the covering involution which defines a complex curve c.2 and from those two formal series you can recover all WKB solutions to the Schrodinger equation indeed, since we are speaking of logarithmic derivative it's just the matter of choosing a primitive of p plus or p minus and then exponentiating it and then you are free to multiply and then factor so we will recover this way when you apply this recipe the first term here is responsible for exponential plus or minus i eta a indeed, remember a is just the primitive of q to the one half that we have chosen the logarithmic derivative of q is easy to integrate and you get to this q to the minus one fourth and then this here will give rise to a series phi plus or phi minus as we had said previously so what is this lambda that we had promised at the beginning there would be one form formal in eta in this problem what is p it's just the h bar extended momentum it's the half of the difference you have two solutions you take half difference and check that indeed it gives rise to one form now each series p plus or p minus is the plus or minus the half difference plus the half sum and it so happens that the half sum is the logarithmic derivative of the half difference this will help us to handle this question of normalization choices the coefficients of p plus and p minus are uniquely determined holomorphic on our complex curve they have previous expansions at the turning points and at infinity and you check that there is no term in the form 1 over x minus xj as a consequence to choose normalization for psi plus or psi minus to choose the factor here is equivalent to select or equivalently a primitive of p curly p so here is what we do so we must integrate this so let's split p curly p there is the first term q to the one half and then there is the remainder curly y with this factor q to the one half and then correspondingly p plus or p minus is plus or minus the first term this we repeat and then we have plus or minus 1 over i eta q to the one half curly y and so we can define what we will call the xj normalize solution xj being one of the turning points or the pole at infinity like this we don't touch the pre-factor the exponential pre-factor here we have curly p to the minus one half we are starting with this we are using this this natural primitive logarithmic derivative and here we must choose a primitive of q to the one half curly y so we choose the primitive whose freezo expansion and xj does not contain any constant term that's a reasonable choice equivalently this is the average of the primitives corresponding to the base point a on the complex curve so something more about this normalization because they play a role in the theory so psi j plus psi j minus you can write it this way using this slightly different notation for the selection of primitive and so q to the one half curly y this is just lambda minus lambda 0 and a of course we know what it is here appear the so-called vos coefficients they allow to connect one normalization to the other how do you go from j to k you multiply by a certain integral and the integral of q to the one half curly y now there is a remark let me do a picture that y is invariant by the covering involution so when you multiply by q to the one half you can use this formula you have xj here and you have x somewhere and so instead of doing this business of choosing the primitive by looking at the expansion you can do that you can say let's do this so there is a cut here so you start from x star the other point for the other branch and you turn around so this is the path gamma j x which is here so it starts from x star it encircles positively xj and then it goes back to x and you compute the path is exactly what you want so this means that our coefficient pi j our integral is just let's say a quantum period i eta over 2 of lambda minus lambda 0 so what you can do in that problem is to maybe I should explain what is gamma jk I mean here we have another turning point jxk and we are supposed to compute the difference between these two primitives so the difference can be computed by saying that we do we do this there is no so we find a cycle appearing in the so this is the gamma jk it's the cycle around j and k and and then we can, then next we can choose, let's say we have our reference point somewhere here x0 so we can choose a basis for the homology and so this would be gamma j gamma k so we can compute once for all these pi j of eta so integral from x0 to xj with the rule that we have set for the turning point following the path gamma j that we have selected in advance ok so this is definition of normalization for each singularity at infinity the situation is simpler because the coefficients decay at infinity and you get integrable function so you can replace this funny selection of primitive by the usual integration from infinity so here is what you can do let us split p plus and p minus like before we have the first term plus or minus q to the one half we have the second term and then we call y plus or minus the rest with this prefactor so we define this way y plus y minus and so you see that curly y is just half the difference and now the point is that the normalization at infinity is just defined like this using the integral of q to the one half this y plus or y minus integrated from infinity to x so the solution normalize that infinity are characterized by the vanishing at infinity of all the coefficients except the very first one we said that this phi plus should always start with 1 y plus it's just a logarithmic derivative essentially of phi plus or minus infinity so why are they resurgent where does resurgence come from for all these objects the y plus the y minus the phi j plus or minus we will deal with phi infinity plus or minus which is easier to define so why is it resurgent let us define a vector field x curve d d is q to the one half d over dx so it's dual to the square root of the quadratic differential it's dual to this lambda zero and this way we can rephrase the Schrodinger equation using the unknown phi plus or minus instead of psi and using d instead of d over dx and here is what we get so we have this function k which is very important k is given by that formula so it comes from the initial q which we have started with and correspondingly we can change unknown in the Riccati equation and again we can use this vector field d this derivative and we get this equation for y plus and y minus ok so y plus or minus in this sense is the d logarithmic derivative of phi plus or minus and here it's not the xj normalised solution it's the xinfinity normalised solution so here we are supposed to invert d so to invert d we multiply by q to the one half and we choose the primitive so it's the xinfinity based rise inverse to d so why will it be resurgent we can compute in a sense phi plus and chi plus so these are the notation the derivative of phi plus so remember that y plus will just be the ratio of these two we can compute as well phi minus 2i eta phi minus using the Neumann series because it's a linear problem so it's a very particular case of parametric resurgence so this you recognize here the Riccati equation for the phi series using d so we insert this chi plus it's d phi plus and so the second equation is this and now we say we have specified the way we solved the equation we have specified that phi plus should start with a 1 and then the other coefficient vanishes and chi plus will vanish in fact we know how to invert this because here we are dealing with operators in formal series in eta with coefficients holomorphic on the complex curve c dot 2 so curly k is just the multiplication operator by k what is the inverse of d plus 2i eta it's given by the geometric series this is a perfectly well defined operator we have eta to power minus k minus 1 so you differentiate all the coefficient and you see that you differentiate the coefficients of holomorphic functions and you do not compensate by any factorial so you expect je vrai growth and you expect divergence yes series in inverse eta always always oh yes you're right minus 1 is lacking thank you so the Neumann series for this linear problem gives you this so phi plus you see you start with 1 and you repeatedly you multiply by k apply this inverse operator and then you integrate from infinity and chi plus there is this difference we have one more factor similarly for phi minus when you do the computation you find this minus 2i eta times something which starts with 1 over minus 2i eta and chi minus is this so you have 4 series of formal series in 1 over eta so we are dealing with series of formal series but each of these term each one individually is a non trivial resurgent function so why is that so it's because the counterpart of the operator that we use d plus e 2i epsilon eta so epsilon here is obviously it's either plus 1 or minus 1 here it's a typo it's minus 1 2 here so we are using plus or minus 1 so in the Borel plane what happens in the Borel plane we divide by the factorials so we get an exponential series an exponential of d so d is the derivation it's a vector field exponential of a vector field gives you the flow so you see here the flow the time t flow map of our d so remember d is just q to the 1 half minus 1 half of x d x so we should inquire about the flow of that vector field and when you know the flow you take it at time minus psi over 2i epsilon you substitute in g so if you start with g not depending on eta if you start with capital G which is a series in eta then the computation is like this multiplication by eta to minus k minus 1 gives rise to convolution by this monomial and then you write the convolution as an integral and again you see the flow of our vector field at minus psi prime over 2i epsilon so we get an integral operator and we see very concrete formulas that we can manipulate and that we should apply to to this so we are taking now we know how to handle the Borel transform of all these series so the claim is that the Borel transform phi hat plus or minus chi hat plus or minus are endlessly continuable and where are the possible singularities, where are they located so the recipe is that there cannot be a singularity at omega unless when you flow along this vector field starting from x until minus or plus omega divided by 2i you hit a turning point so this is the recipe to determine the singularities of the Borel transform so you see that the singularities in the xi plane depend of course on x so the idea is that the vector field that we are considering is straightened by the so-called UV transformation just the action function viewed as a change of coordinate and in that coordinate z the vector field is strengthened so you can compute the flow and in fact the quadratic differential is strengthened too so you can use the local geometry of the quadratic differential if you want to analyze the behavior near xj for instance so in that coordinate z everything becomes simple but be careful that the coordinate z is defined locally and then you follow its analytic continuation so there is a kind of periodicity in the problem but it's tricky because everything is multivalued on the complex curve c.2 so excuse me can you remind me what is omega x so it's omega of x is defined in this claim so what are the potential singular points they are the points such that when you flow until that time you hit a turning point and in a moment you will see that in fact we are determining them we are here reaching an alternative definition because the flow here I'm using this gamma j so from x0 to xj if I start from x0 alpha j is the value of the action one of the possible value gamma j but of course if you follow a different path you get different values you get replicas of this singular point alpha j so alpha j is a singular point for the flow map because we are saying that the vector field vanishes here and so in finite time we hit a singular point of the vector field on the other hand the function k that we are using in our Neumann series is meromorphic but it has poles at the turning point hence in coordinate z k has a less analytic continuation and where are the singularities in the z variable they are located at all the possible values of the action function action function a so alpha j plus multiples of the classical periods so this is the description in variable z starting from x0 but now we are starting from x so we must shift everything by z so the recipe which is here gives us z minus or plus omega over 2i is alpha tilde so omega of x this is the answer to your question plus or minus 2i action of x minus alpha tilde and so why nk is integers yes yes yes I mean it's just the action it's just a representation of the homology group excuse me also among these x1 to xm you count x infinity also no not here not here because you cannot go to infinity in finite time yes of course you can for x4 you can oh maybe I'm mistaken let me think no no no look z is primitive of q to the one half so we are integrating we are integrating the square root of the polynomial so it's integrable at the turning point at infinity this is the non integrable part that we are always skipping when we go to infinity we must always remove that part may I cautiously disagree I mean I discovered that in fact you should not remove it it's nicer you can get cleaner things if you don't remove it I can believe it I mean this is probably not the definitive version of I would be happy to see your comment in detail on that thank you so where does this condition come from so I claim that this is the condition to have possibly a singular point where does it come from so remember we were dealing with these Neumann series of that form so in the case of phi plus we have to invert d plus 2i eta so the Borel transform involves the flow at times minus psi over 2i and so this is why we get the condition like it was written flow at minus omega over 2i should be on the singular locus similarly for phi minus so the next question if we are dealing with the resultant structure of phi plus and phi minus is what are the alien derivatives ? so I'll be a bit sketchy but because in fact it's work in progress so these formulas are written in the literature but our understanding is still in progress but the idea would be the following if you are dealing not with the solution normalized at infinity but with the solution normalized at xj the alien derivative will be governed by the sj that we have computed a moment ago so all everything here is like as if you are dealing only with a pole with a a monomial q of x you can substitute q of x by this mj is the multiplicity at xj and you get that result but that is true only for phi j when you are dealing with the alien derivative at a minus alpha j now if we want to deal with phi infinity we must go from phi j to phi infinity and for that purpose we can use the VOROS coefficients and so now that's an easy computation the alien derivative of that guy should be the exponential multiplied by the alien derivative of that guy so we have the sj and we get twice pi because one comes from here and the other comes when you express phi j minus in terms of phi infinity minus and similarly for the other for the other one so we can go on and get formulas for the alien derivative of the rickety solution because it's a general fact in resurgence I mean this part of the theory of that if you are dealing with resurgence functions their ratio will be resurgence, their exponential their logarithm will be resurgence so and moreover there is an alien chain rule if you want to compute the alien derivatives of these so once you have computed the alien derivative of one set of solutions you can in principle you can deduce the rest of the structure and then the next question is but what about the alien derivative of that coefficient so in fact here I'm referring implicitly to the second bridge equation and the third bridge equation so here this is here that ddp formula would appear because the location of singular points for that factor, that factor does not depend on x it's vos coefficients it does not depend on x we have integrated in x so you can guess that the singular points will be located at those points, the classical periods so I stop here my comments on the resurgence because I want to move on to something different we have said that we would reach the resurgence structure using these representations of the solutions normalized at infinity by means of these series of elementary non trivial resurgence series a way of motivating what follows is to ask can we do the same for y plus and y minus so we have said we are confident that it's a ratio of resurgence series it will be resurgence but can we analyze y plus and y minus in the same manner that we analyzed phi infinity plus y minus so here is what happens we have our four series of series here is a compact way of rewriting them I introduce those series m indexed by plus or minus epsilon 1, epsilon 2, epsilon r these are strings, these are words the letters of which are plus or minus so when the word is empty you have 1 and then how do you define m associated with a word of length r so it's an inductive definition so you invert this operator d plus 2i eta the sum of the letters you invert it using primitive at infinity vanishing at infinity if the sum happens to be 0 then you apply that to b with the first letter b with the first letter is either k or 1 and then this is the function the series associated with the truncated word in which you have removed the first letter so is that clear for you for instance start with phi plus so you have what is this nth term so we have 1 so you multiply by k multiplication by k and applying the operator this is multiplying by b plus and applying d plus 2i eta when there is only 1 letter which is epsilon 1 equal to plus but then immediately after that you apply the inverse of the infinity which is that case when epsilon 1 plus epsilon 2 is 0 so this is why I wrote here m minus plus I started with a plus so that was the and then I have a minus et cetera so these are elementary series this is a family of series indexed by words on the alphabet plus or minus 1 this is the definition of a mold a family of object indexed by words but here we see only a slice of the mold we are using only alternate words of lengths or odd lengths but only alternate words what about the other words what about the rest of the mold what will it do for us so I repeat here the definition of the mold the inductive definition I repeat what we have obtained so far and here is the claim we can represent the solutions to the Riccati equation in the same way so here we have series with coefficient 1 and only peculiar words but here we will use a different subset of words and we have different coefficients these coefficients beta plus and beta minus are integers they are very often 0 to be non 0 beta plus requires the sum of the letters to be 1 and the value is computed by iterating these operators so you have elementary operators d over dy minus y square d over dy you see that there is homogeneity going on here because b plus is homogeneous of degree minus 1 b minus of degree plus 1 so when you do the composition you see that the product applied to y will have degree 1 minus the sum but if the sum is 1 it's a constant and that constant is an integer number so this is the recipe to compute beta plus and the recipe for beta minus is here so now we are interested in words whose sum is minus 1 and so I claim this and I also claim that if you forget the factor minus 1 to the a plus 1 instead of finding y minus you will find 1 over minus 2 i eta plus y minus so why is that true I would like to spend 5 minutes to explain that it's very elementary part of mold calculus as defined by Eckhart so let's rephrase it here in our context the key point is a certain family of quadratic relations so we are dealing with a mold on a certain alphabet which was plus or minus 1 with value in a ring r in our case the ring of formal series in eta whose coefficients are holomorphic on the complex curve c dot 2 so we said a mold is just a family of object indexed by words so we can say this is the collection of values of a sequence of functions so giving a mold v is like giving a sequence of functions and the nth function depends on n arguments these quadratic relations that we will write are called symmetry relations so here it goes v0 should be a constant it's a function of nothing and that constant should be the unit of our ring and then for all pq for all p tupple b for all q tupple c the product should be should coincide with this sum so it's a sum of values of vn n is p plus q so you take the sum of all subset of the set of indices 1 to the n to n so i is the subset of cardinality p, j is the complement cardinality q and what is this this is the n tupple a the unique n tupple such that when you extract the i part you recover b and when you extract the j part you recover c so i wrote the formula in this way because it's reminiscent of something which is well known in topological previous example yes here v is general mould and we will apply this to m so this is just a more general definition a1 a n was plus and minus exactly exactly here is a typical example of mould satisfying this quadratic relation and our m is very much of the same nature so here is the standard way we rephrase this property in mould calculus this is absolutely equivalent to using the shuffling of the two words b and c so our mould m satisfies this so our mould m is defined on the two letter alphabet it is symmetrical and you can check it by induction on the sum of the lengths on the sum p plus q and it's not difficult because you have an inductive definition and you have a characterization it's defined through a differential equation with a boundary condition at infinity so it's not difficult to test this property so our mould is symmetrical what good does it do to us so remember that the Schrodinger equation was written for phi plus in the form of a system so this is the system for phi plus chi plus equivalently we can say this is a vector field or this is a derivation acting on that algebra the algebra of formal series in y1 y2 whose coefficients are in r in the same typo this should be eta inverse so r is our coefficients take value in r and let's consider these derivations so d act on the coefficients d is the differential d over d essentially so we have d over d y2 d over d y1 you remember that b plus is k plus y1 d over d y2 this is what comes from this k phi because phi is y1 and b minus is 1 so this chi here is responsible for this y2 d over d y1 let us call l0 the first part of the operator let's call b plus bar this operator b minus bar that operator so we are dealing with this derivation and this derivation encodes the Schrodinger equation in a certain sense so these are derivations now let's consider this operator so this huge sum sum of all possible words of m epsilon and this product of derivations so this is formally its convergence so it defines an operator and the inductive definition of our moule actually says that it conjugates l and l0 so it's like a normal form problem we are conjugating what is given to us to something simpler we have gotten rid of these two terms but because the moule is symmetrical theta is an automorphism that's a very general principle why is that true it's because you see how theta is constructed using the moule that's coefficient and here we are using product of derivations each derivation satisfies Leibniz rule a product of derivation satisfies generalized Leibniz rule and when you write the generalized Leibniz rule because they do not commute you see the Schroding coefficients appearing and by duality if the moule is symmetrical then this guy will satisfy that the product is sent to the product instead of Leibniz rule some other all possible words including the empty word all possible length it's a huge sum and you don't put a factorial no no factorial here so it's an automorphism what are the automorphisms of the space of formal series they are all substitution automorphisms so we know in advance that theta f bf compose with something and that something is obtained by evaluating theta on y1 or y2 in fact in our problem when you do this you find that this little theta this this formal change of coordinates in the y1, y2 space is this so everything is very simple in that case because the problem is linear but wait a minute I don't do the computation but it's not difficult you obtain that so you can believe it I mean after all what will happen when we evaluate this product of operators on y1 with so elementary vector field it will be very easy to compute you will see on the alternate word it's clear so you get that but remember that operator is an automorphism so I repeat the formulas here bar, b minus bar, definition of theta and this is what we have obtained so far now what is the general solution to the Riccati equation written for the y function it's the logarithmic derivative of the general solution to Schrodinger so the logarithmic derivative of so you see the general solution to Schrodinger is sigma plus and sigma minus are just parameters so since I focus on the equation with a plus I see the multiples of phi plus and the multiples of phi minus appear with this correction exponential minus 2i eta a a is lacking here this exponential minus 2i eta a because I'm switching from phi minus to phi plus so we can introduce sigma which would be sigma minus of a sigma plus and you see that this takes the form of a linear fractional evaluated on this monomial so let's introduce theta for Riccati to be that so this is a formal difomorphism in the y space with coefficients depending on x and eta so and morally this y is just y2 over y1 because theta is an automorphism because of the symmetry when you compute theta of y2 over theta over y1 it is the same as theta of y2 over y1 so you can compute that thing theta Riccati of y as the evaluation of that big operator capital theta on y2 over y1 so previously we were just evaluating on y1 or on y2 but now we are evaluating on y2 over y1 and what do we find well the image of by just one operator b plus is one and the image by one operator b minus is minus the square so maybe you remember the formulas for b plus and b minus they were just as simple as this so this is how we prove the claim so we have obtained that formula that y plus the solution to Riccati is given by this mold expansion with coefficient beta plus which are produced by these operators, elementary operators acting on y but this is a very very elementary instance of mold calculus but what does it get where does it get us we have obtained this we have a recursive definition we have this m and when we solve the induction this is what m epsilon is so epsilon is an arbitrary word and so we are inverting that operator multiplying by b epsilon1 inverting that one multiplying etc and whenever a suffix of the word has zero sum the prescription is that we should use the primitive vanishing at infinity ok let us define the resonance level of the word as the number of zero sum suffixes how many terms here do vanish how many times do we need to use the integration from infinity the rest of the times we use only the geometric series with d you remember but because of those zero sum suffixes our m epsilon will involve an n fold integration from infinity and this is slightly reminiscent of something we've seen in the talk about topological recursion so we are tempted to group together all the words with the same resonance level n so let us introduce w n plus or minus as the sum of the beta so in this big sum we extract the words with resonance level n and we can call that n point component y naught or n point correlation function and then our y plus is the sum of that so what we obtain here is that we have a recursive definition which produce produces automatically a sequence of object whose sum is really the solution to the Riccati equation so afterwards you are just you just need to integrate from infinity to exponentiate and you get WKB solution and you know that it was mentioned in the previous talk that in the airy case topological recursion does produce WKB solution but here we have a different angle a different way of constructing recursions and so let's say this is food for thought we plan to investigate that further thank you for your attention so to prove the claim you need to prove that the sum of terms converges somehow or other you are referring to the last claim or I mean anywhere, the claim and then the well there are claims of different nature for instance that one what we where is it that one ah ok you mean on the resurgence yes yes I skipped that part but indeed to prove resurgence of phi infinity plus or phi infinity minus we should establish that this series of formal series gives rise in the Borel plane to a series of functions and that's a fact no problem but so then we need normal convergence on compact subsets on certain remands of phase yes indeed this is the strategy which was indicated by Jean et Karl in as early as 84 so I mean is that my question is at the end are you planning to say that basically the WNs have some kind of not too many WNs and each one sort of gets smaller and smaller and that's why it converges I mean the the series for Y plus is we can apply the same strategy directly instead of invoking a general theorem which says the ratio between two resurgence series must be resurgence we have access directly to the to a decomposition of Y plus in elementary pieces in the remands of phase so that's the strategy yes it would be to and then after regrouping we would like to see what happens but well it's work in progress as I mentioned at the beginning so you had WNs yes maybe in the next slide on the very end near topological recurrence yes right there so are these WNs supposed to be factorier very soon Y plus is but are WNs themselves yes everything here is resurgent I mean this is a decomposition I mean factorier with respect to N it's convergent with respect to N we have convergence here the divergent takes place only with respect to one of the eta so this is a series of resurgent series which is convergent for the topology of resurgent functions because in topological recursion those terms are factoring divergent so in fact it's a different angle so we are doing something else yes I think the point is in fact that those should be factorially convergent because when you do an N fold integration you get a 1 over N factorial and then maybe there's only an exponential number of terms that is indeed the case yeah but in topological recursion it doesn't happen apparently so here the things, the computations are organized in a different way so this is what we wanted to propose to that audience to see a different way of organizing the computations yes André I would like to ask you I'm very excited by this normalization of infinity because I in my work I was guilty I think I didn't realize that it could be done so later I on normalization and the advantage of normalizing at infinity including the divergent part is that you get something which is translation covariant now is your normalization at infinity explicitly also covariant by translation it's not mine, I mean it's in the literature it's I mean in FAM it's already in FAM yeah and in fact in Eccal's paper in 84 already so is it covariant I mean the trick is that lambda 0 is not integrable at infinity so you have to to remove that part you can't define the integral to infinity by a way a classical version of data regularization so in a way but perhaps finally it might be allowed to what you are doing so I am very excited so I didn't think too much about it because for us the idea is that we fix once for all this x0 and it appears only in this pre-factor it's a totally different business when we are referring to to the integration of that guy because here we have a whole series so theoretically we might change the selection of primitive at each term of the series and then we would get crazy things but here it's just once at the beginning so we didn't pay much attention to that but I would be interested in your comment about the data regularization yes so I haven't quite understood to write a full solution you would have still to write some rich equation for the borrow's coefficient right and the second part was not some method to get around that like you would still have to yes these are two avenues of research I mean I switched at some point from one direction to another