 Спасибо большое. Вчера я поговорил о КП-харархии, и сегодня я поговорю о топологической рекордии. Эти два субъекта, которые я сказал, у вас есть много структурной similarity. У них есть красивый spac of solutions, у них будет красивый группа симметричных актингов, и так далее. Есть еще другая similarity, и также структурно, что люди, работающие в этой субъекте, обычно apresentают результаты, в таком виде, что никого из-за того, не смогут ввести. Итак, я покажу мою собственную версию топологической рекордии, которая является эквивалентом к стандартам. Я не думаю, что мою версию лучше, но я, afortunately, удача, чтобы сравнивать. Например, вчера Дима представила рекордию для хорвитских номер, и я спрашиваю, если после его поговора вы сможете реконструктировать хорвитские номера из-за этой рекордии. Итак, эта рекордия появилась сначала в модели Метрика, и одна из первых версий этой рекордии появилась в пейпе Миронов, Морозов и Александров. Тогда было в пейпе тоже в модели Метрика в районе 10 и Чехов. А потом, обычно, рекордия для всех русских исходят, но теперь это часто refer to в районе 10, в районе 10, в районе 10, рекордия, да. Итак, я буду пытаться использовать репрессивную рекордию, как это, если вы не хотите. Итак, еще одна вещь, что я хочу сказать, что в этой ночи я пытаюсь модифицировать мою рекордию, и это было полностью... после того, что я буду представить только половина моих компьютерных экспериментов, так что последняя часть не работает в момент. Но, дай мне просто дать вам initial part, которая работает довольно хорошо. Итак, то, что типологическая рекордия делает. Так что, у нас много проблем, где компьютерные материалы, связанные с проблемами, собрали потенциал, партийственные функции, коэффициенты. Коэффициентом, это то же самое, как все возможные партийственные. Так что, все эти данные собрали в отдельной гератике серии. И мы видели несколько примеров, как Хурвис серии, Витом Консейдж потенциал, и Петерсон Волюмс и так далее. Их будут многие другие примеры в том, что Петерсон Волюмс. И в типологической рекордии анзат, который производит коэффициенты из серии 1x1 используя какие-то данные, которые используют. Чтобы быть правильным, рекордия не идёт для функции себя. Если у меня есть функция, то есть параметр 1x1, который responsible for генус, и н-стенс, то есть н-стенс. Их будет в индукции гн-н. В правильном капитале n-1, что это значит? В титуле генусы? В момент, я оставляю гену, то есть первые примеры, которые будут выводены к н-1, а потом выводим гн-н. В формуле гн-н на второй линии. Да, это будет подписано. Спасибо. То, что мы actually compute, это не функция, но это деференциал. То есть, коллекция, это параметр диривитов. Так, давайте considерим деференциал, который это комбинация параметр диривитов. И давайте учитываем эту деференциалу т-т-т-алфой, просто, и новой, новой, вибрациал, как всю символу. И в том, чтобы выяснить, что я учитываю просто абстракт символ, я буду выводить хае. Ксай-алфа кладетity-алфой. И мы чувствуем, что уа же для нейтации Uma proportion весeriaNew С путем И этот элемент — это форма-линерная комбинация КСАИ. И линеральная спина КСАИ — это длина Л. И этот элемент — это элемент этой вектории Л. И с коэффициентами в Т. Также... Детали, которые были использованы, — это дюйл-спейс. Я считаю, что дюйл-спейс — V, с дюйл-безисами КСАИ, альфа-стар, с дюйл-безисами. Я копирую здесь некоторые части экрана, потому что я хочу, чтобы вы посмотрите это позже. И потом, я считаю, что дюйл-спейс — это длина Л. И также, дюйл-спейс — это длина Л. Итак, для момента, это может быть сделано для каждой функции. Рекарсия не... не... не упоминана. Но теперь, что используется для рекарсии — это длина. Это случилось, что эта длина Л, с длина Л и дюйл-спейс, должен быть equit с структурой ассоциативной алжебры. Я покажу, как это работает в экзаменах. И потом, у нас есть рекарсия. И это используется для того, что есть длина. То есть, что я делал, functionality can beocide way given, which is between of the two resources. it's engineering so Коэффициент, который зависит от t в нижнем способе, но в любом случае, я считаю, что коэффициент — это функция т. Потому что, в первой линии, она говорит, что есть infinitely many. А, а, infinitely many. Finally many. Верible t, да? T zero infinitely many. Just as yesterday. Ah, it's the same as label by integers. Label by integers, like yesterday, all these potentials here. And there will be also some... Also homogenous in t means t zero has degree zero. No, just usually tk has degree one. Usual homogenous... homogeneity. And also I will need one extra data, which I call shifted origin, which is eta is a particular point of this space, some fixed vector. And the only thing that I suppose that it should be invertible. Okay, so this is the initial date of the recursion. And now let me formulate the recursion itself. And here what do we do? First we apply this operator. So we play this operator to our function u. Sub g minus one and plus one. Thank you. I don't understand this. Later. So how this operator acts. I recall you that this is a point of L with coefficient depending on t. So first we differentiate with respect to parameters. And then we multiply by xi using the multiplicative structure in H. So this is... Here we have a point, means the multiplication in this algebra H. And again the second sum. Again here when we multiply we use this multiplicative structure in H. And also let me add some more term. H is infinite dimensional. Everything is infinite dimensional. So probably you will see immediately in the example how it works. So I take this huge expression. It belongs to H. I use this to... Here you use the multiplication in H. And there is one extra step. I wish to divide this over 2 times eta. So this is still an element of H. And I want to relate this to the newly computed term of our recursion. But the left-hand side is an element of V. And the right-hand side is not. So we just ignore the V component of this right-hand side. And project just to the first sum. And this is UGN. So let me show how it works in examples. I will assure you that after some examples you will see how it works. So the first example is the Witten-Konsehich potential. So Witten-Konsehich potential is the intersection number, a genetic function for intersection number of psi classes on the modulus spaces of curves with marked points. So this is my implementation of this. This is my implementation of this function. Yeah, now it works. This is the actual definition. I need some way to compute it. So for the actual computation I use the Verasor constraints. So this is the Verasor constraints. These are the Verasor constraints. So on the left-hand side we have this differential operator, which involves partial derivatives and also sometimes second-order partial derivatives or linear terms and some factorial coefficients. And the right-hand side is almost always zero, except for the KSM equal to minus one and one. There are some corrections. And yesterday I wrote this Verasor constraint for the exponential of this generating function. Here I wrote for the function itself, so the nonlinear term appeared here. Instead of the second partial derivative, we have a product of first partial derivatives. And also another distinction, that here I use the genus expansion, so it implies that I should put here this formal parameter h. Anyway, these are well-known Verasor constraints for the written constraint potential. Now let us do the following trick. Let's combine all these equations together in a single one in this following way. I introduce an extra new variable x, a new formal variable x. And sum up all these equations with this factor. I put x to the negative power m plus one, multiplied by the left-hand side of the Verasor equation. Then it happens that using this trick one can write these equations in a very nice, compact form. So, for instance, what happens if we sum up this term? If we sum up this term with this coefficient, we obtain an expression, which is exactly our function u, which is this one. So, this sum will appear if we sum up up to the shift of m plus one, and this is just the same. So, summing up this term with this coefficient, we obtain what I denote by u here. Similarly... This function u contains... Yeah. You should have the following say. Yeah, so... Let me write it here. So, let me denote just by this coefficient by xi. Xi sub n is equal to 2m plus 1 double factorial divided by 2x to the power n plus 1. And then u is the sum of 2n plus 1 double factorial divided by 2x to the power n plus 1 times the partial derivative. This is xi n df over dtn. Similarly, here, when you take this double sum and collect all these terms, you have just product of two things like this. So, the left-hand side... This term turns into the function u. This term turns into product of u squared. So, here you have u. Here you have u squared. Then, this sum containing 2nd or partial derivatives is equivalent to applying this delta this delta to the function u. So, u already contains 1st order partial derivative of f. If you apply this delta, you obtain 2nd order derivatives. And also there will be... These factorial factors will appear twice. First, when you take the coefficient xi, which is already in u, and then this extra coefficient. When you write xi n equals something, do you mean it's actually a way to introduce the algebra structure? Because in your notation, xi n is just a... Yes, the algebra will be h. In our case, this will be this ring of Leron polynomial in x. So, L will be the negative part of this, polar part of this. And vz will be the regular part. That's how you get the multiplication in h. Yeah, yeah. This is the algebra structure in h. So, in general, you should have a curve with a smart point in formal coordinates. Something like this again. If you wish. You may interpret it in the way you wish, but this is one of the possibilities, right? So, this is how I interpret these terms. There is also this term, which is probably most interesting. So, it also has this kind of sum. It appears, if you sum up over all possible indexes i and m, this sum appears to be the product of two power series. The first one is u, which contains partial derivative of f. And there will be another series, which contains these linear terms in t. So, let me write here t. So, this stands for cyan star. But if you take the product t times u, it happens that this product contains much more terms. This sum, which is used here, is only some part of this product. And there will be also the terms containing positive powers of x. So, we have some limitation on the summation indexes, which implies that when you take the product, you see that this t has terms of increasing order and positive also degree in x. And this psi has negative powers of x. When you multiply, you obtain both terms of positive and negative degrees. So, here the terms, which we appear in this summation, correspond to only those part, which has negative degree in x. Therefore, the left-hand side is not zero. But there is some contribution of this term in the positive, in the regular part of x. So, the conclusion that if you apply this expression, all lower-hand part of the expression will cancel and also the only part, which remains, is some regular function in x. So, this is how it works. So, I don't compute the left-hand side, the whole left-hand side, but actually I drop the positive part next. So, you may, for instance, look at the next term, that it's really nonzero. So, you see, the positive part does exist, but if you just drop it, then it becomes much nicer. So, this is the equation on U. And there are some several equivalent forms. So, for instance, if I divide by 2x, here... So, there are just formal manipulations of this. I can divide by 2x. And here... Or I can extract the complete square. And I'll get the equations of this type. So, these are two possible ways to represent the same equation. And let me write these resulting formulas here. h bar times delta U plus contribution 1 or 8x U plus t minus... And this is what I denote by eta. This is of x. So, eta, in this particular case, eta is equal to x. And another form of the same equation is this one, that if we take U minus 1 over 2 times eta, h bar delta U plus U plus t squared, and so on. That's all. It's equal to the... Well, up to the term, which corresponds to the regular part. So, these are... This is an equivalent form of the virocyl equations. Now, if you look just at this and take the homogeneous part in G and N, in G, so we extract the coefficient of a particular power of h, and also homogeneous term for the fixed degree in N. So, these are the equations that we obtain finally. So, this is just our... Our recursion relation could be combined in a single equation. So, this is what I called master equation. So, these should wait a little bit. It should work. Still thinking. So, the recursion relation holds for all G and N, except for some initial values, which for G equal to 1 and equal to 1, and G equal to 0 and equal to 3, which are initial terms of our recursion, and should be given explicitly in a different way. I will explain, discuss it later. So, anyway, what do I claim that for our particular case, there are very strong constraints, and just by a formal manipulation, we could represent them in a different form, either using this master equation or this recursive procedure, which is a refumulation of the topological recursion. These are just the same relations. So, the topological recursion, from my point of view, is just a reformulation of erasor constraints. Okay, so this is... Now, this is the equation that I hope you are able to apply, because just compute the right-hand side, ignore the regular part, take the lower part, and this will represent the next term of this recursion. Simple to implement. The next example that I would like to present you is the following genetic function, which is modified from... I modify the written concentration potential, but by adding so-called kappa classes. So, kappa classes are another interesting classes in the model of spaces of curves, which also appear quite often in some applications. So, consider mgn plus 1 bar and consider the forgetful map, which, given by just forgetting, raising, marking from one of the marking points. In the yesterday's Dimas talk, this could be identified also with the universal curve. So, here we have n-mark points, and we have here n-classes, psi1, psin, associated to these marked points, which are first-turn classes of the corresponding associated cotangent line at the isemark point. And here we have also these classes, which are... and there are m plus 1 of this type. Those classes, whose indexes from 1 to n are related a little bit in a certain way to these classes, which are here. Namely, here we have psii in this space, just more or less the same as pullback of the classes psii on the bottom. Up to slide correction, which is divisor. Divisor is here. Divisor in this modular space, which consists of curves, which are singular and has two points, number i and n plus 1, bubbled in a separate rational component, and there are some other marked points inside. So, these classes could be related to this one, but the last one is just completely new. There is no relation between these classes and the classes on the bottom. And therefore, using this class, one can produce new homologic classes on the bottom, which is called Marito-Mampfer classes. Kappa sub k is push-forward of the powers of this class. This push-forward decreases grading by one, so we take k plus first power of psi in order to obtain kappa k. So, we consider all possible monomials in kappa classes multiplied by a suitable and organize them in generating series, which now will be series in both T and S. I use this S as formal parameters, but one can also fix them in any arbitrary way. So, this is generating series, whose coefficients involve all possible intersection numbers for kappa classes and psi classes. But now... Okay, the same... Yeah, kappa k belongs to homolog of real degree to k. And now, let me formulate the statement, that actually this function could be reduced to... ...potential by simple substitution. So, consider there are certain polynomials in S variables, which are written here. And we just make a shift in the written concentration potential. So, the variables T0, T1 are not shifted, but then we shift T2 by S1, shift T3 by this polynomial and so on. And this will be this is how all these complicated power series could be combined. This result implicitly, this is known for many years, but seems that it's never being formulated explicitly. We discussed this with... There are certain special cases. For instance, there is a paper Stograph-Magnon, where this statement is asserted for T equal to 0. Right? But this is just slight modification of the same arguments. Anyway, so the conclusion that this is just the same written concentration potential with a shifted argument. So, in particular... Excuse me, it was mentioned that it's related to one Peterson. So, there is the class which is couple one. So, it happens that this class couple one has a nice interpretation in terms of Riemann geometry of surfaces. This is just the same as the class of the Well Peterson Simplactic form. Up to a factor, which is 2 times p squared. I don't remember. Inverse. So, the computation of Well Peterson volumes could be reformulated in terms of algebraic geometry, adjusted as a computation of intersection indices involving just couple one. So, for instance, the generating series that Dima mentioned in the last talk is just a specialization of this for when we ignore or couple classes other than couple one. And in particular, again, it's the generating series for Well Peterson volumes is obtained from the written concentration potential by a shifter of arguments. Only we put here you use only S1 and S2, S3 are set to zero. So, it's very nice, very simple interpretation in terms of intersection series of modulus spaces. Okay. And as a conclusion, we obtained that these generating functions obeys where are sorrow constraints. The shift of the argument does not change the very sorrow equations. The only thing that here we have the partial derivatives are the same. The term t appears only in this coefficient. It was just t for the written conservation. Here we have this shift by the polynomial q. So, it's a very simple equation. And again, using the same procedure we have all these where are sorrow equations to write them in this form. And it happens that they are just the same. Everything is the same. The same h, the same basis in V and L. The only thing that is changing is just the choice of this eta, which I called the shifted origin. So, we still use h and the rest remains the same. So, it just you apply just the same relations. The only thing, comparing with the written conservation case, you just modify this eta. One more example is what is called another modification instead of couple classes I will use lambda classes. So, I will consider generating function, which contains all possible integrals involving all possible combinations of lambda classes. But let me represent all polynomials in lambda classes in the following way. First of all, instead of lambda classes, I will use компонент of the churn character for the hodge bundle. Lambda classes are churn classes of the hodge bundle and instead of churn classes, I will pass to the churn characters. Besides, there is a result of Mumford, which claims that the even components of the churn characters of the churn character are zero. Only odd components remain. So, we have only durators of Degree 135 and so on. And besides, I will use a slight rescaling of the components of the churn character by this factor. Just some correction. In English, it's character. So, I rescaled the components of the churn character by this scaling factor, which will be convenient for the references for the Mumford formula. So, given any polynomial in lambda, we can rewrite this as a polynomial in mu classes. So, mu is just another distinguished collection of of homology classes on the modular spaces. And I take this generating function, which contains all possible combinations, all possible monomials in these mu classes. So, every intersection number involving lambda classes can be found in this function. By some reason, the obtained generating function is called a potential of n equal to 1 homological field series. So, here n equal to 1 appears. I will consider also the case of n greater than 1, but for the n equal to 1? Yeah, it's kind of in physics n equal to 1, it's supersymmetry. No, no, it's another n. It's another dimension. Dimension of the of the... It's also supposed to be best, possibly. Thank you. I'm happy that you're following. Carefully. Okay, so in particular, if you put instead of these parameters just with these coefficients, then this term becomes just total hodge class. So, the hodge potential for single lambda classes, which appears in the ESV formula and also in the generating function for Hurwitz numbers is a specialization of these for this particular choice of the parameters r. So, now the left-hand side also depends on the r parameters and the t parameters. So, I consider t as the parameters of my manipulation of which participate in the recursion and r just supplementary parameters, which are not involved in the recursion. So, you can fix any parameter r and we have the same recursion. So, this is you may treat it as a family of functions for each parameter where you are, you have the same recursion. But where is t on the right-hand side? This is probably not the best way to write this. This is a Taylor coefficient of this series. So, here n is the degree of this monomial. So, the left-hand side is the number. It's zero. Actually, it's zero. If you wouldn't put n here, you should put here zero. But this is homogeneous of degree n. So, this on the left-hand side you have a constant. Well, you may constant. Well, if you wish if you wish, I can write it this way. If you sum up all these terms with different n, are you happy now? Just the same. Okay. So, this is my implementation. Mu is homogeneous term of degree m of this expression rescaled by these Bernoulli numbers. Be it the Bernoulli number, of course. And here to simplify computations I specify in some random way the constant r. Not the same as the whole integrals, and there is a I mentioned already the Mumford's formula for the lambda classes and this could be formulated in this way that the expansion with respect to parameter r could be described by this partial differential equation. In fact the original Mumford formula was formulated not in terms of intersection numbers but in terms of the classes themselves and they could be written as follows. Take this Mu class in the homology Mgn and this could be written in the following way problem minus Mu to be correct. First of all we have the class which is kappa sub m which is push forward of psi n plus 1 to the power m plus 1 I recall you where p is a forgetful map. Then minus what we have sum of psi classes and plus the class which is will be of this form psi n plus 1 to the power i psi n plus 2 to the power j. So this is this class is supported on the boundary divisor of the modulus space we have our curve split or our curve has a double point it could be either the double point could either could either devise the curve into two parts or it could be the curve might be still connected but anyway this divisor could be regarded as its own modulus space which meets two extra marked points which I numerate by indices n plus 1 and n plus 2 so it has many components and any of the components could be regarded as a modulus space with two extra marked points and so I take this class supported on this boundary divisor so it's just a matter of combinatorics to reformulate this Mumford formula in terms of intersection numbers of classes and it will be will get this form again some small sprint in the middle form it's spawned upon OH bar f lambda is equal to spawn minus operator this is the same guy so maybe it's a 6 it's evaluated when r equal to 0 so here is the function itself and this is the initial function which is of course Witton Konsevich Maxim, and what is J star over there that push forward from where to where so let me denote this divisor somehow sigma and J star is just divisor this is OGM plus 2 No, embedding of some divisor embedding of the divisor, here we have two more marked points coming from the gluing and here we have just 10 points we have N plus 2 points here and N points here Could you please write on the blackboard what is F lambda in comparison with F Witton Konsevich are you just put this lambda classes for any point where is it here I put in front some extra class in front of monomials of psi classes I put some extra class which depend on lambda so r equals 0 gives r equal to 0 it's also could be treated as a deformation of Witton Konsevich which these parameters are so you are right for the initial parameter value it is for when r equals to 0 we have Witton Konsevich but when r is changing we solve this linear equation as usual the solution of this linear equation with constant coefficient is just exponential of this operator so for r equals to 0 we have Witton Konsevich I write here the one for the equation in terms of not exponent but in terms of the function itself as usually the non-linear terms should be added in this way in this case and these are the equations so whenever dependence of r on the parameters I write in this way and you get 0 so when I start to move in the space of r parameters I know how to change our potential so I can go along any curve in the space of r parameters and I will restore our potential so now this statement is that this function also admits topological recursion of the same sort it is formulated in a slightly strange way using so called Laplace transform the word Laplace transform appeared already in the talk of Dima and also I use the same term but let me say what Laplace transform is from my point of view Laplace transform is just a linear isomorphism of the space of Laurent series in the variable X and in the variable Z this is a coordinate isomorphism if I have a monomial Z to the power n here then I just take the same exponential X to the power n here but rescale it with a certain factor to the power n divided by 2n-1 factorial double factorial so just a linear isomorphism of two vector spaces so given a series you can represent it on the left hand side or on the right hand side just by rescaling the Taylor coefficients the same formula by the way could be applied both for positive and negative n because the negative double factorial odd number is well defined and it's just the two formulas are equivalent both both formulas could be applied for positive and negative n I use multiplication when I use topological recursion I use multiplication in this space and this isomorphism does not preserve the multiplication so it's not multiplicative isomorphism Let's speak about multiplicative structure in our space H I will use this presentation of this space but on the right hand side I have a nice operator of multiplication by z here but if you take just multiplication by z it becomes looks very simple here but here you have some strange scaling factors because on the left hand side this turns to be differential operator so you can consider this Laplace transfer as follows you can ignore the right hand side you have this space of Laurent polynomials but z inverse is just treated as a notation for this differential operator so for me I the new data which I introduced for this space of Laurent series is this extra differential operator first order differential operator now the prescripts is as follows the basis that I use is very simple if you use the right hand side of Laplace transform I start with a series r which given by this exponential so r small r are the same is the parameter of my family so I obtain this series r of z and this series which corresponds to the basic vector number 0 the space which I denote by v and on the right hand side this is just a series on z so on the left hand side you should rescale the coefficient to obtain the corresponding basic vector in the x coordinates and then in order to obtain basis you just act by multiply by a suitable monomial in z on the right hand side and respectively I should apply this operator on the left hand side so this basis looks very nice so I obtain the basis vectors on the right hand side which looks very nice on the right hand side of Laplace transform but on the left hand side you should use these strange factorials factorial coefficients and the duality is given by residue as usual I will speak what do I mean by duality later if you don't mind and so I claim that you should apply to these bases so we have these bases x i n which is 1 r of z which correspond to certain series on the left hand side and also we have minus 1 to the power n plus y z to the power z to the power n plus 1 r of z on the right hand side which corresponds to these are regular in X and these are Laurent's so the v v is the linear span of this x i n star which is easy to see just the same as before this space of regular functions but L is different в случае витон-концевич или вилл-петерсон это было только координат саплементный subspace но сейчас это не координат это просто linear span of these vectors which is slightly turned subspace in the space of Laurent's series and also an eta is equal to according to my notation something like x i 1 star which is 2 times x plus high order terms it's also some deformation of the function tx in the viton-концевич and you see it works just apply the same relations this relation these relations they do work so this implies recursion again as usually if you take extract terms of degree g and n in h and t then you obtain this recursive procedure to recover homogeneous component of this potential and in particular if you for the choice of parameter values using these Bernoulli numbers you have specialization of these for Hodge integrals involving single lambda classes or Curved numbers and this is what Dima mentioned in his talk it's so called Buchar-Marini conjecture the Buchar-Marini conjecture claims that certain recursion for Hurwitz numbers or the same for Hodge integrals and if you remember it was related to the it was related to certain spectral curve the spectral curve of this recursion is so called Lambert curve slightly my notation is different from those used in talk of Dima by slightly shift of variables so I use the this is a Lambert curve I use coordinate system such that the critical point of the projection to the x-line is the origin so if you expand the logarithm is what this is y squared over 2 minus y cube over 3 plus y cube fourth over 4 minus and so on so it's something like this and there is a involution that exchanges the pre-images of the projection to the x-line the projection to the x-line is double quantified so there is involution which exchanges these two points and let me denote this involution somehow by by bar this is slightly misleading notation because it's nothing to do with a complex conjugation but however used everywhere in papers devoted to this recursion and also let me introduce an odd coordinate which is given by this equation so in this notation s bar is equal to minus s so this is s is another local coordinate on our curve and this is how the function y looks from the s coordinate x becomes just a square over 2 and if you look carefully at the recursion the form of in R and 10 then it happens that it depends only on the odd components of this function y so you can freely modify the even components they are not participating in the recursion the whole recursion so you can a huge variety of possibilities to change the spectral curve which produces just the same recursion if you modify anyhow the even components the recursion which will remain the same so I treat this slumber curve just as a job or a curve as origin if you write it in this way then it happens then y extends as a global coordinate a fine coordinate on this line but the recursion uses only the germ of the curve as the origin so if you just are interested in computing the coefficients you need only odd part of this and then let's compute the derivative of this odd part over s and it happens that this series is related to the series that I used in my formulation by just Laplace transform so here are two functions so the first one is given by this Bernoulli coefficients and the second one is given by this derivative of the odd part of the y coordinate so I do not formulate the recursion in terms of the spectral curve this will be done probably in the next talk but just to make the correspondence the correspondence uses this strange peculiar identity in this lemma I use the reference b and m if you think that this is Buchar-Marinio that it is not correct the correct reference this one I observed this lemma in a strange paper of these guys from mathematical institute in Venezuela this lemma in this exactly form was proved in this paper this paper is nothing to do with Hurwitz numbers with topological recursion and it happened that these coefficients which appear here appear in a slightly different problem not related to the subject of our talk it deals with asymptotic expansion of gamma function so if you take n factorial then you know the sterling formula n or e to the power n and there are correction terms of the asymptotic expansion which are exactly those and there is a square root of 2 pi n well n here sorry sorry but I this is you know better we should get odd power odd power sterling formula ah sorry it expands to some four powers yeah if not not exactly this but very close to this anyway this seriously appears as the asymptotic correction terms of the asymptotic expansion in the sterling formula by the way put n to the power n to the left hand side you will observe this formula which is close to what was used in the talk of Tima so it's really related but not directly ok so these you have in your case you have the same coefficient as here yes I did the following this correction for this numbers I obtained this series and then I put these coefficients numerators and also numerators in the how it's called the table the library of sequence and sequence and they give me this reference but maybe it is related because of course it's related yeah I think that I finish with my examples and now I'm ready to give more general form of the recursion which covers more or less all the cases treated by another antenna which covers more or less all the cases treated by another antenna and I still have time probably I will start these definitions then probably which one and yeah I will keep this because I just copied this because that remains completely mysterious yeah it looks like so I just think that if you just one of the ways to get use of this form was just to repeat it many many times use this principle would you explain how one is supposed to guess just to speak to the god or how one is supposed to write according to according to the general principles the new ideas come not because people start to believe in this but because comes a new generation which do not pose this question why it works so it I don't know so that's why you're keeping it on the blackboard yes yes get used to it and then yes yes yes by the way so before I go further let me play a little bit with this formula especially it looks very nice in this form there is a strong temptation to denote this u tilde is equal to u plus t minus eta and try this formula in the following form h is equal to u plus u squared is equal to o of x even more compact form so the first remark that this function u tilde combines terms of different nature so this is a point of L this is a point of of complementary space v so what do you use in the what do you you are going to compute the terms of this expansion but these are different nature they belong to different space so I don't like to put them together but if you wish you can also consideration and this is what does correspond in the presentation of DIMA to W0n and you should recognize where is it so so if you compare with the talk of DIMA you should recognize that this is a hidden form of the Bergman Kernel and this is DIMA wanted to add into the consideration to add this to the right hand side to make it in more symmetric form so in my case the formula as it appears in the paper of LNR and around 10 is too huge it compiles many lines so it's it's it's painful to add terms so you can combine these two terms formally if you denote this T formally as you could define this is formally 0 2 and then this term will be included here but I still prefer not to do this for me there are no such terms well another thing is about this summand 1 over 8x let me I did erase let me take the simplest case of within concavage and let me try to compute what is delta of T so when I wish to interpret this is a delta of U I should compute delta of T so by definition this sum from n0 to infinity what do we have xi n which is 2n plus 1 double factorial n plus 1 times the derivative of T over Tn but the derivative of T over Tn derivative of T over Tn is xi star times 2x to the power n over 2n minus 1 double factorial is equal to sum from n0 to infinity what do we sum up 2n plus 1 over 2x right? so we get a divergence and therefore we regularize this by definition this is 1 over 8x so this is just convention so if we add this convention then you can write the formula in this more compact form this convention comes exactly because it should be like this if you wish that it works for the written concage potential but I do not want to speculate this divergent series so I prefer to keep the formulas in this form okay I will erase this okay so now we played a little bit with examples and I think that you are ready to to accept the general definition so now I will give the general definition of topological recursion in the form that I prefer I start with first of all I start with some h with a finite dimensional Frobenius Algebra it means that it is associative, commutative and also with unit and also equipped with a scalar product which is invariant for multiplication which is given by linear function such that is a non degenerate quadratic form by linear form so n stands for the dimension of this vector space and in all my previous examples h was just c one dimensional and now I will consider a more general case the typical example is so called semi-symbol Frobenius Algebra which has admit the basis of idempotence so we have a basis which I denote E1 star E sub n star such that multiplication it looks like EI times squared is equal to EI and for different indices they star star so this basis is automatically orthogonal but it's not necessarily orthonormal and so let me use a dual basis E1 En the dual basis is proportional to this one and EI I don't remember well probably 1 over delta IEI I don't remember the conventions that I prefer so let me let me think where I wish to put the star probably like this so that the unit sum of EI star and also when I use the scalar product you should take this factor and also I will need the Kazimir element which is independent of the choice of the basis which is in this particular case is the sum of delta IEI star it's idempotent so this is a basis so whatever this element it should be expanded in the basis and these are coefficients of this expansion and also all this delta IEI should be nonzero this is one of the typical examples and the case which corresponds to the recursion of another antenna how it appears in in most papers it corresponds exactly to this case and moreover for the choice of these factors all are equal to 1 but it happens that applications usually so for instance the length of the unit vector is equal to sum of delta I's so in application it happens that the norm of the unit vector is equal to 0 so it's the convention when all the delta I is equal to 1 is not physically meaningful another typical example is even of some complex manifold this is the way how it appears in quantum so this is a Frobenius Algebra that I fixed once and for all I will consider the deformations of initial data of the topological recursion and the Frobenius Algebra itself will never be deformed I fixed once and for all the next step is to consider the Algebra that I used in my examples H this is the space of Laurent series in one extra variable with coefficients in H how to put probably it's better to put H in front simply like this space of Laurent series in X valued in H and you can multiply multiply elements using multiplication in H this becomes Frobenius this becomes an Algebra and this space is symplectic there is a plural or symplectic form in this space given by the following formula so if you take AX to the power K BX to the power L it's nonzero only if K plus L is equal to minus 1 plus 1 is equal to 0 and this case is equal to A the pairing of A and B times the coefficient some extra coefficient which is I'm not sure about this factor 2 2 K plus 1 so because of this property it's skew symmetric if you replace K and L the negative sign will appear let me write this in more variant way if you have to learn serious valued in H then you do the following residue can you explain where it's skew symmetric? if K plus L plus 1 is equal to 0 then to K plus 1 and to L plus 1 is different by sign so if make a substitution my favorite substitution instead of X I put a squared over 2 dS G and then I take the primitive of this the primitive is defined up to a constant but the constant does not contribute to the residue so this is the way how it is defined and this is obviously skew symmetric then the next object in this space by the way if you use this this if you use this symplectic form that one can produce many convenient Darbou basis basis and one of the possible Darbou basis is the following so I use the elements with two indexes and new K which is E sub new star times 2 X to the power K over 2 K minus 1 factorial and the dual basis XI new K this is without star this is I new times 2 K plus 1 double factorial 2 X to the power K plus 1 so if you use this basis it's Darbou so the pairing between so here new runs from one to dimension of the space and K from 0 to infinity so if you take the pairing between vectors with the same indices you get one up to a sign and zero in all other cases but you could also use a Z basis yes there is another presentation of this space using Laplace transformer I will discuss it later K so in particular these vectors spend the regular part of this which I denote by V so V subspace V subspace V is just regular part of this H so at the moment I have no data to use for the recursion because all this can be can be used for any V is that I have V is that I have V is that I have it's the regular in X regular in X it's not the one that you underlight oh yes you're right sorry sorry sorry no no no no no no okay no no no this it's important so probably I will I wish to emphasize so this is this particular case this is just complementary coordinate subspace consisting of all around polynomials or polynomials in one over X but now the the data of the recursion that I do need is L and by definition L is a Lagrangian subspace in H complementary to V so this coordinate subspace spent by this vector is just one of the examples but of course there are much many more different Lagrangian subspaces and to be precise I also need a basis basis X alpha is a basis in L and also X alpha star is dual basis in V so the symplectic structure makes these two subspaces dual to one another and so I fix some basis here dual basis here and they together form Darbou basis in the ambient space H actually the explicit form of this basis is not so important but for the matter of computations to make some explicit computations I need also to fix this basis by the way I can change one of the basis and then this produces the change of the other basis for instance I can use this coordinate system such that this is a standard basis the actual basis could be different but if X alpha star X alpha is the same as there U, K а, не-не-не, подожди как сказать значит все subspaces are parameterized by quadratic forms так как в следующем способе так что consider L space V and the complementary space L and also I will use the coordinate space L0 L0 is the coordinate space which was used before which consist of polynomials in X inverse right? so here we use which have chosen the basis XI alpha here we have a dual basis which I denote by XI alpha and also there is a basis here but if you use L0 instead of L I denoted this way then the distinction, the difference between these two basis due to duality should be vertical plus certain vertical vector which is parallel to this one so it belongs to V so this is element of V and what is up there should be XI new K0 no no no XI0 with some basis of L0 no so XI0 XI with upper 0 it is a single symbol and alpha runs over all the basis projection to XI alpha to L0 the basis in V is fixed no no it works also for XI0 you can use different basis but let's for a moment fix the basis here and these are both dual to this one but with respect to different Lagrangian subspaces then actually L is Lagrangian is just a reformulation of the fact that this B is symmetric and again as before I use T is a sum of T alpha XI alpha star that generic element of V and also I will need some eta some particular fixed point of this family this is some specialization of this one and it should be of the following form some invertible element times X plus high order terms so V is always subalgebra V is always subalgebra but your basis XI alpha is the same as this is one of the possible choices but you don't assume that it's the same you don't assume that it's the same so for instance for the Hodge potential it was different for Witten and for Vell Patterson it was like this but for the Hodge cases was different so this is the initial date of recursion now I can formulate the main recursion and actually it's actually written here this is the main equation of the recursion so what we do we use here multiplication in order to apply this multiplication we use ring structure in H this delta uses also multiplication we differentiate the coefficients with respect to parameters and then multiply by this XI alpha using multiplication H and again this equation could be written in compact form like like master equation the only thing that I promised to tell you was the initial date of this of this recursion and the initial date of this recursion require to put here special choice of here and the initial data if you remember was we need to fix U11 and U 0 3 so the actual choice of this is the following one I hope that I will write correctly here I should put C over 8 X C is a Kazimir element and also plus some contribution which actually affect only this U11 and this is written in the following form XI well some XI alpha beta XI alpha star XI beta star XI alpha both regular functions therefore when you multiply you still obtain a regular function if you increase the indices alpha beta then the order of this function is getting greater and greater therefore you have only finally many terms which contribute which contribute to this U11 I will speak more about the reason for this choice of the initial data but let me now formulate the theorem so first of all you may just apply directly this recursion this choice also contains the definition of U11 and U03 then you apply this recursion you find UGN for all GNN and therefore this UGN is uniquely defined by the recursion then the second statement is the following so this UGN is an element of the space V the space L of L and the same they claim that it is polynomial means that there are only finally many nonzero terms here only finally this sum is finite actually only finally many nonzero terms and each individual coefficient of this expansion is a polynomial in T so actually on the right hand side for each GN and alpha we have only finally many terms the next assertion so that this U satisfies the condition of mixed equality of mixed derivatives and so it could be integrated as a differential of a function this property depends heavily on the fact that our subspace is Lagrangian and moreover this condition is also used very much in the proof of this property if you just very slightly the initial condition that this property will fail strangely this is very subtle it looks very natural but actually the proof is somewhat delicate but this is true so actually the recursion could be applied for any L and for any initial condition because they are so the first and the second line does not use the Lagrangian property of L but the third property in the third property this is crucial and so this implies that you can recover FGN such that UGN is differential for instance you can use the earlier formula but this unfortunately holds only if N is positive because having differential of a function we can recover the function itself up to a constant so in the example that I considered before for instance for the Konsevich written potential this constant was always 0 but it happens that we cannot assume that it is always 0 therefore there should be some way to recover the constant term corresponding to N equal to 0 so it happens that this potential has many other nice properties and one of these is that it obeys so called satisfied by so called dilatone equation which says that if you differentiate it in the direction of the vector eta let me check the humidity condition N plus 1 probably because we differentiate once so here this is 2G minus 2 plus N times FGN yeah for holes we already defined F for N equal to greater than 0 so it always holds for those values of N where F is defined and hence since this equality holds we can use this equality to define FG0 is the same way so now we defined this all the components by yes yes yes 2G minus 2 yeah right so for G equal 0 it's still defined and we said by definition so in this by definition we said FGN is equal to 0 for FGN 2G minus 2 plus N non-positive so F 0 0 F01 F02 and also F10 all these are equal to 0 there are many nice properties for this potential let me tomorrow I will discuss how it varies under variation of the Lagrangian plane so what's there is a given tiles quantization of simplectic quantization of quadratic Hamiltonians so tomorrow I will discuss its dependence on the choice of Lagrangian space but now let me mention several obvious properties the first one is the following that how it to what extent the choice of eta is important so the first property the different of eta leads to a shift of the t variables so if you look at this equation the t enters t minus eta and t minus eta is the sum of t alpha minus eta alpha times t alpha star so it's a different choice of eta leads to just shift of the t variables and we observed this this fact when we used the well Patterson generating function shift of the t variables is equivalent to is equivalent to a different choice of eta and the second another property is a what happens if we choose a different basis a different choice of the basis of the basis in v or in l leads corresponding linear change in t again it has more or less invariant meaning and the actual choice of the basis is not involved so the conclusion just follows so the conclusion so we can treat f is a formal function on v and the formal function is well defined independently of the choice of the basis and but if you are interested in particular Taylor coefficients you need to fix a coordinate system so the particular choice of the basis leads to a particular coordinate system on v need to fix basis but if you just consider the abstract formal function it's well defined on v and even is not does not depend on any choice of any basis but moreover we take the expansion of this function not at the origin expanded certain point probably minus eta so the function is well defined but like this so this is a space v and we do not know how to expand this function in the origin but we know how to expand it minus eta a different choice of eta leads to a shift of arguments and this condition on eta where is this condition on eta means that if you change these terms the corresponding variation of the function is well defined the shift of the argument in the direction of these dots is a well defined operation the function everything is polynomial in this direction so actually it means that this potential has an invariant meaning and essentially it's determined by Lagrangian plane so the essential part of this definition of the potential is Lagrangian plane and all the other things are just coefficients which are used for the particular choice of coordinate system so f depends on l of the choice so invariantly and meaning of f actually is uniquely determined by l ok and then the other properties its variation with respect to l and also relation to Gromov-Witton series this will be tomorrow ok now I filled all 2 hours thank you very much we have some applications of gender gender I think that it's better to postpone your question to the next talk the next talk will be mostly devoted to huge huge amount of different applications of this for bigger end, right? No, no, no My next talk is mine I can talk for 1 hour simple example is Gromov-Witton Gromov-Witton Gromov-Witton so you speak now, right? No, I speak after the run ok, ok, sorry I have short questions concerning the setting you said that the initial date is U11 and U03 I do see U11 I do not see U03 let me write here probably U11 is equal to where is the formula ah, here it is here pL of 1 over 2 times eta of this c over 8x plus sum b alpha beta ksi alpha star ksi beta star and U03 is equal to pL 1 over 2 times eta of t squared and that explains the fact that it depends on the Lagrangian explicitly the quadratic form beta depends on L and also projection also depends on Lagrangian are the questions what is the precise this is the subject of tomorrow talk