 So I will start by writing the classical LSV formula. The LSV stands for Ekedal, Andor, Shapiro, and Weinstein. Shapiro and Weinstein. So it's an equality between Herbert's number that I will explain in a moment on the left-hand side and on the right-hand side you have a combinatorial coefficient and the main part is an integral on the modularized space of stable curves. Here you have the total term class of the dual of the Hodge bundle, again I will explain in a moment. And here you have some expression with side classes. So I will try to briefly recall the meaning of all this. So it's not to take the whole hour. So I made a little picture of the modularized space. Here it is. So this thing is the modularized space here of GsG stable curves with n-mark points. As you see points of this modularized space correspond to stable curves. If you take the point at random like here it's going to be a smooth three-man surface. Then there is a divisor with normal crossings and points of this divisor correspond to curves with self-intersections. And then when you have a point on the intersection of this divisor you have two nodes, one here and one here. Here you have one node, here you have two nodes. So that's the component, this is called the boundary. That's the component of the boundary that corresponds to non-separating nodes. So if you normalize this curve it stays connected. Here you have a component where if you normalize the curve it becomes disconnected. There are several components like that. And now on every curve you have n-marked points and you can take the co-tangent line to the first marked point for example. You see this is the co-tangent line to the first marked point on each fiber of the universal curve. So the union of all these curves is called the universal curve. Looks like dungeon. Yes, I cannot draw the co-tangent line. And again as you see now I have a complex line over each point of the moduli space. All together they form a line bundle that is called L1 because I took the first marked point. And similarly you can define L2 and so on up to Ln. So you have n-line bundles and their first term classes are called the Psi classes. Psi i is the first term class of Li and it's in the tucco homology of Mgn bar. And then you also have the hodge bundle. So that's the hodge bundle. And that's the space so that the fibers of this hodge bundle are spaces of a million differentials. So if you take let me call this map Pi for example and then e is the push forward of the relative co-tangent line bundle. So if you take the co-tangent line bundle to the fibers the sections on each fiber are a billion differentials on this fiber. And over each point here you have a dimension G space of a billion differentials. So this is a rank G vector bundle. And in this formula I have its dual. So this is well known since 2000. Now I'm going to write another very similar formula that is still not proved. So it's a conjecture. Before that I should explain the previous numbers. There is also the left hand side and that will allow me to make a transition between the two cases. So hgk1kn is the number of ramified coverings of Cp1 of a sphere. It's actually a complex structure. It's actually not important because it's a topological invariant. So this will be Cp1, this will be infinity. And here I will fix m more simple branch points. And so the conditions for the ramified coverings that I want to enumerate have m per images of infinity of orders K1, Km. So the degree of the covering is equal to capital K which is the sum of the Kis. And then I have as many simple ramification points as needed. I don't know how to draw that, something like that. All these are simple ramification points and their images are all fixed here in the target. And you can, believe me, or you can compute the characteristic and check that this number m should be equal to 2g-2 plus n plus K. If you want the surface to be of g as g, that's the number of simple branch points that you have to fix. So now how do you compute this if you want to actually compute the surface numbers? Let me first of all introduce a generating series. h of beta and then p1, p2 and so on is going to be the sum over n and g and sum over K1, Km. Beta to the m divided by m factorial, hg, K1, Km, pK1 and so on pKm divided by m factorial. So maybe I should make one small comment that this number m here or the genus are expressible right there. They give the same information. So I prefer to write the genus here, but the number m here is expressed from the genus. And I could also write m and express g from m. That would be equivalent. Okay, so now I'm going to write an expression for this function in terms of the representations of the symmetric group. Okay, so first of all let me introduce an notation p of sigma. If sigma is a permutation, then p of sigma is going to be the product over the cycles of sigma, p sub length of the cycle. Let's say c length of c. Okay, so if sigma for example is equal to 1, 2, 3, 4, 5, 6, 7, 8, p of sigma is going to be p3 times p1 times p2 squared. Okay, so now let's denote by T2 the sum of transpositions in S capital K. So capital K is the degree of discovering or the sum of the KIs. Okay, and then I claim that e to the power h, the exponential of this generating series is 1 over K factorial e to the power p of e to the power T2. So beta times T2. Right, this is the correct formula. So where does this come from? What does it say? The T2 is the sum of all possible transpositions. Here I raise T2 to some power m, right? This thing here is the sum of beta to the m, T2 to the m, over m factorial. So I raise the product of all transpositions to the power m. That means that I consider all possible products of m transpositions. And the product of m transpositions is the monodrome, is the product of monodrome is here of the coloring. If I take the monodrome is around these simple branch points, I get transpositions. If I go around all of them one by one, I get a product of m transpositions. So altogether I'm going to get the transposition of the big cycle like that. I've noticed that it's not really good. P, you see this T2, it depends on k, it's a group of symmetrical k. Yes. Yeah, but for every k... No, no, no, because in one of the k-factorials, since it doesn't depend on k, it should be good k as some kind of parenthesis. Here in T2. Yes, but it's better, it's actually better. Yeah, because it depends on k, it makes no sense. P of sigma is well defined for any size of sigma, actually. And T2, you're right. So T2 should actually be... You're right, some of group rings of all symmetrical k. Yes, but T2 actually depends on k and I could write a k here. Because it's maybe better not to actually. It is what is called a stable conjugate class. A transposition exists, the sum of transpositions exists for any k. You can take it for any k. No, no, no, it's possible to understand the formulas. So one sum of k, one sum of k-factorials, I think is number E. And then multiply by sums, which doesn't depend on... No, it's a kind of grammatically formal, that doesn't make sense. Okay, if you prefer k here, I can put it, but it's... Yeah, it doesn't make sense because... Right, so once again, I take the product of m transpositions in all possible ways for all choices of m transpositions. And then I look at the length of the cycles, the cycle structure of the permutation that I obtain as the product. And my goal is to obtain the permutation with cycles of length k1, k2, kn and so on. So if I obtain a permutation like that, this p here will give me pk1 times pk2 times pkn. Right? So in front of pk1 times pk2 times pkn, I will have exactly the number of ways of representing a permutation with these lengths of cycles as a product of m transpositions. And this is exactly the way to enumerate when you fight coverings. And just one thing that is... Well, I had to take the exponential of the generating function because when I take this product of transpositions, I don't know if the ramified covering is going to be connected or not. So if I do it like that, I actually obtain the generating series for possibly disconnected ramified coverings. And this is e to the power h. Is this the ground side formula? This is the what? Ground side formula. This one, yeah. So actually, yeah, I can push this one step further. So this is also equal to sum over k e to the power beta t2 of lambda. So sum over partitions of k, representations of sk. Actually lambda here is a partition of k that represents a representation of sk, times the sure polynomial of e1 and so on, pk. So this is a way to compute the power of this element, right? Now I have this t2. It's the sum of all transpositions in sk. It's actually an element of the center of the group algebra of sk. It's a linear combination of transpositions. So it's in the group algebra of sk, and it's actually an element of the center because they take a symmetric thing, the sum of all transpositions. So now how do I weigh it to the power m? How do I compute it exponential? The answer is that the nice way to do it is to decompose it into idempotence, into the sum of idempotence. So in this center of the group algebra, it's a commutative semi-simple algebra, which means that it has idempotence elements. So for every representation lambda, there's an element in here that acts by 1 in representation lambda and by 0 in all other representations. And then if I raise it to powers, I get the same elements all over and over again. So in general, if I want to raise an element to a power, I should look at its action in every irreducible representation and raise it to the corresponding power. So t2 of lambda is the action of t2 in representation lambda. It's a central element, so in each irreducible representation, it will act just by a scalar. And t2 of lambda is the scalar. And if you know the representation theory of the symmetry group really well, you should know the formula for t2 of lambda. If you don't know it so well, it's a difficult exercise. So it's 1 half sum i greater than or equal to 1, lambda i minus i plus 1 half squared minus i plus 1 half squared, where lambda is equal to lambda 1 greater than or equal to lambda 2 greater than or equal to lambda 3 and so on. So when I have a partition i represented by a decreasing sequence or not increasing sequence of its elements, then I make this small shift instead of every lambda i, I write lambda i minus i plus 1 half, so that this sequence is now strictly decreasing. And this is more or less a renormalized sum of squares of this shifted coordinates. So you can notice that this sequence actually finishes with an infinite number of zeros, right? An infinite number of non-zero elements. After that they are all zeros. And when lambda i is equal to zero, I just have a difference of two similar terms, of two equal terms. So this is an infinite sum, but actually it has a finite number of non-zero terms. It is well defined. And morally if you forget about this part, it's just the sum of squares of these elements. And this is added for it to converge. So now I go in cycles and take it. Maybe I should continue here. So this is just a number? This is just a number. That's a function on all possible partitions. All possible partitions of all k. And because the sum of transpositions exists in every s k, so for any partition of any number, I can look by what scale the sum of transpositions acts in this irreducible representation. This is the number that I get. Okay, so now I want to write a more complicated LSD formula that will depend on a parameter, parameter r greater than or equal to one. So it's an integer. Right, so from now on I fix an integer r greater than or equal to one. And if r is equal to one, I will find the usually LSD formula. So the first thing I do, I will go in the opposite direction. I will start by changing the average numbers. So now first I have to define t r plus one of lambda. So the recipe is very simple. Everywhere you see two, you replace it by r plus one. Almost everywhere. There's this one half, this takes one half. So one over r plus one. Sum for i greater than or equal to one. Lambda i minus i plus one half to the power r plus one. Minus minus i plus one half to the power r plus one. Right, this is once again a function of all partitions of all integers. And the question is, does it represent anything? So it actually, well, it's not such a, it's actually a very good function. It represents what is called completed cycles. For example, I will write it just one. Equality t three of lambda, for example, can be decomposed in the following way. Three plus one one plus one twelfth times one. So that means that t three of lambda is the sum of three cycles. Sum of three cycles plus, so here I have the trivial permutation but with two distinguished elements. So actually n times, sorry, k times k minus one over, times the trivial permutation. And here I have one twelfth times, once again, the trivial permutation but with one distinguished element. So k over twelfth times the identity. So this is again an element in the center of the group algebra and this element acts in this way in every representation. And there's a similar formula for any r but it gets more and more complicated. There's an expression for the things that you should put here but it's not the most beautiful way to see them. So these things are rather mysterious and this is actually the reason why the formula is still not proved because you have to find these coefficients in a geometric way and I still cannot do that. But now, okay, now once you have these t's you can define the new generating series which is for r hermit's numbers h, g, r, k1, km. Let me start with, so e to the power h sub r of beta k1, kn p1, p2 and so on. Write exactly by the same formula. Sum over k, one over k factorial, and I'm sorry, now I have to write the second version, right? So sum over partitions of k e to the power beta tr plus one of lambda times the short polynomial. That gives me a new power series and if I take its coefficient, p1, p2 and so on, its coefficients are called r spin hermit's numbers k1, kn. Once again, so here I should write beta to the m divided by n factorial and here pk1, pkn, over n factorial. So now everywhere I have this parameter r so these are some new numbers. Like coverings, roughly, this should have been preferred. Yeah, so yeah, roughly, yeah, roughly, roughly. These are coverings, so yeah, coverings from the genus G-surface to Cp1 such that, such that df has an rth root, has an rth root. So that means that critical points, right, critical points are zeros of df. If df has an rth root, it means that critical points go by, by collections of r. Pardon me? That's the maximum or is it everywhere? Sorry? All critical points have this rth root. Every critical point has, yeah, so, well, it's, there are some boundary contributions, so let me draw a picture for, again, for t3. So I, right, I wrote this equality 3 plus 11 plus 112 times 1. So now I can draw a picture associated with every term. This thing is just a critical point that behaves like where, so where this map f locally behaves like z cubed, so that's a double critical point. But then I also have a contracted component of genus 0 that is attached to two sheets of the covering. And here I have a contracted component of genus 1 that is, that is attached to one sheet of the covering. So here I have a covering with two distinguished sheets and they're distinguished because there's actually a genus 0 component that, that is attached to them. And here I have a covering with one distinguished sheet because there's actually a genus 1 contracted component. So you see these are actually stable maps. Stable maps can have contracted components. And the image of these contracted components are counted as branch points with some multiplicities. There's a formula that gives you the multiplicity and the formula gives you two in all three cases. And these are exactly the three possible cases. It's called the branching morphism that extends the notion of multiplicity of a branch point to stable maps. Okay. And so now I can write the corresponding ILSV formula. Yeah, okay, let's, I guess I'll have to erase this but it's very similar. So Hgrk1kn is equal to m factorial and this time m is 2g-2 plus n plus k divided by r. So as I said, the critical points now go by collections of r. So the number of different branch points is divided by r compared to the previous to the ordinary case. Now there is a power of a here that I never remember by heart. m plus n plus 2g-2. m plus n plus 2g-2. Sorry, there is also the combinatorial factor ki over r to the power p i over p i factorial. So I'll explain the notation. And here I have what is called the space of r-spin structures. So once again I will I will explain what it is. Here is minus r star p push forward of l. Now there are some new things but so that's the formula. Now once again I have to explain all the letters. So first of all there are these numbers k1km that correspond to the branching at infinity. And I have to divide each of them by r and write the quotient and the remainder. That's going to be r p i plus r minus 1 minus a i. So that's the quotient and that's the remainder that for historical reasons has to be reversed. Instead of going from 0 to r minus 1 you go from r minus 1 to 0. So from each number ki you get the number p i and the number a i automatically. And this already explains the combinatorial factor here and the indices that are here but I still have to explain what the space is. And the space mg1 over r a1an is the space of so it's stable curves of genus g with n marked points and with a line bundle such that its arth power is isomorphic to is identified with the cotangent line bundle twisted by sum of minus a i x i. So this thing is called a north spin structure. Once again a north spin structure is a north tensor root of the cotangent line bundle. So here I have the cotangent line bundle I twist it by some divisor supported on the marked points and then I take the arth root and so the first term class of this let's say the first term class line bundle is 2g-2 minus sum of a i's and it should be divisible by r other ways it doesn't work. If you want to extract the arth tensor root it has to be divisible by r and that's the same condition as this right here in order to get an integer m you also have a condition of divisibility by r it's actually the same condition. So the integral is well defined if and only if this m is integer and that's the situation where the hermit's number exists to begin with. So why do I look at this space? That's once again so if you look at this explanation what you have, so you have a unified covering like that so it has poles f has poles let me write it here of order k i at x i right it has m poles so I call these poles x 1, x m so f has the power of order k i at x i so df has a power of order k i plus 1 right at x i and this is equal to r times p i plus 1 minus a i right, if I rewrite k i like that I obtain r times p i plus another r there's a plus 1 that kills this minus 1 so I just have a minus a i right and that means that if I take a square root so enough root of df it will has a pole of order p i plus 1 and this is the thing that I have to twist the cotangent line bundle with in order for this to be divisible by r right, if I want this to be divisible by r I have to at the beginning it's not so I have to twist the line bundle by this minus a i times x i and then the pole will be of order r times p i plus 1 and I will be able to to extract the rth root okay I want to explain one last thing that's the this thing in the numerator and this thing is the numerator it is as follows, so here I have my moduli space of r-spin structures over this moduli space I have the universal curve and I also have, let's call this pi as before, and I also have the universal r-spin structure since this is the space of r-spin structures it means that on the universal curve I have the universal r-spin structure so here I have a line bundle on each fiber and the union of these line bundles gives me a line bundle over the whole universal curve and its rth power is the cotangent line bundle to the fibers and now I can look at the push forward so the sheaf of sections of this line bundle and take its push forward I mean the push forward this r-star, so the direct functor so more or less on each curve I can take h0 of cl and h1 of cl and the formal difference, h0-h1 glues into two sheaves that are called r0 star of l and r1 star of l right this is for one curve and this is if I do it on the whole on the whole moduli space I obtain two sheaves and so r-star p-star of l is equal to r0 p-star of l minus r1 p-star of l its a rather common construction this is what you see in the Greta-Decremen rock formula for example and you can check that in the case of r equal 1 if r equals 1 you obtain here exactly you obtain actually minus the hodge bundle here you obtain the class of the dual of the hodge bundle exactly as you should because if r is equal to 1 this l here is just the co-tensioned line bundle its h0 is the hodge bundle and h1 is the trivial line bundle c so it has no charm class ok, so I would like to make one remark on this r-spin how it's numbers before I continue that's in order to make them more attractive so these expressions here these tr plus 1 of lambda I learnt them from a paper by Akonkofen-Panderibande where they compute Gromov-Rutten invariance of curves so it actually turns out that this thing is related to the class psi to the power r and I can write completed quality so this heritz number hg rk1kn is equal to a Gromov-Rutten invariant so if I take the space of the space of maps to Cp1 relative to infinity of degree k and the branching at infinity so if it's relative to infinity it means that I have to prescribe a ramification profile at infinity ramification profile is prescribed by k1kn so that's the space of relative stable maps it has a virtual fundamental plus and I can sorry I should add m marked points here marked points and k is the degree of the map degree so now I have m more marked points and I can write r factorial psi1 to the power r times r factorial psi m sorry psi1 to the power r times the pullback by the relation map of a point that's the the block and same thing r factorial psi m to the power r pullback of the class of a point by the relation map so I have m marked points I fix their images because every time I take the pullback of a point by the relation map and after fixing their images I also put r powers of psi plus of each of them and this will be exactly equal to this hermit's number and again the relation between the two is not clear so I mean this is I mean this thing is proved by and but it's not clear why it should be related to integrals on the space of R-spin structures so if you prefer you can reformulate the RLSV formula by replacing this number here by this gram of written invariant okay so now finally I come to the topological recursion can you repeat the connection between the formulas? yes so I mean this formula is proved so this is the R-spin hermit's number that I defined using this completed the representation theory and this is a gram of written invariant and this formula is a conjecture so I should maybe write an exclamation mark here and a question mark here otherwise it's not right so since you know that these numbers are equal you can take this number and put it here it's an equivalent conjecture and in this way you have an equality between two integrals one is the integral of the space of uh relative stable maps with some powers of side classes and here you have a space on the uh an integral of the space of R-spin structures and it's not clear why they're equal to each other it seems like there's a connection between the virtual fundamental classes hiding somewhere I don't know okay so the relation with the topological recursion there are more properties of the hermit's numbers so the first property is called the Buscham-Mainier conjecture that is that was actually proved since which says that the curve the plane curve x equals log y minus y is the spectral curve is the spectral curve for hermit's numbers so more precisely if I write h g n right maybe I should write w g n of x 1 x n equals sum over k 1 k n one over n factorial e to the power sum of the k i x i so and then the hermit's number g k 1 k n and also I forgot the e to the beta to the m over m factorial so here I also should sum over and again m here is expressed from this genus so maybe, no sorry you probably don't need it here because there is just one m once you fix no once again beta to the m over m factorial so these things here well these are functions but you can transform them into you can transform them into differential forms by differentiating once with respect to each x i and then what you obtain are the invariance the topological recursion invariance for this spectral curve so it was a conjecture for some time and then it was proved from the ELSE formula and there is a similar conjecture for these r-spin hermit's numbers the r Bouchon Maria that says that the curve x equals little y minus y to the r is the spectral curve in the spectral curve for r-spin hermit's numbers and once again this is still open so the only theorem I can boast of today is that these two conjectures are equivalent so that's the the main theorem of the talk is that the rlsp conjecture and the arbo-sharp Maria formula are equivalent to each other so if someone proves one of them both will be proved at the same time maybe I should write them down since that's the main theorem let me write it here rlsp is equivalence to Boucharp Maria I still have a couple of minutes so I can I'll try to explain a little bit how the equivalence is proved just to make sure you understand once you have the spectral curve there is the topological recursion procedure that allows you to compute everything to compute all the hermit's numbers right one by one and the erasve formula also allows you to do that if you can compute some integrals on the moduli space so in principle both determine the hermit's numbers uniquely and you can prove that we can prove that it's the same it gives the same numbers yeah that should also write that's joint with Luke Spitz and Sergei Shadrin ok so how does the proof work first of all we have to look at the numerators of these formulas in the erasve formula we have the dual of the hodge bundle and in the rlsv formula you have this the churned class of the minus the push forward of the spinner bundle so these things are actually semi-simple co-homological field theories I'm not going to tell you what that means but the important thing is that semi-simple co-homological field theories are classified and they are determined by one matrix value power series so completely described described by a matrix valued power series matrix valued power series it's the the give and tell our matrix so the power series for the first one here you have matrices one by one so it's actually just one power series it corresponds to the exponential of minus sum of the k greater than or equal to one bk plus one over k k plus one t to the k so that's the Bernoulli number so actually one term out of two is equal to zero because the Bernoulli number vanish the odd Bernoulli number vanish so I'm not going to describe you completely the classification of co-homological field theories it would take too long but I will show you how in this particular case what this formula I mean what this power series tells us about the term class of the Hodge-Bandau so there's an expression called the Manfred's formula obtained by applying the Grotten-Dichry-Manouff formula to the universal curve and it gives us the term character the term class of the of the dual to the Hodge-Bandau so it's exponential sum so minus sum k greater than or equal to one bk plus one over k times k plus one like that times k minus sum from one to n psi i to the power k and plus one half delta delta is the boundary and here psi one to the k plus psi double prime to the k divided by psi prime plus psi double prime so when you have when you have a boundary stratum like that you have two psi classes at the node at the node you have two branches that meet so you have two cotangent lines to the branches you have two extra psi classes you have as usual the usual psi classes attached to the marked points with the cotangent line to the marked points here but on this boundary on this boundary divider you also have cotangent lines to the attachment points of the of the two branches so you have also two extra psi classes psi prime and psi double prime and here is the expression so as you see as you see you can see the power series that appears in front of some expression of classes on the moduli spaces not the moduli space and there is a similar formula I have to stop in like five minutes right? so for the for the space of r-spin structures you have this c of minus star we push forward of L and this corresponds to matrix static power series and the matrix is r times r and the elements here are b k plus 1 so it's sum over k again 1 over k k plus 1 and here have b k of b k plus 1 1 over r and so on b k plus 1 r over r let me write one more term b k plus 1 2 over r so it's a diagonal matrix and here have the Bernoulli polynomials and you take the values of the Bernoulli polynomials at the rational points with the denominator r so once again you see that if you take r equals 1 you just get this term which is exactly the Bernoulli number the value of the polynomial at the point 1 and this is translated by the by a formula by Alessandro Chiodo for the term character of this so that's equal to the exponential of the sum minus sum over k 1 over k k k plus 1 b k plus 1 of 1 over r times k k minus sum from 1 to n b k plus 1 a i plus 1 over r psi i to the power k plus 1 half yeah I don't know if I want to write the boundary term actually r over 2 let me write it down but maybe it's not so sum from a from 0 to r minus 1 delta a there are actually different types of boundary divisors in the space of our spin structures that are indexed by remainder's modular r I think is not so important now psi prime plus psi double prime and here b k plus 1 a plus 1 over r psi prime to the power k minus a plus sorry b k plus 1 r minus a minus 1 over r c w prime to the power k okay it's too long and then I have to close both brackets so once again you can see all the elements of this matrix appearing in different combinations and yeah I wanted to compute one integral but I think I have to stop here so that already mentioned the result that the topological recursion was actually included inside the classification of very simple topological theories by these r matrices so when you have this r matrix when you have a curve you have the remember the b of z z prime or z 1 z 2 that was part of the spectral curve and then you should take its power series expansions at the critical points and take the Laplace transform Laplace transform transform and that gives you this this matrix value of power series matrix value of power series that classifies topological field theories so there is actually a connection between these power series here this one and this one and the expansion into power series of this two form d z 1 d z 2 over z 1 minus z 2 expanded in coordinate y right if you have a rational curve you can take the square if you have a rational curve you have this two form and you can expand it into all the coordinates of the square hood of critical points and the coefficients of this expansion should match the coefficients of this power series after the Laplace transform and so the the Laplace transform is some integral right the Laplace transform of f is something like that d z and this integral when you write it for this spectral curve this integral happens to give you the inverse of the gamma function more or less actually one over the gamma function and I don't know if you know that but this power series both these and this here the actual asymptotic expansions of the gamma function the sterling formula exactly the sterling formula is the leading term equivalent rates to square root of 2 by t t to the power t minus 1 into the t times the exponential of this thing but with a plus sign so when you have one over the gamma function and these actually things appear in front of the integral so they simplify and in the end you get exactly this power series this is the power series that you want the connection between the ELSP formula or the ELSP formula and this actual curve is that what you make when you compute some integrals at the neighborhood of branch points of this curve you get the reciprocal of the gamma function that's whose asymptotic expansion is given by these power series here and here I think I have to stop here