 So I don't know how to call my talk. I decided to call its introduction to everything. I have two hours, and I will talk about, I will tell you more or less everything that I know about the topological recursion. And then Maxim will have six hours to continue. And I will also tell you about the Herwitz numbers and the ELSV formula. So the plan is topological recursion, Herwitz numbers, and the ELSV formula. And I will try to stay as elementary as possible so that you will be able to follow the rest of the talks more easily. OK, so how does the topological recursion work? The topological recursion is a method that applies to many problems where you have generating functions fgn of x1, xn for any g and n. So g is usually the genus and n is some number of marked points or of something. I will give you three examples of problems like that. There are more, but I'll give you three. OK, the first problem is on Herwitz numbers, hgk1kn. There are two definitions, one combinatorial definition and another more geometric. So the geometric definition is that it is the number of framified coverings. Let me draw it here. Here is a sphere, let's say, Cp1. Here is infinity. Here you have m fixed branch points. And you want to find to enumerate branch coverings of the sphere with ramification profile k1, k2, kn over infinity. So infinity has m pre-images and there are branching indices equal to k1, kn. And then simple branch points over all other points. I don't know if I will be able to draw it properly, something like that. And I want this surface to be connected of genus g. So this is the number of branched coverings. And the combinatorial definition is with permutations. So let me denote the sum of the k i's by capital K. And then it will be equal to 1 over k factorial, the number of lists of m transposition. I forgot to say that m here should be equal to 2g minus 2 plus sum of the k i's plus n. You can compute it if you want this surface to be of genus g. You can compute a soil characteristic and you find that the number of simple branch points should be equal to this number. So this is the number of lists of m transpositions, let's say, of lists tau 1, tau m, sigma, such that tau i's are transpositions, sigma has cycles of lengths k 1, k n, lengths k 1, k n, and the group generated by them all is transitive. And I forgot the product. Yeah, tau 1 times tau m times sigma is equal to the identity permutations, the identity permutations. So all this is sk in sk. This definition is equivalent to the geometric one because if you look at the monodramas of this covering, you find m transpositions and one more complicated permutation for the monodrama here. Okay. So this is one Herbert's number and then is it possible to get the top blackboard by pushing some button? No. Oh, there is a stick. Okay, there's a hook. Yeah, I can jump. I was thinking of that too, but since everything is filmed on camera, I should behave properly. Okay, and so now you can introduce the generating function hgn of... Yeah, so I will hesitate. There are two variables actually. One set of variables is most natural for all combinatorial work and the other works fine if you want to apply the topological recursion procedure. So I'll give you both. So sum over n, one over n factorial, sum over k1, kn, hg, k1, km, kn over m factorial, so maybe I'll write it like this, plus n plus sum of the ki's factorial. And here you can write... Okay, let me write e to the power sum of the kixi or product of capital X i to the power ki. So capital X is e to the power x. This is the way that we'll make the topological recursion work properly and this is the usual way to write generating series. Okay, problem number two. That's the end of problem number one. We arrived at a set of natural functions that for every g and for every n, you have a function of n variables. Once we get this, we can try to apply the topological recursion. So the second problem is the problem of Weill-Patterson volumes, volumes of moduli spaces. So if you fix gn and you fix n lengths, l1, ln, and then you can look at the space of complete matrix of curvature minus one on surfaces of genus g and bounds. Yeah, so actually, yeah. Let me do it like this. Then it will be correct. So with n-holes, genus g and holes, 1, 2, n, and each hole has this geodesic here and I want them to be of length, l1, l2, ln. The geodesic that bounds every hole must have a prescribed length. Right, so you get a space of all possible hyperbolic matrix, hyperbolic complete matrix on surfaces like that. Okay, on this space, there is a two form called the Weill-Patterson form, symplectic form. So let me call this Mg, l1, ln. And then you can compute the volume of this space. So the integral over Mg of l1, ln of the exponential of the Weill-Patterson form. And this is called the volume gn of l1, ln. So once again, we have, for every genus g and for every n, we have a function of n variables. And once again, if you want the topological recursion to apply, you have to perform some strange change of variables. Okay, let it be. So actually, you have to write v hat gn of x1, xn. It will be the Laplace transform of vgn. So the integral over l e to the power minus lx vgn l1, ln. So this will be l1, ln sum lixi. And the topological recursion actually works when you apply it to these functions. Okay, and finally, the last example, I called it incrustated triangulations. So here is a picture of an incrustated triangulation. So you have a genus g surface, mostly triangulated, but there are n incrustations that are some other polygons with any number of sides. Here I have a one-gone, a two-gone, a triangle and a pentagon. So t, like triangulations, gn k1, kn is the number of triangulations of a genus g surface with capital N triangles and more polygons, and more polygons with k1 up to kn sides. And here you also can collect these numbers into a generating series. So here you will actually have one extra variable because of this capital N. So fgn of x1, xn and a u will be equal to sum over n and capital N, one over n factorial, sum over k1, kn, x1 to the k1, xn to the kn, u to the n. I think it's over n factorial, actually. Yeah, almost sure. tgn k1, kn. Okay, so once again, for every genus and every number N, you have a function in N variables. U will be a dummy variable. It will not be involved in the topological recursion. Actually, there's one question in great pictures, what should you divide by n factorial as well? Of course you divide them. No, because you... It's... Let me see. In the definitions, divide it. When you take the exponent... Exponent, yeah. No, that's the dimension. No, this is not the same factorial. Now here, the thing is that here you have the sum over all the k i's and here it's just a... Like integral over l i's is the same. I think... I'll have to check. I think it's like this. Yeah, no, it's not related to the exponential of the well-peterson. I'm not sure about your question. I think this is the correct formula, but I'll check. Okay, so now what do you do? What does the topological recursion do with all these functions? So I have to warn you that the topological recursion is a subject in progress. So it is now pretty well understood how it works but not why it works or in what situations it works. So there are lots of examples. Sometimes it is proved that it works in a more or less standard way. I mean, it appeared from some set of examples where it works in a standard way. There are some examples where it is proved but in a much more complicated way. And there are examples where it's still a conjecture. So I'm going to describe a procedure and then for every example you will have to work hard to see if it works or not and why. The procedure is as follows. You take the function f01 of x. So the simplest one. This function will give you something that is called a spectral curve, c. And then using these both things, you will get all the others. So you will actually get n-forms wgn on c to the power n. So you take this curve, you take its n-th power and you have n-forms on symmetric n-forms. Symmetric n-forms on its n-th power. And these will be equal to d1 dn, maybe let me write dx, d1 dn of fgn of x1 xn. Curve is in c-square or abstract curve? So this c is actually in the plane, c in c2. And this x will be one of the coordinates of the plane. So if you look at the examples that I gave, sometimes f01 is really the easiest generating function to compute for the Hurwitz numbers we are going to compute f01 right now directly and it will give us the spectral curve. On the other hand, in other examples, f01 may not be defined at all. For the Wild Peterson volumes, if you take the sphere with one hole, there are no hyperbolic structures at all. So there is no modular space. You have nothing to integrate. And still the theory works very well. This function should be defined in an ad hoc way, a posteriori. So sometimes this is easy to find. Sometimes you may have to guess it by other methods. Okay, so let's do it for Hurwitz numbers. Okay, for the Hurwitz numbers we have to compute h0k, which is the number, so 1 over k factorial times the number of lists of tau 1, tau k minus 1. These are all in sk. These are all transpositions such that tau 1 times tau k minus 1 is a k-cycle. Right, so you have to compute the number of ways to decompose a k-cycle into k minus 1 transpositions. It's the smallest possible number of transpositions to get a k-cycle. And then it is automatically transitive. You can also see it geometrically. So if you draw, if I draw again my picture with the ramified covering, so here you have just one pre-image. Yeah, I'm not sure it's the, maybe, okay. So if you draw a star like that on this plane, let me write, draw it in yellow or in red. If you draw a star like that on this, on CP1 here, the pre-image of this star will be lots of stars. It will be all glued together at the pre-images of these points. So the pre-images are here. So you will get a picture like that with lots of stars. All glued together. No, so only two by two. And then you can erase the unnecessary tails and you get a tree. The tree will have k, k vertices and k-1 edges. So this is equal to the number of trees with k-numbered vertices divided by k factorial or the number of trees, or number of trees with k-1 numbered edges divided by k-1 factorial. So let me tell it precisely since you're asking about it. So when you take a tree with numbered edges, it has a planar structure automatically because around every vertex you can order the edges in the increasing order, right? So when you take the pre-image of this, you can number the points here, one, two, three. You can take the pre-image of this star and you get a tree with numbered edges. Actually your form of a hook should divide h0k, should divide by k-1. Yeah, exactly. Just in front of you have a mistake. Now just look at h0k is equal to 1 over k factorial, no, no, no, this is correct. This is correct. It's just in the generating series. So actually what I wrote here is h0k divided by k-1 factorial. And this is equal to m, 2g-2 plus n plus sum of the k-i's. In this case it's equal to k-1. It wasn't the generating series. So I number these marked points, then I have a tree with numbered edges, and then I divide by k-1 factorial. So in some way, in some sense, I forget the numbers of the edges. And if I want to see it like that, to see it as permutations, I also have to number the pre-images of this point. There will be k pre-images numbered from 1 to k. And then I will get trees with numbered edges, numbered vertices, and I will have to divide by k factorial to forget the numbers of the vertices. I think it will run away. Okay, so now does anyone know how many trees with k-numbered edges, the k-numbered vertices there are? There's the Cayley formula that says that the number of trees with k-numbered vertices equals k to the power k-2. It's actually not a trivial formula. The answer looks so simple that you would think it's an almost obvious statement. There are lots of proof. I mean, it's one of the most famous combinatorial formulas, so there are lots of proofs, there is a bijective proof there are. But none of them is particularly simple. So one of them, let me sketch one of the proofs because I will need the parts of it later. So if I take rooted trees, same thing, rooted trees with k-numbered vertices, 1, 2, 3, 4, and one of the vertices is a root. And I call y of x the generating series. Then I can, let me draw a little bit more complicated, 5, 6, 7. So then I can erase the root and obtain several trees, and all of them will be rooted. The roots will be the vertices where these edges were attached. And from this I will get the quality that this series is equal to x times e to the power y, y of x. So x stands for this root that I removed and e to the power y. So 1 plus y plus y squared over 2 and so on. Every term stands, so this 1 stands for the simplest rooted tree with no other, nothing attached. y is when the root of degree 1 and there is one tree, y squared over 2 is when the root is of degree 2 and there are two trees and so on. So we get an equation like that. And then there is the Lagrange inversion theorem, inversion formula that allows you to find the coefficients of a series that satisfies an equation of this kind. And you find that the coefficients of y are k to the power k minus 1 over k factorial x to the k. So this will be an important power series for us. Okay, so anyway for f01 of x I found that it is equal to sum of k to the power k minus 2 over k factorial e to the power k small x. From this I find w01 of x. As I said it is equal to d of f01 of x. So remember I told you that what you actually find with topological recursions are n forms rather than generating series. So when you have your first generating series you have to take its differential and you find sum of k to the power k minus 1 k factorial e to the power kx dx. So here you can recognize the power series y. And this gives you the equation of the spectral curve. So the equation of the spectral curve is y equal sum over k to the k minus 1 over k factorial e to the power kx. And if you look at this equation you can also write it as x equals log y minus y, something like that. Right, so if I write y equals x e to the power y, then capital X is equal to y e to the power minus y. So this is e to the power small x equals y times e to the minus y and then I take the logarithm and get x equals logarithm y minus y which gives me the equation of the spectral curve. And I have the first differential form that is now just y times dx. Right, if you look at this it is just y times dx. So this will be the starting point for the topological recursion and the topological recursion procedure will give me all the other differential forms only from this information. Okay, so now here is the recursive procedure. Suppose you have c curve in c2. You have omega 01 equal to y dx. And then so for the recursive procedure you also need to know w02. So w02 in many cases it is uniquely determined by this data. So suppose that suppose c is a rational curve with global coordinate z. Global coordinate equal to z. Then I can write the following two form on c2. On c2 I have dz1 dz2 divided by z1 minus z2 squared. And I claim that this is a two form that is unique in its kind. It does not depend on the global coordinate. So invariant under if I change the variables if I replace z by az plus b over cz plus d you can make the computation you will find exactly the same thing. So it's a form on c2. So on c2 you have a coordinate z1 and a coordinate z2. So c is c1 minus finitely many points because your curve is not finitely. Yes, yeah, yeah. So actually in all examples that I personally know the curve c happens to be rational and this is the two form. I know that there are some advanced people who look at examples where the curve c is of hydrogenous and then there is some choice. So in this case you can also require, you can also ask for a two form that has a double pole like this on the diagonal. But then it is not unique and I think it is not very clear how to make the choice. So there is, in the original paper by Aina and D'Anton they make some, they specify some choice of w02 but it doesn't look particularly natural. They have to choose a basis of, of differentials on the curve c. So the choice depends on the basis and it's not very clear how they choose it. So I think it is, I think it's still not an elucidated question. If, if the curve is not rational, I don't know what to choose exactly. And then there is the huge frightening formula that you will see, I'm sure, in Maxim's talks or maybe he will hide it in a, okay. So this is my curve c, let me pick a critical point a. So when I have a critical point a, I have a local involution sigma a of z to x. Yes, this is x, yeah. So I have a local involution sigma a of z that exchange the two sheets. It's not defined on the, on the whole curve, but it is defined in the neighborhood of this, of this critical point. So from this I define a kernel, kernel k a of z and z, z prime. Okay, so it's the integral from sigma of z to z of w02 of z and z prime divided by 2w01 of z. So the type of this thing is something times dz prime divided by something times dz. W02 has a dz and a dz prime, but then I take an integral over z. So dz goes away, right? Yes, you agree, Maxim? Yeah, yeah. dz goes away so I have a dz prime and this is one form with a, with a dz. Okay, now once you have this kernel, I'll take a fresh blackboard. Okay, sum over critical points, residue as z goes to a, this kernel dz prime. Let me write dz on my n plus 1 actually. Okay, here you have a sum over all g1 plus g2 equals g and i union, disjoint union j equals 1 to n. So you split the genus and the set of mark points into two parts. Then here you write wg1, so i plus 1, ziz, wg2, j plus 1, zjz plus sum over, here there's no sum, sorry, wg minus 1, n plus 2, z1, zn, zz. And actually in some, depending on the papers, sometimes you can see, instead of z, you can see sigma of z in some of these places. It does not change anything, everything is, so you can put z or sigma of z. You have to change a sign. You have to change a sign before, okay. No, no, so the left-hand side, I decided not to put any left-hand side. I think it's more beautiful without a left-hand side. So this is equal to zero. And this is actually a triangular system that determines all the wgn's recursively. Because if you look at the, so there is, there is, sorry, there is one main term here. If you take the first sum, in the first sum you can take g1 equals g and i equals 1 to n. So that here you will be left with w01. So if you are left with w01, it will, you see, so here you will have, okay, let me write it, let me write it somewhere. So I claim that this, this choice will lead you to just wgn plus 1 of z1, zn plus 1. And you can convince yourself that this is true rather easily. So if you take this choice, you will have wgn plus 1, z1, zn, z. Then here you will have w01 of z. And then you will have to multiply by the kernel. And the kernel will, in the kernel you have this, right, 0, 2, zn plus 1, divided by 2, 0, 2 w01 of z. So the w01 of z simplifies, right? And you just, you are just left with this thing and with some residue to compute for this w02. So if you look at the w02 on the diagonal, it has the residue. Well, there is a, you will just get 1. Yes? Could you explain where the formula comes from? No, no, I cannot explain where that, yeah. Yeah, so the formula, of course, the formula comes from originally from the matrix integrals, but I think I'm not going to talk about this now. So this is the main term in this formula. Some people put it on the left-hand side. And then they have to do something to kill this term and the summation. So in this case they usually, by convention, put w01 to be equal to 0. Instead of my choice, which is the most natural choice, they say that this is just some special form that is used in the kernel and w01 is equal to 0 in the recursive formula. I think it is the unnatural choice. For Herbert's numbers, for example, you can compute the function of 01 and it gives w01 to be equal to this. So you don't want to change it by force. It's better to write it, more beautiful to write it in this way. Okay, and then you have this term that is the, let's say the main term of the formula and all the other terms have smaller 2g minus 2 plus n, smaller rather characteristic. So once you know w01 and w02, you can compute all of them one by one using this formula. Yes, wgn has poles on diagonal and on the diagonal, yes, as you can see here. No, so wgn, no, they also have poles at the branch points, actually. Yeah. Okay, so, yeah. So, yeah. No, that's actually the only one that has a pole on the diagonal. It's already, yes. Do you have some, I'll show you a quick point. Yeah, they should be simple. So now, again, now there are developments. So now it is possible to generalize this procedure if you have a multiple critical point, but the formula is different. Okay, so the answers to the question, this formula only works for simple critical points and wgn's are meromorphic on c to the n with poles only at critical points. No, I think on the curve itself it doesn't have poles at all. You're writing coordinate x, which is not called local coordinates at critical point. I think it's... So, sorry, what does... I guess it's actually wgn has no, it's a form on power of curve, has no poles at all. Because when you write it's something times dx, maybe, x will be not local coordinate. Yeah, but they, no, I'm sure, they will have poles at the critical points, I'm sure of that. Yeah, you will see in the example of Herbert's numbers you can see the pole. So only at critical points and w02 is an exception. W02 has a pole, has a pole on the diagonal of c squared, but that's the only one. I have 10 minutes left. Oh, yeah, I know, it's fine. Yeah, I was surprised by that. Okay, then it's fine. One hour more. Okay, so now the LSV formula. So it will look like I'm talking about something else, but actually soon after that I will come back to the topological recursion and you will see some, the relationship with the LSV formula. So here is the formula. It's a formula on Herbert's numbers. h g k 1 k n equals m factorial product k i to the power k i over k i factorial integral over m g n bar. 1 minus lambda 1 plus lambda 2 minus and so on plus or minus, okay, minus 1 to the g lambda g. 1 minus k 1 psi 1, 1 minus k n psi n. Yeah, LSV stands for Ekadal, Landau, Shapiro and Weinstein. Okay, so I have already explained what the Herbert's number is. Now there are lots of funny letters here that I have to explain too. Anton asks me to include an introduction to modular spaces, so I will spend some time explaining all the letters here. And this m, again, is 2 g minus 2 plus k i plus n. That's the number of simple branch points. Okay, so this is m g n bar. So it's the modulate space of stable genus g curves within marked points. Maybe I'll even draw a little bit bigger. So that means to every point here is associated a curve of genus g with marked points, let's say 1, 2, 3. So here I take genus 2 and 3 marked points. So this is a smooth curve, but you can also have curves with simple self-intersections. So stable curves are allowed to have singularities, but only one type of singularity, namely simple self-intersections. So here you have some divisors with normal crossings. And if you take a point on this divisor, you will get a curve with a simple self-intersection. 1, 2, 3. See, so in this case I decided to take a curve with a non-separating node. If you normalize the curve at this node, the curve stays connected. And then this is a self-intersection of this divisor when a curve has two non-separating nodes. 1, 2, 3. And then there's another divisor. Here you can have a separating node like that. So there are actually lots of divisors because you can split the curve in many different ways. You can have two components with various distribution of the genus and marked points. And so here the intersection of the two you will have one separating node and one non-separating node. Like that. Okay, so over every point of the moduli space you have a stable curve. All these together form what is called the universal curve. So mgn bar is what is called an orbifold. You can think of it as a manifold. It usually doesn't manifest itself in any way, but properly it's an orbifold. So compact, complex orbifold of dimension, so complex dimension, 3g minus 3 plus n. So this one will be of dimension 3g minus 2 plus n. One dimension more and here have the projection whose fibers are stable curves. There's one tiny piece of yellow chalk. Probably. Okay, I'll try to use this one. Okay, so now over mgn I will define n line bundles. And one vector bundle. So let's take the cotangent line to the stable curve at the first marked point. I can do that in every fiber. In every fiber I have a first marked point. I have the cotangent line. Right, so now over each point of mgn bar I have a complex line. The cotangent line to the first marked point. To the curve at the first marked point. So all taken together they form a line bundle that is called l1. So I have line bundles l1, ln. If you prefer formulas, li is the pullback under the ith section of the universal curve of the relative cotangent line bundle to the universal curve. Okay, and then there is one vector bundle called the hodge bundle. So it's a Frank G and its fiber over a point is the space of holomorphic one forms on this curve. A billion differentials. So ex is the space of a billion differentials or holomorphic one forms on cx. Right, you take a point x, you have a curve cx, you take the space of a billion differentials. It's a vector space of dimension g. So over every point I have a space of dimension g. And they form a vector bundle. Maybe you should say something about first order poles and double poles. You mean about the stable curves? Do you want to know how to extend the hodge bundle to the stable curves? Okay, so the rule is when you have a stable curve it has simple self-intersections. The rule is that at this point you allow simple poles, at most simple poles with opposite residues. Well there is a reason of course for this rule, but let me just give you an example. If you take an elliptic curve you know that there is exactly one a billion differential up to a multiplicative constant. If you take a lattice of parallelograms you can just take the differential dz and it descends on the elliptic curve. And then you can take a stable curve like that. That's just a sphere with two identified points, let's say zero and infinity. Cp1 with zero identified with infinity. So you would like to have exactly one a billion differential here. But on the sphere there are now a billion differentials. If you take a differential one form on Cp1 it will have poles. But if you take this form dz over z it has a simple pole at zero and a simple pole at infinity with opposite residues. So with this new rule you still have a one dimensional space of a billion differentials. And it actually works in the same way for any genus and any degeneration you always get a space of dimension g. And you can actually see that if you take a holomorphic one form and you take a family of elliptic curves that tends to the stable curve you see how the a billion differentials degenerate into something like this. Okay, so now that we have these line bundles and the vector bundle I can take their characteristic classes. So I can define psi i to be the first-turn class of Li. So that's a first-turn class so it lies in H2MGN bar with rational coefficients. This Q here is actually the only place where the fact that it's an orbifold makes appearance. Usually the first-turn class has integer coefficients on a manifold. On an orbifold they can actually have rational coefficients. And then lambda i, let me say lambda j will be the jth-turn class of the Hodge bundle. So that's an H2J of MGN bar Q. And here i goes from 1 to n because they're n marked points and j goes from 1 to 2g because it's a rank g vector bundle. And finally there is one last thing I should tell for you to understand this formula. It's that when I write 1 over 1 minus k i psi i what I actually mean is 1 plus k i psi i plus k i squared psi i squared plus and so on. Yes, so here I have a product of n plus 1 factors. The factor in the numerator and the factors here that are expanded in this way. They look like infinite series all of these but actually the dimension of this so as I said the dimension of this is equal to 3g minus 3 plus n. So when I perform the integral I will only extract the terms of this degree and then it's a finite expression. All this thing is actually a polynomial in k1, kn and it is symmetric with what top degree equals the dimension of the moduli space 3g minus 3 plus n. And lowest degree equal to 2g minus 3 plus n. So I get top degree if I take 1 here and maximum greatest possible power of psi i's and lowest degree if I take lambda g and so the lowest possible powers of psi i's and of k i's. But then if you want to find the actual coefficients of this polynomial this is much harder because then you have to actually study the intersection theory on the moduli spaces and compute the integrals of all these characteristic classes. It's one thing to define them and it's much harder to actually compute anything. So now I'll give you an example. Okay, example. Let's take genus 1 and one marked point. So I rewrite the ELSV formula. It says me that h1k is equal to k minus 1 factorial k to the power k over k factorial. Sorry, k is it k plus 1 factorial 2g minus k plus 1 factorial times integral over m11 bar 1 minus lambda 1, 1 minus k psi 1. So I denoted k, I write k instead of k1 since there is only one marked point. I don't want to write the index every time. So this thing here m11 bar is equal to, well it's a compactification of this modular figure. If you take this thing and you glue the left hand side to the right hand side and this arc to this arc and then you add one more point at infinity to compactify it. You will get m11 bar. So if you imagine what happens you will see it's something like this. So it's homomorphic to a sphere. Okay, so let me rewrite this integral. It's the integral over m11 bar 1 minus lambda 1 times 1 plus k psi 1. And I don't have to continue because it's a sphere, it's of complex dimension 1 so all the other classes will be equal to 0. The next powers will be equal to 0. So that's equal to the integral over m11 bar of k psi 1 minus lambda 1. I drop all the terms that are not of the correct degree. So that's k integral over m11 bar of psi 1 minus integral over m11 bar of lambda 1. Let me denote this by x and this by y. It is possible to compute these integrals but I don't want to do it now. I want to use the ELSP formula that will compute them for me with no effort. Okay, so to use the ELSP formula now I have just two numbers to determine, x and y. If I know these x and y I will be able to plug them in here and get the final answer. So to get x and y I just have to compute two Herbert's numbers. So let me compute h11 and h12. Okay, h11 is the number of coverings of the sphere by a torus of degree 1. Right here I have my sphere, Cp1. Here I have a torus. Here I have infinity and infinity must have a preimage of one preimage of multiplicity k. This is the definition of the Herbert's number. And now I decide to take k equal to 1. So how many coverings, how many coverings there are of a sphere by a torus of degree 1? The answer is 0. Right, it is not possible to. If you have a map of degree 1, this one has to be a sphere too. It cannot be a torus. If you want to use a combinatorial definition, you will have to count lists of, let me see, two transpositions in S1 whose product is equal to the identity permutation. Well, in S1 there are no transpositions, so there are no lists like that. Okay, now let's take degree 2. That's already more interesting. Now you have a preimage of degree 2 and then you have three more marked points that are simple branch points. In this case, if you fix the branch points, you have exactly one double cover of the sphere with branch points here. That's exactly the picture that I draw. Moreover, there is the hyper elliptic convolution. So it is actually counted with weight 1 over 1 half. So I think I didn't tell you that I should have. When you count the ramified coverings, every covering should be counted with coefficient 1 over the number of automorphisms. In most cases, this number is equal to 1, but in this case there is group Z2 of automorphism, so you have to divide by 2. And if you use the combinatorial definition, it gives you 1 half automatically. So it's 1 over 2 factorial times the number of lists of three transpositions in S2 such that their product is also a transposition. So in S2 there is only one transposition, so there is no choice. There is just you have to take 1, 2, 1, 2, 1, 2, 3 times the same transposition. The product of these three transpositions is also equal to a transposition. So that's exactly the monodrome of all the discovering, it's just the permutations of the two sheets. And then you divide by 2 factorial, so you get 1 half. Is that fine? Right, so now I have a system of two linear equations. For k equals 1, I get 0 equal to 2 factorial 1 to the power 1 over 1 factorial times x minus y. For k equal 2, I get 1 half equal to 3 factorial 2 to the power 2 over 2 factorial 2x minus y. The first equation tells me that x is equal to y and the second equation tells me that x is equal to y is equal to 1 24th. You can check, 6 times 2 times 2, that's fine. So the ELSV formula tells me that this integral is equal to 1 over 24 and this integral is also equal to 1 over 24. I got this by just counting permutations in the groups S1 and S2. And now I can write the final result. So I'll simplify k plus 1 factorial and k factorial. I will get k plus 1 times k to the k times k over 24 minus 1 over 24. Or k squared minus 1 times k to the k over 24. So this is the full power of the ELSV formula. It solved me, it computed two integrals that if you don't know how to compute are pretty hard to compute. And it also solved for me a non-trivial combinatorial problem. This is actually the number of ways to factorize a k-cycle into a product of k plus 1 transpositions. If you try to solve this problem by hand it is extremely hard. Once again it looks like the answer is rather simple. If you try to do it by hand I think it's not possible. But here is the answer. And you can see there's 24 in the denominator that is particularly surprising. So now I'll tell you a couple of words about how the ELSV formula is proved. I'm not going to give any actual proof but just an idea. So the idea of the proof is to look at the space of stable maps from genus G surface, genus G curves to Cp1 and then poles of degrees k1, kn. So there is the space of stable maps, mg, k1, kn, Cp1. And then to a map like that you can assign the set of... So you can, if you have a map like that, you can assign to it a set of branch points. So that will be something like C to the power m here. I have to draw a picture. So Cp1 here have the stable map with the poles. So when you have a stable map you can actually have some contracted components. And then here you have the branch points. And there is a theorem that tells you how to... So the contracted components, they also count as multiple branch points. There is a theorem that tells you how to extend this map. So how to assign a multiplicity to every contracted component. So that this map is well defined not only on smooth maps but also on stable maps. So as soon as you have a contracted component its image counts as a branch point with some multiplicity. And now you can try to compute the degree of this map. So this is called the branching morphism. And you can ask what is the degree of the branching morphism. And the degree can be computed in two ways. The first way is to take a generic point in the image. So you take simple branch points. And then the degree will be just the number of smooth ramified coverings with simple branching over these points. This is exactly the Herbert's number. Let me call this BR the branching morphism. So branching the pre-image of a generic point is the Herbert's number. If you take simple branch points the number of pre-images is exactly by definition is the Herbert's number. But then you can also take the pre-image of 0, 0, 0, 0, 0. So I decide to put all the branch points in one place at 0. And then this branching morphism has an infinite fiber. In this case the pre-image, so the maps that have all their branch points at 0 look like that. Here is z to the power k1 and so on, z to the power kn. So you have a bunch of spheres on which the map looks like z to the power something so that there are no branch points here at all. And then you have a curve of genus G with n marked points that is entirely contracted to 0. So that all the branch points accumulate it over 0. And this is an infinite fiber because this curve of genus G with n marked points can have any complex structure. And then in algebraic geometry you have methods to compute the degree looking at an infinite fiber. Here you see some integral of characteristic classes over this mgn bar. And when you carry out this computation you get the right hand side of the LSV formula. So it's a rather common situation when you can compute the degree of a map over a generic point to have a finite number of pre-images. And then over some special point you have an infinite fiber. So you have to introduce a characteristic class and integrate over the special fiber. A small question about this mg bar table maps with the d1kn. What's the generalization of it? Because kind of canonical one doesn't have this d1kn. It's kind of high tangency. It's what is called the relative stable maps. Infinite is the divisor, yes. Okay, I think I can erase maybe. Okay, so now I have a small discussion about this polynomiality. So in my opinion the fact that this thing is a polynomial is the most mysterious property of Herbert's numbers. You see these numbers can be defined combinatorially by enumerating transpositions. It's not a very complicated definition. You would expect that most combinatorial properties can be easily proved from this definition just by studying how transpositions can give you a permutation with given length of cycles. So the property here is that when you take this number and divide by some simple combinatorial coefficient you get a polynomial in k1kn. And for a very long time the only proof of this property was via the ELSV formula. So you had to introduce all these spaces of relative stable maps and modulate spaces and this line bundles and characteristic classes. There was no other way to prove this polynomiality. And this polynomiality is also contained in the topological recursion procedure. And so Maxim wrote a note several years ago, already quite a long time ago, that showed that could actually prove both the ELSV formula and the topological recursion just from combinatorics but assuming this polynomiality property. And this polynomiality property no one could prove independently from the ELSV formula. So finally, several months ago there is a joint paper by five authors, I think. Try to do Dunin-Barkovsky, Luk Spitz, Maxim Nikolaev-Lontan, Sergey Shadrin, right? So they finally found a different proof of the polynomiality that is pretty complicated, but purely combinatorial. I don't feel myself as expected with the proof, even though I am a man of the authors. And then from the polynomiality they were able to deduce the topological recursion procedure and the ELSV formula simultaneously. So now how does this polynomiality appear in the topological recursion? So let me, okay, so let me write once again h, g, n of x1, xn. It's equal to sum over n, 1 over n factorial, sum over k1, kn, hg, k1, kn over m factorial, x1 to the power k1, xn to the power kn. So this was my definition of the power series of the generating functions h, g, n that enumerated the Hurwitz numbers. So now let's use the ELSV formula and plug it here and see what happens. So I claim the following, sum over n, 1 over n factorial, sum over d1, dn, integral over m, g, n bar, psi1 to the d1, psin to the dn, lambda, 3g minus 3 plus n minus sum of the di's. That's the coefficient. And then sum, okay, so it's a product from 1 to n, sum ki to the power ki plus di over ki factorial x to the power ki. It's a trivial statement. I just plugged the expression for the Hurwitz numbers from the ELSV formula to this sum, but I also did something else, so I actually expanded the denominators. You see, if when I write 1 minus ki psi i as 1 plus ki psi i plus and so on, ki squared psi i squared, I can then try to separate all the terms. So let's look at the term where psi i is raised to the degree di, psi n is raised to the degree dn. I fix the powers and then I take the sum over all possible powers. So if I have psi i to the power di, I will also have a ki to the power di coming together with it. So here you see ki to the power di. And then there is ki to the ki over ki factorial that comes from the ELSV formula and the m factorial here simplified with m factorial here. And this thing, it's the degree that is necessary for the sum to be equal to, sum of the degrees should be 3g minus 3 plus n, which is the dimension of mg n bar. So I have to put this, if I want to non-zero integral. Sorry? Minus. Minus, sorry, where? Ah, okay, yeah, there is the, okay, right, yes. Minus 1 to this power, yes, thank you. Okay, so now I want, so now the question is, what is the relation of this power series, right? I have this power series, k to the power k plus sum over k. k to the power k plus d over k factorial x to the power k. And the question is, what is the relation with my power series y of x, which was sum of k to the power k minus 1 over k factorial x to the k. You can see that this series looks similar, but it's not so clear what is the exact relationship. Right, this was the power series that I used to define the spectral curve. And now I said that all the answers should be meromorphic forms on the spectral curve. So in some, so I should be able to express these series in terms of these x, in terms of this y. Okay, so here's what you have to do. First I take another power series, z of x. It will be more practical. Sum of the k to the power k, k factorial x to the k. And I claim that this is equal to y over 1 minus y. Oh, excuse me. Just so you can note that the function of x1, xn is sum over n over 1 over n factorial. n is already fixed on the level. Ah, yes, okay. What does it mean? Yes. Yeah, just, yeah. Yeah, thank you. And yes, so in this case, so in this case there is probably not even, yeah, you see you were asking about the n factorial. I think, I think in this case there is no n factorial. Yeah, I think you should erase all this. Yeah, thank you. Yeah, yeah, that was, yeah. I think I did it all the, every time I wrote this power series. Yes. Yeah, yeah, okay. I did it all, yeah. So you have to do that all every time. Every time, okay. You were asking about the n factorial at the beginning is, you were perfectly right, yeah. There is no sum and no n factorial. So here is my spectral curve. y is a global coordinate on the spectral curve. But it's more practical to have a global coordinate z defined as y over 1 minus y. Right, this is a rational curve. So this is also a global coordinate on the rational curve. It's advantages that it has a pole here. So this point corresponds to z equals to infinity. As you see here is y equals 1. So the pole is here. That's the global coordinate that has a pole at the branch point. And as I said, the, the meromorphic forms that we get have poles at the branch points. Okay, so here I already have the first claim. How do I prove that this, this power series and this power series, right? This is called z. This is called y. And I am saying that this z is equal to y divided by 1 minus y. This actually, so if you write it out, if you write out the power series, you will get the following identity. Sum for p plus q equals n. So n factorial, p factorial, q factorial, p to the power p, q to the power q minus 1 is equal to n minus 1, n to the n minus 1. It is called Abel's identity. And again it looks like a trivial combinatorial thing. Actually it's not, it's not trivial at all. You can prove it only if you know the Cayley formula for the trees. I tried, I gave this as an exercise to an international, sorry? Lagrange inversion. Yeah, Lagrange inversion. I mean any proof that works for the Cayley formula will work here. But it's not a, I gave it as an exercise to an international math Olympiad winner and he was not able to prove it. So it's not a simple, not a simple problem. You can, so if you know the Cayley formula, you can actually prove it by cutting out an edge of a tree, from a tree. You see here you have the number of rooted trees. So here, so here's a root and then you have a tree with n marked, n marked vertices, n numbered vertices. Then you pick an edge so you have n minus 1 choices. And when you remove this edge you get two trees, one of them with two marked points. So p to the power p and the other one was just one, one root so it will be q to the power q minus 1. So if you know the Cayley formula it's easy to prove. And then by proceeding in the same similar way, you can prove that all these power series, so sum k to the power k plus d over k factorial x to the power k is an odd degree polynomial, odd degree polynomial, polynomial in z. So when you raise z to the series z to different powers and by taking polynomials of z you can construct all these power series that you get here. And this is what the topological recursion does. So it starts with a power series like that with w01 and then by some combination of residues it gives you a sequence of polynomials in several variables. So w, g, n, if you look at them in this, in the local coordinate z these are n forms on this curve, here is the curve, n forms on the nth power of this curve you can write them in any coordinate you like. So this z here is the most practical coordinate and then these things are polynomials. So now you have two polynomiality properties that are actually equivalent to each other but in a non-trivial way. Two equivalent polynomiality properties. So the first one that is obvious when you look at the ELSV formula it tells you that the Herbert's number divided by some or multiplied by some trivial combinatorial coefficient k1 factorial kI to the kI is a polynomial in kIs. And the second property is that the generating series h, g, n of x1, xn is a polynomial in z1, zn where zI is this power series. So sum of k to the k over k factorial xI to the k. So the equivalence between the two is actually not a trivial thing it uses the Cayley formula and Abel's identities. But nonetheless it is true that both polynomiality properties are equivalent to each other. And you have to know the second property if you want to prove the topological recursion. So this is the crucial fact about the power series that allows you to prove that the topological recursion formula works. It actually gives you a very strong constraints on the power series if you want them to be global polynomials on this curve. So I am more or less done. You see I am 15 minutes earlier than the schedule but my knowledge of the topological recursion is limited so I told you everything I know. If you have any questions you can ask them now. Thank you. Polynomiality statements for the other two examples? Yes. So for the... So for the... Wilde-Petersen volumes it's actually quite... The statement is very simple. The Wilde-Petersen volumes themselves. So this is a polynomial in L1, Ln. And then you spoil this property by taking the Laplace transform. So in the end what you get is not a polynomial but a Laurent polynomial but the polynomial property is still hidden in this. Yeah, that's a good question. No, no, no, no. So for this one it is something like x equals to sine y, I think. Yeah? y equals sine of square root of x. Okay, so if you know the answer by heart, that'd be very good. No, the parabola is for the endpoint function for mg and bar. So it's a little bit... These volumes also involve the class kappa 1 when you compute this interval. If you remove the kappa 1 you will just get the parabola in this case. And for the triangulations I think I don't know the spectral curve. Yeah, for the triangulation. So there is the matrix integral to enumerate the triangulations and then you have to compute the spectral curve of this matrix model but I don't know by heart what the result is. I have a question also. There was some activity on studying the spectral curve for RULSV. Yes. What's the state of this activity? So there is a RULSV. So there is a formula that is called RULSV that is still a conjecture. It's my conjecture from maybe 10 years ago that it's still not proved. That involves some different Herbert's numbers and called R-spin Herbert's numbers. And here you have an integral over the space of R-spin curves. Space of R-spin curves. What's the definition of the left-hand side? I mean, both are pretty long. So it would be a topic for a different talk. And what is known for R-equal 2? No. So for R-equals 1 you get back the usually L-sp formula. R-equals 1, not R-equals 2, but R-equals 1. For R-equals 2 I don't know. No. Well actually let me think. I think for R-equals 2 I can prove this formula. I actually have a method to prove this formula for any given R by a finite amount of computations. So for R-equals 2 I can prove this formula. And for any R that you give me I can compute many examples and show that it is true. But the proof is... Okay, so anyway, and for these numbers there is also a topological recursion conjecture with the spectral curve equal to x. So it will be log y minus y to the power r. Here you have... So for the ordinary here of its numbers it's log y minus y. Here it's log y minus y to the power r. And so with Sergei Shandrin and Luk Spitz we finally proved that the two conjectures are equivalent without proving either of them. So that's the state of art. And again there is the polynomiality property that in this case nobody knows how to prove. But this topological recursion means that you can compute these constructives. So now we can compute it. Yeah, everything can be computed. There is no problem. If you want to actually compute it it may be a problem, but theoretically all of this can be computed. The Hurwitz numbers are defined just combinatorially so they can be computed. This integral, you know, everything is computable. In the RELSD case you have to upgrade the topological recursion to degree r not just simple critical points. So the topological recursion works for any number of simple critical points. If you look at the formula there was a sum over critical points. I already raised it. But only simple critical points. Yeah, but this curve has only simple critical points. Yeah, there is no problem with that. It has R simple critical points. Yeah. I apologize. I would like to reduce the discussion to a low level. Can you please repeat the construction, the definition of the construction of the spectral curve which you presented for Hurwitz numbers? Yes. Yes, of course. Yes, okay. So the construction is quite simple. So you first compute h01 of x. Then you take w01 of x equal to dh01 of x. And then you decide that this will be equal to ydx. So you just take y equal to dh01 of x over dx. And this gives you a spectrum. This is the equation of the spectral curve. So it looks simple, but you should be warned that you have to choose the right coordinate x. It is not at all clear. A priority y I should take this to be equal to. So it's h0k sum of h0k over 1 over m factorial. And then there is this e to the power kx. So it's not at all clear why you should take e to the power kx rather than x to the power k, for example. And for this I have no answer. If you know the right way to choose the variable, then it's just a simple computation. If you know the first power series equation of c. And another warning, if you want to apply the same method for the vial-petterson volumes, you will fail because there is no v01. For the vial-petterson volumes, you can only define it in the stable range so you don't know what to do. One more philosophical comment on this spectral curve. The original Corvus function is defined as a formal power series next. Yes. When you define the x, when you write the equation... x to the power k here. When you write this equation of spectral curve, you have this y and still this is formal change of variables. But then you consider y as a global fine coordinate on Cp1. And the branching point is far from the origin. It corresponds to the coordinate y equal to 1. This small x coordinate does not extend to the global curve. Yeah, or capital. So you start with a power series, sum of k to the power k minus 1 over k factorial x to the k. This is a power series at the origin. So this series has a finite radius of convergence equal to 1 over e. So if you draw the graph, you get something like that. This is 1 over e, and after that it doesn't converge. And then you can extend it to... Yeah, so in this case actually, yeah. Yeah. And then you can extend it to a global curve with a global coordinate y. This is what you were saying, right? Yeah, we have this Cp1 with a fine coordinate y. So who with numbers of tenders is coefficients of certain expansion in the point at the origin. But the topological recursion place with the behavior of this function is a different point. Yeah, so all these WGNs are polynomials in the variable z that has a pole here. And the zero here has nothing. It doesn't see it as any particular value at all. But if you want to find your Herbert's numbers, you have to take the expansion at this point, yes. Yeah, okay, thank you. Conceptual question. Do you have some reasonable meaning for the... Why do you think of the Laplace transform from this kind of series extension? Probably is that related to the convergence of the asymptotic series, or is it related to the polarized summarized ability of... Why do you have to take the Laplace transform in this example? Well, let me see if... So I have some sort of an answer. I don't know if, Maxim, are you going to speak about your symplectic formalism? Yes, yeah, yeah. So Maxim has a formalism for this topological recursion that looks similar to Given-Tal's formalism for Gromov-Witten potentials. I don't know if you've heard about them. So they are similar, but they act in two... So to go from one to the other, you have to perform a Laplace transform first. So if you want to look at the two procedures as two formal, two big formal machines to compute something, one works well in one set of coordinates and the other works well in the Laplace transform of these coordinates. So this thing here is well-fit for Given-Tal's formalism, if you want to do the topological recursion, you first have to do the Laplace transform. I don't know if you are convinced by the sensor, but... Okay, thank you.