 Hello, this is the third lecture. I will continue on what I started in the last lecture. I started to show you that I recall the general context. We have a certain spectral curve. Let me call it S. To S, we associate some forms, omega g n, of z1, zn. What I was starting to show you is that there is a general formula that expresses omega g n as some integral of a moduli space m g n. With some homologiforms, let me recall that's omega g n. We have already seen that omega g n can be written as a polynomial in the Taylor coefficients of the form omega 0 2 that we called xize. On the coefficients of the polynomial, let me recall, so we had a 2 to the 3g minus 3 plus n, summed over a1, an, among branch points, among ramification points, on d1, dn, that are degrees that are positive integers, non-negative integers, they can be 0, a1, d1, an, dn, on product from i equals 1 to n of xi ai di of zi. These form basis of meromorphic forms with poles at ai of degree 2 di plus 2. Yes, yes, yes, sorry. They form a basis of a certain subspace of such forms, but which are the forms that appear as Taylor expansion coefficients of omega 0 2. So, I mean, because the topology correction takes residues, it can only extract Taylor expansion coefficients. So, everything is necessarily a polynomial of that form. Every residue computation will necessarily give that. A non-trivial property was that this polynomial is symmetric in all its, in all the pairs, ai di's. And, okay, and the theorem that I'm going to state now is the following. So, the main theorem, which is not exactly the one I cited last one. So, last one, I was, somehow, I wanted to show first the case where there is only one possible ramification point. Now, let me state the general theorem. The general theorem is written like that. Omega gn of z1 zn will be, so, we have this 2 to the 3g minus 3 plus n. We will have a sum of our graphs, g. Let me say in a moment what they are. One over the number of automorphisms of g, so, symmetry factor, times, times, we'll have a product of our edges. I'm going to write in a moment what it is. And times, a product of our vertices of an integral over some m bar gv nv. So, some, a modulate space that depends on the vertex of exponential sum over k of t hat ak, sorry, a, a index vk kappa k. So, this is a certain cohomology class times product over half edges adjacent to v of what I will call tau dh. On the product of edges, an edge is a pair of half edges. Let me call it h plus h minus, which are two half edges of the coefficient that I called b a h plus dh plus, sorry, dh plus a h minus dh minus. So, let me explain that formula and let me also say what graphs we are talking about. The graphs we are talking about, so, graphs g, they are graphs made of, so, let me give an example. They are graphs made of, so, we have, so, let me give this example, okay. So, there are a certain number of vertices, there are also some outgoing half edges, and these half edges are labeled by point d, are basically a1 d1, a2 d2, and so on, an dn. So, they are the, sorry, this is not omega gn, excuse me, this is fgn of a1 d1, an dn, that's the coefficients. So, that's the coefficients, and so, the labels of outgoing edges are precisely the variables you are, you have in your coefficients, and each vertex will be associated to a certain modular space, so, each vertex is going to carry a genus. So, those graphs will be somehow dual graphs of nodal Riemann surfaces, but so, each vertex is going to carry a genus, so, the vertex v here carries a genus, g index v, it carries a number of half edges, so, here it's 3, so, let me in, well, so, each vertex carries a genus, a number of outgoing edges, and it carries also a color, av, av, which belongs to the set of ramification points, okay, and v is always the number of adjacent half edges, adjacent to v, so, let me take the example where we take, so, in my picture, I choose 1, 5, so, this one has 5 edges, so, let's take 1, 5, and let's call the color a, this one, in my example, will have 0, 3 on color b, and this one will have 1, 3 on color c, okay, so, each, so, vertices as, so, it has gv, nv, which is the number of half edges, and a color av, that belongs to R, it's stable, we want the stability condition 2 minus 2 gv minus nv, must be negative, and, yes, and, it has half edges that carry a degree, so, h, carries a degree dh, so, for the external half edges, so, for instance, for ai, di, is such, must be such a half edge attached to a vertex of color ai, and, it must have, and, have degree di, okay, so, I'm going to sum of all possible graphs, but the degrees of the internal half edges will be summed on, so, we will, we shall make a sum of all possible degrees inside, but the degrees going outside are fixed, and also the colors, we shall make a sum of all possible colors, but here, the color is fixed. – You don't have any loop in the graph, no? – Yes, you can have loops, here there are lots of loops, yes, so, let me, let me, excuse me. – Self loop, vertex to itself. – Oh, yes, yes, it's possible, oh, it's possible, yes, yes, it's possible, okay, I did not show in that example, but, then let's take 5, yes, that's possible. – Then how do you describe this model space attached to this kind of? – No, no, no, no. – There is no problem, so, let me give, so, so first, so, let me first show what is the dual nodal surface to that graph, the dual nodal surface is something of genus 1, it has 5, 5 things attached, but remember that the edges should be interpreted as nodal points, so, there are 3 nodal points, one is, so, this one is going to be a sphere, okay, more complicated because I draw, okay, but it's a sphere, okay, and this one, as I said, as genus 1, and, so, here it's a1d1, a2d2, a3d3, in fact, in this example, a2 has to be equal to a1, because they are attached to the same piece, and this surface is an element of m3, no, m5, m53, right, yes, there are, so, the number of holes is 1, 2, 3, 4, 5, okay, and we have 3 mark points, okay, so, this is, so, this is just the general formula, let me show you how to use it in practice in a simple example, let me show you, for instance, how to compute f04 of, so, a1d1, a2d2, a3d3, a4d4, so, what are the possible graphs? Well, because of the stability conditions, so, excuse me, there is something that is missing here, you want that, of course, when you do the sum of all vertices, sum of all vertices of 2gv, of 2-2gv-nv, you want this to be equal to the total Euler characteristic, 2-2g-n. And did you explain, as a seamless, like, tau? Also, these are just the written classes, so, these are just the powers of the psi of the charm classes, tau index dh, so, that's the tau class index, the degree of corresponding half-edge. So, but, I'm going to show an example. And where did you articulate a1 equals a2 before? I mean, you said, since they come from the same... Well, since they are, I said that... Yes, I said that for external edges, they have to be linked to the color ai, so, since they are linked to the same vertex, they must have the same color. So, we shall see that in the example. So, here, basically, there will be only two possible graphs, so, basically, we are going to have only those two graphs. So, either one graph of type 04, with some color a, or two graphs of type 03, so, 03 on color a, and 03 on color b. So, here, we have 1, 2, 3, 4. And, for instance, we have 1, 2, 3, 4. But we have also the graph with, let's say, 1, 3, 2, 4. And also the graph with 1, 4, 2, 3. And be careful about symmetry factors. But, basically, these are all the possible graphs. So, for instance, for that graph, what do we have? So, first, the first power of 2 is just 2. So, let me write one half of that equals... So, in that first graph, we have to take a summation of our a. We have to take a summation of our a. There is no internal half edge, so there is no sum of our any degree. And we have... But so, for that one, we must have that all a, i. So, product from i equals 1, 2, 3 of... Sorry, 2, 4. Delta, a, i, a. So, all must have the same color. Times integral over m04 bar of product from... Sorry, exponential sum of t hat ak kappa k product from i equals 1 to 4 of tau di. What do you do with automorphism? This one has automorphism. Yes, yes. Yes, because this one... So, all the edges are labeled, the automorphism factor for that one is 1. Yes, so, yes, with labels. Yes, so, okay. You have to be careful by computing the automorphism factor, but it's usually easy. I mean, it's quite standard. Plus... So, plus, for instance, I'm going to take this graph and there will be other graphs by symmetry. For this graph, you have a sum over a and b. So, that belongs to the set of ramifications points. So, for this graph, you must have that delta a1, sorry, a, delta a2a, delta b1, sorry, delta a3b, delta a4b. Times. And you have a sum over two degrees. So, you have two half edges. You have one half edge here. Let me call that d. So, the degree d on d prime. So, we will have... Sorry. So, yes, sum over d on d prime. That go from zero in principle to infinity. And... And what do we have? And we have the product of b's. So, we have the b, a, d, a, b, d prime. Times. Integral over m03 for the first vertex. So, for the first vertex, we have an integral over m03. So, bar 3. Of exponential sum of t8k kappa k. Times. And what are the half edges attached to this one? So, we have d1, d2 on d. So, tau d1, tau d2, tau d. And times. The other vertex. M03 bar. Expansion of sum of t hat bk kappa k. Tau d3. Tau d4. Tau d prime. Sorry, that's... Right. So, in principle, this is anything in sum. But you know that all intersection numbers that don't match for which the degree, the sum of degree of the classes does not match the dimension of the module space r0. So, in fact, the sum is finite. And in fact, in m03, the only possibility, which is of dimension 0, the only possibility is to keep only classes of degree 0. So, it means that in fact d must be 0, d1 must be 0, d2 must be 0. And here also, in the sum, you keep only kappa 0. So, in practice, in that example, let me say what... So, sorry. Yeah, no, I have not finished the formula. So, this corresponds to that graph. And plus you have the other graphs by symmetry that correspond to this one and this one. Okay? But it's not very difficult. But so, in the end, it's a very small number of terms that you are going to get. Yeah, actually, I don't use what is kappa 0. Kappa 0 is basically a very large characteristic. Well, in fact, up to the sign, it's 2g minus 2 plus n instead of 2 minus 2g minus n. So, indeed, in practice, so here kappa 0 for m03 is 1 and for m04, kappa 0 is 2. So, basically, that will give sum over a. So, a product of delta ai a. i equals 1 to 4. And here, I should start with the other one. So, here, in fact, you will get m04 bar. In fact, only kappa 0 and kappa 1 will appear in the computation, but kappa 0, you can take it out. So, it's exponential, as I said, 2t hat a0 times exponential. So, it's, in principle, exponential t hat a1 kappa 1. You can start the Taylor expansion. It's 1 plus t hat a1 kappa 1 plus. And so, in principle, you would have one-half of t hat a1 to a square kappa 1 to a square and so on. But kappa 1 to a square is a degree 2 form. So, it will disappear. So, it's higher than the dimension of m04, so it will disappear from the computation. So, you can stop the Taylor expansion here. Sorry, yes, sorry, degree 4, I mean, 2 times 2. Ok. Yes, 1t hat a1 kappa 1 times product i equals 1 to 4 tau di. Ok. That's just this number. And this number is going to be 0 each time the sum of degrees does not match the dimension. So, in fact, it's only very, very few terms that will survive. Plus, so plus, as I said, the overgraphs. So, we have delta AB, sum over A and B, sum over D and D prime. And delta a1 A, delta a2 A, delta a3 B, delta a4 B. Times, so, B, A, D, B, D prime. Times, so, this integral m03 bar. But here, the dimension is such that you keep only the degree 0 term. So, let me keep t hat A0 times 1. Sorry? I think D, D prime, already equal to 0. Yes, yes, that's what I was going to say, that in fact, in the sum, you have to keep only the terms D and D prime equal to 0. So, in fact, you can remove the sum here and put this to 0 because all the other terms are vanishing. And also here, you have integral over m. Well, okay, I will also take exponential t hat B0 times. So, the notation is, well, sorry, let me write it this way. Integral over m03 bar, tau D1, tau D2, tau D, which is 0. So, which is tau 0 times m03 bar. I must have a one-half missing. So, probably the symmetry factor of this graph should be one-half. Or, it means that I have forgotten the one-half in the definition. How do you prove it? Okay. It means, okay, I see. In fact, sorry, that would be one-half per edge. So, here, there is one-half per edge. So, we have, in fact, it's there. So, here, it's already here. So, tau D3, tau D4, tau 0. And you know that this integral is just delta D1, 0, delta D2, 0, times the integral of tau 0 times tau 0 times tau 0, which is 1, times 1. And the same thing for this one. This is delta D3, 0, delta D4, 0, times 1. So, you see, in the end, we have computed everything. So, this is the formula in general. For loginus, it's quite easy to use. Okay. Let me mention that while you could think of, well, since all these graphs are, in fact, they just encode all possible nodal surfaces, you could somehow define a modular space that generalizes a little bit mgn bar, but with colors on the different pieces. With different colors on the different pieces. So, there is a kind of way of writing this formula in a more concise way, which is just somehow a notation. But imagine that you define, so mgn bar of r, the set of your ramification point, sorry. I usually write it as a power r. So, this will be the set of remain surfaces nodal stable, sigmagn, with markpoints p1, pn. And this will be stable and nodal. And there is an extra thing, which is a map quoted by isomorphisms, as usual. And so, these are n markpoints. They are labeled. So, I mean p1 is different from p2. They have an ordering. And such that f is a map from sigmagn minus nodal on markpoints 2r, is a continuous map. So, it means that, basically, it's constant on each pieces. The degrees for half edges will be included in the definition of, the sum of the degrees will be included in the definition of the co-homology class on compute on that space. So, r corresponds to the set of ramification points. So, this is a map from the surface to the set of ramification points, which just means, somehow, I put colors on the different components. Do you put topology on the space? For the moment, I did not put topology because this definition is sufficient to write computations. But it would be indeed interesting to understand the topology of that space. C'est compact. I'm not sure. How does isomorphism work on f? Well, I mean, you just pull back by the... So, if you have an analytic map from a surface to another surface, you just pull back the f. This is a question about isomorphism and cost about any mathematical structure, whatever you want. Yeah, yeah. Basically, they must have the same colors. I mean, two such things that will be isomorphic, if also they have the same colors. Okay. That's not very important. And now, I put a topology class on that. The topology class depends on my spectral curve. And it will be basically exponential of sum of k, of t hat of f, let me call that f, so which is somehow the color, kappa k, whatever that means. But it just means that on each component you are going to select the color. Sorry, I'm k. I think you don't need the mark points for... You remove the mark points for... No, I don't. Yes, in fact, yes, it's not very... Yes, it's still connected, but it's just that in the computation there is no really difference so I want to put all of them together, but it does not matter. And exponential one-half of sum of all boundary divisors, so boundary divisors, which in practice mean sum of all ways of decomposing the nodal surface into its components, and decomposing basically MgN into a product of MgVNV that's exactly doing that. So sum over ak, sorry, sum over dd prime b f d f d prime of the nodal points. So... Okay. tau d tau d prime. Okay. I'm a little bit sketchy on the notations, but basically this is going to mean exactly what is written here. Okay. Let me not spend too much time on that. It's just that it's a way of writing this formula in a way that is very similar to the Mumford formula or to the Kiodo formula. And... So in fact what this produces is a formula that looks very much like ELSV formula or... or Marinovafa formula for the topological vertex. It's going to be basically the same thing. And also another thing in that formula is that it looks very much like what appears dans l'informalisme. And in fact Don Bartin, Oranta, Chadrin and Spitz have shown how to identify this formula with Givental's formula for all comological field theories, I think. I have a question. Is this formalism in general Lagrangian subspace in some infinite dimension space? It's symmetric matrix. Here we just have the coefficient of Taylor expansion. So the t-hat ak are the coefficient of the Taylor of omega zero one and the b are the coefficients of the Taylor expansion of omega zero two. And that's fully general. No, no, no. In the usual matrix series for this b will be series in two variables but it's kind of a special type. It's come from matrix series in one. Yeah, yeah. So indeed. So here we have two series. We have a series of a t-hat. So I mean, we can make two series. So let me recall what they are and they are related to Laplace transform. So remember that the series minus sum of our k of t-hat ak. Is going to be integral of omega zero one. So it's a one form times e to the minus u x, so let's call the variable ux of z over a path over a certain which was basically x of sorry x minus one of x of a plus r plus. Basically you want x to be positive real on that path and you have to multiply by some three factors that are exponential minus ux of a. No, I prefer to write the other way around. It's ux of a varies u to the three half over I think two square root of pi. Is it what I wrote? Let me check my notes. No, the two is on the other side. Okay. So that's the definition. So basically the generating series for the coefficient t-hat k is made of Laplace transform of omega zero one and that's why when you take for instance the Lambert curve the Lambert function for so if y for instance is the Lambert function let me call it Lambert of exponential x that means that which means that exactly that it satisfy this equation so if y is given by this equation omega zero one is y dx when you compute that this is the so the Laplace transform of the Lambert function is the gamma function and which means that the coefficient t-hat k are of the Bernoulli numbers this is the steering expansion of the gamma function so these are the Bernoulli numbers and this class with the Bernoulli numbers is going to be the Hodge class. This is my fourth formula. The Hodge class is the exponential of some of Kappa classes with coefficients that are basically Bernoulli numbers This was discovered by by Mumford and there is a generalization which is Kudo formula for our spin for instance which is very similar and which would be the same thing with some I think some power white with power r or something like that it would give Kudo formula so this is the definition of those coefficients and the definition of a coefficient so some of k over b at k b a k b l sorry t-hat a a k k 1 from 1 from 0 so which means that t a 0 is somehow a global scaling factor of omega 0 1 u to the minus k v to the minus l is similarly so there will be a pre factor b of sorry omega 0 2 of z 1 z 2 times e to the minus u x of z 1 u e to the minus v x of z 2 let me erase that so it's a double integral let me call this path gamma a so it's gamma a for the first variable on gamma b for the second variable and we have a power so something similar e to the u x of a e to the v x of b divided by pi I think yes that's it so the pre factor just removes all the terms that would not have negative powers of u and v so this when you expand it at large u on large v there are exponential terms there is square terms and then 1 over u expansion 1 over v expansion ok so this is so these are the questions they just come from Laplace transform now imagine that there is a relationship between omega 0 2 and omega 0 1 so that they are not independent that's the case for give-on-tal formalism you want them to be not independent if you call that if you call this function let's say f of u and this function this function let me call it f of u v I think so basically these are going to be the give-on-tal R matrix something like that in give-on-tal formalism you like to have that f of u v is I think sorry this is f a of u this is f a b of u v I think u so what is the formula but you must have u plus v 1 minus sum over yeah so but I should have a double index somewhere ok well I don't remember but something like that f of v ok no ok I didn't prepare that part so let me forget but so there is a very simple relationship between those two functions it's extremely simple that are needed in give-on-tal formalism but what we see here is that the formula works even if you don't have that property so the formula is a little bit more general than give-on-tal formalism but let me cite that so d Don Bartin Orantin Chadrin Spitz in 2014 I think said no it's is it 2014 ok vague showed that this is equivalent to the give-on-tal formalism special case yes in this special case so relate to give-on-tal so which means that so which implies in particular that for co-homological field theories the generating functions of of well the amplitudes of co-homological field theories will satisfy the topological recursion ok question so this kind of special class of the things that have expressed on series in one variable and so what is geometrically means for spectral curve does anybody clarify this as far as I understand what they did it's not that clear yes but the spectral curve can be written explicitly yeah well in fact yeah in fact yes the spectral curve can be written explicitly and in fact instead of a spectral curve what you re-specify the Frobenius algebra and somehow the spectral curve the way to encode the Frobenius algebra but so let me mention another so before going to the next part let me mention that for instance Grommoff-Wittern theory under this formalism so one application this was my CRM ok but one application of that CRM so I showed last time but among applications of that CRM so applications so we have for instance Mirzharani so Mirzharani's recursion can be deduced from this CRM as I said Mirzharani's recursion so if you take the curve y equals sine so it was 1 over 4 pi sine 2 pi z and x equals z square and if you take omega 0 1 equals y dx and you compute la place transform you find that the times t hat k are just t hat k is just 2 pi delta k1 so the only non vanishing time is t hat a1 equals 2 pi that's just the computation of la place transform it's very easy all the b a k b l are 0 if you choose omega 0 2 equal d z1 d z2 over z1 minus z2 to the square so with that computation you get that all those numbers are 0 and just by applying the CRM you find that omega gn of z1 zn is going to be integral over mgn bar of exponential 2 pi square k1 sorry here omega well product of so sum over d1 dn product of psi i to the di and product from i equals 1 to n so I'm putting back the psi i here which are just dzi dzi over 2 di plus 1 double factorial dzi over zi to the 2 di plus 2 I think that's it and these are the Laplace transforms of the modular spaces ok which you can write also in that way I think integral over mgn bar of exponential 2 pi square k1 plus 1 half of sum of li square psi i integral from 0 to infinity of l 1 dl1 e to the minus z1 l1 ok if you do this Laplace transform this is just equal to that on times product of dzi ok ok this is a very simple computation but these are the volumes of the modular spaces while these are the hyperbolic volumes no, if you fix L yes, you're right no, I integrate over the li's so these are sorry the hyperbolic volume is that and here this is the Laplace transform of the hyperbolic volumes ok I I'm not going to enter the details but I'm just saying that a corollary of a theorem is that the hyperbolic volumes obey topological recursion so which is Mirzhani's result but of course it relies on many other things first it relies on the proof of the theorem by the way the proof of the theorem is quite easy it's just combinatorics of graphs it's just computing the residues in a combinatorial way so picking the good Taylor expansion coefficients at each step so the proof of the theorem is quite straight forward and it's just combinatorics another corollary another application so application is the ELSV formula for that you choose the curve x of z equals minus z plus log z let me check minus z plus log z y of z equals z omega 0 2 of z 1 z 2 equal d z 1 d z 2 over z 1 minus z 2 to the square on omega 0 1 equals y dx if you choose that curve so which satisfies so if you express y as a function of x it's the Lambert function so y is the Lambert function of e to the x which is equivalent to say that's the definition of the Lambert function that y e to the minus y is equal to e to the x ok so if you choose that curve which is called the Lambert curve you compute the Laplace transform as I said the Laplace transform of the Lambert function is the gamma function well here the coefficients of b are not going to be 0 because z is not the square root of x so the coefficients are not going to be 0 but they involve also somehow the gamma function and they involve also Bernoulli numbers so basically the coefficients b, ak al are going to be also Bernoulli numbers and that's exactly what you need by this formula to reconstruct the full memford formula for the Hodge class so what you end up with if you do every step in the computation but the most difficult steps in the computation are the computation of Laplace transform of those Lambert functions so but you still need some kind of origin like geometry requirements that you complete whose numbers satisfy topological so indeed there is another step I'm going to mention it but what you get when you do that to know indeed you need to know that what you are computing is the Hodge numbers but what you compute with this formula is that so what you find with this formula is that omega gn so the theorem implies that omega gn of z1 zn is going to be equal to sum over d1 no let's call me call that mu1 up to mun product from i equals 1 to n of mui to the mui plus 1 exponential mui x of zi dzi zx of zi over mui factorial times integral over mgn bar of the Hodge class let me write it this way product from i equals 1 to n of 1 ok so that's what you get when you apply the theorem it's just a short way of writing so this is just this combination of kappa classes and tau classes and this one you have to think of it as so 1 over 1 minus mu psi is by definition sum of mu to the k psi k ok so it's just that and again every term which is which has degree high degree higher than the dimension of the space is going to be zero so in fact this is a finite sum the integral so this is just a short way of writing this finite sum but this is the famous ELSV formula and so first ELSV proved that this is equal to the Harvitz numbers but one way of proving that this is the Harvitz numbers is now the omega g n's were defined by the topological recursion so they satisfy a recursion and if you see what this means for the coefficients it means that the coefficients satisfy a certain recursion relation which is called the cut and join recursion of Harvitz numbers so here you have two sides of this equality the right hand side is an integral of the Hodge class in a certain module space and the left hand side is something which was defined by a certain recursion and this recursion turns out to be the same as the recursion the cut and join equation for Harvitz numbers so it means that the left hand side equals the generating function of Harvitz numbers and then this proves that Harvitz numbers equal this integral with the Hodge class so this implies the ELSV formula you can do the same thing for the topological vertex now if you choose the curve which is the mirror of C3 you can do the same kind of computation and you will have a Laplace transform of something which will be the Laplace transform of another function which is not the Lambert function but the Laplace transform will be the instead of being the gamma function it will be the beta Euler function which is a product of three gamma functions and in the end you will get a product of three Hodge classes so another application which is extremely similar to the one for ELSV it's the Marinovafa formula for the topological vertex which means that so you get a very similar expression but here with a product of three Hodge classes and again the left hand side is something defined by recursion which means that it satisfies certain recursion formula which are also interpreted as cut and join equation and you can prove that the topological vertex equals a product of three Hodge classes like that and this is a consequence again of that theorem the spectral curve is basically e to the x plus f y plus e to the y equals 1 or something like that so which is the mirror of the manifolds so which is the mirror of the Calabi or manifolds c3 in fact the mirror is really so you see here you have two complex variables the mirror should be a three dimensional complexe manifold which you can write as a sub-manifold of c4 and the sub-manifold of c4 so you need for c4 you need two extra variables so let me call them uv so now this defines so you have four complex variables x, y, u, v sorry u times v sorry u times v so this is a manifold a sub-manifold of c4 yes this is a sub-manifold of c4 which is the Calabi or manifolds c star square product no because I put them in the exponentials yes f is the framing it's the framing factor so in fact yes so this is the frame mirror of c3 with framing f well ok to get the curve so the curve is really when this is equal to zero but the full mirror is with this term but when you just look at the term now just observe where are the branchings and it's very easy to compute it's extremely easy and there is only one ramification point and ok I don't have a formula here but it's something like y equals something like f over f plus 1 or something like that c'est un peu difficult to lografe to add to pi to pi i yes but you no I think so if you want to solve what are the zeros of dx there is only one on the curve ok we can speak of that later which means that on the base curve where x lives you can add to pi but on the curve itself you cannot because the curve is where each of the x lives ah so it's not just yeah it's not what do we ask you exactly yeah so the function whose laplace transform is the order function yes defined by yes well one way to parameterize that curve is to take as a uniformization variable can be chosen as y sorry as exponential y so you choose y as the exponential so it means that you can write x as a function of exponential y ah so yes so if you want to parameterize that curve so if you want to parameterize that curve you shall choose so you shall choose ah I think you shall choose y equals log z and x equals ah log 1 minus z plus f minus f log z I think well you should check but I think that it satisfies this equation ah nearly so which means that if you want to integrate e to the y dx e to the minus ux when you want compute that integral first of all it's easy to see that it is 1 over u by integration by part dy times e to the minus ux it's just integration by part then ah then if you want so it is so it's 1 over u dy is just dz or z and exponential minus ux is something like 1 minus z to the power minus u times z to the power f u and I think now it should be obvious that this is the Euler beta function so it's going to be something like ok maybe I'm a little bit wrong but it's something to be gamma of f u gamma of minus u over gamma of f minus 1 u or something like that maybe I made some mistake at some point ah maybe ok maybe I made some mistake at some point I was expecting f plus 1 but ok it just means that I change the sign of f ok at some point but basically you get 3 gamma functions and that's what's going to give by this formula that's going to give a product of 3 Hodge classes basically you are going to get the Hodge classes lambda of 1 lambda, well in fact it should be I think a minus 1 lambda of f lambda of f minus 1 that's the Hodge classes and it just comes from this very simple computation the Laplace transform of y dx is a product of 3 gamma functions and immediately from the theorem you see that's the class that you have to to take in your mgn is a product of 3 Hodge classes so that immediately gives the Magnove of a formula and the last application is for more general Calabia manifolds so I think it's important to mention the BKMP conjecture historic so what time is it I'd like to go towards the so the last application was the BKMP conjecture so B yes yes I'm going to say it so in fact so this was a conjecture first so this was a famous paper in 2008 but in fact M is Marignot Clem, so this is Bouchard Clem, Marignot, Paspoitis and Marignot in fact already made the conjecture at least well in 2006 he already made the conjecture but only in the case where the curve was hyperlétique because computations are easier so he checked already many examples with hyperlétical curves and then in 2008 with the other people they made the general conjecture so the conjecture is that the generating functions of Gromov-Riton even variants of Calabi of a Toric Calabi-Ao 3fold let me call it X are equal the omega-gn of a spectral curve of the spectral curve which is the mirror of X so I will call it X hat I can't state the full details of that conjecture so it was a conjecture and many people started to work on that in 2008 and the conjecture was eventually proved so now the conjecture is no more or the conjecture is proved and in fact we made the first proof only for smooth Calabi-Ao manifolds and it was a very long proof but mostly it was very long mostly because we didn't know give and tell formalism and somehow we re-derived it on the way by combinatorial methods but then there was a proof by Fang, Liu Melissa, Liu and Zhong they did the general proof which includes also orbifolds and they really used now the result of Don Barty, Laurentin, Chadrin and Spitz and they used all the power of give and tell formalism and they were able to prove this conjecture in a quite general setting in the proof it is essential but in fact we believe that it should go beyond so the conjecture could we believe that the conjecture can be extended beyond but it's proved only for Tauric and yes it's essential in the proof but is it essential in the conjecture we don't know probably not in fact Calabio maybe is not even essential ok, so I'm not going to state the CRM but I'm just saying that it is another consequence of that formula good question I'm not even sure that it's so essential in fact there are some proofs for Calabio's, sorry for grammar free time variance of P1 in this formalism so it's dimension 1, not 3 and it's also satisfy that that's all, so this has been proved and this has been proved mostly by using the CRM which I've erased but which says that well and in fact the proof works the other way around the proof says that ok let's take the omega gn that satisfy the recursion then because they have this graphical expansion in terms of graphs and all that then they satisfy given formalism and then they have to be the grammar free time variance so somehow the proof goes back it's really a b-model side proof and it's mostly combinatorial so now let's ok, it's late to make a break maybe 5 minutes ok now I already gave the definition of topology correction I gave some important properties on one, probably the most important property is that it has a lot of applications there are lots of cases where in enumerative geometry you find this formula and I already said some few properties but there is a very important property is how they can they get deformed in respect for curve how are the topology corrections of the omega gn and that will be related to plenty of very beautiful properties and in particular to the notion of integrable systems the notion of integrability and in the notion of integrability a central object is the tau function so I'm not going to give a lecture of integrability on tau function but I will just be very vague but in order to say that integrability is a tau function it has to satisfy a certain number of equations like herota equations which are basically differential equations which means that you have to deform you have to do some infinitesimal deformations or also it can be defined by other properties but in each case when you go to another curve I mean when you when you move in the space let me give an idea of what we want to do so let me call s I will write it this way is the space and I will put quotation marks because I'm not sure that it can really be called a space of all spectral curves in fact it's more a thing of which the spectral curves are the objects I'm not sure that you can really give the name space to that or moduli space of all spectral curves and I will be more precise later but it's a space of all spectral curves imagine that we have some coordinates at least locally t1, t2, t3 and so on an infinite number of coordinates typically if you restrict to a certain subfamily or a certain subspace it can be only a finite number of coordinates the dimension should be infinite and you have a very large number of coordinates and most of them there will be called the times and the idea is we would like to be able to define a function can we define a function and also I will put quotation marks here because it will not really be a function more like a local section of line bundle a function from S to C and we shall call this function so to a spectral curve we or which is the same as the set of times t1, t2 and so on we shall want to associate a function tau of S or which is the same thing tau of t1, t2 and so on we have to keep in mind that coordinates are just you are making a choice when you speak about coordinates you are choosing bases or something like that you are making a choice of coordinates but you would like better to have something which is completely intrinsically defined in terms of your spectral curve you would prefer to choose something which is coordinate independent and the idea is that we want to use topological recursion to do that we already know that topological recursion is already does already something like that which is that to a certain spectral curve it already associates the fg of S which are also what I call the omega g0 I think I define them only for g larger than 2 but I did define them the definition fg of S which was omega g0 of S by definition is 1 over 2g-2 sum of all branch all ramification points of residue at A of omega 01 sorry omega 1g times a function which I called f01 on such that df01 is omega 01 so you take an integral of omega 01 multiply it by omega 1g take the residues at all ramification points and divide by 2g-2 omega g1 you're right omega g1 it's a one form and the genus is always the same variable the first variable I mean it's always the same notation as mgn so which you can interpret it as you have this is somehow a generating function that counts surfaces of genus g with one boundary this is a disk and you are gluing the disk on the boundary to make a closed surface of genus g so this is interpreted like that, this formula and so what are the properties of the function tau ? I have not said it yet but in order to call it tau function you want it basically to satisfy the keralta equations or Sato relations or something like that you want it to satisfy certain properties and the idea is that we want to construct using tau function something that will indeed satisfy these properties so what we want to do is prove that tau function allows to build tau function in a very natural way and let me say what is the candidate so we already have some functions but these do not satisfy keralta equations why tau functions, why with other functions ? ok, you will see well it's like in physics the partition, well why was it called tau, I don't know who decided to call that tau so but this is somehow the most fundamental object in every integrable system I have no idea who chose the name tau maybe this was the Japanese Sato I don't know no, at the moment yes I didn't define anything I just say what will be my candidate the candidate that I'm going to what I would like to define is tau of S would be but you will see that there are several problems just exponential of sum over all G equals 0 to infinity of FG of S well 1st, this is an infinite sum and this doesn't make sense ok, doesn't make sense at all but well, for a moment we don't care about times we have FG is defined intrinsically for every spectral curve so FG of S is independent of any choice of times this is just a well defined so for a given spectral curve FG of S is a given complex number you can compute it for instance if you take FG of S is in fact 0 for all G larger than 1 the main problem in that formula for a moment is just that we don't know what this infinite sum mean is it convergent or something like that well, in fact it's not but we shall use the fact remember that so recall homogeneity so in fact we shall put a formal grading which is that FG was an homogeneous function of a spectral curve so remember that a spectral curve was, I'm going to come back to that later but the spectral curve of the data of a Riemann surface function X of 1 form omega 01 X of 1 form omega 02 and by definition lambda S so the rescaling of a spectral curve just means we rescale omega 01 lambda omega 01 omega 02 we keep everything else unchanged and rescaling the spectral curve just means rescaling omega 01 and FG of lambda S is lambda to the 2-2G FG of S that's a property implied kind of obvious from the formula How is F1 defending ? I did not say for the moment I did not say so that was the definition for G larger than 2 I didn't give for the moment the definition of F1 and in fact I'm not going to give it but I just ask you to believe me that it exists and I also said that in fact there is not really a definition of F0 and that will be an important point that I want to discuss in a moment how to define F0 there is a specific problem with F0 ok, so we have this property so what are we going to make we are going to so the idea is that we are going just to rescale the spectral curve and make it a formal series and so what we are going to define is in fact we shall rescale by a large parameter which we shall call H bar to the power minus 1 S will be exponential of sum from G equals 0 to infinity of H bar to the 2G minus 2 FG of S so now this is so now this means that H bar to the square log tau of H bar minus 1S is a formal series it belongs to of the formal parameter H bar so now this is well defined as formal series so the first step was to give a meaning to the infinite sum and the second step will be to study the properties of a tau in fact, since it's only the candidate let me call it Z and it will not be really our tau function in fact this is not going to be our tau function except in very few cases so what I will try to show you is that it will not exactly satisfy the heretic equations it will satisfy the heretic equations to a leading order in H bar in an obvious way but not necessarily to all the other orders of H bar and the formula can be corrected and in fact it will be multiplied by something which will be 1 plus H bar I will say later so there is a way to correct the formula to make it solution of heretic equations and another property is that there will be a modular group acting on spectral curves I am going to talk about that later and this quantity alone is not modular invariant this quantity alone is not modular invariant but multiplying by this this one is what I will call Z and the full thing is what I will call tau later but Z somehow is only the leading part of tau it's only what is called the perturbative part of tau so the perturbative part is not modular invariant and the thing that you need to make it solution of heretic equations is also exactly the same thing which you need to make it modular invariant but which is non perturbative so I will speak about that later I don't know what is the relationship between integrability and modularity but it's there I don't know why but it's there I'm not sure but at least I will do something that solves at the same time the two questions modularity and integrability on solutions of heretic equations but first for the moment let's study the property of that Z and so in order to see if it satisfies something like heretic equations so let me recall that well sorry Z of h bar minus 1s so let's also write it as a function of the times t1, t2, t3 and so on so it's a function of all our times so which just parameterize locally on its function of h bar it's a function of the times that are just local coordinates in our space and typically heretic equations heretic equations are typically non linear differential equations but the idea is that they involve some d over dtk in some way I don't want to be more explicit at the moment here at that sorry they involve doing small deformations of your curve so basically we need to understand what is the tangent space so we need to understand the tangent space and cotangent space to the space of spectral curves we need to study that also well there are other equations like SATO relations SATO equations SATO equations typically they involve shifting the times so you have Z of h bar so okay let me call that global variable t and they involve something like t plus well something that will be called x or something like that they involve shifting all the times it's just notation for a certain vector in this space of times so box will be a certain vector an infinite vector and typically it will be 1 over x 1 over x square 1 over x cube and so on it will be this vector so SATO equations are equations that tell that Z is a tau function and it involves to have to complete this kind of things so shifting all the times by certain quantities so it just to say again that we need to we need to be able to somehow to move inside the space of spectral curves so we need to either we need to study infinitesimal deformations or finite deformations of the spectral curves so it just my introduction to say that what we want to study now for the spectral curves or that's the same as saying how to compute the deformations and in particular deformations of fg so one question is to be able to compute the over dtk of fg and in more general way the over dtk of omega gl so that's what we want to compute now finite deformation actually includes this infinitesimal deformation of course on vice versa finite deformations, you just have to integrate the flow of infinitesimal deformation why is it studying too distinctive it's just because in the literature of integrable systems you have hiver basically the hereot equations oversatur equations that are used and indeed you can derive one from the other and somehow one is the infinitesimal version of the other but it's just that we want to compare to the literature and in all cases what we want to study is how to move inside the space of spectral curves and first we shall start by moving by infinitesimal deformations let me recall what is a spectral curve so this was so deformations of spectral curves so let me recall what I call the spectral curve in fact I was not completely précise last time because we have this up to isomorphism which I did not mention last time but I said that the spectral curve is the data of a curve in fact there was also a base curve omega 0 1 and omega 0 2 the base curve will be held fixed in fact almost all the time and in fact let me now let me say that the true definition will be that it will be an equivalence class but I need to say how yes I'm going to say so here the definition that I'm going to make definition so I did already already give a definition of spectral curves in my first lecture but let me be more precise this time so I give again the definition but I explain to I need to explain what means this modulo isomorphisms so here it will be a certain surface and I'm not saying remand surface I'm just saying a smooth surface well differentiable surface smooth surface this will be now a remand surface so this one has a complex structure this one has a complex structure sorry compact non compact so connecting non connected doesn't matter it can have boundaries non connected it can be just a union of disks it will be so all of them in fact will be what I will call local smooth surface in fact what I will need is really that they are defined in neighborhoods of the ramification points so here we shall have a remand surface here this is a map x is a map from sigma to sigma 0 it's a it's making a ramified cover it's defining sigma as a ramified cover of sigma 0 and in fact the pull back of the complex structure of sigma 0 defines a complex structure on sigma so it defines so it defines a complex structure on sigma what is important the subtle point that is important is that the complex structure on sigma depends on x depends on your choice of x so when you move in the space of spectral curve when you are going to move if you change x you will change the complex structure of sigma in a certain way that is given by how you change x then this is a one form this is a meromorphic one form on sigma with that complex structure meromorphic with respect to that complex structure and this one is a symmetric one it's a symmetric tensor product of one forms on sigma cross sigma with poles on the double poles on the diagonal but let's not care about that for the moment and the bracket means that it's that in fact we take the equivalence class modulo isomorphism and we say that sigma sigma 0 x omega 01 omega 02 is isomorphic to another sigma prime we shall take the same base curve x prime omega 01 prime omega 02 prime if there is a map so if and only if there is an isomorphism from sigma to sigma prime let's call it phi and so there is an on meromorphism sigma to sigma prime such that the pullback so such that basically x is the pullback by of x prime omega 01 is the pullback of omega prime 01 and omega 02 is the pullback of omega 02 prime so that's a the graph optimization what says than every defined subsec you already say about the seismomorphism the kind from a symmetric scheme built in if you haven't haven't any okay what I want to say is the following idea it's that when we're going to deform we have to take into account L'invariance modulée de l'homéonomorphisme est que nous pouvons réparamétrer l'invariance de réparamétrisation. C'est juste que le paramétre qu'on utilise sur la surface est irrélevant. C'est juste dire ça. Excuse-moi. La suivante idée que je veux vous montrer est que si vous pliez votre courbe dans l'espace X, Y, si vous remettez, si vous immersez-le dans, je dirais, l'écosystème, donc Y sera quelque chose d'homégazéro-1 divisé par l'exe. Ou vous pouvez immersez directement dans l'espace cotangent. En fait, je dois vraiment dire que c'est l'homégazéro-1. Donc ici c'est sigma0 et ici c'est l'espace cotangent de sigma0. Donc c'est une immersion dans le total cotangent bundle de sigma0. Et ça ressemble à ça. Mais maintenant, vous voulez faire une déformation. Vous voulez choisir une autre courbe. Une petite déformation. Ça ressemble à ça. Si vous voulez... Le problème est que vous ne pourrez pas compter... Si vous voulez comparer l'homégazéro-1 sur le point ici et l'homégazéro-1 sur la nouvelle courbe, comment vous comparez-vous ? Parce que ce point n'est pas sur la courbe bleue. Donc, une façon de comparer les deux est d'utiliser l'homégazéro-1 sur l'idée de ce que nous allons faire, c'est que nous allons comparer les points qui ont le même x. Nous allons comparer l'homégazéro-1 sur les points qui ont le même x. Donc une déformation, une déformation infinitésimale. Donc une déformation infinitésimale, c'est qu'on veut... Qu'on change l'homégazéro-1 sur l'homégazéro-1 sur l'homégazéro-1. C'est une fonction. L'homégazéro-1 sur l'homégazéro-2 c'est une fonction. L'homégazéro-1 sera une forme sur l'homégazéro-2 plus le delta homégazéro-2 qui sera une forme d'homégazéro-1. C'est l'idée. Nous allons décrire le temps et le espace comme un espace qui contient les fonctions sur l'homégazéro-1 sur l'homégazéro-1. Donc en fin de compte, ce que je veux dire c'est que pour étudier le espace, donc le temps et le espace, nous devons étudier le espace des formes. Mais, nous pouvons utiliser cette réparametrisation en variante. Donc, de toute façon, par la réparametrisation en variante, le delta x delta homégazéro-1 delta homégazéro-2 est en fait une réparametrisation qui est équivalente pour, en fait, vous pouvez toujours mettre delta x equals 0. Donc, cela veut dire que vous réparametriserez la curve bleue de cette manière que les points avec les mêmes paramètres correspondent au même x. Donc, c'est delta homégazéro-1 minus delta x times omega 0 1 sur dx. C'est différent de cela. Donc, c'est juste en utilisant la réparametrisation en variante. Vous pouvez réabsorber le delta x dans ces objets et le delta dans une façon similaire. Minus delta x, on va dire delta x1 pour la première variable d homégazéro-2 par dx1. Donc, respectons la première variable minus delta x respectons la seconde variable d homégazéro-2 par dx2. Ok. Ce que je veux dire c'est que nous voulons comparer l'homégazéro-1 sur cette courbe ou l'homégazéro-1 sur cette courbe avec les mêmes paramètres. Bien, dans la réparametrisation aussi, ce que je veux dire c'est que si vous voulez voir la dimension de l'espace vous pourriez penser qu'il contient une fonction, une forme sur un produit de deux formes mais en fait il contient seulement une forme sur un produit de deux formes. Il n'y a pas vraiment une fonction. La fonction peut être réabsorbe par la réparametrisation de l'environnement. Donc, localement le espace tangent donc le espace tangent sorry, de S localement il ressemble le espace de la forme méromorphique de S donc c'est le espace de la forme méromorphique de S sur le sigmar plus m1 de sigmar tangent m1 de sigmar symétrique. Donc, c'est basically ce que notre espace tangent que notre espace tangent ressemble. C'est-à-dire que la loi de l'homégazéro-1 sur le produit? Non, non, je serai plus précis, plus précis plus tard, mais juste, la idée est que choisir un vector tangent est équivalent à choisir une forme méromorphique et la forme méromorphique de S sorry, c'est un vector tangent tu es bon, en fait, je veux étudier le espace tangent donc le espace tangent est fait de 1 forme. La dernière fois que vous mentionnez la forme de S est-ce clair ou différent? Je pense, oui, je pense que vous l'avez étudier dans beaucoup de détails. Ce que je vais vous donner est en fait, de toute façon, une autre façon d'en voir cela, mais ça devrait être équivalent. Juste une question, peut-être, c'est une question locale donc c'est une question locale qui veut dire que la forme de S est un ordinateur de la forme S. Oui. Donc, il y a un ordinateur de la forme S où vous pouvez... Oui, oui, oui. Mais la forme S n'est pas différente pour le moment. Mais on peut parler de la forme Oui. Je n'ai pas donné la définition mais maintenant, c'est juste d'introduire ce que je veux faire. Maintenant, la chose que je vais utiliser, c'est quelque chose qui est très puissant, c'est la forme de la dualité de cycle. En fait, je ne veux pas parler de une forme, je veux parler de cycles. Et je vais utiliser la dualité de la forme de cycle. Donc, c'est ma section 4-2, la forme de la cycle. Donc, en fait, en fait, naturellement, notre espace tendance sera en termes de formes. Mais, je vais vous montrer que c'est beaucoup plus naturel et que c'est vrai de ce que nous savons de la recursion typologique. En fait, la bonne façon d'écrire les déformations de la recursion typologique n'est pas en termes de formes, c'est en termes de cycles. Et donc, l'idée est que nous devons pouvoir, de toute façon, identifier un espace de formes avec un espace de cycles dans un moyen que je vais donner plus de détails. Mais, bien, premièrement, nous avons un point carré, donc, cette formes cycle dualité est aussi, c'est-à-dire, le point carré dualité qui dit que si vous avez un cycle, je vais dire en un moment ce que ça veut dire, mais c'est un élément de H1 de Sigma, typiquement, un cycle integer, mais vous pouvez aussi prendre C ou Z. Pour le moment, ce n'est pas important pour ce que je vais dire, mais il y a une forme Omega, puis vous avez la suivante perring, Gamma Omega, par définition, cette perring c'est juste l'intégral de Omega dans le cycle. C'est le point carré perring, je ne sais pas, mais vous pouvez aussi juste coller l'intégral. Ok, mais ça montre que le espace de cycles est en fait un subset, de toute façon, de formes duales. Donc, c'est un cycle actuel actuel actuel actuel actuel actuel donc, vous devez le voir comme comme comme formes duales formes duales. Donc, c'est un élément de formes duales. Bien, une extension naturelle de l'espace de cycles serait de considérer tous les éléments possibles de formes duales. Mais, c'est trop grand. Nous ne voulons pas le faire. Parce vous avez beaucoup de choses dans le... il y a beaucoup de formes actuelles de formes que vous pouvez définir et ce n'est pas qu'ils n'ont pas tous les bonnes comportements. Donc, donc, ce que je veux faire maintenant c'est que nous avons un autre élément dans notre spectacle curve, c'est Omega 02. Qu'est-ce qui se passe quand vous intégrer Omega 02 sur un cycle? Donc Omega 02 a deux variables et vous pouvez intégrer un de ces. Donc si vous compute intégrer un second variable sur Gamma Omega 02 de Z1, Z2. Donc, c'est un produit transformé d'une forme. Un de ces variables est intégré. Donc, il donne un nombre complexe. En tout cas, ce que vous avez c'est un produit transformé d'une forme complexe de la forme 1. Donc, c'est une forme 1 de la forme Z1. Alors, nous allons l'aider. Omega Gamma de la forme Z1. C'est le problème que Omega 02 a des symbolisations diagonales. Excuse-moi? Vous avez des symbolisations diagonales? Oui, mais c'est juste une classe homotopique donc vous pouvez le faire légèrement pour éviter la singularité. Et Omega 02 n'a pas de résidus. Il n'y a pas de résidus. Donc, ce n'est pas important. Mais en effet, c'est ce que je vais dire plus tard. Donc, vous avez une map. Donc, c'est-à-dire que vous avez une map de l'espace de cycles de la forme. Bon, je vais le dire de cette façon. Cette map, je vais vous donner le nom B-Hut. Et pour un cycle, il sera associé avec Omega Gamma. Et je vous rappelle que Omega Gamma est... donc je vais juste le dire comme intégre de la gamme Omega 02. Donc, Omega 02 permet de définir une map de cycles de deux formes. Bien sûr, c'est un espace de dimension finite. Je veux dire, c'est un curve compact. C'est un espace de dimension finite de dimension finite de dimension finite de dimension finite de dimension finite. Et ici, c'est une dimension finite de dimension finite. Donc, c'est fort d'être surjectif. Ce n'est pas même injectif. Parce que, typiquement, il s'adresse d'une dimension finite. Donc, vous n'avez pas d'un sac. Donc, vous remettez le point? Pour le moment, je ne suis pas très précis. Donc, dans ce que je veux dire, ce que vous pouvez dire, c'est que vous pouvez toujours compter ces intérêts. Je serai plus précis. C'est pas injectif. Ce n'est pas surjectif. Il n'y a pas de 0. Il n'y a pas de 0. Il ne ressemble pas. Il ne ressemble pas. Et vos formes ne sont pas nécessairement 0 résidus. Oui. Donc, c'est pourquoi nous allons générer l'H1 au espace du dual. Même pour des disques, il y aura un espace dual qui n'est pas très dur. Donc, d'ailleurs, l'H1 est 0 pour cela. Mais, nous allons très bientôt prendre quelque chose plus grand que l'H1, qui est un espace dual d'un autre espace. Et, c'est-à-dire, ce que nous voulons considérer, je vais considérer un espace général de cycles. Donc, ma définition de un espace général peut-être être un bon nom. Donc, c'est le espace du dual avec l'image de cette map qui donne des formes méro-morphiques. Donc, c'est-à-dire, et je vais dire que l'H1 de Sigma sera l'inverse de l'H1 de Sigma. Et ce sera un subset de l'H1 de Sigma. Donc, au lieu d'intégrer un cycle, maintenant, je veux faire le pairing avec un élément dual. Mais, parfois, il y a des formes méro-morphiques, mais parfois, ce n'est pas méro-morphique. Excuse-moi. C'est cette map. Donc, maintenant, je défine un espace dual d'un espace de formes. Mais, je considère seulement ces formes méro-morphiques. Je vais juste vous montrer que ce n'est pas totalement trivial. En fait, dans mon papier, j'ai une nouvelle définition de cela. Je défine ce espace d'actuellement construire et j'ai ensuite montré que c'est équivalent à ce que j'ai défini. Donc, je pense que c'est beaucoup plus plus petit. En fait, ce n'est pas très bien défini. Alors, B-hat maintenant, donc, où est B-hat ? B-hat est la map de le M1 de Sigma dual pour quelque chose dont je n'ai pas besoin de dire ce qu'il est. Mais, mais basiquement, c'est ce qu'on intègre sur le pair de Gamma avec Omega 02. Le pair dans la seconde variable. Donc, je n'ai pas donné le nom à cet espace, mais il contient beaucoup trop de choses pour être des formes momorphiques. Je vais juste vous montrer. Il y a un sens que ce fonctionnel doit être un intérieur ou un cycle, plus quelque chose qui peut être expérimenté de nombreux points. Non. Non. Non, il devrait être non. Non. Non. Vous avez Omega 02, vous juste actez par le pair et vous voyez ce que vous avez. Ça n'a pas de sens. C'est-à-dire que pour chaque point que vous appliquez Omega 02, il y a un certain fonctionnel qui est sensorétiquement invariable de... Non. Ok. Ok. Donc, j'ai une définition qui est donc je ne veux pas aller dans les détails, mais ce que je peux vous donner c'est quelles sont les éléments de cet espace. Dans cet espace il y a d'autres cycles en H1. Mais il y a plein d'autres choses. Donc, je vais juste vous montrer une chose. C'est la courante. Qu'est-ce que c'est la courante? Mais il n'y a pas de cycles encore. Donc, je suis juste acté par une forme linéaire. Ok. Je suis juste acté par une forme linéaire. C'est bien défini. Mais en général, ce que vous avez sur la droite est pas une forme méromorphique. Il peut être très, il peut même être non-continue. Non, non, non. C'est bien défini. Si vous considérez l'élément de l'H1 après l'homotopy, il n'y a pas de fonction d'une forme méromorphique au moins il n'y a pas de résidus. Oui. Je comprends. Et c'est ce que j'ai écrit. Oui. Si vous considérez la fonction d'un espace qui n'a pas de résidus, il n'y a pas de résidus. Qu'on appelle donc, donc, ce que j'ai écrit ici n'ont pas de résidus. Il y a vraiment des curaux de Jordan. Oui. Ils sont vraiment des curaux de Jordan. Et ce que j'appelle, donc, ce que nous appelons des curaux, les curaux, c'est le data, c'est un curve, un curve de Jordan, c'est un gamma. Et avec la fonction f, donc, je m'appelle gamma f, c'est le fonction f qui est holomorphique dans un neighborhood de tubula de gamma. Donc, tu prends une fonction qui est holomorphique un petit peu autour de ta gamma. Et puis, tu peux intégrer ça. Donc, intégrer la gamme f de omega02, par définition, c'est intégrer la gamma de f times omega02. C'est la définition. Et c'est vérifié. Mais, en fait, en fait, le résultat est de la forme 1, mais la forme 1 a une discontinuity autour de la gamma. Et la discontinuity c'est juste df. C'est juste la fonction f. Donc, la discontinuity autour de la gamma c'est la fonction f. Donc, ce n'est pas bon. Donc, ces curaux ne belongent à ce espace. Ils belongent à la doule. Ils belongent à ce m1 star, mais ils ne belongent à ce m1. Parce que le résultat est de la forme 1 qui n'est pas même continue. C'est comme si tu dis qu'il y a un résilier, de toute façon, le résilier est df et ça ne marche pas. Donc, qu'est-ce que tu peux faire pour avoir des curaux qui donneront quelque chose qui serait acceptable. Donc, l'idée est que si tu veux mettre un f, il ne faut pas pouvoir passer la gamme. Donc, si tu mets la gamme à la boundary, ou si tu prends quelques points infinitésimaux de petite gamme autour d'un point, donc, une possibilité est de prendre un point p. Tu prends une très petite gamme autour d'un point, ce qui s'appelle cp. Donc, une possibilité est de choisir, donc, une possibilité est de choisir cp. Donc, ce qui signifie une très petite gamme autour d'un point p, times a function, times a function et typiquement, il faut prendre x of z minus x of p to some power. Ok. Donc, let's me call that one a of pk. So, this is the data of a very small circle around p times this function. Or, you can take also exponent of one over this thing as well. You can take plenty of things, but this one, I like it. Ok. By definition, the integral of our a pk of on let's say k positive. Integral of a pk of omega 0 2 of z 1 z 2 and what you integrate is z 2. This is by definition, this is the integral of z 2 belonging to this cp of x of z 2 minus x of p to the power k omega 0 2 of z 1 z 2. Well, if the power of k is I mean, if k is positive, then in fact, you get 0. But 0 is a morphic form. So, that's fine. So, this belongs to our space. It's 0 element here. Yes. Well, a pk is not 0, but it's in the corner of Bihat. Ah, a pk. It's in the corner, but it's a pk is not 0. And in fact, if you integrate not omega 0 2, but for instance omega 0 1, you could get something which is not 0 on a pk. So, basically, what in fact in my paper, what I construct this space by defining it as the space generated by all those things on many others. But another possibility also is to put gamma at the boundary of the surface. If you have a boundary. So, if the surface has a boundary, then you can choose gamma to be at the boundary. And in fact, what this will compute is some kind of Fourier transform at the boundary. But okay. Let me take another example. B pk, now, will be basically the same thing, but with k negative. And for that, I like to put y pi k cp times x of z minus x of p. Well, okay. Non. Let me say k positive and put a minus k. Okay. B pk. Okay. This is another example. Now, you compute the now when you compute the integral, it gives non 0. So, now, b hat of B pk. So, it's a one form. So, it's the one form that I called omega B pk of z one. Is by definition. So, the one over two i pi times the smaller integral. So, it will be one over k times residue at p. z two goes to p of B of x of z two minus x of p to the power minus k omega zero two of z one z two and it is a meromorphic one form. It is always a meromorphic one form. It belongs sorry. Sorry. Sorry. Minus k. So, it gives non 0. It gives non 0 element. So, with all this, so remember that in our space m one, now we have the usual cycles h one, we have h one and we have those elements as well and in fact, there are many others as well that I'm not going to describe but these are the central ones. And so now we have the space m one I should stop. I'm just going to give the definition of the intersection theory on that. So, so definition of so the intersection of two cycles gamma one intersection gamma two is going to be one over and it's going to be a simplectique, it's minus so it will be one over 2 pi i integral over so it will be a double integral z two belongs to gamma two z one belongs to gamma one of omega zero two minus integral over z one belongs to gamma two z two belongs to gamma one of omega zero two ok so because omega zero two has a pole at coincident points the order of computing the integrals does matter and so this is non zero in general and this defines the intersection and you can check that respected on h one this is the usual intersection. gamma one is equals to gamma one ten no suppose gamma one square self intersection ok excuse me so it's intersection on m one yes so it's intersection so intersection on m one sigma ok I will stop here but so we have a space so the idea is we have a space of cycles space of cycles m one of sigma and we have a map and we have a map b hat that sends the space of cycle into the space of meromorphic forms next time I'm going to go to build a map that goes the other way round that somehow chooses so what it says basically also is that m one of sigma is somehow m one of sigma cochanté by Kirby hat so in fact you have an exact sequence I think you have the exact sequence probably which is what is it so if you put Kirby hat I think that's an exact sequence is that right yes ok ok but ok it's just another way of saying that ok so so we but the important point is that we have this map b hat and which allows to identify somehow forms and cycles and that's what I'm going to do I'm going to describe the tangent space of a space of spectral curves not in terms of forms but in terms of cycles and let me ok I will continue next time sorry