 So, this is the second installment of the as I told you seminar, but it's my turn. And as I as we said the other time we do this movie like this that we talk at two parts four to five minutes each one is supposed to be a little bit more generally understandable, although we have to see whether this is accessible. And the other one is, you know, about current desertion. And so, so now that my talk for the part which you are supposed to understand is that it was a point of services and then I do a circle computation. So, okay, if you already we can start. So, we talked about service services. Can you see like this or does one have to lower the light or something. It is okay. Okay, so let's start. So, we look at surface as and it's the human scheme of n point s. So this will come twice the other mental substance, which will be a bonus, so a general element. And the same points on us and when some of these points come together. We have a foundation structure at that point. And so, why this space is more objective of dimension times. So, why should one care about it. One thing by looking at the sheets of the substance when it is an example of the model space of sheaths on the surface and somehow simplest such example what can think of. With this becomes a model case for anything one might want to study about that one last piece of sheaths. And somehow, all the theorems on my phone are the simplest version. In that case, then if I want to look at my life is for instance of higher than sheaths. So the idea sheet of two of them and some things that somehow the building blocks and so on and somehow I understand other model spaces, terms of them. Then, another thing is that it's an example of higher dimensional varieties. So, for instance, if the surface is a key to the surface then it's an is the hyperbola code of your history. You're just going off it's infected. And these are rather such the right it's a rather rare and so we want to make ways of keeping them this. And it's also a very nice chapter which one can use to study the space for you. And then, naturally, people teams of points have a lot of situations. Somehow, we can use them to count points in special configurations but also, and then indirectly curves and other things. So let's know one more thing. So there's some kind of interactive structure, you know, all the people teams together, such as a kind of. So, if you have a zero dimensional sub scheme on and into a point which guys outside the zero dimensional sub scheme to take the union, this is a sub scheme of things and that's why. So that means we have a rational map from S type of the human and points to the human and that's one point that's taking you. And this gives us some kind of interactive structure how one can go from one here with the next and somehow understand them because we need it. So, for instance, from this but also other things, if I want to compute some barriers of the modernized phases of the book, then it's natural to make a general function cell, in terms of this here. And for instance, this because of structure should allow you to find time to hope to have a nice generation. And you could have to open up some other cultures, which put together whatever or the homologies of the spaces. So, let's see. Now, maybe I can go back to get to the mission, but they look more fun. So we have a set of sub schemes degree S that I said the general point, the set of n is important as it's most productive damage. And it's related. Well, there are simple ways how you can prioritize sets of any points, which possibly come together you can just take. Endpoints count the multiplicity. This is what this metric power does just to take the entire part of the surface you divide by the extra Smith's room. This will parameterize. As I said, and points with what this you can also write this as a zero cycle. So some and I XI but the XI artistic points and as the sum of your eyes and you have a channel, which relates to it was the most method how it just says a sub scheme and support with multiplicities. So that means point, which lies in the support of the sub scheme you associate the dimension as a vector space of the local ring of the subsequent. Some of these. This map turns out to be a resolution of some of that. So it's easy to see that she puts the it was the symmetric power, the second symmetric power would be single out on the diagonal and then in this case, this is a single resource, and it has a special property that if you take another class which was mentioned in previous lecture when we're talking about this patient of time dimension varieties, and which plays a big role for the classification. This behaves very well with respect to this so the canonical class of the people seems to pull back that symmetric power. And in fact, this implies that the rational properties of the human scheme of the very close to that of us. This is where the human scheme is, and a codile dimension, which is how complicated the manifold is from the rational point of view. It's just of the human skills. Oh, so now it can give some just the most of it examples. So for instance, once they have a team of zero points. So, zero points is just the empty set. So, doesn't mean the human scheme is zero is empty, but it consists of one point at this point parameterized games. They, they will see one point S, because obviously, as parameterized as the part of the best. And then we look at the two points. Now, is your dimensional sub scheme of things. It's either two distinct points. So it is a, it's one point with a one reduced structure of lanes to, you can easily see that this corresponds to a tension direction for the surface at that point. And so, you get this, that these are given by a protective part of the surface. And if you work this out, you can see that the second thing was being is just you take the product itself, you blow it up along the diagonal, and then you take the portion by the action by this by you know, commuting the two factors as well as to which So, this is, this is. So, So, and it's also, it's a modular space, a fine modelized space or for one parameter space. You see the mental substance. So, and we have the universal family. What do you mean by changing the bank. Before or after. Well, one can do that equivalent. So you can either first. If you change, if you can exchange the fact that the action leaves to the blow up. And you can, but you can also say you take the portion by, by this action, take the single thing if you blow up the diagonal and that you also get. So it doesn't matter which order. So, so you can just say, this is just an incident for the point between points and sub schemes. And this really is a universal sub scheme to get any family of substance in the surface, which is a problem. And then this corresponds. There was some basis course balance for Matt from the base, but they have a scheme, such that the pullback of the universal family via this map will be the family of substance that we're getting. And this is kind of that we may obviously that the fiber for a point which corresponds to a sub scheme is that substance. So we can use this name to find logical she's, which are kind of basically for the funders, which are bundles which are from users for essentially any application of things. So if we're given a vector value on the surface, see a frame s effect, are we can associate with the vector value of rain. And all the different sort of endpoints. So this is done by first pulling back from the surface to the universal family and then pushing down to the other team that means we take a zero of this final. So the fiber of the way. So the fiber of the, this is a logical sheet over the point was when the sub scheme is just somehow a zero of the spectrum of the state to the south. And so in particular in this case, for the trigger bundle is just that, which is the same as taking the structure sheet of the sub scheme, viewed as a vector. Okay. So in particular, we also can take the determinant of such bundle. And this will be a line battle on the labor scheme at least the general funds generated the whole of the car. So we have all the line models. I want to give some results about the most of them are kind of ancient but you know it's just an overview of what to use or what one can do with them so I started my first encounter with them. Which, so these people schemes, they were schemes, as I said, for different and so close related one can expect some life generating functions for the biological that, for instance, we can look at the better numbers. The second number of images of the module groups or the module groups. Some of the backing up. Some of the backing up on the same message. Okay, all of these are important to mine as well. And then we'll be happy. Okay, that's true. I got zero to two times conditions. Well, okay, that's true. I mean, if you have contacts in the dimension of variety. Yeah, yeah, yeah, yeah, that's, that's a bit. I mean, I would say at this moment that they have taken the dimension but that would be cheating because they certainly move them. Okay, so then one gets this somewhat. Generating function so we have the formula for all the better numbers of all the other schemes as this infinite part. So you can see into the product we get to the better numbers of the surface. So this one plus the minus one. And you can see the side function. So we can see if you think so, so can see if you think for instance, the numbers of the of the evil scheme only depend on the better numbers. So this, you know, they're for long years, but you know, but the surface if you apply this out. In the same way you have to follow up on a follow up, which happens by putting the minds on it. But you can also see that you have this incredibly nice structure which obviously in the form of the somehow should indicate that there's something more going on than just to have a nice generating function but there will be some extra structure which type all the parts of all the other things together. And this is in detail. The case. And so, but yes, that's a long one. Yeah, I mean this is very. I can't see it. Yeah. Yeah. I mean, you can also have the size afterwards, it's not like, no, no, no, I don't understand why I hear it. Okay, so that would be a little bit hard because later there will be very many formulas and I think in the second half so if you can copy those you are okay. Okay, so let's see. So we put all these. Together. So, I think for FN, to be the model of the end table scheme, take the data, some of them. And then, like I did 10 years ago that this space F is the, is it your representation of the high demand. Each model on the module of s. So it's possible in pop states representation. And so, what is this, I think that algebra. So it's, it's generated by the, by some operators p and alpha, and the nature of the class. And we have positive and before the creation operators the one with negative and the annihilation operators. And we have this. So most of the times operators commute. When N is equal to minus M. And then what we get as a commutator is to speak this as much. So this is the evaluation of the product of alpha better on the fundamental part of s. So this is this day. And now, each x on it, mainly, I'm not mistaken. So, if I, if I, you apply PM of alpha better. A better life in the homology, and then I get the class and the end of the end. So this is a star. So just the idea is if you, you can think of representing a common class by submit form. And then what this would mean is that if the end, if this end is positive, and this is the class of, we have a certain supplementary code corresponds to A, then we add to, we take classes, we add to the classes, which we had before. The convergent class in the HIPAA scheme corresponding to, you know, schemes of length N, which are supported on the supplementary code. So you take, and the other one is the P minus M is essentially the joint operating. Well, with respect to the intersection there. Okay, so. So, the fact that this is a useful representation, representation means that if we apply the creation operators like this, so for n bigger than zero to the element one, which generates the homology of the zero. We get all the homologies of all the. If one, one can work out from this was the dimension of the code should be, and it is the common dimension that I gave before. I mean, I haven't really said all the information for them. So that's fine. And then one thing on that for instance, and then show some time ago that, you know, when you have the high level of algebra, then in a sense, if there's a canonical way how to get the B or so algebra out of that, in terms of the high level operators. I mean, showed that this year so as well is related to the restructure of the four more to be able to support so that you can use it to study the restructure and so to do computations with it and understand the structure and as a book of paper where they somehow upgrade this to also have some vertex operators. I'm not going to say what that is. And of course, most of what I don't know precisely but anyways, you know, things related points to tension founder can be expressed in terms of the Hilbert scheme can be expressed in terms of vertex of the strategy, you know, one step out of the publication. Okay. So this was this. Now, I want to really talk about this particular. So that thing is not so useful for the is that I'm right there up to three hours or is it not 420. Okay, so now we can talk about one last piece of sheets. And so, I will just review this is a certain interpretation. The space which formifies this will be a sheet so yes, it takes the medical medical variants like the test chair class. But this is such a space. So that is not an algebraic variety. But we have to put the condition, we want the sheets to be the same thing that basically means that the subsheets. You think of the bundles the sub-bundles shouldn't be too big. They shouldn't have too many sections. And so, practically this means that you see the same statement with respect to the end of the age that if we take the sections, a zero of F and the H to the end by the rank of F for in the sub sheet. This is smaller equal to the same things for P, as long as N is large enough. So as we twist with power of the input class, the sections of sub sheets go through faster than those of the sheet itself. And by GAT, there exists a more like a space of same standard Korean sheets with given rank and given turn classes, and our pivoting points can be viewed as a more like space of rank, while sheets with first turn class equal to zero, and the second turn class equal to N. The singularities of the sheet will correspond to the second turn class by just sending C to its ideal sheet. And one can show that every sheet of rank one with these turn classes must be the ideal sheet of a zero differential sub-sheet. And so this is, this parameter is all this, this is the model. Now, I want to just say a few things. So, for instance, set a modelized space. So, a rank R1, 2 will depend on the input class H. There is a system of walls and chambers, so when the rank, the rank is equal to 2, and in the second homology with R coefficients, there are some walls, which are hyperplanes. The autogonal hyperplanes to certain classes in the second homology with Z coefficients, so autogonal will be affected in the section form. And then the modelized space will be constant in the chambers and the chambers will not cross the walls. And what happens if I cross such a wall is that if you move from each class, such as the size and the size bigger than you, which minus that has happened, which minus is smaller than zero, what you do is you take a line of extensions like this, by the sense of a equals to e equals to a w depends on b equals to zero, or some differential sub-sheet, that shed the first chunk of b minus a. This c is replaced by extensions the other way around. So this would be, and so this is basically, it's very related to what was mentioned in the talk last week. So this is the flip or flop of the modelized space, the modelized space, so a certain vibrational transmission of the modelized space. And so obviously there's more complicated things happening at a higher rank and there's large generalizations in the meantime, more cross-importions, but it always can be somehow related to some extent. Okay. So, more generally, I have mentioned this in the previous lectures in the last year, there's multi-tookies formula, which allows us to compute intersection numbers on LF spaces of rank R sheeps on S, in terms of intersection numbers on products, on products of R sheeps of points. There's a way how you can take any formula, I mean any expression for all the classes on this modelized space. And the evaluation of that on this fundamental class would be equal to a sum of a certain n1 for nr of certain other commodity classes which you can explicitly determine in terms of that, I have written down some multiple numbers, which is pretty terrible, which you instead have to go across the pillars. So this means basically any computation on the modelized space of sheeps can be done on these two points. So that's also nice. Okay, so this was as much as I wanted to say about this is just an overview about things. And so then I want to briefly talk about hypercalia reports. So, good. So, it acts as a small project variety of dimension n, it's called hypercalia or very useful in terms of its in tactic, if there exists a unique and we're not a general form of form. So one of the two points so that if I take the the end batch power of it, this comes a trigger but it manages nowhere. Okay. So, for instance, if the measure of x is two, then this means x is a KP service. For instance, if I know a quartet in three. And, and so for all time. I mean, there was even once a theorem that that these are all. These are all hypercalia meaningful so was a theorem. And then, however, it is not true. And we will show that. But anyway, so these hypercalia varieties have a very rich structure. So one thing when said they have is this believable model of the graphic form on the second formology. So this is, this is a product of form on the second model which creates in many ways, like the intersection form on the second module of service. Somehow, this hypercalia varieties have a little bit of a surface paper. And it contains it somehow reports many geometric properties of our hypercalia. So, if ssqp service that is in his hypercalia. More than I showed the ethic is the service and we take any modern life space of sheets where stable is equal to single stable, which doesn't have to be singular. And then this modern life space must be hypercalia. And then, and then, and then, and then, and then, and then, and then, and then, and then, and then, many papers for her with the few others which shows that, if we have such a modern life space, which is more singular of dimension when then it must be have the information equivalent to the helix scheme of n points. So these are all these monolith bases on the D-services, actually up to the information of particular morphings are just given to points. There are a few more examples of hyper-killer varieties even up to the information, I mean there are other constructions which are different, but there are other families, but all of them can be related to monolith bases of sheets on D-services or D-services. Or in particular can be related to given schemes of points on D-services or monolith base of sheets. And if an independent court case is one of the original points, there might be more examples. It's a bit, there's no complete classification, but for the moment. Anyway, this fact that if this is the monolith base and the information equivalent to the helix scheme can be used in many cases, if you want to prove something about monolith base of sheets on D-services, it's often essentially not good for helix schemes of course, then you finish by the information. Okay, so that's a very useful proposition. Okay, then I come to one other kind of ancient interpretation of this, I mean this is a good story. So, if S is a KG surface, and it's a line of under on S, where the seventh sector of the first two in class is 2G minus 2, then you expect that the number of rational curves in the complete linear system L, so which the number of rational curves, which are zeros of sections of L, that number should be finite. The call is number NG, it actually turns out to only depend on G, if you can't guess what this is, and it also conducted a long time ago that if you take a general equation function of this number, this product can be given 0 minus 1 minus x to the end, minus 24. And now, the Euler number of the Hilbert scheme is 24, so this is the generating function for the Euler number of the Hilbert's constant, so this is the remarkable, and the part is actually rules this conjecture relating the pressure of half of the curves to Hilbert's constant, which somehow shows that we can count the curves in terms of the Euler numbers of the so-called compactified covariance, and these compactified covariance, the space of compactified covariance, compactified covariance can be this deformation because we are on a hypercalametic board to a Hilbert scheme of points on the surface, and so you get this. And then, some time ago, I discussed this also, and many years ago, I generalized this conjecture to a conjecture of algebraic surfaces, and this conjecture has, in the meantime, been proven many times, and at least most of the proofs are in terms of Hilbert's constant points. Okay, so this was that. And now, how much time, so actually, faster than that. I kind of left the difficult part to the end, and then hopefully saved by the goal. Anyway, so now I want to, in the last bit, I want to, I want to maybe get a bit more technical, and give you two useful tools for working with the Hilbert's constant. So the first one is what we have called, for more reason, in variance, or something like that. So we have this story, we have this universal sub scheme, ZNS in S times SN, and we know there is this special map from S times SN plus 1, which is the final outside universal sub scheme. And so one can indeed resolve the eternity of this map by blowing up universal sub scheme. In fact, if you blow up universal sub scheme, you get an incidence correspondence, you get if you blow up inside the product universal sub scheme, you get the incidence correspondence of length n and plus 1. And so this will allow us to compute intersection numbers of SN, or SN, because, so you can somehow, you can understand what happens if you apply this law from SN plus 1. If you have a class called the Hilbert's scheme of n plus 1 points, you pull it back to this thing, and you push it forward to this, and you can understand what you're doing when you do this to a certain extent. And so, and so you can build up a recursion of computations. And so we did this, and what you find, for instance, where you work with the spotological sheets that you have to call, we have the following statement, I don't know where I put the sorry one. So we see them again, we have the spotological sheet that we have to call, so she pulled back the bundle, then we have the following statement. So that's just an intrusive summation. If I have polynomial, invariables d i and e j, and then I say, I take a code key of SNB and replace the d i by the general classes of this chronological sheet, and the e j by the general class of the tension problem of SN, the general class of SN. So this whole thing, if I apply this to the polynomial, this would be a polynomial of the Hilbert's scheme, on the Hilbert's scheme of n points on x. Now, we want, we want that somehow something would be nice if it's generating functions. When we do this, so we note the polynomial, if you assume that two disjoint surfaces as one as two. So if you have n points on two disjoint surfaces, what is that? It's n one point on one surfaces and two on the other, such that n one plus n two is equal to n, and all possibilities. And the same is true of the Poisson's scheme. So therefore we get the Hilbert's scheme of n points on this disjoint union. Now, now assume we have kind of polynomials like this for every n bigger than zero. And we assume that whenever n is the sum of n one percent two, if we take this polynomial here, so this is the class of the Hilbert's scheme of, so s is still supposed, s is now supposed to be the disjoint union of the polynomial on s two. So we take this polynomial here, restrict it to this part of it, this component, then it's just the product of the corresponding things on the two vectors. Assume we have that, then it follows from the recursion that the generated function of these polynomials evaluated on the fundamental class of the Hilbert's scheme is given as a product of five universal powers. So we have five universal power series, a zero four three four, in one variable, depending these powers used to take only on the rank of v, and the other American formation that we have here, which is the second standard class of v, the first standard class of v squared, c one of v times a s, a s squared, and this are just taken as powers of v. So this, this part of the recursion by some, I mean, you do the recursion with some argument, and then so that's in a nutshell what one gets. Okay, so this is a, well, start, just don't go away. Okay. Anyway, so this is the, this is the state. Okay, so so basically, as a first step, if you have such a polynomial, you evaluate it on the Hilbert's scheme, it's just the polynomial in the channel numbers for the surface and the boundary, and then on the surface. And then, you know, this condition that we put here ensures that it behaves nicely with this generating number. But there's no state about how it depends on power. So now there are new powers. Yeah, so what one can prove is that, so we're not going to lift it more because actually, so the first step is that if you have a, if you have such a polynomial, if you evaluate it on the Hilbert's scheme, you get a polynomial in the corresponding channel numbers on the surface and the rank. So it's also a polynomial in the rank. So in that sense, it somehow depends in a certain way on the rank. So the coefficients here, if I take the coefficient of this of x to the something, this is a polynomial of degree at most something in the r. Okay, so in that sense, it doesn't depend arbitrary, it depends polynomial. Okay, so that was that. And now I wanted to end up talking about the other thing which I will use in the second part of the talk, which is localization. So that could end up with this. So the back is a small projected variety with an actual, say, for completeness of t, which c star has to start, so some porous c to the n, c star to the n. But I choose this because I don't use the data. And we assume the x and that's 526 points. Then we want to use this to compute intersection numbers corresponding to a grand factor on x. So assume he is an equivalent vector bundle of red r on x. So that means that the action of the, of the stores, leads to x compatible. So then, because of this, if I think that if it's an equivalent of 100 like that, if I have a fixed point, the fiber or the fixed point is the vector space is an action of the torus. So it's a representation of this talk. So, and it's known that if you have an action of such a torus, or the vector space, the vector space must have a basis of eigenvectors or common eigenvectors for the whole thought. So at each fixed point, the fiber has a basis of eigenvectors for the tx. So we can write the i as the sum like this, k equals 1 to r, c v i. And the i values are powers, if we apply t1 to 2 in the torus, they are powers, they are monomials in t1 to 2. So we get, and then we can define what I could solve there. But something from e of pi is the sum of c v i k. It's okay. Well, c is, oh yeah, so it's an i, so i equals, so a is equal to i. No, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, no. So yeah, i's v i k, yeah, yeah, yeah. Yeah, yeah, because there's a mistrint. So it somehow, in t, in particular, now they, I somehow, I somehow have, I don't, and this is also v i k, obviously. Okay, so then, I can go, at some point, I could call the equivalent of the term, plus of the pi bar of the vector 1 left point. This is the sum of the corresponding term, equivalent of the term classes. This is just the product, i equals 1 to r, so the i's should be maybe the k, k equals 1 to r, 1 plus nk, epsilon i plus nk, epsilon 2, where n1 and I mean v smash, this is power. There's a polynomial in epsilon 1, epsilon 2. So the equivalent of the point is the polynomial in, in this case, very good epsilon 1, epsilon 2, but this is this equivalent of what? And so the, the i term class of this, that is, I move the part of v i in epsilon 1, epsilon 2 of this expression. So all the, the term part. So that meant, so to speak, for every point, say what the term class is as a polynomial in this and now, assume we have any polynomial in the term classes of e, okay, then the what does the formula, the following or the different composition formula, that's the problem. If I want to evaluate this on x, so I take this polynomial in term classes, I integrate over the fundamental class of x, this will be another, I can do this as follows. I take the same polynomial in these equivalent term classes at all the points, I divide it by the corresponding thing for the top term class of the tension part, for the Euler class of the tension part at every point. There's some this expression over all the fixed points. Now, if you think of it, this is an incredible rest. So at each point, at, you know, at each fixed point, you have a rational function in epsilon 1, epsilon 2, no polynomial epsilon 1, epsilon 2 divided by another polynomial, and you solve it over the fixed points. And then I claim one should take the, put epsilon 1, epsilon 2, 2 to 0 in this, this equal or seems to be nonsense because there's no reason why I could be able to do this. But the theorem guarantees that this expression action is a polynomial in epsilon 1, epsilon 2, the denominator has to cancel by the theorem, and then we can put epsilon 1 and epsilon 2 equal to 0 and we get a loss. So this is the boundary formula. And, you know, and, you know, this is even useful, for instance, when the answer is 0, because certain expressions have to be 0. Okay, so this is, I don't know what that happened. This is the boundary formula, I don't know what that is. If you are now, one can try to apply this in the case of the Kierburski police. Am I allowed to change the page or? So, so let's possibility assume that our surface is P2. Then the policy starts and she starts with X on it, or P2 in the usual way, by experiencing the coordinates like this. And then the X points will be, let's say, 1, 0, 0, 0, 1, 0, like this, you could be a big point that should have a double point between them, moving the coordinates. And we have to apply, so, the local equivalent coordinates near points to the point P0 with the X, and the action of these coordinates, they are going to be like this, and we can compute what the action of the local coordinates near P1, P2, so obviously this is 7. Exercise. Now, a sub scheme in the river scheme of N points is T invariant. If that's T invariant under connection, obviously its support must be T invariant, so it must consist of a subset of these three fixed points. And so, C will be usually of three sub schemes which supported that one of the fixed points. And so this allows us to reduce to the case that the support of our sub scheme is just one fixed point. So, we have the product of this, which we then assume it's P0. Then, it's easy to see the sub scheme will be T invariant, even only if its idea in say, case C, say CXY, is generated by monomics. Now, you can see that, you know, otherwise, when you apply the action, the idea will move. So, because of this, you can write your idea like this. You can write generate, so that you like this is X to the N0, Y, X to N1, that's all. Write to the R X to the NR, where the, so what's, so to speak, not in the idea is what the O of the sub scheme is. And the dimension of the O of the sub scheme will be the sum of P and I. So, the N0 and to the NR is a partition of X, because obviously, if these are the generators of the idea, these numbers N, I must go down. So, therefore, we see that the fixed point of the action on 2 to the N are in bijection with triplets of partitions. And each of these partitions, the partition of some numbers, so that these three numbers add up there. Okay, now my time does not come to the general thing. So, now, for instance, we want to do this for these top logical boundaries. And for the tangential boundary, we have to know what the action is. So, I just really, really, really want to get out of this. So, it's if we take O and this vector boundary, so correspond to the trigger boundary, it's a background for N, which by that, is just the structure sheet of the sub scheme. And if we have a union, if there is a union of these with the partitions of the idea, the O of this is just the direct sum. So, the total equation is just the product of them. So, let's just look at an example and see that that is the idea generated by this. And the part of this function, so it has an equilibrium basis, one x, x square, and y. And we know how T1 and T2 act on that. So, it affects on x by y by T1 and on y by T2. And so, we get the eigenvalues of the action are these. And so, that means that the eigenvector, the eigenvalues, not eigenvectors, eigenvalues are these T1 to the C of S, T2 to the R of S, where C of S is the row in this thing, starting with zero in this diagram. And R of S is let me see. So, this is the, so C of S is the column, zero, one, two. And R of S is zero, zero, one. And so, this is mean. And so, in this particular case, what we get for the total equation count is this. So, we place the T to the i by i times x to the nine. And for the tangent panel, it's more complicated. I cannot explain how you get this, but what you find is the column for the tangent panel, you get the column eigenvectors for this, the eigenvalues. This is T1 to the minus the arm length of S plus one, T to the vector and here in the other row, where the arm length is, if you are in this diagram, this is how many are further to this side of my thing, which is the arm length of one, of this one here, it's two, of the x, it's one, it's all, and the left thing is how many are below. And so, we get this formula. And so, if one works this out, you get that the total eigenvalue of this thing is this. And so, obviously, one can kind of now compute with this. We just sum up all the fixed points. We do this combinatorial, we have this combinatorial data of all the partitions, we sum it up and we get some expression and that's our answer to any question you might want to ask. Okay, so about the section. So maybe, so this is the end of this half. I went maybe two minutes over, but they are their questions. And I mean, the same as before, this A0 to A4, they're completely universal, few things are. So, can you give us the first two coefficients known as polynomials in R? Are they completely known if R is smaller or what is now? So, you see, they are completely universal. So, I forgot something. No, they're not universal. They depend on R. The universal means, no, no, no, no, no, no, no, no, but I forgot something because here I have the polynomial p. Obviously, it depends also on p and R. But some of this, even some of this, no, no, no, but I made a mistake. The universal means it only depends on, it only depends, so here we have this p, so p was any polynomial. We have a set of polynomials. Obviously, it cannot be that it only depends on R. So that's what my mistake. So it depends on p and R. So the polynomials, you only need a value for n equals 1. This is the suggestion point. Wow. So it only, so it should depend on very little. Yeah, it depends on very little, but it's a, we would see in the second half, we see one example of this, or two, actually three examples, and then one can see what these powers really are. But in there, it's not, yeah, so I really have to work a little bit on the formulation of that. Okay. So if there are no further questions, maybe, you know, so the break is all for the purpose that whoever wants to run away can run away. So you have your chance to run away, with both my eyes, and then, and who doesn't. So you, if you don't take your chance, I will start now with the second half, or if somebody wants to go, I mean, whatever, there's some I think the break should be, that's it. The break should be a break, I think. So these two minutes. Okay, so let's, yeah, 200 break. Okay. But I don't want the long break, but then we'll go forever and ever. So, okay, let's maybe turn it on for you, but that is, and then we can already announce the, you can check in with, I'm not sure, but you know, this was the form of the week, but I think it's more interesting. And now comes the, like he won't support the whole college. Now, he's right, I think it's true, I only need to know what P1 is, the one plus P1, no, no, no, no, I mean, especially, so I think the room looks like too, but I should work it out. This was not how it was called, you know, paper head. If you know P1, the first point, no, all right, and R, then this will turn it, obviously not P0, so I think that. It's a higher run version of the box, right, it's like four range sheets on the tool. It was bought this thing here, but the range sheets, it's not seen here, and that makes sense. But, I mean, there are, there are certainly, I mean, I'm not actually a decisive thing, but there are certain papers for such a particular marriage. Are there any recommendations from DOA in terms of structure? Should we model DOA, there are 100 records that it was, oh, there are some of the chronologies. And I think, I mean, in principle, the sort of work of joint savings, some of that, but in there, the work that they did, and the work that they did, the steps, and so on, and so on, being sheets, but then they saw, so here's the work that we have for, I think, all right, I'm going to say, okay, so let's, let's start again, so by the way, the answer to your question is, this thing depends only on the first point on your, as you said, and our, okay, so now I talk, this is the second part, the joint work, so we talk about the linda and sega formula for evidence gain from points, that's something I would have discussed many times over the years been done, but now the solution is going to come from Reddit, who also wants to go, sometimes go, so let me see what I suit, yeah, I'm going to suit that, okay, so, so let me start again, so we have the evidence points, you know what this is, it's going to be a linda and sega formula, then you're going to copy the same thing, and so we have that, it's again related to the sledding power line, but it's not knowable, and it's a little bit of a notion of the sledding power, and then we have this universal sub scheme, which is in the product of the surface and the payload of the points, and with this two projects, will be achieved, so this is all on stuff, so yeah, I can put it again, so then we have to recall all the methodological sheets, which will be the vector bundles of the analysis, we pull back from the surface of the universal sub scheme, we push down, and we get this universal, it's called oxygen sheet, it's fiber, that's the issue of these, this construction causes, tends to be a group of vector bundles, so the different vector bundles, like this, we push to this, and so therefore we also have in some sense, it's called the construction of sheets, so okay, and you know that the line bundles, we get all the line bundles on the head, but it's in terms of these, so this we have people. Now we want to understand two things, we want to know how many sections do the line bundles have, or down this case what, so all of what we call a characteristic, so the alternating sum of the commode, the length of the commode groups of these degenerate bundles, and another thing is that we might be interested in, we can take one of these technological bundles, we take its top chunk, the chunk has corresponding to the dimensional table scheme, we integrate only to get a number, we want to know what these numbers are, and we can do this also when the length of the bundle is negative, so the top chunk does, my NCN is the saved class of the end, so therefore just that anyone was interested more in the saved formula, and so I call this thing the saved formula, although I always talk in terms of the top chunk class, but there's no difference because we can always, as we can work with different classes, we can always assume the next nature we want, so okay, so this is this formula we want to, okay, so first why should anybody care, but okay there might be many reasons, or whatever, like the amount of recordings there, but they know somehow, one thing that one could say is why are the, for the vanilla formula, why are the correspondence which says you see to the ideal sheet, we have, as you already said, that the Hilbert scheme, and once is a monolith space for brain found sheets on the surface, and therefore if we take the vanilla formula for the Hilbert scheme, this is the rank of one case of what one could call a general vanilla formula for surfaces, so the vanilla formula for surfaces would be such a formula for all monolith space of sheets on the surface, and as we announced there was the corresponding analog for curves, for monolith space of vector boundaries on curves, is the celebrated vanilla formula which was first projected in physics, and then it could prove that also there was something about it, so this is an important thing, so we want to generalize this from curves to surfaces, and this would somehow be the first standard, okay, then what is, why do we want to care about the saving formula, I mean there might be, again the main reasons may be for this there, but the, it has for instance an innovative meaning, it counts, it counts contributions of points in special positions, okay, so if you take a general class of something, it means it's, that's also in so many sections, it holds less conditions, and so on, so in this case for instance if we take a surface in Pn plus minus 2, and we take the hyperplane bundle on this, and we take its restriction stress, then the integral over Sn of the top-side class of this corresponding, not logical bundle, which would be the number, I mean it ages at the number, it would be the number of n minus 2 planes, so n minus 2 dimensional subprojective spaces of Pn minus 2, which are n sequence s, so this means the number of considerations of n points, which lie, which stand only something of dimension n minus 2, and not n minus 1, as we normally should, okay, so this would be an example, but there you could write other formulas for other things, and we could get this, okay, so this would be one thing I work in, so therefore we want to do a formula for this generating function, I mean it's a natural thing that, you know, this depends on the n-tiverse, it depends on n, you know there's inductive structures, so it's natural to make the right amount of generating, so we have the generating function for the identity formula, and we have the generating function for the sigma function, and we would like to know what they are, so the first step is that we can use this component, which I just told you about, we hear the other situation where this theorem that I had just a few minutes ago applies, so we have that this generating function for the identity formula is a product of our universal power series, which I choose, and I could say it's c1 of v square and whatever, but I could choose them differently, but it's just, it depends that all the numbers are there, so I take here the whole molecule, I call it a physical determinant of v, okay, one half pair of s, this expression is, and here for the determinant value, there's one more invariant which plays a role, which is the second determinant of v, but if you look at the determinant value of v, this will be independent of the second determinant of v, so therefore, when it starts, it only sees the first determinant, so therefore we have here only this one, so we have these two, we know if I alright that there are all universal power series for this, and five universal power series for this, and these universal power series now depend on k, which is the rank of k, so we have, it's a now totally universal because we have fixed what the problem is, and so we depend on this, this expression, so to know what this thing is, it's the same as knowing these nine powers, or if you want to solve it in the problem, you have to know this formula, so this is the statement, okay, so I'm allowed to move on, or how is it, but you just move on, so I repeat this here for within the series, in the beginning of the series, so we have these, we know for pair of two powers, this invariant that is like this, so then in this paper where we put the coordinate invariance, we also showed that with this change of pair, this x, which counts on which Hilbert's scheme we are, the coefficient of x to the n is what happens on the n-Hilbert's scheme, if we make a change of variable that causes minus t times 1 minus t to the r minus 1, where r is the rank of b, then we can at least determine two of these power series, meaning a1 is just 1 minus t, and a2 is 1 minus t to the r square divided by 1 minus r square t, so very simple power series, and at least when r is very small, 0, 1, 2, 0, 1, minus 1, these are clearly far harmless there just because power series 1, it holds in general that these power series start with 1, okay, so this is from the beginning, now for the series, there is this odd conjecture by Lin, which I think you also had a bit ahead of me that don't know, I think he wrote in his paper consigned it to find this conjecture formula, and this is an explicit formula for this savior series, so the savior invariant for a line bundle f, so this means the term series for minus, he gave an explicit formula for this in this kind of form, I mean, and this took 30 years to prove, it was proven by Mairi, a player in family family, based on an idea of how was that, which, you know, it is sort of rather a pretty geometric group, somehow, but you are law of a key, so this is a, so more generally, while we are from the family and data makers consider this generally in function for the savior invariants, for general vector bundles, in the key, and what they get is, you know, before that depends on the general function, like this, in terms of these five universal power series, which I think are these powers, okay, which shows the exponents of that kind of grad side, simple, so are the terms to rank, and now are as one was the general, so, yes, so here, yeah, this is not a misprint, this is what it is, I mean, I, couldn't you call the rank but we see in a moment that these two ranks that we have here, they are the rank in the within the case, the rank in the second case, when one wants to look at them together, they are the deeper by one, so there is some, anyway, so, but so now the case, the rank of the three, and we try to still after k minus one, so then we make another change, let's x equal to minus y times one minus power, minus one, and then you get this b zero, this first one is this simple rational function, b one is this simple rational function, and b two, which comes with the entire four x, so it depends on the quality of x, so this question, so, actually, what you see right side, they have a different change of value, and then the formulas obviously don't differ, but this is what we get in our computation and it's slightly simpler than what they have, so just, this is slightly more convenient, so furthermore, they determine these two recent policies as algebraic functions in x or in y, and for small values of k, so k, the absolute value of k is at most two, so I should make a few comments on this, so first, I mean, these formulas go to form for the red in there, and for the same formula, they look rather simple, we have to, the simple formula for b one, b two, but this is after the change of value, the under the change of value just gets quite less, and we are also supposed to multiply this out, if you want to know what happens on an actual given scheme, and then we get incredibly complicated formula, so this is, so even when we are on the surface making s equal to zero, this is actually a very complicated formula, even though it looks simply simple, and second, it is always the case that these b zero, b one, b two, and a one, b two, so the ones which do not contain the canonical charts in the experiment, these are always much simpler to study, and the reason is, is that one can study them by just doing computations on the given schemes of k three services, and as we said, the given scheme of a k three service is a hypothetical variety, and one can, for instance, use the formula for the form or other things on k three services to compute that, and so these three are always much easier to study, when can the others, these a three, a four, b three, b four, that involved k are much, much more difficult, and normally also in other similar problems, you cannot say anything about them, but here they would only say something in this very small, there's one further ingredient to further the mystery, which is the relation between the Belinda and the Sega series, and this relates to the question of thought or its collection, so there is somehow these policies between, so the one for the Sega series and a three for the Belinda series, they are somehow the same of the change of variables, but this if we look at the Sega series for rank k, so for b of rank k, and for the Belinda series for b of rank k minus one, so for b, which I called up, so these two alone power series, this is in just the projector and are related by this change of variables, and in addition by changing the rank in by one, so we did it, so this is a, so this is a kind of, so this is a projector by originally by Johnson or Johnson and used some version of string, very people, together with some static legal arguments, so to show that such a statement passed home, where this change of variables is given by certain intersection numbers, which he couldn't quite determine, only a few of them, and then Mario Opria-Pande-Pande made this formula, I don't know, this is called the draw, so this is somehow the most important thing, and so the conjecture is such a thing by the two, and so he conjectured it by giving an incomplete proof, so I mean it was supposed to be a conjecture, but you know he needs an incomplete proof, and so now how can it be possible to be true, and why do we have this crazy shape between Sege and Verlund, by sending a to k minus 1, that seems kind of ridiculous, okay, so now this is a question, and now I will talk about what we and Anton do, so the first statement is, so this is the result, so our results, and so the first theorem is that in the Sege correspondence is true, so what I said here is this conjecture is correct, without any restriction, this is the first statement, so therefore in order to determine the k and the b series, it's enough to determine a3 and a4, and now the second theorem is that with this change of error, with the same change of error as we had before for the bit in the series, the a3 is given by this explicit expression, so we have here 1 over 1 minus y to the r r's, so where r is, is the red in the Verlund case, the exponential of the minus some y to the n to the n, we take this polynomial here, x to the r minus x to the minus x to the minus x to the minus 1 to the power of n, we take the coefficient of x to the 0 of that, this would be some complicated combination of polynomial coefficients, we take the exponential of this expression, this is our answer, okay, so now you might say this is a very complicated formula, but it somehow isn't, it's much much simpler than what anybody would have expected, and it's much much simpler, so I did suppose some computation before, we don't know whatever for small cases, the individual formulas we get for small values of r are much more complicated than the general formula which doesn't bother, so it's a kind of surprise all these things, but it can also be simplified, but it can also be simplified, and you can work this out, you can work this thing out, and it says that the value of the derivative of a3 of x is the sum the delta 1 over n, and that you can compute in first form immediately because you want to assume that you want the derivative to be the same, right, okay, we can discuss this afterwards, when I'm here, I cannot, I was too excited to, no, no, we can't, I'm just saying it should be very easy, okay, so then, okay, so then we have already an improvement for the, for the later, but anyways, we can see, and so there's an alternative way of saying the same thing, which I'm not sure is related to, okay, so we cannot say differently, we take this expression here, so x to the one half minus x to the minus one half square divided by this, so this power series starting with x to the r minus one, we take the reverse power series, so the composition inverse, now as it starts with x to the r minus one, they are r minus one branches of it, which somehow differ to some extent by doing something without community, so we have, there are therefore r minus one branches of the inverse power series for this, which I call alpha i of y, and then this thing here, after this change available, is maybe like this, I don't know whether this is getting better, but anyway, y to the one half is fine, no, this is the same as here, and here is the product of this, so this is another expression for the same, I have introduced this so that we can talk about A4, so the, the final, the final missing power series, and so here, so we have to get this formula that I just wrote down, where these were these branches of the inverse of this, this version of function, and then now it came, I don't know whether you can simplify that, so if I again take the same change of variables, then you know, these are probably starting with one, so writing it like this is the same as saying what we call it, can be written in this form, in terms of these factors of the AI, so we have this complicated expression, so some simple special function like this, and then product ij from one to r minus one of the alpha i, alpha g, one minus alpha i, and here for the r power, we have this formula, I don't know whatever, so this is a, I'm not precisely, yeah, so, okay, so I hope I was wrong, so this is, but this is just the connection, so we are able to, we know that this is true, but our proof does not extend to that, but, so the first form is all i and j, yeah, the second form is only whether you, yes, yeah, you could also, I mean, okay, and that's just the way it is, I mean, there's actually, if I have done some complication for the higher rank, so what actually happens is that correctly the formula would be also here, the part of all ij, and then you divide by the thing with the i squared and the second power, because that's how generalize is spent, but anyway, this is, okay, so this is the conjecture, and so, so this gives these two things, if I'm releasing the conjecture, it gives a complete answer for the, both for the, within the sake of formula, and the proof, this is proven when a squared to zero, before that term is there, and so this conjecture, we know it's true, because, you know, some years ago, when I was again, he told me how to compute efficiently, this algorithm with localization of the Hebert's team of points, and so I was able to, so we were able to compute with R, this rank, in the variable, until we did the Hebert's, I mean, for the right Hebert's team of points, so if you look at the output, it says so many megabytes of very big numbers, and this formula explains all these numbers, okay, so we know that this, this is correct for any vector bound, for any k theory class, on every Hebert's team, until the 49, okay, so therefore, we, we are sure that it's true, so that, so now let's not get, oh, so this is, as much as we know about the question, our, as I asked it, but there's one more step, mainly, we can compute something more than the Seger and the Belinda formula, we can compute some finite invariant on the Hebert's team of points, which contains both the Seger invariant and the Belinda invariant as a limit, so we have a final formula which contains so what we compute is this crazy thing, so now this is the same thing as I wrote over before, but now without the chair, it's the Belinda, and so we take the sum, just everyone could change the sign, but anyway, this is how it comes out naturally, and so minus x to the n, the whole moment, all the characteristics of, so it's a lambda minus z of this topological bundle, that is the determinant of the topological bundle corresponding to all for the minus one, and we take chi of this, and obviously when I write something, so lambda minus z moves the alternating, the sum of the alternating powers of the bundle with coefficient minus z to the n, where when I write it like this, obviously means that the z just comes out of the equation, so this thing, the x and z powers here is an x, so this is this thing, and so we can look at this thing, and now one can see that this specialized both the Belinda and the chair, so the first thing is kind of easy, at least when we take the chair for the piece, we assume that v is just the vector bundle of rank k, then lambda to the n times, so then this thing has ranked, vn has ranked n times k, so lambda to the n times k of this thing is the determinant of n times k, or is the determinant of this vn, so therefore we see if we take the coefficient of x to the n, z to the nk of this expression, and then for the stupid sign we have this is just the whole of this, but here we have the determinant of this bundle, but here we still have this vector, so we have the determinant of vn, hence the determinant of os to the minus one, which is the same of the determinant of v minus os to the n, okay, so this is one limit of this coefficient, and what is slightly less obvious to see is that if you do this and that thing is right and wrong, then if you put here this, replace x by this expression and z by this expression, then you apply Riemann Roth and some computations in chronology, and you see that the limit is at some of those two zeroes, this term, so this is some computation which one has to do, it's not particular, obviously you can't see it often, and obviously there's a misprint in this formula, then it would be more difficult, but I think it's going to be, and so you can see here already that you have this difference that here if we have i here in rank k, then we get the corresponding Schoen invariant from rank k, but we get the regular formula in rank k minus one, so it's somehow doing this problem in rank by one, so these two are related by one, our series, but you know in a different way, so this is something, and then again we can apply the coordinate invariant, which says invariant set, this thing is again written as the product of five universal policies, one for the policy two, kind of me and so on, and where these five universal policies depend on the rank of k, so we have the same equation, now we have to make our problem a little bit more difficult, we want to compute what these policies are, we know what these policies are, we know what the a's and the b's are, we know much more, because we have the much finer invariant that we both see and develop, and so the answer is this, which is now, I don't know a little bit on the crazy side, so again as usual these policies are never nice in the original, so if we have in terms of x and z it's the total horror, so we have to make a change of variable, so now we have policies with two variables, so we make changes of variables in welfare, so the x now becomes u times 1 minus u to the r, the r k minus 1 divided by this, and z becomes this, and you also keep in mind that we want to call y this combination of u and v, so this is a central function of u and v1, so then because we will find out that many of these policies do not depend on x and y or u and v independently, but they only depend on y, so what we find is that if I take the product of g0 and g1, so this first move, then this is just 1 minus y, so what means is 1 minus this, where x is expressed by this, so it's in the end a rather complicated thing, but in this change of variable something's g0 and therefore g1 cannot be expressed in terms of y, it's this and then you can flip what g1 is, so some rational function in u and v, g2 is also a rational function in u and v, which is a bit more complicated, and then however g2, e and g4, which are the ones, which are the most difficult ones in general, again only depend on y, and in fact they are the functions we already know, if we take this y, then g3 is just a3, so if you take this j3, this is just, it's just the a3 from before, when we have to place the y, by this we divide by this and then act like this, so this is kind of amazing, and for j4 the same, so they depend only on y and the formula is the same as they were for the variable in the formula, this is, this is the same, and so I would say that, I mean although that's the case, is that they didn't say the correspondence is explained by the fact that these powers, e3, e3, e4, they do look more real, so we know that the a3 and b3 are both kind of specializations of g3, and so they are kind of different specializations, but SGT actually only depends on one variable y, it turns out that therefore it must be true that they obtain from each other by a certain change of value, which is the one which was given there, and the same here, so this, if I use this as an explanation, which is maybe a bit crazy, this explains the city of Virginia, and it's clear that the shift by rank, the rank is one, because after all this was how, okay, so this is the theory, so this is the result, and now if there is time, I can give a few remarks about the proof, obviously, you know, this is a full speech paper, and some three or four years to think of the argument, so it's not really that we can explain, but anyway, let's just have a sketch, some of the ideas of the proof, and then basically at some point, so we have again this provodic invariance, so it follows that the formula, you know, because of the provodic invariance, we only need to know the formula for some surfaces and some vector ones, enough to determine the formula, so we can take these surfaces to be toric surfaces, and the bundles to be for vector ones, so that means we have an action of these two torus on S, we find a big spring point such that the action needs to be, and we can therefore use a localization form, and that's maybe also why I'm interested. So again, we have, this is the microbiology, the point that we would like to list for longering, we take, say, we have E6 point of the T-axon S, we take like T1, I, T2 eigen weights, which was a phenomenon that someone actually told that you get from the eigenvalues of the action on the 10-inch space at the iPhone, and in fact this, the weights on the fiber of D at the x-point, each of these is a linear polynomial, and epsilon 1 and epsilon 2 in epsilon, epsilon 2, so we have these next, and then, we know that on SN, the x-points are parameterized by e-tuples of partitions, numbers, and they are for n, and now we write down this, let's say, start with complicated formula. So what do we have here? So it depends on all these variables, x, z1, z2, z1, t, it's a sum over all partitions, of the generative function of something over all partitions, we take that x to the, this partition in that sequence, it's the number which is partitioned. We have a sum over all partitions, and what we have here, so we, we could, if you have a partition and one thing, and whatever, if the partition is whatever, 4 to 1, then we write, as we had before, write a graph like this, okay, and the, so this, this is, so to speak, the start here is number zero, the start is number zero, so this here, we have the, the rows, we have the columns, they, they take an element in this thing, so one box here, the, the row of this would be zero, and the column of this would be one, so it's zero, so the row zero, so this is when the row zero, but then we have here, so the part of this thing is one, and the column is also equal to one, so this is the row, this is the arc, and the c that we have here, and then we have this a, which we also had before in the formula, for the tension part, so a is the, a is the arm length, so if we're going to take this element here, the arm length is how many more, so the arm length of this thing here, so it is a different thing, it's equal to two, and the left length is how many are there below, and there's one below, okay, and so we write down this incredible expression, so product, input product of all elements, so all the boxes in this diagram of the partition, and we, we have this over k, where k is the rate of the bottom, and then, so we have this expression, okay, so this is what we have, so some explicit expression, so it's something which actually comes out, comes up, well this is also part of the thing, so that's what you write down, something which comes up, for instance, for certain kind of the varieties, for all the, of those similar expressions are written down, and so now we take the logarithm of this, obviously that's not significant, now the thing is, we have written down this formula, which is something complicated, but if we apply Riemann Roth, the localization formula to the formula we had before for this generating finite invariance, and the localization formula on the given sum of points, so we have here somehow sum over e2 percent of partitions, so if you might try this out, then this expression is just our generating finite g e s e, so if we replace the z i by e to the v i, so v1 of i to the k of i, where we have this statement that these are the weights of the fixed point, so if we do this, and here this was the weight at the I think point, the tangent point of the service, and so then if we put this together, this gives us, this gives us some expression in terms of the tangent point of the Hilbert scheme, and of the, in the terms of the tangent point of the Hilbert scheme, and the, in the terms of this chronological sheet, and what we actually get is precisely what one gets by Riemann Roth, so this is why one writes this pretty big thing, and so obviously we can take the logarithm, and that's not very, the exponent of the logarithm is this, so in some sense we have therefore, the question is completely answered, so we know what this generating function is, it's this, no, simple, you just write down this formula, that's it, but the thing is obviously the, the power series in x and z, where the coefficients are incredibly complicated, rational function in epsilon 1 and epsilon 2, and it's guaranteed that these incredibly complicated rational functions actually are just polynomials, then we can put epsilon 1, epsilon 2, epsilon 2, 2, 0, but obviously practically what cannot do that was something very important. So, there's our one ray of hope, the first thing one can do, it's not trivial, is that if you look at this expression, and if you have it within epsilon 1, epsilon 2, so we take that into epsilon 2, then so this logarithm has the property that it has only a pole of order 1 in epsilon 1, epsilon 2, so this thing itself, the omega, you can have an arbitrary pole order in epsilon 2, but here the pole order is only 1, so it's almost a polynomial, if I multiply this thing by epsilon 1, epsilon 2, it's actually the power series in epsilon 1, epsilon 2, and then we can do the following, I will say first here, we look at this expression, now here as you can see we have a sum over the fixed points on the surface, and then we can reinterpret what is written here as some equivalent class on the surface, and so then we can we can pull in this limit epsilon 1, epsilon 2, 0 to this, and so then we get some intersection numbers of the surface, and we have the exponential of that, so that's what we do, and so it comes out in this somewhat very way, so we have this expression from before, now we put in what we know, we know that this H has only pole of order 1 and epsilon 1, epsilon 2, and we have replaced at a fixed point we place the epsilon 1 by T1i, T2i, so the weights at the tension space, so we get if we write out what this is, we get this expression, they are higher order terms but they don't matter because when we put epsilon 1, epsilon 2 equal to 0 they will vanish, and so we can pull out 1 over this which is the order class of the tension factor of the surface, which one always has to divide by if one does localization, then we have this expression, we have the H minus 1 minus 1, and this means I take all the dI, so we take this expression here, then we get this, this, and now if you look at this for instance, if you go on, yeah, for instance if you look at this expression, so we have for instance this, then these two will cancel, and we have here just this H0, and so we get, we will get H00 times the order number of the surface, so F would be H00, so we can, we shuffle this, so the point that I want to make is for instance if you take here this expression, for instance you take this one, so if you take some derivative by Z here, we get down some of these Vijs, so for instance if you take the first derivative, we get kind of the first term class of the bundle V, or what was it, yeah, of the bundle V, so we multiply, so the academic first term class of it, so what we get here is the local contribution of the first term class of the surface times the first term class of the, of the bundle, and so if you work this all out, we get that this by localization on the surface, this can be written like this, so C2 of the times somehow of the C2, C1 square times somehow of the C11, and so on, where I can quickly compute what this power C is, what does it mean? Why can't we, okay, so for instance, well let's put this one where we put all, so with K and Z, where you put all the variables here in this H to be equal to Z, and we take this derivative by that, and this F, which is here comes from the order number of the surface, which has to be this H70, the E, which comes with this intersection number of the surface, which has to be H1 from 1 and so on, it's all a bit fast, anyway this C11, which C1, which comes with this intersection number, is just new minus A and derivative of the H minus 1, so the coefficient of the, remember this definition of this, okay, and so on, so therefore if we know these power series here, we know all these power series, and we know our, okay, so we have determined these kind of coefficient, H minus 1 minus 1, or rather some partial derivatives of them, then H minus 1 is 0, H0 is 0, H minus 1 is 1, if we know these, we know it, and so we have to understand what they are, and for this we use two properties that somehow are around here, they are called regularity and symmetry, so a power series into variables like this will be called deregular with respect to k, if the thing that we did before to get from the, from the surprising variable, so the charm thing, if that gives us, you know, something which is divisible by epsilon to the d, the power series in epsilon, so, so power series in x and epsilon, which are divisible by epsilon to the d, so some kind of regularity in this limit, and it's called symmetric, if when we do this gradient change of variables, so x goes to x to the minus 1, and z goes to x, the function goes to itself, so we have these two properties, and now what we find is that these things here, hg1 d2k is minus e1 minus e2 regular, when e1 plus e2 obviously here is more than equal to 0, and secondly, they are also almost selective, so if we add to them something, so here I have some expressions I don't remember, but here we have called only logarithms, just in one variable x on z, then if we add this, so this thing then it becomes symmetric, and now the point about, so, so this is actually, the first part is relatively easy, the case getting followed from the fact that you already know, you know, however, we take more carefully that if we take that the Schoen series is a limit of our refined series, if we do this correctly, it will follow this regularity, and the second one is a deep statement, so it comes from a symmetric function theory, it's a certain, it's some complicated identities of generalized equations, which give us some kind of, some function equation for this omega, and this function equation that they've long written translated for this state, okay, and so with this one is basically done, because there are, you know, we know these things are essentially symmetric and regular, but basically they are no symmetric and regular part, so they are extremely restricted, and so the first thing is, if we have a function which is symmetric and irregular, if t is bigger than zero, then the function must be zero, but they're no such part, so the only cases when they are just zero regular, if they're zero regular, they actually, if are, these are supposed to depend on two variables, but then they only can depend on one variable, and it's precisely the change of variable, so if we change x by this expression, and z by this expression, then they only depend on this expression, there is a unique h such that we must even write it like that, and now these functions we were looking for at e whatever, which are, this is the h1111, so they can be expressed in terms of symmetric regular function by what we know about this h minus one, and in addition, as they are so restricted, this symmetric regular function could be determined by a small part of their positions, so in order to determine which ones they are, we only have to be able to assess what these positions are, in particular, one thing that is there is that if I have a function which is one regular, then if I take the derivative by z, it will be zero regular, and furthermore, if this derivative is symmetric, then it is determined just by the part of this function where we put the z variable equal to zero, and if we look at how these functions work, you know, if you put the z variable equal to zero, then this thing will go away, because it's only, and basically, you only get some polylogs, and so one can easily understand what that is, okay, and so that is kind of the argument, okay, so it's, but you know, as you can imagine, that was, yeah, rather incredible, so okay, so thanks. I think you found the only collaborative book we could have brought this in here, so I, I mentioned at the beginning, this is a special case within the formula for ILA, which I actually wrote. So I had some kind of effort out with how to project for that, and finally I dropped, I worked for a long time with that how to project, and then the special case of that project, or in one case, it was this book, and so at this point, I explained to him what I knew he had been working on this formula, and I told him, I know what the answer is, you know, can you prove it, and so then he kind of came up with this method, and then, I mean, I also did some contributions to the expert, but not, but the main idea of the actual proof is this, but on the other hand, he has this method for many years, he couldn't prove the formula because this method can only prove the formula, you know what the formula is. Can I ask a question? I don't know if you can hear me. Yeah, unfortunately, yes. Sorry. We hear you, we also have to hear the bed, but anyway. Now the bells are gone, I think. Sorry. So this omega function is something that we introduced with Tamash and Emmanuel in our joint work, and we have a conjecture that this platistic logarithm of omega gives the mixed hodge numbers of the character variety, and some of that has been proved. So by Anton and others, and so I was wondering if there is any reason why the same omega appears in what you described, other than it seems like it just falls in the lap that that's the series to consider, but does it go beyond that? Is there any, what's the expectation as to why this same generating series appears in those two contexts? I think it could be the part of the reason is obviously that Anton worked on the other thing, and so he came up with this thing. No, but you told us at the end that the omega was the series that you wanted to compute, I mean in logarithms and so on, specialized in the right way. So it's not, it doesn't seem to be something at hope that he happened to know about. It seems to be exactly what you want. You see, it's a bit strange because by itself, there's no mystery. If you look at the formula, the way it's given, it's not that by some complicated subtle fact, the the the localization formula will give us what we want. It's just obvious. I mean, if you know how to write down such a thing, the formula that I wrote down is precisely this. You know, it's not by some kind of complicated argument. That's exactly what I said. It's not something that it happened to know about and then brought into the problem. It seems to be intrinsic to the problem. Yeah, that's true. Yeah, so somehow this is, yeah, no, but I mean, I don't think I can answer your question. I mean, so that's, maybe I will, I might want to discuss this with Anton whether there's anything I can understand why this thing works. But yeah, I mean, I think, you know, it's anyway, maybe something that we should sometimes discuss. They seem to be very close relations between English teams and your character variety things, which I don't quite understand. I mean, obviously, the English teams are related to traditions. They find this localization as you also, but you have the same formula as Marceline, and so we maybe should talk about it, but I cannot answer it now. Okay, thank you. Yes, if one looks at this formula for omega, that his power is q to the something t to the something z i over the something is the arm of the leg length. Well, not quite, but anyway, something is square as a position. Does that mean that this function satisfies recursion through a z i by q z i t z. So is this something like the q of the non-existent about the third thing, the natural thing that could take the simplest thing product one minus q could be an x and kind of creepily replace x by q x and series just goes itself on this one minus x and it's much, much simpler. So it's an inductive. So is there a relation that we take the value on that with x c one up to so we'd see one of the c k, you change, let's say just c one by the q c one or t c one, then there should be some simple relation. It's so that might be much simpler way to understand this question. I have no idea if that's true. No, I don't know. I mean, obviously, I don't think I can answer that. I mean, there is something that is I thought about and didn't make much progress on, but this same series, if you specialize in a particular way, the q and the t is related to, to representations of a quiver variety, I mean, yeah, of a quiver, quiver Nakajima variety. And there there are these functions that correspond to a veil, veil group action or somehow the cat's middle convolution. And in those, the quiver in this in this setting is a star quiver. So it has one central node and then legs corresponding to the particular to the punctures on the surface. And then the particular is in there's a vital group element associated to the central node, that does something really complicated to the variables. Whereas the the the the value group elements associated to the legs are some simple thing that basically permutes the parts of a partition so they're not they're not they're visible. And that could conceivably be a pretty serious transformation formula for this omega, which I don't know that done is the kind of thing you're thinking of but but that's something that that has been on the back of my mind as to something to to look into. Sounds much more sophisticated. I just meant if tribunary for this formula, or from the definition one sees that there's a recursion of the nation between valued C1 and qc1 or tc1. But I should maybe look at that we should maybe forever more. No, I can look it's my thing that we said, I don't know. So I have that I should look at this but I don't know at least how okay so. So are there for that question? Otherwise, I'm also in source. So then I have one question. So have you tried actually doing any computation for just C2 and not for P2? For C2? Yeah. So like a covariance of Orlando. Okay. So that would amount to looking at this thing, just like this you take the whole power so you take a lot and then you do this. Yeah, negative powers of the weights and positive one, not just the positive like you're getting now. And then it's kind of a mess, right? Yeah, I have not looked at it. I mean, I don't know whether there's a good, I mean, I must admit that obviously it would be interesting to say what are the covariance within the formula of the covariance. I mean, that would be an interesting question, but I don't know. I haven't talked about it. So in principle, it's just if you understand this, I mean, I don't know. I mean, sorry, but I have not looked at it. Have you done any equivalent computations like that? My students start trying to do something like that. Thank you. So I think, I mean, I'm also the chairman, so thank you very much for listening to me. Okay, so thank you very much. Oh, there was something in the chat, so I don't know whether... There's no answers. Well, anyway, somebody asked what we can see about when the surface is singular, and so the answer is we haven't talked about it. And so we, in particular, we can't say the moment any could be solved. Okay, so maybe I would... How do I stop this with the same registers and all that? So, okay, I guess it's many people. It will save it and all that. So it's auto-registering, no? Well, it was registering with a problem, right? Yeah, yeah, yeah. So there's been... In climate, that seems to be...