 projective surface with a positive geometric genus, first protein number equal. We wanted to compute the virtual Euler number of this modular space of stable sheets of rank 2 with general classes C1. But OK. So over this virtual fundamental class of the tangent of the general class in the virtual dimension of the virtual tangent one. And so we had the Mojizuki's formula, which allows us to compute such a thing in terms of Hilbert's Keynes of points. So I will maybe review it. So if E over S times the modular space is a universal sheet, then we can look at classes. And if P of E is a polynomial in classes tau E of alpha k, which is just I push forward to the modular space. Some I've turned class of universal sheet kept with a push forward of a class alpha k, for case in the homology of S. So if I have any polynomial in these, then I can compute this integral of this over the modular space of sheaves in terms of something on Hilbert's Keynes of points. I mean, I will use the, maybe I will not precisely write down the formula, but anyway, we have somehow, we take C1, write it as a sum of two classes. We take the Zyverquitton invariant of the first. And we take the coefficient of S to the 0 of some expression A depending on A1, A2. We have C2 minus A1, A2 is some integer S, where I will not precisely say what this is, but psi. So this thing A1, A2, and S is the sum over all N1 plus N2 adding up to N integral over the product of these Hilbert schemes, points of a suitable expression, which is called psi A1, A2, N1, N2, S, which maybe later I will need to, at some point, to precise form, but maybe not now, anyway, we won't remember. OK, so this, what we, but this was a rather complicated expression, and so it means we can compute, instead of the modelized space of sheaves, we can compute on Hilbert schemes of points, but the price is that the formula becomes more complicated. And, but the point is that we do not know how to directly compute anything on this modelized space of sheaves, so it is an advantage we have at least a space that we understand reasonably well. And so we can try to compute. So if we want to apply this to the virtual Euler number, if you remember the definition of the tangent bundle, here we compute the virtual tangent bundle. This was a certain, essentially, some relative x. So if we, it follows by the Grotten-Dickry-Monroch applied to the projection from S times M to M, that this Cvd of Tm here is indeed, so vd is always the expected dimension, is of the form P of E. As before, it can be expressed in terms of these tau classes. And in the same way, if we look at chi minus y of this modelized space, virtual version, we again find by applying the virtual Euler gives that this is also of the form. So this is also the integral of some P of E over the modelized space. OK, so in both cases, and one can explicitly say, I mean, compute what these things are and express it, we will maybe see a bit more later. So that means, in particular, we can apply Mochizuka's formula to both these situations and also to many more general ones. Now, as I said, this expression is very complicated. So we cannot really directly understand it. So we will try to establish some properties to make it more palatable. And so these are two properties that we want, co-bordism invariance and multiplicativity, which will then allow us to reduce the computation to the case of toric surfaces and apply localization. So we have this thing that we want to compute. So in order to evaluate this expression, we have to compute this expression and then take the coefficient of s to the 0. So the expression is this. So first, if you don't know what to do with something, you put it into a generating function and hope that something emerges. So we make a generating function. So I write, say, z prime of s a1, so generating function, z prime of s, say a1, a2, this variable s and q. This is just the sum over all n bigger equal to 0 of this thing, ns, maybe q to the n, where q is a variable. There's some kind of partition function. And now we want to say something general about this. And the first thing we can say is that it only depends on the intersection numbers which are in this question. So this is the co-bordism invariance. So proposition, there's a polynomial, say, p tilde in all the numbers we have, say, a1 squared. So these are a1 and a2 are classes in the second homologous, or I can consider their intersection numbers. So I just want to look at the numbers a1 squared, a1 a2, a2 squared. And we can have a1 times ks, a2 times ks, and ks squared, and chi of os. So these are all the intersection numbers you can make out of the surface s and natural classes on the surface, and these a1, a2. And then the statement is such that this thing, a, a1, a2, ns, is equal to this polynomial. So I mean, universally there's one polynomial depending only on these, and then this is for all a1, a2, n, and s. So here I should have said, by definition, I write here maybe s to remember that this depends on s, p tilde of a1 squared, a1, a2, and so on, all the things that we have until chi of os. So this is the first thing. So this does not depend in some complicated general way on the geometry of the situation, but only on these intersection numbers. And in what? Well, yeah, I mean, I think there will be some, so if b1 is not equal to 0, I think I have to change the story with the Mojizuki formula a little bit. Also anyway, Mojizuki's formula was under the assumption that b1 is equal to 0, so one would have to first generalize the proof. I don't think there is a very serious problem here with the multiplicativity. I mean, it's something about Hilbert's schemes that's nothing really to do with b1, that you have this fact that it only depends on the numbers in the situation. But you first would have to establish a version of Mojizuki's formula. And I mean, there will be one, but it hasn't been done. And so that's it. And there might be few more terms. No, there could be some kind of something that happens in the picar group of the surface, maybe some intersection numbers, I would have to see. But it will be like this. I think at least for the wall-crossing formula, Munoz had some wall-crossing formula, which also when b1 is bigger than 0. And there are some additional terms in it, which have to do with what happens in the first co-molge of the surface. So maybe it's not just, you don't then just have intersection numbers in the second co-molge, but you also have to see what happens between the first and the third co-molge, and so on. So it's a bit OK. So this is some old, some modification of some old argument that I had with Ellingsoot and Lane. And it's based on some inductive scheme to understanding Hilbert's schemes of points. This is based inductive scheme to compute on Hilbert's schemes of points. We'll just kind of sketch what's going on here. It's based on the following observation. So whatever, it's actually not so difficult, but any proposition, I call it theorem, although it's actually not so difficult. If I take this universal family, so this was this instance variety between points and sub-schemes, such that x lies in the sub-scheme. So we look at this. And we also then, given this, we can also look at another instance variety, namely, we call it SNN plus 1. Some version of it was also related to these Nakajima operators. So this is another instance variety, namely we have a sub-scheme Z and W, where one is in the Hilbert scheme of N points. The other one is in the Hilbert scheme of N plus 1 points. And Z is a sub-scheme of W. And now these are, we have a relation here, namely this thing is the blow-up of, so we are here in the product of S times SN, along the universal sub-scheme. And it is also non-singular. I mean, it's actually not very difficult to prove. Obviously, you can see that if you are outside the universal sub-scheme, if you have a point and a sub-scheme which are disjoint from each other, you can make a sub-scheme of length one more by just adding the point to the sub-scheme. And you find that you can also somehow extend the sub-scheme precisely by going to the blow-up. Now, given this, you can, so this blow-up can be understood reasonably well. You know what the, you can say something about the normal sheaf and so on. So you can understand what happens in this blow-up, also the homology classes. And so therefore, you can make up the following inductive scheme. So we have here our Hilbert scheme of N points. Then we can go to this incidence variety by just kind of in this thing projecting to the second factor, if I call this QN. We can project here to the complementary factor S. And as you know, this was, this thing is the blow-up of S times SN minus 1 along this universal sub-scheme. So we can do the blow-down. And then we can, this was N minus 1. And now we can just leave always factors of S here and do the same to this factor. So we can have here, say, QN minus 1. This goes to S times SN minus 2 N minus 1. And PN minus 2 goes to S2 times SN minus 2. And we go on like this. Here we have S times S, what is it, N minus 2, times S12 maps to SN. I'm not sure I, maybe this is P1 and this is Q2. And so we have this picture. And so we want to somehow compute something on the Hilbert scheme of points. So alpha is a class on this. And we want to integrate it over the Hilbert scheme of points. Then we can pull it back here. Excuse me, the same integral. We can push it down here and we get the same integral. So this is the same. So first we pull it back. So this is S, the map has degree N here. This is the integral for SN minus 1 N of the QN star alpha. And then if I'm integrating over something and over the push forward to something, it's always the same. So this is 1 over N integral over PN minus 1 star QN up a star alpha. And then you can keep doing this. And in the end, you get the integral over just N for product of P1 star Q2 up a star and so on until PN minus 1 lower star QN up a star alpha. And each time you pull back, you get a factor, whatever the number of the Hilbert scheme is. So it's 1 over N factorial times this. So this is how we can compute it. And the point is, so to pull back in this context, it's always something trivial. So it's easy to come. That doesn't do anything. And the push forward is usually difficult because you have to understand the map well. But we understand this blow down well enough so that one can explicitly say what happens to many homology classes after being pushed forward when so one can actually figure out what this thing is. And so then we have just an integral over the Hilbert scheme, over the N for product of some homology classes which we can say of what kind they are. And then you will find it is such a polynomial. So this is this thing. So this is for the Hilbert scheme of points. In our case, well, anyway, we do integrals over Hilbert schemes of points so that some modifications maybe of the argument will be necessary for precisely our setting. But this is how it works in general. So the second nice property. We want one other property. So we get somehow that this thing is some polynomial in all these numbers. We don't know some. This is, in principle, we compute it here. But we kind of just keep track of what terms occur, not what precisely it is. So we have no clue what precisely this polynomial is. It would be very complicated to follow this procedure in detail, and maybe not even very advisable to try. But we can say one more thing which is the multiplicativity. Namely, I had this z prime, which was this generating function of, say, a1, a2, s. So it depended on s. And q, which was this generating function of these, so sum these things with n, q to the n. And so it's kind of, we pull out the constant term, so the coefficient of q to the 0, which one can easily compute. So this would be 2s times to the power chi of s times 2s to the holomorphic Euler characteristic of the line bundle. So I write it additively a1 minus a2 minus a1 times the same somehow the other way around. So these are line bundles. So a1 and a2 are line bundles. So I can take a2 to the, a2 tends a1 to the minus 1. And we take the holomorphic Euler characteristic. And so we get this expression. This is the term which does not depend on q and times something which without the prime now. And this is now power series starting with 1, which I can view as some form of a partition function. It's actually, as we would see, closely related to Nickasov partition function, a certain version of it. And then we have the following multiplicativity result. So there are, so instead of this being just a polynomial in these things for all each coefficient of n, we have that the whole generating function here is a product of power series to these as powers. So there are some power series a1 until a7, because if I'm not mistaken, these were seven numbers, which are power series, which are Laurent series in s and power series in q, with such that we can write this thing as a product over this. So this thing, zs a1, a2, sq, will be equal to a0 to the a1 squared, a1 to the a1 a2, a2 to the a2 squared, a3 to the a1 ks, a4 to the a2 ks. And now it doesn't seem to, so because I start with 0 for unknown reasons, it didn't seem to add up to 7, but if you start. So the canonical class a6 to the ks squared and a7 to the homomorphic Euler cryptos. So we have this description. I can also roughly say what the idea of the proof is. First, we have to remember that this thing here, if we look at this a, this was somehow the sum over n1 plus n2 is equal to n, some integral over product of n1 times n2 of something. Now note that if I take the Hilbert scheme, if I take the union of two surfaces, the disjoint union, and take the Hilbert scheme of points, the sub-scheme splits into one sub-scheme on one and one sub-scheme on the other. So this is just the disjoint union over n1 plus n2 is equal to n of the, OK? So that means that this thing is actually equal to the integral over s union s to the n of the same something, OK? So maybe I write, so let me write s bar equal to s union s. So now instead, obviously, then it also follows that if I take s1 bar, disjoint union s2 bar, this is the same with the bars. So the only thing one has to see, so check that, where is it? Write it anywhere? Yeah. So check that if I take this expression here, a something, so this was psi of something. So if I take psi and restrict it to, so this thing is disjoint union of these things, if I restrict it to such a factor, so here right now this is psi depending on s bar and n. So if I restrict this to s n1, so s bar n1 times s bar n2, then we would like this to be just psi of s bar n1 put back from the first factor times p2 upper star of n2. If this is the case, then it follows that each term in the sum is a product. And if you make the generating function, it becomes a product of the thing. So if you write s as a disjoint union of two surfaces, and I should also say that a1 is a1, 1 on 1 and a1, 2 on the other and the same for a2, if it then splits up like this, then it follows that the whole expression splits up as a product. So it follows then, so if this is true, then it follows that z of s1 union s2 where you have here, so to speak, a1 union a2. So it's a1, a1, 1 union a1, 2, a2, 2 union a2. And then whatever we have, we have sq. That this is just the product of the two factors of each of singular ones. So this is z, s1, a1. So we find that it's a product like this. And now, if I look at the triple, so if I look at these numbers, nobody has forced me that s has to be irreducible. So I can just, so I can use this to, so I can write my given, I can reproduce the numbers here by splitting up s, a1 and 2 into factors like this. And it will follow that it is a product and I can, by some manipulations, write it as a product of the terms where these corresponding numbers are so that one is 1 and all the other ones are 0. And this then will give me this ai and I get a product formula like this. So I said it a bit sketchy, but it is quite simple formula manipulation. And so then from this, so I just say from this result follows formula. So this is this multiplicativity. And so if we have this, then one should notice, well, OK. So in particular, so this generating function is determined by these numbers. So I just, in order to compute it for any surface and any a1 and a2, I only have to compute it for enough examples so that these generating functions are determined. So as we said, this Zs a1, a2, sq depends only on this seven tuple a1 squared, a1, a2, and so on. So in order to determine it for any s, a1, and a2, it is enough we need to determine it for seven examples where such that these corresponding seven tuples are linear independent. So maybe I call this nu s a1, a2 such that the nu s a1, a2 are linearly independent. And notice that I made the assumption that b1 of s is 0 and pg of s should be positive. So rational surfaces are not allowed. But this was for the original question. Now we are just evaluating this integral over the Hilbert scheme. We can take any surface we want. And so in particular, we can just choose examples where s is, for instance, either p2 or p1 times p1. So can choose these such that each time s is equal to p2 or s is equal to p1 times p1. That gives me enough numbers here if I change the line band. And so this is, in particular, s is a toric surface. So a toric surface means s-toric means we have a c-star times c-star action on s, which has a dense orbit and finitely many fixed points. But it may be justified with finitely many fixed points. So p1 to pE. So maybe we can say, but so we want to, now we will want to, in this case, we can compute by equivalent localization. This was already used in some other talks, but as mine is supposed to be slightly more elementary, I will explain it in an elementary way. So if x is a smooth projective variety, so this is actually just a port residue formula, with an action of a torus t equal to c-star to the k with finitely many fixed points, say p1 to pE. Then so if we look at the fixed point, so let p one of them. And so if I have, say, a vector bundle on x, I can, at the fixed point, so an equivalent vector bundle. So let e be a t equivalent vector bundle on x. Then I can, so then if I look at the fiber at the fixed point, the t will act on the fiber linearly. So then the fiber etE of p at p, at this fixed point, is a vector space with linear t action. So it splits, it has a basis of eigenvectors. So we can write ep is equal to the sum i equals 1 to the rank of the bundle of c times vi, where vi is an eigenvector. So if I have t1 to tk in the torus acting on it, this will act by some monomial. t1 to the n1 comma i times tk to the nk comma i, vi, where these nji are integers. So it just multiplies it by the monomial of these elements of c star. And then we can say the weight of this vector vi is w of vi, which is the sum j equals 1 to k of these nji. Epson j for these Epson j's. Epson 1 to Epson k are some variables. So these actually would be coordinates at the le algebra of the torus. But anyway, it's for us, it's just variables. And so then we can define some kind of the equivalent churn class of the fiber. This would be just the sum of the total churn class, the sum of the churn classes. And this will just be the product over all these vi's of the weight, 1 plus the weight of vi. So that is a polynomial in these Epson i's. This actually would be the equivalent churn class of this in equivalent comology. So this is actually, but we can also just do this polynomial. And so then the port residue formula tells us we can compute integrals in churn classes of such equivalent bundles by just computing with these things. So port residue formula, it says that so say we are given some equivalent bundles, e1 to es t equivalent vector bundles on our given x. And we are given any polynomial, so let p of, so I just write c of e1 until c of es, a polynomial in the churn classes of the ai. Then it follows that if we want to evaluate on x, this polynomial, we can do this by summing over the fixed points and looking at the same expression in terms of these equivalent things. So p of c of ct of 1 of p1 pj plus s. So we just evaluate this. This is some polynomial in these Epson i's. We divide by what would be the top churn class, so the dimension of x, of the tangent bundle also equivalently. So this will now be a rational function Epson 1 to Epson s. And for instance, if I just put these equivalent variables all equal to, so this will turn out to be regular in Epson 1 to Epson k. If I put them all equal to 0, then this gives me this expression. OK, so this is the spot residue formula. And so one can apply this to our situation. So we have to see how this works on Hilbert's schemes of points. And maybe for, I mean, OK, this was also used in some form already in other lectures. But I wanted to, maybe it's also useful for those who haven't seen this before. So we take such a smooth projective toric surface. And we have indeed, so that is t equal to c star squared x with fine-term and fixed points, p1. So pE, say. So under the assumption that it's a toric surface, we actually have local coordinates near the fixed point, which are eigenvectors for the actions. So they are coordinates xi, yi at pi for all i, which are eigenvectors. So such that, say, t equal to t1, t2 in tx pi t times, say, xi is equal to, say, t1 to the n1, i, t2 to the n2, i, times, as a bit even more complicated. Let me see. Yeah, why not? No, now I have to see how I get the numbers right. So this maybe uses n1, i. This is m1, i. And then t times yi equal to t1 to the n2, i, times yi. So these are eigenvectors where these are, again, these powers. So we have the weight of xi will be, according to this, n1, i, epsilon1 plus n2, i, epsilon2. And the same here. This was m1, i. And here you have n2, i, epsilon1 plus m2, i, epsilon2. So it's very simple. So we have this action. We find many fixed points. And we can see the action. So now this action will lift to the Hilbert Schemes of points. So action lifts to the Hilbert Schemes of endpoints on s. Namely, you can just say if you have a sub-scheme z, you can just, so if I want t times z, you can just take this as t of z. Because t, after all, it acts on s from s to s is an isomorphism. So if z is a fixed point, then it follows the support of z is a union of fixed points. So it's contained, so is union of fixed points. And so therefore, we can write z equal to z1 until ze, where the zi, so zi, support it at pi. And so that is, zi is given by an ideal in c xi yi, which, or if you want, which has a, which, of finite code I mentioned with support at the origin in this. So, and it should be, and this ideal, i xi. So, and this ideal should also be invariant under the action. So i xi is t invariant. And under our assumption, this can only be possible if it's generated by monomials. So that means we can write i xi as yi to the, say, m0, yi to the m1 times xi, and so on. So yi to the mr, yi to the r, what? Yi to the mrx to the r, and then find the x to the r plus 1. And obviously, these numbers must decrease. So m0, bigger equal to m1, bigger equal to mr. And now the, the sum of these numbers is the length of the sub-scheme zi, or the degree of the sub-scheme zi. So, and on the other hand, the sum of these, the length of the sub-scheme is the, is the number n. So that means that we have a bijection between the fixed points on the Hilbert scheme and tuples of partitions of n, or tuples of partitions, so that the total, the sum of the numbers partitions is n. So we get a bijection fixed points, t fixed points on Sn. And so this would be e tuples partitions of numbers adding up to n. OK, so we can therefore apply. So now in our particular case, we have always a product of two Hilbert schemes. So if we compute, so for us for this thing, a s, a1, a2, n, s, or Sn, I don't remember, maybe ns. We do have the, what we are computing is the integral over something over s union sn of this psi. So in this case, we have to, the bot residue formula will give us a formula, will give a sum over two e tuples of partitions, because we have two copies. OK. And I will maybe congest. And now I cannot really start something new. So the first, so the thing is that the fibers at the fixed points on the Hilbert schemes of all the universal sheets we are considering, so of this tangent bundle of the Hilbert scheme, or this Q of I, a1, I, 1, a1, I, 2, a2, or this taut logical bundle. So the fibers of, so here we have some, in this Q with some extension groups, so and here we have this taut logical bundles. So all can be described in terms of the combinatorics of partitions. We'll maybe explain this a bit next time. So that means we get a combinatorial formula for this expression, this a, and therefore also for this partition function zs. So there's also another thing which I will explain next time. Namely, one also finds there is a, if one looks at the expression, it has a very close relation to the neckers of partition function. And we can, we'll see this next time. I think my time is up. So we see each other. Anyway, thank you very much.