 in terms via Mochizuki formula, in terms of some generating function which I had called ZS A1, A2. So as it was our surface, we should remind you S is an algebraic surface with a PG of S bigger than 0 and first vector number 0. And as this was mentioned before, so this number B plus, which is the number of positive eigenvalues in the intersection form, is always 2 PG of S plus 1. So we see if that is bigger than 0, then B plus is bigger than 1. Okay, so we wanted to compute this. So we expressed this in terms of these, all these numbers in terms of this generating function, which were some integrals over Hilbert schemes, which I will not specify of some expression, Q to the n1 plus n2. And then we had applied, by a universality or cubordism argument, we had seen that it's enough to compute this in case S is a toric surface and A1, A2 are toric line bundles. And so that means we have that T, so X on S with fine-dimensional fixed points and so which I mean we need P1 to PE. And we had seen that then the action of T lifts to an action on the Hilbert scheme of n points still with fine-dimensional fixed points. So with fixed points, parameterized by e tuples of partitions. So we have e partitions and the numbers partitions sum up to n. And then we wanted to apply port residue formula and today it was also mentioned in Pavel's talk at the airport localization which I'm anyway, which is more or less the same but allowing also for the equivalent setting and so forth. So let me just write it so if X, assume we have an equivalent vector bundle on X and we want to compute say something like the integral over the churned class, some expression in the churned classes of it, then this would be the sum over the fixed points. So now we don't know where n is the number of fixed points. So we have the P of this equivalent churned classes at P which were defined in terms of the weight of the action and then I divide by the Euler class. So this would be TPI. So these are not the same eyes as here anyway. So this is 2 if we want to, if we just, if X is compact then we can like this just get this number. If we put the epsilon 1, epsilon, so the equivalent weight of the action so epsilon i equal to 0. Whether X is compact or not, we can also view this as push forward in equivalent homology which would be this expression as an equivalent integral. So this would be in the equivalent homology of a point. And if X is not compact we can also do it then we are, this sum will only live in the quotient field of equivalent homology of a point which would be somehow q field, a fractional function in the EI. So we can always make this expression. And now we want to, I want to relate what we have to compute here to the, to some versions of the Necker-Soph partition function. So let me just review this. So we look, we consider the following modular space. So this is a frame chiefs on P2. So say m of n, I just consider rank 2 case for now. So this is pairs of E, which is a torsion free sheaf on P2 and phi. So E is a rank 2 torsion free sheaf on P2. Whose second churn class is n. I think it was also on the blackboard. And if I take phi, it's a trivialization of the restriction to the line at infinity to the trivial bundle. So an isomorphism. Okay, so this is a smooth and quasi-compact, a quasi-projective, but it's not compact. And this carries an action of the torus. So you have c star, so t equals c star to the 3 acts on this mn. So first, c star squared acts on P2 fixing the line at infinity. So this gives us, and then we get an action therefore. So just if we take the homogenous coordinates, if we put the homogenous coordinates x0, x1, x2 on P2, and we have an element t1, t2, and we act on it, then we can just do this by leaving the first coordinate fixed and multiplying the other ones. And the x star c star acts on this framing. Namely, if we say, if this is the coordinate on c star, and we apply it to such a pair of a bundle and the trivialization at infinity, then this is sent to the same bundle, but we change the framing. For instance, s to the minus 1, 0, 0, s composed with phi. So we just rescale the fibers or the two directions of this trivial bundle in different ways. And so this gives us torus. And now this action has finding many fixed points. It turns out that this is just, the fixed point is just given by the fact that this e splits as a direct sum of ideal sheaves of zero-dimensional schemes and phi is just the identity. So this is i z1 plus i z2, the identity, because where ci are elements in the Hilbert scheme, well, I just say of a2, so this is the open part of the Hilbert scheme of a2 away from l infinity, which are supported in the origin, so of some numbers, ni, where this i zi is a monomial ideal, and obviously n1 plus n2 is equal to n, because the churn class is given by the singularities of the sheaves. And so we see, in other words, we had seen that a monomial ideal is given by partition of this number ni. So that follows that the fixed points of mn are given by pairs of partitions. And now the knickers of partition function is that we want to integrate something over this thing using the formula I just am wiping out. So for instance, the simplest version is the pure theory. In this case, we just say that this partition function, actually the instanton part, but I never consider anything else, so epsilon1, epsilon2, a, and say q, would be the sum over all n bigger equal to zero to the generating function of the following integral to integrate over this modelized space, one, q to the n, where this is meant in the sense before we use localization. We apply this localization formula, so this will be an element in the field of rational functions in epsilon1, epsilon2, a, and then obviously the powers using q, where we just apply this localization formula from before. Yeah, no, actually, so I should say that a is equal to the logarithm of s, epsilon1 is equal to the logarithm of t1, and epsilon2 is equal to the logarithm of t2. We had seen before that to an equivalent variable we associate something such as addition, corresponds to multiplication, so this is what it is. Okay, so this is the simplest case. So now the Neckar's of Conjecture, basically, I mean what I would call the Neckar's of Conjecture for this case, would be, say, an explicit formula for the lowest order terms of the logarithm of this thing. So the lowest order terms in epsilon1, epsilon2 of this. So, for instance, for this particular case, there are several proofs of such a formula, one I know is by Nakajima and Yoshiyoka, and from this, by somehow some argument, one can deduce the formula for the wall crossing of the Donaldson variance in case p plus is equal to 1, so this is related to. So one's got talk, and one can, however, also look at more complicated things, so one can add some, I think, what the physicists call meta, so we can integrate some bundles over it. So for instance, there is a universal, so there's a universal sheave, E over P2 times this modular space, and so therefore we have, so this picture, P2 times the modular space with the two projections, Q to P2 and P to the modular space, and we can consider, for instance, this universal sheave to be, so in the normalization that I've seen, you take the push forward to the modular space of this universal sheave, so R1 push forward, so fibrovised the first homology, tensor Q up a star of E of O of minus L infinity, so the functions, which this is all equivalent, so we look at one more function, vanishing at the line at infinity, everything in equivalent, in equivalent K theory, and so this makes sense because we push forward along this compact direction, so the fibres are P, it's just P2, so this is fine. And then the knickers of partition functions, say with fundamental meta, would be, we, I can write it like this, so it depends again on these things, epsilon 1, epsilon 2, A, some number M and Q, and so here we again take the sum, we take the integral over M, and now we integrate over the Euler class of this dart log to sheave, and we twist it by some equivalent character, M, which I think is going to call it the mass of something, Q to the M. And so the knickers of conjecture for this is also shown by Nakajima Yoshioka, and then from this one can, using this one can prove the written conjecture for algebraic surfaces that the written invariance give the same information as the Donald's invariance, so this was also done by Nakajima Yoshioka, and they allowed me also to participate a little bit. So then we can look at other cases, namely a joint meta, so this would be, we have some other mass, and this would be, we take the Euler class of the tangent bundle of this thing twisted by some equivalent character, okay. But so this somehow should, is something which seems to be related to the Euler number of the modelized base, so one could have hoped that this is what we need in our case, but unfortunately that's not the case. We need both, so we consider both. So it depends on Epsom 1, Epsom 2, A, M, M, and so this will therefore be integral over Mn. So some N of the Euler class of V of N of M times the Euler class of this tangent bundle twisted by this. Now this is actually unfortunately, so now indeed we can show, it's one can show that this thing I wanted to compute, so maybe I just write it in a kind of, who is it? So we can show that this thing that we had, this Zs A1, A2 that we wanted to compute can be expressed in terms of this, actually in a very simple way. So if S is a projective Tori-Serkzvis and we have here the fixed points of the action P1 to PE, then we can, this function we wanted to consider to compute our invariance can be expressed just as a product I, so maybe E, Z, W, so we take the weight of the action on the local coordinates, and then we also somehow have to suitably assign what locally happens to these. So there's the A, I don't write what this precise is, so Ai Mi, Mi Q, maybe with some pre-factor and we again take the limit of Epsom 1 and Epsom 2 equal to 0, this will give me this. And so one can, so why should that be the case? At least we can see that one is summing over the same thing. So if I take the coefficient of Q to the N, so here on this side we are computing on the sum, on the distance union of all Hilbert schemes of N1 points times N2 points such that N1 plus N2 is equal to N and for this Hilbert scheme it's an E tuple of partitions. So this is a sum over two E tuples of partitions and on the other side, if we just look at one of these partition functions, the coefficient of Q to the N is a sum over pairs of partitions and we have E factors, so it's also two E tuples of partitions, so we sum over the same kind of thing and then one can check that term by term you have the same thing on both sides. If you just compare with the Mojizuki formula you can actually see where, which one is the, there's this thing with the tautological classes will be the fundamental matter, the P of thing which is just the P of E is the joint matter and what is in the denominator essentially gives us. So this was, so we were looking at Mojizuki's formula and so the statement was that if one wants to compute something, but if I want to take the integral over this modular space of some expression classes coming from the universal sheaf then this was equal to the sum over all ways how we can split this C1 as a sum of two classes in the second homology, this I wanted A2 and then we have here the Zybergritton invariant of the first and then we have some integral over the, we have the sum over all n1 plus n2 equal to C2 minus A1, A2 of the integral over something on the Zybergritton scheme of points. So that was how the formula was and so A1 and A2 are these classes so in particular A1 corresponds to the Zybergritton class. What? So I explained it the other time so it's just, so we have the notion of weight so the X is a toric surface so that means that every fixed point we have coordinates which are eigenvalues for the action and the weight with respect to T of these is W of X1 at the fixed point is the weight of the action on the coordinates. Okay, so we find this I will not go into, so for these the first two it's there and it has been proven and for this one there's no statement so if you have it with both a joint matter and a joint matter and fundamental matter there's not even a statement what the formula should be and this is also true because from the physics point of view I was told that it has to do with something what is called the conformal dimension which is somehow the expected dimension of what you get when you cut down by this and so when you cut down by the tangent bundle the dimension is zero and then you cut down by this and that somehow is not physical and so therefore the physicists don't want to touch it and so there's no prediction and in particular if there's no prediction there's no proof of the non-existent prediction. Okay, so that's our problem so basically if there was a knickers of conjecture one could try to use that in the same way as before to prove a buffer written formula so if a knickers of conjecture would hopefully, I mean with a big effort so with a very long lead to a proof of the buffer written formula but there is not even so the problem is we don't even know the statement so then we don't know how to prove it so instead it remains a conjecture we can just check it so instead for now until somebody tells us what the answer is we just evaluate the combinatorial expression for the knickers of conjecture and then if you have the knickers of conjecture you could do this you somehow have to compute this and then you are supposed to so you have to first evaluate this and then you have to take the residue at s equal to zero and if you can do all this you would get the answer but remember that anyway so we just evaluate the combinatorial expression because as I said then we had that this is just by counting over partitions so let me just say a few just very few words about it so obviously to a partition we can associate a Young diagram so if nu say s equal to n0 until nr then the corresponding y nu is a diagram of boxes I mean in the positive quadrant so we are with the columns of length n0 to nr so whatever if associated to we would get now which one did I want so we would for instance have that this would be so we would have here first 3 and then ok would be such a thing and so we can think of this also in terms of so these correspond to monomial ideals and we can very easily see how we could say that here this box corresponds to 1 this corresponds to y this corresponds to y squared this to y to the 3 and this would be x x squared x to the 3 and so we can see that if we have a monomial ideal so thus say if I take the monomial ideal i z nu which is the ideal y to the n0 x y to the n1 and so on corresponding to the partition nu then we have that that the for instance the structure sheaf of this thing is just the vector space generated I mean has as a basis all the elements x to the i y to the j such that i j is in this diagram and the ideal sheaf is precisely the set of i x i y to the j in the positive quadrant such that i j is not in the diagram so we see we can describe everything in terms of this so in particular the and so if we have the we know what the action on the coordinate is so we know therefore what the action of a basis of the vector space is because these x i y to the j are bases so these are bases so we know the weights of o z nu if we know the weights of the coordinates and then the other sheafs that we consider we consider both such tautological sheaf twisted by some line bundles and we consider some sheafs of the form x 1 i z i w and these can also be expressed in terms of this so there is a formula so in particular for instance if we take this this will be the tangent bundle and so accurately you can compute this and this is there is a formula for this in terms of what Anton said arms and legs so if you write this thing here you somehow if you have a some box in it then this would be say depending on how you define it say this length here would be the arm leg length of this and this would be the leg length and then there will be there is some formula for this which expresses it equivalently in terms of the sum over all boxes in this z in the young diagram so then somehow maybe I will not write it precisely it would be equal to the sum over all s in the young diagram so if it is of nu so if this is z nu often so this would be a basis of this so x to the something with the arm length of s and something with the leg length and y something with the arm length and something with the leg length minus plus some similar expression but I don't write the I mean if you want actually maybe as I wrote it almost I can also write it so this would be x to the minus leg length of y nu of s a y nu so maybe I just write here l of s for the leg length y to as plus one plus x to the leg length of s plus one y minus the arm length of s and so this means what is supposed to mean is that this space has a as an equivalent basis which consists of as many you know vectors as their summands here and on each summand and on the summand it acts as on the monomial x to the this y to the this and so this gives us the precise description equivalent of all the vector bundles we consider and so we so if one this out with the computer so you can for instance compute this what you had here s a1 a2 s q modulo q to the 31 and so this allows us this gives us a check of the Waffen-Wittgen formula for many surfaces up to expected dimension depending on the surface maybe between 30 and 50 so it's rather well tested and similarly I mean the same kind of computation you can do for the kaiwai genus and you'll get this so let me just so as I said generalizations would be the kaiwai genus the elliptic genus the cubordism class and then we I can maybe briefly talk about the case of modellized spaces of rank 2 also the rank 3 case so maybe I can briefly talk about this as I still have a minute so let's look at the case of the modellized spaces of rank 3 so I should maybe say there's a version of mochizuki's formula for arbitrary rank for any r so it's just the higher r is the more complicated formula comes but it's always similar and so we can also evaluate the rank 3 case and I just wanted to show you the answer so we apply this in rank 3 so we have again s is a bg of s is bigger than 0 d1 of s is equal to 0 is our surface and so we find that the virtual so the virtual dimension of the modellized space will be in this case 6c2 minus 2c1 squared minus 8 times the volume of the Euler characteristic and let's again assume for simplicity that there's an irreducible canonical curve otherwise the result will be in terms of so we again want some expression terms of modular forms there were in the case of rank 2 there was also a theta function in this thing, it was just a standard theta function we also have theta functions here but they are slightly more complicated so we look at theta functions for the A2 lattice maybe just consider 2 called this one theta A0 so this in the moment I just make it depend on the variable x which should be q or some rational power of q if one really wants to view this as modular forms so this is now over rank 2 lattice so we take this pair nm we multiplied with the matrix with the A2 matrix and so we use this to make a pairing so if one wants to write this out x to the 2 times n squared minus nm plus m squared and the other one we is a similar one where we introduce some kind of sign or rather a root of unity so we put epsilon equal to e to the 2 pi i divided by 3 and then this is just the sum over all nm in Z2 of epsilon n plus m times the same so these will somehow be modular forms of weight 1 for some group and we however will mostly want to consider their quotient so these are modular forms of weight 1 and we can consider their quotient so you can find a modular function which is Z of x which is just a quotient so zeta A0 of x divided by zeta A1 of x so it's something very explicit and it's a as we have a quotient of two modular forms of the same weight I mean obviously it's not just this and that so we want to express in terms of this but it's not quite enough we have to solve some algebraic equation so we define two other modular functions for some other group Z1 of x and Z2 of x which happens to be Z1 of minus x a solution of some quadratic equation namely W squared minus 4 Z of x squared W plus 4 Z of x and then we can finally write down the formula so we have this function the generating function for given C1 is 9 times 1 over this 3 eta bar of x squared to the 12 to the chi of 4s which is somehow responsible for the case of K3 surface times 3 times eta of x to the 6 to the 3 divided by one of these theta functions this is to the power KS squared that would be nice but it's a bit more complicated so this is multiplied times Z1 squared Z1 of x to the KS squared plus Z2 of x to the KS squared plus minus 1 to the chi of 4s times just a sine I mean which is epsilon to the first-journ class times KS plus epsilon to the minus the first-journ class times KS okay so this thing will be either one or minus either 2 or minus 1 okay so this is the formula you can see this is we have this Z of x now we make this quadratic equation and this has so Z1 Z2 are the two values of W that come as solutions of this equation yeah okay why do we have okay yeah if that's yeah I thought that was maybe understood okay so this is the formula there's also a version for the chi-y genus which is very similar but with a small extra twist I will maybe not explain it and okay so now I haven't so this was this I haven't made my statement I mean the statement is always the same statement before but I should at least make it namely the conjecture is that the virtual Euler number of this modelized space is equal to the coefficient of x to the power of the virtual dimension of the modelized space of this expression so again this one can do the same argument as before we apply Motrizuki's formula we apply the Kubotism argument to restrict ourselves to the case of Toric surfaces and then one does the computer calculation to check this up to a reasonably high degree so enough so that one is convinced that this is actually correct okay and if yeah maybe maybe for now that is all I wanted to say so okay thank you