 We want to compute them for sufficiently many examples because if you compute them for sufficiently many examples as they are universal, we have become GSA number after one, two, three, one, technology, joint surface, surface, surface with an action of C-star and C-star applied from pickpoints. Pickpoints, historical line boundaries, that's an action in the line number. And actually for the S-alpha, we can take, in each case, we can take just P1 times P1 each time what we vary is mostly line numbers. We do it in such a way, surfaces in line numbers in such a way that the vectors of numbers. We are given here, namely, that of here is in the right way. So these are just, you put all of them together, these are just numbers, and these circles of numbers should be linearly independent. It follows from this formula. I know GSA-alpha for all of these. I know them. I know what these policies are because I get just linear equations for, I mean, for training its back. So I take the linear equation for the logarithm of these, and so then I can use them. So, so, so in this case, if that's such a correct surface, then we have that GSA-alpha, which is an action of C-star times C-star times the fixed point, but this leaves to an action of these programs. So in some form at the beginning, let me just, again, so let S be called surface, that means S is an action. So in fact, it's, I can write it as 1. So, but you can say before, so every fixed point has actually a neighborhood which looks like a 2. And the coordinates of this E2 are activated. So the C-star times star acts on each fixed point. So I assume of E2 with coordinates to the left, but it will be a certain weight. Yeah, I may be interested in a certain weight, which the weights are some, maybe I will not write this, because it's using the first step. And what do I mean by this? T, T1 times Xi is really equal to T1, N1, Xi's notation, like this, these elements are eigenvalues for action, and similar for C4, this is some code that we want. Similarly, we have the weights for WI, so this distracts the action. And then, then T also acts on SF, say that I push forward or back. Yeah. So, and there's still a finite amount of fixed points. First, the sub-steam Z is a fixed point. Its support must consist of fixed points. The action on S moves the sub-steam around, so it shouldn't move the right, because it doesn't move around the support. So then, let me write Z equals N1. And then, secondly, this thing has an accurate table, which is each fixed point, each point that I had to name it, which is as small as E2. So with these coordinates Xi, so in the action of this E2, this sub-steam of E2, which is concentrated in the point PI, must be in there. And I point support it in PI, so that is a fixed point of action, same as the table scheme of NI points S in the origin, E2, this we write to the system, with the, where you have this E2 with these coordinates Xi, this is a fixed point. Now, if you, what does it mean to be a fixed point? It means that the idea to say that the idea has to see this is equivalent to this is generated by normals. Okay, it is generated by normals, because you need these invariants, and if it's not generated by normals, then some are obviously something else. So, but if it's generated by normals, you can also easily see that you can write it in a nice form, you can choose it later, so that you can write it down nicely. You can also see that you can write ICI, to make that now, just X and Y, write X and Y for X, Y and Y, I, to write to the M0, so maybe I can just have normals, and it says in one vector series of the portion, X, Y, by this instance, what's missing is M0 from here and 1 from here, and so on. And you can see it's just, if it's generated by normals, you just take the smallest polynomial which starts with X to the 0, this will be this, take the smallest polynomial in the idea which starts with X to the 1, which is this and this will be generated. So that's N0. Therefore, we can put this together, we have this thing at every six points, at every six points, we have a sub scheme of something MI, such that the sum of the MI is N, if you're under the scheme of N points. So, so that's six point locus, S, N. T is in the direction of the six point and the sum. And in the same way, if we look at this S, N, in the scheme set, obviously, so it's on union over, this in by direction, with E times R. So, we want to compute this, what we are supposed to compute is this integral over S, N, this had a very complicated formula. The only need to remember what the ingredients are, what was in this. It was, was the, basically, the order number of some R on P of some extension of some universal idea sheaths with other ideas sheaths. And then, you know, also, the order number of the divide by the order class of something corresponding for the universal sheath which is also given by such extensions of ideas sheaths. So, what I want to say therefore is the following. So, when we want to compute this, when we have to compute this, this obviously is, so we use localization, this will be therefore, and so this becomes equal to the sum over all of the action, which we know was divided. So addition over, which is gamma, a, we restrict it vector space with an action, or some element in the cave with an action and we just take this, this will give us some number divided by the order class. These things are quite similar. Maybe they are given essentially by X groups of ideas sheaths. So, ways that can be divided. We can see minus. This is basically essentially the, not, not just X one of the ideas. This is plus now. So this is the vector space. And obviously, this always needs to be this equity. And in the gamma, this is the, this is the oiler. We have the following and zero. Yeah, so therefore there are several. We have, we are in this product of these habits. So, there is not just one idea sheaths are different ideas. So, you get like this. Yes, which is somehow made out of these ideas. He is somehow just written as a sum of such things, divided by some of other such things. So therefore, and also here this one is such. So therefore it's enough to understand what the action is on such a space. The idea of one universe of the one. So, one such idea. So there are those 10 or something. And as I said, they're saying to the experts, or there's the form, something simple, then there's the X one, which contains almost all the information and then there's the X two, which is something simple. So, and so, therefore, the following formula, I first tell you what formula. Sub scheme. We can associate partition as we had seen. And for this, the sub scheme constable point and associate partition. You can associate the young diagram and you can put you can put this in terms of this. The sub scheme is a sphere for y zero x y one. And so on diagram. Just is a sentence holds all the elements in the function by this idea. There's squares in the thing. So, this would be this. So, this would be zero zero. The next one is anyone. This would be the point zero zero and three points. This corresponds to five would have it like this partition would be three to one month. This gets this. For an element in this plane. We can find numbers with respect to this. This is the arm length of the leg. To assume our element s here. Then the arm length is how many things are above it here. So that's how many are above it. The other one would be the leg thing. And the arm length is how many years. So in this case, if I take an element outside, then maybe this number might not be next. Anyway, you can see what these are. So in terms of this, it can run. So, I'm going to get six points. And then we put some young diagrams. Then we write down this formula. So just a polynomial. Depending on these two diagrams is a polynomial. This is x and y. This is equal to the sum of all s which lie first young diagram. It's very similar x to the. This is just a polynomial. And the idea is that this is supposed to tell us how something acts on samples in the end and you write down such a polynomial. So now let me. So let's start this. One, two tutorials for the young diagram of, say, C one, which is supported. The idea of it is a idea generated by monomials in x, I, why I, and so it has corresponding. And then we have the formula. So in the representation here is saying that you want to know minus C one, two. This is equal. We have to ask ourselves that this thing is to what happens without the adhesion. So this is minus sections. So this is just a homology of design. And then plus contributions from this. This formula here, which is the number six points on the surface. We take this w. Yeah, place x and y. This is a certain. Essentially, this is a vector space with an action of this force. So writing here is basically the composition of this into spaces. So first we have this day. We have a set the sections, the homology of this line battle. And this isn't a vector space, it will somehow have an action of the tools. We can compute. Then the non-trivial part is this. So here we say, we know. And when we write something like this, you know, as a representation range of T, it means T x itself. We have a vector space of the dimension. These are terms. These terms here, why why does number one by two, the dimension of this vector space. And on each day this vector, it acts like it acts on why to the power of this and X to the power of this Y to the power of this. And these are I vectors, which act, you know, where exactly. Then you finally have my type by this M is an actual line line. So as a fixed point, it's an actual value. Like it's just an actual one dimensional vector space. So it also is the action of it also is a certain way. So it's just a straightforward formula now that I cannot say, I don't think I want to say. There's a different proof. So it's actually this is by having to be very long time ago. One uses somehow a resolution of this idea sheet. I mean the idea sheets are given by these generators. So this gives you some kind of resolution. And in terms of this resolution, you can include these X groups. And you get this form. And the other proof, which they somehow haven't noticed. There is by Jim. Nice. So in this case, they somehow. In terms of representation, representation, so this formula. I mean, what should we say that we have, we have all the ingredients to fly the localization formula for one. So, and so we can write down this very complicated formula for you to just put it together. And on the very start, you want to find the close formula for that. But this we have instead one can just take this as it is and put it in a computer and the computer. And so we have done a bit more, but these four. Compute series is that we have here. G. Okay. In these cases that we needed. And so the result. And then. So, the result and we get. How many. So this is a bit. So I put that. Maybe either show you the slide. We have already seen that, but you know, now. Might want to see them again because a long time ago. What the outcome of this is. But in that case, I don't know, I don't need to hear so then I would might need some help. So, I am supposed to stop. Maybe then it's complicated. That is something. I also put it on. Then I have to put something. The thing is so complicated. I don't know. I also would have to share the screen. I'm not sure it's worth it, but you know. I would have to share the screen with you. It's not going to assume that now how do I put it on that thing. And it says on the. Okay, so otherwise. Yes. I don't know, but I mean. No, no, I don't think that makes a difference. There's something there. That's too complicated. Yeah. The buries for us not working. But anyway, then I. So, okay. So, so let me just, maybe I can do it. Okay. Now I just have to see. So it is on Zoom. So we I just started with the form. Maybe I don't know how this is called. Next page. But then what we're going to test. Thanks. So now I've used all my time code. So we have maybe, just I repeat. So we had seen this before. We have this. I want to have to use the structure formula. So we have this beta function. We have this data function for the AR lattice, which was given by it's not so important what size formula is, but it is different. Yeah. And we have put beta AB for two tasks in the chromology, V1, V minus zero. So this is finished. And now structure theorem of La Raca, which said that what is this vertical of the written invariant was given by this explicit thing to the power pair of S. I just had called it A, but I had said that you can do that. And then you can normalize this thing. So, and then there's something to the power KS square, but I can pull out this term to make this one easier. And then we have the sum left before, the sum delta C i, sum over vector i, times the product of this other thing invariant. Times the product of this, I mean, it was a situation between work. And now the aim was to know what these things are. So, you know, we just have, you know, I already said it, so we have this other side of the written invariant. And the thing that one might want to remember that S minimum general type, then the vector zero is one, the other is zero. Okay. And so we make this, write down this thing, which contains all the long table information of this. So, as you can see, we've just pulled out these factors that we know. We just write basically this lower line here. And this is all like this. And so we need to add, if the answer is enough to know this. And for simplicity, so I write again, so that we don't forget it. And then, so we have contractual formulas for this. So, here I just recall before, we, this is a set of functions we had before, but now we take this function of such data functions function. And I'm going to call it this. And then we can first look at the case of rank two. So then this thing you want to compute. So the thing we need to compute is very simple, but not, so we have to define this thing as the function of these two data functions. So in this case, data is zero. This is just the standard. And then what we get here is just data C one zero plus, so now we assume that in the case of general time. So we deserve a question in variance, but just one or zero and minus one will have a square. This is, you know, just to make the formula simpler. So then we just have these two simple terms. Okay, this is then the answer for what the words are just given by this data function. Now, in the right three case, it is a little bit more complicated. So we have just two with these portions of data functions like this. But for some reason, we have to solve an algebraic equation. So we have the same thing that C one zero and something that C one to S that is one minus to S and something else. And here we always have these data functions. This was not better function, but in addition, we take the solution of this quadratic equation. So X, you know, there is two rules of this question. And we take these, these will be some modular functions in some or some smaller complex equation. And then finally, if I will once, you know, this is only for basically to scare you in rank four case, obviously the problem starts terrible. But it's interesting that you get something as interesting as that. So you look at this function, which is can be written as a certain function, as you can see, social data functions, which also has such a nice fraction development. And then in terms of this thing, you can write a format. So it looks terrible, but it's not, you know, it's not outrageous. You have something for delta C one zero, something for delta C one two K S minus K S. And so you always have these data functions into this portion of data functions. And in addition, you sometimes have this Q and you have in addition is Z, which are the roots of such another quadratic equation. So, you know, the thing is, so I'm going to have thought that, you know, it's nice. So physicists say that we should in terms of model of forms and model functions, this is nice. But what is a little bit surprising is that we'll get something rather complicated. But there's a nice structure. Obviously, the real task will be to understand what it is in general. And then finally, just to show you, this obviously you cannot understand or I cannot understand. So the final one in rank five, you get this order of common new ones, but you need fractions. So you get this start in terms of this hour Q, which is this, or this product. And then, then, you know, you write down this formula, which is against the instructor. But, you know, it gets kind of complicated. You have many terms, which are, you know, you do something with this, what I don't know, you have to solve the number of product equations. And in terms of this question, so this, and so it's, yeah, I mean, it's, by itself, just, I mean, I mean, I don't know. I don't know, precisely, how interesting for you. I mean, it may be originally I had bought something slightly simpler than this, but you know, you get some formulas, which are, I mean, the thing is that you can compute sufficiently many terms, so that basically can be sure that this is correct. But on the other hand, this doesn't come with an explanation of where all these things come from. But anyway, this is the result that contested in the five case until then, and in higher end case, we haven't been able to guess. Yeah, so we've also, for six and seven, but somehow, okay, well, sorry, so I'm not sure how, whether this is for women, but that's, you know, in the end, you get the formula, and the formulas sometimes are pretty, and sometimes they are, maybe not quite so pretty, but that they are what they are. Okay. Yeah, so that I will actually take the next time. I mean, that's basically the next thing I want you to say. I can even check, maybe it might be that I forgot that this actually exists. So if there was another slide, but anyway, there is a formula also. So one can defend the horizontal part, but it's more difficult. So one can compute a little s, it's not only up to right, but there's also by this s duality, one expects a relation between the horizontal part, and that actually gets, finally becomes some rather neat relation, so that you really can see how functions go one into the other. But this I will maybe say in the next time, because it's not s, I mean, that's why, you know, basically, in this particular, you know, you would have a very similar formula where instead of these Cijs that we have here, you have some other functions which are Dijs, and they are related one by the other by placing tau by minus one over tau, and as we all have formula forms here, one can see how it translates, but this I would make the problem. No, I mean, that's actually, you know, it came as a surprise that one needs this for another equation, because we had somehow hoped that, you know, it would just be given in terms of the set of functions that were given, and in fact, that was somehow the expectation, but it turns out this is not the case, and we have to, we need somehow some model function which cannot be expressed directly in terms of this set of functions. I mean, just, there's no explanation, just, you know, by itself, it's just, you know, obviously, you have, it's still all in terms of modular forms, so if you take, if you solve the quadratic equation, you have just the cover of the corresponding modular curve, you have some other group that one has to see, we haven't, we haven't tried to figure out what people precisely did, but anyway, it somehow, somehow means that it's not just, so, you know, just say one thing, so there, this was mentioned there, so there, if you look at these formulas, you have always something with the theta function or with the discriminant, this comes somehow from the fact that, you know, for, it's rather easy to see that for K3 surfaces, you have to have this discriminant function, one over that, and then you have these theta functions here, and you can see that those come from the Bloch formula. There's a formula which relates what happens on the surface and the Bloch of the surface in the point, this requires that these theta functions must occur in a certain way in the formula, but somehow this doesn't determine everything, and the fact that it doesn't determine everything, that more complicated things happen somehow related to these, or not many equations have to solve, but I mean, I cannot do the proper answer. There's actually, there was, I think some, there was some physics paper in this context where they claimed it was only in terms of these, but that's actually wrong. So, you know, it's a, you can, you know, some paper people start from the assumption that it can be expressed in terms of the theta function and this discriminant, and then easily from this term in arbitrary rank, or these above written generating functions, but then it seems to see that there are many health examples. So, instead of writing these