 Okay, so we have introduced the modernized space, of course, this was the termitized communication of the end-to-end and the table-experience classes, basically one of the things. There exists such a modernized space, but it is not compact. We want to still compute some numbers of them. But as an abstract theory, virtual dimension zero. In fact, it's a circular abstraction theory, so the tangent is similar to the abstraction. Now we wanted to get some numbers to it. So we want this abstraction theory, the virtual parameter task, but we still don't have the virtual parameter task because it's not compact. So we use a trick, so we try and compute this as expected dimension zero. So what we want to compute is the variation on the virtual parameter task of this modernized space, because the vector dimension is zero. So we have to make sense of this, because there's no virtual parameter task. And so we do this five equalization formula for this and taking what one gets from the first equalization for this S4. And so let's see how that goes. So I want to recall, before doing that, I want to review the other way how, so this modernized space was satisfied. Because they have another way to describe the same modernized space, which is, in some sense, which more dramatic or related to Thomas and Reyes, which is inspected first. So this is an alternative description of the same modernized space. So we look at this, let's say the total space PS is a total space. So this would be a two-fold, obviously. It would be a non-compact, and now I will see what happens to make it possible to take some. So I just want to see how we associate to such a spare sheet on the scope of space. So just so to spare, we want to associate, so in our modernized space, and I'll see it somewhat roughly, yes, like given as follows in S. Let me just see. So, so we look at the point x of S, the fiber EX. And so this homomorphous space has eigenspaces in addition to eigenspaces, where these i's into eigenspaces, where the corresponding eigenvalues are elements in this. We want to associate to this, so to this thing of the space, yes, the i. So the i is now an element. The other is mesh space project. See that this makes sense. So you have to see that this thing altogether gives you a polynomial sheet, which is not clear. I think it also would not be quite true if one does this. I say here, you also have to see, you know, what would happen that have generalized idea for position this thing has to make sense. Actually we'll find out that. So, so I'm very, I'm very, either be that this thing will be supported the vector space of higher end, but it could also be that. It actually turns out precisely the generalized space, then the support of the subject at that point will be non reduced. So there is a number of things to check here but this is anyway I'm not going to really use it but this isn't a piece not explicitly. So this is this kind of description. And so, if you see. If I release this. So this will be a sheet whose support is a finite number of points in every fiber of KS. Let me just slide the values. So it is a. Yes, I would say of dimension two. That means the dimension of the support is two. You need to, you know, the support. Yes, the support for which life school is right. And so, I mean I cannot. I think I'm not going to take some page. But I mean, you can imagine at least you can make yourself picture. You can see the level and endomorphism is very good. I found the risk of an endomorphism fiber. I can space the position and to get can make a sheet over the total space of the line bundle where the fibers, which lives only over the first one. But obviously they don't really make sense and use together. But they know this somehow ties it up with it. Thomas invariance. I don't consider such a few sheets of lower dimension to study to get something back. So in terms of this, you know, here it's also somehow more clear that one would get one gets also for general reasons. It's a tool of such theory because one is, you know, in a lot of yards, but I will not go really into this. So, in some sense, in the obvious way, you just, you know, you haven't just as a key skills. Maybe for future. When we want to work with this, we actually also want to say, we want to specify that C stars also supposed to act on the canonical bundle. And we wanted to act with the canonical bundle with weight one so by scaling the text. So here anyway the methods just by scaling this. So look at this alternative description with respect to that construction can describe it differently. So in this picture, it's not spectacular for the surface here, but it's, you know, this standard in the context of curves. It's a spectacular surface, but there is, you know, if you have a huge fields on purpose and consider that much more studied for a long time, and one way to study them is by such a construction. And which is then called a spectacular purpose for this support. This sheet will be curved then that would be the spectrum of construction. So the action can be as follows. On the space. Yes. To the action is this is scaling the fibers on the one. So you have a sheet on this space, and you just scale the files you get. And then when I said that we kind of keep in mind that one. Also, we have C star on on KS. So now. So you will see the fixed point of course, this allows us to apply virtual localization to define the event. So, if you, I will review the usual localization formula, which says that if you have a contact right, you have an excellent system, then you can integrate something over the smooth contact thing you can instead integrate the restriction over it. And now, in our case, it's not contact, but you put the same here by justifying this as a definition. So also the virtual situation we have the virtual class we have to use the version of the localization, which works in the virtual set, which is a virtual position formula. So now I just want to do localization you know what it's very boring. I try to do it fast. I think the usual or understand what you know, the homologation, yes or no. You work. You work. Anyway, I'm right. I'm right. Right. And so let's see. First, we have to explain from which I can write like this is each star. So we have so let's see where we have the project space. This is the universe. Look at this thing. So we have here an action of this down to such a portion of this meeting. So now this X, both on the team. And X, so we take the back the portion by the diagonal action. And so this is a, it's a module, the equivalent for module point, which according to this is just a TV by 30. And this will be that team where this is supposed to be. Nice in the second one. I can take this. I can take this T to be the first class of a line bundle from this point. The trigger line on the point with the action, which just multiplies the fiber, you know, times T times T, so with the actual way to see the action that T times. Well, no, we're going to go back to this operation formula. First, say it in general, so if you over X is an acronym for that is an action. X has a C style action. Have an action on me, which is compatible to the actual next. We have a career choices. C. We need to know how this is done. In general, I'm not only interested in the case that the action of T on axis trigger, but it's only on E. I want to know what you think. Also, if they prevent as a fiber wise action, the action of T, one is fiber wise, C star is a cognitive is committed to fence on one can prove that. I'm back to the J. In the most good way, so it. So, on the J, the action is that if T X one element fiber of J. This is just acting by multiplying the iron bundle. So maybe I should like this. But like this, it's not going to be anyway. So G. So I say these J's are integers, but it's not all integers. There's some J's. But it's a finite. Okay, so this is this. So it's a bit like this, and I'm saying in that case, obviously, the. So just the formula. J's. And. So, the terms of J as a bundle. It's a scale from the tensor. By something whose question. And so. And now, here, actively, j, I find it a bit confusing that I use t in two different ways. So this is one t and there's another t. I mean, maybe you can, for a moment, accept that, plus j times t. Maybe make it here and here. Okay, just keep it in mind. So, obviously, if you take the concentration class, I just also want to remind you what the Euler class is. So, the Euler class of the bundle. This would just be the same term class corresponding to the rank of the bundle. When E is an element in the k-loop, the difference of two bundles. So, E t, E minus f. This will, it's not even clear that this makes sense, but it does make sense because one has this t here. So that makes it possible to invert the order. You don't have to, in order to, so if you take the top-turn class, we don't have the one here. But we don't have to invert homology classes in order to invert this because it's a polynomial t. So now, the localization formula says the following. That's the formula so let's say implement homology. So maybe, I think it's the x of some dimension. As I mentioned. And here, that is the homology. That just means you just take class micro-homology so that if you kind of go to the homology, which means if you set t equal to zero, you get this original one. You can obtain this by, you know, if this is, for instance, an expression in turn classes of some bundles, which are equivalent bundles, then you can take it eventually. It corresponds to the technique where it leads up for the others to want. And then the statement is that critique the integral over x is equal. So the overall kind of difference of two is a four series of two. The way I wrote it, it could be power solution t, yes. Yeah. Yeah, I mean, you have a, I would have to see but I think it would be power solution. So there's a, you know, here called localization formats and some consequence of the utilization column somehow tells you how to compute, to express a grand homology class in terms of somehow a grand homology class is on the pixel side. But this involves also localization that is you have to be able to invert t. So anyway, the statement is that then I take one for value on x. And I can, instead, the sum of all the six point local slow side. So this is equivalent x i, six point locals, six point locals, a restriction. This alpha tilde x i divided actually sound amazing if you look at this formula the way I wrote it here on this side of the number. On this side, you have a rational function t. The claim is that they are equal. So that means this rational. Yeah, after we solve it, you know, otherwise individual things. Yeah, yeah, there will be some complicated. You can produce this tool. You will find some interesting way of writing. Okay, now we've come to virtual. Yeah, to be honest. Probably would take a long time but have we just stated in your words. So now, again, we assume that we have M, which is compact. So maybe one should say, I think one has to be very careful what one means by this but intuitively it's kind of. So this is this complex of sheaves, which. And so these are supposed to such a way that is an action on them, which is compatible with the map. The other one. And so everything must be this direction and the maps must be compatible with it. If one has this. So then it follows this. There's some work to say what this means, but you know, and then if one has given the correct definition one can show. And every connective component is also in use of structure. And this is in these papers like a favor. And so that I have not even done in the sense of explaining how this works. And so in particular, we have a virtual fundamental class. So. Now, if we take the virtual tension space of M. So such component art. I is now. So this is now a complex of two vector bundles on a space on which T x trivial. So it's you know, each of these. So it's whatever is your minus one of one minus one. And which are just vector bundles with a C star action on on on the space on which the star activity. So then you can compose. And then we have therefore. And then we have the final set J. DJ minus the sum of a sum. Whatever these are. These are time spaces. So now here we are in the cake. So where that means that. So, complex. And so, so, some, some, using the fiber that T and similar for you. And the same. Okay, so this is. And so, then, and I want to. Yes, I can say what the virtual normal boundary of the fixed point. I mean, you can imagine. If you have. And this is actually necessary for this form of sense. If you are so. And so the fixed point locals. The sub bundle of. Where the action of C star is. Whereas the normal. Where, or whatever. You can more or less write this diagram is the part where the action is not to be with X with it, possibly with some weight. So, therefore, and not causing you to. Where the action is a way from the, from the team, because you actually move on the fibers of the normal. So, and so the same. So the, the virtual normal one. This assumption will be just the part where the weight of the action is not zero. Now, she is different from zero to my shape before we take those. In the papers, they say it's the moving part. Things which are. And then he just can write down. The formula. So again, let's. Alpha. So maybe I want it again. Make it simple. So he was a new. The virtual dimension. The same formula. So the integral over the virtual fundamental class is equal to the sum. So you can see the formula is exactly the same as it's a non-virtual case only in the places where it's necessary one could severe. Same form. And one has to make sure that everything. I was. Besides. I didn't say much about how what it means to have an example of structure. And what the use of structure. Some work meeting, but then one gets the same form. Okay, so that's very simple. And now we want to say, so we want to define. Now, let's just formally apply this form. Good. You could say. Should be this. So we wanted to take this modelized space and one of it. This doesn't make sense because it's not compact, but we just find this. Let's go about things. Integrate over the virtual fundamental class. And then, in this case, you want to integrate one. So, obviously, one is one, so we can put one here is one over the. It's a bit crazy to find something like that. We have a theory which applies when this thing is compact, and we use this definition. It could very well be that the answer is nonsense, but it turns out. And then I said, the other time, there's another definition which looks much better, which is in terms of the so called. I don't know whether anybody knows this, but anyway, there's some kind of. If you have the sample of such you like this case you can associate with construction on the space, and you can just integrate this against. You can basically make a number of some over on the low side where this construction is certain very much like the other number of that number. And this is then you can show that if the things come back to do the same as integrating one over the virtual fundamental class. And so you could do this here to be much simpler. I think it was easier for you, but then it turns out that the answer is also very simple, much simpler than what one wants. And in particular, not related to what it is. So if I want to compute this now obviously one has to understand what the fixed point looks. So we have to want to take these integrals over the connected components of this point locals. Precisely. Because I just think of just two minutes. Yeah, let me see whether something which I can sensitively say, you know, I mean it takes, I mean it takes a while to describe this. Let me just see. I think I can at least it's been one. Which is not what one wants in general. So we will have to describe the point locals in general, but now I just want to look. What happens in the case which we are actually not really good maybe when the geometric genius is zero more precisely when the process kind of negative. And in that case, I think that there's only one contribution. So there's not so many points. So this is. So, so if, say, the degree of economic line. That's the negative. Then, which I'm next one locals. And so follow, I just wanted to do a particular case. I just want to say that very briefly. So, this is by stability. Yes. So, who was that. I don't know anyway so I can answer. So yes, in the paper of Tanaka and Thomas they also compute their function to get something and they. The answer is some, you know, you take the generic function get for something some very simple rational function which is not what one wants it certainly different from what I announced before. In the paper, Tanaka, Thomas, the first paper, there is a computation. Thank you. Okay. So by, so it's by stability so I can. So if you remember, stability was in terms of. So, maybe just, I can maybe roughly say in. So if I call. T, or she, I called T. And the reduce silver for the sheet. So, whatever we are. S. H. And one can also express this stability. So, the statement was that they reduce for not never polynomial of the sub sheet which lies in the kernel of a window which is invariant under. Under the X map. So, it must be smaller than we do silver polynomial. And the community one could also say, if you have an invariant portion, which lies in the open of the state, then it should be bigger. So, then. So, some more that I'm wondering whether. Anyway, so what one would have is so then obviously the kernel of the. The kernel of the. With the sheets, but let's say for the image, invariant and the peace. So, we must have the do silver polynomial of the image of P is a sub sheet of this one. And the kernel is a sub sheet of this. So we must have that P of a smaller than the image of P. And so in particular we have this inequality here. And now, what I actually have more just hope that it's true because I'm not using under my assumption, I seem that this is obviously unless he is equal to the zero map because then there's this is zero for. So they actually say it. And like a Thomas with smaller input. I thought to make it easier for this presentation, I would make this. But now it is, this argument actually just says that this inequality is not possible. So, it's not possible to predict anything because KS degree of KS is negative. So, this will always smaller smaller smaller smaller smaller. The problem with this purpose that I have not assumed that. So we have to maybe can, I can check the next time whether this is true in complete generality, or whether it may be anything. This is right. And the focus more or less like this only I'm slightly worried about the fact that. Much more. The statement. This thing. It's nice to be checked. Anyway, sorry about that. Let's go.