 Okay, thank you very much to the organizers for giving me the opportunity to speak and I want to apologize to the audience for being a person of the last century and not giving a slide talk or whatever, more fancy talk. So I hope you can read it. And so the topic is about the ranked honest Thomas theory of a certain Calabria 3-fold and in some sense it's a degenerate Calabria 3-fold. So in some sense ideally you would want to study Calabria 3-folds which are strict, so which have H1 of your Calabria 3-fold should be 0. And you obviously see that because of this elliptic curve that's not true here. Okay, so this is some sort of, you can say some sort of degenerate case. Of what we should expect in the true sense. So the reason to study this anyway is in some sense that from this degenerate case you might still learn something for the general case and then the second thing is also maybe just the personal motivation that is kind of, in this kind of degenerate case you actually can really give complete explicit formulas and that's kind of very satisfying in this field. At least for myself, okay. And so the second thing I want to say is that the talk will be slightly technical but I thought it fits well because it kind of uses a lot of the techniques that also will be discussed in Richard Thomas' talk. So I hope that is also useful in some sense. And yeah, if there are any questions just start. So I want to just recall a little bit, recall this, what is the standard setting? Okay, so if you have a smooth trajectory 3-fold and we'll just do the really the really classical case of curve counting, Donaldson-Thomas invariance. So we look at what is called this Hilbert Schemes of Curves. Okay, so this parameterizes closed one-dimensional sub-schemes of this x and we'll ask two conditions. So the class of this one-dimensional sub-schemes should be this curve class better and then the structure sheaf, sorry, and then we also put a condition on the structure sheaf and we also fix that the third term character of the structure sheaf should be n. Okay, and so maybe as a warning to, I don't know who follows usually this talk, this is a little bit different convention but yeah, don't worry about it. It's equivalent to just fixing the holomorphic Allah characteristic of the structure sheaf. Okay, and so another way to view this, I mean if since this is a school, it should just say it is the same as the modispace of Kizika, semi-stable sheaves with respect to some polarization of this with Schoen character specified by this data. And so then we learned in Richard's talks that there's a perfect obstruction theory and let me just write it out one more time. Okay, so there's this, you look at this universal sheaf which I call always E in this, in this case. Okay, so you have this universal sheaf on this product of Hilbert schemes times this thing and you look at the map to the Hilbert scheme and then you look at this relative form between the sheaf in itself, you push it forward to the Hilbert scheme, then you still shift it and then you take this trace-free part, then to really get this map to the co-tangent complex of your, of your, of to do a Hilbert scheme you just still have to dualize. So in particular this obstruction sheaf, I mean forget about all of this, the obstruction sheaf of this obstruction theory at the point corresponding to this E would be just this x2 of this sheaf and then the deformation space, so to trace this x2 and then the deformation space would be the traces x1. Okay, and so in particular from this data you, you get this, you get this virtual fundamental class, let me just put it here. So from this data you get this virtual fundamental class of this Hilbert scheme. So I drop this, this data here, so this will be char class in the virtual dimension of your Hilbert scheme and here the virtual dimension is just I mean just first sharing class integrated over this bit. Okay, so if x is Calabria then you can define the dt numbers of your space just by integrating, taking the degree of this virtual fundamental class, right, then Calabria means c1 when this first sharing class vanishes and then just integrate. And really the challenge I would say a challenge in the field is to compute explicitly, computed explicitly, explicitly for a strict Calabria, for a strict, strict, strict compact Calabria 3 and here strict is this condition I put here. Okay, so for example, a Quintic 3-fold compute this number. And so yeah, so let's consider this degenerate case that I discussed. So, and so now I mean that's kind of very unfortunate notation wise but I will not call my elliptic curve E so, but in some sense I have to call it E, so I will call it like this double E. Okay, so I hope that doesn't confuse anybody. Okay, so I think it should be clear whether what is the sheaf and what should be the space but if not then just interrupt. And so on this thing here, if you just use this naive definition I just wrote up that it's just 0. Okay, and the reason for that is I mean there are many ways to formulate it. One reason is that there's a co-section. Okay, so there exists the co-section. Exists the globus co-section co-section of the obstruction space, so of this obstruction space, sheaf. So there's a map sigma from the obstruction sheaf, so I should say hip to space to, so to the trivial sheaf of the hibiscus and we ask it to be surjective. And so how do you define this global co-section? How can I get this board down? Okay, so that's kind of part of this general this co-section stuff I think will be also discussed in this, in all these lectures we kind of play a role. So I want to prove this vanishing by constructing a co-section. In some sense there are two ways to this and let me just sketch them. And the second one is kind of closer to what we will do later and it's kind of maybe more clear what happened. So the first one is really goes back I think to Bufreitz-Flenner, the semi-regularity map. And then I guess there's also some work by Maulik, Pander, Pander and Thomas. And then I guess even farther back you can trace this back to Glock and so on. And so the idea is the follows and the idea is to well what do we have to do? We have to find a map from this obstruction space to a trivial sheaf. And so let's look at this guy here. So this is, yes? Why the existence of global co-section is important? Why do you need this? Yeah, so because it will prove that all the Donaldson-Thomas invariance are zero. And at the same time we'll give you a way to fix that. So actually define your non-zero counts. Okay, so I will just ask you to wait a minute and then I will kind of state this conclusion. So the first thing is to find this co-section. And in some sense this co-section should be some sort of trivial piece in this obstruction sheaf that doesn't play any role for for deformations in obstructions. Okay anyway, so how does it go? So well let's look at a sheaf E. So let's say look at a fixed sheaf with this obstruction space. And so how do you find the co-section of this guy? Right, so how do you, and so the idea is to use a certain gadget that also already was introduced. So I take it, I can say that I can use it. So there was this a tier class and so you just apply this a tier class here to the x3 and then you take the trace. So here's that tier class and then take the trace and you land in a certain cohomology group h3 of x comma omega x. And so now you have to do this community composition in your head because what is this space here? So x is this product. So this will be h2 of isomorphic to h2 of the k3 surface, structure shift of the k3 surface plus h11 of this elliptic curve. Okay, and so here there's a, and so this is in particular, it means unique, there's a unique dual of symtactic form and then the point class. Okay, so that's really just a one dimensional space. So I didn't tell you what is, I mean, I don't know, we didn't discuss much what is this tier class. Just a way to, a certain natural class assisted in this homology group and assisted to do it, you can write down this map. And so the way to show this is really subjective. I mean I said we really want to prove this rejection. The way to do this is some sort of calculation that I want to not discuss too much, but one way to do it is really construct an element in here that you can show goes to zero here. And so the way to do this is you look at, you pick, you pick an h1 homology class in this, in a certain group and you take it in here. And then you can calculate, okay, calculate the image under this map and I will not discuss how to do it, but the image will be this certain thing. And so then you can show that if you take this v correctly, this will be non-zero. So what is the, sorry, it's a bit fast, so I didn't notice it. So this is, w e is the, what is the w e there? Ah sorry, so this is the point class. I mean I want to take a generator here which is the class of the point on the elliptic curve. And I want, I'm saying if you take the class of point of elliptic curve and take the dual of the complex conjugate of your subtractive form, multiply them together, this gives you a generator of this, of the space and so the generator of the space. And this notation, this v in that product, sorry, very left. Yes please. I'm saying you pick any v here and then you take the interior product with this ATIA class. I mean all of this is not important. The way, I mean I feel this is the most natural way to understand this course section because somehow if you look at the model space of K3 surfaces, there's a certain divisor, so this model of K3, there's a certain divisor with this class beta that we started with here. So here I fix it, of course I fix a class beta, let's say a class beta comma d inside the Hilbert's, inside h2 of x, inside, identify this with h2 of your K3 surface plus h2 of your elliptic curve. So here's the beta and the d. And so there's a certain divisor in this model space where this class beta is algebraic. And so here I ask this beta to be, in fact, if beta to be algebraic. So I should say 1, 1 of s comma c under funcratuality. Okay, so I'm kind of messed this up, the order up of these things. So ATIA class is kind of a three form, is it? This one or what? Anyway, so sorry, so essentially what this says is that, so I have your point, so here is kind of the divisor where this class beta is algebraic and somehow you have this point here of this K3 surface. If you take an infinitesimal deformation of this K3 surface which is given by this vector v, then somehow the obstruction to deforming your sheaf along with it is exactly given, I mean, sorry. So then if you want to try to deform this sheaf along, there's a certain obstruction which is coming from the fact that this churn 2 of this sheaf should stay algebraic. And the obstruction to that is really given by this group. So in this way you can prove this is not true. Okay, maybe I did kind of a very bad job explaining this. So let me give you the second way to construct this co-section which is actually much more easy to understand. Okay, so the second one is using the action of the elliptic curve on your model space. Okay, so we have our model space here. And so in particular if you have a subscheme you can just translate it by this action of the elliptic curve. And so in particular this gives you a map from the tangent space into the first x group of your sheaf. So now you can just, I say, you can just look at this in a sheaf way. So, I mean, look at this in a relative sense. So you get the sky, get a really sub-bundle of this sky. And now by seaduality, so we have, so this is isomorphic to the dual one. And so now by seaduality, this produces you this co-section. Seaduality produces you this, or by dualizing you get this co-section. So omega e. And so because you have a sub-bundle you get a co-section. Does that make sense? Maybe that's, that's clear. And so, so why do we do this? So then once you have that, and there's this result of Kim Lee that has the two things. The first thing is that there exists, so the first thing is that this virtual class that we defined with the standard obstruction theory is just zero. And then this is kind of the first statement. And the second statement is that there exists a way to fix this. This is called this reduced virtual class. So this guy here will live in the usual virtual dimension plus one of your models. Okay, so the virtual dimension in our case was zero. So it lives in a, it's now a one-dimensional class. And so, so this class is then really non-zero often. And I mean, in most cases, and you really get a non-trivial dt invariance out of this. So, so the definition here is the following. So you're going to now define the t-invariance of the space. So this x cross e, then you just integrate over this class and you integrate a certain insertion. So, so a certain commodity class, natural commodity class on it. And the way to do this, again, is using this, using the universal, universal, let's say universal sub-scheme in this thing. So there's two maps to x and to the hibiscus. And what you do is, so you have this two projection here. And what you do is you, you pick a class gamma and a homology of your x. And then you define this insertion tau zero of gamma, which is just given by pulling back this tx by, to this thing, taking, taking a cut product with the churned, churned two of the structure sheath. And then pushing it forward to, to, to the, to the hibiscus. And so here what we do is we just take tau zero. And in some sense it doesn't matter really what you take. The, the most convenient thing is to take, you pick a class beta dual in the h2 of your k3 surface, such that the pairing with this given curve class beta is zero. And then you also tensor it with this, this class that I defined up before this, this point class here. Okay. And so in the essential way, I mean before in the usual carburetor three-fold, somehow there was just a single number that govern everything. In this case here, essentially this is the single number that governs all the curve counting in this, for this hibiscus. All right. So the cross-section is rare in this construction? Cross-section comes in here through this reduced virtual class. So, so kind of with the cross-section I know that this vanishes. And then Kim Lee also tell you a way to fix that. Okay. These two cross-sections are the same? Yeah, they are the same. And the construction depends on the choice of the cross-section. This reduced? Yeah, up here yes, yes. I mean up, well, yeah, so in some sense, so the reduced virtual class only depends on the k-theory class of your, of your perfect abstract theory. And somehow it doesn't really matter which kind of trivial piece you subtract from this course. So in some sense, it doesn't matter which one you take. But yeah, it's good to fix one. I mean, the reason we need one is to have this first statement. Okay, so what's the theorem? And so, so basically there are two parts and kind of the first part is, I guess, more exciting one, but the older one, and let me just still state it to, for complete list, these are old results together with Aaron Pixon and then Juniang Shen. And it says, if you take better to be primitive, so better is our class on the k-3 surface. So, so that will be always primitive. So let me just write the picard group here. So, so this is primitive of square, of square 2h minus 2. Then this dt invariant is easily computed as follows, just given as a certain coefficient. So there's a sign here which doesn't play a role. And then there's a certain coefficient of this egusa cusp form, one of this egusa cusp form. Okay, so, so you take this one over this egusa cusp form and then you subtract, you extract a certain coefficient here, free coefficient of this egusa cusp form. So, you take h minus 1, q tilde, d minus 1 and then p to the n. Okay. And so here the q and the q tilde are the free coefficient, I mean, as I said, it to tau and tau tilde. So q is e to the 2 pi i tau and then the same for the tilde and then p is the over z. And this guy here, this fellow is this, this egusa cusp form. So there's a certain Zege model form of genus 2, it's a cusp form. Yeah, some sense as nice as this discriminant model from delta of tau just now in genus 2. So if you, if you have any model form textbook, somehow you can just open it and you find this guy. And so now today I just want to explain the second part to this theme. So this is maybe, and this just tells you what to do in a, in a general case. So in a general case, you have the certain multiple karma formula and it stated as follows. So that essentially you can, so here this better was primitive. And so we can always reduce to this primitive case. Okay. So the, the way to do this is you, you sum up all the, yeah, means it, so, so better is the element in this lattice. And so it's primitive if you can't divide it by any positive integer bigger than one in this, in this lattice. And so, so the primitive case is somehow easier because, yeah, he curves out, you can control them more. But it turns out in this K3 surface case, the imprimitive one is in fact controlled by the primitive. Okay. So the way this works is you sum over the, all the devices of the GCD of this beta and this n. And then you have a certain sign here which is given by this n plus n over K. And then we have a one over K here. And then you just get a primitive invariant here. Okay. So, so for the Euler characteristic, you just divide by this, by this K. And then for the curve class, the D doesn't change. So the D doesn't change. And then here I want to really take better over K. But better over K is not primitive. So I didn't gain anything. So instead, I just assume it is primitive. Okay. So, so let me just say what is this prim here. So, so we are prim over better over K is a primitive effective curve class, curve class on some K3 surface as prime of same square of square given by this of square equal to the square of this, of this, of this class better over K. So you just take your better over K. You assume it's a primitive class of the same square on some other K3 surface. You calculate, take the corresponding DT round determined by maybe part A and then just, just plug it in. Does that make sense? Okay. So, so I'm saying here I take better tilde, which is, so I take, so take better tilde. So this is a algebraic class on some K3 surface. And let's say positive. So there's actually a curve in there, such that, such that the square of this class is the same as the square of this better. And then the prim of this better over K is, I want to define to be this better over K. So is that as prime any other K3 with that property or you really construct this start? Yeah, any other, any other, it doesn't depend by deformation invariance. So, somehow if you have a K3 surface with a primitive algebraic class of a given square then any two of these guys are deformation invariant. K cannot be 2 because beta square is 2h minus 2. I mean if, if K has to divide better here. Okay, so otherwise it doesn't work. Okay, so anyway, so this determines everything. Okay, so what is, for this, for this geometry in, in rank one? Okay, so maybe I, okay, so maybe I can make some remarks about this short history. So the first one is about, yes, about this problem. So, so maybe there are three remarks I can make. So the first one is if d is equal to 0, then, so d is the degree over the elliptic curve, right? So, so we have, okay, three surface elliptic curve here, then we have a s here. And so if d is 0, so d is the degree over the elliptic curve. So if you, then all your curves are somewhat vertical. So just live in, in this K3 fibers. So in this case, this, this statement just reduces to the K3 surface and, and then somehow there are like three results that one can state here. So if beta is irreducible, this goes back to kawai-yoshioka. And then if beta is primitive, there's this, I mean, this was then done by Moly, Panda, Panda, and Thomas. And then beta-imprimitive, this was done, then, by, done by Panda, Panda, and Thomas. And so let me just make also a remark for beta bigger than, so strictly bigger. Then just this part b, so, so part b, b for divisibility 2, was, is also, holds also, was done by follows, by work of Younger and Bay and Tim, Tim Bill on the gramophilm side, okay? So we don't know, for primitive beta, we unfortunately don't know any, we don't know the gramophilm-pt correspondence yet for the space, but on the gramophilm side you can independently prove this. And so then the final thing is the higher rank, you can ask for the higher rank and this is, yeah, still work in progress, but I think this also can be done eventually. Yeah, so, but on the gramophilm side you can just do product formula, and then this b is kind of a consequence of this, of this multiple-cover contractual forgate resource. So, so how much time do I have? I didn't want it. Okay, so for the last half an hour I will just try to explain the proof, and at least the key idea in the proof. And, and sometimes this follows closely somehow ideas that already been used here and also show up, will show up a lot in this, this week later. Okay, so let me just discuss the proof. So, proof only of part b. And so, so how do you prove such a thing? Okay, so, and so the ideas, well we have this s cross e, and you have to break something and what you break is always the elliptic curve here also. Okay, so, so the first thing is you degenerate the elliptic curve. Okay, so, so you have to smooth the elliptic curve and you degenerate it to a non-elliptic curve. And then you can, so this is just, you view this as the p1 where you glue this two points together. And then there's all this business about the degeneration formula. With the degeneration formula, let's say reduced to, then k3 surface times p1 relative to two fibres. And then with some other work, some more work. So, plus, then you can actually just reduce to the case where you serve a k3 surface times p1 relative to one fibre. Okay, so degenerate elliptic curve and reduce to, to this relative dt theory, theory of, of the surface time p1 relative to, to this, to the fibre was zero. So here s0 is five. And so I will not discuss this, but this is kind of standard. And don't worry about the relative dt, it will kind of disappear in a second. Okay, we'll, we'll just reduce the problem in absolute dt in the end. Okay, so this is the kind of the step one. Okay, and then the second step that you do is always you localize. Okay, so you look at this geometry, you look at this Hilbert scheme s, s cross p1, and then you take s0. And then you, you localize. So you take gm. And yeah, so essentially there's three different fixed laws I, one has to consider in this, in this, in this relative Hilbert scheme. So this one I call the extreme one, I haven't told you what it is. This one is what is called pure rubber. And this one is some sort of mixed guy. And in some sense, for us, what will turn out is that the second and third one will not contribute at all. And that's, both of that is not hard to see. Okay, so let me first tell you what are these three components? So what is this A component? A will be the one that is kind of most important for us. Okay, so let me just write it in some more picture schematic ways. So this is a sub scheme for curves inside really surface times p1 of the form. So, so here you have the fiber over zero. That is your relative divisor. And so here's our p1. Here's, I mean, my picture's three dimensionals. Okay, so here's, so here's our k3 surface. Okay. And so, so these are kind of really the, what should I say, the fixed curve that really lives inside just the surface times p1. And so how does it have to look like? So your relative over zero, so, so nothing can happen over this fiber over zero. So here you have really just straight, straight lines, maybe thickened or something. And that meet this, meet this fiber over infinity, then at the fiber of infinity, you have all kinds of fixed, fixed curves here. Okay, so you can have some sort of fixed curve here, or maybe I should use the same color. Fixed curve, fixed points. Localization occurs at the infinity. Yeah, so if you have the scaling action on the p1, so the c star, we use the gm action or c star action on this p1, just scaling the kind of this parameter. And then you have two fixed points, zero and infinity. And somehow if you have a curve inside surface times p1, that is fixed under this action, then you can have some vertical components over zero, some vertical components over infinity, and then just kind of essentially straight lines between them. And the ones over zero, we cannot have in this, in this, in this component, because we have this kind of condition that we're relative to this. And so if you, if you write this out, really you can, I mean, say even more precisely, this picture is kind of bad. So, so this is really the ideal sheaves in this geometry, such that if you restrict this i to surface times p1 without this fiber over infinity, and this is just pulled back from this, from the, from an ideal sheave on this, of ideal sheave of point on this, on this s, okay, so here, so let's say, so where eta is a zero-dimensional sub-scheme, sub-scheme of length d on this s. So just to short form notation, the modus phase is the Hibbel scheme of d points on the surface, and so eta is just the point. Okay, so if you take this ideal sheave of our thing, you restrict to, away from this fiber over infinity, where all the stuff happens, I mean the screen stuff happens, and you just get kind of this straight tubes here, which means it's just the, the corresponding sub-scheme is just a preimage of this, of this fixed zero-dimensional sub-scheme of the surface. Okay, so there's really a lower side that sits inside the Hibbel scheme of this surface times p1, okay, so in this absolute, absolute modus phase. And so yeah, let me also look at the other two, so b, what is this b? So let me also just draw a picture here, maybe c first, okay, and I draw this for the people who know it, this relative stuff, and so in the c you just, everything happens over this, over this, over this rubber component, and, okay, so here you have the original p1 here, and then here you have some sort of rubber component, so where you, it's like a p1 where you, where you identify curves when they differ by a gm scaling, and then you have some sort of stuff here, and then here you, on this side you just have straight lines, so this is this component c, and then component b is kind of a mixed, mixed, mixture of these two things, so you have some, some green stuff here, and then you have this, this straight lines, and then here you have some, some stuff. Anyway, so these two components doesn't matter, and somehow the whole action happens on this a here, on this part, and so this is the guy we have to understand, and try to, try to make sense. Are there any questions about anything? Yeah so, so c, in c I really want to say there's no, the only curves that are allowed in the surface times p1 is just, are just pullbacks, I mean the corresponding curve is just the pre-image of this zero-dimensional sub-schemes on the s, so it's just like p1 times the zero-dimensional sub-schemes, so there's no, yeah, there's no vertical components in any way. On the other end here I ask, I mean allow stuff here and then also stuff here. Okay so, so we have to understand this extremar component, and somehow the techniques are essentially, this is well known in some sense, so and what's nice here is we are on the surface times, well let's say we're in surface times p1, but what really matters here is this, this part of away from zero, so really can just view our sheaf also just restricted to the surface times a1, where a1 is this, is p1 without the, without the zero point, okay, so I mean all the data of this curve is somehow totally determined by just restricting away from the zero point because there's just, and so in fact we can just work on the surface times a1, and so, so the way to do this is then the following observation, so this is kind of textbook edge write geometry, so if you have a current sheaf, current sheaf on surface times a1, this corresponds to a coherent os of t module on the surface, and then if you, and so, so the way this to do this, if you have a sheaf here on this three-fold, you can just push it forward, okay, and so this gives you then this os of t module on this thing, and then if you also ask it to be gm equivalent, this corresponds to, on this side to saying that this sheaf should be graded, okay, so grading with respect to this t is just the same as having this gm equivalent structure, so in particular if you have that you can, you can just write your, your push forward as kind of a direct sum of this, the sheaf, and then there's this feature that we look at current sheaf, so at some point this, this kind of direct sum has to stabilize, okay, so we have some e r minus 1, t r minus 1, but then we have e r, t to the r, and at this point it becomes stable, okay, so, so here's the point of, which I call point of stabilization, of stabilization, stabilization, okay, and so this multiplication by t here, I mean, gives you maps here, between the skies here, t, and you can show that, I mean, now, now it's kind of, that if you flat over this a1, so if you take our ideal sheaf, this will be always flat over a1 because it's torsion free, and so this will correspond to the fact that at t from this e i to e j is injective, okay, so really you have injections here if you start with a flat sheaf, flat over a1, injection between these two things, so in fact everything of here just corresponds to also finer filtrations, filtrations, yeah, on, on this, on this s, finer filtration of current sheafs, and so the way to go about this last is I'm often just sending this to this, this inclusion, okay, so, so this is kind of the, I mean, I guess standard standard two, okay, so how much time do I have now? Okay, so, okay, so now I should, somehow, so let me still say some words about this d equals to zero case, okay, so the d equals to zero case is this, this classical case, okay, and so in this case the, the Hilbert scheme, I'm sorry, this, this component a here that we had is just this nested Hilbert scheme, okay, and so the idea here, so the idea to solve this, determine our invariance in this case, so this is the idea of Pt, is to consider this, our obstruction theory, and so obstruction theory of our Hilbert scheme and restrict it to our fixed component and take the fixed part of this thing, and then observe that this has a second obstructed theory, so should have, so, should have, so, so maybe I put it here, construct a second, second cross-section for, for this fixed part of the obstruction, so you restrict it to this, to this component, and so from this you get that most components, most components finished, and so in some sense it's kind of very natural to expect such a thing in, in hindsight of course, and the reason is as follows, so if you, if you think back to the beginning of the talk, some of the obstruction theory by part one was kind of constructed by thinking about this class better and kind of trying to deform your k3 surface to a loci where this class better is no longer algebraic, so now in this fixed loci we have not just a single class better, but we have the churn characters of all of these sheaves, okay, so we have kind of r plus 1 sheaves here in this, in this fixed loci, and they have all kinds of churn characters, so in particular, I mean we already use this reduced class, so we know that somehow we cannot touch the churn 2 of this, of this sheave e, so this kind of global churn character is kind of, we have to keep algebraic, but we are free to kind of move all of these individual churn characters, I mean, I mean we have, we have no condition on this individual churn characters, so in particular if you have some, some guy here which has a churn 2, which is not linearly dependent to this, to this class better, we can just move our k3 surface to some place where this class is no longer algebraic, so in some sense if you look at this thing, you in some sense expect that you should get r plus 1 different cross sections, and just by this kind of intuitive thing that now to get a non-zero number you just don't have to protect the algebraic city of a single beta, but you have to kind of keep all of these churn characters algebraic under different measures, that's kind of in theory, I mean already this argument allows you to kill a lot of this term, I mean not kill a lot of these contributions, the issue is that what happens, what may happen is that churn character of these guys could be all kind of linearly dependent to this class better that we already have, so Sagan is not quite sufficient, and so instead what they do is they consider the sky here, I mean the structure sheath, so in this case d equals to 0, we have I mean over p1, so in case d equals to 0 really everything happens over this a1, so you have just the sheath here, okay, and so in that case they look at the sheaths, and then because of d equals to 0, at some point you hit this case where you stabilize, and then what comes afterward is all 0, okay, so you just get 0 at some point, and then they kind of do something amazing which is that they construct explicitly deformation deformation of this OZ, so of OZ over this over some field of dual numbers, this yields you a tangent vector, so from this you get a tangent vector conclusion in this x1 pi of ee, and then yeah so I should put a t here, and then again by this by the serduality argument we get the co-section, yeah it should be the torus weight, so I mean in some sense on surface times a1 this is Calabria, but equivalently it's not Calabria, so someone this is this t, so I just forget about it, and so this really very nice geometric construction is gives you additional tangent vector, additional deformation, and then it produces you this, so in our case we have to go to d positive, and then there are somehow two issues, the first one is you really have to work with the deformation theory of surface times p1, so you're not Calabria, and then the second issue you have to deal with is that your sheaves, I mean in our case we don't have the zeros, in our case the zeros still stay the structure sheave of our zero-dimensional sheave eta, so we have kind of this additional terms here floating around, and then you have this non-Calabria, and then so anyway so you have to kind of try to extend this argument to the surface times p1 case, so this is what I try to explain in the last 10 minutes. Is there any reason to take this, why do all numbers, it's why you non-obvious? I mean you have to I mean you have to, I want to construct a tangent vector, and the tangent vector in, so what's the tangent vector of a scheme, it's, I mean in the model space is in the Siebel scheme, it's just the sub-scheme over the x cross the dual numbers, anyway so this is kind of a very nice argument, second tangent vector, but we kind of do something more direct, I mean more stupid. How do you know it's a difficult section? Yeah so this is, I didn't mention this here, so I'm going to do this first, so they prove that this is a different course section if and only if you're, if your OZ here is uniformly stacked, so meaning that, so I should have said this, so this is different if and only if, so yeah so so this gives, gives the same course, same course section as before, if and only if all of these guys are the same, okay, so all of them are the same and then your curve really looks like a fixed curve with in the same structure, and if you, if you, if you have one of them different, then you get a new course section, then you get a vanishing by this Kim Lieben method, okay, so, okay so now, okay so there's kind of, I think two observations here, so let's do the general case, so, so let me just rewrite this line that I just erased, so this is our sheaf here, so now e would be our ideal sheaf in the surface times p1 or a1 for simplicity, okay, so it doesn't matter, so this would be, so we have this the non-trivial stuff, and then we have this stabilization guy, so t to the r, t to the stabilization point, then we stay, stay the same, and here we get this ideal sheaf of this it, and so the first idea is to, I mean, okay this looks kind of complicated, I mean, so you have kind of this non-trivial stuff here, nevertheless you can just view this, you can in some sense just cut it here in this, in this stabilization point, and just view it as a filtration, where you take e0 sitting inside e1, and so on, under er minus 1, then you have er, so er is this, this guy, then you just add a trivial piece, os, okay, so there's a natural unique map here, and this is just os, it's just c, right, this is the ideal sheaf of this, it's just c, and in fact there's a canonical map to os, in case you just view this as a, just add this kind of stupid term here in the, at the end, and so this tells you, also in this case here, this a and beta that I wrote, it's just the union of nested Hilbert schemes, okay, so, and so here the nesting, so we have this, this parameters, so n, I don't know if I should give you a definition, so this is, would be just the sub-schemes inside er, and then inside os, where this ei, ej's will be just ideal sheaf of points of length nj, and then you tensor with the o of minus a beta j, okay, and so then with this convention there's a certain constraint you have, so you have beta 0 plus beta r is equal to beta, and in our case the last one will be just the ideal sheaf of this eta, so this beta r will be just 0, and then you have a second constraint here, the sum over the nr, and well, my notation should be like the summation over ni plus 1 half beta i i square is equal to, I forgot, so n minus r times d or something, okay, and so, so once you have that, so that's kind of the first observation, you just, you don't need to construct a new kind of crazy model space, you just work with a nested habit scheme, and then step two is this co-section, okay, so co-section, and so what you prove is that the following, so now let's look at this, our guy here, so this is our fixed part of our obstruction theory, okay, restricted to this, restricted to this low side, now what you can prove is the following, this is just a cone of the following guy, so this is a direct sum r home of ei ei, mapping to the direct sum over r home of ei ei plus 1, and then you put a trace free part here, and then you have to get the sum and rights, so it matters here, so take the deformations of the first r's, then here these ones you go for, actually 0 to r minus 1, so if I put here r then you get just a rank, I mean the theory for respect to d equals to 0, but here we put r minus 1, because we did this cut, so and so here this 0 here, let me just define this, and so here this r home ei ei 0 is the cone over r gamma of s os, r gamma over ei, sorry r home, and then the same thing also for this ei plus 1 thing, is the cone of the same guy here, mapping first to the same guy here, and then continuing with this natural inclusion with respect to ei, ei plus 1, okay, so here I can attribute this to, I mean Golampur, Seshmani and Yau, and also I guess Golampur Thomas, and then I mean it's really not, I'm epsilon square or something, okay, so now I don't have any time and more I guess, right, so I want to draw a final, I mean the final thing we have to do is just take the, understand this long exact sequence associated to this thing to find our cross-section, once we have that then we're done, okay, so let's just take homology of this, okay, so now we are hunting a second cross-section of this thing, so take a long exact sequence, and it's really kind of a beautiful nice thing to do because you learn a lot about this nested Hebert schemes by just taking this, stirring it, okay, so the trace free part, if you take the h0 of this term, you get home of ei ei, that's just c because the ei are ideal sheet, so they're stable, so this is just c, and then the trace free part just, the identity is killed out by this cone, so this first term is 0 and here you really get the Hormes model of this identity, so you get the first term here is the Horme, Horme of this ei, ei plus 1, Gauchner C times this natural inclusion, let me just write fi on inclusion from ei to ei plus 1, okay, so this is the first term, it goes to r minus 1, so this maps to the x1 here of this ei, 0 fix, okay, and so this term here really just describes you to, I mean, kind of tells you how to move, so the ei and ei plus 1 differ by a curve and some of this group tells you how to move that curve around, so this is this term, then the next term is, yeah, so we're here, then we're going here, and then the next term is here, so this is just a direct sum of this x i's, of this ei, ei, where i goes from 0 to r, and so this is also very natural because this just tells you how to, and when you have a deformation of the nested table scheme, it induces your deformation of each of the innovative members of the sheaves, so if you deform the whole thing, you also deform the individual ones, this map, and then the next one here is the x1 of this ei, ei plus 1, and then here you have the 0, and then you have er minus 1, okay, and what is this map here, this kind of interesting map, and let me just draw you one picture, so you have a deformation, so you have a guy here, let's say a guy, a tupper here, that gives you the deformation of the first sheave, I mean one sheave and another sheave, so let's say you're given a tupper for i, that just gives you collection of deformation of the individual sheaves, and then you have this natural inclusions here, and essentially this guy here just sends you to the corresponding condition that makes this diagram commute, so now I have to be careful, it's like fi, i for i minus this j, or something like that, the guy in the middle is not crest free, I'm just saying I'm just trying to describe you this map, so here, and so finally you end up with this obstruction space, okay, so now comes the punchline of the talk, so where's the co-section, and so this is really by trying to understand this sequence also, so let's just try to understand it, and so here we really took this r-gamma, and here I mean h1 of s or s is 0, so let's start at this first term here, so we have the sequence x1 of ei, ei plus 1, and so you go to this x1 of ei, ei plus 1 to the trace free part, and then you go around, you go to this h2, and then you continue, you have x2, and then you go to the x2 of this guy, which would be 0, and then 0, okay, and so now just, so the punchline comes in this thing to observe that this is just the home of ei plus 1 comma ei by said reality, do it, so in particular this is c if ei is equal to, if you are the same and 0 if they're different, okay, so if they're strictly contained in each other, so in particular if they're different, this guy is 0, and so this means this is subjective if they're different, okay, so if they're different it means you go to 0 here, they're subjective, so then you really have a map to this trivial piece, and so then what you see is that you get a map here to basically direct sum of all the i's where you're different, h2 or s, so these are trivial summands, then you show that this factors to it here, and then you win, okay, so the upshot is that in the end, so the corollary is that this guy here and better, so this, this is 0 if the number of under indices where these two things are different is bigger than 1, okay, so if then more than two steps in this filtration, then you get a 0, and so, yeah, so from this you, I mean you just do some calculation and then, and then, yeah, you get this, and you get, I mean it's kind of beautiful, you get from this geometric argument, you get the sign and you get this factor and then, yeah, I mean it's, yeah, okay, maybe I stop because I'm already over time. Questions? So you get this for the Iguzakas, we have this s3 symmetry, which can live to the derived category, yes, I guess the multiple curve you can express by the hacker operator, yes, you can do that, can you live that to the derived category? You mean this s3 action? No, I mean the hackers. Ah, to the derived category, I mean, no, because somehow, yeah, so this strictly does not come from a derived equivalence because somehow, yeah, I mean the invariants which are preserved, I mean, so for example, this, this, there's certain invariants you can write down for this, let's say beta 1 minus n1, you can write down invariants which are preserved under derived equivalences and which are different, so you can write down a certain number here, a certain number here, and they're different. If, for example, if this one is, if this has a divisor and this one doesn't, if the GCD of the second one is 1 and the GCD of this one is 2, then you can write down an invariant that tells it that there's no, I mean, which is a derived equivalence invariant, so you can never take one to the other. Okay, so, yeah, so derived equivalence will not help you prove the multiple-cover conjecture. Is it easy to describe this invariant? Yeah, it's easy to describe this invariant, sorry for not doing it, so, so the, yeah, so the way you do it is you look at this, look at the integral homology and you decompose it as like, homology of k3 surface times this class of a point on the elliptic curve plus, and then if you have a vector here, so this, I mean, if you have some guy here in the here, then you just take the wedge of this and you take the GCD of this wedge, and at least, okay, so that's, I mean, there's a slide, slide lie hidden here, so it's not, the auto group of autocorons of k3 times elliptic curve is not known. I think it's even not known for k3 surface, so, but under the natural ones, I mean, the ones coming from autocorons of the k3 surface and the ones from elliptic curve, you can prove that this quantity is just preserved under autocorons. And so, so the higher rank is that's, yeah, so the higher rank is not the torque, so, but yeah, I think this is also, I mean, I think that's like, yeah, I think three different approaches to it, so I think one of them would work. I have a question with multiplicity two, because Eugene was also wondering in chat. The fact that these hasted, hasted sheathness ideals come up just like in Richard's heading for the cotangent bottom of this, so is that just a consequence of the fact that you're using the same methods or is there a connection between the geometries? You mean that you get the same, I mean in you have s times e and he has the cotangent bottom of this? Yeah, I mean s times e, I mean, if you think, I mean, we have surface, I mean, we have surface times p1 and we really look at surface times a1 and that is just the cotangent, I mean, the canonical, total space of canonical bundle of this on the k3 surface because canonical bundle is trivial. So, high question is the fact that you can reduce to s times p1 is something that you haven't showed us and then it leads to the same things that. Exactly, exactly, yeah. And this reduction to s cross p1, I can refer to Pander Pander's lecture next week, so. So, then is it's no surprise that with the same thing should be solved? Yes, it's not surprising, yeah. I mean, what was surprising a little bit for me is that this issue with here can be solved too easily, but maybe this was just my stupidities. Could you explain briefly why components B and C doesn't contribute? Yeah, so here really the, if you calculate this, then the x2 really splits as a sum of two x, so for this component, ah, sorry, for this upstairs, really splits as of ee, I mean, maybe splits up to, maybe there's some factor, but let me make a small lie here, okay, so it's essentially the same as the obstruction spaces to these guys and maybe you have to worry that this, they match on both sides, where this is the adieu sheet for this guy. And so then you have a co-section here and you have a co-section here and they're independent, and so then you have again two co-sections, so this doesn't contribute. And here it's really dimension argument that somehow the reduced virtual dimension of this surface times p1 is, is 2d plus 1 and here this rubber just has a virtual class of dimension 2d, so there's some dimension argument that, that we don't need this, so this is actually, yeah, this is kind of, kind of lucky that this doesn't contribute. Any other questions? So you, you have this insertion of degree 1, how did you particularly choose the insertion? Ah, I chose this by tau zero, or you choose why, why I chose this one. Why do you expect much to cover for this insertion? I mean, you can choose, choose a different one and essentially you can re-express it in terms of this one. There's a different way to, you write everything and when you quotient out your model space by the elliptic curve and then, yeah. So essentially it doesn't matter which one you take. I mean, yeah, sometimes this is the, this is the, the natural one from several perspectives, but it's not, yeah. There's nothing particular about it. There's somehow a single number that you can attach to this model space and that's, that's, that's one. Well, to find if you're a reduced DTO fix, you make a choice of beta dual, such that beta, beta dual equal one. Yeah, and it doesn't depend on which one you take. Okay, it punishes when you reduce to the relative theory that beta dual, somehow. I mean, you have, I mean, I didn't talk about insertions, that's kind of also a different story. So you have to, you have to, yeah, deal with, yeah, deal with the insertions that you take. And here the surface times P1 relative to zero, it's not Kalabiao. So this one is not Kalabiao. It has the, this has reduced virtual dimension 2D plus one. In fact, you need a lot of, a lot of insertions here. And I just sweep them under the rock. Do you expect a similar story in the gromovitin side that you can reduce the non-primitive reduced gromovitin theory of K3 times E2 primitive in terms of primitive? Yeah, so I mean, on the gromovin side, you can just use the product formula and just reduce to gromovitin theory, reduce this question to gromovitin theory of K3 surface. And there's a conjecture for exactly that. That, I mean, any gromovin theory of, any gromovin environment of a K3 surface in primitive class can be reduced to primitive class. And that conjecture, I mean, I can refer to as Tim, to Tim, because he proved it in divisibility too. So, but in, in, in the general case, that's kind of an open question. So the expansion for a different point for the inverse. So that's related then to the high rank? Yeah, that's related to the high rank.