 Okay, thank you. I apologize for the remote aspect of this lecture. So, okay, let me, then I'll start, I'll try to share my screen as I write so. Let's hope this works. Okay, no, no. Okay, good. So it's, I hope it's visible. Yeah. Okay, so we'll, we'll just, so we'll discuss two topics. One is something that we started last time. So this is a topic of the representation of the diagonal in essentially a self product of our modular space. This is very, yeah, so so we'll be very specific very soon. So, and the second second topic, I'm going to start to say some things about the homology rings of quote schemes. Uh, sorry. Okay, feel free to interrupt anytime in any case. So let's start with one and here for us the basic fact, let me recall it since it's been a few weeks. The first thing was that if we give ourselves a non singular complete variety M, and we consider the diagonal inside M times M. Let's say we have projections P and Q to the two factors. Well, if we manage to write. The class of the diagonal as as a sum of split split classes coming from the two factors. So something like this. Where is alpha, I paid higher just classes in the Chow of M. So this is a representation in the Chow ring of M times M. If we can write this, let's call this star. Then well let's say the alpha eyes but it goes the same of course for the betas. They generate the Chow of M as a Z module. Okay. And so so, so this is just a it's a very basic fact but it underlies basically this whole this whole discussion so let me so in other words if for any X of course I can write for any class X. I can write X as the sum of some coefficients and these coefficients are of course. So these are some coefficients that say AI. Okay. So you can really represent it as a as a linear combination of these classes of I just because we're representing a diet and okay so so then you can also show further and so if if this relation holds then so if you can represent the diagonal is a split, then in Chow, then in fact the numerical and rational equivalence are the same for M inside and the cycle class map from Chow to homology is an isomorphism. So then this M is actually Chow trivial so it's a very strong condition. No, I mean this is not going to happen easily. But in any case even if it does not if the same represent the in in commonology. Of course you can, you can always split everything using the couldn't decomposition so in for what we are for what we're about to say you know this this might at least on the level of common this will shed a lot of light but okay so so this is so the same statement holds in homology homology once you represent the time you can always do it in homology in in this way then the alpha is will generate the homology. Okay. So, as an example last time we considered the code scheme so let me this is a this will be our first example today we're going to we're going to sort of and look at this a little bit more closely. So we had considered what's while we let me denote it in this manner. But of course so our and then are some fixed parameters are is less than or equal to n. So this is just the parameterizes exact sequences. And here we're in the setting of P one. Okay, I will comment on that later but we give ourselves this exact sequences on P once we're concerned it was him on P one. So, this he has a rank R and degree minus D. And of course F then has a rank and minus R and degree D. Okay. So this this this code scheme in the case in the sort of geometrically more significant case when this is a proper grass money so R is less than an this this compactifies the space of maps to the grass money. So in here's maps from P one to the grass money. Which sits as a note of degree D. Okay. So it sits as an open open subset of the code scheme. Just, and what what is a geometrically well we're looking exactly for maps looking exactly at those exact sequences where F is is always locally free the situation but F might have torsion and when F is also locally free, then it actually comes from a map to the grass money. So, I should say before I go a little bit further with this I should say that basically the subject owes a lot I mean the, the, the, maybe the earliest thorough geometric analysis of this code scheme came in work of strome. Exactly this code scheme so from P one and this is a classic paper from the 1980s on parametrized rational curves in grassman varieties. So a lot, a lot is calculated there as I will describe. As we go on. Okay, so. This is a paper from the, I don't know, I don't remember the exact here but it's, it's back in the 80s. Okay, so. So now, what about a code scheme also we have seen the following that if we consider. So here, if we consider two points in the code scheme. So let's say, given by two exact sequences to. What's nice in this setting and which you don't have actually now in a strictly in a modular theoretic state when you when you're only parametrized sheeps here obviously we're parametrizing quotient of a fixed sheaf. But when you only parametrized sheeps you don't have this. There's a nice total logical map here so let me. So between E one and F two so. So what you can do is you. You know this. You use the, the first inclusion of the one and CNN and then you use the second. Surjection onto F two. So this gives a completely naturally a morphism whenever you have two sections let's say, well, let me actually sigma one two from E one to F two. And this homomorphism is zero exactly when the two points of the course him coincides exact coincide exactly when you have. In fact, the same you're talking about the same portion. Okay, so it's it. Yeah, so you can consider it is a non zero homo homomorphism which is zero if and only if the two sequences coincide. And so what we also saw last time. Was that we calculated. We calculated that X one. So this is calculated on P one between E one and F two in such a setting is always zero. Okay. So this is this uses the fact that we're on P one it does it's not, it's not always true right I mean if we were on a curve of higher genus this would not be true. This is simply H one. One P one of you want to do a left to know just to. That's what we're talking about. So, so we calculated that this is zero. So this of course, I'm using the same sequence. This means so let me also. I also saw that. So this is for all for all to for. Yeah, so for arbitrary points, arbitrary points in arbitrary pairs of points I should say in the code scheme in QD. Okay. So we also saw that if we were we talked a bit about the tangent space. So we saw that this is equal to X zero. This is true for any code scheme in any setting of E into F. And because of this vanishing of X one, we actually obtained that the code scheme is smooth. Since this X one vanishes then. Okay, so we have these smooth smooth projective variety since I should have said I mean they're this is a projective this is a projective variety but it's a smooth projective variety in the case of P one. Since X to one. So anyway in that case as a as a sheaf of course I can write that the tangent sheaf is. This will reacquaint us with the universal sheaves here. So here I'm using the universal sheaves. There's a universal sequence on the code scheme. And this is the projection. I mean I have the projection to QD and roll projection to P one. Okay. So. So now we just to carry through this with these observations so we, we may also consider on the complex. Let's call this w. Well, I mean I can look at. Yeah, let's not be restrictive on the product. So this. So here, the setting is that you know this is to the, the twin this is mean that we're looking at, so looking at a pair of code schemes. And this is the universal sub chiffon QD on the first factor times P one and you have the universal quotient on the second. Okay. And buys the projection to just the product. Okay. Yeah, so then this w is on on. So what X one as we saw will vanish so in fact this is this is a this is a vector is double is a vector bundle. This is this vector bundle. Okay. So, and so, of course it's on QD times QD and what's nice about it is that it has this total logical section that we were discussing. Yeah, so for for a pair of points in the code scheme, you always have this morphism. So this is a homomorphism from this. The fiber over a pair of points of doubly is precisely the homomorphism group from E one into F two, and there is a canonical one. Okay, so the code schemes come with this canonical section canonical section. And as we saw the so we saw what it is on on every we saw the value of the section on each pair of points at every point of the product. So, in particular, we have that the class of the diagonal is precisely the zero locus of this section sigma. And therefore this is the same as the top turn class of w. Okay. So now it takes. So again, we can calculate the class of the diagonal by the remark theorem so well. So, so the remark theorem says that now, if I can calculate the turn character of this double you know because this is just a push forward shift. So, well so the turn character of two is just and minus turn to. Okay. So quotient shift. And then I have the top genus of P one here on plus the class of a point. This is it. So now if I, we may write, of course, if I. So let's think just of homology now explain exactly how. So. So you may write. This is important for us with fixing some notation for us so we're going to write the turn classes of the universal she sub she. Well. So this should come in terms of a as a community composition in terms of the homology of P one so that's just the fundamental class and which is one and the class of a point on P one. And this is for I between one and are so there are our such turn classes. And now a I is in H2I of QD. And of course if I is in H2I minus two. And here you may this notation is classic I'm using it because it's the classic notation that at the end bought used in their study of the endless equation of Riemann surfaces so I'm just sticking with it as well. Okay, so this is for the community composition. Okay, so notice here that this F one so if one is just D. Yeah, so if one is a. Yeah, there are two are minus one classes of positive comological degree. So then if we, if we do this then. Well then, then you proceed with the calculation and it's clear that you know the churn character of W is written as a, you know, as a sum of say poly polynomials of homogeneous degree. I mean of homogeneous polynomials of certain degrees of these are the churn pieces of the, the pieces of the churn character. And you're going to have, you know, the a variables and the F variables corresponding to the first universal sheaf and the second universal sheaf. Okay. That's it. Let's say this is degree. And so then you just, you just expressed the top turn classes expressible in terms of the churn character so we also obtain this so this is the top turn class and the top means. Wow. It's also expressible as a polynomial in, in exactly in these classes yeah coming from the two factors. It is a standard transition from turn turn turn from the turn character to the turn class. Okay, and this is the diagonal and you see how you we've done exactly this we have written the diagonal. In terms of just classes coming from the two, the two factors. So, so then we, we have obtained that so we have obtained in fact that let's call this a proposition, the tautological classes, AI and fi generate the homology of the code scheme. So now, I just want to remark on the difference between China and co homology because obviously this argument as I presented it is just in common you use the community composition right of the of the turn classes of the universal sheaf, which you may not have, you may not have in child. I mean, in this case, you do, but as I, as I would say, but okay, so the point was, let me just, so as a remark, and here this is remark concerns child versus chronology. So it comes with this fact, well, we had to, yeah, so we, we consider this push forward. Yeah, I mean, that that's what got us started. Well, and this is the top genus of P1 here somehow. So this came from looking at M times M times P1 and projecting to M times M. Okay. Also, you could rewrite this a little bit with the, and so just to see clearly where the distinction between child and chronology comes, you can give yourself, we can give ourselves two copies of P1 and really consider things on the universal sheaf on one and then the universal quotient on the other. And I can do the same calculation if I just multiply by the class of the diagonal here. Well, so this is the same as, so, so pi star and now, yeah, so I use the tablet is miserable. Okay, so, so this is the same as pi star, the same thing. And now I have, you know, I have two classes. I have this thought of P1, the first P1 and the second P1. So I have two copies of P1. And of course, I have to put in here the class of the diagonal. And so now you see, it's, it all depends on this class of the diagonal. If the class of the diagonal itself in P1 times P1, which is the underlying variety splits. Well, and it does in our case, splits in Chow is written in splits in Chow as a sum of, you know, as a sum of factors from the two, from the two copies of the variety, then the whole product will split. Yeah, because you're genuinely pushing forward, you're doing two separate push forwards, two completely different worlds. So what you get on M times M will split. So, so then I want to say that it all depends. So if, so if the underlying variety is Chow trivial, which P1 is the diagonal splits as a sum of hyperplanes on the two, on the two factors, then, then the diagonal will also split in Chow on the, on the space M. Yeah. So this is again an observation of stroma. So that goes back a long, a long time. Yeah, so if, but this, this will fail. For example, if we look at, if we look at an elliptic curve, you know, here instead of P1, the diagonal will not split in E times E, which means that, you know, you cannot expect this reasonably on the modular space side. Yeah, so, so, so somehow. Yeah. So, so if the underlying variety, in our case, P1, the simplest possible varieties is Chow trivial, then the modular space will tend to be Chow trivial as well. That's just what I wanted to say. So, so if Delta splits inside one, so as Delta splits, maybe I should say then so does Delta inside the code scheme QD cross QD. Now, this is the remark. So then in particular, this shows that, you know, then the code scheme is Chow trivial. Yeah. So, so Chow and cohomology agree in this case. There's no, there's no odd cohomology. Okay. So this is the discussion of the diagonal in the code scheme on a curve. Okay. So, so now I will, I will talk about a modular space of stable sheets, which sort of, it's more aligned with the, with the, you know, that Lothar has been talking about in his lecture so far. So let's say that M to see, to see what we can, what can carry over in that setting. So, so this is a modular space of stable sheets, stable sheets on a variety X and this is either a smooth projective curve or a smooth projective surface, projective curve or surface. And if we're talking about a curve, there's no C2, you know, but, and I'm going to assume there are no semi, no strictly, no strict semi-stables sheets. Okay. And I'm also going to assume there is a universal sheath just to see how this flows, this argument. There's also polarization. So the stability is with respect to an ample ample line bundle on, on X. Okay. So there's a universal sheath on the X and project to X or to M. Okay. So let's see what we can say about the diagonal in this setting. Okay. And there's a bit less structure floating with a code scheme. We have a lot of structure with a modular space. There is less, but we have stability though, so which we don't have on the code scheme. So here, let me note. Yeah. So I also assume that this is, yeah. Okay. So no strictly, no. So it's a situation when there's no, there are no strict semi-stables. So this is a, is also projective. So the smooth projective variety I'm dealing with here. So if, if I'm trying to calculate homomorphisms between a pair, I consider a pair of sheaths in M. So this is, there are no homomorphism unless, actually, like this, unless the two sheaths are isomorphic. And then they're just the trivial, the trivial automorphism side. You have it. Okay. So the only automorphisms of a stable sheaths are, of a stable sheath are just to see starry scaling, scaling. Okay. So let's assume, and this assumption is stronger than what we need, but let's, let's see what it means. So it's a bunch of assumptions to set the scene. So we're still, we're still setting the stage here. So let's assume also that X2 for, for, for any pair of sheaths, E1, E2, parameterized by M. Okay. And so this is, okay, this is automatic if you're on a curve, of course. On a surface, it's very restrictive, of course. And so, yeah, so this is automatic. This is X, I'm sorry. If X is a curve, and this is restrictive, if X is a surface. And so, but how can we understand it a little bit? So this X2 by, by seriality, this becomes on a surface. This is X0 times the canonical here of the, of the surface. Yeah, so dual, of course. But so when might this be zero, this X0 that we understand better homomorphisms than, yeah, than this X2 group. So, so this is, for example, this is zero. The latter is zero if, for example, if the polarization times kx is negative. So this is by stability then. So by stability, you can never map, and I think Lotta also touched on this last, last time, you cannot map a stable sheath of higher slope to a lower slope to a, to a sheath of lower slope. And this multiplication by kx, if h times kx is negative, lowers the slope. Yeah, so then by stability. So you cannot have homomorphisms from higher to lower slope, basically. Well, so that happens, for example, on take p2, but you know, it is restrictive, so it's, it's not gonna happen a lot. So, but let's, let's, and I'll comment a little bit on that, there are ways you can strengthen this, but this is the, this is the kind of the perfect, and I'm setting just the perfect scenario for this, for this representation of diagonal to work. And if the scenario is less than perfect, then you know, you, you can do, you have to do other things, but sometimes you can, sometimes you cannot. So yeah, so, but, but if this X2 is zero, that that's why we want it to be zero because then you can, okay, so we can consider, well, I'm getting a bit ahead of myself. So, sorry, so I consider as before, not quite as before, in analogy with, with our first example, okay, the code schemes, we're gonna consider the complex, and here I'm gonna stick a minus on m times m. So this is the same, the same scenario as before, you have a universal shift on one factor and universal shift on the other factor. Okay. And in the context of this vanishing of X2, in fact, well, the X2, so this, the X2 sheaf in which appears in W is zero. So in fact, you can resolve the point is that under our assumption, we can just resolve W by a complex of locally free sheafs. So by a two term complex of locally free sheaf. So you can, so under our assumption, we can resolve W by the complex of locally free sheaves. So I will have a bit to say about this resolution. So, so this then, so then obtaining something like this, so let's talk about these as you can do it in several ways, there's a concrete way to do it, which I will point out, but you basically will have the following. So here is what you obtain. Let me write this. So this is an exact sequence on the product m times m. And I wrote this first term just for inspiration, because in fact, this sheaf is zero. Okay. Zero, it's, you know, the, so you'd, you'd expect that you see something along the diagonal, of course, the torsion free sheaf is zero, but it inspires you because then you understand that it, the diagonal defines the, the, the locus in m times m, where this map from A to B fails to be injected. Yeah, which is where exactly this X1 sheaf, the X zero sheaf is, is zero X1 is not a locally, it's not a locally free sheaf. Yeah, so it, and, and, and where it will jump rank, it has a generic number, where it will jump rank is exactly where this map from A to B. Let's see. Yeah, let's say that it's called is fee is, is a fee fails to be injected. Okay. The fee is not injected. Yeah, so the, so then there would be what we would put, would have to check. So this is a degeneracy locus. So then the, this is, this happens on the diagonal, right? Because unless you're on the diagonal, the X zero group is equal to zero. So X1 is constant is given by minus the Euler characteristic of E1 into E2. Yeah. So, so then this jumping locus is precisely the, the diagonal. And so then you would have, you would hope to write if things were not degenerate. So you may hope to write the degener, the degeneracy locus of fee, which is definitely fee. You try to write it as a, as a, as a class as a child class. And in fact, the portious formula tells you what it should be. Let's say the ranks of A and B are little A and little B. This, this is by portious, kind of portious formula tells you this. So, yeah. So now what is, how does this compare to the, well, the diagonal we know is co-dimension M where M is the dimension of the modular space. But this works out very well in fact, right? Because B minus A is the generic rank. So B minus A here minus chi of E1 into E2, okay, which is a constant here. And, and this in turn is the, is the dimension minus one. Yeah, the modular space minus one. Okay. Okay. Because the X group on, yeah. So it's the dimension minus one. And so then, so then this, then, then this works out pretty well. And this is okay in fact, because you, the degeneracy locus is exactly the expected co-dimension, which is exactly M, the modular space. So this is actually okay. It's not just a hope. It's, it's true. You still have to check something. So this, yeah. So you, you have to check the degeneracy locus is actually reduced, let's say, you know, so the, the porous formula tells you, tells you that the, the, the, the, the, this turn class is equal to the, the class of the degeneracy locus with its natural scheme structure. So you have to check that this is exactly the diagonal, but just a smooth structure on the dial. Okay. But, but that's easy. So, so then this is exactly, of course, this B minus A is precisely the, the complex W. Okay. So it's CM of W, because this resolves. Okay. So, so, so then we're back where, you know, we're in the same situation as before where we had to work a little bit harder. I mean, on the code scheme, it was easier. This was a top turn class of a vector bundle. The, the diagonal was the zero locus of a section of a vector bundle. So it was a beautiful setting. This is just a little bit more work, but not, but it's, it's not bad. And yeah. So then, of course, so finally, as before, by Groten di Creman Roch, you can calculate the turn character of W. And so then you can express the turn class in terms of the turn character PC. So in terms of, in turn, the, the QNET components of the, of the universal shape. So whenever this is possible, then in homology, so you can then conclude the QNET components of the turn classes of E, which naturally, of course, they belong to the homology of the product, you know, decompose in terms of the basis for the homology of X, be it a curve or a surface, generate H star when Q multiplicative. And of course, as remarked before, if the diagonal is factors inside X times X in Chao, then this also works in Chao, if not not. So, right. So I should also remark that then, you know, you can, this argument, of course, was given with, in this simply, with this hugely simplifying assumption that, that X2 was equal to zero. So this argument was, ah, okay. So first of all, I should say that this argument is also due to strone, the argument above and maybe Boville, Boville kind of, these are early 1990s, okay. And it was adapted by Markman to work in the case of a K3 surface. It's 2000s for X, K3 surface. Okay. Yeah. So there, of course, you don't have, then the, this, you don't have this vanishing of X2. So E1, E2, or this is, this is the case of a K3 surface. The canonical bundle is trivial. So, ah, so this is then zero. Okay. So this is, so it's zero if E1 is not isomorphic to E2, but of course it's C if E1 is isomorphic to E2. Okay. But in any case, it's a, it's a small variation and then one can deal with it. So the argument, this is just, it's just an adaptation of, of strome's argument. So Boville's argument. Okay. So one thing that I did not say as I move away from this subject is to, yeah, how do, how do you obtain the resolution? Yeah. So this relies on this resolution here. And I mean, you can always obtain it, it's a, okay, but it's, you can actually do it in, and it's sort of instructive to do it very concretely. So it's not, you know, that such a resolution exists. Can you actually write a complete one? Yes. So I, well, let me see if I really want to say something about this or not. Yeah, maybe, maybe I won't actually. In fact, it's a, it's a nice, it's a nice construction, but I see that the time is actually a bit short. So unless I don't know it, I can go either way, depending on what the audience wants. So my inclination is to skip it at the moment. So let me just maybe just write the remark. So you obtain A2B from the geometry of the modular space itself, from the geometry of the universal sheaf. So sheaf. So, so, so, and the key point, then you can sort of think about it. So the key point is that you can choose a large integer, M, so that the push forward of the universal sheaf twisted by a large multiple of the polarization given by M is equal to zero for, so that the higher, the higher direct images are zero. Okay, so this is actually a vector bundle. So if I call this, and moreover, it's a vector bundle so that if I, so this is a vector bundle on M, but if I pull it back to M times S, it surjects onto E twisted by H to the N. So this is true on M times S, M times X. Okay, and this is true. And so this is, this is, it's sort of, it's built in deeply into how you construct a modular space. So sheaf, this boundedness is part of it. Yeah, so the fact that you can, you have this, it's, it has very much to do with how stability plays into obtaining a bounded family. And anyway, so if you call the kernel K here, then this helps you get the resolution, basically. So on M times M, so let me call this, you use star for E2, for E1, let's say, and you take text with values in E2. And then, and then this precisely gives you the resolution. Okay. Okay, so you get, get precisely this, get the desired resolution, resolution of X, of the X complex. Okay. So, yeah, anyway, so this is, this is, so with this, I'm, I'm ready, I think to, so we leave it, I leave it to the audience to finish this, this little argument. Okay. So, so, so now I think I'm done with, yeah, I said what I, you know, was reasonable to say about the diagonal, I think, in general terms. And I would like to go back and time flies, I'm not here. So I'll just, let me at least get started on my second topic here, which much more is to be said, in fact. But so, so this is the study of, now that you have this result of generation, it was a study of, of the cohomology of QDGRM in my original setting on P1. This is instructive, it's sort of a, I view it as I'm sort of toy model for what, for what you might expect to happen in, in a, in a setting which is not as simple as these are after all quotients on P1. Okay. So, chief quotients on P1. So it's sort of a simple setup. Okay. But it's, it's, it's still, it still has some interesting feature. So, so let me start by discussing a little bit the Poincaré polynomials. So, so again, this is, you know, so we're back to this setting. So where we, we're just parameterizing short exact sequences on P1. And here, the degree, is D would, and so what's, what I think of as fixed are these parameters are an N. So the dimension of the trivial sheaf that we take a quotient of and the, and the rank of the, the rank of the sub sheaf which is fixed are, I think I'm going to, we're going to think of this D as moving. Yeah. So this is N minus R D. Okay. So the Poincaré polynomial. So let's say, well, okay, so if we, for every quote scheme, we're indexing them by D. So there's, remember, there was no odd homology. So, so let's write them as like this, the dimension and the dimension is linear in D. Yeah, I mean, it's easy to, so it's, it's N D. And then, of course, you have the dimension of the underlying grass mania here. But anyway, it's linear in D. That's important. Okay, so, yeah, so then we formed the full series. So we include all these. So we formed the, the full series, let's say T. Okay, so Z here keeps track of a homological degree. And this T, the variable T of the, the quote scheme, which, which space you're, you're looking at exactly the maps of quotients of which degree. Okay. Okay. So we sum over all quotes, quote scheme degrees here. Yeah, so this Poincaré polynomial, the Poincaré series was calculated by Linda Chen. And the year was 2000. And it's a beautiful, it's a beautiful expression. So let me write it. So there's a prefactor here. Okay, so here, this P zero is just it's just Poincaré polynomial of the grass mania. It's classically known. So let me write this. So this is Poincaré polynomial of the grass mania. Okay, so, yeah, so we'll do a little, a few, we just get, so it's, yeah, well, so let's, let's take a little bit of a closer, closer look at this formula. So first of all, I want that you can deduce immediately. So what we can observe is that just to get us started a little bit is that B two of QD is equal to two. So for all the RN, of course, I hear these at least one, you know, otherwise. Yeah, so this is just one just says to read off the coefficients of, the coefficient of Z times T to the D, okay, in the above series. So one can do it. And it plays a bit differently. I mean, here, you'll see this, sorry. Yeah, so this plays a bit differently. There are two cases, we'll sort of look at two cases. One case is when we're talking about a genuine grass manian, so R is less than N. And that plays a little bit differently in this case, because, well, I mean, somehow, yeah, so there's a, you know, this Z here appears with this coefficient Z to the N minus R plus I. And when the, when R is equal to N, in some sense, the formula is simpler. And you also don't have this P zero. So the, there's this pre factor. So when R is equal to N, you're talking about a point. Yeah, so this grass manian GNN, it's a point. In any case, in both scenarios, B2 is equal to two. So then you conclude that, oh, sorry, A1 of the code scheme is spanned by these two classes, which, well, we knew this, the only chance was that they would be linearly dependent somehow, but they're not. Yeah, because I'm sorry, A1 plus F2. So remember, we do have, at this point, we have the Poincare series, but we're also looking at it knowing that these classes, A1 through AR and F2 through FR, generate the homology. Okay. So this is the first, yeah, so let me, it's already, my time is already up. Let me just write one thing, which is amusing and sort of, then we can pick it up next time with, because this deserves a more, a more thorough analysis. So, so let me write this as a remark. Remark is that the homology stabilizes as D goes to infinity. Yeah. So more precisely, like we'll see this on, what does it mean on the level of betting numbers? And so more precisely, we see that the stabilization starts to occur for D greater than or equal to K. So if we fix, so this means that if we fix a certain homological degree, then if we look for code schemes of sufficiently large D, and these has to be at least K, then this, these betting numbers with the degree of the homology, I'm sorry, the rank of the homology degree K stays the same, as you move on. And so this is a phenomenon which, is more sophisticated, a lot are analyzed, for example, in the case of the Hilbert steamer points on a surface. Yeah. So some, it's a phenomenon you see there as well. And of course that series is a lot more sophisticated, but this is some sort of, you know, a very simple version in some sense that you can play with and its geometry is much simpler here. So as we'll see, as we develop it. Okay. So how do we see this? Just to, we see this. Yeah. So if we recall it, we have this, we have to, we can read it off in the Poincaré polynomial. Let me explain briefly how, and this will be the last thing I do. So it won't be more than two or three minutes. So yeah. So let's say that we write this, okay, V2KD, this means that you're looking at the betting number in the D-quote scheme. Okay. So if we consider then, so if we write, we can write, to notice this stabilization, we write 1 minus T times PZFT. So this is PZFT minus T, PZFT. The full series. So, you know, there will be some initial, so I, okay. So I'm, okay. So this is the part of degree zero, which I write here. And then starting with degree one, of course I have exactly this difference, which I want to notice is zero. Yeah. So I have B2KD minus B2KD minus one, Z to the K, T to the D. Okay. So this I would like to observe that it's zero. So in other words, what we need to see is that the coefficient of Z to the K, T to the D, in this modified series where we multiply by one minus T is zero for D greater than K. You know, that's what we would like to see. Once D is greater than K, we just want to see that this is zero. But here is, well, this is sort of manifest. If you look at the form of the series, well, look at the original. Let me just point this out in the, yeah, in the series as I wrote it, you see here, this starts when i is equal to one. I'm looking at this particular. So you can, so you have basically two types of two types of factors here. And if I look at the first type that starts with one over one minus T, you know, when i is equal to one, that's precisely what you won't see anymore once you multiply by one minus T. All right. So this is just P0Z. So you have the product of one minus TZ to the i from i is equal to one to r minus one. And then the product from i equal to one to r, one minus TZ to the n minus r plus i. Okay. So notice what's important here for us is that whenever you have a factor of T, it's accompanied by a Z. So it's a very easy point in this multiplication then you see that every T is accompanied by at least a non-zero power of Z. So you cannot simply buy Z to the a for a greater than zero. So that's the argument. So then you're done. I mean, the coefficient of Z to the k, T to the d when T is greater, when D is less than, when D is greater than k, it's just zero. So coefficient of D is greater than k is zero. Right. So there is there is homology stabilization. And so then you can also ask, so what does it stabilize to? So in other words, and then I'll finish on this note actually. So so if you set, let's say, B to k infinity to be equal to B to k of QD for D greater than or equal to k, then the stable Poincaré polynomial of QD is, well, so you, you just form this series with the stable betty numbers. And the fact is that which is easy to see, I will, fact is that is that you calculate this by evaluating a T equal to one, the product of your, of the original Poincaré series with one minus T. Okay. So I can briefly explain this but not now. So in any case, you get the following because now we're running out of time. Yeah. So I will, I guess end here because unfortunately I imagined I was going to cover a lot more than I actually did. But I want to point out one thing that if you set this formula is simplest when n is equal to r, n is equal to r is the, so for n, let me just write this one, this Poincaré series. Well, there's this P zero is one in that case. And then you have just this, I prefer to write it as a, I could write this more compactly, of course, this seems a little stupid what I'm doing, but it's, I'm doing it for a reason. So you have these two distinct products, I viewed them as coming, having a slightly different flavor. So that's why I write them separately. So n is equal to r is the case when you were looking at the code scheme of rank zero quotients. Yeah. So the sub-sheaf has the same rank as the CN as the trivial middle sheaf. So then the quotient is pure, is pure torsion. So it's the geometrically the simplest, most accessible case. I mean in, and so as you, what this suggests, of course we know we have these generators. We have these atia-bot generators. And so let's, so these are a one through a r and f two through f r. And what this formula explains in fact is that these are atia-bot generators actually become relation free. Yeah, as D, as D increases, because in fact this stable Poincare polynomial, if you look at it, this is just, this is just the Poincare series of just the polynomial algebra in these generators. Yeah. So p infinity of z is the Poincare series of the polynomial algebra. So no relations in these generators. Yeah, precisely with the correct degrees, yeah, because, yeah, so, and you can see that the two products, one would correspond to the A classes and the other one to the F classes and the F classes are of course by, yeah, by one, by one less. So yeah, so I, my plan was to explain why essentially, you know, so it's an interesting thing, but I think we'll have to leave it for, for another day. So I'm going to stop sharing. Yeah, so, so yeah, I apologize. I went over time and yeah, it's, it's, it's an interesting this where we left it off is sort of an interesting point and I hope to resume it. I have one question. So first of all, thank you for the beautiful talk. I wanted to ask if the space QD has a cellular decomposition or a fine paving or whatever it's called. And because this would, so the actual question is, does the formula for the Poincare series become a motivic and this would be implied by the affine paving? Yeah, yeah, it's, it does. I mean, it's this in, even in the initial analysis by, by, by Stromed and, and also in, in Linda's derivation of this Poincare series, of course what's used here is a torus action. Yeah, so, so it admits a full torus action. I mean, yeah, so that you can act with a torus on, on CN, but you can also act with a torus on P1. So in fact, you have finitely many fixed points and you have this Bielinisky-Biroudi composition as well. And this is how, in fact, this analysis, I mean, that these were reduced, so yeah, you can sell a lot more. And also I should say that this thing with the torsion code schemes, this analysis works on any curve, you don't have to have P1 there. I mean, I'm, I'm keeping it a bit more on P1. I'm interested in any grass mania, not just the case of torsion sheeps, which is why I, I assume that we're on P1 to simplify certain things, you know, but, but if we are just interested in torsion sheeps, then, you know, any curve would do. Yeah, so, but my point is to actually not just calculate, my point is to sort of understand which I didn't get to, but I will hopefully in a future lecture is to understand geometrically why that is, why this becomes relation free and what, what are the relations in fact, you know, and there are two ways to approach it. And yeah, so to have a basically a kind of a detailed geometric understanding of what this series means. But yeah, there is a, there is a serial decomposition you can, you can study. And it's been studied much from this point of view, but not so much from the point of view I want to emphasize. So, you know. All right. Thanks a lot. Sure. Good question. This tank, Alina, again?