 All right, thanks. So for last several lectures, I talked about one algebra geometric model of link homology using braid varieties and related varieties and some structures that we can understand on that side. So today, I will give completely different algebraic model for link homology for some class of links. And this class of links is called algebraic links. So we start from a plane curve f of x, y is equal to 0 in C2. And we assume that it has singularity at the origin. And for now, at least, I will assume that the curve is reduced. So this equation is reduced, but not necessarily reducible. So you can have several components. And to this thing, you can associate the link. So what happens? So you take a small sphere with the center at the origin, like this blue circle here. So C2 is four-dimensional. The sphere will be three-dimensional. And you intersect with the curve C. This will be a link in three-dimensional sphere. And if the sphere is efficient or small, it doesn't matter what's the radius. The logical type is all the same. And so, for example, if you have just a line, maybe I should start with a line, actually. Like this horizontal line here, this will give you a circle in a sphere. If you have a node x, y is equal to 0, then I would have two circles in three-dimensional sphere. And in fact, these two circles will be linked. That's not completely obvious, but that can be proved. And so we get this whole plane, which we see over and over, where we just have two crossings and two strands and take the closure of this link. But one way to see why you get them linked, because the linking number between these two components, so how they're actually linked, is equal to the intersection number, algebraic intersection number between these two curves at the origin. So there's two lines intersect with multiplicity 1. And so expect the linking number 1. And that's what we see here. Another example is when we have a curve x squared is equal to y cubed. So that's a cusp. And this corresponds to the tree foil T2-3. So this is the closure of this break. And in general, it's not so hard to see that if you have x of the m is equal to y to the n, this would be a tourist link Tmn. So this curve has as many components at GCD of m and n, like here, you have two components. Here you have one component. And in general, you have GCD of m and n components. And this corresponds to a link with GCD of m and n components. And there are lots of useful facts about these links. So first of all, irreducible components of a curve. So if you factor this polynomial of x, y locally in the neighborhood of the origin, irreducible components of c correspond to connected components of the link L. So in the regular if c is reduced and irreducible, we get a node with one component. And in this case, it was actually classically known how to classify such nodes. So not every node appears as an algebraic node. In fact, any algebraic node is an iterated cable of a tourist node. So what happens is that you have this tourist node, Tmn. Then you take a small neighborhood of this. It's a torus knotted in a complicated way in S3. And then you draw another torus, not on the surface of that torus. And you keep repeating this procedure. And so there are some conditions on the parameters of these torus nodes, which are written out in the literature. And there is actually full classification. So if you know what's the expansion of an irreducible link or singularity from Pousso's exponents, you can just read off all these parameters of torus links. And how many Pousso pairs? So the number of Pousso pairs tells us the number of cables that we need to do. And for general links, it's kind of more complicated. So each component of this link will be just an algebraic node, which we just described. But the way the link is kind of tricky and it can be read off the resolution of singularities of C, it can be read off some other things. And there is a nice book of Eisenbad and Neumann on algebraic links and plum tree manifolds where they describe it in detail, how to describe such links, how to classify them and so on. So if you're interested, algebraic links are classified quite well for a while. In any case, we have an interesting class of links, which is very closely tied with algebraic geometry by definition because it comes from an algebraic curve. And so the natural question is then to ask, if I know the algebraic geometry of the curve, can I reconstruct some invariance of the link? And to various extents, it was answered by many people. And so what we're focused today is the following conjecture of a block of Rasmussen and Schender from around 10 years ago. And it says that you take this curve C, you take the Heber scheme of K points on this curve C. So maybe I should write down on C, support it in the origin. And then you take just this, the joint union of all possible case and you have homology of the space and then the claim, which I say is that this homology is actually the same as again, degree zero part of triple grade homology of the link. And so in particular, if I know all these Heber schemes, I can recover this triple graded homology, at least it's bottom eight degree piece and they give a recipe how to deal with other pieces, but I won't talk about this. And of course this has two gradings because we have the number of points K and you have homological degree and this gives two gradings on the space. It's always infinite dimensional because there are infinitely many Heber schemes, but this is also infinite dimensional up to some other things which we'll talk about. And then the claim is that these are isomorphic, at least this vector spaces or bi-graded vector spaces. And you can ask about more structures later and that we'll talk about. And this conjecture is still wide open. So I'll review what is known and what is not known, but it's quite remarkable from algebraic geometry point of view. So if you don't care about income homology again, this, you can read it in two different ways. So originally I think their idea was to say that this is a way to compute the income homology. So at the time, nobody knew how to compute this income homology and they say, well, so here's the explicit algebraic geometry space, outbreak variety. So just compute this homology of this variety and this gives you the answer. And in many cases, as we'll see today, we can actually effectively compute the homology of this space by using a fine paving in different things and other tools in algebraic geometry. And so we know the answer for the income homology. And so in fact, these computations for torus nodes, they were definitely inspired by this conjecture. On the other hand, given all the progress in our understanding of income homology, you can reverse the logic and say, well, so suppose that we know income homology, then we know actually how much of the Hebrew schemes, which is still very, very non-trivial. And I think one of the basic tests which is still not proved in any reasonable form is that the left-hand side only depends on the link. So you can have different curves with different equations which give the same link. So you can slightly deform the curve within what is called the equal singularity class. The curve changes and the ring of functions on the curve changes and the Hebrew scheme of points on the curve changes, the link of course doesn't change. And so you have to explain that why this homology of the Hebrew scheme doesn't change in this equi-singular framework. And as far as I know, this is a wide open question, even except for torus links again. So, and there are other questions that you can kind of import from either side. So in particular, we talked about different structures on the income homology, like action of prelineal ring, action of this astrological classes and you can ask what they're here. And they give a lot of insight on this side as well, which wasn't kind of expected from the beginning. So there is a very, even though the conjecture is still open in most cases, there is a lot of interest in interaction and you can get some intuition from income homology from known computations to get some intuition here or you can get some intuition from here and some constructions, so I wish I will review to get some structural results about the income also on the left-hand side or at least conjecture that, okay? So any, maybe let me pause here and ask for questions. So any questions, how do we build the link from the curve and what does the conjecture say? So this Hilbert scheme is singular. Yes. And so is there any, anything about the intersection of homology? Not as I know of, you can ask what is the intersection of homology and how to think about it. I don't know. I'll give you some examples where they're same with, but I don't know what to say. Any other question? Okay. So anyway, so here is one example, which is very, very concrete. So we'll look at the node example. The curve is x, y is equal to zero. And again, we're looking at the Hilbert scheme of points on the whole curve, just the point supported the origin because that's the easiest one. So Hilbert scheme of zero points on a curve supported the origin in just one point. So you have one point is there, so this is just the ring of functions. This is ideal, oh, could I mention zero. Hilbert scheme of one point is again one point is the maximal ideal of that point at the origin. Hilbert scheme of two points on this curve, because like you can prove it in different ways, but for example, this curve has multiplied to two at the origin. So the Hilbert scheme of two points on the curve is actually the Hilbert scheme of two points on the whole C2 at the origin. And many of you know that this is just Cp1. So Hilbert scheme of two points on the curve is P1. And then there is a computation, which I will skip for the interest of time, but there are some problems in the exercise sheet if you want to do it. That Hilbert scheme of three points on this curve is actually a pair of lines glued at the point. They have two Cp1s glued at one point. And this one point is actually, so intersection point is the ideal generated by X square, X, Y, and Y square. This obviously has co-dimension three. It contains X, Y, so this is really an ideal on the curve. And this is the singular point in this case. And then you have two lines out of the singular point. And then, Hilbert scheme of, question? Okay. So Hilbert scheme of four points is the union of three lines. So the dimension won't grow up. It will be two lines. Here we'll have three lines. Glued at two points. So these two points also correspond to monomial ideals. And in general, Hilbert scheme of K points on C zero will be a chain of K minus one lines, glued in this way. So you have just a chain. And so it's really easier to test this conjecture in this case. So we can compute the homology of this Hilbert scheme explicitly. It is connected, so H zero is C, and you have K minus one components. So you will have H two, which is C to the K minus one. And that's all in this case. And I guess you can also try to compute intersection homology, but yeah. And so in this case, I mean, that's the answer. You just direct sum over all K. If you want to do pancrep polynomial, which many people like. So what do you have? So for zero point, you get one. For one point, you get one. So this is, what is this? This is sum of Q to the I, C to the J dimension of H, J of healed I. So Q degree tracks the number of points, T degree tracks the homological degree. And so if you have zero points, that's one. One point, that's one. Two points, it's one plus T square, that's homology of P one. Three points, it's one plus two D square. So H zero is one and it's two is two dimensional and so on. So you add it up, it's a rational function and you get one over one minus Q. So for this one plus Q plus Q square plus Q cube and so on. And Q squared D squared divided by one minus Q squared. And so this is the answer and you can compare it with the answer for the Hoplian that we had before. So if you remember anything from first lecture, from second lecture, the homology was R plus R mod X1 minus X2 where R was the polynomial ring in two variables. And so the polynomial ring in two variables correspond to this term and the polynomial ring in one variable which you get by quotient by X1 minus X2 is this one over one minus Q. And so these guys are separating kind of a homological degree in the homology and this shift is given by this Q squared D square. So there is some change of variables but up to this regrading, the answers are indeed the same at least on the level of graded vector spaces. And I think that's a very nice example because you see kind of more and more complicated spaces but there is still a lot of structure here. So it's not random like there is a lot of, I mean the fact that this generating function is rational is not the coincidence by any means. Okay, all right, any questions? Okay, so as I said, this conjecture is wide open in general. So besides some cases it's not known on the level of Euler characteristic. So there is a monumental work of Davesh Molek who proved that if you take the Euler characteristic of the Hebert scheme of K points, then you recover the homophilic polynomial. So that was a separate conjecture of a Blomkopf and Schänder about this Euler characteristic and your homophilic polynomial depends on two parameters Q and A. You set A is equal to zero as you would do in the homology. And so Davesh used a lot of machinery to prove these results. So he related this Euler characteristic to the topic of this conference to counting curves and counting stable pairs supported on this curve and various relations between PT and DT invariance and blow up formula. So there was a lot of technology directly related to the topic of this conference which unfortunately I will not review. But let me just say, so this is related to stable pairs. So basically you relate the ideals on the curve to stable pairs supported on the curve. And so you have PT and DT invariance and the blow up formula for this. And so there is a lot of technology related to kind of counting ideals on the plane or in three-fold related to this thing and counting ideals supported on this single curve. And that's very powerful proof which is indication that something is definitely going on. And then a much more kind of low tech thing is that we know the answer for torus nodes. So for torus nodes and I don't think we know the full answer for torus links yet as far as I know. But for torus nodes at least with one component you can compute both sides. So again HHH was computed by recursion from lecture one which I kind of sketched and didn't really explain but the work of Hogan company made it in particular they give very, very explicit recursion and very, very explicit combinatorial formulas for this dimensions of different graded pieces. And on the other hand, hybrid scheme of points on the torus nodes. So torus nodes corresponds to a curve X to the M is equal to Y to the N. This is a curve with torus action. And so as such it has a paving by a fine cells which can be contracted very, very explicitly and combinatorially. And you can enumerate all the cells and I will review this enumeration and slightly different setup. And there is a combinatorial form of a dimension. So I think in this setting, this was explicitly written in this paper of a blog of Rasmussen and Schender. And so they say like what are combinatorial data enumerated in the cells? What is the dimension of a given cell with this combinatorial data for every key? And so there are five and the many cells for every given key, you can enumerate them and you can compute the punk rapidly and all that. So this comparison between these two combinatorial answers solves the problem for torus nodes just because we can compute both sides. And again, like the methods of computation are very different and it would be very exciting to explicitly compare recursions here and recursions here. And that's not fully done. Like to find the geometric analog of these recursions on the left and like what they actually mean, what this K and mean, for example, which appeared in the recursion. And in any case, so like maybe the main outcome of these computations of following up in methods, we can now verify this conjunction. And then besides like actual computations and actual proofs, you can ask for very structural results like the interesting structures on either side, do they match something on the other side? And for example, recall that if you have an R component link then last time I think I explained that triplicate homology has a national polynomial algebra C of X1 through XR. So for each component of a link, you can put the marks point and that gives you a polynomial action. And that is a nice invariant of your link. And so it is important because like usually we're dealing with huge infinite dimensional spaces. So for example, here you had two component link and you have this explicit structure of a module over two variables X1 and X2. And you can ask, well, so on the left hand side do you have the structure of this module? And the answer is yes. And so for one component, there is an action of polynomial algebra in one variable and actually of the Heisenberg algebra constructed by Malik Jung, Milirini, Shende and random also randomly explicitly constructed the Heisenberg or while algebra of one generator. So he constructed kind of the action of X and Z over DX in some sense and the commutator was one. And so he constructed the action of both X and Z over DX. And then for more components, this was done by Oscar Kivenin. So for roughly speaking, for each component of a curve you can add a point. And so if your link has our components, this means that your algebraic curve has IR components and very roughly XI adds a point on this side component. But you need to be very, very careful if you want to say this properly. So all this mod light spaces are singular. So you can't just naively use correspondences to define operators in homology. So the way that it was done and I think all these papers really is that you have to use a versatile deformation of the curve, look at the versatile family and the family of hybrid schemes associated to the versatile family and define some correspondences there and use the fact that the family, the versatile family is actually smooth. So the key fact observed by many people, here, you have C tilde is a versatile family, then the hybrid scheme of C tilde is smooth. Do it fiber-wise for each curve in this family and then the total space is smooth and then you can do some operations there. So, and then you can add a point for each component, one component at a time. And you have some analog of this D over DX as well, which was also constructed by Oscar. And another thing which is useful, which I want to really talk about is that you can also look at kind of global curve. So here I have hybrid scheme at the origin. For some of these actions, it's much better to look at the whole curve C or compactification of C and CP2, but I won't talk about this. And another thing is that you can get a lot more structure if you add, break the symmetry. So recall that our construction of link homology uses the braid. So it doesn't start from a link itself. And here, one issue why this conjecture is complicated is that you don't get a link as a diagram. You don't get a link as a product of crossings. You get a link as like the whole subset of S3 given by intersection with this curve. And it's not clear where are the crossings, where is the braid, whereas anything resembling the things that we saw before. And one step in that direction is to choose a projection of my curve to some line. So this is related to what Richard mentioned. Yesterday that you can think of the C as a spectral curve. So you choose this line again, everything is a fine. So this is just a straight line with coordinate X. And we project our curve onto this line. And there is some degree of this projection, which is N. And so if I take push forward of the structure shape of C, I get a rank N free module over the ring of functions on the line. And again, I completed the origin. And on this ring of functions, I have an action of Y. So the other coordinate, the vertical coordinate, which acts by multiplication by Y. So we'll see some example in a second. And so this is more structure. And as we will see, this more structure gives us more control of the situation. And maybe I think I didn't write it here, but I can say this now. So the choice of projection, what does it mean in terms of the link? So note that choice of projection or C corresponds to the presentation of my link as a break closure. And so this is the picture that I think many of you have seen that you look at the small circle on the base of this projection and you look at the pre-image. So the pre-image of this point will be N points on the curve C. And as this point on the base goes around, you have some monodermin. And so this picture gives you what is called the braid monodermin because for every point I have N point. For every point on the base, I have N points on the curve. And then if I start moving this point on the base, this N points start twisting around and get linked to each other and they behave like a braid or braid closure. And kind of more abstractly, you can think that the fiber is one dimensional complex line. So this is a real two dimensional plane. And then you have N points on this real two dimensional plane. And then as we go around, we have a loop in the configuration space of N points on two dimensional plane. That's a closed braid. And so naturally we can associate a braid or a braid closure or a conjugacy class of a braid to the choice of this projection. And again, for different choices, we'll have slightly different braids on different number of strands because the degree of the projection could be different. But this is additional structure, which helps a lot to kind of understand better what's going on. And so here's a concrete example. So if I start from a curve from the cusp x square is equal to y cubed. So we can write the ring of functions on the curve as well. Polinomials are a series in X and Y mode this equation, but you can also choose the basis in the ring of functions. Namely, you have all possible functions in X and spanned by one Y square and one Y and Y square. And so if I Y cubed, you can express it as X square. So you don't really need Y cubed, but you can always write it in this form. The ring of function is actually a free module over C of X of rank three. And this is the basis of this free module. And moreover, we can describe the action of multiplication by Y in this basis. So one goes to Y, Y goes to Y square and Y square goes to Y cube, which is X square. So the matrix of multiplication by Y is this thing on the right. So one goes to Y, Y goes to Y square and Y square goes to X square. So in this case, the degree of the projection of rank three here for I have free module of rank three. And I have this three by three matrix. And on the biological side, this corresponds to choosing this as a closure of three-thrand break. So here this would correspond to three-thrand, which you can actually write down. I mean, we'll do this one and then it goes. And if instead you project to the Y coordinate, then you can also regard this as a free module over Y. So this C of Y with the basis given by one and X. And again, X square, you can now eliminate and replace by Y cube. And then my capital X is the matrix of multiplication by X. And so here one goes to X, but X goes to Y cubed under multiplication by X. And so this corresponds to the two-thrand break, which looks like this. And there are of course, different descriptions of the same curve. So typologically, you can say you can get the same link by closures of different breaks and different number of strands. And algebraically, you can just project to different lines. And as Richard explained yesterday, knowing this data actually is enough to recover the curve. So if I know this free module of rank three, for example, with this operator Y, depending on X square, I can recover the curve easily. And namely the equation of the curve is just characteristic polynomial of this matrix. So you have this matrix depending on X. If I take characteristic polynomial, I get the equation Y cube minus X square. And then the roots of this characteristic polynomial are precisely the, the locus of the roots is precisely the spectral curve C. And I think in this case, we assume that our curve is reduced, then the characteristic polynomial is the same as minimal polynomial. And so we don't have some of the issues that Richard mentioned last time. And so again, just to sum up, so this is completely formal procedure that we can replace a curve just as a curve on the plane by choice of this projection, we replace it by the following data. So we have a free module over, let's say power series in one variable together with an operator, which depends on this variable X. So any questions about this construction, any questions about this kind of relation to brace and stuff? Okay. So if that is clear, then we can define what is called the fine spring of fiber of C. So this would depend not only of C, but also of this choice of projection. And so we look at all possible subsets, subspaces in Laurent series in X. So here I have power series in X. Now I want to do Laurent series in X. And I want V to be a lattice. So I want this to be a module of maximal rank, which is a technical condition. And more importantly, I want this V to be invariant under multiplication by X. So this is, and I want this to be invariant under multiplication but under the action of this operator Y. And so this is a subset in a fine-gross monion. And you can think of this a fine-gross monion, which Joel already introduced last time. This is just the set of the same V's in CN of X, where just X V subset of V, and again, you have this latest condition. So if you like a fine-gross monion, so you can think that you have this space of all V's invariant under X, and then you put additional constraint that this V is invariant under this matrix. And that is why it's called a fine-sprang of fiber in some sense. And so this is this subset of a fine-gross monion. And so what do we know about this subset? So well, if C is irreducible and reduced, then this fine-sprang of fiber is the same as what is called compactified Jacobian of C, Devonekko. So yeah, maybe it's compactified Picard. I mean, locally everything is local, so I guess it doesn't matter that much. And up to some shift on the latest, which I ignore. And this compact, this is very close, let's say to compactified Jacobian of C, and which is defined as this mod-wise space of rank one to ocean-free shifts on C with some framing because we work locally. And a remarkable result, which was known for a while is that this is actually the same as the hybrid scheme of a large number of points on a curve. So if your curve is irreducible and this is really, really important assumption here, so and reduced, then the hybrid scheme of points actually stabilize for a sufficient large N and this fine-springer fiber is this stable hybrid scheme and it's the same as compactified Jacobian. And so you might have seen this in different settings. And in particular, so this kind of looks different, but in fact, it is the same thing as what we started for the hybrid scheme. And even more is true, so the Merlinian and Schender and one you can even prove that in fact, this homology of hybrid scheme of points on the curve, which were the subject of that conjecture is completely packed in the homology of the fine-springer fiber. So you have your curve, you choose a projection in some way and then you have your, you take the homology of that, that carries so-called perverse filtration, which I will not define, but it's an interesting filtration in homology. And the claim is that if you have this homology of a fine-springer fiber equipped with perverse filtration, then tensor with polynomial ring in one variable and you recover the left-hand side, the homology of the hybrid scheme of points on C zero. And maybe I want to mention here that for example, by this work of Rennemont, so the left-hand side has an action of this Heisenberg algebra in one variable with x and d over dx. And so this is the same x. This is the same polynomial, this is the same x. This is the same polynomial action on the left. And because there is an action of d over dx, it's actually a pre-module. So we just kill this free action and what's left is this. So this is very, very natural thing. And in fact, yeah, you can define, like if you know the action of this polynomial ring on the left, you can use this to define perverse filtration in essentially unique way using this formula and using the work of Rennemont. But they did it slightly differently using again, versatile deformation and the composition theorem and other things, which I don't have time to talk about. And moreover, Molly and you only prove that there is an action of SL2 on the right-hand side, on this finite dimensional thing. Yeah, so Oscar is commenting that if R is bigger than one, we have to replace this P-gamma by this P-gamma question by some latest, which I will describe in a second what is the latest. And maybe we'll come back to this. But like in the reducible case, for one component, so they prove that there is an action of SL2 on this homolog. And this is kind of less known about the paper, but I think it's a beautiful result that you always have an action of SL2 here generated by cup product with some class in H upper tool. The raising operator just the cup product in a cell with a class in H two. And then they prove that this also satisfies curious hard left-hands. So if you remember anything from last lecture, there was an action of SL2 on the homology which satisfied some kind of curious hard left-hands property. And here again, you have this, but this curious hard left-hands is with respect to perverse filtration. So again, this is a singular space in general. And we'll see example where it is really, really singular. And on the singular space, you wouldn't expect the normal hard left-hands and point-correctuality because it's just not true. You don't have it. But if you take associated graded with respect to perverse filtration and shift the degrees a little bit, then it becomes symmetric. And then it actually satisfies this hard left-hands and everything if you do it properly with respect to perverse filtration. So this is really, I think remarkable results. So I think, yeah, this is just proved in this paper of Malikovsky, but this is a very, very nice paper. And this, as I said, compare this with curious hard left-hands for weight filtration on grade varieties from lecture three. Okay, so maybe a couple of examples. Okay, let's start with this example. So if I have the node again, so I want to write the node as x square is equal to y square because I want to choose a nice projection. And I can choose the matrix in different ways. Let me choose the matrix like this. You can also choose the matrix as x zero, zero minus x and that will be useful for us in a sec. And then the corresponding affine-springer fiber is actually bad. So none of these theorems apply. This is not an irreducible curve x square is equal to y square and affine-springer fiber is actually an infinite chain of p1s. So you need additional structure here to control its homogen. But this is probably the most well-known example. Another well-known example is that you have a cusp. So the affine-springer fiber for that thing is p1. And this is the same as compact phygicobin for the cusp. So I think I don't have time to explain this. In this case, it's nice and smooth and the perverse filtration is trivial so there is nothing there. And what is known in general is that if you have x to the m is equal to y to the n, again, this corresponds to a torus node, gcd of mn is equal to one. And then lots and lots of people study this space many more than Hebrew scheme on this curve, but again, essentially this is equivalent. And many, many people proved in different ways that this is paved by affine cells and you can parameterize these affine cells by lots of different combinatorial data and you can compute dimensions of the cells and this is beautifully related to the tier of q to the cut line numbers and combinatorics. So the references, I think the first is used to get smelt and first started this affine-springer fiber and then Piantkowski who studied this as compact phygicobin and then Hikita who started this space and generalization to affine cells. And then our work with Misha Maizen and Monica Zirani when we started kind of more combinatorially and related this to q to the cut line. And so there is a lot of work on this thing which I won't be able to cover in detail. But one kind of great example is that you have the curve x cube is equal to y to the fourth. So this is this corresponding SPY or compact phygicobin is this thing. So this is a cone over here, it's a brook surface. And so this is singular, you have a singular point at the tip of the cone. And so how does this cell decomposition work? So here's the brook surface, have one zero cell, two complex one cells and one complex two cells. Then you take cones of this, so you will have two cell, four cell, two four cells and one six cell and then you have the vertex which gives you the zero cell. So this is really a projective cone, not a fine cone. And so the homology looks like this, H zero is one dimensional, H two is one dimensional, H four is two dimensional, H six is also one dimensional. And you can compute the perverse filtration, it's not so easy, but I mean, you can compute it using the Haber schemes and the theorem of molecule in the dimension, though you can compute it by definition. And it turns out that H four has two pieces in different levels of perverse filtration. So it looks like this. And again, there is a cure as hard left shifts because you have a class in Asia for two, you can composite and then you have the top thing will be like, so this will be maybe alpha and this will be maybe alpha square and this will be maybe alpha cube. And there is another generator beta over here of different perverse debris. And this should be compared with E six picture. So the picture of E six plus the right from last time we're at exactly the same thing for weight filtration in the homology of the braid variety. And so this example and many other examples and also more general work of the Catalde-Hausel-Milirini led with the action to conjecture that this is actually always true. So if you conjecture this also about 10 years ago that homology of braid variety or some analytical braid variety together with the weight filtration is actually isomorphic to homology of this the fine spring of fiber or compact fatty carbon together with perverse filtration at least in this irreducible case. So in non-reducible case, it's much worse. And as far as I know, this conjecture is really wide open even in this case. I mean, you can whenever you can compute both sides it's true. And so I guess that's all what we know. And I mean, there are some deep ideas why this might be true non-nebillion-horch correspondence and things like this but maybe I don't have time to talk about this but maybe what I want to say is that the left hand side is an open but smooth variety. So this is non-compact but smooth and this thing is compact but very, very single. And so there are just two different settings and it turns out that they have the same homology. So this is non-compact and always smooth, at least like in all examples we need. And this is compact. And again, you can probably ask about intersection homology here but I don't know what would be the right question. Okay. And another thing which I want to mention since I mentioned the pathological classes last time is that we constructed or I kind of indicated how to construct the pathological classes on the left and what are the weights and stuff. And on the right hand side at least for this particular singularity X to the m is y to the n, a Blomkov and Eun constructed an action of the pathological classes on this side. So they really come from some vector bundle on SP gamma. And they proved, so this is a big result of a Blomkov and Eun that this homology is generated by the pathological classes like it is here. So there are two pathological classes alpha and beta. And then there are all relations between these classes pretty explicit. So again, like if we believe in this conjecture you can ask, is it true on the left? So we have the pathological classes now. Can we write the relations between them? Can we verify them on the left hand side? Even though we don't know if the conjecture is true but like, for example, is it true that homologies are isomorphic not only as graded vector spaces but as rings? And that's a very good question which one should look at. Okay. And so in the slightly different direction I wanna mention a result of Oskar Kivinen. So what happens if you have a lot of components? So if you have a lot of components like in the extreme case you can have x to the k n is equal to y to the n. So here we'll have, so this corresponds to q n k n. So here we'd have n components and all components are not all linking numbers are k and we saw this example last time actually. So the corresponding n by n matrix has this form. So you have diagonal matrix with roots of unity all different roots of unity of degree n on the diagonal times x to the k. And for n is equal to two this is this matrix x and minus x and zero zero in general we have all roots of unity. And so you can ask what is the fine spring of fiber for this particular matrix? And this was started by many people starting from Guresk equals in McPherson and others. And so in particular you have an action of the lattice which acts on this a fine spring of fiber by translation. So n minus one dimensional lattice acts on it by translations. And you see it here, sorry for scrolling up and down that you have an infinite chain of p ones and it's big and kind of complicated but you have an action of Z by translations let's say one step at the right or to the right or two steps to the right depending on what you wanna do. And you also have an action of the torus because the torus she started then commutes with this matrix y so it stabilizes this and it acts on the fine spring of fiber. And so what Oscar actually proved in this theorem is that if you take just homology of SP gamma it matches the link homology with all the structures. So we have the action of the lattice it corresponds to the action of this polynomial variables up to some subtleties. And equivalent homology of this fine spring of fiber matches wifi or deformed homology of the link which depend on additional variables y one through y one and which are identified with the equivalent parameters. And so I don't want to write these answers again but you have explicit formulas for this in lecture three. And I will return to this problem next time but again, I don't have time for it today but there are some explicit ideals in the polynomial ring involved and so basically he compared this homology and with all the structures to this ideals in the polynomial. And this shows that not only we can identify this homology as a vector space with link homology according to a blog of Rasmus and Chandler conjectures but also we have this additional structure of acting of taxes and wise and we can perfectly see it in this setting. So, and in this setting it's actually easier to work with a fine spring of fiber because you really see this latest action and you really see the total section and for the Hebrew scheme on a curve you don't really see that unfortunately. So you can ask what is the meaning of that but that's a separate story. Okay, and so maybe the last thing which I want to mention today is a very recent work of Gardner and Kivenin which I think is very nice. And I want to advertise as much as possible. So now you can take any curve C and it could be irreducible, it could be reducible, it could be even non-reduced. So for this it doesn't matter. It's so general and so nice that you can have any equation of a curve not necessarily reduced. And then what they proved is that the Hebrew scheme of points on the curve is not in a fine spring of fiber for JLN but it is actually generalized the fine spring of fiber. So there is a notion of generalized the fine spring of fiber depending on the group JLN and their representation N which is in this case the Lie algebra of JLN and together with the vector representation CN. And so this is generalized the fine spring of fiber. It depends on the vector in N of X and on a vector. And we choose this vector. So this is the matrix N by N matrix and the vector in CN. So we just choose it to be this Y and then a vector, let's say 10000 in appropriate basis. And you can see properly what it is. And one consequence of this is that there is an action of a very interesting algebra again we're up to interesting structures in kohomology. And so Braver-Mann Finkelberg and Nakajima for any G and any N, they define a certain algebra which is called the BFN Coulomb branch algebra which Joel started defining yesterday and well I guess defined fully today. So this is really from Joel's talk this is the same algebra, Joel's lectures. One result is under some mild assumptions this algebra for JLN and N acts on the homology of a fine spring of fiber for G and N for any choice of vector really under some really, really mild assumptions. And so in particular there is an action of this big interesting algebra in kohomology of Hebrew scheme of key points on the curve direct sum of the world key. And this unifies like all the construction that I mentioned before. Furthermore, if a curve is glassy homogenous if it needs a cis-direction like our friend X2DM is equal to Y2N then there is a cis-direction on the curve there is a cis-direction on the Hebrew scheme and you can look at equivalent kohomology of the Hebrew scheme. And there is an action of some other algebra on kohomology of the Hebrew scheme which is known as quantized BFN algebra which I'm sure Joel will define today. And this is a non-commutative algebra which acts in this if we're in kohomology. And so it helps a lot to compute this kohomology and gives like very rigid structure to it because it's a module over some very explicit non-commutative algebra. And in fact, Kodere and Akajima described this algebra for us. So they said that this for this particular choice of G and N where G is the group Gila and N is the joint representation of the algebra of Gila N plus CN. This quantized BFN algebra is known as is nothing but what is called as rational Turingeq algebra or spherical actually rational Turingeq algebra if you want to be pedantic. And this is an algebra which people studied and you can describe it very, very explicitly by generators and relations and so on. And so this result says that there is an action of this algebra which people understand in the kohomology that we want to study which is supposed to be in kohomology or if we're varying version of the kohomology. And so this raises a lot of question is there an action of this rational Turingeq algebra in kohomology or kind of deformation of it corresponding to this equivalent parameters is there, can we identify this representation? And in many cases the answer is yes and they identify this representation. And what happens for non-reduced curves? So here you can start doing non-reduced curves. They still have this direction and you still have an action of some algebra. And there are lots of interesting questions here. And maybe one specific example of non-reduced curve is x of the n is equal to zero, it also works. So they think this is the first example where you can understand properly the homology of the Huber scheme on non-reduced curve. So you have ideals in the ring of polynomials and x and y mod x to the n. And you have some interesting structure there. You can compute the homology and the same techniques shows that you have an action of this quantized BFN algebra in this equivalent homology. And you can identify this representation very explicitly. And I think this is very nice and kind of brings all the structure together. And I would expect that maybe other especially non-reduced curves can be started by this machinery even though you don't have a solid composition but geometric representation theory gives you another way to study these spaces. And I think that's all for today. Thank you very much. And again, Oscar is there in the audience so you can ask him after the talk for more clarifications. Any questions for Eugene, for Oscar? Is there like key theoretic version of this last thing? Yes. So you would have a Dacha, some version of Dacha. And I'm not sure like if it's worked out in full detail. So if you would have GLN without C to the N, I think this is really just the Dacha defined by Sheridnik. And this was worked out by Ettinghof, Breibermann and Winklerberg, I guess. And if you have the CN, I'm not sure if the KT or Etting... So there is certainly a notion of key theoretic BFN algebra. I'm just not sure if that was computed explicitly. Yeah, it's trigonometric. Trigonometric. Aha. Okay, so then you have trigonometric Dacha and then, but I think like all this would work in KT area because I think so. I mean, the construction is very general and by some correspondences, so that should work, I think. And would this be related to any like nothing variance like on the other side? I mean, this space, the homology of the hubris scheme is supposed to be the same as the link homology, right? So you would have an action of this algebra and link homology. Here, you would have some kind of deformation of link homology with one extra parameter which might be related to this Wi-Fi homology or not. That's, I'm not 100% sure here, but you would have an interesting action of this algebra there and you can ask, well, so you have this lots of interesting operators acting in link homology. Do they, are they related to the topological classes that we constructed last time? That's an awesome question. Do they give you extra structure in link homology? I don't know, but these are all excellent questions. So, yeah, I wanted to ask yesterday when you discuss this, why deformed homology for things? You had this operator psi. Yes. And do you see, and some DG structures, so do you see them on the homology of spring or fiber side? Yeah, so these sides are casual dual-to-wise. And so there's like a variant parameters here and you can also find kind of, because you have C-stire action, instead of looking at the equivalent homology, you can look at non-equivariant homology with kind of the action of homology of C-star and then size of generators and homology of C-star, but they're really casual dual-to-wise. So maybe I want to mention this. So, I'm not saying, or a casual. And this means that you have kind of interesting degree one operators in here, which you can study and they're not closed, but you can control them. And it, but you can construct some closed operators as well. And maybe what's more relevant is that, I guess, you can take the quotient by the latest action, which you like in many people in the audience, like, and there's still is an action of the torus. And I think there you can see this size more clearly, but I don't know if that is explicitly worked out. Okay, thank you. We also have a question in the Q&A. So is the F-1 Springer fiber related to the Sato-Grassmanian and Cox-Schwarz operators? I don't know, I don't know. Maybe I want to say, so since Oscar mentioned this and I didn't say this and I think it's related to bank's question. So you have this example. So you have this example where x-square is equal to y-square. So we have an infinite chain of P-1s and you have an action of z by translations and you have an action of C-star, which scales all these things. And you can take the quotient. So this is my SP. And if I take the quotient by Z, I will get just one line which glued to itself at a point. So this is a line glued to itself. And here you can see that you have h0, h1, oops, one second. h0, h1 and h2, one second, sorry. Can you see it? You can see your screen here. Okay, great. So this is the line glued to itself. And here I have interest in h0, h1 and h2. And so there are several comments here. So one comment is that this matches the computation for this one plus xy is not equal to zero, which we saw last time. And again, this is a very, very different space. So here I have a line glued to itself. There I have kind of an open subset of C2, but the homology is the same. So this is another argument towards the x-conjecture. And another thing is like how do you actually, and you still have an action of C star on this. And so you can either look at a covariance homology, which is fine, or you can compare it to the income homology where we had this r, r, r, that was zero and that was x1, minus x2. And then if there is no C's direction, so we kill xi and we get just three copies of CCC with zero differential, right? So if I kill xi and r, I get just nothing, you get just C from each of them. So I can take the reference of product with C over r. And so my odd variables, xi that Peng asked about, they would act here. And so you would have an action here. And so I think you would have an action of odd variable in this homology, which you see very clearly, that you should have an action of basically homology of C star here, which would go, I think, from h0 to h1, but I might be wrong. I can't think of it in the spot, but this is certainly in the same space, in the same vector space, that this three-dimensional homology of this quotient by the way is this three-dimensional homology. And there should be an interesting error corresponding to this xi. And there should be another interesting error corresponding to what I called U2 yesterday. And this U2 gives you a class in top-dimension homology. So yeah, homological operations here would give you interesting operations in this homology over here. And again, I mean by essentially Casulto, if you know this deformed homology as a module over x's and y's, you know this undeformed homology as a module of xi's and x's. You have all the structures. And again, this latest action and the third section played the role of different variables in the homology that we saw. Yeah. And there is a subtlety between homology and Braille-Mort homology, which you can ask Oscar about. Yes. Thank you so much. All right, thank you so much.