 So, the mathematical part, I'll just say the statement and in case anybody planning to fall asleep at the talk, so you know, that's the very big, I will say the statement which is not, should not make sense for you right now, but approximately should make sense. So the idea is that if you have a link inside of the F3, I will explain you the way to construct a shift. Well, more generally complex of shifts on the Hilbert's scheme of points on C2, such that if you take a sections, more precisely all of the homologous of this shift, so this is as beta, then it is a knot in there. It's an isotopic invariant. So that's basically roughly the result. So, in particular, you know, there is T, that means you have action of the T is a torus which acts on C2 and that means it acts on the Hilbert's scheme and you have action here and, you know, this space in particular will have action of the torus. It's doubly graded space. So, and I will explain details of this thing during my talk. So, all right. And, you know, things like this were conjectured by many people, in particular there is a work by Shakira Patanage, then there is a work by me, or me and Shende, me and Shende and Rasmussen and Gorski and Eugud, so many people conjectured that now there is a construction. So, and maybe another thing I want to say that for a particular type of knots, like a torus knot, there is very explicit aspect for this shift and I would say, well, Eugud as well, why would you do this? So, the point is that this construction is somewhat, it's much better to compute things with this construction than a usual definition of knots. All right, so let's start from the beginning. So, please ask questions. So, right now, nothing should be clear for you because it's unclear what is Eta, what is N, what's all. Okay, so let me remind you from where we're beginning so that there is a serum of Markov, I guess, beginning of the 20th century, so that if you look at the... The serum of Markov said that if you take a braid group, should I remind people what is braid group? Everybody knows what is braid group? I'll write it out quickly in a second. So, if you take a union of the braid groups on any strands, module or some relation which I'll write, then you'll get that's equal, this set of equations, classes is a set of isotope classes of in R, in S3. Right, so the relations are... So, the first relation is that if you take two braids, you can flip them. So, that's the first one and the second relation is that if you have alpha, sigma and plus minus one is the same as... I'll draw a picture, so the alpha here belongs to the N minus one. So, the picture here is like this. So, the braid group is generated by the simple crossings. So, if this is I strength and this is I plus first strength and everything else goes like this, this is the generators. And they satisfy relations as we all know, the braid relation. So, the Markov theorem tells you that if I have a put in the box, let's say alpha, so this is braid on N strength, then if I take a closure of a braid like this, I'll get what's called closure of the braid. And the point is that this closure, taking closure definitely does not depend on order. If I take two braids next to each other, the first relation tells you that if I have one braid and another braid, I'll just... So, you can definitely flip them around. It's the same as, you know, you just move this bit over here, that's the same thing. So, that's the first relation. And the second relation is more interesting that is that you... If you have a this notation alpha on the first N minus one strands and you added one more crossing and you took a closure. So, you can take this loop could be removed. So, you can... this is the same as, you know, this kind of loop, right? And this you can remove. That's the statement. And basically, Krull and his classes, model of these relations are exactly the same. So, basically, if you want to construct a knot invariant, it's enough to construct some map from this set of braid groups which respect these relations, right? And probably one of the most celebrated, the knot invariant, it's a Jones invariant. And then there is a more general construction which is called homfly PT invariant which is constructed, for example, using the arm matrices from the previous talk. And basically, the recent developments were... Recent developments due to Havanov, Havanov and Razansky and many others were attempts to construct an invariance which would be... Some kind of invariance which would attach not just a number but a vector space to the knot and this vector space is typically called knot homologous. And in particular, there is this knot invariant which is called Havanov-Razansky homologous which I will explain what it is in a second. So, basically, Havanov and Razansky define, give a construction of a homologous of triply graded homologous. So, this is sum over i, j, k to the... So, basically, when they're any braid, they construct triply graded vector space. Sometimes people denote it by Havanov-Razansky, beta. Such that first, basically... So, first of all, this space depends only on the equivalence class with respect to this Markov moves. That means it's a knot invariant. So, in the second, it is actually... it is actually a categorification of the homfly polynomial. So, what does it mean? Second statement is that if you take a generating function for the dimensions of these vector spaces, so then it would be what's called homfly PT invariant of... So, sorry, I forgot to write some numbers. So, it's q power j a power k. So, that's homfly PT of the closure, of the closure of albeta, and sometimes people just write p of albeta. So, I will not tell you what is a homfly PT invariant. It's basically... I just tell you that, for example, previous talk was about r matrices. And if you take the rational r matrix for the SLN group with the fundamental representation, that's the quantity we will get. So, and another remark I want to make that construction of Havana-Fendrazansky. There are two flavors. So, Havana-Fendrazansky used Zergal bimodals. Zergal bimodals defined it. And it's, I guess, for physicists, that would be like a... It would be a symplectic side. Some kind of it's related to constructable shift with symplectic side. And in some sense, I think what we do, it's a kind of mirror-duo for this. What we do, it's more like a coherent side, coherent shift side. All right. So, and the point is that the Zergal bimodals construction is... Computation is very hard. Even if you write a kind of computer program, you basically, like all of the computer program which is available, gets stuck somewhere around like eight crossings. So that's very hard to do computations. And this advantage of the series, which I will discuss, that it's easy to obtain infinitely many examples with exact transfer. So it's in some sense our model is more easily computable. All right. Now I have to justify my invitation and talk a little bit about Gage series. I guess, you know, Hilbert's scheme doesn't qualify as a... Is it okay? Is it a Gage series? Gage group is U1, but it's something. Okay. So it's possible that a lot of this stuff generalizes to other Gage groups. We didn't try. So, you know, that's... Other Gage group means, you know, just go... Instead of Hilbert's scheme, do stuff with the other quiver. You know, the Kajima varieties. And I think it's... All of this is possible. We just didn't have time to go there. So maybe that would be reason to be invited again. All right. So let me remind the definition. So... Field N of C2. So everybody here knows is just set of the ideals inside of the... The polynomial rings of two variables such that the quotient is a n-dimensional space. All right. So... And I wouldn't need to... As I said before, there is action of the torus which is C star cross C star on this... Variety. And let me just specifically say that there is this nice sub-torus which is anti-invariant torus. So basically this guy acts on X, Y by scaling X and Y with opposite weights. It's nice because it actually preserves natural symplectic form on the Hilbert's scheme. So... And maybe I will just introduce some kind of funny notation. There is a smaller subset here, Hilb C2, which is... So that Hilbert's child, symmetric and C2 zero. So what does it mean? So first of all, we always have a map from the Hilbert's scheme of C2 to the symmetric power of C2, which is basically you take an ideal and you send it to the support of the quotient. You take the support of the quotient of the thing, right? So inside here you have a slightly smaller subset, which consists of subsets with a condition that some of the... It consists of n-tuples with a condition that some of the coordinates is equals zero. I'll just do it slightly smaller, you know. Just introduce the smaller subset. It's just because somehow we have chosen to work with SLN group. First of all, I should also say that whatever I will say is available in archive. There is only one paper with me and left so far, and that's the paper which I'm talking about. So a series of conjectures, and basically, you know, the conjectures. You know, I will not talk about the story, the history of the subject, so I will just write a list of names who talked about this thing. So first of all, there is a work by me, J. Krasnitsyn and Shende, and then there is a work, a generalization of this thing by Shende. So then there is also a work by Gorski Nugut. There is also a physicist, also wrote about this thing. There is a work by Aga Nagic and Shakira. So they don't talk about Hilbert's scheme. They talk about McDonald's polynomials, but you can connect it. I would say that they also had some similar ideas, which state the following conjecture. So before I state the conjecture, I need one more piece of notation. So there is this universal bundle over the Hilbert's scheme. So let's denote it like B since I'm talking about this small one. Such that, you know, if I take the fiber of this bundle at the ideal I, it's just literally this quotient. So this one is rank N, and I want to have a slightly smaller one. So basically you have a quotient of B O, because, how to say it, there is always, here there is always a class of constant of identity. This identity, which sits in every fiber, gives you a trivial bundle, and I just want to kill it. So this one is a rank one, this one of rank N minus one. So now I have all of the notations, and I would denote by B check, dual. So then the conjecture, which I think is false, state the right conjecture in second, is the following. I'm not giving all of details who proposed which part of the conjecture, but outcome of, if you put all of these things together, the following state. So that there is the shift be generous and say we just, not even a shift, it's a complex of shifts, which is the covariant on the Hilbert's scheme of N points on C2. So and beta here is the element of the break group on N strands, such that if you take a homologous of this shift, tensor with exterior powers of this tautological bundle, then this is the same as a Havana-Frozansky homologous. So here we have three gradings, and here we have three gradings, one comes with two gradings comes from the scaling action of torus t, and another grading comes from the exterior power. This is exactly these three gradings. So I think this is false, but what is true, so this statement is true, I believe, for the positive nodes. If nodes are very positive, then it's true. If it's something in between, then there is something more complicated, which I will explain. Alright, so what is this here? Now that, first of all, the better object to work with is the Flag-Hilbert's scheme. So let's denote this one. Hilb1.1,N, just a set of flags. So it's I0, I1, IN inside of the Cxy, such that the difference, each of them is co-dimension one. Alright, so one of them is this, exterior power, and two more because we have a torus action, C star cross C star. What's that? Well, you know, there is a lot of kind of bullshit going around that is a fourth grading, but I don't believe in that. So you can say, well, there is fourth grading here. I don't think it's not there, but some people say it is, but I don't believe it. So I think it might be a fourth grading if you go to transfer nodes. We have the whole theory extends to transfer response. It might work for transfer response, but I don't believe it's true for the usual ones. Alright, more questions. Alright, so the paper was left kind of long. It's almost 90 pages, but I think content of the paper could be explained in one page. So the main reason why it's so long because we have to work with this object called like a career and metrics factorization, and the subject was not developed, and we have to develop from the scratch. Push forward pullbacks. We have to define all of these objects. But let's go is actually true. So first of all, there would be always the interesting part of this Hilbert scheme is this Lagrangian piece, which is basically the image of zero times. So basically there's this Hilbert Chalma, and we require that all y variables of the image has zero. And this is like a subspace of a half-dimensional, you know, it's Lagrangian. And actually all of our shifts will be living on this one. And yeah, it's not a coincidence that it's Lagrangian. Alright, so what's even more interesting that there is this, the right object of this study so-called free Hilbert scheme. So what is a free Hilbert scheme? Let's just start with Lagrangian parts. So what is a free guy? It's, see it's inside of the, let's just save a little bit of chalk and say that for now on G is an SL and GN and group G is an SL. Right, so it sits inside of the, it sits inside of the Borel of N times the group GN times mu-potent. With the condition that if this is X, this is G, this is Y. So the condition is that this Y, G belongs here. If and only if, if you take a free algebra generated by matrices X and Y, applied so there is, that there is V, vector V, such that if you apply this free algebra generated by X and Y, you will get whole space. So basically that's a condition that you have two matrices, both of them are protrangular. So the BN is a, it's apotrangular matrices with zeros and N is a strictly apotrangular matrix. So the condition is that you know, you take two matrices of this sort and you have a vector such that you can generate the whole vector space. You have an open set inside of it and that's what we call free Hilbert scheme. So the amazing thing about it that, that this flat Hilbert scheme, which I just talked here, is extremely singular. But this guy is smooth. So you can do a lot of geometries. So then, you know, our main theorem, so that object in this periodic category, which I will explain in a second, so I explain you what is still the thing. So I didn't write you what is the actual one. So this is nested one. It's just quotient of this big space by the gauge group. And this guy is smooth. All right, so then the point is that we construct two periodic complex free Hilbert scheme such that, well, okay, so since I'm continuing it, the two periodic means, the periodic direct category means we're considering two periodic complexes. So they have like odd and eith, I can't say, they would be the ships of the sword S0, S1, and S0, S1. All right, so, and the condition is that, first of all, the property is that if you take a hypergeochemology of this subject, of this guy, then it's, there's an isotope in there, right? And the second, so we can always, you know, as I say, the elements here is just two periodic complexes. And you can always take, you know, the homologous of this complex. So basically you take, they have, you know, the condition is that square is equal to zero. So that means you can take odd and even homologous. Let's denote it, H0 of the complex S and H1 of the complex S. And both of them would be actual shifts on the, not complex, there would be actual shifts on this free Hilbert scheme. So it turns out that actually this, let's denote that this H0 of this shift, of this complex and H1 of S beta are supported on the actual Hilbert scheme, on the Hilb 1n. No free, just usual one. So basically, you know, you have a free Hilbert scheme and obviously the usual Hilbert scheme sits here. You know, you obviously have the usual Hilbert scheme sitting here just by imposing condition that, you know, matrices X and Y compute. So, and, yeah, so anyway, so that's the statement, that's the main statement. So in the easy corollary, you know, which is just statement from homological algebra, you know, you can compute hyper-homologous using spectral sequence. You can first compute, you know, this homologous, then you have some kind of differential on this guy and the corollary of this statement, so that there are two shifts, S0, beta, and S1, beta. Not complex, but actual shifts on the flag Hilbert scheme. So it's flag Hilbert scheme L1n, such that there is a spectral sequence, such that there is a spectral sequence which converges to this hyper-homologous, which is a knot invariant, hyper-homologous of this exterior power. But it's E2 term, so that's kind of important. That's why positive, now you will see why positive knots are good compared to the other ones. The E2 term of this spectral sequence is the following. E2 is that if you take usual homologous, it has a differential, E2 term has a differential that if you take S star, beta, exterior power, things. So it decreases the usual degree by one and it increases the exterior by state. Now, the point is that if you started with, if this S beta was very positive, so then you would have only zero global sections. So this guy would be only non-zero for k equals zero. That means the spectral sequence degenerates in the first term. So when that happens, when p is very positive, then that happens. But then this conjecture, which I was writing before, holds. So basically, if beta is very positive, then hk, whatever is here, is zero if k is not equal to zero. So we get the statement like I was writing above. But if you write stuff in the middle, if something is, if not, neither positive nor negative, then it wouldn't be true. So there is no shift. There is nice shift only on the free-huber scheme. Any questions? So maybe I would say a little bit more about how we construct this thing. Okay, I'm going fast. I guess that's good. Okay, so maybe I'll say why it's interesting, why this series is interesting. So in another series, in another series we can show that, so first of all, there is a special element of the Brazic braid group, which is called, I would call it coxster element. So basically, this is the element where you draw it. So you have this group, right? So when you kind of twist it a little bit like this. So this is this one. And so it's under-crossings everywhere. And in particular, if you take coxster power n, that would be full twist. That means all of the strands go around and come back, but they kind of made it turn. So, and it turns out that this series is nice in the sense that this shift attached to the, and let's call this one twist, T-W. So if you take a braid and compose it with full twist, so then it's very easy to see what happens with this shift. The shift gets tensor multiplied by the determinant of this bundle, which I wrote in the beginning. So basically it gets twisted by line bundle. So that kind of immediately tells you a lot of answers. For example, if you, for example, we know that if, another thing we know that if you take a shift attached to the coxster element, so that would be just the structure shift of the O over pre-image of zero power n, which is called punctual tube. So because of that, these two theorems combined, we get the answer, we get right away, so the answer for the coxster element pre-composed with the twist any power k, which is also known under the name of torus notates, n, n1, kn. So basically in this case, you know, so in this case we know what is this shift. You just take this punctual Hilbert scheme and twist it by line bundle. And that actually was conjectured by all of these people. That's what Hilbert used to get. So maybe I should say a little bit about the construction. Also, yeah, another thing I should say that, another theorem which we show that obviously this thing generally categorifies Homs-Lippenlinon. So also, you know, that if you take the early characteristics with respect to this anti-invariant guy, so then this hyperchromology would be only doubly graded space, so then it's Homs-Lippenlinon. So basically that provides a categorification of the Homs-Lippenlinon. We don't know whether it's the same as Havana-Frasanski. That is unclear to us because it's completely different construction, but we know for sure that it's categorification of Homs-Lippenlinon, and it's not trivial one. Okay, so that was advertisement of the result. Now how we get there? Any questions? So I should say this, you know, this kind of, this statement could be generalized actually. There is more general statement when you take not just full twist, but you know, there's this, you know, there's some called juicy Murphy elements, the ones which you kind of take on the partial break strings and go around. And they also have interpretations twist by line bundles. Basically the point of this construction that a lot of natural constructions, like multiplying by natural line bundles, have an interpretation on the north side by kind of transformation of the break group, some natural change of the break group. What's that? What's one-dimensional? T-A, T sub-A. Oh yeah, it's one-dimensional. So you have one gradient here and another gradient here, it's two gradings. No, it's two variables. One, two. So there is also this plus minus one, which comes from, basically there's three gradings, and in general it would be three gradings, if I would write QT, it would be three gradings. So but I take, like, I take other characteristics of one of the gradings, and I take usually, you know, the grading with respect to anti-invariant one. So actually another interesting thing happens that if a node is positive, so that kind of implies that this early characteristic part is always will have the same parity, which is not obvious from the definition of the polynomial. Actually it might have some kind of physical meaning because, for example, like Gannagish and Shakirov, they kind of talk about torus nodes and don't say how the construction would extend to the other nodes because they have S1 symmetry, and these are the positive nodes, but for them you don't have like this, they are, they hope the polynomials are positive because all of this extra degree always will have the same parity. So that's what we observe here. All right, so the construction. Okay, so here is the method. So we introduce this auxiliary space, which you already have seen, just this one, a potential, which left school, or something called super potential. I think it's just usual potential, but which is this, and the point is that also this object has the actual direction of the two copies of the Borel group, so basically they have two guys, two group elements, so then they would act on this element here by conjugation and so that's the action, and it's not hard to see that this potential is actually endearing with respect to the action of this Borel, and our main object is actually a category of matrix factorizations, which are equivariant with respect to this Borel, these two square Borel. So we introduce this Mf, square Bx, you know, what it is. So it is just... So let's denote this ring of functions on this C, on this ring of polynomial functions on this guy by R. So then the matrix factorization is actually just tends a product R times V where V is Z2 graded and differential, so V is Z2 graded, D is also Z2 graded, so the condition is that D squared is equal to W. So generally this theory of matrix factorization was started by Eisenberg, when he started like... Well, he started it just for a homological reason and later on conservatively adjusted that matrix factorizations would be relevant for the coherent side of the mirror symmetry and then our law developed like some kind of derives categorical approach to this matrix factorization. So it's a long history of this subject, but somehow nobody bothered to study this equivariant, guys. So the one naive definition... Any questions about it? So basically it's a two-periodic complex. So I didn't tell you anything about the equivariant structure. So in the equivariant structure you can say that, well, we can just require that is square equivariant. Well, it turns out to be a little bit too strong, so that's what we call strong equivariant space. Actually to make things work you have to introduce slightly more general theory, so where you require objects to be triple. It will be R as above and some correction. The point about correction term, so what is this D curly? So D curly is... So it acts on this product of... So basically we have to include our more precisely unipotent part of the barrel into definition. So the problem with that, why do we have to do all of these things? People studied equivariant matrix factorization with action of the reductive group. But in our case group is not reductive and we have to be super careful about it. So basically we have to define new homological machinery to work with this non-equivariant which was the covariant matrix factorization. So in the condition here, so I gave a definition that the condition here that D usual plus this correction term plus this Chevalier-Ellenberg differential which would be now acting on the bigger ring. So it's this and times universal enveloping of this guy. So we want this guy square also to be equal to W. So that is our definition of matrix factorization which is the covariant. And we also require and six we require that it's everything T square equivalent. So basically the barrel has a reductive part which is torus and for them you could be like very careless and say well everything is equivalent with respect to reductive part but with a non-reductive part you have to like deal like this. All right, so in particular it's easy to see that there is inclusion. If you have a stronger covariance you would have big one because you can just set D equals zero because if we require D to be equivalent then this Chevalier-Ellenberg differential will commute. Everything will be fine. All right, so that's the most technical part. That's why paper is so long because we have to work with this more general theory. So now we show that actually this category has a natural convolution which unfortunately I don't think have time to explain. So basically we give explicit construction for the convolution. So this MfB square has an associative product. It has associative product. So that's quite involved. So then second we construct explicit homomorphism from the braid group N. So let's call it star to this matrix vectorization thing. So construct is very explicitly. And then basically how do we... So that means for every element of the braid group we attach, let's say, we construct the matrix vectorization of this sort. Let's call it C beta. And well, since I have very little time I'll just say how we extracted the knot invariant. So I should say that these technicalities which I wrote, they're completely unavoidable. So you cannot work with this strongly accurate matrix vectorizations and it's related to the early work of Rosantz. There are some physical explanations but you need to have this correction. So how do we extract the knot invariant? So first of all there is always the shift attached to the identity. So this identity matrix vectorization. And then there is a kind of tool. So you can always change the sign of your matrix of potential. So then what we do, we just take our S beta. It's called prime. This complex, sorry, I just confused it. This is complex. This is complex beta. And you take a tensor product with identity bar. Now this is an element because if you take tensor product of matrix vectorization potentials add. So this guy had potential w. This one has potential minus w. So this guy is a matrix vectorization with potential zero. Which is the same as this periodic derived category. So this is db. Well everything is equivalent with respect to torus. That's how we arrive to the periodic complex. So now I didn't tell you how we get to this Hilger scheme thing. So basically now we have to take this inside of this space. You always have embedding of a b barl times nilpoten guy. So this is embedding here inside of this space x2. So inside here you have even this... Inside here there is this... What is my notation? Let's call it part which is stable part. There's a stable piece. So meaning that we're looking only at the pairs. So stable. So the stable part is if this is x, this is g and this is y we require that again if we take x, c generated by... This one conjugated by g y then it's generated by this whole space. This is an open piece. So now the knot in there is constructed like this. So you just take this to periodic complex and you just do pullback and you're done. That's the construction. More precisely you take this pullback pullback of this s power prime like this and you do one more thing. You take the home over the universal enveloping So that would be the last formula I would write. So take a home over the universal enveloping of the gauge group of the Chevalier-Einberg complex to this pullback from the stable part of s prime beta. And that is our complex s. Because this home is exactly like an algebraic version of taking quotient. Well, of course I forgot to write the quotient. That's it. That's our construction. And we show that if you take... Now we show that if you take hyperchromologies of this guy and it's important that we have to twist by these things because it satisfies Markov moves. Which is kind of quite elementary. There are some statements about global sections of line bundles on projective space. So that's where Markov move comes in. And yeah, that's it. So maybe I'll just stop early and let people ask questions.