 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, more generally complex of shifts on a Hilbert's scheme of points on C2, such that if you take sections, more precisely all of the homologous of the shift, so this is as beta, then it is a knot in there, is 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 Pakanagich, then there is a work by me and Shende, me and Shende in Rasmussen and Gorski in Yigut, so many people conjectured it and now there is a construction. And maybe another thing I want to say that for a particular type of knot, like a torus knot, there is a very explicit answer for this shift and I would say well, you would ask 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 knot. 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. So let me remind you from the very beginning so that there is a serum of Markov, I guess beginning of the 20th century so that if you look at the... So the serum of Markov said that if you take a braid group, should I remind people what is a braid group? Everybody knows what is a braid group? I'll write it up quickly in a second. So if you take a union of the braid groups on any strands, modular sum relation which I'll write, then you'll get that's equal, this set of equivalence classes is a set of isotope classes in R, in S3. 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, n plus minus 1, it's the same as... I'll draw a picture, so the alpha here belongs to the n minus 1. So the picture here is like this. So the braid group is generated by the simple crossings. So if this is i strain and this is i plus 1 strain and everything else goes like this, this is the generators and they satisfy relations as we all know, the braid relation. So Markov's serum tells you that if I have a put in the box, let's say alpha, and this is braid on n strands, then if I take a closure, 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 just move this bit over here. That's the same thing. So that's the first relation. And the second relation is more interesting is that if you have this notation alpha on the first n minus 1 strands and you added one more crossing and you took a closure. So you can take... This loop could be removed. So this is the same as this kind of loop. And this you can remove. That's the statement. And basically, Kuhl 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 the set of braid groups which respect these relations. 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 homfully PT invariant, which is constructed, for example, using the R 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 homologes. And in particular, there is this knot invariant which is called Havanov-Rozansky homologes, which I will explain in a second. So basically, Havanov and Razansky define, give a construction of a homologous of triply graded homologes. This is sum over i, j, k. So basically, when they're any braid, they construct triply graded vector space sometimes people denote it by Havanov-Rozansky, 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 a categorification of the homfle 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 homfle pt invariant of, so sorry, I forgot to write some now. So it's q power j a power k. So that's homfle 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 homfle 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 quantifier we will get. So another remark I want to make, that construction of Havana-Fendrazanski, there are two flavors, so Havana-Fendrazanski use Zergal bimodels, Zergal bimodels, we find 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 constructability with a symplectic side. And in some sense, I think what we do, it's a kind of mirror do for this. What we do, it's more like a coherent side, a coherent shoe side. All right, so, and the point is that the Zergal bimodels construction is a, computation is very hard, even if you write a kind of computer program, you basically, like all of the computer programs which are available, get 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, so now I have to justify my invitation and talk a little bit about gauge theory. I guess, you know, Hilbert's scheme doesn't classify as a, is it okay? Is it a gauge theory? It's gauge group with U1, but it's something, so. Okay, so it's possible that a lot of this stuff generalizes to other gauge groups. We didn't try. So, you know, that's, other gauge 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, 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, Hilbert 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 would 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-taurus, 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 this Hilbert scheme. So, and maybe I will just introduce some kind of funny notation. So, there is a smaller subset here, of C2, which is, let's say, so that Hilbert, Chow, symmetric and C2, 0. So, what does it mean? So, first of all, we always have a map from the Hilbert scheme of C2 to the symmetric power of C2, which is basically taken 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 0. I'll just do it slightly smaller, just introduce this smaller subset. It's just because somehow we have chosen to work with SLN group. And, first of all, I should also say that whatever I'll 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 would just write a list of names who talked about this thing. So, first of all, there is a work by me, J. Krasnitsyn and Tchende, and then there is a work, a generalization of this thing by Tchende. So, then there is also a work by Gorski Nygut. There is also physics, and a physicist also wrote about this thing. There is a work by Aga Nagych 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 talked about this small one. 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, like I said, there is always, here there is always a class of constant, of identity, and this spins, you know, 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 you know, 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. Again, you know, I'm not giving all of the details who proposed which part of the conjecture, but outcome of, if you put all of the 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 equivalent on the Hilbert 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 the shift, tensor with exterior powers of this tautological bundle, then this is the same as a Havana-Frazansky homologous. So, here, you know, we have three gradings, and here we have three gradings, one comes with the, two gradings come from the scaling action of torus t, and another grading comes from the exterior power. This is exactly the three gradings. So, I think this is false, but what is true, so this statement is true, I believe, for the positive knots. If not, the very positive, then it's true. If it's something in between, then there is something more complicated, which I will explain. So, what is this here? It turns out that, first of all, the better object to work with is the Flag-Hilbert scheme. So, there is this, let's denote this one. Hilb 1 dot 1 comma N, just a set of flags. So, it's I0, I1, IN inside of the Cxy, such that, you know, the difference, each of them is co-dimension one. So, one of them is this, exterior power, and two more because we have a torus action, C star cross C star. What 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 response. We have, like, whole series tends to transfer response. It might work for transfer response, but I don't believe it's true for the usual. All right, more questions. All right, 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, is because we have to work with this object called, like, a career and metrics factorizations, and the subject was not developed, and we have to develop it from scratch. Push forward, pull back, we have to define all of these objects. But let's go to the statement. It's actually true. So, first of all, there would be always the interesting part of this Hilbert scheme is this Lagrangian piece. So, 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. So, 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, I saw the coincidence of this Lagrangian. All right, so, what's even more interesting, that there is this, the right object is to 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 sits inside of the, let's just save a little bit of chalk that for now on G is a SL and GN and group G is a SL. So, it sits inside of the, it sits inside of the Borel of N times the group GN times the important. With the condition that if this is X, this is G, this is Y. So, the condition is that X, 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, so just 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 apotrangular. So, the VN is apotrangular matrices. The zeros and N is a strictly apotrangular matrix. So, the condition is that you take two matrices of this sort and you have a vector such that you can generate the whole vector space. It's an open set inside of it and that's what we call free Hilbert scheme. So, the amazing thing about it 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 geometry. So then, you know, our main theorem, so that object in this periodic category, which I will explain in a second, so I explained you what is still the thing, so I didn't write you what is the actual one. The actual Hilbert scheme, 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, the two periodic means, the periodic derived category means we're considering two periodic complexes. So, they have like odd and e, I'll say they would be the ships of this sort, S0, S1, and S0, S1. All right, so, and the condition is that, first of all, the properties that if you take a hyper-chemology of this subject, of this guy, then it's an isotope invariant. And the second, so we can always, you know, as I said, the elements here is just just two periodic complexes, and you can always take kind of 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, they would be actual shifts on the free Hilbert scheme. So, and it turns out that actually this, let's denote that this H0 of this shift, of this complex in 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. So, you know, you obviously have the usual Hilbert scheme sitting here just by imposing condition that, you know, matrices X and Y commute. 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 a statement from homological algebra, you know, you can compute hyper-chromology 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 which converges to this hyper-chromologist, which is a knot invariant, hyper-chromologist 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. So 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, so it decreases the usual degree by one, and it increases the exterior power. 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 spectral sequence degenerates in the first term. So when that happens, when p is very positive, then that happens. Then this conjecture, which I was writing before, holds. So basically, if beta is very positive, then hk, whatever, 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 pre-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 the series is interesting. So in other series, in other 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. This 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 of a preimage of zero power n, which is called puncture tool basically. So, and because of that, these two theorems combined, we get the answer, we get right away. So, 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 puncture tool, it just came and twisted by line bundle. So, and that actually was conjectured by all of these people. That's what you also get. So, maybe I should say a little bit about the construction. Also, yeah, another thing I should say that in others here in which we show that obviously this thing generally, like a categorify is home simple and all. So, also, you know, that if you take the other characteristics with respect to this anti-invariant guy, so then this hyperchromologist would be only double graded space. So, then it's, then it's home simple. So, basically that provides a categorification of the complete PT. 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 home simple PT and it's not trivial one. Okay, so that was the 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 is this, you know, there is 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 grading here and another grading 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 is three gradings. And in general, there will be three gradings. If I would write QT, there will be three gradings. So, but I take, like, I take all the characteristics of one of the gradings and I take usually, you know, the grading with respect to 20 variant 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 Gannagich 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. 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, too. All right. So, the construction, okay. So, here, the method. So, we introduce this auxiliary space, which you already have seen, which is this one, a potential which left school by, for some reason, called superpotential. 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, you have two guys, two group elements. So, then they would act on this element here by conjugation. So, that's the action. So, and it's not hard to see that this potential is actually inherent with respect to action of this Borel. And our main object is actually a category of matrix factorizations, which are equivariant with respect to this Borel, this two square of 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, and then the matrix factorization is actually just... So, the tensor product R times V where V is Z2 graded and differential. So, V is Z2 graded. D is also Z2 graded. So, and the condition is that D square is equal to W. So, generally, this theory of matrix factorization was started by Eisenberg. When he started like a... Well, he started it just for a homological reason, then later on, conceivably adjusted matrix factorizations would be relevant for the coherent side of the mirror symmetry and then our law developed some kind of derives categorical approach to this matrix factorization. So, it's a long history of this subject. So, but somehow nobody bothered to study this equivariant guys. So, and the one naive definition, any questions about it? So, basically, it's a two-periodic complex. So, I didn't tell you anything about equivariant structure. So, then the equivariant structure, you can say that, well, we can just require that 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 object 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? Because people studied equivariant matrix factorization with action of the reductive group. So, 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 equivariant matrix factorizations. So, in the condition here, so I gave a definition that the condition here is 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 w. So, and that is our definition of matrix factorization, which is equivariant. And we also require, and six we require, that it's everything T square equivariant. So, basically, the barrel has a reductive part, which is torus, and for them, you could be very careless and say, well, everything is equivariant with respect to reductive part, but with a non-reductive part, you have to deal like this. All right, so, and in particular, it's easy to see that there's inclusion. That's, you know, if you have a strong, strong equivariance, you would have big one because you can just set D equals zero because if we require D to be equivariant, then this Chevalier-Ellenberg differential will commute. Everything will be fine. All right, so, that's our, 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 a associative product. It has associative product. 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, this. Let's call it C beta. And, well, since I have very little time, I'll just say how we extracted the knot in there. So, I should say that these technicalities, which I wrote, they're completely unavoidable. So, you cannot work with this strongly equivalent matrix vectorizations. And it's kind of, it's related to the early work of Frasunzky. There's some explanation, physical explanations, but you need to have this correction. So, how do we extract the knot in there? And 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 dual. 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. You take a beta, like this complex, sorry, I just confused, 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, and, you know, this guy had potential w. This one has got potential minus w. So, this guy is a X 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 tors. 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 barrel times new potent guy. So, this is embedding here, inside of this space X2. So, inside here, you have even this, there is inside here, there is this, what is my notation? Let's call it the 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. So, this is an open piece. So, now, the node in there is constructed like this. So, you just take this two-period 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, let me write it again. 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, you know, this home is exactly like algebraic version of taking quotient. Well, of course, I forgot to write the quotient by clause. That's it. That's our construction. And we show that it's a, that if you take, now we show that if you take hyper-chemology 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, 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.