 So last time we discussed some relations between braid varieties and maybe Detroit varieties and being homologous. So I tried to outline the general picture. I didn't really give concrete examples except for a very easy one for two-strand braids. So here is like one concrete non-trivial example, which I really like and especially this top picture keeps me up at night for less, I don't know, 10 years. So there is a lot of information here. So this picture is from a paper of Dunfield Gugman-Rasmussen who started this Kavanaugh-Prasantz-Kentripi gradient homology. And they computed a bunch of examples. So rather they predicted based on various tools and methods what this homology should look like. In this particular case, this prediction was certainly confirmed. And there is a lot of stuff going on in this picture. So let me try to explain some pieces. So this picture represents the triply graded homology of 3, 4 torus not. So it's triply graded. So you have a degree, which is represented by the height, by the vertical direction. So this is a degree 0, this is a degree 1, and this is a degree 2. And then for zones of direction, it's supposed to represent the q degree, whatever it is. And then these numbers written there are the t-degree or homological degree, maybe in some slightly weird conventions that I won't really discuss. But in any case, you have three gradings and the whole homology is 11 dimensional. So there are 11 dots in this picture. There are five dots in a degree 0, there are five dots in a degree 1, and there is one dot in a degree 2. And so in particular, as I said many times, we're interested mostly in a degree 0. So these are these five dots here, which happens to be a cutland number, but that may be not so important right now. What is more important is like, what is the corresponding algebraic variety? What is the braid variety that I talked about last time? And that was also studied by a separate group of people, most notably Thomas Lam and his collaborators in the world of cluster algebras and cluster varieties. So they considered the variety corresponding to cluster algebra of type E6. And I will explain why E6 probably next time, but I mean, it doesn't matter really. So you have E6 in kind of diagram, you can build a very specific cluster algebra for it and then the result break procedure, how to build cluster variety if you know what it is. If you don't know what it is, it is just essentially up to some torus, it's a positroid variety. So this is this open stratum in grass mining 37, which I denoted by P37 last time. So some miners again represent every point in the grass mining 37 by three by seven metrics. And you require that cyclically consecutive three by three miners are all known zero. And up to some torus, this is this variety. And so these people also computed the homology of this variety, even before understanding the relation to homology. And so in particular, they found out that the homology is five dimensional. It's concentrated in even degrees, H0, H2, H4 and H6. So H0 is one dimensional, H2 is one dimensional, H6 is one dimensional, but H4 is interesting in two dimensional. And this perfectly matches this bottom row here because they have zero, two, four, four, six and there's exactly the same numbers. And again, like maybe I didn't say it clearly last time, but it's not true that this variety has a fine paving or anything like this. It's pretty complicated, non-compact variety. So it's not completely clear, as we speak, why this homology is all given. And another feature of this homology, which was phrased differently in these two worlds. So in this world, in the original paper done for Google and Erasmuson, which goes back to 2005 or 2006, they observed that there is a perfect symmetry of this picture. So there is a vertical axis of symmetry. And if you flip it, of course, you get the same picture up to some regretting which they wrote down explicitly. But it's clear that all these dots are symmetric. In this picture, it's less symmetric, but what they observed is so-called curious hard leftist property which was also observed in many different settings which are closely related. So there is a unique class in H2. So there is a unique algebraic two form here. And you can look at powers of these two form and this would give this class in H4 and this class in H6. And what they observed is that this multiplication by this class in H2 actually gives an action of SL2. But it's centered slightly weirdly. So these two rows that are written here correspond to the weight filtration and co-homology. And so this action would preserve the weight or rather the difference between the weight and homological degree. And so there is an SL2 representation here which is four-dimensional and there is a separate SL2 representation here which is just one-dimensional and trivial. But in general, you always have this action of SL2. And if you center things properly you would see that the picture is symmetric. And so this table 46 is actually by graded again by homological degree and the weight filtration. And if you do a proper change of gradings you get this picture which is clearly symmetric but also there is an SL2 action here and it's natural to assume that there is an SL2 action here. And so I think like again, I personally think that like these two pictures are very nice and illustrated like the first non-trivial example that people can compute and study in all possible details in both languages in many settings. And the air computation of homology is completely independent of not homology just uses some recursions in cluster world and some laboratory sequences and things like this. But here we got completely different computation and matches. So I think it's really cool and nice. And another thing is that like, where is this SL2 action? So for them, one property of cluster right is that there is a pretty canonical two form which is very, very important for building this cluster structure, whatever it is. And so these two form is not random. And so I guess what I'll try to explain how to build this two form and not homology and how to relate it to various structures here and how to actually prove this symmetry at least what is the idea, okay? So this is kind of illustration for the previous lecture but it's also, I think, announcement and advertisement for what will follow because that would be a bit more abstract but I'll try to keep it easy. And again, please ask questions if you have any. So any questions about these two pictures? So I just borrowed them from the picture of Dunfield Book of Rasmussen and LaMageberg. Okay, so if there are no questions let's discuss about different structures. So I wanna talk about not only the SL tool which will appear but more generally like what kind of homological operations can we construct or what would we expect in the homology? And so just very roughly what happened in previous lectures. So we start from this polynomial ring R there's a polynomials in N variables which was think as kind of associated to strands of our braid. And then if you have a braid and N strands we construct a complex of RR by modules which is denoted by T beta. And these are equivalently modules over C of X1 through XN, X1 prime through XN prime where this X is index prime is going to left and right action of R. And so natural question, what can we say about this complex of modules or by modules like from the viewpoint of pure commutative algebra or homological algebra. And so the first observation which is very easy is that for any semantic function F in N variables the left action and the right action on this complex is actually the same. And this is true for a single crossing and this is true for any braid in fact. And this is just again if you have seen Zorgil by modules this is true for any complex of Zorgil by modules just by definition. So this is just true. And more abstractly you can say, well consider this algebra B which is the quotient of polynomials in N variables and N variables with primes and quotient by the ideal generated by F of X1 through XN minus F of X1 prime up to XN prime for all possible semantic functions F. So of course it's a vision to take I don't know elementary semantic functions if you want. So this is clearly an algebra and the action of my polynomial algebra in X is in X primes on TB on every term of TB if you want factors through this algebra B because the left and the right action of semantic functions are the same. And for people who have seen Zorgil by modules this B is usually called BW naught. So this is an decomposable Zorgil by module for the longest element of SN but if you don't care about it, it doesn't matter. What matters is that if you have a random complex where semantic functions on the left are not equal to semantic functions on the right it never comes from any break or anything resembling a break. So this is one series restriction. The second series restriction is that the action of X size on the left is actually homotopic to the action of X primes on the right with the exception that you need to twist the action on the right. So any braid corresponds to the permutation in SN and the left action of XI is actually homotopic to the action of XW of I prime where W is the permutation corresponding to beta. So here is one concrete example, some random braid I don't know what it is. So how do you get the permutation? You just start here, label it by one and then trace your strand until you get here. So this permutation would send one to two, two to one and then three goes back to itself. And so what this fact and this property says is that X1 is actually homotopic to X2 prime and X2 is homotopic to X1 prime and X3 is homotopic to X3 prime. And this is true again for any braid and in general, it's not true for random Zorgi band module but this is pretty special for braids. And somehow way to remember this as people in not homology community like to say is that this action of variable slides through the crossings. So you can imagine that you put some mark point here and the action of X corresponds to this mark point. We'll draw this mark point, let's try it at least. And so you can slide this mark point over here and you get another action of X in here and they're homotopic, they're not the same. And then you slide it again and you get here. And so one way to phrase it is that the action for different mark point is not the same but it's at least homotopic. So in homology, it's all the same. And then after you close the braid, this means that you identify X1 in the X1 prime. And X2 and X2 prime and X3 and X3 prime. And so before X1 was homotopic to X2 prime but now after closure, X2 prime is the same as X2. And X1 prime is the same as X1. There's a one here, but whatever. And X3 is homotopic to X3 prime which is the same as X3. And so if we close the braid, we actually have an action of sampling of an algebra except that some variables are identified. And so I think some people asked this question in the first lecture, like, do we still have an action of R? So in some sense, yes, but you need to be careful. So a better way to phrase it is that you really have this mark points and what you can say is that the action of polynomial variables correspond to the choice of this mark points. And then if you slide this mark point around, the actions are homotopic. And if you close the braid, the action of the mark point on the top gets identified with the mark point on the bottom. And so in particular in this example, if we ignore all this homotopia and stuff, what we can say. So the action of X1 is the same as the action of X2. The action of X2 is the same as the action of X1. And the action of X3 is separated. We don't know anything about it. And so effectively we have an action of two variables, polynomial ring in the homology, in X1 and X3. And so if we phrase it like this, so you can say that if you have the closure of beta, it's some link with R components. And triply graded homology of this link is naturally a module over a polynomial ring. And you have one variable, one X variable per component of this link. And of course the components of the link correspond to cycles in the permutation W corresponding to beta. So in this case, the permutation is actually just the transposition one two. So there is one cycle connecting one and two. There is three separately. And so you have two in three different cycles and you have two interesting variables. Set differently, you have equivalence relation on axis given by the cycles and given by the strands of the break. And so this is quite well known. And so you have an interesting polynomial algebra action. And again, I didn't write it here, but if you have a node, all axes are the same. So maybe let me write it here. So for nodes are equal to one and all X1 and X2 and Xn, that's the same. And so effectively you have just one variable action. And it's actually free over this one variable polynomial ring. And you can ignore it and that gives you finite dimensional vector space, which we discussed below, this is free. But if you have a link, even with two components, usually it's never free. And you can study the module structure over this polynomial ring and you get a lot of interesting stuff. And we will discuss it very soon, okay? And again, this is just very, very general property which happens here. And we'll see these actions in lots of different settings later tomorrow and on Friday, especially on Friday. Now you can do a little bit more of homological algebra and try to study homotopy between axes and X primes more closely. So we don't just say that they're a homotopy, we need to say like, what do we know about this homotopy and what kind of homological operations we can build from this homotopy. So let me just try to spell out what does it mean that to have a homotopy between X1 and X1 prime and if we're going to put prime here. So there is some operator X1 and the differential of X1 is X1 minus X2 prime. That is the definition of chain homotopy between X1 and X2 prime. Similarly, there is an operator X2 and the differential of X2 is X2 minus X1 prime. And there is an operator X3 and the differential is X3 minus X2 prime. And now we close the braid. So what happens? So differential of X1, so we identify prime variables with non-prime variables with the same letter. So differential of X1 is X1 minus X2. Differential of X2 is X2 minus X1. And differential of X3 is X3 minus X3. So this is actually zero. And differential of X1 plus X2 is X1 minus X2 plus X2 minus X1, so it's also zero. And so we see that X1 plus X2 and X3 actually give you closed operators on this triplicated homology. And so these are non-trivial homological operations of degree minus one, which are interesting. So in a sense they're kind of monodromes of these marked points. We're saying that you have a marked point here. We use X1 to move it here and then we identify this point with this point with the closure, which I don't want to draw. And then you slide it here and then you go back to here. So X1 plus X2 is the monodroming which we use to come back. And this monodroming turns out to be closed. And this gives you interesting homological operations in lean homology. And for each cycle in a permutation corresponding to a braid, you would have such a monodroming. And so you have one of these monodromes per component of the link. So one monodroming. And well, so it could be maybe not so interesting, but like one abstract thing which you can do, which is if you want to say it in hydro with some kind of causal duality, but like you can just do it and you can use this Xi variables. So this monodrom is to deform the differential on the chain complex. So you start from your chain complex Tbeta which is associated to a braid. And you tensor it with polynomial rank and in zero variables, Y1 through Yn. And then on this thing, you deform the differential by taking the old differential and adding Xi, I, Y, I, the sum of this where Xi's monodrom is from above and Y's are some formal variables. And again, you can check that at least when you close the braid, this query is equal to zero. And maybe I didn't say this, yeah, I didn't say this but this is a Lincoln variant. So theorem that we proved was not Hogan-Comp that this deformed homology is a Lincoln variant. So it satisfies all the market promotes it satisfies all the braid relations and you can define this deformed homology. And again, you have one monodrom per component and effectively this means that you would have one Y variable per component after all this identifications. So the number of deformation parameters is again the number of components in your link. And so here is one concrete example which might be interesting to some of you. So you have this hope link with two components so you expect two deformation parameters and two X variables which are alive. So the old complex is this complex from R to R to R with differentials given by zero and X one minus X two. And that we discussed in lecture one how to get this complex by first considering the complex of Zorgil bimodals and then taking home from R. And then the Xi, there is only one Xi or like took signs up to a sign which goes from the middle R to the left R and this is this map and this is a chain map because if I apply this guy and then differential I get zero or if I apply differential then there is no Xi so I get zero as well. So this is honestly a chain map chain and the morphism closed in the morphism of this complex. And then we can deform it by this rule so that you just put Y one minus Y two times this sign this red arrow. And again in this deformed complex this square is equal to zero. That's easy to see. And you can actually compute the homology. So this is R of Y which is just polynomials and X one, X two, Y one, Y two. You have two generators Z and W which leave in this degree and in that degree. And there is one relation that Z times Y one minus Y two is equal to W times X one minus X two because this, the differential of this guy in the middle is precisely W X one minus X two minus Z times Y one minus Y two. Maybe I'm off with some signs but it doesn't matter. And so this is something concrete which you can compute. And again, you see that the homologies even degrees but you can ask like why this is interesting and that I will explain in a second. Maybe let me actually explain this and then ask some questions. And so like one application which we found is the following. So you have N, K, N, torus link. So this is a link with N components. For example, this for K is equal to one you have T and N. So all components are actually on nodes and the link in number between different components is equal to K. And so we just computed was deformed and undefined triply graded homology for this link. And so how do you describe the answer? Again, like if you like combinatorics you can do it recursively as I sketched in the first lecture. So there is some really complicated recursive description and there are some more explicit combinatorial formulas but it doesn't give you, for example, the module structure for this axis and it doesn't give you some other interesting structures. And instead we deform the homology and say that the result is J to the key where J is the following ideal. So you have the polynomial ring in two N even variables X1 through XN and Y1 through YN and N odd variables theta one through theta N and you look at the ideal generated by XI minus XJ YI minus YJ and theta I minus theta J and you take intersection of such ideals for I not equal to J. So if you don't like these odd variables you can just ignore them. So again, if we restrict to degree zero part what we have is the ideal generated just by XI minus XJ YI minus YJ intersection of false such ideals for I not equal to J and this is just ideal in the polynomial ring in X1 through XN and Y1 through YN. And so this thing has a geometric meaning. So this thing because you can consider N points on C2 which is of course related to Hebrew scheme of N points on C2 but that we will see a bit later I guess on Friday but for now just consider N points on C2 with coordinates X1, Y1 and so on up to XN, YN. And then you look at the place where it is at the diagonal where some of the two points coincide. So where do I's point and J's point coincide? Well, this is a co-dimension to hyperplane which is given by equations XI is equal to XJ and YI is equal to YJ. And so this is the ideal of that hyperplane. So maybe I's point is equal to J's point. So this is precisely the equation that XI minus XJ is equal to zero and Y minus YJ is equal to zero and so if I have the union of all this co-dimension to hyperplanes they take the corresponding intersection of ideals and this is it. And so this is RG and this is kind of super analog of this which is not so important for now. And then we take the powers of this ideal which also make a lot of sense from the Hebrew scheme point of view and from the work of Mark Heyman especially but that we will discuss properly on Friday. But for now this is some ideal in this polynomial ring. And so the claim is that this ideal as a module over polynomials and X's and Y's is actually this deformed homology of Torus Link and KM. And to me I would say this is, I mean let's get a lot more structures. So maybe we don't know or like the combinatorics is hard to describe in the actual dimensions of graded pieces of this ideal but this ideal is clearly interesting and important and the fact that it's related to ring homology I think it's quite remarkable. And if you want undefined homology without Y variables you just kill them. So you just quotient by maximum ideal in Y's and then a separate result of Mark Heyman plus a little bit of work if you have all these status says that J to the key is free of our Y variables. So actually you don't lose any information. And so it is kind of a paradox that you want to describe some module over X variables. So as I tried to explain in the beginning for any link with N components you expect an action of polynomial ring and N variables. And you want to describe this module over this polynomial ring but instead you introduce N additional variables Y1 through YN describe this deformed object using X's and Y's and then just kill Y's. And this turns out to be the right answer because in this case it's flat over Y's. And it's interesting that it's much easier to describe this deformed homology before describing undeformed because somehow, I mean, I don't know any good commutative algebra description of this thing rather than J to the key and Y J to the key. And yeah, I mean, in examples you can see this but yeah, I think that's all what I want to say here. So any questions about this theorem about the deformation and about anything. Maybe it's a remark. So this is kind of one similar thing which is not exactly right in this setting but the motivating is that if you have a variety of store section and you want to describe its homology sometimes it's easier to describe a covariant homology by localization or anything and then just kill a query environment. So roughly speaking, that's what happens here. But it's not exactly the same thing. Okay, so let me still pause for questions. Any questions here? Comments, no questions. Okay, so anyway, so if you don't care about this example you might see it again on Friday but for now this is just an example. And then the next thing is really what I want to talk about today is this symmetry and SL2 action like how to get it from this X's and X's and for this I want to do some homological algebra. So I want to rephrase what we've said in parts one and two and three using slightly more advanced homological algebra language. So you have this algebra B. So these are polynomials in X1 through Xn and X1 prime through Xn prime with the relations that symmetric functions and X's are the same as symmetric functions in X primes. And you can resolve R. So R is naturally a module over this B and you can resolve R over B by three modules. And so you can denote this resolution by curly A. And so concretely this curly A is what? So you have B, you have the variables Psi1 through PsiN which we saw before. So the differential of PsiI is XI minus XI prime. And these are the homologues that I talked about. And you have additional variables U1 through Un and the differential of Uk is more complicated. So the differential of Uk is H PsiI times the complete symmetric function HK minus one of XI and XI prime. Maybe kind of good way to write it is HK minus one of XI, XI prime is actually XI to the K minus XI prime to the K over XI minus XI prime. That's kind of easier to digest. And then you have this thing, you multiply by XI and you sum over all I. And a very good exercise to understand what's going on is to show that D square is equal to zero. For example, if I take D of Uk, I get this thing and I apply the differential again, this will be zero. And this is not completely obvious but this is not hard to see. And slightly harder exercise is that this is indeed a resolution. So there are no more relations. And so you can think that like in R, you identify left action and right action, obviously. So you need this size, but then you want to describe seizures between sides and these are given by these things. And then there are no more interesting seizures after that. Anyway, so this is some complex and this is, you can think of it as a digital algebra if you want. And so the theorem that we've proved in a recent paper with math, a plug-in component on Mellet is that this digital algebra over here acts on any Rukia complex, T beta for any braid beta. So you have an action of all this size, which we saw before and the action of this use, which is more interesting on any complex of bimodials for any braid. And so let me try to sketch the argument for this. So the argument is the following. So first of all, you need to construct this action of size and use for every crossing. And size you construct explicitly, that's not so hard. Maybe I don't want to write it, but it's really, really easy. And like this is why the left action and the right action are correct. And the action of U is just zero because you have a homological degree too. There is no rule for it. And you still need to check that this relation is satisfied. So for example, if U is equal to zero, then this right-hand side is zero. But this is easier in this condition. So I don't want to explain this, but that's true. And then you want to extend it to the product of crossings, to arbitrary braid. And here is an interesting idea. So you, if you wanna extend the algebra action to the tensor product, what do you usually do? You build a coproduct on the algebra. And so you construct the coproduct delta on this algebra A. More precisely, this coproduct would go from A to A tensor A over R. And A has some Xs and X primes. The second copy of A has X primes and X double primes. And the coproduct is given by these explicit formulas. So maybe it's not so important. I will comment on geometric meaning of this coproduct in a minute, but the coproduct of Xs is just Xs and X1. So Xs on the left, coproduct of Xs primes is one tensor Xs prime, which is Xs double prime. So this is kind of, if I Xs prime, it will be the right most. And then coproduct of Xs I is Xs I tensor one plus one tensor Xs I. That's not surprising. What's more surprising is that coproduct of Uki of this new guys is Uki tensor one plus one tensor Uki plus an extra correction term. And this correction term involves complete symmetric function in Xs I, Xs I prime and Xs I double prime times Xs I. So there is some formula. Maybe it's not so important. I just wanna say that if you don't put this correction term, it doesn't work. And this correction term actually is super important for what we'll follow. And so one of the exercises I think is to check that this is indeed a chain map from here to here. And I mean, you can explain it more abstractly why such a map should exist, but maybe it doesn't matter. And so if N, M and N are two modules over this algebra, you have an action of M, A on the left on M. You have an action of A on the right on N. And you just have an action of A tensor A on this thing. And then using co-product you can translate it to an action of A on the tensor product. And so because we constructed the action on single crossings and we can extend it using co-product to arbitrary products of crossings to arbitrary braids. And maybe let me say that like because of this correction term over here, it's really non-trivial. So although U's are equal to zero on single crossings, when we start multiplying things, these correction terms will accumulate and they become non-trivial. So even if you have two crossings where U is equal to zero, once you multiply them, these correction terms will give something non-trivial. And the action of U will become non-trivial pretty soon and interesting. So there is some question. Some question in chat, Eugene. I'm sorry. The product that bi-algebra hope for algebra. It is a bi-algebra, so co-product is an algebra of homomorphism if you want. So we defy it on generators and then we extend it to products. The only subtlety which I'm really swiping under the rock is that this is not associative. So maybe I should say this properly, although again, I guess nobody here would care. The delta is co-associative up to homotopy. And so it's not actually, oh, sorry. I'm sorry. That's a good time to ask questions. I don't know why it crashes every time I give a talk here. Usually it doesn't crash. I'm sorry. So I was saying that this is co-associative up to homotopy. So if you want to be really careful, you need to keep track of this homotopies and higher homotopies and they give you some kind of infinti structure, but you can actually avoid it in this case by some tricks. So we can write something about this higher stuff, but it's really, really nice to end. And instead, we just prove abstractly that this higher homotopism which give this homotopy the associativity and some kind of higher associativity relations, they exist on abstract rounds and then it doesn't matter how we choose them, the result will be the same. But in principle, there is something interesting. And again, it's like something really, really basic. It's all about this algebra B and resolution of R over B. There is nothing else. And again, maybe it's actually a good time to say that because it's resolution of R, you know that R tensor R over R is R. And so this is a resolution of R. This is in some sense resolution of R tensor R over R which is again R. And so you know that there should be a chain map which relates these resolutions and we just write it explicitly. And extend, yeah. So this is an algebra homomorphism. And then co-associativity of the homotopy, I mean, it's immediately follows from what I just said but there is a trick like how to prove it very clearly and that's what we use. Anyway, that is definitely too technical. But in particular, why this co-associativity is important because if you have a triple product of three things, so if you have, I don't know, M tensor N tensor K, a priority, there are two ways to define an action on this thing and they're different and they're honestly different but homotopy. So let me try to sum up things. So there is an action of this algebra on any read and there is an action of interest in operators size and use on any Rookier-Kohm books for any read but they're not closed. And so instead, so solution to this issue is you just deform homology as I did before. So in the form homology, you deform the differential and then you can also deform use by saying that instead of use, we consider use plus we gain some correction terms involving partial derivatives in Ys. So this is again some formal homological algebra thing but this gives in operators in this deformed homology. And so what we proved is the following thing. So this deformed use, so this FKs, they're actually closed under D. So this is easy to see because somehow the differential of U is not equal to zero. So these two terms don't commute and these two terms don't commute because Ys don't commute with partial derivatives. But somehow this is designed so that they cancel out each other and this commutator is zero. Now it's more or less clear that they commute between themselves FK and X sub I also commute that's clear. And then FK and Y sub I, they don't commute but you know how they commute with each other. So the commutator before closure is something like HK minus one of XI XI prime. The commutator after closure is just K XI K minus one. And so these Fs are what we call the pathological classes and they act in deformed homology of every link. And that is kind of the main result is that this F2 corresponding to U2 is an interesting operator and that satisfies this kind of hard left so curious hard left condition. And that operator actually leaves to an action of SL2 on deformed homology for any link L. And so as a corollary this deformed homology has an SL2 action. And so it is symmetric as this explains the symmetry that we saw in the bottom. And this resolves this conjectural look of Donald Rasmussen that for knots you always get a symmetry because for knots this deformed homology is essentially the same as under form. And so we get it like the bottom line if you don't care about the symmetry is that we have a lot of interest in commuting operators acting in link homology. And they're very new. You can ask tons of questions about them and how do they act? What are the other relations? Can you, how do you interact with this SL2? Like what is the structure of homology in known cases as a model over this operator? So they're all very interesting questions. And this is very, very, and so geometrically you can ask, well, okay. So we don't care about all this monological algebra. We like algebraic varieties. What happens there? What happens for this great varieties or this droid varieties or whatever. And so we need to find some interesting operators acting in co-homology of these varieties. And of course the natural guess would be that there are some kind of the astrological classes in co-homology. And we just multiply by this the astrological classes. And of course, this is the case. And that is actually one of the motivations for us to build this digital algebra story. And this goes back to the work of many people, Atya, Bot, Shulman in particular, and more recently I guess Lisa Jeffrey and others. So how to build the astrological classes on character varieties and more recently on braid varieties. And so this FK would correspond to some algebraic K form on this braid variety. So how can it be built? So we'll start from the group JLN and Bot and others give you a family of differential forms on the group and on some related spaces. So given any symmetric function of degree R in N variables, one can construct a characteristic class which I would call phi zero of Q in two R's homology of the class find space BG. So this is probably familiar to many people. If you have a degree R symmetric function you get a characteristic class. What is also familiar to many people is that you have another interesting class which is, I mean, really it's a form in this case. I mean, I should write a form instead of H. So this is an algebraic differential form on the group itself. So, and the form will be two R minus one form and this is closed. So it represents a class in H to R minus one. That's why I wrote it. And in fact, if you think about the respectful sequence for the universal, for the vibration EG over BG this form, this class would kill that class by a differential. You have seen this, but in any case for any symmetric function you can build an interesting co-homology class of degree two R minus one on the group itself. And this is also quite known. And what is less known is that you can continue this procedure. So you have a two R minus two form on G cross G. So this is just the product of G with itself. And these two form, two R minus two form, sorry it won't be closed, but it would satisfy some kind of co-cycle condition that the differential of phi two, the differential of this guy will be two R minus one form and it is related to two R minus one. Okay. Yeah, so what I was saying, that you have a G cross G and there is some differential form on this. It's not close, but you know the differential. Here is a concrete example of this. And now what you can do, so if you have two algebraic varieties X and Y with the map to the group or alternatively you can think either we have algebraic varieties with the map to the group or we have just matrices kind of depending on parameters which live in this algebraic varieties, X and Y, then you can build an interesting to an interesting form or two R minus two form on this. So suppose that you pull back this form phi one of Q from the group to X and you pull this to Y and they turned out to be the boundaries of some on the X and on the Y. So you have some form here and you have some form here. And the important point is that you can glue them together by taking the form on X, the form on Y and this correction term which is phi two of Q which appeared over here. And maybe that was too fast, but it doesn't matter. There is a procedure how to build interesting forms and product of varieties. If I have an interesting form here if I have an interesting form here I take the sum of these forms and just add this form correction term which lives on G cross G. And I claim that this is very similar to the co-product on my DG algebra that you have this kind of element or the action of U on the left, action of U on the right and then do you have these corrections or which glues them together. And so classically people have used this construction and this specific form phi two of Q to build a two form and subtract two form on character varieties. But Anton Manit, for example, proved this that one can use it to construct algebraic two form on this braid variety. And this two form on braid variety is constructed exactly like this. What is the braid variety? You have a bunch of matrices for each of them we put the two form to be zero and then you glue them by this procedure inductively. And this gives a non-trivial two form and this turns out to be almost symphactic up to some torus action up to seven things. And this two form is closed and it gives you a cohomology class and it does satisfy Q's hard left shift theorem for this braid variety exit beta. And so you can use this to build an action of SL2 on the braid variety and you can use this to prove this Q's hard left shifts by this machinery. So actually these people also observe this Q's hard left shifts in this case but they use very, very different machinery. But this is exactly the same two form which appears here and which represents the class in HF or two. And so maybe just to conclude I wanna say that in principle this works for any symmetric function. And so the fact that I had this F2F so this gives me an analog of F2 but I can build F3, F4, F5 and so on. It has lots of interesting things and this gives you, in this case it gives you lots of interesting forms and cohomology classes on X of beta by the same procedure you have something on matrices and then you keep building them. So it might be less explicit but you might not need it to be more explicit. And so for example, the next optological class would correspond to sum of XI cubed, I guess and this would leave here. So this would give you four forms. So if R is equal to three, where is R? So if R is equal to three it would give you six form on BG it would give you five form on the group which is there if the group is at least JL three and it gives you a class in H upper five of the group. Maybe I'll write it down actually so if R is equal to three, you have H six of BG and you have H five of G and you know that there is a generator of degree five in the group and this gives you a class it's not closed by the class it's in a four form on J cross G and then by some reasoning you can check that this four form has an interesting weight and this actually gives you a generator here. So this is kind of the form which you can build from the cubic symmetric function and it's there. And in this case, you see that cohomology is a ring is actually generated by this class in H upper two and this class in H upper four and it's a very natural question which is as far as I know, we're wide open is it true that cohomology is generated by this theftological classes for torus nods? So maybe I will ask this question. Some problem slash conjecture. Is it true that for co-primes whatever braid variety is generated by the theftological classes or maybe equivalently as H H H is generated. You see that all this theftological classes they are in even degree and we just proved that cohomology is in even degrees. So maybe it's not so surprising. And one reason to believe is that in slightly different setting there is a space with very, very similar homology and their homology is generated by the theftological classes. And another question is whatever. So suppose that it is generated by the theftological classes. So can you describe this co-homology? So again, here co-homology would be a ring because it's the co-homology of a space. And you can describe this, can you describe this ring by generators and relations? And here co-homology is not really a ring. There is no ring structure, but it's a module over F2 and F3. And hopefully this module has one generator or one kind of co-generator. And again, what a, how to describe this module explicitly? What are the generators and relations? These are all wide open. And I think that's an excellent problem for someone to work on. So at least I try to convince that, like besides these combinatorics and the fact that we know the dimension of homology, this opens up lots and lots of questions. Like what are these theftological classes? How do they act? And what can we say about all this? And I think I'll stop here. So again, I'm very sorry for all these interruptions and technical issues. Any questions? Can you say a little bit about which symmetric function gives rise to this element that does this curious hard left shift? Yeah, sum of XI squared. So if you have sum of XI squared, this would give you, so again, you can write it down. So if R is equal to two, then you get two form on G cross G. And this is precisely that two form. So you'll get three form on G, but you'll get a two form on G cross G and this is these two form. And so the recipe is that you just take this. So for this particular queue, you take this correction term, this two form on G cross G, which is very explicit. And whenever you see a bunch of matrices, you do it inductively. So you do it for the first matrix, the form is zero. For the second matrix, the form is zero. For the pair of matrices, you have this correction term. And then you have the third matrix. And then you add correction matrix, corrections are for the product of the first two matrices in the third one and so on. So this, you build inductively these two form. And this gives you a class. And like whenever you close the braid, this gives you a class in H upper two. And this class in H upper two will give you hard left shift. And maybe I wanna add that, like Anton actually used this theorem over here to prove curious hard left shift for more general character varieties because they can be somehow stratified by this. So this is like really, really powerful thing. And like most of what I explained in the beginning was motivated in kind of trying to translate it to not tiering more algebraic language, more kind of homological algebra language. Yeah, now I can see, sorry, I crashed again. I don't know, maybe my network or my sum or something, I am very sorry. Were there any more questions? He didn't have any more questions. One more chance for a last minute question. Well, if not, let's thank you, Jean again.