 Okay, so welcome back everyone. So let's start the second half of Eli Kasby's talk on the representations of affine groups and Ikevarian homology of affine respondents. Please take it away. Thank you. Thank you so much. So yeah, now I'm going to explain I'm going to explain what this categorification is. And yeah, so we're going to provide natural interpretation of D-bar in terms of this categorification. And we will see how this allows us to answer questions one and two. All right, so quantum affine algebra is or affine quantum groups. So you can view them in two ways. So it's this UQ of G-hats. So the first way you can view it is you take an affine deformation of G-hats. So a loop deformation of G-hats. And then you take the quantized universal enveloping algebra of this. Or alternatively, you take the quantum group UQG. So quantized universal enveloping algebra of G. And you can have a loop version of this. And it's a famous theorem by Drinfield that these two things are actually the same. So anyways, we don't need this for us. It's just, for us, it's just going to be some algebra which is a hopf algebra. And because it's a hopf algebra, its category of finite dimensional reps is monoidal. So if I take V and W, V tensor W is not only a UQG hat tensor UQG hat module, it's actually a UQG hat module. Because, yeah, because I have this co-product here. So yeah, for us, so it's like a monoidal category of finite dimensional reps of this guy. It's a very huge category, infinitely many simple objects. It's not semi-simple, so it's complicated category. But at least it has something nice. There is a notion of character. So that means there is an injective ring homomorphism from the Grottenly group of C. So Grottenly ring of C because it's a monoidal category. So actually it's a ring. So from the K0 of C to some torus. So basically for each object in C, you can write down a Laurent polynomial in a bunch of variables which records the dimensions of weight subspaces. So any representation, finite dimensional representation of this guy has a little bit like this L of lambda that we saw earlier. It's just more general, more, yeah, high-tech. So there is also a notion of weight subspaces and you record the dimension of these weight subspaces into some Laurent polynomial. So that's really what this guy does. So this is due to Frankl and Reshetikin at the end of the 90s. And yeah, so as I said, there is a monstrous quantity of simple objects in this category. So there are as many, up to isomorphism, as many simple objects as monomials in these generators. So really monomials, not Laurent monomials, yes. And so for example, you can take a monomial with only one guy, one variable. So this is what we call fundamental representation and somehow they generate this category C, okay. But of course, you can consider more complicated monomials. So it's really like an analog of highest weight, yes. So these YIAs are analogs of the omega i in classical theory. So analog of dominant weights. It's just now you, for each i in one comma n, instead of having just omega i, you have as many weights as non-zero complex number. So it's really something massive. But you should think of this as omega i twisted by some complex numbers, roughly speaking. Okay, and this classification result is due to Shari and Presley. So yeah, so for us, we have a nice, so we have a monomial category, it's Grotten group, it's just a nice ring. And we have, so it's called Q character, but everything is commutative here. Okay, this is purely a billion setting. This is a commutative ring. This is a nice Laurent polynomial ring. Okay, so what Hernandes and Leclerc did, they looked at some kind of discrete skeleton of this category. So some kind of Z version, some kind of integer version of this. So now instead of looking at any non-zero complex numbers here, we're going only to look at integer powers of the indeterminate Q. So recall that for me, G is simply laced. So I can choose an orientation of its dinking diagram, so-called dinking quiver. And we're going to look at the subcategory of C generated by these fundamental reps. So we choose a certain collection of monomials in the Y. So if you want, we choose only the YIA, such that A is an integer power, so it's Q to the P where P is like this. Okay, so we look only at the simple modules whose label involves this YI, Q to the P. Okay, so this is what we call CZ. So it's really the same as C, but in the discrete version. And then we have some kind of truncated version of C, which depends on the choice of height function, where basically we take like one half, if that makes sense. So we take only, if you want roughly speaking, only negative, yeah, only like one half of CZ. Okay, both of them are monomials subcategories. And so what they proved in, yeah, 2015, 2016, what they proved is that our coordinate ring here, C of N, is isomorphic to the grotting ring. So extended scalars to complex numbers is isomorphic to the grotting ring of a certain special monoidal subcategory inside CZ or inside C lower XI. So we have a nice monoidal subcategory, so which depends on our initial choice of Q here. So if you change your Q, you'll get a different subcategory of CZ, but in the end, the grotting ring will be the same. So yeah, so once we fix a Q, we have a nice, so Q tells us which fundamental reps we should choose, finally many of them. And this CQ, so now the CQ is going to be a small category that we can handle a bit better. And it's grotting the ring is isomorphic to CZ. And moreover, this isomorphism maps bijectively the dual canonical basis constructed by Loustic and Kashiwara to the classes of simple objects in CQ. Okay. So the dual canonical basis is not the same as the Mirkovic-Vilonian basis that we saw earlier. It's two different basis, but they have a significant overlap. So yeah, so here it's dual canonical basis mapping to classes of simple objects. But what's good for, so in particular, what we'll be using in our construction is that, so we have this isomorphism, and this category is a subcategory of CZ. So we have an obvious inclusion here. And then we have this Q character morphism due to Frank-Helen-Rachetique. So that means now that we have a way, so this representation theory of affine quantum groups, we can take it as a black box that tells us that for each regular to each regular function on N, we can associate a nice Laurent polynomial in whatever bunch of variables. And this is what we'll be using. Yeah, so concretely, do I have an example? Yes, I do have an example, fantastic. So my running example was SL3. So I'm choosing this orientation, okay? So in this specific example, the category CQ is generated by three fundamental representations. So this one, this one, this one, okay? And so believe me that there is a short exact sequence. I mean, there are many more of course short exact sequences but let's look at this nice short exact sequence in CQ. So if you tensor these two guys together, so these guys are simple, okay? They are both irreducible representations but if you tensor them together, it's not simple anymore. So it's kind of big question in this area to know when is it true that the tensor of two symbols is still simple. Okay, I'm not gonna go into that. In this example, the tensor of these two symbols is not simple. So it has a sub-module which is actually this guy and the quotients of this map is also simple. It's not a fundamental but it's still simple, okay? It's a monomial with two guys. So this guy is not simple but it's of length two in other words, okay? So it has irreducible sub-module and irreducible head, irreducible, yes? So we have a nice short exact sequence in this category. So in the grotnik ring, having a short exact sequence means that the class of this is the sum of the class of this and the class of that and because it's a ring, so the class of this is the product of these guys. So the product of the classes of these guys, the product of these two classes is equal to the sum of these two classes. So maybe that slightly rings a bell from the beginning of the talk. So this is really the x, x prime equals y plus z of slide number one or number two, right? So Hernández Leclerc isomorphism in this example maps this guy to x, this guy to y, this guy to x, sorry, my bad. This guy to x, this guy to x prime, this guy to y and this guy to z, yes? Right, so this x, x prime equals, so we started with x, x prime equals y plus z which when you look at some minors and matrices is completely trivial. We have a first level of interpretation which is, well, some mutation from the cluster structure of CA. This is another level which is the categorified level where we say, well, this mutation sequence, this identity in the Grottenegring comes from a short exact sequence in this category, okay? Yes, so what we are going to do is we're going to, we're going to, so now we have regular functions in CN such that x or y or z or x prime and now we know that they correspond to classes of modules in the CQ and these modules in the CQ, they have a Q character. So I can map them to get nice lower on polynomial. So this is what we're going to do now. So, Franco-Rachitikin defined so this Q character morphism and Hernandez Leclerc for reasons related to cluster theory that I will not tell, they had to truncate it. So they used what they call truncated Q character. So basically, if you have a module in CQ or more generally in this half of CZ, well, you look at its Q character so you get a Laurent polynomial but then maybe this Laurent polynomial involves variables that are in the other half in the half we're not considering the upper in the other half of CZ. So you just killed it. So it's still a morphism, it's still injective and yeah, this is the reasons related to cluster theory. This is, okay, I will not go into much detail okay, so you take the Q character or some guy in CQ and you kill the terms that live in the other half of the category. So for example, if you look at these representations this representation which corresponds to X it's as a vector space, it's of dimension three. It's Q character has three terms and the truncated Q character kills these two guys and keeps only in this one. And similarly for this one which corresponds to Y Q character, the truncation kills these two terms. This guy is more complicated so the truncation also kills everyone but the leading term. And it's in that case which corresponds to X prime the truncation kills only the last term and these two guys survive, okay. Any questions on this categorification business? So if not, I can state our results, our constructions and results. So unfortunately I need a little, one more piece of notation. So it's something called quantum carton matrices. So we are in simply lay style. So you have the carton matrix of G. So two on the diagonal and a bunch of minus one outside of the diagonal. So you take this, the very same matrix but instead of two on the diagonal you replace your twos by quantum two. So Z plus Z inverse where Z is a formal variable. And this matrix is invertible. So you look at its inverse. You get a beautiful matrix. And the entry of this matrix are, so our functions of Z of this formal variable that you can expand into formal series and the integer. And this is what we want to extract as information. So this is a family of integers. So the coefficients of power series expansions of these functions obtained in this way inverting the quantum carton matrix. So these integers have nice properties. You have some periodicity property. You have nice recursion relations. So very nice, nice properties. They are very nice, very nice. So what we did with my collaborators, Jean-Rongly, with our construction goes as follows. So we construct a map from this torus. So this torus contains all the, it's really this torus. So it contains all the truncated Q characters of whatever objects in CQ, whatever. So this is the torus, the torus containing truncated Q characters. And we're going to map it into our field of rational functions. So for each generator of this torus, we associate a certain rational function, which involves here these coefficients of the inverse quantum carton matrix RG. And this is, so it's really a positive root or a negative root, it's a root. Viewed as a linear function on, yeah, like. Viewed as an element of T star. So the dual of the carton. So this root is specified, is prescribed by some classical Outlander Heiden theorem. So I will not go into details, but there is some nice combinatorics. So if you like representations of finite algebra, finite dimensions of algebras, this is a big deal. So you have a whole theory that tells you how to associate in a natural way, a root to each of these tuples YP or JS. Okay, so I'm not going into details. The only thing that matters is that to each generator of this torus, we can have an explicit rational function defined in this way. Okay, so now this is an isomorphism. Okay. And our result goes as follows. So here we recall we have this categorization isomorphism due to Hernandez and Leclerc. So our coordinate ring CN is isomorphic to the Grotten degree of this little subcategory CQ. And the truncated Q-characteromorphism allows you to associate the Laurent polynomial to each of the guys here. Okay, so you have this compose map. And if you plug in, so if you take a rational function in CN, you look at the Laurent polynomial given to you by this affine quantum groups representation categorification business, and you plug in this Laurent polynomial into our map D tilde, into this map. Then what you get is exactly D bar, the same rational function, like it's the same map. So it's really the map defined by Bohr-Math-Cameter. And so using this, we can prove, so the conjecture I mentioned that for these specific cluster variables called flag miners, the D bar is one over blah, blah, blah, it becomes, yeah, it becomes clear when you go through this. Okay, okay, I still have a little bit of time. So this is, yeah, what we published, this was published last year. And I can mention what we are doing at the moment. So ongoing project. So here we are using Hernández Leclerc truncated version of the Q-character morphism. But a natural question is, after all, after all, we can decide that we don't want, we can, yeah, it's a very natural question to ask. What happens if we don't truncate? Okay, so this upper side of the picture is the, yeah, the previous slide. But now we can also say, well, I don't want to truncate. So I really take Frankel-Rechetikin Q-character. So I get more complicated, Laurent-Palinamiel's now. But I can still plug them in into detailed R because for some reason detailed R lives on the whole torus. Okay, so I don't need to truncate to apply this. I can plug in whatever I want to this. So what happens? If instead of plugging in truncated Q-characters, I plug in the whole Q-character. And the answer is you get zero. So this is what we are proving at the moment. So if you take a fundamental representation, you take its whole Q-character, the detail of this is zero. And if you take an arbitrary object, it's gonna be a constant, maybe not zero, but at least a constant. But it doesn't really matter. It's fundamental reps are sufficient because they generate our category. So in other words, a purely, if you want technical way to view this, would be that from this representation theory of affine quantum groups, you can get nice complicated Laurent-Palinamiel's. And each term which will give you a rational function when you plug it into the detail down. So this tells you that you have a very complicated sum of rational functions that somehow magically simplify and you end up with zero. Okay, so this is, so I still have, I think I have a couple of minutes left. Yes, do I have a, how much? Yeah, that's good. Yeah, okay, good, fantastic. So now we can maybe go beyond this purely technical approach and ask what kind of interpretation, how can we understand this statement? And so what I have in mind is some geometric interpretation. Yeah, so classical, so classical facts that I didn't mention earlier when I talked of, I mean, I may be slightly mentioned, when I talked of equivalent homology. So again, very classical facts going back to Attia-Bott, Berlin Bern. So if you have, yeah, under a mild technical assumption you have this famous integration formula. So if you take an homology class, oh, there is a T missing here, I'm sorry. So if you take a class in the equivalent homology space of X, you integrate it over X, then this is the same as the sum over the fixed points of, so here it's the pullback map to this point and this is the inverse of the Euler class in equivalent Euler class at P. And if X is compact, then the sum is gonna be zero. Yeah, X is compact of dimensions, strictly positive. Then the integral of the fundamental class is zero. And this is the interpretation I have in mind. So it's somehow speculative at the moment to be very honest, but this is, yeah. So the interpretation that we have in mind is the following. So for each representation in CQ, there should be an algebraic variety, a compact algebraic variety XM endowed with natural action of the maximal tourism G. Such that the fixed points of XM, the T fixed points of XM should correspond somehow to the terms of the Q character of M, so to the weight subspaces basis of M. And what happens, what's going on when we plug in every term into D tilde, every term of the Q character into D tilde is somehow must correspond to the inverse equivalent or the class at P. And then because this variety is compact, then when you plug in the whole Q character, well, what you're doing is that you are summing over all fixed points. And because of Attia Bot integration formula, you are, that means you are integrating your class over X. And yeah, this class should somehow be constant, or yeah, and so you get zero. So this is the interpretation we have in mind. So I said it's speculative, at least on my running example, it's not speculative. So this is nice. So let me present it and then I'll be done. So on my running example, yeah, so this is my, this was my X and my X prime and my Y and so this is the guys we already have. They correspond to, yeah, so I should have called, yeah, sorry about that. This should be X and this should be X prime and this should be Y. And the corresponding MV cycles in that case are, yeah, they are fairly simple. So X is a P one, X prime is a P one, Y is a P two. And the tangent weights, this is a P one with tangent weights alpha one. This is tangent weight alpha two. Of course, they're smooth, obviously. And this has tangent weights alpha one and alpha one. So when I say tangent weights is at the fixed point that Beaumont-Canisse and Knudsen need to consider. And so yeah, this is typical example. So if I look at this guy, this middle line here, X prime. So as I'm, so the whole Q character of X prime is this. So this is what I get if I plug in detail that to each term. So as you can check, it's a, this is zero. Yes. And this is exactly what you get if you consider a P two with these weights and you get, you consider the class alpha one, alpha one, alpha one. So constant class alpha one everywhere. And you can check that this is the, this is this fixed points alpha one and alpha two, alpha one, alpha two. This is this fixed point and this is this fixed point. So this is really the right-hand side of Atiyah-Bott integration formula on a P two with this. So you get zero. And now our result of Tian Hongli with that we obtained with Tian Hongli that if you plug in the truncated Q character, you obtain the same as Beaumont-Canisse and Knudsen's map becomes clear on this example because the truncated Q character kills this guy, which corresponds to this guy. So the taking the truncated Q character means that you consider only these two fixed points. And the MV cycle is here, Beaumont-Canisse and Knudsen fixed point is here. And then it's a classical fact of equilibrium homology that if you have a map from X to Y, then the equivalent multiplicity here is the sum of the equivalent multiplicities on the fiber. So this would be the geometric interpretation of all this story. And there might be more to say, but I really need to stop now. So thank you very much. Thank you very much. Thank you very beautiful talk.