 OK, thank you very much for the invitation. So let me explain the kind of general structure of the talk. So the main purpose of the talk will be to describe certain conjecture sort of within local geometric langans. But the bulk and some sense maybe the biggest part of the talk will be reviewing some local geometric langans is. So I'll send and to discuss some applications maybe of this conjecture or corridors of this conjecture, applications and evidence. And I should say from the very beginning this conjecture, the conjecture, the way I'm going to describe it is essentially maybe slightly differently formulate. So it's essentially due to David Gallotto to Gallotto. So my kind of contribution to this is just being able to translate what he says into a slightly more mathematical language. So secretly, the talk is about computing or realizing explicitly as duality for certain specific boundary conditions or interfaces, boundary conditions in, let's say, 4D n equal 4 super young meals, topologically twisted. I should say, topologically twisted. But I'm going to only say these words once here. And I'm not going to mention them in the rest of the talk for two reasons. One is kind of lack of time. And the second is that definitely in this room, I'm not the right person to say these words in the first room. So in particular, I should say that the main conjecture, when we get to it, so I mean, David was able to produce it from sort of looking at this setup. And the way he was able to produce it is completely mysterious to me. So it requires some strength. I mean, the conjecture eventually will be completely mathematical. But it will be pretty surprising for experts. And he was able to use somehow some kind of string theory arguments to produce this conjecture. And I really sort of don't understand that at all. OK, now the actual plan of the talk is as follows. So we'll part 0, which will be extremely brief recollection of, so to say, usual local correspondence. Then we'll part 1, which will be sort of brief, but not extremely brief recollection, local, geometrical English correspondence. And so here, this will come into two disguises. So they'll be classical, both classical and quantum. And somehow sort of the emphasis will actually be on quantum. And then there will be main conjecture in its corollaries. And then some theorems. That's what I want to try to do. OK, now let me do part 0. So well, let's assume that we're given some field k, which is a local non-alchymidian field, for example, biotic numbers all around power series with k-efficient of finite field. And then usual local language correspondence deals with the following things. We'll also need some group G, which is a reductive group. And for the purposes of this talking, you can think about the group G ln, because eventually the main conjecture will be about G ln. And then the usual local angles deals with irreducible representations of the group G of k, well, in the appropriate categories. Some kind of smooth representation. And claims that this is closely related. Not quite, it's not quite a bijection, but somehow, well, let me make a dotted line here. Some kind of almost a bijection between these guys. And homomorphisms from the Galois group of the algebraic closure, actually separable closure of k, if it's positive characteristic, of k into G check, which is the language dual group. And again, for the purposes of this talk, you can think about G ln. And then the dual group is also G ln up to conjecture. And it is well known that there's a lot of problems with the exact formulation of this. So somehow, first of all, I have to be careful about what you mean by this. And also, even if you formulate this in some way, then somehow it's not quite a bijection. It's slightly bad, actually, for G ln or for general group, but still somehow, so we have to, I mean, there's some problems with this. And also, it's kind of not a very natural statement from this point of view. Because somehow here, you only talk about the reduced representations. And there's no way in this framework to formulate it for just all of us. For example, if you want to study the category of all representations, then somehow it's probably impossible to realize it on this side. And this problem is actually cured in geometric langlines. So now, local geometric langlines is about the falling. So here, I'm going to say what it does. And then it will involve some notions which I'll have to, well, if not defined, but then at least discuss what they're about. So main thing is that, first of all, we replace k by, say, Laurent-Pauwes series with coefficient complex numbers, so any algebraically closed field of characteristic zero. And then we look at the group G of k. And this is some infinite dimensional in-group scheme. And then, so the analog of this will be, so, well, here, I'm going to deal with an analog of all representations, not just necessarily of useful representations. And everything, the categorical level of everything will be one. So we should move one level up in the categorification of everything. So somehow here, we should look at categories. And then here, I'll also discuss in a moment what they mean by categories for the purposes of the store categories with, OK, let me write it strong, and also explain what strong means, G of k action. And this should be also closely related. Well, this is actually a two category. And these two categories should be more or less the same. And, well, there are some problems. Sorry, I want to say that what I'm doing now is kind of about classical local geometric langos. And then there was a quantum, and the quantum somehow some subtleties, which appear here will actually disappear. But the roughly structure is that this is more or less the same. And there's actually no correction on how to make it really precise. As categories, as the two category of categories leaving over, well, the stack of local systems. So I'm going to define all the players in the moment. So here, d star is a spec of k. This is the formal puncture disk. So let me talk about what all these words mean. So first of all, let me look at the left-hand side. So let me explain the meaning of this. So let me do it first, not for the group G of k, but just for, suppose that, let's say, a is some algebraic. Could you mention what's replaced left-hand side? If on the right-hand side you replace disk by cure, then what happens? No, that's a different story. I mean, you don't do that. So I don't think there is any sensible statement. So he's looking at surface operators. He's looking at surface operators. You're looking at surface operators. Absolutely, yes. So if I have algebraic group, well, there's a notion of, what, if you just have any group, there's a notion of group action in a category, which is pretty clear what it means. Now if you have an algebraic group, there's an action of algebraic group action in categories. So algebraic action, and so by a category, so let me just remark is that, well, I'm going to ignore this kind of problem, but basically category usually means some kind of derived category of something. So category, well, can mean, well, actually, for some application, it will be enough to work with the beginning categories, better to work with some kind of triangulated category, or if you want to be really careful in category, so you have to actually work with DG category. But let me, this is something I'm going to ignore completely in the stock, otherwise somehow I won't be able to get anywhere. And actually the main conjecture when I get it to, somehow, in the main conjecture, you can ignore this kind of category. The subtleties will even hold for some, because point AB line categories, but in principle should be thinking about categories, some kind of derived category of something. And so if you have an algebraic group, so there's an algebraic notion of algebraic group action in the category, which you can easily invent what it means. And so algebraic, so in fact, and one way to think about is that algebraic action of A on a category C, it's equivalent, this data is equivalent to the action of the monoidal category of quasi-K and sheaves on A, which is a monoidal aspect of convolution, so the monoidal structure has to do with the group law on C. And so in fact, DG category, if you want to work with co-complete DG categories, so DG categories which have arbitrary direct limits. And so, and well, this is what is actually called, I'm going to call weak action. And a strong action is the following. So intuitively strong action means it's a weak action such that it is infinitesimally trivialized. And one way to think about it is that strong action on C is the same as action on C. Of the monoidal category of D modules on C. Is it the same if we make kind of odd cotangent but we'll make DG group? Probably, yeah, probably. No, actually I'm not sure. Okay, I can't figure out on the spot, but somehow it means that, I mean, intuitively we should think about this one way. So what does mean in action? In action means that if you start with an object, you can make a family of objects over the group, which forms something like quasi-cagarenship. And a strong action means that this family will have a flat connection. That's, but it will be convenient for me to think about it in this way because, and maybe for future purposes, let me give you immediately some kind of generalization as construction. So suppose that you have now let A tilde be a central extension of A by, let's say, C star. So it fits into a short exact sequence, one C star A tilde A. And then for any complex number, there's a notion of strong A action or A tilde action depends on how we just noted. Of on, let's say, on level C. And this means that, so it's easy to define in this language. So somehow you can consider D modules on A tilde now, but now you can consider D modules with an index C. So it means C twisted D modules. So it means that D modules, which sort of have monodromy, you're given by this number C along this. So it's D models with on the space of A tilde, but which kind of monodromic with respect to a C star with monodromy given by this number. And that's again in monodrome category. And that means that this guy acts on C. And this notion sort of kind of doesn't change if you shift C by an integer. So this two category of categories with a strong action of level C is the same if you shift C by one. I have the same differential operations and C spirals of this line bump. It's, well, yeah, oh, you mean, what do you mean by that? Well, yeah, it's the same thing, but it's the same as to say D models on the total space, which are kind of monodromic with respect to C star with monodromic. But, yeah, it's the same as, it's the same as models over differential operators in the C power of this line bump. So there is such a notion. So it sort of depends only on image of C and C nodes. Okay, so that's, well, we don't even need, well, we'll need this guy in a moment when we discuss quantum local angles. But at least somehow now I can have told you what left-hand side is. So we have to apply to A being this group G of K, which is an infinite dimensional group in scheme. But it's fine system, it can be rigorously defined. Now about the right-hand side, so what I mean by this, so this local system, so by this kind of abusing of language, so local systems means the local system of the round sense. So this is the stack of, this is the stack of the round local system. The round local system, so it's a stack classifying, let me say, G check bundles with the connection, which is automatically flat. So it's not really an algebraic stack, but again, it's fine system. Now what do you mean by categories leaving over this? So let me just say that if S, let's say, as a stack, then there's a notion of category leaving over S. So kind of the most simple-minded example is if S happens to be an affine scheme, so if S is specter of some ring R, then life over, life of some category C over S is the same as sort of action of the ring R on C, so it means that it's the same as by definition as a map from R to the center of this category. This is the same as action of classical chemistry? Well, yeah, exactly, this was my next sentence, so that, but so, and that's actually definition in the general case. So in general, life over S is by definition same as action of the category of quasi-korean sheaves. And again, somehow, since we actually want to live in the world of DG categories, this quasi-korean sheaves should also be in the student sort of derived sense in a DG sense, but let me ignore all these problems. And so now, this is a monolithic category just with respect to ordinary tensor product. And so now, so this is what local language studies, so somehow this is almost true as stated, so in fact, there's some, I mean, but it's not completely true, it's stated, so there's actually, so there's a way to slightly change the right-hand side due to our ink and which actually, at least, can actually make something to actual equivalence of two categories, so it can actually formulate a precise mathematical statement. Now, I want to also discuss quantum, quantum local geometric languages, and so you know, is it visible if I write here or I should not write here? Because I want to keep this stuff on the board for some time. I'm sorry? So quantum, local geometric languages is the following, so this has to do with the fact that the group G of K comes with a natural central extension, so here you're gonna have to be a little bit careful about what the central extension is, prime first by, so in fact, if G is simple, so if G is simple, then the central extension, then there's kind of one parametric family of central extension, basically there's one canonical central extension, if G is not simple, if G is general reductive, then in fact, you have more parameters, but I'll kind of notationally pretend that there's just one parameter, so in fact, central extensions are primetized by integral even in varying forms on the Lie algebra, so maybe notationally I'll pretend that the group G is simple and then we have sort of G hat, which is a central extension of G of K, so again, there's not much choice here of G is simple, if G is not simple, then you have to be a little bit more careful, but let me sort of again not deal with this right now, and then quantum, local geometric languages says something which is kind of more symmetric with respect to G and G check, yeah, and sorry, before I say this, let me just introduce some notation here, so therefore we can talk about G of K actions on a category on level C, which is a complex number, which is just a special case of what I talked about for over there, and in fact, notation it will be convenient, so it's convenient, so we shift, convenient to shift C by H check, which is dual coaxial number, so we shift them by some integer, so which in fact doesn't do much, because of the marker over there, it doesn't change much, but somehow, well, you'll see that anyhow, so I want to change the just, when I say that level C, it means that the action is not by that number, by that number I shifted by the dual coaxial number, and you will see why it's important in a moment, well, although this is, well, approximate this is not very, very important, but just to be precise, let me do that, so then the statement is that the injector says that G of K actions, so let's say categories with G of K action of level C are really the same as categories with G check of K actions of level minus one over C, so, well, it's kind of an exercise to convince yourself why what's written over there is a special case of what's written here, namely, so if you want to take C to be equal to zero, then you get, I mean here, to get the left-hand side we had originally, it just should put C equal to zero, and then minus one over C becomes infinity, so what you have to convince yourself is that when you consider this notion of categorical action at level equal to infinity, then it's natural to think about it as just being a category living over the stack of local systems, that's, I'm not going to do that, I mean, it's kind of not a mathematical statement, it's a psychological statement in a sense, that somehow sort of you said, you can convince yourself that it's natural to think about this as a limit of this as the level goes to infinity. Okay, this is all kind of well-known stuff that I'm talking about, at least to some people, but so the main purpose of this talk will be to provide some examples of it. Yeah, and maybe let me just make a remark that if you define as things carefully, sort of, if you define as categories, DG, co-complete, DG categories with this section and so on, then at least if C is not a rational number, then it is expected, a statement that D is true as stated. Otherwise, you need corrections, and particularly in this kind of zero infinity case, you need, now maybe say that I mentioned this kind of boundary conditions and of course, I promise not to mention that anymore, let me mention one more time and that will really be the last time, that somehow if you can see that this four-dimensional and equal four-supremian meals, then it's known that it has a P1 worth family of topological twists, which are called Kapustin-Witton twists and if we can say the boundary conditions in this Kapustin-Witton twists, then boundary condition corresponding to, well, the twist given by number C is should be more or less the same as this category with GFK action of levels C. That's an expectation. In other words, one other way to say this is that if you have a four-dimensional topological field theory, then to a circle, it's supposed to associate two categories and the two categories associated to a circle should be this one. I have a question. Is this supposed to be kind of algebraic and C or transcendental, so in this complex numbers? No, algebraic, I mean, it's, I mean, you can replace complex numbers by any field of characteristic zero here. Is it known that the two monoidal categories are not equivalent? Does it just mean? Which two monoidal categories? Oh, you mean the corresponding category of D models? Well, it's probably not known that they're not, well, I don't know, but I don't think it's known they're not equivalent, but it's not known that they're equivalent, let me say it like this. Okay, now let me first do a couple of examples of how this works or how this is expected to work. And, well, this is kind of too abstract, but let me, I want to work out one example from which I can actually get some kind of equivalences of categories, which you can then test. And then I want to, so I want to do some, first of all, some kind of known examples and then I want to explain some examples which come from this Galliot's argument, which I kind of already completely knew. So, okay, so examples of how this works. Well, let's even forget about langons for a moment, let's talk about how in principle can you produce categories with action of A or maybe with action of A of sub-level. So, let me write here. So, how to produce categories with a strong A action or AC action, whereby AC action, I mean action of A on level C, just for general A. Well, there are gonna be two sources of examples and both will appear. So, source number one is the following. Suppose they have some X, which is some space and some variety or scheme or stack. So, well, let's say scheme with A action. Then we can consider the category of G modules on X and so this has an action of A in a natural way. That's one example. And if you want A sub C, then you need some additional structure in X. So, if X has a line bundle, which is kind of equivalent with respect to the corresponding central extension. So, if X is equipped with a line bundle, just a tilde equivalent, then we can consider DC modules on X. So, somehow, because we're in twisted G modules on X and here we have sort of, let me just, abusing notation, right? We have an action of A sub C. That's one source of, that's kind of a slightly sophisticated source of examples. Less sophisticated source of examples is the following. So, we can just consider, let's consider little a to be the lead algebra of A. And then we can just consider the category of modules over this lead algebra. So, then we can consider A modules has the category of A modules has a natural strong A action. And again, if you say have a central extension, let me consider the category of AC modules, which means that it's action, actions over corresponding central extension on which the central element X by C has a natural strong AC action. So, that's one source of examples. Well, it's two sources of examples. So, let me then do some specific, well, coming back to Langlands, let me give you a few examples of who goes to what or who goes to who. So, example one is the following. Example one will be still slightly abstract, but it will be important in order to, you will actually, this example will be important and to get some kind of concrete statements out of Langlands correspondence. So, example one, well, this is what I'm going to say is not a theorem, it's an expectation because this Langlands does not have a construction, it's a conjecture, it should exist some kind of correspondence like this. So, somehow we can only guess what kind of properties it should have. So, one thing is the following. Let's consider the fine-grass model of G. This is G of K mod G of O, where O is the field of Taylor series. Then it does have a natural line bundle, which is a covariant with respect to the exponential extension. So, we can consider the category of DC modules on the fine-grass model. And that guy is expected. So, let me denote this correspondence as by C going to C check. So, if I put check here, it should be actually a category of similar nature. So, I should write minus one over C here, modules on the fine-grass model of G check. I'll explain some importance of this result in a moment. So, again, you're welcome to do an exercise just to think what this means when C is equal to zero and this guy is equal to infinity. Then we get some kind of interesting statement. But, I'll ignore this. So, let me keep C in minus one over C finite. Then let me explain some interesting corollary of this. Corollary is the following that it says that if you consider the C, well, it's a category of G of K action. And inside G of K, we have the group G of O. And over G of O, the sample extension always splits. So, in this context, it makes sense to consider G of O invariant. So, this is the category of Geo-4 equivalent objects. At least when working in this DG setting, somehow this is all completely well-defined. And the corollary is that that guy should be equivalent to this guy. And so, why is it so? Well, if you believe in this thing as equivalence of two categories, then it should, well, it should match objects of this two categories, but it should also match morphisms. And so, one way to think about this, this is home over G of K from exactly from this guy to C. I should maybe put G of K, C here. And so, if you believe that this thing is an equivalence of two categories, then this statement immediately implies this. Although C and C check might be completely different categories, but if you take their Geo-4 equivalent objects, you should get the same thing. So therefore, any time we have a statement that one category is London's dual to some other category, we get automatically also some statement about two categories being equivalent, because if we take the Geo-4 equivalent objects, by the way, we can also replace Geo-4 by a Wahori group here, but I won't do it in the store. So for a Wahori, the same is expected, was also expected, but if we consider some smaller subgroups of Geo-4, then it will not be true. Only for Geo-4 and for Wahori, this is true. So in particular, it's actually nice to, you can actually do it for these guys themselves, then we get twisted Geo-4 equivalent objects, which are also Geo-4 equivalent on the left, both here and here. And we get actually some non-trivial statement, which is known, but with really non-trivial, in this case. Yes. I think something, the general discussion you said that C was different to an integer, the London's case, it's not- Well, it's actually, yeah, it's a good question, but see, you can actually ask this question here. So here, if you change C by an integer, you get, well, get just on the nose the same thing, but the effect of changing C by an integer on the number minus one over C is completely different. So therefore, we get some kind of, and this is absolutely non-trivial, so somehow it's very difficult to see it. So here, you get some, so in particular, this statement implies that, implies a lot of equivalences of these two categories, one sort of, you don't change the group, and they're absolutely non-trivial. I mean, knowing them is known, this is more or less the same, so it's, I mean, but you're right, you get some really non-trivial sort of equivalences just on one side of London's correspondence, and it's very difficult to see that, actually, whole. Question, if you want to have a quiz, you probably need some kind of bi-module, or two categories. Absolutely, yes, but, and the bi-module is just the following, it's like, I won't try it because I won't use it. You just take G-modules on G of K itself, and this has two G of K actions on the left and on the right, and then they just apply the slang ones to one of them. I mean, so it's, I mean, yeah, no, no, okay, sure, but, right, sorry. Do these two categories have any extra structure, like representations of a group can be tensored or something? Is there like a symmetric middle structure on them? Well, they do have, you know, you can't answer them over G of K for, well, no, you can, well, there's no issue of tensoring two categories of G of K, but I mean, I don't think this equivalent, I mean, I don't know any structures which should be respected by this equivalence, let me say it like that. Okay, let me proceed because otherwise I won't get, I mean, I have only 20 minutes left, and there's one example. So now I want to do one example of this sort, and the main thing that I want to do eventually will be generalizing that example. So now it's natural to take C to be this G hat, well, G hat C modules. This has an action of G of K of level C, and you can ask what is its dual, and there's an answer. So, well, expected the answer. And again, so this expected answer can be tested because I said anytime you have a, anytime you say one category is dual to another, you also get some equivalence of categories like this. So, expected answer is the following. So I consider, so the only thing I always get confused about, yeah, so I can see that the group G check of K, and then I consider D modules, well, let me write it, and then I'll DC modules on that, sorry, the D one over C modules, and then I consider Wittaker of that. So what is Wittaker? Wittaker means the following. So if we inside this G check, we have the corresponding, which have some Borel subgroup, and some Borel subgroup, we have the corresponding important radical, and then we consider, then we, so Witt in general means considering N check of K comma Chi equivalent objects, where Chi is some, well, non-degenerate character of N check of K. So I won't go into what non-degenerate character means, but in some kind of generic character. So Chi is actually, well, it's a character in the sense that it's a homomorphism to the editing group. So, well, the central extension here, well, here what I mean by this is that I do this on the left, and so if you can adjust G modules on G check of K, it has an action on the left, it has G check action of level one over C, and it's also known that on the right, it has an action of G check of K of level minus one over C. So somehow the levels on the left and on the right are negative of each other. So here when I do this stuff on the left, when I apply this, this kind of equivalence on the left, then on the right, I still have an action of G check of K, which will be of level minus one over C, so at least the sense will make sense. And this is expected answer. And as a corollary, we get the following statement, which is actually theorem of Gaetz, Gore and Luger, that, so let me actually introduce some notation before, introduce notation here. So if one considers, so I want to consider the category of GC modules, and then I want to consider here geofold covariant objects, and in other words, it's modules over this, sorry, a fine algebra, which are integrable with respect to geofold, which this group acts, and this has a, this is sometimes called the Kashten-Lüste category of G check level C because it was started by Kashten-Lüste, and Kashten-Lüste actually, well, they introduce some tens of structure here, and actually it's a tens of category. So let me assume that for simplicity G is simple, and let's say simply laced, otherwise I have to be a little bit careful about how I formulate it, then Kashten-Lüste constructed an equivalence of this, well, for almost all C, so let me just say that it is equivalent to the category of finite dimensional representations of the corresponding quantum group, finite dimensional quantum group, not a fine work, U is e to the power two pi i over C, and here we have to exclude some C, so maybe we should write that C is not a positive or non-negative rational number for any C, which is not like this, this is going to work, and so the core I want to format, I want to take here this geofoic variant object, so here by definition I get the Kashten-Lüste category, sorry, this is modules. So I get this Kashten-Lüste, so this is called Kashten-Lüste equivalence, so if I take the Kashten-Lüste category of level C, then this thing should be the same as, well, and the equivalence is also, well, I mean, for this to work nicely, I have to make some assumptions on C, like best assumptions to assume this, but I think this equivalence I'm going to say right now is to work always, if you will format it precisely, so here I should write, sorry, I should write wheat, and so here I should write Dc, sorry, D minus D1 over C modules, and I should take geofoic variant objects here, but that's the same as to just consider G-monus on this affine-grasmina, which is geofk-mon-geofoic, so if you can see the Whitaker D-model on affine-grasmina, that's the same as Kashten-Lüste category for the dual group as if C is, if C and one over C, if the levels are matched in this way. So that's actually a theorem in this case. I guess, well, at least it's written maybe if C is not rational, for rational somehow there's some, it's more delicate, but I think it's proved that this one C is not rational. With C is rational, sorry, I mean it's proved I think in full generality, but written maybe only for irrational C. Okay, so, and again, thesis I have some, I think have some ways to explain why I should expect, why this is actually the right expected answer. And so in the remaining 12 minutes, I want to generalize this. So what I want is to slightly actually generalize this. And for this I need to slightly generalize actually this part of the board. So I mean I said that somehow here, one way to construct categories with this action of my group is just to consider the modules over the Lie algebra. Now I want to do something a little bit more general than that and again from, turns out that from the 4D Young Mills point of view, it's also natural to do that. Let me do the following. Let me assume, so assume that we're given a Lie super algebra. Let me call it, maybe it's a bad notation, but it wasn't come up with anything, with any other notation G prime. So G prime is G prime zero plus G prime one and such that G prime zero is my G. Then I can try to play the same game. I can try to consider this morning loop algebra. So I can consider G prime and I can consider it's central extension if it's sufficiently nice super algebra. And well, let me also put number C here. Well, you have to be a bit, again a little bit careful by, but by what the central extensions are parameterized. But again, let me ignore this, but you can consider just modules over this super algebra. And this category has an action of, well, let me write it G heads of sort of group G of K of level C. And so idea that if G prime is some nice super algebra, for example, if it's a simple super algebra in some sense, then this logless dual guy should be computable. Let me say it should be computable. There should be some explicit answer about what this thing is. And if one reads carefully the paper of Gaiot and Whitton from 10 years ago, then apparently there's actually an answer there for all, for at least for all classical super algebras. There's probably an answer, but today I'm going to concentrate on the simplest possible example. The simple possible example is the following and it's already pretty interesting. And again, for people working geometric long on some how it was completely unexpected that you could do what I'm going to say in a moment. So I want to take this G prime to be GLN slash M. So now what it means is that my group G is GLM cross GLM. It's also the same as G check in this case. And I want to compute, so I want to formulate an answer about what the longlist dual is. And this will imply some equivalence of categories as I explained by, if I take the corresponding geofoic variable. Before you have A, the algebra was simply dimensional plus Casmury. No, I mean, I said that, so here I started, suppose I have, I'm saying that here suppose that my fine dimensional algebra has some super extension, then I can consider the finization of that. So in particular, there will be a statement that the Casdan-Lousty category for the least superalgebra GLM M will have some kind of realization. So the question, so now, so I take this GLM M hat C modules, and I ask what is its langons dual when I dualize it with respect to both GLM and GLM. This is by the way also what Jesus would call interface between sort of young mills for GLM and young mills for GLM. I mean, that's an example of an interface. Okay, let me produce an answer. Now, and this answer is supposed to generalize this one. So this is the case, well, if G is GLM, then in particular, this is a special case corresponding to say M equal to zero. So, well, first of all, I'm going to assume that N is greater or equal than M. I mean, everything's symmetric with respect to N and M, so I can do that. And so somehow the easiest answer will be in the case when M is equal to N minus one for some reason. So, okay, so the structure and answer will be the following. It will be one type of answer when M is strictly less than N and slightly different answer when they're equal. So let me first do the case M less than N and it should generalize this, but I first do the case M equal to N minus one and it will not look like this at all. But then I will do the general M less than N case and it will be clear how it includes this one. So, okay, so, for example, M equal to N minus one. And let me do this for, so then the answer is the modules, well, of level minus one over C, sorry, minus, yeah, minus one over C of on just GLN, but consider as what? So somehow this has GLM of K, GLNK action on GLN of K. On the right, and on the left it also has GLN of K action. Well, I'm ignoring the levels now, but I'm kind of ignoring that. And I'm just, so I can naturally embed GLN minus one into GLN, let's choose this embedding in a natural way. And then on the left I can see there GLN minus one K action on the left. So on the left there's also full GLNK action, but I just ignore that and I only can see the action of that guy. So the claim is that then this two guys are supposed to be, well, sorry, I don't need minus here, sorry, that's just one over C. Because the minus will come because I can see the GLNK on the right. So that this is the langos duals in particular quarrel array, well, well, this is not a mathematical conjecture because we don't know what, we don't know a construction of langos correspondence, but the conjecture becomes completely mathematical. Is that the cash done loosing category of level C for GLM slash M is equivalent hat is equivalent to, well, I should take here objects, which are equivalent to respect to GLN of all on the right and GLN minus one of all on the left. So somehow one way to think about this is that, so let me write D minus one over C modules. And here I should take the affine Grassmannian of GLN. So this is thinking quotient by GLN of all on the right. And I should on the left also quotient by GLN minus one of all. So in other words, I can see the GLN minus one. So usually when people do geometric satake, you would consider just GLN of all equivalent objects here, but here you can see the equivalent aspect of smaller group and the claim is that, sorry, this is M equal to N minus one. This is N minus one. So this should be in equals like this. Okay, so I have three minutes left. So I probably won't be able to discuss any theorems, but let me still finish the formulation. So I should say that, again, the formulation of this is essentially due to Gallota. In fact, well, he didn't, of course, formatted this in this particular way, but somehow basically 99.9% of the formulation is due to him. So my only contribution is being able just to translate it into some kind of mathematical language. Okay, now let me do the general case. So let me do the general M less than M example. So M, sorry, less than M, general. Then the statement is that what you should do is the following. Then if you consider this GLM slash M hat C modules, and if you consider it's Langlund's dual with respect to GLM times GLM, that thing should be equivalent to the following. So now you should do the, you should, if you consider D modules on level one or C one, GLM, okay? And now I should consider it's with a reduction with respect to GLM minus M with respect to any important group in GLM minus M. So the two extremal cases, one is when this number is equal to one, then you have GL one here and the important group is trivial, so you don't do anything and that's what we did before. Another example is when you can see the M equal to zero, then I should consider the full Whitaker here and that's the theorem up there. So, well, no, I mean, that's the thing up there. So, but I mean, the theorem becomes a few. So in particular, GLM all times GLM all equivalent objects, the categories of GLM objects should be equivalent. There's also an interesting case when one of this number is equal to zero and the other is equal to infinity. Actually, when this number is equal to zero and this is equal to infinity, then it should also, this equivalence should also hold and now maybe let me conclude with a few things. So first of all, the equivalence up there, it's known that actually holds in a lot of abelian categories, don't need any derived categories and the same thing should be true here. One other thing is that let me also say what happens when N is equal to M. If N is equal to M, then we have here GLM, N have C modules and if I langons dualize it, then the answer is similar but slightly different. And the answer is that I should consider G1 over C modules on the following thing. I should consider GLM of K times K to the N. And so here this has two actions of GLM of K now. One is just say the left action just on GLM of K itself and the other is the following. So I have the right action on GLM of K and I also have the action of GLM of K and K to the N and I consider the diagonal one. So I can say these two actions and again if I take GLM of O times GLM of O equivalent objects, I get some equivalence of categories. And so maybe if I can have two more minutes, let me make a few remarks about this and formulate. Just tell you a few cases when this is known. So first of all, I should, one thing I want to say is that I never, until recently, kind of psychologically, I never admitted that the importance of, and then we acknowledge the importance of least super algebras, so to say. But that's kind of the first time in my life when something in least super algebras play some kind of a role. And now when I understood the formulation of the conjectures, then I started kind of researching both the literature and asking experts and turned out that this, for example, this casualistic category for least super algebras has never actually been studied. Although there are like a lot of people who start representations of least super algebras, but there's no, for example, analog of casualistic, well, not that there's no analog, there's no known analog of casualistic equivalence for this, I mean, nobody. Although, I mean, there's no obstruction here, just nobody has ever proved this. So I think that it might actually be sort of a good thesis problem to generalize casualistic equivalence to some kind of least super algebra like that. I mean, I think it should be relatively straightforward, but somehow nobody has ever done this. So, some remarks. The corresponding equivalence, so if I can see the g ln of all times g ln of all equivalent object, equivalence should hold for a billion categories. This is not something coming from a general, from any general expectations, but it's just effective. But I mean, if you work with examples, you see that that should be true. Also, in this case, this, again, I'm talking about g ln of all times g ln of all equivalent objects, somehow the corresponding categories, they're not just categories, but they can, what's called factorization categories. So also it should be equivalence of factorization categories. Okay, and the last thing I'm going to say is that there are a few known cases. The first known case, which is relatively easy, but not completely straightforward. So the first case is when n is equal to m is equal to one, so you can set gl11. It's not completely true in this case. So this is known except that I don't know how to prove at the moment that it's equivalence of factorization categories. So I can, the corresponding equivalence of categories I can check, but not with factorization structure. And by the way, I can also check the corresponding characteristic equivalence in this case. So the characteristic category for gl11, it's equivalent to representations of the corresponding quantum. We can define the corresponding quantum group and you can check that they're equivalent. So that's one thing that, the other thing is that when n is equal to two, m is equal to one, and c is equal to infinity, one over c is equal to zero. That's also, well, in this case, by the way, the super algebra degenerates to something much simpler, but still in this case, it's known and it's written in a recent paper of myself with Finkelberg, where it was studied for completely different reasons. Funny enough, it was also kind of an attempt to mathematically confirm certain conjecture of Gallota, but a completely different conjecture of Gallota. I don't see any relation between these two conjectures. So, but, and it's not, I mean, this is actually not very easy. I mean, it's not super difficult, but, and the other thing is that independently before, even these whole conjectures were formulated, Finkelberg, Ginsburg, and Trafken started the case, n is equal to n, and again, c is equal to infinity, one over c is equal to zero, and actually formulated the right conjecture in this way. So, I never actually formulated carefully what the conjecture says then in this limit, but that's easy to do. And so, they formatted the right conjecture, and so recently, just a few weeks ago, Trafken told me that he thinks he can prove this. So, probably, and I think that's the full list of known cases right now, at least known to me. And maybe I will conclude by saying that you indeed can actually, GLNM is just an example. So, just for any classical e-superalgebra, there's an analog of the statement, but that's just, that's just as simple as possible. Okay, thank you. What's the factorization category? Well, it's a form, it means that it's a following structure, it means that it's a category, it means that if you have any algebraic curve and any power of an algebraic curve, then you have a category leaving, sort of which leaves over any power of any algebraic curves, and with some compatibility between these. And so, the original category is the fiber, if you can see the just first part of the curve, then you can see the fiber of that over a point, then this is your original category. So, it's, if things are kind of finite enough, then that thing is the same as a tensor structure in the category. But, but, but. So, it's kind of first or actually, you have to put some kind of curve? Right, right. No, I'm saying that there should become some compatibility means that somehow if you have maps between curves, then somehow everything should be, yeah, sure. So, there is such, no, I mean, when you do this Germanic language stuff, this is an extremely basic notion. And so, for example, this equivalence of Gaetz-Gurion-Lury, it's actually an equivalence of factorization. It's kind of trying to be in equivalence of tensor categories. But, I mean, can they actually are, I mean, cash-analytic, they actually define a tensor structure in this cash-analytic category. But the tensor structure is kind of transcendental because if I were to define this tensor structure, you have to at some point compute monodrome of certain connection. Sort of an algebraic world that doesn't exist. What does exist in an algebraic world is this factorization structure. And so, this, both of these, if you know what factorization structure is, then both of these categories have a factorization structure on the nose. And the claim is that it's an equivalence of factorization categories, which is actually a much stronger statement just in equivalence of categories. So, I mean, it's really kind of more difficult to prove this equivalence of factorization categories than that's an equivalence of categories. So, this equivalence is really very basic for local geometric language. So, Geyser calls fundamental local equivalence. So, he has a precise formulation of what local geometric languages should do, and it's based on that equivalence. So, you mentioned, sorry, maybe I'm just asking a physics question, but so you mentioned that this N and M are the Lanko gauge group. Yes. Between, I mean, two different gauge groups with interface. Yeah, but that's the same, I mean, it's kind of the same as the boundary condition for the product. Interface between two gauge groups is the same as a boundary condition for gauge theory when the gauge group is the product of the gauge group. So, yes, so, and so the point is that if you, just, I mean, where do these super-algebra appear from this point of view? So, somehow, if you want to leave where at the level being infinity, so you ask yourself, when can you construct naturally boundary conditions? So, somehow, the claim is that if you start with the representation of your group, of your gauge group, then you can construct a boundary condition at this topological, for this topologically-twisted theory at infinity. This you can do for any representation. Now, you can ask, what does it take to sort of extend this boundary condition to this P1 family of topological twists? Now, the claim is that the structure you need is the, so, if you start with G and the representation, then you can consider Li-algebra plus V. And so, again, this just gives you boundary condition to infinity, but if you want this boundary condition to leave everywhere, then you need to extend this to a Li-superalgebra structure on this sum. So, any time you have a Li-superalgebra structure on this, it defines the boundary condition for any point of P1. And so, this is just kind of the easiest Li-superalgebra you can consider. And it's only good to n, so it's just the product of the same theory, and then you just consider the boundary condition. No, no, it's the boundary condition for the product of GLN times GLN. So, it's an interface between GLN and GLN. And so, and actually for any physics argument, that's actually the most basic example. So, somehow, well, again, I won't be able to repeat this. So, somehow, this really requires some kind of string theory arguments, but no, I mean, the question is, where do we get this from? I mean, how do you know that the S-dual of this boundary condition is given by this? So, David has this kind of argument with, you know, brain construction of all of that, and so, for him, it's a kind of natural statement, but I don't see any kind of mathematical. I can form it mathematically what he says, but I can form it mathematically the eventual statement, but I cannot form it his arguments mathematically, even approximately.