 So, I'm going to talk about paper we wrote recently, it's 3D partition functions in Kivara Sora, which has been written together with Antoni Delin, who is doing his first post-doc in Milano, and with Fabriz and Yuri, who is doing his post-doc in Upsilon. So please, you know, unfortunately, I have to leave tomorrow to US, but please feel free to ask the many complicated questions and, you know, don't ask me, okay? Good. So, let me try to give you idea first, and then I will try to give more pedagogical talk because the subject is rather technical, so I will only outline the some technicalities, but I will concentrate on the idea, but the thing is the following. So, thank you. The idea is the following, that in principle, typically when we discuss, so we do localization, so for example, we discuss partition function for S3, I don't know, for land spaces, et cetera. So, for simplicity, I concentrate on 3D partition functions, and maybe toward then I will make some comments about 5D. So we calculate these partition functions, and typically, you know, there are functions of very few variables. So there may be functions to be of, I don't know, of rank of gauge group, there may be a level, maybe some kaoplan, phalaeolopoulos term, but it's a finite. So in a way, to a large extent, it's a number, and you know, it's a bit boring number, so it's a bit boring to study. Of course, some people concentrate on this number quite a lot, there is typically some matrix representation, and this guy depends from very few parameters. At the same time, it's not the only quantity we can calculate localization, we can also calculate typically some set of wilson loops in some, you know, representation. So in matrix model it corresponds to certain insertion, so in a while I will make it very precise. But typically in all discussions we talk about either partition function or we talk about wilson loops, et cetera, but what does make sense, it makes sense to put, to take these two objects together and put them into the generating function. So eventually my zeta t now will depends, of course, on some finite number of parameters which is related to the theory, but then typically it will depend on infinity many guys which people typically call times, it can maybe one set of times, it may be two set of times, et cetera. And then the claim is the following, that's what I will try to explain, that these guys in terms of this infinity many constraints, such as for infinity many, many PDEs, very complicated PDEs, but these PDEs, they have very clear geometric algebraic structure and they're related to cuverasaur. So this is idea and this is, this is general pattern in all other dimensions, again in some dimensions people just didn't study, but, you know, for example in 5D necrosse of partition function, it was a result by Pastein from December 15, you can do the same thing and it will satisfy infinity many algebraic PDEs. So in a way the idea is better not to study this or this object separately, put them together and then you write this. So that's what I will explain and the basically idea would be that this object has very clear interpretation in terms of CFT and this is BPS-CFT correspondence. So before I go there I assume that you guys don't know much so I will just outline your idea of what's verasaur constraints and then tell you some basic stuff about cuverasaur constraints, et cetera. So typically all these guys are matrix integrals with different measure. So when I'm giving you integral typically, right, so if I write for you the integral, so I have to calculate the number, okay. This is very cool, you know, you sit down, you calculate it by some means and typically for this integral what people typically write there is word identities, tts, which is basically an insertion of derivative under the integral. I mean it's a true statement but it's a useless statement because you cannot really calculate anything. So typically this is the whole idea here that instead of calculating this integral or calculating many integrals you can basically try to reduce the problem to generate a function. So namely for example I can, again, I'm here very few mathematicians because they ask me a lot about convergence so let's ignore these issues but I can basically park infinity by integrals in the following object, let's say t1, t2, et cetera and I will write something like this minus 1 over 2g x square plus n from 1 to infinity t and xn dx. So in principle if this is a reasonable function and there is not much analytical things so what I can basically translate this purely analytical problem to this, calculating this, then of course the question how do I understand this integral but typically the most natural thing to say that this is just Gaussian weight and then these guys expand formula and, you know, I can do the formula and write everything explicitly. So then there are two ways, I mean for this, this is very nice, for example, it's good for students to do but you can basically do this thing in two ways. So one way you can actually sit down and calculate this integral, for example, it's, it's, you know, the only thing to calculate this you need to know x and power 2p Gaussian what is equal and then you expand equally but you can do something else. So you can trade this guys for, you can write infinity many PDEs. So one of the things which I can try to insert, there it's a fallen operator. So as you can recognize it's almost verasora but the thing is I put this guy up inside. So of course this guy will satisfy the classical verasora. Then you take this guy, you put inside and then you get of course identity, very trivial identity e minus 1 over 2 g x square plus. So if you have questions please feel free to interrupt me, let me put k here. And this is dx equal to 0. Well now what you do, you just simply differentiate with respect to x everything and then whatever extra terms you have, you can generate them by taking tk derivatives and then what you eventually get if you, yeah, are you a mathematician? Then why do you ask? I warn you that, you know, I mean it's, the thing is of course it's convergent, depends how you understand this integral. I'm told you all the way you need to understand this integral. So you treat g as actually positive guys. So this is Gaussian weight and these guys you just expand formally. For example, this is one way of understanding integral. Yes? Yes? No I'm not there. This is, I mean, this is that level. I mean we come to three manifold, nothing to do with this. I'm trying to explain your idea. So typically, you know, the idea is in a way trivial but, you know, as you just make this idea algebraically more and more complicated becomes, you know, much more meaningful. But the idea is very trivial in a way. So instead of calculating integral, so I have integral, I have word identities. This is an empty statement. Now what I'm doing is I'm writing for instead of integral, I'm generating function for all integrals. Let's forget convergence, let's forget all this nasty mathematical questions, let's believe everything works. Okay, and then what I'm telling you that I can just tell you right, infinity many these guys. So for example, if I assume that n more than equal to minus one, then this guy is equal to zero. Now what I'm telling you that I can do this exercise, I can actually differentiate on the integral, I can write I will have only powers here of x, and then eventually what I will get out of this if you want it to be concrete, I will get some differential operators, ln written in t's and derivative of t's, acting on zeta of t equal to zero. You know, if you want that just to make sure that I'm not cheating, I can write for you this operators. Again, this is extremely trivial exercise, but in a way so n plus one factorial t, tn minus g n plus two factorial dTn plus two plus, sorry I forgot here, I just decided for convenience because would he k factorial? Again, it's not of course important, so n plus s factorial, and then s minus one factorial ts s plus n something. And then you can write, let me not write it, you can write the same for l one, there is operator l minus one operator. So what I'm telling you is the following, that this generating function of infinity many variables, just killed by this first order differential operators, and of course next thing you can just check that these differential operators of course satisfy exactly the same algebra. So you can check that these operators satisfy the same algebra. Now what you can do again, the thing is why do you do it? Because as I told you, either you try to solve this PDE, so you can actually calculate integrals and model some analytical details, what you can get, you can just get this thing. So this is generating function for integrals in one variable. So this is called Bell polynomials related to Schur polynomials. Okay, so the whole thing is a following, you take this integral, either you calculate it, in this case you can calculate it, or you can prove, you can try to derive the value of this integral from verisoroconstraints, again, there is obviously verisoroconstraints not feel fixed normalizations, because this is PDEs, I mean, this does not fix the constant value of the integral. But this is idea of verisoroconstraints just in very simple example. Any questions? So the next level of complication, this is our favorite standard Hermitian matrix model. So Hermitian matrix model is the following, again, this is like 25 years old stuff, but so what I can do for you, I can write for the following model, again, I'm introducing, so I will do the integral over UN le algebra. So it's a vector space of, so this is a space of Hermitian matrices. So I'm writing Lebesgue measure here, and then I'm writing here minus sum over S0 to infinity, TS trace of MS. Again, if you're worried about convergence, you can put Gaussian factor, I mean, you can sort of make things, so you can in principle try to look at the algebraic things. So what is obvious is that, I mean, the Gaussian measure over le algebra will be invariant on that joint action. So if I sent my mu to U dagger mu U, and U now is in UN, so this is invariant. So there is very standard Texas side that you can just translate this integral to something like this. So this is again my infinite times, and then I will have here a product over Rn. So by any Hermitian matrix, by unitary matrix can be diagonalized. So I can basically diagonalize this measure, the only thing I have to keep track of things like Jacobi, et cetera. So I will have dn lambda, and then here I will have i minus j lambda i minus lambda j square, and then here I will have S from 0 to infinity TS, and then I replace here i from 1 to N lambda i S. Then again, here what you can do in this simple integral, you can try to repeat exactly the same game, and that's what people did. So it was one of the speaker who refused to come. I did the Russian one, so Morozov and Mironov, I mean, they did this version of the trick, but you can take this object, so you would like to calculate, and one of the things you will insert his operator, ln, which would be minus sum i from 1 to N, d d lambda i lambda i N plus 1. So obviously this is just classical verisor, so I have N variables, I just look at the sum of these guys, so you can easily, well, there's nothing to check. You know that ln, lm, N minus M, ln plus M, so this is classical verisor. Okay, now I'm not going to repeat the thing, but I wanna basically bring you the message, so this guy, this is matrix integral, it has nothing to do with conformal field theory. But then what you do, you do the following exercise, and that's what Mironov is Morozov did in 9091, so in principle you can put this operator here, you can actually differentiate, you will get some terms, you use some simple combinatorial identities, and eventually you will get that ln on zeta, t is equal to zero. Any questions? Okay, and then, so you do this exercise, and then you write this operators. So let me write for you this operators very explicitly. So now this is not first order operators, this is second order, so they will be tk dtk plus second derivative over t0, and then finally this term ln k from zero to N, second derivative of tk dtN minus k, so this is the second term, et cetera. So this comes just from word identities, so this matrix integral. So typically you'll put these guys, it's very obvious why you put here this condition for these powers and more than equal minus one because you don't wanna mess up this convergence, for example. So now, if you stare for these guys, whoever did a free boson and stare for these guys for a while, then you would realize the following thing. You basically will realize that what you have there, it's a free boson representation. So if I look at my Heisenberg algebra, and I can, for Heisenberg algebra, I can pick up some representation and the most obvious representation to pick up. Again, in my discussions, I'm a bit ignoring zero modes, et cetera. So basically what I can do, I can choose these guys to be something like square root of N tN. So introduce infinity many variables. So this is my creator operators. So my N is positive. I'm sorry to confuse you, but this N has nothing to do with that, okay? And then my creator operators, this is guys which would be just proportional to derivative, something like this. So this is the representation of Heisenberg. Now if you pick up these representations up to some numeric, stare at these formulas, what you realize is the following, that this is nothing else. So what you have there, it's the following formula. It's one over two M from minus infinity to infinity of normal ordering of A M minus M A M. So basically this is free boson. Again, at the time it was not that obvious. So this, I mean, this integral, you take a integral, you know, just got integral over the algebra and eventually this integral, the very sort of constraints related to free boson. So if you now try to think a bit more what you have, so this is what I'm telling you, this is pretty standard thing. I'm just, I will do it more and more complicated in a while, okay? So basically one of the way of looking at this integral is the following. So I introduce my free boson. And then basically I would like to construct some guys so I can think about these guys as an element, element in a folk space. For example, I pick up this representation. So for example, you know, I can construct all my states, polynomials, et cetera. So all this, you know, theory of symmetric functions, I mean, can be recast it. I mean, it's related to representation of Heisenberg. So it's in this folks space. And then I basically have to construct your state which would be annihilated by these guys, et cetera. So in a way, this problem already, it's related in somehow to singular vector fields. But the very canonical way of doing it is the following. I basically can introduce for you a screening current which up to the fire of lambda. And the thing is that these guys with these guys behave in very concrete ways. So it's typically ln is s of lambda. This is some derivative of some operator. So it means that if I integrate this guy, so if I do integral of a correct contour so that derivative vanish, then this is zero. So this is typically called screening current. So this is current, this is screening charge. And then what you can think about zeta t. So zeta t for Hermitian matrix model can be thought as a following, that you just take your object, which let me denote, this is let me denote q. So I just take a qn and act on the vacuum. So in this representation, vacuum is just identity. And of course, this is by definition, we'll satisfy Verasorra constraints because Verasorra heat in this, they commute the heat to vacuum, then he later this is zero. And in a way, the matrix model, if you will just try this integral of this, you have to do the normal ordering. And exactly when you will do the normal ordering of these guys. Sorry. Where here? Right. Yes, exactly. I mean, it's a representation, right? So what I'm saying to you, for you, for example, it's obvious that I can, what I'm saying is as a following that I can write for you a minus one, a minus two. This is the same as a polynomial t one, t two, right? That's the sort of statement. It's a representation. It's true in a concrete representation, I choose for Heisenberg, okay? So this is in a way, one of the way of looking at Hermitian matrix model. And the thing is that whatever matrix model we get, you just have to modify the story somehow and make it more exotic. But this is the idea of Verasorra constraints. Any questions? Just final comment is that when you order these guys, so when you do normal ordering, what you will get, you will get exponent of a sum i not equal to j log of lambda i minus lambda j. This is standard calculation in the CFT, which is of course the same as a wonder month. Okay, I mean, right now when I'm presenting it, it's very obvious. 25 years ago it was not obvious. So now what you can do, so our goal is the following that our goal eventually to reproduce some measure. So this is where it's canonical, simple measure. In, for example, in localization, we have more and more complicated measures. So the idea, I will first explain one basic example and then I will spend time explaining the main result of the paper. But one of the things is the following that we'd like here to get a different function, something more complicated. And one of the way of doing this is basically try to deform Verasorra. So you look at different deformations of Verasorra. So let me write this down explicitly. So I will now start to talk about Queverasorra or Sorra. I'm not going to give you any general theory, I'm just, and mainly I concentrated on a free field realization, free realization. But the whole idea, we keep this philosophy from Hermitian matrix model, but we deform Heisenberg, we deform guys, we deform these guys so they become, now not quadratic, they become something complicated, but they still satisfy some algebra and then you get eventually the same thing. So these things still will be true and you will get something annihilated by these operators. So let me talk about Heisenberg. So now I will deform it by some complex numbers, which I will tell you in a moment. Let me be, so PQ and just this is zero mode, I guess it's what, and then so in general Q and T can be complex numbers and there are the following relations. So it depends which numbers you discuss more fundamental. So P is equal Q, T minus one, and quite often you say that T is equal in power Q beta. So either I pick up Q beta or I pick up QT or I pick up, et cetera, QP. This is just standard things. Okay, again the whole idea is now if I try to form my guy like this, so just think if I take this formula and I deform my Heisenberg, I start to calculate something like this. Again there may be some, I mean modifications, but roughly speaking, I automatically will not get this measure. It will be more complicated. Okay, so let me tell you about free-filtralization of Q verasaur and what's how it's related to verasaur. This is just zero modes. It's a momentum Q, so I have to have A zero. So the formula which I wrote here, this is for N and M in integers minus zero. So typically we have to keep a zero mode, so we, I mean for zero modes we have separate guys. This is some technicality if you wish. Right, so now, Hi, Q verasaur. So this I just take, I postulate and then what I can do, I can do to write for you the following. So what is Q verasaur? So this is basically deforming verasaur not within the algebras, but within associative algebras. So this has been done around 95, group of Japanese a lot at all. So let me write the thing and then tell you in what sense it's a verasaur. Oh, deformation of the verasaur. So you can write this algebra. So I fairly some structure constant, so they come from expansion of concrete functions. But the thing is that you can fix this function by associativity, okay? So this is fixed, fixed by associativity. And then if you now take your Q and try to say EH bar and try to expand this in a up to quadratic order, then you can see that sort of only, so if I write for you the formula, TN, it will be two, some constant term plus H bar square beta, ln plus Q beta square over four, delta N zero plus terms of all the H four. So Q beta is just some variation. So what I'm telling you is the following that you can write this algebra sansats, you can fix it by associativity. But another requirement is that you would like in quadratic order in H bar, these guys will be actual verasaur. So this is called verasaur. So you deform verasaur using associative algebras. This is basically a W algebra, okay? So of course you can ask the following question, how to write these guys. So you can write concrete sansats for these things. And so this is L plus of zeta plus L minus of zeta. And this L plus minus zeta is, so I'm just writing to give you idea, zeta N minus N one plus P minus plus N, A N normal order in Q plus minus square root of beta to capital P, P plus minus one half. So this is free field realization. So again, it's a computation which has been done by Avat at all that if you take this Heisenberg, you know, you write these guys, you take these guys, of course you expand them over sum over TN zeta minus N. You can see that TN satisfies this relation. Of course, this relation can be written in many different forms. And I'm not writing for it, but this is a coefficient of expansion of very concrete function. Okay, it's a quverasaur. Then next thing you can mimic in Hermitian matrix model, you can write also the analog of a screening current. So screen current is given by the following formula. So just one comment for you right away that you realize. If I wanna now look at these guys as a differential operator, so all of this Heisenberg is deformed, I still have a canonical representation of Heisenberg. I mean, in a way, it depends on your test. I may say that I'm not deformed in Heisenberg. I can put all these coefficients inside of this A N and then my complications will pop up here. So I mean, it's a folk space. It's the same folk space. I just have an action of totally different object there. But up to these factors, which you can put in definition, I can still write A N for positive N as a derivative in T and this is just a multiplying with negative N as a multiplying T. So now if you look now at these guys and you put here your operator, so you put your representation in derivatives, et cetera, you can see that the operator is horrible, but of course it's well-defined. But in principle, it's exponent of differential operator. So these guys in the representation of Heisenberg is up only quadratic guys. This becomes now a very complicated operator, but it's totally well-defined. There is ordering problem solved, et cetera. So now I'm writing for your screen in charge. So again, it's some expression. Beta Q, W, beta. The whole thing is done because I would like to mimic exactly what I had before. So if you do the calculation, then the thing is that T and this SW is equal to, now it's a difference operator. So I'm not specifying this. This is always a sum operator, but this is almost, it's a Q derivative. So you can calculate this, so we don't care. Important thing that if I take a correct contour, sorry, W, so if I take a correct contour, so what is very obvious is the following, that if I take a correct contour for integration, then of course this is zero. Simply in this integral I can change instead of QW, I multiply here, I change variables in this integral. So this is the same integrals. Again, the thing is that you have to think about the contour, but let's say we take a contour, which is, so it means that if I integrate these guys, so if I integrate these guys of appropriate contour, then I will get that this is equal to zero. So from this point of view, I can repeat exactly the same game I did before. There is some subtlety, zero modes, et cetera. I'm not going to discuss them, but now what I'm telling you that I can take my integrals, roughly speaking, so in power n, put on the vacuum and get something, okay? So I get something and then I'm saying that this something, if I call it zeta, so this something you'll be annihilated automatically by this positive. So let's ask what is this, so you can actually calculate. So this is, I'm going to write for you the thing explicitly. So my zeta of t looks as follows. And that's when three manifolds and gauge theory will enter in a moment. So this is just some complicated, but straightforward calculation. So what I have, I have some contour integral over a region. I from one to n dwi two pi iwi then I have here and from zero to infinity. Okay, let me write this in these notations and then w, so i not equal to j, w i divided by w j q infinity, w i divided by w j t q infinity and then here I will have my t k, k rounded from one to infinity, i from one to n w i k. So how do you get this integral? What you do, you write your screening charges. So this comes from the measure, so this guy is identified is exactly with my screening charges in power n acting on the vacuum. So this comes from normal ordering and this is just my generating function. This is creator operators in my conventions. So you still have them there. So I mean, these guys still act on the vacuum. I mean, this is a generating function and this comes from the normal ordering. So is everything clear? The symbols are all just clear. Do I have to define them? Everybody understand what's written here, right? Let's assume so. Okay, good, but now the thing is that when you stare at these guys, what you realize is the following, that this is in fact a partition function for u n theory for n equal to u n theory plus a joint chiral with mass on the following geometry. So t, so it's a disk, which is you currently rotate. So it's a two disk time circle. And the parameter t, it's roughly exponent of the mass. So this is, so in a way what one can say right away, then if you know, again, it's identification, but the statement is the following. If I would turn things around, if I will put Wilson loops along s one circle and put them all in generating functions here, then I will get a sum object which would satisfy infinity many constraints. So there's a lot of relation to McDonald's polynomials, it's, I'm not discussing it here. Okay, what is important here, it's identification of this object with this object. Again, when I'm saying partition function, I might actually mean a generating function. Okay, so now let me explain the main result here and that's what I'm doing. So I'm interested in a compact manifold. So let me start with s three. So for s three, if I take the theory, so I'm taking n equal to u n theory plus adjoint karel of mass m. So it's generalizable to marginal theories, I'm just looking at simple examples. So my partition function would look as follows, it's i r m and I will have a d n x and then I will have here i not equal to j. Then there is this double sign. So my s is squashed. So what I'm thinking about my guy is omega one of zeta one square plus omega two, sorry, omega two zeta two square equal to one. So xi minus xj w one w two. Then there is s two m plus xi minus xj omega one of omega two. And then typically here, so let me be more pedantic and try to do it explicitly. Yeah, nothing special. The thing is you do what you can do. So this is a thing. I mean, I'm going to, so unfortunately, you know, there is, if you want to go to concrete things, if just let me make comments. So the results were explicitly has been calculated. There is closed form, it's s three, it's a land space, it's s two times s one. So this is on the market exists concrete formulas and you can take them. And the statement is that I will tell you that all this generating function will be killed by two copies of commuting, whereas our constraints is appropriate way of identifying parameters. Now what is up to the catch? And this is for any postdoc speech students or even senior staff. I mean, n equal to theorists exist on Cyford. So it's a result by Fistucci, Demetrescu and others. So you can put n equal to on Cyford, but there are no formulas here explicit. Because one thing that you have a supersymmetry, another thing is to calculate a matrix integral, take into account all flat connections. So in principle, we would believe that we should be able to do something here, but you know, nobody. So s three is because people calculate it. We can sit down, there is concrete formulas and do it. But in principle, I mean, this is up to catch. I mean, one should be able to produce formulas for Cyford up to some orbit fold. I mean, it's locally free, you on action. Absolutely. So this is huge class of manifolds and people prove that it is identical to supersymmetry, but nobody dared to write the concrete matrix integrals, well, maybe exception, Simons. But again, there is a lot of issues here, et cetera. So, no, they're much better than Rouschi-Tichin to arrive. So this is all the stories secretly related to complex Chin Simons. So this is, I mean, this is some upgrade. And but again, this type of theory is not yet well defined. It's up under discussion, but secretly it's related to complex Chin Simons. Absolutely, yeah. So I don't want to discuss the subject because it's outside of my scope, but secretly sort of adding the mass, you know, having U.N., this is somehow should be identified with the complex Chin Simons, et cetera. So there is this discussion in the literature, but it's not very conclusive. I mean, there is no clear cut answer. So is that equal to? Absolutely, yeah, it's specified. So the rank of this guy tells me how many integrals I have. So then if I take another, you know, theory than I have integrals. So this is like this. And then what I wanted to write, I still have at least to have time to explain your result. But, and here of course, what you will have, for example, you will have standard this terms, minus i pi k2 of omega one, omega two, x i square plus two pi i k one, omega one over omega two, x j. So this guy's Chin Simons level, this is Philly and Lopoulos, for example. So now what we want to do with you, we would like to put Wilson loops, but we have to think how to put Wilson loops because we would like to put a supersymmetric Wilson loops. So it's important in this logic that I'm choosing my omegas to be generic. So there is, we'll be lot of algebraic degeneration when they're rational, et cetera, et cetera. So I choose them generic. If I choose them generic, there is only two supersymmetric Wilson loops I can put. One Wilson loops when zeta one equal to zero, and it will have this periodicity. And another Wilson loop here, it will have this periodicity. What I'm saying is it's not true if omegas are rational. I can have torus nodes, for example. But if I choose them generic, then that's what I will have. And then it means that for example, if I wanna put a Wilson loop, then I have to insert in the integral correspondingly the following guys. I have to insert, for example, trace in some representation of my exponent of two pi i x over omega two, for example. No, no, well, it's basically, if I think about S3, S1 vibration, it's only two loops which sits over north and south polo S2, right? So this is the only thing. But this is what supersymmetry is telling me. So either I put this guy or I put another guy. And I would like to write the generating functions for both of them. So either I put these guys or I put, so this is representation one, I could equally put in another representation two pi i x over omega two. And of course, these guys is nothing as a short polynomials in dependent from the exponent. R1, R2. So if I wanna get a Wilson loop, I just put a short polynomial here. And then there is this standard formals which I'm going to use it. If I have a sum of all source, let me call them S, I think this is more canonical. So if I write this at T hat S Rw, this is equal to exponent of n more than one Tnj Wjn while these Tn's are related by formal change of Tk hat n over n, sorry, T like this. So that's what I'm doing. Now what I would like for you to introduce a generating function. So my generating function will depends on two set of times T and T prime. So how I define this? I define this as a sum over two set of representations. And then I will have S of R1 of T hat S of R2, T hat prime. And then I will have my integral which I had before with insertion of, so we'll have d and x, I will have my measure. So this guy is exactly this measure constructed out of double signs. Then I have my key exponent, okay. And then I have corresponded the Wilson loop R1, e to pi i x of omega one. And then I have S of R. So I would like to have a generating function for Wilson loops, supersymmetric Wilson loops. So if I do it and I use this formula then eventually what I will get I will get the following integral. So my measure, my e w x j. And then here I will have correspondently exponents of n more than zero Tn. Then I will have j from one to n e to pi i x j of omega one. And then I have tn prime. Now j from one to n e to pi i x of omega two. So I have to have two times if I wanna have keep track of all these loops. So this is my object. So let me now tell you the answer, the main result of this object. So let me for simplicity choose my chance simons level equal to zero. It just was simplicity result looks simpler. So in principle one can generalize it appropriately. So now I told you that I can write for you these operators as I wrote before in some representations for t and t, t prime. So I will have to free field realization for quiveras or I'm now looking that these guys is more than equals zero. So these guys by construction are commuting because this is used different times. The statement is the following that these two guys, tn will kill this function. So this is a priority. Let me also put here prime. This is different operator. So again, I'm repeating that this guy, there is no really obviously symmetry because this enters with this periodicity, this enters with another periodicity. Now it is important that this is not identical algebras. I mean they have a different parameters. So let me write for you the parameters. So if I'm writing for you the thing, so I have two copies. Copy one and copy two of my verasaur. And in gauge theory I have the following parameters. I have my squashing parameter for this here. I have a mass, et cetera. So then correspondingly copy one, this q one will be the following. It's a e two pi i omega of omega one. So omega is just omega one plus omega two. And t is equal t one is equal to q one and power b and it's equal to two pi two pi i capital M of omega one. And then q two is equal to two pi i omega omega over omega two. t two is equal to q two beta two pi i M over omega two. So this is your glutuk verasaur and this is the following relation of guys. Again I'm repeating what is quite unique here. This is just one integral. It's not two integrals. I mean the statement will be obvious if there will be two integrals. So if I took my result from before, when I got verasaur then it will be, the thing is here the trig works because these parameters are gluten very, very particular way. So if you start to think geometrically, so if you think that q one is equal to two pi i tau, so tau is just here, it's a ratio of omega over omega one. Then my q two is just equal to the following things. It's a minus two pi i tau, tau one minus tau. So this is a sum element of s l to zeta. And this is exactly element which you will use when you will glue s three out of solitary. For example, if I will, I mean, the same statement will be true. The measures will be different. The integral is different. But the same things, for example, true for land spaces. For land spaces you will put here r, et cetera. Now what is important, the trick that these two commuting guys lives together nicely, it doesn't work. So I cannot really take two results and reduce two integrals to one integral, et cetera. So the whole trick, it works nicely only when the parameters of verasaur of two q verasaur glued by s l to zeta. Otherwise it doesn't work. So I have no time to explain for you the details, but this is the idea. So this is the main result. And of course I skipped for you a lot of technicalities. And what I would like to stress at the moment that it may appear trivial result, but it's not trivial result. And the thing is because of screening charges. So when you prove it, of course, you will start to prove it by basically by gluing two free bosons and thinking that if I take one free filterization, I take another free filterization, everything works. So I can basically repeat for your story. I can take TN. I will write for your screening charge. One, I will take another TN prime. I will write a screening charge two. And then somehow what is obvious is that you would like, for example, now to construct S1 tensor S2. But don't forget they would depend from the same guy. So when I will act this by TN, so of course TN will act on this. So I will get D of S1. So this is my Q difference operator. Q difference operator. TN tensor S2. On S2 it does not act. I will use totally different free bosons for representation of this guy. But the thing is this is not good because I cannot put on the integral because this is not equal to D on S1 times S2. So I mean, there's a standard stupid trick of just gluing two copies. Well, it works, but I cannot put this under the one integral. Of course, if I will have here X, here Y and integrate, then there is no problem. It's really a copy. But I would like to do one integral. And because of this property, it's not good. But the magic is the following, that if the parameters of this theory are just in such a way that this D acts trivially here, then this becomes true. And this exactly happens when it will be glued by SL2Z transformations, OK? So I have to finish, but let me just say a few things. So it can be generalized to these cases, LR, and some two versions of S2 times S1. Why we don't do other manifolds? Because there are no explicit formulas for this. So for example, we can maybe come up with some ways of gluing currents, et cetera. But we have no measure to compare. So it's simply there are no gauge theory results. So this is one comment. Second comment, you can generalize everything. So in fact, the theories which one can do, one does not have to do your own theories. If the theory is which is extensively considered, it's there are 3D closings considered by passed on by Nikrasov. It's a skewer type theories. So in principle, you can have a UN node, and then you have a marker. The only thing we always have to decorate our UN theory by some adjoint to have an M. And this is responsible for T parameter. In principle, you can go to the limit when T goes to 1. So there is rounds here, a lot of things that will correspond to certain degenerations of the algebra. Some of them studied, some of them not studied. So for example, one can generalize this to ABG mode. The only thing is if you would like to see full algebraic structure, you have to take two nodes. So you will have these guys by fundamentals. But they also have to decorate by adjoint. If I don't want this adjoint, there is certain value of M which I can do it. But in principle, so they will be this diagram, your diagram. So Peston constructed for this diagram, the w-algebra. So there will be these two copies. So if I construct this default ABGM for general values for scores here, some partition function, generating function will be annihilated by two computing copies of this clear w-algebra, which was constructed by Peston, et cetera. For all the theories, it works. It's first generalization, second generalization. Of course, everything, all the theories can go on to all these three type of manifolds. You can include, of course, Chen-Simon's level, which corresponds to insertion of some vertex operators. So the rest of our constraint becomes a bit more complicated. OK? So I have to finish. But basically what we see now here in details is some remains of 5D picture. So in principle, the screening charges which I wrote, it's the same screening charges which were used by Peston in his December paper. So in principle, one can write a 5D microsoft partition function for this keyword type theories, roughly speaking, as ejection integrals. So this is symbolic things as ejection integrals of infinity-many copies acting on the vacuum. So that's a Peston result. So this is a representation. The thing is that in principle, what you do now, you talk about generating functions, and it will be annihilated by infinity-many constraints. So of course, this is an R4 times S1. So this is the result with the Peston at December 2015. So for example, one would expect here that you can glue some similar things. We sort of, it's under consideration. We don't know. But one can also, I mean, for the compact manifolds, you will glue things there also. The only thing that apparently, if you think for a while the correct notion of modulo double for q-verasora, it should not involve not acyl-2-zeta. It should involve acyl-3-zeta. And the thing is, because verasora, q-verasora automatically has a symmetry over q and t. So it means if you take this picture and you impose symmetry about qt, you have to generate full acyl-3-zeta. And that's exactly how you, in general, five manifolds you glue partition functions, et cetera. But again, it's something very speculative and remains to be understood, et cetera. But for example, the result, which is sort of obvious for Peston, that a partition function on S5, a generating function, will be annihilated by three commuting copies of q-verasora with the parameters which you can read from geometry. So I have to stop here. Thank you very much.