Higgs Bundles, Gauge Theories and Quantum Groups

0:00 I have a reference to this as M&A, and the salary that we introduced, sometimes I use young meals, as opposed to young meals. This salary, which I will describe in detail during this talk, is based on the ideas which are very old, and the job it does, this salary, computes intersection for on what's called a HIT bundle, a modelized base of HIT bundles, which was introduced by Hitchin also like 20 years ago or so, and this is for any Riemann surface It depends on one parameter, which is everywhere and five meters, I call it C, the U1 action of X1 that was introduced by Hitchin in the same paper, so this is the one parameter associated to the C-direction, and in the paper, with Moore and Necrasso, we exactly sold the cellar. So this cellar is exactly sold. I will claim, and I will consider this cellar for Gates Group UN, but it can be considered for any Gates Group. I claim that that's already the same as two-dimensional non-linear Schrodinger quantum severing in the n-particle sector

2:30 So I basically argue that these two things are the same. Now this theory also depends on one parameter which is called coupling constant. C, that parameter would be positive in both cases. Now, people who study all best-measure systems related for, well, whatever, this subject is very old, know that this salary exists for two values of coupling constant, c more than zero, c less than zero. This one corresponds to the one which does not have a bound state. So for c goes to infinity, this has the salary of pre-permium, and for c goes to zero, this is the salary of pre-bord loss. Second thing is, this theory also is at the top of it, by many methods, so the first one probably I would call coordinate right-hand side. I think it goes back to Eliot Leib and the papers of Terezi and so on, so this is very old, early 60s. It also was sold by algebraic batanda, Leningrad school, my colleague Leningrad also called it Fanto method of inverse battery, part of inverse battery method. And it is classified, basically, now that theory, like I said, is defined on any Riemann surface, sigma. Of course, the people who did this other thing, in that sense, had in the case of the plane or the disc or something like that. So this theory I will talk about only for the cases like cylinder, disc or R2, once that is for any sigma. And it is described in terms of, when I mean it is described, I mean that the complete universe states, eigenvalues of Hamiltonians and so on, are

5:00 described in terms of something called the Antians, equivalently double affine Heckel algebra but they generate one and slight modifications of this statement will link to all of them sometimes better words is to use quantum groups that's why I put in the title quantum groups as you see this every whatever I just said has a two parameters uh the generation means that it's only one parameter which is at c that's over the one parameter but this server can be also deformed to something called generalized g divided by g gauged west domino-witten model then it has also two parameters the second one of the algebra and then okay so as i said this is the work with gerasimum on which we extend i mean basically the few words of our history moon across of an eye introduced this gauge salary in order to learn how to integrate over how to define equivariant integrals over certain moduli stages which are hyperkeller quotients and that parameter actually there is a regularization in the compactification for the hyperkeller or the equivalent parameters there and derived formula for partition function for arbitrary sigma in order to put it in a little bigger perspective so by the way I should have said that if this is a quantum cellular and particle sector of quantum salary in Northern Schellinger equation then that one which has all quantum groups will be related to the X X Z model and maybe sign Gordon like what I want to emphasize here is that there are many gauge salaries in many

7:30 dimensions this is a topological salary it's a four-dimensional two-dimensional related I'm talking about basically how to find a piece in the big big class of the gauge salary some little piece of it that exactly matches the integrable salary and if it is like that then using the relation between coordinate betanzas to algebraic betanzas which itself is this one basically identify everything in in that theory with the things in this theory. For example, there are Hecke generators here, the simplest one in the center of the double-alpine Hecke algebra, the generate one. And that center will turn out to relate to the cohomologists in this theory, and actually the old Hamiltonian here will be related to some very simple objects here. And that's its job. But in order to put in a little perspective, I want to make two statements before. talking about here, mathematically, is a precise formulation, but maybe does not have a mathematics proof, I don't know how to say that, but in physics language it is established fact. I'm just saying that this conference is mathematics, physics, and many other things probably. And so mathematically I would not claim that I have a proof, but I think I do. Oh, there is some pass integral involved. I can't, I mean, look, I'm doing some localization which starts with pass integral and ends up with a finite dimensional integral. No, you can take it as an action. Thank you. Well, I mean, okay, let me be precise. This statement from the paper I said from was later in the equivariant integration of the X bundles and so on was claimed as a result of this thing. It has been proved and some people were prepared to prove it. So I would

10:00 say that the model of that it is a step is fine and anyway so now I want to link the examples which was targeted in that paper ten years ago and this actually will be helpful later in terms of understanding so the The examples of hyper-catellar quotients Considered there were ALE, which is a four dimensional monoclonal, which looks like Walker-Linn, which is like C2-Mont-Gamma, where Gamma is a subgroup of S2. also ALF technically trampled for the model like space of instantons now here also integrations are very nice and reduced to the simple finite dimensional integrals Then we combine this to consider basically theories on arbitrary finger manifolds. For example, this one later leads to the exact production and compensation of the black producing like in Hida, the Hida, after, and also K-theory version of this, which we called the 4d berlinde formula and i heard the talk by george recently when when he calls that the four dimensional berlinde formula is a case of reversion of donaldson invariant and then And we wrote that the equivariant localization leads to the answers for the case of pitching system as a sum of a solution of that time that's coordinate, that's equation for, for northern intersection. so basically we computed partition function and so that like like here for

12:30 example partition function here would be the extension in the here for the expansion in the backhand for the no-linear setting the equation so now I want to so there was Gerasimo we decided to establish complete equivalence because happened that the sum over the background of the equations here could have been accidental fact and it happens at this by the way the second background of equations is solutions is classified completely by partitions of integers so I mean partitions of integers comes in many different places okay so now I want to define the left hand side and I want to define the right inside and establish the relation, mathematically precise. Well I hesitate to improve because you know in both sides there are many things which were not static. So when I claim they are exactly the same, I have to identify just everything. So on this side for example the wave functions were static, the Hamiltonians were static, so those are rule identical. But there are on that side the same, there's a lot of homology here, but there are also with some lines and stuff like that. So what I will argue, the simplest version of this equivalence, is that the space of the wave functions in this series is given basis of diagonalized homology completely coincide with the wave functions here. Okay, so now... This is the first time I'm giving this token in a more mathematical audience before I get into this token, so I try to be as modest as possible. You will see whether I have it or not. But this is the correct one. But, from what I know, okay, so now I want to define the, okay, before that also, the very interesting thing about this parameter C. I want to list those properties before I go to the very technical details.

15:00 The salary we defined for C goes to infinity, when this parameter C goes to infinity, from the quantum field salary side we will see that A, this will be a two-dimensional Yann-Mill salary for compact group. uh this thing but and this one uh will be established this correspondence will be established in this limit because uh this salary has a description in terms of prepermions those in fermions that in this salary for c goes to infinity is a preferred point you know the same one then the second for C goes to zero which is a bad limit in in both service like that C was a regularization parameter if I would not see in this story everything but after I put C I can computes the answer in this end if it's C to zero and see what happens. So in two-dimensional Yann-Mills theorem, in two-dimensional Yann-Mills system, this theorem at C goes to zero becomes two-dimensional Yann-Mills again, but for complexified rules, which are known in C. That's the reason why everything is emergent because the first thing you can't use as i said was the covariant volume and here it's a complex group so it's infinite and the the second thing will be is that it could be described also in terms of three bosons from earlier scheddinger's side now for c not zero again it's like what happens in the conformal filter the partition function one thing is is defined there is a definition of partition function but in another language it is only partition function can you tell you the volume which is defined but you see the asymptotics and you pull the information so that same thing

17:30 happens here for C naught equals zero and infinity I don't know what I wrote here is true for both and the answers are easily constant so this by the way suggests that the theorem is better viewed as a deformation at the point C goes to infinity rather than at the point C goes to zero because at the point C goes to infinity it's a lot of fun theorem now for C not equal to zero at infinity it interpolates between these two limits an interesting thing from the point of your nonlinear Schrodinger equation we are talking about representations of basically spherical functions or GLN in the limit when P goes to 1 so the answers of this over here turns out to be spherical functions here corresponding to the GLNZP for P goes to 1 that's a bar so when you look at this you take the Magdoros function or how it was polynomial and coincide with the other one I did not know it's equal to the algebra now from the point of the Young Mills as the way I described it it should be related to the representations of the Youngian as i just claimed the the partition function was sum over the solutions of the cornet beta and that's the question so this is generalities now i started so sorry i spent 20 minutes on general statement but i think at least what what i claim is explained now we have to describe this model in that sense Anyway, from the point of view, my understanding is a good move from that considerations can be in follow. First, this subject in mathematics is now where people are very much interested in studying this model like this for various reasons and so on, and try to find some properties there. This is a very developed subject. This exists for 40 years and more and lots of things must have been done here.

20:00 So it looks like this can be used, this equivalence can be used to understand that better. But actually I think that the opposite phase is nicer, although someone who works on language correspondence and geometrical language correspondence seems like that would think that's it. Why do I need to study anything more complicated than that? If all we take out the glass and put whatever is entering into the geometry length, but here is obvious, here is absolutely not obvious, and I'm telling you that they are the same. I think the input wanted to have the most general function groups in a picture, and I explained it with that. Okay, so, definition. So what is the theory that we have introduced in the space of 10 years ago? Okay, so this is the gauge theory, now this is the quantum field theory we're talking about. So I have to introduce the space of fields. And space of fields here is starting the same as in two-dimensional topology in this case or you consider because they're in a surface sigma you consider mass from remember the sigma of any genus I don't know genes like H sigma H to a group which can be any group then consider the fields first the gauge field a which is nothing more but connection in a principal g-model and to make sure we are here so the g takes taking the compact group here we also take it to the compound the field five which i'm sorry we can't if you read oh i cannot move this up sorry So I'm basically describing the content of the fields in Pudimental-Yan-Mill theory and this theory has more fields. So I will first introduce this one. So A is basically Yan-Mill's field, the big field, on the remote surface of genus H. Phi is a scalar, so it's a section of the function, it's a function, and it takes values in the adjoined representations of the algebra, which is a real algebra to that.

22:30 And the third one is a sign, so they're noted like sign A, which is a one-point. But in the salary of Young-Neill-Higgs, we had other fields to introduce, and those fields are following. So there is a pair of the phi and the psi, I call it psi of phi, and they take value of both in A1, so these are one forms. but you need two more fields I call them phi plus minus chi plus minus and there are colors everything is in a journal so after this field search so phi plus minus sometimes I use phi 1 plus i phi 2 plus minus equals phi plus minus similarly for chi's and from this field content we write of gauge theorem, let's call it Young-Neill's case, which is later it's clear to show that since it's related to the Fitching equation, and Fitching equation is the dimensional reduction of four-dimensional self-loathing conditions into two dimensions, this will be also possible to write in four-dimensional terms, but I'd like to make an example, so this is integral over the human surface, sigma H, I'm given to extra. Trace, which is trace in adjunct representation of all of the objects. So I have phi, I call it now phi zero, that phi which was in ordinary and mid summary, I will call it phi zero. Phi zero times the real moment meant for this hundred-color reduction, which looks like Carbature for this connection A minus 5 with 5, now this 5 for the one-fourth, and that's plus minus C times 5 with star 5, again if I get two-fourth, so this is the first term. so this is how this parameter c enters here as a master so for physicists is nothing more than this since phi is a vector this is a mass for the body by the square times c so c is a mass for

25:00 it then there are two more terms which are related to the complex normal mass so i have the if phi zero was like an amount for this equation the phi plus minus move before other one so i have to take So phi plus covariance unit with respect to connection A is one zero component, I chose the complex structure of the C1 surface, I take one zero and zero one component for one four, and take the covariance unit of phi which was another one four, zero one part of it like z plus complex conjugate which will be phi minus times nabla A of zero one of phi 1, 0, so this is a bosonic part of the action, plus fermions, and these fermions are, now I will mark, this is a fermion, this is a fermion, so anticomitic variable, and here this was not anticomitic variable. Now, just very briefly, if I put C equals zero, and take the phi zero, phi plus, and phi minus, enter linearly, so I get the equation as a critical point of this one, so part of the equation, F equals phi phi, and the number of phi equals zero. these are kitchen but this is some of the right so now on the space of fields i just introduced i think everything is introduced any questions please about what is this I want to get done quickly with this whole paper with Greg and Nikita in order to start studying the complete solution for the partition function as I said so now on this space of fields there are certain operators that can be introduced one I call Q, which is related to the gauge symmetry

27:30 this is a DRST of our return then there is a linear derivative of a gauge transformation with a gauge parameter phi zero so as you see, I have a gauge field I have one form, but I have a scalar phi zero so now L of phi zero will be the linear derivative well, gauge transformation with respect to phi zero and introduce the action of the u1 c star action on the moduli space and it comes with its own derivative and I call it L of b so basically if you don't want to know any of this there are three operators which act in the space of the fields and I give you some of the formulas for example Q acting on A gives Psi of A this is the gradation there so this is an odd this is even and this is even okay so Q acting on A gives me this is odd and this is even so this gives me the odd and there are other formulas for example Q acting on chi plus minus gives me pi plus minus. An interesting one would be let's say Q x in one side of pi let's say component one zero so that's pi. We have to distinguish this big pi into small pi. So I'm acting now with Q of this one and here this zero enter so this is equal as it was supposed to do the phi zero of that plus C times five so in this operate definition of Q and all of them C will be entering somewhere so similarly L by zero acting on A is a covariant derivative with respect to A of five zero and let's say L of B acting on let's say five plus minus would be minus plus five plus minus so it comes from the chart. Now I can't write the big formulas here so what what matters here now that Q basically squared to zero. So this operator Q squared to zero, but not here. So properties of this Q are that Q squared is equal to gauge transformation with respect to Y zero, I think it's the second operator, plus C times action of the vector field of each integer.

30:00 So Q will square to zero as long as I'm acting on the gauge invariant objects which are also c-star invariants. So if LV kills the object, L phi zero, then Q square. Now in two-dimensional, which is a limit of this, as I told you for c goes to infinity, this term is not there. and you have an equilibrium localization there, but here so then one can ask the question how can the action be written in terms of the operators and the action S turns out to be action of trace 5, 0, F psi A with psi A plus Q and this part of the action is the same as intradimensional numbers this one also can be written as Q-exact but it will be Q-exact of non-gauge environment so one can introduce basically here in Q of something two parameters of interesting objects I will denote them by tau one and tau two for example tau two tau one in the section tau two would multiply integral chi plus five minus plus chi minus one plus like introducing something like a So I can add another term to the Lagrangian, which is Q inside, like that, and then take tau 1 to 0, remove this one, and tau 2 to 1, and answer will not change, because this object, the time field is here, makes the space of integration basically compact. So in the beginning, it's a non-compact space, so it's difficult to use the methods we know in the aquarium localization, but that parameter makes it basically compactified.

32:30 So, anyway, so we continue to answer, I want to describe the answer, from this paper with Stradmore and from Grasso, for the limit C is not equal to zero and not equal to infinity. And then we look at the limits of this. So I have actually prepared this talk very consistent in a sentence that at this moment I had to compute that only for the limit c goes to infinity. And then I had to compute that for only limit c equals zero. I'm showing you something. And then compute for the arbitrary c but I'm running out of time. so I give you the answer for arbitrary C and sketch how it's related to the image so this is a partition function for the human surface sigma H of this theory I call YLH so it's proportional with some irrelevant constant in front of you which might depend on C but irrelevant in the that it does not depend on a genus of the surface and other characteristics which will become clear now. It's proportional to the following integral over Rn, product over j equals one to n, d of lambda j, sum over Nk integer, so there are several integers I know by Nk, k grows from one to big N, was the rank of the gauge group of exponential 2 pi i lambda m nm sum over n from 1 to n so what's happening here? I have integration variables lambdas they are labeled by j's which is from 1 to n and then I have a set of integers which multiplies those lambdas. I have a sum which is a vector then all this stuff multiplies the product of I not equal J lambda I minus lambda J to the power N I minus N J plus one minus H H is the genus original surface and then this multiplies products a not equal J no

35:00 actually all of them, ij, it can be, I might speak of j, lambda i minus lambda j minus i times z-parameter c to the power n i minus n j plus 1 minus h. So that's basically it, but of course this is not a final story, I will very strictly simplify that, but the reason I wanted to write this actually will become clear very quickly. because if I would consider the same salary but in quantum mechanics which means I would instead of Riemann's surface Sigma H consider trivalent graph what is called I mean basically take and reduce this salary to the one dimension on circle s1 turns out that I will get exact same formula except this factor which is I would denote now by the project so this happens to be there only for remote circuit but the same theory can be defined in quantum mechanics and that factor will not be there and we will see the reason is that those integers and i i wrote there are basically the integrals of the u1 part of the gauge connection over the pi watts so this is Well, there is a gene, it is empowered, you will see it's a second, this formula gets simplified. So you can get it in some units? I can get one from another, I will explain this, this is an old hawkies actually, but basically I mean these U1 factors are very important for me. So this formula, when it's written, should be compared to the formula we derived for the instant non-modelized states for the case of the Donaldson invariant, where it was left like that. So this is Abelianized theory. Now I have only Abelian left us. In theory, it was not Abelian, but as you see, I have n integration variables instead of n squared. And in Donaldson theory, we left it like that because we thought it was like whatever, polynomial times the Fourier factor. big deal. But as later turned out, computing is related to the partitions of integer and of the tableau. And then Nikita wrote a paper with OpenCore where they actually did this to the answer. We will see. It depends on the, so you have to take only basically U1 part

37:30 of the gauge group and then you are considering u1 to the power n principal g-bottom that one has a monodrome so you have to see how it's actually sits here because after i go to the leg length dual you will see the table there will be tricks related to the very slight modifications of this but first let me let me simplify this formula to the level that everything is trivial Okay, so now, as you see here, I have to integrate over Rn and sum over n. So it's like a Poisson resumiation formula now I have here. I can first sum over n, then integrate over n. Absolutely. And this is where the integrability now will be hidden very beautifully. So if I first sum this, then what will happen? at the time, now I have to watch the integers here. I have N here, I have N here, I have N here. So let me now remove these guys here, okay? And write those products I just removed as exponential of log. Okay, so I removed them and I write exponential of NI times log of whatever it was, right? Then it will change this. So instead of lambda, I have now something I call alpha m of lambda and I give you a formula for this alpha m so I put everything which was summed up with this integer together in this exponential I take it over and that alpha turns out to be oh exponential of i times alpha j of lambda so this is all dependent as I said and I just collected here equals something very known from the C and Young the first derivative of something called Young's function and this is just notation is exponential of 2 pi i lambda j times product k naught equal j lambda k minus lambda j plus i c 1x5 minus under J minus under J minus IC so this alpha that I collected together exponential of it is this horrible object and everything else then as Alan was mentioning

40:00 almost doesn't depend on genius or whatever but it still depends on genius it's the power of 1 minus H now we have the exponential of this sum it's a sum of delta functions of zero so five okay so this integral luckily for us becomes easy to take because this Poisson resumation factor gives us delta function of zero so five but then we easily integrate over lambda okay so turns out that's the answer that after we do this is what I would say integration in sum? No, not for every, every, everything. Maxime and I had a fun formulating for which alpha set, so what conditions is actually true. And this particular one can solve two is one of them. Okay, so here is an answer after I do that. So partition function on sigma H, equals sum over lambda, which I will call now the set of BN. Set of BN will be those alphas for which this is equal one. Right, because I said only zeros of alpha will contribute, have to multiply that. And so then this has to be equal to one. This is some set of lambdas. I call it lambda belongs BN. of certain functional which I denote by D of lambda to the power of 2 minus 2H and now I make one slight modification in that theory which is defined there I can add to it with some coefficients and pieces like the generating functions very much TK times sum over K times the polynomials of degree 2K zero okay i can take that five zero square to the power four and so on and start adding two okay and it turns out that i can compute all of that not only for t equals zero but for all of them this answer it's an exponential of minus sum of pk k equals one to n of symmetric polynomials P2K of lambda this is basically for K plus 1 and so on

42:30 where P of lambda so the set Dn is a set of exponential 2 pi I lambda equals product I less than I naught equals J lambda I minus lambda J minus I C divided by lambda I plus I C and this is called the coordinate theta-android harmonium linear equation with d lambda is the product i less than j lambda i minus lambda j c squared plus lambda i minus lambda j squared squared so d of lambda is this product then I have to take it to the power of 2 minus 2 8 multiplied by that this is complete answer question is what is this set? so Jan long ago proved that for every partition of integers p1 more or equal to p2 and so on more or equal to pn any part integers ordered integers there is exactly one solution to the equation so then this set is lambda to the part so some and then lambda is a complicated function of those there is a better now this is a corner but that's the equation for our inner Schellinger and young proved that for every integer there's exactly one solution I mean, I have 15 minutes, I have not even finished the definition, sorry. No, even for two lambdas it's not easy to prove. This equation for two lambdas is exponential of lambda equals like a minus lambda divided by a plus, sorry, a minus lambda divided, and this is the imaginary. It's a transcendental equation, I mean, even that is not easy to do. By the way, your question, Maxime and I tried to write a condition, general condition for which this works like Poisson-de-summation. And there is an interesting condition, I think this is called in physics a completeness group, that the set of solutions is complete. Then the sum of something equals zero, it defines it. anyway so okay now there are questions so there are two ways to compute this so

45:00 one way we propose was the way I just described well I wanted to describe here do the localization with those, what I call hyper-color localization. So take a fixed point set of those things and then, so turns out the fixed point set of those that you are in action is enumerated by those solutions. Another way to write it as an ordinary portion, not a hyper-color portion, but a unitary portion, and then what you get is an integral over fixed point set, which now has an Euler, a current Euler class in it and it's complicated. So the algebraic geometers are having that fixed point set. What I claim here, you can further localize and get to some of the partition. So this result was very strong and we will see now that this result actually allows to go farther. So the first thing I wanted to say here in my way of proving it was two studies or two links and we will see at least maybe we do that and it's anything else so if the real proof goes like this i have a partition function i know the partition function is a trace of certain operator in the complete hubert space okay this partition function turns out like has almost no Hamiltonian it's just some of some wave functions and so on question is If we lift this from the periodic one, so I have these pi ones I said, and I go to the R2, what will be the wave function, conditional wave function, that after we compactify and include the monodromies of the H connection, and sums the wave function, that we get this partition function. And it turns out that condition of having a wave function is a function of some phase space, which I will describe in a minute, hopefully, as the cosycle condition of wave function beautifully matches these representations of that else. So that's the way to go. This is cosycle. You can't even guess what that cosycle is. This cosycle is something that I could feel, this projection.

47:30 So normally, you would have, when you go around the circle for the wave function up to the permutation of the coordinates, it acquires a phase, like a Fourier Fx. That would be an ordinary and mill series, and in this series, it has to require that phase. There are not that many functions, which when you have linked it to the R, have that phase. So we wrote a general solution, and then when we went down to the S1, we got the solution written by Berenzian and others for this. okay so now let's study the two limits okay this is very very very easy when c goes to infinity let's look on the definition when c goes to infinity by the way uh this theory where it's written has a gauge symmetry of compact group only but when c goes to zero something changes fair to study c goes to infinity this is a mass term so this drops up so you have to remove c goes to a delta function phi equals zero, it has the dumping factor exponential minus c x squared to send c to the infinity. So the all integral localizes from zero phi. So this guy disappears and whatever is left is to the young mean. Now we go to the answer. What happens when c goes to infinity is the answer. You have to consider solutions of the lambda up to the multitudes of integer and up to the permutation. So this, if you wish, let's take, this example is good. When c goes to infinity, this becomes minus 1 to the power n. so this becomes minus 1 to the power n then that tells me that lambda is some integer plus n minus 1 over 2 then these integers actually are those ordered set of integers which define highest weight of the irreducible representation of our compact group U n and the sum is over the highest weight of representation of U n and what is standing here, D lambda you look under what happens when C goes to infinity it with d lambda and you see that this d lambda becomes the product of lambda i minus lambda j. Sorry, I had SUM or U.

50:00 Now, what happens with d? d becomes a dimension of irreducible representation for that high school. And this is a known answer. In two-dimensional Young-Mills, the answering partition function of two Young-Mills is the sum of dimension of irreducible representation labeled by the weight mu in the positive, what I was talking about, to the power 2 minus 2H and sum over all mu's in positive chamber. so we will cover that's after the number for c goes to infinity now let's look on this c goes to zero when c goes to zero this term disappears and what I have? I have phi zero equation of motion f equals z which was not the same for phi equation of motion with c so now let me claim the claim is that it's easy for anybody who studies I'm trying to filter that this is a partially gauge fixed to the Young-Mills with a complexified gauge group. And the reason for that is that those equations, F equals pi pi and number pi equals zero, can be collected in one equation to find the use complex connection AC equals A plus I pi. And then I write the flat connection condition on the AC. So in the real part, it will give me F equals zero. In the imaginary part, it will give me number of I equals zero. No, you are. So I consider V plus AC squared equals zero. This is my equation. I take the AC, which is the algebra of G, take V plus AC squared equals zero, and you will get those physical points. F of phi phi and another phi equals zero. Which is not the case for C on zero because the phi equation and so on will start mixing C. But in this case, it is that guy, when you fix the partial gauge fix, what I meant, when you fix unitary part of the gauge transformation, gauge fix. So now, since it becomes a complexified gauge group, Young-Wilso, let me mention Young-Wilso, For C equals zero, you expect, you expect something like that.

52:30 But the problem is that this complex group, the representations of the complex group, even dimension is not well defined because it's infinite dimensional representations and there are continuous and discrete series of them. So which sum you mean, what do you mean? Now, what I claim is that for every C, this is well defined. so so my answer for c goes to zero gives you one very particular answer so the limit asymptotics of this formula for sequence it goes to zero is very specific but that formula if i would copy nikdo's computation for complexified gage group is unclear completely what is the dimension so what you do you take some old books and if you're russian like russian educated like i am He opens the books like Gelfand and Jelob-Banko and Schterr and so on and learn about the representation of the complex groups. It turns out that there is a notion of the formal character, there is a notion of the formal dimension which is called degree and so on. So you look and see what your answer is. Is that C? Is that C? Or square integral. Yeah, or square integral. So what we are looking now. See, C introduced the square integral. Yeah, okay. So you take this picture. It actually happened. So since Alan knows everything, I'll give you a quick answer, right? So I have to say something else. So what you get is that our answer turns out to give only part of the representations in the sum. And these are discrete series representations unitary, which are classified in orbital classified by numbers real number s and integer k you get s equals zero and you have sum over case and then what enters here for c goes to zero this formula for c goes to zero becomes sum over only those representations when d of lambda becomes what they call degree Or something like the formal degree which was introduced, I mean, I shall go into the formal character and prove it's actually not generalized function but a regular function, introduce degree and you get that. So finally, this formula that we derived, turns out to integrate nicely between set of irreducible representations in complex case group and the real. But in the middle, we suspect it's a quantum group.

55:00 Actually, we suspect it's a quantum group because we got better answers. Okay, so now, finally, we go and... So that's why, I mean, you know, when C goes to zero, this is one, so lambda are integers. now now if you allow me i say two words only about northern setting their equation because in a modern uh literature the representations is the wave function on a disk and even after is described in terms of the double algebra introduced by and independently and the version of it on the circle which were are interested to consider a Riemann surface with the boundary of the circle, co-interviewed by Genevieve, okay? So, the way that that guy goes is that you have the Hamiltonian, second Hamiltonian for knowing the Feddinger looks like this. One half D by F here, D by star BX plus C of five stars. C is positive for us. so this is a Hamiltonian with a Poisson structure of 5x this is, by the way, non-relativistic cell it's very funny that we are trying to keep the non-relativistic cell so this is a Poisson structure there is a quantization of it it turns out that the particle number is conserved it is a dx 5 star x 5x so we can consider this cell restricted to the n-particle sector an n-particle sector wave function function and i don't know whom to i mean let's say eliot leave and baris these are two papers and while i check the same time one written in russia that eliot's paper is earlier actually one year earlier so from i because one two n minus the over three x i squared plus two c some time less than j that x times j acting five and this is interesting now we have some set of

57:30 and set of X's. So, we have set of X's goes X1 to Xn and lambda goes lambda 1 to lambda n. Now, I wanted to explain to you that if I look on this theory in Hamiltonian formulas, I can prove, although it's not obvious, that the phase space of this theory phase space of Jans-Mills is T star to Cartan divided by Weyer's group and not a complexified Cartan you would think that the Fitchin model would enter here HC but the Gerasimov and I prove that this theory is a deformation of the theory at C goes to infinity where T space is Z and basically what it says the phase space of this theory and two-dimensional are the same, but Hamiltonians are different and real functions are different. And that's the statement from equilibrium homology. I just thought it was, interestingly, when C exactly equals to zero, there are extra elements in equilibrium homology. Suddenly they pop up. But if you consider for C non-zero and that takes it, they don't. Exactly like in the computation I just said. So for C non-zero, these spaces are the same of Cartan of course the Cartan eigenvalues all classified will be our wave functions will be wave functions of Cartan so there are X1 to Xn and we have periodistic condition and then there are momenta lambdas and as these guys prove to this Hamiltonian in a particle sector becomes just the lambda I square acting on phi lambda X but the funny thing is there are many Hamiltonians this is the integral of several as many home components as degrees of freedom and all of them are the powers of something called the dunkel operator which enters in the description of the center of the headcount now i don't know how elliot libel the people who were doing that before i was probably born managed to prove such things and i'm not expert of that side but the statement is extremely powerful they know everything so they saw completely discernment and well I learned modern description of which has

1:00:00 this but nowadays of course they did know it they actually wrote everything but it's on the business for us it will correspond for the case of this now from the gates of a point of view I'm looking for a week's thing that's how we also look at quantum field so as i just explained its phase space is final dimensional it's a wild rule i have to make sure that i have symmetric functions on t's power of h h is a carcanta group and so it's new one to the power n that's how you want count but now i have to look on cycles on the human surface with those who want and then i have to pick the right wave function and from the quantum field or point of it will correspond follow I have to take a disk and not a human surface I can take any human surface if I want but that's now it's getting more interesting and put the operators from the spectrum of this equine to homology in the middle then can't you pass integral results in certain for every operation okay I'm not even in the middle so but I will wrap it up in one No, but I, look, my teacher, I think I have more than 10 notes prepared to make, I have 26. I think our question is the other way, maybe if we could spend more time. I don't want to see that people are living in this room, it's so depressing. Anyway, anybody can leave from now. So what I have to compute, I have to compute plus integrals now. So basically what I said, this paper of Moon and Grasso and Man, the starting point, and unfortunately there is a long walk, and it took us 60 pages to start from there and get where we are now. I have to put every operator here. First I have to prove that the cohomologies are completely the same as in two dimensional yard mills. So that Q equivalent ones are traced by zero to the power of N. So we're telling you this was the ordinary one else. So we prove it for arbitrary C, it's correct for C equals zero, there are extra ones, but we don't care. We define C equals zero severance as a limit of C goes to zero. Then we have to take this operator and insert here and compute the wave functions,

1:02:30 as a function of boundary condition and we argue, if you wish, that you get those bytes that this guy throw. So it's first of all, the first thing to check is that there had to be as many variables or such. And as I told that, n here and n there. Then there had to be exactly n momentum which is trivial over there because it's this time. And here there n momentum. Then, you have to show now that these wave functions are eigenfunctions of this Hamiltonian, this one. That's difficult. To derive the Hamiltonian is difficult, but you can check that those guys solved this equation. And this solution that the people wrote, the great guys wrote long ago, I mean, good books on this are, I mean, I looked on Gaben's book, and I looked on, and they have all the description, actually, this function I had here, the alpha one, is first derivative of the Young's functional, which counts the critical points of the charts, the solutions of the account, but there is some very nice stuff. So basically, once we prove that the, not only spaces are the same, but the basis of wave functions coincide, then you have one-to-one identification between upwards. The funny thing now is that if we have two-dimensional Yang-Mills, that some of the people are experts, like Sanjay there, who wrote, I think, thesis on the two-dimensional Yang-Mills, the trance phi zero to the power n in Young-Mills' series in x of Young-Mills becomes just the nth power of that. This operator becomes very important in ordinary in x of Young-Mills. In this series, from the solutions we see that they get the problem very specifically like this. For example, someone would say, can we deform some different way? I don't care. This particular pass integral, they form it like that. one for example one operator which is called under the name of dunkel operator i'm sure there are which is described completely in terms of the symmetric group and differential operator the di is d over dx i plus symmetric x i less than j so j is from one to n of a step function x sin minus x sin minus 1

1:05:00 and this is a basic operator which sits in the center of this Hecci algebra its n powers completely coincide with the operators we just diagonalized so these Hamiltonians are what they call the center of the Hecci algebra and then there are this algebra is very big this is a way of describing Yangian so we have too much evidence that the Everson comes from and the mathematical proof would be the good mathematical proof would be I take that operator and prove that its quantum version is this instead what I prove is that that partition function with the data and cohomology has no other choice but the path integral solves this equation not very direct proof but it's still a proof okay so now I don't just write for you the solutions exact solutions and normalization so my first conjecture actually was at the top this thing that I called B of lambda which was entering the sum and the dimension of B I first thought that it was a norm of the wave function of the of the norm you are shedding yourself because they are normally not normalized there is no reason why should we think that this plus integrals will give us a normalized wave moreover we know that they're not normalized because if i take this another disk and i glue them i get a sphere and then sphere partition function i already completed i completed it for everything for example for a sphere so moduli square of that wave function summed up should not be one or n whatever it should be something which is not trivial and so that suggests that the wave function there are not normalized so first I thought it's a no I looked in the Gaden's book and the norm is written there it's almost the same one up to subtlety and that subtlety that Gaden's one doesn't satisfy limiting procedure so this is kind of big stuff so lastly okay so the statement I made lastly good question would be what about the whole quantum world? So I get here only Andrian, which is a degenital. So what you have to do, you have to consider the rest of the linear within conformance in theory. It's g mod g and deform it with introducing the parameter

1:07:30 like c, such that when the level of cosmological goes to infinity, theory becomes this. That will introduce two levels into the business. Turns out you can do that, and we wrote that survey, which is generalization of these 40 vessels a week, and then you get the whole, the sequential of beta-anzas here gets displaced by beta-anzas for x, x, z, which has whole quantum group. We have proved that the wave function match, but we believe that they do. The lastly interesting thing is that, see, now we are prepared, we have the oldest equivalence, why don't we just do what's called in this business a non-duality? Because it's known that the reduction of the four-dimensional zero-dual equations down to the two-dimensions is one of the particular cases of Berners' non-duality. And in that duality, the group gets replaced by a language duality. So let's do it. We have complete answers. So it's very sketchy, but it's completely true and simple to do follow. Believe me, we would like to follow more in much detail. So the non-duality is about following the duality. Consider 4s1. okay s1 times s1 times s1 take this s1 for the radius r0 this for r1 and that takes them these two okay the story we started was when r0 go to infinity and r2 go to zero so then this becomes this is a story that we studied but these guys who told the better who told the norlinear will prove that they are the same This becomes R times S1, okay. This is the story I described for the trading curve. But this one, solutions of the moduli space of that just simple thing, is the same as S1 of 1 over R0 times S1, and this is not like S-dollity or something which you have to do 100 times before it's true. This is a number of simple stuff we know from many places. one over R1, S1, one over, I have to reverse all of them. One over R2, and now we take same limit,

1:10:00 so this guy disappears, because R0 was going to infinity, it disappeared, and I'm left with S1, one over R1, times R times R. These are called periodic monocles. Okay, so the integral we're studying is equivalent to this quantization of moduli space of periodic monopause. But the funny thing is, that in this case, what has to happen, the group has to be replaced, let's call it level-run duality. What happens is that the run here of the standard hormonal vector balance we're starting here was N, and instant-on number of the moduli as we're looking was K, the difference in this mass group. And also, when you replace it, like when so i actually think that the all ingredients which enter in the geometric language conjecture enter here i mean i'm not saying that i understand it i actually even don't know the whole ingredient enter here very well and the uh one side is completely established big subject of mathematical physics another side is a very established subject of people who do things out there, and they are completely matched in this example. Thank you very much. And let's have the next quote at 335. Sorry, excuse me. that's okay

1:12:30 I know that's the face It works. You want to study the phase, then you will start. Because hydrogen is a space space. OK, we can prove other spaces. Space of classical solutions. So it's just half a lot. That's why. And I'll see you next time.