OK. So, yesterday we were discussing n equal to two three-dimensional theories on these backgrounds, sphere, and the circle beta square. And the other crucial ingredient is a background field for the asymmetry of the theory. OK. I could replace this guy here with a generic Riemann surface. I will not do it just because the metric is ugly and so on. But everything goes smoothly everywhere if you replace the sphere with a Riemann surface. I show you the Lagrangian. And I say that it was pretty boring. And Samir made a lot of jokes about it. It's pretty boring because it's almost like the one in flat space. But there are a couple of things that happens. And one is there is this coupling, for example, for the scalars to the curvature of their symmetry. That's typical in curved space. You get the phi square multiplied by something. This is my, actually it is a very mild modification. But there is something hidden in the Lagrangian. When I say that it was the same as in flat space, I meant that it looks the same as in flat space. But now we are in curved space. So first of all, whenever there is a covariant derivative, maybe not for the scalar, but certainly for the fermions, you have to put the spin connection clearly. But not only the spin connection. The point is that you turn on a background for some symmetry. So if you have a chiral field scalar, which is charged under this guy here. So if this guy has R charge, R phi, this derivative is not the mu of phi. But it has an extra piece, which is R phi, the charge, times this guy here. So all covariant derivatives are also covariant ties with respect to the background.  And in a background that looks like a monopole for their symmetry. And if there are also flavor symmetries, we will discuss later, you have a similar picture. So for all background symmetry acting on the phi, you put a background that looks like a magnetic background. And so the important point of, well, this is very similar to what Francesco was discussing yesterday. Well, yesterday or the day before. I don't remember now. Most of the term that I show you are actually q-exact. For example, you can up to coefficients, write the Yamlis Lagrangian in terms of the q-q tilde of something. This is lambda dagger lambda over 2 plus d sigma. And similarly, also the matter part for chiral fields and so on is q-exact. And there is a similar expression here with bilinears in the fermions and something else, 2i phi dagger sigma phi. So most of the Lagrangian is q-exact. And I can use the combination of q and q tilde as a localization supercharge. And since the square of each of these guys is 0, this is the q tilde of q-q. Well, I don't know how to call it. Let's call it q tilde just so that I can also pronounce it. It's the q tilde of something. This is very similar to many other examples in localization. The only term that is not q-exact in the Lagrangian that I show you yesterday is the term Simon's term, which is roughly e d e plus completion. And then there are terms like lambda dagger lambda plus sigma d. So the only place in the Lagrangian, which is not q-exact, is the term Simon's term. So the only piece that will contribute also to the classical value or the classical part of the localization is the term Simon's term. Yesterday, I haven't written the interaction for the chiral fields in n equal to 2 supersymmetry. You can have a superpotential because, again, you can check that superpotential terms are q-exact. This is a general feature of most of the localization in gauge theories in various dimensions. The superpotential, the explicit form of the interaction is often of no interest for localization. It's important that there is some interaction because it tells you what are the charges of the fields and so on. We will see an example. But the explicit form, if it is cubic, the coefficient, and these kind of things, disappear completely in the localization business. This is quite common. So to localize, what you can do is, well, you have this Lagrangian, Young Mills, Matter, Charon Simon's. Since these two terms here are q-exact, you can put coefficient in front of them. Well, let's call this one over g-squared Young Mills coefficient, just a second. And you can put an independent one here, or even equal. It doesn't really matter. And then you send the coefficients to zero. This term of q-exact, the pat integral is the same. And the crucial point is that everything will localize on the zeros of this term here. Yes? No, usually, well, in this, it could be that in some kind of localization, you need to be careful. I'm not sure we would need Francesco. We can ask Francesco later. Perhaps in what they call X localization, you need something. But in general, well, in the one that we're considering, the background for the chiral fields is zero, so that the superpotential is zero. Most of the time is like that. Could be that in some kind of localization, you need to be careful. But usually, it's not a constraint. Maybe it enters in some coupling. OK, so that's the logic. I'm not doing all the computation. I'm giving you the result. You can find the explicit computation in this reference with also many examples. Let me take it. And on arbitrary genus, you can find here the details. OK, but instead of giving the details, I will tell you what is the structure of the result. As usual, localization reduce an infinite set of fields to some matrix model, roughly. So you need to specify what is the relevant matrix model for this particular twisted compactification of a three-dimensional theory. Now, the result is of this form. I will write the ingredients, and then I will explain all of them. So there is a sum here. I will tell you about what. There is an integral. As usual, it is an integral of something. There is some integrand, which depends on a variable over which you integrate. Yeah, let me write this way. Where? This is not particularly important. This is the order of the vile group of the gauge symmetry you have. Let's call g the gauge group. This is a sum over a set of magnetic fluxes on the Riemann surface. I will be more precise later. This is a particular integration contour that I also discuss in a moment. And this is, in this case, is a meromorphic function of a variable, of a complex variable, that I'm going to define in a moment. And also depends on this magnetic flux, use a mover. Let's analyze all these ingredients in order, starting with the variables that remains after localization. So this is usually the integrand here. And possibly because this is often the case, essentially typically here you have a remnant of the gauge degrees of freedom. Very often scalar fields in the gauge multiplet remains as a zero dimensional object you have to integrate over. And in many different situations, as in Francesco example, as in Pestun localization, you have to sum over topologically nontrivial gauge bundle, meaning monopoles on the Riemann surface in this example, instantons in Pestun localization, and so on. So there is a sum over topological sectors. And these guys are specified by the localization locus. So the first input is what are the saddle point of in this localization. As Francesco was discussing, if you are good enough and you find a good deformation term for your Lagrangian, and this is the case, everything boils down to the zeros of the variation of the fermions. So this is the BPS condition. You take the variation of the fermions. In this case, it's enough to look at the gauge fields. The other are simpler. That gives essentially zero for the chiral field. The nontrivial one in this example come from the variation of the gugino that I showed you yesterday and is of this form here, where epsilon I was using yesterday, a spinor that has only top component. So you need to set this to zero. And OK, you easily discover that with this spinor, you have to set to zero independently. This guy, this guy here. And essentially, then you analyze the possible solution of this equation on the sphere. You discover that, essentially, the result is that sigma, you see that there is a derivative here, should be a constant. Then remember that we are on sigma g times s1. S1 is pretty flat. So nothing is going on on s1. So whenever you have a circle and you have a gauge field, you can turn on a enronomy, or a Wilson line, call it as you like. If you want, it's the time component of the gauge field. Looks like a scalar. And again, this is not constrained by this equation. For the simple reason that here only the derivatives appear in the gauge curvature. So the Wilson line is always in this game. If you have a circle, it's typically a zero mode. And then you discover that you have to set to zero this. And this, here we are in three dimensions. So this is along the sphere as component 1, 2. This is gamma 1, 2, which are sigma matrices. So sigma 1, sigma 2 is sigma 3. You use this condition, and you discover that the curvature here along the sphere or Riemann surface must be equal to the d term. And as usual, these are gauge degrees of freedom. So if you have a non-abillion gauge group, our matrices. But typically using gauge transformation and solving the equation of motion, you discover that you can diagonalize all these guys. So all these guys belong to the mutually commuting. So I can restrict these guys to the Cartan subgroup. And this is, again, typical. And this also explains why then you have to divide by the vile group. You know that if you, with a gauge transformation, bring the fields in the Cartan subgroup, then what remains of the gauge transformation is precisely a discrete group, which is the vile group of g.  So what's about these guys here? These two guys are real. I can combine in a complex guy. I call it to u. And this is this variable here. I put a factor of the radius of s1. This is the only place where this beta enters. Once you put beta in the definition of your guys, it will disappear everywhere. As I told you yesterday, there will be no trace of the metric or radius of the Riemann surface and also of the circle. So this guy, this is the variable that I will be using. And whenever you will see a u today or tomorrow, is this particular combination. And what's about the second condition? Well, this is a gauge field on the sphere or Riemann surface. So this guy is quantized because, again, Dirac quantization conditions. If you integrate, let's call it f, let's do on the sphere, which is simpler. On the sphere, you divide by 2 pi, you should get an integer. Well, to be precise, what you should get is, so essentially, you integrate the guy over the sphere. And if you want to be general, but we will not need it, the Dirac quantization condition in group theory term tells you that this guy should belong to the corout lattice of the group. It is not really relevant for us. In the case of u n, what is the corout lattice is z to n, which means that these guys are the diagonal element of the joint of a matrix in the algebra of u n with integer entries. So the logic is that this is an integer. The right condition for the quantization is that when you exponentiate this guy here, tote as a element of this lattice, you should get the identity. And to be precise, this depends on the matter content. This is in the corout lattice. So you apply this operator to all the matter that you have, and the result should be integer. There are subtleties, depending if your fields are only in the joint in the fundamental and so on. But none of this will matter for us. OK? Yes? I think, let's see, Leo, where is it? Did it? Yes, it works pretty much in the same way. Yes? So in an uncomfortable case, you do not have this condition, because it's compact. So you have, so m is in the continuous integer. M in principle is continuous integer. But then with the spectrum, you will get a conditional m. But that's later. OK. So you also have to speak the spectrum in the continuous part. And there you will see. I think I can start. Yes? I think if you fix them, then it's pretty much similar to what I will get. So the other technical point here is that this is a Wilson line. So it's along a circle. Wilson line are periodic. And, again, the right condition is that this is 2 pi times root. OK? Roughly speaking, it should be an integer up to 2 pi. The reason is that the physical object is the olonomy, not the Wilson line itself.  Let's do a u1 just to understand what's going on that you can generalize to other situation. Sigma is non-compact 80 is compact. So this is a cylinder, essentially. So most of the time I will work with this variable here, the exponential of this u, with this exponential map, essentially you can transform the cylinder into c star, OK? So x essentially lives in a complex plane, u on the cylinder. OK, so this is the BPS locus. So you see that there is an integer, m, and a complex scalar, u. And these are the guy in the integrand. Now, what's about the integrand? You do the localization. You compute the value of the classical action on the saddle point. You compute the one loop determinant. And in this particular case, there are various subtleties that I'm not covering in the lecture. You can find in the paper. The subtlety will tell you what is the contour. It's a delicate story. But luckily enough, as usual, for what I need to do, the large limit, all subtletly, usually in the large limit, goes away. So I will not discuss all these fine details here. I give you the form of the integrand. I give you for a generic sigma g. But let me start. Well, let me write it. Well, let me write the back. Then I will correct term by term with subtlety. So the main thing to know is the following. This integrand has a classical part. There is a one loop determinant for the vector. There is a one loop determinant for the chiral field. On the sphere, this is the end of the story. At arbitrary genus, then I will put something else in a moment. I will tell you what it is. But just not to complicate the original formula, I will tell you this in a moment. So the classical part comes from the chiral simons and is roughly this one. So x, remember, is from now on x refer to the exponential of u and u is the variable in the integral. So if you have just a v u1, this is the result. xk is the chiral simons level and m is the magnetic flux. Why is that? Well, it's easy to... Well, it's easy. You can understand what's going on. This is the chiral simons roughly. So on the sphere or on the surface you put a flux, this give them. Once you have selected the coordinates on the sphere, this is on time, and on time there is the Wilson line. Then supersymmetry is intelligent enough to complexify in the right way everything. There are other terms that transform at, remember that at enters in u this way. So there is also a term with sigma. So you see that you get at times m. So you get something e to the u times m.  For a generic group g, you will multiply the same factor. So for each entry in the cartan you will exponentiate, you will multiply this. So this is the only possible classical contribution. Only chiral simons. Remember that he amels and the matter part of the Lagrangian arc is exact. The vector one loop is the following, product over the roots of the gauge group. Then we write in a explicit example later on. And actually one way of writing it is thinking of this as a meromorphic form and putting the integration variable here. So that you are integrating a form. What is this? As we saw in Diego Letcher at a certain point is what is called the Van der Mond. It is very typical when you have a non-habilian transformation. So we will write probably later today or tomorrow in the case of UN explicitly. But it is easy to construct in terms of the data of the theory. Every chiral contribution at one loop, every chiral brings a rational function. So suppose you need to write the expression for a generic representation. So this is the same notation as Francesco was using. So there will be some representation. I denote with raw the weight of the representation. These are just the charges of the fields of the gauge group. And each component essentially give you a rational function of this form here. So essentially for u1 fields of charge 1, u1 with charge plus 1, the expression is the following. The rational function that appears something like this. There is the gauge variable square root 1 minus x. There is a power here, which is the magnetic flux. And if you have representation in the joint and I will show you an example, you need to put the right representation for this. And there is a contribution that comes from the R-charge. So there is a particular exponent here that I am going to explain in a moment. And remember that in this game as you see here, it's better that your r symmetry is integer, so that you have a rational function to some integer power. What's about the term that is missing here? Well, there is another term that you need to be careful if you are on Riemann's surface of genus greater than 1 because Riemann's surface is of genus greater than 1. So there is an extra set of fermionic zero modes that you have to deal with that are not there on the sphere. And essentially, all this contribution can be rewritten in terms of these guys here. Let me call this guy Zeta tilde. You need to take the second derivative of the logarithm of Zeta tilde in respect to u, the variable u, and the variable m take a nice determinant and put it to some power. So this is the contribution of fermionic zero modes to your tilde. Third contour C. Now, supersymmetry tells you where you have to integrate. Here the story is actually slightly more complicated than this. When you do localization, you start with an integral which is an integral on the complex you plane. So it's an integral over u and u bar. Then if you look at the integral and you put all the pieces together, you discover that the integral is the del bar of something so that you do the first integration over, say, the variable u bar and you get something holomorphic. But in this operation you need to do it with care because the integral of del bar can be zero. So you need to regularize your construction. And when you do everything with care, the final result will be just the integral over a contour in the complex plane and supersymmetry will tell you where the contour sit around which singularity of the contour sit. Because the integral is a holomorphic function, then you have a line integral to perform and you will do the integral by residue theorem. So it's important to know where the contour is. You need to know which poles are inside the contour or not. And the result is very interesting because we have no time to discuss it because the result is something that is not a mathematician. It's the so-called Jeffrey Kirwan residue.  Typically, when one sees a metric, if you have some experience with metrics model, the Gaussian metrics model that Diego was discussing and so on, you have similar structure. You have a Van der Mon integral and then again you have to do an integral where you do the integral in standard metrics model where you take the unit circle is the natural guy so you have a bunch of variable, complex variable and you do the integration of the unit circle. Well, in this case it's not that. It's not so simple, even if there is a Van der Mon. If you want to look at the details of the Jeffrey Kirwan residue, I refer you to the paper and I'll show you an example later. I will tell you how it works in practice. Now, well, actually maybe it's better to write this an example to start with and then discuss the interpretation. Yes. Integral should be an integer. Yes, you can see. I'll show you in a moment what I will actually compute is a generalization of this of fugacities. In that case it's not an integer but it's a generating function, yes. But, well, you can check that this generating function is a rational function when you compute, so you expand in power series and the coefficients are integers. It's not obvious, but it's the case. So let me give you an example before proceeding and the example is super QED. Super QED you can realize in N equal to 2 in three dimension as the theory with a gauge group U1 and two chiral fields electron and positon if you want and partners with charge plus one and minus one like electron and positon. So in this theory you have two fields you have the gauge symmetry and these guys have opposite charges and what kind of are in global symmetry you have? You need an r symmetry an integer one, I told you. Well, how many r symmetry there are many because super QED there is no super potential is just electron and positon to the gauge field and fermions so there is no super potential there is nothing that tells you what is the symmetry you can choose whatever you like for example, I can assign the same charge to Q and Q tilde and define an r symmetry with integer entries the charges of all the fields are integer this is an r symmetry which means that if Q as charge one is partner which is a fermion as charge zero r symmetry in the addition there is a global symmetry that in this table looks a little bit weird because it has the same entry there is an r symmetry which is just the symmetry that transforms with the same charge electron and positon in this notation with chiral fields and this is a global symmetry meaning that I transform Q with charge one and this partner same charge so the difference is not visible in the table but is a difference one is an r symmetry and the other is a flow symmetry with this ingredient we will come back to the flow symmetry later with this ingredient the object that you have to compute is the following there is a sum over magnetic flux sum over an integer in z you perform an integration over one variable a contour integration and here you will put an expression like this suppose that we consider the case g equal to zero I will take this expression here m I will set g equal to zero I will set r phi equal to one so for convenience so this guy goes away so the only thing that survives is m and this is the contribution of q and the contribution of q tilde is pretty much similar is this one so instead of x you put one over x this is the role of the rho here is the charge of the field and again or even at exponent instead of m you have minus m because the charge is opposite and then you need to select a contour a contour of integration and in this simple case essentially jefriki one in this simple case is again pretty boring because it tells you take the poles of the fields roughly speaking is more complicated than this but if you have an intuition of what jefriki one tells you you look at the charge the gauge charge of the fields and you took the poles of the fields with positive charge and not the other so you need to do a contour of integration here is is not immediate to see how you need to do it but essentially you take the residue of this and not the residue of this one and you can perform this computation we will come back to the example in a moment ok so I won't briefly to discuss two interpretations of this result so interpretation number one among sigma g times as one sigma g is compact you may think that you are dimensionally reducing to a quantum mechanical problem what kind of quantum mechanics while in principle it is simple you have a three dimensional theory on the sigma g there is a magnetic background so you will give a lot of Landau levels for your fermions and boson you have a complicated theory in a background magnetic field you can reduce to some quantum mechanics and what is I will tell you later when I put also a flavor fugacity but the result is when you put the flavor fugacity I can anticipate this is a very simple model and there is a duality symmetry in 3D that tells you that this theory is the same as the so-called XYZ model without gauge field with an inter cubic interaction so the result is that of essentially three fields is the product of the contribution of three chirals ok I can write it for you when I put fugacity is more transparent so yes, the point is that you start with a two dimensional integral but then you discover including everything including fermionic zero modes that this integrand is the bar of something and then you eliminate the integration over the u-bar and you discover that the final result is allomorphic in u so it is a contour integral this one that you have to take so let me put a circle here to be more clear no XYZ is a model of three chiral fields no gauge interaction and the cubic is called XYZ because this is the superpotential it is a Vesumino model with the superpotential XYZ it is not a spin chain so in this picture what are you computing well as we already discussed we are computing the with an index of these complicated quantum mechanics with many Landau level and as you always mentioned this is supposed to be an integer you will tell you what are what is the number of vaquas of this quantum mechanic and then to make samirapi again let me add that this is pretty boring it is ok but it is not particularly interesting things become interesting for what I am doing when you add flavor symmetry remember for example you have a flavor symmetry transforming the fields the simplest way of thinking about flavor symmetry in this game of localization is the following you can always think as a flavor symmetry of something that has been a gauge field which is not gauge so if you have a flavor symmetry you can always gauge it the flavor symmetry so if you gauge the flavor symmetry it becomes just a gauge field you already compute things for the gauge field the only point that is a flavor field symmetry has no kinetic term so it is a background gauge field that I wrote before is still valid for any gauge type of object so the point is that you think of a one flavor symmetry as a one background gauge symmetry and then since since it is essentially if you want every flavor symmetry has to do with a vector multiplet this is the same spirit of say you couple to some background field then you can turn on something in the preserving supersymmetry some component of the background gauge field preserving supersymmetry and what component you can turn on well, since it is a vector multiplet what you can do is for every background flavor gauge field you can turn on an integer again a magnetic charges this is background not gauge means that is not integrated over in the pat integral which is a magnetic flux and again this is essentially in simple situation is just an integer and similarly to the gauge field you can introduce a combination that I call UF of two parameter let me call it delta F a Wilson line for the flavor for the flavor background field and there is also the other guy which is the parameter sigma for the background flavor symmetry remember that yesterday I stress the fact that sigma appears in the Lagrangian as a real mass so it give mass to to chiral field boson and fermions and whenever you have a flavor symmetry a background value for this sigma F is just a mass for your boson and fermion in the Lagrangian so this is a real mass so what I have to do when I introduce this this background so the only thing that I have to do in the previous formula is to take the chiral one loop determinant and write it not only as a function of y of x but also as a function of a new variable which is conjugated to this so again I exponentiate this UF and I call it the result y and x is the exponential of the gauge variable gauge and background but the result is mostly obvious is like having two gauge groups so this is raw belonging to some representation of the gauge symmetry you have the gauge part as before but then you have an extra background U1 you put an extra factor of y for the background you put actually one over half one of here you put a y here and you exponentiate with the following exponent before there was the magnetic flux the gauge one now if there is a background flux you add it for the background gauge field and the rest is as before g minus 1 r phi minus 1 so this is the only thing that you have to do for example in this example here I use the gauge variable I can use a background variable and the difference is just the choice of charges so the gauge symmetry give opposite charge to the two fields the flavor symmetry give the same charge so the replacement that I have to do is simply this one I introduce a variable y here is at numerator here is also numerator because this charge is positive and I also introduce a background flavor symmetry here and here you see that the gauge enters with opposite sign because of this condition the flavor symmetry enters with the same sign and now you have no more an integer but the result will depend on y which contains this online and real mass and on the magnetic fluxes and tomorrow for the black hole this y will be the chemical potential associated to electric charges this n will be the magnetic charge of the black hole the other variable x and m are integrated over these are the gauge variables and now the result is numerator and integer is a function of y and n the interpretation in terms of the with an index well essentially what I'm introducing with y is a Wilson line so the effect in the trace of a Wilson line is to introduce if you have a global symmetry and you call the charge the operator associated with the charge j you couple to the Wilson line you see that this is really a chemical potential and you get the function of delta f here and what is this this function now is no more an integer and is no more an integer but as we will see in example this can be expanded in power series in y y remember is this exponential of e to the delta f plus i beta f is a complex variable and what's the meaning of this sum when once you expand in power series it's pretty much in your lecture this is just a number of vaqua with charge n ok so the point is that you have now quantum mechanics you have a global symmetry you have a set of vaqua you are counting all the vaqua but you can specify what is the charge under the global symmetry of each of them and you get a generating function each coefficient will tell you what is the charge of the guy I shouldn't call it this n because I already use n for something else and I call it p ok no p I was using for magnetic charges q I was using for electric charges let's call it a name for a charge ok and you can see explicitly I have no time to discuss it looking at the super algebra this object doesn't obviously look holomorphic but you can check explicitly using the super algebra that the result is holomorphic in y ok and certainly this guy here is holomorphic in in y ok so so this is the quantum mechanical interpretation let me add something again I know I am not entering in detail so I am quite beyond schedule but if you are interested you can check in the paper there is a natural interpretation in terms of quantum mechanics of this function here and exponent this function here essentially if you reduce to quantum mechanics and you look at the type of multiplet that you have, you discover that you you can define two type of multiplet they are called chiral multiplet and Fermi multiplet this is typical of two and one dimension Fermi multiplet is a funny guy that only contains fermions fermions and auxiliary field you can have those in two dimension and one dimension while the chiral multiplet is the standard one and essentially if you evaluate this trace for chiral multiplet boson plus fermion you will find an expression that is roughly this one if you try to evaluate this guy here I call it why the final symmetry because this guy intuitively is summing over a fox space of fermions ok, this is very typical in two dimensional field theory if you do the same for for a fermi multiplet that only has fermions this guy goes up and up to the square root of y this is just the vacuum and the presence of a fermion ok, so you can get these two explicit expression from evaluated this trace here in a quantum mechanical system with this symmetry and you see that this guy here is this guy essentially if you set y to zero this square root just an overall charge that you put in your Hilbert space doesn't really matter and this guy here is the inverse of this and the point is that in this picture the exponent come from the following reason you start in 3D multiplet, some are bosonic some are fermionic and the number of multiples that you get you can compute with some index theorem you use Riemann rock and indeed this exponent here come from the Riemann rock theorem you have a charge on the sphere on the Riemann surface you use the index theorem you count how many zero moles you have and you will discover that you get this object here this guy can be positive or negative you get a chiral field when it's negative you get a fermi multiple so there is an operator operatorial interpretation of this form if you want more details you can look at the paper so this is interpretation number one quantum mechanics interpretation number two is also interesting and it's the way in which actually we will do the computations tomorrow and interpretation number two is somehow the reverse so let's take another point of view I have a circle and I have a Riemann surface I can also go down to two dimension there is some interesting two dimensional physics that you see in this game yes there is and maybe some of the talks next week we will deal with it there is something that can deal with this general structure but let me show how it works what is the relation in a simple example let me go back to this guy here you see that here you have an integral which is a rational function you have to sum over all m's but you have to take the poles of only of this piece here this one obviously when you find poles here well m should be large enough otherwise the denominator goes to the numerator so the pole is the zero of this object here so we will find poles for m greater than minus n for m large enough the point is typically is always like that half of the m those that are less than minus n doesn't matter in this integral n2 doesn't contain them and half of them instead you have to take this is a very simple example you can do the computation for every m and resum but another way of doing the same thing is to first resum over m look at how m it appears it appears in this way so it's just an exponent so you can rewrite this integral in the following form there is a sum over m there is x over 2 pi ix and if you put piece together you can rewrite as follows y minus x divided 1 minus xy to some power xy divided 1 minus xy x minus n to another power ok this is the same expression now you want to resum this is easy because you need to sum the guys that are smaller than minus m doesn't matter, you throw away the other one you perform in geometric series resummation geometric series resummation tells you actually you can do a geometric series resummation taking n number larger than n and say I sum some cutoff and I sum over all the m's greater than minus m the one that are smaller than minus m doesn't really matter because they have no poles so the result is the same so if you sum this guy obviously you will get something of this form the precise result doesn't really matter here but call this guy say p of x this guy q of x and perform in the geometric series starting with capital capital m you will get some powers of m from the geometric series the geometric series itself which is one minus this guy here and rest I just need the structure not the actual value of this result so the computation and Jeffrey Kuhner is always like that it tells you that in this complicated integral typically you have to select n number of fluxes that typically are sub lattice only half of them say the positive or the negative or some complicated lattice in the bigger one so you can always do this trick and resumm them and get an illuminator and then the result of this is obvious you need to find zeros of this function p and use residue theorem you can do it yourself in this case it's very simple and the result is obviously sum over all the zeros of this function p there is q evaluated at the zero and there is possibly a factor that comes from the illuminator because if this function is not trivial you need to take the first derivative it will be something like this but the computation can be done explicitly in terms of residue theorem and this is an example of a general situation because in the last couple of minutes I will tell you this because all this matrix model can be written in the following form so there is always this variable that e to the i, u there are many of them if you have the carton of the gauge group of dimension greater than one there will be an integral over these variables I'm suppressing the factor of 2pi i and so on just to give you the structure of this guy and there will always be something that is linear in the magnetic fluxes because these are always exponent and I can call the coefficient I can give the coefficient a name and I choose to introduce a function such that what multiply the magnetic fluxes is called the derivative of w you can do that and then there is some other stuff like in this example as always M you call it the exponential of this guy derivative of w the interesting point is that this derivative of w you can reconstruct from what I told you before the explicit expression can be written explicitly this w for just one field is something of this form there is k u square over 2 then there is a boring factor again pretty boring just to do is parity anomaly and so on so it's important but not for what we want to discuss here and then the crucial part of this guy that come from this you exponentiate this, you get a log the integral of a log unfortunately is a polylog it's called polylog 2 l2 of x is the sum over x to the n in general l of n of x let me use another variable k 0 infinity n to the k this is the definition generalization of the power series of log in particular log l1 of x is minus log of 1 minus x k to the k to the n yes on the right hand side of the blackboard, sorry sorry yes so there are the zeros of this denominator and what is this guy when we already encounter in Francesco Lecture you can check explicitly that this is the twisted superpotential of the 2D theory meaning what and then I conclude I can still couple of minutes to finish this meaning what well you see an example in Francesco Lecture here what is the two-dimensional theory is the compactification of a g times s1 so you have s1 you have many Kaluza Klein modes the idea is that if you integrate out all the matter field you get the superpotential for the parameter u that parameterize the gauge the gauge multiplet and how do you see that this is the twisted superpotential for the compactification of a theory well there is a some representation of the little function which looked pretty much like a one loop contribution you can rewrite this as the sum of u plus 2 pi n log of u plus 2 pi n over 2 pi minus 1 now this is not enough to give you a full explanation but typically the point is that when you integrate out a massive field you get log of u as a contribution and then you have many of these depending on an integer because you have all the possible Kaluza Klein modes on s1 so the final point of this story is that inside the matrix model this exponent here that you can always write sum and write as product over i1 minus e to the derivative of w with respect to i your matrix model can be always rewritten in this form and can be always evaluated in this way you take the residues of this denominator which means that this condition here can be written in a invariant way in term of two dimensional data of your theory in this way so essentially your matrix model can be evaluated by finding the solution of this equation so you take the twisted superpotential of the compactification to two dimension you find the solution of this equation here and essentially you take one pole, this correspond to the pole of the integral and the final result is a sum over the solution of this equation of a certain integral and that you can explicitly evaluate and there is some determinant of the second derivative of this guy here I'm not very precise we will not need the explicit expression but roughly speaking the structure is always of this form so the matrix model can be evaluated by taking residues these residues as a physical meaning these are the vacua essentially because these guys here is setting the derivative of the is setting the derivative of w is set to i not to zero but essentially to two pi times an integer but if you look at the literature this is the right way of looking for vacua in two dimensional theory there are ambiguities so these two pi's are just ambiguities and just to conclude let me tell you that this formula maybe is familiar to some of you because it's very old in the end of the eighties beginning of the nineties and then the solution of this equation has been used quite a lot by Nekazov and Shatashvili in their game of beta gauge correspondence goes back well the twisted super potential maybe not super symmetry is from the 70 it's the beta equation well the interpretation is that the vacua of the twisted super potential have the beta equation the set of zero corresponds to the beta equation of some integral system associated to the gauge theory integrability will play no role in the tomorrow computation but there is this interpretation you can relate to integrability you cannot solve it exactly no yes no no I don't want to solve exactly I don't want to take the larger limit of this set of equation so these are the two interpretations quantum mechanics with an index is one interpretation reduction to one dimension another possible interpretation reduction to two dimension is related to this integrability business it tells you that you can express everything in terms of the solution of this equation by the way last comment to finish there is a paper by where they show not only for the partition function of sigma g times s1 but they were able to use the same construction this beta vacua to give an integral to give a formula for all, well not all but many partition functions in 3D including the tree sphere and so on there is some trick that you can have the look to the paper of we'll let maybe closer we'll give some details in this seminar next week ok, tomorrow we'll discuss the black hole case it will be a generalization of this in the larger limit and sorry for being late


Yes, Nils. Mr. President, you, I don't have to stand about our problems here in New York as you know about them. I'm taking every step I can with state authority to help the little business man and the individuals who are really up against it because they're losing their business. And a little fella can't do anything. He doesn't get in cash and a lot of them are going bankrupt. And an awful lot of people are not getting paid even though his honor the mayor asked him to pay him. So I would like to send you a wire, but I wanted to speak to you first. And in the wire I would outline the steps we're taking in the state. And then I would say, in view of this major catastrophe, I request you, Mr. President, that you declare in New York City a major disaster area, that you take all necessary steps to make available to individuals and small business federal assistance including low interest loans to small business by the Small Business Administration. Two, guarantee of outstanding loans to veterans on both homes and businesses by the veteran administration. And three, highest priorities to loans to individuals and small businessmen in New York City by the Office of Economic Opportunity. Nelson, I think that what we ought to do is let me have the people who handle that to get back in touch with you and review what you want to send because it'll sure get out. And when it does, it'll create a problem for us if we're not able to meet it. It'll hold out hope that we may not be able to meet it.  Let me get our best man to call you after we get our small business, see if they have any possibilities of doing it. I would question that when we have a strike per week that we can write a new rule because if we did, we'd have had it out and never know our last week and would have it some other place tomorrow. But we want to help anyway and every way we can, but I don't know that we ought to do it by wire back and forth. Let me get to our best person. I'm sitting here now in a security meeting and they just buzz me and I knew nothing about the nature of the call and what it's about. I want to answer as quickly as I can. Let me get my best person and I'll have them call you as soon as the lunch hour is over. And see what facilities we have here that would be available without legislation, which we don't have, the Congress is not here. And then have a kind of understanding of what we can do without getting up people's hopes and you're recommending something that would make you look bad if nothing existed for it. I understand what you're saying. We have researched your federal laws naturally very carefully before security. I haven't though and I'm going to be the one that has to make the decision. My own impression is there's no funds in the small business. My own impression is that they're exhausted. And I don't know whether your research shows that or not, does it? That does not show, although I assume that emergency funds could be used. No, I think we have to wait until the Congress gets back in and they go through the appropriation channels of the House and the Senate and their hearings. I think the New Orleans thing was an unexpected emergency of such huge proportions. And they had to have legislation and I think the funds are pretty well depleted. But as you know, I'm not an expert on these individual veterans' loans and small business loans and just the extent of how much we have in each pocket. But I will, if you'll give me a... If I don't just have to... At the time it's after one, within the hour I'll have somebody going into it and I'll get back to you this afternoon. Well, now let me just make one more point on this. We have got a situation... You probably got a report from Willard Woods already. I got a thorough one from him. I've had it every morning and every afternoon. As you know then, as a result of that, at this moment there is no break in the clouds anywhere. Unless Willard's got something that I don't know about. Well, I think they had some things that they were uncertain about what would come. They had plans but they are in the process of being discussed and evaluated and he made a report to me, I guess, at 11 o'clock. And they have some plans and they have some questions as to procedures they'd like to get inaugurated and it's conceivable it could be agreed upon and if they got those procedures why that would give us a little time. The problem we face is with no really clear evidence of the course leading to a solution that we are facing the people that is not just the 8 million in the city, there's about 16 million people involved and this thing is a very... Well, I'm not telling you because you know it but it's extremely serious as far as what's happening and there are a lot of people who are beginning to get very upset, little people who are innocent victims of a big situation that's unfolding. I quite agree with that and it's been distressing and disturbing to me every moment and it hadn't been a day, it hadn't had an oral report and a written report. And when the mayor, we have said to the mayor, I don't need to review all that we've said to him that we would do this, that ABCDE, he goes back to his panel, he calls back, says I don't think this ought to be done now, the panel doesn't agree with it. So we said whenever you're ready, whenever there's anything we can do, whatever resources we have that might be helpful to you in reaching agreement, we'll do it. So he called me at 6.15 yesterday, said that he would like to discuss some of the strategy involved with the Secretary of Labor and I said well he will be asked to leave within the hour. He caught a 7 o'clock plane, he went up, they did go into it late in the evening. I think before he left they had made some arrangements about some of the parties getting them available for the conference and things of that kind and out of jail and so forth. And words tells me that they carefully explored various possibilities and one of them is in motion today. Now what comes from it, I don't know but anyway the mayor is carrying it out as he thinks it ought to be and I want to give him any assistance that we have here and I want to give you any but I don't want to exchange him wires on front pages of papers when there's not any hope and I sure don't want you to move a situation in here that can be better handled by the responsible and selected and elected officials there. We want to cooperate with him everywhere in the world we can but I don't want to as a kind of a conservative, moderate Democrat I believe in your Republican principles of state's rights and local self-government a little bit and I want to cooperate with him and supplement him and not supplant him. Well that is exactly why I called you instead of wearing you. That's right and I'll get my people busy and we'll get back in touch with you. And we will take all the steps we can at a state level and you know of all the governors there's nothing that has got to criticize more for doing things and there's a certain senator from this state who keeps criticizing if they're not getting enough money out of the federal government. Well don't get me, don't you think I've got enough problems with it now? Without getting in your New York personality but what I will do is this I'll get our people see what it is that we might do and I'll have words or somebody get back to you within the afternoon and we'll keep it absolutely under the cover so we don't build up people's false hopes and make them think it's Sanies down here that can take care of these things but we'll let you know what can be done and then we'll decide who our people ought to meet with to do it. If I could give whoever you designate to this if I could give them some thoughts on the phone it would be very helpful because I can point to the areas that we've explored that might be useful. That'll be good. I'll do that and I'll have somebody call you after. Wonderful. My best happy. Same to Mr. Johnson. Thank you a lot.