 So I will continue the lecture I gave before. It's about this one, four-day necromancer S2-S1. It's called Supercoma Index. This is actually the overlap thing with Guido's lecture, but maybe this is the last lecture in this week, and you are very tired, so maybe it's okay anyway. Okay. So I will start. Okay. So I wrote down the suji algebra on S3 cross S1, and of course this manifold, so this manifold, S3 have SO4, so there is SO4, SU2 cross SU2, so it's SO4, and there is S1, so there is a U1 direction, and this SU2 contains J3 and this also J3, and I will call this L and R, and for the U1 direction, a genetized maybe H is Hamiltonian, it's more appropriate, but I will denote it's a D here. Okay. I will choose the generalizing spinners are satisfying this equation. I mean, it's related to this three direction. Okay. We can choose that. Then R charge of this one is minus, this is the same, but then this is a spinner, so I mean delta, spinner, so it's have a one half momentum, I mean angular momentum here, and delta epsilon bar is R charge one, and J3 have minus one half. The interesting things are good stressed, is this both of them are right one, not left. Okay. This may be strange, but actually we construct a Suji on, I mean, for example, I mean, we can consider some u-be limiter of the this theory, and then Suji algebra becomes a flat one. Okay. So then for that case, SO4, I mean, this charge have left and right angular momentum, but here it's different. This is for the SOS3. Okay. So this is why we have this. We can check that. Okay. Now, because DT is called delta T minus, so this Q is R charge of the field, okay. Then, okay. So this means minus delta T becomes D minus, okay. Here, I mean, in a previous lecture, I write down the Suji algebra and this one, I mean, from the flat space to here, we just rewrite this delta T to this one, this combination, that is the only thing. That means that this is like a twist, and this is R charge, so sorry, I will call this R charge is capital R. Okay. So then the dilatation, this U1 direction generator is not T itself, but delta T becomes D minus this one. Okay. Is it okay? Then, partial function S equals S1 becomes this one. This is the user, just the equivalence between the first interval and the operator formulas. Okay. Here, this minus phase because we, sorry, impose the periodic boundary condition for the fermion on this S1. Then, this partial function is equivalent to this one. Here, usually, some of the partial function, it's here is a Hamiltonian, so D itself, but because of this one, this twist, this becomes, okay. So this is just a relighting for the first interval. Of course, there is some factor, because of the ambiguity of the percentile measure and overall constant, that is called Kashmir energy, but here, we think it is absorbed in the definition of this percentile measure. Okay. Then, we can do the further twist. It's delta T2. So this is possible because first, this combination, this psi and the eta is a parameter. Okay. M is something flavor-similar to generator. For the vector, much there is no such things, but anyway. This is zero. This is from this assignment. I mean choice of this one. From this assignment, delta is commuted with this combination. Of course, delta commutes with JL3, because it doesn't charge. Okay. Only the right one. Of course, delta commutes with flavor charge. Okay. So then, this is a constant matrix, a constant generator. Okay. So if we think the suji transformation and the supersymmetry reaction and replacing this dt to this twist, this then, this twist, I mean this replacing is consistent with the closure of the suji algebra and the supersymmetric action. Okay. So then, from here to here, then still we have the supersymmetric action on this one. Okay. And then, this function becomes changed because this one, I mean, this is the S1 and there is the Hilbert space and exponential minus H or something to here to here. Then, there is some like Wilson charge here. Okay. Then after the here, I mean and we give the trace here, then there is some insertion by this element, exponential of this element. Okay. Expansion of dt. So then, the partial function, I mean partial function with this twist is different from here to this one. Okay. So this is insertion and so x, q, so here small process d or Hamiltonian. Okay. Is it clear? Okay. So then, I mean, partial function is equivalent to this one in operator formalism. This is just a rewriting. And then, this is called super-conformal index. Actually, here we do not consider the super-conformal field story, but it's called super-conformal index. So why? Actually, if we consider the first super-conformal on flat space, then there is a user conformal map, conformal map to the S3 cross R. Okay. This is a user cylinder map. Okay. Then, there is a delta epsilon, delta epsilon bar here. Okay. But what is this corresponding in this one? Actually, commutator of these two includes this D. I mean, S1 direction, it's a Hamiltonian in this here. But D in this R4 is a dilatation because of this conformal map is non-trivial. Okay. So these should be, if this is a q and q here, then it doesn't give the dilatation, that gives a moment here. So this should be S here, because of S is a super-conformal generator. Super-conformal generator and user fan curry, suji generator, gives dilatation and special conformal transformation. So this map relate to this S super-conformal and super-conformal charge and user super-charge to this epsilon and epsilon bar. Then, we can calculate the h. Okay. So then the commutator of this one, I mean, we can compute it from here or some. Then, we can think, I mean, from this super-conformal R4, we can construct just with the index, from this Hamiltonian and this charge, like a q, q, like a q, q bar. Okay. So then this is called a super-conformal index. Okay. This is just a generation of the index to this super-conformal philosophy. Okay. But here, from the localization point of view, or philosophy on S3, S3 cross S1, we don't need to impose the theory is conformal. Okay. But if it's conformal, it's related to the disk-conformal index. So this is called a super-conformal index. Then, okay. By the change of the parameter there, we can rewrite it as, we can rewrite this one into this form. But here, of course, there is some, I mean, there's only a parameter p prime, but here, the q, x, y, but some combination becomes a proportion to h, and this doesn't contribute. So we can neglect that. So this is called super-conformal index, and this was given by the grid. Okay. So this is super-conformal index. Okay. Then, of course, there are two ways to evaluate this expression from the percent error or from this operator homoism. In operator homoism, we just count the number of the state. Actually, the theory is free theory. I mean, because of the localization, becomes free. So we just count the free gauge theory. So it's in some sense trivial, but it's, of course, not so simple, but we can count that. Or using a partial function, then that case, actually, the theory is free, but the geometry is three-cross S1. So we can, even at a free theory, there is a Wilson loop for S1 direction. Okay. So we need to integrate over the Wilson line. So it's called A, and of course, there's a gauge symmetry. So we take some A t equal constant, T is S1 direction. So then the Gaussian, around this one, we can do just a Gaussian integral. Okay. It's a free theory, so it's trivial. Okay. Then the expression is what it's like, it's like what Francesco wrote down for the two-sphere. So this is for the fermion, for the lambda, this is linear, so this is like a Dirac equation. Okay. This is the Dirac operator on S3. This is for the gauge field. Okay. So this should be for the vector Laplacian. You're fine. This is a, she is because we fix a gauge, so we need to cancel it as a ghost here. And prime means the zero mode is eliminated, because we integrate it over this A direction, like S2 case or S3 case. So actually, this is very similar to the N equal 2, so you see only on S3. Then it becomes, sorry, here, A hat equal beta A. Beta is the length of the S1 and this becomes, so this is just a fundamental determinant to fix this gauge. And I vector, this is come from this one. Of course, there are many, or my huge cancellation between the denominator and denominator. And the final result is this, this same. And for the chiral mass specter, there is the same procedure and we have these things for the psi and phi in chiral mass specter. It's just a scalar and vector, our spinner. So it's, I mean, spinner is this same, but the presentation is different. So then it becomes, gamma is called the elliptic gamma function. So this is what we do, right there. So this is just, I mean, sort of for the computation of this, this, okay. Then it becomes same as the index computation. I mean counting the number. It's used up pretty solar exponential or something like that. Then it's coincided with this expression, okay. So from this, we can compute a super-common index. And for the cyber duality case, if we choose the R-charge appropriately, then this is not a simple. This complex integral is a coincide for the cyber dual pairs. Actually, so this integration, number of integration is different. Actually, this is a rank. So it's an electric cell, it's Nc, but the real cell, it's NF minus Nc. So this integration is completely different, but there is some identity found by the mathematician and we found actual coincidence. So it is a very strong support for the cyber duality. Actually, cyber duality, there are not so much support before this index, I think. Actually, there's only two-foot matching condition or some string cell radio or something like that. But here, this exception gives infinite number by expanding the P and P prime, but it's coincide. So it's nice, okay. So this is about, so this is about the super-common index. And then I will move to the different one. I mean, computation of the gauge in a condensation. Okay, no question, okay. So this case, as he said, oh, this is for, this is dB is the whole root cause and madness. So this is, yeah, okay. So first of all, for the, I don't say about chiral mass spread, chiral mass spread V equal, if we choose this delta psi dagger as, I mean, satisfying this bound, then delta V bosonic is like, so the subtle point is the same. I mean, three-wheeler one for the chiral mass spread. But with this choice for the vector mass spread, the subtle point is unsafe their connection. But actually, I will move to the S1 cross R3, I mean, flat space. This case, V becomes the n phi squared plus f squared. So there is no, this term here. So actually, not phi cos 0, but dn phi, I mean, phi is cover-oriented constant is required, okay. And since the comment about the homo-phi, but so there's no super potential, our associate super, super field. There are no notion of super field, but actually we consider the, and okay, sorry, okay. Okay, consider this delta exact term. Then because the delta psi bar is epsilon bar, f bar for this choice, you can see this act on this one, because if we choose the eta bar as this one, then satisfy this combination, okay. So it gives just a del w bar, del phi bar times f bar. So it is just super potential term, okay, in a component, okay. So this, so this is like delta phi bar i, delta w bar times the f bar plus fermion, okay. So this is. This is what should be behind me. This is the bar. This exact, the delta exact term would go up there. Should I, if you keep it there for one hour? I don't know, I just, maybe it's related, but sorry, I don't know in the relation, yeah, okay. Okay, so this is a super potential term. So this is delta exact. Actually it's almost trivial, but here this eta bar where epsilon breaks low-range symmetry or rotation symmetry, but this result is not, I mean, rotational invariant. And if we increase the fermion also, it's a rotational invariant, okay. I mean. Can I ask you a question? To do this, my theory needs to be understood. Don't have one? I don't know, yeah. I mean, just to consider some function of phi bar and to consider this combination, then it's a delta exact term, okay. So this is the form of the super potential term. I don't need to start with the super potential to reach my equation. I'm not gonna get a one or two more. Yeah, yeah, yeah, two, two, yes, yes. Yeah, so this combination is, yeah. So this is, yeah, I mean, almost trivial, I'm not sure, but yeah, actually we can construct this one, I mean. So actually, so the homomorphic combination, I mean, anti-homomorphic combination doesn't give any contribution for this localization computation, okay. What do you mean by homomorphic here? A homomorphic, I mean, this is just a, yeah, I mean, phi bar, I mean, this combination. This only includes phi bar and no phi or something. Yeah, actually, yeah, I mean, it doesn't, I mean, anti-homomorphic variable, anti-homomorphic coupling or anti-homomorphic variable cannot contribute to these types of the computation. It's almost, in a flat space, it's almost, as a user, I mean, actually, yeah. Phi is in which representation of the column local? And this one? You have to be Gage invariant, I guess. All right, all right, all right, all right. Phi is in which representation? Which, yeah, any, any, yeah, it's a contracted, yeah, this is contracted. My most general Gage invariant operator I could only use phi. Right, right, right, right, right, right, okay. Yeah, so this is like a user super potential, again, but yes, yeah. But yeah, with us, yeah, usually we use a super field or something, but we don't, yeah, this is delta example. Are you saying that the anti-homomorphic parameter is in the special? Uh-huh. And only to, in that term? True, true, true. Yeah. I think that we can change such things. Yeah, that is a user's comment, but yeah, here, we- That's what you mean by form of form? Right, right, right, right, yeah. Yeah, we just wanted to understand the statement, so it is not part of the homomorphic. Yeah, yeah, yeah, yeah, statement is just, this is a delta example. So this type of the, this type, this type of the deformation cannot change the result. For the, for the, okay. So this is, yeah, user statement of the homomorphic. So let's move to the Gage invariant conversation in N equal to one. First of all, we consider super young males on S3 cross S, R, sorry, R3 cross S1. So this is, I mean, the computation is almost same as this David's at all, but okay. There's some, something we need to care about. Okay, so Gage invariant conversation is this one. Here G can be anything, I mean, any semi-simple algebra, but for definiteness, so I mean, for the simple notation, so in this case, Nc minus. We choose S, N, C, and then what I said, localization is just adding this terms, plus family. Okay, so then the theory is weak coupling limit on instant, or ASD, or instant, answer the connection. So this is, yeah, actually the classical computation on this background. So it's trackable, okay, and then this is by the Gross-Yafupitovsky configuration on R3 cross S1 are classified by instant on charge. So this is, of course, and Wilson loop on this S1 direction, we call it A0 direction. And finally, it's a monopole charge for U1R. So this U1R is because of this Wilson loop. This breaks G to the U1R, okay. So this is top. So you probably have a relatively simple expression for V right now. Sorry, R delta V, for the fermion, I do not write down, but yeah, of course it's a simple one, but yeah. Actually fermion for the fermion is lambda, lambda bar coupled. So it gives something lambda bar equal to zero condition or something like that. So yeah, okay. So we need a two zero mode, okay. A zero mode means a fermion zero mode because there is, I mean, it's a lambda, lambda, okay. But for example, instanton has a two Nc zero mode. So instanton cannot give C this one. But so we need an anti-safe dual connection with a two zero mode. It's called the fundamental monopole on S3 cross S1, sorry, R3 cross S1. Okay, so then it's a known, there is a R plus one fundamental monopoles. These are actually very simple, I mean, well known one. One is a T-der over PPS monopole, I mean, Tostopoliakov monopole. Or more precisely, it is fractional instanton. T-der means the PPS monopole is defined by R3, okay. R3, and there is a scalar, I mean, for the Tostopoliakov monopole. T-der means the phi can be considered as this A zero, okay. Just changing, I mean, that is called T-der of PPS monopole. I mean, S1 cross R3, I mean, yeah. PPS monopole is a gauge field, a mu equal one, two, three, and phi, okay. In a scalar, okay. Tostopoliakov monopole has a scalar. And we can, this is a non-trivial some configuration. And we interpret this as a zero. Then it's called a T-der over PPS monopole, okay. It's a, I mean, satisfies the PPS condition. I mean, T-der, I mean, this T-der is a, I would say, okay. Just an instant on an arctic of this one? Sorry. Is just an instant on an arctic of this one? Not actually, it's a fractional instant. Okay, I will explain, sorry. You mentioned the instant source. Did you just solve all the equations for the A mu in four components? You know, okay, so. Sorry, sorry, I'm just speaking. An equation of the, and so, have you just used all the equations for A mu? For A mu, yes. I can't do it. Yeah, for A mu, yes. Because if an average of the A mu. Yeah, yeah, it's, yeah, yeah, instant on, but it's, actually, it's R3. So it's non-compact. R3 cross S1, yes. R3 cross S1, so, yes. The typical dependence on S1 direction, coordinate, so. Right, right, right, it's. Because, as you said previously, it sounds like there is no dependence. Okay. In the way you said it before. I mean, this is for this one? Yeah. I, yeah, actually, yeah, there is a, and there's a, there's a R1, this is, and there's another one. Cause the Karatsa climate monopole is found by the, between E and the community. So actually, sorry. So TDR means, there's a brain picture. Maybe this is a, for SU3 case. And, for this is identified as S1. And there is a, for example, 3D4 brains. That thing, this one, okay. Then, instanton is a D0, instanton. This has a charge one. I mean, instanton charge one. So then, take a TDR. Then, it becomes like a D0. So D3, D3, D3. Okay. Because, oh, sorry. And, phi equal, for example, phi1, phi2, phi3. This is a Wilson line. So, in some coordinate, it's on a D0. Okay. So then, this D0 of TDR becomes D1. D1 wrapping this dual circle. But, because of this D3, we can move this D1 to the far away in R3 direction, or somewhere. Then, okay. So this is called VPS monopole. So this segment is a VPS monopole, okay. In this picture, actually, this gives a monopole configuration, user one, in a 3D, because of D3. We forget about the time direction. Okay. So this is because of monopole. This is also monopole. But, this D1 is called, this laps this S1. This is called the Carlta-Klein monopole. Okay. For 3D picture, it's, I mean, this is a non-compact limit or something like that. So, there is no this Carlta-Klein monopole. But, if it compactified it, so there is a Carlta-Klein monopole. Okay. Okay, so then, this instanton have charge one, but this VPS monopole have fractional instanton charges. Okay. And the solution, XP solution of this segment is given by just a replacement of this five to the A0. Okay. It's clear? Not clear. Okay. Okay. What? So, forget about this one. Yeah, we can consider this D1. Okay. So, this is a VPS configuration. And there is a TDR of this one in this picture. It is called here, VPS monopole or fractional instanton. And if we consider some bound set of these three, then it becomes instanton, but, or Kerala here. But we can consider this fraction. I mean, just part of this one only. Then it's called VPS monopole. Okay, maybe I can, yeah. I can write down the solution explicitly because this is normal. I mean, this is like, use our TOFT-POLACO monopole. We can write down the explicit solution in a textbook. I'm just replacing phi to A0. But for the Kerala-Climb monopole, it should depends on the X0 direction because it's a Kerala-Climb monopole. But it's a bit same. But actually, from this picture, Kerala-Climb monopole and other VPS monopoles are not distinct. I mean, it's same, you know? So, it should be same. Actually, if we take a different gauge for the TOFT-POLACO monopole, then we can think we can have this monopole is same as this one. So, I mean, all of them are actually same class of solution. So, what do you think was the biggest question? Okay. The TOFT-POLACO monopole, the solution depends only on R3. And Gory is on R3. So, I filled phi, but it's all found is on R3. Yeah, for this one, yes. But your picture is clearly, the things are not invariant on the translator. Okay, okay. Okay, actually. Something localized. Right, right. Yeah, actually, for the one VPS monopole, if we consider just one VPS monopole, by the gauge transformation, it's a singular gauge, it's called HOH. Then, it becomes X0 independent. Yeah, but this gauge is not, I mean, for, yeah. But if we consider several one, then there's no gauge that makes them are X0 independent. Only the one monopole. Okay, so that is what's here. That is strange, but that is, yeah. Actually, yeah. This color of the monopole was found from this picture. I mean, then there should be something. It's related to the gauge here, dependence or something. Actually, this color of the monopole is X0 dependent in a user gauge dependence. Okay. Okay, then, okay. So, that means, ah, that means so. Yeah, of course, NC or something. Then there is SU2 embedding. Related to the simple root, one to R, and also, sorry, this is, so then V is this one. This is a VPS monopole. So, it's clear from this picture. This segment is correspond to the simple root. Oh, sorry. So, I did not write down the explicit form, but it's, then the magnetic charge, I mean, magnetic charge is this alpha i. I mean, this is the level of the magnetic charge. And, sorry, instanton charge. Instanton charge is Q equal, so it's proportional to this combination. And the classical action, S equal minus i tau alpha i. But, so, then the Carlta Klein monopole, this is, as you expect, magnetic charge is alpha zero. It's this, for this SU and C case, given by this one. Okay, so, I think root, I mean, it's, I mean, lowest root. Okay. So, as you see in the picture, this alpha zero and alpha i's are all, I mean, equivalent. And charge Q equal one plus alpha zero phi, and S is, so this looks different from this one, but this is just, I mean, coordinate dependence in the algebra. You put one more beta times the graphic point and alpha i. Sorry, V, this one. Sorry, beta is the length of this one. Okay. What is this one? Ah, V, ah, sorry, sorry, we define this one. I mean, sorry, I will use, maybe, sorry. Define this one, sorry, I have, yeah, sorry. Actually, I, I, I plan to write down the explicit solution, I mean, so it's a political monopole. It's, I include this V, but sorry. I didn't do that, so. Okay. Then, we, here, we can identify the background, which contribute to the, this case in the conversation. Then, we, then what we should do is just evaluate the lambda, lambda in the background classically, or if we need it, including one loop collection. Because R3 is non-compact, and the three direction, so we need to find the vacuum of the theory first, okay. So, there should be, for, for this case, we need, yeah, there is some notion of a vacuum, so, okay. But computation is a weak limit, I mean, semi-classical. Okay, so then, just consider the effective action for massless fields. Here, it becomes just a U1R, vector mass field. So, this is free. Okay, free means, there is only a joint matter. So, there is no coupling, okay. This is because by this Wilson line. And actually, with zero culture crime, momentum. Do the localization S1 times R3, I thought that is, S1 times R3, okay. R3, yeah. So, you first should specify the boundary condition at the infinity of R3, at the left of the path integral. Right, but it's, yeah, here we take the user one, I mean, something dumping at infinity. Right, like a user instanton in R4, something like that. It's a user feel-self, like boundary condition. Okay. It's okay? Yeah, actually, yeah, of course, if we, yeah, this is, I mean, yeah. What do you mean by need to find back here? I thought that fixes the boundary condition already. So, yeah, actually, there is a choice. Actually, this, yeah, yeah, this, yeah, sorry, this Wilson line is a modular, but. There's a choice of the subtle point, but there is no choice of the boundary condition. Yeah, sorry, so this is a modular. I mean, five is a modular. I mean, modular of the back here. A classical modular of the back here, okay. Oh, okay, so need to find back here means need to find the subtle point. Right, subtle point, yeah, yeah, yeah. Yeah, actually, some of them are lifted, and it actually, there is only the NC points remains. So, we need to find, yeah. I was a little bit confused because it said non-compact, and hence, you have to find the back here. Ah, if it's compact, actually, yeah, we should integrate over the back here, I mean, some over the back here, I mean, back here. It's not called back here, but, about here, it's a non-compact. So, for the field theory, we should satisfy me, specify as back here, okay. If it's R1, then it's okay. So, we need to integrate over everything, like integrate over this A itself, like a S3 function, or S4 function, it's okay. But it's a, this is non-compact, so we need to specify, I mean, we need to fix what is A, I mean, back here, okay. Not okay, I mean, it's a user, I mean, general field theory, I mean, not suji or something. I mean, if we have a two or three or four non-compact direction, then we need to specify the back here. But in low dimension, over compact case, we need to integrate over system. Sorry, does it impose the condition to the case use dump of the zero or the infinity? Yeah, but A is okay. A does, I mean. So A does, so what boundary condition impose for the A? Okay, or maybe it's a dump, I mean, dump means action is finite or something like that. Or A is a constant, it's okay. It's like a Wilson line in a user sense, I mean, for S. So A goes to constant infinity. Right, right, right, right. But then you have to choose the value of the constant to fix the boundary condition. That's true, that is what we want to do here. Yes, okay, okay, I'm not okay, maybe anyway. Okay, then the, actually it's zero-calculation climb momentum. Actually because of this weak coupling limit and this, yes, U1 vector multiplied, actually there is no coupling between the machine and the massless one. So we can forget about the machine means, I mean, non-zero-calculation momentum state. It's a machine in a 3D sense, okay. So we can consider it as a 3D theory. So it's direction to 3D. I mean, machine mode decoupled and T goes to infinity limit, okay. So then there's a 3D N equal to U1R vector multiplied. Actually it's, in literature it's N equal one, but I think in a user sense it's N equal two. I think so, but anyway. Then five is this one, I mean, this Wilson line, constant mode in C. And this is the real photon. Actually by using a beyond-guidance or something like that by, I mean, adding some terms imposing a beyond-guidance, we need, yeah, we can do, I mean, actually this is U1, so we can dualize explicitly. Okay, this is a Wilson loop. So this, yeah, can be a classical module, okay. So this is maybe a classical module of this space. So this is a 3N equal to Kyla. Actually there is a supersymmetry, so it should be this combination. And then, original action. Original action is just a Yang-Mills action plus, yeah, plus some topological term. It comes, oh my gosh. So this is just a gauge coupling constant or something like, I mean, it's a dual gauge coupling constant because it's a dual photon, it's Kyla, okay. So original Yang-Mills action is a Kyla potential, so it's no potential. So it's a flat. It's of course, of course, because the Wilson line is modular. So this X is a classical module. Sorry, T goes in infinity limit, yes. Actually, yeah, here first we consider the T goes to infinity limit. Okay. You asked me, the circle size is finer. Ah, circle size is finer, better. Why do you locate all the colors at the same time? Ah, because of T goes to infinity limit. Because T goes to infinity means, I mean, so we, externally weak coupling limit, okay. So if we consider some loop collection to the machine mode, it becomes some beta times something, but it's translated by this beta times exponential minus T because of the coupling constant. So in T goes to infinity limit, that vanishes. Always, okay. So actually there is only the mass dimension beta and the beta times exponential minus T or something. One over beta times exponential minus T or something. It's very small in a T goes limit. So it becomes bigger. So machine mode is decoupled, okay. Yeah. So R can be fixed. Actually, R goes to limit, R goes to beta goes to zero limit is discussed by this Davies. But here it's fixed, but modify the action. Okay, then to find the back here, we need the scalar potential. Yeah, it's a, here I explain it with the back here, we need the scalar potential. Yeah, it's a straight forward computation where we can do that, but it's for the, I mean, I mean, some, something easy way. I mean, in principle it's okay, but easy way is the scalar potential. Ah, sorry, this is scalar potential is related to the super potential. I mean, fermion bilinear term because of a suji, okay. Because of supersymmetry, phi is related to the scalar bilinear, okay. So if we compute the scalar bilinear, we have found the super potential or scalar potential. So we find the back here, okay. So this is a fermion bilinear. So we need the two fermion zero modes. So we can, we should consider the fundamental monopole. If we consider the, if we want to consider the scalar potential, we need a different one. I mean, different configuration, but to consider the fermion bilinear in an effective action, okay. We want to compute the fermion bilinear in an effective action. It's, I mean, I mean, necessary to compute the two fermion zero. Okay, and then the procedure is the user one. I mean, user instanton computation or monopole computation. So pass integral measure of the zero mode on this fundamental monopole is like this. This SJ is a cluster action with the fundamental monopole including, I mean, yes, how does that monopole here as I wrote down here. Actually, it's becomes a, not five, but it becomes a Z. I mean, including a real photon. It's because of the boundary term given by the, sometimes imposing Bianchi identity to realize something. So this is correct one. As expected, it's actually a form of a combination. And sorry. By the fundamental monopole, do you mean the embedding that you put monopole by a simple one? Yes, yes, simple one and also a character crime monopole. This, yeah. Fundamental monopole means that these VPS monopole plus character crime monopole. So I want one to R and zero, alpha zero. No, no, no, not superimpose. If you superimpose, it's because bound state, this becomes an instant one, but The fundamental monopole means either VPS monopole Yeah, right, right, right. Then we have a two. Filling of zero. Okay, so this A is a position in R three, a position with a monopole in R three. And this is a user, I mean, to put a monopole measure. Omega is a U one phase. And this is a position in R three. And this is a Filling of zero mode. With that. Okay. This is a bit strange, but cut off scale, mu and gauge coupling constant, gauge coupling g at the cut off scale mu, which is defined, which defined at T equal zero. Okay. That's because actually we modify the action, but we do not modify the measure. So this is a measure comes from the measure of the original theory. So this measure should respect this T equal zero. Okay. So what, yeah. Here we consider the T goes to infinity limit. Action is different, but then, fermion zero mode is given by the supersymmetric transformation of the fermion and explicitly given by the, okay, here, sorry, here is X infinity, spatial infinity, okay. This tensor mean, just mean that we do not write down the gauge indices. So gauge indices like this one. I mean, okay. Specialized by alpha g here, I mean here. So this is a abbreviation. So it's not, I mean, this doesn't mean important things. Okay. Here, actually here, we consider the different back, I mean, different background. I mean, different BPS monopoles, the cuts and crev monopoles and we already summed here. Okay, so this is okay. Okay, so this is a free fermion propagator here. From the instanton to the fermion. Instanton to the fermion. It's in X goes to infinity limit. This is a user in a monopole or instanton computation. Then, superpotential, which reproduces this correlator. I mean, that means reproducing, I mean, reproducing this one, we need some terms, fermion binary terms in effective action, which is given by this one. Sorry, I'm getting close. So I thought that you were doing the localization. So you should identify the subtle point and then do the one loop determinant around this. One loop computation, okay. You seem to be doing something rather different, so. Yeah, actually, this is a classical, I mean, this is a classical result. I mean, just plugging in the classical configuration into the field. It's the same. Actually, I do not write down the one loop determinant around the VPS monopole. Yeah, that is true. But actually, it's a bit difficult, but we can evaluate it and then the result doesn't change this one for the superposition. So one loop determinant is one? Sorry? One loop determinant is trivial? Not trivial. Actually, it's a, it's a, it changes the kera potential on this. It doesn't, I mean, doesn't change the potential. I mean, the kera potential, something like that. Let's see, but I just want to compute lambda, lambda, so right, sorry. Then it's just one expression. I don't even know why you're talking about the effective potential. I mean, in the localization, you specify some correlator or something. Correlator, yeah, yeah. So that's lambda, lambda. Lambda, lambda, yes. That's the other key exact term and the rest is all the computation. So I don't know why you begin to talk about superpotential, et cetera, in this way. I mean, why do you want to use more of it? Yeah, actually, I mean, here, just, I mean, localization just means consider the theory with different action with, which depends on t. So here we just consider or compute the correlator, I mean, fermion-balena in some strange action, depend on t. Okay, so this is just a user computation, except that the action is very strange, but it's, so it becomes a weak upper limit or something like that. So, okay. Sorry, you're fine, but I thought that the point you send t to infinity and then you can restrict it to the subtle point locus, but you should still take into account the one root correction and everything. Sorry, maybe, okay, maybe please, please continue. Maybe I can see after. Okay, the subtle point is the AST connection. It's a BPS multiple. So, yeah, and actually with these terms, actually there should be some correction by the one loop, but it only affects the care potential. So this is, actually, this is just a classical result, but it's okay. I mean, one loop is good. That's if there is something that you are using the homomorphism in some way. Actually, yeah, if, okay, okay. No, yeah, here, yeah, I use homomorphism, but yeah, of course, yeah, this is just a trick, I mean. We can compute this care potential directly in this theory, I mean, with some, something, okay. But here, yeah, it's just a trick. I mean, trick was some shortcut to compute the potential. Okay. I see, because if you use the homomorphism, that's what essentially what Matt called the door. Other thing, no, actually, yeah, yeah. That was the weakest smoke in this competition, but then, yeah. Analytical continuous radiance, that's what it is. Yeah, yeah, yeah, but yeah, actually, analytic condition of R is, I mean, is different. R is not analytic, so. Yes, sir, maybe, maybe we'll go ahead. Okay. How did you get the film? How did they come about? This one, actually, the film, zero model is given by the Suji transformation with this background. What transformation? Suji transformation of the lambda, the lambda, it's become the zero model, is that it? Okay, around the bucket. How did you get the propagator? Yeah, no, no, yeah, actually, this is just not propagator. I mean, this is a zero model, and it becomes this propagator type in a large X. I mean, why? It's a, yeah, I mean, if you compute it, it's like that. Yeah, actually, it's, yeah, I mean, delta lambda, Suji transformation is given by the something, by F or something. I mean, field strength, something like that. It's like something dumping X, one over X square or something. Okay. So, this film of zero model also dumping in this term, and it becomes just a free propagator. Free propagator means just one over X times gamma mu or something like that. Okay. So, gamma mu is in a film of zero model. So, this is, yeah, I mean, yeah, okay. So, this is just asymptotic behavior of the zero model. So, asymptotic behavior becomes just one over X something. Okay. So, it's not, I mean, yeah. Not special, I mean, not special meaning. Okay. So, actually, so, so this is X goes to infinity. So, maybe there is a derivative, but to determine the potential, it's not, it's okay. Then, so the vacuum is determined by logic. Mm-hmm. Logic. Logic. Yeah, actually, yeah. We compute some correlated, okay. In a, then in a low energy effective action, low energy effective action, a very sunny effective action contains something lambda lambda because of the, to give this one, okay. So, this, this is determined. So, we can specify what is effective action. I mean, this coefficient in the effective action. Okay. So, lambda, I mean, film in violin in effective action. Then, then, because of the supersymmetry, we don't, we don't care about super potential and everything, but there is a supersymmetry. So, we need, there should be a scalar potential in the effective action, right? Is lambda lambda ideal for you to get? Lambda lambda identified, yeah, with X, yes, yes, yes. Yeah, yes. It's contains, yeah, sorry. Yes. Yeah, I would say that you go to three dimensions, three dimensions, they go to break supersymmetry and then you go, then this is the political potential, political mechanism of confinement, where there is a cosine for the, your bottom. But this is a supersymmetrize, for example, and also a little bit of a finite Okay, then the vacuum is determined. So, you have a, you assume some potential coupling in the effective action. Compute the two-point function of film that was called the super potential coupling and then match it with the classical computation at the surface of the ground. That's the way to get super potential. Yes, yes, yes, yes. Yes. But maybe you have one fiber diagram. Sorry? One fiber. Partition of one fiber diagram? One fiber diagram, yeah, yeah, two, yeah. So, you have like a vertex induced for this term? Right, that too, yeah, yeah, one vertex, yes, yes. You want to reproduce the classical computation? Right, yeah, from the low energy effective action, I mean low energy limit effective action, yeah. So, yeah, we can compute the scalar potential directly, yeah, it should be, but yeah, it's, yeah. But yeah, this is a trick, but yeah, it's easy. Then, sorry, C2 is, sorry, sorry. C2 is a real coaxial number, sorry. And W is a fundamental weight. Fundamentally it means WI times alpha j equal delta ij or something like that, okay? Alpha zero is the sum of the minus sum of the alpha, one to alpha, oh, okay. So then you can check, actually exponential is same for this computation, I mean, this exponential is same for this part. And yeah, it solves equation. Ah, sorry, flatness condition. Okay, so, two over two. Okay, then plugging this back to super potential, so this is low energy super potential. Okay, here we consider the S1, so there's a beta, but as I explained in the first lecture, if we found this super potential, I mean super potential in a low energy limit, then by derivative with lambda, or I mean, I mean by taking lambda to the chiral super field, we can compute lambda lambda from this one. Yeah, maybe so it is. So this becomes correct one. Here, because these are defined at t equal zero, so it's okay. And actually this, yeah, this doesn't depend on the beta. Okay, that is the result. So we can take beta goes to infinity limit to the R4. Then, okay, so then in a weak limit computation, we found this, six. Fundamental weight, fundamental weight, yes. So, so now, consider theory at t equal zero. So what we consider the t goes to infinity limit. It's different. So first, as a creator, original theory, but okay. So this means the action is, action is the same, I mean original one t equal zero, but the back here, if we, I mean, choosing the real, I mean, correct back here for the action with t. Then, use our localization computation or something. There is some terms remain in this one. Actually, because of the back here, it changes. So this doesn't vanish in general. But actually here, back here are discrete. There is only, I mean, NC back here. Only NC back here, because of this fraction. So if this is this change, this divergent, I mean, it's a, I mean, if this change, it's a, I mean, sudden change, okay. If there is a modular, then this t dependence, I mean, means, can mean the dependence on the modular or something like that, but it's back is discrete, okay. So we, so this should be zero. I mean, as a user, I mean, assumption for the index or localization computation. Actually, actually, if there is something like, asymptotic behavior changing, then may, I mean, this may change like a written index can change. Wow, localization doesn't work anymore. But here, we assume as a user, I mean, like a user written index in a field theory case. There's no such things happen here. Okay, so because of the back here are discrete, so back, I mean, vacuum doesn't change. That is the conclusion, okay. So that means in this T goes to infinity limit computation can give this gauge in a condensation. So you're already using the n-bock yards input? No, this is, yeah, no, no. This is in a T, this is a T goes to infinity limit, okay. Okay, then it becomes T, yeah. Of course, if there is some sudden something happens, I mean, it becomes one vacuum or something like, it's happened like a wall crossing, I don't know, or the written index jumps. Then, yeah, of course, this computation doesn't work. So vacuum, the T equals to infinity theory on R three times S one? Right, right, right, right, right, yeah. And then we changed, yeah, one over T to the zero to one. Discrete, yeah, at least at the T goes to infinity limit, okay. We computed. Okay. I thought that you need that to compute it. No, no, we computed it for the T goes to infinity limit. The action, if action is a change, then we can compute this one because it's a weak coupling. Okay, then we found it's a NC vacuum. Okay, as it changed the T goes to infinity to finite one over T to zero, zero to something. Okay, yeah, yeah, yeah, yeah, yeah. If we change, it's a drastic change like a jump over the index or something like that. It can happen. But it's usually assumed in a user computation, I mean, user localization technique, it doesn't happen, that is assumption, okay. Martin, okay. If it happens, yeah, if the modular, I mean, yeah, the backyard is not discreet, then if we change the T, then the vacuum can change in the modular space or something like that. It can happen, but it's discreet for this case. So it's okay, I think. Then, yeah, here we consider the localization in S1, S1, sorry, S1 cross R3, but we can directly use this one to the R4 case. Then maybe it's more simple, but this doesn't work. Actually, there's a two scale for this case. And coupling is weak for this region. This region is weak because here, the coupling is, I mean, scale is here. So above this scale, I mean, much of the discreet is weak, but here, this region becomes strong. And even if we take T goes to infinity limit, then there should be some region strong. So, theory never becomes strong coupling, but okay, for this case, this region is weak because of this factor, but this region is also weak because in this region, theory becomes three dimensional. So it becomes weak, okay? So this case, we coupling computation is okay, but for R4, there is no this scale. So theory never becomes weak. Even if we take the T goes to infinity limit, okay? So, localization doesn't work here. So we need to compactify this one. And S4 is not good, as I explained. Okay, so finally, consider N equal to one 4D vector. And the general action is like this one. Here, this is a Yang-Mills term, but we can consider the general action. Then we can include this, the other terms. Then following this computation, okay. And yeah, some work, then we can show. This is given by the, this is given by the, sorry, this one. This is called the Benetsian-Yankirovitz potential. So, okay. So for this action, the nonpartuitary production is given by this one. That is, yeah, with some computation, we can show that. And then, consider the chiral multilay state, matter multilay state. And then we can add the matter multilay state for, I mean, chiral potential of the matter multilay state arbitrarily. Then it becomes a weak coupling. So we can integrate the matter arbitrarily weak. So in a part of it, we can integrate as a matter field, okay. Then we get this action, but because of the, we should integrate as a matter field with background field, S or W alpha, okay. So then it becomes this one. Then, the gauging conversation is given by this one. If we know this one, I mean, part of the result, okay. And if the matter cell is, I mean, adjoint case, it is given, I mean, by this F is given by the matrix C or one matrix model computation. Okay, that is called the digraph buffer. And this means, there's some proof of the digraph buffer theory by Kacchia's digraph survey written on the Xanon digraph buffer, something. But they assumed adding this term by hand. I mean, no part of the potential is this one. Actually, we can show this is okay for the non-Suji breaking case, G to G, then it's okay. But if you consider, for example, SUN2, SUN1 times SUN2 or something, then we cannot trust adding this one is okay. But here, this type of the computation can show this is okay. This is just a straightforward computation of the gauge in a condensation. So this is okay. So that means, yeah, for with any chiral matter or any action, we can compute this gauge in a condensation by the part of the integrating as, then fixing A. So this is some proof, I mean, no part of the proof of the digraph buffer connection. So that takes, thank you very much.