 In the first lecture, I introduced to you 2D QD formula melts. In the second lecture, I introduced to you the super conformal index of SU2 gauge theory with four flavors. I didn't tell you, I haven't told you, the precise relationship between the two. And this precise relationship is the topic of today's lecture. So just to remind you about what we did at the last minute of my last talk, I rewrote the horrible expressions for the super conformal index of SU2 with four flavors. Here it is in its full glory. So it's a product of tons of infinite products, but it's a bit too complicated, right? And in fact, Rasteli and collaborators wrote it down in the summer of 2009. And they already noticed that this is related to some kind of 2D mill theory at that point. But it took them two years to find out exactly how the correspondence goes. So after two years of struggle, they found the crucial simplification, which was to take q equals t. So why was this so difficult to find? Well, as I answered you, some one person in the audience, so these p, q, t are the fugacities or the exponential chemical potentials of the super conformal generators, which commutes with the chosen super charge, capital Q and capital Q dagger. But there are no intrinsic nice bases in it. So they are just three dimensional. You can take whatever basis of it. So Rasteli et al was using various different conventions, bases, vectors in every paper of theirs. So in the first few years, their papers are very, very confusing to read for me. But after two years, they found the correct basis, which is very nice. So this is q equal t. So let's see what happens. For example, so I should say that this is the Gallotto's notation for the S duality of SU2 with NF equals 4. So let me quickly remind you that this half corresponds to NF equals 2, which is a tri-fundamental chiral multiplet. So they transform as a doublet of SU2A, doublet of SU2B, and doublet of SU2G, similarly for this. In the S dual side, the combination of the flavor symmetries are exchanged. Therefore, we should see some kind of invariance of the super conformal index. And the super conformal index of this side is written here. So these three terms correspond to the contributions from the gauge multiplet SU2. This has eight infinite products. This corresponds to the contribution from the left-hand side tri-fundamental. And the last one is the tri-fundamental contribution for the right-hand side. So the equality you expect that is mathematically proven. And I showed you mathematical computation last time is that this is invariant under the exchange of B and D, say, right? So how do we see that? Yeah, you're right. There are also another duality frame. Anyway, so let's set this to this particular component. So what happens is that if you consider PQT1 half of Z, this becomes, well, I just plug this in, 1 minus T inverse Z. But now we are putting this, right? So we have, so I should write Q here, right? So this combines in this form. And this is inverse. And you have also this contribution. But you see that in this combination, because it is a hyper-multiple, always a fugacity and its inverse conjugate appear together. So there's an additional factor of Z inverse, which is basically this. But inverse placement of the exponents are reversed. So you immediately see that this part of the infinite product and this part of the infinite product almost completely cancels, right? The only difference between this and that is that here the product runs from 0 to infinity. Here the product runs from 1 to infinity. So there's an almost complete cancellation. And therefore what happens is that this product is just a single infinite product where it might not look like much of a simplification, but nonetheless it's a simplification, much better. So here I explicitly deduce the computation, but it is a general property of n equals to super-conforma index that if you set Q equals to T, P dependence automatically drops out. This is due to the enhancement of the supersymmetry preserved by the R. Yeah, that's I can remind you, but that's not going to help you much. So the definition was that the super-conforma index is something like minus beta QQ dagger P j2 plus j1 minus 2R. Q is j2 j1 minus 2R. And T was i3 minus R over 2. So this is a funny combination. And whatever that is, we are setting QT to be Q. So what happens is that there's only one combination here. So this becomes Q's j2 j1 plus i3 minus R. So well, you need to do the computation of the super-conforma algebra a bit carefully, but you can rewrite this into epsilon minus beta prime Q dagger prime in this particular case and check that this linear combination of the super-conforma halter algebra element commutes not just with Q, Q dagger, but also with this Q prime and Q prime dagger. Therefore, using the standard argument of the invariance of the written index, this SCI is not only invariant under the change of beta, but also by the change of beta prime. That's why I said there are twice super-charged preserved in the background. And that's why P dependence automatically drop out. Thank you for the question. Right. So let's see what kind of simplification we have. So there will not much simplification, you might say, but it is, in fact, quite significant. So just as I told you, oops, no, no, no, this is fun. This will become plus, minus, plus, minus, plus, minus n over 0 and 1, n plus 1 half, z plus minus, a plus, c plus minus, d plus minus, inverse. And so I'm going to erase. And this part becomes similarly, maybe I shouldn't bother writing all of the equations here. But sometimes doing this computation very explicitly helps you understand what is going on. So these are the contributions from the matter fields. And there are contributions from the vector multiple. So they become a certain term of this form, which I write as 1 half, again, dz over t2 pi i z, 1 over z squared, inverse, and kz inverse squared, where kz is still an infinite product. So kz here, inverse, is this following, again, single infinite product. So what do we get by the simplification? So here is the crucial step. This infinite product is the matter contribution. But there is a formula, which we write this, into an infinite sum instead of an infinite product. So let me write it down. So plus, minus, plus, minus, plus, minus, and over all positive and non-negative integer, z plus minus, a plus, minus, b plus, minus, it's really horrible. But this can be written as the following way, k. So this same k, but without minus 1, kz, ka, kb, divided by k0, which I introduced shortly. And sum is over in positive integers. And ka in q1 half, ka in a, ka in z, ka in a, ka in b. You might be familiar with this expression. Haven't you seen something like this a few days ago? So I'm not going to give you the proof of this equation, but you can prove this equation by just checking where the poles are on both sides of the equations very carefully. And I should say that ka in a is just n minus 1 plus n minus 3 plus dot dot dot plus a1 minus n. So this is the character of the element a inverse in su2 in the irreducible n dimensional rep of su2. So this is the equation. And finally, I need to tell you what this k0 is. The k0 inverse is just an infinite product like this. So now I'm going to plug this in to this formula. So let me erase this general case, which is very long and hard to remember. So I'm just going to rewrite this using that expression very explicitly. So let's move very slowly so that you can take the notes if some of you are taking notes. kz inverse, and I plug in that expression. kz ka kb over k0 summation over n ka in z ka in a ka in b over ka in q 1 half. And again, we have, yeah, this is horribly complicated. But you just need to be patient. You have kz kc kd divided by k0. And summation over, let's use different index m, which is over 1 of ka mz ka mc ka md divided by ka mq to 1 half. So now, why is this good? This is good because the dependence on z is localized in the expression. I mean, the only place z appears here, and this kz inverse, kz here, ka in, and kz there, and ka in. Now, kz cancels out because here you have minus 2. This is plus 1. This is also plus 1. Now, the only z-dependent factor is just this, this, and that. But there's a certain formula of this form, ka in z, ka in z. This is just chronicle delta. This is because this is just a measure induced on the Carlton torus of SU2. And these are the characters. So this is just the orthogonality of the characters. This means that in this horrible expression, the only remaining term is where n and m are equal. So you can just perform the integration very explicitly. And the final answer is ka kb kc kd divided by k0 squared summation over all positive integers. And you have ka in a, ka in a, ka in b, ka in c, ka in d, divided by ka in q inverse squared. So now, in this form, the invariance under the exchange of b and d is completely manifest. In fact, the complete symmetry under a, b, c, d is manifest. And if you remember the content of the first lecture of mine, you recognize this as exactly the partition function of q, d, form, d, and melt on a four punctured torus, sorry, sphere, with holonymy variable given by a, b, c, d. But with the area sent to 0, please remember in your note, I mean, in the first lecture, the summation is of this form a and quadratic casimir of the representation times ka, ka, ka, things like that. Here, we don't see anything like that. So you need to set a to 0. So this is the 0 area limit of q, d, form, 2, d, and melt. And as I already told you, there can be some strange renormalization factors involved. And indeed, we see that associated to the Euler number here and also due to a strange, from the 2D point of view, strange normalization of the measure on the SU2 manifold. But this is exactly the partition function of q and melt up to this bulk renormalization and the wave function renormalization. So this is the explicit for the 2D correspondence. I wanted to show you. And this is very explicit. So that's the good thing. I mean, better thing about this S3 times S1 case compared to the S4 case. So four or five years ago when I gave this lecture in the spring school here to yesterday, I talked about the case of S4 partition function. In that case, you need to compute conformal blocks, which is very complicated. And on the other side, you need to compute necklace of partition function, which is also very complicated. And you cannot see the full equality. You cannot just prove that. Yeah. Pardon? That's right. That's right. Well, this depends on a very explicit embedding of SU2s into SO8. So I don't think you can see what happened to the other SU2s. Well, what's that to your question? Right. That's right. So in this sense, in that sense, with this expression, we only see that SU2a times SU2b times that time SU2d within SO8 are properly permuted under the stability. But because SU2 to the fourth is a maximal rank subgroup of SO8, by checking that, you automatically guarantee that all of the SO8 acts correctly to the fugacity is ABCD. But that's indirect. So yeah, so why is this the case? So the rest of my talk today will be spent to explain why this is the case. Simone last week told you that answering why is often a very difficult question. But in this case, I try. So why is this the case? So this picture is already something is suggestive about this relation. So the natural thing you try is that somehow this trivalent vertex corresponding to the tri-fundamental field corresponds to three punctured sphere, similarly on this side. There's some connecting tube. So this will correspond to the QGCD here. And this should correspond to SU2 gauge field. So the reason behind this correspondence is basically that there is a certain six-dimensional, n equals 2 comma 0 SCFT. And labeled by G, which is either AN or DN or EN. So this is SUN plus 1. And this is SO2N. And N is 6, 7, 8, such that its compactification on S1 in the IR is given by 5D, N equals 2, superion melt with gauge group G. And I'm going to describe how instead of compactifying on S1, if you compactify the same theory of type A1 or type SU2 on this three punctured sphere, you get this four-dimensional hyper-multiplet QABG. And if you compactify that on this tube, you get N equals 2 vector-multiplet. And similarly with the other three punctured sphere. But the basic fact is much more deep that there is this nice something in 60. So before getting to the physical content of this statement, how do we know this fact? So you need to use string theory at this point. Existence proof. And I need to put the quote around the proof. But the existence proof is that if you consider N M5 brains on top of each other and take the low energy limit, and this gives you the 60 N equals 2 gamma 0 theory of type A N minus 1. And you can do some similar construction with D N and EN2. So, pardon. You need to use type 2B in the EN case. So if you think that super-string theory and M theory exists, then that already guarantees the existence of this 60-dimensional theory. But in the rest of my talk, I don't use much else about string theory or M theory except for this particular fact. This particular fact is very powerful in deducing various properties in 4D. So before getting to discuss what this fact tells us about 4D N equals 2 field theory, I'd like to first discuss 4D N equals 4 theory. When did I start? 20, okay. So I have a lot of time. So let's start from the 60 theory, which is a super conformal field theory. Let's put it on some S1 of radius R6. And from this fact, we know that this is 5D N equals 2, a superion melt, with gauge group G. So at infrared, you can write the grandstand of this form, G5 squared, trace F mu nu, F mu nu, plus lots of super partners. So what this gauge coupling would be? In five dimensions, this is dimension 4, and this is dimension 1. The original SCFT doesn't have any mass scale. Therefore, the only sensible thing to write down is that this G5 squared is proportional to R6. I will fix the coefficient later. Let's do another compactification on the circle of radius 5. And this becomes 4D N equals 4 superion melt, with again G as the gauge group. And what would be the gauge coupling constant in this case? This can be seen in a following way. So as is usual in the Caltz-Klein expansion, you decompose this five-dimensional integral into the integration along the five direction and the rest of the four-dimensional integral. And you have this G5 squared, trace F mu nu, F mu nu, plus that of that. But this total combination is the one of G4, I mean four-dimensional Yang-Mills coupling squared. And if you compactify on a circle of R5, size R5, this will give you a factor of R5. So this is roughly of the order R5 over R6. So this is what you get. So what we did is to put the 60 theory on the torus. So you identify the opposite size to get the torus of the size R6 and R5. And we obtained 4D N equals 4 superion melt with coupling constant R5 over R6. But remember, we made a choice of first compactifying along this R6 direction. And then this is the compactification along R5. You can, of course, switch the order. So if you compactify first on R5 and then R6, you can repeat exactly the same argument and conclude that this is 4D N equals 4 superion melt with coupling, which is instead of R5 over R6, which is R6 over R5. So these two coupling constants are inverse to each other. So this is the estuality of the N equals 4 superion melt. This is a really surprising thing about this simple postulate or fact. So estuality in four dimensions is a very non-trivial operation. However, from this six-dimensional point of view, it is really just a part of Lorentz symmetry, Lorentz invariance exchanging the two sides of the torus. And in the first, sorry, in the second lecture, I told you that estuality exchanges W bosons and monopoles. And how does that arise in this setup? So is there a colored chunk? Yeah. So I will talk about the culture of client modes. So that's part of the postulate. So this fact includes the behavior about the KK modes, too. So the 60 theory has a string-like excitation. So in four-dimensional theory, basic excitations are point-like particles. Here, you have strings. And if you have strings on T2, that can either wrap this direction or wrap the other direction. Or if you prefer, you can wrap in various other directions, too. But let's just consider these two cases. And let's say if you are in a, I didn't explain, but if you go to the generic point on the modular space of the supersymmetric vacuum, these strings are tension-full. So these are massive strings. Then this excitation looks like a point from the four-dimensional point of view. And the mass in 4D will be tension times R5, of course. This string excitation wrapping around this cycle will have different mass, which is given by the same tension times R6. So the ratio of the mass of the excitations, ratio of the mass of the excitations is actually given by this tau, or tau inverse, depending on your point of view. And I please remember, recall, that in the last lecture, I told you that in n equals 4 super-miles, the ratio of the mass of the W boson and the monopoles are given by this coupling constant tau. So you identify W bosons as this guy and monopoles as the other guy. So W boson and the monopole come from the same object in 60. So this is an interesting point. Many people try to write down a nice, useful Lagrangian for this 60 n equals 2 comma 0 theory in six dimensions. But suppose somebody finds such a Lagrangian, which is very useful. But then four-dimensional estuality of n equals 4 super-miles will become completely trivial. The 4D estuality exchanges perturbative excitations like W bosons with almost semi-classical excitations like monopoles. But they should be completely manifest if you have a completely nice Lorentz invariant Lagrangian in six dimensions. That shows that writing down such a Lagrangian would be extremely hard. This doesn't tell you that it's impossible, but I think it's very, very hard. So let me come to this analysis of Karlsruhe Klein modes. There's another peculiarity of this relation between the 6D theory and the 5D theory. So before getting to the peculiarity, let's recall what's the standard behavior under S1 compactification. So let's say you have, let's just consider a very simple case of free massless scalar phi in capital D dimensions. And let's put it on S1 of radius R. And I guess everyone has learned somewhere that in that case you can decompose this d-dimensional field in terms of the Fourier modes, which depends on the d minus 1-dimensional directions and exponential i n x d over R. When this is massless, these components will use phi of n. We will have mass given by n over R. So this is called the Karlsruhe Klein Tower. So in this fact, I'm just erasing. What do we exactly mean about Karlsruhe Klein towers? So interesting thing is the following. Just consider 5D n equals to a superion melt with gauge group G on a flat space. So it has four-dimensional flat space-like direction times the time. And as you know, superion melt has an instanton configuration. So it's a solitonic configuration of the gauge field, which is point-like along R4. But in five-dimensions, this extends along the time. So using the word instanton is a bit misnomer, oxymoronic here. But this becomes instanton particle. So it's a particle-like excitation. The mass is given by the action in the four-dimensional sense of the instanton. So if the instanton number is n, the mass is in the standard convention 8 pi squared over G5 squared times n. And using this relation, this is proportional to n over R6. So the mass of the instanton particles behave exactly like the mass of the calzacline towers. In fact, assuming that they are indeed calzacline towers, you can fix the relation between G5 and R6 by just demanding the equality. So from this analysis, you can fix the proportionality coefficient here. R6 is G squared 5d is 8 pi squared R6. So what does that mean? Usually, if you consider lower-dimensional theory coming from higher-dimensional theory under compactification, you need to add calzacline towers in the lower-dimensional description. But however, in this particular case, from 62 comma 0 theory down to 5d super-yarn mills, at least some part of the calzacline towers is already included in the five-dimensional yarn mills. That's a surprising fact. And there is an empirical fact here. As far as BPS quantities are concerned, supersymmetric quantities concerned, 5d super-yarn mills computation, including instantons, gives all the expected KK models. So there are many impressive computations of five-dimensional gauge theory using instanton counting. And if you look at the final result, without adding KK models by hand, the final partition function looks like 60 theory put on S1. In particular, if you add, say, KK towers of 5d yarn mills fields, you get the wrong answer. You get double counting. I don't know why. At the current level of my understanding, this is as much as I can say. So this is an empirical fact. So this puzzled many people. But again, this means that writing down six-dimensional 2 comma 0 theories Lagrangian in a manifestly Lorentz invariant way is an extremely hard business. Because suppose you have such a six-dimensional Lagrangian. Let's compactify that on S1 and do the analysis like that. You always expand six-dimensional fields in the Fourier models. And then you have tons of KK models in the Lagrangian. But the computations people have done so far, at least as far as the BPS quantities are concerned, means that you shouldn't have those KK models. So therefore, if you have a 60 fully Lorentz invariant Lagrangian, then that Lagrangian should have a huge new type of gauge symmetry which should gauge away those KK models and transform them into the instantons of the zero models. I'm not saying that it's impossible, but I think it's really hard. So I know there are many colleagues of mine who is trying to find the six-dimensional 2 comma 0 theory Lagrangian. Please continue the hard work. I really love to be surprised. Honestly speaking, before ABJM theory came, I felt in a similar way about the Lagrangian of the M2 brains theory, everybody thought that it's almost impossible to write down a useful Lagrangian describing multiple M2 brains. I mean, maybe most of the students here are too young to remember what surprise, what shock it was when ABJM theory or Bagger-Lambert theory first appeared. But that was a shock. So if some of you write down the Lagrangian of six-dimensional theory, I would be the first one to use that to localization to get something interesting out of that. So please do the hard work. Yeah? Yes? Yeah, it should be doable. So in the Abelian case, something is strange. In the Abelian gauge theory, you don't have this instanton. So there is no contradiction about having an Abelian Lagrangian in 60. In that case, there is no instanton in 5D U1 gauge theory. So it's perfectly OK. And it's perfectly necessary that you have KK towers if you reduce 60 Abelian self-deal to come as a Lagrangian down to 5D. It's necessary. That said, if you turn on a non-commutativity a bit, then there is something called non-commutative instantons even in the U1 case. So in that case, you can compute the partition function of 5D and equals to super-U-Yanmills with U1 gauge group, including the non-commutative instantons. Surprisingly, that computation exactly reproduces the standard U1 Yanmills computation plus all of the KK models. They are equivalent. So there are nice papers showing that, not only on R4, but on a more general four-manifolds, so there seems to be really a kind of gauge symmetry which can transform a KK tower into instantons of the zero models. At least for the U1 case, this was done very explicitly. I mean, we don't know what the symmetry is, but you can write down both sides and compute in the both sides, and you get the same answer. So it seems to be working. So the problem is how to realize that with non-Aberian G. Any questions? Yeah. Pardon? In this case, there is no Chern-Simons term. If you want to have any equals 1 super-Yanmills in five dimensions, it's compatible with Chern-Simons terms. But with this maximum supersymmetry, you cannot have Chern-Simons term in 5D. But these days, we are slowly generalizing this relation between 60 and 5D. And now we know that we can start from 61 comma 0 theory and compactify it down to n equals 1 5D theory. And in that case, Chern-Simons term plays an important role. Yeah. All right. So this is the discussion for n equals 4. And let me just finish my talk today by discussing the n equals 2 case. So yeah. So let's take 6-dimensional 2 comma 0 theory of type A1 or SU2 and put it on this type of surface. So I forgot to say that I often say that 60 2 comma 0 theory is of type G. But I do not mean that there is a gauge field with gauge group G in six dimensions. I do not mean that. So nobody is sure what is going on in 60. If there is something, there will be some kind of non-Aberian two-form field. So there's definitely not the standard gauge field. So let me just stress that you need to always say it's just type A1, not gauge group A1, but whatever. Let's put 6-dimensional 2 comma 0 theory of type A1 on this kind of Riemann surface, a two-dimensional surface. So these two parts are rather complicated. But let's just consider this cylindrical part. So let's say this direction has length R6. And these two parts have something like length R5. So if you look at this part, just this part closely, what you find is something like this R6. This is R5. And something is going on around here. Something is wrong on the two sides. Let's use the fact that if you compactify 60 theory on a circle, you get the five-dimensional superion melt. So this is equivalent to having a 5D superion melt on a segment of length R5 coupled to something here, coupled to something there. And gauge coupling constant was like R1 over R6. Therefore, if you go further down to four dimensions, this is 4D gauge multiplied with tau given by roughly like R5 over R6. So that's what you get. Here, you have this maximally supersymmetric 5D ion melt on a segment. But you have something going on at the boundary. So there can be many things going on at the boundary. But you cannot keep the full supersymmetry on this setup. At most, what you can do is to keep half of the supersymmetry. And suppose you are doing that by carefully choosing the metric here or the R-symmetry background here, which I don't have time to discuss. But let's assume that there is half of the supersymmetry preserved. So originally, we started from n equals 2 5D superion melt. But remaining 4D gauge multiplied is 1 half of it. So it's n equals 2 vector multiplied due to the boundary condition. So what is this boundary condition? From the 60 point of view, you cannot really say directly what exactly is this boundary condition. But there's one nice class of boundary condition in 5D superion melt, which is the following. You realize that this boundary is R4, right? So you can have some 4D theory with SU2 flavor symmetry here living on the boundary that couples to the bulk. So there's some theory here and there coupled to this SU2 gauge multiplied. At this point, there's not much. I don't know how to directly argue that this theory, which should be there, the boundary, is this tri-fundamental hypermultiple, ABG and GCD. But at least you can see a rough structure emerging just by the basic fact. So let's try to determine exactly this three-point, three-function sphere is from the 60 perspective. To do that, let's consider the super-conformal index of this system. So please pretend that the final answer is not yet known. Let's just consider the super-conformal index of 4D theory obtained by putting 6D theory on this, right? What was the SCI? SCI is trace over the S3 Hilbert space minus 1 to the f times in the particular choice of the variable that was basically something like this and various flavor fugacities. So this was the Hamiltonian way of writing down the SCI. But if you think of q as exponential of beta, this is essentially partition function of the system on S3 times S1, where the length of the time direction is beta or log q with some additional chemical potential put in. So let's apply this to this six-dimensional theory. So what is this SCI of this 4D theory? This is a partition function of the 6D theory on S3 times S1 times this thing by definition. Now you realize that there is a S1, which proves to be very useful. Here we invoke the fact that the 6D theory on S1 is equal to the 5D theory. So you get 5D theory on S3 times this Riemann surface. Now this 5D theory has a Lagrangian. So this is something you can really compute. And there are brave people who really computed this partition function. But roughly, you can see what's going on. This is just 5D super-Yaml on S3. So basically, you should get some version of 2D Young Mills on this Riemann surface plus corrections that come from the fact that you have this S3 of a finite size. And after employing the localization along this S3, which is exactly like what Marcos told you last week. So in that case, what was done is to reduce three-dimensional supersymmetric theory down to zero-dimensional non-super symmetric matrix model. Philosophically, you do the same reduction from supersymmetric one down to non-super symmetric model pointwise on the Riemann surface. So you get starting from five-dimensional supersymmetrical Young Mills down to two-dimensional non-super symmetric Young Mills. And then there are lots of corrections coming from the KK models around S3. And people have confirmed that these corrections exactly gives you the Q-deformed Young Mills. So just by starting from this general feature, general fact that the 60 theory in question on S1 becomes the 5D Super Young Mills, you can compute the super conformal index of this 4D theory without knowing exactly what it is. So here I'm using the computation in the logically backward order in some sense. So starting from 60, you do this computation. And this has a certain explicit form as a Q Young Mills. But by looking at that, you can read off exactly what is the super conformal index of, say, well, let's just consider the super conformal index of this three-function sphere. So if you follow that argument, the SCI of this thing has this infinite sum from KKK over K0, chi, chi, chi over chi. But you realize that this can be rewritten as an infinite product showing that it is the tri-fundamental hyper-multiplet. So that's how you see just from the basic feature of the six-dimensional theory that it becomes 5D Super Young Mills. That 60 theory on the three-function sphere is the tri-fundamental hyper-multiplet. So now that we know that, S duality can be derived in a nice way. So I don't know where to erase. So I will finish. So now we are pretty sure that this compactification of the 60 theory gives you tri-fundamental and another tri-fundamental coupled by this vector-multiplet. And unfortunately, I erased. Ah, I didn't erase. Coupling constant is this, and this is 1 over g squared. So the coupling is very weak when this sitting is very long. Let's make it very strong, right? So slowly you try to merge these two things and make these two very, very short. In that limit, you can instead split the 2D surface in a different way. So this is a strongly coupled limit. Now you consider the dimensional reduction along this direction. And then you see that there's dual SU2 emerging, coupled to dual Q fields. So that's how you see the n equals 2SU2 S duality from a six-dimensional point of view. Thank you very much.