 So this is going to be the last lecture from me. So in the last three lectures, I described how the s duality of n equals two supersymmetry. I guess I should wait a bit more. More people than expected are coming back, thankfully, to my talk. So, so far I talked about the s duality of n equals two to 40 theories of type SU2 and how they can be understood starting from 60 theory with two comma zero supersymmetry, again of type SU2. All the essential features of the duality, I think, was already in this type SU2 case I talked about for three hours. And in this last lecture, I just want to generalize that to SUn, that gives you a lot of technical complications and, but that's inevitable. So, yeah, somewhat SUn is rather difficult compared to SU2, so please bear with me. So let's remind ourselves what we learned yesterday. So if we start from 60 n equals two comma zero theory of type SU2 and put it on three puncture sphere, each puncture carries one SU2 flavor symmetry. I explain to you how you compute the super conformal index of this system. So let me remind you the definition of the super conformal index. So in this particular case of just having the variable Q, I forgot to mention, but the exponent simplifies, it's just a difference between the scaling dimension and the SU2R charge. And you can insert flavor symmetry element. And I told you how without knowing in the first place what this theory is, you compute this SCI using the fundamental fact of the 60 theory that it becomes 5D super Yang-Mills if you compact the final S1. So the SCI is given by this nice formula, K a, K b, K c divided by K zero, summation over all of the representation, K a lambda, K lambda b, K lambda c divided by K lambda q to the one half. So this is the basically partition function of 2D Yang-Mills on this three puncture sphere, Q D form. And for today's purpose, the only thing you need to know is that this K a has this form, K a joint of A, Q plus OQ squared. And K zero is basically just one plus OQ squared. So not so difficult. So how do we know exactly what 4D theory this three puncture sphere is? In my lectures, I started by telling you what this is, but pretend we don't know that, but we can expand this to some order, right? So what would be the leading contribution? Of course, the leading contribution is one. The next contribution just in fact starts at order Q to one half. So, and for simplicity, let's just consider the expansion until OQ. So you can just forget about K a and K zero, right? So all it has to do is to expand this part. The trivial representation give you this one. Next contribution comes from the doublet representation. So what you have is that, well, I guess I should first explain how it works. So doublet, doublet B, doublet C. And you need to consider this. So this is Q minus one half plus Q one half. But in any case, this is just Q minus one half one times OQ, right? So for our purpose of expanding things up to order Q, this is just becomes Q to the one half, dot, dot, dot. So this factor, Q to one half, is the telltale sign of having a free hyper from the representation theory of N equals to S, supercommon algebra. If you have something like this, the theory is guaranteed to be free hyper. And you see how these free hyper transform under the flavor symmetry. So these are in the so-called tri-fundamental. So thus we see that this is corresponds to a free hypermultiple of this one, Q A, B, C, right? And once you understand that, now we can understand the exteriority of two tri-fundamental coupled together, G, C, D, in the following way. So you can continuously change the shape of the Riemann surface. You'll contactify the six-dimensional theory. So you get dual objects and dual gauge group. Yesterday I also told you how to read off the coupling constant here. So the coupling constant of this guy is basically given by R5 over R6, where R5 is this length and R6 is this radius. So if you want to make this very strongly coupled, you need to make this ratio very small. So you need to crash them together in this direction. And in that limit, you can split them on the vertical side so that you again find the dual weekly coupled gauge group. So that was how we understand the exteriority of N equals two. So SU2 gauge theory coupled to four flavors because this counts as two flavors. So far so good. So all I want to do today is to generalize everything to SU3 and up. So the natural thing to do is to change this to SUM. So I'm sorry for doing that. This is very easy to do on the blackboard, but for those who are taking notes in a traditional note pad, this is very bad. If you are using tablet, then you can just all copy the content and modify. I'm very sorry about that. So I'm going to change everything to N, this. I'm so sorry. The only thing you need to change is the argument here. So as I told you, the argument becomes a certain diagonal matrix of this one, QN minus three, dot two, dot dot dot Q one, two, right? So what is this theory? So from this on I can rewrite everything because things are significantly different. I might reuse this though. So let's just consider the next case N equals three. What would be the expansion of the super conformal index? It starts with one. The next term is in fact, so let me just write down what this object is. So in the denominator, you have chi lambda diagonal, Q plus one, one, Q minus one, right? So clearly there's no fractional power of Q. So the leading contribution is OQ. So let's compute the super conformal index up to OQ squared. Then order Q term comes from two contributions. One contribution from here, from the case actors, and from this term when lambda is a fundamental or the anti-fundamental. So let me just write the contribution from case chi adjoint B plus chi adjoint C. So these are the contributions from the case. And also you have this contribution just as before. Q plus minus one plus one plus Q to the one chi box A, chi box B, chi box C. And similarly, contribution from the anti-fundamentals. I'd like to keep it, so I'm so sorry for going over. So this is it, but as I explained in the case of N being two, this is at this order you can just approximate as Q, right? So this is the matter content. Now, we don't see any factor of Q to the one half. So this theory is no longer a hyper-multiplet. It's something much more complicated. And for those of you who love exceptional groups, this appearance of adjoint, three adjoints and product of fundamentals is a telltale sign that this has E6 symmetry. What is going on is that E6 has SU3A times SU3B times SU3C subgroup. So this is rank six. This is rank six two because SU3 is rank two and there are three copies of it. And E6 is adjoint, it's 78 dimensional, which everyone knows. This is the composition into adjoint of one, adjoint of another, adjoint of the third, and product of three fundamentals and product of three anti-fundamentals. So, and this structure persists to higher order. It's a fun exercise if you know how to use mathematical or other computer algebra system to compute higher orders and check everything works. So when N is three, people have found that this particular three punctured sphere theory is no longer a free hyper-multiplet, but this theory is known as E6 theory of Minahan and Nemesh Chansky. So this theory was introduced in, I don't remember, 2009, sorry, 90, 97. It's completely different. So it's known for quite some time, but it's a non-trivial super-conformal theory. So using that knowledge here, we now know some strangest duality. So I just changed this to SU three. This is Minahan and Nemesh Chansky's E6 theory, and this is also Minahan and Nemesh Chansky's E6 theory. So you start from two strongly coupled SCFTs. You couple them to gauge group SU three, and this SU three gauge group is exactly marginal. So you can choose its tune is coupling constant. You make it very strong, and then you have dual Minahan and Nemesh Chansky's E6 theory here, and dual Minahan and Nemesh Chansky's E6 theory coupled to dual SU three. So this is the natural generalization of SU two, four flavor, a sterility to SU three. This is the most natural one, but from the perspective of an old guy like me, this is not very satisfactory because this is a mysterious theory, another mysterious theory coupled to something which we know this is gauge group, and you make it strong, and it is self-dual in some sense. So it's a new knowledge about strongly coupled N equals the theory in four dimensions, but this is just too mysterious, right? So I'd like to understand what happens when, sorry, two Minahan and Nemesh Chansky's E6s, but I'd like to understand what happens when you just take as N equals two SU three plus NF equals six. So this is a more conventional gauge theory, and I'd like to understand what is the dual description when the gauge coupling of this theory is made strongly coupled, so that's what I'd like to do. But as you will see, analysis of this system, which has a simpler Lagrangian description is much more complicated than the analysis of this sterility involving two strongly coupled CFDs. So one lesson we will learn today is that Lagrangian theories in 4D are not always easier than non-Lagrangian theories. So as you will see, the analysis here would be very, very complicated. But let me go on, but before getting there, let me just say a few words about exactly what happens when N is not three, but some general N. In the general N case, if you do this computation, it doesn't fit with anything known before 2009. So for general N, this three-function sphere theory has a name TN introduced by Davide Gallotto. I asked Davide what T stands for. He told me it stands for theory. Yeah, but this is the name we use. So in general N, again, you have this sterility relating two TN theories coupled to SUN gauge group. And the dual is again, SUN coupled to two TNs. Very symmetric, very understandable, but you don't know exactly what this TN theory is. Well, since 2009, we have been working on the analysis of these TN theories for quite some time. So we know quite a lot about these theories, but still they are not Lagrangian theories. So we would like to understand what happens if you start from SUN with two N flavors. But again, the analysis would be very, very complicated, much more complicated than this nice case. So Lagrangians are not always good. Right. So to do this, I need to introduce another concept, another technique, which is very often used these days by people who does the analysis of, non-parameter analysis of supersymmetric field theories, which is called partial closure of punctures. So that's a funny name. So for the next 20 minutes, I need to introduce to you this concept. So, and also there's something called complete closure, but the complete closure is a case of partial closure. So I always call them partial closures. So what's going on? So in general, suppose you have a given N equals to SCFT, let's call it curly T with flavor symmetry, GF. What kind of properties do you have? Of course you have current operators for this flavor symmetry, but thanks to the supersymmetry, you can have more operators, you have more operators acting by Susie once, you have Fermion operator, and acting Susie twice, you have in fact scalar operator, labeled by our joint. But because we are using N equals to supersymmetry, in fact you have SU2R index, and two SU2R indices, and this is symmetric. So this is SU2R triplet, GF our joint, scalar. So this is a very general property of SCFT. And in the jargon of class S series, I've been talking about, a puncture on a Riemann surface corresponds to SUN flavor symmetry. So a puncture means flavor symmetry. So, but before getting there, let's give an example. So consider bifundamental hypermultiplet. Bifundamental hyper consists of N squared fields, Q twiddle ai, both ai are one to N. Then there's a flavor symmetry acting on the index a and the index i, as always. So in this case, GF is SUN times SUN. Well, for simplicity, let's just consider UN times UN. Then there should be three scalar operators, right? So they can be written very explicitly, mu plus, mu zero, and mu minus. And if you consider UN acting or a, say, you just perform a contraction. And again, you perform some contraction. Finally, I'm getting, obviously. Iabi minus Q twiddle bi, Q twiddle dagger ib, right. So in the N equals one notation. So as you see, this component is a chiral super field. I mean, lowest component of the chiral super field. This is the standard D term of the flavor symmetry. And the third component of SU2R is this anti chiral field. So this is, you get the idea. So in the case of free hypermultiple, which is a special case of SCFT, you can write it down explicitly. But the good thing about this is that the existence of the mu's are guaranteed by just super conformal algebra. So they are guaranteed to exist for any non-Lagrangian theory. So what you do, suppose GF contains some SCVN, you do the following operation. So take mu plus, which is a scalar field, which is a N by N matrix. And you give a wave of the following form. So it's just almost zero everywhere, except in this form. So this is a Jordan normal form. Of a new potent matrix. So in this case, I'm using an example N is six. And as you know, Jordan normal form of a new potent matrix is formed by fundamental Jordan blocks. So in this case, you have three by three Jordan block, two by two Jordan block, and one by one Jordan block. So such a wave is characterized by this decomposition of six into three plus two plus one. You can encode the same information by drawing a young diagram consisting of columns of height three to one. So this is a standard convention. So you can give a wave like this. What will happen in general? So you start it from a SCFDT and you gave this wave mu plus. So you get some different theory. I don't know what to, nobody has decided the notation, precise notation, let's say three plus two plus one. And in the IR limit, you get something, but there are some fields you are guaranteed to get because here we have SCN flavor symmetry, right? And we are breaking that symmetry by giving a wave to some fields. So definitely you should have some number of gold stone molds, right? So from the procedure, you know you need, you definitely get some number of gold stone molds. So you take away that from the IR limit and you define the rest to be the partially closed theory. So this is the definition. And as I will tell you, a complete closure corresponds to giving a biggest wave. So mu plus is given by this biggest Jordan block. So this corresponds to N equals N. So the good thing about complete closure and partial closures is that we can start from this mysterious TN theory and perform this partial closure to get three hypermultipleths. So I'm going to tell you that if you start from TN theory, so they have three punctures and undeformed punctures are called full punctures. So full punctures are punctures before the partial closure. And as I emphasize in there are three SCNs each associated to each puncture. You pick this SCN, one of this, and perform partial closure by this particular young diagram, N minus one, one. This means you have this young diagram and do this operation. So you start from this TN, do this operation and in fact you get exactly this N by N free hypermultipleths. So I'm going to explain to you exactly how this arises. But let's assume what this happens. I mean what this means for estuality assuming this fact. So assuming this fact, we can start from considering this partial free puncture sphere so in my notation today, this star stands for partially closed puncture or using this data. Partial closure is just this giving of a wave and stripping away of all the number goes to almost. So this is, this is N by N cues, right? And you connect this with another free puncture sphere. So this is again N by N cues. This part gives you SUN gauge theory, vector multiple, right? So on this side, you know that you get SUN gauge theory plus NF equals two N matter content because this counts, if you consider A as gauge index, I counts as flavor indices. Therefore this counts as N flavors. This also counts as N flavors. Together you have two N flavors coupled to this SUN gauge field. So this is the kind of system you want to analyze whose strongly coupled limit. And its SDOR is of course, obtained by crashing this together. You split them on the other direction so you get this system, right? So all you have to do is to analyze what this is. And I will analyze this system by considering an intermediate step. We start from this intermediate step. You know what exactly this is because this part is just three punctured sphere theory. So this is TN theory, right? This part is a dual SUN gauge group. And finally this part has two full punctures and a puncture of type star. This is the bifundamental hyper multiple. So this is N flavors of SUN, Q and Q twiddle, right? So this final SDOR form of this SUN plus two N flavors is obtained from this funny system by partially closing this puncture. So we are going to partially close this puncture to this type. This way you can understand what is going on on the SDOR side of the SUN with two N flavors. So that's what I'm going to do. So I still have 30 minutes. So hopefully I can get to the end of the analysis. So this is the rough idea of how you understand the SDOR of SUN with two N flavors. So this will be more complicated than just analyzing the duality of TN couple to TN. Good. So let's perform one example of closure in the simplest case. So let's consider SU2, right? And suppose you have a theory like this and pick a puncture. This has SU2 associated to it. So you have mu plus field. And let's give it a wave, 0 1 0 0. In the case of SU2, there's no other way to do something non-trivial. So how do we analyze this? In this case, we know what this is as a gauge theory. So this can be analyzed using this trick. So you have this puncture connected this way. So let's call this puncture B, puncture A, puncture A and puncture B. And this is a gauge puncture, additional puncture. So this part is just Q, A, I, U, say. It's a tri-fundamental. Let's say SU2A acts on QA, SU2B acts here and gauge acts here. In this case, mu fields can be explicitly written. So mu AB with indexed down is given by QAIU, QBJV, epsilon IJ, epsilon UV. This is a standard result just following the super perform algebra. And because I used two epsilon symbols, A and B are automatically symmetrized. So this is in the adjoint of SU2A, which is as it should be. And so this is written in the way one index is lower, one index is upper, but I'm now using both indexes lower. So to set this VEV, what you need to do is to set mu plus one, one to be one and the rest to be zero. This can be arranged by giving a VEV to AIU in this quite obvious way, delta A1, epsilon IU. So I need to give a VEV to the hypermultiplet. So what this does is that you need to use this funny operation of using epsilon symbol relating two different flavor symmetries. So what happened is that SU2B times SU2G is broken down to diagonal subgroup. Another thing is that, let's analyze what happens if you act to this VEV, this SU2A flavor symmetry. We are breaking the flavor symmetry by giving a VEV, therefore there should be a number of gold stone molds. So what would happen in general? Well, A appears only here. So what happens is that this delta AI, A1, just becomes some general vector, QA. So this is the number of gold stone mold. So what happens is that in this operation, SU2G is broken and Higgs mechanism eats three hypers and there's a number of gold stone mold. Number of gold stone mold, there's one. This original QAAU counts as four hypers. You might say that if you multiply two, two, two, you get eight, but this is a half hypermultiplet, therefore there are only four full hypers. So what happens is that out of four hypers contained in this Q, three gets eaten. One is a number of gold stone mold. So I declare that in the process of partial closure, I just remove them. So you don't get anything left. This means that after the partial closure, where should I write? So this guy just becomes, yeah, I have no place to write. This guy just becomes this. So original puncture B and the gauge puncture G is identified because the flavors image is broken down to the diagonal subgroup. So end result of this operation is that we started from this Riemann surface. I gave this VEV, I just ended up erasing just puncture A. That's why it's called closure of punctures. Puncture is just closed and it disappears. So this is the complete closure. So how do we analyze such a partial closure or complete closure in the non Lagrangian case? Because this analysis was possible due to the fact that we knew exactly what three puncture sphere theories is. But here I'd like to apply this procedure for something we don't know the Lagrangian description. So to see that, we need to analyze that process of SU2 in a slightly different way. So let's write this there again. Look about it for a minute. Well I cannot spend that a minute just thinking, but this is not just SU2A triplet, but it is also SU2R triplet. So this plus sign means that it's the highest weight of SU2R and this VEV clearly means that this is also the highest weight of SU2A. This means that in fact a diagonal subgroup SU2R prime is unbroken. This is similar to the procedure Nathan explained to us yesterday. He gave a VEV to phi which had both scale charge and the ghost number. And after giving a VEV, the new ghost number is the combination of the scale charge and the ghost number in the original theory. Just as in that case, I'm giving this funny VEV which carries both SU2R charge and SU2A charge. So this breaks SU2A completely, but SU2R remains. And the new SU2R charge is a combination of the original SU2R charge and original SU flavor charge. So what does this do super conformal index? So remember in the super conformal index you have this structure K, A, K, A and various other terms. So I'm just concentrating on terms which depend on SU2A. So this A variables are taken from SU2A and remember I just erased that super conformal index has this form Q minus I3. So this Q, Q is basically in some sense SU2R. So at the level of SCI, at the level of SCI because the new SU2R symmetry is a diagonal combination of original SU2R and old SU2A, you need to set Q to be, sorry, A, A inverse to be Q one half, Q minus one half just because of this diagonal identification. So what this does is that K lambda A, sorry, K lambda A and K A should be replaced by K lambda Q one half, K Q one half, so far so good. But this is not the whole story because I declared when I defined the concept of partial closure, we are going to strip all of the contribution from the number goldstone nodes. And you can check that this part K has contributions from number goldstone nodes and you need to remove it. So it's helpful to do at least exactly one example. So for SU2, this K A has this form pi over one. So this is the infinite product form of K and let's set to A to be Q to one half. So what happens is that this becomes one minus QN minus one, one minus QN, one minus QN plus one. And you see, this gives a divergence because N starts from one. So this is a telltale sign of an appearance of number goldstone node and because number goldstone nodes in N equals two theory also comes in a hypermultiple form, in fact these two components, two factors in the infinite product corresponds to number goldstone nodes. So this is the reminder. And if you have taken the nodes, this infinite product is exactly the thing I called K zero. So what happens is that at the level of the, at the level of the super conformal index, let's say you have genus G, N puncture theory, so this is G equals two and N equals two, then let me just explicitly write the contribution for this. So this is K A, K B and the K zero to the power, yeah I don't remember, this is two G plus two minus N. So when N is two, this is four and sum over lambda, chi lambda Q one half to the fourth and chi A lambda, chi B lambda, sorry chi lambda B. So let's say you perform the complete closure of this puncture. So let's close this, close this puncture. Then in that part of the blackboard I said, by closing this puncture, you just lose one puncture. But what this, let's check that at the level of the super conformal index. At this level, what you need to do is to set K to Q one half, replace one factor of K A with K zero. So this cancels against one factor. This cancels against one factor in the denominator. And this is exactly the general form of the Q one mils amplitude for the punk, for the Riemann surface with one less puncture. So this characterizes the procedure. You need to be, so thank you for the question. So this K of A is basically the contribution from operator's muse. This is an adjoint in SUM, right? So what happens is that mu takes values in the adjoint of G, complexified, because this is a complexified mu. We are taking some particular wave. So in order to count the number of goldstone nodes, all you have to do is to act on it with SUM. This gives you some kind of a cone. So this is a subspace of GC, which is whose Jordan normal form is the one I gave. So just using the SUM representation theory, you can identify what are the number goldstone nodes very directly. Yeah, that's the method. So I'm almost done. So as a generalization, let's consider the type SUM theory. And let's say we have a puncture, then you have SUNA, and accordingly there's this mu plus. If you give the biggest wave, this eliminates this puncture completely. Instead, let's consider the second biggest wave. So this has N minus one, and this is size one. So this is the partial closure by this young diagram N minus one and one. Now the level of the super conformal index, all you have to do is to replace kA, kA by the following. Again, a linear combination of original SU to R and SU, some SU to subgroup of this SUNA remains. And also, there's some U1 symmetry remaining, right? U1 symmetry generated by one, one, one, one minus N still commutes with this wave. Therefore, U1 subgroup of this SUNA remains. So if you use that, so the rule is this and alpha to one minus N. So alpha is in this unbroken U1, which is subgroup of SUNA, and some factor alpha. So this is the replacement you need to do. But for today's purpose, I don't have to write down exactly what this is. This is just OQ, or one plus OQ. So now I can show you that this puncture where one puncture is closed, partially closed to this type, is a free hypermultiple. So SCI can be written in a standard way. So let's say this part has carry label B and label C, then this has label alpha. And from the general formula, we have KB, KC, and this KN minus one alpha divided by K zero, summation of a lambda of chi lambda QN minus one to one, QN minus three, dot, dot, dot, Q one minus one over two. And on the upper side, I have chi lambda B, chi lambda C, and chi lambda that. I don't want to repeat it. This part is exactly this one. Let's expand this up to Q to one half. So I mean, these parts just don't give you anything up to at the order Q to one half. And the reason you have Q to one half from here is as follows, if you take lambda to be a box, right? Then what you have is you have box B, box C, and this funny thing. So what this is? So it's the character of this diagonal matrix as the fundamental representation. So you just sum over them. So the most dangerous part is this, right? So the most dangerous part is this, plus dot, dot, dot, divided by the trace of this, which is Q one minus N plus dot, dot, dot. Then this part, two part cancels giving you Q to one half. And similarly there's a contribution from chi box bar B, chi box bar C, and alpha inverse. So as I told you at the beginning, the fact that you have super conformal index starting with the power Q to one half means that this is a free hyper-multiplet. And you can read off exactly what is the matter content. So this is a fundamental of this SUN, fundamental of the another SUN, and has U one charge one under the baryonic symmetry. This side is anti-fundamental, anti-fundamental, and baryonic charge minus one. This exactly fits the spectrum of Q, AI, and Q twiddle, IA, right? You need to relabel the fundamental and the anti-fundamental of the second part, but you see that this gives you baryonic charge one, fundamental fundamental, and this gives you baryonic charge minus one, fundamental fundamental, anti-fundamental, anti-fundamental. Right, so I still have five minutes to go to complete the description. So why have I been doing this? This is to understand the duality of Lagrangian gauge theory. But as an intermediate step, let's just consider this duality, so now you know how to read off the matter content. This side is TN theory. This part gives you SUN gauge group. This part I just show you gives you N flavors, right? And this length controls the coupling constant, makes this very strong, goes to the dual frame, you get TN theory, coupled to dual SUN, coupled to NF equals N. Good, we still have nice self-dual behavior of this mixed Lagrangian, Lagrangian, non-Lagrangian QFT, but that's not the final goal. Our final goal was that we want to make this part also this type of puncture, so that this becomes another in quark fields. So to see that, to analyze that, you need to close this too. But the problem is we haven't analyzed this type of fully punctured sphere with two partially closed punctures yet, so we need to know exactly what this is. But the good thing is that we already know this part is a Lagrangian theory. Therefore, we can just do something similar to what I did in the case of SU2 to see exactly what is going on. So let's do that. So this part is QA, I and Q2, AI, and let's say this guy carries A index, and this guy carries I index, right? So we need to give mu plus, again, VEV. Mu plus is QAI, QB, thread line, let's see, say. And we need to give a VEV of this form, ah, which is written here. One way to give this VEV is just to take this itself to be Q twiddle IB, and QAI to be this, 0, 0, 0, 0, 0, 1, so this has n minus two columns and this is two, this is two, this is n minus two. So if we do that, mu multiplied in the opposite order, contracting A instead, IJ, sorry, contracting A, so in the opposite order becomes this. So this is a puncture, I mean, VEV of this form, n minus two, one, one. So what happens is that if you partially close this puncture, sorry, partially close this puncture, sorry, I'm completing. This is the A index and I'm giving I index, sorry, I index is SUN gauge. So this means that I index is broken by this VEV. So SUN gauge is broken to the subgroup which commutes with this, which is SU two. So because of this, this SUN is broken to SU two here. And this puncture is no longer full, but this is closed by two that type. And again, you need to carefully count how many multiplets are eaten and how many number goal storm modes needs to be split. But what happens is that only NF equals one of SU two remains. So that's the outcome of this analysis. Sorry for getting over time. But the final answer is that if you start from this SUN with two N flavors, this is dual to, I don't know, some other punctures of this type, couple via SU two gauge theory to one doublet. So this is the dual frame you finally find. This is very complicated. So not all strong coupleness disappear. You still have this funny object remaining, but this is the fact of life. You cannot get any easier. So the final minus two minutes, let's just consider N equals three case. So in this case, this part becomes one, one, one. So what is a puncture closed by this one, one, one? One, one, one corresponds to Jordan block of size one. This means it's just zero. So in this particular case, you are not doing anything. So closure by one, one, one, one, one means you do nothing. So this is T three theory itself, which is a Minahan-Nemechansky's theory of E six. So in that case, SU three with six flavors is dual to Minahan-Nemechansky's E six, couple to SU two gauge theory, couple to one doublet, which is what Argyres and Zeibach found in 2007, which inspired the Davide to come up with the general method to do this analysis. So that was what I wanted to say today. Thank you very much.