 I would like to start by reminding a little bit about the PQ-Webs as duality in type-2-V and what happens when you place these three brains in the mix. And then I will discuss in detail a billion mirror symmetry in three dimensions and a stepwise approach. And then, can you hear me now? And then an approach which is called stepwise dualization. And then if we have time, we will try to generalize to the non-Abelian case. So these are the brain setups I would like to talk about. Let me introduce, first of all, the standard Honeywitten brain setups. The main players are D3 brains, NS5 brains, and D5 brains. And usually you draw something like this. You have an S-brain, and then you have D3 stretching between an S-brain. So here I explained where the various brains are stretching. So all of the theories we are discussing will be three-dimensional. So all the brains stretch along the first three directions in type-2-V. So here will be X6, will be the direction where the D3 brains are going. Then there will be NS brains, will go vertically. So this will be X4, 5, and 6, 3, 4, 5. And then we can add the D5 brains, which will go along the remaining three directions. So these setups, if we just stop on the first three lines, we'll have N equal 4 supersymmetry. There will be an SO3 times SO3 symmetry acting on this coordinates, which is the SO4R symmetry of the N equal 4 supersymmetry in three dimensions. So what is known is that if you do S-duality in type-2-V, you replace NS5 brains with D5 brains, and D3 brains stay D3 brains, you will have a different gauge theory. And this is related to 3D mirror symmetry. And this part of the story with N equal 4 supersymmetry is quite well understood. What I would like to do is try to generalize to the case with half the number of supercharges. So one way to do is to rotate some of the brains. So we will have NS5 prime brains, where instead of stretching along 4 and 5, they will stretch along 8 and 9. And similarly for the D5 prime, instead of stretching along 8 and 9, they will stretch along 4 and 5. So once we do this, we have half of the supersymmetry. So the theories which live at this type of brain setups will have just N equal 2 supersymmetry in two dimensions. In this case, even drawing a generic brain setup with NS and S prime, D5, D5 prime, one does not know what is the gauge theory living on it. So what we would like to do is just take one example and try to work out the gauge theory. So one example where we know is the following. Instead of having just an S5, we put NS5 and then we put many D5 prime brain, one on top of each other. So this is called the PQ web. If we just do this, this is a five-dimensional theory, because if we take the NS and the D5 prime, you see that they have 0, 1, 2, 4, 5 in common. So they have five directions in common. So this is a five-dimensional gauge theory. In this case, it's an easy gauge theory. It's just free. But what we would like to do is to add a D3 brain going like this, transversely. So this would be now the sixth direction. So this is known how to do. Let me just do the case where we have two D3 brains arriving on the PQ web, which is what I drew here. And then we also add the two NS prime brains just to stop the D3 brains. So what is the gauge theory over here? We have an S brain, an S brain, and then we have D3 brain. So this will give us two U1 gauge theory if we just put one D3 brain. So I draw it in this way. The circle means gauge node in the theory. Then we will have, by fundamental matters, which will be something like this. These are massless strings, which have 0 length, so they will give rise to massless state, massless multiplets. So these will be chiral multiplets. I use arrows. There are two of them. And then I have to add the D5 prime. So if I add just the D5, this will give a fundamental field for this gauge group. If I add the D5 from this part, it will give a fundamental for this. But if I put the D5 prime on top, then I have many massless states going from D3 to D5, which will be fundamental. There will be some fundamental of this group and some fundamental of this group. Then the result is this. So we have K arrows going from one to this global symmetry group. So the square means it's a global symmetry group. So I have an SUK global symmetry, another SUK global symmetry. And they have all these type of lines. This was well known. This was known from the 90s. Now there is also super potential here, which is a cubic super potential. I just wrote it down. It's related to loops in the quiver. Now the question is what happens if I do NS duality? What happens is that instead of having a PQ web with one NS and many D5, I will have a PQ web with many NS brains and one D5. In this case, the answer is not known. What is the gauge theory living over here? It's probably going to be some billion gauge theory because I have just one D3 brain. But the answer was not known. So the purpose of the talk is to work this gauge theory out, and along the way we will find some interesting things involving monopole super potential, which means we have gauge theories where in the super potential there are monopole operators. These are the monopole operators which were discussed yesterday. In the case discussed yesterday it was n equal 4 supersymmetry and the super potential was never involving monopole operators. But with n equal 2 you can add the monopole operators to the super potential and you don't break the supersymmetry. So first of all let me review again 3D mirror symmetry. The simplest example is just one gauge group U1 with one flavor, one anti-flavor. And let me add also a singlet. So this is P and this is Q and then I add a super potential which is sigma PQ. So this is actually n equal 4 super young mills U1 with one flavor. And this is the super potential which ensures me that we have n equal 4 supersymmetry. The mirror here is just a free hyper. This is the simplest example of 3D mirror symmetry and this was given by Interligator and Cyberg I think back in 96. If you draw the picture, the brain setup over here we have 2 NS5, 1D3, 1D5. When I do the S duality I get NS5. So this is the free hyper. These are the massless things that give the free hyper. And this is U1 with one flavor. So we can read off the duality very easily from the brain setup. What is important here is that what is the mapping of the operators. So here we have two free fields, let's call it P and Q, free chiral fields, n equal 2 notation. And what are the fields which are mapped over these two free fields? These are the monopole operators of this U1 with one flavor. So if we call them m plus and then minus, we have that this is mapped under the mirror symmetry in this way. We also have another gauge invariant operator which is sigma here. So where is sigma mapped? If you remember from the talk for yesterday, the sigma satisfies some quantum relation which is the following. Sigma is equal to m plus and minus. So sigma is mapped to the map to the product of P and Q. Okay, are there questions about this? Okay, so let's try to do, this is the very basic example. The next example would be U1 with n flavors. But then we can do it for n equal 4 and then we will have some generalization. We will have the sigma to the k equal to plus and minus. This is what was explained yesterday. Moreover, let me modify this duality a little bit to go to an n equal to dualities. What we do is that we can flip some field. What does it mean? Flipping a field. It means that let's say we have a gauge theory with some operator O. We couple this operator O to a new field sigma. And we get, so we have some theory with some super potential and some operator O. We change the super potential in this way. And we add a new field. So in this case I want to flip this operator sigma. So the super potential will be sigma PQ. This will go in sigma PQ plus sigma times, let's call the flipping field X. So in this case you can see that sigma and X become massive. So I can integrate them out. So I have to use the question of motion of sigma and X to integrate them out. What I get is that I get W equals 0. And I lost both of these fields. So I started from U1 with one flavor in n equal 4. I flipped this singlet and I just get U1 n equal 2 with one flavor and W equals 0. So basically I lost this singlet. What happens on this side? I have to flip sigma, but sigma is P times Q on this side. So the duality now, the mirror duality becomes a theory with three fields, which I can call sigma PQ and W is equal to sigma PQ. So this is an n equal 2 duality, which is the simplest one you can have. And this is just the equivalent to the n equal 4 duality. Just you can go from one to the other flipping at one field. This is called usually X, Y, D model. It's a Vestumino model. There is no gauge symmetry over here. But now it's in the duality between two interacting field theories, the simplest one. So now we can try to do something more general. We can go and do U1 with K flavors. So we start to get something similar to this. You see now here we have the U1 with K plus 1 flavors. So let's try to do U1 with K flavor. One way to do it is to use this duality in this direction, or the duality we had before in this direction. So we have P and Q, which are two free fields, and we can replace them with U1 with free fields and the superpotential. But now I want to do it for each flavor over here. So I will pick the first, let's call this small. I want to pick the pair P, I, Q, I for each I, and replace it with the U1 with the flavor and the sigma I. So once I do this, what happens over here? I have a U1. This is what I call stepwise dualization. This type of approach was advocated by Kapustin and Strassler, always in the 90s. We are now the superpotential. I replace every pair of fundamental fields with the U1 with one flavor. I do this K time, so I have K object like this. So now this theory is not the U1 theory anymore, it's the U1 to the K plus 1 gauge theory. And there will be a superpotential which has K terms, sigma I, P, I, Q, I. But now there is something missing in this picture, and it's the following that when I replace this fundamental field here, these fundamental fields become the monopole operators. So these fundamental fields were charged under the U1. So this means that the monopole operators of all these K gauge theories must be charged under this original U1. The way to do this is to add the BF couplings. So what is a BF coupling? This is like a Champ-Simon coupling. So a Champ-Simon coupling is usually for a billion theory, something like ADA. But if I use different gauge groups, gauge fields, then I will have something like ADB, or BDA, let me put it like this, BDA. But DA is usually called F, so this is a BF coupling. So I started from a theory which is U1 with K flavor. Now I have a theory with many U1s, the same number of charge multiplets. I add the K singlets, sigma I, and then I have to add many BF couplings in the Lagrangian. These are equivalent. I just use the basic mirror symmetry K times. But now something happens here. We see that this U1 doesn't have flavor anymore. This means that I can do exactly the pati integral over here. Once I do this, this pati integral gives me a delta function because this gauge field, I can exactly integrate over this gauge field U1. This will give me a delta function that tells me that a function of delta function, that the sum of all these field strands is zero. So I can think of this theory as the product of K U1 with one flavor, but then there is the relation that the sum of all the field strands must be zero. So a different presentation for this theory is something like this. Take a linear quiver. Instead of K gauge groups, I have to put K minus 1. And then I have many singlet fields, which are flipping to zero all these bifundamental fields. So this is the statement. This is the statement about 3D mirror symmetry for a billion gauge theories. U1 with K flavor is dual, is mirror to a linear quiver. What we did is just assumed that the original symmetry was correct, and then we proved this. So if you believe the first original duality, then this just follows. One way to see this through brains is the following. We can start from here that we have an s-brain, a d5, an s-prime brain, which is rotated. So we have n equal to 2. Then here we put many d5, and then we do the s-duality. What happens? The d5 become an s, and then here we have a d5 prime, and here we have a d5. So we see that also in this picture we have a linear quiver. So many U1s, one connected to the other. And then there is a flavor for the first one and the flavor for the last one, which is precisely this. So again, all this goes back to 97, I think. But this type of procedure is some kind of procedure that we can try to apply to this theory. So here we have k plus k plus 1 pairs of fundamental fields where I can apply this basic step. When I do this, I will create many U1s. So we can already see that here we will have a theory which even if it looks like there are only, you know, the Dennis brains are all on top of each other. We will get a theory with many U1s, which is kind of strange. Anyways, let's first do the case k equal 1, because in the case k equal 1 you already know the results, and also we learn something. So the case k equal 1 we have from one side. We have this theory, U1, U1. There are six fields. And from the other side we know what is the theory, the theory here, because we have just one and s5. It's basically pretty much the same theory, but it's not gaged. So there are six fields, which you can see them as massless strings going from d brains to d brains. But there are no gaged fields, because there is just one and s brain. In order to have gaged fields you need two and s brains and these three that can slide upon them. And we have this type of super potential, the cubic super potential. So this would be s and s, d5, d5. And here we always have the same 1, 1, pq web, which upon s duality goes into itself. So this is the same on both sides. OK, so we can try to understand this duality, but one way to understand this duality, first we would like to prove it, usually always using the basic Abelian mirror symmetry. So all this talk is based on a paper that I wrote with Sara Pasquetti. Which is 16, 0, 5, 0, 2, 6, 7, 5. And in the paper we prove this duality. But along the way we found the third theory, which has just one gauge group. And so this duality actually becomes a reality. So one way to find a theory with one gauge group is to basically start from here. So let's put here k equal 3. So you see that the matter content is the same as this. We have a linear query 1 cross u1. The difference is that here we have three singlets with this type of super potential. Here we have zero singlets and a different super potential. So here the super potential is, if we call this p and q. The super potential is Apq plus At tilde, p tilde, q tilde. So what can we do is that we can try to modify these to arrive here. But we know that there is a mirror on this gauge, which is one with three flavors, without super potential. So let's try to add here, we can deform and add sigma i xi. So we have three singlet fields and we flip the sigma i. If we follow the map, the mirror map, the sigma i's where, so this is like pi qi, this was where like pi qi. So there are three sigmas and there are three mesons that I can write. So what it means is that here I should add a super potential, which is xi pi qi. So three times in the super potential. So I can write, let me write this differently in this way. So now I break the su3 global symmetry to u1 to some power. And here again I can integrate out sigma and x and I get w equals zero. So this is a different presentation of the basic duality, but this goes, now we are getting something similar to here. Now we have to add this type of super potential, Apq. So what was the map before? When we had u1 with k flavors, the map of the mesons, so we had pi qi is a matrix of mesons. So let me do, let's do the case of three mesons of three flavors. Then here I have p1 q1 p3 q3. And this is a non-trivial aspect, which is not discussed usually. So the diagonal one, they were going to see to the sigmas in the basic ability of middle symmetry. The off diagonal one, they're going to monopole operators. So for instance p1 q2 goes to the monopole operators which charge one and zero. So here I have two gauge groups. So when I say a monopole operator, so I have to assign the charge under both gauge group. Then I can have the monopole operators which charge one and one. And this goes, this is mapped to p1 q3. Then I have to charge zero and one. And this goes to the other one. On this half, on this triangle is the same, but with negative charges. So this is the map for the basic middle symmetry for u1 with three flavors, and it's possible to generalize. Here we also have mesons m with charge plus one. Here is mapped to the operator product of all the pis. And then with charge minus is mapped to the operator product of all the qis. So we have a linear quiver. We take the product of all the operators going in this direction. We get the gauge invariant. This gauge invariant is mapped to a monopole on the other one. On the other hand, here we have mesons, quadratic gauge invariant. These are mapped to a monopole on this. This is a generic feature of middle symmetry, which is mapping normal operators to monopole operators. So if you have some relations, which are satisfied by normal operators from one side, they will become quantum relations for monopoles on the other one. Okay, so here we want to add the apq to the story, which is the product of all the p in this direction. So we want to add the super potential. This is p1, p2, and p3. We want to add to the super potential p1, p2, p3, plus the other p1 tilde, p2 tilde, p3 tilde. So we get, now this theory, if we do this, becomes the theory describing the pq web. But we saw that what are these two terms mapped? These are mapped to the monopole. So here we have to add monopoles plus one, plus monopole minus one. So, say it again. This is a u1 cross u1 gauge theory with some super potential, which is precisely the one that we know describes this pq web. This is due to a u1 gauge theory with three flavors, three singlets, and monopole super potential, which is not obviously related to the brain picture. So this is a duality. Now it is also possible to show that this duality, one way to do is, I'm just going to sketch the discussion, one way to do is to think about this theory as u1 with two flavors. So we have two incoming arrows and two outgoing arrows, but then I gauge these two u1s together. So basically I take u1 with two flavors and find these flipping fields. Then I gauge and I get this theory. But if instead of starting from u1 with two flavors and the flipping field, I start from its mirror, which is just u1 with two flavors, I get some theory which if you inspect a little better, you see that actually it becomes... So from one side I have this, from the other side I have u1 with two flavors. This is the mirror symmetry for u1 with two flavors. Now I gauge something over here and I get that theory. So these are u1 cross u1 gauge theory. When I do the same gauging here, I get the product of two separated theories, which is u1 with one flavor, two times. But this again is equal using the xyd. This is equal to xyd model and this is equal to another xyd model using duality again. So I see that the product of two xyd model is equal to this theory. But this is exactly what we have here. We have six fields with two superpotential coupling, which are xyd. So basically what we found is that there is a reality here. This is one presentation, this is another, and this is obviously related to the brain system. This is another and this is related to the dual brain system. So what we want to do in the next half an hour is to generalize this reality in two different directions. One is to add more NS5, and the other is to just look at this. So one is to generalize this part of the reality and one is to generalize this part of the reality. Here we have u1 with three flavors and here we have just the vestumino model. So what we're going to do is that here we can replace u1 with three flavors, with u and c with an F flavor generically, add the monopoles and see what happens. There will be another dual. This will be a Aaroni cyber type of duality, but for the moment let's just try to generalize in this direction. Are there questions? Okay, so let's apply this stepwise dualization to this theory. So here I have many, I have the two u1s, then instead of taking A and A tilde, which are despite fundamental, let's replace it with the usual big A, big A tilde, and sigma A. Then instead of having PI, PI tilde, I replace it with u1. So I have sigma I, PI, big tilde, big and so on. So I have K of those. Same story from the other side. PF couplings, which are connecting me, but then I have a PF coupling like this. So I just use the 2K plus one times the basic a billion bill of symmetry on this case theory. And I get this strangely looking object. Now I can do the integral of the two u1s. I get two deltas now. When I implement the delta, I get a linear quiver exactly like the beginning. Then I get this u1 in the middle here. So these are A and A tilde. Here I have K minus one. And then I get another linear quiver from the other side. I have all the singlets. I'm not drawing the singlets, but there will be many singlets. 2K plus one. OK, so we see that what we are getting here is a theory with many u1s gauge groups, which should describe this brain system. We should be careful about the super potential because here we have all the singlets and all this type of super potential which will be sum of sigma i, pq, pi, qi, pip and so on. So we have all this type of flipping super potential, but then we also have the super potential coming from this cubic. So what is this going to become? It will become some monopole operator super potential. But the monopole operator here is not local on the quiver. So for instance, the terms p, i, q, i, a will be a monopole. So the super potential here will be this. Then we have sigma a, a, a tilde. Then we have something similar to this with the tilde, with the qs. But then we have a monopole. So the monopole will be something like all zeros, 1, 1, 1, all zeros. This will be one monopole term associated to p1, q1, a, p1 tilde, q1 tilde, a tilde. This is just the first half. Then we have the same with the minus, all zeros, minus 1, minus 1, minus 1 and so on. Then we can do p2, q2, a and this will be this object. 0, 0, 1, 1. Then we have 1 in the middle. So this notation is the charge. These are all the charges under all these ones. This one is the one in the middle. 0 and the same with the minus. And the last one will be m with all 1 all the way. So we have many monopole operators, 2k monopole operators, half with the plus sign, half with minus sign. But they will involve groups which are not attached to each other. So this is not the local in the quiver. Usually everything was, when you have a quiver with an equal force supersymmetry, every node will just talk with the nearest neighbor. This is a little bit more complicated. Another thing we can do, we can use the result from before to replace this one with three flavors in the middle with the square. So before we remember, if you remember, we had this duality. So we can apply this duality inside here and we get a theory with one less gauge group because this is one gauge group, this is zero. And we have six fields. So this is a presentation which in the case k equal one reduces to what we know. The other one didn't reduce. So we have four different theories. We have this theory, then we have this one, which would be the dual of here. And then we can also use this duality inside both to change the presentation a little bit. If we use this presentation, we'll use this term, the sigma tilde a tilde, but we get, if we call this is p1 q1 p1 tilde q1 tilde. This is xx tilde. We have a cubic term, px p1 q1 plus x tilde p1 tilde q1 tilde. And here we lose the one in the middle and it's just like this. So again, we have many super potential operators. We have the cubic terms and then we have flipping terms. So this will be the result for this question mark that I drew before. And the interesting thing is that there are many ones more than you would get if you spread this k NS5 brains because we have k NS5 brains. So in n equal 4 you would expect k minus 1 u1, but here we get 2k minus 1, so almost twice as much. And then we have all these monopole super potentials. So this is the result. So I think one thing one can do is to map the chiral rings from here to here. This is not easy. We just mapped the generators of the chiral ring because it's easy to study the mesons. It's not easy to study the whole set of monopole operators. So for instance here we have many mesons which are mapped into many monopole operators that you can construct here. And then we have a few monopoles, just the one you can construct out of these two u1s. And these are mapped to many mesons of many type of normal meson operator you can find over here. But you can check that at least at the level of the generators of the chiral ring there is a perfect matching between the two sides. OK, if there are no questions, I will go back to this duality again. So let me write it again. So from one side we have u1 with three-flavor and the super potential is, let's call it this Pi, Pi tilde. The super potential is sigma. There are three flipping terms, sigma i, Pi, Pi tilde. And then there is the monopole with charge plus one and the monopole of charge minus one. This is this side of the duality. From the other side you just have six fields, A, B, C, A tilde, B tilde, C tilde, and the super potential is A, B, C, plus A tilde, B tilde, C tilde. So this presentation of the duality is not SU3 invariant. So what we can do is to get rid of all these of the three singlets in such a way that we have SU3 times SU3 global symmetry. So what we have to do here is to add the sigma i, xi. Now we have to look at the map. So here we will be introducing some more cubic terms because the sigma i are mapped to quadratic operators over here. The end result is the following. We have u1 with three flavors and just a simple monopole m plus one plus m minus one is dual to this object. So what is this? This is xij. This is a bifundamental matrix of SU3 times SU3. So there are nine fields. We started from six fields. We added three fields. And the super potential is just the determinant of x as a matrix, as a three by three matrix. So there are nine terms in the super potential. In this way you can see immediately that the SU3 times SU3 global symmetry here is preserved on this side. One thing is that the duality, this duality we discussed at the level of the field theory, but if you put the field theory on S3, so this is what in the previous talk we heard about, these S3 partition functions for N equal 2 on S3. You will find some partition function on both sides. And this will be something like dx. And then we will have three times. We have Sb of x plus Mi, Sb of minus x plus Mi tilde. So this will be a U1 gauge theory with three flavors and three anti-flavors. And from this side you will see that there is just a product of nine singlets. So basically you are projecting down the duality from the level of the full gauge theory to the level of a three partition function. You will find some identity. This is the identity. It doesn't really matter. But this identity was already present in the literature. So for instance already in 2002 there was a paper that was discussing this identity. Of course they didn't know about the relation with gauge theories and this was called ultimate integral identity because apparently for their purposes which was some integrable system story this was a very basic identity. Otherwise it was also known as a new pentagon identity. Okay, so what we did in a paper which we didn't publish yet with Francesco Benin and Sara Pasquetti is to try to generalize this story to the case where you have UNC with an F label. So I just tell you the result. The result is you have NC, then we have NF flavors and the superpotential here is just M plus plus M minus. So these are the two basic monopole in UNC with charge. So M plus is the following. The charges are if we go on the carton of UNC there will be plus one and all zeros. This is M plus and minus will be something like all zeros and minus one. So these are two different monopole operators we have to add both of them to the Lagrangian. And once we do this we are left with the global symmetry which is SU and F square because the monopole operators if you don't have the Lagrangian here you have SU and F square but then you have a u1 topological symmetry and then u1 axial symmetry. But these M plus and the minus are both charged under both of these two you want symmetries with different charges so you're breaking both of them. And the dual is okay let me write U and F minus NC minus two is the two end flavors the NF flavor sorry and the superpotential is the usual cyber superpotential so sum i and j M ij Q i Q j so this is very similar to the Aroni duality but the Aroni duality doesn't have the minus two. And then here we have the other superpotential is again M plus plus and minus where this M plus and the minus are the monopole operators of the dual gauge group and here is another different the Aroni duality there is a flipping of the two monopoles here we don't have any flipping but they both appear in the superpotential and again here the global symmetry is SU and F square so one way to argue for the symmetry one can check that this duality makes sense for instance one can compute the modular space of acqua on both sides and it matches another way to argue for the symmetry which also provides a ultraviolet completion for the theory is that because the one question you can ask here is whether starting from just UNC with NF flavor and adding this superpotential is relevant so generically if NF is too big the superpotential is not relevant the boundary case is NF equal to NC plus one anyway but we can start from a duality in four dimension so this is a 3D N equal to duality we can start from a duality in four dimension which is a USP duality USP 2 and C with NF flavors usually it's called 2 and F with the 2 and F flavors so this is not anomalous in the four dimension and there is a dual which is the interligator pull-out dual which is USP 2 and F minus 2 and C minus 2 with 2 and F flavors and here there is a superpotential which is MQQ this is very similar to can you read here ok so basically you are this is like the cyber duality for USP gauge groups if you take NC equal to one USP 2 is SU 2 and so this is the notation I'm using so this you can reduce to three dimension you find the same duality except that when you reduce to three dimension you are turning on a monopole superpotential so in 3D let's just go directly to 3D you have this duality but now the superpotential here is the monopole which is there is just one monopole for USP and again here we have the monopole now there is a way so this theory is SU 2 and F global symmetry there is a way to turn on real masses in such a way that the USP is broken down to U and C and 2 and F is broken down half of the flavors become massless make up massive half of the flavor but now the superpotential gets doing this again you are going you are giving some expectation value and giving some expectation value so this is the mechanism which was discussed by Bollywood back in the 70s generates monopole superpot can generate a monopole superpotential in the infrared so here we get monopole plus plus monopole minus so we get precisely this and if we do the same on the mirror on the dual we get precisely this dual so this is another generalization of this reality I was discussing at the beginning one nice thing is that if you plug F equal 2 and C plus 2 which is the in some sense the self-dual point where there are charges for instance of the basic fields will be exactly one half so if you are familiar with the cyber duality in four dimension if you have SU N with two end flavors it's a similar situation this theory lives at the S duality wall of a four dimensional gauge theory which is SU N with N equal 2 with two NF flavors so this is N equal 2 in 4D so this theory can have a duality wall this was discussed by LeFloch and he found some sort of duality like this but he didn't show that there is a monopole superpotential so some things are not really working well so anyway so last comment I want to make is that if you start from this S duality wall you have a co-dimension one you have a wall in four dimension on the wall there is a three dimensional theory which is precisely this type of theory with the monopoles so this is another example where going from higher dimension to lower dimension you generate monopole superpotential the other example was when you compactify on S1 you generate one monopole superpotential and also here I think I can stop you