 So, thank you for coming, and it's a great pleasure for me to be back at Trieste Spring School. This is my third time. The first time I came here 11 years ago as a graduate student, and I still remember very nice talks back then. Marcos Narinio was one of the speakers. I remember him talking about topological B-model and integrability, and I really fondly remember that. So, let me start, but let me start very slowly, and let me first give you the overview. So, the title of the talk is Estuarity in 4D N equals 2 theories. Can you read it? I hope you can. Right, so my main question, even bigger than this one, is this. How does a very strongly coupled gauge theory behave? It's a very difficult question, fascinating question, and it has fascinated so many physicists in the last few decades, not just the last few decades, but more than half a century, but it's a very difficult question. So, we'd like to start by starting from a simple example, and first you notice that there are two types of strongly coupled gauge theories in four dimensions. One is, one type is where coupling becomes strong via renormalization group flow down to low energy. Another thing is, another type of strongly coupled theory is that UV coupling constant doesn't run, so UV coupling is a parameter you can choose. You choose, you make it strong, so there are two types. The first type is rather difficult because the system becomes strongly coupled all by itself, so it's very hard to analyze, but in the second case, you can, it's you who are making problem difficult in some sense, right? It's a tunable, controllable situation, so in this case you can start from carefully analyzing the weakly coupled case and slowly making the problem harder. So in the four lectures I'm going to give, this time I'd like to focus on the second type of the theories, and in particular I'm going to impose lots of supersymmetry. So with Susie, this problem is quite tractable, thankfully. So there's a long history of such a study of the strongly coupled behavior of Susie gauge theories. It all started in a paper by Montonen and Olive from 77. So there they studied, well it's not quite true, but in any case, they considered any equals four superion mills. Simone has been talking about scattering amplitudes of this theory, but we consider any equals four superion mills with SUN gauge group. This is clearly, you can compute the beta function, and it's one loop finite. It is believed to be, it's known to be finite to all orders, so this is the second class of theories, and we believe that if you crank up the gauge coupling of this theory, eventually this is mapped to itself. So this is the strong weak coupling duality of any equals four superion mills, and that's called S duality, and this is from 77. The next example found in the literature is by Cyberg and Witten from 94. So this is any equals two. Super symmetric gauge theory with SU2 gauge group with four flavors. Again this is one loop finite, believed to be finite to all orders, and in this influential paper where they introduced the concept of the Cyberg-Witten curve, to begin with they also showed, they gave a convincing evidence that the strongly coupled limit of this particular gauge theory becomes basically the same theory, but with a different flavor charge assignment. So with four flavors, I will explain it in more detail, but with four flavors, this theory has SO8 flavor symmetry, and in the original theory, in the original theory the matter fields is in the vector representation of this SO8, but in the dual theory the matter fields are in different representation, so in the SO8 flavor symmetry, so it's a eight dimensional spinner representation, so that's in 94. So a natural question you ask is that, well, you can find a very similar example to this. Yeah, SO8 symmetry? Pardon? Ah. So this SO8 symmetry, so let me say a bit more carefully, so this jargon any equals four, means that you have four pairs of quark and anti-quark fields, and i is one, two, four, i is one, two, four, so this one, two is the gauge indices, and there are four flavors, but you know that in SU2, doublet and anti-doublet are in fact the same thing, so it's more convenient to write this as SU2 doublet with eight components, and the interaction term in the Lagrangian in fact preserves SO8 rotation of these fields, so that's the standard charge assignment of the original SU2 theory. So what Zeibergen Witten found was that if you crank up the gauge coupling of this theory, eventually you have some dual gauge theory with basically the same gauge group with the same matter content, but the assignment of the SO8 flavor representation needs to be changed, so if you know the Dinkin diagram of the SO8, you have three nodes corresponding to the eight-dimensional vector representation, eight-dimensional spinar representation, eight-dimensional conjugate spinar representation, and it's your choice of assigning the SO8 representation to these eight matter fields, so that's completely up to your choice, but what they found is that if you go to the strongly coupled dual description, you need to simultaneously change the flavor charge assignment, so this will become very important in the third lecture I guess, so I will come back to this. Thank you for the question. So everybody thought this should be generalizable, right? If you know the one loop beta function, you can easily compute that N equals two gauge theory with SCN gauge group with NF equals two N. Gauge theory is also super conformal, I mean the gauge coupling is tunable, and that is a special case where capital N is two, but for quite some time people couldn't understand exactly what happens when you crank up the gauge coupling here. The problem was, just as we can see from his question, this duality in the simplest case involves this accident, some mathematical accident which only exists for SO8, so it's not quite clear how to generalize this concept to this more general case. So the breakthrough came in a paper by Gires Zeiberg in 07 where the SU3 case was treated, so what they found is that SU3 with NF equals six in a very strongly coupled limit is dual two, so it has a weakly coupled description with different gauge group SU2, coupled to just one copy of the matter field, together with some mysterious strongly coupled sector known as Minahan and Mishanski's E6 theory, I will come back to that, but so there's a qualitative difference between these two cases, in this case if you crank up the gauge coupling eventually you find a dual description where everything is weakly coupled, but that is a bit too good to be true in the general case, what Gires Zeiberg noticed in 07 is that in this case if you crank up the gauge coupling you can go to a dual frame where you have weakly coupled SU2 gauge group, but you still have some strongly coupled mysterious matter system coupled to this SU2, and quickly Gaiotto in 09 generalized this to all SUM, yeah general case, so what I'd like to do today and in the four lectures to come is to explain these dualities from very basic point of view, so Gaiotto's great insight is that he gave a very general perspective point of view to understand these dualities in one go, so what he found is that the dualities in, so I didn't have to erase this, but N equals 2 can be understood by compactifying, putting something called 60, N equals 2,0 theory on a Riemann surface, so Riemann surface can be split in various ways and that corresponds to various weakly coupled limits, so that's what he found, and that explains a lot of thing about this mysterious occurrence of strongly coupled matter field in the dual frame, so in this SU3 case we had some strange matter content in the dual side, that's inevitable because I mean this 60 theory itself, so this doesn't have useful Lagrangian so far, Lagrangian description, many people tries, there are some, but it's not useful yet, so if you start from a very strange theory in 60 and compactify that on 40 to 240, you would expect you get something very strange and that's the generic situation, but sometimes even if you start from this strange theory, you get very understandable thing in 40 and that allows us to understand the strongly coupled behavior of the gauge theory better, even in four dimensions. Another line of development started by Davide Gayotto is the following, so let's consider this six-dimensional theory on some four-dimensional manifold times a Rayman surface, that's right, so you need to use various indirect methods to study these things, so sometimes you know that the resulting 4D theory is a Lagrangian theory, sometimes you don't, typically you can compute two point functions of currents and such things, and you need to guess exactly what these 4D theories are, but I will come to concrete examples later, thank you for the question. So let's start from this product space situation and consider computing some partition function of this, and let's name, let's give a name to this object, so let's call this 60 n equals 2, 0 theory, just theory S, 60, so this is a partition function of this S6D theory on this product space time, if the Rayman surface is very very small, you can first compactify along that small Rayman surface and consider this system as a four-dimensional theory, so in that case you have some 4D theory determined by the Rayman surface, and you are considering that theory on X4, so that's one way to analyze, there's another way to analyze the system, so if suppose X4 is smaller than this C2, then you can do the same thing, so you first compactify along X4, then you have some two-dimensional theory determined by this X4, and you consider that theory on a Rayman surface, right? Of course in the general situation these things are all different, you need to take limiting procedures, but suppose you consider a nice, if Z60 is a nice Suzy partition function that doesn't depend on the size, so this is a big assumption of your choice of precise background, but suppose this is the case, in that case and the size doesn't matter, therefore I guess I should use a colored chunk, in this case the three things I just wrote on the blackboard are the same, so this becomes the same because this shouldn't depend on the size of C2, this shouldn't depend either, but because these are the same, we should have some relation of a partition function of some 4D theory on a 4D spacetime and a partition function of some 2D theory on a 2D spacetime, so these days this four-dimensional theory determined by the Rayman surface, there are tons of them because there are tons of Rayman surfaces, you can even add various punctures to it, and these are called class S theories, I don't know who introduced the terminology, it appeared around 2012, no, 2000 already had the name class S theory, oh thank you, okay yeah that's an easy choice, so until now I didn't know what this S stands for, great, great, so that, yeah, alright, by knowing the origin of the name it seems so less mysterious, which is too bad, so apparently this stands for six, alright, such mundane choice anyway, more interesting thing is that if you compactify 6D theory on S4, we now know this is 2D Liouville theory, in general Toda theory, and another famous example is S2D and if you instead consider S3 times S1, you get something called 2D QD form, so before getting further I would like to mention that these equations are amazing to me, usually in physics when we write equations, both sides of the equations are numbers or functions, here in these equations both sides are quantum field theories, so we are now, I mean QFT knowledge of the human beings as a whole has progressed to a certain stage where we can now write down equations between quantum field theories, so I feel good about it, but we need to proceed, so I already said that this is my third time here, when I came here in 2010 or 11 as a lecturer, I talked about this particular equation, so I don't want to repeat that, so this time I'd like to explain to you about this third equation, and in my perspective, in my subjective way of viewing things, this is much more understandable than this one, because this is simpler, tractable, contains lots of information, easily extractable, so I really like this version better, so that's a very general overview of what I'm going to do in the rest of the lectures, and before getting further, let me mention that I kind of have a lecture note for this, it's not exactly the lecture note for this set of four lectures, but a friend of mine forced me to write short introductory notes about these two relations, and that's available as a preliminary draft, so let me just say the URL, so member.ipme.jp, ug.touchcour, and temporary file, and 2d40.pdf, or maybe it was 42d.pdf, I don't remember, you can try either, and you will find a file, so you don't really have to take the notes, but yeah, right, so this is the very general introduction, and since I have a nice blank space here, let me just give you a general contents of lectures, I would start as an appetizer, talking about 2d, qd, from the armills first, I mean we'd like to study 4d gauge theory, but as a toy model 2d gauge theory is a very, very instructive example, and then I move on to the very basic facts about 4dn equals to Lagrangian's, and then I will cover SU2 NA equals 4, and it's super conformal index, so this super conformal index is another name for S3 times S1 partition function, and I will, once I get to the point, I will give some 16 interpretation, and then I will discuss the generalization to general SUN, generalization to SUN, so I'm not sure how far I can go, I'd like to explain various things concretely, so I might not get to this generalization part, but it should be okay, SU2 examples are instructive enough, so any questions up to this point? Yup, so in order for this kind of analysis to work, there should be a way to add the background fields that couples to the theory that preserves some amount of supersymmetry in the total space, that gives you some constraint on the choice of usable 4-manifolds, and people have been classifying possible types of 4-manifolds you can use in this way, so typically easy thing you can do is to take all before either of these things, and you get various complicated things, and even that case is not completely understood. More questions? Alright, so let me start with 2D young males, pardon? So sorry, sorry. S4, I don't think it's a complex manifold, so it's not required. So there are series of works by Zeiber and Festutja and Dormitor SQ, where they discussed exactly which manifolds allow some number of remaining supersymmetries, so you can typically take one of them, plug this in, and do the rest. So I'd like to discuss 2D QD-formed young males, but before getting there, of course we should discuss undeformed 2D young males, right? So let's just discuss 2D young males theory. It's such a simple theory. In 2D, there's no physical propagating degrees of freedom, so let's take G as the gauge group, some non-Abelian V group. So everything I say can be found in a great review of 2D young males, either by Blau Thompson or by Cordes, Ramgulam and Mua, they are cited in my lecture notes. So if you are interested in a very detailed structure of 2D young males, you can find everything there, but let me introduce to you how this thing works. So this is a really simple theory, so if you consider this in a flat space, you don't get much. So instead, let's consider this in a general two-dimensional manifold, in a euclideanized sense, and Lagrangian is, so this is the coupling constant, I could use G, but I cannot use G because I'd like to use generic metric, which is denoted by G, and you have the standard kinetic term, so this is Lagrangian, nothing spectacular, but in 2D something interesting happens. In 2D, the only non-zero component of F mu nu is F01, right? So you can rewrite, rewrite, trace F mu nu, trace F mu nu very explicitly as G00 G11 minus G01 G10 trace F01 squared, and as you know, this is determinant, inverse, inverse, right? So if you plug this back in, what you find, what you find is that S is just given by D squared sigma, a DG squared inverse, trace F01 squared. So you see that explicit components of the metric never appear, G mu nu only enters into Lagrangian in this combination. In 2D, giving determinant of G means that if you have a two-dimensional surface and given a very small patch, you can measure its area, because it's given by determinant of G times dx dy, but in 2D Lagrangian, there's no need to specify exactly what is the metric. All you have to do is to specify the area element. So from this, you can see that 2D, young males, only depends on the area. So this is a good thing to know. And this coupling constant can be absorbed into a redefinition of the area and just set in the scale of the area. So from now on, I can just ignore it. So let me just give you a Venn diagram before proceeding. So there are tons of 2D QFT, many of which are very complicated, right? And many of you know there's something called CFDs. So CFDs are some very special QFTs, which depends only on the complex structure of the Riemann surface. So that's a special type of 2D QFT. But we just learned that there is another type of special QFT, which doesn't have a nice name, but we can call it areaFT, where the theory only depends on the area, total area of the surface. And you can have some intersection between the two. If a 2D QFT only depends on complex structure and also only depends on the area, then this is purely topological. So this intersection becomes 2D TQ, topological quantum field theory. But it's important to remember that 2D young males is an area field theory, if that's the word. So how do we solve it? Solving it using path integral is also very instructive, but due to the lack of time, let me just do it in Hamiltonian formulation, which is very instructive again. So the easiest space time is of this form. So it's a cylinder. So let's say this is x0, and this side is x1. This periodicity is L. This length is T. And so this is the time direction. Maybe I should have drawn this vertically. So in any quantum field theory, so this is a constant time slice. Given a constant time slice, you should have a Hilbert space of the system. So what's the Hilbert space of this system of the 2D young males? So classically, what characterizes the gauge field configuration on S1? So what you can do is to pick an origin here and define a group element, which is path ordered exponential from 0 to L of a1 dx1. So this is a nice almost gauge invariant object defined by integrating the gauge field around S1 once. But of course you need to remember that there's a residual gauge symmetry at this origin. So if you perform a gauge transformation by g, u is sent to g, u g inverse. So this is the residual gauge invariance. Therefore, the Hilbert space of states of this theory is given by wave function psi, depending on u, which is invariant under this gauge transformation. So this is the space of wave functions. So such a function is known as a class function on g. What are the class functions? Typical example is the character in some representation r. So this is just trace r, a trace in the representation r of this matrix u. And it's known that chi r for all irrep are spans hs1. So this is a mathematical theorem. In a product can be nicely given by a standard integral over g. So let me just give it to you. So if you have two such things, so you would like to take the inner product and you integrate that over the space of u using the standard heart measure. And this is given by just a delta function of the labels. So it's a very nice orthonormal basis of hs1. So now we know the Hilbert space. What's the Hamiltonian? So I'm using a canonical approach. So we just write down the Hamiltonian. So this is an integral from 0 to L of electric field squared up to some constant. I mean, in 2D there's no magnetic field. You just have f01, which is the electric field. And in quantum theory, this is given by the standard canonical quantization. Electric field is the conjugate momentum to the gauge field. So you get this. So you can really directly act this h on, say, this chi r of u. You can really compute it. So this is trace of r and px and 0 to L and a1 dx1. So we can take a gauge where this a1 is constant along the x1 direction. But this h just hits a twice. And in order to see the process more carefully, I guess we should introduce a joint indices a. So this is a sum of a. And it's better to write it as ta a1a. So this t is the representation matrix in the representation r. So if you hit this u by one variational derivative, you just get one factor of ta. So this is basically trace r and integral 0 to L ta ta dx1. And you have original px0 to L ta a1a dx1. But you know that in a irreducible representation, this is just a value of the quadratic cashmere. So you found that this is equal to L. So there's an integral. So this is L times quadratic cashmere of r times chi r of u. So we know that this representation basis is, in fact, the eigenstates of the Hamiltonian of the 2D young mouse. And the Hamiltonian is given by quadratic cashmere. So we are talking about this situation. But for example, we can further identify this part and this part identify. So you get the torus. You can compute a partition function. And this is given by trace over the Hilbert space of exponential of minus ht. And because we know the Hamiltonian and the basis here, you can very explicitly write as a sum over the irreducible representation of exponential of minus t times h is L times c to r. So this is the partition function on the torus. As I already told you, the partition function only depends on the area. Here we arbitrarily introduced length L and length t in the system. But the final answer depends only on the area, as I already claimed. So now you see that. Next step, let's consider the following geometry. So this is a disk with area a here, disk. So it only has one boundary. And in such a situation, you get, so you still have some Hilbert space associated with the boundary, and you get the wave function determined by doing a path integral of the system with given boundary condition fixed in holonomy to be u. So we'd like to determine this thing, psi a of u. These are called Hartl-Holkin wave functions, originally introduced in the context of cosmology. So the world starts from nothing and pops up, and that uniquely determines some nice wave function. And 2D-Yaml is simple enough so that we can know this creation from the nothing. So this is Hartl-Holkin wave function. How do we determine it? So there's a first nice trick of changing the area. So adjoining a cylinder of psi area a prime gives you a disk with a plus a prime, and we know this cylinder contribution. This is just the action of the Hamiltonian. So this means that psi a prime a plus a prime of u is given by exponential minus area times h. I guess I'm using some wrong normalization of h, but you get the idea, right? So you can easily change the area. Therefore, it suffices to know the zero area wave function. So what's that? The zero area wave function can be understood in a path integral point of view. So we have a very big space time, right? And you have a zero area disk cut out of it. You might not be able to see it, but you get the idea. This is very, very small. You see? Very, very small. Therefore, the homonomy around it, u, needs to be basically one. Therefore, zero area limit of this disk wave function should be just the delta function. But as always, in the case of delta function, this is not square integrable. So natural normalization is not very clear. So let's just put the arbitrary constant alpha. So this is the result. Of course, you can write it in representation basis. So let's say this is given by CR. C is already used too much. FR, chiR, u. Now, how do you know chiR, f of R? All you have to do is to multiply chiR prime u and integrate over u. So on this side, because of the orthogonality, you get just f of R prime. On this side, because this is a delta function, you just get chiR prime 1 times alpha. And this is a dimension of this representation. So you know that this has an expansion of this from alpha, dimension of R times chiR, u. So very explicit. So now we know the discomfort. And finally, let's determine this geometry. Let's consider this geometry. So this is a pants with three holes with area A. So in this case, you have three Hilbert space associated with it. Therefore, this should determine a vector in Hs1 tensor, Hs1 tensor, Hs1. Or equivalently, if you assign three holonomies, u1, u2, and u3, the wave function psi of A, u1, u2, u3. So how do we determine the three boundary wave function? It is, in fact, very easy by using the following trick. So if you have a three boundary pants, you can glue a disc. And I already told you repeatedly that the final result only depends on the area. So this becomes a cylinder. We know the wave function here. We know the wave function here. This is the exponential of the Hamiltonian. Therefore, you can determine this part. I'm not going into the detail, but let me just write down the answer, which is very beautiful. So this psi A of u1, u2, u3 is given by the sum over r of exponential of area times c2r. And you have chi r, u1, chi r, u2, chi r, u3 divided by the dimension of r and times r alpha, which is a normalization constant. So it's a good exercise for you, if you haven't done that, to plug this in to this equation and check the equality, which is a very useful exercise. Now that you have the wave function for the three punctured, three boundary sphere, you can glue that in various ways. You know my drawing is terrible. So you can take one, two, three of these wave functions and glue that one, two, three points to get, say, g-nus-1, surface with three boundaries. You can easily find the wave function by multiplying. So the corresponding wave function is, again, given by a sum over r of a, c2r, of product over the old boundary of chi r of ui divided by dimension of r of 2g-2 plus n, where n is the number of boundary. And I forgot to include this important normalization factor. So this is basically what I plan to tell you today, but let me say a few words about this constant alpha. What is this? So originally, this came from the fact that the disc amplitude when the area is zero is a delta function, which is not square integrable. Therefore, there is no natural normalization and that introduces this. Another way to see that from the path integral point of view is that on a g-nus-g Riemann surface, sorry, g-nus-g surface, if you do a quantum field theory, it's very easy to generate a counter-term of this form. This is defaeromorphism invariant, combination of the metric, and in renormalization, typically every term which is allowed by the symmetry is generated through regularization and renormalization. So if you do a path integral, typically you get some number, some amount of this term is generated over the Riemann surface. But if, so, suppose such a term is generated on a closed Riemann surface, this evaluates to 2g, sorry, 2 minus 2g. Therefore, this multiplies, this multiplies s, I mean, z by exponential of beta times 2 minus 2g. So this has exactly, and if you introduce the boundary, you get a similar factor. So you can understand this as an effect of possible finite renormalization, which is inherent in quantum field theory. I should also mention that we took a very nice basis, chi alpha of u. This is orthogonal in Hs1, right? That's why we get this nice formula, but if I am very stupid, if some strange person might choose a different basis, I mean, how should I say? This choice was possible because we introduced the inner product. This is because inner product was just psi u, psi prime u bar du, right? But if you choose a bit strange inner product, say it's totally possible to choose some strange additional factor in a different inner product, orthonormal, in this case, basis is changed to k u inverse chi r u, right? So I'm just redefining the inner product in a trivial way, but that just introduces some superior factor, but that would introduce some superior factor k of u, or maybe inverse in this convention, in the amplitude for the 2D amulet's partition function. So this alpha is a certain renormalization constant, and this additional ambiguity k is a choice in the inner product, but remember this is independent of r, so this just comes from the inner product. So I have three minutes left. This is undeformed 2D amulet. We need to discuss q-deformed amulet. What is it? So there are many ways to introduce q-deformed amulet, but the most conceptual way is the following. So far, we consider gauge group G to be the standard regroup. You deform it to something called quantum group. So in general dimensions, gauge theory with quantum group as the gauge symmetry is very hard to define. So quantum group is kind of a non-commutative group. I mean, group are non-commutative in general, but what happens is that the entries, matrix entries themselves become non-commutative. So that's the kind of deformation. Let me just give you an idea. So SU2q in 2 by 2 representation has this entries alpha, beta, gamma, delta. So if there's no deformation, the entries themselves are commutative, but after the deformation, you get alpha, beta becomes q, beta, alpha, and alpha, gamma becomes q, gamma, alpha, et cetera, et cetera. So you try to define a 2D amulet in a lattice formulation, say, where all of the link variables are replaced by the quantum group elements instead of regroup elements. You can try to do that in any dimensions, but because of the non-commutativity, you need to specify exactly in which order various link variables appear in the path integral, and that's only doable. I mean, it's known to, at least in two dimensions, how to consistently do that. So that determines a gauge theory where the gauge group is literally the quantum group. You can do, if you know a bit of quantum group theory, you can almost exactly follow what I discussed. The end result is very simple. You just replace this dimension of r into something called quantum dimension. So at this level, all of the change is just, this becomes dim q of r, and when g is SUn, then dim q of r has a very explicit representation, which is chi of r, and you have some diagonal elements and q. So you need n by n matrix, right? So you take a diagonal element. So this is the q-deformed dimensions. So you can check that this becomes standard chi r of 1, which is the dimension. So that's the q-deformation, right? So that's what I wanted to say for this first lecture. Thank you very much.