 a few minutes. I think it's one minute, so I'm going to wait a minute. The clock is in his computer. It's 10 minutes ahead. I don't know how he managed to do this. I don't know what he would have been saying. Now I start. All right, guys, welcome back. Now we have the last talk of the day, the last lecture, by Diego Tancanelli, who's going to continue to explain which moves and this biography. Just before Diego starts, one quick announcement, the session after this talk, the discussion session, will start right after Diego's talk, OK? So it won't be a break. So right after Diego's done, we will start the discussion, and then the way we finish the lecture, OK? So, Diego? OK, so I suspect the last few minutes of yesterday's lecture were too fast. So let me go over this giant Wilson-Lew business one more time and just repeat the main ideas. So the crucial thing is that, OK, so there is something that weighs order n square, which is the n equal to 4, Super Young-Mills, action, or equivalently the Wigner semicircle distribution, or equivalently the ADS-5 times S5 background, OK? So this is something which is fixed at this order in the n counting. And then, OK, we inserted something that goes like order, sorry, first let's go order one, which was the Wilson-Lew in the fundamental representation. And correspondingly, we had this trace e to the 2 pi m in the pat integral. And correspondingly, we had a fundamental string in the back, OK? So in particular, this fundamental string in the back, I haven't really emphasized this yesterday. I said it a couple of times, but maybe I should really emphasize this more that the induced metric of this object, if you compute it, is ADS-2, OK? So there is a surface, which is a minimal surface, which it has a disc topology. It ends on the circle, and it has ADS-2 metric, OK? But now, OK, we can ask, OK, what happens if we insert something heavier? And so the next thing that you can think of is something which scales like n. And in the gauge theory side would be with the loops with a representation, which is either a symmetric or anti-symmetric of SUn, where k is of order n. So we have either a row of k order n boxes for the symmetric representation, or we have a column with k order n boxes for the anti-symmetric representation, OK? So in the matrix model point of view, this was the topic in one of the exercises. And you can think of having just an isolated point outside the vigorous semicircle distribution or having a missing point in the distribution in case of a symmetric anti-symmetric representation, OK? But anyway, so there is, again, some other insertion of this matrix. And then in the background, we have either D3 or D5 brains. But the crucial point is that both of these things are probes. So both the string and the brains are probes that do not change the background geometry, OK? And OK, so how we understand that the D3 or D5 brain, or how do we understand, for example, this ADS2 factor from a symmetric point of view? Well, simply from the fact that what we have, so as an operator, there was some loop insertion. It's either a line or a circle. It's actually simpler to think about the line, although the result doesn't change if we consider a circle. It's just that the way the symmetries are realized is different. But let's think about the line. It's a line in R4 times one point on this S5 internal space, OK? So we have, as I told you yesterday, SL2R times SO3 coming from this part, SO3, times SO5 coming from this other part. So we have ADS2 times S2 times S4, OK? So again, this is just the remnant of the conformal group, which is left unbroken by adding a line, which has three generators. I can do three transformations along the line. I can translate along the line. I can do a dilatation along the line. And I can do a special conformal transformation along the line. So it's three generators, SL2R. Then around any point of the line, I've got a sphere, SO3. And then since I fix this coupling to a scale has to be 1, 0, 0, 0, 0, 0, I break the SO6, our symmetry, down to SO5, which explains this. And again, in terms of isometries of the background, I preserve this thing. And there are associated supersymmetries. Half of the supersymmetries of the background are preserved. And you can really write down even the supergroup, if you want, is this OSP 4 star 4. OK, so now the important point is focusing on these higher representations is that we have an extra parameter in the game, which is this k, the number of boxes in my young tableau. And so the natural question is, how do I see k holographically? So what does this become? And so if you think about it, what happens is that there's going to be some fluxes treading either VS2 or VS4, depending on which brain you consider. And this flux is going to be quantized and is going to tell you how many boxes you're putting in your representation. So this is the computation I've done the other day. So there is some, in the DBI action, there is some gauge field on the volume of the brain. And this gauge field, so I got pi mu nu, which is the conjugate momentum to this gauge field. And if you do the computation, if you compare how the brain couples to the B field and our string compares to the B field, you discover that the condition is this one. So you discover that A and B, if you want the VS2 directions. So the gauge field is turned on along the VS2 direction. And if you integrate the momentum on the sphere, so this will be VS2 in the case of a D3 brain, or it would be the S4 in the case of a D5 brain, you get precisely this information about the number of boxes in the representation. Good. OK, so now what we can do, we can actually do computations and we can check this from the, so we can check this idea, this derivation with the matrix model and localization. So let me start with the D3 brain. I'm not going to do the full computation because it would take me too long, but I'm going to give you the gist of the argument. So we have the tension, and then we get a four-dimensional volume. You've got a determinant mu nu g mu nu plus 2 pi over the square root of lambda f mu nu. So this is the DBI action. And then you have the sumino term, which is one of the four factorial, epsilon mu nu lambda rho c mu nu lambda rho. So this is the tension of a D3 brain. This is one of the tension of the fundamental string. And essentially what you want to do, you want to find the solution. And so we know what was the, and think about the circle because this is the one that we want to, I mean, which has a non-trivial expectation value that we can compare interesting to the matrix model. So if you want this was the minimal surface for the string, now you can think that what's going to happen is that I'm going to add a cycle, this S2, which is going to be at any point of this minimal surface. So it's like this sheet of the string, which is puffing up and it's becoming fat. So this is why these objects are called giant objects. And it pinches back to the boundary. So this becomes like, well, I cannot really draw in dimensions, but think that this would be something like this all around the circle. And it pinches off. So the cycle shrinks when you go to the boundary and you land again on the Woodson lobe. So how do you get the solution? Well, this is a more complicated problem than what we did for the string. But in the string, we learn a trick. So the way you do it, you first solve for the line, which is simple, and then you do an inversion, and you get the circular Woodson loop. So start with the line, and then you have a line along, let's call it the x direction. Then you have an AdS2 times S2 for volume. So this is along the x and z directions. Now, you don't get the same answer in the expectation value, because when you evaluate the action, indeed, you get either 1 or something on trivial. But if I want to find the solution, I can use this method, of course. And then I got some sphere with radius r of z, such that r at 0 is equal to 0, so the sphere shrinks. So you can do this computation. I've wrote down some more details. I mean, there's many, many. I cannot really do it in real time. But the important thing is that there is a C4, which has this form, dr times the volume of S2. And then you have the gauge field, which is 2 pi over square root of lambda fxz. And then, OK, so if you do the computation, you discover that there is a natural parameter, a natural combination of the parameters that appears, which is this kappa equal to square root of lambda k over 4n. So this is a parameter that controls your solution. OK, so you first find the straight line, the straight line is simple. Essentially, the solution of the straight line is this r equal to kappa times z. And this f is equal to i over z squared. So this is the solution for the line. It's very simple. So you see it's a line with a sphere with a radius that grows like this linearly in the bulk. Anyway, so after you have the solution, you do that inversion that I was discussing yesterday for the string. You get your solution for the circle. You plug it in. You have, again, to be careful about boundary terms. So you have to add these Legendre transforms. But after you do everything carefully, and so all the details are in this paper 05, 011, 09, you discover that the action of the d3 is going to be minus 2n kappa 1 plus kappa squared plus arc sinh of kappa. OK, so it is some result. And if you've done your exercise, you're going to find that this is precisely the answer from the matrix model in this representation. OK, in this, yes. Yes, the definition of SE4. In which gauge? In which gauge you take this? It's not really been sorted out. But OK, so you can couple with three form that acts like that. The circle that the gauge transformation is, what they find at the boundary, and that's what induces the harmony in the expectation boundaries? I don't know if it is right to the same thing. No, because if everything is taking variance, then the actual action of the line and the circle should be the same. All you did was do a coordinate transformation, right? We would not know if it would have been between the circle and the line. Even within the circle. Even within the circle, yeah. In one case, in one case or the other, you have to write the best middle term. Right. Yeah, so here is written in a sense respecting all the symmetries and the most natural choice that you can have, which is this essentially volume form. But yeah, so there could be some harmony. OK, so anyway, so the thing is that if you take now, so this is your result. So if you take kappa going to 0, so you find that the expectation value of the symmetric representation goes like k power, so the kth power of the result for a single fundamental string. So this would be the result for a fundamental string to the power k. And so the interpretation of this result is the following. So from the matrix model, you have this. So the trace of this insertion is one of the k factorial. So the trace e to pi m to the k. So this is the term with the largest number of traces plus the expectation value of some other term, which has k minus 1 traces. OK, so something which has less traces than this. OK, so this is essentially, as I was mentioning here, so this is the leading term in this limit is just adding k fundamental strings one on top of each other. But then when you go to higher kappa, which is higher k, so this is k this parameter that I can control, I start deviating from this situation in which I just have some coincident strings if you want. And so the terms with less traces are becoming important, and they start combining in an untrivial way. So and this interpretation is, sorry, if you write down, this is actually n to the k, k factorial, w the fundamental to the power k plus orders, and k minus 1 if you include all the normalization and you're careful about normalization. So this is, you start seeing how increasing k, you start including non-planar corrections, one over n corrections. OK, so you have this extra parameter k that allows you to explore these extra corrections. And so the statement is that this action knows about all the one over n corrections in this particular combination kappa and resum測m. OK, so the complete result is a resummation of lambda k squared over n to the n corrections. Good, so now I'm running a little bit late, so let me be fast for the defi brain. You do the same thing, actually. This is a simpler situation because you have an AdS2 times S4, where the AdS2 is AdS5, and the S4 is in the S5. OK, it's a simpler situation, and as I was mentioning yesterday, you see right now you have some polar angle on the S5, and this polar angle controls the location of the S4. So you can write d omega 5 as d theta squared plus sine square of theta d omega 4 squared, of course. So you have this angle theta, which is some polar angle. And now this angle theta is determined by this equation. So now you see that the important parameter that appears in the game is not this kappa, but it is this guy. It's just k over n. So there is no factor of square root of lambda. And you can understand this because in terms of just tension of d-brain, so the defi brain is heavier by a factor of square root of lambda than the d-t brain. So you have this particular combination, so k enters here. And so now you understand that theta cannot be larger than pi, and otherwise you don't have a solution. It's a polar angle. And so theta less than pi implies that k has to be less or equal than n, which is consistent for an anti-symmetric representation. You cannot have more boxes than n. But anyway, so if you compute the initial action, and again, you compare it with the matrix model, you find agreement. And again, you have a resumption of these powers. Finally, a very interesting situation. So let's. The spring chart by the definition that you gave. I mean, I guess you have to generalize it for the d-5, right? I did. So k is an integral of. Yeah, yeah, on the sphere, on the sp minus 4. So that's going to be an integral over the s-4. Yes. And that integral, the result of that integral is that condition there, or? Yeah, more or less. So this is, yeah, yeah, so. Right. And that integral is 0 for theta equal to 0. That's how you see that the charge, which you're saying is that the b-fibrin cannot have charge more than n. Yes, yes, yes. Precisely. So let's cement. So we had the fundamental representation. We had an f-1. And then we go to the symmetric representation. We have a d-3 brain. Then we go to the anti-symmetric representation. We have a d-5 brain. So you have this situation here, this situation here. But of course, now, and OK. So this was order 1 insertion. This was order n insertion. But then now it's natural to consider, OK, what happens if I got an order n-square insertion? So for example, if I take a large young tableau, which has order n on this side, order n on this other side. Of course, now, you don't have AdS-5 cross S5 anymore. Because these weights in the same way as the background. So there is a back reaction. And you create a different geometry. So these geometries are called bubbling geometries. And they are very nice. They are very, so I like them for various reasons. But let me tell you also what is the bubbling geometry. So first of all, this is a half-EPS geometry. Because it corresponds to a half-EPS operator. Asymptotically, AdS-5 crosses 5. But then when you go deep, well, when you go inside of AdS-5 times S5, it becomes something complicated. And however, it has to preserve the same symmetries of a loop. So I can do an ansatz, which is dS squared. So it has to contain AdS-2. It has to contain omega-2. It has to contain omega-4. OK, so now we have 2, 4, 8. Direction plus remaining 2 ones. And then I got some fiber over this base. So I got f1 squared, f2 squared, f4 squared, where this f1, f2, f4 are functions of sigma. So this sigma is the upper half plane. So how do you get this solution? Essentially, you have to use killing spinors. So there is a very nice, powerful formalism developed by Gauntlet and other people, and then used extensively by UCLA group, like Docker, Gottperle, and other people, as this, in which they computed lots of bubbling geometries for lots of different symmetry groups. And so I mean, there is an algorithm. It's a very complicated algorithm, but it's an algorithm. And anyway, so what you find is that, OK, you have this solution. And in order to have a regular solution, you have to impose boundary conditions on the boundary of sigma. And there are essentially binary boundary conditions. You can have either 0, 1, or 1-1-1-1 for some fields. And so you can essentially define intervals in this boundary of sigma. So for example, in this region in blue, you get that f4 goes to 0. The regions in white, you get that the f2 goes to 0. So now you can consider, for example, a sphere like the s4, vibrated over an interval connecting two blue regions. And this will give you ns5. You can consider the s2 vibrated along two regions in which the f2 goes to 0. You are going to give ns3. Anyway, so you have cycles popping up. And this is why it's called bubbling. It's like bubbles of spheres in this game. OK, and then you can solve everything. So the whole metric is encoded in an hyperalytic curve. So essentially, you give me a distribution of white and blue regions on this line. And I produce to you a half EPS solution which respects ads2 times s2 times s4 and preserves half of a supersymmetry. So now, of course, the question is, what is the relation between these blue regions and white regions and the Wilson loop that I'm sorting? And this is the picture that I tried to draw the other day in which you have this Wilson loop that you write down as a. So this would be the representation that you write in this oblique way. And then you project down these sides. OK, missing something. And then, OK, this would be like the blue regions or the other one is the blue regions. But anyway, there is a very precise mapping between. So on the matrix model, the point of view is would be a multi-matrix model with multi-cuts. And so you have very precise mapping between the spectra curve, the resolvent that appears in the matrix model with these particular cuts and the geometry with cuts given I mean with this boundary conditions given by those cuts. OK, so there is a very, very precise formula. And you can think of it as an example of emergent geometry. So you have an auxiliary structure, a matrix model, which has nothing to do with geometry. But it knows about some half EPS geometry with ADS 5 times S5 boundary conditions. Yeah, OK, I guess isn't it the same? Yeah, I think it is the same. So you have a big matrix M, but now you can divide it into blocks. And every block is non-diagonal, but after you. So every block gives rise to a different, I think it's called multi-matrix model or multi-cut. I think it's the same thing. Yes? Give me a tableau. Yeah, so I give you a tableau. I can compute the matrix model. And then I can compare it with the super gravity with a solution that has precisely the same structure of cuts. OK, good. So I think I'm more or less on time. So I want to do now more. So OK, again, I like this picture a lot because of this emergent geometry thing. So this is not just ADS 5 cross S5. This is a very complicated geometry. And all the information about this geometry is encoded in a resolvent of a matrix model. So this is interesting for me. Yes? If you use cases of decode. It would be the case in which you have only one cut. So it's like having just one, for example, just this line. And then you just project down this guy and you have only one. OK, so well, you don't have a bubbling geometry in the case. You don't have a back reaction on the geometry. So yeah, let me say, originally, so we had large edges of this giant tableau in order. So you want this edge to be large in order to have a thermodynamic many eigenvalues here so that you can have a continuum limit. And then you wanted this to be large so that the distance is large and they don't interfere much with each other. The interaction between these distributions, it's simpler. And this is what we compare to supergravity. But then from the matrix point of view, then Leo had a more recent paper in which they can solve this thing for generic edges. OK, so now I'm going to briefly review some other contours, some other operators which are all related to each other and can also be computed using localization. So this is going to be the extra. So time permitted. So I'm going to do the extra. So more examples. So and for this part, I'm following a nice review by Zarembo, which is in this big review on localization and there is a chapter on with all loops. But OK, I have to say that many of these things have actually been done by Diego Correa always sitting here. OK, so the first example that I'm going to consider is a wavy line. So this is no longer a supersymmetric operator because we know that the operators are either lines or circles unless they do something peculiar to the scalars. So now let's consider just a line. And then consider some waviness associated to this line. So let me parameterize this waviness. So I got the parameter s going along the thing. So I got x i of s that tells me how this wavy line differs from the straight line. So for example, something like this parameterization. So I got the line along the forward direction plus some waviness from some small deviation. And now I can consider the combined propagator. So remember, I'm going to write down the definition again. So this is delta AB GM. OK, so this is the combined propagator. Of course, it's no longer 0. But it has this expression. Of course, I can do much more if I don't tell you what x i is. But even so, at small lambda, I can compute, for example, for the fundamental representation, this thing. And it becomes 32 pi squared integral between minus infinity and infinity ds1, ds2, psi dot 1, psi dot 2 squared. OK, so at this point, I'm just doing this computation. And I'm doing this computation because I want to define what is the pre-factor in front of this integral. So if you do this computation in QED, you are going to find LaRmov's formula for the radiation, for the energy radiated by an electron. So I'm going to define this pre-factor as B over 2. And so this is a Bram-Strahlung function. Again, so if you compute it in QED, B would be equal to 2 alpha over 3 pi. So you have this LaRmov formula for the energy of radiated by an electron. OK, so this is something that we can consider. We can define. And this is the first example. Now I'm going to move to the second example and then the third and the fourth. And then we're going to see that we are all related to each other. So the second example is called CASP, Anomalous Dimension. So now I'm considering, so again, I'm in Euclidean signature. And I'm considering, so this would be a line. But at some point, I introduce a CASP. And this angle, I call it phi. So the line goes like this. So we know from quantum field theory that if you compute, so usually in quantum field theory, you have a linear divergence in the Winsor loop. But we saw that this linear divergence disappeared once you consider the gauge field plus the scalars together. But we know that if you consider a CASP contour or a self-intersecting contour, you get on top of the linear divergence, you get a log divergence. And the log divergence survives supersimid. I mean, survives the coupling to the scalars. So you're still going to get a linear log divergence. And so let's regularize it by introducing two cutoffs. So one is L, which is an infrared cutoff, and epsilon, which is a UV cutoff. So you can think of like, OK, L is the cutoff at which the CASP line starts deviating from being a straight line. And epsilon, you can think of smoothing out the angle, the sharp angle, at a distance epsilon. So this is some epsilon here and some L here. OK, so now I can compute the, so let me parameterize this loop using the letter S before I reach the CASP and the letter T after the CASP. So I can compute this at weak coupling, and I got 1 plus lambda over 8 pi squared, the integral between 0 and L dS dt, 1 minus cosine of phi, S squared plus d squared plus 2 Sd cosine of phi. So 1 plus lambda 8 pi squared phi, 1 minus cosine of phi, sine of phi. So this integral is going to give you a log divergence that is expected. And that actually contains interesting physical information. So now we define this to be equal to some constant over epsilon. So gamma CASP, which depends on lambda and phi. So in general, this will some loop. You can, with this CASP, we can write it down like this for any value of lambda. So at weak coupling, it would be this particular expression. In general, we introduce a function of lambda and phi. So we see in particular that at weak coupling, this gamma CASP becomes lambda over 8 pi squared phi tangent of phi over 2 plus orders lambda squared. So we just computed it. Yes? Which one? That's fine. Yeah, so this goes here. So now you can do an analytic continuation in which this phi becomes i theta, becomes an imaginary angle. And this is like going to a Minkowski contour with a kick. So you get a particle which stays at the origin and suddenly gets kicked, and it moves in a theta direction. So now gamma CASP lambda i theta, when theta goes to infinity, now you can take the limit of very large theta. And this is minus 4 theta f of lambda. And this f of lambda is called the CASP anomalous dimension. And you have now connections with integrability, where you can compute this object using integrability. OK, so now if you take phi going to 0, so you have a very small CASP angle. So you stay in your clear case. You take phi equal to 0. And then, of course, you recover in the strictly phi equal to 0 limit, you recover that the Winston line has trivial expectation value at that gamma CASP at any value of lambda in 0 is equal to 0. There is no log divergence. And you can think of it as a wavy line. OK, so this is, of course, a consistency check. And then you can think of this wavy as this CASP contour as a wavy line with a deformation psi, psi dot of mu, which is a step function times phi times a unit normal to the first segment. OK, so if you do that, then you can compute the Bremstrahlung function associated with this wavy line. And then you find that for small phi, this becomes 1 plus beta phi squared log L over epsilon. And so you find that there is a connection between the gamma CASP and the Bremstrahlung function. So the Bremstrahlung function that was introduced from waviness is actually connected to gamma CASP in the limit of small CASP. Actually, there is another possibility that we haven't really considered. Namely, remember that we had the coupling to the gauge field, but also we have the coupling to the scalars. So if you want, you have a geometric contour with a CASP in spacetime and a CASP parameterized by phi, but also you could have a CASP in the internal contour. So you remember it was a theta i, but let's say it was 010000. But then you can consider, OK, since I'm allowing for CASP, I'm allowing for discontinuity, let's say that I've got one coupling to the scalar on this half line. And then after I jump, I couple to a different scalar. So this would be a CASP in the internal space. And you can also do that. And also there are expansions. You can introduce a dependence in this internal angle in gamma CASP. You can expand again, and this is very similar. You are going to have phi squared minus the internal angle squared. So there is a very similar structure. Good, so now we introduce Bremstrahlung function, gamma CASP, you have this. Now you can think about a q-q bar potential as a third object. So the q-q bar potential, you usually think of it as a rectangular contour. So this is how we learn it in quantum field theory. When we study lattice gauge theories, we study Wilson-Lupes in quantum field theory, you think about a rectangular contour with a very long side and a very short side. And so it doesn't seem like we are considering rectangular contours. But think of the CASP actually as a quark. And so think about a CASP that now is pretty big. And think of it as a quark and anti-quark being generated from this point and now propagating. So if you take phi, so this would be phi, so if you take phi going to pi, you have essentially collinear, well, in the limit phi going to pi, you have collinear quark and anti-quark pair. OK, so now you can compute this. So for example, the log. So you can compute this. And then you can introduce the log of w phi, let's say for particles in the fundamental representation, when phi goes to pi, you can introduce some function alpha of lambda pi minus phi log L over epsilon. And so if you compute it at weak coupling, this alpha of L would be like lambda over 4 pi at weak coupling. And it is actually also related to this. And this is essentially the quark anti-quark potential. And it is also related to this gamma CASP. So you can see that alpha of lambda in general is the limit for phi going to pi of pi minus phi gamma CASP lambda phi. So there is this gamma CASP. It's a very interesting function that packs lots of information about different interesting physical situations. So radiation, gram-stallung, and quark anti-quark potential. So you can compute this at weak coupling, but actually you can also compute this at strong coupling because it's a simple surface. Probably I'm not going to give you all the details because. OK, so weak coupling is a trivial computation. You just essentially use the formulas I wrote before and you get this result. At strong coupling, essentially, OK, what do I have? I've got this CASP like this. So this is the z direction. So this is a CASP in the z equal to 0 plane. OK, now I got this infinite, this wedge. And I have to compute a minimal surface associated to this wedge. And I mean, you see how the minimal surface is going to be. It's going to be something like this, right? So it's convenient to use polar coordinates, r and bar phi. And it's clear that this surface has to be self-invient under rescaling, right? It has to be self-similar. So you have a self-similar form. So the answer for this surface is going to be something like z equal to r, some embedding function u of bar phi, right? So this is at the origin, right? And then you tell me, OK, you tell me at which distance you are from the origin. And then depending on phi, I can tell you how deep you are in the interior of the surface. So this is the parametrization. OK, so you write down the Nambu-Goto action. You get the square root of lambda over 2 pi integral dr over r d bar phi 1 over u squared, 1 plus u squared plus u prime squared. So you solve it. There is some integral of motion. You have to compute what is the highest point in the surface. Of course, it's going to be the point halfway in the cusp. It's going to be the highest point. So this is your integration constant. Anyway, you can compute it. And so compute action on shell, subtract the divergence and everything, and then you get that gamma cusp is equal to some elliptic integral. So square root of lambda pi u0 square root of 2 plus u0 squared. So u0 is the highest point. I mean, it's this integration constant that comes from integrating over the highest point. 2 plus u0 squared, some elliptic integral minus 1 plus u0 squared, some other elliptic integral. Anyway, you can do that. And OK, you can take the limits now. You can take various limits. So this is valid for any phi. So remember, phi was this angle that I got in the cusp. And phi was the polar angle. So if you take phi going to pi, which corresponds to u0 going to 0 limit, you get that alpha of lambda. It goes like 4 pi squared square root of lambda over gamma, Euler gamma 1 quarter, gamma to the 4 of 1 quarter at lambda equal to infinity. And if you have done exercise number five of the problem sets, you should have gotten this structure from computing. So this was the original papers by Maldesain and Suyam Ray and Yi. They compute this quark-anti-quark potential at strong coupling. But anyway, so you see that everything is related to this gamma cusp. Gamma cusp, as I said, packs lots of information. Then you can take the limit of phi goes to 0, or u0 going to infinity. And then you see that you get the Bremstein function. And this is square root of lambda over 4 pi squared, again, for lambda equal to infinity. So this is a very nice object to consider, this cusp with some lines. Very good. So now, what is the relation of all this study with localization and supersymmetry? Because these are non-super symmetric operators. So should I really only hope that I can compute them in particular limits like I've done so far? And actually, as Diego Correa and collaborators found out, actually, there are exact results that can be obtained from localization for all these quantities. And these exact results have to do with the fourth example I want to talk about, which is this 1 quarter BPS latitude that I introduced in the first lecture, and then it was a subject of an exercise. So the idea is that this B of lambda can be obtained from the quarter BPS latitude. And so a quarter BPS latitude, let me write down explicitly what the contour is going to be in the internal space. So you have a latitude. So there is, you remember, you have the sphere in R4, and then you have the S5 of our symmetry. So you have this latitude, and then you have a dual latitude on the sphere parameterized by this angle theta 0. So you have this angle. So what is called T0 theta 0 here becomes theta 0 here. I guess, well, this is something that we always have to double check 10 times. OK, I'm not sure. Maybe, yeah, it's like this. It's the complementary angle. But I guess it's the other way around. Sine cosine, maybe OK. Maybe I want to write down cosine theta 0 here and sine theta 0. No, I think it was right. Anyway, so you can check that this is 1 quarter BPS if you've done the exercise. And now the fact that it's 1 quarter BPS gives you something very nice, that this width on loops can be computed by localization. And in fact, it is the same matrix model as the usual width. So it's going to be a Gaussian matrix model. The only thing that changes, if instead of adding lambda, you have to replace lambda to cosine square of theta 0. So you can import all the results that you computed already for the half BPS circle to this case of these loops on the S2. OK, so for the particular latitude, you have to do this replacement. For generic loops on BPS2, you have to do another replacement. But it's very simple. So now you can compute this. This, as I said, you can import all the results that you had before. So in particular, we can compute exactly the expectation value of a latitude at an angle theta 0. And for n going to infinity, this is cosine theta 0, some Bessel function square root of lambda cosine theta 0. So it's the same Bessel function. So now, when theta 0 goes to 0, so yeah, actually, yeah, it was the opposite, right? So this is theta 0, and this is theta 0. Sorry about that. So when theta 0 goes to 0, you'll go back to the half BPS circle, OK? And you can see, so the latitude has a wavy deformation of the circle. And so now you can compute the Bremstrahlung associated with it. So you can compute the expectation value of the circle minus the expectation value of the latitude normalized by the expectation value of the circle. And this is, if you expand it for small theta 0, this is going to give you minus 1 over 2 pi squared a Bremstrahlung function theta 0 squared plus dots, OK? So you see from this analysis that the Bremstrahlung function can actually be computed exactly and is given by 1 over 2 pi derivative with respect of lambda of the log of the expectation value of the circle. So this is square root of lambda, Bessel function 2 over 4 pi squared, Bessel function 1, OK? So this is valid for any lambda and n going to infinity, OK? So there is a connection between these four operators. So a wavy operator, a wavy line, a casped line, the quark-anti-quark potential line, so anti-parallel lines, and the latitude. And there are very nice connections that also connect to integrability. So this is an arena where lots of computations can be done, lots of insight can be obtained, OK? So thanks.