 First of all, to organizers for inviting me to give us lectures. So I was asked to give lectures about an introduction to supersymmetic width on loops and their role in localization and holography. So this is what I'm going to do. And before I start on the blackboard, let me just give you a roadmap and other view of big conceptual ideas using slides. Also, if you don't want to, you don't need to take notes because my notes are already online. So you can just maybe, I don't know, print them out and follow or whatever. OK, so a width on loop is a crucial central object in quantum field theory and holography. And to remind ourselves, we had the first encounter with a width on loop in quantum mechanics when we study Bohr-Maronoff and QED when we study these gauge compensators. And then you go to lattice theories and then you see that width on loops are important tools to diagnose the presence of confinement. And if you study mathematics, you know that there is a relation with not invariance into dimensions. This is how Whitten got his Fields Medal in 1990. And also, going back to quantum physics, they are very important for testing non-perturbative dualities like S-duality or electromagnetic duality because there is a relation between this electric object, the width on loop, and its magnetic counterpart, which is the top loop. So you can use the two computations of the two objects to test S-duality. But also in string theory, of course, they are ubiquitous. They appear all the time. But of course, this is a school on localization and holography. And of course, width on loops have to do with supersymmetrical localization. And the way they enter in this picture is because they can be computed exactly using localization, and in particular, using matrix models, which is what the quantum field theory pat integral reduces to after localization. And so they give you exact results. So having exact results in a quantum field theory is something very rare, very, of course, non-trivial, and very powerful. Because now, with these exact results, you can really test holography, which is, again, weak-strong coupling duality, like S-duality. So you really have to have control over one side of the duality very precisely, or even exactly, in order to be able to test non-trivial these dualities. And so in holography, there's lots of different aspects of width on loop that are important. So OK, they can give you precision tests. And also, you have many interesting objects that appear. So you have minimal surfaces. You have deep brains. You have bubbling geometry, so fully back-reacted geometries that can all be used to test holography. And they have a duality with width on loops. OK, so of course, there's also another big part of the literature having to do with computing scattering amplitudes through strong coupling using width on loops. And of course, this is also very important. So what we're going to focus on is these two blocks. So there's going to be a part one, which is width on loops in supersymmetric gauge theories. And then, so this is going to be the subject of today and maybe tomorrow. And then we're going to move to the gravitational side of this holographic duality. And we're going to see what these results that we obtain here, how these results can be checked in holography and what they imply in holography and so on. So this is everything I wanted to say with slides, so that you have at least a big picture of what we're going to do. So we're going to start with this. And then the last two lectures, we're going to do the holographic side. And also, I'm going to maybe in the last lecture, if time permits, I'm going to talk about some other kinds of operators that I'm going to leave for last. Anyway, so if there's any question, please interrupt and ask. So before we venture into supersymmetric territory, so it's, I guess, always nice to review a bit what we know. So review width on loops in gauge theories. And as I told you, so the first place where you encounter width on loops is in QED, where they appear as, well, width on lines more precisely, where they appear as gauge compensators, or if you want to think more in geometric terms as parallel transporters. OK, so you have some objects that, let me call it, U gamma of yx, which is the exponent of IE and the autonomy of the connection, well, the integral of the connection along gamma. So gamma is some contour going from x to y. And so this guy is nice, because it allows you to define a covariant derivative, which transforms nicely under a gauge transformation. So if you have some field transforming, like e to the i alpha x phi of x, you can also, you can define a covariant derivative, which also transforms in the same way, because this guy transforms as e to the i alpha y U gamma e to minus i alpha x, OK? So this object depends on the path gamma that you're taking to go from x to y, unless the field A is a pure gauge, so unless f mu nu is equal to 0. OK, and what is the physical meaning of this object? So if you take the expectation value of this object, essentially you discover that this is the phase acquired by a particle which has charge e going along gamma. OK, so this is just the interaction term between the particle and the gauge field, the background field. OK, and this phase is, as you know from quantum mechanics, it's observable in some situation, like when you have a non-simply connected spacetime, like for example you have some solenoid with some magnetic field, like in the Baum-Aronov experiment, you can detect physically this phase by doing some interference experiment, OK? OK, so this is UAD, then you can go to Young Mills, and we want to define a non-Abelian extension of this guy. And so we have U gamma, y, and x, which is defined as the path-ordering. So now let me call g v coupling constant. And now this guy is some field with mu A x ta, OK? Some generator of a group, so you have some group g. So let's say, for example, g is equal to most of the time we're going to think about g as being SUn, OK? And OK, so now we have matrices in the exponent, and so we have to worry about what it means to take the exponent of these matrices. This is why we introduce this path-ordering symbol, which you can think of it this way. So first of all, let me pick some parameter along the curve. And now I can think of this path-ordering as the products f0 to tau, 1 plus ig, c dot mu s. So under a gauge transformation, now we have the same structure, except now instead of having just exponents of functions, we have some gauge transformation omega. We have this u gamma of x and yx transforms like omega y, u gamma yx, omega dot of x, OK? OK, so you see that this guy is not a gauge invariant, but we can find some gauge invariant object out of it, which is just take a closed loop, so take x equal to y, and take the trace. So if you do that, you get a gauge invariant object. OK, so I need more time. I think that's enough. OK, so now let me write it down. So now let me define this guy as trace, path-ordered, exponential, ig, integral over a closed contour, this mu, a mu of z. So we see then that this guy, it depends on the contour c that I'm considering. And also I could do something more general actually than just coupling the theory, Young-Mills theory, to some fundamental particle. I don't need to do that. Well, this is what you usually do, because you always think about quarks. But you could take this in a different representation of the gauge group, and then you could take this trace in this other representation. So we see that this also depends on the choice of representation that you are considering. So this is not something that you usually do in Young-Mills. But actually, I can give you an example of a paper where, so this was done. For one example, this is some work by Schiffman and collaborators, 0307, 097, where you say, OK, so for SU3, the rank 2 anti-semitic representation is like a fundamental representation. So you could think of quarks transforming in this representation, and they have done it. So this is Armoni, Schiffman, and somebody else. So again, it's not something that you usually do in quantum filtering, but this is going to be actually very nice for us in holography, because this is going to be an extra parameter that we have to explore interesting things in holography, like these D-brains and these bubbling geometries that I was mentioning here. They are explored precisely by changing this representation. And also, the nice thing is that supersymmetric localization does not depend on representation. So you prove it for the fundamental representation, for width-allupian fundamental representation, it carries over for any representation, and so you can use the same techniques also for higher representations. Of course, as you know from lattice theory, some contours are particularly interesting, like these rectangular contours, which can give you indications of confinement, the confinement transitions, and so on. OK, so this is what I wanted to say about a review, just essentially to write down this formula so that we have a starting point. Very good. So now, let's move on to part one, to the supersymmetric discussion width-allupian supersymmetric gauge theories. So let me start with some generalities and conventions. OK, so I decided to do one example with details rather than doing a survey of many examples with less details. And the example, which is the easiest to talk about, is n equal to 4 super yam males with gauge group SUM in d equal to 4. But essentially, everything I'm going to talk about has counterparts for many other examples, most notably in three dimensions, in this ABJM theories or transformable matters theory with various degrees of supersymmetry. You can carry over many, many of the things that I'm going to say. But this is the example, which is, I guess, more pedagogical, because it's simple and we can do computations in great detail and compute things exactly and so on. OK, so let me remind you, so what are the fields of a theory? So we have a gauge field, a mu. Then we have six scalars, phi i. So this is a singlet of SO6R. This is vector of SO6R. And then we have alpha equal to 1, 4, and a equal to 1 and 4. And this is a 4 of SO6R. And OK, I can write the action, but I don't need to, so I'm going to be in Euclidean signature most essentially always. And so it's particularly convenient to recast this. So we have four spinors that is nice to recast in terms of a single Majorana vial spinor in 10 dimensions, OK? So these are 16 components. And now, so if you write down the action, you find that there are some gamma matrices, gamma m, that can be, so which are 16 by 16, which are basically for the algebra, gamma m, gamma n equal to delta mn. OK, so now the n equal to 4 action is invariant under the following transformations. So we have two sets of transformation, the q. And then there is transformations for the fermions that I don't need. So this plus the transformation for psi that I don't need. And also there is another set of transformation that I call delta s, which is epsilon bar s. I've got the same thing with x mu, gamma nu. Plus a corresponding transformation for psi. OK, so we have now epsilon q and epsilon s are two constant Majorana vial spinors in 10 dimensions. So they have 16 components each. OK, and I can combine them into a single epsilon, which is an epsilon q plus x nu, gamma nu epsilon s, which is a super conformal killing spina. OK, good, so now it simplifies computations a little bit if I actually change a little bit this convention, this notation. And actually, so let me think about gamma mu as something which acts on alpha only, not a. So it only acts on the Dirac index. And instead of using gamma i, let me use gamma 5 rho i, where rho i acts on a, not alpha. So, I mean, another way. So gamma 5 is minus gamma 1, gamma 2, gamma 3, gamma 4. So gamma mu are the SO4, the uB and SO4 cliford algebra. And rho i, the uB and SO6 cliford algebra. And the two guys commute. So gamma mu rho i is equal to 0. So this simplifies the analysis, OK? OK, so this is everything I need about conventions. So let me propose an answer for a supersymmetric Wilson loop. OK, so I don't know, in general, how to define a Wilson loop in a supersymmetric theory. So I have to think a little bit about possible answers, something that has a chance to work. And the obvious idea is that we need to include phi i's in the exponent. So in a non-supersymmetric gauge theories, we had the definition that we just erased into the i integral a. So there was only the gauge field. Now we have, in the same gauge multiplet, we have the gauge field, we have the scalars, and we have the fermions, OK? OK, so you could say, OK, maybe I also need to include psi a, because we are in the same multiplet. And yes, maybe in principle, you should. But including phi i in this particular case is enough. And now let me try to motivate why this is. So let me justify this strategy. Well, so the first point is that phi i is in the same multiplet as a mu. So I should treat them more or less on the same footing. And then I'm looking at these transformations here. And then I say, OK, of course, if I have only a mu, I'm going to produce a variation of a width of loop, which is non-zero, but proportional to this big psi. Of course, if I include also phi i, maybe I've got a chance to cancel the two variations. OK, so delta epsilon of phi has a chance to cancel delta epsilon of a mu, right? And then there is another justification, which is the fact that d equal to 4 and equal to 4 super young males can be seen as a dimensional reduction of d equal to 10 and equal to 1 super young males. So in a sense, we already had a taste for this reduction because I already arranged the fermions in terms of a single Majorano-Weil-Spiner in 10 dimensions. But also, you see that these two guys, a mu and phi i, originate from a single gauge field in 10 dimensions. But after I do the reduction, this remains a gauge field. This remains a vector field. And these phi i's are just a bunch of scalars, OK? So since I got, so if I think of AM like this and big xm as some x mu yi, so I got some contour in spacetime and in some internal space, then I got the natural starting point in 10 dimension would be AM d big xm. And so this would be a mu dx mu plus phi i dy i. So I see that there is a natural, so this is, including phi is something very natural. It's something that you do very naturally. So you think of enlarging your contour from being purely spacetime to being spacetime plus some internal component, OK? So with this justification, let me write down the ansatz. So now I got, I will solve it. It's going to depend on the representation r. It's going to depend on the contour. And it's going to depend on this internal contour. I call it theta. So this is one of the sum normalization factor. OK, so this is my ansatz. I, so up to here was the same thing as before. Now we included some, we included the scalar. I just write down this x dot explicit to make it clear that this is reparameterization invariant. And then I call the couplings of the scalar theta i. And again, this is sineoclidean signature. OK, so essentially the supersymmetry analysis is independent of the representation. So if you prove that something is supersymmetric for the fundamental representation, it's going to be supersymmetric for any representation. So for now, let's just fix the representation to be a fundamental one. And so we have two unknowns, which are c and theta i. So we need to solve for these unknowns in order to have a supersymmetric width on loop. So is it clear? So it's going to be clear in a moment that r doesn't play any role in the supersymmetry analysis. Only c and theta is what you need to determine. Good, so now let's see the supersymmetry variation of this guy. So this is the dimension of the fundamental representation is just n. Try spot-ordering tau epsilon bar, I guess that's psi. OK, so I'm going to tell you what I've done. So I took the variation of the width on loop. So these two variations, they act on the exponent and they bring down this piece. So this is the variation of the gauge field and this is the variation of the scalars. But I already write down with these new notations in which instead of having the gamma i, which anticommute with the gamma mu, I take this gamma 5 explicit and then I got rho i instead of gamma i. OK, this is what I explained before. Yeah, so this part is in spacetime and this part is in some internal space. I still don't know what theta i is at this point. Let's see. Oh no, OK, so s is the common parameter for the two. So if you want, you have x mu s and theta i s. So it's the common parameterization of the two contours. I don't know what they are, rho i, right? So rho i is what I... So instead of using gamma i, which was a gamma matrix in SO10, Clifford, obeying the Clifford algebra for SO10, I use gamma 5 rho i, where rho i now is an SO6 Clifford. All of them are 16 by 16 matrices. But it's nice because now I got that rho i and gamma mu commute. Still, gamma i anticommute with gamma mu, but because there's an extra gamma 5. Good, OK, so this is the variation of the Wilson loop. OK, so now the simplest strategy, and we want this to be 0, right? So we want this to be 0. So in this particular example of n equal to 4 super-miles, a strategy that works is just to require that this is 0. So if this is 0, of course, this is enough to be 0. But so in other theories, it doesn't work this way. So in other theories, you have to do more complicated things. So things have to mix in a complicated way. But in this particular case, this is enough. So the strategy is make this guy to be 0. So for example, in ABJM theory, things are more complicated. First of all, it's not so OK. You could do that, but you don't get the most important Wilson loop, which is the most supersymmetric Wilson loop. You don't get it this way. You have to do another construction, where you also include fermions in the exponent. And then you have to expand the exponential in some Taylor expansion. And then the variation of the bilinear of the fermions cancel with the variation is a more complicated construction. But in this case, this is enough to get everything. So we want this to be 0. The first observation is that if theta i is a unit norm vector, then we get that ix mu gamma mu plus, so we get that this guy squared is 0. So this operator is squared to 0. So it is degenerate. And I can use it to build projectors. OK? You? Yes. No, I think that's enough. But yes, yes. Yes, yeah. OK, I'm not sure. So there is a classification by Damascus and Perthstone. And I don't remember if I use this idea. But yeah, that's the most generic classification that you could. OK, so now we have this operator that's squared to 0, so you can see it immediately, because essentially you get an x squared from v. So the cross terms cancel precisely because of the commutation between rho i and gamma mu. And then you have this ix squared, so that gives you minus x dot squared. And then you get a plus x dot squared multiplied by theta i squared. And if theta i squared is 1, things cancel. OK, so given this, so you can easily check it. So let me define projector p plus minus to be given by this. OK, so you can easily prove that p plus minus squared is p plus minus. And p plus p minus equal to p minus p plus is equal to 0. And now we see that this particular combination that appears here is essentially proportional to p plus. So you can prove. You can see easily that this is proportional to epsilon bar p plus x dot mu gamma mu big psi. OK, so and we want this to be 0. So this is the condition. So then it's clear that any epsilon bar, which has the form of some epsilon bar 0, p minus is going to be a half rank solution. So this is a project. So half of the components of epsilon are going to be preserved in this way. So if we find an epsilon of this form, this is going to preserve half of the supersymmetries. OK, good. So now there is a crucial observation. Namely, so these projectors, they depend on x, x dot. So they depend on the point in the contour in which you are. So if you move from one point to the next point, you change the projector and you change the equations. And so at different points, you are going to preserve different components of epsilon. So you're going to preserve different supersymmetries at different points in the loop. And this is not a symmetry of the action of n equal to 4 superyamides, which is a rigid supersymmetric action. Rigidly supersymmetric action is not a super gravity action, OK? So this would be only local supersymmetry, but now we have to preserve rigid supersymmetries. And so we have to, so but we want rigid supersymmetry. And so we want the same condition at any point of the loop s, OK? So we want to preserve always the same components of epsilon. OK, so let's go on and find the solutions or some solutions. So a first solution is a straight line in the four direction. So let me take x mu equal to 0, 0, 0, s. And so x dot mu is equal to 0, 0, 0, 1, OK? So now this becomes constant with projector, but I still have a problem that this guy theta i, sorry, when I said that this depends on the loop, I also meant to say that this guy depends on the loop in general. Theta i depends on s. So OK, this is enough, of course, to get rid of the problem here. To get rid of a problem here, let's take theta i equal to constant. And just by our symmetry, I can take this constant vector to be 1, 0, 0, 0, 1, 2, 3, 4, 5, 6, this guy, OK? So when I do that, so p plus and minus become s independent. And so I've got a Wilson line, which is 1 over n trace but ordering exponent. So now the line has to be infinitely extended. Otherwise, it wouldn't be a gauge invariant. So this is what we find. And this guy is half EPS. So by this language, we mean that he preserves half of the epsilon. And so let's check that this is the case. So we can take separately epsilon bar, epsilon bar q, p minus. So this preserves eight components of epsilon q. Or we could take epsilon bar equal to epsilon bar s x nu gamma nu p minus. And this preserves eight component of E s, the way to see it. Because now, so this guy actually is just p plus x nu gamma nu. So it goes through, changes sign. But that's always a still half rank projector. So we have eight epsilon q's and eight epsilon s's. So we preserve in total 16 supercharges. And this is why we call it half VPS. Because the vacuum has 32 supercharges. Well, I didn't really do the counting. But early this morning, we were more complicated countings than this one. So I assume that you can do this count. What? Yes, I put the formula in p, and then I commute the gamma up here in here through this gamma. And it's still a p. Yeah, I always want to use, so epsilon bar has to be of this form, where this is constant. Right, I specialized, yeah, I didn't write it down, sorry. Yes, so p plus minus in this, sorry, yeah, I was supposed to do it. p plus minus is 1 over square root of 2, 1 plus minus i, gamma 4, gamma 5, rho 1. So you see it's s-independent, very simple. Pino, you have to know x-nu, gamma-nu. So it has to be, so epsilon, it has to be an epsilon q plus epsilon s, x-nu, gamma-nu. But for this very simple case of the line, I can turn off one or turn off the other separately, and I find that they are separately solutions. But now, as I guess Leo was alluding to, this is not always the case. Sometimes you have to consider all of them at the same time. Yes? Because in 10-dimension, I probably only have supersets. But in 10-dimension, it wouldn't be supersets. OK, you mean? Seems that everything comes from the reduction, right? Like having a line in 10-dimension. Yeah, but I guess in 10-dimension, it would still be locally supersymmetric only. And this is really a question of having it rigidly invariant. I think, yeah, it inherits some local supersymmetry from the 10-dimensional one. But I think that's not enough. You still have to impose rigid supersymmetry. So Q and S, Q and S, they are 16 components, constant components each. And P plus minus, they kill half of them, right? So you go, you get, I mean, Euclidean. What do you mean? Intent Euclidean. Yeah, so I don't know. So the question is, if it is really 8 real plus 8 real, or if it is complexified. So OK, maybe you can switch. You can re-rotate in terms of just, I'm not sure. Yeah, the straight line you can do in Lorentz and put it on a space direction. And I think nothing changes. Yeah, so you don't see the difference. OK, so this was one solution. It's the most obvious solution. But now, if you think in terms of conformal transformations, so the second solution, you see that the line is conformally related to a circle via an inversion. So let's take a circle in the 3-4 plane. So x mu is 0, 0 cosine of s, sine of s. And theta i is the same vector. Now x dot, so a is equal to epsilon a b, x b, a b are 3-4. Now gamma a gamma 5 becomes minus epsilon a b, gamma 1, gamma 2, gamma b. So if you write down your projectors, you discover that they have this form, 1 plus minus i, gamma 1, gamma 2, rho 1, x b, gamma b. So now you see that these projectors are no longer s independent because you have this piece here that didn't go away, still depends on s. But they have a very specific form. So if you take epsilon bar equal to epsilon 0 bar b minus, this is epsilon 0 bar 1 minus i, gamma 1, gamma 2, rho 1, x b, gamma b. And you can write it as, so this becomes epsilon bar q plus epsilon bar s, x mu, gamma mu. So even though the projector is not as independent, so you get a super conformal kilispine with the right form, where now this guy is just epsilon 0. And this guy is epsilon 0 minus i, gamma 1, gamma 2, rho 1 constant. So now, again, you get a half-rank condition. So you preserve half of a component of epsilon. But now they are mixed. So you can turn it off epsilon 0 bar. Otherwise, you cancel all of them. So this is, again, 1 half with s. So these are the two simplest examples. So they preserve 16 supercharges, either 8-q's and 8-s separately or a combination of them. But OK, this is very simple because essentially we are using constant t-tas. So we take a very simple contour in this internal space, which is actually just a point. So in this internal space, the loops sits at the point. It always couples to the same scalar. You can go to s4, and then this would be the equator of s4. The line, not really, because the line would have trivial expectation. I think on vs4, you have this half-EPS operator, which is the equator of the s4. You can think of it as coming from the line in R4, I guess, but yes, the variation of the exponent has to cancel. No, it's not necessary. It's sufficient in this particular example, as I said. In some other cases, it's not sufficient. Well, yeah, I mean, it's just like I have this theory, and I have to propose some operator. It's not written anywhere what is the algorithm to get the operator, so I have to do an ansatz, and I have to check that it works, and this is what I've done. Then you discover a new theory, like ABJM. You try to do the same operation, and it doesn't really work. So you get an operator which doesn't preserve half of the supercharges, preserves less than that. And then you start thinking, OK, how do I get the one that preserves half of the supercharges? And then you have to find a different trick. OK, so I got two minutes, so now let me give you a third example. So so far, theta i was constant, but this is not necessary. So I can take a theta i, which depends on s. So the scalar coupling changes along the loop. For example, let's take, so there are many operators that you can build this way, but let me focus on a particular class, which is let me take a sphere, s2 in R4, and now let me, now this loop is confined. So this loop, c, is confined to beyond the sphere. So let's take x4 equal to 0, xi equal to 1, 2, 3, as a unit factor. So this is a function of s. So xs squared is equal to 1. OK, so of course, well, this is not in general supersymmetric unless I do something with this. So I have to tune them in a particular way. And this is something that you can do. So now you see that the scalar coupling is non-trivial. It's not a constant theta. As I go along the loop, I change the coupling to the scalars. So there is an exercise 2 in the problem setting, which you can check with this guy is, genetically, 1, 8 pps. So the way to see it, you're going to see that, OK, you try with the answers epsilon 0 times the projector. But then you see that this is not genetically in a super conformal killing spinor form. So it doesn't look like epsilon plus epsilon gamma x and new gamma nu because there are extra terms. And you can cancel these extra terms if you impose further conditions on the epsilon 0. So instead of just killing half of the supersymmetry, you killed actually 1, 8 of them. So you preserve only 1, 8 of them instead of preserving half because you impose three conditions in general. So you impose the projection plus two extra conditions which are also half rank conditions and you reduce by 1, 8. And this is going to be interesting for the last lecture. There are cases. So you give me any contour C like this on the sphere. If it is built like this, this is genetically 1, 8 pps. But you can consider, so this contour, of course, is irregular. It doesn't have any nice symmetry. You can consider nice contours. Like, for example, instead of taking something generic, you can take some latitude. So you have a latitude at an angle theta 0. So this is more symmetric, of course, because there's some azimuth angle that is there's invariance under this thing. And you are going to see that actually one of the extra conditions that you have to impose drops. So instead of having 1, 8 pps, this is 1 quarter pps. Of course, you can go back to the circle by taking theta 0 equal to pi over 2. Then you go to the equator. And again, you have some cosine of theta 0 multiplying some constraint. And when theta 0 is pi over 2, it becomes 0. And then that constraint drops. And you go back to the 1 half pps width on loop. Or some other. So this is going to reappear during the last lecture. So this is called the latitude. Another nice contour is if you take two meridians, two semi-circles, meeting at some angle. So this is like a wedge, like an orange wedge. And so this is also 1 quarter pps. And there are a few operettos which are 1 quarter pps. In general, they are 1, 8 pps. And so this is a construction in exercise 2. That's also an exercise 1 in which you have another construction with the loops called Zarembos loops, which can also be obtained from this class by some limiting procedure and are also interesting. And it's very nice. Actually, it's very easy to check to understand how the counting of supersymmetry works in this case. So let me stop here. And so tomorrow we are going to do, we are going to evaluate these objects, the expectation value of these objects.