 Okay. Thanks for coming again. Are there any questions that you would like me to address before I keep going regarding the material yesterday or anything related? Yes? I'm not sure if it's going to be covered. I'll be repeating the questions. That was my agreement with their conditioner guy, that I'm repeating the questions. So we saw that we had this non-HoloWordPick. Does this come from a perturbative part or a non-perturbative part? The question is about the capping constant dependence in the extremal correlation functions that we haven't yet discussed. I just mentioned that as an aside. So the question is whether that's perturbative or non-perturbative and the answer is that it's both. There are both perturbative and non-perturbative contributions. As they're typical for any observable in quantum field theory. That there are both kinds of contributions. Any other clarifications before I proceed? Okay. Very good. So yesterday, I'm just going to start by summarizing the discussion yesterday. So we discussed extremal correlators in N equals 4 as a historical motivation, and then we switched gears to N equals 2 supersymmetric field theories. In particular, super conformal field theories. We started the discussion by talking about the two famous half-BPS sectors of those theories. So there are the one half-BPS sector is the Coulomb branch sector of operators, which are annihilated by all the Q bars, and they conjecturally have zero spin even though, as I said, the fact that JR vanishes is not proven, and they're vanishing SU2R spin, and their U1R charge is related to the scaling dimension. So this is the U1R charge. Then there is the Higgs branch operators, which are annihilated by a different set of half of the Qs. For them, the U1R charge vanishes as well as the spin, and their dimension is fixed in terms of the SU2 spin, the SU2R symmetry spin, isospin. So this is the SU2R isospin. So as I said, it turns out that at least in Lagrangian theories, this sector of operators is classical, so it does not depend on the coupling constants. We also mentioned that it's typical for N equals 2 super conformal field theories to have exactly marginal parameters, which we'll discuss in much more detail today, which are comprised of the Young-Mills couplings and the various theta angles. It turns out that the observables in this sector do depend on those parameters, but those do not. So these are classical, and therefore in Lagrangian theories, they're not interesting. Well, the observables in this sector do depend in a non-trivial fashion on the coupling constants and that's what makes them interesting. In particular, the cyber-guitand solution pertains with the Coulomb branch VACWA, which are closely related to the Coulomb branch operators. Well, the Higgs branch is not corrected, and that's more generally true for. So we will focus on those guys because of the interesting dependence on coupling constant that they furnish. Okay. So the outline of today's lecture is that I'll define extremal correlators for Coulomb branch operators. I'll give you a quick proof that they satisfy similar properties to those that we've seen for N equals 4 maximally supersymmetric Young-Mills theory, and then we'll describe some special cases of these extremal correlators that are particularly interesting and they measure distances in field space. So we'll discuss a little bit conformal manifolds and anomalies on conformal manifolds, and then we'll make contact with extremal correlators. So that's the plan roughly for today, and I'll continue that same line of development tomorrow, and then we'll talk about the applications or analogies with QCD that these power series allow us to make. Okay. So I want to define now extremal correlators. So this is the definition. So an extremal correlator is defined analogously to what we had in N equals 4. So now we'll take a bunch of Coulomb branch operators. We'll just put them at some points, let's say. So we'll take some Coulomb branch operator O1, put it at some point X1, and then we'll have N of those. We have N Coulomb branch operators, all of which are chiral. At the end, you remember the idea is to put another operator that soaks up the whole dimension. But clearly, these guys carry an asymmetry charge according to this formula delta equals R over 2. So we need to put an anti-chiral operator to soak up all the lines that come out of the chiral operators. So this is the analogous construction for N equals 2. So we put an anti-chiral operator O dagger at Y in such a way that the sum of the dimensions from I to N is the dimension of this guy, which we may call just O. So this is what extremal correlators. These are extremal correlators in N equals 2. They're a straightforward generalization of those that we studied in N equals 4. Each of those operators is a half BPS, but they're not a half BPS with respect to the same sub-algebra. Those are annihilated by the Q-bars, while this is annihilated by Q. So I would like to show a few properties of those correlation functions. Most notably, I want to establish the property that we found for N equals 4. I want to show that there is some function of the coupling and the theta angles, and perhaps it's a function of N. It's like SUN gauge theory or something. So it's a function of various exactly marginal parameters and N. But then the dependence on the coordinates is like we had in N equals 4. So it's just that. So that's what I want to prove, that the dependence on the coordinates is straight, is easy and it cannot be renormalized. So that's an exact result. The main subject is of course the computation of this pre-factor, which depends in a perturbative as well as non-perturbative fashion on the coupling constants. So this is the first thing that I want to show today, that this dependence on the coordinates is correct, making an analogy with N equals 4. Okay. So let me show you the proof. The proof looks a little bit tricky. But if you think about it, it's correct. So how do we prove that this is true? So we are going to use the fact that Q bar annihilates O that we will use this fact. Of course, Q annihilates O dagger, since O and O are chiral and O dagger are chiral and anti-chiral respectively. Then another fact that we will use is that in supersymmetry, the derivative is the anti-commutator of Q and Q bar up to some pre-factors. So these are the two properties that we will use. That's not all. Another thing that I'm going to use, this is where the trick comes in. So this is a somewhat nasty trick that allows this proof to be much simpler than what it could have been. So the trick is what I already alluded to yesterday. We will use the conformal invariance to put the last anti-chiral operator, this operator at infinity. So that's the essence of the trick. So we'll put O dagger at infinity. According to the formula that we used yesterday. So this is when y goes to infinity. So this will be an important little trick that will simplify the proof quite a bit. So how do we proceed? So now we have to study O at x1, O1, or then On at xn, and then we have O dagger at infinity. Now what we want to prove, as I already explained, is that it's independent of the coordinates. So this is equivalent to saying that when the point y is at infinity, then this is entirely independent of the coordinates. So it's some function of the couplings, but it's independent of the coordinates. So that's what we want to prove. This is a little bit easier than proving this for generic y, although of course they are the same statement by conformal invariance. So we take a derivative with respect to x1 of this correlation function. So we want to prove that it's independent of x1, and the same argument would be true for any of the xi, so that will be fine. So we use the fact that this is proportional to q, q dagger, and then we use the fact that q dagger annihilates the operator at O1. So therefore what we find is that the derivative of this correlation function, okay, let me just write it down explicitly. So the derivative of this correlation function with respect to x1 is given by a q dagger of q, sorry, this is an anti-commutator, q dagger of q with O1, and then you have O2, On, and O dagger at infinity, okay? So so far so good, or is it clear? Okay, very good. So this is what we got. And now the idea is of course to observe that q bar annihilates all of those guys. So q bar should really act only at infinity. So we make another step. We see that the derivative with respect to x1 is the same as a q at O1, then you have O2, On, and then you have q bar acting on O dagger at infinity, okay? And now how do we proceed? Here comes the little nasty trick. So this is a new operator that sits at infinity. We just take the operator that was defined at infinity, and we act on it with q dagger. So we generate some descendant, super conformal descendant of that operator. So this is now some q of O1, O2, On. So this is some fermion operator. So this is like some fermion bar associated to the operator O sitting at infinity. Is that okay? Now the point is that, so now this operators can be all oped. Now the op is non-trivial because there is this fermion here, but they can be oped in some way until you get something on the right hand side of the operator product expansion of these guys that has a non-trivial overlap with the fermion. But what is the dimension of this fermion? The dimension of this fermion is no longer delta. So here the dimension of this guy was delta O. But the dimension of this fermion now is delta O plus a half. However, the power of y that multiplies it is still the same as it was before. So this wasn't corrected, which in order to have well-defined correlation functions with a fermion that sits at infinity, we need to correct the power of y by adding another plus one here. But we haven't done that. And therefore all the correlation functions are going to decay to zero. So this is just identically zero because we have not multiplied it with a proper factor of y. So all the correlation functions are going to decay like one over y. So that's a quick argument. So it doesn't matter what's in the OPE here. As long as we realize that the power of y that we've multiplied the operator here is not sufficient to make the correlation function non-zero. So it just vanishes because it decays too fast. Let me just do it a little bit more. Maybe it was too quick. It's a tricky argument. It seems too quick. Just do another step maybe to explain how it works. So this is a statement about any conformal filter. So suppose you have a bunch of operators. It doesn't matter whether they're descendants or primaries, it doesn't matter. And then you have some operator that is at some y. And suppose this y is very, very far away from the axis. So the axis are here. So suppose we have one point that's extremely far from all the other points. So from general principles, essentially from locality, we know that this has to decay like one over y to the power two delta O. That's a general statement, okay? That's because we OPE these guys and we must get something which has dimension delta so that it has a non-zero overlap with this guy. So it must decay faster than this. It must decay at least as fast as that. Because any other operator that has an overlap with this primary must decay at least as fast as that. So therefore, if we now have, if we now multiply this by some power y, whose power is not sufficiently large, then we'll just get zero in the y going to infinity limit. So that's why this vanishes. Because the power of y with which it was multiplied was not sufficiently large. So this proves this assertion that the x, y dependence of this correlation functions is restricted to be just it. And therefore, the interesting observable is the pre-factor, not the x, y dependence. So it's a tricky argument. It seems too quick, but it's actually correct. Are there any questions about it? Good, so now we'll try to keep going and try to unravel some interesting special cases. So to try to understand these extremal correlators in general n equals two theories, it's useful to try to look at some special cases first. And one particular special case that would be very interesting is the two point function, which is a special case of this endpoint function of some chiral primary of some Coulomb branch operator and some anti-chiral operator located at x and y. Such that the dimension of the operator i is the same as the dimension of the operator j, and it's two. As we have described yesterday, those correspond to exactly marginal deformations of the theory. So such operators can be used to deform the theory by some coupling constants lambda i. And therefore, this two point function has an interesting interpretation that we're going to discuss now. So this is a special case of an extremal correlator, which is an entry in which the pre-factor has a physical meaning that is more clear. So here the pre-factor is a little bit more, is a little bit too abstract to parse. But in this special case, where these operators have dimension two, and they can be used to deform the Lagrangian, and you get exactly marginal operators in this way. This coefficient function is known as the zamologic of metric. So this is some function of their couplings, and it multiplies, of course, one over x minus y to the power four, which is a special case of that formula. And this function now has a physical meaning and some interpretation that I'm going to explain. This is called the zamologic of metric. So the zamologic of metric in theory space So the zamologic of metric in theory space is an interesting observable in many conformal filters, and in particular in this setting equals two super conformal filters. So this will be the first extremal correlators that will learn how to compute exactly using localization. And the more general case will follow by some procedure that I'll explain tomorrow. So today we'll focus about understanding what is the zamologic of metric, and what it's good for, and how to compute it. Are there any questions? Okay, so now, what I'm going to do now is I'm going to forget about supersymmetry we'll forget about supersymmetry for the next couple of minutes, and it will be like a general discussion of what are conformal manifolds, what are these exactly marginal parameters, and what is the zamologic of metric. Okay, so that's what we're going to do next. So the idea of the following, suppose we have a conformal filter, a conformal filter in D dimensions. So we're not even going to specify the dimension to be necessarily four. So we have an abstract non-supersymmetric conformal filter in D dimensions, and suppose it has an operator, it has an operator, well, it has operators. OI, whose dimension is D. Here one does not need to get confused because here the dimensions were not D, they were D over two, or D minus two, but that's okay because the operator, we need to integrate it over D for theta. So the marginal deformation is not given by the Coulomb branch operator, but rather by a descendant of the Coulomb branch operator such that the dimension of the actual deformation is actually D, not D minus two. So the general story is that we have marginal operators of dimension D. So now it's a tempting idea to try to deform the action. We can try to deform the action by adding a sum with some coupling constants. So these are space-time independent. I could have put them outside of the integral, which let me do that. So we have OI, X, DDX. And I'm gonna assume that these operators are her mission. So these are real operators, or her mission, so that I don't have to worry about adding complex conjugates, which I can do with it. So it's tempting to try to consider this kind of deformation of a conformal field theory. And an interesting question, which is an interesting natural question is, does the theory remain conformal? So does the theory remain conformal? That's the natural question to ask in this context. So what are some examples of this phenomenon? This is a very common, I mean this kind of thing is very common in statistical physics or even in simple examples of QFT. So an example that you could keep in mind if you like and is free field theory in D dimensions, let's say in four dimensions, free field theory in four dimensions, that's a good conformal field theory, and it has a marginal operator. Who knows which marginal operator does the free field theory admit? Good. So an interesting deformation is to add lambda five to the four, sorry, D4X. So does anybody know if this remains a conformal field theory or not? Is this a conformal field theory? No, right, indeed. The beta function of lambda is non-zero. So in the case of the free scalar field, we have a conformal field theory which is free, but there is the operator five to the four, it turns out to be marginally irrelevant, right? So if we add this coupling lambda, it flows back to zero. So this is like the lambda coupling. So you don't generate new conformal filters if you try to deform the theory, you just go back to the original conformal field theory by doing the RG scaling. So you don't get any new conformal filters, but it could be that there are examples where you do get new conformal filters in this way. Does anybody know of an interesting example without supersymmetry or the beta functions cancel? Does anybody know of an example with supersymmetry? Yes, n equals four is a good example. So with supersymmetry, one famous example is maximally supersymmetric and super Yang-Mills theory with gauge group SUN. So this theory has a Yang-Mills coupling as well as a theta angle, and these are two exactly marginal perturbations. So they do not have any beta function. Does anybody know of another example with supersymmetry, perhaps? Yeah, with an exactly marginal perturbation, one in conformal field theory. Very good, that example will study very much. So this is an n equals two example. We have SUN gauge theory with two NF hyper-multiplets. So this is, oh, sorry, two n, two n hypers. So this example will study in great detail. In fact, this will be the canonical example to run these questions on. This example was constructed in the 90s and analyzed in great detail. Does anybody know of examples with n equals one supersymmetry where there are exactly marginal perturbations? Very good. The most well-known example with n equals one supersymmetry, perhaps, is the beta-deformed and supermax, beta-deformed maximally supersymmetric Yang-Mills theory. So there is a certain deformation of that theory that preserves only n equals one, and it's exactly marginal. And in fact, this list is extremely long. But without supersymmetry, this phenomenon is not very common. However, it does appear in some, there are several examples where it does appear, even without supersymmetry. Does anybody know of an example without supersymmetry? No, it was a Witton model. The parameter K is quantized. So it's not a real infinitesimal deformation. So, yeah. So I think the Liouville is another interesting, so you can change the linear dilaton coupling. So you are talking about changing the linear dilaton coupling. That's often, that's like the most confusing example, because it's an infinitesimal deformation, but it actually changes the central charge. So it's not a good example. It changes the energy momentum tensor. So it's not within these rules, actually. It involves some coupling to curved space, which it's not exactly the same thing. So the Liouville deformation is not a good example. So I'll give you one example, which is very famous. This is called the Ashkin-Teller model. We call it just the C equals one model. It's an example in D equals two. The Ashkin-Teller is how the condense matter people call it. So it's an example in D equals two, where we have a compact boson of radius R. Compact boson of radius R. And it turns out that the radius of this boson is an exactly marginal parameter. Equivalently, you can think about a quadratic fermion interactions, which is called the Thuring model. So there is a quadratic fermion interaction, which is exactly marginal in one plus one dimensions. So these things appear in condense matter physics, especially in two dimensions. And with supersymmetry, it's very common also in higher dimensions. Although the construction of these kind of things without supersymmetry in higher dimensions seems to be very hard. So the question is if I know of any attempts to prove that there aren't conformal manifolds in higher dimensions. So I recommend you to read some recent work by Bashmakov et al. This is one thing. And then there is also recent work by Behann et al. This is Connor Behann et al. So I'll tell you more or less what they are trying to do in a second. So these are some attempts to say something about that question. So the question of whether these parameters are exactly marginal or not, you can answer this question sort of in some perturbation theory. You can try to add them with an infinite, you can try to add them with a very small coefficient and then compute the beta function. And I won't try to teach you how to do conformal perturbation theory, but what you find is that the beta function for this coupling's lambda i, which we can just call beta i, take the following form. So the beta function, the coefficient of lambda i, so the coefficient of lambda, sorry, this index should be up, yeah. So the coefficient of lambda i is zero. Does anybody know why? Why is the coefficient of the linear term zero in the beta function? It's because the dimension was chosen to be d, right? Then there is a quadratic piece with some tensor and then there is a cubic piece with some tensor and there is like an infinite series like that. So for these perturbations to be really exactly marginal, meaning that they would define an actual space of conformal field theories, which is often called the conformal manifold. So this is called the conformal manifold. This is the space of conformal field theories. It's not, maybe the word manifold is misleading because this space could be slightly singular, yeah. Yeah, sorry. Yeah, let's say A, B, A, B, then C, D, well, A, B, C, then A, B, C, then A, B, C, and so on. So the space of conformal field theories is often called the conformal manifold. It could be just a point, but it could be a more interesting space. In all of those examples that were listed above, this space is very interesting both topologically and geometrically. And also even in the Ashken Teller model, this space is interesting. It's the upper half plane divided by some discrete group. But it could be empty, like in many of the non-supersymmetric constructions. Or it could be just one point. So in any case, for this M to really exist beyond just the one point that we started from, all of those coefficients have to vanish. So C, A, B, I has to vanish. C, A, B, C, I has to vanish. So there are infinitely many conditions on the underlying conformal field theory. The origin, in some sense, that would be required if you wanted to extend the conformal field theory to a whole continuous space. So you can try to compute these constraints order by order. And they were computed in the literature, and you can find them reviewed in these papers. And then you can try to prove that maybe supersymmetry is necessary for all those constraints to be magically satisfied. For example, this constraint has an interesting interpretation. This one is really easy. This constraint means that if you take the operator product expansion of two of those marginal operators, the coefficient of another marginal operator vanishes exactly. So the OP of two marginal operators cannot spit out another marginal operator. If they do, you get a beta function that's non-zero. So this is an easy constraint. This is just a constraint on the OP of the original conformal field theory. But this one is already harder. This implies some integral of the four point functions has to vanish with some kernel and so on. So the constraints become more and more cumbersome. And it's unclear how they could all be satisfied without some magic. So this magic could be supersymmetry. And in two dimensions, there are also ways without supersymmetry as I've remarked here. So now we'll suppose that this space does exist. We'll suppose that this space of theories does exist. So let's suppose that M conformal exists. So we have this picture. And the coordinates on this space are the coupling constants. So the natural coordinates on this space, that parameterize the space of conformal field theories are this lambda i's. So the coordinates are naturally the coupling constants. And you can ask, okay, so is it a Riemannian space? Is there a natural metric on this space? But is it just a manifold? So the observation of Zamalochikov, well, I think it's kind of misattributed. He didn't really discuss that question, but he did some related work. So the observation is that you can define a quantity that looks like the metric by, let's say that you are at some lambda. Let's say that you are at some lambda zero. So this is like a point on the conformal manifold. So let's suppose that you are at some point on the conformal manifold. And you wanna ask, okay, what is the metric? 10, sorry, that point. So the point, so what you do, you do, you do, you just define the two-point function. Let's say you put this at zero. This one is at infinity, just so that we don't have to deal with the X minus Y dependence. Sorry, and the indices are importantly downstairs. And this defines a certain two-point function. It's a bunch of coefficients. They don't depend on anything other than lambda. So this correlation function is computed in the conformal filtering at lambda. And this is some function of lambda. So the idea is that this defines a metric tensor on this space. So the two-point functions at each point define the metric tensor. This is called as a logic of metric in theory space. So this gives you some metric on the space of conformal filters. So in many of these examples, we actually know the metric now. And of course, in this example in two dimensions, the metric is also known. So one has to, it's not a trivial statement that this defines a reasonable choice of a Riemannian structure. Because as you remember from differential geometry, the metric at any given point is meaningless, right? You can just take the metric at any, like you can just choose the inertial coordinates. So you can choose the metric at any given point to be just one, delta ij. So similarly here, what is the physical meaning of choosing the metric to be delta ij? Physically, if you have operators given at some point, you can always redefine the two-point functions to be one, right? You just normalize the operators. You choose an orthonormal basis. So physically, when we have a conformal filter, we are often compelled to choose an orthonormal basis of operators O i. And then the metric would be just the delta function. But you cannot choose such a basis everywhere. So you can make some basis choice. This basis choice depends on the coupling constants. But you will not be able to set the metric to be one everywhere. You can also argue, and that is non-trivial, that there is no obstruction to choosing the first derivatives of the metric to be zero. This is also something that you can always do in differential geometry. In conformal filtering, this is harder to see. This amounts to be able to choose some contact terms in three-point functions. And to show that, you have to use this condition very, very importantly. So this condition comes in in the proof that this can always be done. So you can always choose inertial coordinates. And therefore, the physical object on the conformal manifold is not the metric. The metric is not physical. The physical object is the curvature tensor, the Riemann curvature tensor, which is invariant under, well, which is still coordinate dependent, but you can form various scalars, like R, R squared. And so this is the physical object that you can compute from the metric. So it's not obvious, but it's true that you can define such a metric. And that will be what we'll try to do for these n equals 2 models to compute the metric exactly. So the question is, what is the physical meaning of the curvature? So if you are, I mean, this description is not entirely satisfactory, because what I'm telling you is that you have to compute a two-point function at every point and then observe that there is no orthonormal choice everywhere. And then, by taking derivatives of the metric, you could extract the Riemann tensor. You could say, OK, I don't want to do it. Can you just give me an observable that spits out the Riemann metric? And it turns out that you can do it. So there is a way to, if you, there are some, well, there is a lot of literature on that. But certain integrals of four-point functions anti-symmetrized in some indices over some funny regions of the cross ratios spit out exactly the Riemann tensor. And they're independent of orthonormal basis choices, independent of reparameterizations. So there are some combinations of four-point functions that spit out directly the Riemann tensor, rather than the metric. But there are integrals of four-point functions with some funny regions and commutators. So this is a little bit easier. Sorry, I cannot hear because of their conditioner. Oh, yes, yes. I remember the formula in two dimensions. Actually, the exact formula for the Riemann tensor depends a little bit on dimension, because there is some change of variables involved in the space of cross ratios. So in two dimensions, I actually remember the formula. It's just an integral of all the possible cross ratios of the four-point function, where you put one operator at 0, one operator at eta, one operator at 1, one operator at infinity. And then there is a funny weight, which is a logarithm of eta squared. So you can argue that this integral converges. For this integral to converge, you got to use these conditions just to show that it indeed converges. And you can show that this leads to the Riemann tensor. So this formula is only true in two dimensions, but there are similar formulas in higher dimensions. So it is an observable. It's an integrated four-point function. This is something that's actually very natural from the Bootstra point of view. These kind of integrated four-point functions in Euclidean signatures could be studied with the Bootstra. OK, so now I want to describe a slightly more advanced idea. This idea is more advanced. It doesn't appear in any textbook or something like that. But it's crucial to understand how to compute. So our main goal is to compute this in any pool stone, right? That's the main goalpost of these lectures. So to actually, well, one can explain how to do it in several ways. There are several derivations in the literature. One derivation that I like is through some anomalies on the conformal manifold, because that derivation has some features which are also true without supersymmetry. And so it may have other applications. And then there are derivations which are a little bit more suited to supersymmetry. So I want to, for a second, go back and introduce these coordinates x and y. So we have the zamologic of metric, and then we have 1 over x minus y. And since the dimensions of these operators is d, so the power here is 2d. OK? So an interesting idea is to try to write it in momentum space, which seems weird, but just tag along for a second. You'll see that this is useful. Let's try to go to momentum space. So this idea is actually very useful, and there are many papers about various aspects of conformal filters in momentum space. So let's try to write it out, write it in momentum space. So we have OIP, OJ minus P. And then we have the zamologic of metric G. So I need to put some twiddles here because we do a Fourier transform. So then we have the zamologic of metric, and then we have the Fourier transform of this thing. So what is the Fourier transform of this function? Let's do it here. So we need to do e to the ipx, 1 over x to the power 2d. And we need to integrate over x. So just from dimensional analysis, this is P to the power d. Is that OK? So we write P to the power d, but actually this is not correct. Let's say for them, well, there is a mistake here. We have actually two cases. So this is almost correct, but it's not entirely correct. If d is even, this is P to the d times log p squared. If d is odd, this is just P to the d. How do you see that you need this log squared? Sorry, how do you see that there is this logarithm? So the intuitive reason is let's take d equals 4. So if I didn't put the log, you would say, OK, the Fourier transform is P to the 4. But P to the 4 is a polynomial in momentum space. This is a polynomial. A regular polynomial. What happens if you do a Fourier transform back to position space of a polynomial? You get derivatives of delta functions. So the Fourier transform back would have been just a box squared of a delta function. That would have been wrong, because what we wanted to do is the Fourier transform of a non-local function, like 1 over x, not something that has only supported coincident points. So for even dimensions, this is a polynomial, so it cannot possibly be the right answer. But for odd dimensions, this is not a polynomial. P cubed is not a polynomial in P squared, so it's fine. You cannot write it as a differential operator acting on a delta function. So you actually need the logarithm in even dimensions. Otherwise, it's wrong. So here, I'm just assuming that d is equal to 2n. So d is even. And then we need the logarithm. So on the one hand, so we see that we need the logarithm. Is there a question about why we need the logarithm? Or it's fine. OK. So people don't like logarithms in conformal filters. The reason is that a conformal filter is not supposed to have any scale. So how can we put a scale? Like, what is the scale here? Seems meaningless, right? Because it's a conformal filter, it's not supposed to have any fundamental scale. So how can there be logarithms? So the point is the following. This is actually very common in conformal filters. In many conformal filters, correlation functions in momentum space would have logarithms. And you might get confused. How can there be logarithms in momentum space? So the point is that, suppose you try to see if this logarithm has any important role to play in this story. So suppose you just rescale p by some mu. And you try to see what happens when you rescale the coordinates or the momenta. So what would happen to this p to the d log p squared over lambda squared? Well, this will transform to p to the d log of mu squared over lambda squared. But as I have just explained, if this even, this is a completely local object. So this is only supported at coincident points. So this is only supported at coincident points. So conformal invariance is not strictly violated. Because conformal invariance is a property of a separated point correlation functions, not of coincident point correlation functions. So conformal invariance is not violated because the conformal worded entities are only true at separated points, not at coincident points. So in fact, these logarithms in momentum space in various conformal, in various correlation functions have many applications. For example, you can also look at a paper by Pimentel and Maldesena about some similar story in the context of correlation functions in cosmology where similar logarithms appear in conformal filters in the sitar space. So they pop out in many applications. So it's good to know. So actually, logarithm squared is not allowed. Does any, maybe I'll just ask the audience, why is logarithm squared never allowed? Like, could it be logarithm squared, or that would be not OK? Does anybody see the answer? So logarithm squared would be bad because if I rescaled the momentum, it would not be a pure contact term. It would not be a pure polynomial. So this can never appear in conformal filters. But this is fine. So when you see such a phenomenon that the conformal invariance is preserved at separated points but it's violated at coincident points, that's the smoking-gun signature for a conformal anomaly. So a similar anomaly is the central charge anomaly in two dimensions, which is very well known, the Virasora anomaly. It's exactly the same story, that there is some logarithm in momentum space. So, well, I should have said the central charge in four dimensions. This is exactly the same story. But in two dimensions, it's a little bit different. So this is a conformal anomaly, and there is the mathematical way to treat it. So the mathematical way to handle such an anomaly, the mathematical way to handle this is to take the coupling constants and promote them as two functions of space-time. This is the common trick that allows you to understand these anomalies a little bit more systematically. Because then you can ask, what happens to the partition function? So let's define the partition function to be an integral over all the fields, and then there is an exponential of the action of our conformal filters. Then we add coupling constants lambda i, which are functions of x, and they multiply these operators. So now we've promoted the coupling constants to functions. It's a useful trick, because now the partition function is a function out of some functions. So it has more information than just the partition function as a function of some fixed couplings. Now it's a sensible question to ask, suppose these are exactly marginal operators. So these are exactly marginal, meaning that the beta functions vanish, and you have a conformal manifold. It's a sensible question to ask, what happens to this partition function if we change the scale? So as we've learned from this little exercise, when we change the scale, you might get some contact terms. So a mathematical machinery that allows you to encompass these contact terms in some useful formula, is to think about the partition function as a function of the couplings, and then perform a conformal transformation on this whole partition function and see what you get. So let's perform a conformal transformation labeled by some function sigma of the logarithm of the partition function. So this is an axiom of quantum filtering, or let's say, of conformal filters, that if you try to do a rescaling of the partition function in a conformal filtering, you'll only get a function of this classical objects lambda. So there won't be any quantum operators, because the theory is genuinely conformal, so the violations of conformal invariance can be only due to background fields, not due to the physical observables. So the idea is that this is an integral over d dx of some local action that is made out of these functions lambda ix. So that's the general expression for a conformal anomaly. In any conformal filtering, you could also introduce a metric, which I'll do. So you can also couple the theory to some background metric. So it's not a dynamical metric, it's just a background metric. Then the partition function is a function of lambda and g, and you perform a conformal transformation of all the coordinates, or equivalently, you wire rescale the metric. So this is just, I mean, I'm trying to say some general things about how anomalies work very, very quickly. The details might not be clear if you've never seen anomalies before, but at least you can follow the rough logic. So the point is that using these logarithms, you can, the point is that using these logarithms you can determine the anomaly. Now I'm just going to tell you the anomaly in four dimensions. It has a different expression in any even dimension, but I'll just write down the answer in four dimensions and then you'll see what we can do with it. So in four dimensions, this is a hard computation, but I'll just write the answer for this computation. It's proven in the literature. So the anomaly in four dimensions takes the following form. I'm just going to write it and then step out so that you could see. Sorry. So the zomological metric appears statistically in the anomaly since it appeared here and this is the coefficient of the log. So the zomological metric appears here. There are some operators here and many other terms. This box head is the ordinary box plus some Christoffel symbols. So when it acts on lambda, well, there is some expression that is not going to shed light on anything, but I'll just write it in any case. So there is some modified box operator and there is a mess of terms here. So what you find is called in the literature sometimes panacea fratkinsatling operator. I think it's wrong with the spelling. So the details here are not important. What I wanted to say is that just from these logarithms you can determine a certain anomaly polynomial and supersymmetry comes at this point. So all this is true without supersymmetry and there are many applications for this thing. But supersymmetry comes at this point because you should also require that this anomaly polynomial is appropriately supersymmetrized. So this is like saying physically that these logarithms cannot just be isolated. Logarithms, they have to obey the word identities of supersymmetry. So this is where supersymmetry comes in. Up to here, this is very general and does not depend on supersymmetry. So supersymmetry comes at exactly this point, and that's how we're going to enter the subject of extremal correlators. So supersymmetry in particular n equals 2 supersymmetry, implies various restrictions on the anomaly polynomial. Most interestingly, one finds the following fact. So when you try to supersymmetrize this anomaly polynomial, this is kind of a complicated exercise, and it's not very illuminating. But there is one thing that I do want to tell you, which is that if you supersymmetrize this anomaly polynomial, there is one term that pops out, which is particularly interesting. This term is a half integral of d4x, then some function k of lambda-lambda bar. These are the exactly marginal couplings times box of Delta Sigma, box squared of Delta Sigma. This term is a particularly interesting term that pops out from supersymmetrizing the anomaly polynomial. Now I have to tell you what is k. Delta Sigma, it's the scale variation. So you know what is Delta Sigma. I'll just tell you what is k. So it turns out that in these n equals 2 supersymmetric theories, the zomologic of metric, so the first thing about these n equals 2 supersymmetric theories is that the exactly marginal couplings, which are in general some real couplings by which you can deform the action, it's natural to think about them as complex parameters which come in pairs of some complex parameter and it's complex conjugate, and the intuition here is that these are like the angles coupling in the theta angle and this is the complex conjugate. So if you have cyber-guitand theory in mind, this could be 1 over g angle squared plus, let's say, i over theta then some 2 pi, and this is the complex conjugate in some convention, and this will be the complex conjugate. So what was the minus sign? So the first thing you need to know about n equals 2 supersymmetry is that since the marginal deformations appear in front of chiral ring operators, you remember this expression which we wrote already many times, so the exactly marginal deformations of n equals 2 supersymmetric conformal filters come in complex parameters because we need to multiply chiral ring operators which are complex operators. So that's the first thing, so it's a natural complex manifold. So n-conformal in the context of supersymmetry is a complex even-dimensional space. So it's always even-dimensional because these are coming complex pairs. The other thing that you need to know is that the zomologic of metric has only mixed components. So the metric components which are i, i, and j, or i bar, i bar vanish identically. So the metric is only between lambda and lambda bar. Like on the complex plane, z, z, z bar. And it turns out that it can be written as the second derivative of a scalar function. This is called the scalar structure. So when the theory has n equals 2 supersymmetry, the conformal manifold is not just a generic Romanian manifold, it's actually a scalar manifold, a scalar space. So in particular, it's a complex space, an even-dimensional space with a scalar function. So exactly this scalar function appears in the anomaly polynomial, amazingly, okay? Are there any questions? Yeah. Can you just speak a little bit louder? The question is if this anomaly polynomial can be obtained by anomaly inflow, right? Yeah. Right, so the ideas of anomaly inflow have been very successful with chiral anomalies, but conformal anomalies have never worked out. Like nobody has ever been able to obtain the conformal anomalies from anomaly inflow in any useful way. So it's not known. I see. Yeah, I understand. Okay, so we can go over that. Okay, good question. So the question is what are the geometric structures of conformal manifolds? Let me summarize it here. I've been a little bit fast with that point, so let me just do a more pedagogical summary. So without supersymmetry, this is the generic case. What do we know about the conformal manifold? It's just a Riemannian space. So M conformal has a Riemannian structure. So we are not aware of any additional structure on this space without supersymmetry. Just seems like a generic Riemannian manifold, and it could be odd dimensional in principle. Then you can try to label the properties of this conformal manifold like for any given number of dimensions with any given number of supersymmetries. So I won't try to do it like extraneous systematically. I'll just give you a few examples. So D equals four and D equals three. Without supersymmetry, there is always nothing more to say. It's just a Riemannian space. But if you have N equals four supersymmetry, the conformal manifold, as far as I know, is just scalar. Well, it's a... So N equals four tier is essentially unique. It's maximally supersymmetric Young-Mills theory. So it's a scalar manifold, but we also know the metric, and it turns out that it's a constant curvature space. So it's a constant curvature. And in fact, it's just a... Well, it's just the upper half plane, mod SL to Z. So it's essentially just H mod SL to Z. If the gauge group is SUN, this is the upper half plane. So when N equals four, we know everything. For N equals two, as far as we know, it's just a scalar manifold. There might be additional restrictions in the future. Maybe somebody in the future could prove that the curvature is non-positive, or maybe somebody in the future could prove that it's a quotient of something. At the moment, we don't know of anything beyond just scalar. And in fact, for N equals one, it's the same. So there is no more structure that's known on the conformal manifold besides just being a scalar space. Now, in three dimensions, with N equals one supersymmetry, well, presumably here, it's just a Riemannian space. With N equals two supersymmetry, you get scalar. And with N equals four supersymmetry, you cannot have a conformal manifold. That's the situation in three dimensions. So with N equals four and up, you cannot have conformal manifolds in three dimensions. So, hmm? You can prove that there are no marginal deformations. Exactly marginal deformations, yeah. Yeah, so there are many interesting questions about these spaces. You can ask, are they compact? So as you see here, it's non-compact. In this example, it's non-compact, but it's a funny non-compact space. It's like the fundamental domain. So it has finite volume, but it has like some directions which are infinite distance away. The Caspian infinity is at instance this infinite distance away, but the volume of this space is nevertheless finite, even though it's like a thin throat, like a very, very thin throat that goes to infinity so the volume doesn't blow up, but the distance goes to infinity. So a natural conjecture is maybe the volume of the spaces is always finite, okay? Another natural conjecture is that maybe all the non-compact directions on these conformal manifolds look like thin throats. That's also a very natural conjecture consisting with what we know. So there are many questions about these things and well, not all the questions are answered. So the metric can be computed as far as we know in this case, exactly. This case is known, this is trivial. This will be a special case, but what I'm talking about here now is the computation of the metric in this case. For this case, there's no progress. And also for this case, there is no substantial progress yet. Yeah, that's a great question. So actually this is something that was at the time we struggled with. So a priori this connection could have been anything in that anomaly polynomial, but there is something that's called the Weslumino consistency condition. You know what it is? But there is something that's called the Weslumino consistency condition. That turns out to imply that this connection has to be Levy-Civita. Yeah, yeah. Yeah. So the question is whether the fact that there are conformal anomalies means that the conformal symmetry is broken, right? So there are two types of anomalies in physics that are often confused. When you learn like quantum filtering, they're often confused. One is called the ABG anomaly after some people. And then one is called the Tuft anomaly. These are completely different things. The ABG anomaly means that the symmetry is actually broken. An example is the electrodynamics in which there is a unaxial symmetry that is broken by instantons. So this is an ABG anomaly. So this looks like a background gauge field, not a dynamical gauge field, then a triangle, and then two dynamical gauge fields. Dynamical, dynamical. So this type of anomaly is called the ABJ. I think it's Bardin, Jakiv, and Adler, right? Yeah, so this is called the ABJ. That means that the symmetry is just gone. So that's not an anomaly. That's just explicit violation. So nowadays, I just think about it as explicit violation of the symmetry. It's as bad as just adding like a mass that would break the... So you could add mass to the electron. In masses electrodynamics, you could just add the mass to the electron. That would also break your unaxial. These two things are as bad. There are many examples of duality in which on one side the symmetry is broken by anomaly, and on the other side is broken by a mass. So there is no fundamental distinction between an ABG anomaly and just explicit violation of the symmetry. A tuft anomaly is something much more subtle and much more useful. This is not useful. This just means that the symmetry is gone. A tuft anomaly arise when all the external legs are background, not dynamical. And then the symmetry is not broken. Rather, this is the anomalies that you have to match. So these anomalies need to match in our G flows. But the symmetries are unbroken. So the symmetries act on the Hilbert space. The charges are conserved. Everything is good. But these anomalies need to match, which is a different concept. It's not broken. These ones do not need to match. So there are many examples of dualities where these anomalies don't match. But it's fine. They don't need to match because they correspond to symmetries that don't exist. Here the symmetries exist, and these anomalies need to match. So the conformal anomalies that the wife described here are of this type. And that's when it's useful to write an anomaly polynomial, when the anomaly is of the tuft type. In this case, there is no... Another case that is left out is what happens if you have background, background, dynamical. This actually did not appear in the classical literature. So you can ask, is it the tuft type, or is it the ABJ type? So this is much more interesting. What does that mean? But it's one more subtle, but I'm talking about this one, which is a more conventional anomaly. Okay, any other questions? Okay, so I wanna get back to this. This is the supersymmetrization of the anomaly polynomial. So the next idea here, and that's what I'm gonna finish the lecture with, then I'll just have 10 minutes for questions. That's the last thing I wanna say. I wanna show you that this leads to some fantastic connection with localization, which is something that we haven't discussed yet. By the way, this is not true in n equals one. There isn't that for n equals one, supersymmetric theories in four dimensions, there isn't as much progress yet. It's because this term does not appear in n equals one. When you supersymmetrize the anomaly polynomial with n equals one, this does not appear. And this is the key player in this game, turns out. So suppose you wanted to try to understand the connection between R4 and S4. It's well known that they are connected by a stereographic projection, namely a conformal rescaling. So there is a stereographic map here. There is the well-known stereographic map, which means that R4 and S4 are just connected by conformal transformation. So it preserves all the angles. So a conformal transformation is this parameter delta sigma. Delta sigma was a small conformal transformation. Well, by some finite delta sigma. So the idea is that we can use this term to learn something about the partition function in S4. Even though we haven't specified the base space yet, we can now use this term to integrate the anomaly by doing successive small conformal transformations until we get to S4. So integrating the anomaly polynomial, the following claim comes out. So integrating the anomaly polynomial, what we find is that the partition function on S4 is just e to the k. We integrate that, that becomes one, and then we get e to the k. And there is a factor of one over 12 here, if you do it carefully. So that's how this business of extremal correlators gets connected to localization. It turns out that the force-phere partition function is just the exponential of the Keller potential on the space of theories. So miraculously, the force-phere partition function was computed by Peston, but when he wrote the original paper, he did not say what is the meaning of the force-phere partition function. He just computed it. And this is the physical meaning of the force-phere partition function. It's the exponential of the Keller potential in theory space. So this is the physical interpretation of the force-phere partition function. So therefore, we can write the following useful formula. This is the first extremal correlator that we can now determine using the force-phere partition function. So we see that to compute the zoological metric for n equals two super conformal filters, sorry, so this is Gij. Gij is that. And this is obtained through this reasoning. We can now relate it to the force-phere partition function. So we can write two formulas that are exactly the same. And this is the homework exercise to check that they're exactly the same. So this is a trivial thing. When you take the logarithm of z, you immediately pull out k, and then you take two derivatives of k, and by definition, this is G. So this is trivial. There is another way to write the answer, which is actually more useful for what will come soon. So we put the partition function squared in the denominator, and then we have a determinant over the following matrix. Zs4, d over d lambda, Zs4, d over d lambda bar, Zs4, and then the second derivative. d over d lambda, d over d lambda bar of Zs4. These formulas are the same, you can check. So say again, it looks like a fantastic question. So the comment here is that this, the gentleman in the audience does not like this formula because he knows that k is not a well-defined function. So the metric is actually also, the metric is not a great observable. It's also not completely well-defined because you can always choose an inertial frame, but k is much worse because k can also be shifted by a purely holomorphic function of lambda i plus a purely anti-holomorphic function of lambda i bar. If you shift k in this fashion, it doesn't affect the left-hand side since the derivatives annihilate those two pieces, right? So you might be worried that this means that the four-sphere partition function makes no sense. I mean, this formula cannot be possibly right. So the first comment is that when we make contact with the zoological metric, this ambiguity disappears. But the second comment is that indeed, in Pesto's partition function is not universal. He just chose a convention. No, you can think about this as a, you can think about fixing f and f bar as in some sense a generalized choice of scalar frame. So you can think about it as a choice of scalar frame. And it's a completely ambiguous process. It turns out that there is a similar ambiguity in the computation of Pesto, which is in one-to-one. So when you compute four-sphere partition functions of physical theories, it's always ambiguous. The ambiguity is given by a holomorphic function. And he just made the choice. The ambiguity is that there is a counterterm, which looks like a holomorphic function of the couplings times the curvature squared integral over d4x. And it's fully supersymmetric and it's all good. So whenever you compute the four-sphere partition function, you can always add a purely holomorphic function of the couplings of the chiral multiplets times curvature squared appropriately supersymmetrized. Alternatively, when you regularize this determinants, there is a small ambiguity, which corresponds to a subtraction of some divergence. And that ambiguity is given by a holomorphic function. And in the original computation, Pesto just made an arbitrary choice without saying it. But he did. And that choice corresponds to a choice of scalar frame. So this formula is true. Both sides have the same ambiguity. So it's a covariant formula. So it's a very good question, indeed. In fact, one of the consistency checks that you can run on this formula is to see that there is a corresponding ambiguity on the left-hand side. Otherwise, this formula would make no sense. OK, so we've made a connection between the two-point function. Sorry, this must have been Degger. So we've made a connection between the simplest extremal correlator, which is the zomological metric correlator, and the four-sphere partition function. So next time, we'll analyze this in SU2, or maybe even SUN gauge theory. I mean, I'll show you that this receives perturbative, as well as non-perturbative, corrections. And we'll see what is the physical meaning of that. And then we'll write more general formula for any extremal correlators and make some connection with resurgence. Any more questions? The question, if I want to say a little bit more about this derivation, which one, this one? The idea was to just integrate the anomaly polynomial. So the idea is that if you know the partition function in some space, from the anomaly polynomial, you can find the partition function in any space that is related by conformal transformation. Yeah, yeah, yeah. So the full anomaly polynomial is the four lines or something. And there are many terms. And several of them contribute to this computation. And you just have to do the algebra. And that's what you find at the end of the algebra. I can give references where this is done carefully. But it's not illuminating. It's just you have to write the supersymmetric version of this anomaly polynomial, which is a mess. And then you need to integrate it. But the advantage is that if you know the partition function in any space M, you can obtain the partition function in any space that is conformally equivalent by just integrating the anomaly polynomial. So yeah, that was done in the literature. And that's the result of this computation for S4. Any more questions? Yeah, so I believe, yes. So for example, people have computed the question is whether there is a similar formula for other various computations, right? So first of all, literally the same formula, just with a different coefficient, is true for two-dimensions. So since you already asked, I'll just mention that. So there is also a two-sphere partition function for two comma two supersymmetric theories in two dimensions. And literally the same formula is true, just with a, it's like e to the minus k, literally the same. And the derivation is, in fact, exactly the same. And these anomaly polynomials are a little bit simpler, because they are in two dimensions. So in fact, one way of presenting this subject would have been that I would only discuss two dimensions and then say that four dimensions is similar, because all the technicalities in two dimensions are easier. So this is actually, this is exactly true. And this is useful to compute the distances between two-dimensional conformal filters. The conformal manifold is, again, a scalar space. And this has been used to check mirror symmetry, compute new gromm of width and invariance, and you know. Now, there is an interesting question. What about this? Is there a similar formula for this? Here, I don't know the answer. But the speculation that I've never checked was that if the theory is n equals 2, then at least in the limit of small enough s1, this should be the exponential of some line bundle that was discussed by, I believe, Andy Niteski. So I can explain that a little bit more precisely, but that's a very specular. I've never checked if this makes any sense. That could be true. In three dimensions, the three-sphere partition function also has an interesting interpretation, where people write it as e to the minus f. And this f is the f coefficient of the underlying conformal filtering. And it turns out to be independent of the exactly marginal parameters. So three dimensions is in one way different, that it turns out that the three-sphere partition function is independent of the coupling constants. And it's just some constant. But this constant is interesting. This constant has some interpretation in terms of entanglement entropy. It's called the f coefficient, and it's monotonic energy flows. So three dimensions is in one hand easier because there is no coupling constant dependence, but on the other hand, it makes some contact with entanglement entropy, and it's nice in some other ways. So yeah. And for the fives, I mean, there is also, obviously, you can ask this question about six dimensions and five dimensions, and then I don't know. I have no idea. Right. Right. That's a very good question. So there is a paper about this. I'll tell you what I know. So the question of Maxim is, what is the interpretation of this formula from the point of view of gluing blocks, right, polymorphic blocks? So there is a paper by Bakas about something that's called the Calabi diastasis. Maybe I'm not spelling it right. And so the idea is that indeed he's trying to understand what is the interpretation of this formula from the point of view of cutting and gluing. And so his idea is that the hemisphere partition function measures some kind of scalar potential for the boundary degrees of freedom. And there is some invariant that's called the Calabi diastasis. And well, I cannot fully reconstruct the argument, but if you want to learn about it, you should just look up Bakas's paper on Calabi diastasis. I won't be able to fully reconstruct the logic of this. Very good. Yeah, I think. Yes. Yes. I think it's called the, you have in mind the canonical bundle? Yeah. Yeah. Typically, you know, the standard store. It might be. I've a, it could be that the hemisphere partition function. So indeed, the exponential of the scalar potential, you can view it as a section of L times L bar, where L is the line bundle that you mentioned. And it is, of course, a natural conjecture to say that the hemisphere partition function is a section of L. And the other hemisphere partition function is a section of L bar. And I think that's consistent with, I don't know. I don't know. Any more questions about? Yeah. Say again, just a little bit louder. Right, so an open question in the field, which is interesting, is how to compute the distances or the zymological metric in three dimensions. Unfortunately, the three-sphere partition function speeds out a number that's independent of the conformal manifold coordinates. And therefore, it's just some kind of invariant of the whole conformal manifold. But it doesn't depend on where you are on the conformal manifold. It's the F coefficient of the conformal filter. And nobody knows how to compute the scalar metric in three dimensions for n equals 2 theories. So that's unfortunately the situation. The cases in which we can compute this extremal correlators or scalar metrics are 2,2 in two dimensions and n equals 2 in four dimensions. Yeah, yeah, yeah, yeah. So you can all trade it. You can understand all these things from the properties of the supersymmetrized anomaly polynomial. That's why I presented that point of view. It seems a little bit baroque and complicated. But it's a stream line. It's a machine that allows you to answer these questions without new conceptual work. You just need to understand what happens when you supersymmetrize the anomaly polynomials. And it turns out that in three dimensions, you don't get anything. And in four dimensions with n equals 1, you don't get anything. But you do get something interesting in four dimensions with n equals 2 and two dimensions with 2,2. And I think also two dimensions with 2,0 actually. But I'm not entirely sure. Yeah, any more questions? OK.