 And so today we're going to get into the heart of the matter with this 4D2D correspondence. So given a 4D super conformal field theory with n equal to 2 supersymmetry, we define a map chi onto the space of vertex operator algebras. So first let me describe this map at the qualitative level. So this theory clearly lives in R4, and we are going to choose a plane in R4, which we parameterized by coordinates z and z bar, whereas of course we have a transverse plane, which will be parameterized by coordinates w and w bar, for example. And the main statement, which I'm going to justify and at least sketch the derivation of the claim, is that a suitable set of local operators inserted on this plane, a priori of course this would be a function of all positions, both having a dependence on z and z bar, but for suitably chosen operators, this is in fact just a meromorphic function. And the operator proud expansion of the four-dimensional theory will then descend to a purely meromorphic operator proud expansion defined on this plane, and then this is just a matter of language, but this will be interpreted as the correlation function of a 2D chiral CFT or a mathematical parlance of vertex operator algebra. So that's the main claim, yes. So are you worried about wide mannexioms and operator value distributions and all of the jazz or are you worried about something else? Okay, that will not be the case, and so I'm sorry, the question was whether the restriction of the operator to the plane could perhaps be too singular or there could be any pathologies, and the answer is no. I mean this is of course something you can do. You have a correlation function where the positions of the operators are a priori anywhere in R4, and it is just your choice, if so hard so desires, to put all these insertions on the same plane. Nobody can ever stop you from doing that. Now in a conformal field theory, of course what the correlation function will depend on is conformal cross ratios. So up to four points, in fact there is absolutely no loss of generality in taking the operators on the plane because by conformal transformation you can always put them on a plane, but from five point onwards this is a true restriction, just choosing to study correlators of operator inserting on a plane. But of course generically for arbitrary choice of insertions, this is of course going to be not a meromorphic function, and the assertion I'm making is that for special choice of operators it will be meromorphic. And I'm going to go through this in some detail, but the way this will work is by a variation of a rather familiar trick in supersymmetry, which is we will choose these operators, these special operators are selected by the principle of being homology elements or representative elements of some fermionic operator that I'm going to call square q that obeys square q equal to zero, and so these operators will be, again they will be annihilated, they really should be understood as homology classes, so we identify operators that differ by exact terms, and the statement is that the derivative in the anti-homomorphic direction z bar turn out to be q exact. And so by the usual type of argument, this means that if you translate an operator in the z bar direction, that doesn't matter because you will pick up something which is q exact and then by integrating by parts, given that all the operators are q close, the correlation function will be independent of the z bar translation. You may be familiar with a version of this argument in n equal 1 supersymmetry, which is the tail of the chiral ring, so you will recall that if you choose operators which are in the homology of one full-fledged Poincare supercharge in n equal 1, the correlation function, those of you that are completely independent of positions in the full space because in that case, all translations are exact. In this case, the z bar translation is exact, and so we get something richer than the chiral ring, we get something that depends meromorphically on the coordinates, and that is of course a richer structure, okay, that's the basic idea. Now the only slightly unusual features of these operators is that the familiar operators, composite operators of conformal filtering, we saw example yesterday, you know, things like trace phi square, rq, q tilde, etc. When we compute correlation function of these gadgets, we just translate them everywhere in R4 and compute the correlator. Here we're going to do something slightly exotic in the sense that we will introduce an additional position dependent of the operator as follows. So it will turn out that the operators that will be relevant for this purpose carry a representation of SU2R, so remember that i's are the SU2R indices, and so here I'm looking at a spin k over 2 representation, so these are symmetries, fundamental indices of SU2, so just to be very concrete, this would be a doublet, this would be triplet, etc. And so what we will do, we will take ordinary composite operators that transform in some finite dimensional representation of SU2R, and contract them with some explicit Z bar dependence where this U is a doublet of SU2 that is just one Z bar. So hopefully this is clear, so if I expand this out, this would be O1111, that's the highest weight of the SU2 representation, plus Z bar O211 plus Z bar squared O221 dot dot dot Z bar to the k O222222. So every time you lower one SU2 index from the highest weight to the lowest weight of the doublet, you pick up a power of Z bar. So the operators that obey these meromorphism properties are the operators which are dressed in this way by these additional coordinate dependence. But of course, if you expand them out, you're going to get just a large sum of terms, each of which is a perfectly legitimate and standard canonical correlation function of your original theory. Okay, is the claim clear? I haven't proved anything, but I've just asserted what is true. And so it turns out that this, for suitable choice, for suitable choice of this sort of seed operator O1111, which is the highest weight of SU2R, if I carefully pick this one, and I will have to pick it as living in some suitably short representation of the full superconformal algebra, then it turns out that this resulting operator will be q closed and then the claim will fall. Okay, so let's get to work a little. So what is this q? Q is a certain linear combination of the fermionic generator of the superconformal algebra which in our conventions is this choice. Now you can make different choices that are related by conjugation. So the indices here are not terribly important except that once you have decided the indices of the Poincaré q, the indices of the conformal q are uniquely fixed by the principle that this object has the property that I want. And different choices of indices which are, again, related by conjugation are related, something you could have asked, well, what decides which particular plane over four I'm picking, well, that's totally arbitrary. With this particular choice of q that selects a particular plane, if I make a different choice that amounts to rotating the plane but all configurations are, of course, equivalent because they're related by conjugation. Now we can look at the Dagger operator which remember that the q and s are dual to each other. The s1 minus plus q tilde 2 minus dot and compute the anticommutator, a 1, 1 finds is this combination of quantum numbers. So remember this is also what is sometimes called delta, this is the scaling dimension, this is the carton of the non-Abelian piece of the Earth symmetry and j1 plus j2 are the cartons of the SU2 1 times SU2 Lorentz. In fact, you can also think of this j1 plus j2 as the angular momentum in the plane, in the preferred plane that I chose where my BoA lives. L is just the angular momentum on the plane. And the other combination would be the angular momentum in the transverse plane. Now this object is actually greater or equal than zero from unitarity. And when it is zero, then this is the argument that I quickly reviewed. Yes, then we are talking about an operator that lives in a special short-term representation of the super conformal algebra. And so operator that precisely saturates this bound are generalized. So yesterday I gave you a very quick introduction to super conformal representation theory and you will recall that I had a single out three classes of multiplets. This one obeys E is equal to 2R, this one obeys both taking together the obey E equal the absolute value of R and this one is something like E equal, well this particular one is just E is equal to 2. So this condition clearly generalizes this one. If I take a scalar operator I am in this class but more generally I'm generalizing this condition. Now there's an important point I need to emphasize. When I was labeling representations yesterday I was labeling them by the quantum number of the highest weight of the full super conformal representation whereas here I'm not necessarily doing that. I was looking at the quantum number of a state that need not be the super conformal primary. And so I have to be a little bit more careful why it is clear that these representations contain operator of this kind because I just look at the highest weight and I immediately see by inspection that this condition is obeyed. It could be that as I build my representation acting with Q's I could find somewhere not on the highest weight state but somewhere above. Sorry, let me perhaps go here. So I'm making the statement that it could be that some of the other representation was considered yes or perhaps some even more general representations contain such an operator but not as a highest weight state. So you really need to do a more careful analysis and search for where such states appear. And so the complete list turns out to be that these operators obey this condition. So let's give them a name. So operators that obey this condition are called sure operators for historical reasons. And actually four dimensional unitarity implies a second condition that is a little bit less obvious but it's true that the transverse pin is equal to little r. That comes for free by the conditions of four dimensional unitarity. And so out of the five quantum numbers that a generic operator is labeled with the energy, the two spins, big r and little r, you have two linear relations. So the generic operator in this class will be labeled by three quantum numbers. So the generic sure operator will be labeled by, for example, e, r and little r. Such an operator one easily shows is necessarily a highest weight of SU2r. And so necessarily the index structure is the one where the SU2r indices are all in the highest weight representation. So let me conclude this line of thought and then I'll go back to the enumeration of all possible such operators. By the usual homological arguments, if I want to compute the homology of q, it better be that the homology of q, all the homological elements must obey this condition because otherwise if they don't, I can use q dagger to invert and I find that the operator, if it's killed by q, is also q exact. And so the homology of q is really identified after a little bit of further analysis with precisely the set of operators that obey this condition. So sure operators at the origin of my plane are in homology, but it turns out that as I translate them in the plane, I need to be a little bit more careful because genetic transitions in the plane do not commute with q. So this object at the origin, if it's a sure operator, is in homology of the square q. But the translated operator is not, necessarily, and the reason for it is that while the z momentum commutes with q, the z bar momentum does not. And so if you want to insist that you have something which is annihilated by q, you need to perform this funny dressing that I have just deleted. You take the sure operator, which is the one the highest rate of SU2R, and then you lower its SU2 indices in a way that contains these additional z bar dependence. And now this operator is q closed and then the homology class of this operator by the argument. So, and so this operator does not come, sorry. By the argument I gave earlier, the homology represented or this operator defined some operator, how should I call it? I should call it, what is a good name? Little a of o, which is purely meromorphic. Okay, is the structure of the argument clear? So to recap, given a 4D super conformal field theory, n equal to 2 t, we have this map to the VOA. So there's chi that maps the theory t into some VOA. And the space of states, which you will record the same thing as local operators at the point of the VOA are identified with the space of sure operators of the 4D theory. You have to play this funny game of which we call twist translation. But given a sure operator, you can find a twist translated operator. So for each sure operator you're gonna find a two dimensional gadget. So each sure operator will give rise to a meromorphic field in two dimensions. Okay, so I'm gonna do a little bit of representation here but not too much. And then, given that this might sound a little bit abstract, we'll illustrate how this happens in free field theory, which we'll hopefully convince it's true, and then take it from there. So again, so we have this condition. Where do I find operators obey this quantum numbers? Well, the highest weights of b at r multiplets obey this condition, so they are good. It turns out that there are additional multiplets that I didn't quite have time to discuss because there's too many multiplets in two little time. Which generalize the stress tensor multiple. The stress tensor multiple you recall takes of this formula with r equal to 0. And so there are generalization of the sure multiple where this is true. And you will find a sure operator at level two. I think with qq tilde on the highest weight of this guy will give you a sure operator with obeying this kind of condition with where however the r that is there is a little shifted from the r of the sure operator, so let me put an r prime here. And then there are additional sure operators in some other type of multiplets which are called d and d bar. I don't want to belabor the story too much because it gets complicated. And this operator will have, this will be a level one on the highest weight. And this operator will have u and r different from zero. So for the purpose of not confusing you with too many representations, in the rest of this lecture I will try to focus on vertex operator algebras where all the states have little r equal to zero. So that we need to all worry only about two quantum numbers, the total energy and the spin. Which given this relation that will also fix the value of big r, okay? So yes, the short answer is no. Yes, yeah, after you go through this map, so you take the four dimensional operator, you do this funny linear combinations and then this, well, and then up to exact terms that don't matter in a correlation function where all operators are of this type, this defines a, as I wrote here, a purely metamorphic object in two-dimensional. No, but in conformal field theory, in two-dimensional conformal field theory, it's true that we have at least conformal symmetry, but nobody can stop us from having additional quantum numbers if you want. So from the point of view of the two-dimensional, okay, so thanks for the question. So for example, this little r quantum number, it is some additional quantum number that the two-dimensional operator carry, but it is realized non-locally. So there is no affine symmetry related to this little r. This turns out to be some sort of some outer automorphism symmetry of the theory. It's not related to a, it's not a local symmetry. And, okay, okay, so, okay, so you see what, this is saying, in the words of BPS-ness, one is often used to consider relations of this type, where the energy is equal to some charge. It is slightly more involved to also consider operator carry spin. And this is the universe of Schur operator. So it's some generalization of the operators of this type, but that also carries spin. And that is of course necessary if I ever want to get a holomorphic Z dependence, because taking Z derivatives adds quantum number. So it better be that I have a holomorphic angular momentum as part of my shortening condition. So perhaps some of you were not at this discussion section yesterday. I explain in some detail the physical meaning of these operators. I explained that the vacuum expectation value of these operators, what parametrizes the Higgs branch of Vakia of the theory. I also explained that the other half-BPS operator I had, the Epsilon R operators, parametrized the Coulomb branch of Vakia. But these operators do not participate in my homology, okay? So the vertex operator algebra I'm constructing is some huge souped up version of the Kyler ring on the Higgs branch. The Kyler ring on the Higgs branch will obey this condition. And now we are relaxing it to also consider some more exotic semi-short representation that also have spin. But there is some moral sense in which what we are doing is Higgs branch physics. In particular, the Higgs branch physics is fully captured by this VOA. And the Coulomb branch, at least naively, is not at all part of the story. Now, here, as a special case of this kind of multiplets, there is a case R equal to zero, which as I explained yesterday is the Tristanage tensor multiplet. So let's look at that in slightly more detail. So the highest weight of this operator is equal to two and R equal to zero, and it's some operator x. But as I was explaining yesterday, you go to level one and you will find some good things such as the SU2R current, the U1R current, and then eventually you get to level, so this is QQ tilde. And then when you get to Q square, Q tilde square, you will find the stress tensor. And the Schur operator in this multiple happens to be the highest weight of SU2R of the SU2R nether current in this is plus, plus dot. This is Schur. You can check it because it has E equal to three being a current. It has R equal to one being the highest weight of a triplet. And it has L equal to one, so that the Schur condition is obeyed. So let's look at this. The Comology class defined by this operator, which again, if I want to be super explicit, it would be this. This operator defines a meromorphic object in two dimensions. And another important relation that I haven't written down, and I really should have, is what is the homomorphic scaling dimension of this Comology class. And that relation is, this is the L0 eigenvalue, the purely meromorphic dimension that is big R plus L. And so this operator defines an object with homomorphic dimension two. And, well, there's a famous object with homomorphic dimension two is called the stress energy tensor. And it's not obvious, but it's true that if I do a detailed analysis of the consequences of four dimensional super conformal word identities obeyed by the arcymmetric current, which we know very well, you get to deduce that this T is not just a homomorphic object dimension two, but it obeys the operator product expansion expected from a two dimensional stress tensor. And so we really declare that this is the stress tensor. So you see, the way this is happening is kind of curious. The two-distress tensor in some moral sense comes from the four-distress tensor, but not quite directly. It's not directly coming from T. It's coming from a superpartner of T, which is predicted to exist because of super conformal symmetry. And so the statement that the four dimensional theory has a local stress tensor. I reviewed this. Yes, there is a local stress tensor with associated central charges. This predicts the existence of a local stress sensor T of z in my VOA. And you can compute, again, using the details of this map and super conformal word identities, the two dimensional central charge is minus 12 times the C anomaly coefficient. OK, this is a nice surprise. Because up to now, this construction could have looked a little bit artificial. I'm slicing the four dimensional space, taking this funny plane. Of course, by construction, given that I'm embedding this plane in a four dimensional conformal field theory, conformal transformation that preserve the plane, which is just a usual SL2 times SL2 bar, are guaranteed to be present, of course. It's just a subset. It's a subalgebra of the full conformal algebra. But what was not at all guaranteed to happen is that the local SL2 generators, which are L minus 1, L0, L1, in fact, enhance to the full Virazoro symmetry. This will be negative because the four dimensional central charge is defined from the two-point function of the stress tensor. And so four dimensional unitarity imposes that C4D would be positive. And so necessarily, the two dimensional theory is has negative central charge, in particular, nonunitary. And so we are going to get nonunitary vertex operator algebras, which, however, are far from generic because the details of the map from 4 to 2D, although unitarity is broken, four dimensional unitarity is still hidden somehow. Here, in some sense, is the first indication of it. Well, C is not generic. It's negative. And the story will become important later. So we can play the same game with the other thing that I explained yesterday in the discussion section, that BHAT1 is the multiplet that contains, if you go to level 2 with q, q tilde, you will find the conserved current, which is the flavor current. We interpret this as the flavor conserved current. So necessarily, this object must live in the adjoint of some group GF that commutes with the super conformal algebra. And so if it is the case that the four dimensional theory enjoys some continuous global symmetry, then it must have a conserved current. And yesterday, I also went into some detail explaining how, necessarily, at the generic point of the conformal manifold, that conserved current must, in fact, belong to a BHAT1 multiplet. So I'm going to run an argument similar to a run for the stress tensor. I have a stress tensor. Superximity predicts the distance of an arc current. The arc current is a true operator. Here is the same story. I have a conserved current. Superximity predicts that it belongs to a BHAT1 multiplet. But the highest weight of the BHAT1 multiplet is a true operator, because it obeys this condition. So BHAT1 highest weight is this object mu ij. And so mu 11 is sure. And so this will give rise to a holomorphic object. So this object, mu 11, has r equal to 1 and l equal to 0. It gives rise to a holomorphic object with holomorphic weight 1, which we are going to identify with an affine cut smoothie current for the symmetry GF. So the similar phenomenon happens here. You have a global symmetry in four dimension that becomes local in the two-dimensional carol algebra. Again, you can use word identities to convince yourself that this is truly a affine current that obeys the correct OP. And then what I need to tell you is the level of the two-dimensional cut smoothie algebra, which, again, is an interesting minus sign, is related to the level of the four-decurrent, which I defined in my previous lectures. OK, so these are two very universal fissures of this correspondence, that the global conformal symmetry becomes local, and the global flavored symmetry becomes local in this two-dimensional vertex operator algebra, with the precise correspondence between the central charges and the levels. OK, let me comment on an important point. So as I said, we start with, let me do it slowly. So the generic Hilbert space of the four-de-theory is quintuple graded because it depends on these five quantum numbers, and then possibly, in fact, there could be additional three of quantum numbers that I'm ignoring for now. And then the Schur condition imposes these two relations. The second one is the concept of the first using for the unitarity. And so the space of Schur operators has this, I could decide to label them just by E, what should I choose, E, little r, big r, and little r. Or by using this other relation that, let me write this here, I could also, so this vector space of the chiral algebra, this is a 2D statement, will then have a gradient by little h, by big r, and by little r. There are three quantum numbers, just by taking linear combinations. So little h is obvious, it's just a scaling dimension. Little r is less obvious, and I'm going to ignore it in the rest of the talk. And big r is also not at all obvious, but it's predicted to be a quantum number that must be hidden if I exactly know the map. However, there is a major distinction between how these quantum numbers behave under the operator product expansion. Whereas little h, and in fact also little r, are preserved by the operator product expansion, big r is violated. Why is that? Well, it's obvious, because I took this funny linear combination of objects which had different big r quantum numbers. And so the big r quantum number is not preserved by the operator product expansion of the vertex operator algebra. But nevertheless, if I really fully know the details of the map, I can unambiguously assign this quantum number to the operators of the VOA. So this is a subtle point, but it's essentially in more recent developments. Under the operator product expansion, you see the way this correspondence works, I start with the highest possible values of r, which is my true operator. And then I decrease it by z bar dependence. And then I do OPs. And so what you actually learn is that, although big r is not preserved by the OP, it can at most decrease. Big r can never increase in the operator product expansion of the VOA. It can at most decrease. So this defines in fancy parlance a filtration of my space. The filtration is what is preserved by the OP, but secretly there is a true way to define an actual grading if I really know the details of the map. I'm insisting on this because what we will often do is to use very general properties of the Ford theory to just guess the VOA. And the guess is correct, but the guess does not come equipped with the details of the map. So the guess will automatically know about little h, but the dependence on big r will be, in fact, a bit hidden. And so that's additional information that you only know if you know the full details of the Ford to the map. Why is that important? It's important because knowledge of this big r is what determines the sign in the two-dimensional inner product. So the four-dimensional theory is unitary, so norms are positive. The two-dimensional theory is not unitary, but there's a very precise sign that I can predict in the norms, which is correlated with the hidden value of little r. And so for very simple operators, I know the sign. For example, by detailed construction, there are these two-dimensional operators, the stress tensor and the affine current. And there, I know exactly the sign has to be negative. But as I start taking normal-order, proud-of-olomorphic derivative of t and j, for example, or the more complicated operator I may have, the big r doesn't add up. I have to very carefully take special linear combination of this operator that depend on the details of the Ford to the map to predict the value of big r and once I know that, I can predict the precise signs that must appear in the inner products. So knowing the full details of the map amount to having some version of the funny unitarity of the 2D theory, which is not the standard unitarity because there is this r quantum number that predicts what signs I should get. OK, so I'll hopefully come back to this point later on. Questions about this? OK, so now to make, yes. I can have an act. So I kind of can, in the sense that if I'm a mathematician, I can just postulate that the VOA had this additional r grading and then tell you what the norm should be. That I can always do. But it's not very useful. Because if I give you a VOA presented abstractly, I say I give you the set of generators and the operator product expansion, a priority that could be multiple ways or perhaps even no way to assign this r grading. And so the full structure is, in other terms, there's additional structure on the VOA that is not something that is standard. So yes, I can just say that what I'm really interested in are VOAs that have this additional structure. But if I don't give it to you from the beginning, then you have to struggle a bit to either find it and show that it's unique or know some additional information to identify it. OK, I can be precise if you really insist. So the additional structure is the VOA must admit involution. Well, it's not really an involution, a linear map, sigma, which is actually not any, OK, first of all, the first statement is that the VOA admits this grading, this triple grading. In terms of that grading, sigma square is minus 1 to the 2R. So it's only an involution for operators with only squares with only 4R with integer R, which for simplicity we could consider to be the case. And then if I work in a basis with definite values of R, I have the following statement that the two-point function of A and of sigma of A times Z to the 2H, this is just a number, times crucially minus 1 to the H minus R is positive. That is the additional structure I have. Actually, I should perhaps say strictly positive for non-zero A. If I find 0, it means that I should set that this is true. It means that A is an outstader should mod out by it. That is the additional structure. But it's not standard. So if I give you, sorry, say it again? In the sense of, OK, we should talk about it. You probably know more about it than me. But it's not, if I give you something as trivial as the Virazoro algebra, what is this r-grading? You don't know. And so that is something that you only know if you have carefully tracked this map from 4D to 2D. But this is the additional structure that you can axiomatize. OK, so some trivial free-field theory example. Let's do the free-hyper to demystify this whole story. The free-hyper, you will recall, there is this. This is not curly Q. Not to be confused with a supercharge. It's uncurly Q. And so the Schur operator here is just Q1, which has r equal 1 half and h and e equal 1. I can also define the tilde version of this, which is Q tilde minus Q star. And so here I have Q tilde 1 equal to Q tilde, which has the same quantum numbers. And of course, the 4D non-trivial OP would tell me the Q of zz bar with Q star of ww bar. Well, let's call it. Let's do this at 0 for simplicity. This is just, of course, this operator is dimension 1. And so this is z times z bar. And similarly, Q tilde zz bar with Q tilde star. OK, but now we construct the Twisted Translated Operator, compute the function with the Twisted Translated Operator of tilde type. OK, well, I mean, this is kind of boring. But this piece is clearly 0 because I am at the origin. Z bar is equal to 0. And then, OK, so this is a trivialized size. We see that this is just Q contracted with Q tilde. Well, Q and Q tilde have non-singular OP. So the only singular OP terms is here. And you see it's a triviality. But that's just what happens. I got a holomorphic correlators for you rather trivially because this bar dependence just canceled out. Yes, exactly. So what has been asked is that the OP of what you really need to compute is the OP of these Twisted Translated Sure Operators. That contains a lot of junk. But that junk is the coupled in this correlation function by the homological argument that I gave. So what we're really saying, of course, is that within this subset of operators, you have this enhanced infinite dimensional symmetry. But this is not a symmetry of the full four dimensional theory. It only holds if you carefully choose the operators inside this subset. OK, so you see how the homomorphic arises rather trivially in this free field example. And so the statement is that the free hyper-multiplet gives rise to Q, Q tilde of what are sometimes called symplatic bosons, which each of them has. This is like a beta gamma system. But you can think of this as a beta gamma system, but with equal conformal weight. It's a special case with equal conformal weight with the OPEQ of Z, Q tilde of W is 1 over Z minus W. This generalizes rather easily to the case where you have multiple hypermultiplet. In fact, I could, in one shot, do the case of the half hypermultiplet in some pseudo-real representation R. And then here, this would be pseudo-real representation R with the omega is the anti-symmetric of the representation. Now, free vector multiplets work similarly. And there, you get your shrew operators from the gay genus. So for the shrew operator there is this gadget, as it turns out. And this is something that has R equal minus 1 half. So I promise not to talk about OPEQ with non-zero little R. And I will not. But for the free hyper, let me just make the statement that what the free hyper will lead to. And then there's also lambda 1 plus lambda. This lambda and lambda tilde will translate into a BC GOS system of dimension 1, 0. But where the zero mode of C has been removed, this is the small algebra of the BC system. So what really happens is that lambda gives you B and lambda tilde gives you delta C. And now these are fermionic objects. But again, so the central charge here of this BC GOS system is minus 2. And you can check that that agrees with my general formula, that C2D is minus 12 times C4D. And for the free hyper, you get QQ tilde whose central charge is minus 1. Again, agreeing with the statement that the free hyper multiple has C anomaly coefficient equal to 1 over 12, 12 in four dimensions. OK, so free theories, not very surprisingly, give rise to free vertex operator algebras. But of course, as I emphasized yesterday, we have ways to always insist that we're talking about non-free theories. And that's what we are going to do. And then it's going to become much more interesting. It's probably a good time to stop.