 So that's it. So whenever we last class, at the end of the last class, we were discussing how we use, you know, we gave a Hilbert space interpretation to the path integral that defines the formality of the area on the two-pronged area. And at the end of the last class, we were discussing how we can compute the commutative relations between the generators of the Bayesian graduate. So let's remind you of the limitations. We have 3 of z with the sum over n, ln over z to the power of 2. This defines ln when we make this expansion in the z plane. If you want to know what t of z is in terms of these ln's in the w of ln, then we'll just apply the usual conformal transformation determinant. In particular, you pick up a minus sign, apart from the usual shift of ln's. You pick up a minus sign to remember that you picked up a factor of, but you could have an expansion, up to a shift, you could have an expansion like t of w, even minus sum over n, ln into the power of i. I think I put a plus minus wrong. The additional factor of minus z's where it came from the conformal transformation that worked even for a primary field. And then there's also the fact that here in n0, we would have an n0c by many four or five, but that would be the same as we discussed. So these ln's, in everything we're doing, are defined by the ln, the expansion coefficients in the z plane, and they turn out to be whatever they turn out to be entirely. This is not a convenient way to work, because now once you've got this definition, you don't even know what they were in the landing plane. You just work with this definition, never refer to the landing plane, except for concession questions. Because for the answer, when we eventually use this, we may want the insertion to the landing plane, in which case then we'd have to work out what the insertions were interested in there, and that was at least this. But at least one can, you know, all intermediate results, you know, if you get it. Then we set up the calculations for ln commutative. And we said that this thing was equal to integral dz, dw, z to the power n plus 1, w to the power n plus 1, 2 pi i, okay? T of z, t of w. The first term had n acting before, and therefore at earlier times, the n. So n was associated with w. So the first term we had this being the w and z, the z counter. Minus w counter that is unchanged, and the z. So just one more, one more, and before we get to the end of the calculation, just one more concept in one's mind. Remember what this expansion meant? As we discussed last class and this expansion has a greater expression in the Eisenberg picture, it's just always correct. Expansion is not always correct for correlation functions of t. If you have correlation functions of t, this expansion asserts that t is an analytic function whose only singularities are 0 and infinity. Okay? But correlation functions of t are not analytic functions whose only singularities are 0 and infinity. In fact, the analytic functions of singularities are at points of insertion of other operators. Expansion is a great way of not crossing other. Okay? So in computing l, m, n, that's why we have something non-trivial in this calculation. In computing this first graph, l, m, n, we have this, where this is an active representation of this quantity in that vector. Similarly, this is an active representation of this quantity in that, in this order. Had it been that t of z was everybody completely analytic, you know, in the beginning, regardless of w, then these two contours could have been, had they defaulted to each other, cancelled. Okay? So this is something that's of course very important to keep in mind. That correlation functions of t will not have this form. You know, we have, are not always analytic functions. They have singularities at points of insertion of other operators. Okay? So, so, so let's go. So what we want to do is, we're still doing this calculation, this combination. But then we talk about our strategy of doing this, this calculation. Our strategy of doing this calculation was to fix a particular point in w, let's say this point. And then, default the z contours, sorry, it's about to be done, default the z contours so that this one goes to, you know, surrounding here and then line of the boundary function. And this one goes to surrounding here and line of the boundary function. So that this calculation was the same thing as the calculation of the z contours going like that. That contours, it picks up the residue, cancelling one factor of 2 pi i. And so we have finally that lm, lm is equal to the integral dw over 2 pi i of something. What's the something? The something is the residue as z goes to w, the residue in z, in the variable z. As z goes to w of z to the power u plus 1, w to the power u plus 1, g of z, dw. This quantity is a residue in z, so it's a function of w. And that function of w has to be, let's carry this through. Let's compute this, this rest. So in order to compute the residue, we know that z to the power u plus 1 is a completely analytic function of z at w. And so the only possible places of non-analyticity, the only possible sources of non-analyticity are non-analyticities in this, in the two-point function between d and z at w. Any contribution to this two-point function that is analytic in z minus w will not give us support here. So we can ignore this. So it's only the non-analytic pieces that are of interest to us. What were the non-analytic pieces? The non-analytic pieces were d of z, d of z equal to c by 12, sorry, c by 2, z minus w from the cube, plus 2t of w, z minus w from the square, plus del t w, that's 4, c by 2. You might think, I'm the only pole in this piece, but that could not be accurate because we've got a double pole here which will give us a pole when we take the expansion of z to the pi n plus 1 about z equals w. So let's write that data expansion. So z to the pi n plus 1 is equal to w to the pi n plus 1, plus n plus 1 into w to the power of m plus, in time, z minus w, plus m plus 1 into m by 2 into w to the pi minus 1, z minus w equal to the square, plus m, m squared minus 1 to m by 6, w to the pi minus 2, into z minus w, thank you. We don't need to go further than this order of the data expansion because the maximum singularity was z to the 4, so this is the largest term that can give us a pole. So now let's multiply this with this and isolate the coefficient of 1 over z minus 1. So we get 3 terms, we get this multiplying this, this multiplying this, and this multiplying that. So the 3 terms are c by 12, m squared minus 1 into m, this thanks that side. So 2 m plus 1 into t of w, plus the original power of w, right? So this was w to the pi m minus 2, this was w to the power m, w to the power m, and we also had overall, right? We have w to the power n plus 1 into t by, so multiply everything by w to the pi, because this was going wrong, just w to the power n and then, plus f t of w into w to the pi. So that is this, right? This is the quantity that we have to multiply by d w by 2 pi i and integrate over the power, what is this stuff? Of course, what we do is, we consider integral of either constant or t, so this can be written once again in terms of l, okay? So let's see how we do that, so let's first take the constant. So the constant term is w to the power n plus m minus 1 into c by 12, integral d by 2 pi is the 0, unless n is equal to 0, in which case we have to contribute to the power, then we want, okay? So this is equal to delta of n m equal to m times c by n minus n, that's called n plus m, actually both very similar and can be converted to each other, but that's what I'm trying to do. This is w to the power n plus m plus 1 to t of w plus w to the power, and this is into 2 into n plus 1, plus 1, plus w to the power n plus n plus 2, then integrated d w by 2 pi i. Okay, now we should really substitute each of these expansions, each other in the power series and there is, this is a little trick that helps, that little trick is an integration by parts, okay? We've got a derivative of d, we should, we put substitutions in the derivative of d in the expansion, but it seems similar to just integrate this by parts, okay? By parts, what we get is a minus side here and derivative of this. Now, this minus side of the derivative makes the power the same as that, but makes the constant, which was here 2 into n plus 1 minus m plus n plus 2, which is w to the power n plus n plus 1 into d w by 2. No, so the 2 is cancelled, so the question is 2 and 2 cancelled, so we get m minus n into w to the power n plus n plus 1 d of w d w by 2 pi i. Now, this is simply m minus n times l of n, is the thing you multiply by, you isolated out the coefficient here, remember that d was equal to l of m divided by z to the power 5 plus 2. So, if you wanted the coefficient of z l m, you multiply by z to the power minus 1, okay? So, this is z to the power n plus n plus 1, then it goes in. So, we've got both pieces in this expansion, so let's sum up what we concluded, we've concluded that l m, committed to l m, is equal to m minus n l m plus n plus delta of m plus m of 0, 0, d divided by 2 is for the biggest. Okay? It's a very important, these combination relations are very important, because as you know, the l m's and l m's are conserved charges, l m's and l m's are conserved charges in our theory. The combination relations between the conserved charges defines the symmetry algebra of the theory, just like the combination relations of the SC2 generators defines the symmetry algebra of any theory with rotation. Okay? So, this is telling us what the, you know, this is telling us a lot about what we can say about the theory of the symmetry, there is a very important relationship. Okay? It's a physical point that I'll bring up now and I'll also be talking in the end of the lecture, I'm a little clear on that, but anyway, let's, let's talk about it. And that's this. Some of you might be wondering about the formula. You know, when I was a student, there's a kind of thing about formula. Look, the formula transformation is just a formula. It's just, you know, one of these analytic coordinate changes. And so if you want to robot the algebra of these coordinate changes, it should be very easy to do. You rather than make the field to generate support, we know how to do that. Without the before, it's like z to some power times z is the power, z to the power n plus one or something times z times z is the generator. Let's, well, we'll get all factors right. So, something like z to the power n plus one times z is the generator for n to the power n. So, if you wanted to sit and compute the combination relations of these symmetry things, it should just be computing the combination relations between these two vector fields. And then you get whatever you get. So, instead of doing the fancy calculation, why don't I do that simple calculation? Okay? You get an answer and the answer doesn't have to be exactly the same as this but without this fact. Okay? Now, this might seem like something money. What's going on? If the symmetry in question was that these coordinate changes, how can the final symmetry algebra not have the algebra of these coordinate changes? Certainly, this thing did have rotations. The generators of rotations are going to be the same symmetry algebra, but there's no special information. We demand the variables to surprise. So, what's going on? How can it be, how can it, you know, all of this sounds very like a lot of fancy mathematics, but how can it make any sense? Okay? And the important point here that's missed by this news argument is that the coordinate transformation is not just a coordinate change. It's a coordinate change plus a wide change. Okay? If you make a coordinate change, if you make a coordinate change, you also change the metric. Right? So, you make one of these conformal coordinate changes. You also change the metric in theory. And then you have to rescale your metric in order to put the theory back into a theory with flat space. A conformal transform, remember conformal transform in theory is, you know, when we're making these transformations, we are also changing the metric of which it is. So, a conformal transformation is the product of these two things, these two operations. Okay? We were assuming that the wide transformation did nothing. Okay? It's just a product. It's just a product. It's just two operations. One of which is a coordinate change, the second of which is a wide change. And you don't want to say something about the global structure of the group when you think about it. So, you see, the stress tensor, the oh, you see, this part of the commutation relation that came from the most singular piece of the stress tensor, the GTOPE is actually telling you how T transforms under wide transformation. This part is just the transformation of the stress tensor under a coordinate change. Okay? This part does not have a coordinate change. If you just make a coordinate change, that's not how T transforms. So, it must be, since this is related to how T transforms under a conformal transformation, must be this part is telling you how T transforms under a wide transformation. What's happening here is that the wide transformation that is needed to put the theory back in the flat case method is not acting truly linear. That's this additional bit that is giving us this additional value. Okay? So, this thing is completely physical. This part, this transformation is not just a coordinate change. It's called the change for something else. The combination of all the changes is giving us this, the combination of the wide transformation. Okay? That's what's going on. We will see this in more detail in probably many of these videos where we will use this by view of ethics. Okay? Well, there's no contradiction. It's all perfectly sensible and makes sense if you think about it correctly, cleanly. Not losing it. So, this is the data sort of algebra. Any questions or comments? Let us use this data sort of, sort of a value problem. What was special about primary states? Now, you know, whenever I define a term for an operator, I would often, without telling you about it, use the same term for states. And the understanding is always this is the inheritance under the state operator. So, we define the primary operator as an operator that has only singularity in the TTOP. It is that of the square term and the linear term. The particular forms of the, sorry, T operator operator. The particular forms of the, of the rest of the corrections. Okay? Now, we, where any state that is due to an operator that obeys such, obeys the fixed-end state and the operator is going to be called the primary state. Okay? I want to understand what I can say about the action of these N operators on primary states. Primary states. A primary state by definition is obtained by, is the way it comes to that as obtained by inserting a primary operator of at the origin. And we do the bottom table after some points. Let's say Z equals 1. And we get some state. Understand what I get when I am on the state. I should act with the operator, act with the operator T of Z, Z to the pi n plus 1 by 2 pi i on over 0. This one, this, this, this this one, this integral is regular over this part you will see. What I see, I don't care what it is. It's just some quantity that's set up to 0. You know the answer doesn't depend on what quantity it is because everything's empty except it's 0. I could know on such issues that apartment state is, what is lm on 0? lm on O. Let me learn some notation. Suppose I got an operator O, the state corresponding to that O. Put that operator in. Yes. So I am going to compute lm on O. So lm on O corresponding to that O was a primary. So we'll see what we get if it was not there. Let's assume for a moment that O is a primary. So we get T of Z times O is H of 0 divided by Z squared plus delta O divided by Z times Z to the pi n plus 1 DZ divided by... I should be picking out the pole. I should be picking out the pole. I've made an assumption here. I've made an assumption this calculation is correct. This calculation is correct provided that m is greater than or equal to minus 1. Because m was less than that then I get contributions to this operator problem. I am going to get contributions that involve analytic terms of the operator problem. Cancel the non-analytic insertion. m greater than or equal to minus. This calculation will only receive contributions from the non-analytic states of the operator problem. These non-analytic states are under complete control. So that is what we get. This term clicks if m is equal to 0. So we get delta m 0 times H times O is 0. This term clicks if m is equal to 1. So we get plus delta O if m is equal to m O times delta m minus 1. Can we just using the primary operators for this case? Yes. We will see that in just a moment. The statement is that as day is not true but a relationship is true. So this calculation was true for all m greater than minus 1. This is non-zero only for m equals minus 1 and m equals 0. So let's this work. So this 1 on O is equal to m. Actually this statement is not true just for primary operators. It's true in complete generality. So as we said, any local operator the whole part of the T operated expansion is definitely operated by an operator. So this one will always work. We've also found that L is 0 on O is equal to H times O. So this is true in greater generality than the primary operator. This will be true for any quality operator. But this is rather generally true. These two statements are quite generally true. But this state L m of O this is O not 0. This is the state O is not 0. O is equal to 0 for all m greater than that. Crucially it was the fact that there were no singularities in the OP beyond inverse quadratic. In order to understand what its physical implications are I'm going to ask you to remember what these L m's were on the cylinder. Remember that we had T of that to be equal to L m over back to pi plus 2. Plus 2 was taken away by the conformal transformation business and we got a T of w which after some shift was even pi m into w L m. The important point is the following L 0 is the zero mode of derivative and we integrated over the cylinder and we added to L 0 pi actually which gives you the energy of the theory on the cylinder which seems to perform many times. Now let's look at what we got for the common. Remember the physical interpretation of L 0 was the energy of states or the left moving part of the energy of states. So the energy plus moment. Now let's remember what we got from Virasov. We got the computational reaction to L m L m is equal to m minus n L m plus n plus m squared minus 1 into m by 12 is it? c. By 12 c delta m plus n 0. n equals c. When we get n equals 0 we get L 0 L m is equal to minus n L m that tells us that modes L m have energy minus state of energy e with L m. The result state is energy e minus n. Therefore L m's with all n's positive are operators that lower the energy of states. Whereas L m's with all n's negative are operators that raise the energy of states. Is this correct? This is just telling us that operators that we face correspond to the driverly operators are good. So what does it mean that L m or 4 is equal to 0 for all positive energy? Now let's think a little group theory. You see whenever you have any group implemented in the quantum field theory you can take the Hilbert space of the quantum field theory and decompose it into irreducible representations into irreducible representations of that energy. If you have a question that does not commute to the energy but has not the operators and the aspirant questions that we can see don't commute to the energy operator out there but non-trivial computational representations, the energy operator in both ways as well as other energies. In the quantum field theory, in a sensible quantum field theory I mean we often demand, certainly in an initial field demand that quantum field theory have a lowest energy state. That we can act indefinitely with the energy of lower operators on any state. We have a class of distinguished states in the theory such that these states have energies that cannot be lowered further by the operation of the symmetry. We have seen that these states, some states are exactly prime presidents. But we have not seen that in the words we have done we can easily run them through the algebra. You see, suppose we had any OP coefficient let's say it's like a 1 over z to the k that would give us a non-zero action of some error with something positive. I have not said it in these words but it's the extension of what we've done to see that. So that says that if it is a quasi-primary state it should be a primary state. But we've already argued that it's always possible to use a base. So you're right. We could take two primary states with different weights and take any of the combinations of them that would technically not be a primary state. I would still have a condition that is evaluated by ordinary operators. But what we have argued effectively is that it's always possible to choose if you take the states of space of states that is evaluated by all positive elements then it's always possible, that's a linear space. It's always possible to find a basis in that linear space such that basis elements are because it's always possible to find a basis which is quasi-primary and then that's fixed. But can we just take the proof that we can always choose a quasi-primary state? In this space, because we've already seen that it's always possible to find a quasi-primary basis in the whole space. So you now want to know whether it is or whether that's it. You're right, though I'm sure since you've made the proof let's see if we can do it on this one. So you're asking, well of course I've got a set of states that is evaluated by all ordinary operators. The question essentially is can we diagonalize the energy operator in that set of states? Is it quasi-primary? Yeah, it's more or less here. You see, what does 0 mean that the OP between T and O would not have a higher term than one-wise that's correct? That's correct. So that's correct. So what he's asking is couldn't be, let me guess, what he's asking is that well, suppose we put that condition then all that we know is that the maximum or singularity of OP is one over 6. So any such operator has the property that T of z on O is equal to let's actually make the OP. It's some other operator. So some other operator of O tilde by z squared plus del O by z. And but in that definition of an operator to be primary we require this OP to be the same as OP. You see I'm getting the wrong kind of basis in such states that how do you know? Well, you see firstly firstly notice that this is to be true it must be that O tilde has the same weight as T. Same dimensions because T is dimension 2 or has some dimension. So these two sides have the same dimension O tilde has to have the same dimension as T. And because this is the dimension of this side of the OP is dimension of O plus 2 we've got the 2 from 1 over z squared. So this has that same dimension as OP. So O tilde is some other, the state corresponding to O tilde is some other state in the Hilbert space of the theory. The same energy. So what you have here now is now let's try to make a basis change in the space of L0 the state O will be the state O tilde. But L0 the state O will be the state O tilde. L0 of the state exactly. Exactly. And at 4 you're definitely going to get more dimension. Then we're just diagonalizing the Hamiltonian in that. Right. So let me say again what we've concluded. We've concluded that there must exist a distinguished set of states that are evaluated by all energy lowering operators of the theory. Lm is greater than 0. And it's possible to find a basis of these states such that in this basis the states are due to primary operators. Now in any state in the Hilbert space of the theory given any state in the Hilbert space of the theory let's act on it with lowering operator. So run out the energy minus infinity or you'll reach one of these states. That every state in the Hilbert space of the theory can be reached. It is possible to take any state of the Hilbert space in the Hilbert space of the theory and act on the sufficient number of generators of the symmetry algebra in the distinguished states. Let's draw what we've concluded. Suppose we draw the primary states of these special states which are evaluated by all the operating operators by these horizontal lines. We set all the states that we can reach from these primary states by the action of the symmetric states in the symmetric energy. What do you get from these states? You can act on the energy by some energy raising operator. You can then act by lowering operator. That is redundant. It's redundant because you can then take the lowering operator through the raising operators until you annihilate the primary state and we let the only terms in terms of raising operators by using the commutation detection of the theory. So what we do is we do it with J minus and J plus by computing commutators. You can generate a whole set of states by taking this primary state of the Hilbert space state and acting by all the energy raising operators in all possible ways. This may not by itself be a hundred straight things to do. Certainly you can do this. All of these states will give you some powers. Take these states here and by acting with lower operators you can go back to the primary unless the state is null. Unless something very funny happens which we already discussed you know recently genetically that you get an idea of these states go back to the lower states and get to the primary. So this is the structure of it. Also it's clear that there's no limit to how high the scale goes because you can act with N1 as an index, N minus 1 as an index or N minus million, N1, N minus million at all. You just keep going. Yes, that's an assumption. The structure of the Hilbert space of the theory has this structure. There are these distinguished lower state states, primary states and then you build up modules all of them for the group. Module means just means an irreducible representation and irreducible representation that you get acting on this lower state state by all of the raising operators in any way you want. All of these objects are related by action of the symmetry. These states we have played from one another by action of elements of the symmetry algebra so in the same representation of the algebra. Things that you get from different priorities are not related to each other by action of elements of the symmetry algebra and form different representations. The structure of the Hilbert space of the theory goes as follows. The Hilbert space is divided into a bunch of modules or representations each of which is lower and limbled by its primary state. So the question you would ask is that it's not true that you have races in the Hilbert space which is spanned by primary operators. But it is true that you can find a basis in the Hilbert space that is spanned by primary operators and use for all positive elements. Just from symmetry that's good enough for the ordinary purposes. Any state that is not primal is called a descending element. It's a descending to perpendicular primary state. If you know everything about that primary, you know everything about that for instance, if you know the energy of this primary state and you want to know what the energy of the descending is, well suppose this descendant was L-20 acting on the primary we know that there was an exact statement that L-20 has energy of primary plus 20. The energy of L-20 of the primaries, 20 plus the energy of the primary. So we know the energy of L-20 for instance, if I do a full spectrum of the theory. If we do a spectrum of the primary states, that would be enough. Everything has a dimension. It's also true of some statement like it's also true of correlation functions. You know these correlation functions of these guys can determine the correlation functions of descendants by acting with differential operators as some part of what we discussed. But how do we know that we can never have a state which is actually two modules and then in both the modules how do we know that this thing you see the basic point is that we can go up and come down. But maybe you can come down to some other primary operator. No, no, no. You see coming down is the reverse. There is actually one subtlety associated with null states which we've not yet gone through and which is related to the question. But ignoring that, let's give a nine thousand first. So see, suppose this state was related to our primary but we actually have some elements. So let's say first L-M1 is L-M2 after L-Mk acting on some primary. Is there a way by acting on this state with lowering operators so that the resultant state is primary? So now this question is completely algebraically answerable because what does it mean for the resultant state to be primary? It means that all raising is annihilating the state. So I'm going to ask the following question. Can I act with some L-M2? So that the next result is that L-M2 for all positive K, that K is just no, okay, this is that L-M2. So there is just one way of doing it. And that one way of doing it is as follows. Suppose we choose lowering operators to kill all these so that we just get rid of all of these. That will do it. But in that case we just reach for primary P. Now I want to ask you, there is no other way to do it. Basically as follows. Suppose you've acted with some raising operators, lower operators, to kill some of these else but no. You know, you can always, suppose you've acted with lowering operators, we always take the lowering operators through. When they reach for primary, they kill it. So whatever we have in the LKP written as some number of raising operators on the prior, some raising operators are left or they are none left, then we reach the primary question and we act. But if there is something left, let's suppose as an L-M2 left, okay, then I'll act with an L-M2 on this state. Now according to the, according to the V-R-M2, L-M2 and L-M2, then I'll take the commutator, this thing I had as a primary. So L-M2 and L-M2 is something very concrete. What is it? It's 4Lg, exactly, plus some C into yeah, some number. So I say it was some number, yeah, 3 by 2. Acting on this state, okay, that allows this number to be the same as this number, then unless that is true, okay, then what we would, what we tell is that it's impossible, right, that this is not true. Now these can sometimes happen. They relate it to the existence of other states. And what the structure of the different space becomes then is more complicated. Okay, well what happens is that there are some states here that, then what, what actually, okay, let me just tell you the answer of what happens then, and we'll come back to study in a little more detail in a little bit. You see what can happen exactly for this piece that we will create this class, is that there is a state that is both a descendant of a primary as well as a primary descendant. The reason that although you reach it by acting with raising operators on some primary, you cannot reverse that that ascent because the state that will reverse it actually gives you zero. But it turns out that whenever this happens actually we probably see that when you say, but I won't try to do that now. So whenever this happens, this state has zero this state and all its descendants have zero more. Okay, so if you're really interested in a unifying quantum field theory, but no states, no physical states allowed to have zero more, you should delete these states from the spectrum. Those states are unclearly there. Okay, but I even take two points from your ideas on an unacceptable spectrum or that it's a pneumonic for a spectrum which does not include this in the concept. Okay, in that situation what happens is that the structure of the interviews of the representations of the bigger sort of algebra, at special values, see there's a special values of L0 and C, you get representations that are different from the generic ones. These represent, you know, the generic representation you've been able to act with all the LNs which are raising in an unconstrained fashion. For these representations you have to predict yourself having to get mouse patients who are supposed to be single. And these states are different. Okay, so the next conclusion is the same. What happens, there is a single primary for every real state in the theorem. Because these guys could have ended here, they would exist. So this may not have been terribly clear. We will get back to this at some point, but any questions at the moment? Just to clarify. I mean, for a generic representation without this outstakes, is it true that the primary is unique? There's only one state for one real state representation. Yeah. But for higher dimensions? Ah, you're wondering about representations of the ER. What happens to representation of SOD? Now, in this case, these two do you have a representation of SO2? Good. Other questions about this? Partho, are you... Okay, some good. Other questions about this? Before we go back. This is a very important. You see, the way you do it is that you know, generically, you get some state by acting on a primary state with enough number of elements. However, for these special representations, you know, in a repair, these states are deleted, by acting with these elements in that state, you just get zero. Now, we can make the same statement about correlation functions of operators. How elements act on operators, just by some differential operators. Okay? So some combination of differential operators and any correlation function of these special operators must be zero. So I can see differential equations for correlation functions of these short operators, which then and always gives you great control over the solutions of the correlation operators. Okay, so that's a throwaway remark. Okay, so any other questions or comments? So that's my primary operator's perception. There was a perception in the sense of representation. Let me now demonstrate something we use in the dedication of the stress tensor, stress tensor over here. And there's something, while this became, in any unitary theory, the dimension, the scaling dimension, the scaling dimension of every local operator is strictly greater than zero. And that's a unique operator of dimension zero, namely the identity. That demonstration is very recent. Okay, suppose I have an operator given a scaling dimension. You don't care what it is. It's always possible to choose the basis of operators in there. Given a scaling dimension. Okay? Let me call that operator O. And let me act with L1, L minus 1, and let's say that O is a private operator. Let me first study private operators. So we have L1, L minus 1 acting on it. This is equal to 2L0. There's no central charge term here, because remember there was an M at any square minus 1, then square minus 1 is 0. Let me sandwich this with an O on both sides. Now this how do you take that term? It's just exactly, it's just O, L1, L minus 1. Obviously, the other way, L1 and L is private. And now we bring something important to that. Let me first make a statement and then we have a discussion. L1 is the permission conjugate of L minus 1. Let me go ahead with the statement and then we'll come back to our discussion. Okay, because of this positive element, with this 2H and OO, this is positive element. In theory, it must have a positive scaling dimension. Of course, our proof of the statement is 0, but A is over the private operator and B, that is 1 and we should conjugate it. Let's address the first thing. Since we proved that every primary operator has a dimension greater than 0, it immediately follows that every operator has a dimension greater than 0. Because if there is an operator with even negative dimension, then there must exist a primary, because you can keep lowering the energy of that operator. The assumption was not restrictive. What about this L1 being the permission conjugate of L? Manifestly true. This is manifestly true because of the way we are doing our organization. Remember that in the expansion of the stress tensor in terms of W, what was that? Let's go back to Minkowski's case. What do we have? We have G of sigma and tau. The stress energy operator of the classical stress energy function of sigma and tau was taken and just fully expanded. It was taken and just fully expanded in the modern fashion. So e to the power i m tau, will be handed as e to the power i m sigma m plus e to the power minus i m sigma minus e to the power minus i m sigma minus is the sum over m in the expansion. In this classical expansion, we demand that t is a real function. That demand of reality lets us the complex conjugate of t is itself. But in complex conjugate, this function we get minus i m sigma plus. Sigma plus is just tau plus sigma. Similarly, here in complex conjugate, this function we get minus i k l k. So how do we ensure that the stress energy tensor as a function is real? It's real provided that l m is equal to l m star. Sigma plus does not become sigma minus. No, first motion. Sorry, l. Similarly, l minus k is equal to l k star. Now this statement was about classical functions but of course it becomes promoted to a quantum operator. The condition of the cylinder is the same. In the quantization of the cylinder, this condition was manifest. What we are doing on the plane as we have seen repeatedly is implementing this quantization in a more linear path. Defining norms of the condition conjugate on the cylinder. None of this changes. That's why we have to be very careful to define t as equal to l m over z phi plus 2. This plus 2 is what relates this expansion to this expansion. Of the transformation, the primary path of the transformation between the w and z plane. You might have thought why not take t and expand this l m over z to the phi m. You can do that with an l m and n minus m. It will be a pain in the neck. Working with the dimensions that we have been working with, whether doing things carefully, these things are all omission transients of each other. In addition, we have one of the statements I used before, namely that in the eutriquant with every local operator as positive norm by the way, let's understand something a little more clear. Suppose h is 0, that implies an l minus 1 on over z. What is the l minus 1 on over z? At the beginning of this lecture, that l minus 1 on over will be 0. If and only if the operator is at the top, then 0. Both 0 and 0 are at the bottom. If and only if the operator is a constant. It's the identity operator. If you reach the most sophisticated conclusion, there are interesting local operators in the theory, these ones. Which although operators, but always have an antihologmophic way to see. Moreover, the only operators that have antihologmophic way to see are operators that are above. Any operator that has a weight of the form h comma 0 must remain the equation. It's not that operator. So it's worth saying many times. And of course the crude version of the statement is that an operator with both weights equal is necessarily constant. It's only one such operator. There is hope to start on new discussions. So the plan for now is, in today's lecture, I'm going to complete every general bit of discussion about conformity in the next one. From next lecture on we start discussing specific performance details. We'll start with just a free scale theory in detail in this language. And we'll try to see some of these abstract arguments work out in more concrete way. But today we must really complete our general discussion. So there are two things that I wanted to say about conformity in theory in generality. Both of which have to do with the importance of it. I'm not even trying to do some of these exercises. I'm going to suggest exercises for you. I request you really to do these exercises. It's the only way we can get to anywhere real. First thing, I want to suggest one thing. This is effectively what we need here. Remember we said that we have primary state at the age. There's a primary state and then there are states obtained by acting on that state with the n minus 1s. Also states that we can get. This has energy 1 and 1. What about states with energy 2? The two kinds of states we can get. These are l minus 1, square on primary and l minus 2 on primary. First we take an arbitrary level to state. First we scale. What I want to do is try to understand the structure of inner products between states that are descendants of the same product. We already understood the structure of inner products at every moment. We found that there's only one state. Firstly, there are no inner products between states at different levels, vanishes. Why? It will become l positive. It will become l positive. There's a combination relation with n positive. Acting on primary operators. Of course this is a very general state with the general rules of quantum mechanics. Now we can prove a solid. Suppose you got something at level 8. So this energy is h plus k. And we are looking at something whose energy is h plus k prime. k prime is not in the k. Then insert l0. l0 is a mission operator. This way it produces h plus k prime. This way it produces h plus k. Unless h plus k prime is h plus k, this inner product is c. This is standard application of quantum. So this product is inner product. Let's say that we work with the normalization such that the inner product is primary with itself as 1. This inner product of l minus 1 on the primary with itself is it's 2h. And we've got a constraint on h from the requirement of this p positive. Now let's say we go to level 2. Suppose we go to level 2, we'll have two states. We'll have l minus 1 squared and we'll have l minus 2. And we'll have 2 cross 2 matrix of inner products in these two states. There's 2 cross 2 matrix of inner products in the computer. The bit is c relations to the l's and the wave as a branch. I want you guys to do this. It's a little calculation. Compute this 2 cross 2 matrix of inner products of states. Add this computation in your hand. Suppose you have this computation in your hand. What requirement about what requirement does unitarity impose on the state? But unitarity requires that any linear combination, any particular state you take has positive value. The state that you take is some linear combination of these two vectors. The fact that every such linear combination has positive norm is a statement about what this matrix is. Technically it's a statement that the matrix of inner products is what's called positive. Which means that it's all positive like linear combination. It's a positive and some parts zero. So what I want you to do is to take this thing, compute the output values of this matrix and demonstrate that positivity requires that the central charge of the theory is greater than 0. It's impossible for both eigenvalues to be positive unless the central charge of the theory is greater than 0. It could be equal to 0. Perhaps one of the eigenvalues will be able to see. Let's take that possibility out. But once you've done this calculation we'll discuss it a little bit in class next time. What this calculation shows you is that this business has had, you see, we had this bad beginning C from in the TTOP. We just said why is the most general possible TTOP? It put that C but C could have been 0. But this calculation shows you that C equal to 0 is essentially a very trivial thing. Actually I didn't even show that if C is 0 then this theory can have at least one state or something like that. When you do the calculation we'll exactly answer it. So this possibility that we talked about by adding this extra term to the TTOP wasn't just a mathematically non-sensitive in order to get something physically sensible we must have a non-zero and actually positive central charge. So first exercise that I want to suggest to you which is strongly suggestive. Now this is the second exercise I want to suggest to you which I also strongly suggest to you. I'll count the exercises in ideas and what do you fit in the numbers. And that's it. The exercise is let us try to see in any fact we can understand well the end result of this exercise is a very deep result. It is the fact that the dependence of the partition function of a reformative theory. So far we put studies in a reformative theory in flat space but we can study it on any dimension. The dependence of the partition function of this reformative theory on the metric of the theory is completely fixed merely by a central charge. I'm going to wrap by the end of this exercise we'll write down a formula which tells you what the value of the partition function is given the metric of the map. By partition of every part of the table of the reformative theory on this map, on this dimension. This is an extremely important as you can see it is totally non-trivial statement. So let me try to start out planning how I want you to go about this. It's an important exercise. It's dealt with in Pochinsky. I want you to do it yourself because you can't get trouble with discussing it. But let me start by our planning. The first step in this exercise is to prove what you can see by planning. So what's going on? We've got a reformative theory. And a express tensor in flat space stresses. We argue for that and we use it a lot. Now the claim is that if this reformative theory is non-zero central it would be inconsistent for the stress tensor to be zero when the failure standard is ending up. While the stress tensor is non-zero, what its value is the stress tensor, what its value is is independent of the state in the theorem. Independent of the particular solution of hand. It's the same for the vacuum in any other state. And it's given just by the coverage of the manifold equation. This is the claim. You understand what the claim is before we go on to try to explain how we can try to prove it. This is a local statement actually. As long as it's a manifold in the middle of your point, you'll gain about what's happening. It's not. How are we going to prove this? How are we going to do these dimensions? The first thing we're going to do is say this is a statement about a reformative theory. It carries it in flats. It ain't not going to actually go in a manifold. Weirdly, the overall reformative theory was in flat space. So came up with this statement, in your statement about the behavior of CFTs in flat space. It can be the definition of stress tensor. So let us remember that when you go back and trace through all the factors that we used, our definition of stress tensor was as follows. Suppose you have some correlation function, O1 and On. Because we take integral dx into power minus s and some operators now. And we vary a bit. So the change in this correlation, upon a smaller variation of the metric is given by the following. It's given by integral dx minus 1 by 4 pi integral dx into power minus s the same O1 and On times square root g delta g du nu times dj. This is the definition of the stress tensor. We have this definition of the stress tensor at the moment, but there are better than no ones. Just to the polishing function. The definition of the stress tensor was delta of let's call it z is equal to integral dx e to the power minus s times square root g delta g du nu. In order to do this, I don't want to apply this to. I'm going to apply it to the insertion of one stress tensor. So delta of expectation value du nu. For the next step what I'm going to do is to choose a particular form is to choose a particular form of this delta g. I'm going to choose delta g du nu from flat space and choose delta g du nu to be equal to e to the power 2w times delta g du nu. That's 2w. We'll change my metric to e to the power 2w times delta g du nu. So the change is 2w times eta. I plug this in here, and I find the formula of delta of du nu mu at x. So the change in du nu mu at x, when I change the metric from eta to eta plus 2 omega n. It's going to get this formula. It's equal to, and now I want to leave the order in omega. I'll drop this way to g. This is dx, this is e to the power minus s. Times this back to 2. I've dropped by 1 by 4 pi, so that becomes minus 1 by 4 pi. Times t of y, t mu of y, t nu nu of y, t mu mu of x. Integrated omega of y, integrated omega of y. Back to leaving order. So the leaving order of omega of t mu mu, the claim was C r by 12. We just have to expand our leaving order of omega. And we're going to have to look up the factors. If we were working just infinitesimally near that space, then the demonstration of this statement is true. Would be the same as the demonstration that infinitesimally near that space. So pi t mu mu of x, t mu mu of y is equal to C by 6 square of delta x. This is the thing we have to do. At least this statement being true requires that we can also argue that when this is true, we can show that at least this statement we want. We want it first. So we want to show, we won't want to show I hope this factor of 2 and sign is right. That depends on this this thing which we have some number but I can pass. So we want to show that t mu mu of x, t mu mu of y is equal to minus pi by 3 z. We said nothing about the trace of t. We just said that was 0. Nothing was just 0. We know a lot about t z z, z bar z bar, but we know nothing about t z z bar. Certainly the expectation value of the operator is 0 but could it be somehow that there exists such an operator? That there is 4 states of such an operator, non-zero, 2-point function even in transverse. Well, let's see. Let's see. The case is that yes, there is nothing. You see, the stupid function is very small. It's non-zero everywhere except the contact points. Such things are very subtle. These things are associated with curing divergences or things associated with regulation that we want to take into account. And they are very subtle to get right. So of course we have got some nice power law form of never been in a cross violation of everything that we do. We can talk about it but it would just be inconsistent. But all we want to get is contact. It's very hard to get right. So it could be that when we do things that we find correctly there should have been this contact. Now I'm going to argue for you that we are forced to have such a look. Okay? And the argument... Oh yeah, I'm going to outline this for you soon and then we'll... I'll try to fill it up. Next time in class maybe I'll give you a number where you're thinking about it. Okay, we won't finish everything. We have to say goodbye.