 As preliminary to our study of the formal field theory, we began studying what identities associated with symmetries in any quantum field theory, as seen from the point we go through our bottlenecks. So I went through the main idea of derivation of the water identity last time. This time, I'm going to go through it again by keeping factors in consonance with Pojinski's dimensions, which is going to be a very painful task. So this is, of course, what you're trying to keep factors in. So let's see. So let me quickly remind you of the idea and then get quickly to the point where we want to be added. In fact, so if you remember, we had a path integral bx into the power minus s to the power of x. And then some operator insertion, lower sigma 1, lower sigma 2. And then we define the symmetry operations. The symmetry operation as x is equal to x, right? An infinitesimal symmetry operation. Plus epsilon times epsilon del tau. This operation was a symmetry. If and only if, this is our definition. If you plot in this change of variables here, the integral, and we write as an integral over x prime, we get the same functional form for the integral. This was the definition of the symmetry where this was true without any escape to look at what the operator is doing. Now, under the symmetry operation, operators in general transform. So it was symmetry if the statement was true without any operator insertions. But now let's define the change of an operator as 1. That's the essence of O1's in our way of thinking. Let's say that O, delta of O n, is a 1. O expressed as a function of x prime, O n is a function. You take this object, wherever it is, and take it as x prime, when you make this symmetry. This thing will be O as a function of x, when it's a function of x prime plus epsilon delta. This would be O as a function of x prime plus something that's first thought in x, doing a test. So plus epsilon times, whatever we get, we'd call delta. When the operator x itself, delta O n would be just delta x prime. If O n would be the operator x square, we would have got 2 x prime times delta x prime. And so this is what we need by the change of an operator under a symmetric observation. Whenever I use the symbol delta O n, I'm going to express this thing. Then last time, we went through this little calculation. We said, well, we know that if we make this change of variables without any operator insertions, and if epsilon's a constant, the integral measure and the action don't change. Let's look at varying epsilon's. Let's look at varying epsilon's. So if we make epsilon as a function of sigma a, then what would the change be once we make this change of variable? And so now let me put it back in my packed dimensions. Straight. But last time, the integral, ignoring operator insertions for a moment, we've done that in Egypt, right? But ignoring operator insertions, the measure and the action have to be of the form plus the first order in epsilon, something that denotes the derivative of epsilon. And now we choose a convention factor, which says plus i by 2 pi. Plus i by 2 pi integral j e. And if we're working in that space that has a longer real metric, we put a factor of the square root of metric. We're going to make everything a little less. And mostly, we'll be working in that space. We might forget these factors of square root of g. We're all going to go back to these coordinates every day in due course. All right, so this is our definition of we do things upon a change. The physics that's drawn to this is the statement that the action of the measure have not don't change the epsilon's constant. Any changes can be promoted to derivative of epsilon. There's more than, if there's more than to the absolute function of the derivative of those derivatives. There, okay? So this statement is only made to first order in epsilon. Okay, so that's the physics. And then there's our convention dependence, which is what factor you put out here. Of course, it can be changed by refining this gv function, but we should be saying nothing. This is going to be our convention for what? What is g? This will be our definition. G is the word g. Word g. Okay, so this is what we did when we performed this variable change. With our update, it was completely, it was entirely in irons, okay? So, suppose we do a set of operate insertions. These operate insertions, yeah, yeah, yeah, yeah, yeah. Okay, let's, the first thing we do is choose a region and choose epsilon. That's support only, that actually, okay? Though it's slightly singular, but useful when I'm saying to choose epsilon to be one in that region and zero in that region. Okay? In that case, this change of variables does not affect the operator insertion variable. Because the variable changes identity. And the operators are local functions of the x fields. Over sigma one depends only on the value of x and sigma one in the derivative. That tells us that if we do that calculation, we take all these operate insertions, they haven't changed. And then we use the fact that the integral, of course, doesn't change under the transformation. Subtract the two. And that gives us a statement that integral n a epsilon ga, square root g, or any region that does not contain an operator insertion. Then as we did last time, we could take the statement integrated by parts, they move to del a, you know, the derivative to ga. And we land up with the conclusion, but since epsilon then is completely out of the function. Okay? We land up with the conclusion that, we conclude that del a of ga over sigma one, over sigma, over one sigma one, when sigma n is equal to zero, provided sigma is not equal to any of the sigma. Okay? This is what we talked about last time. The missing statement that we will use a lot. Okay? When we allow our region, our region over which this epsilon is non-zero, then, of course, just for simplicity, one of the operator insertions. So that's non-zero insertion that includes one operator insertion and other things. Okay? And as we said, we choose exactly one everywhere in the region and zero. So now, let's focus in the liquidation and see what we need. The first thing we get is that, the first thing we see is that when we do this change of variables, we get that the additional equivalence, let's follow this quantity, C, C equal to C. So we also have to see now the equilibrium integral of the x prime, exponential minus x sub times, right? Then we've got this term, that's the same as before, by the g that is a epsilon g h. Okay? Well, and let's take the operator which is included as the first operator. So this is all one, we'll just stand it out of one. There's an epsilon, yeah? Okay. Yeah, so we choose this epsilon function to be epsilon in this region and C equal to C. We want a small band of meters so that we can expand everything. Okay? So that's what this epsilon is. It's the constant value of that function inside this region. Is that clear? Okay, good. And then, O2, that's not going to affect it. O2 of sigma, O n of sigma n, now this is going to affect it. Okay, so C, which was equal to this is also equal to this. Now, we subtract the two and keep terms at leading order next. Okay? So what do we get now? We get i by d phi. And then, we take this and integrate it by mass. So this derivative goes against this j. And, okay, so let's do that first. Minus i by two del A of, that would be A of sigma, O1 of sigma one, O n of sigma n. And this is integrated over, over sigma. And this is going to take us over the region. Over this region of epsilon, it was not zero. So let's follow that region R. Okay? Plus, we get the correlation function of epsilon. Yeah, well, that's what we, this is the coefficient of epsilon. So that's O1 of sigma one, O2 of sigma two, O n of sigma. And we use the Stokes theorem. Well, I think the text of that, integral of the total divergence to integral of, at the boundary. All right? The Stokes theorem you can do, the i by 2 phi, of j dot enemy, the normal vector to the boundary of the region. Output of the boundary of the region. And this integral is taken, this is a point sigma. And this integral now is taken over the boundary of the region. Yeah, this symbol is often used for the boundary of region. Yeah. All right? Then, O1 of sigma one, da da da da, is equal to, correlation function, is equal to the correlation function, delta O1 of sigma one, da da da. Next step we do some algebra. We each one of them for, let me do one sort of my, just to get a factors of two steps. I think this is the first time I'm thinking that I was trying to get a factors of two. Let's see if I can. Okay? So, so, now the next thing I want to do is to take this j mu and mu over the contour and write it in terms of complex variables. We decided to use complex variables on our worksheet. Okay, so what is j mu? Well, suppose we've got a little different. We're getting j, the contour along which, you know, the little tangent factor along the contour is the vector dx mu. The integral j mu and mu is the, okay, what's the, what's the normal vector to this tangent? That has this, that's what's the vector that has the same length as it. Okay, so it's very clear that it is plus or minus dy minus mu. It's the vector that has zero dot product first time. It's normal, right? It's clear, it has the same length as it. Now, which side do we choose? Well, we want the outward pointing normal. We want the outward pointing normal. So, if we're going here, we're doing the integral and the clockwise as we were always choosing. If we just had dy, we should get the normal and the dx direction with the positive sign. Okay? So, this thing's correct. Okay? So, j mu and mu is the same thing as jx dy minus jy dx. Jz, jz dz minus jz bar. Jz bar, dz bar. This is the same thing as jz dz minus jz bar dz bar. Well, of course I need to make, well, I need, what is jz? Well, you see z, where this is to be thought of as adding an upward index, is equal to x plus i. So, for instance, dz is dx plus i. That's a usual dimension. Okay? Upward index, okay? So, that's the jz with an upward index is jx plus ij. We should, of course, put x and y in the extramaritons because upward is z. But, jz. Okay? How does that add? Dx is the same thing as the derivative. Okay? So, j little z. So, let's remember that dy dz is equal to dx by dz by dx plus a similar term with y. And x is z plus z by two. So, jz is half of jx plus ij. Which is the same as half of j and upper x plus ij. Because j is a given. And it's not a convention with which level. Whenever we see a jz with a lower z, if you want to convert it to Cartesian coordinates, it's this. If you want to convert it with, you know, if you wanted to write it with, wait, am I not going to customize this mic? Let's check that. Let's check, that's from this point of view. Let's work it out completely. So, we have dy dz is equal to dx by dz d by dx plus dy by dz d by dy. Now, x is equal to z plus z power two. But y is equal to z minus z power by y. Jz will have a one value. This is what jz is and jz power z is half of jx plus ij. This is the same as this. Okay? Let's, you know, once in a rhythm. Of course, the main one's specific in the way you see it. It's basically the transformation of the epsilon symbol that we want to be on here. Okay? But, but it's, that thing's also straightforward here. Okay, so, so this is half. This, let's do it in the same way. This is half of jx minus ijy into dx plus ijy minus jx plus ijy into dx minus ijy. Now, you can see that the jx, dx don't cancel. The jy, dy don't cancel. What remains is the jx, dx and jy, so jx, dy and jy, dx don'ts. And, so I'm just, I've got a factor of r, right? Yeah, good, good, good, good. So that's what it looks like. That's the thing, so this is jx, dy, i times jx, dy minus j. Okay, good. So this divided by i, so one by i, jz is n minus jz by dz. Is equal to jx, dy plus j. Okay, there's somebody who's gonna have to remind me to sign the equation. We had an equation that was i by two pi j dot n, del, or one, will do the delta of the one. Oh, what was the sign, please? Was the sign correct? I'm just gonna do it. Okay, good. So, ij dot n is jz, so this thing is one by two pi jz, one, blah, blah, blah. Minus one by two pi, this is dz. That's five, four, one, blah, blah, blah, blah. Is equal to, is equal to delta of one, make a special additional simplification. And a simplification that will turn out to be true in every case that people use. You see, we know that this function, whatever it is, apart from singularities that couldn't occur when j hits o one. Okay, we know that the insertion of, we know that del z bar jz, del z jz bar, insertion in any path integral is equal to zero, apart from singularities where things collide. Now, there's one simple way in which this constraint can be met. And that simple way is that this insertion in the path integral was zero, and this term was also zero. Nothing we have derived requires that so far. But of course, in a special case, it's a special case that might sometimes be true. And will often be true, and that's why I'm looking at it. In that case, the function which has a j insertion is entirely analytic in z. And the function that has the jz bar in insertion is entirely anti-analytic in z. This integral here is a contour integral over z. The equation function that we have just concluded for the special case is analytic in z. This is a contour integral over z bar. Over a quadruple equation function which we have just concluded is anti-analytic in z. No, we did not conclude, just assumed. In this special case, these integrals can be done very simply. Listen, this integral is zero unless there are poles. Okay, so you remember what we have to get is delta O at the point of insertion at all. So we probably get something that's non-zero, but only at the point of insertion at all. But from complex analysis, okay, the way that happens is if these integrals have poles at the location of all, and if one of these two integrals, integrals must have a pole at the location of all. And when that is true, the contribution of this integral is two pi i, out of two pi i here, this minus is cancelled by the fact that it's anti-analytic integrals. See, you're actually with minus i, you pick up an sd. The next conclusion is that you have a history to monitor our assumption that j is, the j by jz by itself is analytic and jz bar by itself is analytic. Okay, we meant that it's true. It follows that high times the residue, residue of the pole in jz, jz bar with four pi is equal to, I don't agree with my notes, so that's good. So this is the conclusion that we drew. We reached very hardly enough strength to explain the more, take more time. It's the same argument, just a little bit more time. This is completely clear. What we have done is that in this particular case of the situation where the symmetry currents and analytic parts are, now the jz is analytic and jz bar is anti-analytic, we have related the transformation properties of operators to polarities in the correlation functions between this operator and the current that generates the situation. The second meaning, the second means that if you compute the correlation function of j with an operator open, there must be a singularity in this correlation function. The singularity function must blow up as the insertion point of j approaches the insertion point of open. There may be many kinds of singularities. The only part of the way sensitive to this is the portal. There must be a portal and with a determined risk going to apply this analysis to a particular symmetry of great importance, namely the symmetry under space-time calculations, okay? And we will apply this analysis to a conformity in the special case. Okay, this is a bit of an aside and really useless, right? We want to understand that there is a second, but what is the, how do we generate the con that generates space-time translations? We use a usual method procedure, one performing infinitesimal translation and then you make that space-time dependent. So let me first start with the explanation. Let me define that the, suppose I perform a translation by epsilon mu. Okay, in the direction epsilon mu. Then I will get a symmetry problem. Let me define, and now for some i approaches, we put several reasons, a reason of respect on the problem. So let me get this, I can make this. Let i, in the name of the union, i direction in another way. This is why it transforms like a vector. Instead of having a vector of columns, we have a tensor of columns. So we've got two indices, index of them, which is the index of the column and which direction we translate it in. Okay, this is the precise connection between the tensor and the vector. We've got tensor, which is what we call the stress engine tensor. Translation in the mu direction, in the epsilon, we call a mu, the normal vector. We take the stress tensor, multiply by the i times m mu, and you've got some current. That current is the local current, corresponding to translations in the end product. It's with the particle. So, another expression, independent of an, oftentimes more useful expression, for this translator of 40 mu, then the expression of, then the implicit operation of just acting with the normal procedure. In order to do that, basically what we're going to do is use a clever trick to implement the normal procedure. Okay, so first let's start. Suppose we wanted to implement the normal procedure. And then we'll make n mu space for that. So what we do is to make a change of variables, delta x mu is equal to x. x mu is equal to x, prime plus epsilon mu of x. And the current in question will be the coefficient of the derivatives of epsilon. I could do this in many steps. I could say epsilon times a mu n was a constant vector. And then that would give me the current for that particular translation. But it's faster just to do it all at once. Okay, if you get confused, you can do it that way. Okay? There is such a change of variable is to see how the bottom table changes and identify the coefficients of the derivatives of epsilon mu as terms of the stressors. But, we know something about, we know something about is what this is. It's just an infinite number of variables. A small change in the coordinates of our base math. Now, we know of an operation under which the action, and presumably the measure to define the measure that is from what we've made in a way, the action of the theory does not change. And this actions perform the diffeomorphism not just on coordinates but also on the metric. If you make this coordinate change and you change the metric appropriately, appropriate to this coordinate change, okay? Then as a function of the new metric and the new coordinates, your action will be the same as it was as function of all metric, you know? Because just statement of general coordinate. It's just fair. If you know the procedure, that's not what we're supposed to do. What we're supposed to be doing is just making this change of coordinates, not touching the metric, and seeing how the action changes. Since we know that there's an operation under which a combined operation of two things which leave it unchanged, instead of making this change of coordinates, we can make the change of metric, but you ask the change of the action. So, what we know is that under making this change of coordinates, we want to ask how the action changes. The answer is the change. So, suppose we've got the action which is equal to, okay, we'll do it well. The answer is simply minus of del S by del G mu nu. The change of the metric that would have been induced by this coordinate change. Now, what is the change in the metric induced by such a coordinate change, okay? It's easy to work out. It's, also, once you've worked it out a couple of times, you generally remember this. It's got a fancy name. It's the lead derivative of the vector field. But how flat-spaced is the leading order? The answer is simply del mu X around mu plus del mu X. This is the change in the metric induced by this coordinate change. So, you know, the law of the procedure asks us, the law of the procedure asks us to find what the change in the metric was when we made this coordinate change. Using the science-eject, we know the answer. This is the answer. Now, of course, G mu nu is the symmetric tensor. So, this contraction with this and this contraction with this is the same thing, which is active two. Okay, so this is minus two delta S by delta G mu nu del mu X around mu. If we had written this S as an integral over a local quantity, this would have been the integral. This is more accurately. So, the integral of L d mu X around mu, where S would be equal to the integral. So, we've got to answer the rest is the Pinkfield-Gracken-Crow connections. Okay, according to the procedure, we should take the coefficient of del mu epsilon nu. Let me just remind ourselves of what we wanted. We said that we'd start with e to the power minus S and that goes to e to the power minus S plus i by 2 pi del mu epsilon... Well, del mu epsilon times G. In this particular case for epsilon itself carries an index. This is... Okay, this is true for any kind. Now, by epsilon itself carries an index, this is equal to epsilon nu times i because G. Now, let's take the minus sign out overall. So, this is a G, so i times i is minus. There's a minus in the S. So, this is a delta S, which is equal to plus of 1 by 2 pi del mu epsilon nu times T nu mu. You see the delta S is minus of this? And so we conclude that T... Okay, of course, we can raise the value to see that with it. So, T mu nu is equal to del n times del, del times G mu nu times minus 4 pi. You see, because what we wanted here was this tiny square root of G. Okay, so, maybe we end up with a square root of G over 1. Let's say we need a new T here. Okay, so, thank you if we put a full definition of the square root of G here, we would have value at this point. Okay, value at this point. Okay, so now we're working with Purchinsky's conventions. I really cannot figure out why I put i for a general covenant and remove it for stress tensor. It's just painful. But anyway, it's probably like he's trying to match with other people's conventions. No, she's just rewriting conventions for example. But anyway, so, yeah, so this is what we got. So, the stress tensor has this nice, had this nice interpretation, has this nice alternative formula. And the alternative formula is that it's the derivative of the Lagrangian with respect to the metric. And this Lagrangian mind here includes the square root of G. So, whatever the entire thing, this is integral n, not the integral of square root of G. Yeah, the differentiated square root of G is n. So far, anything which says to be totally general, it applies to everything, it applies to the other dimensions. Okay, now we're going to specialise, now we're going to specialise to two-dimensional conformity. Okay, so what can we say about a two-dimensional conformity? Well, let's take the Lagrangian flat space for a moment. In a flat space, there is conformity in there. So, the fact that the theory is conformally inventant implies, conformally inventing in particular implies scaling. And that implies that the partition function of the theory, in the partition function of the theory, does not change under the scaling of coordinates. But the scaling of coordinates, if it compensated for by the scaling of coordinates, you could rescale them back by the coordinate change. What that does is rescale your metric. So, the theory being inventing and scaling of coordinates is the same thing as the partition function being made under the scaling of the metric. Is this clear? Because, just to say again, because if the theory is generally coordinated in there, and if you scale your coordinates, you can undo that stage of scaling by coordinate change. What that coordinate change does is change the metric. The derivative of the Lagrangian, you know, is inside the partition integral. Insertions of this object inside the partition integral. Derivative of the Lagrangian with respect to the metric must vanish for those changes in the metric that are proportional to the original metric. What you can do is that del L by del g mu nu times g mu nu. And therefore, g mu nu times g mu nu is equal to 0. That either can form the field theory. You can build that either can form the field theory. The metric in particular, the stress tensor in particular, stress is g mu nu is equal to 0. This general argument might sound like it applies to, you know, beyond the flat space. We will see that there is a quantum subtlety in this argument. And that cannot be both true at flat space and at all spaces in any quantum field theory. We'll review that very soon perhaps, I think. I just want to warn you that this conclusion is not going to be too strongly true. It's the 90th conclusion that you reach when you try to understand what's going on. It's the 80th conclusion. Okay? We'll come back to this something. So we've concluded that if we're dealing with a conformal field theory, then so far we've not specialized in two dimensions, any conformal field theory, and see that the trace of the stress tensor is flash. But when we make this argument at various levels of distribution theory, this is good enough for now. Okay. But now, but now, let's specialize in two dimensions. Okay. So what is this equation like? This equation. Oh, also, of course, the stress tensor is conserved. That is a constant. So we have d nu, d nu, nu. The statements mean, by the way. Both these statements mean, what these statements mean, really, is that any correlation function, involved in this quantity is another operator, 0, at least a vector of coincidence. We know that this statement is not true at coincidence. We found the correct version, that's it. That's what I mean. We'll use that very soon. Okay. There could be trouble with this statement also at coincidence points. We'll come back to that. But certainly, if you insert this operator inside the correlation function and you stay away from other operator insertions, we can see it. That's just the conclusion we come to. Now, let us see what we can say about correlation functions and therefore, involve in the stress tensor. Stay away from insertion points at the moment. We'll start approaching insertion points in a very small way. Okay. So, let's work out, so, let's do dimensions in complex coordinates. This thing, this quantity, in complex coordinates, is the statement that tz, z power, it's something that we do like. tz, z power, z power, this can work out, z power. Yeah. So, what is the statement? This is tz, z plus tz, power, z power, e to 0, which means tz, which means t alpha z, g alpha z, plus t beta z power, g beta z power, e to 0. That's what the statement means. We need non-zero component of the automatric with the z index. That's the one in which the other index is z power. This thing is a z power. And similarly, you get some factor times tz, z power. Okay. So, all the three components of the stress tensor, tz, z, tz, z power, and tz, z power, z power, one is just set to 0 by statement. Yeah. Fine. These equations are dz, tz, power, z, plus dz, power, tz, z, e to 0. That's one equation. It's something, I think, how I got this to happen. If I want to write the equation as del, mu, t, mu, mu to 0, what value do you think this equation has? What can I choose mu to be given? Z. Z. Right. So, this is the x mu. And this is the same contractions. This is that one. Okay. But this equation can be simplified because we know that tz, z power is equal to 0. Since it's 0 at each point, it's derivative of 0. Why do we stay away from insertion parts? We have to be careful about insertion parts. Why don't we stay away from insertion parts? This is just 0. Okay. So, we read the conclusion that del, z, power of tz, z is equal to 0. So, inside correlation functions, the correlation function of tz, z is an analytic function of the point of its insertion. As long as you stay away from it, at least as long as you stay away from it. Yes, this here. We're going to let the free index be del, z, power, tz, z power, tz, z power. Right. In fact, del z. So tz, z, z power, z power is an anti-analytic function of the position of its insertion. So, you see, I need one last homework. We've learned a lot about correlation functions or stress tensor insertions in a conformal way. We've learned that the correlation functions are 0 if you're inserting a tz, z power. that we insert a TZZ or a TZRZ bar, respectively you have an analytic or anti-analytic function of the point of insertion. It's good. It's a scare. And now we're going to learn more because now we're going to scale, you know, over our pure approaching operators. Try to approach these operators and see what we can do. It's useful before we do that to react to the currents for the infinite other symmetries we have on the problem. You see, we're dealing with conformal theory. Those translations are symmetries from them. But you remember that there are a whole infinite class of symmetries in this theory. Namely, what are the three definitions? By any analytic function of Z, by any anti-analytic function of Z. And I'm sure you've said this, but this is just the analytic definition statement we've made. Namely that if you take sigma plus and make your function sigma minus, and sigma minus and make your function sigma minus, that induces a conformal change. Let's just check that here. The metric is dz dz bar. Is that to any function of itself? Is that part of the complex analytic function? So, here what you'll get is, in the fact that it's dz dz bar times some function of z. And therefore it's conformally related to the original metric. So since we're dealing with a conformal theory, this is a quantum theory in which all such coordinate changes, remember with no compensating change of a metric. This is crucial. Any coordinate changes symmetric, you also change the metric. Very much. It's changing the metric, it's changing your Lagrangian. It's meant to be as a, this is, if you've got the problem, you'll get all these changes of coordinates without any compensating changes of a metric. Let's see why that's true. And we're dealing with the north of the left, corresponding to these, to these coordinate changes. So we can do it more generally, and it's essentially clear to do it one at a time. So let's make a change of variable z to be equal to z prime plus epsilon z prime to bar n. And z bar, z bar prime plus epsilon bar first. Now remember we're dealing with complex variables. So epsilon here is a complex value. So now I've written it like this. The way I've written it, it looks like there's one symmetric, there are actually two symmetric. There's a symmetric orthogonal to the real part of epsilon. Symmetric orthogonal to the imaginary part of epsilon. So this is two symmetric transformations, immediate. Instead of dealing with the real part of epsilon and the imaginary part of epsilon, you deal with epsilon and epsilon bar as the two independent ones. Okay, I want to emphasize a clarification. There is no, it's not true of course that epsilon and epsilon bar are independent ones. It's in the sense that if you know epsilon, you know what epsilon is. That's of course true. What I'm doing is the form of the thing of changing variables from two independent things like that. The real part and the imaginary part. Two real plus i times, imaginary and real minus i times. Symmetric, yeah, and parallel matrix is metrizing it, but real plus i times imaginary and real minus i times imaginary. That's what I mean by saying that we're very dependent. You know, quantum. Symmetric. The number is true at parallel matrix at any time. So, as you will see, it will always be convenient to think of analytic and anti-analytic because independent to the extent that we can. It's in the sense that the same is true. So we're looking at the symmetric. It's a thing of real power of epsilon and imaginary power of epsilon is two independent parallel matrix. Think of epsilon as two independent parallel matrix. It's by making the appropriate linear approximation. Okay, good. Somebody is, for example, when we are actually doing the operations, so one is independently varying z and z bar, so it might seem like there are four independent parameters. But one is not independently varying z. Finally, where is what it is? You see, because when you say you change z, you also automatically change z bar. You've just done a change of the complex parameter. So this often leads to confusion. A lot of people say things that are just thorium. Variations z and z bar never really depend. Just that there's a complex parameter. It's the only, only complex principle. Everything I do, I'm not going to remind you symmetrically correct. But the symmetry comes for varying epsilon and the symmetry comes for varying epsilon bar. Okay, what that really means is I've found the symmetry comes for varying the real power of epsilon, which is some of these two, and the symmetry comes for varying the imaginary power of epsilon. It's a different system. But the point is once I've got some symmetric power, say the linear combination of it is also symmetric, with varying epsilon and epsilon bar. See, that's all that we need. There is always a complex conjugate of z bar. There's no sense in which it's not. Okay, this coordinate change. Let's take this coordinate change. See, I don't have to write a symmetric pattern, but you can also proceed. Now we've got very smart. See, we said, ah-ha, whenever we have a coordinate change, we can implement another procedure by performing the compensating change on the, on the metric. Right? See, this is a symmetric transformation. So in order to get something with the, with another procedure, we must take i epsilon and make it a general function of coordinates. We must take i epsilon and make it a function of z and z bar. And here, this is z bar, which is a function of z bar and z. A general possible function of coordinates. There are four functions of z, z, z, z. Now we've got this coordinate change, and we want to see how the action has changed. But the action has changed minus, minus the amount it would have changed by making the metric of it. Then add by delta g mu nu, times the change in the metric, and we use that to go to formula. Okay? Times, ah, let's call this thing delta z, and call this thing make, ah, rotations of group. So, times, ah, delta mu delta x, ah, mu plus delta mu delta x. Wait. But let's work always in flat space. We don't have to get this way of doing that, this way of doing that. Okay. Ah, now this guy, where they finish. Minus the mu nu of delta mu delta x mu, plus delta mu delta minus, oh, so that's plus, and we're going to replace it with 4.2. That's 4.2. Now, let's see if I can get the final answer. Okay, good. We know that delta x mu is a delta x mu, ah. Ah, just since we know in the upper, let's write this as plus 1 by 4 pi, and t mu, ah. Okay, now, this is twice the, this is symmetric, ah. So, I don't need to keep trying to go to the base. I'm going to make this one over 2 pi, right, I'll use that. I'll just put this in a nice, you know, way over 2 pi. So that, so this I can write as t mu nu delta mu delta x. Okay. So, now I'll, this is 2 pi. So, this is plus 1 by 2 pi times, ah. Ah, t z. Now, t z z, we both go down to the 0. Which means t z z, one up, one down, or t z, pa z pa, one up, one down. Okay? I'm both safe. Okay? So, the only, if I put a z here, the only thing I can add here is a z bar. So, t z z bar, there's z delta of z bar. Plus, t z bar z. Ah, t z bar z, delta z bar delta z. This business is a danger there. Only what we know about what delta z is. You see, what we come here is an analytic derivative of delta z bar. But delta z bar has an anti-analytic function times the amplitude function. This derivative does not hit the z part of the body. That's why you have a symmetry. You see, the change in the action is 0, 8 is x and all is constant. So, we see. So, we verify, the first thing we've done is a verification that this actually is a symmetry. So, let's work this out. Yep. So, this is plus 1 by 2 pi. Now, t z z bar. This is z bar to the power n times delta z of epsilon. t z bar z z to the power n delta z bar of epsilon. Each of the numbers which asked us to identify the coefficient of the derivatives of epsilon and epsilon bar as the common scores point into the transformation induced by epsilon and epsilon bar. So, we see. Let's first do the guy with epsilon. We see that the guy with epsilon has only a z bar derivative of epsilon. There's no z derivative of epsilon. There's z bar derivative of epsilon and z derivative of epsilon bar. So, if we're interested in the symmetry corresponding to the epsilon transformation, which I remember we'll remind you just needs the symmetry for the real power of epsilon plus i times epsilon. The first thing you can do is that j z for epsilon. The symmetry for epsilon. Because it's j z with the coefficient of delta z. There's no delta z. And j z bar of epsilon, because we have to say what our definition of j z bar is. We put in the i, we'll put in the same i that we put in for translations. So, I'm going to define my j z bar without any. You see, if we have a 1 over 2 pi that we had in the general stress tensor thing, you remember that going from the general definition of the current to the current for translations, we just keep putting an extra i. Since translations are special case of conformal transformations, you want to put in the same i for everything else. So, when you put that i in and not be the model for the genitals. Now, I'm getting the right sign. That's just a statement that this is j z bar z times z. It's still z bar z times z to the power i. No, there's no i in the picture. Had we been defining our current, the utility, there would have been an i. But remember, in defining the stress tensor, we put in an extra factor of i. Let me say this. Let me say that what I'm going to do is to define, let me try to get the same. What I'm going to do is to define the current with this for conformal transformations to be such that to be such that No, you see, in the usual definition of the current, the change in the action had an i. Delta s, we put i by 2 pi. It was minus i by 2 i by 2, l a x is around j. This was the usual definition. So, if we use the usual definition, we have an i. Because we put 1 by 2 pi, but without an i. Just on these translations, it was scaled by an extra factor of pi. Let's remember, let's remember. I think that i was introduced because this i was there. So, the i was introduced in order to make the definition effectively for translations without an i. The most general definition had an i, and then we put an extra factor of i for the transition. The general definition of current is that the change in the action has i by 2 pi. That's a plus minus. Times this minus. Identify the corresponding j. I just gave that. If you identify the corresponding j, you get the i. Exactly, exactly, exactly. But you see, now let's just put it at this point. You see, the variation in the action has no i's in it. And that stress tensor has no i's. So, the stress tensor is related to the variation without an i. So, if we use the same convention, because the variations with no i's, according to these figures, like the stress tensor should have no i's. Shut up, shut up, shut up. There are two options here. N is equal to 0. When N is equal to 0, this should go to the last positions. This should go to the transitions exactly. And there is an i there. No, there's no i. You see, with the translations, now we're talking about the transitions that are really unimportant. We use this as a matter of conventions. I'm just trying to follow the functions of these conventions which, unfortunately, are confident. But if you use some conventions, there's no i. Okay, fine. But, so now let's, let's do both ways with lower vertices. What we call is j epsilon z is equal to t below z, z, z, z, fine. j epsilon z bar is equal to 0. Then we've got j with epsilon bar z is equal to 0. j epsilon bar z bar is equal to z bar by t z bar z. So now we see that conformal transformations with respect to epsilon and epsilon bar are one of the special class of symmetry operations that we dealt with earlier in this lecture. Because j z, t z z is empty, z z bar z is empty. Therefore the exertion of this is actually empty. And j z bar is an anti-antibiotic function. It happens to be the anti-antibiotic function c. So epsilon transformations, this is an anti-antibiotic function, falls in, the conformal transformations with respect to, you know, the antibiotic conformal transformations fall into the class we talked about last time. So we do the anti-antibiotic. So conformal transformations, the word identities, terminate into a statement about in this trick of you converting word identities into statements about residences of poles using Cauchy's theorem, books. That's the principle reason we introduced this trick. Because the conformal transformation is very important. Is this yet? Let's use this trick. Let's use the trick that we introduced analysis was that the residue of j with the operator plus the residue of j bar with the operator has a singularity which changes the operator times 4. Okay? In our general analysis, we also did an i. But since our convention depends on the general analysis by an i, we won't have a i. Okay? So the net result is I'm not going to try to justify the sign. I'm not it will take years. Okay? Then the residue of z to the power n and now, you know, I've done for z to the power n. Since it works for z to the power n, it works for any analytic functions. It could be any positive or negative. So the residue of epsilon of z times anti-z, let me just say more clearly. We conclude that epsilon of z times tz acting on o inserted at any point let me choose t0. tz is tzz. I'm changing the notation of tz is tzz and when I want tz bar tz bar z bar quality bar so epsilon of tz times o tz times o. One of the z minus one of the z and the residue of that pole is delta o under the photogram. We reverse that epsilon bar on z bar and z bar on z bar o of 0 to 0 is equal to okay? There's no ambiguity here, it's not like we've got some conclusion on some of you things because the first one is the second one of us to epsilon bar is the second one of us to this general statement this general statement about similarities and correlation functions and the name, it's called the part of what's called the operator product expansion which studies the operator product expansion in detail on the right. I just want to introduce the name so that it works. Let me now see what I can deduce about the operator product expansion the similarities and the correlation functions and the operators approach each other. What can I say? More more concrete what delta o is for operators under the photogram we don't yet, for a general operator we don't know how it transforms we don't yet know how it transforms and what it transforms because any local operator under a translation changes by an end so before that let me do some formal integration so I'm going to do all these configurations with the top line every configuration I do with the top line can be repeated for the same I'm not going to do that so let's suppose that epsilon z where epsilon z is anything it has some data series expansion and del and epsilon by n factorial z to the power n no, there's no choice you see because here is the z bar that doesn't work that tells me that the current corresponding to epsilon has only a j z bar and what is that for so there's no choice I'm just reading off the epsilon and i k again epsilon and i k so I'm specializing in those epsilon and i have a j z bar I could look at any thing that have singularities but we'll deal with those later so I get the epsilon and i k and suppose d z with o and the following singularity structure it's sum over n a n by z to the power n what this equation means what this equation means we will understand that later in the next lecture see what this equation means I'll let you just say the words for the last time you say this is a statement about correlation functions now do you tell me about d and o but there may be other correlators in certain way about them if a remarkable property one appears here that if you have some operator in certain so far over 8 we'll see what it means and if there are two operators nearby then the dependence of this correlation function on the distance between these two operators is independent of what other operators you have in certain and therefore this becomes an absolute operator a statement that is true you don't have to say about which correlation function in any correlation function it's not that these two operators come near to each other but it will be true inside the property instead of inserting in this sense you can insert these policies this is a remarkable feature of quantum field theory it's called the operator product we'll understand that later in the next class there will be more conformal field theories but for now we're not trying to be so sophisticated let's just assume that in some way true you'll understand this let's combine these two statements let's combine these two statements to do and this statement to see what we can find so what we're going to do is the pole the pole comes when we hit it because we've got one of those a left over that delta o must be equal to del n epsilon by n factorial times 8 and if the operator product expansion between T and O has this general structure there is no consistency to this statement it must be that the change in the operator under an informal transformation by n epsilon is given by this formula there is a certain speciality system let's first take epsilon to be a constant so in that case all the other terms of epsilon except 0, that's 0 so under epsilon being a constant delta O must just be a 0 but any orbital operator under a translation shifts by itself to conclude that del O for any orbital operator a 0 is equal to 0 this statement is true for any operator which under an informal transformation parameterized by epsilon changes by delta now while this is a true general statement it's not that useful it's that we don't know how operators change under how they transform the transformation let's look at those informal transformations which we do understand we don't know how operators change let's look at simplest informal transformation then we have translation under the translation how does a local operator transform but it just translates so in an informal translation it changes by its derivative O at Z goes to O at Z plus delta Z it's a constant so it's just a trip under epsilon being a constant must be delta O must be a must be the derivative is when N is equal to 0 so it is del N of epsilon is 0 it's a constant that delta is del let's say in slightly more invariant terms what we found we found that when we take TZ and take any operator home there is a fault in this operator that's always true and the residue of that fault is universal it's a derivative operator because I was at work in the office so girl, you should know what I said general statement which is true of my problem which is in this fancy setting implementing the fact that we shift things if you're integrating over outer space okay, in theory so we know more than operators change under translations we know how they change under rotations because they transform in a certain spin so let's say we take every operator in the theory and find a basis of operators such that the operators have different spin we're working with the following theory we can also ask for the operators to be idle states under scale transformation now this is very intuitively plausible why I made it however, next time we will prove it in a true dimension problem we will figure out if you use the state operator map that's the isomorphic statement that in any other space you can always find a basis that is diagonal that is diagonal for the Hamiltonian okay, so let me say what I said the first statement about the theory the T with O the derivative of O would say true for any local operator now the next statement I'm going to make will not be true for any operator however, in this space of all operators it is possible to find a basis such that the statement I made will be true for every basis element as a statement that why not every state in the Hilbert space it's an eigenvector of the Hamiltonian so I will also remind the basis of states in the Hilbert space that is such that the basic factors are eigenstates what I'm going to say is that this is why the beautiful theory about leading with two dimensions would probably be the theories all these deep general statements would want to be theory that after 30 years of experience people have learned to trust but don't really know why it's true why are true if you think it's true or not you can just see why it's not true this is really a beautiful simple example and we'll move on to the next class see what I'm telling you I say you don't believe but we're going to the next class and let me just repeat that so what I want to say is let us suppose that you believe me for a moment that it's always possible to find a basis of operators that are eigen operators and and eigen operators understand the second one I think might be quite plausible but what is the coordinate what is the coordinate change that generates that generates speed the coordinate change that generates speed is rotations rotation is x dy minus y dx and we've already checked algebraically that this is the same thing as z dz minus z bar dz that is z dz minus z bar dz is very clear because what a rotation does is rotate z in one way and z bar in the other way every time you see a z you replace it by e to the power i alpha times z every time you see a z bar you replace it by e to the power minus i alpha times z bar and in the first class version of the statement is the operator z dz minus z bar dz so rotations are generated by z bar dz these are rotations first scale transformation taking every coordinate and uniformly scaling it what generates that something like that exactly, x me there will be one which is z dz plus z bar dz so let us deal with that basis of operators I think we will prove that it is possible to find a basis always which are eigen states under these two operations that is transformed homogeneously under these two kind of formal transformation now you have already seen by experience that is always there to deal with the analytic and anti-antiality thing so it is a deal with this and this we just take deal with the sum and influence scale transformation it is transformation of weight h bar and the transformation epsilon z equals z epsilon bar z bar epsilon bar is equal to h times o you change the coordinates inside plus you change by h times z so let us do that z again let us do it at zero delta o at zero is equal to this similarly delta o is equal to h bar this is the definition if you know what h and h bar is the split is simply h minus h bar and the weight of the overall scale transformation is h plus this is the physical interpretation of all so now the next question we are going to ask is what else can we say about the operator product expansion between the stress tensor and not an arbitrary operator but an arbitrary operator that is positive so last example you see we now know how it should be when epsilon is there we do it all at zero so the only element that is non-zero is the first set so that is what a1 is that is what a1 is so when the operator was the primary we conclude that d of z is equal to is equal to h o by z squared plus delta o by z plus delta plus h bar o z bar squared plus delta this is only true, it is not true of all operators only true of quasi-primary operators that are definitely h and h bar let's see any questions or comments what I am going to do is to work out what the operator product expansion is not between the stress tensor and any operator but the stress tensor and the stress tensor so everything that we said in general was true so we wanted to compute d z of d zero we would find this is two times d of z it's two because both weights of the stress tensor the weight on the scaling dimension scaling is two because if you take a stress tensor and you integrate it over space which is the energy which is dimension one the integral over space is dimension minus one so this thing was dimension two so stress tensor must have weight two or just spin two usually the famous reason that you know but let's just spin two objects symmetric two things this is z squared plus delta o by z similarly an equivalent statement for the end now what I want to do is see what we can deduce about plus that facts that you may not be able to believe but which we will provide in the next video well