 It's OK. Yeah, it's OK. It's OK. Yeah. OK. In terms of, can you put it on standby right now? Just a minute. OK. Let's do it. OK. Bye. So, do you have any more discussion on the honest programming computation of the string as matrix? In particular, we'll more carefully write the last table that we were interested in. What about, of course, in more detail, and then by the end of this class, we'll begin to study a symmetry of this action, namely, PLS. OK. So, let me quickly remind you what we started talking about in our class. Let me remind you that what we started talking about was that our limit of all of the moments which calculate string as matrix elements. OK. So, string as matrix elements, at least intuitively, is related to amplitudes involving long tubes going off to infinity somewhere in space, right? And going out to infinity a particular way, that is, the vibration on this tube, once you go far enough in each of these tubes, that is done with particular states of string, a particular string of this. OK. Now, inside, there even has many holes in the ground. OK. The number of holes would be related to the loop order of the relevant process, in terms of treatment of this. And then we use the state operator map to replace each of these long tubes by an operator in section. Which operator was the operator corresponding to the state by the state operator? OK. So, our final goal in the end was to take some compact treatment surface. OK. The simplest would be a sphere that we need. There could be some holes in it. It could be a ball or something. Two-handed. OK. And we compute the correlation function of operators. This is our goal at present. We make another video to do more. Calculate the equivalent of offshore amplitudes. If so, we get lucky. It wins the other. Basically, we are good. We are good to do more. But I'm the most responsible if I do. And we'll see why it becomes more difficult to do more. And that's because we understand the answers. OK, great. So, our goal was then to write it about the table and did the competition. Yeah. So, that's trying to be systematic. So, let's say that the number of insertions, that the number of insertions of operators has got it down. We calculate the I. The I will be the insertion. So, we are calculating an I point function. As Captain goes, I will be the I point function. And we have decided that on whatever genus surface we are working, we will choose some fiduciary metric, some class of fiduciary metrics, which have the probability that Bayer or defumonism plus Bayer transformation, any metric on that topology, can be related to one of the metrics in this class of metrics. You know, about Bayer, this structure. OK. But then the class of metrics we are dealing with will be labeled by some number of real parameters. These parameters we call TI. OK. And they have a name they call module I. And let's say that I is equal to 1 up to 2. Is that not that? As we see as we go on, that the number of these modules are equal. So that's a complex module I, so yes. OK. We also want to balance the fact that just asking that our metric, you know, it could be that there are some wireless between operators and transformations that don't that the standard form of the metric. It leaves the standard form of the metric in there. For instance, on the plane, we're really interested in the plane, all these conformal transformations which are combinations of wireless, if you want, wireless transformations and if you're an officer, leave the magic, the flat magic on the plane and that. OK. So I assure you that the number, while the number of conformal transformations is infinite, we will always be interested in compact treatment services. OK. Most conformal transformations blow up in infinity. So if you compactify the plane, most of these conformal transformations will not be about singular transformations. And understand what we're just going to tell you as we go on. But the one point about this that I wanted to show you about as we've seen in detail as we go on is that the number of these wireless between most of them transformations that aren't fixed by requiring that we stick to a particular form is always finite. And so let this number be 2 j. Let number be 2 c. Number of unclipsed. Since we've got... This will also always allow me... See, that's why I'm doing that. So since we've got 2 c unclipsed while in decomorphisms, just requiring the metric to be of a particular form has not completely changed. So what we decided was that we will, in addition, take the positions of C of the insertions. C of the operator insertions. So we will assume that m is greater than c. As we would see, it will turn out that anything would n less than the wireless seeding quality doesn't calculate anything for c. So the assumption will not be of zero restriction. OK. Now, since each operator insertion is an insertion at two points, it's really very great in a two-dimensional surface. OK. If we fix the positions of m operators at c-specific locations of the two-plane, to be at c-operators to be at specific locations of the two-plane, we fixed an addition of two c-specific locations. OK. So what we do is we use the un-fixed, if you want to do this, if you want to do this, that's why transformations. To put the positions of c-operators at specific points of the worksheet. OK. So we choose these points of insertion to be sigma 1, 0, sigma 2, 0, sigma c. Some given points, which could be anything. So we see the number of un-fixed, but each operator has a position in two-dimensional surface. So the sigma is a two-dimensional set of four things. OK. So these are the coordinates that we will fix anything. OK. So that's what our basic set. So then, what we want to calculate? What we wanted to calculate was the initial thing we started out wanting to calculate was path integral over all the tricks, path integral over all the fields of the, of a field theory, which is the x-muse. OK. Then exponential of minus the OYACO action associated with operator insertions. And you remember that these operator insertions we would always choose to be integrated over the worksheet. So what we wanted to do was to fix the cubes in space-time way to some specific point. But not that they started off at some particular point in the, some particular worksheet kind of transition. I mean, clearly that would be something that broke if you wanted to make it. It's nothing. What a cheat, people. It's nothing special, but a particular point in the worksheet. OK. So the insertions of the two sigma, let's say one, square root g of sigma one, that's v one of sigma one. It's some local operator, you know? And we discuss a little more about what these local operators can be. Is this clear? So this is the thing that we want to, we start off by taking action. Now, of course, the problem with this path integral is that there's this huge h integral. And there's more important problem with the path integral. That is, we're doing this integral over all, all metrics including the white factor. But as we saw in our initial discussion of the stroke theory, the thing that we physically require to do is to do the integral only over, only over the, you know, the white factor should be treated as metrics that are different, because different, by at least, different up to rescaling by a white factor should be treated as physically accurate. OK. So this path integral will end up calculating what we actually wanted to calculate. Only if the integral over the white factor is actually trivial. It's a symmetry, which is some overall factor which we can then throw out. Because the tree holds the metric, you know, the geomorphism as well as the wild tree looks as it gets. OK. And check in the end that whatever we do doesn't depend on what we do. OK. So that's what we do. So anyway, this is what we want to do. We want to calculate this partition, this correlation function. Now how do we do that? This correlation function is difficult to calculate because of the gauge. See? So we use the usual polypropagic in this context to simplify it. It says, well, insert identity to the path integral. Now insert identity in the following fashion. We say, well, consider the metric of your system and that's what we do. Subtract from it the fusion metric, GTI, transformed by the geomorphism of omega and then wild tree. This is one gauge friction. The gauge friction delta function we want to insert. There's another gauge fixing delta function we want to insert. If product over i is equal to 1 to c, delta of sigma i minus sigma i0, that fixes the insertions of the first c operators to be at points sigma 1, sigma 2, sigma 1, sigma 2, sigma 2, sigma 0 and so on. OK. So we want to put these two delta functions into our architect. But this was the problem technology we wanted to calculate. So we can't just put these delta functions in my hand. What we do instead is to say, well, put these two delta functions and then I take this and integrate both and this is, say, the diffeomorphism of the royal transformations at omega and sigma i. Only diffeomorphism of that and sigma i. Difffeomorphism of change points while transformation. We say, well, what we're going to do is to insert identity into the path integral. So as not to have changed what we, the quantity, we originally stuck on our cat. So, what we do is to put an integral over omega i and integral over 2 phi here and integral over t, over ti. This is an ordinary thing. Ti. And then we insert this quantity delta fp which can depend on g and on sigma i 0. We demand that this thing is equal. So we will define delta fp g sigma i 0 by this equation. And we saw last time, as you know, intuitively what's going on here is since, there is exactly, if we be careful to make sure, there's exactly one value of these diffeomorphisms and royal transformations for which all these delta functions simultaneously for some choice of p. We'll give us something, some jocopia associated with the delta function and we must have that jocopia for whatever it is. It is this delta x. Now, let's see, integral here is over omega phi at ti. Whereas this thing's a function of g and sigma i 0, this thing can be another variable of us. So, equivalently we could say that our definition is that delta fp inverse is equal to g omega d phi d ti times this delta function. Now, the g ti is the class of confusion. g is the real dynamical metric that entails this. Now, before we actually calculate delta fp, let's see what we get by getting the stuff and putting it into the example. We add, you know, we can write this as this times one and one is the same as this integral. So, you just, enhance the positive integral here to be a positive integral over the g x mu d omega d phi d ti and with all this other stuff. Now, all this other stuff includes delta functions. We take it very carefully to choose to be a linear engine. You know, g is one of the arguments of it. Sigma i, omega j is one of the sigma i's here. It's not the fiducian form. So, similarly this delta function here has sigma i as one of the arguments. In this positive integral here, we have a positive integral over g and we have an integral over sigma 1 to sigma, sigma n. These two delta functions just heat up the, heat up the positive integral over g and heat up the integral over d 2 sigma for i is equal to 1 to c. So, initially you had an operator in sessions, right? We had, I think we called it m operator. No, sorry, what was it? i operator in sessions, yes. m was the, I mean, model. Sorry, I said it wrong. The number of in sessions was i greater than equal to c. So, the number of in sessions is i. All of these inter-each operator in sessions come to the d 2 sigma and c of these integral are equal to delta function. And the fourth part of technology is heat up the positive integral. We can get this delta h p whatever it is. Okay? What would we have found? We would have found that the integral that we want to do is simply an integral over dx mu. When the integral, well, then we have exponential of minus s or plus log of delta h p and the exponent. And then we will how many c of these operate? Yeah, c of these in sessions that are fixed to be given about itself. Then the i is equal to 1 to c square root g tens of v at sigma i 0 and then the remaining operator in sessions continue to be integrated over on it. Okay? So, we have a product of j is equal to c plus 1 to i. Square root g to v is the scale. The key point here is that this is a much simpler part of the integral than this one. Because this part of the integral involves integration over matrix. You know, such as the part of the integral to dimension of quantum gravity. It's a hard thing to do as various gauges can do. This part of the integral, as we will see, but at least the part of the integral is just the part of the integral of an ordinary component. Okay? And we have a given metric. Okay? In S-Pol, we just put in the metric. We put in the metric g omega of t i. We put in the metric. And we will also continue to have an integral over the module. What we have to do, some part of the integral for a particular model will be there in the given metric. The metric will be some g related by t i. Then in the end, we have to sum over these parts of the integral for all metrics in other interesting things. We'll have to do the integral over t. We'll have to do the integral over some correlation function in some ordinary fixed metric component. Okay? In a good case, all the things that we have to calculate is the thing that we will have to calculate is a correlation function of some operators which are given positions and other operators which are integrated into the worksheet. As we will see, as we go along, we will be able to prove that the final answer will be independent of what positions we choose to insert the operators in, which operators we choose to insert. Now, we'll check out my next question. Okay? But at this end, all that remains is for us is to compute delta i of t or even delta i. Let's look at it. Okay? It's just kidding. Any questions or comments before we turn to the application of delta i? So now, let's go ahead and compute delta i. So, let me clear it. Now, we strike not by that, but one of them has t, which is function of g and sigma 0 is equal to equal to d g i omega d of phi in delta of g minus g d i omega i d phi d phi d phi in delta of sigma i, sigma i, 0. And we, we want to show is that this delta i at 3 does not, is a gauge invariant function of g and sigma i. Of g and sigma i. Of g and sigma i. g and sigma i. By which I mean that delta i, if you, if you calculate it for one set of g and sigma i, and you calculate it for another set of g and sigma i that are related to each other by, by, if you want to understand why, why transformations, they are the same. Okay. Now the proofs that, that we did is basically very formal because it relies on operations that are correct in some, for finite dimension in textbooks, in textbooks of lead group manifolds, but may or may not be correct for this huge amplitude in the textbooks that we had. Yeah. So it should be taken more as a motivation than proof. Now in the end the attitude that allowed people to take is that the final result of this answer is the starting point of the definitions. Okay. So but, but you will never be able to reach the result without having some reason. So let's, let's understand you just, however realistically what the, you know, how, how these have been described. So, the argument says that we don't require these, you see, we haven't really specified what we mean by the omega. What does it mean? What is the major of the space of difference? We haven't specified what we mean by the delta part of that precisely. Okay. But what we would require is that the, the major over, over the space of different organisms has a property that is left and right in there. That is an ingredient under, you know, if you take one different organism and compounded by, you state different organisms, compounded by a fixed different organism. So that's a variable of change. And then the Jacobian of this variable changes like that. Okay. So what's the equivalent of this problem? Okay. What's the equivalent of this for a fixed group, for a, for a lead group? For a lead group, you know, suppose you are doing an integral over all group elements, which is all the unitary, you know, all the unitary matrices. Okay. It's possible it's probably wasn't for a lead group to define, it was for the garbage, which is a, which is an integral over the group manifold. So with a measure that is such that if you take Ukraine and say Ukraine is U-times and not, then d of U-times is equal to d. For the group, yes, when you take it over a circle, that this, this whole measure is simply the usual measure over the circle, d theta. And the actual of the group is translation. D theta is a translation in that, in measure. It's always possible for a finite group is done by Mr. Harbour. This is why it's called the Harbour. Okay. So we assume that in these much bigger groups of symmetry transformations there exists such a measure. Okay. And we assume that this delta function that, that, well, that there exists a delta function that respects the symmetry. By which I mean that we have delta function delta of A minus B and you symmetry transform both A and B by some fixed element. Okay. Then the delta function is invariant under the subtraction. Again, the average you keep in mind for the U1 group is delta of X minus Y is the same thing as delta of X plus A minus delta of minus Y to C. Okay. Again, for the groups it's possible to find such a function always. We assume that it's possible to do it. These things have to move forward. Okay. Now once we have we have these two statements the gauge and variance of delta of B is obvious. It's obvious because suppose we wanted to calculate what, you know, so what do we want to check? We want to check that if we replace G is equal to G prime times omega times 5 absolutely with the same i then we get the same answer. You see by the property of delta function if we make a replacement and then to constant diffeomorphism from both entries in B in the delta function the delta function is empty. So let's do it. What that ends up doing is taking this quantity and multiplying by inverse diffeomorphism and inverse inverse i transformation. But then using variance of the measure to change dummy variables in the integral of the diffeomorphisms to absorb the inverse diffeomorphism and inverse i transformation into and take a look at redefining diffeomorphism and redefine i transformation. So the end product of those two operations is the definition for the fabric of determinant of G prime. Is this clear everyone? So so we demonstrated using this formal requirement of the measure delta function of the fabric of determinant is invariant is gauge invariant. Okay because it's gauge invariant we can compute it at any particularly useful point of the gauge orbit. And what's a useful point? Well we know that there is early age transformation that sets G to be this fiducian metric for some value of T which is let's call that T i T i 1 and sets this sigma i T i 0 So it sets sum to 5 0 it sets G and sigma i we put this G of T i 0 and sigma i 1 over delta x p G of T i 0 and sigma i which we compute before is these two our relation we demonstrated that delta x p is patient. Is this clear? Yes So you said that we were let's say transforming G G omega and then sum it and then the heart ratio. Wait so there are two steps first suppose G was equated G prime let's go into 5 just for conversational simplicity let's say G was equal to G prime times gauge transform by some gauge transformation omega okay let's not call it so let's call it omega delta then first we use the property of the delta function that delta of we use delta of a omega minus b omega is equal to delta of a minus b so what we do is how we use so what we have here is delta of G omega minus G minus G T i and we have 3 to the power omega I'm just systematically ignoring the fights next one what we now we'll take this omega and write it as omega is equal to omega prime times omega G what we do is it's a group we multiply by omega okay so what is this this is the delta function G prime omega minus G T i omega prime omega now that's property of the delta functions we cancel out this so if I say up here what we have to see integral over the omega what we have here is omega prime but that doesn't matter because it's a between omega and omega prime so we change the integral variable of integral to omega prime and now this now the definition of the value for what we have then for the metric G prime and that this it's possible to do it's an it's an assumption to it's an assumption it's an assumption if once again if we were doing this for just finitely groups the integral over all matrices belonging to a group then that in such cases not in such you can prove these but you know when you're doing these things yeah let's just say it includes it's an assumption certainly mathematicians want to do that yeah that is an assumption so we decided to calculate one over delta Fp of the metric in the form of special metric that our metric is gauge equivalent to and the points at which we want to fix our special insertions by construction we know that it's always possible to use a gauge transformation to put the metric into this class and to put the points in these points so we can do that now let's do the gauge you see what we're going to do is do this integral to find one over delta so this is equal to d omega d phi delta omega e to the power of 5 and then another delta of sigma i0 minus sigma i0 use one more assumption again it's also true for these integral over finite groups about the nature of the measure of this part you see we can't do this calculation unless we know something about what we need about the integral because then what happens is despite my integral you don't say anything about the integral you don't know you know number of groups is true that is if in the table we're going to find the integral of non-linear factor reduces simply to an integral over the basis elements of the first you take let's say you had sq2 and you take the group n would be pi theta theta a t that's e to the power of 3 I don't know either small or region around identity this measure is simply d3 theta so this is you know we don't get about proportionality we just make sure that it's not d3 theta times theta square but this is true a few we are analyzing the state is true of the part of the network you'll see how it is you're just expanding this and making it less yes but of course but it uses the property of what dU was you see what we're doing is taking the our measure and expanding it so if this measure was something very odd then this would go great actually it's almost guaranteed to be true for the following years that in the lea algebra you know in an infinite as a network you know group multiplication is simply translation you have to go to the next order to see the difference between these two so a gauge a group invariant measure has a translational invariant pressure in the lea algebra and there's only one translational invariant right now by our first assumptions this is not a news in order to calculate delta activity which is some Jacobian we only need to know what the integral looks like to first order in the neighborhood of where the delta function kicks because it's a Jacobian of first day so we have to get the measure right to first order in the neighborhood of identity because we do things so that the clicking of the delta function happens at node of your in the same case so we only need to know what the measure looks like in the neighborhood of the lea algebra if your if your organisms are generated by kidney vectors kidney vectors what's the lea algebra for the transformation for some friction that's fine that's simple okay so the next important step is that for the concentration we can take this v omega and defy and replace that time integral over over eight of these this one by integral over this is just an integral over five okay so this becomes d2v alpha the set of all vector is generally the set of all infimorphisms and this becomes dv over five and this becomes dv over five and we still have dv over five okay, great great so that's that's the dimension then the next step what about the delta functions now because we know how infinite infimorphisms act on a measure how infinite has no formal transformation factor in a measure okay, so what goes into the argument of this action well there's a part we should we should get when we get the identity so that's just cancelling this so what what changes is you have delta of the change of the metric because of an infinite infimorphism which is del alpha v beta plus del beta v alpha the leader of that of that okay plus plus two five times gf g what g factor in this t itself definitely take it over t so this clicks at ti you can see now so we should add plus del g alpha beta by del ti times del and so now we have correctly obtained what this what the argument of this delta function is in the data order of in omega finite d space around the key was this guy so once again we know how diffulorphism acts on a quadrant this action is the vector filter that generates the difference okay so when omega identity is cancelling that the remaining thing is simply v alpha of sigma i naught product i into three steps getting from but once we got this now we because now we were in our formula every term on the right hand side every term made six seconds so now we can actually it's not like what we after is some explicit what we want to do is take this answer process it put it into a form that is used for moving calculations like on the straight so the form that we put it in is the form is to use the famous identity from delta function that tells you that delta of x is integral over k into the pi kx up to two pi's that are absolutely no elements at the level of which we need this calculation drop it all constant factors here's a delta function in the space of matrix okay so we use this identity twice once here once here so so the first letter we use this identity is we have one so the delta meaning is equal to 0 now we have many elements like integral v v alpha d phi dti and then we have let's call it v alpha beta this will be the the electron for this and then we have d and that's why that's something else r r let's call it r okay r alpha sigma naught the Lagrangian multiplies this delta function and this i is i is equal to 1 to c and then we have an exponential of n's Lagrangian multiplies there is this stuff so v alpha beta times this stuff so whatever it is is in this delta function plus i times times this so v alpha of your mind that's not call this function it's just some discord and we like it with an integral expression to the next function actually one more thing that we have to be careful about here you know we want everything in the game to be a few more for clean so when I write this it's of course an integral here and I'm going to square root g I'm going to square root g so that my integration there it will be now charged also like an ordinary case so it's like if you're more but clean very much about it if we were able to carefully define our delta functions and so on this will automatically come out because everything will keep taking trouble in the game okay great so now to get this line you see at this point let's do some of these integrals let's do those I actually just want to do one of the integrals that we'll make guys the integral that I want to do is the integral over 5 you see 5 couples very simply it just multiplies this v alpha beta by g alpha beta and you do the integral over 5 so what does 5 couples do? the trace of v alpha and the integral over 5 that will be the ground to apply 5 sets the trace of g sets the trace gives you delta function that tells you that the trace of v alpha beta does it is the step take care to ensure that we only do pass the integral over traceless v alpha so one over delta fb is equal to integral over d v alpha node d d i d r i an exponential amount that's right I'll give you that explicitly exponential of i times of i times integral square root g v alpha beta since b is traceless we don't have to worry about this g alpha beta done times delta alpha v beta that will break it down into alpha plus i times this r alpha v i of the grind out and the v where the path integral v here is now only over traceless only over traceless it's a useful technical expression for one over delta fb however the thing that we want to insert into the path integral is not one over delta fb it's simply a quadratic integral an integral that's quadratic in its arguments we know that an integral that's quadratic in its arguments now if we want the formula for the determinant rather than the inverse determinant if we want formula for the determinant from the inverse from the inverse determinant all you have to do is write down the same path integral but just make sure that every key in the path integral commutes anti commutes now delta fb now delta fb now delta fb anti commutes so I just rewrite the names of the variables to make them more standard actually b is already already so that's dc alpha dti dmi dv dv alpha beta that's the explanation dpi square root dv alpha beta del alpha c beta del beta c alpha del alpha v alpha c9 every field in the path integral now is anti commuting now the path that's the determinant I should have said it's more clearly this so we should have had that good so that goes in this in this thing so we have thank you that's important last del g alpha beta by del ti delta dmi and here this integral now is an integral over d of delta right which is itself replaced by an anti commuting some let's call this delta dmi and some zti see because it's part of the determinant something quadratic in the unit is this clear so now we have to be this is what delta is it's possible to simplify its expression you look for the first thing we do is to do the path integral over zti and okay this is very simple because because r and zi appear so simply at the path no the modularity itself is a quadratic integral and when we get right to the final expression you know the integral that we do includes an integral over d right this is just a contribution of the modularity of the value of probability it's the contribution of the nature of it so in the s matrix there is no integral over modularity there is an integral you see because you remember expression that was identity had integral over modularity as well I see okay so the delta function is the integral only is the integral over symmetric and the integral over quarts but it then theta of the integral over over modularity and in the s matrix we also have an integral over modularity because there are also we have to integrate with the symmetry and from there we have an integral over modularity because do you remember we replaced identity with some integral that integral was an integral over dT d omega d phi okay now the d omega and d phi integrals have just come out they are overall integrals but the dT integrals are there and the phi integrals are there it was because the delta functions and the major answer one was because we proved the age of the aliens or you see what we found in the alien was that it did not depend on omega and phi but it does depend on this dI okay whether it depends on phi or not on the satellite navigation because the procedure been completely consistent had we been doing things correctly but it does depend on those things but it does depend on you so the final integral omega and phi is just an integral over one see because we have that integral left but those things don't enter the answer however dI does enter the answer because the metric is replaced by g of dI in this one the metric that we used for integrating this data function is basically the function of the dI also yes and then we have to do the calculation in one of these spectacles and in the end integrate the answer over all such factors all such classes all of them exactly okay now let's do the integral over dI and dI the integral over dI and dI is very simple you see because these are anti-community coordinates so we take this exponential and we expand it the terms that are of order you know this bigger expansion of the exponential ends at first order this is first regard something like e to the power a b where a and b are most accurate okay let's just leave for the 1 plus a b plus a b square by 2 plus blah blah so this is 0 if we take a to b after a minus sign and then we get a squared which is c actually it can be replaced by well this integral over 1 which you see is 0 remember that the integral over of a constant over an anti-community number is c okay then this integral over half times c the integral is not the r but it is the c and here then the integral over c c times the key point that I am using here is that there are no derivatives of r or derivatives of c times that end of the action if there are derivatives that's just a difficult couple problem that's why I come to the same thing here since there are no derivatives you can just use the integral once and for all okay so what this gives you an explicit insertion of c of alpha at sigma is 0 and this gives you an explicit insertion of whatever is left so so we have delta fp as you put in table dc alpha d zeta i d b alpha beta then exponential of i times square root g b alpha beta delta alpha dc beta del beta c alpha then this is gone away what we have to say is product i is equal to 1 to c c alpha of sigma i naught and is equal to 1 to m that I want to modify and I will write some notation the notation will be b alpha beta del g alpha beta by del g i this is a notation that is often used in Gochiski it denotes the inner product between two tensors of the same rank and what it means is take this tensor contract from this tensor multiplied by the metric integrator b alpha beta then this thing times square root g integrator this is our final expression and remember the integral b is only over traceless traceless symmetric matrix it is symmetric because the non-multiply matrix is symmetric it is symmetric okay it looks simple it is even simpler than you think it is even simpler than the thing for the following let us now choose I assure you this is already possible that every name answer of this let us now choose our conformal class of matrix to be locally always conformally flat a special class let's choose that so that g alpha beta of g i takes the form g alpha beta of g i takes the form into the power then at least in some coordinates you know patchwise in the coordinates system in the simplest case of the torus of the sphere that you did do it globally otherwise you might have to do this patchwise better but this pie is also better so we are choosing our its global issues that will be done you might say that all of these things are equivalent to each other some of them will be equivalent to each other under bi-transformations but there might be global obstructions to why we escape all of them to our identity as we will discuss in detail as we look at it there is a specific remuneration okay but now or at the moment we are just for interest in a local analysis okay so locally the matrix is conformally flat in this situation with such a choice okay with such a choice of of matrix this actually becomes even simpler than you might have thought the reason for that and it becomes simpler once once you choose the index structure so what we are interested in is integral square root g b alpha beta del alpha c beta plus del beta c alpha now what can we say about that the first thing is that we are only interested graciously but in this conformal coordinate system tracelessness is very simple right it simply means that the only non-zero components of b now we use the same z the z is in our coordinates the only non-zero components of b are bz and z and bz plus z as we see in the details okay so let's work out what this is in coordinates this is integral square root g bz z okay integral square root g bz z del z c z what's up actually 2 because these 2 things are the same plus something else with z with z in space back to 2 which okay now let's work this out what is del z del z is equal to gz z bar times del z bar because the metric has to go del z okay therefore therefore we can rewrite this as integral square root g gz z bar times let's work out a real first zw bx similar bz z del z bar of c z part of the polynomial metric metric of this form this quantity is again inverse metric if z z part of the polynomial metric cancels the determinant first thing is simply bz z times del bar del bar c what can we say about del bar c okay so what is del bar c z of course this quantity is equal to del bar c z plus what well z z bar alpha of cf what is that and now what can we say about this constant see because when we lower this this will become a z bar this is proportionate to that thing it's proportionate to gz bar alpha z bar minus z plus g z bar z bar alpha plus gz bar alpha z bar this is cap i lower this this becomes a z bar and then i have to complicate this this times this metric derivative of respect to that that's this one plus this times this metric that goes to that that's this one plus this times this metric minus this times this metric derivative of respect to that but these two are identical so they cancel cancer. And this is zero because G says Z bars. Z bars, Z bars, Z bars, Z bars. So you only want zero component of G, that's a G says Z, Z says Z bars. So what do you need? And you could make the same amount of parts like you could Z out, like you could Z bars. Yeah, what else can I explain to you all about this? And therefore the crystallism, this is just zero. So this thing is simply D through sigma, Bzz, Zz bar. So in a locally conformally flat coordinate system, this, this, the polypropagodermity, if the polypropagodermity is incredibly simple, it's just, just the socket. Now, you see this action, as we've seen, you see the action is independent of this part, which in particular tells us that it's widely maintained. It's widely provided in regards to C, Cz and Bzz bar and wide scales. Okay, if under a wide transformation, this is neutral and this is neutral, we've already seen that all other factors in the wide factor cancel out. So conformal transformations are special. Conformal transformations are those, those are the transformations that leave, you know, metrics in these forms unchanged. Now conformal transformations made up of a wide transformation plus a difficult one. We've seen that B and C have way to zero on the wide transformation. So a conforming wave is given completely like that if you're opposite. But the if you're opposite, why does it make an exception? A lower Z index has the same if you're opposite weight as derivative, therefore weight one. So this quantity, okay, is weight two, is weight minus, that's scary. This is the BC conformal theory. The value was the vector, the weight of B to be minus two and the weight of C to be one, B to be two and C to be minus one. It's the special value that we worked out, the state of greater man. Okay, that was not right, that's right. V was lambda, weight lambda and C was weight one, right? Exactly, okay, so weight lambda, weight one, right, lambda and lambda equals two, this matches. So if we chose this fiducian metric to be of this local conformal form, four, then the action of the final part of the network can have to do for string theory is very nice. Okay, so now I'm clearing. Maybe I have to do it with the bar and C bar. But we've chosen this to be conformal class, simple. Okay, my first, I'm going to be getting a V del bar, del bar C plus V del bar, del C, we also have the integral of mod unite and we had, we had a V z z z del, it was this square root g ten, this state product of all i, where we should, where V z z is the same thing as B to the left, V z bar z, so it's B to the left, and where this is the same thing as B z z, everything wrong. Okay, it was so crucial that, you know, to get this nice, simple form, it was crucial that we could B to have lower indices and C to have upper indices. Nothing good would have happened to B, because in the attitude of the same answer that I feel, you see, covariance derivatives wouldn't have become ordinary derivatives at C and lower indices. The whole thing wouldn't have been a violent meditate, by itself, had, yeah, so everything would, would have got messed up, so this is really convenient choice there, so we much about B to have lower indices and C to have lower indices. Okay, and then, then that is the product I think would want to see, C of sigma i naught times the insertion of the vertex operatives, square root g ten times V of sigma i naught, V i, and then the remaining vertex operatives, j is equal to 1 to C integral of sigma square root g. Yeah, you got it. I mean, this was there, this is the path we were going to get, right? Yes. Now, there, there, if you remember, we converted this integral of T's into xi's and then read the integration. Right. Now, what I have a little bit confused about is, there, this was multiplied by 1 xi and therefore since you did, did the integral, did you got the coefficient that is, which, each of the, yeah. So, why can't, this g's are in general functions of D, right? Right. Then, derivative of this with respect to T i's will give you in general some function of T i's. Right, right, right. And, when you, when you are replacing those T i's in that definition by xi's, then you should replace all those into xi's, right? No, no, no. You, you see, with the, the, the thing that got converted with the integral over, over xi's was the integral over delta i's. Remember, that's just been linearizing our software of delta, for the Jacobian of a delta function. The fact that the things that enter here are the derivative of g z, z, g z, z bar with respect to T i's tells, tells you that you couldn't get something interesting, non-trivial there. Well, you can, you can, you can, you can get enough, non-zero answer. Only if the complex structure of metrics with different module i are different. You know, the coordinates that you have to choose, the coordinates that you have to choose in order or at different values of T i, you put it in this nice and formal form, can't be conformally related to each other. Otherwise, these things would just be zero. Come this complex, free definition, as these things would be zero. So, in a way, the integral over T i's is like the integral over complex structures of our network. And we will see that, you know, all these things are nice, particularly, you know, of nice mathematics. Okay, so this is now the final answer, in our final answer. One last step, sometimes we need to put a minus here into plus i. It doesn't matter at all because quadratic integral is a term. You put the i, you get some extra actors in the determinant, we're throwing away complex, you know, constant factors. Okay, and it's just another way of saying it is that you could clearly define c to i c. And when we do this for monolithic takers, you have to worry about whether something converges or not. But for monolithic takers, there's no issue. Yeah, so this is the final problem we work in. Okay, so this is a problem, four problems, let's see. So this is our final answer, our final answer for, you know, our final starting expression, the basic expression we use again and again in the game of interest, you know. We enhance it when we, when we deal with the superstructure. The starting expression we use again and again in the game, for the scattering amplitude for states. Okay, so just let's look at it once more for you. See, the answer has integral over the x field and integral over the bc posts. Okay, and it has an integral over bc. It also has some insertions. Some of these insertions are integrated over all of the machine. Fixed to special points. Those of us fixing fixed to special points also have a factor of c associated. The answer is given by the fact we have this insertion. As we will show, this insertion implies that the integral over t is invariant under co-ordinary definitions of the space formula. So you integrate over modular with a particular major on-space modular given by that thing there. And, you know, this is a completely well-defined thing. Over the next few lectures, what we're going to be doing is to study this formula in various ways. Okay, we will try to understand various consistency properties of the space formula. We will try to see that it has the properties one expects of a consistent S matrix. Okay, first we have to lower our properties like it doesn't make sense under co-ordinary definitions. We will also check that it's invariant under an interesting symmetry of this pathontation because b r is just a symmetry. And we will check many things about it. We will also, in many examples, guide it just to get a feel for how these things go. Okay, so a lot of the next few lectures, not a lot of the next five or six lectures, will be devoted to studying this formula. And various aspects. Okay, so it's very important formula is a sort of central formula. It's a very beautiful and simple formula at the end. Okay, so we're going to start by studying this formula and soon before that we have a question. Now, let's first ask, let's look at various consistency aspects. The first thing we want to do is to ask this troubling question of why you can use it. You see, what we gave us, we started with this number order action in which, which was equivalent to the Boolean factor provided, the y factor wasn't an issue. So we decided that the Polyakov action with number, with y factor gauged was the system we wanted to study. And we went and implemented our procedure. But now it's time to come back and check whether our answer affects our assumptions. So answer was that we were gauging y invariance. You can only gauge y invariance if the positive integral has y invariance as a symmetry. Okay, so let's check. Now you might ask, how do we check now? Because we've already assumed that it was, you know, it was a gauge invariance and done all these manipulations, so how do we check? Well, we do it the way that we usually do engagement areas. That is, if we're done everything right, then if we choose different ways of fixing gauges in the same answer. We have chosen a particular way of fixing gauges, but it's another broad particular way that we require that our metrics be conformal, right? This gauge choice is invariant. I'm going to go into these things. You choose one metric, I'll choose a well-defined function over all of space, why I always get my metric with respect to that. And I'll ask if the formula that you get and I get are the same. I'll ask but what is the structure of, you know, we've got various insertions. But apart from the insertions, what is the structure of the path of the integral? We do a integral of the integral over an informal field theory on a given background metric. So we are asking, where is it true? Is it true? And if so, where is it true, that the path integral of this conformal field theory over different background metrics that are related to each other by wireless gauges? To choose the same answer. Does anybody have any answer? Where is that key? Which is ever true, so far. Well, one way to answer this question would be to make an infinitesimal biodegradable scaling and check that this path integral doesn't change. What insertion in the path integral is an infinitesimal biodegradable scale? What particular property? The trace of the stress tensor. We've used this before. We're asking, this path integral is, in every correlation function, which involves at least one copy of the trace of the stress tensor function. But we have a formula for trace of the stress tensor. Now, put in the formula for the trace of the stress tensor, we're adding the formula here. It's minus C by 12 times the curvature of the boundary. The curvature of the boundary doesn't get you to be zero. So the only way this thing can be zero, the only way this thing can be zero. Either all the quantities that we're interested in calculating are zero, then the curvature of the boundary will also be zero. Or it must be true that C is zero. Do you see? So none of this makes any sense. None of this makes any sense unless the conformity of theory we're dealing with is zero. If it didn't have zero in the central church, different ways of doing the j-fixing would give you different arm sets. What is the total central church of the whole of the tenor? The total central church of the whole of the tenor. Because when we change the dimension, we change it. No! Okay. So let's see. What is the central church of the whole of the tenor? Well, we know the central church. Remember, everything is new in this case. We know the central church of the XP, those are each one. So if we were to get a factor of D from the x-fixing. But if you remember, for the BC system, we calculated the central church, the father and the daughter equals two and the father is 26. So C is equal. So this whole procedure is consistent if and only if. Do you see what's going on? Basically, what we did was to treat why are we scaling as an invariant. Because an invariant is a classical theory. But why are these scaling? And really, invariance is a quantum theory. Because quantum theories need a lot of defined sense of it. And you rescale the lens gauge chip. Even though you define the theory to be nice, conformally, on an arbitrary curved metaphor, it's not easy for this thing to be wild. You can dream on a whole procedure a little differently. What we can do is gauge-fix only the cure of this. Left the integral over phi, the scaling factor in the world sheet, explicit. And let the ideal answer, in terms of integral over phi as well. And the action for phi then would have turned out to be the value of the path integral of string theory, as function of phi. This is something we at least discussed computing. We didn't compute it in class, but here is one of the exercises. At least told you that the answer is del phi, del phi. Do you remember this? The part of the equation of the change in the action, the change in the path integral, is curvature under the variation of phi. And then to treat that as function of the dimension equation and integral. So it's possible to do. If you guys haven't done this, this is an exercise I want you to do. Okay? And so there you go, you are a god. How do we jump? How do we delve a little differently? I'm not going to do this. How do we delve a little differently? What people have got is path integral over x and phi with the longitudinal action of phi. And see not in, you know, the critical direction. So phi would have become another physical field, like the x field. Except that it would have been different from the physical field. Because this, as we will see, is the way I've written this. It looks like it's got the same actions. There's also a card finder in this theory, makes it different from the other physical fields of the dimension. So we would have ended up perhaps with a consistent strength theory, but not a relative theory. Where we would have got more degrees of stream fluctuation that we started out thinking we would have had. Flatuations were somehow in these phi directions in addition to the transverse. In addition to the transverse, they're in those ways. Now that, people don't have a greater point in this course when we talk about activation. To theory of this slide. And this introductory point in the post, they're not interested in such complicated things. We want to study skin theory, propagating under the X and B, and that sort of things. And if you want to know that, phi has to disappear as a nuclear field. You take an over phi to be trivial, just a constant between, and just track the bottom. So at this first class, that's what we're asking for. And that works only when you do it. It's not trivial that it works at all. And there's any value of D, any integer value of D has to be D equals square root of 3. Okay, it makes no sense. It's not trivial that it works by any integer value at all. Of course, it's a big integer value. We'll come back to that. What would the connection with real world be like? Okay, so that's the first thing. So this calculation gives us a second pass on the critical dimensions. Okay, any questions about it? Now the next question we're going to ask, the next question we're going to ask is, what about the operator insertions? You know, what operator does not be allowed to insert in here? Such that we get sensible answers. Okay, now the answer to this question is given by raising it very similar to the answer to the question, and it says D equals 20 cents. You say, the final answer must have been, for any illegal question, cannot depend on the choice of fiduciary metric, cannot change under why are the rescales of our gave choice of the fiduciary metric? Okay, there was one place in which it could have entered in this basic public action. This is in which it could have entered, namely the insertions. Okay, now why are the insertions? So the insertions here, okay, the insertions here are d z d z bar, okay, times square root of g, so that's e to the power 5, 5 times v. This quantity, this, in order for this to be well defined, you know, insertion, and insertion that doesn't depend on the file that goes into the distribution metric, it better be that this v transforms under 5 in such a way as to cancel this explicitly to the bottom 5 dependence of the sketch. Okay, now a necessary condition for this, a necessary