 Equation 6.5.6. I think in the last two lines, the s and t should be in the way. I think that's what I call that. I haven't gone to check this website, but maybe I called it. But I gave it. OK. Now, let me read this. I checked the s and t. 1, 2, 3, 2, 4. OK. And 1, 4, 2, 3. OK. And the others. OK. So, whatever the shift that we get in the typo, there's a difference. OK. So, very good. So, just to remind you, we will discuss the legacy on my afternoon class. Now, we discussed how we had budget operators in certain areas of different orders. And given the particular order, if a particular order is associated with a digital search, will students, by the way, students, right? Who is associated with the digital search? But what went in here, whether it was sb or su or u? Now, I don't know. This is selected in 180. T. I'm going to vary to change x to the 1 and 6. OK. So, which variables do we have? Basically, adjoining momentum is 4 and 1 and 4 and 2. Now, you know, you might wonder, what's special about the momentum 4? Suppose we looked at this diagram and said, let me give you three years of the momentum 2 expression. You can do that and you get the same answer. Because 2, 4, will be associated with 1, 3, and sb. And 2, 3, will be associated with 1, 4. That's it. See, you know, we chose to have 4 as the adjoining momentum, but we couldn't do anything. Anything else? OK. There's another thing that I didn't describe much in detail later. We couldn't have enough time. We just, you know, the way this blindly-using comparison would get this, would involve two integral of a white. An integral of minus infinity to the digital infinity. One in which 1, 2, 3 are over there and the left of the board. And left of the board, one in which 1, 2, 3 are over there and the right. If we get all these three regions, we might think they don't look quite like the same integral. And one of them is an integral from minus infinity to zero. Yeah, it's an integral from 1 to infinity. So how did you, how did all those integrals become the same? Well, they had to, of course. Because that's the symmetry. But if you want to see it explicitly, you can just do a change. You can convince yourself by changing your variables, that these integrals are the same. For instance, the integral from 1 to infinity will become the integral from 0 to 1 and the value of a change takes those two away. And more and more important, they're going to change the dimensions. Okay? So it had to be me, but you can explicitly check that it is. It says, okay. I suppose we're happy with this formula and now we want to understand this system. So let's go on. Suppose if you want to understand what this ASC means. Now, ASC is very simple. It's a simple thing to understand completely. Once you use this classic formula, the beta, the beta of A3, which is, by definition, equals to gamma of A plus B. Now, gamma of A and gamma of B divided by gamma of A. It is given by the integral formula by y to the power b minus 1. 1 minus y to the power b. Well, all of you have seen this formula before. All of you have already seen it. But it's very easy. What is that x? Let me use it. You can imagine it all the way. The best integral we have is of this formula. Once we identify A with minus x minus 1 and b with minus b. Now, ASC minus 1 minus s gamma of minus 1 minus b over gamma of minus 2 minus s. Now remember that s plus b plus u was equal to minus 4. Now, this minus s minus b minus 2 can equally be written as u plus b. These are the various different ways of doing this formula. What I'm trying to understand is, in fact, what I'm trying to understand is, what the CG-dialysis are approximately. So we get this search. Just immediately from the structure of this answer, see it's a product of the other functions. Now, gamma functions, the singularities of gamma functions, are four of those. Add zero days of values to the good of the argument. Immediately from the structure of the formula, we see that you see where they get fours. Where are they going to get fours? So let's look at the very less. Where are they going to get fours? Where are they going to get fours? Where? I need to read it to one of you. We get fours with minus 1 minus s is equal to negative n. n is equal to 0 over 1. Which means that s is equal to n minus 1 for non-negative. Particles are like this. When you get fours, you get this whole three letters. You get fours with the collision introduced, and you can make it possible to propagate and you can do the other part with the air. You get fours when you sit on the mass of this one. The fact that you can get fours of these values of s tells you what particles can be produced in strictly like the lattice in that case. They're particles that have masses that have s z to n minus 1. Remember how in our convention, we absorbed an alpha prime with s. So if we restore an energy with s z equal to 1 by alpha prime over minus 1 plus n. So this is m squared of the particles that can be produced. And this is exactly the same mass formula we had for this case. Okay? Minus 1 will be the vacuum. n equals 1 to the mass of s over 1. And higher n will be the mass. Okay? So first, let's just see. The particles that can be produced from the scattering of the particles are the same masses as the states that are present. Analyze the scale a little bit. Now that we've understood this, we know that there's only one source for the mass of the energy at, for instance, s equal to 0. We need to make the particle not be attacking. For that, a consistency check and an opportunity to clear up some of the readings. The consistency check follows because we know that the three points catching are free and attacking. That means that we can build that. So the structure of this inequality must be exactly the form of two to three point vertices. Okay? Times the points of the vacuum, which we all understand. Okay? And the opportunity we have is that there was some, some things that we let the videos in the scattering rule of calculations. Namely, what factor we put in these two readings as we put, as we insert them into the thing. The other things in the scattering analysis, we'll do a clear back up. As follows. You remember the other side of your stream and the other side. We'll just do another thing. Okay? So first let us let's come in the residue of the city that we want to take. Let's look at this one. Okay? I said s is equal to minus 1. So s is equal to minus 1. This kind of idea becomes minus 1, minus 2. Okay? And, that is this. Gamma of minus 1, minus s is equal to the limit that s goes to minus. x as x goes to 0 is 1 over x plus correct. Now, you see that, you see that from the formula that gamma of 1, 1 plus x is equal to x times gamma of x. But we all know gamma of 1 plus x to the limit x goes to 0 is equal to 1. So that is the gamma of x to the limit x goes to 0 is 1 over x. This thing, you know, on the side is the same thing as s plus 1. Okay? It's 1 over the limit, but it goes to 0. 1 over the output. I'll put this in the side. I'll point to 1. Because the residue of this form is just 1. At least it is less than 1. It's just here. It goes to 1. But it's here. I'm going to place t over u into the same thing. So a, s, t, and a, s, u both have a pole. I'm going to place t over 1 at s equals 1. At s equals 1. Okay? So the city of architecture, where I t equals minus 1, is what? Now we're going to be ahead of me. I need it. Okay, I need to take this approach. I need the, yeah, it could remind us, I'm not giving back of science. There are also some ideas also. Okay? We just understand this approach. Okay. Yeah, yeah, yeah. Okay. So, okay. So now we have to give back approach. At least what we get is 1. That is all the trace structures that are built in a, s, t, and a, s. Okay? So now what are the trace structures that are built in a, s, t, and a, s, u? In a, s, t, we have 1, 2, 4, 3, and plus 1, 3, 4, 2. This is from a, s, t. And from a, s, u, we have 1, 2, 3, 4, plus 1, 4, 3. These two terms combine together into 1, 2 times anti-commutator 4. Two terms are written. So we take the 2 down this side. We go to these two terms. And these two terms then combine together as 2, 1 and anti-commutator 4. So the whole answer is trace of anti-commutator of lambda 1, lambda 2, anti-commutator of lambda 3, lambda 4, and x, that's what we have. That's very effective, you know. Yeah, you may be effective. Actually, you may be effective but it should be in here. But it should be in here. What do we expect from the second module of three cations and then cations run in the middle? So the second module can still run in the middle. So, okay? Remember that we need to compute the last class that the three cations that we have to be able to do between 1 and 3 was anti-commutator of lambda 1, lambda 2, and lambda 3. That is a fact. And no fact at the moment for the structure to see that. What we would expect from the numbers we will work out in the next one, what we would expect is that that goes all the other way. Okay? So we'll have trace of lambda 1, lambda 2 times lambda a times trace of lambda a, lambda 3, lambda 4. Some of them. lambda a lambda a is the is the basis of the space of n for 10 meters. You choose the basis to be having one element as zero, non-zero and everything as zero in all the n squared places. Okay? And now you can easily look to yourself basically from the observation that you know, all n is complete of the basis and n is equal to right n. Okay? Think of these lambda a as the basis of the space of n squared and n to the matrices with some obvious inaudible trace, the trace of quantum matrices generated in the inaudible. Okay? You can look to yourself that this is the same thing when you do the sum over as trace of lambda 1, lambda 2, lambda, lambda 3. The structure that we see up here is beautifully introduced by sum over all intermediate colors. The remaining conditions are the one and then we expect to see of course this divided by the all x plus 1. So that's exactly it. What is the value? We become a exactly the structure that we expect. In full detail, we become exactly the structure that we expect from our knowledge of 3.5 degrees from the form. But now it's any question about the standard we can use to do it knowledge. So let's try to use it knowledge. Now that's going to be a little more systematic of the three factors. I still don't think number 1 and 2 by look up would be the same for that. Okay? We're just uniquely important unknown factors in human behavior. Okay? So as we can unknown factors are what? You see, anything that is that you know with x plus 1 you do something that unknown with some coefficient. So let's say we call that x. Okay? And in addition to everything that we've got, you see, if you remember our rules for computing the correlation functions of the worksheet with this, they were rules for computing correlation functions. So these rules actually gave us you know, we need the whole function of this to be very excellent. So there's a product that the partition function also gave us. But it says what is the three-point function of the backyard proportion of the group? Apart from the dollar factor, it's proportion of the GPU times this function of the disk. The budget function would know where the property is. It appears very hard to compute these out. It's much easier to compute what we've done in 20,000 ways. But, you know, but suppose we were trying to cheat. We didn't want to have to go through the wall of computing this series. But if we say, this is what three-point function is going to do. So when you do this analysis, more carefully, we get an act of listing squared of two three-point functions. But from the four-point function we get gp to the form that we see. Remember, it's the same cd because the partition function on the disk is the function of the disk. Okay? So that each c does us that gp squared is equal to cd is equal to 1. So it relates normally. The unknown normalization is not exactly normal where it's all right there. So this is the same systematic analysis. But every particle, we can find all we can find all gt's. Okay? In terms of cd. So all the issues of these unknown eternally unknown normalizations of the next couple years are determined. Okay? The absolute value of the normalization is not determined. Because I'm not getting determined because very rarely has the data down to you whose expectation value of continuity is a coupling constant. Did they any value of it? So nothing will determine this absolute value. But all ratio. So I just made a assertion that every other field is normalized to constant value of it. But from the same magnitude, let's check that that's working in the deck. Okay? So that's what's the same magnitude. We didn't look at the factorization. On the let's look at the same amplitude. The same amplitude. Okay? So now that's the same. Okay? First question minus 1 gamma minus p minus 1 divided by gamma minus s minus p minus 2. Okay? And look at how it factors on the most. The most standard of the rest is 0. So when we say 1 gamma minus 1 is equal to gamma 0 of your factor problems. Using the identity that gamma of x plus 1 is equal to x gamma of x. Okay? Now if you get back the sign of the game, then we get the same type of hope, right? So this now we conjugate the identity gamma of x plus 1 is equal to gamma of n. And that gets us a factor of minus p minus 2 we have minus p minus 2 divided by x. So minus p minus 2 is still useful for whatever you have in a game. So remember that we have s in general s plus u is equal to minus 4. But s is equal to 0 of this problem. So as far as this recipe is concerned t plus u is equal to minus 4. Therefore minus t so let's multiply this by minus 0 minus u minus is equal to 0. So minus t minus u is equal to therefore minus t minus 2 is equal to minus t minus 2 plus u plus 2 over 2 at that u plus 2 is equal to minus 2. So t is equal to u minus 2. So to realize we can prove useful in everything else that's as such. What if we can do this? We can do this that from ASU from ASU to by 2 into what? The recipe will u minus 2 by 2 times whatever we find. The question is what do we get from ASU? Well it's here the symmetry of the change of u and t. So we will save it for ASU but from ASU we get minus of u minus so t minus u over 2 times whatever that is. I am not going to talk about what about we have 1 2 3 4 2 1 4 3 2 3 4 minus of u plus 1 2 3 4 1 4 that's it. So this minus is 1 2 times anti-commutator 3 4 and what about this minus this thing does 2 1 then anti-commutator 4 3 then this thing does 2 1 then anti-commutator 4 3 then this thing does 2 1 2 3 4 5 4 3 4 3 4 4 3 4 5 4 3 4 3 4 5 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 39 50 1 29 30 31 30 31 31 32 32 33 34 34 35 36 37 37 39 39 40 40 40 50 50 50 50 50 50 50 The four point function, as we just seen, has gp to the 4, then cd, and that is an additional factor of 1. So now the gp to the 4th data is cancelled. So we get gp is equal to 1 power of cd. gp and cd is equal to 1 over 1 power of cd. What I'm very careful with all the coefficients and so on, is that g photon is equal to 1 over, oh, and this is the g photon squared. So you get the g photon is equal to 1 over square root of 2 alpha prime, then g times. And the interesting thing is that this is the same relative normalization that follows from the state of order. So let's take a minute to just understand what that last state of the day, for the last state of the day, may be. So remember that what are we really doing in this way? Well, we're starting with some open string here, and taking it at, you know, starting with some open string, some other open string, some other open string, some other open string. There's some word sheet which has the long tues in the script and then you're going to operate this. And then we use the fact that a long string like this can be component related to a little patch like this with the insertion of an operator, namely the state operator that. To replace this annihilation, the calculation of this link is the calculation of these long strings, they're not related, but the calculation of correlation functions is corresponding to this. Now, the states that we set, they belong to each other. So we don't really have any of physical immunity in what the factors that go behind each other. We have to stop, we're supposed to stop the normalized states and determine the correct way to operate, as we normalize it. That will be able to normalize it. That is an XML that's not able to do it, that either we or you in an exercise deal was supposed to do last time for closed strings. This relative normalization, starting with the state operator and starting with normalized states and then determine what the operator of the last one is that states. Okay? So that gives you relative normalizations directly so there are two different ways to get these relative normalizations. First, sort of from logic, from the state operator. But second, just by taking the answer, if it's described as a quantum field theory, that's unit actually and so on, it must be the difference. And these two ways give you the same answer. Anyway, this relationship that I'm having about the numbers but that's C, D, 10, G, D, the whole difference is 1 or 1 over 2 by whatever the right answer is. This can also be verified with a careful calculation of C with careful normalizing, you know, keeping track of normalizations. Okay? When you define this correctly. I won't go into that. So that's why I wanted to say that every element that we've used by assuming unit attitude and assuming that we get the same structure of singularities as dictated, you see, the way I presented this and I said, if you write a quantum field theory with exchange of particles like this, this is what you do. But that, you know, doesn't require you really to have quantum field theory because in the end, the structure of singularities being determined by lower order graphs follows directly from unit attitude. Quantum field theory has in it even the structure of unit attitude. So it's convenient to analyze these things by referring to how to use normality quantum field theory. But what we're really doing is using unit attitude. So the idea of quantum field theory is to put it as a new model. Okay? So what we've been doing the last 10 or 15 minutes is checking that the structure of the amplitude is consistent with unit attitude and we're fixing some normalizations to ensure the unit attitude back up. But all of these normalizations could have been a big way of learning as well. So it's a check. Okay? Also it was a structure. Right? So for instance, the fact that we had the stream of this unit was very significant. If we just got one in the rest of the view of the quantum amplitude then quantum is consistent with unit attitude no matter what constant it is. It has to be the product of the 3-point amplitude squared divided by the pole up to some number. That's, you know, that's stringent variables. Is this good? So the last thing I want to say about this, well the last thing I want to say about the energy element to do is, well, go to 3. Okay? So you come out minus s minus two, come out minus one come out minus two, one come out minus s minus three. Because you can remember that we have to pull out every particle of s to get the energy. We have to pull out the s equal to n. What? It's a kind of s equal to n. But if we have to pull out the s equal to n then this gamma function is gamma of minus two minus ten, not n. But this gamma function is gamma of minus two minus one. So that's n plus one in the teachers separately in this class. This guy is larger. This is the initial negative number. So we're going to get the factor p minus minus two into minus three minus three all the way out through minus p minus two minus n in the rest of the orchestra. First we take over energy to be very large. Then the c is very large. What we're thinking is this is about a c to the power n. This is telling us that though we haven't computed it the scattering of two tachyons to some state at mass level n the amplitude must have n plus one out of the moment. So that that thing's going to be very large. That's very interesting. You see, the tachyon the tachyon to these increasing high mass particles grows at like increasing power of the tachyon. Now if I say, wow, this is a nice picture. We take the energy we go to very high energies and the amplitude is going to grow up very fast because where the amplitude to produce a mass zero particle already is high amplitude to assume a mass one particle. So I'm going to just write that. But that is not the case. That is not the case. This is not the case. This is not the case. So for instance if we take s to be large and t to be large but even the ratio of s and t find it what is this answer? We know. If we take s to be large and t to be large even the ratio of s and t find it this answer becomes exponentially suppressed e to the power minus s then e to the power minus t and it seems lost. Lost time. For closures. Everyone remember this? It's just beautiful practice. The formula gamma n is sort of like n to the n times e to the power n plus something. Using that formula and cancelling all things to cancel you will find exponential suppression. You find exponential suppression and large energies and you work out for close places. It's not true that amplitudes are going to have that high energies. Even though things are coupling to any individual particle that can hide the power of energy. That's the miracle of strength there. You got all these auto-particles there. And if you isolate the contribution to one particle running in intermediate section it's going to be very dangerous. However, if you stop it or not what you're seeing is that what's probably in the middle is a soft, puffy screen which has size a large form factor and that one that you find exponentially are exponentially suppressed. That's the miracle of the formula that all of the individual things come to the other end of momentum the formula is the dice of the event there. It's the softness of strengthening what you see with all this. Is it a game of the formula? That is the light of the heart that is actually lighted and questions about space is one of these comments so if you look at the photon you see from this structure that you can believe even with you one theory there's no absolute, there's no singularity because it's complicated to manage but of course it's consistent with the fact that the photon is uncharged under the work of any other theory Okay, good. Any other questions or comments? Otherwise, I think I'm going to stop my discussion and I believe something this short. Yeah, pardon me. What's left of today's class is not much given that way or you're going to look at the asset of the token. What's left of today's class and in Wednesday's class what we're going to do is to discuss the loopback of the usage. Now did I I could find, I found I could remember how much the token loopback is the question. Did we stop on loopback, please? Could you remember doing it? No. No one remembers it, so that's a follow-up. Okay, so I'll start with a brief discussion of loopback and then we'll go on to the discussion of the loopback that seems to be open. You know, measure modular space to the token. Then I come to the actual computing and the learning of the completion of the measure of modular space. Sir, recall. Recall that I pointed out the general structure of the theory. Every modulus appears in the mean search. And the mean search in policy in general spending G is alpha, beta G is alpha. My and del by del D of G is equal to modular space. So the actual we didn't have a computer actually. Did we? So what does function function as a? Yeah, we did, we did, we did. It's quite, quite, quite well shared with you. Okay, share with you just in the memory if you can, then it's a quick reminder. I didn't remember none of you did, it's clearly not a question of you. Okay. So, a very quick reminder. Basically, important is reminder because most of the gymnastics we do here we're very similar to the gymnastics we do here. We have a more familiar context for remind you of what we did and then move on to the discussion. Okay. So, so first thing about the regular model analysis. Does anyone remember the every factor of a modelist? The modelist in this case is T. Okay. A pureness with satellite. The basic backhand that you call that. Okay, but we definitely have an exact response. Maybe we should have a discussion of T duality itself. Basically, and maybe in a moment we're going to finish a discussion of increasing the body of one of the students. Never going to have an exam increasing the body of one of the students. Okay. And I expect you all to do brilliant things. Okay. For every fact, you remember every body has had a real session. But it's not like say real session, say real session. He was the modelist, because the panelist tries not to overwrite by the body of the panelist at that time. So, the torus was a space with lectures d z d z bar. And the identification is d z z plus 2 pi and z z is equal to d z plus 1. Or maybe we said this one is a strange way of panelizing the modelist of the torus because we changed the domain over and the space was back was located in it around the change in the domain. But if you want to remember as we discussed last time a change in the form, you can undo that. So, suppose we have two torus which has taught non-linear panelism at that point. And the second one is the non-linear panelism at that point. What what power does that power what space this is? We can equate quantum identification the same as they were where the model of panelism was found. The major variables have the same identifications z equals to z plus 2 pi z equals to z plus 1. But that would change the matrix. You want to see what the matrix would look like. Okay? So, how do we do that? So, we got a new w-quad so let's suppose that we have a model z which has this identifications z equals to z plus 2 pi and z is equal to z plus 2 pi tau is z. We have some new w-quad so that identifications w is equal to w plus 2 pi and w is equal to this value plus 2 pi. There you can change the dosage matrix too. So, we suggest that z is equal to and in this space the matrix is z to z. Okay? We have z is equal to a plus b, that means v. Okay? So, we want to determine a and b. There are two equations to determine. Firstly, if we put z is equal to 2 pi that should consist of w. So, we get a plus b is equal to 1. And secondly, if we put z is equal to tau plus 10 is equal to 2 pi that should consist of w so this identifications should consist of w. Okay? So, 2 pi is equal to so we get a into tau plus b into tau is equal to tau plus 10 is equal to tau plus 10 is equal to the space with matrix z to z plus and those identifications are very equal to z plus 2. Let's get you what we call let's call it a problem. So, z is equal to z plus 1 and z is equal to z plus 10. Okay? Now, we get two different spaces. One with tau and one with tau plus 10. So, you know, I have thought about the patches of the plane. These are different patches. This is a fine way of categorizing this into a right. However, in this formula here what we're assuming is that we work with some of the same coordinate range that are different matrix. What we did initially is choose the same matrix from both spaces. I mean, in fact, we did. But you have to change the coordinate range. Now, these two things are equal out of the coordinate range. Okay? And if I have to find the coordinate transformation, that makes a matrix here. As these two have been taught, one with tau and one with tau in the center. We put this identification in the new coordinate. The new coordinate is all out. So, what I mean is that normally it equals tau. I get z is equal to tau plus 10. I'm guessing that the coordinate range is equal. z is equal to a w plus b times and I'm going to try to find that here. Okay? So, I have two other ones around here. I need two equations. So, the first equation is the one and the second equation is that tau is nothing but the one. This is tau plus 10. That's the first equation. That's at 1, 1, 1. Second equation, tau plus 10, is that tau? 8, tau, 1, tau, 1. Okay? No, I'm not even going to get too far. This is negation. But here, it's just nobody remembers where this is going to go. Okay? Yeah. Okay, now let's solve. Okay? Let's multiply this equation by tau and subtract. So, we get b tau bar minus tau is equal to tau dot answers delta. So, b is equal to delta tau over tau bar is tau. Okay? And I'll also multiply this equation by tau bar as a fact. So, we get a into tau minus tau bar is equal to tau minus tau bar plus 10 tau. A is equal to 1 plus delta tau over delta minus tau bar is equal to w plus and then there is tau bar xw plus delta dz bar bar in z space because something is out of this space. We don't know what that something is to first all in delta tau so that we can defend it. But also, we don't care about components. If we only care about those components then we have non-zero over that with b. And remember that b was clearly over object or clearly over object. There's only a bz set of components or bz set of z bar z bar but no b star. What we're going to get from this since the ambient space is just enough. So bz z and delta g z z z z bar z bar by by doing some consideration delta tau r bar plus bz z z z z bar z bar and delta g z z and delta tau then we want to tell we want to compute and then the second term we want to compute delta w z z z z z bar by doing some consideration plus second term with tau z we need to know g z z bar and g z z z in a very easy function of tau. So we need to compute the omega k. So we need to compute the pattern that has d w squared and d w bar squared to first all in tau and get that out. So the pattern that has d w squared clearly has one term over here d z bar. So from here minus d w times delta tau bar over tau bar minus tau okay and then the possible d w bar clearly has one term for this d w bar the second term is d w so the list minus delta tau over tau z star d z z z bar d z z z bar and really in terms of w bar this not even too much of an element like that out not even a fraction of the d w d w bar part of anything okay so we conclude that d g w w by d tau is zero but d g w w by d tau bar is equal to minus 1 by tau bar minus and similarly d g w bar by d tau bar is zero we can do that again in this but d d g w w by d tau is equal to whatever we have here now minus 1 over minus tau bar is a scare this one we find that the two insertions that we need we find that the two insertions that we need are this one is non-zero we get b times z bar z bar b times minus 1 by d tau and we have a second insertion just b bar two insertions that we have we can write as b times b bar times d tau bar over b square times factor we need to just figure out the factors and then we have one fourth we do that we just keep the focus of the movement this is what we have inside our eyes we do not know what we have what we need to do in this scene you remember that the donors had two conformity vectors named the vectors which were just constants and two modularies this is the beauty of the facts so we get these insertions that come from two modularies we have to get these insertions that come from two conformity vectors these two conformity vectors are one in the energy and one in the energy so we can deal with that those two conformity vectors one vertical and in the energy energy we should have a question so what we would get then is c of z c bar of z bar then the vertical operator and z of z bar now we use two facts the first fact is that this c operator as you know the insertions of these these operators are only the zero mode we take the zero mode otherwise it can be zero and the zero mode in this case has constants so these c operators can since you get this since you get the constant anyway it doesn't have a value so don't where c is inserted is limited to where the vertical operator is inserted in logic all the colors you can move c and c behind such as anyone because everywhere you get the contrast to zero mode the value of zero mode is the same thing everywhere okay oh there okay too much more okay so the first thing is I move this okay I move this both of you move that then I get the verdict the initial insertion of the vertical operator is that any so instead of inserting the vertical operator at a given point we can add the vertical operator insert that point to the second one and integrate d2 to the second one divided by the integral of d2z normally that would be quite complicated because the insertion position of the c operator is limited to where you insert your vertical operator but all the facts we have to talk about so the list we've mentioned that's saying that you convert the fixed vertex operator once you put two over the insertion of c at the already if you convert the fixed vertex operator to the integrated vertex what if you also make a price of dividing true by the volume of the operator that's what everything is like so as far as the model system is concerned what we get we can't and I remember also that b is a constant and b is a constant so we can think of this integral over b as b of 0 and 20 so what what's going on? b of 0 c of 0 c of 0 we get this piece then we get a factor of 1 by 0 the whole thing squared from the explicit factor we got time to prepare a factor or that's the situation down the integral of this by b of 0 times all of the top is equal to the power of all this we have to divide it by one factor for the volume of the torus due to the strength of converting the vertex operator into the integrated vertex operator so we get this integral factor of 1 by e to the power from that and the next result is d tau d tau d tau and now c of 0 c of 0 b of 0 b of 0 and now after this all of the vertex operators are treated as an integrated vertex there's another way to measure it it's through this derivation you know all of these derivations by fixing an operator as another fixing it just because we wanted to use the general results we had the key point here the two conformal derivatives that we had left unfixed we didn't need to fix anything because the remaining selection that was left unfixed was just the volume of the torus integrated over the coordinate changes that generated the unfixed gates of athlete just gave us the volume of the torus so instead of fixing those unfixed gates of athletes you can just done with the general results of the finite volume okay so for the special case of the torus we could have thought of the age fixing in a completely different way than fixing a particular object and we could have got this answer directly if we did this far enough from the beginning the only point we want to emphasize is that this answer is correct even better know it even though our derivation meaning the finite answer is correct even though we don't have this we are not trying to derive this from you but if anyone is uncomfortable we could have a little session on doing it including the vacuum ant if the vacuum ant was no vertex operator to fix fix the conformal gate but that didn't matter we need to fix so this final answer is correct if you look at the vacuum ant another thing is a starting point of our analysis of the vacuum ant in ten or three minutes before going on to discuss before going on to discuss objects okay so let's break now but that's the exercise we want to suggest to you guys try to derive from the for about the dollars to do they send out the signal okay the the shape that we are that will be useful for one-two amplitudes always things and see if you can derive the correct major factor the correct major factor for I guess we can add on this formula for amplitudes we will do it in class next time thank you