 In fact, so today's class, so we will mainly be giving you a scattering answer to this in a little bit later on. But before we do that, I just want to complete this discussion with you. This is about the modern era. So this reminds you what we were talking about. You know, even regarding a complex category, that's a bunch of patches, on each of which we have a coordinate system where the metric takes form d z d z. Okay, on overlaps of these patches, we have one coordinate being a function of the other. We want to match this up with some situation where these, we're dealing with metrics, modulo, conformal transformations, and since analytic coordinate changes give us conformal transformations, the kind of functions that we are interested in are just analytic functions. Okay, so z is a function, n is a function only of z and batch n, z bar and batch n is a function only of z and batch n. Okay, so then in that situation, and we're trying to write down the factor that we get, it's a measure factor for modulo that we get in this language. Okay, we had the other language that is in terms of how the metric changes if you keep your coordinate matches fixed, and we're trying to transform it into this language that we want to use. So let's do that. So we have a few formulas in the model class then, which I have to write into them. The first was that the measure factor in the language that we had before was integral b z z mu z bar z plus b z bar mu. And we have this definition of mu z z bar, the definition of mu in general, which was mu other for beta, using the g beta teta del k. All of this should have k in the text, this mu should have k in the text for the kth module, but I'm so pressing that. So del k is a very, very tk value to respect the kth modules of g beta of mu. Okay, and then we applied, here we work in the special case where we were starting about the metric that we said that we worked them out was flat. There were changes as we changed the pattern. That allowed us to simply erase and lower things. So recently. Okay. Now, what we were going to do was to try to rewrite this in terms of changes in transition functions between matches. Okay. So suppose we have the nth patch, we have the nth patch, and we've got some functions z, z. Okay. Now, as we go from t equals 0 to t equals 0 plus delta k, our quadrant changes from z to z plus z. Okay. So the new coordinate z prime is equal to z, okay, plus delta tk, but that's the rest of the k index, times v. Okay. Let's solve this vz. We also have the zr. That prime r is zr plus delta t times vz. And this coordinate changes we chosen somewhere up to a conformant transformation. So that up to a conformant transformation, the change in the metric is this change. It's the change generated by this element. That's the condition you want. Okay. So now, in the new coordinate, the metric is dz prime, dz prime. So now what we have to do is to take this, this metric in the new coordinate and rewrite it in the old coordinate and equate that metric with the slightly modified old metric up to conformant transformation. So what we get here? You see, what is dz prime to start with? dz prime is equal to dz plus delta t into del vz by del z delta z plus del vz by del z, delta dz. And similarly, dz prime power is equal to dz power plus delta t into an isomer. Del vz power by del z dz plus del dz by del z power. Now we're working the first order of the delta t. So that's what I find with the new metric. So the new metric, dz prime dz power prime to first order delta t are two kinds of things. There's a filter of dz dz bar and that has one plus some stuff. Now some stuff is this del del vz by del z plus del dz bar by del z power both into delta t. Okay? Last we have a term that is dz dz. We have a term that is dz dz and that comes from this with this. That comes from this to this. So we have del vz bar by del z and plus we have a term that's dz power dz power delta t plus del vz by del z power delta t. So let's count the total of four terms for the delta t, two, three, four. And that should be this to this term and this to this term. So we've got it. Okay, good. First question is what was the first term? Now about the first term, that's very simple. You see, that term here can be absorbed but that term is an informal term. That's because it takes the metric because dz dz bar and rate is dz dz bar times something. We are only interested in changing the metric module of the form. So this term doesn't work. This is simply the statement that if we have a coordinate patch and we perform a change of variables on a coordinate patch, that is analytic. Because that doesn't change the logic. So this part we're going to throw. This part we have to equate with the change. So this part here is to be identified with delta tz z. Del vz bar by del z but delta tz z by the definition of this was equal to mu z z bar. Because this is delta tz z and then we get a z bar there will be this. So delta tz z bar is this time delta t. So we've identified it. We've identified mu z z bar is equal to delta v z bar by del z. And of course, symmetrically mu z bar z is equal to delta v z. Let's proceed. So what we're interested in doing is evaluating this object. So all we have to do is replace this mu z z bar and mu z bar z with this. Okay. So the insertion pattern that we did last was the order of the object part and the order of object part. As usual, I said that this is 1 by 2 pi and we know 1 by 2 pi is d z z. Then we have to del z because v2 z. Okay. So we do the z pi integral what's left behind is a z contour and integral surrounding the edges. Okay. So this is equal to integral over a contour surrounding the region of interest. Okay. Integral over a contour surrounding the region of interest v z z v z. Just a step. Okay. We'll do the practice. We'll do the problem to pi i integral. Okay. Practice. Except. So now we have this expression here. But we set it up, you know, in the region of interest. Now we have to be a little more clear about that. So that's suppose we've got three coordinate fractions that come as m, n and q. First we'll do this integral over the whole map. Okay. Because I'm never talking to you two points of the manifold. Okay. So we can choose to do, we can choose the boundary of the nth manifold to be the same. What's actually here will be better about this pattern. We need some other patches. I mean, no. Okay. So let's need that. I'm saying. We can choose the boundary of this category here. But then that gives you the boundary of the qth manifold is here and the boundary of the nth manifold is here. So the nth manifold is this boundary. The qth manifold is this boundary and the nth manifold is this. We can choose to do, to regard these integrals as partly in the nth manifold partly in the pth manifold partly in the qth manifold, side to p by some of the integrals. Let's look at what we're going to get. Let's look at what we're going to get on this part. On this line. Line the line to the nth. From n we have to be integral which gives us this line integral. To be missed. But we also have to do a similar line integral from n. We also have to do a similar integral from n. Okay. So what does that mean? Well, now we have to be integral not integral in the negation. So how we've been dealing with just one patch. So we've been allowed to be blaséed. But, you know, on different patches we have to do different coordinates. And the coordinate changes are just, are just with respect to different vector fields. I'm going to put two indices here. The first one is m which tells you that this is a coordinate change happening in the negable patch. And, the second one I'm putting here is m which tells you that once a nice coordinate change has got a vector field. Now if I've got a vector field I can view it in any coordinate system. v n nz is the vector field of the mth coordinate change viewed in the coordinate z. That the vector field is this and z by z is z. Is this clear? Now, what are we going to get from the other side? What are we going to get from the other side? Well, let's see. You see, okay, before I work on any first word now, what, can you work on one of the following words? What is the change in the transition function on this line? Okay? So what is the transition function? The transition function is what z m is as a function of is z m of z n and some particular value. The transition function is the nth vector coordinate between the nth and nth batch and this is the function of d but the transition function is superior to particular value. So the change in the transition function is z z m by z z n at constant d. Is this clear? Let's see what that is. You see, when I change d to z m is this function and so what I want is how much this function changes. How much then I want to evaluate what z n m is by z n m. At 6 z n m. So this is the thing that I want to do to evaluate what change in the transition function is. It's how much this function changes and a fixed value of the coordinates in it as t changes. Now how do I evaluate it? Well, that's very simple. Because I know how these functions change. So the functions z m has changed how? It's changed like this. Maybe I should just give you the formula then. Formula and then it's like this. So this is equal to minus z m by del z n by that constant v n. This is my n. Now let's understand the formula. This formula. How does this formula work? You see if I change t a bit and I fix to a particular point on that then z m changes. How much does it change? It changes by exchange is v m m then delta m. That is the change z m. But z m is not the only thing that changes. z n also changes. How much does it change? So delta n is equal to this. Then delta m is equal to v m n by delta t constant. Say formulas. Say m m. So I, you know, what I'm asking is I'm fixing a part of the manifold. I'm asking how does the value of z m change? So it depends on two things. t and z n. Because you're looking at the line. No. I'm fixing a part of the manifold. Okay. So I stick at the same point and I ask at that point as I change t how much does z m change? That is completely intrinsic. That's intrinsic to the Mx batch. That's nothing to do with z. Then I fix at the same point of the manifold and ask how much does z n change and I change t? That's also completely intrinsic. It's got only two to the nth batch. Nothing to do with the nth batch. Then it's given by this form. Now what I want to know is how much does this function change? I mean, you don't understand. I mean, things are different. m and This function is this function. This function is going to be evaluated momentarily. Now how much does the function change? Firstly, z n changes because the value changes. So here this is a part of the change of z m which is just v m m kind of like that thing. You see, the argument of the function has changed. This is now to be equated with delta of the function z m not evaluated at z n but evaluated at z n plus v n m delta t t plus delta which suppresses I'm suppressing the b effect. Everything is now at t plus delta. Is this clear? So the change in z n is this much? But that is the change in the function not at z n but at z n plus v n delta p because the argument inside the function is also changed. So the change in the function is z n delta z n z n is equal to delta t into v m m minus what you get by delta t minus delta m by delta z n equals to t times v n m This quantity gives you this option which is delta z n m delta t So this function is equal to that That's a problem Is this clear? Look at this This quantity there multiplies a vector field by a transition function and therefore just give you the vector field to the new set of coordinates This quantity is equal to of a negative Sorry This quantity is equal to v m in the other coordinate minus v m in the other coordinate That's exactly what we get by doing this in delta You see You saw that by doing the integral over the nth batch we got v m in the nth integrated over v z n Yeah If you visit over the nth batch you get v n in the nth coordinate integrated over v z n which is the same thing as v n in the nth coordinate integrated over v z n This is just a coordinate change Basically you want to evaluate everything in the same coordinate So vector field you can evaluate in the same coordinate because one of these goes clockwise there's a minus So if you do this thing on this batch on this line the path that you get from this batch and the path that you get from this batch combined together gives you exactly this option which is the change in the transition function to size sizes d z n by d del z n by del t We see that this will answer here I mean the whole homomorphic path is 1 by 2 point that goes b z z del z m by del t comes from z n Change this transition This is over the m n along this line then this will take over del z m by del t at constant z q And so the whole thing will be closed so it is up closed So this is the answer This is the answer for what the insertion in the path of integral has to be when viewed as a function of the transition function in terms of the transition function The topology is fixed by how the batch is and how to choose this Precisely it does not depend just by the whole homomorphic path This is the whole homomorphic path You see the original thing was completely independent of how you chose to slice things up That is my sort of nothing to check but it has to work See one of these steps here is that if you take this shape this thing and bring it to the right it can't change it because of the whole homomorphic path You see this is the homomorphic function You move it around you get the same answer You may have to think more about moving with this spot No So Now let's try to work out Now let's try to work this only out in a very simple case This simple case is going to allow us to output all the moduli of our scattering amplitude So if you remember what we had we had the our remote surface remote surface at some point and then if you remember long, long ago when we actually looked at the path integral tried to fix all the additional degrees of freedom and so on and so on We found that we had to integrate the vertex operators Some of the integrals of the vertex operators we can use we use the unfixed homomorphisms to fix The other ones we still had to integrate the positions Now you know you might think we are thinking this is think of the the the the the the space of remote surfaces with a set number of properties and as you see with each puncture there is a new modulus which is basically the position okay So instead you know what that's the way we didn't view the scattering We were going to view the scattering amplitude and amplitude that has an integral of this if I am teaching a class at the moment I will go back in Okay So we got this this is modulary space of remote surfaces as many punctures as there were vertex operators and integrated into the whole modulized space of the remote surface how do we associate this modulus with the position of the puncture You see the position of the puncture for example is just where in a particular patch the puncture happens to be So everybody has a patch and we've got a puncture so that puncture can be okay so what we're going to do is to surround this puncture by a whole patch make the corner say this is z I'll describe some new corner system let's say z alpha okay so that z alpha is simply equal to z n plus some shift z how much you shift it tells you where exactly the the incisions so let's say this again in a new coordinate system the z alpha coordinate system our incisions always are the same the fact that it can be wherever it wants on this patch is important in this coordinate in little case the this coordinate system is entirely in the bigger this coordinate patch is entirely within the bigger coordinate patch so this shape is very simple the spin line is just also del z m delta delta pz it's a parameter how much you shift how much the modulus has changed del z m and delta p is just a constant because this is linear okay then we call this t it's different from the parameter no it's the same because it's a modulus it's a modulus associated with this particular puncture system that's why I'm calling it it's not a modulus it's a Riemann surface it's a modulus it's a puncture it's one of the modulus of the puncture what I was supposed to do was evaluate del z m by delta p evaluate del z del z m by delta p in this case it's an alpha by delta p I get a constant okay and so the insertion is integral d z d z z with nothing and this integral goes over this coordinate goes over a coordinate that's around my insertion over a point where z has the insertion the insertion has some operator so the insertion operator now let's see how this compares to the old rules for scattering activity clearly because this is one of the problems this is one of the integrated vertices it's one of the integrated vertex operators and therefore according to our old rules we shouldn't have had a z insertion in an operator should have just the conformal primary the natural conformal primary conformal prime but now we have to recover that old rule so what should this of be it's easy to see what it should be if of is equal to over time c that's called c time c time nowhere over c conformal primary the natural conformal primary then what would we get we'd get d z and then we'd become over of the b or p but always we are in the matter sector so there are no singularities so the only thing that we can see is we get up to some signs we'd get o by z plus plus automatic and so we would get over so we would have integral over you would have integral d t that's position of insertion and over we could use old old rule says that you should insert just the conformal primary and integrate it over the world sheet without the scenes this tells us that if we think of insertions in this okay insertions as association with the module of the function of the surface then the operation that we have to insert is sequence though that is always the rule is that for all operations you insert the operator for all scattering activity insertions you insert the operator is dual to the to the given state but you have to integrate over the moduli all to the full moduli space of the theory so some of these operators those that correspond to the moduli are surrounded by the appropriative modulus weighting factor that appropriate modulus weighting factor is their integrity which kills us therefore recovering the unit so a new rule goes as follows write down the integral integrated version of the operator or write down sequence of the operator with the P insertion with the appropriate weight insertion for the modulus but treating all moduli now equally the modulus of the Riemann circuit as well as the moduli as associated with the functions so this in some sense is a match the nicest and most geometrical involves an integral over a moduli space which is the moduli space associated with N punctured remonstration every moduli is in this integral that is associated with that in some way of parameterizing this moduli space you can get rid of some of these moduli things rather easily at the expense of it the most feasible line strength or is the conformal primary that corresponds to the state of the circuit so the operator that you have to insert here is P kind of a popular primary so that when we do integral over B that gives the C if you just write now we should have all of this discussion a year ago we did because we're going to look for music in some place but now it was useful to have this discussion this all the way things go with abstract stuff you know people with a certain inclination and that includes me then they think oh this is just some fancy way of saying something they know so I'm not interested however that fancy way is often useful for regular situations where you don't that's often the use of fancy okay so this is one use of this situation this fancy way of learning the Buzonic strings has a natural generalization to the state okay and this natural generalization I will only talk about next class because otherwise your environment will have time for this exposition but I'll just give you an address the natural generalization will include you know always putting all operators in the natural of the community okay and introducing some some factors for corresponding to the integral over super conformal modularity we will see what they mean and just like this fancy way reduced to giving us integrated vertex operators by dropping the C that fancy way will reduce to our old rule of inciting appropriate picture changing of actors once we once we work it out correctly giving you a reasonable justification of this whole rather ad-hoc picture changing a rule also apparently allowing you're giving your principle to resolve sort of subtle alright in question about the stuff next lecture we got to that super conform the super conformal version yes and I mean this insertion this insertion well you may want to take up sort of the B point you are in some sense I mean this insertion yeah in some sense I mean no I mean for example can we just write this in terms of the JPRS oh that I don't I've never seen that I haven't tried to do it but next it's like you mean should we have to do it in that no I don't my guess is you know the whole JPRST business depending on the fact that in the Vittagama if you try to rewrite Vittagama or the integral in terms of phi and zeta Vitta there's a left over insertion you know zeta that you must introduce and that was very important to this whole picture I don't know any analog of that I suspect there's no clean analog at this level there will be a clean analog in the supermarket maybe there will be a clean analog of this formula we've got to integrate over all modular but I don't know I love the Vittagama other questions sir problems okay so our plan is the next lecture I'll go to the super formal version of this I'll I'll ask also for our volunteer for the subsequent lecture to tell us about the one new vanity which is being discussed this is the one new vanity that verifies that the normal is ducal appropriately appropriate F meter okay that we have completed our study of the chapter okay so we've completed the formal part of the study of and produced then we move on to the more you know more immediately physical things about in chapter 13 and chapter 4 and then we talk we have some discussion space then the discussion okay so we've got from our you know a regular street day where they do this in my sense that is that's something that goes through every idea touches on every idea with some help um has a part that talks about okay just how do you say apart from one important addition that is the sigma how how did I realize the straight theory last phase we will soon declare you know we will soon declare that we have completed our study of that stuff of course unfortunately that study is just a prelude for doing anything we don't care about the completed application most of the time we need to know the rules in order to do anything just okay completed the background so that when we start doing anything really interesting really interesting stuff I don't have to say it can be shown that as we have seen that it doesn't seem to have a necessary background for doing the interesting things in the straight theory so it's sort of a strange situation going through a year and a half of sports just to set the tone for doing anything but still that's the state of the subject and now you've suffered through the the technical the technicality and it is becoming more immediately interesting things very soon so it's pay off pay off okay that's that's going to be pay off pay off start up by the class after that we complete the second three-point functions in the last class we completed the three-point function of in the of three k-posons in the type one theory after that we complete the type one so there was this Chen Chen which was left over just complete the discussion completing this this this is a picture these are the what is operators and since these are all k-posons I consider two what is operators in the minus one picture and one what is operator in the zero picture and so for the for the this the insertions I also need to consider the one for for completeness and I have corresponding corresponding matrices here in the what is operator so let me write what is operator which I am leaving for this discussion for the correlation function would be without considering this t as now if you consider this t as then this certain would be considering trace over t a t b and the other diagram 1 to 2 would also contribute another trace over t b t a and p c and when I exchange this to a vertex operators together same same object then I will get assigned with the other c so that is minus and this would give me trace the result that we obtain in the last class we need to multiply that by this factor and then we will get 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Now, let me write down this, this boson gate boson vertex operator we have already seen in the minus 1 picture, it is just e to the power minus 5 times sin u and we got i p x. The permanent vertex operator would be e to the power minus 5 by 2 times theta alpha q alpha, where this theta alpha is the stream field that was introduced earlier. This was defined in the following way exponential i times sum of vertex a h p, where h p are the boson lines portion of the size. However, the h a is the h a is the of the corresponding values under the 5 part terms of what values can I take, they can take plus minus half of this and so each h a is like a plus minus half right and the sum of product of sin is this. Yes, however in this definition Lorentz's inference is extremely good because we have to pair up the things, but in this way of describing this we find it convenient to keep Lorentz invaders. Therefore, we would consider this theta as to have Lorentz in this alpha the permanent Lorentz. And it would be I mean it is not straightforward, but we one can you know obtain what exact is vector it corresponds to in this way of writing. Ok, it is quite straightforward because yes and you just look at what alpha means which sin vector and then you learn the Lorentz here to obtain. Idea values under the various patterns is manifest. Right. So, however here would be explained in the Lorentz's variance, so we keep the alpha in this. So, now we split this portion, now we split this correlation function up into different CFP's. The c the cc and p is obviously it is the same thing that we did in the last class. So, we have to give us a x 1 2, x 2 3, x 3 1 and then we have to worry about the pi's. So, we have to calculate something like e to the power minus pi by 2 e to the power minus pi by 2 e to the power minus pi by 2 e to the power minus pi by 2 ok. Now, in order to compute this we recall some formula which is like e to the power product of e to the power i k x I mean some formula some other formula is the case of Poisson theory which we have used earlier but there is some a's let's say this goes as some factors x to the power i k take x i k to the power e to the power minus 2 to the power of alpha i not b ok. If we do that then we will find we have to incorporate actually i's and we have to find this goes as x to the power of 1, x to the power of 1 4 half of the x's. This is so I am giving this is 1 2 k, x's are the right this 2 x 1 2 and x 1 3 and x 3. Now, could you just argue out that the 1 4th and the half power. Yes. So, this is this is ok. So, here we have to identify k as i times half in fact minus i times half because in the bottom formula we have alpha prime equals to alpha prime equals to half x. So, here I use k equal to minus i times minus i times half minus half. So, k 1 by k 2 is going to give me minus of 1 4 and k 3 is i just i. So, k 3 with this is going to give me a half with the minus sign. So, yeah. And just a way to remember you know without remembering the formula to remember the master formula is e to the power a times x times e to the power b times x and if you have many of such things it is just sum over e to the power a b times the 2.4. This is always true the product of the formula to follow so that we see integration. Now, the 2.4 function is minus log. Right. Yes. So, the half over the half will give you 1 4th times minus log. So, x to the power 1 1 over x to the power 4. But the half with the 1 will give you 1 4th. Now, the next thing that we are going to do is we have to look at let us look at the ok. So, we have to evaluate something like theta beta 1 2 and you have the 2 theta from these 2 body suffered and you have a sign from here. So, you have to do that let us first write down the formula given to theta. So, this will do by arguing about conformal integration of x and towards integration. Now, this should be proportional to c alpha beta where c is the charge of integration matrix. And so, you know the dependence on z let us find out the weight of theta ok. So, the weight of theta can be found out from this expression this being 1 and e to the power e to the power minus 5 by 2 has weight minus half e to minus half whole square minus plus half which is equal to 3 by and therefore, the weight of theta is 1 minus 3 by 8 which is equal to 5 by 8. Since, we have 2 theta's here. So, it is 5 by 4 and therefore, the dependence is x to the power 5 by 2. So, we also argue that x is from the formula theta it has a weight space ok. So, the we can argue the we can of course, I will argue for the x dependence, but this should be written. So, theta's in terms of h is e to the power i. So, i is some number times h into e to the power i times some number times h. So, this should go as actually this number it is important. So, the number is less than it is half ok. So, the weight of 1 by 2 is less than it is half ok. What is the formula for the weight of e to the power 8? So, it is x 1 2 to the power e to the power 1 by 2. Just the dimension versus that power. Oh, I see you are trying to talk I see. But what is that just what is scaling dimension of e to the power 8? So, i times k times k. So, it goes as k square. k square by 2. By 2 yeah, k square by 2. Now, we evaluate the weight of theta the scaling dimension of theta by 8. Oh, you are saying you are trying. So, this is k square this is plus minus half. So, this is 1 by 8. And how many h's? And there are 5 h's. So, that is 5 by 8 just the same place. Thanks. Second relation and also I would like to write down the relation the O B of theta with the psi. Just if I suggest when you clear what you do not need I read it. I need these relations and I am even reading this. So, let me clear. You can clear what you can do here. You need the relation to leave that. Just clear the left-hand side. So, I want to evaluate this and I have this expression. Just look at the O B of theta and it is not right. I will write it also double 3 point after you type on this. The atmosphere. We will argue through conformal ingredients and the weights that this should be proportional to theta. I am sure that you are interested in. Wait, wait. Give us the answer. Firstly, why do some x's fit by the weights? How many others of theta are there? So, this is in the let us say one of the tidal vectors and this is a vector and we are taking a product of them. So, it should have something like an anti-guidal vector here. I mean, anti-guidal is finite here. So, that fixes this dependence and then the weights are determined by. In both the language, suppose I was looking at a particular vector. So, I have written a theta with all the classes and psi with the minus. So, suppose the theta have all the classes. So, write that down into the power 8 by 2 and psi goes into the power minus h1. That particular psi. Now, what would we get from this? So, it would be only when this is h1 and that would be like it is to the power. Firstly, basically what I should do is expand it out and use the log. Yeah. There are four pages going out to the right. Right. So, it is just to get our sum over 8 by 2 to 2 to 4 h. h? Times up again. This is the notation. So, what? Now, we are going to write out the path that is just h1. Okay. So, this is the OP between h3 power i. So, this is the OP between e to the power half h1 i minus h1 e to the power minus h1. Minus i plus the psi goes. Right. The dot is e to the power 5 psi. Then, it is just h3 power minus half. Because it is. h3 minus half times half. Oh, h3 power minus half times 4 power. I mean. So, if i is h to the power minus half. h3 power minus half because of the thing that I wrote down is h3 power k dot k1 dot k2. Here, j is half and here, j is minus 1. And the rest of the thing is e to the power normal order e to the power i half of h1 with what psi? Minus i. Half. e to the power minus i. Oh, you were writing. Right. You were going. Yeah. e to the power minus i h, which is equal to this. Equal to e to the power minus half. Yeah. The most. They keep interested. Yeah. So, if you look at that. Exactly. So, this is an example. Yeah. For me, that is where it is. And you get anti-calibration. Right. Yes. Okay. By the way, the thing that you get on the right-hand side. Yeah. And that would be, is it, you know, gs of protection? It doesn't matter. It's just, to me, is that correct? Yeah. But, does it need to be predicted out? Then, okay, then we'll probably be out of use. I'm sorry. I'm sorry. I'm sorry. Okay. Fine. So, okay, then this is, so this object now, theta alpha, theta, theta. Psi mu. I will use these two relations. This relation and this one. So, write this as c times lambda alpha. Yeah. The singularity is, so as I take these two objects here, I should get a singularity which is like, which would say half. So, if I take level 1, 2, 3, I should get a x2, 3 to the power half, and x1, 3 to the power half. And, if I take these two together, I should get, no, the rest of the thing will be derived by, so that's basically the boundary data. So, the rest of the thing, we're dependent to determine better conformal ingredients, which is, in this case, this is 5, 5 by 4 plus half. Oh, right. Plus half, minus half, minus half. But, 2 minus half comes from this. So, that's basically 3 by 4. So, this should be equal to minus 3 by 4. x1? Yeah. It's fine. Okay, now just check the, have you checked that all, all OP is the same. All C, D are the same, correct? Yeah. I haven't written down the OP between theta and theta. Okay, I haven't written down. So, that I will use from these two relations. Okay. So, the OP between theta and theta, and therefore, okay, this side, don't need it really, but I'll leave it at some time later. So, there'll be a loop. Yeah, there'll be a loop to this. So, the OP between theta and theta, let me write it down since I'm, let me include the total vertex operators, which I will be automatically to write. So, which is 2 to the power minus 5 by 2, 2 to the power minus 5. But why would you give us theta and theta first? Okay. Theta and theta would be, from this relation, we can see, gamma mu alpha beta, psi mu, divided by, this is half, minus 5 by 4, 5 by 4 minus 2 by 4, which is 3 by 4. 2 by 4. Half the same as 2 by 4. Which is what you have there? Which is what I have there, and there are short things to do, obviously. Of course, the OP also has the identity here. The thing you wrote down first, 2 points. Theta alpha, theta theta also has the difference proportion by identity, divided by x to the power 5. Identity mean. But, right. When I take expectation values, I need this expectation values. Okay. And one more entertaining question. Suppose we were to use the bosonized form, could we directly see something like this? So, So, it's like the product of two spiners, giving a vector. Okay. Suppose I do, you are not giving a scalar. Okay. So, it is only true that, this is a spiner and this is an anti-spinner. And therefore, in the product of the spiner and the anti-spinner, there will be a vector. There will be a scalar. Scalar. There will be a scalar spin and a scalar spin, and a scalar spin and a scalar spin, and you do all forms. Right, right. Oh, right. So, a scalar spin and a scalar spin should give me what forms and vector is one. Let's take an example. Let's take all classes and four minuses and one pluses. Both. Okay. Clear everything on the right-hand side of the board? Right-hand side. Right-hand side. Right-hand side. And let me just write it down. Some more. Right-hand side. H1 plus H2 plus H3 plus H4 plus H5. That's H5. And put four minuses in any search engine. Firstly, what is a numerator? Before working out the problem. What is the operator that's left over? That will be left over in the normal order of these two. Which is basically one zero. Right, so right-hand? Okay. So, e to the power i times no half anymore. Half less also. Is that so-called same thing? Okay. Now, what do you mean? Now, as you say, what you mean should be worked out by put a search engine. Okay. Right. Okay. Suppose it should complicate your life. But suppose I had five pluses and two minuses. One plus H2. So, then you will have got much more complicated operator than the right-hand side. You will have got something with H3 plus H4 plus H5. Wouldn't have been a side. But, what? I mean some combination of sides. It will have been some combination of sides. Why would you have done that? You only had the side. The government was the problem with the side. You gave us a P. And you know, there was a some right-hand side. If you had seen the harmony of the computer there was some difference. But, I believe that you have this proportion of the side. Firstly, if you have you could have one side or you could have what is the next number of sides? Can you have two sides? You can't. The sky doesn't matter. She could have three sides. Okay. What is the way to three sides? Three sides is three and two. What is the way to three sides? Five by four. What is larger? Three by two. So, there is no singularity. The only thing that is singular is the one side. The other thing that exists is the OP. But, they are not the same. It's an example of the you take two kind of speakers and you get all odd forms. The odd form that you produced is the vector. There is a three form which is side, side, side. But, it's not singularity. Okay. Oh, just one more thing. Where was the final form for me? I asked that. Now, I want to ask a question. Before they take one to infinity they take one much larger than two or three. What is the dependence of this coordinator that you have? Okay, so it will be twice the weight of one. Let's read it out from there. So, if it's divided plus half which is side by four. Five by four plus half is side by four. Divide by two is five by eight. So, basically when I take it far away the only thing that will continue, you know, if I think of bringing this and taking an OP with this composite operator or whatever is the part of this OP where this object is the operator itself and so that will determine the way to be. Yeah. You take one far away you take the other group. Now, because you can expand everything else in the other group. It doesn't have a 3.5, 3.5, 3.5, 10.5. Everything else in the other group that the the only part of the OP that will contribute is the part where the OP gives you the same operator because two point functions are there. And so the general rule that in any such correlation function, you take any given operator of the infinity to be everything else by name and the dependence of the coordinate of that thing will be one over X to the part twice that of that. You can see that very explicitly in this formula if you take one to infinity X to the power of 5 by 4 which as we said was twice of 5 by 8 if you take 3 to infinity you get one over X and the dimension of size half so half of it to do this. Yeah, this will be a feature of all the OP's. All of the correlation. Now, let me enter the result. The result of 3 point function of 2 now why can't I have added anything analytic to this? You know, you showed us that all the singularities are increased but if I add some analytic thing to this, it won't have any singularities but I can't. Yeah, so I mean one way would be to say maybe you are saying that this would be protected of the GSM protection and so. Yeah, there is a correlation function to perform it. And then it would be like so basically it is an argument of following of the city. Actually, I can't. But from one general ground as you said, you know, OP determines singularities even some variables. So one way to say it would be like I know what are these objects and therefore I can determine the X dependence. I mean what the X dependence is completely in this OP because I can calculate the OP d bar h, d bar h, tan c. You could do it by explicit calculation but there is a general argument that this allows analytic dependences additional analytic dependences. We just said the level argument to this. Explain. Suppose something is analytic then how does it grow? We have reviewed we had a general argument that the dependence on X1 has to be 1 over X1 to the power 5 as X1 goes in. The dependence on X2 has to be 1 over X2 to the power 5. It has to be 1 over X3 to the power to the power 1. Anything that is analytic that is by which I mean it has both of these singularities must be a polynomial. And so it grows at least as as it was. Which is not good. So the behavior that is being separated out basically clustered is actually you can't add anything. Now you can see whether the X dependence cancels because you fix this coordinates and you check that. So here I have X1 to the power 3 by 2 and with that I need to combine X1 to the power 1 4 X1 to the power 2 that cancels with the C contribution and similarly with X3 so this is half plus half that is 1 by X1 3 and cancels with X3. So what is the final answer would be we have to worry about I k X would be but it turns out to be just the given amount it is only that amount we have to we have to terminate this and it doesn't be just one for the same reason we talk about mass transfer. You can also calculate explicitly between the values. Because K square is ours. It's a X to the power 2 by K1 dot K2 and