 OK. So we continue with the next episode of this interview for a good cause. All right. So I just remind you what we discussed last time. Last time we discussed that there was an interesting and useful way of thinking of super-conformist transformations in super-space, OK? We discussed that super-conformist transformations could be thought of as three definitions of the z-coordinate and the p-coordinate in a way that causes the derivative, the super-saturated derivative d theta, the scalar component d, this way. And we discussed what in some detail was or how that worked and what that implied in general in particular for the transformation of the super-conformist transformations of super-conform and primary fields, OK? There's an interesting story I want to develop in more detail then. The only aspect of it that I'm going to use today is the aspect that I'm reminding you is this. That suppose you've got a super field phi, OK? Suppose you've got a super field phi, OK? So that's some function of z, z-bar, d-bar, theta-bar. So everything's sensitive or amorphous and active, or amorphous, I just think of z and theta or whatever it's z to you. Suppose you've got a super field phi and the super field is a primary super field in the sense that we define it. That is, it transforms homogeneously with a factor of del theta and bar to h under super-conformant transformations. Then in the last class, we noticed that this implied that the zero component of the super field was a super-conformant primary operator in our own sense, OK? That is, that the zero component of the super field was an operator such that it obeyed the OP that we, in the earlier lecture, the factor of it. That was a conformant primary and was also annihilated by the supersymmetric gender vectors in the appropriate fashion, OK? And the second thing that we understood about this is that if this phi had an expansion phi plus theta times psi, then psi was equal to g minus half, OK? So if we have a super field, that is a super-conformant primary in the sense that we defined in the last lecture, that is, it transformed under super-conformant transformations in an appropriate fashion, in an appropriate dimensionally fashion, then it's zero component, it's part that's independent theta, is an ordinary super-conformant primary. And the super-conformant primary is what we talked about in previous lectures. And this object is a descendant of the super-conformant primary in particular g minus half, OK? So this, of the discussion that we had in the last lecture, that would be interesting in other contexts to continue more fully and develop more fully for various reasons. The only part that I'm going to use in today's lecture and in the next few lectures, if there are at least a few left to remind you of any questions or comments about this. Before we proceed, for the reason that we're going to use this, let me tell you about certain, let me tell you how the x and psi CFD can be combined in this nice geometrical fashion. Can be combined in this nice geometrical fashion into a super-conformant field. So as you know, as we've already discussed, because the super-conformant field theory made up of free boson and the free pervian, or left-moving, right-moving, OK? It's a super-conformant field. I'm going to explain how that works geometrically in the sense of super-sets. So define a super-multiple x, which is equal to x, plus i theta times psi, plus i theta times psi, and plus theta times psi, this object. This object can easily be used to write the whole three, or if the interesting equation can be written in the theory of i-free bosons and free pervians can be written interestingly in terms of this super-conformant. Then let me start explaining how. Let's start with the action to be written as follows. The action can be written as 1 by 4 pi theta of capital X and v theta bar. Here I set alpha prime equals 2, or equivalently every scale in the scalar field in a way that is set alpha prime equals 2. Just to be a formula for this. So this, OK, so that instead of 1 over 2 pi alpha prime, the scalar part of the action becomes 1 by 4 pi. We can restore that by re-designing this x to it, or the scalar part of the x to be as next. OK, let's see how this works. Let's see what this action is. So capital X was this object. Now that's evaluated d theta of capital X and d theta bar of capital X. By the way, something I said last class that is obvious. Suppose you've got a primary super field in the sense that we define it. Then we take super symmetric derivative since thereof. It remains a primary super field. In more or less, it remains a primary super field, but its H increases by half. And this you can easily convince yourself just from the fact that this d theta transforms in a nice way under super component transformations. So for x transform in a nice way under super component transformations, the d theta bar, the two things add up the way it acts. So this d dx here is only a primary super field. And we make up an action taking d theta of x and d theta of x, which gives us other super super super component super fields. And although I'm going to present this to you, one can check how this major term transforms into super component transformations. And how these guys transform into super component transformations. This x by its end is a bigger point 0. Its lowest term is this guy x, which has a 0. And one can check just geometrically that this object is geometrically super component. By just looking at how the volume factor, the transform transform transformations, and how these two. The transformations of these two are obvious, because x does not transform into super component. That statement that itself weighs the object. And so that this d theta makes it transform in the appropriate fashion. And that's exactly what it can solve. In purely dimensional analysis, in terms of purely dimensional analysis, d theta of x is an object of weight half. So this is an object of weight half from a half under holomorphic and geolomorphic weight. Now, d2 z is weight minus 1 comma minus 1. But this d2 d-cell here has weight half from half, not minus half from a minus half, as you might begin to do. Because a fermion integral is actually a fermion integral. It removes the coordinate. So that's what the weights are. And the important point about the measure is that it's not always its weight half comma minus half comma minus half, but it transforms into super component transformations like a primary of weight minus half of weights. So it actually cancels the transformation. So the general rule is that you don't build a super component very naturally. You take this measure, this d2 theta and d2 z, and multiply it by a super field that's super conformal primary of weight half from a half. And this is an example in which we built half from a half. We built a half from a half. So let's proceed. Let's see how this works. So this is there of del theta of x bar of x dot. This is equal to plus. Remember what d theta was equal to del theta. And we have convention to the plus or minus. Somebody can remember what it reminded me. Plus plus theta and del theta bar was equal to del theta bar plus theta bar. Let's evaluate this object on this object. x plus i d theta psi plus i d theta bar plus theta d theta bar. So the first object becomes what? Let's do the del theta bar that gives you zero on x. Here it gives you i psi. This is zero. And this part is plus theta bar. Now let's look at this part. This part is zero for any term that contains a theta. So it's non-zero only in theta and del zx. And it's non-zero on plus i theta d theta bar del z. So this first term is the subject. Let's also evaluate the second term. So let's say simple, you get the i psi bar plus theta theta d theta bar. And del z by del z plus theta plus del z bar del z plus i plus i d theta bar d theta del z bar. So that's the product of del theta x and del theta bar of x. And then we're supposed to take this guy and integrate over d2 d theta. Now the integral over d2 d theta just picks up the term that is the coefficient of theta d theta bar. So let me get theta d theta bar. So one place is if something that has theta d theta bar model, these terms are multiplied as a constant. So from those terms, you get the terms that are of the form psi bar del z psi bar plus psi del z bar. Which is the action of the big one. The action that says psi becomes a morphic and psi bar. I'm skipping i. Well, where else do you get theta d theta bar? Another option is to multiply the theta term of the theta bar. And that gives you del z x del z bar. This is the action of the three bosons. Or do you have any other options? Well, there is another option. And the other option is to multiply this guy or actually, how do we get it? Or it will end up as theta. That would be great. That shouldn't have worked yet. Yeah, that's theta minus square root of theta. Exactly. Good. So what remains is an x square root of omega. That's the reaction of our three bosons. This part is very familiar. It's the reaction of the three bosons. But what's this term? What's del x? Well, this is what happens when you write a write in the symmetric elements of a space. In order to get things to transform nicely in a supermarket, we have to introduce what are called auxiliary fields. Well, the dynamic strategy that these auxiliary fields have no dynamics. In this case, what's important here is that there are no derivative terms in the subset. It's not a propagation. And the equation of motion is also very simple. And the equation of motion is very simple. The equation of motion tells you that I am simple. Firstly, it's free. So that's an exact statement. But more generally, when you've got a field that's not propagating, it's infinitely massive. So it can always be integrated. Yeah, giving you a local direction. In this case, we're retrieving it. It's like just omitting it. So anywhere that's an x. In the full model, there is x would be set to 0. x is just not there. This is usually the situation with auxiliary fields. There are interesting situations in the theory that's not free. If they're not going to be 0, but they're not going to be something specified in terms of the problem. That's the way it happens. In super symmetric theories. The term is called the F term. They're equal to the supermodation. The derivative of the supermodation. But in this case, it's already simple. So you see that this, rather than back to the expression, reproduces the action for both the free goes on and the free goes on. And goes into the way we introduced, could have been one way to introduce this idea of super component, the super component x i. This is one other quick comment about this exercise. It's super basic problem. We won't use it. But it's sort of a, sort of a neat word thing. The thing we really need about this is that, suppose you have x theta. I'll only present Yeah, okay, that doesn't do it. Let's go with theta 1 to say 1. And I'm not writing in the rotation, but it's also function of theta 1, 1. That is x of theta 2. Okay? So suppose I try to work out the OP. Oh, then let me try the expectation value, okay? In super space. What update do I get? Well, I'll tell you what update you get and then I'll tell you how to do it. Okay, let me first tell you the answer and then we'll understand. The answer is I'm just going to put in the dimensions in which we set up alpha prime 2. It's minus of log of z1, minus of z2 plus theta 1 to be theta 2. A whole new pass. Two things about this answer that I want that we should do. Okay, firstly, you see that this answer has let me check the systems. Let me check the systems that we have. So suppose you take the path that is independent of all the things. That should give us the free goes on. Answer? And that's the familiar answer that we know when we set up alpha prime 2. Fine. If we take a term that has a theta here and no theta that should be the opinion that we accept psi. Like if this theta is the coefficient of psi and no theta is coefficient of... So there should be no term that has a theta on this thing and no theta on this thing and that is everything that has the term that is theta here and theta here. Oh, by the way, this mod square should be written as minus log of z1 minus z2 plus theta 1 theta 2 minus log of z1 bar minus z2 bar plus theta 1 bar. So the scary truth also is the separation between holomorphic and anti-holomorphic in that process. And then the last thing to change So the coefficient of theta 1 and theta 2 should be the opinion between psi between the two sides. So how do we get the coefficient of theta 1 and theta 2? We take the x-landers so take x-landers minus theta 1 and theta 2 by z1 minus z2. So this log is equal to minus log of z1 minus z2 minus 1 over z1 minus z2 in the theta 1 is equal to this. Why is it equal to that? Because any function of bosonic or somnonic variables should be thought of by definition because of the data expansion in the somnonic variables. And in this case the data expansion comes because theta 1 square is easy to get that to square is zero. And we can see real writing this formula and this part of course with the reproducibility of these components. I think this is a neat way of packaging it. But now I want to explain that we can guess this answer in super space. We can guess this answer in super space. There is some... Yeah. And I will explain how that works. People remember that we were discussing actually long, long ago. We deduced the form that correlation functions have to be functions of z1 minus z2 and one of the other things about the correlation functions. Meal of it from... Get actually powers and the two-point functions would vanish unless weights were equal. Now, you can deduce the function independent of correlation functions in particular with these two-point functions on coordinates and super coordinates by significant changes. So we can ask that something has to be a function only of z1 minus z2 was translational. You move z1 by a, you move z2 by a, and then you change. Okay? Now, there is a similar consideration. So, clearly things have to continue because translation in values continues to be a subgroup of super coordinate functions. But in general, you might need to get arbitrary functions of z1 and z2. So the condition of invariance of these functions under familiar transformations. Now, you remember the last class we deduced under a super coordinate transformation parameter I say this function delta z. Let me set v to z. So just the formula requirements. Delta z was equal to epsilon into minus i epsilon times eta and eta. This is this i that is indicating the range of the functions. So we deduced the last class. Let's write down the difference over here. And delta eta z equals to epsilon times eta. So under delta z under delta z is equal to under delta z is equal to epsilon theta eta. And delta theta is equal to epsilon is equal to epsilon times eta. This whole thing was effective. And this whole operation must be effective. So let's try to see if we can find delta z to theta 1 and theta 2. Let's satisfy this problem. The first combination is obvious. It's theta 1 minus theta. Because each theta shifts by eta and that cancels in the difference. So this is the closest delta of delta 1 minus theta 2. It's obviously in many kinds of operations. However, there is another more interesting combination. And that is delta 1 minus delta 2 plus delta 1 theta. This is a more interesting combination. It comes from the fact that z and theta both shift under under one another. Let's see. Let's take delta of this object. So let's take delta of this object. So delta of this object with z1 gives us a theta 1 epsilon integrated with z1 theta 1 times eta z2 gives us minus eta 2 times eta and theta 1 2 we have to get. The first one gives us plus eta of eta 2 plus of theta 1 eta 1. We can reverse the internet. So this can be written here. It picks up a minus sign. So this can be written here. Oh, did I do it the wrong way around? It adds actually. So next up is something that messed up either my transformation law or maybe I just decided to do it. You see that if I just flipped this around and decided to do it. So I messed up on something and then I messed it up. So in this case, we should say that z1 minus z2 minus eta of eta 2. Minus theta 3. If we will notice if we will notice if we will notice that I noticed by the way that z1 minus z2 by itself is not independent of the situation. z1 minus z2 transformed by this object which needed to stick out in that reaction. So there are only two objects that are invariant under both translations. The first is z1 minus z2 minus eta of eta 2 and the second is eta of eta 2. So any correlation function that you write down has to be a function of only these two objects. Okay. So now we could have without any work written down this answer. The way we could have done it is that what we could have said well you know it goes on into things. It's this. The only way it could be invariant is by fixing it up like this. Okay. Now there could have been in addition some function of purely theta 1 minus 2 or some function of theta 1 minus theta 2 multiplying some other functions z1 minus z2. But then you could argue quickly using a knowledge of things. The right one later. Okay. So super component invariance in this way unifies various aspects of the system. Okay. Just again we're not going to use it very much but just because it's sort of Q-tent the usual things we talked about let me let you go to the BCC. Okay. So it's a BCCF thing or any lambda thing. You remember we had a super conformal theory of the BCCF which combined to be Kagama for any value. And the way this was in super say so in terms of B you define as beta plus beta times and C you define as C plus beta times you might wonder how do I know which which factor it was beta. You can deduce that from weights. If you remember our BCCFG had weights as it goes lambda and 1-1 2-1 2-1 2-1 2-1 2-1 So beta times lambda is 2 at related weights. How do you remember that? So this let's think of this as a weight of lambda is 2. Let's think of this as the weight of C so lambda was minus 1. And the beta-delta is 2 and had weights lambda minus 1 1-lambda I think it's possible. Let's check whether that's true. Okay, so this whole object has weight lambda by itself. Because this has weight by itself so this object has weight lambda by itself On the other hand, this whole object has weight lambda by I think so, has weight 1 minus 1. This object has weight 1 minus 1 and this object has weight 1 minus 1. So if you were going to put them together in this overview, this is basically the way to do it. And now putting that in the same overview what could the correlation option be? Well what could the action be? I'll let you check the integral b d d t c integral b d d t c works so let's quickly see how that works. Remember the general rule we want to get something that is of weight half we want to get something that is of weight half we want to get something that is of weight half then have the code left and right The d t c bar makes the thing of weight half under the right hand the only thing that had any weight under the right hand was because we can see purely the transport purely under the left hand Now let's look we said that this guy had weight lambda minus half and this guy had weight what? he had weight half that's the thing we call formula because this d has this weird thing like the leading thing has a formulaic nature so generally we can assume that's only an extension we have formulaic supervisions you would have a formula and an example is w1 there is a free strength supervision there is w alpha which starts with a g so not only can you can you do yes you are right this supervision c is anti commuting whereas v is a commuting is a commuting yes you are right about that and this d t bar it makes it always commuting because I think we have only one you will try to work out what the action looks like you see all the derivative with respect to d t bar you see but there is the term that is d t bar times anti-order walkout that's not zero and in fact adds to the density ok so you have to keep that's the only term that can come to you and then you have to keep one of the two d t bar you can answer the d t bar because that goes from there but you need a d t bar to answer this so you can either keep the d t bar as a v expansion which will give you a v c or you can keep the d t bar in the c expansion which will give you commuting ok so you can easily check that this will give you the way that we used to and now it's all nice that the v c you can utilize all our discussion of v c and v t bar to my functions by the following device v of z 1 v t bar 1 c of z 2 v t bar 2 this is equal to v t bar 2 divided by z 1 and z 2 actually this I've written this z 1 minus z 2 minus v t bar 1 minus minus v t bar 1 v t bar 2 to be fancy just but if we expanded this v t bar 1 v t bar 2 it's zero so that's saying that we don't need this term ok but this term is equal to v t bar 1 minus v t bar 2 over z bar 2 but if you want to bring out the components just drop this in your mind and then the term that is that is that is proportional to v t bar 1 is v with c and that gives you a z bar minus z 2 for b and c the term that is proportional to v t bar 2 is v t bar 1 minus z 1 that gives you a z bar minus z 2 for b and c you can check all the various signs and so on that's very beautiful sorry involving writing involving writing super component super symmetry super symmetry super component in super space and this is a very good development to talk a lot about especially when we add to the amount of super symmetry in the world in many interesting ways we're not going to pause to do that here make any difference so now we end our formal discussion of the development of super space and then to use it in this to discuss the stacking aptitude for the world shape of the string but any questions at the moment ok since we've built this all began in the past there's one thing that I will immediately that I will need to use and this useful will be useful but then you wind it ok now let's return to the computing scattering aptitude ok we're interested in computing scattering aptitude to start with just three levels scattering aptitude so that we don't have to deal with issues of volume now at the end last class we already discussed some issues that came up in the study and the first issue that came up in the study when it follows we discussed how on the sphere just the anomaly of the zero mode current of the fight so for this discussion I think it's useful to remind everyone of the results of the quantization of the of the beta gamma system so you remember we got this beta gamma super conformal first system and you remember that we both organized the system beta was equal to e to the power minus 5 beta was equal to e to the power minus 5 times del the symbol and gamma was equal to e to the power plus 5 times eta where zeta was always 0 eta of which one and these two together zeta and eta form the basis system also this final formula is theory because the three goes on theories but with a certain background charge okay so the translational current of 5 not of the operator del 5 was not a genuine primary field there was an anomaly in the OP of t with del 5 and that anomaly led us to conclude that del view genuine for the translational current was not 0 it was equal to some number times integral r and that number worked out for the wages that were relevant to the super string that number worked out to be equal to 2 when integral r was what it is of the sphere so del view genuine was very similar issues in the bc where this del view genuine was related to the number of 0 and 0 was of b minus 0 was of c this is very similar to the system you can think of it directly in both the diagram you have this mosaic current it computes the stress it gives you something not 0 and then it needs to be developed since the beginning it would translate back to an anomaly okay and the same thing as I want to remember the other thing I want to remind you del 5 of 0 is equal to is equal to plus I think like 3 to the moment I think plus is equal to minus yeah but it was the wrong side it was the wrong side wait plus minus I get it was the wrong side it was the wrong side but then we adopt that and write in vertex operators e to the power 5 e to the power minus 5 rather than e to the power i to the power i okay good all of this sufficiently familiar that it doesn't seem mysterious I know you don't remember the problem as I did today let me know when you look it up do you know what you have on this okay good so we first came up and that that way the spot look as in a development of static amplitudes in the boson extreme we are motivated by the spot by the following time period we've got some we've got some worksheet extreme and there's a little cube coming in here some more cubes coming in here going out okay that represents states coming that's basically okay we want to compute the amplitude for this object and our idea was that this whole thing's been formally read so we've made it map this just to picture with the vertex operators so whether everything here will be mapped with the vertex operators this intuitive picture will tell us that the objects that we should put at the origin are the vertex operators that correspond to the states now we know what the vertex operators that correspond to the states in the in the super stream we worked this out a great detail in the last 2-3 months and the vertex operators that correspond to the states are appropriate lattice super conformal primaries that's equal to the power minus 5 in the second or equal to the power minus 5 by 2 in the roman sector because e to the power minus 5 was the vertex operator corresponding to the vacuum of the beta gamma cft in the ls sector and e to the power minus 5 by 2 was the vertex operator corresponding to the vacuum of the beta gamma cft in the roman sector as we worked out in the beginning okay if we're very infinite this can't really work it can't really work because suppose we're interested in 4 ns sector fields we'll get a factor of e to the power minus 5 in each of these times whatever comes from that now we've said that this but that correlation functions vanish unless the net charge are inside of insertions on the sphere is equal to 2 as of the lower volume the translation could have come that 2 wouldn't be 0 it's not 0 it's something mixed okay so if we use the rules that we just insert vertex operators as we played some state operating map we use that rule then the boundary gets 0 unless we're interested in 2 guys with 2 scattering 2 is not a scattering it's just something wrong how do we cure this problem okay I'm going to first give you the technical answer you get 0 because the net charge with that whole thing with k scattering with k ns sector object would be k whereas the only way to get down 0 operators is the net charge of the sphere where it's 2 you think of this way suppose you think of it like initial time final time you've got some operators inserted in the initial time if there's no anomaly in the gun then the charge of the operators inserted in the initial time so the charge inserted in the final time because the outgoing time limit that gives you net insertion 0 for the charge of the operator if there's an anomaly we don't get 0 but we get something fixed that's equal to 2 so the only way to get down 0 operators is if the set charge is equal to 2 or perhaps more formally suppose you add some into the power n1 power 5 power 5 had translations as 5 then let's translate this subject if that's translated by a then we can prove that this object is equal to itself which means that this is not traction it's 0 this translation is not quite reasonable because there's an anomaly fact which reduces the secret power 4a even the power 4a minus 2 there must be that we just put the vertex operators due to the states and under the state operator map inside our inside our coordinates that doesn't work okay now at the end of the last lecture I told you the way out you see even in the bosonic stream we didn't really put the vertex operators due to the states under the state operator map always because some of the vertex operators that we integrate with and those guys have no scene search the scene search is what we get by the state operator on the other hand there are the guys that were not integrated and they had a scene search you see the careful analysis of the path that gave us string theory to understand what's wrong to think of these vertex operators all being as the guys as the guys who got into state operator map some of them were but others were of this integrated and so in the course of that they did not come from the state operator map so you might think well well it was just very kind to do these operators let's say we're dealing with 4 nsx 2 of these operators but the other 2 operators remove the other 2 operators that's like removing the seas and the integrated vertex operators in that you might be tempted to think oh there's nothing to do but if you did that nothing would work what would not work but for instance the answer would depend on which operator you remove each of the power that's fired from and so on and you get all guys again you won't get any kind of negative you just do random things in string theory it doesn't work so what is the right way of doing this well at first telling them the words understand this more deeply the words go as follows the words this is so much as an analogy the analogy goes as follows you see in the wasa history a part was typically in the sea what we did was integrate the operator over the world shape now integral integration here is replaced by the idea of integration you see integration of the world shape makes sense for things like you were fixing world sheet translational and so on so next the beta data have to do with super space so you might think what you should do is to take is to strip away to the mic strip away strip away this guy but then also take oh and integrate the opposite same with the g minus half g minus half that is replaced or by the integral of super space and then strip away to the mic that would be all way of making an analogy with what you did in the wasa history not with the correct answer which as for this reason that I introduced the super space business they take it both way however you see that it's the correct answer you see that it makes there is a sense of the answer but I am dying to do that but frankly you understand the answer so the answer is for instance we are just capturing only a roman sector sorry never a german sector so we are just capturing four we would choose two to be able to use a picture with the e to the power minus five and two players might know e to the power minus five but there are four as replaced by e to the power that is g minus half g minus half that's the rule of course if you had to go scattering n you would choose two to be in the usual picture and then you might get minus two to be in this picture but as we have shown now for instance it will not matter which one is this will not mess up we will see you see that you need to choose it because this object is not a super conformant right this object was this object is so then you need to choose about where this all makes sense okay this is the this is the rule and you understand this sir if you are on the other side so basically the we need to we need the two operators and for the rest of the lines you should be able to choose minus one e to the power minus five g minus half of course is not an object it's an operator the same replace all by one replace e to the power minus five o by a particular same now let me try to explain to you why this rule makes sense I will keep this on the board okay now in order to see why this rule makes sense let me start from a side rk in the place and the rk in the place is it clear enough clear enough a loose end from our discussion of bosonization of the vita okay we can so the vita down our system was bosonized into this where zeta e to the power phi that was our claim and the object of that claim is that every correlation function computed in the first theory can be computed in the second theory as well how we can make the appropriate replacements and you remember that our proof of the statement was essentially to say that the balls and zeros of of op is assure us that correlation functions of the two sides would be equal just because they have the same organisms up to some overall normalization okay however there is one serious issue and that is that overall normalization might be zero if you look into that series issue when studying this theory what do you think see suppose we were interested in computing the kind of correlation functions that arise in string theory okay the kind of correlation functions that arise in string theory that is we have insertions of e to the power minus 5 that is the matter stuff discussion insertion of e to the power minus 5 plus 5 but no other insertions are the beta-bamasic of this eta zeta-intelsic correlation functions computed in the goes in x language but it is zero but let me simplify the simple reason is that zeta is weak to zero and so has a zero modern input in particular just the constant and you take a look at zero more because the insertions that we are interested in have no factors of and similarly if it were computed just an insertion of beta and gamma beta and gamma the way we can do that is by contracting zeta and zeta contracting 5 5 so that leaves nothing to soak up the effect of the zero I tell that insertions of making this replacement that you get and just putting it into the zeta-beta field that we give you the same correlators that can't be quite right at least in examples where we know that the beta-gamma calculator is now zero it gets zero what is the right way to do it what is the derivation derivation if we have got these non-zero if we have got these non-zero correlation functions they have to be proportional to each other so is there a way of salvaging this derivation it sounds bad it sounds like I am leaving you down the government path I mean a very formal argument but it doesn't make any sense how much the formal argument is salvageable and the salvaging when it is taken through is very important and this formal argument tells you that it is salvaged if you are correct for the following proposition that is for any beta-gamma insertion you make the corresponding insertions of these but also add by hand a single insertion of zeta because we are not interested in the beta-gamma or the formative theory for itself let's restrict that action those correlation functions then we need the correlation functions in which correlation functions are totally independent of zeta they only matter okay? we can't get those correlators but we also put by hand a single insertion of the operator zeta at some point in the world's history now you can ask at which point in the world's history and the answer is it doesn't matter because the only rule of that insertion is to soak up the integral of the 0 whole so that integral gives us factor of the function but it gives us the 0-mode function evaluated at that point but because zeta is constant it is weight 0 the 0-mode is a constant function so the constant function is the same evaluated at any point this is a very important point when we are computing correlation functions in string theory we are supposed to compute the correlation function in the phi cfd and the itas zeta itas zeta is almost no rule it is almost no rule in anything except for one thing if you just did the integral over the itas zeta you get 0 unless you also have the insertion of zeta so let's just put that there and that insertion does not it continues to give you translation invariant correlation functions because the only rule in life is to soak up the 0 there is more elaborate discussion there is just more elaborate because invariant not real options come we understand that the kind of correlation functions that we introduced in the algorithm in string theory have the insertion of zeta now it might be valid that this insertion messes up many things okay and the kind of thing that might vary you is the bottle the kind of thing that might vary you is the bottle look suppose I put some vertex some vertex operators and I put it on the insertion basically now this vertex operator is brsq closed so so well suppose I add something brsq trivial so I put a brsq trivial insertion so some operator with a brsq come a q brsq now now this is always 0 this is supposed to be 0 actually brsq trivial obviously is supposed to be 0 the way it usually works is that you this is a sphere actually you get this circle unwinded around the other sphere that means little loops surrounding each of these guys they are all brsq closed so those loops vanish oh what about this what about this is that guy there what's going on so this is supposed to be studying the following object it supposed to be studying the object that you get by integrating j with drsq understand clear your interest to understand the object if you want to understand it so with an understanding actually because it's not an interesting object for the reasons we described earlier that sounds like a monotone we're going to study the monotone that you might think is necessary so let's go ahead and do it okay now the thing that we want to do is to compute this so what we have to do is to make we have to compute the pole in the brsq talent with z we want the pole we want the particles like that of course there's a straight forward way to do it a straight forward way of doing this to take the expression of the brsq talent that we derived a couple of months ago it doesn't make a comment more than either expression and then build a combination now it's straight forward and painful and you want to do it that's not the approach that we're following we're following approaches to give you a strict generation of this this this residue which is sort of quite neat okay so the strict derivation of the of the object and I'm interested in one more thing that should write down here write down the structure of this this file okay firstly I'm just going to go to what the real estate talent is I'm quoting 10.5.2.1 which is the book there's nothing written down on the book cc plus gamma this is the matter bosonic part matter bosonic stress tensor matter bosonic stress tensor plus bc del c plus 3 by 4 del c beta plus 1 by 4 c del beta minus 3 by 4 c beta minus beta minus b gamma this is the expression for the brsq talent this horrible huge expression for the real estate talent okay so it's this object that we have to take with zeta and find okay so how are we going to do it so the way we're going to do it is to note two things um the first thing we know it is that it is something so important it's important to keep the gamma real and look at the op with any function of vtac this function is a function of vtac gamma for the value of 0 it's not dependent on what we get so of course I try to compute this over here I take gamma with any function of vtac gamma note that it's just vtac gamma what are we going to get what are the contractions gamma contracts and contracts gamma, gamma, gamma op I'm looking at the whole part but what are these vtac gamma for beauty but it doesn't matter they expand it and then what vtac so it's supposed to be gamma with vtac so one will say what are gamma with vtac for two vtac exactly two vtac over z what about vtac at the end would be n vtac n minus 1 so this is going to be there with respect to vtac now plus what do I get in this statement did they get any gamma or any function of vtac gamma and that function gets the part that is irregular the part that is singular it's just the differential of that function but the second thing that is a little dodgy I don't understand is the following let's remember you might remember that you might remember that let's put let's let's evaluate let's evaluate the op between beta and vtac plus minus 5 let's evaluate between gamma and e to the power let's evaluate this op now e to the power minus 5 let's see gamma is e to the power 5 times eta eta goes through e to the power 5 e to the power minus 5 e to the power 5 to the power minus 5 so it's simple you use fact log that the 2 y function is minus 1 so this gives you 7 kind of normal order eta of 7 e to the power 5 of 7 e to the power minus 5 of 0 and this happens here e to the power gamma of 7 e to the power minus 5 e that starts is Z it wasully onboxing we were very familiar with such operation how many things remained familiar with such operation? for instance c with itself at an orp they would start with z just try any content 0 x 0 0 0 there exactly is that's it It's only unusual that you lose money cause! What does oscillate! so the question I'm going to ask is suppose I was to try to rewrite this oscillator And some function, some operator purely in Vitta Gama. Suppose we try to write it as some insertion in terms of Vitta Gama. What kind of insertion, what kind of insertion can I write that would have the property that if I took a gamma field in the end, this object would go to 0 like so. If this one is approached. If we support, you see, what this insertion is, it's some insertion that forces the gamma field to go to 0 where you insert that. Where, of course, a local function that I have, it's the object that will come. An insertion of this object into the path. It will be integrated, not over all fields. But only on those three configurations such that gamma, the value of the gamma is 0 at the origin. So then, isn't that part integral? If you bring, if you try to evaluate an insertion of gamma near 0, this thing would appear to be a data expansion. Would be, would suffer, that's it. It would give you an interpolation. You know, there are a lot of things you would want to check for this. This is quite a signal as I've been searching. So, I'm not sure if you think of it as a literal formula, a way of thinking. But anyway, there is a way of thinking of this formula, at least since I've said it before. Of thinking of e to the power minus pi is the insertion of the delta function of gamma. Notice it's not the delta function of the function. Because we take e to the power minus pi, we take the overview of beta, and that will give you a whole lot of the right of z because it goes the other way. So, it's, and there are various things that you might want to think about and check here. But let's just move on, accepting this rough interpretation. If no, if for no other reason, just as in order to motivate this quick and dirty derivation of what we're going to get here, you should think of this as perhaps motivation. This serious derivation would be much more involved. Much more straightforward. That is, both are nice to be doing the other thing. If you don't like this, many things to explore. You might be able to make this interpretation solid because there are many things to explore here. We're not going to try to do that. Let's just go ahead with that. But any questions about this? What is beta? What? What happens if you have such a thing as beta? When the beta gives you one over z. We get one over z. What? There's a, there's a, the other, you see, it looks like, it looks like it should get delta. What kind of approaches is it? I'm not sure. It's worth it. Suppose I try to approach this object. Suppose I take beta of z and approach it to five minus five. Okay. Then according to this formula, what I'll get is delta times of z times, divided by z, divided by z times e to the power minus five of z, e to the, where? Minus five of z. Everything. Now the pole is the power minus two five. So why don't we expect a general process that we can have with this object? But you give me the derivative of that function with respect to z. Okay? But by z. So this interpretation would work if you would argue that for some reason the e to the power minus two five was the operator that was derivative, delta prime. e to the power minus two five gives you a higher order zero. You see, it gives you square. So it's sort of like it sets even the first derivative to z. Okay? So it's not obviously, it's not obviously consistent. Okay? But as you said, you know, this is the kind of exercise that if this interpretation could be made completely cultured, and perhaps it can, there are many exercises of this sort in what you want to do. Okay? But it's not obvious that it's not working. The Puritschitzky asserts it's true. He generally talks such things through. Okay? So it's probably, I mean, it's probably true that if you do these exercises, you find that it works. Okay? In fact, there's one such exercise that would work. Okay? We don't have a talk today, so we want to do it in a second. Okay, great. Let's, let's, let's move on. Okay. So, now what we're going to try to do is to take this, this instruction and use it to find, along the lines that Joe Benoit has suggested. Use it to find, use it to find other sports. So, how do we do that? It's, it was to use this formula that, okay, what we're going to claim is that this object, zeta, has the inductive, it was getting a little weird. But it's okay. You see, teta's function that makes sense only in one language. Right, makes sense in all languages. But it's a local operator. It's a line of operations. Okay? But essentially in one language, because everything's that. Everything's constant, say. So, teta's, teta is an insertion in the part of the table that sets veta to zero everywhere on the left of the insertion. And, you know, leaves veta obituary. The right. It's zero. If teta's right inside the problem. Okay. We should confuse the, the identity of teta. Teta obituary makes sense for a real function of teta. I should confuse the direction. It's not teta in position spaces. It's teta in space. Sorry. Sorry. We'll draw about it. Okay. So, it just says, you know, it's doing the integral over only positive value of teta. Because the local operator, it's very valid. It's like, but everyone else should do the integral over all the value of teta. But at this point, you have the integral over only positive value of teta. And I'm going to claim that this, this teta object can be interpreted as inserting teta. Now, let's see. Let's see how, how we did this thing. I'm going to make this claim by, by doing the modeling thing. I'm going to compute gamma of z to the teta of teta at z. At z. Okay. According to our general formula, I should get this option. The delta of the teta function. So this should be equal to delta of teta. Equal to delta of teta. Divide by z times z. Right. Ba, ba, ba. Okay. But let's see what it is if I make this, this integration. So what I'm going to do is compute gamma of z times, but I'll replace gamma. So that's equal to half phi of z, eta of z times zeta of zero. So the eta, zeta, cfk is a one over z. So as far as the pole goes, this is equal to half phi of zero over z. That's easy. I got it. And we had that e to the power minus five was delta function of gamma. But by exacting the same dodging, e to the power plus five was the delta function of eta. Okay. So if you take this, so this is equal to this. If you take our earlier interpretations. That's correct. So that forces you to interpret. That forces you to interpret zeta, this object, zeta, zeta, zeta. It's very interesting. Okay. It could be eta of eta plus any function of gamma. But just from the fact that the eta, eta, cfk is completely non-singular. That's a function. And that's completely non-singular because eta is bigger than z. So gamma eta, what do you mean? The gamma, the eta, zeta. It's bigger than z. Zeta is bigger than z. Because the eta, cfk is completely non-singular. Okay. Great. So we conclude that we really want this delta function, doesn't it? We conclude that this zeta field should be interpreted as eta. So now my mind is coming to the point. Because zeta is what we want when we are going to be, you know, under this, beyond this. Okay. It's fine. Okay. So now let me immediately compute what JBRSTM on zeta is. This whole exercise, all the last 15 minutes, was in order to express zeta as a known function of gamma in zeta. Because then we can use just the unbosanized beta gamma expression for JBRSTM. I'm not going to put that. Okay. So now we've achieved this. So we just have to proceed with them. So the gap. So the gap in this. So the gap in this is all the follows. We take this guy and we act on eta of eta. Firstly, it's a function only of eta. So the only non-singular, the only thing that we're going to get is the delta from gamma. So we connect all the terms in both gamma here. We have to connect all the terms in both gamma here. And act. Okay. So we've got gamma here. We've got gamma here, gamma here, gamma here, gamma here, gamma here. This term has two bounds. Okay. So contracting into the view to derivatives of the delta, of the eta. So give you a term that is of form beta and delta prime of beta x, beta and delta prime. And this object here has many terms. This object here has many terms. The term that we're going to be interested in involves the matter piece. So if we get a term in this object, we get a term here of the form gamma, just tf, tf times delta. We also get other terms involved, let's say for instance here. We will get a term that is density beta times delta beta. So that is zero because beta times delta minus delta. Anyway, this is really a little bit of a problem. Okay. So the claim is that when you work it out carefully, perhaps you should think of this just as a good one. No, unless you get the time it doesn't matter. The claim is that in the end, you can work out the BIS term and you get the answer. I mean, it's OP. And you get the answer JBZ times theta times zeta of zero. It's equal to minus 1 by z squared V of z delta prime of beta zero plus 1 by z tf of z delta. Finally, we've got this thing that is of interest to us. Right? In particular, we find that the whole part, the part that we were interested in involves the f times the delta beta. Delta beta of zero, we know what that is. The other formula is in modernized language, it's e to the power i. Now let's go back to the thing that we started off with. First thing, what about the BIS to come? What if we take this BIS to come and surround this fake insertion of zeta? We don't get zero, which looks like we're just computing what we get and we leave it there. However, if some obvious was going on there, although we don't get zero, wherever we get has no zeta left. This is purely function of five. And so it will vanish. The path of delta will vanish because the integral of zeta is zero. Precisely the reason that the zeta was needed in that path of delta there.