 So today's class, I'm going to finish off all the leftover items and I'll start talking about the calculations and the attitudes. But maybe since we won't have the time, maybe depending on how much time we take in the beginning, we'll clear up on the results and maybe I can ask you for questions and comments. Because in a way it's the end of one part of course. Maybe we can have a little discussion. Okay, so let's start. So in the last class, there was one outstanding check we wanted to do of the entity, the scattering entity. I'm not going to write it down. You know what I'm saying. And that check was there of BRS. Okay, BRS. There are two things you might be worried about with BRS preparedness. You see, we know that C times V is mirrored. This was the content of our lecture a couple of days ago. Okay. But however, our insertions also include things like D2, D2 sigma square root g times V without the factor of C. Okay? Now this is not by ZPR. V by Z is not reality. All we know about it is that it's a 1-1 that makes up regular. Okay? So the question we're going to ask is what we can say about the reality transformation problem. Is it this one? This is one more check we have to worry about. Okay, so let's remember. So J BRS E was equal to C times D plus R lambda. But lambda doesn't matter to this operator because it's made up of the matrix. Okay. So now let's take this BRS E operator and act with an LV. So let's look at the OV, C of Z, C of Z times V of Z. Okay, what we're going to do is compute O of this OV. That will give us the reality transformation property of the operator here. So what do we get? You see, we know that the operator is 1-1. So that tells us that the double point is C of Z times V of 0 by Z squared. The C is just going on to the right. H V of 0 by Z squared by H is 1. Okay? And then plus that's the other number to C of Z times L, like I just said before, plus C of Z times L Z G by Z. Okay. And be careful. So because it's obvious now we have to C of Z, Z, Z, V. But there's another term that comes from del z times C. Okay? So when we group that term, it's what? It's del C. So when we group things together, what we get? We get L of C V divided by Z. This is the full point. Let's work it out. So this guy, the contribution to the pole, you know, the del z times this, this contribution that is Z del z C times V. Okay? So the two pieces of that are 1 by Z. C and del z V plus del z C times V. So we're using the chain load of possible density of the derivative. This is del of C V divided by Z. Inclusion of the change in V under the VRSV transformation. Okay? Yeah. So now that we're working, you know, the form of the H, say, that's locally transiting. Okay. So the change in V, the change in V under VRSV transformation is simply del z of C. It's taken an integrated of all space. That's the integral of the total derivative of space. In the end, it will be doing some integral of the component. That's just C. Okay? Okay? So the lack of VRSV invariance of the insertion of V. So V is not VRSV invariant, but it's integral over the whole map of this. That's the point, because the variation of the VRSV transformations of this object is a different one. Okay? Now we might wonder if there's any other term in this category I have with you who has VRSV invariance question. So this is, these are the insertions of V and del H. Okay? So VRSV transformations act on V. VRSV transformations act on V. And what's the change in V? Well, the change in V is D. Right? We repeatedly use the fact that J, VRSV, say, 10V over 0, was like D. It's not before and about. It doesn't really care about it. This change is of the form D times del H. There's a summary and it's not real. So you might be worried. Any ideas about why this is done? What is the definition of the stress density? The stress density is the variation of the path integral under a change in the metric. Okay? So now we take this stress sensor and take the rub rub rub with this change in the metric. What is this? This definition is delta by delta of the key of everything else. Everything else that went into the definition of the stacking amplitude excepting this one particular insertion. So all insertions except this one and take it to the boundary space. Is this clear? Since the expression for the scattering amplitude involves an integral of a moduli space, provided there are no contributions from the boundaries of moduli space. Provided there are no contributions from the boundaries of moduli space. This thing will be much easier. Now the question of whether the boundary of the moduli space will contribute or not is actually a subtle one. And I'm not even sure we have enough to look at it in this course, but when we go to complicated three-month services, it's a subtle one. Generally the answer is they don't, but there's no general one. That's something we address in the search case by case. No, moduli space can have some of those natures. Let's postpone that sample data when we do it. If you're really in your service contributions, then that will make sure that we are at the end of this insertion. So at this general level, that's all I wanted to say about the invariance of the amplitude. There's actually one more thing, one more discussion I want to have about the invariance and the diffeomorphisms, moving the points of insertions of these fixed operators. But I think the best place to have that is when we start doing our first calculation. Because the arguments are a little subtle when you say abstractly. It's much easier to say in concrete context. So I'll postpone that discussion to make it to either the end of this lecture or the next lecture. So please remind me when we do scattering amplitudes, treat that as scattering amplitudes, we have to have in that point a discussion of the invariant field that you have under motion, the points. But while you're at that, we're done with our general discussion about scattering amplitudes. Any questions about it? We also saw that it doesn't matter. So is it possible to write down all the vertex operators simultaneously? No, it's not possible to write down all the vertex operators simultaneously in this form, unless you also put some B insertions. You see, because we're not possible, then you would be changing the number of C insertions. And you saw that the only way we got an anzine or path intake, you saw that that would not be, but by itself, the system would get an anzine or path intake. But we also had an anzine and a path intake. That is a very good question. Yes, there is a way of writing the path intake, in which you put a free insertion as a sequence, and then have a B inserted on a contour around the remaining C's, the ones that would have otherwise been connected. Now, this is a way of doing it that... Fine. I don't want to talk about that in Egypt. I haven't prepared it, for instance. And it's a slightly complicated mathematical discussion. It's a discussion on... Well, it's slightly complicated, but I can point you to reference it, right? And if it's a pressing question, if you feel very echoing... Okay, let's say in the next lecture, I'll give you a little discussion on this. There's a way of addressing this. I will have to remind you this. But yes, the answer is yes. That is a good way of looking at these amplitudes, where you put extra B insertions. It's actually the most elegant way to write it, this way that you're doing it, because it's the most elegant. It's also a mathematical way to write these things. It makes clear the basic objects in some things are the insertions with C times B, which is very nice, because that's the way... That's the way... That's what the state operator can have to use. If you want the state to have a spot in there, matter. Of course you can't state it. This is the operator constant. Okay. Good. So... Okay, so I'll go on for another 20 minutes or so, and then I'll talk for a little discussion on everything we talked about so far. Okay. So let me now go on to a discussion that I want to start. You know, we're building up now to actually calculate between the strings. Now, you remember that the standard number of views were evaluated upon three months. Okay. So I said that we should be getting this one and sum it over all of your months. Okay. So there are... We have this abstract discussion involving the number of C0 bonds and the number of B0 bonds on the Riemann surface, whatever that happens to be. Now, it's an actual question to ask. Can we calculate... Can we calculate how many C0 bonds there are in the Riemann surface? How many B0 bonds are there? And the answer is yes, of course. So my very bad study question. But part of the calculation can be done very easily using physics rather than actual things. And that's the part that I want to... that I want to... I want to review for you and then about the remaining part that I'll just take to you on some... of your thought. Okay. So, the part that I want to review for you might remember that when we were discussing the VCCFT, we found that on general curve manifold, then the Uj mu are equal to 0, but was, in fact, equal to... was, in fact, equal to some number times the curvature of that. Do you remember this? There was an exercise. I'd give it an exercise. But I don't choose the point. At least I've never did it. Not sure anyone else tried to solve the exercise. We worked this out for trace of the stress tensor in the derivative enistability. There was 2 lambda minus 1 over some 4 or something like that. I'm going to give you the discussion without keeping track of the numbers I'll ask you later. But the key question is, what does this mean? You see, it's not suggested that we get the curvature on the right-hand side, because if I take this equation and I integrate it, if I take this equation and I integrate it over a particular manifold, what I'm going to get here is a quantity that is proportional to the eigenvector, because the integral of R of Riemann's service gives me some number that is the eigenvector. If we conclude that integral d t u j mu over the manifold is something, some number which I'll quote for you at the end, that is the eigenvector. Or from the point of view of category that I'll give you the skills kept in mind. Or from the point of view of the eigenvector functions of the BCC edge today on a particular, on any Riemann's service. Now, let's remind ourselves, let's remind ourselves what the definition of the count cost on the intersection was. Okay? What we did was to take any action, so we had some d phi in the power of s and we said, well, suppose we have some entry, so suppose we have some symmetry transformation and delta phi is equal to epsilon phi is equal to phi is equal to phi tilde that's epsilon times delta phi, phi tilde dot. We got some symmetry operation. Then we said that we know then we said we would make the epsilon function of x of the virtual signal. We said we know that it was a symmetry. Then the change in the measure of the action will involve no terms involving epsilon. But could I in general really involve terms involving the derivative of epsilon? We said that we have some entry so the change in this was was was well, under this transformation this thing goes through d phi tilde e to the power minus s by tilde plus del u epsilon alpha del alpha phi g. And this was our derivation of the term. Because these two this is just a variable transformation of this it follows the the integralism. It's value is unchanged. Since these two were equal we could pull out one power of this if we got the insertion of this object inside positive papers was zero or if there were other quantities here we could consider we got the water to remind them of it. Suppose it turns out as we've seen that del u j mu is not seen. What does that mean? Well actually we need to question this of del u j mu because del u j mu is not what does it mean? What it must mean is that the original operation was not actually a symmetry near the symmetry of the classical but it's not a symmetric measure of x times e to the power x ok, so suppose what the situation is is that when we perform such a variation we also get the term which is plus epsilon times plus epsilon in terms that the action and measure don't remain invariant even when epsilon is an offset. We can still form this integration by powers operation here and still bring down powers of epsilon that's not going to change anything ok, so during the integral over all of space what we conclude we conclude that the insertion of epsilon times the integral number times the integral of del u j mu is not a symmetry we don't learn the equation that del mu j mu inside the path integral is zero however we define another equation and in particular if we choose epsilon to be a constant ok, with constant over all of the manifold we find that the integral over this integral of j mu it is plus the integral of del u j mu is equal to zero but in our situation here we know the integral of del mu j mu inside the path integral it's kary that reflects the non-immediates of the symmetry whatever it is, it must be directly kary in which we get the kary kary all that information many is minus 1 by 2 times i is this one and not this one so, what we can do since we know that the integral of del u j mu over the whole manifold is we know this path integral changes when we make a constant symmetry transformation because when we make a constant symmetry transformation, this is the only difference epsilon can be pulled out and we just know the answer just with u this is zero isn't that operative as we are the operative is we didn't need to do it I'm like that would be completely that's completely right but this also they are that's great but I don't even need that all I'm interested in is what is the change in the path integral under a constant symmetry shift you know I just take the symmetry shift which is c goes to c into e to the power i alpha b goes to b into e to the power minus alpha alpha is a constant and I want to know what the change in the path integral is okay that constant symmetry shift we have concluded that integral now let's look at the case that we have just written db times dc and d to the power minus s dc goes to e to the power so this is under what operation? under the operation c goes to c e to the power i alpha b goes to b into power minus i alpha this goes to some number e to the power i alpha times chi times some number that we have to get in that unit times integral db to c e to the power minus is this clear? suppose we had a path integral with no no other insertions we were just in the db dc e to the power minus vc path integral as long as this this number here was non-fear we have just proved that the path integral is 0 because we have proved that the path integral is equal to some phase times itself because remember all of that is a variable change when I say goes to and evaluates this path integral but when the evaluator changes again another path integral of the same form so the same answer times something so I have z is equal to e to the power i alpha chi times z so the only variable that could be correct is a path integral to 0 if this number it must be also clear to us that the only path integral of this form that are non-zero are those with insertions of c and b you must have an operator inserted here such that it is charged under the c goes to c to the power c times e to the power i alpha and b goes to b times e to the power minus i alpha is equal to cancels this variation we get some number here m path integral of the vc system that are non-zero are path integral where let's call the number of b insertions b and b and the number of c insertions c that only when b minus c is equal to m will the path integral be non-zero? yes it is firstly I should emphasize that things that we get in string theory are not correlation functions of that form you know that's we weren't dividing by b this argument holds all of the bs just for the bc system that's a few that's a few correlation function of the bcc entity on an arbitrary manifold is just 0 you see basically you're correct about this and basically it's the statement that if you want proper Hilbert space interpretation of this theory the inner product will be defined with extra actors of b0 system let me try to see um independently try to explain that you see because we also saw the ground state of the theory wasn't a right ending yes it's a holiday to that exactly that if you weren't really doing the part of the technique of the ground state you would have extra insertions right there is a better way to say this there's something better to say about this but I'm not going to get into it again that's a good question but it's just true it's um this is a situation though whenever you have whenever you have anti commuting fields it's a path path integral involving the 0 but it's just 0 you need some insertion you need some insertion in order to get something non-zero so just by the statement of a path integral it's true and to translate that into some statement of the hidden space you often have that's a good question I should give you a better answer but getting back to our purpose so we have this path integral is non-zero only when the number of b insertions minus the number of c insertions is some particular number let's see when we make as many b insertions as number of b zero words the same insertions as number of c zero words okay so that m b minus mc is which is equal to m is equal to the number of b zero words that why the number of b zero words remember with the number moduli of the numbers conforming killing vectors un-fixation vectors so we conclude that the number of moduli conforming killing vectors number of un-fixation vectors is equal to to this stuff because in standard normalization we say that the integral of r but we normally say that the sphere has chi equals zero but the integral of chi over the spheres you know some non-zero number so this was better in the case of 9 minus 1 it's proportionally 9 minus 1 it's zero for the torus it's non-zero with one sign for the sphere it's non-zero with the other sign for the first genus 2 sorry so in standard definitions this is 9 minus 1 so now we go through this whole thing and carefully track through what this factor is something from the coefficient that we got here and then tracing through the definition of the element to see exactly what this change can't find is you can calculate my angles it's a painful exercise involving taking cracks of 2s and 4 bars and i's and I'm not going to do that to you though I strongly encourage you to do it yourself which excuse me and the answer to that is the number of v zero modes minus number of c zero modes I'll give you the answer for the answer for the number of holomorphic guys this is because of doubting because there's actually no holomorphic set here so this turns out to be equal to 3 of the element minus 3 and this is called and we're working this chi chi equal to 0 in the sphere chi equal to 1 of dollars and to calculate is the number of the difference between these number of the number of moduli and number of the log okay we have to manage to calculate them individually so let's see if for for low manifold we can just directly do the calculation in a net okay so now we're going to try to do the calculation this way what is the sphere the sphere as follows we can think of the sphere as conformally equivalent to the complex plane by the usual projected map so you've got this projected mapping that maps points from the sphere to points of the complex plane this projected mapping maps points to the sphere and points to the complex plane including the point at infinity so it provides really an honest covering of the sphere okay we should do it with two patches we could have one projected mapping from north pole another projected mapping from the south pole and these two patches overlap everywhere except the north and south and you know I'm honest on how of course this is the same relationship between that and z is equal to 1 by x okay so we've got two different patches one is all of the finite region of z plane the second is all of the finite region of u plane and there is a conformal transformation that goes between them that identifies points that are everywhere except zero and infinity and this conformal transformation is there to our point so now it's really most purposes for most purposes we can think of the sphere just as being the complex plane however we must keep track of the fact that anything that's supposed to be well defined all over the sphere must also be well defined at the point that infinity at the sphere now we have to understand what that means what it means is that when you take that object and transform it into a real coordinate that is well defined that you can see that means for something to be well defined at the point that infinity at the sphere let's try to understand let's try to understand so in our parameterization we know from the form of the BCCFT you remember that the model i was zero modes of the operator p dagger these conformal hearing vectors were zero modes of the operator p but in this particular on the complex plane the operator p and p dagger were words just simply monomorphic derivatives and there of course the action was b del c b del bar c plus b del bar del c so zero modes of this zero modes of the operator p or p dagger are simply infinitesimal vector fields or infinitesimal two tensor fields that are entirely monomorphic and it is the number of all well defined we want b z z monomorphic because the operator p dagger is just the operator del z bar that should be well defined on the whole step what the transformation property what the transformation property of this object b is under the map that takes us to the u z is equal to one by u so and u is equal to one by z is equal to minus one by u squared and u and so something that transforms like del z del z which is what b z transforms so we get b z z is equal to one by u to the fourth and b u u so we really put it like this z to the fourth and b u u u but now b z is said to be monomorphic and we want it to be well defined at z equals infinity which means it should be well defined at u e now it should be well defined at u equal to zero sorry z z z is equal to one by u so d by d is the n because if you put it up sorry d by d of one by two okay this is equal to u squared minus u squared d so this guy now goes to the u to the fourth one by one by z to the fourth okay that is equal to z to the fourth times b z we want b u to be well defined at infinity but there is this explicit factor of z to the fourth which tells us that b z z must die off at least as fast as one over z then must be an analytic function which dies off at infinity like one by z to the fourth how many such analytic functions are there right? functions of the plane how many analytic functions on the plane exist which are everywhere analytic but die off at infinity like one by z to the fourth at least as fast as one yes that is the famous theorem right that the only only function that is bounded and analytic everywhere is the constant that satisfies this problem number zero goes to the operator p dagger number zero goes to the operator p number zero goes to the operator p okay so suppose we are in the operator p so that is the number of vector fields that is the z of b z to the fifth does not say magic z is equal to one by u so that tells us that c z which transforms like d z so d z is equal to minus one by u so c z is equal to minus one by u c one by u squared now we want all more elements c u is equal to is equal to c z by z c u is equal to u squared times c z which is c z by z what we are looking for is analytic you know analytic vectors okay that c up is specified because now index is up rather than down now that is not true so what is that c up here all we need is that z c z grow at infinity no faster than z any functions are there that grow no faster than energy than z exactly now the only functions we need that every one that it can have polynomial growth are polynomial systems okay so the number of vector zero modes it is 1 plus plus 8 plus so there are three conformity vectors in the sphere and no more okay every vector in the sphere can be put into a standard form by conformity by y by y automations but once you are put into that standard form there are three there are three n going 3 right 3 analytic the homophones that remain unfixed are generated by c z is equal to this is the vector here c z is equal to 1 plus 8 and 7 plus and a similar one is this one on the right 1 plus 8 bar times so 1 can be made up of there are three parameters any constant any coefficient similarly with constant bar will be in bar 5 okay so there are three complex numbers if you want to let that fix now let's do the counting so what we've done is sticking to the homomorphic side we've counted that the number of conformity vectors is 3 the number of moduli is 0 okay let's compare what we have in the bar the number of moduli minus number of conformity vectors minus 3 so I'm going to do this about three times you know what run is coming okay that's great what we've learnt from this analysis is that if we were sticking to the sphere all amplitudes of the sphere are going to have three fixed vertex operators so that we have three C's of those okay okay and we will have none of those horrible insertions with B and N those B del G insertions so the only part of the of the ghost particle that will bother us is computing the ghost particle in the presence of those three three C's plus the fact that there are these three C's what have given us the extra playoff that we have to do three less integrals than you might think because three of your vectors are fixed now I emphasise a conformity vector is actually great very convenient why is it convenient? because it's like unfixed gauging you can use it to fix the remaining gauging in a convenient way that's how we did our derivation we fixed three vertex operators insertions to be at convenient points not fixing vertex operators at convenient points is much easier than integrating the answers once you have them so we've already understood the calculation of three point functions of the sphere I'm going to give you guys now somebody tell me what are two point functions actually Logan and I come down to go when we first asked this the doctor was kept in aptitude asking this question take a compute two point from aptitude with the demands of the sphere something possible right somebody didn't voice why it's not impossible the positions that we can fix is three m and then we need to fix the sphere and the number of operators we have in our hands is two so we just don't have enough to fix gauging variables we don't have enough we won't actually see zero just get zero what's going on so the answer about here is you're designing of apt zero that's correct what are we doing we're computing a two point function on shell we're computing an estimate of a two point function on shell what a propagator on shell if it's honestly the propagator one of them is split where it's just infinity think of neighbors the propagator is zero there's no sensitively defined question involving two point functions on shell there's no well defined question about two point functions on shell if you want to think of the two point function as being a propagator with propagator stripped off anywhere you can zero if you want to think of just propagator you're getting clear because no question is there any interesting non zero answer this is distinct from field theory because we calculate every propagator from field theory but we calculate them off shell things that we've done in India has been stream theory set up on shell there's no interesting question about classical two point functions on shell what the propagator situation is a little more interesting compared to you know the mass shell condition has to be corrected what the propagator we'll understand when we get to help that there is classical there's no no interesting question about the propagator first interesting question has to do with three point functions three point functions after are totally trivial they're totally trivial to calculate because they were just doing a CFT correlation function calculation with with vertices in such a particular point four point functions and I am more tricky because why do we insert three of the the fourth one has to be something something in some sense three point vertices these have been the basic interactions of field theory whereas four point interactions could include basic but also include particles running in channels can be compounded from from two to three point vertices we will see the the four point amplitude the open stream management that's but we are so happy to do that the closed stream persons finish then it's the other should be our the guessing of this amplitude before stream theory was invented was in some sense the birth of strongman these guys in the 1960s were studying the amplitude properties by state noted that there was this really cool analytic function that satisfied all the analytic properties of one of the basic matrices and that made it it was later realized that this four point function was the four point function of string theory but all these analytic properties are too interesting to come out of just just a correlated function of the four point we came out of doing the integral of this one vertex operator over a four point function correlation in the component as we will see in that stage for everything in the next lecture what we are going to do is to start calculating as you said we will start calculating a bunch of 3.5 minutes which can be a graph or a graph to an amplitude check to see that we get what we expect of my string theory but we will also calculate a four point function but first view to this answer I think we should probably probably stop here until Ajay comes to take the camera although we have back 3-4 minutes of discussion other any general questions either general or specific about hey you know this we are getting to the point of the course where we have done a bit of formal development and now we are starting to do calculations so it is just about the formalism or just structure of string theory and the question is not about it yes from the boundary right see these are two different boundaries the first thing with the boundary of modularity yes you see it was a total derivative on modularity oh this lens oh sorry so you see it may be important for the string boundary conditions we will make sure that the boundary conditions are such that we have not done this yet but we will make sure that the boundary conditions are such that all of these terms are zero that will be one of the we will get you know we won't allow absolutely any boundary conditions for open string theory we will make sure that the boundary conditions are such that all of these terms are vanishing you see you will see how that works for people you see the point basically the point basically is that there was a scene in that thing now a scene was something that mimics the properties of of the allowed refusal of string theory now if your morphisms are such that they respect the beginning of boundary search then they must vanish in the boundary now you see the set boundary conditions for C for the open string that will cause that del C we will see this in detail in detail as when we start making the open string that's a good question any other questions? we had this cylinder strings corresponding to those strings and that ended on the spreadsheet then we used the operator state map to say that we are inserting an operator state into the operator yes why is that? yes it does in that case what we will be doing is taking these strips strips once again there is some analog of the operator state which tells you that these strips is the same as inserting an operator at the boundary of the white sheet that's a good question only computers who are having amputations turn to discussions of the open string basically we do that with D but see so you remember that for the closed string we had these cylinders that went out and these didn't work for the open strings we had these strips and they didn't operate in the open string so open string states will correspond to insertions of the boundary of the white sheet and closed string states are insertions of the boundary the integration of the position of the white sheet will be one dimension exactly every one of the open strings is constrained to 9 yes we have not talked about that yet but we will let's get a thorough understanding of closed string theory first we will then discuss it other questions can this operator in the sense that can non three what is the operator any local operator ok good question in a compact let me give you the legal answer in a compact conformity by which I mean for instance a conformity can be thought of as a sigma model on a compact then every local operator is a vertex operator the definition is an operator that is due to a state ok however in conformity theories that are sigma models are non compact out in space that's a danger it's a danger we can discuss so I just have a question there's a danger and the danger is that when we do a power map you remember what a map was converted from operators to states a map was put the operator inside the path and then we got some function this is the way I do the function this is state but remember quantum mechanics not every function of your variables what's the state what are the what are the condition do they need to satisfy normalize very good these functions need to be normalized now this is generally a problem not for like those of the harmonic oscillator but for the baseless degree massively afraid of the function no finite energy function goes out correlation function is centered around zero but for the zero model there's a potential problem here see you are depending on the wave function on the zero model has to turn out to be normalized now if we do a sigma model on a compact out in space that's all I think but if we are looking at a sigma model on a non-compact out in space that's not all I think and in some cases it can fail a simple example is e to the path kx where k is over there this is formally an operator versus zero model dependence of the wave function it's sort of clear just maybe I need to find kx not just close up statement that so in these theories in theories like it's not that every operator is a vertex operator in the sense that corresponds to the state it's true but you know these normalizations actually the simplest example of an operator doesn't correspond to the state it's the operator x if the free force of the property corresponds to the state what's the scaling dimension if it was a state with enough energy zero the reason once again is these diverges x essentially has an operator with x not which blows up you see del x removes the zero model out so that's totally justified e to the path kx has the zero model but if this delta function normalizes the sense but if you look at the x it's like the state x the wave function this is what it begins that's not a wave function the shoddy curve wave function side of x and del x blows up uncontrollably so the answer is this is the answer now there are some conformity in theories which I like to see more when I was in contact matter of course so many of the abstract theorems that we derive for conformity in theory generally speaking apply to these speciality conformity theories that are rather perfect they may be there and another way of saying what conformity in theory is that it has mass scan see these two things are more or less equal if you were a non-compact artist then you have a continuum doing a zero-mode path wave function so you have no mass scan so when you go more abstract conformity in theory conformity in theory you can't think of it as a sigma model at any time the question is whether that mass scan when you take it up it's closed thanks I guess