 OK. Good for you. Has anyone tried the exercises we talked about last class? Yes, Kertima, you tried them? Sure. Nobody tried the exercises? OK, I'll get you one. I'm going to demand submission of exercises. So by the next time we meet, it's going to be some time. How many of you guys have been in Bangalore to explain? Kertima, you should have done me. Um, this question should be continued to the next look. So what are your thoughts? If we continue, then you look them up on video. Would that be OK or would you prefer your way? Well, that's half the class of you. You will see that. OK, fine. So let's say that we continue on. Everyone's back at many times, right? So that's good. 24. So what is that Wednesday? The Wednesday of that week. So 15th is a Saturday, 22nd is a Saturday, 26th, OK. So the next class, then, will be the 26th. OK, OK, so next class. Sometime today or tomorrow or the week end, I'll send you an email with a list of all 10 problems. And I want you guys to solve and submit to me. OK, fine. OK, fine. So I'm going to assume, though, that in this class we're all not going to be comfortable with the formal thing. So we can use that as a tool to bring us up to your string. OK, so how are we going to do that? It's a connection. But let's remind ourselves when we work with, you know, our discussion of string. Remember, we said that the string word sheet will describe with the action that's space. When G is the word sheet, the metric in space, the isentropic space is just a fillet. OK, what are we going to try to do in order to study strings here? OK, so in order to study strings here, we're going to try to solve this path integral. So what do we want? We want a question. Compute this path integral with what are the conditions for what are the conditions? So let's start discussing that question. As you would see by the end of this lecture, OK, you might think you want to do this the wrong way. What are we going to do with quantum mechanics? Or it's superdub version of field theory. We ask questions like the following. Suppose we've got some particles. And we compute a path integral over all particle word lines with the constraint that these virtualized particles are specified points. This gives us the amplitude for the particles to propagate. This gives us the amplitude for the particles to propagate from some set of points where it doesn't go out of mechanics. The propagation just happens like this. In quantum field theory, we allow also various other such the split things and joint things in these particle parts. OK, so the most natural kind of question that you could ask in string theory, you might think, is the antelope of this question. So you might think that, well, so let's back up. From the quantum field theory, all the parts that go that are disconnected. So suppose we have two parts like this, you know, parts like this, and then parts like that. These are disconnected finite diagrams. And so if we just compute all connected finite diagrams and then exponentially answer, that generally some of them are all disconnected. So why I often allow you to compute something physically, you need to include both connected and disconnected graphs. The interesting computational question is how do you compute the connected graphs? So that's the question we'll look for the antidote of the string theory. What is that question we want to address in string theory? You might think that natural question is the following. Consider, a natural question, I think, is to complete the architectural overall world sheets of the string that end on a given set of curves. We want to think of this as the amplitude for closed strings to begin somewhere and propagate to something else. We think these curves within the surface. So let's think of that. If I think of this as a natural question, we want to integrate over all string world sheets that have the property that they pass through some given sets of curves, and these things are important to the string OK, so such a string world sheet is this, with some hole in it out here, or something very complicated. You might think this is the second part of the technique. This then would give us the amplitude for a string starting in some place, propagate for a set of strings starting in some place, to propagate and become a set of strings from any given part of the piece. Roughly speaking, that's what you might think we want to do. And that if we knew how to develop this computation, it's the computation of all Green's functions, maybe now a quarter-fifth here. If you know all Green's functions, you know everything the rest of them. So yeah, you can do this calculation. You know everything the rest of them, but the string. You may have to think about how to extract the information you want, but it's all there to do. OK, that is interesting. I have a calculation of what you can do. But I just have to explain to you at the end of the lecture, we do not know how to build the amplitude of this calculation, it's everything. If you give me some arbitrary set of curves and demand the amplitude for strings to begin and end at this set of curves, we do not know how to do this calculation. The technical reason for our lack of knowledge of how to do this calculation, we can't clearly, hopefully, make it to the stage. However, I'm going to spend two hours talking about this physical possible, it could be, there are two possibilities at this point. It could be that there is a way to do this calculation and we have here to be smart enough to figure it out. Or it could be that there is a fundamental restriction which tells us that we can't do this calculation because it doesn't have a way to find out. It's a little like asking for what is the position and momentum of a path in what mechanics. It's a stupid question to ask. It could be the same. We don't yet understand what the gravity will have to be totally sure of what it will be saying now. But I would say that there are some indications that the second thing is working. That it's just not a good question to ask. And the indications why then it really go roughly as far as. You see, strings theory, as we've already seen, is a theory of gravity, okay? So now, gravity in particular is a gauge theory. It's a gauge theory from the age of the few molecules. No, not this one. Loading the right side of the equation is what I get. Okay, so in any gauge theory, the kind of questions that you, the set of questions you allow to ask, questions to which there is a gauge theory comes. Okay, now, if my is a quantum field theory, the question of what some green functions at finite fixed points of field insertions are, is probably a good question. In the theory of gravity, that's not manifestly where the thing is. It's not manifestly where the thing is, because if you ask for phi of x, phi of one, phi of zero, what does that mean? What is y in the zero? These arbitrary labels for points which take that into your answers. Okay? So arbitrary correlation functions of fields should not be modified in the theory of gravity. It goes with the answer. These things are not ethemophically related to observance. It's like asking for correlation functions for a mu in a gauge theory. And it's not modified. Different gauges give different answers. You're okay. It's not a good question. So once you've said this, you might say, okay, that's all very well. But then, if that's the case, one of all the good questions you can ask in the theory of gravity. So the questions that you have to ask have to be gauged in there. Now, there's some class of gauged in there questions and I emphasize this is not the only set of gauged in there. It's not at all. The people sometimes say it is. No, it's not, I think. But anyway, some class of gauged in there in question you can ask is the following. You can ask for these correlation functions once the points of these, at which you insert these correlation functions are taken off the infinity. Basically, because when you're in the path of gauging of any theory, you do it subject to some boundary conditions. Why do you allow yourselves to make arbitrary deformisms in the bulk? You free what's happening at the bottom. Basically, roughly speaking, because fluctuations of everything, we should be looking at metrics that fluctuate. But fluctuations of the metric and infinity generally are infinite action. That fluctuate over a very large range in order to reach infinity. Okay, so it's usually in quantum theory in our setup of quantum gravity. You see this very clearly, actually, when you start doing semi-classic in quantum theory, it's the state of the S-space. The statements I'm making, some play-making around everyday, may be precise in some things. The way you do the quantum experiments by freezing the metric and boundaries allow you to fluctuate in the bulk. So once you've done that, insertions of things and boundaries are well defined. Nice and easy to learn because you're not allowing the metric to fluctuate in the bulk. Is this clear? Now, insertions at the boundary, correlating functions that are inserted at the boundary are essentially isometric elements. How do you have an isometric element? You have an isometric element by summing over all insertions with the weight factor pointed by kx. But that weight factor goes over all x and is dominated by what happens at infinity. So insertions of operators at infinity, roughly, actually more than roughly speaking, but I'm not going to try to make any of this precise in this class, I'm asking for questions. Is it related to estimate it? Okay, so the kind of questions that you might want to ask, the kind of question that you might want to ask, the kind of question that will have a gaussian, but if you've always been there, answer it. In a path that can describe fluctuating gravity, provide you with a sense of the boundary conditions. These boundary conditions are almost insensitive space. It's forced on you by dynamics. The kind of questions that you might, one second question you might want to ask about is about estimate results. What is the amplitude of things to come in from infinity and be a certain weight that is specified at infinity, interact and go all the way to infinity and reach a certain state that once is even specified because nothing is happening in it. Philosophically, there's a big difference in a theory that has to do more with equivalence. There's a big difference between correlation such as finite points and correlation functions and infinity are more precise. Yes, that's correct. Okay. Diffusorisms, such as the lack of observers should presumably be well defined. And it's these observers that we, now all this is going to ask for the non-sequence, but I have a short technical statement that we appreciate by giving this class is that technically, it's only the end of this matrix that we have yet understood how to compute instruments. So it could be that this philosophy of blah, blah, blah that I told you about, is the reason for this technical background. You should always be where you believe in philosophy. It could be that we've just not been smart enough. Many things are possible, but there's some reason, some general reason to believe that perhaps it's not just trustability that we haven't been able to compute more than S-matrix elements because at least most of the other things that we would like to compute, it is not very funny. It's just bad questions, the kind of questions you should ask, okay? So all of this was just to motivate the fact that instead of asking this question, the question we're going to ask, okay? The question that we're going to ask is questions about what the same part of the technical that these legs explain about. Okay, any questions about it? We can ask, I hope, what happens to that finite element? Totally. And if we think that this quantum theory is going to come out as a little bit theory of a cheat theory, then why does the theory exist then? Yeah, it's a very good question. So the question is, well, you know, suppose you take a little theory of quantum theory which we thought it would not be. You have a quantum theory without gravity but there are calculations in a clear way to find it. The question is does the quantum gravity would make it confusing if you find there's an approximate category, something that reduces the probability of completion. I don't have a very good answer to this question. I mean, I have one or two thoughts about it. Not everyone likes to go into a way that makes sense. You know, to say that, to say that just the stupidest thing you could do is just asking for correlation functions and fixed points. But first, first it has a practical answer to this. And the practical answer is this, okay? If gravity is very weak, you know, then why is a manifold fluctuating? It's fluctuations of our given classical background are very simple, okay? So the ambiguity about what we mean by five zero five one is very small for practical purposes in the situation, you know. So it could just be that. You can do the calculation that isn't terribly well-mined but the error is very small. A little like classical mechanics is a good approximation of quantum mechanics. You know, you can't do all the positions and the velocities and things in certain situations, okay? It could be that's the answer to your question. However, there's more to say about this. And the more to say about this is that, you know, if you think a little bit, you certainly can create observers that are local, that are more local than expected, you see? And that are different of a clean bed. For instance, you can ask for the two point function between two operators that are separated by geodesic distance one. Now, this is different from saying point zero and point one because some average of all directions. But it's a clearly dated there. The few more are clean bed in question. And there's information about how operators behave when they come sort of near each other rather than infinitely, okay? So, you know, sort of, it's a bit. Okay, I mean, what does it mean to say geodesic distance? I mean, basically, it will be a quantum field. Well, the quantum field. So you can put in this, you can put in, suppose you had to be this far, some scale of a quantum graph, right? Suppose you want to compute phi of x, phi of y, with an extra function, okay? That says, gives the expression for geodesic, give it a metric, you can compute the geodesic distance to be x and y. Right, so let me call that geodesic distance, g of x and y, which is also a function of the matrix. Again, mind the game. This is an insertion that I can do to make it a path vector. So, you don't have to know the matrix, you see, because this is some function of the matrix, and then you do a path vector over dg. In fact, that's, yo, with any set, you take any matrix and parameterize it with some coordinates. What coordinates does it make? Okay? The question is, is this, you know, where you found that it might not allow some coordinates. Oh, okay, yeah. So, if you, okay, it could be that there are global constraints like that. It could be that, something like that. It could be that this is not good enough. All right, it's possible that if you, and I really want to use my calculations, change topology in some, but not in my calculations. Okay, such problems, such problems. I don't really understand that. Now, even though this is, you know, an ancient, dead-end insertion, you know, provided you fix the topology of your, of your, of the matter towards over which you need it. Yeah, I suppose all of the matter was at the end of the reading. There's a one-to-one map between them. So you can parameterize them by the same class four. Then this looks well-behaved. Maybe there are a lot of different issues. Okay, you see, such questions are hard to answer until you know something about the theory of gravity. Let's leave it at that. Back to the point of view, now, maybe there are only computer-experienced experiments in street theory. We'll see why, we'll see what goes wrong if we try it. Okay, good. Now, so what are these estimated experiments? These estimated experiments involve these two who's going all the way to infinity. What is an estimate? An estimate tricks has a particle that is non-interacting for almost all of its life. Okay, that comes into some small, minor-intensive interaction, it goes out and becomes non-interactive for all of those experiments. But these non-interacting particles, these things we have got to control. We can certainly describe. This is just particle-propagating of the cylinder. I think that we solved for the beginning of our lectures today. We know all the possible states that could come in and go out. Those are the states of free string of propagating of the cylinder. Next thing we know is that in order to do a path integral like this, okay, in order to do a path integral like this, instead of doing, okay, go. So, so far we've been in mid-task space. Now, the next thing we want to do is the problem. We want to try to turn this calculation into a calculation in the ukrainian space. We want to do that because ukrainian path integral is much less defined than mid-task path integral, okay? And so we want to convert this problem for which we have some intuitive sets of what we want to calculate in the problem which has a precise problem. So what does it mean? Now, what does it mean to take a mid-task and convert it to a calculation in the ukrainian space? So let's try to answer this question with the simplest theory of gravity, the theory we started out in our first lecture with, then in the theory of a point particle, that was the point particle, in the Kolyakoff property. First of all, we have the action s is equal to i, and then we have x dot square, x mu dot, phi eta minus n squared, sorry, s is equal to i in part of the problem. Not in this part of the variable subject, some boundary conditions, something like that, okay? But in the end, the thing that we're doing is path integral. A path integral is product of several integrals. One integral for eta of x, sorry, eta of t and x of t at each time. So what we've got is product of several integrals. Okay, so now what I'm, now these integrals aren't terribly well defined. Basically, because the modulus of the integral is always, since the modulus of the integrin is always one, when we take the, when we go to extreme values of the variable we integrate over, let's say eta of x at some time. It's not like the integrin is dying off. It also takes very fast, that's the reason, conclusions are correct, but it's not like it's dying off. So it's hard to get meaning to this integral. I'm gonna say some sort of infinite set of things, which are not very secure, I didn't think, I didn't think. Is can we take this path integral and change it a little bit? So I say give me, okay? And the answer is clear. You see, suppose we do the following. Suppose instead of integrating over all of your eta, eta by eta prime times, e to the power minus i theta by, in terms of eta. In terms of eta and amounts to change, think of the integral over eta at any given time. What I'm saying is that, do it any given time. Think of the integral of eta as an integral on the real axis of the complex plane, okay? What this amounts to doing is to say that instead of integrating eta like this, I'll integrate it into a set of states, okay? I'll integrate eta, so eta, for instance, goes from zero to infinity. So I'll integrate eta along this line. You see, at the moment, we're not even talking about rotating back, I mean, eta's something infinite. This thing is going to give meaning to my object. So we can think of the other thing that is the limit of theta goes to zero of this integral where eta is given by eta. That's the real number. Times e to the power i theta by 2, okay? Why does this give meaning to this integral? This gives meaning to the integral because now you see that the imaginary part of this thing, that the imaginary part of this thing is always positive, you see, because I divided by e to the power negative i theta by 2 here. So when we're in numerator, it's e to the power positive i theta by 2, and you expand that, you get a positive imaginary part, which when it is this i gives you a negative real part to the argument of the exponent and the integral. Therefore, when the integral comes very large, when you go to extreme values of the integral, of the integral, of the integral integration, the integral would probably be that much. But you can think of this as an ordinary integral, okay? You see, because of this minus sign here, this term also has a positive imaginary part. If you expand this, you get a minus sign from here, but that cancels minus sign here. Yes, we've done it in such a way so that every term in the integral is there. There's one more thing we have to discuss, and that is, I've been hiding the fact that this is the extra minus sign here because we've got the x v dot, okay? So, x v dot is xi dot squared minus x zero x. This would dampen the part of the martian integral coming from xi dot squared, but not the part coming from x zero squared, okay? We also have to rotate the continuity integration by x zero in order to make it smooth, okay? So what you do is, once again what you do is say, x zero is equal to x zero prime times e to the power i theta, okay? Now you see that the term that was minus x zero dot squared gets weighted by not minus n. The term that was minus x, since that's square, you don't need, sorry, yeah, this way. The term that was minus x zero dot squared gets a factor of e to the power minus i theta, which is once again a positive value, so it's very simple, let me just write it down. Let me write all terms in detail. So this is xi dot squared by theta minus x zero dot squared by theta minus n squared theta. Under this shift, let's look at the phase factors for each of these. Here, what do I get? I get e to the power i theta. Now what do I get? I get e to the power minus i theta, but it comes with a minus. I get e to the power minus i theta, but it comes with a minus. The integral, including the i, the thing that appears in the argument of the exponent, including the i, has a negative real part. I have an integration in this way, okay? It's the same thing, but it's the positive integral, okay? So you should think of the positive integral as the limit of this part of the integral, limit as theta goes to c. Yeah, but you should have the virtue of being well-defined just to reduce to what we mean to intuitively want to get. Suppose you're doing a complex integral over a certain contour. Suppose you're doing a complex integral over a contour starting from zero to infinity along k. And it's true, suppose it's true that the integral is strongly zero on this curve k. Then, if you reform the contour of integration to this thing, you get the same answer. I provided the integral is analytic. I provided the integral is analytic, which is certainly true of everything we know. Nice exponential functions, nice analytic. I provided everything is analytic here and we get no contribution to the contour of infinity. Whether you do the integral here, here or here, you get the same answer. But that, the circular infinity if you do it's nothing is clear because we're saying that as you go out of infinity in this direction, the integral goes to zero. Because it's a possible demand. Okay? So once you're agreed to define the department integral in Kastli's case by reforming the contour a little bit like this, okay? Well, you find yourself all the way. It makes no difference. And all these arguments, of course, heuristic. Heuristic because this is not really any integral. But you know what I mean, regulate your path integral. Reduce it into a finite set integral and so on. I think these arguments are basically precise. They're just arguments of analyticity in complex analysis. All the subtleties become integral and so on. We will forget about all of that. This is not precise. Okay? So instead of, what is that conclusion? That conclusion is that instead of making this mild change of variables, it can make stronger change of variables by replacing this by minus i. Sorry, I think that was minus i. It was minus i? Okay, great. So we replaced it by minus i and we replaced this guy also by minus i. Okay? Now once we do that, what happens to i? Okay? So we have i here. This guy picks up i, so we get a minus. This guy picks up minus i and this is minus i squared. So once again, so when we do it, what we get? So the path integral turns into and it goes half when we forget about that. Except for this being much better if I just manage it. It's clear, right? What we're doing is now we're integrating all of our phases. We've got negative real, the argument of the exponent is negative real numbers. You know, go to some extreme value of integral space if they're very large, and the real number greatly damped out the multiplication of the integral. Okay? So taking this point of view, the Euclidean path integral, okay? It is simply a definition is simply the analytic continuation of the Minkowski path integral. So Euclidean path integral is simply the analytic continuation of the Minkowski path integral. Once you've taken the problem to make a lacker of integral space. So this Euclidean path integral that we've been studying, we've been studying the formal fields here. The reason we started here is that it's just a technical trick to compute the Minkowski path integral. You know, unless you do some binary calculation, you're often not interested directly in the Euclidean path integral. But if you know how to compute the Euclidean path integral, you know how to compute the Minkowski path integral. That's the reason for your interest. Okay. Great. So that's what we're going to do. We're going to start doing the calculation in Euclidean. Okay. I have to say two more things. Though I won't try to show it to you. Firstly, if you take the string theory path integral and replace the integral over matrix by integral over ion bounds. And then you make the ion bound corresponding to the time lag direction. You make the same change of variables there on the ion bound corresponding to the time lag variable. And the same change of variables at x dot. Exactly the same argument at x dot. Okay. So this is an exercise. This would be one of your exercises. Okay. Show that the, if you take the problem to make the string theory path integral in terms of binding lines and x fields. When you bind in Minkowski space, when you use complex analysis, you just get the Euclidean path integral that we've discussed. Okay. This is the way. It's on the bridge. Space management is fine. Okay. Great. So, so, so we've done the argument in the simplest context. In fact, it's the same argument. One additional term that goes along with the right space there is the do not thing. The time there is exactly there. But I want you to go through that. Okay. It's actually exactly the same argument. But please go through it. Okay. Great. One additional comment about this. Now you might think, while this is exciting, every time you have a question in graph in Minkowski and gravity, you can convert it into a question in nuclear gravity. Unfortunately, in higher dimensions of space, if you try to do the same thing in three or higher dimensions, you encounter new features that we have in the entire company here. Namely that the action innovation to the square root of G and magnetic terms to scale up also has a curvature. And in higher than two dimensions, it turns out this trick of analytically continuing our material doesn't, why is it, it damps out all terms and say the scalar field action more or less the same way that we talked about before. It doesn't damp out all terms in R. In fact, there is a famous part namely the conformal model. The part of the metric that corresponds to conformally scale you just write down R for conformal factors. There's a famous part in the nuclear space becomes negative definite rather than positive definite the way that we find it. In fact, to my knowledge there is no known way to perform an analytic continuation of the Mikhailovsky part integral of gravity to Euclidean space in higher than two dimensions in a way that makes the part integral manifest you like a friend. This is one, actually this is one important reason for suspecting that gravity is not going to be defined by some emphasize version of the Einstein part integral which is forget about subtle issues of renormalizability and so on just defining the path integral at this very crude way no one will have a difference. Everything that we've talked about is Euclidean continuation while it's completely sounds completely reasonable in one and two dimensions. If you try to do the same thing in three dimensions the state is what you're doing instead of here. Okay, because it's not like this continuation gives you a better defined path integral than four. In fact, if you're worse with a defined path integral than four. It's when you have just phases or moduli will remain one when you're ridiculous with moduli, in some directions. Okay, so this Euclidean continuation is a trick that sometimes is used when you should examine case by case. It's not always guaranteed that it's a good thing. But in the process of looking at it it is a good thing to do. At least that does not happen. Okay, fine. So now that we've done Euclidean continuation of a path integral what are we going to try to calculate? So we can try to calculate these world sheets. You know some overall world sheets that go off infinity. I'm happy to so happy about you you can do whatever you want. Please. Sir, actually like suppose we did not do the Euclidean continuation then even though the integral would not be well defined but we know that it's not that the integral is vanishing but it's cancelling of because of the oscillation. Because of the oscillation. Yeah. And in the other case that you mentioned there also you've seen that the integral may not be well defined but actually it might be finite because of such a thing. No, I think that just as an mathematically integral once you are able to continue it like this it's just unifying because there are some directions in this case where divergence and then unless you know you know say it's some of a positive number because some of which are infinite unless there's some conspiracy or some dynamics to give to give you measured effects that suppress the region of field space that gives you a divergence enough so that it's not you know it could be look at some but naively it's just not well defined you know it's naively integral that may or may not make sense of real access but certainly it does not make sense as which eternally okay I don't know what this is saying about gravity but it might be an indication that during the pathogenetic rover the gravity is not the gravity because we don't have to make sense of the concept at all and then naively I'm not talking about you know subtleties about renormalizability and all that I'm just talking about giving a basic definition of the pathogenetic appears to a high school student to look at it to make sense okay so in fact so anyway we're going back to computing these some of our euclidean worksheets now in euclidean space we've taken a lot from in our study conformal conformal field theory to show the equivalence between the bottom between computing the path integral on the cylinder with some less boundary conditions at minus infinity and computing the same the path integral bottom integral on a disk with the insertion of an object the operator determined by B the operator that is due to the state that define the boundary conditions at minus infinity so what we want to do is to start scattering with some states okay so we want to find a cylinder with some state propagating on that cylinder we want to know the path integral but that we know the same the path integral with a disk with an operator with the corresponding operator insertion the question about the question about having stream worldsheets going all the way to infinity in some particular states nice sir looking mathematical take the stream worldsheet insert operators dual to the states that you want to scan on this worldsheet let this worldsheet do whatever it wants and at the end of the process what you are going to do is end up creating a scattering attitude for the states due to these operators scattering to each other so that is the goal that is going to be a goal in stream worldsheet we are going to try to compute in the Euclidean formulation the path integral over all stream worldsheets will insertion to some operators on the stream worldsheet what kind of operators we will discuss in the next video now let's let's proceed more systematically more mathematically so now I am going to start with more mathematical analysis of trying to formulate more precisely this thing that we want to calculate that we said in words in journey so what we want to do we want to sum it over the worldsheet trick and we want to sum it over the x mu remember now everything is repeated remember that actually it is quite a beautiful thing if you did the analytic continuation of the worldsheet then it changed the worldsheet metric from being mid-class space-time metric for being mid-class remember both rotations were going to get over the path it is enough to make both the worldsheet and the space-time metric are now completely all d dimensions of space-time are equal no distinguished time okay so we want to do this integral e to the power minus 1 by 4 5 where the g then alpha x mu then alpha x mu everything inside is equal okay we want to do not just this integral we also want to do the same integral with the insertion of certain factors with the insertion of vertex operators vertex operators insertion very much look there is no natural point to insert them in fact we can violate if your model is an invariance as we talked about we inserted them at any given point whatever okay so the only way to do you know which is basically saying that why we want the state to go to a particular location at infinity we do not want to say where on the worldsheet that is located that thing is located in fact it is not even well different models of invariance worldsheet okay so the only way to make that precise is by saying that the string should go up infinitely but where on it we don't get so the insertion we want is an insertion of of some vertex operator integrated over the whole string is he ready say suppose in some parameterization of the of the string at some point sigma you got the cylinder coming out okay or another point sigma you have a cylinder coming out as well as it goes to the same point space time model that gives the same physical explainings okay so the insertions that we are going to deal with are of the form let's say that number of insertions is literally so we have sum over n i is 1 to n integral square root g integral square root g and then v at sigma i what these operator insertions are going to be allowed to be we will see more clearly as we go at the moment let's just keep that we are trying to calculate it sources yeah it is sort of analogous sort of but exactly yes it is but as we will see there will be a restriction as to which kind of sources that would be the key point about being restricted on if we were allowed to add any source then we would be able to calculate the amplitudes you see if we were allowed to effectively be making the path integral indietely allowed to be so the answer is both yes and no it is a subset of what you are allowed to do so you will see when we understand what the restriction is these allowed sources so we are going to do this path integral we are going to do this path integral modular difumofism in variance of the order of the state and at least in our first round understanding strength theory we would be most interested in path integrals that do not depend such that the answer at the end you know once we formulate the answer dynamically the integral over 5 is strictly where 5 is the formal factor of the motion of the string because otherwise it was doing the same integral path integral as we should have been doing from the non-linear direction right we were interested in looking at this theory at least to start with fourth theory is in which though the formal factor on the worksheet of the string is not an integral but fine bottom line would be gauge theory so what we do is to use the standard trig location variance and we find it for modifications so let's review how we do that in the context of this example okay so suppose we are working on a certain manifold as we will see we will not maybe I should say this immediately right but maybe I should say one you see suppose we were doing this calculation on the sphere then this would be the analog of this sitting there is going on like this and if you take this the time slide still what you can see okay let me time slide set the map let me make the make the diagram better if you took this time slide you can't see two strings becoming one string going over to two strings the final diagram analog of this on the other hand suppose we are doing the an accuracy of the torus then what would be the right let me bear this torus maybe this time this is an accuracy what would be the you see one string and two strings going over to one going over to two going over to one what is the final diagram we will be followed by beer theorist as a loop diagram this would be thought of as a trigger clearly we want some over all to 1 to the text we want some over all loops so we can't restrict ourselves to working on now first this quantity once you replace all these tubes with the operated insertions is topologically the sphere with operators insertions whereas this final diagram once you insert replace these tubes with operated insertions is a torus with final diagrams insertions and we will see as we continue I studied the most general kind of manifold which we will do in general what's called the Riemann's which is classified by the number of whole trophies we will discuss that in detail so we want to do this path and take a look without to be prejudiced about the topology of our manifold because if we demanded that we do the path and take only our spheres we would be calculating only the anode of three level diagrams but we must also be able to calculate loop diagrams this path and take a loop manifold of certain topology who wants to specify what it is and well we want to go ahead and do that path there so in order to do this path and take a look what I want to do is to gauge things so suppose it was true already seen from parameter counting and it appears to be true that while plus if you want this of invariance would be enough to take any metric to any other now this parameter counting is counting a function so it may have some discreet amenities may have some parameters as we will argue soon that's the case different manifolds and different topology will have some but in abstract discussion now in whatever manifold we're looking at suppose we will take this possibility in account so let's say that every metric that you can write down on this manifold can by a while and if you want to do transformation we put into the form g mu nu of t i where i is equal to 1 to m labels what I call the modulite the modulite so modulite that you have to add to your space of metrics such that with this kind of meters of metrics you can really cover from while in the thermomorphism transformation every metric on that map t i for instance on the torus the modulite will be the stubble map it doesn't matter we study all these things in detail and the moment all we're saying is that it may be not entirely accurate that you can take a standard form metric on a particular topology and through while in the thermomorphism transformation get any other metric it may be that you need a standard class of metrics parameterized by some number of parameters let's call them t i's i run from 1 to n in order to achieve that this will turn out to be at least let's keep that possibility now there's something else that may also happen and in general it will happen and that something else that may also happen is the following it could be that there are some while in the thermomorphism transformations that acting on a metric don't change it we see the examples of this while just if you offer the transformations that are acting on a metric don't change it in flat space conformal transformations actually one of the changes plus the appropriate while transformation gives you back the same metric okay so it could be that there are some populations of while in the transformations which acting on a metric give you back the same metric we'll be careful about the overall transformations once we go you're most another thing the conformal transformations we talked about blew up at infinity we will be only looking at things that are well defined everywhere in space when we do that as we will see we will always refine our conversations okay so let's say that moduli let's call it c let's say that j is equal to 1 to c represents the set of what we call conformal key parameters why is all that we see okay it's the number of in the set of while in the transformations that do not change but what is the importance of what is the importance of the number of moduli well the moduli that you what parameters and metrics you have to integrate over in order to be able to relay any metric to this the number of conformal key parameters that you that if you demand that you gauge you fix gauge by demanding is put in in one of these standard forms then we don't completely fix the gauge there will be some gauge redundancy of what you're doing and the amount of gauge redundancy of what you're doing is the number of conformal key parameters is the number of conformal key parameters because those set of gauge transformations may be while in your office of transformations which do not change the metric and therefore if you put a gauge condition in doing the math the metric it fixes all of the while in the if you're except this if you we are yes so I mean I have a lot of what do you mean a lot the sense okay if you are on the sphere we see that there's six and every higher two and every other higher zero but we just thinking of general possibility we're seeing detail how many there will be which kind of insurance I mean all what and so on you'll see all that though if you just try to have a general formulation not making any assumption about how many conformal key parameters or how many okay so let's say the number of conformal key parameters whatever it is okay now actually let me call it juicy so good yeah we now want to set up we now want to find set up you know so we use the standard what is the standard for the standard for the other for involves inserting identity okay so we say well I take okay suppose I take a standard set of metrics okay parameters like the eyes then why transform and then if you want to transform later by some abstract if you want to that metric can have the metric let's call it just this and I integrate over and I integrate over all overall five and overall okay then because I've been as as good as I as I have I I it's guaranteed it's guaranteed that this is definitely click exactly once for everything actually we have that discrete not continuous discrete we we will be a little more careful within a couple of pages but after some discrete which we have taken we could take that some patience yeah you have included this TI and then some samples or some body armor some samples we we thought some of the what I mean is that in terms of you may need to I mean the radius exactly exactly okay we deal with such possibilities okay so because the way we've been so careful the guarantee that this thing will click exactly once okay was one but that's not right because the delta function it integrated is one only when the argument appears as the argument of the delta function okay otherwise you have the jacquoq factor you know I've got at least the delta factor x of x integrated over x is not one but it's one over x triangle 0 so this is not going to be exactly not going to be one but it's going to be something non-zero and whatever that thing that stops it from being one is we just by hand define you know so this also where there is some function of the metric and we define what's called the fabric of determinant but choosing the function of the metric G such that that integral is one but this thing depends only on G so it can be taken outside the integral is an integral of a T phi and omega doesn't matter okay that's is equal to integral of the T the phi the omega is part of G delta of this is this is a definition of that part this part of the determinant is defined so as to be one okay it's defined so as to you know so that this integral is one so we can take this object and bodily included inside the boundary which is seen this this will allow us as you know to to fix most of our as we see that as we so but because it's a problem getting better will not allow us if it's all because there's some which this condition doesn't fix so we want to go back so we will also insert into the part integral we will also insert into the two-dimensional delta functions so I see two-dimensional delta functions and the the way we do that is to choose c of these vertex operator insertions okay and to put a delta function for the position of the insertion or c of these vertex operators so we will use the remaining coordinate freedom remaining wire to set the position of insertions or c of the vertex operators to be c specified points that we choose that's very important then in which case the part integral will not be we can't let us assume we have at least if we don't become but it will turn out that we just can't get things and there's always differences there are in the case except for the two-dimensional okay basically this question that c is 6 but c in the way of in fact is 3 of the sphere and is one of the torus now the torus has a special way the special exit room to understand which we will discuss in a bit but if we forget about the torus which is an exceptional case the only place where you can have these conformance operators in the sphere the answer there is 3 and basically basically the point is okay so strictly if you compute a one-point function we can see it all because because it's a one-point function you know because the background will be the equation motion if you compute a two-point function because the only shell what's the two-point function of something one shell and one two p of it what's the two-point function at the moment p of of a scale of p okay so probably I need my property but think of it better p squared or 1 over p squared p squared think of it okay but now we put the thing on shell again as a zero infinity you see so on shell two-point functions have no information on shell one the first kind of function that has information in it is required and that's the maximum conforming operators are always the reason where you can't do the calculations there will be no physical because what you're doing is asking for the vacuum expectation value of an operator of a fluctuation operator around the solution you know it's the the usual thing one-point functions these things are fluctuation yes okay so that's where you can be the answer to your question so it will always where it's basically they're always okay fine so what we will also do is insert in addition to this set of functions we will also insert i is equal to 1 to c sigma i transform minus ai sub-chosen numbers you choose it with c one of them will be 0 the other will be 1 sub-chosen yes yes so that is that that is a function not just of g but also it's also even the the diagram yes in any even calculation you can you know everything consistent into a given operator as you will see what we do is all diagrams are the same that's you can just be in a different relation if you want but you could do it if you want it I mean there's no if the what you do that's that's diagram by diagram field theory of first quantized operation if you do something very simple you know for each diagram there's a path and a loop so what we're doing essentially is doing stream theory of first quantized operation which is why we're saying this thing I don't think there's anything terribly okay let's go most the important part of this lecture once we made this insertion now what is this part taken well the first thing that happens is that you can replace the you can use the delta function to do the integral over g okay and replace g by this replace g by this replace g by this by this fiducian okay there should be some impact so there's also an integral here over v to sigma i per every insertion that's how you what this sigma is to move to we want to use our the few opposites to move sigma i to point against okay sigma i is some given number and we want to move it to be pointed but this will be equal to one yeah sigma i is you know the factor we have the delta function here in sigma i we have here and out there in the world I mean ah yeah yeah where's the sigma in this no no no this is equal to one this is equal to one and we like this sigma yeah ah it's a problem this is equal to one with an insertion in the path sigma i is equal to one so that means I say yeah there are there are there are some over that sigma i that is said to given that this will fit into it yeah no you are integrating over gauge transformations you are choosing a gauge transformation to make a i gauge transform so this is a delta function on gauge transformations I mean this is true when you integrate over on gauge transformations so this delta function here helps you to set some of your gauge transformations in 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 to set an distribution method equal to other that uses up most of the the sum that you want from the one phase backend we'll use these unfix different Mambo to choose some specific point sigma i 0 0 and make them particular oggi what Privilege because I forgot I just like your to out I The three usual. Is this clear? Two steps. The first one step, which is the trivial step, is... Okay. Many are just the non-trivial things. Let us first now compute what this... You know, this always sounds very unsatisfactory because you put all the lack of information into some function which you're not saying anything about. Okay? So let's first, before we start using this, let's compute that. Let's compute that IFP of GDS. Well, we do it using the depth. So let's take clear this to write the depth. So the definition was that one over the depth IFP of G and sigma i is equal to integral dt d phi d phi d omega of delta of d hat d e to the power 25 gauge transform minus g. The product i is equal to 1 to c delta of delta... This is 2, actually. Sigma i minus c. That integral. If we compute this integral, we know how to compute that IFP. But remember that this is an integral derivative of delta function. We need to compute the integral of it. Only in the neighborhood of where this delta function takes. So remember that we want to matrix the first delta function. So you only care about what this function is like in the neighborhood of where the delta function takes. And then it will work for last. Okay? So, in order to perform this calculation, well, what do we do? The thing we do is... Firstly, we have to say a little bit about this delta function. This delta function, as it stands, is not terribly well defined. Now, we're going to... You know how this is very formal. But we're going to demand that this delta function has certain properties. In particular, we demand that the delta function is diffeuble. If you get the delta function of some metric minus some other, and if diffeuble doesn't transform both things that enter, it's invariant under that process. Is that clear? It's like saying delta A minus B is invariant in translation. On the real life. Okay? So we're going to define this formal delta function, sort of like the delta function of group methods all the time, where we have no such definitions. This is a little ill-defined. It's an internal dimension and so on. We're going to assume that there is a definition of this delta function that has this property, but it's diffeuble clearly. This measure is also not terribly well defined. What is the integral of all the few models? But whatever the definition is, we're going to demand that each be diffeuble. That is, you take a diffeuble function, you define a new diffeuble function variable, as your only one compounded with a fixed diffeuble function. Then this variable change has unidirectional. It's such a different one. I can understand if you want both of them together, the delta function and itself, No, no. See, I've been working with finite variables. Such delta functions do exist, and such measures do exist. There is a hard measure, and there exists hard, you know, hard compatibility delta functions. Okay? Now, what we're doing is thinking of the diffeuble function as the limit of some finite variables. Okay? So in that sense, it sounds reasonable that maybe it exists. Now, this is the kind of question that will make a mathematician tell us that everything we're doing makes for sense, you know. How do you know that? Okay? We're going to work with this heuristic here. Okay? If you write, what you can do is, we got the end result of what we get as the starting definition of strength. Because what we're going to end up with in the end is some manifestly well-defined integral, which we know how to compute and then make a check on properties. So that would be one point of view. However, that's very unsatisfying. Because, you know, where do you come from? This is the kind of thing that motivates us. So it's good to see it, even though it's completely clear, it's completely right. We're not working at any level. Any number of the things we're saying might be wrong. This is always true when you deal with these parts of the relations of this sort. You know, you could ask the same question from ancient eras. It's very similar kind of issues that are either. Now, if you know how to manipulate the theory like a lattice case theory, then you can make all the arguments that you use there. You can make them really nice. We don't have such a regulator here in any theory with your argument there. So, could be that everything we're saying. So, this is the kind of reason why we put this in the slide, which is people call it what they say now, the heuristic derivation. Think of it as, think that sort of makes sense, but maybe wrong mathematically. Okay, it's just to motivate me in research. I don't think I'll ever answer such a question. You'd have to really study in great detail in the space of all the algorithms and try to define the nature of what we're doing there. I doubt that you won't understand it. Maybe so. But we're going to make this assumption, motivated once again by the fact that for finite groups, what we're saying is true. You know, how it makes us beautiful as all the problems we're going to have to stop in two minutes. Okay, let's, let's, let's, it's particularly bad point to stop since we've taken that two-week break. Let me just try to go on for two minutes. Let's see. Okay. You see, you see that the two, these two problems, the gauge invariance of the delta function and the gauge invariance of the delta allow you to prove a simple but important fact about the formula of probability. Namely, that this function wherever it is is a gauge invariant function of G and sigma. Okay. Now this, if you compute this for some value of G and sigma, right? And you compute it for another value and set up another value of C of C and G and sigma, right? Which are related to the other value by different models. Then you get the same answer. Okay. How, how do you do that? You see, suppose, suppose you had G1, which was G2 omega. Let's say omega. It's some geolocation. Geolocation. Okay. Then, by probability gauge invariance of the delta function, this is the same as this delta function with G2 omega naught replaced by G2 and this gauge transformation replaced by omega minus omega. Omega, omega naught inverse. By the variance of the measure, you can change integration from omega to omega, omega naught inverse. So now we change your variables, which gives you back the same, which gives you back the same table. There's no beton here because the variance of the measure and so we've got an expression that is the particle of the, of G1 or G2, okay? So we take in the particle of the determinant of a metric and prove that it's equal to the particle of the determinant of a metric at a point. Of course, that is related to the value of your variables. Now, you have to study the proof of determinant. It's a gauge invariant function of its argument. Since it's a gauge invariant function of its argument, we might as well compute it for the values of its arguments. Okay? At the values of its arguments, where the delta function kicks. We know that there always is a gauge transformation, such that the delta function kicks. Okay? So there's some, so that is at when sigma i is equal to ai and this metric here is equal to a metric of this form without the omega. And there we do this calculation. Now, so this thing is something of this form without the omega. Now, remember that in order to compute this integral here, we only need to know how this delta function, the argument of this delta function, behaves in the neighborhood of omega e to the identity. Because at omega e to the identity of the diffeomorphism, the delta function is kicked. Okay? And, and, let's say that there's some, at that point we've got some e to the power 5 naught times g of t naught. So this, this metric is gauge equivalent, this diffeomorphism equivalent to e to the power 5 naught of g hat of t naught. Okay? So we only need to compute this thing in the, for diffeomorphism, they go to identity and let's take this 5 to be equal to 5 naught plus delta 5, g to be t naught plus delta 3. So we need to know how the argument of this, this delta function changes in the neighborhood to first store them delta p delta 5 and infinity delta 3. Okay? If we know that, we can compute the integral of this integral. Knowing that is very simple. You see, all we need to know, as we said, is how the metric and the coordinates change under infinitesimal diffeomorphism under an infinitesimal revolving transformation and infinitesimal change of this monitoring. So what's the change of the metric? g alpha beta is equal to delta alpha v beta let's find out whether it's equal to delta diffeomorphism. delta alpha v beta plus delta v alpha that's the change under diffeomorphism. Plus 2 delta phi times g alpha beta g hat how much of that? You know, the metric of which it takes. Okay? That's the change in, plus the change in field of modules. delta pi times del i of g hat at the end. That's the difference in the change in the metric. That's the coordinate. Delta x of sigma i is equal to what? Well, the vector field parameterized in diffeomorphism tells you how much it changes. So it's simply v i. v mu so this is alpha v alpha x compute delta fp we can replace this minus this. 1 over delta fp is equal to integral d d r d delta pi delta pi d delta v we call it v v alpha d v alpha times delta of this stuff times delta of this in the next step what we do the integral expression of delta function simply says that delta p pi omega x integral mu hat puts up to pi to treat your game. What does that mean? That gives us that this pi integral of okay, we stop at just one. Now, for each of the arguments here for each of these alphabets we need some new field. So let's call it v alpha beta times this stuff this we need some. Let's call it z alpha v alpha we have to integrate over we have to integrate over delta t delta pi v alpha in addition we have to integrate over this that we don't have to do this beta times z so the integral in here is d delta t d delta pi d v alpha d beta beta and d z. That's the expression for the inverse of the particle. What we insert in the particle is the particle not the inverse of the particle. But you will notice that this is a quadratic integral. This integral in that makes it a determinant rather than an inverse vector. What the determinant rather than the inverse of the determinant all we have to do is to replace all variables of integration inside here that's what we are going to do. Then we are done. There are more the variable of integration being treated as one. Inverse so now we are computing the determinant of the operator itself and that's what we get when everything is permanent. The path integral representation for this for this value of integral again, the path integral representation simply is this path integral with every variable in the terminal. Actually it's a little more convenient going to do it this way it's a little more convenient to first process this path integral. The processing is the variable of phi appears in a very simple way. The variable of phi multiplied by g r for beta doing the integral of phi gives you a delta function of the trace of phi first. Again, just to process this integral a little bit. What we are finding is B is restricted to be a traceless object and phi integral is like yes, it's a traceless object. So let's first just remove this thing make B traceless and then then replace then replace everything that's a little more convenient. That is the expression for our output. I'll be putting it for you. We'll discuss more.