 So, let's continue. So, just to remind you, last time we worked out the mathematics of the suck line deduction. So, we moved to a study of strings on pop-axe pieces. We worked out the study field theory, which is an ordinary gravity and we discussed how the action reduced on a circle. You've got a redefined interval, a redefined Einstein field. You've got an extra scalar corresponding to the size of the circle. And you've got a gauge field corresponding to a g mu that. And we worked out the action for that. We also discussed how objects carry in charge, after it's getting momentum in the extra direction, the charge of the gauge field. And we discussed how the b mu field decomposes into a gauge field and a lower dimension of the g mu field. There's an exercise that I asked you guys to do, which all of you know. Now, we're going to move to the study of strings theory. And we discussed also that the gauge field that came from the dimensional reduction of the b mu field, namely b mu 9, was different in field theory. From the gauge field that came from the gravity, in the sense that our obvious modes never charged under the gauge field that came from the momentum in the extra direction. But it was no, in field theory there were no modes that were charged under the gauge field that came from the union. And we promised that this would change when we looked at string theory. Okay? So now we're going to start studying string theory with the next one on that circle. The first one that will eventually change. Okay? So let's say that we're going to x25, but we'll soon drop the subs from just because we're still talking about something. So whenever I write x, we need to develop it. I mean x25. And when x is equal to x plus 25. Okay? Now, this object is a field where x is equal to 0. Okay? So what does this identification mean? It means that the field x is identified with the field x plus 25. So doing the more expansions identification affects only the 0 more. So typically there's us that this is, all analysis of the conformal invariance and so on is low on the word sheet. You know, low in the word sheet, low in the space. Okay? So nothing is going to be affected, although the properties of the opiates and so on are not going to be affected by that. So we used to get conformal fields there. So some identification of the old conformal fields there. This is going to be a good background for a string property. Okay? All that's going to be affected is the 0. So we're not going to work on everything that we got from the oscillators again, but we're just going to focus on the 0. So that's it. So how is the 0 more first? So what is the action? We have to do 1 over 4 pi, alpha prime, sterile pi x, sterile pi x. If we substitute in just 0 more here, we've got something very simple. We've got, we've got the integral over 2 pi into 2 and we've got 2 alpha prime, integral x dot, okay. So what we're going to do is to substitute in the 0. Now let's work this out from two different points of view. The first point of view is that we'll work out from two different points of view eventually. But what, let's think of this object, you know, what's different than the 0 one second? The difference between the 0 one second is that a string, okay, with this x being compactified, as you go around the sigma circle, so what we're going to do is first work out the spectrum analysis. Imagine we're doing a lot of light cone quantization once again like the first few lectures of our course, we'll be working on the cylinder of the worksheet and we're going to work out the spectrum. Okay, so what's different? The difference between the 0 and the 0 one second here is that as we go around the sigma circle, it's not necessary for x to come to be periodic. So what we've done is taken x previously and expanded it as x 0 over the function of time plus sum over i n, sum over n to the power i in the sigma times some out of i n. And that procedure did not work out. Okay, this cyclic function, you know, works with this periodicity. It's not necessary for x to come back to itself as you go around the sigma circle, which is the boundary condition we proposed. All that's necessary is that x comes back to itself up to a multiple of two pi r. So the most general expansion will be this plus two pi r sigma r x of sigma r tau will be, we'll do m sigma r, okay, for something. Okay, so let's call this x m sigma tau will be m sigma r plus x 0 over the function of time plus number. Is this clear? This m sigma r is taken because sigma goes from sigma to two pi. This quantity changes from x wherever it was to two pi r times n, which is there out here. So the field space that we're working with splits up into different sectors, labeled by winding number n. So this winding number tells us how many times the tautic space field wraps as it goes around the word sheets. Yes, only one question. It's very different questions. We'll return the questions. So sorry, I should have said that, but we didn't understand. Yeah, this is, and that, and all of this will be deeper into that. Okay, so we've got this kind of expansion into sectors. Yeah, fields, let's say the sum of sectors, in each sector we do such an expansion. So the quantization of this part is unaffected, and this is the part of which can separate. So in each sector we have a bunch of harmonic oscillators, whose we can see is, goes from one to another eight, both to the left, both to the right. So let's go ahead about this one, and why just about this one. In addition to this system, what changes? So once again, you know, once again, fine. So there are two things that change. The first thing is that when we plot this kind of configuration into this action, we get a contribution in the action. We get a contribution in the action, and there are more important things, a contribution in L0 and L0 column, from the two space derivatives in this way. We see that more clearly as we go. The second thing that's more important for our cognitive purposes is that the quantization of this x0 thing, which is just the quantization of one to another x0 sum squared, what I did with this action, looks like it's not changed, but that's not true. It's always functioned to be bigger without x0 goes to x0 plus 2 pi r. So the quantization of this is a free part of the moving in a circle, which effectively just quantizes the momentum in the x25 direction. The momentum of the x25 direction, the x25 is restricted to be equal to n by... Look at the solution of this x field in the nth sector. So in the nth sector, xm of 1000, we've got this new piece of what I'm going to call it, this nth segment, and that. I'm looking at the golden piece, which we find, if you remember, alpha prime times p times tau. Why is the alpha prime here something different? What fixes the fact of this, that if we want momentum, this thing to be identically momentum into an alpha prime, we have the correct combination. Exactly. So the canonical momentum corresponding to x0 is 1 over alpha prime times x0 dot. You can't actually argue that it's alpha prime or alpha pi prime. Okay, the canonical momentum of x0 dot divided by alpha, yx0 dot here with alpha prime times p, divided by alpha prime. Now this p is n by alpha. So in the expansion of the field, we get n sigma r plus n by r of prime times tau. We also get this, which are unchanged, but what we have. So let's rewrite this whole thing, even if it's useful to rewrite this whole thing. It would be useful to rewrite this whole thing in terms of sigma plus sigma. So let's remind ourselves that sigma plus is equal to tau plus sigma. Sigma minus is equal to tau minus sigma. So tau is equal to sigma plus plus plus plus sigma minus by 2. Sigma is equal to sigma plus minus sigma minus by 2. So that's the whole idea. So we get sigma plus into, so we get x is equal to, sigma plus into, well from here we get n by r of prime. We have n by r of prime and the whole prime. n by r of prime, so that's it. Maybe it's very different from alpha prime. n by r plus alpha prime by 2 sigma minus into n by r minus n. Terminating n versus n. Terminating n plus 1. Oh, you're right, sorry. Now this quantity here, this quantity that appears behind sigma plus, we get alpha prime by 2 taken out. Yeah, we give it a name. We call it bs of prime br is equal to n by r minus m. And bs and br both have been equal to the momentum. And quantity is the clear role of momentum multiplied by the left moving coordinate. This has been periodically. Then the whole expansion of x in the left, you know, the combination of x is the left-movers, right-movers, both are oscillators, but not for the zero. For the zero mode, if we try to make such an expansion as we did it somewhere, of course, the coefficient of this would have been b and the coefficient of that would have been b. So for the zero modes, left-movers and right-movers were linked in the sense that their coefficients were forced to be the same. Now we see some sort of relaxation on this. The left-moving momentum, that is the left-moving contribution to the zero mode and the right-moving contribution to the zero mode need not be the same. And in fact, the nth set are not the same. Left-moving contribution to the momentum will be of the total momentum is bm plus pr. The total momentum is n by r. The total momentum is n by r. Pn reduces to n by r. If n is equal to 0. If they were all halves out, such that Pn would have been the ordinary momentum, and one thing you can see. Similarly, Pn would have been accounted for in normalization. If I made a manifest, that's why it took this algorithm to get an object that would have reduced the momentum. Physically, as a momentum carried by the momentum... This is not quite what... as a momentum carried by the momentum. For example, if I write down the mass function. That's right. We do that just now. It's a pr... In the mass formula, pn will appear exactly and p will appear before. Because mass reduces to the normal mass formula with pr. So, perhaps the formula that... what we're doing physically is clear. We're adding whiteness in this case. Because it's something interesting that has to have a formula. At the formal level, it's making zero modes behave more like the oscillators. In the sense that we're allowing what's happening on the zero mode side of the... Let go of some zero mode sides to write for us to become different. We're allowing ourselves to become different. For these two things to be different. Really, that's not right. Let's immediately write down the flow manuals afterwards. And leave the dispersion relation. The formula for the mass-related particles that arise is called activation. So now we'll do that if we want to know what L0 is. So L0 as before gets convolution only from PLs. And with the normalizations that would have said you know, equal to p. P L is equal to p. L0 is equal to alpha prime by 4 is the usual one. P L squared plus 4 p squared. Okay, so this is... Now, the rest is in the long-term direction. So we've got the mass. Minus alpha prime by 4 minus, because space life... No, make your part. Mass plus nl minus 1. The number when the minus one comes to the blue will be energy product. Caffeine energy product. Okay, and as you know, mass is equal to minus alpha prime by 4 p r squared minus alpha prime by 4 m squared plus nr. No, no, sorry. This is mass. I'm sorry. This is the mining number. That has nothing to do with this. This is mass squared. Because we've got a hold. Okay, it carries some momentum in the non-compact directions. This mass squared is k squared minus mass squared is k squared in the non-compact direction. Mass squared, because we are showing that I equate it with this value of the mass. This is the mass if we are... Okay, this is the mass if we are going to reduce. This is the mass if we are going to reduce. This is the mass if we are going to reduce. You see, mass is well defined notion when we've got a Poincaré action. I've got no Poincaré action like the direction in which we dimensionally reduce. So we are looking at the mass square of the effective particle in 24 n, in 25 n. So by mass square, it just means k is minus of k squared which is also equal to p r squared plus 4 pi alpha prime into n r. So in any way, with the x coordinate, there are no identifications. p r is equal to p r and the equivalent is to now p l is not necessarily equal to p r and n l needn't be equal to p r. What I'm trying to do is that these two quantities are the same. I'm going to write the n l and n r. The relation between m and n here because n n minus n r would be an integer. Yes. So we will not be able to satisfy the mass definition for arbitrarily m n. You know. So for crazy value to the radius, we will not have physical particles with the... Yeah. So logonite comes like this that if we subtract these two... So let's do that. So logonite comes like this that if we subtract these two, what would it be? Okay. So what we get here is p l squared minus p l squared which is equal to 4 times n by r. And that must be equated with n r minus n l 4 by r. So the equal to the level match of addition is that n r minus n l is m times n. So if you fix the value of n, then you must satisfy this definition. You must choose your oscillators by x and x. Fix them n, fix them n. Then you have to choose your oscillators by x and x from the right and left such that this difference we have translation. I don't understand. So we should get some condition out of that. Some condition out of that. I mean, conditional... Discharge here. This n by r is the momentum corresponding to translation at the end. I have understood. On the string, the fact that lz have to be equal to lz over r. Again, apply this system. It reduces in the case that n is equal to 0 to the condition that n r is equal to n. From the condition n r is equal to n l and m 0 n l. Notice that you have to be in a new region. So it's a filterable condition. It's not something that depends on r or depends on our uprights. Given any m and n, in our mind that that's how it works. Quite formula and a little motivation. Okay? So let me look at this... Before going into the detail of that system where you go into the moment, notice one thing in which this squared and the other squared have an invariance problem. A certain formula invariance problem. So suppose I do the following. I set it to alpha prime divided by r. It's equal to n prime and n is equal to n prime. Then p n as a function of r, m and n is equal to p l as a function of r prime, m prime and n prime. p r as a function of r, m and n is equal to minus p r as a function of r prime and n prime. Can you see this? You see I exchange r by alpha prime divided by r. So I get alpha prime times r prime. And I exchange n by n. So this term becomes that term and vice-versa. In terms of r, minus r can be written as this one. Something very interesting. It tells us on the circle of radius r has exactly the same spectrum as a string on the circle of radius alpha prime. But in all the math states to each other you need to flip labels. This flipping of labels is m is equal to n prime and n is equal to n prime. Now this is from the point of view of quantification. The point of view of just gravity of a circle seems very non-inclusive. You see, just in gravity of a circle you've got a bunch of massive modes coming from the Kalooza climate states. And these states are more and more massive if the circle gets smaller and smaller. Circle leads not true that a small circle and a big circle have the same effect. So what's going on? You see, what's going on is that in string theory in addition to the Kalooza climate states, you have states that practice the winding states. As the Kalooza climate modes get heavier and the circle goes smaller and smaller. The winding states go wider and lighter. The circle goes smaller and smaller. The heavy Kalooza climate states of a small circle that have exactly the same mass as the heavy winding states of a big circle if we change r to alpha prime. We have physics from the point of view of a low energy observer, you cannot at least at the level of spectrum you'll soon see that this is a more general thing than the spectrum. You cannot distinguish between string theory compactified in a circle of size r and string theory compactified in a circle of size alpha prime. The spectrum is identical. And the way it works is that the winding states in one picture that's the momentum states of the number that we need to change. We've only seen the subpoena level of the spectrum. We'll soon either end with this lecture or the next lecture. This is a full non-potable symmetry of string theory. This is the first glimpse of the symmetry of string theory called T-duality. We're going to go on with our technical analysis but just a little word about T-duality. T-duality is a phenomenon that we can see in the years. But in string theory it's hard to go beyond the minimum length scale. It sounds like something. You see, you might have thought that you were looking at a compactification that dealt with very small length scales by compactifying a very small scale. But in general we get exactly the same spectrum as if you're compactifying lots of things, you know in a more fundamental way we could say that there's no way to distinguish a string compactification on a very small circle from very large scale. There's no way to distinguish string compactification on a circle of size other from string compactification on a circle of size other than that. So the very notion of whether something is bigger or small once you get that string length at least as far as the size of the circle is concerned seems to become more difficult. This is one of the first indications of using many of these as we go on. But that somehow the notion of space and time must get modified in string theory. You know these the absoluteness of big and small must get modified in string theory. How that plays out in the end has not yet been worked out properly. We don't understand exactly what replaces these notions. But we are getting into this notion not yet at least. Is this true even if we let you be formative is this true of which like deformed and awkward meaning? You see you can't do it in a way that solves the equation of motion. So now in many situations there are generalizations of this to work. So let's look at the string theory on compactification on a circle of size other than that. It's a complicated they're not analogous it's phenomenal mirror. We discuss much of these these things as well. Is there a reason that you expect this to be true about arbitrary meaning? I want to take this too far. It's not a claim that basically the point is basically have level the break-off of that. The main point is that you know the winding was in a moment and you need to change it too. As long as this statement continues to be true you have a hard time distribution between big circle of winding and small circle of the moment. Let's say even like I take some minor strings so you definitely have some gravitation because there's a gravity in the sphere and the circle will react to that. Because this is part of this question. Yes. So in what sense are we saying that we can't distinguish between the two? In the sense that if you give me a small circle then give me a big circle can I deforming this circle to make out the winding backslide? Like using three probes. You see anything that has a small deformation out of this cannot be distinguished because any experiment to perform in one picture has to come What is true is that if you prepare a probe that you know is winding then you can check whether that that state is winding or momentum. So if you have some way of saying this is winding and that is momentum then you can distinguish between these two. But for any experiment you can perform in one picture there is a corresponding experiment you can perform in the other picture probes which will get the same result. I mean obviously my probes don't know this circle but conformal fields here are just the same. To the extent that stream perturbation is determined completely by the worksheet these two things are the same spring compact. Any question you have in one picture has a cause of any experiment you have in one picture has a cause for any experiment you have in the other picture then I will give you the same answer. And why do I need states left all over the circle? Both of them can in the other picture. But obvious that that you can distinguish in some invariable way and to the extent stream perturbation is very important. It's not true. There is nothing you can do as a small perturbation on this picture, the distinguish between these two. We haven't shown that but I will show you that these are just identical as conformal fields. Stream perturbation field is determined by the conformal fields here in the machine. To the extent that the conformal fields are identical. This is the statement that they are identical. Depends on the fact that this thing is a circle. Both of them are the same. So if we take something as a small decoration around this it's a small deformation in one picture or the other. So any small deformation here is written in either picture. Because any small deformation is just some vertex operator put on this thing and put this vertex on in one picture and now it's the massive vertex operator here. If I start here and start before me using vertex operators then I need to make sure that the sigma derivative is valid. Yes. You will not be able to distinguish this from starting with the other picture putting on the corresponding deformation and doing the corresponding. Ok. We are going to get back either at the end of the lecture or in the next lecture the vertex operators corresponding to these states. At this point it was just to make sure that let's now go back to the analysis of the spectrum. Ok. Let me start by analyzing the spectrum and now first you know as in our study of Hong Kong accident theory particularly interesting question is what are the masses states in the land? Let's first do this arbitrary value of the radius and then special value is the spatial radius. Ok. The radius at which this under this operation changes. Maybe r is equal to square root of r. Again. But let's first just do this as arbitrary values. Ok. So both we want to understand what the massless particles are in the theory now. You want to understand what the massless particles in the theory are and some arbitrary values. Now because of our interest in complicated fashion in this expression the only way to the only way to satisfy this relationship is to have this equation and then we're back to the other analysis. So we have m, n equals 1 n power e to the 1 and so what are the modes? What are the states? The states are alpha minus 1 alpha tilde dot minus 1 m, n acting on 0. But remember these m, n's could be either 25 or mu. So now there are many options for the point of view of the lower dimension here. Firstly if m to n equals mu what we need to know is the diluchon, the gradochon, the b mu in your field of the lower dimension. If m and n are both one of m and n is mu and the other one is 25 okay? Let's for instance say alpha minus 1, 25 alpha tilde dot minus 1, mu this has a single space time index so it's a gauge gain. We can recover a master's gauge gain. There's a second master's gain which is minus 1, mu alpha tilde dot. Actually it's a symmetric combination of these that are most natural. And these two gauges these two master's gauges are what we saw in our analysis because it's a kind of amplification class. Maybe the master's gauge gain that came from having one leg of the gradochon in the 25th direction and one leg of the video. Finally this gauge where alpha minus 1, 25 alpha tilde dot minus 1, 25 okay? And this is the this is the scalar of what you get from the size of the circle. Remember we had in addition to the low dimension we did it on an extra scalar to the size of the circle. Among the gauges you see from the point of view of field theory one of the gauges came from the direction reduction of the gradochon the gradochon is a symmetric combination whereas the other one came from the direction reduction of being used and this is the radius we recovered what we saw last time from here. There's one little point I wanted to make here which goes to form. Let's look at the vertex operator corresponding to the gauge field that came from here so alpha minus 1 maps to del x alpha tilde minus 1 maps to del yx so the vertex operator will be of the form del x, 25 del bar x mu times minus del x mu del bar x mu times del bar del x where the gauge field will be in the non-combat. The vertex operator corresponds to the gauge field the length of the dimension reduction of being. Now we are interested in measuring the charge. If you are interested in measuring the charge the next charge of the system you couple the derivative of a Lagrangian with respect to a mu gives you the hd a mu and x gives you the count at x and if you are interested in a genuine integration of ground space you take the derivative with respect to the 0 more of a mu in momentum space so we are interested in measuring the next charge under the gauge field that comes from the coupling period we should be interested in this vertex operator and k equal 0 k equal 0 that is to this vertex operator and it is very near to be triggered it is very near to be triggered as you can see in the following way this thing we can write as using the equation of motion of exactly a second data x 25 del bar you know I subtracted an extra term here which is del bar which is 0 and then we could also write this term here as minus x 25 why I want you to know that each of these terms is a total derivative so this object if we were working on combatability this quantity was single value insertion of such a quantity under the worksheet it would just give you something that is 0 over the equations of motion for instance if this was the integrated vertex operator calculating sub-rotation if the other operators that we insert into the worksheet don't create you know this vertex operator all these single value things then we just get 0 all the other x 25 need not be single value the other vertex operators that you put into the your put into your worksheet could force x 25 to wind around and then this vertex operator is not strange what we are seeing is the fact that only states that I know about in x 25 can carry charge under the gauge field that comes from B union now in fact the states that I want to know about x 25 do carry charge under the gauge field that comes from B union and to convince yourself of that I would like to suggest a linear calculation okay I want you to calculate the three-point function three-point scattering of the gauge field now with some of it the gauge field that comes from B union with some momentum and two tachyon operators which have equal amounts of momentum yeah you could basically what you are going to get is because the charge under the gravity the radius from gravity is basically N so I mean it is the exchange that you work out there the charge under that that's absolutely that's absolutely as we will see this duality symmetry is the symmetry and do it x L goes to is equal to x L whereas x R goes to minus x R in 25 and under the symmetry momentum gets in the shape of the winding the gauge field from B union so that follows but now that's I would like you guys to do this calculation again so what calculation let's do this you know what the vertex operators are actually I haven't quite told you about the vertex operators for state of the catapult sorry just to go on this slightly I'm going to tell you about that in a moment once I told you about that you know what vertex operators insert and you know what the vertex operator on this B field is that there is simple generalization of the many calculations we did before for 3-point functions in the Gloucesters do the calculation just to get your hands away from this stuff and a check that you will find with coupling only with states that I am not here already as what I am going to say is that I am many more structural in general reason of understanding this good so why don't I bring this up in a moment I brought it up because already at the less arbitrary radius we see one phenomenon in string care that we did not see in string care then the gauge field that comes from this B field of mu carries states that have charge okay and these states have states that are long enough arbitrary value of the radius which we are interested in early in the mass respect no? that's it there is nothing else but all these states are very big which are the states I mean the case of the charge and arbitrary values of radius that's true okay not an arbitrary value of the radius now just the answer kind of question don't matter let's focus on the particularly interesting value of radius the particularly interesting value of radius of picot cosine is r is equal to square root of r this is the self given point of the r goes to our parameter the point in which there are two different descriptions of the same theory of the same theory without taking the radius and that example is like true here as a warm-up to understand the physics of r so I am interested in the mass respect so we have that n squared is equal to n by r plus w r by r we have the whole thing squared plus nl minus 1 it was also equal to n by r minus w r by r plus n r this is true now if we said now if we said r equals to square root of r what happens what happens is sorry that was a 4 by r this is what we get each of these things has one of is proportional to 1 over square root of r so we get n squared is equal to 4 by r by r into n plus w the whole thing squared by 4 plus nl minus 1 is equal to 4 by r into n minus w the whole thing squared by 4 plus nr at the special value of the radius and now let's search for solutions to the for solutions to the problem of mass resistance let's see we have to cancel this minus 1 we can do that i equals to nl over this n plus w so let's see how we can do it with n plus w so I'm going to write nw in a bracket like this to deal with what happens in this area so what we can do it is if we choose nw to be 2 to 0 you see in order to cancel minus 1 this minus 1 we need n plus w to be equal to 2 so what we can do is to choose nw to be 2 to 0 that way and that way nl and nr go to equal to 0 and give you a mass resistance you see because this will give you 2 squared so that's 1 minus 1 plus 0 but once we've done this there are clearly 3 other solutions to this problem each of them with nl mass resistance with both nl and nr we also try n to be about 1 and 1 let's also solve this so we could try 1,1 and then if 1,1 works minus 1, minus 1 also works okay then we could say nl equals 0 but now nr is not equal to 0 because 1 minus 1 is 0 so we have to say that after 1 oh we could try 1 minus 1 and then minus 1 to 12 and nl equals 1 and nr finally there's 0 to 0 and nl equals 1 enough now we can see the list of the ways in which we can get mass resistance particles at the special level of the range so the interesting exercise to take is mass resistance particles and the group that we with symmetric properties are under the range group of the non-compact range okay let's first come to a group together the same come down the same there are 4 scalars here that's clear where nr is equal to 1 the index to buy will be 25 or 0 so 2 is 25 we get a scale so each of these options has the option of putting all of these guys 25 and each of them is now implemented each of these options is a scale okay so we got 4 plus 2 plus 2 plus 1 so nice what are the gate codes? a gate code is an object with a vector index to forget if one of these could be 25 the other one mu and which one does not and then what about higher dimensional things like the tensors and so on so that's exactly what we had there before in a part sorry this is 9 scalars yeah so that's non-counting like a trace of there's also the idea of a new mu the trace part of alpha mu, alpha mu that's the little jump of the low dimension that's enough 9 scalars plus the little jump of the low dimension you know what else? everything in the field here you would have given mu for the dimension it reduced to 0 and 5 these gate codes are these 6 gate codes here are useful you see two of these gate codes came from here came from taking a mu index here and a 25 index here these were familiar they also come from actually reducing the graviton as the beam which from non-mining is like all of these from w we don't need for 0 as a purely stringy object there's no field here there's no graph there's purely gravity here I'm going to show you as we study the vertex operators corresponding to these guys in the middle of the gate I'm simply going to show you that these 6 gate codes are grouped together into an SU2 and SU2 gate codes these are 6 U1 gate codes they are in fact gate codes are of non-mining gate codes you can see this immediately actually you see these guys have a momentum in mind and therefore are charged under the gate codes on that comes from dimension reducing the graviton and dimension reducing beam this is a theory of gate codes on such that gate codes are charged can be charged under other gate codes the only consistent way to do that is to not be released exactly a and w and I will also tell you that these 9 scalars transform in the fundamental by fundamental of this SU2 gate so these guys also can recharge under momentum of winding these scalars so they also charge them in this SU2 gate to SU2 and they will examine what way that happens I've shown you all of this but what I'm going to do is it looks very interesting you see what what have you seen we've seen that you take string theory which was non-compact was relatively boring in the sense that the fields had no non-immediate structure the graviton is in the fundamental by fundamental by you see you have to say how they're charged under each of the two resistors so there will be like this by fundamental fundamental of these of this SU2 gate so it's very fundamental there's like three times it's in the three times right but these I and J are SO3 vectors the four we are seeing here does carry momentum and winding and they have to recharge under the and they get less 1 and minus 1 but every plus plus there's a minus we'll see the sound much more think this is more detailed as I promised but even before that I wanted you to see how interesting this is you think string theory and you put it on a circle when it's non-compact you put it on a circle and a momentary science the mass of this vector is just that of U1 gates the U1 gate the two U1 gates goes on it came from something of no non-abidance structure parameter in the video okay then you tune this this circle size to a particular size and suddenly you see the emergence of non-abidance okay we'll see that that you might ask how does it go away how when you move from this particular size to another smaller size just go away so we'll see that that basically will be like these two gates because the non-abidance the vector will be broken to U1 gates U1 by the vector these two gates please see all this as we go okay you know it's quite quite interesting you get there's a new source of non-abidance gauge symmetry this new source involves winding in the bush winding strings in the bush okay there's more symmetry than you would have thought the same situation which is very small about the symmetry U1 by U1 in fear therapy actually has an international international symmetry from quite a few years ago which has been his domain when he moved to Larson he's seen some of the non-abidance gauge just give you the layer of land there are new awesome states there is a good approximation when every other string of state every other string of state is well-centered that's another thing new awesome states okay it will be actually very similar to you know looking at a an S2 to S2 theory with the matter and the three types and then you slowly turn on a wave to that matter okay at the origin there's all the non-abidance structure when you go far away you just have U1 by U1 there's an effect of U1 by U1 theory breaks down as the wave of this thing goes to zero goes out of mass you would see it restore I mean some of them right no I think you would see it restore some high energy state here as well you see minus the if you go to energy such that you can ignore the mass difference between these winding states see these winding states always exist and every time it breaks it's just that they're not masses every time it breaks if you go to energy such that you can ignore the mass difference between these massive W-ball sounds and the mass difference you will see the approximate S2 to S2 it will be the same amount so there's simply some partition which is there's a literature which is almost useless which yeah if you have a generic value of the circumcision you know you go to energy such that or whatever there you go so this was to show you how in the presence of mind you can get new interesting low energy physics new massless states and new pedometer we're going to come back and discuss this in more detail but in order to do that in order to do that it would be useful to understand the vertex operators of this of this theory so let's get to some partition function sorry right at the beginning of this lecture you should have calculated okay I'm sorry I'm going to take a 10 minute break before we go into vertex operators you can this is just a little spectrum once again just to compute at an arbitrary value of the radius the partition function the one loop partition function this theory on the on the daughters just to show you how modular in there makes this requirement in this this lecture has become a little happy I have to go on and leave you all with that but let's move on we need to discuss the partition function of this theory and then I will bring back the operators okay also there it goes it's unchanged so we've got this x component in theory we're going to do this daughter's partition function so if you remember we had to define no one should have defined as no function for you so we want to define it so there's a part to come from the oscillator including the zero white energy just totally unchanged in this okay so what are we doing we're computing the daughter's partition function we're computing the daughter's partition function of the part of the conformity theory that comes from this x just this one loop okay so we have this term that comes from the oscillator and nothing has changed because this which by the way in Batman I would like to write as eta of tau more squared and we can write it as eta of tau more squared just because it saves us a lot of white energy okay so so the partition function is this business times what times whatever we get from the zero let's discuss what we get from the zero in the case of non-compact in the case of non-compact theory remember the way we did it we said that well what we have is the is the is the spectrum of a particle on a circle of size tau 2 the spectrum of particle on a circle of size tau 2 and and the particle has effective mass for the daughter okay and then we said well and that is tau e to the power minus tau 2 times k squared plus n squared dk integral then we did the Gaussian integral and we got the answer that is portion to 1 times tau all of this here but there are two differences differences are firstly that firstly that we have winding and secondly that momentum is discrete okay so let's do it again so what we are going to do is compute the trace of q to the power n of 0 q bar to the power n of 0 from the 0 1 sector from 0 the 0 of contribution of that when we have this we come back now n of 0 is simply equal to 4 by alpha prime k n squared the contribution of 0 and n of 0 bar is the 4 by alpha prime k r so what we are going to do is compute trace of and the trace now is just a sum over n to the power 4 by alpha prime k n l squared q bar to the power 4 by alpha prime k r well remember k l d r we are going to do that so we are going to do that we are going to do 2 equal to the power 2 times tau and q bar and tau 2 tau 1 now you can take all these definitions blind here and get some expression as an exponential involving m n tau 1 and tau 2 okay we are going to do a little thing for the power n be done I am just going to go through this this thing is the initial of minus pi n squared by power squared plus 2 squared by power 4 by alpha prime this thing this thing is equal to this thing of course if somebody give me interpretation of this tau 1 we multiply by n times w does it remind you something we say what does tau 1 double do we just come then tau 1 then operator 0 bar is an integral so the integral over tau 1 in the in the integral of the modular space of the word chip of the torus will enforce that L0 minus L0 bar oscillate L0 minus L0 bar is equal to n times something resolved there is a new part of the portion of tau 1 to better what we have just from the oscillation okay integral over tau 1 enforce that L0 is equal to this sensible thing and then this is where it is this is essentially the energy of this we call w in my books I have a game I was looking at police escape 20 times w x m and you actually here is multiplied by 1 over eta tau and written this way is far from obvious it's far from obvious this expression gives us a modular invariant one new there are two sources and not all but firstly mod eta square by itself is not modulated it means an extra factor of square root of tau 2 exactly square root of tau 2 into mod eta square is modulated okay so first we need a factor of that somehow and then the sum has to be modulated as ready and it's far from clear that's the case however it's a trick here an interesting technical trick that allows us to demonstrate modular invariance and has a physical interpretation so let's first look at the technical trick and then look at the physical and then you take the sum over n that is the sum over all the momentum modes along this circular and do it after and close on reason okay let me remind you about that first you've got some function some echo which has which has which has Fourier which is equal to f of k f tilde of k h bar 2 pi i k like the following the sum you know provided this ratio of all of these functions the sum of f and n is is equal to the sum of f tilde and n the 2 pi that's something you all familiar with but that's pretty easy but here you need to check that this is true or you don't know this is true yeah, I mean you know basically for support suppose you take this thing say what do you want to do you want to you want to take this and sum it over all so that's what you want to do is to calculate that's the same thing as integral dk for k and then sum over n e to the power 2 pi i k m you know what this is now this is a little unifying doesn't stand but one way you want to understand is to put you know to discretize space time and compact it put the values of m to be periodic which periodicity came to discretize the possible value of x and then take allow k to take allow values then the m divided by n the respect to periodicity is the problem we have periodicity of this function or as a function of x or with periodicity capital that's the allow values of k Then they allow that k would be k is equal to m divided by n. And of course, because we only look at these discrete points, there would be an effective period of this k in k. m plus n would be a period of this k. So now this is very well done by mathematical function. Let's call it. Let's suppose we do some over it. This is e to the power 2 pi i over n would be like n. n runs from 1 to n. This is very clear. This of k is 0, sum over the various roots of unity. m root of unity. So it's 0 unless r is either 0 or n or 2n. OK? So this is equal to, and then it depends on what we've done is n times delta r plus delta r n. Under this discretization, they go back to continuum. So this sum over n becomes like an integral over n. But we were very interested in integral over k. And the difference between k and n is the factor of n. So if we want this to be true, we want to replace this by an integral. So these get replaced by delta functions separated from that. The only tricky part in the soil process is taking that normalization. So this is correct. So once you find things this way, this is correct. And now we can apply this formula to that object, with x being taken to be the object n. So what we want to do is to compute integral into 2 pi i. And now we replace n by, let's say, r. So 2 pi i, no, that's not the best. We want to keep it by 2 pi i rx. Then minus pi tau 2 alpha prime by our square x square plus 2 pi i tau 1 into xm. xm. There's also this part. But since that doesn't involve our variable n, which are called x, it just goes on to that. So let's do this integral. Firstly, you see this part and this part are very similar. So this is integral into pi pi. And then there is r plus tau 1 times m times x. And then minus pi tau 2 alpha prime x square pi i. So when we get into this integral, we get the factor of square root pi that we get from the integral. Then we get square root of this stuff. So we get square root pi square root alpha prime r, and then a square root of tau 2. And then we get what we get by computing square. Times the exponential of, so computing square, we remove this 2. We take this as square, i squared, r plus tau 1 in the whole thing, square, and divide by whatever's here. So pi tau 2 alpha prime, where's the half? This 2 has cancelled the fact that it's 2 n's. Twice 2 n's. 2 n's squared. Because the standard deviation is the way you can do it. Ah, you want to do it here. So 2 n's here, factor of 4. And this is the way you can work it. Very good. Very good. That's completely squared. And that's completely. So this thing, no, I think this is OK. You see, what it does, we go here, and lift up to 2 n's squared. Yeah, you see this one. Yeah. So this is r by square root of tau prime tau 2. Note the number tau 2. This is the thing that we wanted to do. Take the model and there is eta. Times the exponential, in other words, pi n's is this, we have. And now we add what we actually need. And there's minus sign by the way, which is the i's squared. So minus pi, minus pi being 2 tau 2 n squared r squared by r prime, plus r plus tau 1 m dot squared. But there's a r squared, so that's right about r prime. R prime. R prime. R prime. This has an r squared. That's an r squared. That's good. And then there's some tau 2. There's a tau 2 there. And here there's a, let me make this, tau 2 squared divided by 2. Later, it can be written as exponential of minus pi m power. This is something over all that is for r. This is a much neater form. From the point of view of modular mathematics, than what we had before. But as you can see, there was no hope for modular mathematics in the original thing. There were all these factors of r that appeared to be in different terms. Now all the r's have been found out. So there's no hope for simply modular mathematics. And let's check. Under tau 2, tau plus 1. That's obviously modular mathematics. Under the real definition, n goes to n minus 1. Under tau 2, tau plus 1. This is obviously modular mathematics. What about tau goes to minus 1? Minus 1 by tau. Minus. Yeah, let me look it up here. N goes over. His m. His m is how the r goes to minus n. This is modular mathematics. That's transformation. Oh, we can change. Yeah, can I include you in change? Minus equal to minus 1 by tau. Okay. So what's the imaginary part of this? So this is equal to minus 1 by tau, tau bar into tau bar in tau, right? This is equal to in tau divided by mod tau. This is modular mathematics. That's correct.