 So after this long gap, we continue our lecture in the theory. We go on to the end of the semester and probably continue through next semester, providing very sufficient. So in this semester, the things I want to talk about is the date. What's the date today? So the plan for this semester is to discuss open strings. So let's continue with the bononic string. Let's start with open strings to discuss to avoid compactifications and de-viewatity, de-brains, or one of them that we use in string theory. You know, we don't want to hate the quotas and so on. This thing will go on for about four or five or six inches. And we go on to discuss local physics and the world sheath of the string to derive the beta function equations in the world sheath of the string and to connect that with the equations of space-time, science-time, ice-time equations. This discusses string theory in curved states. That will be another three lectures or so. And so that's eight or nine lectures, maybe ten lectures for all of this. And after which we will continue with the discussion of the super string. We'll try to do a sort of harder discussion of whether we should carry with the super string. We'll discuss what's called the R&S formalism, which is what's given in Pulchinsky. But I was thinking we should also discuss the bulk of this formalism, which is a new thing in the developed city's Pulchinsky's book and seems to be proving it in the useful. I don't understand that formalism yet, so that's the motivation for us to discuss it. That will probably take us through to the end of the semester. I mean, I'm very excited. I think so really well in the next semester we start trying to discuss space-time. So brain solutions, super-symmetry and space-time and so on. And the ABA safety course. Okay, fine. So now let's start. So our mission for now is to, if you remember last semester we discussed the bosonic string reasonably far away. You know, the closed bosonic string, we understood it in many ways. That's of course a drawing model for most of real string theory. We didn't stick with drawing models and so on. Just so as to avoid large complexities as you can see with me. Without being consensual as you can see with me. But anyway, so let's start. Stick with this one. So what we want to discuss is the open string. Okay. Now what do I mean? If you remember the way we started our lectures, 70-80 lectures ago or whatever, was to look at world sheets that propagated like that. The little cylinders that were propagating. Then we tried to write them an action that described the propagation of these strings. You see, we assumed that the strings, and then we did a lot of work with this, but we assumed that the strings we looked at were these little loops that were propagating correctly. Now you might ask, are there any other possibilities? Now that we've done all the work, we've understood the local physics and the world sheet of the string very well, you might understand that you might wonder, the same local physics might describe other motions. And that's the thing we wonder about, you know, if you've got a bit of string, well, I mean, many things you do, so that if you think in space-time, but intrinsically, what you can do, there's basically two things. You can keep it open, but you can close it. So we've discussed closed strings so far. Now you might wonder, what's this story? So could we discuss strings that are a little lying, rather than little loops? Now I understand they're both virtual and special. That's the question we're going to address. Okay. Why didn't I assume, you know, much of our analysis has been local on the world sheet of the string. And so suppose you've got a world sheet like this, a world sheet like this, we take it as a local patch. Yeah, it doesn't care. There's no other difference. So world that was lower on the world sheet is string mechanics problem. So all we have to do is to worry about new effects coming from the high-previval objects. Okay? So let's start in a low-brow way. Let's start as if we were starting with the first lecture for closed strings. Okay? And try to progress from there. So you remember that the way we tried to quantize what was our basic action? We had 104, 5, 5, right? Minus somewhere right again. Square root g times g alpha beta del alpha x mu del beta. Okay? And then in both our initial low-brow as well as our more sophisticated ways of working with this action, we effectively fixed a gauge in which this g was some standard metric. If we were working with a cylinder, it's just a flat metric with the theta coordinate periodic with the data we do by and the tau coordinate. And then we worked with that metric for much more than what we would have. Okay? So now what we're going to try to do is the problem. We're going to try to see if we once again fix a gauge, in the flat metric, in a space barometerized by 0, by sigma and tau, but sigma will run from 0 to pi instead of 2 pi. And we will not impose a periodicity on the string. It will not be a closed string. It will be a string. Tau runs. Okay? Now, having said that we're not going to impose periodicity on the string, in order to make it well defined in the dynamical system, we have to impose so let's see how that goes. So the condition that we're going to impose well defined dynamical system, the system from which you go, well, if you take the action of the system and you vary the action, you get well defined equations of motion. So let's see. Suppose you start with this action, yeah, and you vary the action with respect to x, what do you get? Well, the usual procedure of getting equations of motions, it puts one delta at this x, yeah? And then you integrate by parts. Okay? And the equations of motion of what's left once you throw away the surface. Now that we've got boundary of our state sandwiched model of the surface. So what's the boundary? What's the surface? So the surface level is proportional to integral of its del alpha square root g, g alpha beta x mu del beta well, okay. So this is total derivative. So it could be integrated to something of the boundary. So what is it on the boundary where we use stocks here? Right? So it's the dot product of this vector with the normal to the boundary. Okay? So you can use stocks here and rewrite this as proportional to effectively this vector whatever it is this vector whatever it is dotted with the normal to the boundary. Now what's this vector? This vector is let's move to flat space clear? This vector is essentially del x mu that's del alpha of del x p. So this vector is suppose let's call it the alpha okay? If we want equations of motion to follow from variation of the action this vector on the boundary should be a z. Okay? So that we can do this in different parts to recover the equations of motion. But in natural ways, they had this vector dotted with the boundary dotted with the normal to the boundary was 0. The first way is to import del x mu that is to say that x is some fixed value at the boundary. So the boundary condition of the x is some fixed value. Okay? That would certainly do it. This is del x mu of transition. But there's a second way and the second way is to demand that del alpha by x mu dotted with the boundary. Boundary normal is equal to zero. That will also do it for all the three fluctuation fields del x mu. Now these two different names the first boundary condition is called Dirichlet boundary conditions and the second boundary condition is called Neumann boundary conditions. Okay? The first guy is just saying that x can't change a boundary. The second guy is saying that the normal derivative of x at the boundary is zero. It's sort of like reflecting boundary conditions. We continue the spirit of of the earlier languages and search the solutions of string theory that are forensic. Okay? So in principle you could do something with x1, something as with x2 or so. But let's just start with search the solutions of string theory there are full residents of 26 dimensions. So you're achieving that by adding that you're choosing this boundary condition or this boundary condition for all the 26 dimensions. Now let's see what will happen if you choose this boundary condition for all 26 dimensions you will have zero more of the string near some place in both space and time. Because the end points of the string will be stuck to some point in space time the middle of the string can fluctuate away a little bit but not very far. This will not describe particle proper conditions particle lives and all kinds of things. As we go on with our string theory we will find that actually this does describe something interesting it describes an instant dawn in a path integral in string theory. It's what's called a D instant dawn. But at the moment we're just discussing this this is not what we want for now. So the only lens in the boundary condition in what we discussed so far is that this boundary condition is closed for all 26 dimensions of space time. Is this clear? So now let's take our boundary condition seriously and deal with it. So once again we're working on the string we have sigma goes from zero to pi and we've got various fields of x. So I'm going to now discuss the quantization of any given field x so that we ignore the label of x there will be a new label of x they all behave equitably as you know. So we've got some field x which is x of sigma and we have that del sigma of x at zero is equal to del sigma of x at pi as you can see. Let's solve this theory and then I'll let you have a convenient way of thinking of this that makes the open string look like half of the closed string. So what do we do? We modify x into modes that obey the boundary conditions. Okay? So one of the modes that obey the boundary conditions is actually cosine cosine of x sigma of x at this boundary conditions but we can need to work with exponentials so let's write it as e to the power so we've got x of sigma is equal to sum over n okay? e to the power i n minus sigma plus e to the power minus i n sigma okay? This would be then let's put an an again I'll use some foresight to write numerical constants so square root of my alpha prime I'll put an an and I put in all possible time dependencies so the only possible time dependence is all the equations of motion plus minus i n sigma i n tau so I can just write it as e to the power i n tau and sum from minus not including c okay? So this is a classical solution mostly in the classical solution I should also deal with the zero so there's some x constant plus some number let's call it q when we change it to p with it we can put the right again use some foresight the right constant and then we just write it as this is your familiar with response solve the equation del squared of x is equal to zero this is the most simple solution which respect the right boundary conditions that's a scare now to overcome that you might have thought I should only sum from 1 to infinity because the cosine is same but I sort of have a different time dependence same space dependence this is very the quantum solution it's a free theory classical solution determines what usually all we need to know is what the right term so each of the a n's will be promoted these a n's, this x and the p will be promoted in operator all we need to know is about the right combination okay? so let's try again working in this slightly more sophisticated way than as usual just because it's so convenient okay? so what we do as well is the symplectic form okay? so the symplectic form is simply 1 by 2 pi alpha prime into delta of x dot where delta of x integrated over okay? whatever I written delta x dot by 2 pi alpha prime is the momentum conjugate x so what I written is delta of p of sigma where delta of x of sigma integrated over the over the g of the string and does everyone remember what this means we discussed it when we discussed the closed string example do you remember what this means? this is a simple way of including commutation relations okay? so the coefficients okay? so space that has omega alpha beta delta x alpha wedge delta x beta then is a space whose commutator is between x alpha and x beta are the inverse of this omega after some factors of 2 and so on that we find in the north okay? so I am just including the commutation relations in a simple fashion and basically what I have done here is quantization the fact that the basic commutation relation is p commutator x as 1 now this formula is convenient because it allows you to change variables so this is the symplectic form in the space of x fields but we want the symplectic form in the space of a's x and p all we do is take this general extension here and put it in and that will give us the symplectic form in the space of a's x's and p's okay? so let's move let's do the plugging now firstly the symplectic plugging we get things that have particular sigma dependencies and the integral from 0 to pi we kill the we give us 0 integral unless the sigma dependence is exactly x because we got a cosine and you integrate with it sine, sine of 0, sine of i up with 0 okay? so what we need is exact encapsulation of these explanations okay? so let's see how that's arranged firstly of course 0 more can only click with a 0 okay? so let's work on first the 0 so what we get from here x dot gets no contribution to x because there's no time dependence so x dot is 2 alpha prime integral delta p which delta x integral delta p which delta x there's no delta p which delta p because that's 0, it's exciting so once we chose that to be delta p then if I have to be at x and we have this integral we have 2 pi alpha prime and we have this integral in the sigma the integral of sigma goes from 0 to pi so we get 2 alpha prime divided by 2 pi alpha prime which is equal to dp wedged to dx so this is the standard sublactic form between 2 variables that awaits commutation relation x commutator p is equal to the i is equal to the compressive component okay? so this is what we started with in the field dp wedged to dx this is the standard that's why I chose this 2 alpha prime you know so as to get this to be the standard but I didn't know what to start with I will put an arbitrary constant that shows it okay? now let's move on and do the oscillators so for the oscillators once again you only have to worry about what you get when n's are either equal or opposite but n's equal don't contribute because delta a n wedged to delta a n is equal is that symmetric? so you only have to worry about what you get when it's opposite okay? so let's do that so what do we get? so let's take a del x dot and I act from here so you get delta a n okay? and then we need to dot this so we get now both of these things have the same time dependence i n times e to the power i n sigma plus e to the power minus i n sigma to a particular time dependence then we're going to multiply by a minus n that has the lowest time dependence time dependence answer and so we get delta a n which delta a minus n and e to the power i n sigma plus e to the power i n sigma and this thing has been integrated this whole thing has a 2 pi on top right and it has to be integrated on one side this integral is 2 pi 2 over which conflicted that moment yeah square root of that there's a factor of 2 over 2 over 2 over 2 over so what do we get? let's use this integral this is 2 this is 2 pi okay? this is 2 cosine so it's 4 over square cosine which is the house that's 2 times pi okay? so what we get? we get 1 by 4 pi times 2 pi times d a n where d times i times a this is what we got when we chose a in here a in taking the dot and minus a in the n dot n in taking the dot a n in taking the dot and a minus n where we didn't take the dot there's also a term with the other guy with the other guy that term is also 2 pi times 4 pi into delta a minus n where it should delta a n but with a minus i because it should take n by minus n it's exactly the same thing but that's good because this thing is anti-symmetric so when we put this factor that cancels the things so the next net symmetric form measures up to half so it's sum over n a n delta a n which delta a minus n with an i in n so that's an even clearer answer in our expansion we should have been putting it in sorry we're just if you want to stick to standard string theory conventions sorry better better standard string theory convention we expanded with an n so this whole thing comes with a n here and now we're going to see what is this what is this symmetric form it's clear that that up to this factor of n and up to some possible minus that it's the symmetric form between the creation and annihilation operator and if you just add delta p which delta x you get x commutator p is i now we've got an extra factor of i so we get a n commutator if we didn't have the same n commutator a minus n is 1 plus minus 1 with the n the n is down to this so when we invert omega to get the commutation relation it will go up to this so it's clear that we should check that we're either a n a minus n is equal to plus minus the painful thing is to work out plus minus and I'm going to ask you to check that it will be properly divided by i the way to do it in our n is that n goes down you know what we're going to get is i times n by i you see instead of n we can n by i and other than that we get the anyway I'm going to ask you to check there's no n equal to 0 there's no n equal to 0 n runs from minus n to infinity excuse me fine? no no it's the integral over sigma perfectly and you get a n commutator a minus n is equal to this is the quantization system this expansion that we wrote down is the correct quantum mechanical expansion of the operator at sigma where a n is excess of p's operators there will be a commutation relation we just derive it what are the commutation relations a n to the positive ends are amygdational operators and annihilation and creation operators so we've got an infinite number of harmonic oscillators plus we've got the 0 the quantization of 0 is showing the wave function of a freely moving particle okay and the quantization of the harmonic oscillators it's just half the constant okay this is exactly the same structure with one important distinction notice that while the closed string had had two sets of harmonic oscillators we only have one set of harmonic oscillators what's the physical reason there's a boundary condition to tell us that what's going left bounces back and comes left over right over right there's the same degree of freedom you know what's happening to the left you know what's happening what's going to change significantly that's taking one detail what is the difference between this expansion of the 0 rods for the orchestrated closed string there's just one number if you want to examine the commutation conjugate what will this number but if you look up it knows it'll be 1 you can divide it'll be 1 exactly so the 0-mode structure has changed in one time but otherwise it's the same but the oscillators are doubled in size if you have left over in oscillators you'll know what's going to change okay now that we understand this what what no it's making a difference in the mass once we start computing the yeah because momentarily is what becomes dealt by the x so it's simple this is the right thing to preserve the x it's been omitted we're still going to talk about what this means in space time what spectrum we get but before doing that let me tell you about a little trick that people find very useful and that little trick is what's called a doubting trick and the base of doubting trick is a very simple observation so let's consider any function y which is defined from 0 to 2 defined from 0 to 2 pi it's a periodic function y is a periodic function defined from 0 to 2 pi okay, periodic let's define an auxiliary function starting with y which we call x I thought in x we define sigma and range 0 to pi let's define a periodic function from 0 to 2 pi let's see why this x what boundary conditions this x satisfies and it's bound okay that's very simple because del x is simply equal to del y that's sigma minus del y at 2 pi because of this so the function del y evaluate it at 2 pi and del y are periodic functions if you set sigma equal to either 0 or pi this becomes equal to 0 that's sigma equal to 0 or pi that means you just get function del y at 0 minus the function del y at 2 pi if you set sigma equal to 0 similarly for pi then I define with these normal boundary conditions from 0 to pi is the same as the space of periodic functions okay that's not quite true there's a map from the space of some functions from 0 to pi do the functions on the line this will not be a 1 to 1 map basically it's halving there will be different functions here which will not be important to us but that's another side so actually that's a map from the space of periodic functions in 0 to 2 pi do the functions on 0 to pi with the boundary conditions that's very interesting the next thing we are going to do is to ask let's see if there is a y function for our solutions a nice y function defined in 0 to 2 pi that is nice and periodic that maps to our solutions okay and I want to claim that the answer is yes and that this y function is this map so for any x we can find a y but it won't be that's right yeah there will be different y's that give us the same x but we won't bother you know that we are not going to use this in any way except for convenience this is going to be a mathematical trick that will be useful so for the x that we are considering we will find a y okay that's what we want that's what we need yes we just want the most general x that's what we are going to do and we will find a y for it so I am going to illustrate okay now I want to claim that the right y function is y is equal to a prime prime 2 sum n is equal to minus infinity to infinity again excluding 0 infinity to the power i n tau plus x plus 2 alpha right 1 over okay why is this correct this is correct basically because if I replace this by 2 pi minus sigma the 2 pi doesn't do anything is there where this part is not affected by this is not a function scheme okay this whole doubly trick is only useful for those of you you see this is not an economical quantization okay then we are defining x plus y plus oh okay good we probably won't be able to find it no we don't want to find it so we are just saying this should be divided by 2 I get this because the rule says that we should add y of 2 pi minus sigma but the 2 pi doesn't do anything because of the n here so it's just y of minus sigma under this map right I will emphasize that this is a mathematical trick I think this is an unmotivated thing to do but it's a mathematical trick that will prove useful in many contexts only prove useful for the oscillators okay so we will see what works you see the key point is the following the quantization that we achieved of our of our system is exactly the quantization you would get this expansion of y from the closed for the oscillators if you remember for the closed string what we had was a left moving sector and a right moving sector that didn't talk to each other there was a left matching condition we not get what is the point of imposing constraints we are just working out the quantization of this world chain the right movers didn't talk to each other the quantization we have quantization of a n and a minus n with the left movers and a n tilde and a minus n tilde with the right movers and we have exactly this expansion for the left movers to get the standard and then commutation predictions okay so suppose we form the thought of the action as come you know being a closed periodic closed string but we expand the solution for the only left movers we would get exactly the same quantization as for the open string now in successive steps we use this observation again and again it's a mathematical trick but sometimes it is slippery in your mind but it's very useful so we use it again and again okay now let's use this to okay now there is one more so far we have only been discussing these examples but let's talk also about let's talk also about the stress stress okay so remember that there were two important stress tensors in the game that was so there were two important stress tensors in the game that was d plus plus and t minus minus t plus minus is just the, you know so, the tracers, all the stress okay now remember that d plus plus is proportional to 10 plus h is ten plus h and past minus is proportional to So, let us look at this in a little more detail. You see, let us look at what the boundary conditions impose that the x fields have a boundary, what that implies for the boundary conditions on t plus plus and t minus. All the other, I mean the definition of plus and minus and all the same, all the same. It is plus is tau plus sigma minus is tau minus. The condition behind the boundary was del sigma of x is equal to sigma. S sigma is simply its proportion to del plus minus del x. So, in terms of plus minus, the boundary condition is the same as del plus x equal to del plus x and del minus x. We get the condition that t plus plus at 0 is equal to t minus minus at 0. t plus plus at 5 is equal to t minus minus at 5. I would suggest to define a t, again form as form of t, but t plus plus, you know, extended, which is equal to t plus plus of sigma and sigma belongs to 0 and pi and is equal to t minus minus, what is it, 2 pi plus sigma and sigma belongs to pi 2, yeah, pi 2, 2 pi. It is a very similar thing to this x and y. I am going to leave, it is a simple algebraic exercise and I leave it to you to check that with this definition, t tilde plus plus is simply equal to del plus minus del. So, whatever is the constant, exactly, let us not keep track of the constant, you know, what do you have some overall constant with that, okay. We just can get del plus y and del plus pi. Okay, so please work on the constant. Yeah, yes, this is how it works. Okay, that is please me if the word does not teach that. Okay, good. So, you see, that has doubling tree as a counterpart for the stress tensor. What are we supposed to do in canonical polarization? We were supposed to impose both t plus plus equals 0 and t minus minus equals 0 at every point in the string. And in terms of what is the same condition as imposing just t plus plus equals to the point of the string. None of the great cases that this reduces to a problem that we already, we already saw before. Yeah, why is t plus plus 0? Remember, this is the only one. Remember, we start off with an action that has both stress tensor as well as x. Okay, then we fix gauge to set the stress tensor to be this conformal gauge. But there are the g equations that we have to impose. One of them is automatic, stress, t plus plus and t minus minus and to be imposed. And that is how we first do that quantization. We have to do the same thing for your object. Okay, so that requires us to point twice set t plus plus and t minus minus equals to 0 from 0 to pi. But that is the same thing as setting t plus plus till I am equal to 0. It will be quantized as a closed string. We set both t plus plus and t minus minus to 0 on the whole circle. But t plus plus care only about the x plus of that. So, this is exactly the same thing as a quantization in the closed string. Because now we have an expansion in terms of which we have the same commutation relation as one half of the closed string and the same constraint as one half of the closed string apart from the same one. Right. Okay, so, exactly. So, now let's leave that as an additional factor of 2. So, the only complication is the additional factor. Now, let me remind you what we got when we quantized the closed string. Okay. What we got of course, you remember the first thing we did was, and let's repeat that for a moment, we did a live voltage. So, we set two of the transverse fluctuations, times the one space we got 0. But then we still had to impose m 0 equals 0. And the l 0 equals 0 condition gave us, we got 0. Now, in the closed string, the closed string with the alpha prime p squared by 4 plus sum of n is equal to 1 to infinity alpha minus n, alpha n replace all the which x in this is. Okay. So, this is the full vector, p mu squared and the i squared. And then we got this equal to d minus 2 already. I'm not going to derive this again, because we did this in great detail. You have this in your notes. You have notes. Okay. So, and remember that we had various ways of understanding that d had to be equal to 26. Okay. So, let's just see what changes. The only thing that changes, you know, is that p's are expected, as you see. So, this becomes the alpha prime p squared plus this. Because 0 is actually, it was different by that factor. The full quantization of the open string including all some strings gives us one half of the closed string. Okay. But this one changed. Now, if we want to go through the logic, let's examine the spectrum. Okay. Of course, we have twenty thousand ways of seeing the d's equal to twenty-six. But we had one nice way of seeing just like that. Can anyone remind me what that way of, way of also seeing it like that when you did closed strings? Okay. But why does it have gauges in it? Because you can, let's do like that. You have, you have, you have, you have, you have figured out what is phenolic about it. That is the, go back. Okay. That's one way of doing it. Just let the spectrum do something easier. So, we have this datum and at the first level, we have to, we have to ask this particle and for that to be a particular representation of the, of the maybe group and that. Very good. So, what Josephine is saying is that at the first level, there are d minus two particles of that energy. Just to be in the right, if we are maintaining Lorentz's energy, it has to be the representation of the little group of the theory that's d minus two dimension. But there is no representation of s of d minus one. There is d minus two dimension. And there is the other one. Ah. But there is a representation of s of d minus two that is d minus two dimension. In the way. Particles have to transform into representations of the little group if we maintain Lorentz's energy. Now, there are two options. Either the particle is massless or it's massless. It could be datum. That's not going to change. Okay. So, either, if it's massive, it has to transform into representation of d minus one, s of d minus one. But there is no representation of s of d minus one that is d minus two dimension. So, it can't be massive. So, the only way to be consistent is if it's massless. Okay. Which, when it's massless, if this is equal to one. Okay. So, d is 26. So, we replace it. Now, so, we have the same logic for the other string. Actually, in the closed string it's more complicated because we have to take a thing from the left and a thing from the right. So, this is the logic that I expect. That was really the logic for the other string. And it's massless. So, what is our spectrum? So, the lowest level we applied to attack the arm when this is zero. m squared is equal to minus one. p squared is minus n squared. Okay. Next, we are going to get more and more because the mass is at the field. The only way to make consistent theory of mass is to get more. And that's the high level that we got. Can somebody remind me why this problem of having representation of s of d minus one rather than d minus two does not affect the next second in science. Now, it looks like everything is in representation of s of d minus two. Why d minus two, by the way? Because I have to say it's in life on quantization. Run from one to d minus two. Okay. So, why doesn't this argument also tell us that the second in science has to be massless? Which of course would be a contradiction. It could be the first and second. So, it's good if we're in the open string now. This is a review of everything we did for the closed string. So, can we list all the articles that appear at this level? Can it help with us? What do we get? I think standard notation is called alpha. I've called it a. Okay. Tell me. So, at the next level, at level two. Yes. What do we get? So, you get alpha, i, d, alpha, i, alpha, i, alpha, j. What should I do here? I'm going to say n. I want level two. My understanding is that minus is racist. That's okay. But you don't want to raise it so much. One minus one. Shall we have exams in this course? Maybe it's worth it. Let's have an exam in this course. Okay, I'm going to send an exam, a big-term exam, and do it in a month. We should be examining everything we learn for closed string theory because whatever we learn. This you shouldn't remember. Come. This is minus one, minus one. How else do we make up two? That's one. I can go back here. There will be an alpha minus two. Alpha minus two. I. Acting on the back. Now, get something. This thing is a representation of S of D minus two. What representation? Describe another representation. Symmetric tensor. Let's just call it a symmetric. Listen to what representation. Vector. What can you build a representation of S of D minus one by combining these two? It's the symmetric, traceless tensor of S of D minus one. See, we've got a symmetric tensor of S of D minus one. We'll decompose it into a D minus two. It's a symmetric traceless. Because it's traceless, let's say that we effectively get rid of this vector. We get this vector, same as this vector. It's one vector. And then we've got a full symmetric tensor that's not necessarily traceless. So this is the state of content of our representation of S of D minus one. So it's possible that things prove together in representations of S of D minus one. You work at a higher level, you find the same where it all happens. At every level, and then there's formal proof that it works. Okay? So it's only at level one, but there's only one state. So you could combine together with something else to form a representation of S of D minus one. At every level, there's many kinds of things that can combine together. So it's all working nicely. The only thing that's changed is that we don't have to tensor the left movers and the right movers. So at level one, we genuinely have a gain to both of them. What if we have a level of one in the closed string? In these agarotrons, we have agarotrons that's great. What else? Agarotron was not a level one, it was a zero. What else do we have at level one? Well, very good. B mu, the anti-symmetric eye, and the phy, the dilatron. The dilatron there, anti-symmetric me, and the gravator. How do you see that? It's whatever you get by taking a vector and a vector, this will be. So that's symmetric traces, anti-symmetric, and the trace. Okay? Here everything is simpler. You don't have to take anything, anything is whatever you get. Just the aft. Very good. And well, how do we get high and high levels? Okay, so that's it. We've completed our canonical quantization of the open string of light on the edge. And we've understood already a lot of physics. A lot of physics that we've understood is that in the same sense that the closed string was gravatron's interactive with many other things, the open string gives us gauge bosons. We could like to put many other things. You know, I had of course lots of masses up. And there's always this irritating tacky on that. We don't know how to do it, but we'll go away more sophisticated constructions. And there may not be physics. There was a talk in this year's strings conference trying to clear that they understood that at any point in the tacky on compensations. Not sure it's entirely inclusive, but it was interesting. I hate it. I hate it. Let's try it. Okay, it's so great. But not now. So now, you remember, what else could we do for open strings? The next thing we did when we were trying to understand open strings was to understand this quantization of the mosquisticator manner. Okay? Well, I'll give you two things. We first understood the word sheet theory of the string of the mosquisticator manner that I need to conform with what you said. And then we got to the quantization of the string, this integration of a matrix in a mosquisticator. Okay? Let's follow that from that part. Just things that we did in trying to understand things in conform. So first, our first word is now to try to understand the word sheet theory of the open strings, like the mosquisticator. Okay? We understand the equilibrium space, like we did for the closed string. Okay? So the first thing that we did was to remind ourselves that so we analytically continue with the equilibrium space. And then basically a formal transformation from the cylinder to the plane. We're going to do the same thing. We're analytically considering the equilibrium space. And then you can follow the transformation from the strip to whatever we get. So how do we decide what we get? Well, let's remind ourselves of that and what the continuance is. Omega, which is equal to sigma 1 plus i sigma 2. These were the coordinates of the strip. Sigma 2 is i times tau. Okay? And then with the line z is equal to e to the power minus i. e to the power minus i sigma 1 plus sigma 2. Okay? So you remember that what I said was to take where lines of constant time on the cylinder were constant surface on the strip. Okay? The size of the surface was determined by the number of times. T equals infinity was the infinite size circle. T equals minus infinity was the origin. Zero size circle. T equals zero was the universe circle. Okay? Now what do we get? So let's look at how this, all life is correct when sigma ran from 0 to 2 pi. But now sigma ran from 0 to pi. So we get the same thing first before that. Except that we run out of the whole plane by the half of the origin. Lines of constant sigma 2 are circles that are ever bigger smaller. I mean, all of this action is happening on the half of the plane. What's the z? Z is the half of it. X, the one that we had in the beginning of this world in terms of z. Keep your wits about if you get the i's of it. Take the expansion that we had in terms of w. Take, maybe I'll give you a continuation of tau. So tau goes to, how does it go? i sigma minus i. The way that you always analytically continue in an action. Okay? And then replace. So the thing that was e to the power i tau minus plus sigma will go to some analytic combination of tau and sigma. And then under this map, we'll map to some power of z. Is this clear? So that's where we can do this. And then the other one will be in the same path. Okay? So this is just a rewriting of the same thing that we had in terms of these new z variables where it's actually the same rewriting for the closed string, except for the closed string, this, the coefficient of this guy and the coefficient of this guy will be independent. Here, they're the same. Each, where is each? Elephant. This is tau. It's a fancy way of writing 2 tau. You have this nice, this nice expansion of x in terms of these analytic coordinates. But what about y? Okay, we also have this. Yes, yes. Right here. Yeah, this i can do. You see, let me just check. I think there was an i in the previous expansion. Where could it have come from unless there was an i in the previous expansion? So I think the whole previous expansion should have an i. Now the way to check that is to check whether the commutation relations I claimed worked. Let me check that on the fly. Sorry, so let's back up. So we had an expansion which had alpha prime by 2. We started with no i. Alpha prime by 2. Then we had alpha prime by n. And we had e to the power i n sigma plus e to the power i sin sigma times e to the power n. Okay. Now, let me work this out as well as we can. So let's do that really carefully. We get i in... Okay, so the i is in the new one. That's the new one. So we get d alpha n d alpha minus n into i by n. This is what we got. Now, the standard guy is dp where he takes. Okay. So this tells us that alpha minus n commutate to alpha n is equal to i divided by n by n. So we got the wrong. You see, when you have dp by dx you have x commutate to pz. I've got there d alpha minus n by dx. Here you have d alpha minus n by dx. Let's check that. Let's check that. We have dx dot. We integral dx dot by dx. Okay. So we get... If we take alpha n from here we'll get a plus i. So we get d alpha n by d alpha minus n but then you get a minus 1 by n. Without minus n. So we get plus i by n. No, no. But we always write it by... plus i by n into minus 1 by n. Okay. You're right. You're right. So we write it in the i now. Let's take this. That's probably... No, no, no. You know... You know the way that... Yeah, I think the i is from back. That... Okay. We have to be careful, right? Because that's the... Because... So what's the right way to transform x from one side to the other? You don't transform x. You transform del x. Because del x is a good conformal factor. Now del x is a conformal operator of weight 1. Okay. So the transformation property, the transformation rule that takes you from x of w to x of... That has an i in it. That's the key. You see. Yeah. Let me never clear this. The transformation... Basically, that's taken from the derivative. Let me try to clear that with an i for x last time. Well, okay. We probably did it correctly for the closed stream. If you look up your notes there, we probably have the i's correctly there. It's the same story. But I'll look it up. I'll clear that. Sorry, sorry. I'll find this next time. Probably the right way is the solution. Okay. Actually, since... That's the i. I agree with this. Mysterious. No. Let's just forgive this i for the moment. We'll keep it there since we should see how this is properly arranged. Okay. And... Okay. It may have to do with the translation between x of w and x of z. That may be somehow the hard part. It's a little bit... But... Is to try to understand what the animal of this y field was. What the animal of this y field was on the z field. Now, you remember the way we got this y field was that we just took the... You see, the left-moving part of x was all of y between 0 and y. And the right-moving part of x was all of y between y and y. Okay. So, what does that correspond to here? That corresponds to saying that y is a function first thing. y was defined from 0 to 2 pi. So it's defined now the whole complex set. Okay. And it corresponds to saying that y of z is equal to x of z in the apartment. By x of z, I mean the analytic part of x of z. Because that's how it works. You remember? It's just left-moving part in from 0 to pi. But the right-moving part of the apartment. Okay. It's equal to the x of z part in the apartment. So this has to be... This has to go... This will be... Sorry, one more. One. The z part in the apartment. So, y of z, the corresponding... What is the corresponding point? At z star. Right. So you just want to change the... Yeah. At least star. Sorry. Okay. So suppose I want to find out what y of z is at some point in the apartment. Okay. So I want y of z. So I take z and star. That gives me some objects. Okay. And then I evaluate x and z. Okay. That's an analytic function of z bar. That's this object. Okay. This is the same thing as this reflection to pi in the sigma. In the z bar. Yeah. So can you say that it is... It's upset because people drop that term. Drop that one by z one. Yeah. So it's this term above and this term below. But it's this term not. You know. So suppose we're interested in what... what our field is here. We take the anti-amplity power of what it was there. See, what do we do on the strip? If we want the... What the y field was here, it was left here. But what the y field was here was the right moving part at the corresponding point here to pi by the sigma right. So we do the same thing here. We say if we're interested in what the y field is here, it's the anti-amplity. Okay. If we're interested in what the y field is here. Okay. It's the anti-amplity part. But at the corresponding reflected point. Where the... Where the... Where the imaginary part is reflected. Okay. Now, the fact that we have this nice periodic function in Y when we did that before. Okay. Translates to the fact that this is just some nice well-behaved analytical function on this whole plane. It's basically the fact, you know, you would have to check that derivatives here match. But they do it from the boundary equations. It's very basically the... You know, the boundary equations in the boundary world, remember del plus is equal to del minus. Okay. Which is del z is equal to del z bar at the boundary. Right. Because del z of x is equal to del z power of x. Which tells you that this way of doing it gives you matching derivatives at the boundary. Zero more will give you a branch. Yeah. Zero more is going to be a tree. Right. So, uh... Yeah. Yes. It's a good device. So, for real, this will be useful only for the field del y. Then they solve it. Then they solve it. As usual, as we did with the boundary theory, the field x of the field y is not a good field. But the field del x exists. So, we never use these tricks for the zero. The zero more doesn't... The strict doesn't do anything nice with it. But it does for all the oscillators. A clear way to separate the oscillators from each other. Right. So, this nice W trick that we describe at the level of this field of the cylinder and the strip. Because the U of y is the same thing. Defined from zero to two pi means defined from no pi. Just translating that into this language. It's the same thing. But it's very nice. All the component field theory tricks that we played, analyzing the closed string go through for the analysis of them of the open string when we use this double field y. Particularly, for instance. Yeah. So, one of the various things that we should do. I'm going to suggest three exercises. The first exercise is to compute the combination relation between alpha n and alpha a using the component. What do I mean by that? Write down in terms of field y write down a contour integral expression for alpha a. Okay? And then use the OP, the usual OP between x and x. Okay? To compute the combination relations in the way that we did it 20,000 times in the previous class. Okay? I want each of you to do this exercise on x and y. Just to remind ourselves of all these techniques. It's the same thing as for the open string. But please do it. You'll also have to check whether this eye is connected. It's directly in the set space. Okay? So, please do this. Okay? Secondly, compute the combination relations between p and p. I mean, l and n. The usual expansion of t of z is equal to sum over lm by z to pi plus 2. Okay? And once again, compute the combination relations between lm and ln using this, using the, okay, of course, the t, p, p remains unchanged. Because the same expression stays in terms of p. In terms of p. Using that, compute the combination between d between lm and ln. What I'm asking you to do is to just want to be everything you need for the closed string. It's the same exercise. Just identify. Ask me to do it anyway. Just to remind ourselves of that. So please for the next class, just do these. Okay? And lastly, lastly, how does the state operator work? Okay? So, okay. So, how do we work with closed strings? So first, let's work in the w-plane, physical w-plane before we can do that. How do we work with closed strings? Close strings behind the whole circle. The whole circle boundary conditions in that circle. For the path integral in that circle, define the state. Now we can use scaling to shrink this thing and just to make it as, you know, as small as we want it, that defines an operator insertion in traffic there. Okay? How do we work? How does it work here? How does it work? It works once again with a state here. We shrink it, and we get as near to this point as we want. What we get? We get an operator insertion at the boundary of our machine. Is this clear? Clear? Let me just remind you about the closed string discussion. In that situation, state was defined by the boundary condition of path integral here. But by scaling, we could shrink that to as small to as small a circle as we wanted. And then, in fact, there was a local insertion somewhere at the word sheet. It happened to be the origin of the word sheet to not a special point. It was like any other point. Once you learn to define an insertion at any point, you know, you define it everywhere. It's a local point. So you have to find a local operator somewhere at the bulk of a word sheet. You hope that string, it works as what? A state. Where does a state define it? A state is defined on a line that comes in that indomitable way. Some say, this is where we give our boundary conditions to the path integral to find a state. Scaling brings it down here. So we get some local insertion there. Now, this point is not a special insertion. It can move anywhere along the boundary. But clearly, it comes with the bulk. It's a boundary point. It's different from the bulk. The state operator acts for open strings. It establishes an isomorphism between operators that you can write down on the boundary of the word sheet and states on the string. Is this clear? Now, how are operators on the boundary of the word sheet different from operators in the bulk? You see, what were the operators in this x-scaling? What were the operators in the bulk? Effectively what you had was any number of holomorphic derivatives on x or any number of anti-holomorphic derivatives on x. Times each of the five. These were the same operators. Now, on the boundary, you have fewer. Why? Yes, one of the boundary conditions doesn't have x. Now, what does that mean? Is that it? There's an x and there's an z by x. So, suppose you try to insert, there's z bar to the power n of x. The same thing is inserted in z. So, instead of adding two sets of operators, all the various derivatives will put on z. Z derivatives will be good. And products, they're not. And what is the most general operator we can have at there's z to the power n1 of x times a n1 times all power to variable power. Times there's n power of the state operator map. State operator map works. There's z to the power n1 of x back to a n1. The holomorphic a n1, a minus n1 acting on the back. This operator map back to a minus n1 acting a n1 times. What's holomorphic derivatives of x? Anti-holomorphic derivatives of x. The only thing you could have, you have the false base of both. Now, there's now the state operator map worked. For the closure. Now, however, you don't have two sets of derivatives of x to the power n1. You only have holomorphic derivatives. Or ideal derivatives. And that maps to the fact that on the stroke we have only one set of derivatives. Let's call them holomorphic derivatives. My third exercise for you is to work this out in detail. So, what I want you to do is to repeat the arguments that we had to deal with. Same thing. Remember the way we worked was to was to say suppose we had the insertion of del z to the power n1 in x. And then we worked out what state that was followed. How did we work that? We worked it out by surrounding it with the A operator. And finally what we did and we acted by A. And that would be completely different. So, work this out again for the open stream. You can work it out either the y-language or the x-language, but somehow work it out and find the size of all of them once again. I just sped out the answer. You know what you're going to get. But worked out the open satisfaction and if there's something confusing coming up. So, all of the formalities I've only by the way reviewed the parts of the formalities that we discussed in the hand. If you remember we went through this in five or six lectures. It was a long one. But I only reviewed the parts in different ways. Anything that for instance involved deriving an equation of motion and what that meant inside the path integral of no third star. So, all of the local properties of the spring bulging part of them or the only thing that we discussed was how to have them out. We changed the things. So, that ends my brief review of effectively chapters one and two of which is the path integral of the open stream. Okay, let's quickly quickly quickly go on to chapter whatever this is. Three. The path integral of the open stream. That's three of them. So, that's the part about honest quantization of the stream. That was chapter four. Chapter three is called chapter four is transfect. Maybe we have different issues. No, no. I want to understand what happens to BC coasts when we have that. Okay, so again, we're not going to review, we're not going to do that old quantization procedure. Okay, we've done it. Just remind ourselves about the elements that are necessary to understand what we know. So, what was the end that, you know, what do we have? We have gl capita, some world sheet. And we had x. And then we went to some conformity. Alpha, beta is equal to some fiducial vector. For instance, you see that with some g hat, alpha, beta, and into the body. We used the thermos as a means to do that. And then we worked out what the particle of determinant for this gauge is. Now, if you remember how we worked out, we said, a deferomorphism change. How does a deferomorphism change life? How does a deferomorphism change life? So, what is the particle of determinant? So, you're supposed to look at the variation of your gauge condition, and then compute the determinant of that. Our gauge condition was that g was equal to this. So, we have to look at the variation of this under our symmetry transformation. So, the variation of that was del alpha vibrita, class del vibrita, this was under deferomorphism. So, we had a conformal transformation that made this traceness. The other part was traceness. So, divided by q minus del v g hat of vibrita by 2. So, this was deferomorphism of this particular conformal transformation. We'll do it in the second term. But let's take the trace. I do, right? Because this trace is 2. I would have been by d, generally. These 2. Okay? So, we have this change. And then the particle of determinant was generated by vibrita vibrita. And there is the reduction of vibrita. So, we had a vibrita vibrita. And del alpha vibrita del vibrita 2. Of course, though it turned out to be most convenient to write these b-winder lower frequencies and c-winder upper frequencies. When we did that, it was in a local analytic fact, this action always reduced to b del bar c plus b bar s bar. And this sees in these bits. The c was simply the infinitesimally deferomorphism parameter converted to an anti-interactive material. And the