 the morning session, so we are going to have the lecture by Matias in a few seconds. Okay, so, okay, so very good. So last time, remember we looked at ADS-3, cross S3, cross T4. And the way we describe this was in terms of this SO2R, N equals to one, vessel mean a written model at level one. And then the SU2 supersymmetric vessel mean a written model at level one plus T4 on the world sheet. And then we know that we know that the SO2R subalgebra here is to be identified with the Möbius symmetry of the dual CFT. And that's to determine the space-time spectrum organized by conformal dimension. And what we saw was that the space-time conformal dimension coming from the W-cycle twisted sector of the, sorry, coming from the W-spectrally float sector of the world sheet theory. So in this factor, we have this spectral flow. So what we saw was the answer was equal to W squared minus one over 4W plus N over W. So I'm assuming here that H0 rest is equal to zero and we'll come to that later on. And so that was sort of the upshot of the last lecture. I think you were meant to be excited about that, but maybe you are not, but the idea is that this looks exactly like the spectrum of the symmetric orbit fold. So I want to start today's lecture by giving a brief interlude for those of you who are not familiar with the symmetric orbit fold, what the symmetric orbit fold of T4 will look like. So to make life simple, I'm going to discuss the symmetric orbit fold of S1 of one free boson. So let's look at the symmetric orbit fold of one free boson. And then it will be clear how that will generalize to the case of T4. I mean I don't want to clutter notation with all sorts of stuff that's irrelevant in order to explain the gist of it so I'll just focus on the key element. So what do you mean when people say the symmetric orbit fold of S1. Well, so what you mean is you have X n boson fields. So this makes a funny noise right. But let's hope it'll be better. Not sure. Anyway, so there are n boson fields which I'll denote by DXI. And each of them has conformal dimension h equals to one and I runs here from one to N. So the coordinates on the N copies of S1. And then you have what you mean by this is you take the theory that consists of N copies of a single free boson ID S1 theory, and you divide it by the symmetric group. You take an orbit fold with respect to the symmetric group where the symmetric group acts on the end bosons in the most obvious manner it just interchanges the end bosons. I mean, there's no natural other action that you can think about. Okay, so now this is an orbit fold just like any other orbit fold. So how do we describe an orbit fold, but an orbit fold theory has an untwisted sector. And the untwisted sector is sort of naively what you would think you get. So for example if you think about it in terms of a fox space, you would take the fox space of this theory. What does the fox space of this theory look like? Well, if the modes of this I denote by alpha, the fox space of this theory will look like something like this. Right, you take all the modes of the different bosons and apply it to the ground state and then you write on all the possible combinations you can write down. And then you take the fox space of the end fold, the free boson theory. And now the idea is that in the untwisted sector what you do is you project onto those combinations of states that are invariant under the SN group. So what you do, the untwisted sector consists of all of that subject to a singlet condition. It's invariant under the SN group action, where the SN group action acts on the upper indices only. It exchanges the copies of the fields. So that's the obvious untwisted sector like you always describe it in any orbit fold theory. Now as you know, the untwisted sector somehow doesn't describe all of the orbit fold theory, because you're thinking of closed strings living on this quotient manifold. In this quotient there are closed strings that are originally closed subject to this constraint, but then there are strings that didn't used to be closed, when thinking of it in terms of this theory, but only become closed, once you identify the end points under the action of the symmetric group under the action of the orbit fold group which is a symmetric group in this case. Now, generally you know that the twisted sectors are labeled by conjugacy classes of the orbit fold group. Now people are probably most familiar with the billion orbit folds. If the group is a billion, then the conjugacy classes are in one to one correspondence with the group elements, and the correct generalization to a non a billion orbit fold group is that you label the twisted conjugacy classes conjugacy classes consist of all those group elements that upon multiplication by h and h to the minus one from left and right, get mapped into one another. So the untwisted sectors are in one to one correspondence with the conjugacy classes of SN. Now what are the conjugacy classes of SN SN is a permutation group, and you can write any permutation as a product of cycles. I mean, for example, you can write any permutation. For example, as 12473512345 and then six, for example, that would be an element of a seven and any permutation in a seven you can write in terms of its cycle you start with the first element. It goes wherever it goes, then this element goes wherever it goes this element goes wherever it goes at some stage you come back, and then you start with the next one that's left over and so on. In terms of permutation you can write in terms of products of cycles. And if you think about it, under the conjugation, the structure of the cycles doesn't change the only thing that changes. If you conjugate with this with an arbitrary element in a seven are the specific numbers that appear here because if you think about it, if you conjugate it you just read this rename 127 up to buy something else, then you do the original permutation and then you re unnamed so therefore this is, it's clear that the conjugacy class that's different permutations with the same cycle shape, sit in the same permit conjugacy class because they're clearly conjugate to one another. And you can prove that that's basically all the all the identification you get from the conjugacy classes. The conjugacy classes of SN, I want to run permit one to one correspondence with cycle shapes, and therefore in one to one correspondence with permutations with with partitions of N of writing and in terms of a sum of positive integers, and the positive integers are just the length of the cycles in which you decompose any given permutation. When, when you have standard or defaults for a billion groups, the, the twisted sectors are labelled by the element of the groups which of course are the same of Newcastle classes. Can you clarify why in the only a brilliant case, what levels the twisted sector the Newcastle classes are not just the element of the group. But suppose you start with an individual group element in the non abelian or default, and then you have to make it invariant under the or default action right so in a billion or default, but you do the twisted sectors are labelled by group element, but then within each twisted sector you have to make it invariant under the whole or before group. If you start in a non abelian case if you start with a twisted sector associated to a specific group element, and you try to make it invariant, but you're going to do you have to sum over all the conjugacy, all the elements in the conjugacy class associated to the specific group element, as you make it invariant that the thing that stabilizes the specific group element is just a centralizer. And that produces for you the copies in the conjugacy class so each the or default action links together the different elements in the conjugacy class. That's the reason I mean, this is not something I invent I mean this is standard folklore about or defaults but that's why for non abelian or defaults it's labelled by conjugacy classes, rather than even individual group elements. So that's this general structure of a of a of a twisted sector it's labelled by a partition or if you wish a cycle shape of a permutation of order and. Now, if you think about what we're trying to do we're trying to describe the CFT dual of string theory on ads tree but we're looking at perturbative string theory, and we're looking at states that are made up from one string. And if you think about it what you expect say in the context of n equals to four super mills is you expect to find the single trace operators, the single particle stage you're not expecting to find the multi particle traces from a single string analysis, the multi trace states will come from multi string states. And likewise here, the idea is that the single particle states, the things that are described by a single string when I look at the interpretive string theory on the single string should correspond to the states that come from a single cycle twisted sector. And the single particle states are those that have a con that whose conjugacy classes of this time, where I use the usual convention that you don't actually write cycle of length one cycles of length one always make up the number everybody you don't write out sits in a cycle of length one. And the idea is that a single particle states are not all the twisted sector. There are those twisted sector that has corresponded to a single cycle of length arbitrary length, and nothing happening in the rest of the world. So this will be a two particle state, because this consists of two non trivial cycle, and correspondingly the multi particle states come from the sectors where you multiply the cycles together. That's the closest analog to multi trace versus single trace in n equals to four. And that's, and if you think about it, if you go to string field theory, then this will exactly do this multi particle for you and it will produce all the other sectors of the symmetric or default. So that is the correct part of the spectrum that we are going to see from a interpretive bullshit string theory analysis. So we're only going to look at a single cycle twisted sector. I mean if you're interested in the theory of symmetric or defaults obviously you have to look at all of them, but in the context of relating to string theory a single interpretive string, we'll just be focusing on the single particles part of the spectrum, which come from the single cycle twist the sectors alone. Okay, so what we have to understand is, what does the particle spectrum what does the spectrum, what's the Fox space in the w cycle twist the sector. So let's think about what this w cycle twisted sector looks like. So suppose, Sigma is the ground state of this w cycle twisted sector. Then what does it mean to be in this w cycle twisted sector. Well remember the intuition about twisted sector twisted sector describes those solutions that are not closed in the original theory, but they're only closed after the permutation by the permutation. So what this means if I start with my field say DXI here, and I go around. Once I come around, this has changed to DXI plus one. If I is equal to one up to w minus one right so this is what this permutation does for you. If I come around with this field, it ends up with the image of this permutation applied to this field. That's what it means to be in the twisted sector. I mean it's a solution that it's not closed in the original description, but it's closed up to a permutation. And this permutation says exactly up to what it is closed. So it's up to up. It's closed up to the action of this permutation on the labels of the individual fields. The closed zone will become the second, the second will become the third, the third will become the fourth, the w minus ones will become the Ws, and the Ws will become the first. That is the, that is the structure of the w cycle twisted sector that's what it means to be in the twisted sector, namely you don't close, except you only close up to a group. Okay, so how can we describe the excitations in this w cycle twisted sector. It's simple. Yeah, the idea is, so you see, you could try to find modes for the individual fields, but you won't have any luck, because you see suppose you try to write modes down for DX one, then it's not going to work because somehow the modes of DX one will mix with the modes of DX tool will mix with the mode of DX three. So I can't just write down the mode expansion for each of the fields, as I did originally in the untwisted sector. So I have to find a smarter set of fields that have a good mode expansion. And if you think about it for a moment it's clear what this smarter set of fields is, this will be linear combinations of the first w many fields. So what we're going to look at are the linear combinations of the form. We're going to sum from one to up to w e to the two pi i j times L over W, the x j. Therefore, so I choose L to be zero up to W minus one. And for each such L, I look at this specific linear combination of fields. I'm obviously free to look at any linear combination of fields. This is all linear this vector space. I just decide instead of working with DX one DX two DX three, where I mean, with the curly bracket I now look at DX straight bracket one, which is a linear combination of all the curly bracket ones. Now if you think about it what happens with DX L as I take it around. So if I take DX L around the point Sigma what happens, well it will become the sum. This is equal to one up to w e to the two pi i j L over W times DX j plus one with the understanding that I periodically identify by w, which is fine you see this is periodic in w so whether this is w, this is zero the difference shifts by w drop out because they have the same face. Right, that's what happens. If I look at this linear combination I take it around this field sitting in the w circuit with the sector, it comes rearranged because every, every of the field has moved one up one in the list. So you see this is easy to rewrite. So this I write as e to the minus two pi i j over w, sorry, e to the minus two pi i L over w times the sum from j is equal to one to w e to the two pi i j plus one L over w DX j plus one. Multiply by e to the two pi L over w and e to the minus two pi L over w so I haven't done anything, but now you recognize that this is just the original field itself. Right, I just, I've just relabeled my summation running now from one step further but doesn't matter because it's periodic in w. So what you see is that these linear combinations of fields. They have a simple monodrome as I take them around this point and the module is simply the e to the minus two pi L over w. So they pick up a face as I take them around the twisted sector field. Now what does this mean, if I write them in terms of modes. Well, remember, so this is a spin one field. I should still be able to write as a sum of R, the alpha. Straight bracket L R z to the minus R minus one right I mean a spin one field has a mode expansion where you have a minus one here has been H field has a mode expansion where you have a minus H here. So this is, I haven't changed the conformal dimension. This is just a linear sum, a linear combination of spin one fields. So it still has been one field. So what happens when I take that around. Well, when I take that around I just replace z by e to the two pi is that because I just go once around the origin. So this will go to the sum of our alpha L R e to the minus two pi I R times z to the minus minus one. So what happens. So here I'm taking that around. And what this means I multiply is that by e to the two pi is that which is what happens if you go around the origin once. But now you see this has to be equal to what we've just seen e to the minus two pi I L over W times the sum of our alpha R L. So this has to be equal to the minus R minus one, because it has to be equal to the field up to this phase. But now if you compare the two sides, you see what you learn is that this phase has to be equal to that phase, you see the rest of the sum is exactly the same. So what we have to do is that this linear combination, if I think about it in terms of modes has modes that are not integer mode it anymore. But their numbers are such that e to the minus two pi R is equal to e to the minus two pi L over W. So what this tells you is that alpha L R has modes where R runs in L over W plus integer. So these are exactly the modes that I mean if I was deliberate vague I just introduced this label I didn't call it didn't call it n for a reason because, as you see in general it's not an integer it's whatever is required to have the right monotron me and what's required to have the right monoprene monotron me is that R has to be equal to L over W mod integer. So what's happened in the w cycle twisted sector. Originally we had w many fields that sort of were part of the game. You can always ignore the, the rest of the fields they're just going along for the right they're not doing anything we are just concentrated on the first fields. So originally we had the first w fields, all of them were integer mode it. And in the L cycle twisted sector, sorry in the w cycle twisted sector, what we end up are new fields, where L runs from zero up to W minus one, but they look like the original field like one original field except that they have a fractional mode number. And if you and you see in some sense it's now redundant to introduce this label out here, because you see whatever. I mean another way of saying this is that you now just have modes alpha R, where alpha R is an element in Z over W, right, any value of the form Z over W will appear. I'll just tell you which L you're talking about you're talking about the L for which the arm or do you've written down is of the form L over W plus it so what has happened is, instead of having w many integer modes, you have one mode, which is now w fractional. I mean in some sense you still have the same number of modes. Originally you had w many things all integer mode it. Now you have one thing but it's much finer mode it, it's noted in units of one over W because that's exactly what this gives you and I'm accounting for all the fields and just have really reorganized them so that they have a good mode Yes. Sorry, maybe I missed it. Could you repeat why L should be an integer at least I mean from if I start from there from the definition of the XL. I mean, I would use any L in. Well, I mean you see you have w fields to start with and you should write down that w linear combinations. So what I propose this they write out these w linear combinations where I take out to be an integer. Right. And what I'm claiming is and I haven't shown this is that the set of the set the lead these linear combinations are linearly independent of the original. Okay, so I can either write it in terms of the original w field or in terms of the small else then I would introduce linear relations among them. Okay, so it's just convenient and then one matches the bigger is a freedom. So this is just a convenient rebinding I may have a w dimensional space, and I just use a different basis but obviously I still have to use w many faces like this because I mean think more can't be linearly independent. I just choose them smartly and anything like that. And then I see they have simple mode expansions and the mode expansion of that form. And now if you ask yourself what does the spectrum look like by the spectrum where normally you would say L zero gets lifted up by integers. But now, in the spectral if in the w cycle twisted sector, it'll get lifted up by integers divided by w. And you see bingo this is the first explanation of this factor. You see this is exactly what this chapter is telling you see you see and are the degrees of freedom of say that he for theory. So there's nothing twisted about this this looks like a single copy of a T for theory. But what we see is that with respect to the space time conformal dimension it behaves this though, the mode numbers are w fractionally moded, and thereby, they behave exactly like the modes that you would see in a w cycle twisted sector of the symmetric orbit. So this is the, the first big hint that we are doing the right thing. And is and that I'm not going to explain it I could explain it in detail but this will probably take too much of my time is, you could ask what is the ground state conformal dimension of the so Sigma is the ground state in this w cycle twisted sector. And you could ask what is it conformal dimension what is the conformal dimension of Sigma. What you find is that. And I mean, on a certain level you are familiar with this because you see when you go from the never short sector to the remote sector. That you're doing basically the same thing right instead of having half integer mode it feels you have no integer mode it feels. You can ask what's the, and you know there's a Kazimier energy involved in shifting the energy up from an average board sector to a moment sector. And likewise you can calculate here, what's the energy that's what's the energy shift that the ground state energy that you pick up from from being in the w cycle twisted sector and what you find is for each the shift is w squared minus one over 24 w. I mean a different ways in which you can derive this formula, probably the easiest way is to use a modular techniques, you write down the character with the insertion of the, of the obi fold element of the w cycle twist and then you use the s modular dimension, and then you read off the conformal dimension in the sort of you in the twisted sector. And when you do this what you find is that if you do this for a single boson, you pick up a factor that the ground state energy is w squared minus one over 24 w. Now, that's what you get for a single boson so for the example I've discussed here, but if you think about doing this for T for you have four bosons. So, you'll get four times as much. And in fact fermions, but there's a small subtlety here but let's, let's say w is odd then the subtlety disappears but I don't want to explain the details of it. Then the fermions behave exactly like half a boson, then this just scales with the central charge so in general you would expect H of sigma to be the central charge times w squared minus one over 24 w. But the central charge of T for is exactly six. So this will give you w squared minus one over four w. And that's exactly the first time in this formula which we found here so this is the Kazimier energy that you would expect for the symmetric orbit fold of T for because in the w cycle twisted sector the grounds that is shifted up by w squared minus one over 24 times the that reproduces exactly this ground state energy, and then on top of that you have all these fractionally modes, which come from the modes as I explained to you here and that's, that's exactly the structure as to see over there. So that's the reason why when you see this formula you say, ah, this looks exactly like the symmetric orbit fold of T for there is a question here. So you also think of this Kazimier stuff by going to covering space. Yeah, you can. There's various ways in which you can derive this. I mean you can derive it by the modular transformation can go to the covering space you can write down the stress energy tensor in that. So I mean the simplest way of deriving this is to write down the formula for the stress energy tensor in the sector. So this will be like this. And then you can simply work out L minus one on the ground state this will not be zero anymore, and then in order to extract the eigenvalue you can just apply L one. I mean the reason you look at L minus one and L one is then you don't have any normal ordering ambiguities then it's unambiguous. And then if you work this out. You don't want to make a mess of it, but then you, you can work it out and you arrive at this result. And I'm not attempting because otherwise. Another thing. So maybe. So in this studio, like a lot of Veltex operators to know, of course. So what are they on the string in the S3 times S3 and so on. Well, so you see them. I mean, at this moment we are identifying states right so we identified so we know for each. Yeah, so the state here is characterized by which fractional modes you excite. And then according to this dictionary that just tells you which modes where you sort of remove the fractionality you multiply the mode number by W. Well originally excited on the T for the construction of vertex operators is an interesting one and I'll come to that probably tomorrow. I want to explain how the correlation functions match, but that requires a certain number of preparations and I'll try to explain. Okay, then ask tomorrow. Please ask again if I don't explain it satisfactorily. I think there's another question over there. I mean, I understood I misunderstood the starting point but I understood that W, if you want to take the SN, somebody called before the W can go from one to N. Correct. But yeah, so here in in the strings that W can take any integer value. Right, but remember that G string is proportional to one over N, and in perturbative string theory that means we take these string to zero. So what I'm doing here is always in the large and limit. So we are always in the larger and limit so we've taken N to infinity. But as I'll explain to you the one over N corrections which will correspond to the G string corrections will be able to reproduce from the world sheet the world sheet does not reproduce the finite and answer. The finite and answer would correspond to sort of non-perturbative effects from the point of view of the string theory, what you're going to do is perturbative string theory so we are always taking the one over N goes to infinity expansion first, and then we systematically keep track of one over N corrections. But there are no one over N corrections to the spectrum, and therefore this formula is exact as it stands. So what should I think about this, I mean, if you think N to be finite, it seems that you are truncating the spectral flow to not to take all the possible integer values, but there should be some mechanism that I mean, this description I don't know how it'll work at finite and I think that's, I mean this is a finite and isn't something it's finite string coupling right then, then you're, I mean when you want to do perturbative string theory you first start off by the string equal to zero and then you think of switching that's the spirit here. The spirit here is you start with the theory where N is infinity, and then you systematically include one over N corrections in all the calculations and they should come for higher genus world sheets. How the non-perturbative effects arise from the point of view of string theory and finite and effects would be non-perturbative from the string world sheet perspective is an interesting question but one where we don't have an answer yet. But look, I mean, this is already pretty good if you can account for all the one over N corrections. It's like all the non-planar corrections of N equals to false supang mills. You're also not, I mean the finite N supang mills is also totally out of reach from the point of view of ADS CFT. Sorry, another question. Is it completely under control? This may require revolving in the larger limit. Absolutely. So, and maybe I didn't explain this, but this has a very nice large end limit. So there is a sense in which you can just take the, and the idea is that you see all the copies that are not involved are just in the vacuum. So therefore, things stabilize. So if you fix W and you take N to infinity, then everything stabilizes because you just add more of actual in the additional copies where you don't do anything. So therefore, the result doesn't really, once N is bigger than W, nothing ever will change again. So therefore, this has a very stable, you can write on the interpretive spectrum and it has a, it stabilizes order for order. I mean, there's no problem with taking the large end limit. Any other questions? Okay, so, yeah, so but, so, so, so this looks right, but obviously I told you I cheated in a number of ways. I said that J was equal to a half. I didn't really explain to you why J had to be equal to a half plus I times zero, but I didn't explain this. And then the other question you would ask is, why are you just looking at the T for degrees of freedom you have all this other gunk in your world sheet and you have this s to two at negative level what are you going to do about that. That's a legitimate question. And within the context of the NSR description at this moment we don't have a satisfactory explanation I can give you a heuristic explanation of it but in view of time I'll probably skip that. It's, it's not totally satisfactory. It's just this answer smells to be the right answer but the NSR description at level one is a little bit problematic. It has issues, the Americans would say. And so, so we have to, so we have to do a little bit better if you really want to find a satisfactory answer. And the idea is that we use the fact that there's an alternative description for strings on ADS three and that's the so called hybrid formalism. So I mean this wasn't invented to solve our problem. In fact this was invented to solve an entirely different problem this was invented to solve a way of giving access to world sheets that also have a month on one background. That's what it was really invented for, but you can just take this theory and evaluated for backgrounds with pure never Schwarz never Schwarz flux. And in that case, the hybrid formalism involves the world sheet theory that has a resume new written model based on PSU one at level K, plus topologically tested T for theory, plus suitable ghosts. And what has been checked is that this theory if you take K to be any number bigger than one, this theory seems from what one is able to check to agree exactly with the NSR description of Maldesino or glory. I mean this is not the proof but this is has been checked to the extent that one can check these things. And there's very good evidence that this is a perfectly equivalent description to the NSR description for any value of K greater equal than two. And then this theory as I'm about to explain to you also makes perfect sense that K to one. So our proposal will be that in order to make sense of the cables to one theory, we should rather work with this description, rather than the NSR description where I have all these little issues I have to be to be fighting against. See from your eyes that you're a little bit scared so let me explain to you what he is you one comma one slash two is, and why it's natural that it appears, and then let me explain to you how you analyze this world sheet theory and you can analyze it by very similar methods than what I've done. And the reason I spend so much time on the NSR description is that here we can be extremely explicit and you can see everything how it works. But the answer will effectively work more or less the same way for this hybrid description. Okay, so what is PSU, a one comma one slash two, and why should it appear. Okay, so PSU one comma one slash two. If you think about it, it looks like a four by four matrix. It hasn't. So, so that the convention has you have an SU one comma one and an SU two. You have one subgroup that is SU two, and then you have one subgroup that is SU one comma one, and as many of you probably know SU one comma one is a different name for SL2R. These are the same real forms. This is just the identically same the algebra, not just the complexification of it as it really is about every the same so it's a different name. Whenever you see SU one comma one you can just replace it by SL2R is really exactly the same. So this is a superly algebra. So it has a person except algebra that's SU one comma one plus SU two, and then it has fermions sitting here. So you have a fermions, and the fermions sit in the two comma two with respect to the SU two and the SL two because they sit inside a two by two block on which the SU two and the SL two act from either side as a tool. And because we have guys here and guys here we have two of those. So P is your one comma one slash two is a lead algebra that has eight fermions that namely two pairs of fermions that sit in bias by spinner representations, and then it has six bosons the bosons are just SL2R. And SL2R is a three dimensionally algebra SU two is a three dimensionally algebra. So this will have six bosons. So it's a superly algebra. So what a superly algebra mean, it's basically a lead algebra, except the instead of commutators of the fermionic generators you write anti commutators, but that's it. I mean for us for physicists that's not a big deal and we are used to anti commutators or commutators so it looks like a lead algebra except every now and then you have to write anti commutator, but then it satisfies the Jacobi identity and blah blah blah all the rest of it so it's not, not much more complicated than a normally algebra it just has some anti commuting generators rather than commuting that's not that's not such a big deal. Why does PSU1,1-2 appear in this context. Well, that has a very simple reason, because you did the dual CFT as n equals to four super conformal is a super conformal field theory in two dimensions. Now, maybe you also don't exactly know what that means so what this means is it has a very thorough algebra because it's conformal. It has four supercharges plus four supercharges so these are fields of the form g plus minus and g prime plus minus. So these, this is a field of conformal dimension h equals to two. These are four fields of conformal dimension three halves. And then it has an r symmetry and the r symmetry is SU2. And that is our fields of conformal dimension one. That's what the n equals to four super conformal algebra looks like I mean I can write down all the commutation relation that easily fill one of these blackboards, and you'll copy it down probably make a typo and, but you can also just look it up on the web. I mean it's, it's a standard algebra. That's what you expect that you will conform a field theory to be like because that's what the symmetric of T4 is because that's what the T4 theory has the T4 theory has this n equals to four super conformal symmetry. Now, from the point of view. So remember, in the case of the Bosonic theory, we had the SL to our symmetry on the bulk, and that became the mobius symmetry in the dual as conform a field theory right remember the transformations of our ADS space became L0 L plus minus one on the boundary. Now what characterizes L0 and L plus minus one inside the virus or algebra, but these are the global conformal transformation these are the transformations the generators that kill the vacuum, and it killed the out vacuum. These are the generators that are really well defined on the sphere. There's an analog of SL to our for these other generators, and the answer is always simple you take all the modes with mode number is strictly less than age. Less than the conformal dimension because these are the modes that will kill the in vacuum, and that will kill the out vacuum. So how many modes do we get. So if we look at the global modes of this, what are the global modes of this. So we're going to get L0 and L plus minus one. And we know already that this gives us a copy of SL to our. What are the global modes of SU tool. Well, this will be the zero modes, the minus one mode and the one mode, don't annihilate the in and out vacuum but the zero modes tool. So from these guys, you get just the j a zero modes are actually I'm going to call them k a zero in the future, and they'll generate a lead algebra of SU tool. So these guys, they're going to get the generators g plus minus, but since conformal dimension is three over two, the allowed modes will be plus or minus a half. So you will have these modes with generators plus or minus a half, and you will have the generators g prime plus minus with plus or minus a half. These are the global super super conformal transformation the super current transformations. So if you think about it, you see, you have a three, you have an SL to our, that's this SL to our, you have an SU tool. That's this SU tool. And these guys, the upper index tells you that they transform as a doublet of SU tool, and the lower index tells you that they transform as a doublet of SL to our. So these guys sit in the two comma two. And then we have another copy that sits in the two comma two. In fact, this precisely account for these families. So when you take the n equals to four super conformal algebra, and you restrict to the modes that annihilate the in vacuum and the out vacuum, what you discover is that the lead algebra they generate is PSU one comma one slash two. That makes a lot of sense from the worksheet perspective because you see the zero modes of PSU one comma one slash two. These will be the global transformations from the point of view of the dual theory, and they should precisely reproduce the global part of the n equals to So this is the supersymmetric generalization of what we saw earlier that the SL to our vessel mean a written model gives rise to the mobius symmetry in the space time CFT. And this is the souped up and equal to four version. Now you don't just have SL to you also have SU tool for the asymmetry, and you have the appropriate supercharges. So the fact that PSU appears is just the way of this theory to making sure that space time supersymmetry is manifest. That's exactly what this factor does for you the generators of PSU one comma one slash two are exactly the n equals to far transformations of the dual CFT. That's why I sometimes say this is like the green Schwartz formulation in the green Schwartz formulation space time supersymmetry is manifest. And here it is manifest by virtue of the factor that the zero modes of this factor, precisely, the global n equals to fall transformations of the dual CFT. Okay, so, so this is why this PSU factor appears. And now I have to explain to you how we're going to make sense, how we're analyzing the representation theory of that at level one. And that is actually not so difficult because in particular we already understand now the structure of this super the algebra. So now, let's ask what are the possible highest rate representations of PSU one comma one slash two at level one. Well, so how would you analyze this remember so the highest rate representation the highest rate states will form a representation of the zero mode algebra. So the highest rate states, they form a representation of PSU itself right without the affine beats so I mean the zero Now, now you know the structure of this this has a bosonic algebra that's SL2R plus SU2. So every state we can label in terms of say, and say we label them in terms of a pair of labels so this are the states that transform say in the continuous spin J representation with the parameter alpha. And this is the dimension and representation with respect to SU2. So suppose I start with one state in the highest rate space and I say it transforms in that manner. So here the first label is a representation with respect to SL2R or SU11, which is the same thing. And this is the dimension. So n is equal to what you would call 2J plus one as a representation of SU2. So we can write down all the states that appear in the highest rate space in terms of representations of SL2R plus SU2 because that's the bosonic algebra that just acts. Now we have fermionic zero modes. Well, we have eight fermionic zero modes. They sit in two copies of the two comma two. So what do you do when you have fermionic zero modes? Well, you make creation and annihilation operators out of them. So there will be four creation operators and there will be four annihilation operators. And without loss of random generality I can say I start with the state that's killed by all the four annihilation operators and I apply the creation operators to it. So the fermionic creation operators and they sit in the two comma two with respect to SL2 plus SU2. So what happens when I apply one of these creation operators, one of the four. Well you see as you apply one of the four, you're going to generate states. And because the fermionic generator sit in a representation of SL2 plus SU2, you will change the representation with this vector, the bosonic algebra. So if you apply this, what will happen you get a term that looks like J plus a half N plus one. That's when you take the plus plus component where the spin shifts up in both directions. Then you get C alpha J minus a half N plus one. Then you get C alpha plus a half and minus one. And then you get C alpha J minus a half and minus. Is it clear what I'm doing here I mean, I'm starting with the state and I'm now applying the four fermionic generators. And since these generator transform in the two comma two, what I have to do is I have to take the tensor product with respect to SU2 of the n dimensional representation with the two dimensional representation. So I'm going to get either N plus one or N minus one. And then I have to do the same thing for the SL2 bit and the SL2 bit it does exactly the same thing so J will be shifted up by a half or down by a half. So there will be four terms, because I can shift J up by a half and down by a half and I can shift N up by one and down by one and the two are not correlated with one another so I get four summons. At the first level when I build up my Clifford representation, starting from the highest rate state and applying the fermionic generation creation generators once. Okay, so now I've created the ones now now let's do it again. Now, when I do it again obviously I have to be careful, because I can't apply this exactly same fermionic generators again. So in the first line, I should expect to get four summons. In the second line I should expect to get four choose to equals to six summons. So which six summons do I get. Well the six summons that I get. And where I get C alpha j plus one. So when J get shifted up yet another time, then I can't shift up and at the same time so this has to go down. So this will be one term. Then there's a term where I do the opposite, where I shifted J down so I go back to CJ, but then I go up to n plus two. So there is a term that goes like then they're actually two times that of the form CJ alpha, sorry, G alpha j comma n because there are two ways in which you can get it if you think about it. But it's actually not important for the following and then you have the mirror image of that, which is G J alpha and minus two. And this is plus C alpha j minus one. So so there so there are 123456 state at a second level, and then at a third level you get another copy of this, and then the fourth level you get another copy of this this is what the sort of Clifford algebra representations of super algebras look right. I mean this is like standards. This is good to you. I mean this is a standard for me on the generators on some on some Clifford. Now why am I belaboring this. But remember, we are sitting at level one. And what do we know about the highest rate representations of SU two at level one, as you to at level one has only two allowed highest rate representations. Now just think about the SU two factor SU two at level one has only the n equals to one representation, and the n equals to two representation, the only possible representations at level one. Any other representation is incompatible with the null vector in the back in the vertex operator algebra. So the only allowed representations of SU two level one is the one dimensional and the two dimensional representation. So now look at this diagram. And let's see whether I can find color that one can see. So, you see, we start with n whatever it is, but then halfway down the diagram we find n plus two. Now this is pro this is a problem right because suppose we start with n equals to one here. We are going to reach n equals to three here. So number three is not an allowed representation of SU two at level one. So this means the generic representation is not compatible with the representation theory at level one, because it produces an SU two representation. That's incompatible with just looking at the SU two bit. I mean, in particular it has to satisfy all the constraints that just come from SU two. So I know that this term can't be there. So this is one. Obviously if I said at level two. No problem I can start at level two I also have the three dimension representation then there will be a term that starts with the one has a two here three here to one everything is fine, but that K equals to one. Generic long representation is not allowed. The only way to get away from that is that the representation theory at PSU one comma one slash two at level one. The only allowed representations for the highest rate states are short representations representations that are shorter than a generic Clifford algebra is some fermionic generators have to vanish. Otherwise you run into trouble, just with the representation theory of SU two. Okay, so then. So you see that something special happens. And then if you if you analyze more carefully to discover two things. You discover the only possible representation is ultra short representation. So the only allowed representation is of the form that you start with the representation. See a JJ in the doublet. So I'm writing this upside down for reasons that a little bit more convenient. And then actually I was I think I was a little bit miss here I should have shifted the alpha the alpha values also shift by a half. So, actually here there's always a plus a half, because I mean when when you when you add the J three eigenvalue also shift us by half integer so the alpha parameter also shifts by a half. And then what you find is that you get alpha plus a half. What you find is that it's this representation. It's this. But these are the only representations that appear this is a short representation so it has only three factors. All the other fermionic zero modes vanish. And then, as is familiar from BPS representation. You see this, what you have to arrange is that a certain product of fermionic zero modes vanish. It's generically not the case. So in order for this to be the case you get a constraint on what the spin is. And what you find is that a constraint on the spin is that Jay has to be exactly equal to one half, and no other value of Jay is allowed so in particular. If you try to do this for a half plus IP, then the representation is necessarily long. So it's short is demanded by the fact that you are not allowed to have this state on grounds of the representation theory of SU two, and therefore it must be a short representation but that fixes the spin to be exactly equal to one half. That's the special thing that happens at level one, the continuum that we had gets frozen out because it's incompatible with the structure of the super the algebra the super the algebra insists on the fact that the spin is exactly equal to one. These are the only representations that are compatible with the structure and except for the vacuum representation but the representation of this kind and I started here with a continuous representation. If I started with a discrete representation, the analysis would have been completely identical, and I would have ended up again with a representation of this kind with Jay is equal to a half. Similarly, at Jay equals to a half. If you choose alpha to be equals to a half, then it is actually the discrete representation. I mean a continuous representation for alpha equals to a half Jay equals to a half. You can't distinguish from the discrete representation because I mean the discrete representation is basically half of it, and because it has all the same values it's just, it's essentially the same reps. I mean, this is the indie composable representation that contains the discrete representation as an irreducible sub representation. Now I've said it and I'm not going to say it again, but basically the discrete representation of j equals to half is contained in there. And if you do the analysis starting from the discrete representations again you get exactly the same. This is the BPS constraint. This is the BPS constraint, and it comes from the shortening which is an intrinsic property of this level one representation but this is the easiest way of you can also see it in a, in a variety of other ways you can study the null vectors of some virus or algebra generator. There are various ways in which to see that this is the only consistent representation of this level one theory this is the easiest way of seeing it because it appeals to something you know namely that the SU two level one representation theory is simple. Okay, so that that explains why, why this is this is the explanation of this fact. And in this context it really just comes out of the representation theory of the super the algebra and I'm not making any assumption here I'm just analyzing it from first principles and that's what I get that there is nothing, nothing else I can do. And then what you can do is you can. In fact there's a free field realization of this theory but maybe I don't really have. I don't really have time to explain it, given that I've only five minutes left so maybe I'll try to sketch it briefly at the beginning of next time, but what's important is now then you can do the counting and accounting here you see. It's a very special representation, and you can calculate its character. I mean the representation is easy one comma one slash to level k. There is some math literature on it this is not uncharted territory so you can look up the books, and you can work out what the character looks like a K equals to one. And what happens is that it only has two Bosonic degrees of freedom. You see generically, you would expect six was on the dissentance if you look at a generic representation of this algebra. So if you look at a boson should should go like one over ita to the six because each boson you can basically excite independently, but you know already that that's going to miscount badly, because at level one you know that as you to at level one is really equivalent to a single free boson. So as you to at level one, instead of having three boson has really a single boson. In some moral sense the same thing happens for so to our. So instead of having three plus three bosonic excitation modes, you have one plus one. If you just. So if you calculate the character of this representation. It just grows like one over each squared if you look at the bosonic degrees of free. If you look at the physical state condition, the physical state condition is going to eat up to bosonic degrees of freedom like it always does. And what you read off from that is that there are no dissenting degrees of freedom, coming from the PSU one comma one slash two factor of your world sheet theory because if you calculate this character. In this case, it grows as fast as though it has two free bosons, and then once you impose the physical state condition that will eat up the bosons from here and the only bosons that will survive come from the T for, and that explains why the spectrum really looks like only stuff made up from the T for. Remember, I said it was N over W but I didn't specify what N counts. Now I'm telling you, in this way of thinking about it a physical state condition removes all the dissenting degrees of freedom from here keeps only the dissenting degrees of freedom from here and then you match exactly the degrees of freedom as you would expect from the T for. And this calculation for how what the conformal space time conformal dimension of this data goes through, not identically but in an analogous fashion, and to then end up on the nose with the single particle the single trace spectrum of the symmetric or before the T for, and now there is really very little room for maneuver I mean we have honestly understood why this is very close to half you honestly understand the physical state account counting, pretty honestly understand the physical state counting condition, and we get on the nose, the partition function of the same at the single part the single cycle sector of the symmetric or before the T for. I mean, this is the sort of more honest version of the NSR description it requires a little bit of technology, but you see it also tells you why there was a problem with the NSR sector. You see what this tells you is the three dimensional representation of SU two is not an allowed representation for the highest base states, because it's not compatible with destruction. So if you think about it in the NSR formulation, the fermions it in the adjoint representation of SU tool in the three dimensional representation of SU tool so in the NSR description you started with degrees of freedom, which you discover from the hybrid point of in the first place so therefore the NSR description screams at you that you have to remove this fermions they won't you started with something you shouldn't have started with because at level one. They're not compatible with the actual symmetry underlying the problem. So the reason why the NSR formulation leads to this SU two level minus one factor the minus one factor tells you that you have to remove some degrees of freedom. And the degrees of freedom you have to remove other degrees of freedom you shouldn't have introduced in the first place, because they're not actually allowed by the representation theory of the super conform. So this is a short answer for what why the NSR description is a bit awkward, and how this comes out cleanly in the hybrid formulation. So, I think my time is up so what I want to do next time is I want to, I probably briefly mentioned the free field realization but then I want to concentrate on how to reproduce the correlation functions of the symmetric or before so the correlation function of the symmetric or before you can calculate in terms of covering maps. So I explained to you how you can calculate correlation function this metric or default. And then I explained to you how they arise from this specific world sheet description and they arise in a very transparent fashion, and you really reproduce exactly the structure of the correlation functions of this metric or before coming from this world sheet theory, but that's what I'll do next time. Thank you. Questions. Yes. Hi, I have an elementary question about symmetric or defaults. Typically in or before