 Recording in progress. Okay, so very good. So last time remember we looked at the ADS3 cross S3 cross T4 and the way we describe this was in terms of this SL2R N equals to 1, vasomino-witten model at level 1 and then the SU2 super symmetric vasomino-witten model at level 1 plus T4 on the world sheet and then we know that the SL2 are subalgebra here is to be identified with the Möbius symmetry of the dual CFT and that allowed us 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 and what we saw was the answer was equal to W squared minus 1 over 4W plus N over W plus, so I'm assuming here that H0 rest is equal to 0 and we'll come to that later on and so that was sort of the upshot of the last lecture and then 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 now in order to make life simple I'm going to discuss the symmetric orbit fold of S1 of 1,3 boson so let's look at the symmetric orbit fold of 1,3 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 n boson fields somehow this makes a funny noise right so let's hope it'll be better not sure anyway so there are n boson fields which I'll denote by dx i and each of them has conformal dimension h equals to 1 and i runs here from 1 to n right so these are 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 i.e. the 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 n bosons in the most obvious manner it just interchanges the n bosons so 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 well 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 so 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 that describes for you the fox space of the n fold 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 by being 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 think you have closed strings living on this quotient manifold and on 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 endpoints under the action of the symmetric group under the action of the orbit fold group which is the 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 abelian orbit folds if the group is abelian then the conjugacy classes are in one to one correspondence with the group elements and the correct generalization to a non abelian orbit fold group is that you label the twisted sector by 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 so I mean for example you can write any permutation for example as one two four seven three five one two three four five and then six for example that would be an element of S7 and any permutation in S7 you can write in terms of its cycle you start with the first element this element goes wherever it goes this element goes wherever it goes and at some stage you come back and then you start with the next one that's left over and so on so any 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 S7 are the specific numbers that appear here because if you think about it you name one to seven up to by something else then you do the original permutation and then you re-unname so therefore this is it's clear that a conjugacy that different permutations with the same cycle shape sit in the same conjugacy class because they're clearly conjugate to one another and you can prove that that's basically all the identification you get from the conjugacy classes so the conjugacy classes of Sn are in one to one correspondence with cycle shapes and therefore in one to one correspondence with permutations with partitions of N of writing N 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 yes when you have standard orbitals for obedient groups the twisted sectors are labelled by the element of the groups which of course are the same of conjugacy classes can you clarify why in the non-Abelian case what labels the twisted sector are the conjugacy classes are not just the element of the group well suppose you start with an individual group element in a non-Abelian orbifold and then you have to make it invariant under the orbifold action right so in an Abelian orbifold what you do the twisted sectors are labelled by group elements but then within each twisted sector you have to make it invariant under the whole orbifold group and then with the twisted sector associated to a specific group element and you try to make it invariant what 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 the thing that stabilizes the specific group element is just the centralizer and the rest produces for you the copies in the conjugacy class so each the orbifold action links together the different elements in the conjugacy class I mean this is not something I invent I mean this is standard folklore about orbifolds but that's why for non-Abelian orbifolds it's labelled by conjugacy classes rather than individual group elements okay so that's the general structure of a twisted sector it's labelled by a partition or if you wish a cycle shape of a permutation of order n now if you think about what we are trying to do we are trying to describe the safety dual of string theory on ADS-3 but we are looking at perturbative string theory and we are looking at states that are made of from one string so if you think about it what you expect say in the context of n equals to four supermails is you expect to find the single trace operators the single particle states you're not expecting to find the multi particle traces from a perturbative 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 perturbative string theory on a single string should correspond to the states that come from a single cycle twisted sector so the idea is that the single particle states are those that have a conjugacy classes of this type 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 that sits in the cycle of length one and the idea is that the single particle states are not all the twisted sector there are those twisted sector that correspond to a single cycle of length arbitrary length and nothing happening in the rest and then you see the multi 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 that's the closest to four 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 for you all the other sectors of the symmetric orbit fold so that is the correct part of the spectrum that we are going to see from a perturbative world's string theory analysis so we are only going to look at a single cycle twisted sector I mean if you are interested in the theory of symmetric orbit fold or perturbative string we will just be focusing on the single particles part of the spectrum which come from the single cycle twisted 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 twisted 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 by twisted sector well remember the intuition about twisted sector twisted sector describes those solutions that are not closed in the original theory but they are only closed after identification by the permutation so what this means if I start with my field say dx i here and I go around once I come around this has changed to dx i plus 1 if i is equal to 1 up to w-1 so this is what this permutation does for you as I go 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'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 the action of this permutation so the second the second will become the third the third will become the fourth the w-1 will become the w-th and the w-th will become the first again 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 element so how can we describe the excitations in this w-cycle twisted sector now the idea is very simple you could try to find modes for the individual fields but you won't have any luck because suppose you try to write modes down for dx-1 then it's not going to work because somehow the modes of dx-1 will mix with the modes of dx-2 will mix with the mode of dx-3 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 for a mode 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 you take the sum from 1 to up to w e to the 2 pi i j times l over w dx j so I choose l to be 0 up to w minus 1 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-1, dx-2, dx-3 where I mean with the curly bracket I now look at dx straight bracket 1 which is a linear combination of all the curly bracket ones and now if you think about it what happens with dx-l as I take it around the point sigma what happens well it will become the sum j is equal to 1 up to w e to the 2 pi i j l over w times dx j plus 1 with the understanding that I periodically identify by w which is fine you see this is periodic in w so whether this is 0 or w doesn't make a difference I have the same phase that's what happens if I look at this linear combination I take it around this field sitting in the w-circle-twisted sector it comes rearranged because every of the field has moved up one in the list but now you see this is easy to rewrite so this I write as e to the minus 2 pi i j over w sorry e to the minus 2 pi i l over w times the sum j is equal to 1 to w e to the 2 pi i j plus 1 l over w dx j plus 1 right I just multiply by e to the 2 pi i l over w and e to the minus 2 pi i 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 just relabeled my summation running now from so what you see is that these linear combinations of fields they have a simple monodromy as I take them around this point and the monodromy is simply the e to the minus 2 pi i l over w so they pick up a phase 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 1 field I should still be able to write the sum over r d alpha straight bracket l r z to the minus r minus 1 right I mean a spin 1 field has a mode expansion where you have a minus 1 here a spin 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 1 fields so it's still a spin 1 field so I can ask what happens when I take that around I just replace z by e to the 2 pi i z because I just go once around the origin so this will go to the sum over r alpha l r e to the minus 2 pi i r times z to the minus r minus 1 right this happens so here I'm taking z around and what this means I multiply z by e to the 2 pi i z 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 2 pi i l over w times the sum over r alpha rl z to the minus r minus 1 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 this tells you is that this linear combination if I think about it in terms of modes has modes that are not integer moded anymore but their mode numbers are such that e to the minus 2 pi i r is equal to e to the minus 2 pi i l over w so what this tells you is that alpha rl has modes where r runs in l over w plus integer because these are exactly the modes that I mean if I was deliberate vague I just introduced this label r I 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 monotromy and what's required to have the right monotromy r has to be equal to l over w mod integer so you see 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 w fields so originally we had the first w fields all of them were integer moded in the same sector what we end up are new fields where l runs from 0 up to w minus 1 but they look like the original field like one original field except that they have a fractional mode number and you see in some sense it's now redundant to introduce this label l here because you see whatever I mean another way of saying this is that you now just have modes z over w any value of the form z over w will appear and that just tells you which l you are talking about you are talking about the l for which the r mode you've written down is of the form l over w plus z 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 I mean originally you had much finer mode it's moded in units of 1 over w because that's exactly what this gives you and I'm accounting for all the fields and just have reorganized them so that they have a good mode expansion yes excuse me sorry maybe I missed it could you repeat why l should be an integer at least I mean if I start from there from the definition of the use any l well I mean 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 is I write out these w linear combinations where I take l to be an integer and what I'm claiming is and I haven't shown this is that the set of these linear combinations that I write it in terms of the original w field or in terms of d if I were to introduce more else then I would introduce linear relations among them so it's just convenient and then one matches the diggers of freedom so this is just a convenient rewriting I mean I have a w dimensional space and I just use a different basis but obviously I still have to use w many basis vectors simple mode expansions and the mode expansion is of that form and now if you ask yourself what does the spectrum look like well the spectrum where normally you would say l0 gets lifted up by integers but now in the spectrally 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 n this is a freedom of say the t4 theory so there is nothing twisted about this this looks like a single copy of a t4 theory but what we see is that with respect to the spacetime conformal dimension it behaves as 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 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 the conformal dimension of sigma and what you find is that and I mean on a certain level you are familiar with this you see when you go from the Nervous Schwartz sector to the Ramon sector that you are doing basically the same thing instead of having half integer moded fields you have no integer moded fields and you can ask and you know there is a Casimir energy involved in shifting the energy up from a Nervous Schwartz sector to a Ramon sector and likewise you can calculate here what is the energy shift the ground state energy that you pick up in the w-cycle twisted sector and what you find is for each boson the shift is w squared minus 1 over 24w I mean there are different ways in which you can derive this formula probably the easiest way is to use modular techniques you write down the character with the insertion of the orbifold element of the w-cycle twist and then you use the s-modular transformation and then you read of the conformal dimension 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 1 over 24w 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 T4 you have four bosons so you'll get four times as much and in fact fermions and 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 1 over 24w but the central charge of T4 is exactly 6 so this will give you w squared minus 1 over 4w and that's exactly the first term in this formula which we found here so this is the Casimir energy that you would expect for the symmetric orbifold of T4 because in the w-cycle twisted sector the ground state is shifted up by w squared minus 1 over 24 times the central charge 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 has exactly the structure so when you see this formula you say ah this looks exactly like the symmetric orbifold of T4 there is a question here good can I also think of this Casimir stuff by going to covering space like this as well as you know there's various ways in which you can derive this I mean you can derive it by the modular transformation you can go to the covering space you can write down the stress energy tensor so I mean the simplest way to derive this sector so this will be like this and then you can simply work out L minus 1 on the ground state this will not be 0 anymore and then in order to extract the eigenvalue you can just apply L1 I mean the reason you look at L minus 1 and L1 is then you don't have any normal ordering ambiguities then it's unambiguous and then if you work this out I'm probably going to make a mess of it another thing so maybe so in this theory I like a lot of vertex operators too of course so what are they on the string in the ADS3 times S3 and so on can you see them so I mean at this moment we are identifying states so the state here is characterized by which fractional modes you excite and then according to this dictionary so which modes where you sort of remove the fractionality you multiply the mode number by W were originally excited on the T4 the construction of vertex operators is an interesting one and I'll come to that probably tomorrow because I want to explain how the correlation functions match but that requires a certain number of preparations and I'll try to explain it please ask again if I don't explain it satisfactorily I think there's another question over there maybe I misunderstood I misunderstood the starting point but I understood that W if you want to take the SN symmetric orbit fold W can go from 1 to N correct but here in the strings W can take any integer value right but remember that G string is proportional to 1 over N and in perturbative string theory that means we take G string to 0 so what I'm doing here is always in the large N limit so we are always in the large N limit so we've taken N to infinity but as I'll explain to you the 1 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 N answer the finite N answer would correspond to sort of non-perturbative effects from the point of view of the string theory what you are going to do is we take the primitive string theory so we are always taking the 1 over N goes to infinity expansion first and then we systematically keep track of 1 over N corrections but there are no 1 over N corrections to the spectrum and therefore this formula is exact as it stands and how should I think about this I mean if you take 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 yeah so this description I don't know how it will work at finite N I think that's I mean finite N isn't something it's finite string coupling right then I mean when you want to do perturbative string theory you first start off by G string equal to 0 and then you think of switching on G and then you can automatically include 1 over N corrections in all the calculations and they should come for higher genus world sheets how the non-perturpative effects arise from the point of view of string theory and finite N effects would be non-perturpative 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 And another question, is it completely under control? This may require revolving in the larger limit. Absolutely. So maybe I didn't explain this, but this has a very nice larger 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 vacua 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 perturbative spectrum, and it stabilizes order for order. I mean, there's no problem with taking the larger limit. Any other questions? OK, 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 0, but I didn't explain this. And then the other question you would ask is, why are you just looking at the T4 degrees of freedom? You have all this other gunk in your worksheet, and you have this SU2 at negative level, what are you going to do about that? And this is 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 not totally satisfactory. This answer smells to be the right answer, but the NSR description at level one is a little bit problematic. Let's put it that way. It has issues, the Americans would say. And so we have to do a little bit better if we 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 ADS3 cross S3 cross T4, 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 Raman-Draman background. That was what it was really invented for. But you can just take this theory and evaluate it for backgrounds with pure Neville-Schwarz-Neville-Schwarz flux. And in that case, the hybrid formalism involves a world sheet theory that has a Resumino-Widden model based on PSU1,1-slash-2 level K plus a topologically-tisted T4 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-Auguri. I mean, this is not a proof, but this 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 or equal than two. And then this theory, as I'm about to explain to you, also makes perfect sense that's K equal to one. So our proposal will be that in order to make sense of the K equals 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 fighting against. Now I see from your eyes that you're a little bit scared, so let me explain to you what P is, you 1,1 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 P as you a 1,1 slash two and why should it appear? Okay, so P as you 1,1 slash two, if you think about it, it looks like a four by four matrix. It hasn't, so the convention here is you have an SU 1,1 and an SU 2, so there is one subgroup that is SU 2 and then you have one subgroup that is SU 1,1 and as many of you probably know, SU 1,1 is a different name for SL 2R. These are the same real forms. This is just the identically same, the algebra, not just the complexification of it as it really, the algebra, they're really the same. So it's a different name. Whenever you see SU 1,1, you can just replace it by SL 2R. It's really exactly the same. So this is a superly algebra, so it has a personics subalgebra that's SU 1,1 plus SU 2 and then it has fermions sitting here. So you have a fermions and the fermions sit in the 2,2 with respect to the SU 2 and the SL 2 because they sit inside a two by two block on which the SU 2 and the SL 2 act from either side as a two and because we have guys here and guys here, we have two of those. So PSU 1,1 slash two is a lay 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 SL 2R plus SU 2 and SL 2R is a three-dimensionally algebra. SU 2 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 lay algebra except instead of commutators of the fermionic generators, you write anticommutators, but that's it. I mean, for us, for physicists, that's not a big deal. I mean, we are used to anticommutators or commutators, so it looks like a lay algebra except every now and then you have to write an anticommutator but then it satisfies the Jacobi identity and blah, blah, blah, all the rest of it. So it's not much more complicated than a normal lay algebra, it just has some anticommuting generators rather than commuting generators. That's not such a big deal. Now, why does PSU 1,1 slash 2 appear in this context? Well, that has a very simple reason because the dual CFT has N equals to four super conformal, it's 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 zero algebra because it's conformal and then it has four supercharges, plus four supercharges, so these are fields of the form G plus minus and G prime plus minus, so 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 these are 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 will 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 a standard algebra. That's what you expect that dual conformal field theory to be like because that's what the symmetric orbital 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 SL2R symmetry on the bulk and that became the Mobius symmetry in the dual conformal field theory, right? Remember the rigid rotations of our ADS space became L0, L plus minus one on the boundary. Now what characterizes L0 and L plus minus one inside the Virasora algebra? Well, these are the global conformal transformation these are the transformations, the generators that kill the vacuum and that kill the out vacuum. These are the generators that are really well defined on the sphere. So you can ask what's the analog of SL2R for these other generators? And the answer is always simple. You take all the modes whose mode number is strictly less than H, 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? Well, from Virasora we are going to get L0 and L plus minus one and we know already that this gives us a copy of SL2R. What are the global modes of SU2? Well, this will be the zero modes, the minus one mode and the one mode don't annihilate the in and the out vacuum but the zero modes do. So from these guys you get just the JA0 modes or actually I'm going to call them KA0 in the future and they'll generate a lead algebra of SU2 and then of these guys we are 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 conformal transformation the super current transformations. And now if you think about it you see you have a three, you have an SL2R that's this SL2R, you have an SU2 that's this SU2 and these guys the upper index tells you that they transform as a doublet of SU2 and the lower index tells you that they transform as a doublet of SL2R. So these guys sit in the two comma two and then we have another copy that sits in the two comma two and in fact these precisely account for these fermions. 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 PSU1 comma one slash two and that makes a lot of sense from the well sheet perspective because you see the zero modes of PSU1 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 four super conformal algebra. So this is the supersymmetric generalization of what we saw earlier that the SL2R as a mean of written model gives rise to the mobius symmetry in the space time CFT and this is the souped up n equal to four version. Now you don't just have SL2 you also have SU2 for the r symmetry 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 fact that does for you. The generators of PSU1 comma one slash two are exactly the n equals to four transformations of the dual CFT. But 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 fact that the zero modes of this factor are precisely the global n equals to four transformations of the dual CFT, is that clear? Okay, so this is why this PSU factor appears and now I have to explain to you how we are going to make sense how we are 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 superly algebra. So now let's ask what are the possible highest rate representations of PSU1 comma one slash two at level one? Well, so how would you analyze this? Well, 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 without the affine bit, so I mean the zero modes. Now we 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 these are the states that transform say in the continuous spin J representation with the parameter alpha and in the dimension n 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, right? 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 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 there are four 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 the representation of SL2 plus SU2 you will change the representation with respect to 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 N minus one and then you get C alpha J minus a half 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 generators 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 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 two equals to six summons. So which six summons do I get? Well the six summons that I get, so there's one term where I get C alpha J plus one. So when J gets shifted up yet another time then I can't shift up N 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 shift the J down so I go back to C J but then I go up to N plus two and then there is a term that goes like then there are actually two terms that are of the form C J 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 N minus two plus C alpha J minus one. So there are one, two, three, four, five, six states at the second level and then at the 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 like, right? I mean this is like standards. Is this familiar to you? I mean this is a standard fermionic generator on some Clifford algebra. Now why am I belaboring this? Well remember we are sitting at level one and what do we know about the highest rate representations of SU two at level one? SU two at level one has only two allowed highest rate representations. So if I 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. These are the only possible representations at level one. Any other representations incompatible with the null vector in the vertex operator algebra. So the only allowed representations of SU two level one is the one dimensional and the two dimensional representation. But 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 you find N plus two. Now 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. But N equals to 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. 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 if I sit at level one. Obviously if I sit at level two no problem. I can start at level two I also have the three dimensional representation then there will be a term that starts with a one has a two here three here two one everything is fine. But that K equals to one that generic long representation is not allowed. So what you learn from that is that the representation theory at P is you 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 are some fermionic generators have to vanish otherwise you run into trouble just with the representation theory of SU two. So now you can ask okay so then so you see that something special happens and then if you analyze more carefully you discover two things. You discover the only possible representation is the ultra short representation. So the only allowed representation is of the form that you start with the representation Z, A, J, J in the doublet. So I'm actually writing this upside down for reasons that's a little bit more convenient and then actually I was, I think I was a little bit amiss here I should have shifted the alpha values also shift by a half. So actually here there's always a plus a half because I mean when you add the J three eigenvalue also shift us by a half integer so the alpha parameter also shifts by a half and then what you find is that you get alpha plus a half and actually what you find is that it's this representation, same as this. So 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 and that is 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 J has to be exactly equal to one half and no other value of J is allowed. So in particular if you try to do this for a half plus IP then the representation is necessarily long. So the fact that 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 and 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 this superly algebra. This superly algebra insists on the fact that the spin is exactly equal to one half and these are the only representations that are compatible with this 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 J is equal to a half. In fact, you see at J 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 at J 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 essentially the same representation. It's not strictly speaking I mean this is the indie composable representation that contains the discrete representation as an irreducible subrepresentation. Now I've said it, I'm not going to say it again but basically the discrete representation of J equals to a 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 theory. So this is the easiest way of, you can also see it in a variety of other ways. You can study the null vectors of some ReroZero 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 SU2 level one representation theory is so simple. Okay so that explains why this is the explanation of this fact. And in this context it really just comes out of the representation theory of the supali algebra and I'm not making any assumption here I'm just analyzing it from first principles and that's what I get. So there's 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, it's a bit, 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 the counting here you see so this is a very special representation and you can calculate its character. I mean the representation here of P is 1.1 slash two 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 at K equals to one. And what happens is that it only has two bosonic degrees of freedom. You see generically you would expect six bosonic descendants if you look at a generic representation of this algebra so if you look at a boson it should go like one over eta 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 SU2 at level one is really equivalent to a single free boson. So SU2 at level one instead of having three boson has really a single boson and what happens here is in some moral sense the same thing happens for SL2R. 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 eta squared if you look at the bosonic degrees of freedom. But then if you imply the physical state condition the physical state condition is going to eat up two bosonic degrees of freedom like it always does and what you read off from that is that there are no descent degrees of freedom coming from the PSU1,1 slash two factor of your world sheet theory because if you calculate this character it really behaves, 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 T4 and that explains why the spectrum really looks like only stuff made up from the T4. I mean 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 the physical state condition removes all the descent degrees of freedom from here keeps only the descent degrees of freedom from here and then you match exactly the number of degrees of freedom as you would expect from the T4. And this calculation for what the space time conformal dimension of these states are goes through not identically but in an analogous fashion and you then end up on the nose with the single particle the single trace spectrum of the symmetric orbit fold of T4 and now there is really very little room for maneuver. I mean we have honestly understood why this is J equals to half. We honestly understand the physical state counting or pretty honestly understand the physical state counting condition and we get on the nose the partition function of the single cycle sector of the symmetric orbit fold of T4. 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 SU2 is not an allowed representation for the highest weight states because it's not compatible with the structure but if you think about it in the NSR formulation the fermions sit in the adjoint representation of SU2 in the three-dimensional representation of SU2. So in the NSR description you start it with degrees of freedom which you discover from the hybrid point of view are not allowed in the first place. So therefore the NSR description screams at you that you have to remove these fermions. They want, you started with something you shouldn't have started with because at level one they are not compatible with the actual symmetry underlying the problem. So that's the reason why the NSR formulation leads to this SU2 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 are the degrees of freedom you shouldn't have introduced in the first place because they're not actually allowed by the representation theory of this super conformal algebra. So this is the 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 orbital. So the correlation function of the symmetric orbital you can calculate in terms of covering maps. So I'll explain to you how you can calculate correlation function of the symmetric orbital. And then I explain 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 the symmetric orbital coming from this world sheet theory. But that's what I'll do next time. Thank you.