 Yeah, maybe let's wait one more minute since people are still coming back Okay, I think we can start now. So Matias is there It's the microphone. Okay Okay, welcome back. So we now have the Lecture by Matias about strings in ADS 3 Please okay. Well, thank you very much. So let me remind in progress So let me remind you where we got to last time. So we are we are describing the strings on ADS 3 cross S3 cross T4 and so far we've been trying to understand how you describe strings on ADS 3 and we've concentrated so far on Bosonic strings on ADS 3 and The idea is that ADS 3 is geometrically a group manifold Corresponding to SL2R except we have to be careful about the periodicity in T Which we will have to undo and I'll come back to that so ADS 3 is really the universal covering group of SL2R and That fact that we can describe in terms of an SL2R vessel mean a written model if we are in the Situation where we have pure nervous words nervous words flux. So it's a vessel mean a written model based on the Lee algebra SL2R and as I explained to you Once you add the vessel mean a term you get you get conserved currents these conserved currents are defined by this expression and these and there's a corresponding formula for the left-moving version of those and They give rise to an affine cut smoothie algebra, which is this infinite dimensionally algebra that I wrote down And that's characterized by a level k And then the idea was that instead of trying to describe all classical solutions Given the fact that this theory has this enormous symmetry We can use this as a way of organizing the space of states of this theory and we argued that it should be given The Hilbert space well Hilbert space is the wrong word because it's definitely not a Hilbert space It has a non. I mean it's not positive definite It's the infinite I'm a vector space of of physical excitations And it has to organize itself in terms of representations of this affine cut smoothie algebra the left-moving version and the right-moving version and therefore Schematically it should be of the form that you have some representation on the left and some representation on the right and Initially you would naively expect that these representations will be highest rate representations And what I mean by a high state rate representation is that it's Fox space is generated by terms of the form Negative modes acting on some ground states and The ground states are annihilated by the positive modes So these are conventional highest rate representations so they're killed by the positive modes and the negative modes freely generate some Fox space and Normally you would expect your spectrum just to be described by highest rate representations on the left and highest rate representations on the right And then the question is what should this sum over J run over and what the sum over J labels are These highest rate states and what I mean by that is that we have an action of the zero modes on of SO2 are on this highest rate states and specifically I choose the conventions that the plus mode just shifts I mean you should I mean you have obviously seen bracket J M before if you know the representation theory of SU2 So J will stand for the spin and M will stand for the magnetic quantum number. So J 3 0 on J M will be just M and J plus will move you up one step and then J minus will move you down one step and in these conventions Obviously, there's a I mean sometimes you write this funny square root bracket factor here I've decided to rescale my my my states So as to absorb this square root factor and then it bites you here There'll be a factor here and the factor that appears here then is M into M minus 1 Minus the Kazimier Applied to J M and what is the Kazimier the Kazimier operator is the generator J a 0 J a 0 And if you write it out in terms of this plus minus the generators, it'll be of the former half times J plus 0 J minus 0 Plus J minus 0 J plus 0 Minus a half times J 3 0 J 3 0 So just like for SU2 what you show is that this combination of generators commutes with all the zero modes So it's what's called the Kazimier operator Therefore it takes a definite value in each irreducible representation and what you take the value of this to be in The conventions and so all of these things look like SU2 except there as a random Number of minus signs scattered throughout so for SU2 There would be a plus sign here for SL2 R There's a minus sign here for SU2 This would be J into J plus 1 and for SL2 R. It's minus J into J minus 1 So there's always a 50% chance that a plus sign turns into a minus sign Well actually judging by that is bigger than 50% because there are more minus signs than plus signs here but that's I mean okay you have to work a little bit to fix these signs, but it's basically SU2 with some Some small bells and whistles giving rise to signs. So then the question is what is the so So so what are the values of M that run runs? Oh, I mean when I specify J what I mean by that I specify the representation the representation will be given by specifying what value J takes and therefore what value the Kazimier takes That specifies the action of the zero modes then uniquely and then I have to specify which sorry which values of M Mod integer will appear because the generators move the values of M up and down by integers so I Was mumbling words like p-tavale theorem So if you look at the situation where k is large you can think about this geometrically and then you can ask What is the L2 space of SL2 R? And it's described in terms of tensor products of representations of the finite dimensional of the fine dimension group SL2 R and then the logic is that in the quantum theory in this in the string theory You should sum over the same set of representations So what are these representations for the case of SL2 R? What you should there are two classes of representations that are so-called discrete representations and They're characterized by J being a real number and Then M J M minus J must be an integer In fact M minus J must be a positive Or the discrete representations of this kind as another family of discrete representations But I'll just look at those and in fact in order for this theory to satisfy The no-ghost theorem that is one to impose the physical state condition you end up with a positive definite Space of states you have to restrict this J parameter to a finite range Namely it has to satisfy that it's bigger than a half Well bigger than a half is sort of That's that's it goes for free because I mean this is a quadratic relation So for a given value of Kazimierz you can always choose if it's real you can always choose J to be bigger than a half So it's it's bigger than a half But then the no-ghost theorem tells you that it also has to be less than k plus one over two But you see in the k-ghost to infinity limit you don't care and this is the analog of what you know for SU2 For SU2 the spin runs from a half to k over 2. So this is just the sort of SL2 analog of that So these are the discrete representations. So here I have to sum over all of them But you see this is now an integral because J is real J is not quantized because SL2 R is non-compact and Therefore you integrate over all the J's that run in this in this range So this is really some sort of integral or integral sum and then the second class of representations are the continuous Representations and these will be the real heroes in our story So we will treat this a little bit We will sweep them under the carpet and I'll give you an explanation for why we are allowed to do this later on today We'll just pretend that it don't exist But we'll come to that later and the continuous representations are characterized by the spin formally Well, it's been taking the value a half plus IP and then if you plug this into the Cosmere What this tells you is that the Cosmere is a quarter plus p squared Where p is a real number and then You see I mean this condition guarantees This condition guarantees if you stare at this formula that's a J minus Zero on the state J J is equal to zero So what these discrete representations look like is they have a state here where m is equal to J and Then J plus acts freely and J minus stops here So it basically looks like an infinite line and you have a J plus a zero going to the right Right and you have J minus going to the left So you you increase the eigenvalue of M with J plus and you decrease it with J minus But there is a there is a place where it stops Then the even m is equal to J this pre-factor goes equal to zero and you see this is sort of like the SU 2 representations you are familiar with Except for the SU 2 representation you would look at this range and here you look at this range So it's basically that's what it is and for SU 2 you would choose J to be half integer Whereas here we choose J to be any number and you just look at what happens to the right Whereas the continuous representations So so they are they are sort of semi infinite right they run to the left at Infinite them they never stop, but they stop to the right There is the smallest value of m, but there's not the largest value of m and then the continuous representations There are unbounded in both directions They run all the way because you see once C is of the form of a quarter plus p squared or rather One J is equal to a half plus IP since M is equal to real this pre-factor can never become zero So therefore this this sort of a row of J minus operators and J plus operators will never stop You will go as far to the left as you can go to the right So this extends infinitely in both directions and then because it stop never stops There aren't any preferred values for M. So then M can really take any value So the way you the way you say is that M is in the set Z plus alpha So these continuous representations are actually labeled by two parameters They're labeled by the spin which in this language corresponds to the real parameter p and they're labeled by this parameter alpha which tells you the quantization condition of M a mod integer and Remember I told you that the folklore is that strings on ads 3 can't put at the pure never Schwarz-lewitz background Which is what I'm describing here can't be due to the symmetric orbit fold And the reason I gave was that they have this continuum coming from the long strings out at infinity and this continuum is exactly this continuum So this p parameter is basically the momentum out at infinity and the fact that this is continuous means There is really a continuous spectrum here even after you impose the physical state condition And therefore this looks totally different than your symmetric orbital theory which has a discrete spec So that's the folklore and obviously we will have to break the folklore in order to identify the world sheet theory That's exactly due to the symmetric orbit fold because the symmetric orbit fold I mean on the face of it It has no chance to match any of these theories Okay, so this is this is what you would naively expect But that's not quite the right answer and it's not quite the right answer For reasons that has to do with the fact that we're really interested in the universal cover of SL2R Rather than the group manifold SL2R itself namely what this part of the spectrum describes if you think about it in terms of Solutions, I mean so you can ask what are the equations of motions that I I mean there's a sort of two Dictionaries here you can either write down the vessel me know written Model and you can look at the classical equations of motion on G And you what you find is the solutions for G Functions of the form sigma and tau that you can write as functions of G plus of X plus times G minus of X minus This is the this is the family of of Solutions of the vessel me know written model equations of motions Right, so so you could try to start with this and then quantize the phase space produced by all of these solutions Now here we've sort of gone down a different route We've said okay We know that this theory has this symmetry and therefore we know that this space must organize itself in terms of representations of The affine-cut smoothie algebra so you can ask how does this language translate? into writing down this fox space of physical states and the logic is that the states that are described by this fox space correspond to those maps that have the property that G plus of X plus plus 2 pi remember X plus and X minus At how a secret how plus or minus sigma so these are the world sheet light current coordinate and It obviously has to be the world sheet has to be periodic in sigma goes to sigma to 2 pi So therefore this function has to be periodic in sigma when sigma goes to sigma plus 2 pi And how do you arrange for that? Well you arrange for that by saying that G plus of X plus goes to Under rotation by 2 pi goes to G plus of X plus times time fixed matrix and G minus of X minus minus 2 pi Goes to M to the minus 1 to G minus of X minus Then obviously the product will be periodic because you see this M will kill this M to the minus 1 when I multiply them together So this will describe Periodic solutions and there is this sort of mental dictionary that the matrix here is fixed up to an element up to Conjugation so it lifts in the carton Taurus and the different values of J You can effectively think of describing the different matrices here So roughly speaking I don't want to just explain this in detail But the way you should think about it the different values of J Describe the different monodromes this solution describes, but they all describe strictly periodic solutions for SL Because this is like the analog of what you do for SU2 and for SU2 you want there is nothing else But for SL2 there is something else so this doesn't account for all the interesting solutions because the the The T parameter here is not periodic in in 2 pi I mean when you write it as a matrix in SL2R it appears to be periodic in 2 pi But it shouldn't be periodic because we want to look at this covering space And therefore we have to include additional solutions and the additional solutions. We have to include They are of the of the of the form that you so suppose you have a solution of this kind I'm going to engineer a new solution that will Account for the fact that T is not periodic. So what I do is I define G left of WR of X plus To be given of the form of e to the i times WR over 2 You'll see in a second why this is a smart thing to do X plus times sigma 2 times G What I call G zero left of X plus So this is a solution of this kind. So let's call them G zero here These are and they satisfy this sort of periodicity Where M is a fixed matrix Fixed matrix in SL2R and now the the new solutions I take an old solution of this kind and I multiply G left WR by this factor and G right of W Sorry, this should probably this actually an R and then this is an L of X minus What you do there is you multiply it from the other side So I should have called this right So this is left X minus and then I multiplied e to the IWL over 2 X minus times sigma 2 So okay, so what I'm claiming is suppose you have a solution of that kind I propose let's look at the solution that comes from this function for G plus and this function for G minus What's the difference? Well the difference is not as Monkey can see the difference is obviously the Factors of e to the I is times something over there But now remember and that's the reason why I wrote out these equations over here How we parameterized the group manifold to start with So you see on the left we have e to the u times sigma 2 and the right we have e to the v times sigma 2 and The way I'm modifying these solutions is by multiplying them on the on the on the left by Yeah, I must admit this is the world's most stupid convention to call the thing that stands on the left right and the thing That's on the right left and as you see I've confused myself But so so this is acting on the left and this is acting on the right despite appearances so what this mean is I'm take a solution and Relative to the solution I have before what I do is because of this factor You see this will effectively shift you as a function of X plus so you will go to you goes to you plus W are over 2 You see it's sigma 2 so it's W are over 2 times times tau plus sigma And it will shift V Which is the factor standing on the right here? You see this is the V So V is what stands on the right and you is what stands on the left so you get shifted by By this factor and V gets shifted by the corresponding factor Wl over 2 times tau minus sigma and now why is this a smart thing to do? Well, remember that you and V are the light cone coordinates in target space So if I translate them back into T and Phi T is the sum and Phi is the difference so what this means is that T gets shifted by V Plus the sum so there will be a term proportional to tau which will go like a half times W right Plus W left and there will be a term proportional to sigma times a half times W right minus W left and then Phi gets shifted by Phi plus Tau times a half and now I have to take the difference so here there will be W right right minus W left and Here there will be a half times W right plus W left Okay, so suppose I have a solution that's described by this G zero and once I modify it in this way I get a solution whose T and Phi dependence I mean previously it was periodic and now it picks up these additional pieces and Now why is this interesting? You see we want T not to be periodic So we have to make sure that this factor is equal to zero Right because when sigma goes to sigma plus 2 pi that describes must describe the same point and In order to avoid T being periodically identified. I have to make sure that this factor is equal to zero So this tells me I have to choose W R is equal to W L So when I do this then this term goes and this term goes and what I see here This just becomes W R equals to W left equals to W. So what I find is T goes to T plus W times Tau who cast Tau is not periodic T is not periodic and and Phi goes to Phi plus W times sigma and sigma is 2 pi periodic and Phi is 2 pi periodic So therefore I've included now solutions that are not periodic in T Even if I go around the sigma by 2 pi So I've made solutions that were periodic in T into solutions that are not periodic in T And what I see is it's not obvious that this accounts for all the possible things but you at least see that you are producing arbitrary winding in the phi direction and But you have undone the fact that this would have identified also the T direction Okay, so that's The proposal of Maldese in Onaguri was that what you have to do is you have to look at the solutions corresponding to this But now I've translated it in terms of classical solutions So in terms of classical solution I also have to include these new solutions But because I want to work in that language I now have to explain what does this modification do on the level of the representations I've written over there Sorry The winding in Phi does have to be integer so like W is an integer or it can be Oh, yeah, sorry. W has to be an integer. Absolutely. Sorry. I forgot to say this. Yes, absolutely So W has to be an integer. Otherwise I would yeah, yeah Thank you So so here I've explained to you on the level of the classical solutions what I have to do But I'm an algebraic and CFT type person. I want to work in that language So now I have to translate this description into this language So what I have to do is I have to calculate what happens to the currents when I do this Remember the currents are defined by the formula that are also wrote down here You have to take the trace of the le algebra generator and that's the convention I'm picking times d plus gg to the minus one and similar for the other current So let's work out what happens when I modify my solution in this manner Okay, so what we do is we take a solution that What's of the type g zero I a periodic solution? I or rather fixed by this fixed monotomy and now we have to include this factor So now we what we want to calculate is that the j a of r It's now k times the trace of ta and now I'm writing out a d plus of g my of d of g and So, okay, so this stuff doesn't not do okay, so I'll write it as Let me write it What happens so I have to apply this to e to the i times w right over 2 x plus times sigma 2 Times my old g right to 0 of x plus And I have g left of x minus and because I'm interested in the plus derivative I don't have to write it out and then I have here g inverse which is g l x minus to the minus one g right 0 x plus To the minus one and then I have to write the inverse of that e to the Little bit more blackboard space here e to the minus I w over 2 x plus Sigma 2 and I should write now w because W right is equal to w left is equal to w isn't integer Okay, so now I have to evaluate that now. Obviously you see The dx plus I can stop the dx plus the action here because this doesn't depend on x plus So this term obviously cancels against this term That makes it already less to write and now they're obviously two derivatives that deep the d plus can hit here So what am I going to get I'm going to get a term that looks like k trace ta And now I'm going to get two term if the d plus hits here. I get an i times w over 2 times sigma 2 G 0 and then the g 0 r will cancel against this g 0 r So that's all there is and the other term is Plus if that the derivative hits here, then I get e to the i times w over 2 x plus times sigma 2 d plus of g r 0 x plus Times e to the minus i sorry times I'm so g 0 r x plus to the minus 1 so this is this term times e to the minus i w over 2 x plus sigma 2 Right and then so that's what I have to calculate Now if you think about it, you see this is basically the old current, right? This is the original current and it's now conjugated by that But the country I can either think of this conjugating or I can think of the conjugation Conjugating the ta so I can write this as k times trace of ta times i w 2 over sigma 2 plus k times trace of e to the minus i w over 2 x plus sigma 2 times ta Times e to the i times w over 2 x plus sigma 2 times I'll just write it DJ gr gr to the minus 1 the old thing the one I had before and Now now it depends a little bit on which component I'm looking at so If you so in the convention so the conventions are I mean Somebody should have worked a little bit better on these conventions because they're the sigmas and then the teas So the tea tea 3 to confuse absolutely everybody and this has confused me for a long time I misread their paper many times is proportional to sigma 2 There you go and then t plus and minus is proportional to sigma 3 plus and minus sigma 1 But that's what it is the sigmas are the Pauli matrices so everything is totally explicit So now we can just work this out. So let's work this out for J3 of R So when you pick ta to be t3 You see t3 is just equal to by this funny convention minus i over 2 Sigma 2 Then you have a minus i over 2 times an i over 2 gives you plus 1 over 4 Sigma 2 squared has trace equal to 2. So this term just tells you This will be equal to k times W over 2 Plus so this comes from this term and then from this term if sigma 2 Since t3 is also proportional to sigma 2 obviously sigma 2 commutes with itself So this time goes away and you just keep the old current So what you read off from that is that the three component of the current just get shifted by a constant term and What happens to the plus minus component of the current? What you have to do for that is you have to ask what is this for this being equal to plus minus and That's a little calculation, which you probably better do in the privacy of your room than Seeing me struggle on the blackboard. I mean, it's not the rocket science You I mean e to the sigma 2 sigma 2 is an explicit matrix So you can write e to the i sigma 2 as an explicit matrix It's basically cosine sine sine cosine and then you just multiply this rule and what you find at the end of the day and it's a Very elementary calculation, but I'm not going to do it for you here Is that this is equal to e to the minus plus i times w times x plus times t plus minus So the t plus minuses go back to themselves, but they pick up a phase e to the i e to the minus plus i w x plus So what this tells you is that j plus minus r you see for them This time is 0 because the trace of sigma 2 with sigma plus minus is 0 because the traces of the of the dig With different sigmas are 0 so so this term is absent and this term You just pick up this phase and then you have to again the old current So what you learn from that is that this is equal to e to the minus plus i w x plus times j plus minus of x plus That's what you calculate. So that's the that's the sort of algebraic Description of having introduced these additional solutions, right? I'm taking these additional solutions and I'm translating it What it means from the point of view of the affine cut smoothies symmetry and that's what it means There's a bit of confusing point here for me because you introduced the these extra solution Because you want to describe universal cover but even if you want to describe just SL to R You should include them, but WR is not equal to WL or not No, but I mean if I'm looking at SL to R Then I have to be Then Then it has to be periodic in in T, right? So you're saying it's also periodic in T Yes, you just need that WR minus WL is a multiple of four pi Well, so in this convention Yeah, so you would have to use something like that. Yeah, maybe maybe you have to do this for SL to R as well Yeah so it seems to me that Going from SL to R to the universal cover is not adding new solution by extricting Yeah, you you're right. Well, I mean but but then that's not compatible with this, right? Are you saying WR minus WL has to be equal to a multiple of Pi or 2 pi or pi maybe 4 pi whatever. Yeah, so maybe maybe you're right Maybe you're right. Maybe there's also additional solutions for I think you're right That's probably also additional solutions if I don't go to the universal cover But if I go to the universal cover, I have Disconstrained on this class of additional solutions and in fact, this is I mean you see for SU 2 you may ask Why don't I have to do the same for SU 2 and for SU 2 if you were to do the same thing You would actually land on the same spectrum You're not introducing new degrees of freedom because I mean this will turn out to be spectral flow and the spectral flow maps The SU 2 representations again back to standard highest rate representations So normally you're probably right one has to include the sectors as well But for SL to R they generate new representations as I'm about to explain and you're right The universal cover is not responsible for adding new solutions. They're there anyway. It's more responsible for removing some Okay, okay. Thank you. You're right Okay, so let let me explain to you what this means in terms of the affine cut smoothie algebra Remember we have this melt expansion j a r of x plus. I Said you can write as a sum over J a n and I think I forgot a minus sign here yesterday There should be e to the minus i and x plus otherwise. I'm going to run into trouble now So that's the consistent solution with everything else. I'm doing so when you translate this you see what this means is that j 3 So j 3 n the new j 3 n is the old j 3 n and Then the constant term means that the zero mode term gets shifted. So plus k times w over 2 Delta n comma 0 and what you find for j plus minus n is that this is the old j 0 plus minus But now n gets shifted up and down in terms of w because you see this factor basically just shifts the mode number Because I mean this just multiplies to the exponent and thereby you shift the mode number and when you find this that this is the Set of transformations you get and what this means is the new solutions are described You can think of them as being described on the Fox space of the old solutions except that the SL to are now acts in a modified way So so so you have the old Fox space on which the J zeros act like they acted before But now on the same Fox space I define a new action where the new modes are defined in terms of the old modes plus corrections and In fact what you can check is that this is an automorphism of the affine cut smoothie algebra associated to SL to are I These guys satisfy the same commutation relations as the original guys if you redefine them in that manner So this is an automorphism and whenever you have an automorphism and you have a representation Then the automorphism produces for you a new representation Now there's no guarantee that it's a genuine new representation It may be the old representation in disguise But at least potentially it's a new representation normally would think it's a new representation if the automorphism is outer Then generically you get a new representation if the automorphism is inner then you won't then you're just relaping the states Now you can check that this is that this isn't in general an auto automorphism and the reason for that is you see That it shifts the mode number in this funny way. So if we start with the highest rate representation And so on the highest rate representation You remember we had j 0 plus m on the state j m is equal to 0 for n bigger than 0 That was it what it meant to be a highest rate representation But now if I think about it in terms of the action of j plus m J plus n acts on these states By saying that this is the same as j 0 plus n minus w acting on these states So this is only 0 if n is bigger than w So what this means is that there are some negative j plus modes that will not anneal a the highest rate state And as a consequence, it's not a high straight representation. It's a representation where there are certain negative modes that you can apply as many many times as you want and they will always be non-trivial and so So so these are generically not highest rate representations You can also see it in terms of the a very several algebra. So if you look at it in terms of the if you ask What does the very server generators do under this modification now? There's sort of an an abstract way of doing this because you know the commutation relations of the of the virus or a generator with the currents I mean, you know that the l n generators as To with the j a m generators have to satisfy this commutation relation and Then compatibility with this will tell you how l transforms and what you find is that l m transforms as l m 0 minus w times j 0 3 m Minus k over 4 times W squared delta m comma 0 this term You only see if you insist that they still satisfy a very several algebra and this term you get by simply demanding That this is also an automorphism under this under this transformation and you see because of this term and The fact that the j 3 0 spectrum is always unbounded to the positive line if I take w to be positive Then the zero mode of L 0 will be unbounded So so so these representations are not included in what I had before because before I had a bounded L 0 spectrum Because I had the highest rate state and the positive modes killed it and L 0 move you up and Now once I've included a spectrally float sectors. I've genuinely produced New representation with an unbounded L 0 spectrum But this hinges on the fact that these representations are infinitely extended to the right and therefore that this eigenvalue can become as negative as you want in Particular if you think about doing the same thing for SU 2 you see for SU 2 you have finite dimensional high state representations And what you find is that after you have applied this spectral flow to an SU 2 representation You just get another SU 2 representation and for SU 2 what you find I mean for those people who are sort of familiar with this Sigma will map the J's representation to the representation Associated to K minus 2 over J So it'll just flip around the so it's all the tool so Sigma squared will be trivial will map a representation to itself Sigma 2 is Sigma squared is inner Sigma is outer and in fact This has to do with the fact that SU 2 is the double cover of SO 3 and what this is really implementing is the quotient going down to SO 3 But this is just as a side comment but what's important here is that we genuinely get new representations and therefore this spectrum that I wrote down before was Really too small because it didn't include any of these resolutions So I have to include my spectrum and now the correct answers for what's my my Fox space of this Worldsheet theory should be is that there should be a sum over these winding sectors and Restrict myself secretly to W bigger than zero Then I have this integral over J or sum over J and then I will have and I write this like like such So by this I mean the the representation induced by this automorphism Sigma to the W and I apply this Simultaneously to left and right move us because W right times W left is equal to W So this is basically the Maldesina or guru a spectrum and as I explained to you the spectrally flowed sectors come from the fact that you have to have this periodic solutions that are periodic in fine, but not periodic in tea Well, I mean so I mean that you have to these additional solutions because SL 2 are has these additional solutions and then the fact that We are we are in the universal cover means that W right is equal to W left rather than the difference is something funny Okay, so this is this is the spectrum we have to work with and now the idea is that now we've identified as well Sheet spectrum now the aim of the game is that we are going to work out the the The physical states that satisfy the physical state condition But before we do that there is one more thing we have to do No, I'm losing track of my notes. Oh, yeah so so far we have done everything Bosonic and Bosonic is fine, but it kept just the essence of the spectral flow But in order to really get the symmetric orbit fold we have to deal with the super string So what's the super string version? You see so far I really only concentrated on the ads 3 fact done only concentrated on the Bosonic spectrum now I have to look at the super string version So what's the super string version? Well, so The the super string version is actually relatively easy to describe So what happens is you see we had this SL 2 are Bosonic algebra level K And if I'm look there's a natural supersymmetric generalization of it and a natural supersymmetric Generalization consists of writing an upper index and writing a one That means it's n equals to one super conformal and what does this algebra consist of well This algebra consists of the generators J a m that satisfy an SL 2 are Level K f I cut smoothie algebra with the commutator as I wrote down before but then in addition you have fermions That also transform in the adjoint representation of SL 2 are so there are three fermions Psi 3 and sub psi plus minus and so the J's with themselves give you an SL 2 are and the J's with the with the with the step size Give you just the they sit in the adjoint representation. So they just transform as somebody who sits in the adjoint representation So I hope this is so these are the structure constants of SL 2 are so so 3 with plus gives you Plus times plus 3 with minus gives you minus times minus and so on so it's It's just that they transform in the adjoint representation and then the size by themselves just satisfy their free fermions So the non-trivial Anticommutators they satisfy of the form Psi and again, there's a funny minus sign Because we are in SL 2 are so there's a minus. I'm sorry. There's a K over there's a K There's a factor of K here and there's a factor of minus K over 2 here So so going to the super string is basically means you replace the bosonic algebra SL 2 are by the sort of Super-symmetrized version and the way you think about it. This is like NSR Normally functions. I mean these are the bosonic degrees of freedom of your target space and for each Boson DX mu you add the fermion Psi mu so the fermion has the same labels It's also sits in the adjoint representation It transforms under the bosons in the adjoint representation I can't really do anything, but and then it's a free fermion just because there's nothing else you can do Okay, so that's basically the Susie version for ADS 3 So that's what you have to do for ADS 3 and then you have to do the analogous thing for SU 2 For SU 2 you also enhance the SU 2 level K prime affine cut smoothie algebra by adding in fermions in the adjoint representation of SU 2 and they satisfy The analogous commutation relations that's that and then the T4 is the T4 Now now we want to so the first thing I have to explain to you is why the level of the SL 2 R is Quantized I was asked this earlier and the reason for that is that it must be equal to the level of the SU 2 So let me explain how this comes about Now when you look at this super conformal Super affine algebras you have to you can ask yourself. What's the central charge and the central charge? There's a clever way of calculating the central charge and that's sometimes useful You see this is a coupled system So I mean how do you calculate central charge? You try to separate it into blocks and then you just add that Paravise commute with one another and then you add the central charges of the blocks now here You can't do that because the stupid fermions transform under the bosons So you can't just say it's the bosons plus the fermions because they're not you have to disentangle them now in fact you can disentangle them and That's what goes under this decoupled current. So this is meant to be a different letter than this J This is a more curly version of this letter J And what you do is you define new generators new currents Which are the old currents and you correct them by terms that are bilinear in the fermions and they look explicitly like that So for J plus minus You just do that and then for J 3 You take these J 3's and then the way you decouple is that you add to it the term J minus J plus I Mean these are to be understood as normal order products of this size So what you do is you take your currents and you add to them suitable fermionic bilinear terms and then what you can prove and that's a little exercise is that these currents then Commute with the fermions. You basically remove the fermionic piece of the currents so that these guys Commute with the fermions and they still satisfy an SL 2 are Affine-Cutts-Moodie algebra, but their level has been shifted. So they now satisfy SL 2 are Not at level K, but at level K plus 2. I Mean for people who know what's secretly happening here is that the free fermions themselves build an SL 2 are algebra at level 2 And you are basically sort of taking the coset You're sort of taking them out and then a rather minus 2 taking them out and then you get this term plus the decoupled fermions so So but now it's very easy to calculate the central charge that comes from this factor so when I calculate the central charge of ADS 3 So you see for ADS 3 I have this SL 2 are Super-symmetrized level K, then I have S3 then I have SU 2 Super-symmetrized level K prime and then I have a T4 So let's calculate the central charges Well, so I have to calculate the central charge of this decoupled SL 2 are now the central charge over of an of a Bosonic SL 2 are at level K It's 3 K into K minus 2 So now when I shift this what I get is I get a 3 K plus 2 into over K This comes from the from the decoupled SL 2 are currents because they are level K plus 2 Then I from the fermions I get three halves because three free fermions Give me three halves each fermion contributes a half to the central charge Then I do the same thing for SU 2 so for SU 2 what happens is for SU 2 level 1 Level K is the same as SU 2 Bosonic at level K minus 2 plus 3 fermions So here for SU 2 level K the central charge is 3 K minus so if this is at K prime so this will give me So this is at K prime So this will give me 3 times K prime minus 2 over K prime plus three halves and Then the torus will give me six because the torus is for bosons Which gives me four and for fermions each fermions gives a half that's another two So all together six and this has to be equal to 15 That's the critical dimension of super strength theory, right? I mean no ghost theorem tells me that I have to live in the critical dimension So that has to be sequence to 15 which corresponds to a super conformal theory in ten dimensions Ten bosons and ten times five fermions, ten times fermions, ten times a half fermions give you 15 And if you stare at this you see you get a three from here plus six over K plus three halves plus three minus six over K prime plus three halves Six and now you see three plus three halves plus three plus three halves plus six is equal to 15 So therefore what you learn is that six over K minus six over K prime has to be equal to zero and therefore you learn that K Has to be equal to K prime. I mean this is also familiar from a supergravity perspective That's just saying that the radius of the ADS space has to be the same as the radius of the three sphere That's the supersymmetric supergravity solution But as the CFT version of it is that in order to get a critical strength theory the level of the SU2 and the level of the SL2R have to be equal to one another And therefore because the SU2 level is quantized the SL2 level has now also turned out to be quantized Because they have to be equal to one another. That's the question Sorry, I have a question on the shifting of the level. Maybe I'm confusing myself, but With the different conventions, but also in the bosonic In the bosonic double zero model you have a shift of the level by the dual cocceder number Well, that only appears in the central chart the central charge formula of S of G level K Is K times the dimension of G? Divided by K plus the dual cocceder number. So that's for example what for SU so for SU2 level K This gives you three times K because the dimension of SU2 is 3 and it gives you K plus 2 Because the dual coxed the number of SU2. Yes, and if I'm not wrong for SL2R is there is a minus sign So for the SL2R, exactly for SL2R I basically you can think of the dual coxed the number being equal to minus 2. So is it true that by adding fermions? I have an opposite shift. Absolutely. So this is actually a very important point So you notice here you get a factor of 1 over K and 1 over K prime here As in the bosonic case you would have had a factor of 1 over K plus 2 and 1 over K prime Minus 2 or whatever. Now, why is that important? Remember K is the radius So if you think about it like you weren't smart and you took this vessel mean original model you turned it into Feynman diagrams and you would just do a Feynman calculation then the powers of 1 over K is basically the higher loop corrections as you would calculate it from a Feynman diagram and then in the bosonic case You would have something like 1 over K plus the dual coxed the number and the way you think about it Is that this is 1 over K times 1 my 1 plus H dual over K To the minus 1. So this is 1 over K times the sum from n is equal to 0 to infinity of H bar over K minus this to the power n, right? So what you would see is you get infinitely many corrections in perturbation theory if you treated this Naively in perturbation. Now what happens in the Susie theory? You see The Susie theory is sort of renormal not renormalized and that's manifesting itself that the central charge involves just the unshifted level and therefore it's one loop exact you don't get any higher order correction So that's the reason why this is exactly opposite and that's a hint that Susie is doing something good for you. Okay. Thank you Okay, so so this is so this is basically so now we know that we have to live at the same level and now we just have to put everything together and and Analyze the physical spectrum Now I'm out of my many many right. So so now what we have to do is We have to So now so let's write down the spectrum and Actually, so here. There's some small subtlety. You see I explained to you spectral flow on the level of the SL2 I assume in a written model, but now you're going to ask yourself So should I spectrally flow the J's or should I spectrally flow the curly J's? Right, I mean they're both SL2 are with some in a written models and I told you how to flow them So which one should I flow and the answer is it doesn't matter? So why does it not matter? Well, because you see spectrally flowing the J's and spectrally flowing the curly J's Differs by whether you spectrally flow the fermions or not There's an analogous way in which you can spectrally flow the fermions But for the fermions if you spectrally flow them, you're just rearranging the states You're not generating any new states in your in your fox base Just like with SU2 level K where the spectral flow just maps the jth Representation in the k of a 2 minus j representation with the fermions. It just re-assembles the states So whether you spectrally flow the fermions or not is up to you. You're not describing a different spectrum You're just describing them slightly differently And it's more convenient to actually spectrally flow the coupled SL2R generators because the coupled SL2R generators Are the geometric ones because geometrically the fermions will also feel the mobius symmetry So they should also transform under say the zero mode So decoupling the fermions is just a trick for the purpose of calculating the central charge But the real SL2R is the coupled one because the fermions I mean these are the tangent vectors They live in the tension space to the manifold They also feel the rotation of SL2R and to remove them would be artificial So what we are going to do is we're going to spectrally flow also the fermions I expect to be float the coupled generators rather than the decoupled generators And what this means is that the k that will appear here is really the k I'm talking about here rather than k minus 2 or k plus 2 for the case of SL2R and SU2 So I'm flowing the full generators And then on this resulting fox base, which will look so there's the SL2 factor So again, this will look like the sum over W Then I will have the sum over J and then sigma to the W Hj, but now these are the super conformal representations But that we are happy with and then I have a similar factor for SU2 This is like the covariant description of string theory, right? This is This involves the time direction ads3 contains the time direction So in order to describe the physical string states I have to impose the physical state condition So this will be the condition that gr of phi is equal to 0 for r bigger than 0 And I'm thinking of here working in the Neville-Schwarz sector There's always a Neville-Schwarz sector and a Raman sector But the Neville-Schwarz sector is the interesting one And then you have ln on phi minus a half times delta n comma 0 On phi is equal to 0 for n greater than 0 So these are the the physical state condition This is just like if you're open green Schwarz-Bitten You do flat super string That's what you have to impose in covariant quantization, right? It's the it's the usual n equals to 1 super conformal symmetry You have to impose and in the Neville-Schwarz sector the mass shell condition L0 has to be equal to a half That's the mass shell condition in order to get the no ghost theorem going So now what we have to do is we have to evaluate We have to look for the states That satisfy these conditions We have to enumerate all of them And then we want to find out what their space-time charges are And that will describe for us the space-time spectrum of this specific world sheet theory So what does so the interesting condition here is that L0 has to be equal to a half So what does this mean? Well remember we have to work with the real L0 And in the spectrally float representation L0 gets shifted in that way But what I'm going to do is I'm going to spectrally flow Yeah, so this is I'm going to use exactly this formula So the L0 condition will amount to the following It will amount to the H0 of SL2R So for my for my level for my Susie theory at level k Plus the H0 of the rest So the rest will be SU2 So the rest is SU2 Plus the torus So this will be the ground state conformal dimension Then there will be some excitation number This is the total excitation number coming from everybody And that has to be equal to a half Right, that's the mass shell condition That I will have to impose on this spectrum Right, you have the SL2 factor and the rest The rest is SU2 and T4 This is the conformal dimension of the ground state of SL2 That's the conformal dimension of the ground state of SU2 That's the excitation number And I'm I'm thinking of looking at states that are in the in the vacuum of the T4 I could also include that doesn't really matter for the analysis It's just to be simple And this factor for this factor I now have to use the fact that this is spectrally flowed So this will be the ground state energy before spectral flow And the ground state energy before spectral flow is the casemia So this will be minus j into j minus 1 divided by k Right, because in the Susie theory I get this shift It's the the the Shuka-Vara construction will have a level k down here For the same reason that we discussed before And then it'll be minus w times m So m I will call the eigenvalue of j03 before I apply the spectral flow So this is the eigenvalue of this guy And then I have to subtract minus k over 4 times w squared So that is what this term is And then I have plus h0 rest Plus n is equal to a half So this is the mass shell condition I have to solve And now I'm going to solve this mass shell condition for you And explain to you that we get something interesting So let's let's do this This is a maybe I can do this Yeah, perhaps you can take another five minutes Is that enough? Yeah, it'll be five minutes It's very it's very simple I mean it's very very elementary Except it's quite key So I would quite like to explain it today And then we can discuss its consequences So what I'm going to do And at this stage I'm just declaring this So I'm going to set k is equal to 1 That I've told you before that you are used to And now I'm going to say that j is equal to A half times i times zero So I'm going to look at the continuous Only contissuous representations I'm only going to look at the one for which p is equal to zero Now the justification at this moment is non-existent But I'll promise you next time I'll begin to explain to you that this is what you have to do If you look at it from the hybrid formalism In this language it's not obvious Why these are the right representations to look at But grant me just look at these representations And see what happens Okay, so what's the casimir? Well the casimir is a quarter So what does this equation become like? This becomes a quarter minus w times m Minus a quarter times w squared Is plus h zero And I stop writing this rest stuff now Plus n is equal to a half Okay, so now this equation I have to solve And remember in the continuous representation The eigenvalues of j three zero are up to me Because the continuous representations Remember they are labeled by cj alpha And m has to be of the form alpha plus a dot But I can simply take this equation to solve for m And then declare that that means I'm looking at a representation With a corresponding value of alpha And because all values of alpha appear inside this direct sum So if I'm looking at the continuous representations I would sum over all values of j and alpha There will be one term in this sum Where this mass shell condition will be satisfied So I have in mind I take a specific dissentant And I ask for which term in this integral Does it satisfy the mass shell condition And I'm saying okay I'll just take this equation I solve it for m Then m tells me which alpha And that picks out the term in this integral Where it matches the mass shell condition Okay, so solve this for m Okay, that's not difficult I just bring wm to the other side So what is this? Wm is then equal to minus a quarter A half minus a half A quarter minus a half Minus a quarter w squared Matthias, sorry just to say that The equation should be equal to zero right So Say again A question equal to zero So the the quality The equation means it's equal to zero I'm here It's equal to a half Ah there is an equality sign I mean this is in the Navier-Schwarz sector Of a Ramon Navier-Schwarz sector string It was the minus and the equal Because I can't see from here I'm sorry Sorry sorry no this is This is equal to a half And this is this a half This is the usual a half ground state energy In the Navier-Schwarz Yeah sorry Okay so this is the equation Okay But now remember That we are not interested in m m is the j30 eigenvalue before spectro flow And we are interested in the j30 eigenvalue After spectro flow Because that's the real j30 Right? The other thing is just a way Of describing these representations So the real j30 eigenvalue Remember that j30 is equal to j030 Plus kw over 2 And this is This eigenvalue is denoted by m Remember that was the term that appeared In the shift formula for the conformal dimension So if I'm interested in this And remember this is the scaling dimension Of the dual CFT So this I can also call h And think of it as the conformal dimension As looked at from the point of view Of the spacetime theory So for that I should not look at m I should add to m k over w Well first of all I should divide by w But that's not hard And then in order to find out the state The spacetime conformal dimension Of the corresponding state What I have to do is I have to add kw over 2 Which is just w over 2 But look there's w over 2 And this term is minus w over 4 So what I'm going to get is w over 4 Minus 1 over 4w Which comes from here Plus And let's set h0 to 0 for simplicity Plus n over w Now Now it depends whether you've seen The symmetric orbital before or not If you have seen the symmetric orbital Before you should say aha If you haven't you should not say anything And from your reaction I assume You haven't seen it before So that's what I'll explain to you next time But this is exactly The conformal dimension spectrum Of a symmetric orbital Where this is the casemia energy Of the ground state And this is what you expect For the symmetric orbital of t4 And the excitation numbers In the spectrally flow In the w-cycle twist detector And I'll explain this next time Are not integer-moded They are fractionally 1 over w-moded So this exactly reproduces the spectrum Of the symmetric orbital So obviously I have So what I'll explain to you next time Is the other way of getting at this formula From the symmetric orbital And then I have to answer All the nasty little questions Is why was I allowed to do that And what happens to all the other degrees Of freedom and all the rest of it And then the proper answer to that I'll give you some hand-waving answer And then the proper answer Will be the hybrid description Where there's a very clean answer Where this comes out of the Representation theory of the super-lie algebra And where all the degrees of freedom Do exactly the right thing To match then exactly the spectrum Of the symmetric orbital But here you see the first sign That you are on the right track Did this looks to somebody You see the symmetric orbital spectrum Before like the symmetric orbital spectrum So now you just have to dot the i's And cross the t's To make sure everything works out And it does work out But that requires a little bit more work But since I'm over time I'll stop here Yes, thank you Recording stopped