 Ok, zato v zeločenju lečenju smo zeločili, kako se zeločilo vzeločne teori, supersymmetrične teori in vzeločne spasje na vzeločne superspasje vzeločne supersymmetrične teori. in, če je to, da je pošta, da je vzelo, pošta je, da je začnila vse supersymmetrične teore na supergravite in potem, pa je prezete supergraviti vse na zelo, da da se daje vzelo začne, in pošta je vse supergraviti klasike obježenje. Zato si smo zelo, da je izgleda, da pozavljamo neko izgleda, in da imajo nekaj supersimetry, kaj je obježeno, načo je boj, da je nekaj, da je nekaj pozavljamo, nekaj zelo nekaj, in to ješnijo, da je dobro dosveteljne nekozavljamo, pozavljenje barametri, kar trenička, sega da lahko dodeni drugih začinitih. Ten vseč je eno kako svojeza izgleda, zapravk vseč, na prorovu moželo na 0. Pozolim som je nekako, zelo se dodenil, One example, which was for theories in for dimensions, which possess a Ferrara-Zumino supercurrent multiplet, then we can couple them to what is called old minimal supergravity. And this generalized giving spinor equations that you get in this particular instance. I have the following form. I'm only going to write one. And there is another one for that bar. And here in this equation, m is a complex background field in the supergravity multiplet. And b mu is a complex one form, sorry, real one form, or you can make it complex again in the gravity multiplet. And the other way in which the fields in the gravity multiplet appear in this equation, well, I mean, they appear implicitly in the sigma muze because there would be factor of the frame and also in the covariant derivatives and in the spin connection, which enter in the covariant derivative acting on the spinor. So then, in order to find backgrounds which preserves some supersymmetry, you have to find backgrounds for which you can find solutions to this equation and the other one for the zeta bar. Yeah, m is a scalar, and the b mu is a one form. So for instance, I can give an example. So we can set b mu to 0, and we can take the metric, we can take the manifold to be the force sphere, and we can take the metric to be around with radius r. And then we can try to see if we can solve this equation for some value of m, and the other equation, which will have an m bar. So maybe we can just write the other equation here suppressing the piece with b. And indeed, you will find out that there is, this was one of the exercises. If you have done it, you have found out that there are four solutions to this coupled equations, which means that you have four different couples of zeta and zeta bar that solve this equation for b mu equals 0. And you have to set m equals m bar equals 3i over r. So this is indeed one of those cases in which you have to allow the background fields to be complex. And in particular, m and m bar are not related by complex conjugation, because they both have the plus i. OK, so, well, then you can figure out, once you have the spinor variation parameters, you can check what the algebra is, and what you find is that, in this case, you get OSP1 slash 4. And indeed, in OSP1 slash 4, you have an SO5, well, SP4 corresponding to the rotations on the S4, and there is no r symmetry. OK, so then you can, so let's comment a little bit about this background that I have to turn on. So if you remember the expression for the deformation of the flat-space Lagrangian that I wrote down last time, or if you open, like, Weston-Bagger and check the equations for all minimal supergravity coupled to matter, then you will see that this m and m bar coupled to certain operator appear in the Lagrangian as coupling to some certain operator, which appears in the current multiblet. So this was this operator x that we talked about. And this operator, if the theory is not super conformal, will not be 0. And therefore, there will be terms in the Lagrangian, which have funny factors of i, which coupled to this operator. And these terms would go down as 1 over r, as advertised last time. Now, one thing, which is interesting, is that if you, so these terms, which, like, somehow spoil reflection positivity on S4 because of the funny reality condition on the background fields, they go away when this operator x vanishes. And that, indeed, as we discussed, happens when you can shorten the multiplet from the Ferrara-Zumino multiplet to the super conformal one. So for super conformal field theories, these terms are not present. And indeed, like, you know that you can place a theory on the force sphere without having to add any 1 over r terms in the Lagrangian by just doing a conformal map. So that will introduce a coupling to the curvature, which goes, like, 1 over r squared. But it does not introduce any imaginary couplings. And this, by the way, also, so you might have wondered, why is that I can write down, people write down theories on spheres, and they are supersymmetric for, like, say, some non-conformal n equal to theory on S4, or, in this case, a non-conformal n equal 1 theory. But no one has ever written down anything in the sitter. So the sitter is supposed not to be supersymmetric. And the reason is that, indeed, like, you could imagine doing the sitter, but then you would have some background fields, which are imaginary in Lorentzian signature. And, like, that would correspond to some theory, which is not unitary. And people are not interested in theories, which are not unitary, not interested in writing no unitary theory in the sitter. So that, I guess, is why you don't see this. OK, so this concludes the comments that I wanted to make for now about the old minimal super gravity backgrounds. I will maybe come back to this at the end of the next lecture. OK, so, but I wanted to make a general comment before continue, which, I think, is a good slogan, which is the following. So we had this diagram last time. I drew this diagram where we have a set of, this is the set of all n equal 1 theories. And all of these theories, we argued, have a supercurrent multiplet that we call the S multiplet. And then inside this, there are some more special theories, which will have a Ferrara-Zumino supercurrent multiplet and some other set of theory, which have a conserved r symmetry, so they have this r multiplet. And then in the intersection, there is an even more special set of theory, which are super conformal. Now, so first, a question that I got by a few of you, so maybe I should clarify, is that, indeed, this is a proper subset. So there are theories, which have both a Ferrara-Zumino supercurrent and an r-current, but are not super conformal. So that means that you do have improvements available that make the S-multiplet, both an FZ-multiplet or an r-multiplet. But the improvement for the r-multiplet and the FD-multiplet are not the same. So you cannot improve all the component of the supercurrent multiplet S that will be needed to be improved to 0 in order to get an SCFT at the same time. So the simplest example of such a theory is just the free chiral field with mass. That's clearly not a super conformal field theory, but it has an r symmetry. And it also, you can check, has a Ferrara-Zumino supercurrent. Yes? So will any mass deformation of an SCFT be both FZ and r? Would any mass deformation of a, so you take an SCFT and just add some masses? Well, no, because that will presumably break r. So I don't think that's true. That means you can have twisted mass. So OK. Twisted masses are for two dimension. Yeah, because there was a comment about. Yeah, I'm not sure what the comment exactly refers to. But OK, so let me. So there is a slogan that I wanted to say, is that the following, basically the more special a theory is, the more backgrounds it can be placed on. Which I think is, I mean, it does not require an explanation. Because if a theory is more, sorry, a theory which has an r-multiplet is clearly also an s-multiplet. Therefore it should be, if you solve the conditions for having supersymmetry stemming from 1616 supergravity, clearly any theory which has an s-multiplet, which is any unequal one theory, can be placed supersymmetrically on this background. So in particular, the theories which have an r-multiplet. So that means that there is a dual, if you want of this diagram, which are all the backgrounds on which you can put a super conformal field theory. This is the most special class of theories, so they have the biggest set of backgrounds on which they can be placed, preserving some supersymmetry. And then inside this there will be a smaller subset for the theories which have an f-c-multiplet and a smaller subset for the theories which never r-multiplet. And there is gonna be an intersection and in this intersection, like, well, there will be the theories which only have an s-multiplet. So actually, even if naively you would think that the equations that you get to find supersymmetric backgrounds stemming from 1616 supergravity should have more solutions just because there are more background fields that you can play with. Actually, they do have less. And this comes because of this general argument. So you don't need to solve equations to get to this conclusion. Any question? Okay, so if not, I will continue by talking about theories which instead of possessing an f-c-multiplet possess an r-multiplet. And this is the case that we are gonna spend the most of our time on, mainly because it's actually the one where the classification of supersymmetric backgrounds is more uniform and nicer. So you can repeat this classification also for theories with an f-c-multiplet, but there are more sub-cases and it's a little bit more complicated. And not as interesting. So, okay. So, again, I will remind you that, so we consider theories with an r-symmetry. So our supercurrent multiplet contains the energy momentum tensor, the supercurrent, and it also contains, as its bottom component, the conserved r-symmetry current, J nu. And then, as we have seen, it also contains some string current, C mu nu. Or, well, which is the dual of some closed two form. And so this multiplet can be coupled to a version of minimal supergravity called new minimal supergravity. Okay, so let me write here the string current which is proportional to the dual of f. So this is automatically conserved because f is closed. And so, what are the fields in the new minimal supergravity? Well, there is gonna be the metric. Well, gravitinos. And then there is gonna be an auxiliary field, which is a connection for the u1 r-symmetry and we will call this field a mu. So this is a background gauge field. So it has a gauge invariance. And then we will also have another background gauge field, which is a two form. And how do they call it in my notes? B, maybe, yes. And again, so this is a gauge. It has also some gauge invariance. B mu nu plus B mu omega nu. And these two form couples to the conserved string current. So indeed, the coupling of B mu nu and C mu nu will be gauge invariant because C mu nu is conserved. Alternatively, in actually most of what I'm gonna say, I'm not gonna use this two form gauge field, but I will instead consider the dual of its field strength, which I'm gonna call V mu. So that's gonna be, so by its definition, it's conserved. So this satisfies a conservation equation. Okay, so these are the fields in the, well, and plus fermions, I mean, there will be gravitinos that couples to the super current. Okay. It's not without it. Oh, I already put it there. Okay, sorry. Good. Okay, so to continue, I have to say that, I mean, I'm gonna make a switch of reference because I discovered that I've wrote my notes actually using the conventions of a different paper than the one I gave last time. So actually the conventions for what I'm gonna write now come from 14 or seven, two, five, nine, eight by Sir Klosset, Thomas Lumitresko, Zor Komargoske, plus myself. Okay, so let's write down what this generalized killing spinor equations are for the case of new minimal supergravity, which I think will also make it clear why these are nicer than the ones that I've wrote over there. So we have an equation for d mu zeta and this, again, has the same structure as advertised in general, so it will be some covariant derivative of the spinner is equal to some object, which is linear in the spinner and it's complex and zeta bar. Okay, but now I'm gonna write it in a better way. I over two, d mu sigma mu sigma. Okay, in the following I'll try to be to use tilders instead of bars because in Euclidean signature, the dotted spinors are not related to the undotted spinner by complex conjugation, so I will emphasize that by using tilde instead of bar. So this is one equation and then there is another equation for the right-ended spinner, zeta tilde. So here they are. So let's make some comments. So first of all, from this you see that the spinor zeta and zeta tilde, they are not just spinor, but they also have u1R charges, so they actually live in the u1R line bundle times the spin bundle. So these are left-ended spinor with charge one under the r symmetry and these are right-ended spinor with charge minus one under the r symmetry. And in particular, the charge spinner can exist on any spincy manifold, so you don't even need your manifold to be spin for these objects to be in principle well-defined. And in four dimension, any orientable manifold is spincy, so that is nice. So that's was comment number one, zeta and zeta tilde have R charges and correspondingly the super charges which correspond to zeta and zeta tilde will be, will like, so there will be, we know that there is an r symmetry and this r symmetry acts on the super charges. Okay, so that's first comment. The other comment is that with respect to the equation that I brought there in all minimal, you can see that here zeta appears by itself, there is no zeta tilde in the first equation and vice versa, there is no zeta in the second equation and so that makes things nicer. If you want, this is also due to the fact that none of the fields, well, except for the gravity, but the bosonic field in the super gravity multipet are not our charge, so there is no way of writing something here which would involve zeta tilde. Okay, so then other comments is that, I already made, but they are worth stressing is that in Euclidean space zeta and zeta tilde are not related by complex conjugation. That, and the other comment, so in particular there is no real good reason to say that a mu and v mu have to be real. Now, you, as we will see, we actually, so a mu is a connection, so it doesn't, it's not just a well defined one form, so all the examples that we have, like the imaginary part of a mu will be a well defined one form, and it's only in the real part, which is not, and hence basically we know the examples that we have, we always only have to do real U1R gauge transformations in going from patch to patch. We don't have to do, we don't have to invoke complexified U1R gauge transformations. And the other comment, which I've made yesterday, but to repeat is that because we are considering some fixed background, v mu, it's all the auxiliary fields in the supergravity, namely v mu and mu, don't need to satisfy equation of motion. So is there any question so far? Yes. This theory also brake the passive reflections, similar to the whole theory? Which theory? The one with R, my point. I mean, if you have a background with, so that depends on the specific background, then we will consider round sphere similar to the one. No, the round sphere cannot be, it's not in this class of backgrounds, da lebo je nivak nesetratnja solušnja, je. Zato if it's not spin, then you have to choose a twisted bundel where the r-charge is loaded appropriately so.. Yes, it has to be... Zeta lives in an honest bundle. Some other question? No. OK, so let's find another blackboard here. Never, nothing but here. OK, so now that, the other comment that we can make is that, from the super gravity, from the way that supercharges act in super gravity, you can, from the super gravity transformation from the algebro of the super gravity variations, you can find out what would be the algebra of the supersymmetry variations that remain. So suppose you have some solutions to these equations, then you can start varying the fields using this spin of variations parameter zeta and zeta tilde, and commute this variation and find what algebra do they realize, and so we can write down what that would be. So suppose that they have some variation, which is parametrized by a solution zeta, and some other one, which is parametrized by, well, first let's do another zeta. So let's say zeta prime, then you find out that these commute, and well, similarly for two zeta tildas, and if you have one delta zeta and one delta zeta tilda, then on the other side, well, and let's suppose that you act on some field phi of r charge r, so this is any field that you want, then this is going to be 2i times the leader derivative along a certain vector field k, with a prime, I'm going to explain what the prime means, of phi r. So, first of all, what is k? Rk is what you would imagine, it's zeta sigma mu zeta tilda, and you can actually check for your, for your edification that indeed, if zeta and zeta tilda satisfy those equations for some v and a, then this vector is killing, so this is a nice, is a translation along some isometry of the manifold, and then I have to still explain what this prime means, so the prime is there because the leader derivative is twisted by the u and r gauge field, so that means that the leader derivative prime k of some field phi of r charge r is equal to the usual leader derivative, but then there is a piece, which is just what you would imagine, times phi r, but not quite because of the convention that I am following here, there is also 3 half of v mu. Okay, this is not, this is just a question of definitions, because you could always redefine the u and r gauge field by absorbing a factor of minus 3 half of v mu, v mu is a well defined one form, so there is no problem, times what? k, no, that's v, it's the other background field, the conserved vector. Oh, the indices, oh, oh, times, yeah, yeah, sure, yes, that's obvious to be there, great, so again, this, I mean, you can expect, and actually it's not just that, I mean, if my fields are charged under some gauge group, then on the right hand side there could be appropriate gauge transformations, and so on and so forth. So, okay, good, so now we can write down what the Lagrangian of some theory would look like, so what is the coupling of some flat space theory to this background, so again, so if you start with some theory with some Lagrangian, then in order to find the Lagrangian of the theory in some supergravity background, so either you open your supergravity book, well, that's what you have to do, and then just look at how this thing is made, but I will just make some comments, so you can divide these into three pieces, the first one is just the original flat space Lagrangian minimally coupled to the metric, then there is a second set of terms, which we called L1, so these include couplings of the, of various operators in the arc current multiplet to the, to V mu and A mu, so again, so what can this be, so this will be in particular, there will be a coupling of the conserved arc symmetry to the arc gauge field A mu and then there is also a coupling, as we discussed, of the, of this gauge field, the gauge formed V mu nu to the conserved string current C mu nu, or alternatively, we could write this as this field V mu, which is conserved, which couples to some connection giving rise to C mu, so this twidled A mu does not have to be well defined, it could shift by the mu of something, but because V mu is conserved, it's not a problem. Okay, then finally there are single terms and this we called L2 and these include couplings to curvature and also to products of two of these auxiliary fields. So this is consistent with dimensional analysis, where you find that the background gauge field V and A, or B and A have dimension 1, so they scale like 1 over r, where r is some overall scale of your manifold, overall radius, while the curvature scales like 1 over r squared. Okay, so I'm actually gonna write down some explicit Lagrangian so that you can see these actually born out in practice. Okay, so indeed let's write down an example. What did I have here? Oh yes, I can get rid of this. Okay, so let's consider chiral fields. Let's consider a chiral field of charge chiral field with r charge r. Okay, this might be confusing if I also use r for the size of the manifold. It shouldn't be a big deal, but I could call this q. I don't like that, keep it r. So we can write down what the supersymmetry variations will be in some putative background. So the chiral field, again, this should be known that there is a complex scalar phi, a vile fermion psi, and an auxiliary field f, which is again a complex scalar. And the variations are gonna be delta z of phi is square root of 2 times z of psi. So that's exactly of the same form as in flat space, and indeed you could have guessed this because it doesn't involve any derivatives. Then we have delta psi, delta z of psi. So this is square root of 2 zeta times f. Again, this is the same as in flat space. And then you have, I'm clearly missing some, okay. So there is this, where this covariant derivatives means that the derivative is covariantized in respect to the U and R gauge field. But actually to be more exact, d mu of some field phi with r charge r is, well, you know what it's gonna be. It's gonna be the same as it brought somewhere else, I guess. It disappeared. Anyway, it's gonna be the covariant derivative times minus i r v mu. And was it plus, yeah, plus 3 half, no, a mu plus 3 half v mu times phi. So there is a reason, actually, why this particular choice has been made. And that's because this A is actually the field which couples to the, so if the theory were super conformal, then that field will couple to the super conformal r symmetry. Okay, so that's the, so you get this extra piece. So here you see that there is some deformation of the variations in flat space and that's just obtained by covalentizing the derivatives respect to the U and R connection. And finally, we have to write down the variation of F. And this is gonna be, so this is actually somewhat different than the one, this I can write d mu zeta tilde, sigma tilde mu psi. Okay, so here you see, again, that there are several differences with respect to the one in flat space. First of all, the derivatives are covalentized, but then here when you expand this, you also get a factor where the covariant derivative acts on the spinner. And then if you want to be explicit, you have to replace this using the equation over there. And then you will get a bunch of terms. Okay, so the variations are modified with respect to the one in flat space, but again, the modification dies off as the radius of the manifold becomes larger and larger because the fields A mu and V mu scale like one over the radius of the manifold. Should you have another correction on the covariant derivative? Sorry, should I have what? Correction to the variation of the formula. No, that's it. The correction is inside the covariant derivative. Okay, any more questions? Is there an intuition for why the deta tilde has to be inside in the variation of F, but not in the formula? No, this is just that it's particularly nice to write it this way. I don't have an intuition why that happens. Maybe somebody can come up this, but it's... Yeah, I don't know. But it is true that this is nice. But you could just... Yeah, I could have written the more complicated one. Okay, so now let's see some example of some supersymmetric Lagrangian. So, for instance, we can take... You have already set to zero the gravity, right? Yes, yes. This is assuming that there is a solution to those equations. Then this would be the supersymmetry variation in the background corresponding to this solution. Okay, so our Lagrangian will be... I have to be a little bit careful because I made a little bit of a mess in my notes. In the usual term with the auxiliary fields. I could have put these two last pieces together, but... Okay, so this should be it. So, let's make some comments. So, first of all, this has the general structure, advertised some... Is it here? Yeah, okay. So, there are terms, which are just the minimal coupling to gravity. Then there are terms of order 1 over r, which include the coupling of the auxiliary fields in the supergravity multiplet to various currents in the theory. So, in particular, from this Lagrangian you can read off what is the operator which couples to V mu and what is the conserved arc current. I guess that's not particularly interesting. And check that that is indeed what you get from the r-multiplet for a free chiral field. And then there are terms of order 1 over r squared, which depend on curvature and on squares of the auxiliary fields. And notice that all these terms do depend on the value of the r-charge. So, the formation of the theory does depend on the r-charge of the field that you are considering. So, now, one thing that you can notice is that let's suppose that we consider some value of the r-charge which corresponds to, well, the super conformal value of the r-charge which is two-thirds. If the r-charge of a chiral field is equal to two-thirds, then you can have a cubic superpotential. Then we can check that indeed this will just correspond to do the conformal coupling. So, let's see if that's indeed correct. Well, if r is equal to two-thirds, this term disappears. So, this goes away. Then this term becomes r over 6, which is indeed the conformally the value of the coupling to the rich scalar, which is appropriate for the conformal coupling. And then you have to check what happens with this covariant derivatives. So, what is this? So, this will be d mu minus i r times a mu plus 3 half v mu. And then there is plus... No, plus that. And then there is minus i v mu. So, indeed, if r is equal to two-thirds, this v mu cancels with this other v mu. And what you are left is the coupling to this mu, which as advertised is the object which couples to the super conformal arc current. So, this coupling would expect even in conformal supergravity. So, indeed, the coupling to v mu, which is the operator which has to disappear, does disappear. And you can also check this for the fermion that they leave you as an exercise. So, is there any question so far? So, the theory with a cubic superponential is classically supercomboment. Well, okay, this is a free. We are considering a free chiral mode. Yeah, but you set the arc to third. And I thought the motivation was... Yeah, so, that is the value which would correspond to... So, you would expect for that value to be able to couple the theory to conformal supergravity. B, because the superpotential... Because for that particular value you could add a superpotential, which is... So, suppose I take article one, then I would be able to add a mass. That is not conformal value for the arc charge. If you include quantum effect, the theory is not really supercomboment. If I include what? Quantum effect. Oh, yeah, yeah. And then there is a question of... Yeah, these are just classical considerations. So, presumably in the quantum theory, if the theory is not conformal because of quantum effects, then the coupling to V mu will reappear at one loop or higher loops. Okay, so, now, are there any comments or questions? I don't like to comment. So, really, was there a question, I think, about why the variation of past is so derivative? Was there such a question? Yes. So, I don't have a good reason. It's just easy to write it that way. But isn't there intuition that generally when you want to write in Lagrangian, you require that this... I thought it was a general principle that the variation of the top control in the curve is supercomboment. Oh, okay. Yeah, that may be true. Yeah, that may be true. Because I'm sure that when you write after a contribution, it's still Lagrangian that they're super symmetric. Right, except that I'm not... It's a little... Yeah, it's... Well, but, okay, you can still think. I mean, maybe, like, if you can do all this in, like, actual curved superspace, you might be able to, like, show that that is the case. Okay. So, I would want... How much time do I have? I have another half an hour, 40 minutes. Okay. So, okay, so I will want to give an example before going into more formal considerations. And so this example will actually be considered by Terashim and his lectures. So, I can be glib, maybe. Or, hopefully, I will say things that do not completely overlap with what he's gonna say. So, the example I want to take into consideration is that of S3 times R. So, that's a cylinder. So, again, if you have a super conformal theory, then it's obvious that you can couple it, you can put it in a cylinder preserving supersymmetry because that's just a conformal mapping. But here we are trying to find if this background is supersymmetric in the more less restrictive set of theories, which just have a U1R charge. Okay. So, the isometry the isometry of the cylinder, well, okay, there is the rotation of S3, that's an SO4. So, there is SU2 left times SU2 right. So, these are the two rotations of the S3, not to be confused with the frame rotations. I don't know, we can call them S2 plus and S2 minus, which corresponds to dotted So, there are two different set of SU2s. And then, well, then there is R corresponding to the translation along the axis on the cylinder. Okay. So, we would like to find a background which is as symmetric as possible. So, what we are going to guess that because we don't have general procedure yet, is that the auxiliary fields are going to be just along the axis of the cylinder, which put a coordinate tau along the axis of the cylinder. So, v is going to be little v times d tau and a is going to be little a times d tau. Okay. So, now we can write down what those equations look like. And, okay, so, if I've not done any mistake, okay, so, I choose a frame which is adapted to the cylinder. So, that means that one of the frames is just d tau. I mean, you could make a bad choice and then your equation would be look very ugly, but you don't want to do that. And there is a similar equation with the opposite signs for zeta tilde. So, this is along the time direction and then you have along the s3 you have some equation which looks like this and also here. Okay. So, these are the equations that you have to solve. So, now I'm not going to solve them, but for exercise you can do it. But, so, the solution to this is, okay, let's forget about this equation for a moment. Let's just look at this equation. So, well, if you are familiar with this kind of equation it's clear that the solution is, you're going to find solutions for v equals plus or minus i over r. And this correspond to s2 right and s2 left invariant spinor on the sphere. Okay. So, let's choose a sign, minus. And so, the solutions are going to be spinors zeta and zeta tilde that are invariant under s2r. And vice versa, if I choose the other sign then they will be invariant under s2 left. So, some comment noticed that the sign here in this equation is the same as the sign in this other equation. So, both the zeta and the zeta tilde, which is undotted indices and the other one is dotted indices but they are both s2r invariant. So, indeed, these two s2s have nothing to do with the other two. And, okay, so once you, once you have said what, so this you can check for exercise. It's actually pretty easy. You just have to choose a, I mean, if you choose the right frame on s3 then it's easy. You have to choose s2r invariant frame and then these equations become very nice. Okay, so now let's look at this other equation and here it's clear that I have possible, different possible choices because any constant value for a will, like, give rise to a solution for zeta. It just changes how zeta depends on the Euclidean time along the cylinder. Okay, so, for instance, I can choose a equals 0 and then if I choose a equals 0 I find that commutator of h, which is the generators of translations along the tau with supercharge generated by zeta, let's call it qzeta, well, that will be equal to a half of q. So, this is the case that corresponds to the conformal coupling of the theory to the cylinder. Indeed, as we said before, if for a conformal field theory all the couplings disappear except the 1 to a, so if you set a to 0 then that's exactly the conformal coupling that you know and love. So, this is the conformal coupling. However, this is not maybe the nicest value to consider. It's nice to take a to be equal to i over 2. So, these are r is actually the radius of the s3. I guess, maybe I should use this. Let's call it l. And then with this choice h commutes with q and with q bar. So, an important comment is that I really need the u1r symmetry to do this. Actually, if you look at this equation over there and then you look at the old minimal equations that I've wrote in the last lecture or even at the beginning of this lecture, you might wonder that actually you can obtain the same by just you can obtain equations which look very much like this by just setting m to 0 and just adding b mu in the old minimal equations. So, in old minimal you can indeed find a solution on the cylinder exactly in the same way. But the difference there is that you don't have the freedom of adding the background gauge field for the u1r symmetry and therefore you are stuck with some commutator between the time translations and the supercharges. So, in particular it is impossible in old minimal to add this background u1r gauge field in the same way that the supercharges are independent of time and when they are independent of time then it's possible to compactify this cylinder on, instead of being s3 times r it would be s3 times s1 preserving supersymmetry. Well, if the supercharges are independent of time this is clearly possible. So, that means that s3 times s1 in the compactified case is only possible for theories which do have a u1r symmetry. So, this requires u1r. So, in particular say, we take superium mills. This has no u1r symmetry because it's broken by an anomaly and therefore it cannot be placed on s3 times s1 preserving supersymmetry. So, you're saying that Römeres-Berger index rekevajas arktuči. Yes. So, in particular if you were to try to compute the Römeres-Berger index for superium mills I think you will find something but the something cannot be interpreted as an index. Like, the coefficients would just I don't know either be like negative or like not even integers. I actually don't know which of these things happens but I'm sure one of these things will happen. Okay. So, okay. So, now let me write down a little bit better what the what the superalgebra that results from this actually is hopefully still slightly visible. So, I'm going to be sketchy but you can check all these statements by yourself. It's not too complicated. So, let me introduce the following notations. So, we have supercharges q bar. Those are the ones which come from the zeta tildes and that they have this index A which is an index under s2L the left part of s2. So, this zetas are invariant under s2R and they transform a spinors under s2L. So, the corresponding supercharges will also be s2R invariant but they will transform a spinors under s2L. So, this A is an s2L index and then we'll have qB's again also the q's transform under s2L and this commutator ends up being h minus R over L times delta IB plus q over L jAB where jAB are the generators of s2L and as we said before qA the q's will commute among themselves and the same is the same will be true for the q bars and ok I already said that the q bars and the q's transform doublets under s2L but let me skip this one so then there is the R charge and so you can also check that the supercharges have R charges so this is for q and then there is another one for q bar with the opposite sign then as advertised we will consider the case where h commutes with q and it also commutes with q bar you can check that indeed the value of A which makes one of the the A minus V disappear there it also works for the other one ok so the rest of the commutation relations have the one that you would know and love for s2L so we can define j3 to be equal to j12 and then jplus will be equal to j11 and j minus is going to be i times j22 and ok so then you have the usual commutation relations j3 with jplus minus is equal to plus or minus jplus minus and the commutator of jplus with j minus is equal to 2j3 and then you have commutators with the supercharges I'm not going to write these but for instance jplus with q2 is equal to minus iq1 and j3 with q1 is equal to q1 with q2 it will have the opposite sign and jplus with q1 is equal to 0 et cetera so these are the saying that q is a doublet under s2 left so what is this superalgebra? this superalgebra has a name and it's s2 slash 1 but it is centrally extended by h h indeed commutes with everything and it just appears in the anti-commutator of q with q bar so if you consider the entire theory supersymmetric filter that you will write on the cylinder this is going to have a symmetry group which includes s2 slash 1 but then it also has a u1 which corresponds to the translations along the cylinder or well if you are in the non-compact case I guess to br and then there is another symmetry which is there for free which is s2r because everything is independent of it no supercharge there are all singlets and the rest you are ok, so is there yes? wait so in the product there um yeah, this is not quite yeah yeah, this thing yeah it's like a symmetry yeah, it's a symmetry actually it is a product provided you do the right definition of h and r yeah, ok so if it commutes with q you can actually use the r symmetry r minus lh or something yes sure and the rest will be there yeah it is an actual problem but ok it's clear what I mean um so ok, so now the the other thing that you can check is that if you were to put the theory in Lorentzian sej, so at the time instead of being will be Lorentzian then you could you would like to check that the theory is actually unitary and it is like because when you change the um when you change signature like this uh d and a will pick a factor of i so they will get real and then indeed uh you can check what the hermiticity conditions on the supercharges are gonna be and so the hermiticity conditions on the supercharger is follow from the Euclidean continuation to uh Lorentzian signature are just what you would expect which is that q bar a is the dagger of q a so now we can study the representation theory of this object are there any questions before I get into that? ok, so if not let's see what I can erase maybe I'll pull down this very upper so the generator of that you want is h yeah so then if you say it's a direct product then it's not a central expansion just no, what I'm saying is that h enters into the commutator of the q's that's not sorry, but that's not a representation relation of s2 s2 s1 that's centrally extended by h yeah it's not a product that's ok, let's just erase it is a product that it's abstract it's abstract so if I call it artyndis r-lh and then you can call the u1 generator whatever you want it is a direct product s2 s1 has different commutation relations it doesn't have h it doesn't have h that's the difference it's centrally extended yes I agree I should erase this line but the commutation relations are correct hopefully ok so now so let's study the representation of these guys so because of the because the q-bars anticommute we can just choose highest weight state which is annihilated by all the q-bars so let's consider some state which has spin j and r-charge r this is maybe not the best so and let's suppose that this thing is annihilated by all the q-bars so by q-bar 1 and 2 but then from the commutation relation of the q-bars with the s2 left generators you can see that a commutator of a q-bar with j will only produce a q-bar so that means that this commutation is going to be true for all the states in the multiplet that you can obtain by acting with s2 on this on this state so I can just take this to be the highest weight state and then all the other ones are going to be obtained by acting with s2 left so you have a bunch of states already annihilated by the q-bars and this will form an entire multiplet under under s2 left so let's call this object j which j is the top spin component and r charge r so for instance we have j plus on j r so maybe this will call m and then this will be 0 and then all the other ones are obtained by descending and this will form this part of the multiplet so far I have not used the q's yet so now I can ask what happens if I take this multiplet and I act on it with q how do you define how are you defining that you're imposing those equations but how are you defining this bar j of r is like so you take a state which is annihilated by all the q-bars so those will lie in let's take some state so now like the comment is that because of the commutation relations with s2 left then these states will lie in complete multiplets of s2 left ok so then I can just choose one with a stop spin j so we start with that state and then the other ones here are obtained by acting on this state by s2 by j minus ok so far I haven't used the q's yet so now I can act with q so what happens is this multiplet with q so then I will obtain 2 complete multiplets 1 which has corresponds to j plus a half and r minus 1 because acting with q lowers the r charge by 1 but q has spin half so I will get 2 I also get the one with j minus a half and r minus 1 so actually I get both of I don't get the second multiplet if j is 0 so this other multiplet I only get if j is greater or equal than a half if j is 0 I only get the top one just by following additions of angular momenta ok and then I can act with q again and that will be it because I only have q1 and q2 so I can act with both of them and that gives me another complete multiplet but with at spin j but with r charge r minus 2 so that's the structure of a generic multiplet of this of this superalgebra now let's so now we can compute the norm of the highest weight states and to check conditions that these charges have to satisfy in order for the multiplet to actually exist so so for instance we can take so what is the highest weight state inside here so the state with top with highest spin so that is clearly obtained acting with q1 on the top guy here so I can consider q1 acting on j r where j is the state which is annihilated by j plus and then I can compute its norm so this you can do for exercise and what you find out is that this norm is positive or equal to 0 if h is greater or equal than r minus 2j then you can take the highest state inside here so that's a little bit more complicated because you have to construct it so now it's I mean it's gonna have a piece which is gonna be given by q2 acting on j r but then this I mean you have to work a little bit harder to find out that in order to find the state here which is annihilated by j plus you have to add a piece with the following where q1 acts on j minus 1 r so here you have to take also the state which is the lowest which is obtained acting on j with j minus on j r and then you have to act on that with q1 so that gives the highest weight state here and then you can so let's call this guy psi so the claim is that j plus psi is 0 and now you can compute the norm of psi and impose that it's greater or equal than 0 and you find the condition that h has to be greater or equal than r plus 2 j plus 2 very good so then the other thing that you can check is that the last guy here has also positive norm so that's easy because that's obtained by acting with q1 and q2 on j r so just check that the norm of this thing is positive and that gives a further condition which when folded in this one just ah, ok there is I have to make a comment so this condition is clearly there only when this multiplet exists so this is only there for j greater or equal than a half so now you have to consider this guy and when you fold it together you find that it gives the same I mean and you intersect that you find that you get the same condition but now it's also true for j equals 0 so again this ok so to be precise this gives either h less or equal than r minus 2 j which however like is not compatible with the rest or h greater or equal than h but now this is also fine for j equals 0 ok so now you have found what the conditions are for the norms to be positive so we can study the possible multiplets that that to arise so what are these so you have long multiplets so the long multiplets are there whenever h is greater than r plus 2 j plus 2 ok so if j is greater than equal to a half then the structure of the long multiplet is what we have above so there is this plus j plus or minus a half at with r charge r minus 1 and then there will be j with r minus 2 but there is also a long multiplet when j is equal to 0 and then you only have j r and j plus a half at r minus 1 sorry j is equal to 0 so we can just put 0 here a half and 0 at r minus 2 so these are the long multiplets so we can call this long j and then we have short multiplets and these arise when the norm of these states become 0 so when you iterate the bound so for j ok, so let's see for j greater than a half you get that h sorry yeah h is equal to r plus 2 j plus 2 and then the multiplet contains j r and j plus a half r minus 1 but not the rest and instead for j equals 0 then h will be equal to 2 plus r and the multiplet contains 0 r and a half r minus 1 so we call these multiplets which have the same structure s j standing for short j but then there is one final very short multiplet which is just a singleton so this happens for j equals r equals 0 and then the multiplet is just one state and we will call this s at 5 minutes oh 12 30 alright ok ok, well I think I can do it in 5 minutes so we have the structure of the short multiplets so now we can study how this how it can be that short multiplet recombine into long multiplets so suppose you have a long multiplet at spin j and now h goes to r plus 2 j plus 2 from above then as the long multiplet hits the bound it splits into short multiplets and it becomes s j plus s j minus a half with the with the corresponding r charges that you can find out because they are related and and in the same way when you have l zero sorry so this was for this was for j greater or equal than a half and then for the zero case when h goes from above to 2 plus r then this splits into s zero plus the singleton so you would want to construct some quantity which counts the short multiplets up to the combination so that is some chance of being an index so let's build up the most general quantity we can imagine so this will be a sum over all the spins of some pre-factors which can depend on the spin time the number of short multiplets spin j and then we can add some other factor let's call it beta times the number of singletons sorry but in order for the singleton to be contained in l zero you need particles to pool but the singleton has zero or not sure yeah but the singleton will be the singleton will be the state here right so it will have zero charge so it's inside this multiplet? yeah it's so this multiplet splits into these two so that this thing is the so the long multiplet which okay so this is this long multiplet the one at spin equals zero it contains three pieces so when r becomes equal to two like this two become a short multiplet and this becomes the singleton yeah but your condition is that of r being two yes it does h has to oh what do they do here no okay so right right so but but the singleton will have the r corresponding to whatever it has to be yeah yeah they can have no zero h oops so okay so now let's see what conditions so we want this to be invariant under these recombinations so the first one tells us that alpha j plus alpha j minus a half better be zero and the second one tells us that alpha zero plus beta better also be zero and so the index that we get is we can raise this put minus one to the two j plus one and this we can just put to be one and that would be our index and you can check that this is the same as the trace of minus one to the two j three or also called the trace of minus one to the f on your multiplets exponent sum over j minus one minus one to the two j three or which is the same type two j three or which is the same the trace of minus one to the f well you can define f to be this okay so now I'm done but we'll continue with comments about what this means in the next lecture