 Thank you very much. So as the title says, I will talk about 4D N equals 2 supersymmetric theories on S4. And then I'll talk a bit about our localization. But the bulk of today will be to put an N equals 2 theory on the force sphere. So 4D N equals 2 supersymmetric theory is, of course, just a quantum field theory, which is invariant under the N equals 2 superpoincare algebra. So the superpoincare algebra is generated by the usual momentum generator, where in my notation, M is a space index. So I'm already in Euclidean signature. So I label the space indices by 1 to 4. There is also the usual rotational generators. And there are additionally eight real supercharges, which are organized in two spinners, which carry an additional index i, which runs over 1 and 2. So alpha, alpha dot are the standard spinners. Although, again, Euclidean signatures, so alpha and alpha dot, they're really independent. They're rotated by independent SU2s. And these two are not quite complex conjugates. Instead, each supercharge separately satisfies some symplectic reality property. And to be actually precise, there is also a central charge and its complex conscience. So this is the set of generators. And their algebra for this is just a standard poincare algebra. So let me indicate what the algebra is of these supercharges. It's just something as an extension of the n equals 1 poincare algebra, where we have an anticommutator that looks like so. But I mentioned there is also the central charges. They're somewhat similar to the central charge that appeared this morning. But they don't carry extra indices, so they really do act on point operators, on local operators instead of on extended objects. And they appear in the anticommutation relations of supercharges with themselves, or at least Q with Q instead of Q with Q tilde. And similarly for Q tilde, where you have a complex conjugate sitting over here. And the central object, given that the name already indicates that it's central, it commutes with anything. So this is the superpoincare algebra in four dimensions. And for most of my talk, I will really not care about the central charge. I will mostly just set them to 0. But just know that, in principle, they're there. Now, when the central charges are 0, you can easily see that this algebra has an automorphism acting on it. Namely, these indices i and j, they can be acted on by SU2R rotations. So I already indicated that we will call this R symmetry. And Q and Q tilde can be acted oppositely by a U1R symmetry. So this superpoincare algebra really has an SU2R times U1R automorphism, which is the R symmetry of an N equals 2 theory. But again, when Z is present, you immediately see that the U1R symmetry is really not present anymore. So you would see you have the SU2 still, or does it break the whole thing? No, you can see that. This is an invariant answer on the SU2. So we have an algebra. Now, let's look at which multiplets we can consider in this algebra. In particular, I will just consider the massless multiplets. In fact, the massless matter multiplets, which are relevant to built Lagrangians. And just as a side comment, as soon as you restrict yourself to considering massless multiplets, you necessarily have to set Z and Z bar equals to 0. It's a consequence of the fact that you're massless. So there's two such multiplets. First of all, we have the vector multiplet. Its field content is, of course, the gauge field. It contains one complex scalar, so phi in its complex conjugate. And it contains a bunch of k genie. So the k genie, they are rotated by this SU2R symmetry. The phi and phi tilde, they carry charge under the U1R. And A is, of course, neutral under it all. And given that this is a vector multiplet, I'm considering this all to live in the adjoint representation of whatever gauge group I would like to consider. Now, in fact, if I want to consider an off-shell version of the vector multiplet, I should add an SU2R triplet of auxiliary fields. And you can see that this multiplet is actually nothing else than a copy of an n equals 1 vector multiplet and an n equals 1 adjoint kara multiplet. So these guys and half of the fermions is the kara multiplets. These guys and the other half of the fermions together, well, together with one field here is the vector multiplet. And the two other guys actually sit in the off-shell kara multiplet. So that's one massless multiplet to build Lagrangian's width. And the other one goes under the name of the hyper-multiplet. And in fact, let me immediately consider R, such hyper-multiplets. Then the field content is an SU2R doublet scalar fields, which also carries an index A, which I'll explain in a second. And there are two fermionic partners. So this index A is a USP2R index, because it's just a fact of life that our hyper-multiplets carry in the USP2R flavor symmetry. Let's unpack this a little bit. So notice also that I just restrict myself to an on-shell representation here for the time being. So let's unpack this a little bit. Let's just look at a single hyper-multiplet. A single hyper-multiplet, in fact, is nothing else than two n equals 1 kara multiplets. And let me denote these n equals 1 kara multiplets in super-field language as q and q tilde. But in abuse of notation, I will also use these notations for the scalars themselves. Then you can organize the scalars in these two n equals 1 kara multiplets, so q and q tilde, and their complex conjugates in a matrix like so, where in this matrix the rows are exchanged by the SU2R symmetry, and the columns are exchanged by the USP2 flavor symmetry. So another way of saying it, we really had in two n equals 1 kara multiplets, we had four real scalars. In principle, you can think that they are acted on by an SU4 symmetry, and SU4 is, of course, nothing else than SU2 times SU2. One of these SU2s acts as an R symmetry. The other SU2 is isomorphic to USP2, which acts as a flavor symmetry. But more generally, these USP2 gets enhanced to USP2R. And given such a collection of hypermultiplets with this type of flavor symmetry, which acts as the index here indicates, we can try to gauge some subgroup of this flavor symmetry by identifying it with the gauge symmetry that is carried by the vector multiplets. If you do so, then however you embed the gauge group in this flavor symmetry group, there will be some kind of a commutant, and that commutant will be the flavor symmetry of the remaining of the theory that we're actually looking at. So more concretely, let's imagine that we have n hypermultiplets transforming in some representation R of the gauge group, then what is the remaining flavor symmetry? It depends on the properties of this representation R. So it's really an exercise of embedding things in USP2R, but let me immediately give the answer. If R is complex, then these things carry a flavor symmetry, SUn. If R is real, then there is a flavor symmetry of USP2n. And if R is pseudo-real, there is a flavor symmetry, SO2n. So I went through the three cases, because it will be somewhat important in later talks. So it will be given to have the three cases in mind. So maybe you can do this as a little exercise to show that this is indeed the correct flavor symmetry, depending on the properties of this representation R. Maybe let me do one example of this. Imagine that you have one adjoint hypermultiplet. Then according to this statement, the adjoint representation is always real. We expect, since we only have one of them, a USP2 flavor symmetry. And indeed, that USP2 flavor symmetry is simply this flavor symmetry that is depicted here. Q and Q tilde, and well, all the fields here transform the same real representation. Q star in principle would be the complex conjugate representation, but since the adjoint representation is real, this flavor symmetry is not broken by introducing the coupling to the gauge field. So this USP2 is really the flavor symmetry of a single adjoint hypermultiplet, which means that if you have an Ennecos IV theory described as an Ennecos II theory, you really have an SU2 or a USP2 flavor symmetry. So good, we know now what the algebra is. We know what the massless multiplets are of that symmetry algebra. And now I would like to describe how the algebra acts on these multiplets. But to save me some writing, let me just introduce the notations I'm using. So we had over there Q i alpha. And when I write the object delta, I really mean that I have some type of a keeling spinner, which contracts with the supercharge. And similarly, for the Q tilde, I will have a situation like this, where in this talk, and presumably in all localization talks, you will hear psi is not treated as a fermionic object, but I keep it bosonic. So maybe that is not standard, but that's what I will do. So delta is still a fermionic object. And this delta acts on these multiplets in the following way. So there's no need to copy these all. This is just to indicate how things transform. At one point, we will need some of these transformation rules. So I will just keep these things on the blackboards. If you would like to play with these transformation rules themselves, I just copied them from a paper of Hamai and Hosomichi. So good. We have multiplets. We have transformation rules. Let's now consider actions. So the Young Mills action starts off, of course, in the standard way. We have the kinetic term for the gauge fields, the usual f squared. We have a kinetic term for the two scalars over there. The auxiliary fields enter quadratically. We have some commutator term among the scalars. And then we have a bunch of fermionic stuff, which I will not quite write. And similarly, you can construct an action for the hypermultiplet. Again, you can read the concrete expression in this paper I mentioned by Hamai and Hosomichi. Now what I would like to point out is that in the variations I wrote, I added two somewhat special terms. Or maybe I haven't added it yet over here. I want to add here a somewhat funny term. Because at the moment, we're just doing Poincare's super symmetry in flat space. So these spinners over here, they're just constant. They just label which supercharges am I talking about? They're just constants which tell you which combination of supercharges. But now I'm adding a derivative of that parameter, which really is just 0 for the time being, since I'm just talking about super Poincare. So the derivative is 0. Here is a similar term. And over there, I already included them. Now the reason I included these derivative terms is the following. If we were to take this action and we try to transform it, say we take the Young-Mills action, we try to transform it with respect to these transformation rules. Of course, it's an invariant action. So at worst, it should become a total derivative. Which then, if you integrate it to form an action, you actually get 0 up to boundary terms, which we ignore. But let me now try to compute the nurture currents associated to these super symmetry variations. So the standard procedure to compute nurture currents is to take the transformation rules under the symmetry we would like to consider. But consider position-dependent parameters. If you take the parameters to be instead of constant position dependent, you should expect that these rules change a bit. You will pick up transformations terms in the transformation of the Lagrangian, which are proportional to derivatives of the parameter I'm considering. So generally, you should consider, you should expect that the transformation will look something like this, where I pick up terms proportional to this now position-dependent transformation parameters. So again, if you made them constant, we got invariance. Now I take a position dependence. I pick up derivatives, and the object that multiplies these derivatives is the nurture current for the symmetry we're considering. So these are really the super currents which already featured this morning in Guido's talks. OK, this is all fine. But now you can sort of start feeling why I felt like introducing these extra terms containing derivatives in the transformation rules. I introduced them such that the terms that pick up proportional to the derivatives have a special property. These currents will have a special property. And a special property is, well, first of all, the not-special property of these currents is that they are conserved on shell. These are just the consequence of the fact that they're nurture currents. But the special property, which appears as soon as I add these somewhat empty of meaning terms from the point of view of the Poincare super algebra, as soon as I add those terms, the super currents all of a sudden have a second property, namely, that classically, their sigma traces are 0. So by adding these extra little terms proportional to derivatives in the killing spinners, we found that the super current that appears here has an extra property. Now, this extra property is actually quite useful, because now you see that if we give a very special position dependence to these objects that sit over here, we're going to find the invariance of the Lagrangian or of the action even with position dependence. And that extra special property is given that the sigma traces of j is equal to 0. If this object were actually proportional to a sigma matrix, we would find that this object is 0. So if the derivative of the killing spinner is, in fact, let me introduce some extra minus i's just because people like these conventions, if I choose a position dependence of the spinners such that this property holds, then you plug that in here. You use that. Well, you just throw the sigma matrix to the other side. And then you find sigma times j, which is 0. And similarly, for the Tilda spinners, if you demand that a property like this holds, you can say something similar over here. So that's interesting. We started off with actions over here, which we designed to be invariant under super prime carrier symmetry, invariant under these super prime carrier super variations. When we add these extra terms, proportional to the derivatives, we all of a sudden find that there is more supersymmetry in town that we had hoped for. Not just constants over here, but also position dependent objects as long as they satisfy this equation. So these equations go on the name of conformal killing spinner equations. And you will hear a lot more about these types of equations from Guido. But I've derived them here in a somewhat simple fashion. So the solutions to these equations on flat space are particularly simple. Of course, we still find, or I should have mentioned, these prime spinners, they're just all the arbitrary spinners. They can be anything, because the only property I really need is that there is a sigma matrix here to use this property to make this 0. So these prime spinners are anything, but of course, they're not really anything. If I contract this with a sigma matrix, I find that this really is sigma u down u xi i up to a factor of 4 and some factors of i. But fine, in principle, you can take them anything they need to be to satisfy this equation. And the solutions then are just this constant which we started off with. Of course, we didn't lose the constant solution. So this is some constant plus a manifestly position dependent piece. So we find a solution like this, where both xi hat and xi tilde hat are constants on the left here. No. I'm solving for xi. And this guy can be whatever it needs to be. So in this case, it would be this object again. So and similarly, I can solve for xi tilde. It takes a similar expression. There's two constants and an x sigma piece in between. What is the character that you've written between the sigma and the minus i? An x. It's position. x m. And the hat of the sigma tilde? That's just to indicate that it's a constant. Oh, where? On the sigma tilde? On the xi tilde. Oh, sorry, sorry. Yeah, xi tilde, the hat is just to indicate that it's a constant. Maybe I should have done it all along. So these were also constants, if you like. So stupid question, but sigma matrices are always invertible because they have got non-zero g values, yes? So I can always find such xi prime to satisfy this equation. Sure, you're saying that xi prime is really something like up to factors. Sure, that's good. The point is that the different derivatives of xi i are related to each other because you have to be all equal up to a multiplication by some sigma matrix. Exactly. So you see the index, yeah. So the other solution simply looks like something like this. So we already understood these constant pieces of these solutions. These constants just describe my Poincare super symmetry with which I started off. But these additional constants seem to add another eight fermionic symmetries to my system. So the question is, what are these eight additional symmetries? And the answer is, as already indicated by the C, they're conformal supersymmetries. So we're discovering slowly superconformal invariance of four-dimensional n equals 2 theories, at least classically. So I should maybe re-emphasize this. All the words I'm saying here, they're classical words. As soon as you have quantum effects, these kinds of equations may or may not hold still. Well, this equation in particular, this one will still hold, but this one may be violated as soon as you have a quantum mechanic or as soon as you allow for quantum effects. So we slowly discover classical superconformal symmetries of n equals 2 theories. So let's explore that a little bit more. Let's explore the additional symmetries by simply trying to compute the supersymmetry variation squared. So let's compute the algebra of the supersymmetry variations. But now take into account that these parameters can be position dependence as the solution indicates over there. And in fact, I have explicit derivatives in the variation rules, which will talk to the position dependence. So let's, for example, take that multiplet and you just act twice with a supersymmetry variation parameterized by some psi and some psi tilde, which you can think of as explicitly looking like the solution I wrote over there. If you do that, what will you find? Now, if these things were constant, we should, of course, recover the superpancare algebra, which I just erased. So if these things had been constant or were still constant, we would have expected abstractly that the square of the supersymmetry variation is just a lead derivative in the direction of a vector v, where this v is a particular bilinear in the killing speeders. So given these two killing speeders, which parameterize my supercharge as over here, if they were still constant, it would be like over there. Then I find a particular vector, which, of course, will just be the translations in the four directions. As indicated by the algebra, I delete it over here. But now I have this position dependence. So this thing actually picks up more types of transformations. The first transformation that you additionally find is a scale transformation with some parameter w, which I will not specify in detail. There will be a u and r transformation with some parameter theta, where both w and theta are, again, some bilinears in the killing speeders. In fact, they also contain the derivative of the killing speeders. So indeed, they're 0. As soon as I go back to my old case of superpancare, you will find SU2R transformations. And of course, as always, if you don't act on gauge invariant stuff, you also find the gauge transformation. So again, these parameters, all three of them, they are parameterized in terms of sine xatilda and their derivatives. And they are 0 as soon as you set to 0 the derivatives. So this is the general transformation or the general algebra given this position dependent sine xatilda. And in particular, I'd like to focus on this object, this vector over here. The first property that you can easily verify given my conformal spinor equation over there is that this vector is, in fact, a conformal killing vector. So what does that mean? On flat space, it means that we have an equation like this. So if the right-hand side had been 0, then we were talking about actual killing vectors. But now on the right-hand side, it's not 0. So I'm talking about conformal killing vectors. Conformal killing vectors, of course, specify conformal transformations, which are, OK, I'm still talking about flat space. So conformal transformations are transformations which keep the metric, in this case, 8IMN, or since I'm really doing Euclidean, I'm here to write delta MN, which keep delta MN invariant up to a vial rescaling. So that's the definition. And if you impose that requirement that the metric remains invariant under the action of this type of a vector, then you will find that this equation needs to hold. So this is a conformal killing vector equation. And its solution can be found in all generality. First of all, you find constants, which parameterize the translations. You find a matrix of constants, which parameterize the rotations. You find yet another constant, which parameterizes scaling. So dilatations. And finally, you find something somewhat more complicated, which is parameterized by a vector. And these are the special conformal transformations. OK, so we already knew what the constant piece was. That just corresponds to the translations. We already knew the rotations. This guy corresponds to dilatations. So three scalings. And this guy, as I said, are the special conformal transformations. Now, in principle, you can have some fun given these explicit conformal killing vectors to figure out what the transformation rules are or what the algebra is of these generators by just studying, well, essentially, the commentators of these things. But let me already give the answer. Schematically, the dilatation commutes with, well, gives as a commutator with PM, PM again, which, roughly speaking, just says that P is an object of dimension 1. On the special conformal transformations, you find the opposite. It's a minus sign. It's an object of dimension minus 1. The more interesting commentator is between the special conformal transformations and the translations, which contains, first of all, first of all, the dilatation back, and second of all, the rotations. And now, the more interesting, mn, yes, sorry, mn, mn. The more interesting transformation rules are, of course, among the fermionic objects. So maybe I should have, well, written over there. We now have these two constant parameters. And delta, really, the delta I'm using here, parameterized by sinx at tilde with that type of position dependence, it's really the one constant talking to the q and the other constant talking to a new fermionic generator, which I will call S. And then, similarly, for the tilde objects, sorry, it's just S for supersymmetry and C for super conformal symmetry. So I have the q's. I already understood their algebra. I already wrote it before. It is just a translation. Similarly, you can convince yourself, given this transformation rule, that the anticommentator of two S's is the special conformal transformation. And finally, the anticommentator between a q and an S, let me now just be schematic. Let's not write all the indices. It's some epsilon. Well, it's roughly speaking a d plus an m plus an r plus a little r, where d is a dilatation. So to soak up the indices, you have two epsilon symbols here. m is, again, the rotation. r is the SU2r symmetry, and little r is the U1r symmetry. So this algebra can be just derived from the explicit expressions we found on this multiple, for example. Sorry, if we go back to Poincaré's supertrade case, then what we want external that? For Poincaré's supersymmetry, all these are absent. Poincaré's supersymmetry, of course, just has on the right-hand side the conformal, well, the Killing vector corresponding to translations. But now we pick up these extra terms, which are reflected in these additional anticommentators. In particular, the k corresponds to v being precisely this object, and then the d, m, and r, and little r. Well, the r and little r we had explicitly sitting here. The scale transformation was the d, and the v soaks up the m, and the spatial action of d. So far, we have found that, classically, n equals 2 supersymmetric theories have actually some additional symmetries. They are, in fact, invariant under a full supermaforma algebra, namely, that algebra. Maybe I should mention, at this point, that to keep these statements true quantum mechanically, you should really ensure that the beta function vanishes. Examples when the beta function vanishes are, of course, pure n equals 4 super-yang-mills, or n equals 2, and n equals 2 theory, with k h group s, u, n, and 2n fundamental hypermultiplets. And there are more examples, but these two will do. Now, if the beta function is 0, then, well, then scaling symmetry is restored. Scaling symmetry, in particular, implies that the trace of the stress tensor is equal to 0. The trace of the stress tensor actually sits in the same multiplet as these sigma traces. So you see, this is what Guido was mentioning this morning as well. There is a multiplet of anomalies, which, in particular, contain these. The trace of the stress tensor, the conservation of the u1r current, and a bunch more. And if the beta function is 0, this sub-anomaly, sub-multiplet, can, in fact, be consistently set to 0. And then you have all the statements I mentioned just now. OK, so at the moment, we found that n equals 2 theories classically have that symmetry quantum mechanically, still if the beta function vanishes. But in fact, there is even more symmetry that we could consider. We could not just consider conformal symmetries, which, as I said, are, say, for the flat space case, there's symmetry such that if you act on a metric, you do not get back the same metric, but the same metric possibly with some position dependent rescaling. Now, vial transformations, they have a similar flavor. Vial transformations are transformations that send the metric to some rescaled version of itself. And while the metric rescales also local operators undergo some rescaling, so this is a vial rescaling of some theory. So the metric goes back to itself up to a vial factor. Local operators transform into themselves up to the same vial factor, which is now raised to the power minus delta O, where delta O is the scaling dimension of the operator we're talking about. So in my algebra, which is deleted now, I had this scale piece. There was a scale W. The coefficient of that piece will be precisely this dimension. So good, these are vial transformations. And let me just to be clear what the difference is between vial transformations and conformal transformations contrast the two cases. Vial transformations are defined over here. Conformal transformations are transformations such that the metric also transforms into itself up to a pre-factor. But now, this transformation should be such that this vial rescaled metric is, in fact, defiomorphic to the original metric. So this is, in fact, g prime mu nu at position x, where g prime mu nu at position x prime is the standard transformation rule for a metric. So this is the standard transformation rule for a metric under defiomorphisms. And conformal transformations are such that the metric goes back to itself up to a vial factor, which is defiomorphic to itself. So said differently, they're just all defiomorphisms such that if you apply them, you can back with the same metric, but with a vial factor. The vial transformations are more general. They're just transformations that scale the metric but without the requirement that it should be defiomorphic to itself. So in particular, you can immediately see that if you have vial invariance in any arbitrary background metric, then, of course, you will have conformal invariance. And in fact, the opposite is also true. If you have conformal invariance in flat space for a unitary theory, in fact, you can prove vial invariance. But this is a complicated proof. If you want to read it, it's in the paper by Lutti and Frantz. So very good. I would like to study vial transformations of my Nicos II theories. And the first observation I can make when trying to study that is that the conformal killing spinner equation, so these equations over here, these killing spinner equations, that they are, in fact, vial covariance. So let me write these equations in slightly more covariant form. So replace the partial derivatives with covariant derivatives, which I can do if I have given any remanumetric. So this type of an equation is vial covariant. Maybe you should check that as an exercise. So the vial weight of the killing spinner is 1.5. So if you take the metric and send it to omega squared times itself, you take the killing spinner and you send it times 2 omega to the 1.5 times itself, then this equation picks up an overall factor. What is nabla? Nabla is just the covariant derivative. Or you want it, how it acts on the spinner. That's what the question is. The forward says that psi is not a spinner in the beginning. No, I said psi is a spinner, but it just, sorry. Psi is, of course, a spinner, but it's not anti-commuting. That's what I meant, sorry. So just to be totally explicit, this thing is the usual partial derivative plus 1 quarter spin connection term, something like so. Yes. So you know how the metric transforms. In particular, that also implies you know how the wheelbinds transform. If you know how the wheelbinds transform, you can figure out how the spin connection transforms. If you know how the spin connection transforms, you can figure out that this thing is covariant and the wild transformations. So that's good. Second observation you can make is that the covariant-type transformation rules are also wild-transform, wild-covariant. In fact, you could have figured out which derivative terms to add to the transformation rules precisely by imposing that they are wild-covariant. So that's good. We know killing spinner equation and the transformation rules are wild-covariant. Now what about the actions? I did not quite write them completely, but I did write the bisonic terms, although they're not on the blackboard anymore. But you should expect that the action for scalar fields should be modified a bit, simply because the box operator on curved space is not wild invariant or wild-covariant in and of itself. The combination that is, in fact, transforming properly under wild transformation is box minus the scalar curvature divided by 6. This is in 4D. So if we add terms that reflect this curvature dependence to the action, so we should add terms to the action that go like scalar curvature phi phi for the vector multiple action, scalar curvature, qi, qi for the hyper-multiple action with appropriate coefficient, like 1, 6, times whatever coefficient was already present in the kinetic term. If you do that, then also the action will be wild-covariant. In fact, the action will be wild-invariant. So everything seems to transform very nicely under wild transformations, which is a useful observation because my aim as the title used to say is to put supersymmetric theories on S4. And S4 happens to be related, the metric on S4 is related by precisely a wild transformation to the metric on flat space. So given that I explained everything there is to know about super conformal symmetry on flat space, I can just apply my wild transformation to everything I said on flat space and get statements on S4. So we get S4 almost for free from all the work we did in the previous, I don't know how long, because the metric on S4 using the stereographic projection, if you like, can be written as follows. This is my value factor. So this is my flat space metric. This is my value factor. And then I get a metric on S4. It's four times r squared, where r is the radius of the S4. So it's some constant radius I chose. And the x squared that sits here is just the sum of the xi squared. So good. Essentially, I have put now an n equals 2 supersymmetric theory on the force sphere because these transformation rules, they were well-behaved under wild transformations. The actions are well-behaved under wild transformations. The metric on S4 is related by wild transformation. So these rules and actions, the model law having to add these extra terms are precisely supersymmetric transformation rules and actions on the force sphere. Are there any questions about this? So in particular, on the force sphere, we can realize, at least classically, a full SU2,2 slash 2 super algebra. I haven't mentioned this name yet, but the algebra I described over there is described by the super algebra SU2,2 slash 2, where here the SU2,2 is isomorphic to 4,2. I'm a bit sloppy with my signatures, but this is the correct as a morphine between the algebras, but I was really talking about Euclidean algebras, so really the Euclidean conformal algebras is of 5,1. So then here you need to, well, on either side, you need to put some stars to make this a correct statement in Euclidean theories. But anyway, so SU2,2 is a conformal algebra. Then the SU2 that sits there, this SU2 is just the SU2 r symmetry that our theory has. And then there is a U1 r symmetry still in town as well. So these are the global, well, these are the symmetries that this, the bisonic symmetries of this super algebra, and the fermionic symmetries are precisely parametrized by Qs and Ss. So this full algebra can be realized on the force sphere. That's good. Now, on flat space, on flat space, the massive algebra, by which I mean the algebra which you can preserve when you turn on masses, was just super Poincaré. It's the algebra which I started. But now on the force sphere, well, let me first say, though this flat space algebra in particular, you can also think of it as the algebra which closes onto isometries of space. So flat r4, the isometries are rotations and translations. And super Poincaré, indeed, in the algebra, you only find translations and rotations. Now, on S4, we would like to find a similar object. We would like to define an algebra which closes onto isometries of space. Now, the isometries of S4 are, of course, S of 5, just rotations. And we would also like to make sure in my, yes, in my algebra over there, the scale transformation, this parameter w, we would like it to be 0 because dilatations are obviously not isometries of S4. And since I also like to think of it as a massive algebra, or as just a generic n equals 2 super algebra, I should not expect a U1r to be a symmetry. Typically, U1r symmetries, except for super conformal theories, they are anomalies. They're not really symmetries in the quantum theory. So that symmetry I also don't like. So there are a bunch of symmetries I would not like to have in my super algebra that I would like to describe as this massive superalgebra on S4. And if you impose all these conditions, you will find that the massive algebra is OSP 2 slash 4. So the SP4 is really isomorphic to S of 5. S of 5 were indeed isometries of the force sphere. And the O2, that's really like a U1. And that U1 is the carton of SU2r. So the U1r we didn't like. But the carton of SU2r. We do like, and it appears over here. So notice this algebra. We think of it as the massive algebra, but it does contain r symmetries explicitly. On flat space, this is not the case. Massive algebras on flat space never contain r symmetries. We started off with the pancari algebra. The r symmetries were isomorphisms, but they were not part of the algebra. Here, the r symmetry is explicitly part of the algebra. That is a special feature of having super symmetry on a curved manifold. r symmetries are or become part of just super algebra. And of course, they're always part of super conformal algebras. But here also, super algebras. So at this moment, I have put the theory on the force sphere using this vial transformation. And next week in Zohar's lectures, you will notice that there is, in fact, a really interesting relationship between correlators on the force sphere and correlators on flat space precisely using this type of vial transformation logic. But you need to be careful to take into account all kinds of vial anomalies. But OK, Zohar will talk about that. Excuse me, sir. We'll play the theory on S4, because vial transformation, you have the same g that I asked you, 2,2. This is a super conformal symmetry on S4, yes. Super conformal. So far, your theory is super conformal? So far, how do you think it's super conformal when I impose that the beta function is 0, so I select super conformal theories to start with? Or the statements I made are classical. A masculine classroom. Classically, you have this algebra. How do you make it a method to be? No, it's something you need to, well, it's by hand, essentially. If you, well, it's either you say it's by hand or you could just say, OK, I'm going to turn on masses, say, for the hypermultiplets. And then you will manifestly see that you cannot possibly preserve all the symmetries, because the masses introduce a scale, and the particular scale transformations are violated then. Anyway, so if you introduce masses on the force sphere, you will notice that this algebra is definitely not there. But what is still there is this thing. Yeah, my question is, once you have the supergenome theory, master supergenome theory, you use vial transformation and place it on S4. Then this is still classical, a supergenome theory. So in the sense, but you have to introduce a scale, right? Because the sphere has a sphere down and off, which is like in a mass. Sure. Then the symmetry should be like glass. No, the sphere still has, OK, the sphere has fewer true kiln vectors, but the number of conformal kiln vectors on the force sphere in R4 is just equal. And they're both, in both cases, they're described as a 4.2 algebra. How many square chargers do you have in this sense? Where? Here? Here, 8 Poincaré and 8 superconformal. And here I have 8, which I can maybe call Poincaré, or just massive. Just to introduce some notation, the constraint that takes you from a generic pair of xi and xi tilde satisfy my conformal Keeling-Spiner equation. So satisfying the equation I had before, and similarly for the tilde version. So you have such xi and xi tilde satisfy this equation, then the projection down to OSP 2 slash 4 is implemented by declaring that this is not some arbitrary spinner, but that this spinner is in fact related to this spinner in a very specific way, namely as such. Whereas it's some arbitrary matrix, sigma mu nu is the standard product of two sigma matrices and symmetrized in the two integers, and xi is, of course, this guy. So if you impose that this spinner is not arbitrary, but instead satisfies this type of equation for some s, then you're not describing the most general transformations anymore, but you are describing transformations that lie in this algebra. I just wanted to introduce this symbol, s, because it will appear later in the talk. And if you're interested in how the solution now looks like, so after vile transforming, after vile transforming, the conformal killing spinner's picked up this vile factor, which I described over here. So a square root of the vile factor, the vile factor was this thing. So after just vile transforming, we have 1 over square root that factor. So notice there is a square here. So it's really the square root of just what's in the brackets times just the solution we had before. So this is the solution on s4 without imposing this condition. Now, if you do impose this condition, then this guy is not generic anymore. But in fact, it gets replaced with 1 over 2R times xi tilde s. So xi determines, let me get rid of hats. They're just annoying. So if there's a sub superscript in brackets, it means it's these constant things. And now I have over here that this guy is not arbitrary anymore. But in fact, it's related to the xi tilde constant that's it over here. And here in this dot, dot, dot, you do the same. But then the xi that appears here is really 1 over 2R, this guy. OK, this is just to be totally explicit. It's not all that relevant for what I'm trying to tell you. It's a very nice question. But this SU2, tomato slash 2 is a particular case of sp2 slash 4. So this is a sub-algebra of this thing? Yes. So what you're saying, in this particular case where you choose xi, well, the coordinate of xi really relate to xi, this case corresponds to OSP2 slash 4. So if you want a transformation that lies in here, you should impose that this is true. Why do you have to include an x-bar constant? You have SU2 slash 2, 2, comma 2, slash 2 is a particular case. No, no, no. This is a sub-algebra of this, not the other way around. This is the bigger algebra. So to get to a smaller algebra, you need to constrain something. And the constraint is this correct like so. Is it clear? It's not clear. So over there, we had this W parameter and this theta parameter. We also had transformations which were not isometries of the force sphere, which are not S of 5 rotations. We would like to make sure that all those are 0. And to make sure that all those are 0, what I can figure out, it's not obvious because I didn't tell you, for example, how W and theta look like. But anyway, if I had told you, then it would have been obvious that this is indeed the constraint that manifestly makes sure that, for example, W and theta are equal to 0. It's not all that obvious that it also just restricts the conformal killing vectors that appear there to be just as of 5 rotations, but that's also true. So this constraint is imposed to get to the smaller algebra, OSP 2 slash 4. Excuse me, why can't we have to impose this formula? Can you do localization with such a bigger algebra? Well, it depends what your goal is in life. If you want to study massive algebras on S4, or massive theories, theories which contain masses, then you simply don't have this algebra. You only have this one. Yeah, but if you consider conformal. If you have a super conformal theory, then you can place it super conformally on the force sphere, and you can try to work with this algebra. That is the form of the matrix SMU? No, SMU, it is whatever it is. Yeah. I mean, eventually, when you plug this back in, you can plug this over here. You will figure out what this is, and then you will figure out what S is. So S has a specific form. But to describe the constraint, it's enough that there exists some S that does a job. OK. And the second question, there's dot in the plus and minus, this missing half? And this is also proportional to the F? Yeah, it's proportional to this guy. Yeah, yeah. OK, not enough space, but any more questions? So maybe this was all a bit fast, but the bottom line of the entire discussion is we know how to put theories on the force sphere, preserving super symmetry. And in fact, we can do a little bit better. We cannot just preserve it on the force sphere, but you can also put any costue theories on a squashed version of the force sphere, where this manifold is really an ellipsoid defined by an embedding equation. So I introduce two new parameters, L and L tilde, where this squashing parameter, as it is often called, B is defined as the square root of their ratio. And this is some manifold, some foreign dimensional manifold. And it is possible to preserve super symmetry on this manifold. But everything I said clearly hinged on the fact that the round force sphere is vile related to flat space. This beast is not related in such a way. So to show that there is, in fact, super symmetry on this space, you really need full machinery of coupling to background supergravity, which you will hear about from Guido in his lectures. So let me not try to even describe how you would do that. Let me just state that it is possible. And if you really want to see the details, I can again refer to this paper by Hamai and Hossamichi. So they describe which supergravity background will do the job to preserve some amount of Susie on this manifold. So I have put the theory on the force sphere. And this is, of course, a school on localization. So let me now start localizing. So Francesco, this morning, started explaining how the localization argument works, not quite yet in quantum field theory, but just in symplectic geometry. So let me quickly sketch what the argument is in quantum field theory. But let me also refer you to Francesco's lectures for more details. So imagine we would like to compute this path integral on the force sphere. Of course, I have just e to the minus s, whereas it's the action. And o is some insertion. Say I wanted to compute some correlators. I can insert whatever fields I need to actually compute a correlator. So say I would like to compute these objects. Then very much like Francesco did this morning, you can, in fact, show that this object is equal to the following object, where I have deformed the action with some q exact term. So here v is some fermionic functional. It's some arbitrary fermionic functional at the moment. And q is such that three requirements are true. So this equality is only true if, first of all, q is a symmetry of s. Second, the operator insertions I have over here are also annihilated by the q I choose. And thirdly, so this you can think of as equivalently closed, but we need to make sure that it is, in fact, well. We need to make sure that q squared, which is not necessarily 0, just as it was not 0 in Francesco's lectures, we need to make sure that it is also invariant under q squared. So those are the three requirements. If those are through, if those go through, then the argument you can make is very similar to what we heard this morning. You can take the derivative with respect to t of this object. You will pull down this integral, so it's here. Now, because q is a symmetry of both s and the operator, you can put q outside of everything. Well, it's both symmetry of s and o. And q squared on v is equal to 0. So really, maybe I should write the argument. So I'm going to take the derivative of this object. So this will be equal to the same thing, but with this extra insertion. Sorry, I should have an OK. So I have q integral of v, maybe with a minus sign. Additionally inserted, s minus dqv. Sorry, you know what sits here. So now, because o was killed by q, s is killed by q, and q squared on the integral of this fermionic thing is also 0. I can rewrite this as an overall q variation of something. And total q variations in the path to integral of 0. They're like total derivatives. So this thing is really 0. So this is the localization argument in quantum field theory, sketched out somewhat quickly. And once we have this statement, there's a few choices we need to make in order to get to some useful expression for path to integral we would like to compute. So at the moment, this was just any fermionic functional. And the argument, if these three statements are true, guarantees that this is equal to the thing I actually would like to compute. But if I now choose v, if I choose v such that it's bisonic part, it's positive semi-definite. So this is really the equivalent statement of what Francesco had this morning in terms of the norm squared of the vector that generated the isometry. So if this fermionic functional, sorry, if q on this fermionic functional is such that the bisonic part is positive semi-definite, then I can take this expression, send t to infinity. I will create infinitely steep potential wells for the bisonic fields. And they will all localize to the localization locus, which is described by precisely the equation here, but when it is saturated. So this statement, for such a v, this equality can be further refined if this v is true. To a sum or an integral, whatever it takes of configurations, which are zeros of qv, where the operator insertion is evaluated on this configuration, where we evaluate the classical action on this configuration. And of course, we're still doing quantum mechanics. So even though I have this infinitely steep potential well, I still need to take into account little fluctuations around the volumes of those wells. And that translates into the insertion of additionally a one-loop determinant. So this one-loop determinant is really the quantum field theory analog of the product of eigenvalues that Francesco had this morning. So when doing localization in quantum field theory, there's really three things you need to figure out. You need to figure out what is the localization locus, what are the zeros of the supersymmetric variations of this fermionic functional that I chose. You need to evaluate the classical action on those configurations. And finally, you need to compute a one-loop determinant of quadratic fluctuations, where the quadratic fluctuations should be computed for the deformation action, since that is obviously the dominant piece. The deformation action is the thing that creates infinitely steep potential wells, and which is where the quadratic fluctuations happen. So three things to do to perform localization computation and really two choices to make. The first choice you should make is the acid of some interval that's a Q or an O. Is it? No, above it. That's being restricted. Yeah, it's the O. Yeah, it's the O, the operator insertion, or whatever insertion I'm considering. So two choices to make. The argument I gave you is true for any choice of supersymmetry variation. So you need to make a choice of supersymmetry. And you should make a choice of V. V is really not all that constrained. It just needs to satisfy condition three over there, and you should choose it such that the bisonic part of its supersymmetry variation is positive definite. So for the localization on S4, I will make my choices. I will choose a supercharge, which in fact lies in a massive algebra, such that I can study massive theories. And I will in fact choose a supercharge, which is preserved also on this quashed force sphere. It's clear that not all supersymmetries in the OSP2 slash four are still preserved on this quashed manifold, just because you don't have a full SP4 of isometries present. What you do have, you can see it from there, are two U1s, the U1 that rotates X1 and X2, and the U1 that rotates X3 and X4. So I should in principle allow for, for I could consider supersymmetries as square to these two U1 symmetries. And in fact, I would choose Q such that it squares to B, where B was the squashing parameter square root L over L tilde, J12 plus B inverse J34. So J12, J34 are the rotations of coordinates X1, X2, X3, X4 in that notation, plus B plus B inverse R, where R is the SU2-R symmetry. The carton of the SU2-R symmetry. So these factors B and B inverse, if you compute, if I would have given you the precise expression for the conformal Killing-Spinners, or in this case, just the Killing-Spinners to satisfy and generalize Killing-Spinner equation, then you could just compute this square explicitly and you would find all these coefficients. So it's surprising that I was preserved on the splash case four, called Massif and Massif series, do you remember? And it sounds subversive, doesn't it? Maybe SU1 slash one times SU1. Yeah, it's a bunch of SU1 slash one, so. I would think it's two, just because there's two. Two assumptions, maybe it's only just one, that may also be, I'm not sure. It's just Bruno says, it's a bunch of SU1 slash ones, maybe just one, maybe two. Okay, so for this particular Q, I'm not going to describe what xi and xi tilde should be, but I just want to emphasize one property, namely xi vanishes at the north pole, and xi tilde vanishes at the south pole, equal to zero at south pole. You'll see in a second why that is an important observation. So okay, I chose some Q, I could have chosen, well, on S4B, it's not very easy to find other Qs, maybe there's just one other, maybe that's not the case, I don't remember. On the round S4, of course, I had a lot more choice. And in fact, if you make these other choices, you find dramatically different localization results. If you're interested, I can tell you more, maybe Balt will talk about it next week, maybe not, I don't know. So the other localization result that you could find is you could, for example, localize onto what goes under the name of the car algebra associated to 40 and equals two theories, or at least you can do so for free hypermultiplets. So anyway, so I chose some Q, it's not the only choice on round S4, it's pretty much a unique choice on the squashed sphere. And now I need to choose V, I need to choose this fermionic functional with property number three over there, such that the bisonic part is positive definite. So the canonical choice for such a V, it's easily described, you just focus on this condition over here, QV should be positive definite in its bisonic subsector. Well, if I then just take the sum over all fermions, all fermions in the theory, I take their supersymmetric variation, complex conjugates times the fermion again, where of course implicitly, I'm contracting all the relevant indices. If I write this down, then of course, if I act with Q, the bisonic part will be Q psi dagger Q psi, which of course is positive definite, that it also satisfied property number three. Well, maybe you should do that in exercise, I don't know. It's pretty obvious by just realizing that whatever you will have written here looks like a singlet under everything, even the killing spinner is treated as a field. Of course, the killing spinner is not a field, so the killing spinner is not part of this Q squared transformation. Q squared doesn't act on the killing spinner, but in fact, you may as well consider it as acting on the killing spinner because Q squared acting on the killing spinner is equal to zero. If you act with these bisonic symmetries, what they would do on the indices described, or ascribed to the killing spinner, you will find zero. Said differently, nevermind. So this is the quick argument. Maybe you can flesh it out in some exercise. So this is a good choice. It's the canonical choice. And given this Q and this V, we can start localizing. So let's start with step number one. We should find what are the zeros of the bisonic sub part of QV. Now, given that I chose V over there as the sum over all the fermions in the theory, it's clear that the bisonic sub, well, these zeros are described by just reminding that Q psi is equal to zero for all the fermions subject to the reality condition that defines this dagger over there. Subject to reality condition. Okay, so the transformation rules are still around. We have transformations on fermions over here. We have transformations on fermions over here. I'm probably not going to describe what are the precise reality conditions because I'll simply tell you the answer. The answer is that for the hyper-multiplet, all hyper-multiplet fields are set to zero. For the vector-multiplet, things are a little bit more interesting, but I should first make one small comment. As the OS over here indicates, these are on-shell transformation rules, which is not good for the localization argument. The reason is, okay, it's erased. The reason is that Q squared on V should, well, you want to make sure that Q squared on V is equal to zero, but if Q is not closed off-shell, you will find equations of motions in that thing, and you want to avoid that from happening. So you should really make sure that over here, you also have an off-shell description. Now off-shell descriptions of hyper-multiplets, they don't exist for all supersymmetry variations in the supersymmetry algebra, but they do exist if you make specific choices, and of course we made a very specific choice. We chose this particular Q over there is just one supersymmetry, and one supersymmetry we can put off-shell also for the hyper-multiplet. So you will introduce a bunch of auxiliary fields with their own transformation rules, and they will also appear in the transformation rules of the fermions such that this thing actually closes off-shell. It's just a side comment. If you do all that, then you reach the conclusion for this particular choice of Q that all hyper-multiplet fields are equal to zero. Now for the vector-multiplet, things are a little bit more interesting. So what we're really doing is we're solving this equal to zero and this equal to zero subject to a reality property. And the answer, let me just tell you, is that the field strength should be zero, which I can take to mean that in some choice of gauge, the gauge field is actually zero. And additionally, these complex scalars, phi and phi bar, they are equal, and they are equal to some constant where, okay, I put some minus i over two for no good reason. So phi and phi bar are equal and constant and the auxiliary fields, the triplet of auxiliary fields is also constant and proportional to some particular tensor, Wij, where this Wij is really defined as the following object. It's a volume object. So remember the definition of S that related psi prime to psi itself. You contract that with this bilinear, you divide by what is essentially the norm of psi and that is what this W is. So this thing is symmetric in its indices i and j because sigma is symmetric in its spinner indices and psi i and psi j were chosen to be grassman even. So you want, well, you need to have the symmetry so i and j are symmetric. So this is a symmetric tensor in i and j and the ij is proportional to that thing with proportionality constant, this constant matrix A, which also appears here. It's just zero, yeah, well, you know it. I'm just trying to follow, anyway, some notational conventions. So this is the vector multiplet localization locus or at least it is the smooth part of the vector multiplet localization locus. So let me indicate that. These are the smooth solutions. But I remarked earlier, it's probably still written there, psi vanishes at the north pole and psi tilde vanishes at the south pole. So if you look at these transformation rules, we don't care about anything else, we know that that satisfies the smooth solution, but then additionally, if psi vanishes at the north pole, sigma mn fmn can be anything it wants at the north pole. And similarly here, if psi tilde vanishes at the north pole, sigma tilde mn fmn can be anything it wants at the south pole. So in particular, that means that at the north pole, so using that sigma is really anti-south dual, that means that at the north pole, the constraint on f is a little bit weaker. It's not that you need to set the full f to zero, but you just need to set the self-dual part to zero. And similarly at the south pole, you don't need to set the full field strength to zero, but just the anti-south dual part. So these equations are, respectively, the instanton and anti-instanton equation. And, well, they're part of the localization locus, so we will have to take them into account when performing step two and step three. Okay, so your bundles are actually just that first, including these ones, I think. I'm just going to ask. Okay, I think I can at least finish step two and step three. So step two, we're supposed to evaluate the classical action on the localization locus. I've described the localization locus there. It's clear that the hypermultiple actions will not contribute anything because the localization locus is trivial. The vector-multiple part does contribute something which is purely from phi's and these. Of course, the action is gone, but you remember that D entered quadratically, which is essentially the only location from which we're going to get anything. Anyway, when you do it, you just find something quadratic with some pre-factor. So here I introduced a hat. A hat is just the same as a, well, a zero, really, with a factor of L and L tilde absorbed. So I'm a bit ambiguous whether I'm doing S4 or the squashed force sphere. The supercharge I chose was valid for the squashed force sphere, so I may as well think of doing the squashed force sphere. But of course, on the squashed force sphere, I haven't told you how to write down an invariant action given that you have all these supergravity background fields present. So anyway, I'll be a bit ambiguous, but if you like, you just think about round S4. So okay, this is it. Just evaluate the classical action on that smooth configuration. Of course, the instantons will also contribute their own classical action pieces, but I will get to that in a second. And the third step is to compute the one loop determinant of quadratic fluctuations of the deformation action. Now there's a lot of technology that goes into actually performing such a computation. So this step, it's easy enough. You just take the configuration, you plug it in, you find the result. Computing one loop determinants, it's, well, you can do it in a low brown way. You just stare at the deformation action. You write down the operator that describes the quadratic fluctuations. So this is the field configuration. You consider small variations around that field configuration. You find the operator which describes quadratic fluctuations, and then you just compute the eigenvalues of those operators. And for the bisonic fields, you get a determinant of the operator describing quadratic fluctuations in the denominator. So roughly, you get a determinant of some operator describing quadratic fluctuations for the bosons. Similarly, you get some determinant, maybe some Gaussian, whatever, fermions. And you're supposed to compute these kinds of ratios of eigenvalues of these operators describing quadratic fluctuations. Okay, this is a hard problem, especially when you're not on S4 but on the squashed force sphere. On S4, you can try to diagonalize the operators you find by introducing some spherical harmonics. No one has actually done it because there's a better way. And in any case, writing down spherical harmonics would not work as soon as you're on the squashed force sphere. So the better way is by using indexed theorems, which I will not explain at all, but I think next week you will have four lectures on precisely this topic. So you will learn all about it. Long story short, I will just write the answer for the one loop determinant of quadratic fluctuations around these smooth configurations. It takes the form of a product of overall the positive roots of the Lie algebra associated to the Lie group where the Lie group is really my gauge group of an Upsilon function and the second Upsilon function. I'll define in a second what the Upsilon function looks like. It's some regularization of an infinite product, but let me first write down the one loop determinant for the hyper-multiple. It's a product over all the weights in the representation in which the hyper-multiple is transformed, under which the hyper-multiple is transformed. Of again, some Upsilon function. So this B is again a squashing parameter. If you like the round sphere better, so just set B equals to one and Q is B plus B inverse. I already explained what I had to us. These are the weights. I think I explained all symbols except for Upsilon B. So Upsilon B of X, as I said, it's a particular regularization of an infinite product and the infinite product looks like. So a product over M and N, positive integers of MB plus NB inverse plus X times M plus one B plus N plus one B inverse minus X. So if you had done the computation in terms of these differential operators described in the quadratic fluctuations, you would have found a ratio of eigenvalues. If you simplify that ratio, and for example for the vector-multiple, you would see that only modes of the fermions still survive at the end of the day. And for hyper-multiple, since there is an inverse here, you find that only modes of bisonic fields contribute at the end of the day. But the fact that you should expect a lot of cancellations is, of course, the consequence of supersymmetry. And this is exploited in the index theorem approach, which I will not talk about. So okay, I will stop here, but you can see how to put these ingredients together to get the answer up to an important, up to the instant non-contributions. Yeah. Is it possible that you'll give different weight on the X one to five to have different squashes? Because in the previous case, for this question, you'll only keep the Q one plus Q one, as I mentioned, right? But what if you have this question that might make, for example, X one, two, three, equally weighted, and we'll have a bigger... Sure, you have a bigger isometry group. Then I will do this... Pardon some, please. Do some of this. No, you can still choose the same supercharges. The supercharges I chose will be present. It's present for the squashed force sphere and anything that has more isometries. So in your case, if you set R and L equals to... I think there are two different squashes, right? Right. One is sensible and the other is much higher than the other. I mean, it's easy. Yeah. So... Right, but I think... I think the only squashings I've seen can always be reduced to this squashing, but maybe you're more inventive. But you're right, in... Because the difference squashing, and also, even if it's the same geometry, sometimes there are different multiple ways of... Right. There could be different squashes, but the question is really whether the answer is different. But I don't think it will be different.