 So I'd like to start by pointing out a few references for what I was talking about yesterday. There are a set of excellent lecture notes by David Tong. You can find them on his website. There's also a big review by Van Doren and Van Nieuwenhaazer, which has a lot of computational details, in case you're interested in the more computational side of this subject. And then there are these two papers, which I mentioned as a reference for the original brain theory construction by Michael Douglas. So let me briefly review where we were yesterday, where we left off. So we were trying to study the instant on equations, which if you solve these equations, you've automatically solved the equations of motion. And moreover, you find a finite action configuration where the action is proportional to the topological charge of the object. And for a given topological charge, k, we found that the solutions to these equations, they're not unique. You have various continuous parameters that parameterize your solution. And these parameters we called collective coordinates. So in particular, I wrote down the solution for an SU2 gauge group with instant on charge 1. And we saw that there were eight collective coordinates. More generally, there will be 4k n bosonic collective coordinates when you solve this equation. Now, we also saw that these collective coordinates, which are these continuous parameters on which the solutions depend, were in a one-to-one correspondence, which so-called zero modes. So the zero modes, they were really normalizable solutions to the linearized field equations for the fluctuations. So what we did is we wrote the gauge field as some classical solution, whereby classical solution I just mean a solution to this equation. It depended on these continuous variables. And we studied a fluctuation around this solution. And then there were two conditions for these fluctuations. First of all, it needed to satisfy, as I just said, these linearized field equations. And second of all, you want to prevent that this is a pure gauge transformation. So you want to make sure that this is orthogonal to all possible gauge transformations. So this problem is defined by these two equations I just described. You find what goes on the name of these zero modes. And they were in one-to-one correspondence to the collective coordinates. Moreover, there were not just bosonic zero modes in the problem. As soon as a theory you're studying, in which you're studying these instanton configurations, also has fermions, there will be also fermionic zero modes, which are defined in essentially the same way. They're again solutions to the linearized field equations around this background. But now, not for a fluctuation since fermions classically don't have a profile. So it's just for the fermion itself. So in general, when you have some generic interacting field theory, you'll have these two types of zero modes. And in fact, when the interacting field theory possesses supersymmetry, these two types of zero modes, they're really paired up by supersymmetry. So we showed that in an instanton background, part of the supersymmetry is preserved. And that preserved supersymmetry will act on these zero modes. It will organize these zero modes in supersymmetric multiplets of the supersymmetry algebra, which is preserved in this background. So Suzy will pair the bosonic and fermionic zero modes. So in particular, if we are studying n equals 2 supersymmetric theories, the supersymmetry preserved in an instanton background was two-dimensional n equals 0,4 supersymmetry, but dimensional reduced to 0D. So this was all still within the four-dimensional quantum field theory. But then we made something of a conceptual jump. We tried to start thinking about a configuration in R4. So we have some instanton configuration in R4. The gauge field in this R4 is essentially looking like this. So it depends on these collective coordinates. And of course, it also depends on the position in this R4. And then what I suggested to do is we add an extra dimension to our space. We still look at an instanton configuration in R4. But while it moves in what we may call time, I'm going to let the collective coordinates depend on that corner time. So in every slice in time, we consider an instanton configuration. But while we move in time, I'm going to allow these corrective coordinates to depend on where in time we are. So if you write the action for such a configuration, so now I'm in 4D plus this extra 1D direction, we have the time integral and the usual space integral of the Yang-Mills term plus a lot more. If we take this action and we plug in this configuration, it is clear that as far as the four-dimensional integral is concerned, you will just notice that at every point in time, you have an instanton configuration. We already know what that integrated with integrated, essentially, to this topological charge. But we will pick up various time derivatives because these collective coordinates are dependent on time. So if you do this exercise, you will actually find that this five-dimensional integral reduces to some integral along time, where the thing that enters here will contain various derivatives or will contain dynamics of these collective coordinates, which now depend on time. So this I wrote yesterday as well. You should generically expect that the result here is some sigma model as its target space, precisely the modelized space of instantons. So let me somewhere here. I should say that these collective coordinates, they define for you some modelized space. So both bosonic ones and fermionic ones, they come together, define some modelized space. And this is the metric on a modelized space. We defined it yesterday. And in general, you should expect if you do this procedure to find the one-dimensional theory that tells you what the dynamics is of the motion, essentially, in this modelized space, while this instanton configuration moves forward in time. Sir, the initial five-dimensional action was also an action of five-dimensional n equals 1 theorem? Well, yeah, let's just think for a second that I'm just doing the pure Young Mills case again. But you can, in principle, complete it or try to complete it. And what is the new fields, the five-dimensional fields, which was absent in four dimensions? Yeah, I'm doing 40 n equals 2. So there is a completion to n equals 1, if you like. But there are other fields which was not present in four dimensionals. No, if I do 5d n equals 1 in 5d, well, 5d n equals 1 reduces to 4d n equals 2. Well, maybe we have d4 to d4x, dT, dQ. Sorry? You have dT and d to the fourth n. Yes. So you do in 5d n? Yes. Oh, OK. Yeah, I think this does not mean something. In principle, no. Right, right, right. But the completion, well, anyway. For example, what's the potential to obtain five components in 5d n? Sure. And what is the fifth component? Choose it to be 0. OK, it's done. So through this procedure, we land on some quantum mechanics describing the dynamics of the model I. And we would really like to study this object a bit better. It's clear that this description as a sigma model into the target, into the model I space, is not all that useful. So instead, we started thinking about brain pictures that maybe provide a more convenient description of precisely this one-dimensional theory. So let me not repeat all this brain theory stuff, but let me just tell you the answer. So if you want to study the world volume theory of an instanton in a four-dimensional n equals 2 theory, it will be given by a quiver gauge theory. So k is the instanton charge. I have a UK gauge group. I'll define what dotted lines and solid lines mean in a second. So we find this type of quiver. I drew it using two-dimensional n equals 0,2 notations. But you should really think to get to that picture, you should dimension-reduce it by one dimension. So in these two-dimensional n equals 0,2 notation, the solid arrows are chiromultiplets. So correspond to chiromultiplets. The dashed lines are Fermi multiplets. And of course, this corresponds to a vector multiplet. So let me remind you of the field content of these things. In the chiromultiplet, you have a pair of complex scalars. And a pair of left-moving fermions. In the Fermi multiplet, you have a pair of right-moving fermions and two auxiliary fields. And finally, the vector multiplet, which sits hidden in here, contains, of course, a gauge field and a pair of right-moving fermions together with some auxiliary field. So this is the two-dimensional description. If you want to reduce to the 1D description, it will look essentially the same. But in particular, this gauge field, one of its components will become a scalar. And if eventually we may want to look just at the zero-dimensional world volume theory, then you pick up two scalars over here. So we would like to study this world volume theory. This is really the object that describes the motion in the modular space. And as I mentioned yesterday, one of the branches of the vacuum manifold of this thing is precisely the modular space we're after. Of course, for the specific case of an n equals 2 theory, with gauge group SUn, and in this case, I chose two n fundamental flavors. So these correspond to the two n fundamental flavors split in two groups of n. If you have different matter content, you can try to change these numbers. Are there any questions up to this point? This was essentially the review. OK, so let me then go back to my original problem. We were localizing on the force sphere. And we had chosen some localizing supercharge Q, which had the property that it squared to two rotations, plus an SU2R, carton generator, plus a mass, plus a gauge transformation. So here m dot f means the flavor symmetry acts on whatever field you're acting, and the coefficient of the action is a mass and same over here. So this was the supercharge we chose. Let me just remind you that we're on S4B, which I gave coordinates, or which I parametrized at least as follows, where the B was defined as a square root of L over L tilde. So this J12 is the rotation of these two coordinates. J34 is the rotation over here. And as I said, R is the SU2R carton. So this was our localizing supercharge on the force sphere. In particular, this is how it acts close to the north pole where the instanton lives. And in fact, part of this preserved supersymmetry that in the instanton background, I mean, we know that the instantons are solutions of the localization procedure with respect to this Q. So certainly this Q is preserved in the instanton background. And in fact, it acts on precisely this world volume theory, or it's one of the superchargers that still is preserved in this quiver. So in particular, I should be able to think of this supercharge at the level of this quiver. So in that quiver, there is a supercharge which squares to accept exactly this thing. Let me slightly rewrite the square just to have more conventional notations. So I didn't do anything. I just took the sum and the difference of these two terms. But now it's conventionally, this thing is conventionally called JR. This thing is called JL. These are really, but it doesn't matter what they are really, this thing is called J. They're all just names. So at the end of the day, on this world volume theory, I have a supercharge that acts like so. But now I also added an extra term which corresponds to gauge transformations with respect to this guy. So Q is cool. There is a Q in this theory which squares to this thing where in this zero-dimensional theory, this J and this J left are really thought of as flavor symmetries. And this is still a flavor as well. This is also flavor from the point of view of the zero-dimensional theory. So this G is really corresponding to this flavor symmetry. And now finally, we have the UK, which is represented over here. So the task we have to perform now is to assign the quantum numbers under these flavor symmetries to the various multiplets which appear in this quiver. So we have the various 0,2 multiplets. Let me try to keep it somewhat close. So we have in the brain picture, we had these 0,D4 strings. With these 0,D4 strings, they gave, for example, this Fermi multiplet. Maybe I should do it like this. Let me call this Fermi multiplet number one. It comes from over here. So this has charges under this J of 0 and under J left of 0. Now I should mention that in the brain theory picture we had, all these charges act geometrically. So you should be able to figure out what exactly are the charges. And I will just give you the answer. So we had two chiral multiplets that came from the D0,D4,2 strings. So those are really these two, chiral multiplet one, chiral multiplet two, Fermi multiplet two, Fermi multiplet three, chiral three, chiral four. So chiral multiplet one and chiral multiplet two. You can verify half charges, a half, a half, and 0,0. Fermi multiplet number two is similar to this guy. And finally, we had a vector multiplet sitting there. We have Fermi multiplet number three and chiral multiplet three and four. Let me just give the answers 0, 0, 0, 1, 0, a half, a half, a half, minus a half. OK, so this was not very illuminating. I just gave you the answer. These are the charges of the various multiplets under these two flavor symmetries. Since the flavor symmetries, all the elements of these multiplets should have the same charge, at least when I think of it as a zero-dimensional thing. So now we have all the information we can possibly hope to have about this quiver in order to localize it with respect to precisely this supercharge. So this is the localization of, well, pick your dimension. We can do it, say, in 1d n equals 2, where the n equals 2 is thought of as dimension reducing from the 0,2. Or we can do it in this dimension and then we reduce it to 0d. Well, you do it however you want. I will just tell you the answer. I think I hope that Francesco will tell you more about localizing the elliptic genus. That computation is the most similar to the computation you're supposed to do to compute the partition function of that world volume theory. So let me just give you the answer. So the procedure is, of course, the same. We have a queue. We'll study the BPS configurations. We evaluate the class collections if they give any contribution. We evaluate one loop determinants. And we integrate over the entire BPS locus. So the integral, in this case, is over an element valued in the cartile of that UK gauge group. So we have K integration variables. And while we integrate is a whole slew of one loop determinants. So in this computation, the classical actions, they evaluate to 0. In fact, they're all Q exact. So we shouldn't expect any contributions from there. And indeed, the result just contains one loop stuff. It contains one loop stuff from the various elements in this quiver, which I again denote by their brain names just to be brief. So we find one loop determinants for the D0, D0 strings, which is really for this vector multiple and, well, all these adjoint stuff, essentially. This is all the adjoint stuff. This is for one of the Fermi multiplets that sticks out to one of these ends. This is for that other Fermi multiplet. And this, finally, is for the pair of chiral multiplets that connect to the nodes, which is really to be identified with a four-dimensional gauge group. So that's why this guy depends on the gauge parameter. And these two depend on flavor symmetry parameters of the four-dimensional theory. So in this picture, these m's coupled to the four-dimensional flavor symmetry. And these are represented by this m and m tilde, which are really the flavor symmetries rotating here and here. While this phi over here, capital phi, couples to the four-dimensional gauge group, which is really that box n up there. And finally, this little phi, as I was mentioning, is the parameter of this UK gauge group. So this is a computation you can do. You can compute these one loop determinants one by one. You multiply them, and then you perform the integral, where I wrote some funny letters here, JK. It's a particular residue prescription that you need to impose when computing these integrals. So these various ingredients have various poles. And this JK prescription tells you precisely which poles you should pick and which poles you should avoid. It will be somewhat useful. So sorry, I should have written this before the lecture to actually have the explicit expressions. So let me quickly provide you with these expressions. And in fact, then you will see why it was useful to have this entire table of charges. At least intuitively, you can understand where these ingredients come from. So the zero, the zero strings, as I said, it's everything that is adjoint that is this entire list. For the vector multiplet, we add charge 0 for both of these slaver symmetries. But of course, the vector multiplet is adjoint valued under this phi. So we get phi ij, which is sort-hand notation for phi i minus phi j. So this is like the root structure that appears when you act on this adjoint value thing. And the prime means that in this product over i and j, you should omit the terms when i is equal to j. OK, so you can really see this 0, 0 appear here simply because there's nothing else. We evaluated the one loop determinant. And we have the 0 here, 0 here, 0 here, 0 here. We just got stuff from here, which is reflected here. Now for this guy, this guy comes from the Fermi multiplet. The Fermi multiplet carries charge 1 under j, charges 0 under jl. And you can indeed again see that reflected here. This B plus B inverse comes precisely from here. j is 1, so we pick up a B plus B inverse. Again, it's an adjoint value thing. So we pick up this phi i minus phi j from over here. Similarly for these two chiral multiplets, they carry charge half, half, and half minus a half. So we get this times a half plus this times a half. That is this B over here. And similarly, half minus a half gives you this B inverse. So you really see that this one loop determinant is not all that mysterious. And in fact, the pattern recognition I'm doing between this one loop determinant and these charges is if you actually do the computation, you will see that this is precisely how these things arise. So now we understand the pattern. We can easily write down the other contributions. For example, for these chiral multiplets, they carry charge a half under j. That means that we should expect a B plus B inverse. Sorry. So I'm writing the Fermi multiplets. For these Fermi multiplets, we don't expect anything. But these Fermi multiplets, they are charged under this type of flavor symmetry. So you get phi i because they're fundamental under the gauge group and they're fundamental under these flavor symmetries, which gives you this type of expression. And similarly for the other ones. And finally, we have these two chiral multiplets. The chiral multiplets contribute bosonic fluctuations, whereas the Fermi multiplets contribute fermion fluctuations. That's why the Fermi multiplets give you stuff in the numerator. But the chiral multiplets will give you stuff in the denominator. So this comes from these two chiral multiplets. You finally get, again, they're bifundamental under the gauge group and this top node over there, which was identified with the gauge node in the four-dimensional theory. So we get a result like so, where you see the charge one half reflected in this extra term. Right? How about you have d0, d4, 1, d0, d4, d4, d4? Oh, I'm sorry. This should be, this is 3. And this is 2. 3 is the one we're buying. 3 is, well, OK. Sorry, the names are not optimal. But these two Fermi multiplets, they correspond to these two contributions. So these two, the two Fermis sticking out are these two guys. d0, d0 is all the adjoint stuff sitting over here. And finally, this stuff comes from these two chiral multiplets. Yeah, of course, it will be the equivalent. So let me, maybe I should have drawn the brain configuration. So we had n d4 brains, another n d4 brains, a third set of d4 brains. We had k d0 brains, this stack I called 1, this 2, this 3. And we have all the strings connecting things. OK, very good. Maybe you should have more time. OK, so it's just when you do the localization, you're going to compute these one loop determinants of quadratic fluctuations. There's a lot of cancellations between bosonic mode and fermionic modes. But in particular, you can see that the Fermi multiplets has two fermions and some auxiliary stuff. It's these two fermions that will contribute fluctuations. They're fermionic fluctuations. So you get stuff in the numerator. The fermionic one loop determinant sits in the numerator. Similarly, here for these chiral multiplets, there are two scalars and there are two fermions. But it turns out that contributions are such that only the scalars have a leftover contribution. That sits here in the denominator. You do the computation. It's not supposed to be obvious. But it has these two fermions sitting there and those guys contribute. I mean, the modes coming from the fermions are not canceled by modes coming from the bosons. Not all fermionic modes are canceled by bosonic modes. There are some leftover modes and they give you this. Well, you have to know how. So in 2D 2.0.2, as Francesco was explaining at some point, the vector multiplet gives rise to the field strength lies in the twist and chiral. Could it be that the vector multiplet field strength here lies in the fermion multiplet? Would you explain? Yeah, I think that's true. Yeah. That's a better explanation, indeed. The question was whether it happens to be so that a 0.2 vector multiplet is equivalent to a 0.2 Fermi multiplet. And indeed, that is the case. And that's why you can understand, in another way, why the vector multiple contribution gives fermionic, looks like a fermionic determinant. OK, with some effort, we wrote down a final expression for the partition function of the world volume theory of an instant of charge k in an n equals 2 super conformal field theory. So at this point, we are really done. We have that object. I wrote it as a zero-dimensional computation. Well, the result I wrote is for a zero-dimensional computation. We have this zero-dimensional object, this point-like instanton sitting at the north pole. There's a similar expression for the point-like anti-instant at the south pole. And the only thing we need to do is add up all possible topological charges with a proper weighting factor, which comes from the classical action in four dimensions. So this is a total contribution from all possible topological charges. We have such a contribution, and we have a similar contribution for the anti-instantons. So this is the extra ingredient that we didn't know quite yet when we were writing down the answer for the localization. So we had this integral over the smooth localization locus. It was parametrized by what I have called over there, capital phi. So let me keep that notation. We have capital phi, which is the integration variable. Then we had, well, now we have these two instanton contributions. We had a classical action, which I think had coefficient like so. And then we had one loop determinant from the four-dimensional fields. We had hyper-multiple stuff and vector-multiple stuff. These two guys were expressed in terms of epsilon functions. And this is really now the final thing. This depends on phi. This depends on phi m and m tilde to keep notation from there. And these two guys, they depend on the dependence here is on capital phi, capital m and m tilde. Because, of course, we integrated out the little phi. And everything depends on b and b inverse. q is the instanton counting parameter. I got it by evaluating the classical action. And it is really q is e to the 2 pi i tau, where tau is the holomorphic coupling. So it is theta over 2 pi plus 4 pi i over g Yang-Mills, I think, g Yang-Mills squared. So if you take an instanton and you want to evaluate the only terms, let me do this over here, so the only terms relevant to evaluate the classical action on an instanton configuration are the Yang-Mills action and this topological term. Well, there's traces. This thing had coefficient i theta over 8 pi squared. This guy has coefficient, I think, 1 over 2 g Yang-Mills squared. And now we know that this integral precisely gives the topological charge. So this is precisely k. Now star f, since we're looking at an instanton, is, of course, the same as just f. So I can kill the star here. Then this thing will be proportional to k, but not quite. You need to take into account these factors of 8 pi squared. So this thing is 8 pi squared and then you put it all together and you should recover this combination as a counting parameter where the k keeps track of the topological charge. OK, so we know in principle the answer. Let me just mention that you will typically see this answer expressed in a slightly different form. Let me just mention how that comes about if you try to compute these integrals. So if you're familiar with instanton partition functions, you typically see them expressed as a sum over Young diagrams and then stuff. So where does a sum over Young diagrams come from? Well, it comes from actually performing this integral I wrote up there. So the JK prescription, I will not explain it in any detail, but for my purposes, it's enough to know that it will instruct me to consider the poles, consider poles of negatively charged stuff, negatively charged fields. So in particular, what does that mean for that integral if you look at, oh no, yeah, I'm sorry, I removed the integrants. It seems. Oh, excellent. OK, so if you want to look for negatively charged stuff, you can see that one of these chiral multiplets, it has a minus sign in front of its phi, little phi. This minus sign is precisely the gauge charge and we're looking for negatively charged objects. So that guy is one of the poles we should take into account. This is a pole that comes from a negatively charged field and the JK prescription tells you take that pole. So we have a pole from that equation. Let me just write it. So the pole from that equation from this guy gives me, well, you have a pole if this equation is satisfied. So if one of the phi i's happens to be equal to this, you will have a pole from over there. Now from this adjoint stuff over here, this is phi i minus phi j, this is phi i minus phi j. So you see again, minuses pop up. So from here and here, we will also find poles of positively charged objects just because at the level of patterns, you see a minus sign in the expression and you find two more pole equations which look like so. So if you want to compute this integral, we have to pick up all poles which solve these equations. We need to solve these equations such that we have k linearly independent solutions for these phi's. So OK, let's try. Let's try. So the first thing, these guys, they need as an input. They eat as an input already one expression in terms of little phi. So we don't have a phi yet when we're starting to construct the pole. So we should start with this guy. So step number one is start with an equation. Start with equation A, which in particular means you should choose one of these capital A indices for which you take phi i to be equal. So this starting point, let me denote that just with this little box. In here, I chose phi i to be equal to i capital phi i plus 1 half b plus b inverse. So that's always step one if you are looking for a new pole of which you would like to compute the residue of that integrand. So step number two could be, well, I choose this equation again, and then I would start drawing what I call yb. You'll see in a second y. And I would have a new box. And then OK, so this is not, it's possible, but let me be a bit more adventurous and try to use poles b and c. So every time I choose a b, I need, as an input, already a phi expression that is already on some pole location. So we already have that. We have this phi expression. It sits on some pole equation. I can show that in here. Then I add a b and I find another expression for a new pole. When I do that, let me just draw a box over here to indicate I took this one and I used equation b. So every time you use b, every time you use b, I'm going to propose as a rule, add box to the right of the input. So this thing needs input. So that means that input already had a box and you put a box to the right of that guy. So that's what I did here. I added this box to the right of the guy I already had. And then this guy will, of course, be precisely this expression plus an extra b. So at the moment, we picked two variables phi on the pole location. We should do this game k times. Every time we use b, we're going to add a box in this picture to the right. Now what happens when you choose option c? You're going to add a box to the bottom. So if I choose option c, I take again this pole as an input. I add b inverse. This defines a pole of that integrand. And this guy then sits at a location, defines a pole location as follows. And you keep doing this game. But it's not an unrestricted game. You cannot get any arbitrary shape of this box configuration simply because you see over here, this thing becomes 0 as soon as phi i is equal to phi j. As soon as any i different from j is such that phi i is equal to phi j, you get a 0 from here. Similarly here, as soon as phi i is equal to phi j up to b plus b inverse, you're also getting a 0. So that means that if you had found in this list of pole configurations things which are either equal or differed by b plus b inverse, you really didn't have that pole because it was canceled against the 0. So what does that mean in these pictures? It means that if I delete this box, say I didn't choose this box, I just chose those two boxes. And I try to add a box to the bottom of this, meaning that I try to input the thing over here, which was i phi a plus I have b plus b inverse plus b. Say I try to input this pole in equation c, that would mean that I need to add another b inverse to define the pole location in that integrand. So this would naively define a pole location like this. But this is precisely equal to this up to b plus b inverse, so it will cancel against that 0. So this is really not a pole configuration. So if you think a little bit, then you will figure out that the only configurations of boxes you're really allowed to have look like so. They look like young diagrams. So this was a young diagram based on choosing this pole once, where I chose the index a. And then I used these other poles to complete. But you can, of course, choose multiple times number a. So you'll have a young diagram that is labeled by a. You will get one that is labeled by b. And so forth, you keep doing this game until you have defined precisely k linearly independent pole locations, which do not hit 0s, which is implemented by having this kind of a shape. So at the end of the day, when you actually try to perform this integral, you will find an expression that looks like a sum over vectors of young diagrams, such that the number of boxes in the total number of boxes in this collection of young diagrams is precisely equal to k. If you have precisely k boxes, you have precisely k poles, and you're doing a k-dimensional integral, so that's good. That will define you something. And then you can compute the residues and fill in these dots. So this is perhaps the more standard expression of the instant on partition function. It's a sum over all boxes with overall young diagrams of some integrand. So over here, this sum over k, if you like, I can refine it. Internally, it will be a sum over young diagrams restricted to have k boxes of this thing. And then this integrand is what we were just defining. Or I can just say, I have a big fat sum over all type of young diagrams. And then you need to think a little bit more what this integrand should be, but you can figure that out. Or you just look it up in Nikita and the Krasov's paper. So I think at this point, we have achieved localization on the force sphere in all details. Well, all details that I was willing to mention. I haven't computed for you these one loop determinants in great detail, but you can do so. We know the instant on partition pieces. We know the classical action. So this is the S4 partition function. Are there any questions about this computation? Yes? Somehow this can be translated to the page could work with you and not S4. So S4, because in the workflow by HEP, you point out that you are wrong. I don't understand. OK, that's true. And maybe we can talk about later. But there is a trick that you can do at the level of that integrand to take care of the difference between un on SUN. Or you just do like AGT and you pull out some U1 factors. The trick has to do with this thing is, well, it's not really wall crossing invariant. And you need to make sure it's wall crossing invariant. If you do that, then the claim is that this thing actually does produce a SUN result. Yes? Sorry to say again? I mean, young diagrams, you don't see them with permutation groups. They're also in the permutation terms of permutation groups. You see, it doesn't coincide with these young diagrams. I wouldn't say it's a coincidence, but I'm not sure if I have an answer off the top of my head about permutations. Oh, I'm sorry. The question was that we ended up with a sum over young diagrams. And young diagrams are related to permutations or partitions. There's a lot of things that are labeled by young diagrams. The question was, does that have a meaning in this context? But at the moment, I'm just going to say that it's a convenient way to characterize the pulse of these integrants. Next week, you should hear a more field-theoretic understanding of why this young diagram should show up when you do the actual integration of the modelized space as defined by the ADHDM constraints. If there's no more questions, let me start with some new topic. Sorry, say again. Jeffrey Kerwin. It's two names, Francis Kerwin. And I forget the first name of Jeffrey. OK, so the new topic I would like to talk about is the following. So just now, I had the result for a localization computation of the S4 partition function. We did localization on S4. But to do localization on S4, there were really two choices we made. The first choice was which super-charge to choose to localize. We chose a particular one that had these two rotations, this SC2R carton generator. And the other choice is which localizing fermion, or fermionic function, how we choose. So in this computation that we just finished, I chose the canonical one, which Francesco also chose this morning. It was a sum overall fermions of the variation of the fermion dagger, where the dagger is defined with respect to some good choice of reality properties times the fermion again. So really, there's two things I could try to do to get perhaps more useful, well, not more useful, other results for the S4 partition function. I could change the choice of Q and I could change the choice of V. Let me stick to the same Q, although it's very interesting to consider different choices of Q. And if you choose right, you will be able to get chiralgebras, which you will hear about next week from Baldt, at least the chiralgera part. I'm not sure about the localization part. So in this next topic, I would like to change V. So recall that all we're doing to the path integral is we have this integral over all the fields, we have the action, and we're adding something Q exact, such that the bosonic part of the Q variation of this thing of this V is actually positive definite. So here you see that manifestly this positive definiteness. But whatever V I choose, as long as it's entering in the Q exact way, it should not change the answer. So whatever I will do now or modifying this thing or just choosing something new altogether, I will not change the answer. The answer will always be the same, but the representation of the answer will change, perhaps, if I choose V cleverly enough. So let me choose such a clever V. I'm going to keep this term, but I add something new. I add to this expression, the trace over the gauge indices of some H, which I'll explain in a second. So I add this to the deformation action, where here lambda and lambda tilde are the gauge genie. So they're elements of the n equals 2 vector multiplets. I contract the spinor indices with the keeling spinors. So these guys, they describe my particular Q. So Q is described by this pair of keeling spinors. I use them to contract the spinner indices. Both of them carry an SU2R index. I symmetrize that SU2R index. And then I take this expression, and I multiply it with an object H, which is a triplet of SU2R. Set differently, these indices are also symmetrized. And H is just some arbitrary functional. At the moment, I'm not going to specify what H is. It's just something depending on bosonic fields. So this is some function of bosonic fields. So good. I chose this to be my V. If I act with a Q, I'm making it Q exact on the nose, because after all, I'm acting with a Q. So this is a legal thing to do. The only question I should answer now is if I can actually run the localization machine, because as I just mentioned, you also need to make sure that the bosonic part of this QV is positive semi-definite. Yes, I can. So well, you see it in five minutes. If it's not clear why I chose this thing, then I'll get back to your question. Good. So I should also mention that S4, as we already noticed, is a somewhat complicated localization computation when I'm doing what will turn out to be Higgs branch localization. It's not going to get any easier. But fortunately, Francesco already announced that he will perform a different Higgs branch localization computation on the two sphere, I think. There, the details will be easier to follow than what I'm doing, but I'll do my best to at least sketch how the computation works in this case. But we'll notice that somewhat complicated configurations appear in intricate ways, and that complication is not present when you do this game in lower dimensions. So as I was saying, we would like this property to be true, but if you look at it, if I'm going to act with Q on my new V, it's the sum over fermions of something that is manifestly positive definite. But then I have this sum. So I'm restricting to the bisonic piece. Then I have this trace over this thing. H was something bisonic. So if I vary it, I make something fermionic, which is not what I want. So I should really vary the Gagini over there. So OK, this is the Q of V, and we would like this to be positive definite. Now, it's obvious that this is not quite the case. But given that I refuse to specify what H is at the moment, there's no reason why this should even be positive definite. So it seems like I'm in a little bit of trouble if I would like to use this for a localization computation. But that is actually only superficially so, because remember that the variations of the Gagini, it has a lot of stuff, but in particular, it has this auxiliary field term. This triplet of auxiliary fields appears in the variation of lambda. And similarly, in the variation of lambda tilde, I get a lot of stuff and the auxiliary field again. So that means that when I use that variation here and here, this term will be linear in the auxiliary field. There will be Dij times some stuff. In particular, it will, roughly speaking, look like Dij times Hij with some fermion by linear coming from the contraction over here. This thing, when psi is actually also the Gagino, you see that this thing will be quadratic in D. Since both variations will contain these types of terms, so you can get quadratic terms from here, you also have some linear terms, and you have some linear terms from here. Now, in all this, the auxiliary field, of course, doesn't enter with the derivative. It's just a pure Gaussian integral, and we can perform that integral. So we can perform this Gaussian integral exactly, which essentially means we solve for the equation of motion of Dij in this thing, and we substitute it back. So the equation of motion is this. Dij is equal to minus 1 half times this object, minus i times phi 1 w. So let me remind you what phi 1 and w mean. So we have two complex scalars, phi and phi bar. They're complex, so I can try to parameterize them in terms of two real scalars, which is what I am doing over here. So phi 1 and phi 2 are two real scalars, and I expressed phi and phi bar in terms of them, and one of these real scalars appears over here. And now, if you are paying a lot of attention, you notice that this is quite a funny reality property, if you like. The complex conjugate of this is definitely not this. But yet, when I define this dagger, I will precisely say that phi 1 and phi 2 are real. So you see this is what Francesco is also mentioning this morning. The reality properties you impose in doing these types of integrals are not necessarily the natural ones. So phi 1 and phi 2 are real, and phi and phi bar are not each other complex conjugate, but there is an extra minus sign. So that is what this phi 1 is. Now, this w, concretely, on S4, you can think of w just as being some 2 by 2 matrix. It's symmetric, and its entries are simply this, where r is the radius of my round S4. So on round S4, you just have this type of a matrix. It's very concrete. But you remember from last time, we had a definition of w in terms of this S. So roughly, but then I need to define S. So anyway, look at last time's lecture to see what w is in terms of abstract symbols. It's OK. I can do this. I can integrate Rd. At the moment, that doesn't seem to have bothered me all that much. But the miracle, yes, please. The question is, OK, I integrated our drj, and the question or the worry is that when I integrated our drj, my transformation rules went off-shelves. That's the worry. But really, I know how they went off-shelves to speak. Drj is just this, so I just substitute this in the transformation rules, and I'm still fine. So maybe a different answer to your question is, before doing all this nonsense, I have guaranteed you that this thing will not change the answer simply because it's a Q-exact information, whether it's positive definite or not. That doesn't matter for the argument that there is no dependence on this parameter t over there. Once I am at that point, then I can start fooling around with performing integrations like this. And I know how these integrals will affect my transformation rules because I will just substitute this in the transformation rule. Yes? Are you happy? Sort and initial action is also depending on the. Yeah, but OK, that's true. So in principle, here I just looked at, I should have this parameter t floating around. Well, no, Qv is this, but in the action there is an extra parameter t. So everything I wrote here is dominant over whatever you get from the original action when I sent t to infinity. So you're right. In principle, from the original action, there will be extra contributions here, but they will be suppressed by 1 over t. But if I substitute that contribution back into Qv, I will get some contribution which would go to 0 or the wrong quantity. This back in here? Yes. Oh, you mean the excess stuff? Well, maybe it's 1 over t squared. It works. It's probably 1 over t squared because of the argument you just said. OK, so I'm doing this. I'm integrating out this dij, plug it back in here, and then there is a lot of work to be done. But at the end of the day, you will find almost miraculously that adding this weird extra term after integrating out dij still gives you a positive definite answer. So you get still a sum of squares or mod squares of stuff. So the squares are typically from the vector multiplets stuff because they are more real than hyper multiplets, but anyway. So you get a sum of manifestly positive definite things. Of course, I didn't specify all the dot dot dot, but you'll have to believe me. If you don't believe me, I can give you the reference. You can look at this paper where we spell out all the details in this particular case. Or you can look at the S3 version, or the S2 version, or the S3 times S1 version. It's always the same moral story where you find that this particular type of choice gives you a sum of positive definite things. Now, let me come back to Eli's question. The question was, why do I choose this thing? Now, the reason is now that you see in this equation of motion for dij, like a dij is equal to hij. And let me ignore this stuff for a second. So really, if I'm going to choose hij, which roughly speaking looks like the moment map, then I'm imposing vacuum equations. And this is exactly what I want. Is that clear? Sort of. It will become clearer later. So if I chose hij to be essentially what goes under the name of the moment map, ignoring this piece for a second, I'm imposing vacuum equations. And that is where I want to go. I want to get rid of this integral we had in the original computation. And I want to replace it with a sum over vacua. And this type of equation will precisely tell me, or already hints that that is what we will achieve by choosing this particular deformation. So the trick is really to make sure that if you integrate our d, that you get d is equal to what you can set equal to a moment map. It's OK. This is good. And now let me specify a bit. Let me try to specify what are these ingredients that enter there. Of course, it's an annoying computation, which would only waste time if I do it live. So this is the result. So it's very explicit. That's basically the message you should take home. So you see here, hij enters in these rules. If I hadn't included hij, if hij were 0, then these pieces would be gone, and then you would just be staring at your original localization equations on S4. So now I added this hij, and it enters contracted with this tensor theta, which I defined over here. It's some bilinear in killing spinors. Yes. So as well. So again, phi 1 and phi 2 are the real scalars defining phi and phi bar. And we already sort of remember this equation. We know that on S4, the localization locus was constant values for phi 1. And this is the equation that precisely imposes this constantness condition. So phi 1 needs to be constant in the original computation. So in the original computation, that means I ignore these two terms that are manifestly proportional to hij. This equation just constrains phi 2. It's not obvious that phi 2 needs to be 0 just from this equation, but if you later combine it with this equation, you will see that phi 2 needs to be 0 again in the original localization computation. If I want to phi 2 need to commute, that's fine. And these are the juicy equations. Let me ignore the hij pieces for a second. Then in fact, you can simplify these equations a bit more to find precisely the localization locus on S4. So if you ignore this piece, you can play with Fierce Identities to simplify this little mess. And then you find that phi 2 should be 0. The field strength should be 0. And that was the localization locus for the vector multiple, which you found in the original localization computation. Again, you've noticed this feature here. As in S tilde, they're proportional to psi squared and psi tilde squared. These guys vanished at the North Pole and South Pole respectively. So at North Pole and South Pole, we find these weaker equations. Because the coefficient is already 0, you don't need to manifestly impose at North Pole and South Pole that these guys are separately 0. So at North Pole and South Pole, we find instantons and anti-instantons. This is all still for the original computation. Now for my new computation, I have these two extra terms sitting in these transformation rules. And they mess up the last part of the argument I just gave. You cannot do these Fierce rearrangements anymore to argue that phi 2 should be 0. And that field strength should be 0, just because you have these extra pieces here. So what we really need to do now is solve these more complicated equations and see what comes out. I notice there's one symbol, which I did not define. Kappa is just the one form dual to the vector field V, where V is defined over here. OK, so this is this set of equations. And over here, we have a similar set for the hyper multiplets. In particular, you see that the auxiliary fields are equal to 0. So I didn't even bother introducing an optional description of law. It's somewhat necessary. And then you get three other equations. Essentially, phi 1 is sufficiently generic. Q needs to be 0. So all the hyper-multiple fields are 0, roughly speaking. Now you find also these differential equations, which are obviously solved when the Q is equal to 0, which was the original localization of locus. Now, because over there, we are changing a lot of things, you can imagine that a lot of things will change here. So we need to combine these two sets of equations and find new solutions. So that is what we should do now. The problem with the sum of the square, I'm sorry, but it's there after inequality. Is there no easy way to see that? Or is there a sort of guarantee other than try and error? I'm not sure. I think I did a try and error. The structure is the same, but it didn't seem to be always the same. Yeah, it's always like this. So probably there is some deeper reasoning. On the left side, the hyper-multiple is reduced to this very simple thing. We're saying the things up there that look complicated are reduced to the very simple things down there. No, no, no. These are four equations you need to impose. All I was saying is that if Q is equal to 0, then these things are obviously satisfied. And if you set f equal to 0. So in Coulomb branch localization, I claim that all the hyper-multiple fields have to be set equal to 0. This is the solution to this set of equations on the Coulomb branch locus, where the Coulomb branch locus was defined by essentially only having phi 1 non-zero and all the other vector-multiple fields equal to 0. But now we changed the bit, the equations over here by having these extra terms. So it's not guaranteed anymore that over here I will find nothing that I feel will find only the trivial solution before I'm modified. Yeah. The statement was. These equations have trivial solutions. So first thing we can do, or which we can observe, which I already said in words, if we turn off this extra term I introduced in the deformation, we get back the original equations. We get back the original localization locus, which, let me just write it explicitly, was phi 2 is equal to 0. The gauge field is equal to 0. Phi 1 is equal to some constant, which I call small a. And the auxiliary field was equal to minus Iawij. But now at least this relation you can easily understand from over here. If hij is equal to 0, then dij needs to be minus i phi 1 wij. And that's indeed exactly what you find here, which, of course, had to be true. I can always impose determinations, and this should always be true, even if I don't have this extra term. And on top of this, at the north pole and side pole, we found instantons and anti-instantons. It's OK. Let's go to a situation where we actually turn on hij. As I was telling already in response to Eli's question, we really like to set hij such that this dij equation is morally the same as imposing the D term equations in just the original quantum field theory. The D term equations set the right hand side of dij equal to the moment map operator. Let me write that thing explicitly, maybe just in the example of u1. You can look at the paper, or you can ask me for the most generic thing you're supposed to be writing here. But let me just stick to a u1 gauge group. For u1 gauge group, I would choose hij to be equal to minus zeta over l. Again, this w matrix. So here I introduce a parameter zeta, which you should think of as an fi parameter. So when I impose D term equations in a u1 theory, you can imagine that there is an fi parameter around. So in this q exact term, I'm going to have or impose that there is indeed such an fi term. And then I have minus i. Let me just split it, or maybe write it as a matrix. So the 1, 1 component is essentially q, q tilde. The 2, 2 component is minus q tilde star q star. And these diagonal components is q, q star minus q tilde, q tilde star. And the same over here, because this thing is of course symmetric. So this is what I will choose for hij. Again, the motivation is that if you just look at this right hand side, you should recognize these as vacuum equations in Enneco's 2 supersymmetric theory, where this object, if you like, I could have called it uij, like a moment map. If you're not familiar with vacuum equations in Enneco's 2, I just took q and q tilde, which are the scalars of the hyper-multiplet. So imagine, or recall actually from my first lecture, the hyper-multiplet is really equal to two chiral multiplets in complex conjugate representations. These two chiral multiplets, they contain complex scalars, q and q tilde, and these guys that enter here. In fact, belaboring this point, perhaps a bit too much. But you remember from Guido's exercise, maybe that he also had a structure like this in his vortex string problem, precisely for the same reason as we put it over here. So these are really ubiquitous expressions for vacuum equations. OK, so this is my choice. And actually, I will split this FI parameter in a piece, which I will call zeta vacuum plus zeta sw, where the sw will stand for cyberquit and eventually. So zeta has a specific sign, and I will choose this split such that both pieces have the same sign. At the moment, this seems like a little bit ill-motivated, but you will see why I do this eventually. So I will choose Hij to be this thing. And then with this choice, I'm going to have to solve all these equations all over again. And hopefully find some nice localization locus and then eventually find a new expression for the S4 partition function in terms of this new localization locus. You are thinking that the gage with the u1? Well, for the example I wrote here, I just thought chose u1, but the final result I'm going to leave with u1. You can do it for any gage group, but I'm going to insist though that there is a u1 factor somewhere. So other than insisting that there is a u1 factor somewhere, any gage group will go. How about the matter representation? The matter representation can be anything it wants. You just replace this simple expression with, as I was saying, with the moment map. So I could write it in all generality, but it will just be a bunch of meaningless symbols. I can show you after the lecture how I would change it. What is this? OK, I'm supposed to be finishing. Well, maybe I just finish here. Since the next step would be to take this choice of h, plug it in the d-term equation that is trivial, but more importantly, plug it in here and see which solutions pop out. But that's like a whole spiel, so maybe I'll just stop here.