 So, let me start and so I got a feeling that people at some point got confused but I will try to remove the confusion but I don't know, we will see. But the things will get more complicated so I basically recommend you that you understand the general structure, details you have to understand by yourself. So what I started yesterday to talk about index, so my goal is now, so today I want to do two things so I want to tell you about index calculations and also beat some of geometry and then try to apply this for 3D theory which is basically a toy example of 5D theory and tomorrow I will tell you about 5D, so full scale calculation. So we are discussing with you index theorems and the main thing is that what I told you that we are interested that in the operator D, so which is elliptic operator and then we are interested that for elliptic operator, so I am reminding that on compact manifold this is Fred Holm. So then for operator you can define the index of operator which is the same as dimensions of the kernel minus dimensions of co-kernel of D, so just that you know this is the same as dimensions of kernel of D dagger. So the magical thing is that about this operator, I mean typically this index that this is topological quantity, this is not but this is it is. So let me suggest I came last night, I mean just some good exercise to do, very simple exercise but sort of very nice to see. So consider S1 and then consider the following operator, my lambda is a complex number. So please calculate for this operator index and also calculate dimensionality of kernel of D and dimensionality of kernel of D dagger. So in this example what you will see that as you are, so the answer what is magical is what is good in this example that these things will be always the same for whatever lambda is. These things will change as for different lambdas because you have to find the solutions and of course because you are solving this in the circle you have to impose periodicity and it may not work for arbitrary lambda. So these things may change as you change parameters of your operator, this does not change, this will be always the same number. So that's why it's topological environment. So in a way what is important that this thing depends more on topology of this thing and it does not depend very much on this operator. So this is one important thing when I talk about index theorems, so this is one exercise. So basically if I'm talking, so typically the index theorem, the standard thing is that it's called ATIA, both theorem and again, I'm skipping details. I will give you just a rough idea, but roughly speaking is the following that this guy, so this is typically called when you solve it, it's called analytical index, analytical index. So in principle like in this problem you're supposed to actually solve your problem and count solutions. This is equal to some quantity, let me not explain this very much. So basically this is generalized Gauss-Banne theorem. So this if you want, this is some characteristic classes. So in principle if you want to calculate the index of operator you have to construct some objects out of curvatures, et cetera, and calculate it. But this quantity is actually independent of concrete connections you do, et cetera. So in a way this is invariant, I mean this is topological invariant. So for example it may depend only on the bundle and things like this. So this is part typically called topological index. And the whole beauty of index theorem is just saying that analytical index equal to topological index. So the magical thing is the following that if you actually want to calculate the difference between dimensions of kernel and co-kernel, then eventually it depends only on topology. So actually there are many variations of the theorem with many names and different variations. So in principle, I mean this you should not understand very literally because there are different classes, et cetera. And depending which operator you put, you put here Daalbu, you get one version of index theorem, you put here Dirac, you get slightly different, et cetera, et cetera. So in principle in old days when this theorem was proven, what you basically wanted to do the calculation, you need to know how to manipulate these characteristic classes, et cetera. So just let me remind you because I will need this for generalization. So characteristic class is the following concept. So I will need the equivalent generalization of this. So if I have invariant polynomial, so I have invariant polynomial. Which I can denote as just something on the algebra and it's invariant. So it's invariant about joint action. Just let me give you example. So example of course, example, I mean if I take, it depends for which group I'm doing. But for example, definitely Fathian would be a good polynomial if I'm looking for just orthogonal group. So under orthogonal group it will work. Or different versions of determinant or trace. I mean, this will all give me a polynomials. Then typically what I supposed to do, so I'm supposed to take this polynomial. And then if so, I have my bundle, I have for example some vector bundle of M. And then I take this polynomial and I put some curvature in there. So pick up any connection, put the curvature, put it in. And then basically why you call characteristic class? Because you can identify this with element of Dirac homology. So as element of Dirac homology, this is independent of concrete connection you are choosing. So there are standard theorems. So this is typically called churn while isomorphism. And this guy as element in homology is independent. So you have to prove first of all that it's, for example, closed, okay? And next thing is that you have to prove that if you change your connection. So in fixed bundle you change your connection by anything else. So you go from connection to connection. Then this guy will change by view of something. So this is standard theorems. You can look at Nakhahar how to prove it. It takes a couple of pages. But the important thing is that this is a way to produce something which is invariant. And then this object you take and you plug it in, etc. What I'm skipping, there is a whole list of classes depending which group you take. I don't know if you change characters, change classes, etc. Okay, then there is one important thing which would be important also in applications. So there is this notion of elliptic operator. There is also notion of elliptic complex. In fact, this is a special case I will tell you in a moment. So here typically what you're interested in, so your operator, in here what you're interested in typically you have one bundle, you have another bundle. And you look just some operator which acts from section to sections. So for example, I don't know, example. So think the operator which acts from zero to, I don't know, a one zero, right? So this is example, for example, operator. So in most of the examples, it's important that you write this. It's natural to continue the things. That's where notion of elliptic complex will typically arise. So you start and you write one bundle, you write another bundle, etc., etc. Yeah, so now if you wanna talk about ellipticity of bundles, so you have operator. I mean in principle it can be different operators. I don't d1, d2, etc., d whatever, m minus one. So you have to actually look at the symbols. So you have to look at symbols of these guys, etc. And typically you have to see that with respect to these symbols, this is exact sequence that the kernel coincide with the image. Again, it's a matrix calculation. I mean, what I'm telling you, it maybe sounds complicated, but it's not. So let me give you example. Again, this example, typically whenever I have just one operator in F, I just always can make a complex by just writing these things. You can ask, why do I talk to you about this nonsense? It's actually not nonsense because in gauge theories we actually need a complex, not this. So let me give you example and actually I suggest you to calculate. So I hope you did exercise, right? Yesterday you did this as a following exercise. It's from yesterday. I ask you to prove this thing that this is elliptic operator, right? So this is these two equations, it's elliptic PDE. And again, here I've just looked at a billion case for simplicity. So in principle, if it's not a billion, we're interested in linearization of the problem. So now that actually what appears in gauge theory, this problem, I mean, when we will do calculations to more of a 5D theory, etc., then actually there is associated elliptic complex. So let me write for you the things. So this is exercise for today you can do. This is a bit more complicated, but still it's a linear algebra. So I have a one form on form manifold, then I have, sorry, zero form, one form, and then cell dual two forms. Again, when I'm writing complex, I can put all this put here to zeros. So here it goes to DRAM and this is goes to D plus. So D plus is just the following operation. So I'm just taking a plane one over two star to DRAM. So I apply first DRAM and then I project. So this is definition of D plus. So show that this complex is elliptic. So typically the thing is that these two problems is related the following way. You're dualizing these things, again, sorry for confusion I realized. So this is plus, this is plus, this is dagger, okay? This is plus, so dagger it means conjugated, so this is actually plus. So the two problems is related, you just fold this complex and you're doing like this. But engage theory actually, I mean, you have these things because these guys where your ghosts sit. So in future, this is will be the field C, which is called ghost. And then it maps here, this would be a connection and that maps to here, this would be typically the field chi, which I will tell you later on. So please show that this is complex elliptic. Basically what you have to write symbols of these guys and show that this is exact sequence. If you're confused by any definition just Google it. Yeah, it's just it's a definition of elliptic complex. That's how you define things. But the conversation of the dagger together with the- I mean, you have to do the compositions a level of symbols. So if you will do it, I mean symbol composed one guy with another guy will give you zero. I think you can just use the d plus as projector followed by d. So you get projector d squared, which is zero. Right, Google it, okay? The answer for everything is Google. So let me keep going because I'm afraid that I would not have time, okay? Now what else I wanted to tell you? So please do this exercise and that would be important for tomorrow. And again, it's not that complicated what you may be confused. I mean, how to do it, but actually write the symbols and you have to prove that kernel of one operation is coinciding with the image of another. So there is some linear algebra and some properties of cell dual forms. Okay, right. So why do we need this stuff, etc.? So actually this is very nice, but I mean, to work with these things, it's a bit complicated. What we actually need, we need a better version of this. Well, not better, but we need version when on the manifold we have some symmetries. So we would like basically now to go to a equivalent version of index theorems. Again, let me give you a warning that the subject, if you do it properly, it's a complicated subject, there are a lot of subtleties. So what I will do, I will give you a rough idea how it works approximately. So if you wanna study, you study this by yourself, you read books, etc. What actually for me will be important, I will work out some examples. And that's exactly examples which we use in physical theory. So right now my presentation, I'm not giving you anything systematic. I actually give you a rough idea how it's gonna happen. So now in our story, that whenever we would have a bundle. So the key thing is that now we'll have some action of the group. Typically it will be torus. So we'll have some action of the group here, G, okay? So then I would like always that this section will be lifted to the whole bundle, so it will be lifted here. So we'll deal with what I told you if you are in bundles. So typically again, I told you that you lift it, it should be compatible with the typically projection and it should sort of act nicely on the fibers. And if it's principle bundle then two actions should commute, etc. Okay, so then as I told you here, again, the story is the following. Then whatever you do, usual geometry. So here what you had, you had a notion of Diramka homology. You had a notion of connection, curvature. And for example, the way of producing characteristic classes was taking a curvature, plugging in an invariant polynomial, and getting an element of homology, and then you're doing. So now you do exactly the same. You just attach to every word, adjective, equivalent, that's it. So it sounds simple, but of course it's not. So you have to, the bundle, associate a coherent and connection, typically I don't know, I will use, for example this thing. So typically it's a usual connection plus some formal guys, so my VA. So I have a vector fields V associated to this. So this would be a current connection. And then my current curvature will be F, G. So in localization when we discuss gauge theories, I will tell you more concretely what's actually appearing. So what is important here to realize that, for example, curvature typically, I mean if you look locally, it's a usual curvature, it's a two form. Now in a current store it will be always two form and zero form. Okay, so this is formally if I write a current curvature and then the story is the following that, so I have now my invariant polynomial, invariant polynomial. And then I will produce this so I have to put an invariant polynomial basically this object, okay? Now the statement is that whole thing is this churn-wild isomorphism follows. It just becomes now element of a current homology. So this is no current homology. So again, my intention here is just display for your schematic things. I suggest that you study this yourself. But the story is very much similar like this. You will see in a moment what's the catch while you are doing. Because right now it's not a big deal, okay? The thing is what is important and this is very, very crucial that it's element of a current homology, right? So if I would like to write some integral for your M on M, and then I put here something whatever M F of A G, right? Or some combination of them. The catch is the following that if I write this, this will be killed by this guys. Because it's element in homology, right? And then during first lecture, I told you that if I have a form, which is equivalently closed, then this just becomes nothing. So I apply now another formula, ITIA, what is ITIA, I mean ITIA is everywhere. I mean, it's also both. Sorry, this is actually called, I mean, this is called ITIA singer. So index theorem is typically called ITIA singer, ITIA boat, it's a localization formula, sorry. So ITIA boat, what happens is the following, that it becomes a sum of fixed points and then, well, you get here whatever you have to evaluate your form at Laus degree and then you get something here. So then why we like with index theorems, actually we get a way to calculate things much faster. So what we actually need, we need to know, instead of calculating, so, yeah, question. I was just wondering if there's any, how things will depend on epsilon, epsilon A? They will depend on epsilon, absolutely. I mean, here I'm supposed to be very pedantic, I'm supposed to put this, yeah, yeah, yeah, yeah, yeah. As I said that I'm not giving mathematical lectures would require too much. I'm giving you idea, so the whole thing is that this is complicated, typically. We have to know a lot of topology. Here, with the Q-variance, actually everything will be reduced to fixed points. And then actually I will give you example that what you need to know, you need to know to calculate index theorems, basically information around fixed points and life becomes very simple. Again, to arrive to this result, you need to know to do a lot of math, but this is the idea why we like them, okay? So this is, sorry, I may be missing the obvious here, but F is a curvature, is a field strength, so it's essentially two form, isn't it? So how can a polynomial of a two form have like a zero form component? No, no, but this is what I'm saying is that curvature anymore is not a two form. Curvature has two components now, it's a two form and zero form. One of the thing is that you cannot solve, so what is important that if you think I told you how do you solve these conditions? I mean, the way to solve these conditions, it's only to have combination of two forms. So in supersymmetry, I don't know if you saw already, there will be always this localization locus. In whatever dimensions you have, you will have this. So the statement about this is a following, that this is actually the following thing. So this is correct notion of a curvature now. So your characteristic class of, for example, if you wanna do the things of, for example, your characteristic class would be a trace. So first-gen class, a current first-gen class will be trace of F plus five. And now the thing is when you will calculate everything, so in principle, integral of these things will depends on the values of five at fixed points. But whatever theory you take, you will always have this locus as a one law of localization. Two, three, four, five, six, seven, whatever. I mean, it's always will be a part. I mean, the current bundles everywhere in the supersymmetrical localization. Okay. Right. So let me say a bit of more general nonsense and then try to go to examples because it's in a way more productive. Right. So basically, if I wanna give you the version of index theorem, so the analog of that formula, so let me write you the following. So index of D bar operator for the twisted bundle because all my examples will be about Dalbo operator. That's because we need for three and three and five decalculations. So this is basically evaluated at the group element. This would become one over two pi i and then there will be integral over tot class, equivalent Tm and chen class of my bundle. Again, everything now becomes equivalent. So you replace everything by equivalent. Now, if you apply next further on iTabot theorem, then actually what you get, and that's what we actually need. I will get this. It's a sum over fixed points, trace of e x at the group element of g, determinant of T of x10, one minus g1. So what you typically do, index you will evaluate at the group element always. So now I will tell you that if you put this element to trivial identity, you'll just arrive to usual index. So in a way, this is much more powerful formula. So I will give you example. So I'm pretty well aware that I wrote for your formula. This formula to use more or less nonsense. This is nonsense. I mean, you just simply, you know, I didn't define for you. The whole catch is the following. If you ask me to calculate the index of operator on the manifold with a action, actually what I need, I need to do a linear algebra at every fixed points. I have to collect certain data and then I can write my disinformations and then, for example, I can put my group element to one. If it makes sense, I mean, and then get answer. So from now on, I will do the example. So I wanna, this is a general theory. I think it would no sense for me just to tell you this stuff. So it's much better to do example and then you will have a pretty good idea. Okay. Yes. So let's concentrate on the examples and, but again, let me still warn you. So I will consider two examples. Of course, CP one and CP two. But don't get misled that if you wanna actually apply for module, manifolds, et cetera, there is a lot of subtleties. So you actually, if you're ever going to write a scientific paper about this study subject first, don't refer to my lectures, okay? So CP one, which is the same as S two. So I actually advise you to fill the CP one. So actually try to do different coordinates, et cetera. So this is defined in the following way, right? So I take a zeta one, zeta two. Right? So this is which belongs to C two minus origin. And so CP one is defined. So CP one is defined as the following thing. It's a zeta one, zeta two identified with lambda, zeta one, lambda zeta two, where lambda is a C star. So C star, it's typically a notation C minus origin. Now, so there are two patches. So you can cover two by two patches. So first patch is the following. It's, I can introduce four coordinates. So let me see which coordinates I wanna choose. So psi, which would be zeta two over zeta one. And zeta one should not be equal to zero. It's one patch. And another patch is, so again, this parameterized me well, because if I'm doing this, I can always parameterize my things because it means that if zeta one is not zero, I'm basically saying that this point is always the same as zeta two, zeta one, one, right? They cannot do both simultaneously equal zero, but they can be equal to zero. So this is lambda is zeta one over zeta two. So zeta two is not equal to zero, okay? So this is what you are doing. This is one, zeta two over zeta one, and this is zeta one over zeta two. And of course what you can see here that on intersections, on intersections, u one with u two is psi goes to one over lambda. So this is standard, you know, gluing two things, okay? So this is would be one of very important examples. And actually this was related to S three. Now remember that I have a u one action on my CP one. So this, if you wanna write your like pictures, you write this is just a rotation of this guy. So rotation of the sphere and there are two fixed point that is north and south pole. If I just cut this in this coordinates, so I can take a parameter. So u one action in this coordinates is just the following. So this is psi to T psi, and then lambda to T minus lambda. So again, since it's u one, I will typically assume this, right? Okay, so this is the action. So then let's do the index theorem. So for example, let's pick up my e to be zero zero and CP one. And then my another bundle, let me say this will be zero one on CP one. And obviously my operator, so my operator is del bar, it maps e to f. Okay, then the formula which I wrote there, what you have to do now. So let me first write for the formula and then try to explain every term. So the index, the statement is the following. The index of del bar operator is one minus T minus one, one minus T, one minus T minus one plus. One minus T, one minus T, one minus T minus one. So if I actually will put T to one, then upon the thing, so this would be equal to one. So if T is equal to one. So what are we doing? So this is contributions of two fixed points. So actually you have to look locally and you have to stare at this formula a bit. So what you have to write, you have to write, so I will do it now, but then I will skip all the things, et cetera. But so I introduce for you the things. So what I have to now remember how these guys locally act on different forms. So for example, side bar obviously goes to T minus one side bar, dick side bar goes to T minus one dick side bar. Dick side goes to T dick side. So lambda bar goes to T one. So this is just lambda bar, d lambda goes to T. So all the things of course, follows from this perspective things. So here for example, fix this is just, I mean follows from my change of coordinates, otherwise it's inconsistent. But the whole data eventually to calculate this contribution just need to know things on C. So whole data is associated to C basically. So then this formula, which I wrote over there, so these things is understood. So d lambda bar goes to T, d lambda bar goes to T, right? Thank you. So actually what I have here, so my index, if I take this formula and write it better, so index of del bar is equal to sum over fixed points, which is two here. And this is just, I have to take a trace of at E at my fixed point of T minus trace of F at T and then hit determinant of a tangent bundle, one minus row of T. So let me, I don't know how much it's obvious. But what I actually have to do, so for example, everything boils down basically for calculating, I mean my index and this quantity is locally. That's it. So everything is done here at fixed point. Again, here's example, I mean the fact is that if I do, so these things cancel, these things cancel, and then I can, if I don't try this, I will just get a current guy. Again, the thing is that if things cancel, this is very obvious what you get. So in a way, the fact that your index is one over T. I'm writing for you things a bit more complicated than they are just to confront this formula. But in a way, these formulas are not that complicated because this is nothing, it's a one over minus T and this is the same as a K from zero to infinity, TK. That's exactly example I was giving you last time. And here's the same thing, but here everything will be expanded T minus one. So in a way what you are doing, you're just gluing two indices incompatible way of Dalbo operator on C and C. So this is beauty. The thing is that you have to put this thing just to confront more general formula. But in principle, this formula is simpler than it is. So this is example number one. It's not actually what we need to calculate. For calculation, we need something more. Sorry, one question. Yeah. You're evaluating the formula T equals one. Say it again. You are evaluating the formula for the index at T equals one. Yeah. And then, is it like it will define? It is it will define, yeah. But you have to expand it and then evaluate it. What I'm trying to tell you that, so one, if you would do without TQ variance, you can get one. That's the only thing I'm saying. But so let me see how to, well, that's one. I can maybe during the exercise class I can tell you this. I mean, in a way what's the only thing that I'm trying to tell you that there is a way to get. So this is non-accurrent index. Non-accurrent. Well, it looks like it's just one without any limit. Yeah, it's, yeah. Right, so it's one over minus T plus one minus T minus one, right? So this is, yeah, exactly. So this is, so this is T minus one, T, right? Right, just one minus T minus T, right here. That's one. But this is exactly non-accurrent index which you can calculate right away. So it's a representation, whatever you choose representation is acting in your forms. So it's representation. So here it's very trivial. I mean, this is exactly what you get here. But sometimes when you have a bundle, you may have something more complicated. So this is just obstructive. Okay, let's move any other questions. Yeah. So what exactly at EX and FX is the matrix given the action that you're taking the trace and determinant of? Because it's acting on the, I mean, because if you wrote one minus T and one minus T inverse, I'm guessing it's some diagonal matrix of the form one minus T, but I don't see where that, how you got that from this transformation. I don't understand the question. And I don't think I want to answer your question because I mean, I'm telling you, I feel very stupid. I'm not giving a lecture on index theorems. I just need this a tool. Those guys ask me, et cetera, but I actually have to give you like 10 lectures to make a sense what I'm saying. And I will now systematically avoid any questions because I will not calculate for your partition function of S3 today. If I don't, if I start to answer your questions. So just admit that you don't know what I'm saying and then you study. But I told you right away, I mean, it's, I don't know, I feel actually stupid because I have to go and explain through this formula. I have to explain you how you deal with the things because typically this is distributions. You get delta functions, et cetera. So typically the index, a current index is not a function. It's a distribution or group value distribution because next thing you will start to ask me how to expand the things, what I'm start to do for you, these things. And the more I start to calculate, I mean, the more I will, there are more, I have to go to distributions. I don't wanna do it actually. So I feel a bit stupid, so I admit it. So let me go to another example which we actually needed. So Google it and I would be happy to answer your questions during exercise class. So actually what I wanna do another example. So what we need, we need with your O and bundle now for Cp1, okay? So the way to construct it is the following. So I will tell you the definition and I will give you some exercises later on. So the definition is the following. So now, so this is a line bundle over Cp1, okay? And it's constructed in the following way. So let me take now three copies of numbers. So zeta1, zeta2 belong to C2 minus origin. Well, zeta3 just belongs to C. And then I will identify them in the following way. Lambda, one, lambda, zeta2, lambda and zeta3. So if you do this identification you will get a total space. It's a line bundle over O and bundle over Cp1. So what you can do here, you can do similar things what I told, I mean, so I have to introduce coordinates. I mean, you can convince yourself why it's line bundle, et cetera. But so the coordinates is the following. And xi1 is equal as before zeta1 over zeta2. And xi2 is equal zeta3 over zeta1n for zeta1 not equal to zero. And then u2 is lambda1 is equal to zeta1 over zeta2 for lambda2 is equal zeta3 over zeta2 and zeta2 not equal to zero. So on the intersections, this would be the following xi1 is equal to one lambda1 xi2 is equal to lambda2 lambda1. So I hopefully I'm not just to be faster. I don't want to actually derive this formula. So hopefully I'm just copying from my notes and I'm not making any mistake. So now this is an already example of HQR and bundles. So now I'm having this u1 on. So I have my u1. So this is u1 on Cp1. No, it's not correct. Sorry, here should be zeta1, right? Sorry. So now what I have to do, I have to postulate now how the section extends on a whole bundle. So xi1 goes to txi1, xi2 goes to xi2. And then by transformations, I have to, I mean, this line I can postulate. But next line I cannot postulate its derivable. It's a t minus one xi, sorry, t minus one lambda. So this is what I wrote over there. But then for this coordinate lambda2, it will t minus n lambda2. So this is my line bundle, okay? So let me write the index theorem and then I will give you a bunch of exercises to do and you will understand why we are doing Cp1 and why we are doing On. So here my bundles will be, it's two comma, sorry, zero comma zero on S2 with a values in On. And F would be zero, one with a values in On. So my Dalbo operator now, it's a twisted Dalbo. It will act from E to F. So the index will be given by the following formula. One minus t minus one, one minus t, one minus t minus one, plus t minus n, minus t, one minus n, one minus t, one minus t. So then eventually, we will evaluate this. So we will get one minus n. That's what I'm interested. So let's do the following. Don't ask now questions. Try to understand these formulas and tomorrow we have exercise classes then you can torture me, okay? Because actually my goal is to show that why this formula is useful using today. So let me do a number of exercises and that you understand why, I mean what we are doing is useful for us. So I'm going to do it. So it's a lot of now will be facts and I ask you to check them. They're very explicit, very nice, et cetera. So I have S3. So S3 is the following things which is written in C2, right? By definition. So then S3 has a T2 action, okay? So T2 action corresponds to the following. Zeta one is to e alpha zeta one, zeta two to e beta zeta two. So you rotate the things and this is of course preserves this equation. So this is T2 action on S3. If you take diagonal one, so when these numbers are coinciding, this corresponds to diagonal one, diagonal one. This corresponds to U1 acting on S3. And this is exactly the action corresponds to hopf vibration. So S3, two. So what I suggest for you to do the following thing. It's actually, I mean the best thing to understand what I'm telling you is just do a very concrete local coordinate thing. So just there is a very wonderful coordinates for parametrizing of S3. And this is adapted to the hopf vibration. So the coordinates are the following. It's one, so what did I use? I used one plus xi, xi bar e theta zeta two is to e theta xi one plus xi xi bar, right? It solves this condition. So now what I ask you, this is coordinates adapted to hopf vibration. So what I'm asking you, go to lambda coordinates when lambda is equal to one of xi. So I'm using these coordinates which I'm used over there, okay? And I would ask you how theta tilde will change. So this is not globally defined coordinates here, right? So there are two guys. So locally you can see. So locally this is actually C times S1. Then you do this change of coordinates. You will get another C times S1. So basically theta will get a shift related to argument of xi. So that's why when you glue this sphere, it's not actually S2 times S1. It's a twisted thing. So as you go around, you will get this twist. So this first exercise, please do it because this is very nice. And quite often people say a lot of this formal nonsense but if you didn't do ever in your life this type of calculation, this is bad. Then you can ask why do we talk about ON bundles? And that's the whole thing. Again, this is you do it yourself. So my interest is the following. My interest to take a P form on S3, right? So if I look at this thing, right? So if I look at trivial vibration. So if I look at trivial vibration. So I'm preparing grounds for calculating determinants. I will not tell you all the math, right? So you remember I told you this example. So for example, if you take a P form on S1 times S2, right? What you will do, it's natural. You just take every form and you expand for your modes here, right? For example, I mean, what would be much more interesting we would be interested in horizontal forms along S2 which doesn't have any legs. Then you expand in modes. And for you it's over that this is equal to sum over N and then you would have something like, you know, P form, I mean, N's component over S2, right? I mean, it's a very trivial statement. So just take every form, expand for your modes, et cetera. What we're interested in here is the following. We take any form on S3. So locally it looks like C times S1. So you would like to expand it. The main thing that S1 is not actually fiber trivially. So the correct statement here, what you will do. So for example, when again, you will take a horizontal forms. It's the following. This would be a sum over N and this would be a P form horizontal on S2 while you know N bundle. Please prove it. So this is exercise, prove it. So the way to prove it is the following. Of course, first you write these coordinates, okay? And naively you would say, okay, I can expand in the things I can expand with the mode. And then you have to go around and come back and see how things are changing and they will change. And then change will be the same as related to N bundle. That's how you identify that this is correct thing. So this is a generalization of Fourier expansion here. And so whatever in future, well, we will do core, I mean, we will do today, hopefully, we will do the determinants of operators on S3. The trick is the following. Whatever we do, we have to reduce on S2ON, then we basically are operator, what I call Big D will be Dal Bu. We will have huge constellations and we actually need to know this N minus one. And this will give me answer, okay? So I suggest for you all to do all these exercises. Are there any questions? So now I will switch to physics. So remember this N minus one and remember this fact so that we actually can do Fourier modes on S3 with respect to your one action and any questions. Okay. Are you going to describe index on S3 or not? I don't wanna talk about index anymore, so that's it. So you just, right now what I'm asking you to remember the following facts, that there is some magical thing I calculate index for you, it's whatever I raised here, right? Then what I'm asking you to prove this fact, oh, to understand this fact, okay? The question is, if I've raised this formula then, if I want to calculate some kind of index on S3, I just have to sum up over this. I'm not interested in the index on S3. Okay, so why is this fact is useful for us, if you're not interested? Why this fact is useful? Because what I will do, you remember at some point I wrote for you this formula on some kernel of D and I can write it kernel of D dagger, right? So now I will write for you now a gauge theory on S3. I will have really derivatives here of U1 and this will be just Dalbo operator. So I have to actually know what's a mismatch between norms. So here typically there will be two spaces. One will be zero forms and other will be zero one forms. I have to know mismatch between how literary effects on the kernels of Dalbo operator and the mismatch is given exactly by one over N. So, you know, that's why I was basically telling you maybe message didn't come across, maybe it would be much better than I do calculations if they tell you the whole structure, but we always will calculate this. So in a way I don't care about S3, eventually I will reduce everything there. So for me it's important, right? I mean I have this lead derivative. So with respect to lead derivative which U1 I will have, I have to do this mode expansion and then I have to see how things are. What's a mismatch? So let me try to now introduce for actual physical theory. Okay. So as a toy model, not very toy, but anyway toy model for 5D, I will now calculate for you a Chen-Simons partition function in not the way we tend it. So it's a very nice exercise. So I would like to calculate. So the goal is to calculate 3D Chen-Simons partition function on S3. So my money fold, a big one, I mean. So I will have now, so the setup is a following, that I have some principle bundle. So this is now gauge group, gauge group over my S3, right? So my theory would be the following. So I will have a Chen-Simons. I choose a connection. This will be kappa over 4 pi trace of ADA plus 2 over 3A cube. And what I would like for you to discuss, I would like to integrate the thing of space of connections, right? So let me call this a space of connections. So in this example, there is very little topology because this is simply connected. I will typically choose the most semi-simple, simply connected group. So this is bundle over the S3. So there is nothing to worry about. Okay, so this is space of connections. At the same time, what I would like to do, I would like to try to construct my thingy equivalently. So right, I would like to doing the same way I did. So what I will do is the following. I would like, so in this story, what is important that I have this section of U1 here. So for simplicity, I would assume that my action is just hop fiber. So this is diagonal action, just for simplicity. Then you remember what I told you, right? So this is our thing for atia-bot, so something V mu of X. So for example, if you're very, very pedantic, this is where, I mean, I will have here parameter. I don't remember how it was denoted. Let me write phi A V A mu of X. And that's, so if I write everything, so in usual atia-bot, we ignore this parameter. So let me try to mimic this. So I have my connection and I will go to psi. So psi is odd one form, odd one form. Delta psi is equal to, so now what I would like to do, so which symmetries I would like to make it a current. So I have a space of connections on S3. So on S3, there is a symmetry action, for example, rotation. So obviously this symmetry will survive on my space of connections. So what I can write for you, I can write LV A, plus I will be allowed to write some gauge transformations here. Gauge transformations. There's some parameters, so let me be very explicit. So this will be D A phi. And then I would say that, so that's what I would like to try to do. So there is number of problems right away here. So this is sort of very natural thing, I mean, to write. But of course there is number of problems and you have to protest because if A transforms a connection, how derivative transforms, it does not transform well. What you can do actually, you can redefine slightly, I mean, fields, so you can pull out some things from phi, et cetera, so you can rewrite these transformations in a bit better way. But this is, in a way, nothing. So this is rewriting, let me see if I can, so you can check that. So F is just field strength for A. And phi is just i sigma minus i r A. So now sigma, so sigma is a zero form on S3. So there is one thing just for conventions, of course they're all the algebra valued. And all fields, connection will be connection, all other fields will be in a joint representation. So I will skip writing this G, but I'm assuming that there is a algebra here. So this is my model. Now, you can ask, can I write the extension of Chin Simon's section, which is in the variant under this guy? And the answer, you can do it. So the supersymmetric Chin Simons will be the following. So I take this Chin Simons over there, and I will write A minus i kappa sigma. I will define for you the things minus kappa over four pi kappa by chip psi by chip psi. And again, here where we trace as every single algebra value, et cetera. So you can prove the following very simple fact. Again, let's say it's exercise, you prove it yourself. That delta of this supersymmetric Chin Simons is equal to zero. If the following things are satisfied, if kappa is the one form on S3 with the property that IV kappa is equal to one, and IV d kappa is equal to zero. So this is observable. Sorry, I'm having these bad habits of using V and R. R is because it's contact geometry. It's very, in many places, people use R because it's called rib vector field. Sorry, it's just because I use different conventions. Okay. So now actually if you start the section, if you play with this section, it's a current extension of Chin Simon section. In a way, you could start from this term and just reconstruct Chin Simons by requiring to be invariant of this symmetry. So in a way, you can do things other way around. Again, remember, so what I'm having here, so my fields A and psi, this is coordinates on this huge space. One of the things is that when we go to infinite dimensional spaces, I mean, we cannot talk very much about smooth geometry, et cetera. Although this space is not so bad. This is a fine space. I mean, space of connections are contractible. There is not much topology. But anyhow, what is important that we have T1 naturally defined here. So this is why I was stressing to you. So why localization works because there is a canonical measure here. Again, it's sort of hard to comprehend, but I mean, we impose on this geometry here. Okay, and this section is actually a current extension. So this statement is exactly what I told you in finite dimensional case. Okay. So now the problem is that you would like to calculate something in the theory. And unfortunately, it's not enough to calculate and you need more fields. And you need more fields. There are two ways of explaining the things. And of course, they're more or less the same. First way is a physical way because if you look at supersymmetry in 3D at vector multiplet, there are two fields are missing. One boson, one fermion. Mathematically, the way I explained to you, it's important when I would write this sort of Birst exact terms, et cetera. It's important that operators which I get there either elliptic or transversely elliptic. So I will not be able to do here. So actually the full set of fields which I have to add, so I have to add more fields here. So the full transformations will be the following. Then I add chi is H and delta H is L, V, A, chi minus i sigma. So here's the conventions is that L, V, A. So in this sign, there is a coherent version of literati which is just a D, A, I, V plus I, V, D, A, which is the same as a derivative plus I, V, A. And I'm not using physics conventions. I hate those size. Okay. So this is a full multiplet and mathematically we actually need otherwise we will not get good theory. Now, physical thing is that, so this thing is actually one to one. You can map to N equal to vector multiplet in 3D. So there is exist a map which you can construct. So for example, if you take S3, you look what Kapustin et al did, I mean they constructed killing spin or et cetera, you write all this ugly supersymmetry, then you take bilinear and you map to the forms. So it's one to one map. Again, the thing is that actually the magic of supersymmetry, so supersymmetry is smarter than we are. So many people do supersymmetry, but they don't understand why it's valuable. I mean, in this supersymmetry, we guarantee to get correct operators. So all our operators will be nice, elliptical, reality conditions guaranteed, et cetera. But I think I'm trying, you know, I hope Guida told you everything about supersymmetry, so I'm a free person not to talk about it. Okay. So if you don't know supersymmetry, you need to come up with some, I mean, you need a guess where it requires a guess and what kind of use it has to obtain the transverse elliptical operator and things like that. Well, it requires guess, but I mean, there is very systematic procedure. So this is basically theory comes from the lift. So the story is the following, I mean, this you can understand. In fact, it was pointed out in 90s by Nikita, Losef and Bollet, I mean, it's not very cited paper, but they were saying that you take elliptic problem and dimensions and this is just a natural lift in D plus one dimension, et cetera. So there is totally systematic way to do it. So I will, in a moment, I will write for you some operators and actually I will confront what I told you during first lecture why our operators, et cetera. So in a way, everything works and works beautifully, mathematically and physically. The main thing what I'm trying to stress that, I mean, I used to work in, you know, in supersymmetry with people like Rocha, et cetera and he would always say his phrase was, it's super feel smarter than we are. So this is incredibly good thing because you learn a few tricks and supersymmetry, it tells you how to do things. But eventually if you write everything, wow, everything works. I mean, operators elliptic determinants, a calculable, et cetera. You try to deviate from this framework and everything fails. And supersymmetry are smarter than we are. That's a statement. Okay. So now I have to write for you bearstick zhakturm, right? So this is again, always confront the things against what I told you before, right? And the answer was actually, so you get something. So evaluate a zero form, et cetera and this is determinant of dv. So let's do it. So let's write this bearstick zhakturm so delta w will be delta of psi mu star delta of psi mu bar plus chi, sorry, I didn't tell you that chi is odd zero form and h. So this is odd, so fermionic. And this is even zero form. So now if you count odd forms here, this is one form, it's a three component. This is zero component. So in total four components, this is exactly a Dirac spinor in four dimensions. So that's why there is a map. Then you write this and then you will write star h minus f horizontal. So I didn't tell you. So of course in this setting, I will presumably discuss more tomorrow. So I have these conditions, IV on kappa equal to one. So I can decompose any forms to vertical plus horizontal guys by just applying a projector. This is kappa of HIV. This is one minus kappa of HIV. So just now let's think about, so this is a whole catch is that I'm looking at two forms in three dimensions, right? So then I would decompose a vertical part of two forms plus horizontal part of two forms. So how many components vertical part will have? I'm asking students that you don't decouple. Okay, let's give you a simple question. How many components f has in three dimensions? Be brave. How many? Three, good. Then you have to take three. So what's the decompositions here? No, two plus one. So it's two, it's one. Because here I take f and contract with v. So the only index is one free. It's like a vector, but it cannot be a long v. So it's two components and this. So it mean that f horizontal, it's actually one component. So you can think this, I mean, actually you can map it naturally to zero forms. That's why I'm writing this condition. It makes sense, okay? So now moreover what you can see, so here always exist. If you look at this example, there is a compatible metric. So every actually round metric, if you take v and contract, you will get kappa. So actually these two spaces are orthogonal with respect to scalar product. Prove it, exercise, exercise. If I choose for your metric, I choose this decomposition, they are orthogonal. So the whole catch here, so if I start to write the things again, so I'm looking at bosonic things and I combine, I will get f star f. Again, this term gives me f horizontal square. This term gives me f vertical square. Plus da sigma, h star d sigma plus the stuff related to odd guys. Okay. So here's the my localization locus will be f equal to zero, sigma equal to constant. And this is just a one isolated point because the only flat connection on S3 is just trivial connection. You can do this calculation more complicated, manifolds and then life becomes a bit more complicated. That's why I choose this, okay? So I actually have one fixed point and then I have to do it. So now there are some subtleties which I'm not going to discuss, but actually what we do, you have to apply this formula directly there. So the partition function will be the following. It will be integral over constant sigma. So my, I mean, it's isolated in these directions here. I have, of course, sigma is a constant. Then I have to evaluate my action at my fixed point which would be E minus some number trace of sigma square. And then I have to write for you these guys and I'm writing for you these guys, right? So the first line it's mapping from even, so everything is linearized level. So at linearized level, this is actually one forms. So I'm mapping one forms even to odd one forms. So then what I will have here, I will have here a determinant over one form on S3 of Lv plus adjoint of sigma. I'm a bit skipping how I get the adjoint, et cetera, but roughly speaking, if you assume that, okay? Now another part of thing is that this goes here. So this is zero form. Now it goes from here to here. So then what I will get, I will get a determinant of zero form on S3 Lv plus adjoint. Now that's a story which I didn't tell you yet, but as you know very well as a physicist you have to gauge, fix it. So there are three fields I'm missing. In fact, this field has a super-partner. It's related to the ghost. So there will be more fields. There will be fields C, C bar, B. And actually, so C is related to sigma then these guys as a super-partners, et cetera. So actually these two, I mean fields. So there is purely systematic procedure to get them. So I told you there is this mataiqueline formalism which is built in, et cetera. But basically having C, C bar, it contributes me the following thing. It's a determinant of zero form of exactly the same operator. So that's what we have to calculate. That's what gets from localization. Again, I did not explain to you this term, but I will try to draw maybe next time diagram. Again, I didn't have enough time. Actually I wanted to spend for this more than half of the lecture, I mean in the theorems. We're just small deviation. So the whole goal is that this is supposed to be a partition function of transform on S3 and according to localizations, that's what I have to get now. So now we actually have to calculate. So there is, I'm going to do it in a smart way but on S3 you can do it in a not smart way. The way Kapustin did. He's a smart guy, but yeah. What you can do actually, and that's, you write this thing, right? You expand everything up to quadratic order. You get Laplacians and you calculate the ratios. So you can do the way exactly as I told you at the first thing. You can actually calculate this determinants of these things. Because it's actually group manifold, you know so much about harmonic analysis, you can do it. That's the way Anton did the original calculations. So this is totally fine. Since you have a localization, you just can put a parameter and only quadratic terms are contributing. This is Laplacian. So you actually can do these things. So this is in fact, it's another step when you actually go to this formula. I'm contrasting to you the things because sometimes in localization community people tend to do these calculations, which is in principle much harder. But if you know about manifold too much, then you can do it. So S3 is one of these examples. Okay, so now the whole idea here, what we have to do, and that's why I was telling you about in the theorems is the following. That you have to actually start to expand these guys. So for example, so one form, right, you can expand on the following things. This will be a vertical guy, one form. Plus it will be a horizontal guy. But horizontal will have one zero plus horizontal zero one. So vertical guys, it's just one component. Five minutes, okay? Less. Less. Oh, you're already for two minutes showing me. Okay. Right, so this has a zero component. So this is actually zero forms. So now if you start to look is that here I have one operator zero forms in power three over two. Then one of them will cancel. And here this is more or less the same spaces. Let me ignore the face. But face here is very, very important. You have to take a square root in a clever way. It's responsible in the quantum level for the shift of a level of chance assignment. So let me ignore the face. But so actually if I take square root of this and my answer will be the following. So I will have a D sigma E minus some number trace of sigma square. And then here I will get a determinant of zero form LV plus a joint. And then determinant of horizontal one comma zero form LV plus a joint of sigma. So now actually because of all nonsense I told you before and it didn't come across presumably. But the thing is there is this DH operator here which is transversely elliptic. Okay. So one of the things, for example, what I told you that I can take these guys and expand in the zero in the exercise I was giving before. So I can expand. So I have to finish. We'll talk tomorrow. So one of the ways is to do the following. You can expand these guys, okay? Because when you expand in words, you know very well what's the value of this operator will be, right? Because LV on my form n will be two pi i n omega n. The only thing is now this guy takes values. This is a form on S2, these values in O n. So it means that this guy, when I expand in the modes, we'll have a value two pi i n plus sigma. Now what I actually have to go to the kernel of this operator and see what's minus much of the modes. And roughly speaking, I have to go to this operator I have to finish, let me say one thing. What I will get here, I will get one two pi i n plus a joint of sigma in whatever n, n minus one. Something like this. Again, if you want, I have to finish now. You can play yourself with the things and convince. But the steps is the following, that you take these guys, you can expand in modes and then you can go to kernel. There is a huge consolation. This is the only thing which is left. And this is overall n I have to sum up. I mean, I'm a bit sketchy. I have to actually think differently when it's n positive and negative, et cetera. But this is the whole idea and that's the answer. Now if you take this function, you plug it in here, you calculate everything, you get what we can go to end of 80s. So this is exactly the same result. And yeah. So I will stop here. So I'm sorry. Again, I think I suffer the same problem. I didn't have time, but I spent too much time on the index theorems, but this is much more important. So tomorrow I will comment about this. I will tell you this result, write some diagram, gain stressing for some sequence and then give you 5D theory again, very axiomatically. Just tell you briefly how it's related to supersymmetry, but I will not actually talk about this. Okay, thank you.