 So let me, first of all, start by summary what I was doing and what I would like to remember from last lecture. So you were doing it about theorem. And again, as I told you, this is just integration. I mean, we are using this quite a lot. So it's integration of the currently closed form. So I have a manifold with U1 action and then I have this formula. So a fixed points, pi D over two, alpha zero at fixed points and then fixed point. And then again, remember that my supersymmetry was this guy V more of X and then the requirement for my form is this. And again, actually what we are doing is, I mean, this is integral of our whole manifold. So you actually integrate the top form, but this is a large component because to solve this condition, it involves all degrees of the forms. And then what I would like that you remember when you were constructing this guy, right? So we were doing things up to quadratic order. So minus T H mu nu X mu X nu plus S mu nu psi nu psi nu plus high order terms. And I told you that the answer can be written in two ways. Either it's a square root of determinant of S or square root of determinant of H. And then if I use my linearized symmetry, right, then I know that H is equal to dV of S and then I get something much better, the level of determinants. So I mean, this is important to remember because of infinite dimensional setup. So this is something I would like to, that you remember. Then basically I told you another setup. So I was discussing, I mean, this thing when we have, you know, vector bundle. And so for example, I was telling you about this model, so integration over this model. So this is in principle formalism and there are two forms. So I'm just mentioning, so if you actually wanna do things properly, I can explain some of the things during exercise classes. So there are different versions of Mathai, Aquiline, et cetera. I mean, if you wanna actually understand geometrically what we are doing in physics, it's good to study. So there is a normal thing about, I mean, non-accurant version of Mathai, Aquiline class representative. Then there is a QR inversion, version. So that's what basically I was writing for you last time. Very sketchy. I can give you more details. And then there is something which is basically related to the quotient. So when I have to have a principle bundle times V, quotient, et cetera. So it's the same representation of this, but this is actually what we have to do in gauge theory. So there is this another thing. So if you're interested, there is a paper by ITIA and Jeffrey, which is discussing. I don't have time to do it, so but basically at some point I have to introduce for you ghosts and stuff like this. And of course as a physicist, you know that you need ghosts to do the gauge fixing, but there is a proper mathematical explanation for this and this has been done at the end of 80s, et cetera. So I mean, there are these different levels of the model. I'm just mentioning this in case you wanna do it and learn it properly. I simply don't have time to explain. So what I was doing last time, we actually replaced whatever problems we write here by some linearized problem. So what I was telling you that I can write for you this linearized problem, sorry, X. So again, this is even odd, odd, even. So this is just some operation. Then I would have delta chi equal to H, delta H equal to R1 chi. So if you want either I construct the exact term so I can equally write in transformations here. I can also add these terms IDX plus IDPSI. I mean, it doesn't matter either I put these things because this is actually a field definition how you define your H. Either you define your H by itself or you write like this. You can go back and forth. So you write baresticks exact term. So D is another operator. Okay. And then what I did last time for you. So we were calculating determinants. We write baresticks exact terms. We do whole things. And then what do we get? We get the following. That's zeta up to some terms is proportional to determinant one over two. I zero, I over two, D dagger minus I over two, D term one. And then here we have determinant of one over two. So here for everything to work. So for this algebra to work or another thing is that I have to require the following thing. I have to require that R one D is equal to DR zero. I mean, it depends for example, this can follow from this requirement. Okay. So if you want all the supersymmetry I wrote at linearized level you can think of also some diagram. So I have my field. For example, I can write as a bundle E zero. I go to E minus one. So this is field X. This is sky. This is bundle F one. This is my field psi zero. This is my field H. So you see this is fermions. This is odd guys. So I have here, for example, minus ID, then ID, then I have here I zero, then I have R one. If I add simply this errors from here to zero I get what people call in mathematics complex. It's basically by complex. Okay. So now we would like to discuss how all this operations and what we actually would like to have an infinite dimensional setting and the whole thing you would like to make sense. So this formula, as I told you last time you could equally collapse to the following formula. So determinant one over two on the kernel of DD dagger of minus R one. Determinant of one over two at the kernel of D dagger D of minus. I mean, I don't actually need this minus. It's not important to R one. You remember I told you formulas this here but it was one over fourth here so I can pull it back and forth. Okay. I have zero for this. All right, so zero. So for example, just for future it's very easy to remember what is going on. If your symmetry goes from bosons to fermions then it goes downstairs. If it goes from fermions to bosons this is comes upstairs. So this is always will be the thing. So now all this operation. So in infinite dimensional setting infinite dimensional setting all these guys are zero R one D differential operators. So we have to discuss actually what do we need from these operators so the answer to make sense and also discuss different opportunities. I mean, what we can do, et cetera because there are for example if you go back to 30 years and typically operators like R zero, R one used to be zero. So I would like to explain you these things. And the main thing what we have to do now I would like to discuss with you it's different notions for operators. So I would like to discuss for you. So let me remind you elliptic operators. Elliptic operators. So elliptic operator, it's a linear operator on my manifold. So I can write as L of U. So this is some, if I write so it's a local notion. So it's enough to write for me in local coordinates but of course I can have operators acting differential forms, et cetera, et cetera but I'm not concerned here with that. So in general operator will have some coefficients and then I will have something like this. So by this of course I understand this is a multi index. So this is I don't know D one and some power, I don't know alpha one, et cetera, D and in power alpha and it's so this is derivative. So I look at my operator, et cetera. So for every operator, differential operator I can correspond what's called a symbol. So symbol is the following thing. So the operator, this is what you will call mth order operator here but I'm looking only at the leading part in derivatives and what I will do, I will basically replace this guy. So symbol is the following object. So symbol of this operator, let me call it D. Okay, so symbol of this operator. So I'm looking at the top thing in derivatives and then I'm basically, so I'm looking at alpha equal exactly to m. So I will call this operator for the ram and then I will look at alpha x and then I introduce formal variable xi. So xi is just a vector which depends to our n. So instead of every derivative I have here just put xi, okay. And the thing is that this should be non-degenerate away. So this should be non-degenerate away from origin. So again, for every operator I can identify the symbol until symbol satisfies this property that an operator is called elliptic. So let me give you example. So example, so your favorite operator for example, Laplace operator on the scalar function. So if I would have a function then I can write for operator i from one to n, second derivatives of d x a on u, okay. So the symbol for these guys is that's a symbol for Laplace operator is just some i from one to n xi one square xi i square. So it's definitely positive away from one vulk xi is equal to zero. So this is elliptic operator. Now what is important here is that ellipticity if I add here the terms in a law order I don't care about it if it's, it will not modify ellipticity. Well, exercise you can do it yourself. So if you look again what you were reading last night on the Kahara, I hope. So there is the things on if I have a manifold, right? I have a differential form. So from p form I can map to p plus one form. I have a diram. Then if I put a metric on the manifold I can then try to use d dagger. So which will map in opposite direction. So it will have also d dagger. And then they will be on p forms. I mean they will be correspondently law plus operator which is defined d d dagger plus d dagger d. So I'm asking you to prove that this is elliptic operator. Well, in fact, if you're reading a Kahara it's, it's proven there. Okay. So the question is why do we interested in these operators and another exercise? Okay, another exercise. It's important. So dirac operator. So prove ellipticity. I mean there are a lot of details because you have to think in which dimensions you are, et cetera. Again, I think it's done in a Kahara if you look it up. So I mean there aren't two questions. I mean questions why do you like elliptic operators and they do IP in physics. So there is this famous mathematician. So there is, so I have operator. Somehow I'm hesitating to tell you too much math but I have operator d on m, on manifold. So there is another notion of for the operator it's called Fredholm operator. So this is a guy who got a PhD in Uppsala by the way. So the operator you call Fredholm if it has a finite dimensional kernel and copernal. So if you look at dimensionality, so again I'm talking of course only about linear operators. So if dimension of kernel d is less than infinity and dimension of copernal of d is less than infinity. So copernal you can think is just a kernel of d dagger. So for example by the way these operators are self-adjoint, so d, okay. So if they are finite dimensional, so basically the set of zero modes for given operator then it's called Fredholm. Then for compact manifolds why we like elliptic operators because actually so there is a theorem that on m which is compact if d is elliptic then it's Fredholm then d is Fredholm. So of course this is not true for non-compact, for non-compact line it becomes complicated and we will discuss it later on. So but what is wonderful is the following if you give me operator I do the simple check of the index or sorry of the symbol, I would actually know that the kernels are finite dimensional. So in a way when I will write for you just as example, so if you would write for you the action, right. So for example for the scalar field on m dimensional manifolds, right. So this is just this action. So this is a, phi is a scalar field. So what is very, very important why we do like, why we would like that he's standing elliptic operator because this guy we can actually invert. So here what we are having in the case of scalar, right. So this would be exactly, and if I write more abstractly, this is phi Laplace phi. And because the kernel is finite dimensional I can control it very clearly. So I can actually know exactly how many zero modes. So it makes sense to talk about this. So for example, if I would replace just, if I go to some other manifolds, et cetera, and my operator, I don't know, not elliptic, I don't know, parabolic, whatever. Then I mean, I will not have any possibility to define in any meaningful way the things. That's basically the reasons. And supersymmetry is perfect in this sense. So I will tell you a lot of stuff. I will comment about supersymmetry towards the end. So I hope you heard enough about supersymmetry last week, so I will not repeat it. But supersymmetry is basically wonderful in a sense that it picks up elliptic problems because always for bosonic fields, your operators are always elliptic up to gauge fixing, et cetera, for the gauge fields you have to take extra care of. And then for fermionic fields you have Dirac operators. So if you look at your actions and for example, look up to quadratic order, all your operators will be always elliptic and supersymmetry is automatically taking care of this. So if I would, for example, try to do it more abstractly, I have to think about it, okay? So this is about elliptic operators. So let me see what else I will not tell you. That is the ram laplacian. One can have also the tensor laplacian with a current river. One does not work with one. You can do many things. The question is that elliptic too? Your question is not well posed. Let me explain why. No, no, no, no, wait. The thing is what is important here, so you're right operator, you have to decide on which object it acts, et cetera. So first of all, covariant derivative you don't care about covariance because covariant derivative, the connection, it's allow order in derivative so you can throw it away. So it has no effect on ellipticity. So ellipticity, this is what is amazing about the things. And this is, I mean, this theorem is not that trivial but the thing is that I don't care what stuff you add there. And what I was warning you why your problem have to well pose because for example, when you have covariant derivatives, you will also have kinetic terms for A's and then you have to think a bit harder. So I will tell you more about this later on. But in principle, always, you know, I mean your covariant derivative, for example, if I add D, A, D minus, let's do, let's be physicists and try to, although I hate it. It drives me crazy to put this size but just for your different, this is difference between mathematicians and physicists. That's what Nikita was mentioning yesterday. Physicists put IS everywhere, although it's nonsense. So for example, what I'm telling you from point of differential operators, these guys, or if you start to calculate Laplacian from this, this is allow order guy. So you can throw it away. It has no effect on ellipticity. But gauge fields are by itself important. So now let me give you a bunch of the problems to solve that you actually get a hint what ellipticity is. And this is very important. Right. So exercises, these comments. So typically if you write the actions, for example, for Bosonic guys, then your operators, what appears in the action, it's second order elliptic. So for example, engage theory. So when you write the engage theory, this thing. So again, I will now do non-abelian gauge things because I'm interested only up to quadratic order. So whenever I linearized my problem, I don't care about being non-abelian. So this is formula is this. If I write more abstractly, this is not elliptic operator, of course. And you know very well because these guys has a too much degree, I mean too, you know, it has a gauge symmetry. So actually, of course, when you do gauge fixing and you introduce goals, et cetera, so the terms which you actually considering these ones, right? You have a D-dagger A star D-dagger A. So this is gauge fixing. This is standard Florence gauge. So now if you write this abstractly, this is exactly this. So again, here what I'm doing is just when I'm discussing forums, omega. So I mean my scalar product defined this way. So when you discuss ellipticity, actually you have to see what for the gauge field you inverted there. So you have to look at the gauge, charm plus gauge fixing. And of course you add, I mean, when you do actual calculation, you add goals, et cetera, compensates certain things. But when I'm actually looking at elliptic operator for gauge field, it's Laplacian in one forms. Okay. So in actions, as far as bosons concerns, even fields of concerns, you always have a second order elliptic operators when you do the gauge fixing. Now there are a number of very important elliptic problems which is first order. And that's exactly what I would like you to, for a reason. What I would like for you that you do this exercise. So please prove the following things. Exercise number one. So in 2D, F equals zero, D dagger of A, it's elliptic problem. So by the way, first check when you do elliptic problems, just check the conditions you do. Because for example, right here you have connection A. So when you will write a symbol, so the exercise you do the following, you write this again, every single linearized order, I don't care about being non-ibilite, you write this, you write this, and then you replace derivative by psi. So you will get two by two metrics on A1, A2 equal to zero. And the important thing that this matrix when I write psi should be invertible. So for example, this object, you obviously write psi one, psi two, right? And then, well, this you can figure out yourself. Okay, I'm basically going to write it for you. Psi one minus psi two. So now you calculate the determinant of this guy. I did it. Right, sorry, yeah. Otherwise I would not get what I want, right? So this is, right? So the first term, what I'm doing, I'm doing the following thing. I have D2A1 minus one, two. And then I have D1A1 plus D2A2. And then I'm just writing this guy. So and then determinant of this thing is psi two square plus psi one square. And this is positive away from zero. So it's elliptic. I solved it for you. But now I will give you something more complicated to problems, which I'm going to use ellipticity. This would be important for discussion of SO5D theory. So exercise two. So in 3D, so I'm in 3D. So I have a gauge field and I would have a joints color. So transform and adjoint. So please prove that this problem is elliptic. Exercise three. So now I'm in 4D. So I just have a gauge field. So I will have a metric. So I defined F plus, which is equal to one plus star F equal to zero, I can put one half. And D dagger of A. So this is instant on a question. So this is again elliptic. Well, finally, since Nikita was talking about this yesterday. So if I have a Riemann surface to the manifold, let me say this is complex manifold, complex. So then this problem is also elliptic. So all problems we are dealing in, this is basically our ideal world that you would like to ever introduce to elliptic problems. So please do it. I mean, this is a relatively simple exercise. In a way, maybe it's better to start from this. This is just dimensional reduction of this problem actually. Okay, so let me tell you something more about operators. And then I will go through some examples, what I'm doing there. There is another notion which is apparently, in old days, we were ignoring. So in old days, basically till, I don't know, till end of 90s, et cetera. I mean, the whole pattern was that everything should be elliptic. So either you do second order problems, they're elliptic, they're typically hard. But for example, you know, you cannot calculate exactly whatever young mills, you are trying to do something simpler. So for example, you go to this problem, it's elliptic. I mean, wonderful thing that ellipticity basically guarantees you that the more or less space of this problem find a dimensional, then you can do quite a lot, et cetera. What we actually learned, one of the main, at least mathematical conceptual thing or what Vasily did is the following that actually supersymmetry likes not only elliptic operators, they like what's called transversely elliptic operators. So again, the catch is that for elliptic operators, the kernels find a dimensional so we can calculate, you can write determinants of operators, we can do quite a lot and more less spaces. But basically there is other situations. So let me try to give you idea. So for example, I have this operator D on S2, which is just Dalbo operator and this is nice elliptic operator. So what I can try to do, so I will give you definition, formal definition in a moment, but first I will not try to tell you the idea. So what I can do, so this is a nice elliptic operator. So I can lift my problem to something three-dimensional so I can do S1 times S2. So here I would have coordinate zeta, zeta bar and then here I would have coordinate t. So then if you ask me to write a formula, whatever Laplace operator here, well this is not a rocket science, you will write something like minus operator t square plus d d bar. So I'm writing this minus because this is a then positive definite operator because if I expand in Fourier modes, I will have IN. So then it will be positive. So now if you stare at this example, there is a following things happen, see that a priori, if I now, right? So formally my operator D is defined on the whole space. So I can take for example now my form, so I can, in this very simple example, I can discuss, so I have my vector field T, I have one form DT. So for example, I can talk about horizontal forms. So the forms which has a legs only along as two and I can have PQ decomposition, okay? And of course I will have for your operator which I can define DH. So it's operator for example, which map a horizontal, I don't know, zero zero form to horizontal one zero form. So it will do the standard thing, but the only thing is that these forms in this form depends on T, right? So this operator is not elliptic in any way. I mean, there is no possibility to have a finite dimensional kernel simply because this operator acts only on zeta coordinates and T dependence is not controlled. But the funny thing is the following in this particular problem, this is very elementary problem, what I can do, I can basically take and expand everything in Fourier modes. So instead of looking at this problem, I can basically reduce this problem to the following thing. So I look at my Fourier modes. So I would basically would say that L of T on whatever my form is equal to I n of, I can put label n here, right? And then I can do it with a zero forms H. So now I can put label n. So it's a Fourier mode expansion. And if I add by D, I will get to H one zero n. Again, very trivial statement. So it acts on Fourier modes. But now this is elliptic guy. Now from point of view for presentations, what I'm doing is that this is basically, I'm taking my omega and I'm decomposing. So I have my omega H PQ and I'm decomposing in the representations with respect to U1. So I'm just expanding in Fourier modes. At every component D is elliptic operator. So this is basically the notion of transverse ellipticity. Trans elliptic operator. So this is the idea, this is very important. And that's what supersymmetry is also teaching now that this operator is important. So of course, this is very trivial case. Before I give you a formal definition, let me just give you much less trivial case. So for example, if I have S3, S3 is just this example. If I'm writing as part of, you know, in C2, right? This is my S3. So again, if you read Nakahara last night, you should know that this is famous hop vibration. So you can take S3 over S2 over S1. So now in this example, in this trivial example, it was, you know, trivial. So here's very easy to expand in Fourier modes. In fact, I will also do this exercise later on, expanding in Fourier modes here. But the thing is what is important, there is U1 action. So in principle, I can, like here, I can introduce my vector field for this U1 action. So again, hop fibers. And then I can have a kappa, which is connection with a property that IV kappa is equal to one, IV dkappa is equal to zero. So then I can decompose my forms. I can decompose my forms to vertical and horizontal. So this is done by the following things, is this is kappa of HIV. This is vertical thing. This is one minus kappa of HIV. So this is a standard thing which people use for principle bundles, et cetera. Okay, so what I can do further on horizontal guys, I can easily introduce, because it's effectively like forms on S2, I can introduce PQD composition. So there are basically horizontal complex structure. I can define for you the complex structure there. And then I can define for your operator DH. So, and this is again exactly the example of transversal elliptic operator. Now the story of Fourier expansion in this example, et cetera, it's a bit more complicated. So we will talk about it. So let me give you a definition of a transversal elliptic operator. Then you will see how you actually check transversality. So basically what I have to do, so I will define my cotangent bundle of the manifold. So I have an action of the group on my manifold. So I mean, in this example, it's U1. So I will define the elements such as way that this is elements, so this is whatever my xis, et star of M such that xi contracted this vector of field is zero. So V is associated to the action. So of course, if my G is not U1 I have more vector fields. I will satisfy everything. So I'm looking at one forms which is transverse, which are contracted with fundamental vector fields. So then I check transversality elliptic so it should actually, I have to check ellipticity on all these directions. So if I have operator D and I'm looking sigma. So in this setup, what I have to do, I have to put here guys, so alpha, so I put a alpha x, xi alpha. But I have to check this invertibility only on xis, which is in TG of M. And again, this is point like statement, this is linear algebra. Because what is important in general, this vector fields can vanish. So I mean, this is not a regular thing in any way. And in fact, most of the cases, for example, for two dimensional manifolds, et cetera, I mean, so this is a free action. This is example of free action, but there is a lot of actions are not free. So it's important that these conditions check point wise. So when your xi belongs here and then xi of course is not zero, so not all component zero, then you have to check that this is basically a guy invertible. So this is called transverse ellipticity. So supersymmetry actually is telling us that they appear all the time. Now the question is, why do they appear all the time? So there is in fact a very simple criteria there. And this is already boils down to this formula. So let me tell you abstractly and then I will go to examples in field theory. How do they appear in supersymmetry? The effect is the following basically. So imagine I have my operator. So I will write what we have d, d dagger for example. Something like this. So typically imagine I have u one action. So this is typically will be some le derivative of v square minus one. So typically there is a conspiracy between this being elliptic operator and this being transversely elliptic. So if this is a second order elliptic operator, then this is part exactly corresponded to the derivative. Then whatever is remaining, this guy typically will be transversely elliptic. So in all days when you think about instantons, instantons is the following. It's basically about trying to take a smart way square root of Laplacian. And then taking square root of Laplacian, you get actually this problem. So you get instantons. Now if you have a u one actions, you can do things in a bit different way. And that's exactly why supersymmetry leads us. So you actually can take your second order and you in a way try to take a square root only in directions which is transverse to the group action. And this type of problems find a dimension. Now again, the kernels of these guys are infinite dimensional. But the problem is that the kernels, so the kernel of D typically its dimensions will be infinite dimensional. But kernel of D has basically its decomposed in representation. So it's a sum of a reps of whatever group you have typically its combination of u ones, some spaces. And this is all spaces as finite dimensional. So this is very powerful concept. So now instead of actually counting, so before the K thing was dimensionality of the kernel, it was number. So as you know, there is a smart word is called categorification. So number is not good. So if it's number, it should be value of a function at some point, et cetera. So I mean, in a way in this approach, it's much better because this would be a function typically here. So this would be certain whatever character stains because I have to keep track of representations. But you see the problem is not now bad. So I actually can deal with the things and exactly this way. So I mean, transversely elliptic, it means that kernel can be decomposed in irreducible finite dimensional representations of this. So just maybe let me give you a very trivial example but very instructive, which we are going to use. Now, also this statement is very good even for elliptic operators, but then when spaces is non-compact because when spaces is non-compact ellipticity and being thread-holding is not the same. So sometimes I have to use it. So let me give you this example, which we are going to use. So imagine I have C and I ask you to deal with this operator D bar, right? And for example, I would like to ask you, can you calculate me? So this is the type of situations we like. Again, formally, it's elliptic operator in this setup but I mean, it's already important. So I ask you to calculate for me a kernel of D prime. So typically if I would be on compact manifold, this would be a very small space. I mean, for functions it will be typically only constant, et cetera. Because my kernel is equal, what is this? Well, you can know how to do it. So I would have one, zeta, zeta square, et cetera. So I will have basically all homomorphic functions. And if it's regular, then they have expansion, et cetera. So in a way, if you look at dimensionality, it's infinite dimension, it's not very, very useful. But at the same time, what I can encode my information, I have a U1 rotation of plane. And I wrote all this guy there exactly in reducible representations under rotation. So instead of this, I can encode my index in the following way. I would write my index of D bar operator is equal to. And then I basically will write a character for every representation one, plus t, plus t square, plus et cetera. So this would be basically k from zero to infinity tk, and that would be one over t. So this is index of DeraCoperator. So you see, now you understand this is basically as a sum of characters, so I keep track of the things. So I carry this information. Again, I cannot do it if I wouldn't have a U1. So I think this is all this formal stuff I will go through examples. Any questions? It's a good time to ask questions now. So everything is crystal clear. I'll come back to you, if operator is a flip phone, would it automatically be particular? I think so. I don't remember these details of math, but I think so. Typically you like to, I mean, ellipticity it's very easy to check. You give me operator, I mean, within 30 seconds I check ellipticity. Well, Fred Holm is hard to prove. Any other questions? Okay, so I will go through some, so my plan is a fallen, I will just give you now. So remember, I had this formula, which is unfortunately erased. So I will maybe put it here, and I will go for your first three examples now. These will be two and three dimensional examples, just pinpoint for your operators, and it's not a gauge theory yet. So then later on I will switch to gauge theory. So for gauge theories, I will introduce certain things, and how much time do I have? I'm lost, more than half an hour, okay. Right, so let me just, I'm sorry, I'm erased, but I would like to write for you this, because now I will give you a different, and so then I had, sorry for copying this, I'll be fast, but then here's appropriate kernel of operator, okay. So I wanted to show you which situations actually, I mean, we have in physics, et cetera, and I'm considering examples without gauge fields, just simply life is simpler. So when I will switch to gauge theories, I have to lie to you, because I actually don't wanna, I don't have time to explain to you why all fields are there, et cetera. So this is explained in this paper by Tia Jefferies about Gauss, et cetera, about wire model. So this part I wanna skip. So for now I don't wanna complicate life, so let me start with the following problem. So it's a field theory. So I'm looking first at two-dimensional field theory. So again, I'm using very similar notations. So imagine I have a map from S2 to CN. So again, this is very similar what Nikita was discussing yesterday. I can map it to complex manifold, but let me do it simple. So then my space would be, is this. So now what I will do, I have a complex coordinates here. So what I will do, so this is X bar I equal to H zeta I bar. So in this example, first of all, you can see that R0 and R1 are put to zero, which we can do, so it will square just. And then if you want, I can of course modify them by putting appropriate conditions. So here I will put minus D bar X I, and then here I will put basically plus I D bar X I. So it does not modify the algebra. Instead of writing them in burst exact terms. So first of all, this fields, my fields here. So this is fields, we just take value in zero one form on my S2 with the values in T. Well, I mean a pullback of T one zero. If you wanna be pedantic. So it's exactly this structure. So my space, my manifold now. So my manifold, which I have the manifold, it's maps on from S2 to CN. Then when I introduce psi, it's actually, so this psi, this becomes a odd tangent bundle and this is just a bundle. So these guys happen to be a vector bundle of a maps. Again, it takes a time to get used to this language, but anyhow, I mean the fact that geometrically everything works, it's exactly this. So the thing here, why do we need, in this particular example, we allow to have R0 and R1 to be equal to zero. It's simply because the operator del is elliptic. So del, del bar, sorry, del is elliptic. So in principle, if you will write your delta of w, this would be the following. It will be delta of x zeta bar i g i j bar h zeta j bar minus d x j bar. So this model, I mean, in fact, just to let you know what exactly I'm writing, this is another rewriting of chiral field into dimension. There is, you can actually take a physical chiral field on S2 and rewrite in this form. Okay. So if you write this explicitly, what I will get, I will get d x i g i j bar d bar, sorry, d bar, d x j bar plus chi zeta i g i, okay. So in this case, what we're actually getting, so if we calculate determinant, so first of all, this would sit on holomorphic maps. So the maps, which map, I mean, preserve complex structure to complex structure here. So the fixed point, so fixed point, so the fixed point, so the fixed point, so the fixed point, so the fixed point is just d bar of x i equal to zero. So now if I'm looking at determinants, now my determinants are, I mean, I assume that they're around isolated point. So my determinants are basically one half d d dagger, determinant of one half d d dagger. So if my holomorphic map would be actually isolated, so I would just get one. So typically it's not isolated. I mean, the thing is that because this guy's non-compact, there are too many of them, et cetera. But in principle, I mean, this is example where, I mean, determinants is very, very easy. So this type of calculations people were doing, you know, this is what Witten introduced end of 80s and then keep going, et cetera. So this is a model. So this leads to Gromov Witten invariance. Again, the complication is that actually you have to understand this one because there is a model of space, et cetera. But as far as determinants, it's just this. Sometimes it may be one, maybe minus one, if we start to discuss subtlety relating to square roots, et cetera. So this is first model. So you see here why I could put these guys to zero because it was important to me that this is what we transfer, sorry, this is elliptic operator second order. And then I could actually take a square root and I just get again elliptic problem. So it looks nice. So I mean, in a way, the things are canceled out. So the next level you can do, you can do the following thing. So next example to D. So my S2, again, on CN, but my S2 has a U1 symmetry. It's two fixed points. So I will have a vector field V for this. So then my transformations actually can be written as follows. Then I will have again here the same object as before. So again, I can write here minus del bar X, I. And then here delta H of zeta, sorry, bar bar I goes to the derivative of V of chi. Delta bar I minus plus del. So here my D is del operator. Then my I0 is LV on zero forms and R1 is LV on either one zero form, so zero one. So now the fact is that, so remember that what was important to me, it was important this property, otherwise algebra doesn't work. This is obvious because derivative commutes with Dalbo. It commutes with diram, it commutes with Dalbo. It's just a result of carton calculus. So this is guaranteed. So this is example what people sometimes called, some version of the current version of Gromov-Witten. Again, for example, if you would put actually a supersymmetric theory with the carton field on the sphere, that's algebra you will get, so you can map it. This is a rotation, so this is more close to supersymmetry because it squares not to zero, so this algebra will actually, so if you calculate this algebra, it squares to lay derivative on all fields. So of course, if you start to write here in this example, delta w, then this is, you can write two terms, psi mu g minu, literati for v of x, mu plus, I mean the same term I wrote before, chi zeta bar i g i j bar x zeta j bar minus d zeta. So if I look at the Bosonic operator, then what I actually will get, it's lv square plus d d bar. It's still okay operator. So by the way, if you would have a symmetry, so otherwise it wouldn't make sense. I mean, you could also replace this by c by plane. So you can actually deal with this operator. It's okay, it has infinite dimensional kernels, but you basically can parameterize the kernel. So this part using the representations of this, that's what you actually gain by introducing this equivalence. But you can also do it on S2, okay? So that's what you get. And again, we can deal with this. And for example, now the problem becomes very meaningful because formally what you will get, you will get basically the following thing. So if you start to look at determinants, again, I'm not discussing if it's my localization locus isolated, not isolated. I mean, you have to do certain subtleties. Now let's look what I will get here. So now if I start this symmetry, I actually will get here by localization if the following. So upstairs I get fermions. So whatever from here to here. So I have to calculate my operator lv on these objects on one zero forms. These are values in T01 of my manifold. I can take a square root because I have to copy. So if I'm ignoring phase, that's what I will get. Okay, this is just comes from these fields. Downstairs, I will get a determinant of zero zero forms of lv and that's it's a question if it's one, not one, et cetera. So this type of determinants will become non one. I will discuss how to calculate these determinants. I mean, for this case, I will not do it. I will tell you exactly explicitly for S3 and for S5. But this is concrete determinant. So what I have to do actually, I have to take S2 and then I have well-defined forms here. I have this guy, I have zero forms here. I have these guys, they will be presumably at big constellations but that's what I have to do. So then we have just to calculate which we don't have enough tools for the moment. So that was a problem number two. So problem number three. Let's erase this thing. And that's where we relaxed even more things. So you see, first I started to having things purely elliptic and then the problem number three, 3D. So I will have a map from S3 to C. So in fact, carol field can be lifted from two dimensions or three is the same field content. So again, I have a rotation now, U1 acts on S3. So in principle on S3, I have two T2 action but for example, I can think of this for simplicity as just a hopf vibration. So it's the vector field corresponds to hopf fiber. And then I can write for your introduce. So I can, as before I told you, I can introduce vector field, I can introduce one form and I can decompose all my forms to vertical and horizontal with respect to hopf fiber. And then moreover, horizontal will admit PQ decomposition. So there will be almost complex structure on this subspace. In fact, it's integrable. So it's what mathematically called society manifold. Okay, then what I will have here, my fields will be the following D is equal to psi mu. Delta psi mu is equal to LVX mu. So here's I told before I can introduce this operator which is horizontal. Now delta chi of zeta bar i equal to, okay. Okay, and then here I will have delta LV of chi i my plus DH bar psi. Okay, so now what is important that my field chi zeta bar i and the same field H zeta bar i, they in horizontal zero one form in a values again to be pedantic, I have to write a pullback one zero. Now, if I write for you a burst exact term, so I have to, so I feel that everybody got decoupled and bored to death. But you know, life is not easy, it will just get worse. This is a baby game, I mean, gauge theory is harder. Okay, so now I write this term. Now if I again write my exact term, so what I will get, I will get the following. So quadratic term I will get minus LV square plus DH DH bar. So in this story, so if I identify this operator, so I zero, I one, this is just the derivative and then my D operator is this DH bar, et cetera. So now this is almost transversely elliptical operator. So now if I wanna calculate determinants, so downstairs, this is zero form, zero form. So I will get a determinant on zero forms, one half of LV. So I'm writing this, I mean, again, actually it descends not only zero forms, it descends further down, it descends to the basically kernel of Dell operator. And up there, I will have a, this is a horizontal one zero forms LV. So here I have to write this square root, by the way, sorry, I forgot to write downstairs square root. Here I can get rid of square root because formula I will have one zero plus zero one. And up to the face I can take a square root. Here I cannot take a square root. So I made a mistake, let me correct it. It looked different yesterday. And again, so this is actually, I get omegas, they descend to Dalbu. I mean kernels of Dalbu operators and then I have to analyze there what's going on. Now the kernels of Dalbu in this problem is infinite dimensional. But still I can do it. So because they are transversely elliptical operators. So you see this is different stories you can have and different things. So all these operators in principle possible. And the sort of weakest possibility is when this is transversely elliptic, this is just some operator corresponding to your one action. But I always have to require that this is second order of elliptic operator. Any questions? Because I feel, as you know everything, you're just questions. So how does this horizontally operator define the length? Do we start from this splitting into horizontal forms? Right. And then you have to introduce so. So for example in S3, so you introduce vector field, introduce one form. How you do it, you're written a caharaman. This is one of the basic exercises. This is connection one form for hope vibration. So then you introduce these guys, omega V plus omega horizontal. Then the thing is that horizontal subspace, it's two-dimensional subspace. So there is exist almost complex structure. Most complex structure exist. Well, I mean, it's just some map which squares to zero you can define. So further on, these guys you can define as a PQ forms. Is it clear or not? I mean, here we enter the length that either you know a lot of differential geometry will be lost because index theorems, nothing I can help you about. That's, I'm not wasn't joking about this. I mean, you have to know naka-hara quite well. I mean, this is not, I mean, this is general fact for any contact manifold. I mean, there is this decomposition and horizontal forms can be decomposed further on. Then what I have to do, so the definitions of my operator is basically, so D has horizontal length and then I can basically decompose this as usual doing like this. For general contact manifold, this does not square to zero, but for this guy, I mean, I don't know how to explain better for you, but this guy in case of hope vibration actually will square to DH to zero. So is it just composition with projection for the horizontal part? It's a composition with horizontal part and then you use almost complex structure. The thing is that this property is not guaranteed. So this is, this manifolds for which it works. I mean, I can write for you the name, Sasakian manifolds. I mean, this non-trivial property, before it's algebra, you take DRAM, you decompose, and then you have almost complex structure, decompose, et cetera. Any other questions? I'm pretty honest. If you were lost already, then it would not get better. So please ask questions. I actually have no idea what you know, what you don't know, et cetera. I was told to tell you about index theorems. So that's what I'm about to do. Okay, you're lost. Good. And at least it's not my problem. I gave you a chance to. What we have to study from these examples, I just don't understand. You've written a few examples and we didn't calculate anything. You don't have to understand. You have to understand that there are different possibilities. I mean, what you have to understand, I mean. Yeah, but you're asking questions. I'm asking you, do you understand what I told you or you just don't? Because the thing is that you don't have questions, it seems maybe you don't follow. I'm just telling you, giving you examples. I'm saying is you can have elliptic, you can have elliptic operator plus U1 action and then you can have U1 action, transverse elliptic operator. All these examples are well-defined and calculable. So in all days, you will do only first example. I mean, now with supersymmetry, we can do more examples. So I'm confronting with the things. The main thing is why I'm doing this slowly because at some point I will write you 10 fields, many, many bundles and I will write you for you. I mean, a lot of determinants. And what I'm saying is that underlying principle it's always very simple. So it's always will basically fulfill this structure. So the only thing I may have more lines, et cetera, but it's always has a structure. I always will get this as answer. It's always will be super determinant of some operator. So I'm moving so complications because one of the problem is that when I write a gauge theory, I will have 10, I mean, not 10, but three times more fields. So that's why I'm going through these examples. Okay. For many forms of a boundary, do you have some gender? So you understand everything without boundary already? Roughly. No, I mean, there is versions of different theorems with the boundary. With boundary, it's much more complicated to do the things. Because with the boundary, for example, to guarantee that your elliptic problem is well-defined, I mean, you have to put a certain boundary conditions, et cetera. It's a complication. People study this problem, they didn't care about this rigorous mathematical foundation. I mean, how do they know that they get correct answer? What's the question? Some physics criteria, I don't know if, who has to go through this rigorous math. If you want to calculate something, understand you have to go through the rigorous stuff. Otherwise, you know, I mean, I don't know how to answer. Okay. But again, if you have more questions, it's time to ask. Maybe just comment on the previous question that shouldn't be confused between the boundaries on M, which is basically the field space that we're going to do, and the boundaries in space time. Correct, yeah. Yeah, by the way, all this operators I'm considering, they're from, you know, from, well, world sheet. I mean, that's, I'm not discussing anything. I'm looking at field series. I'm not discussing the CN, et cetera. So everything on S2 or S3 in this example is where. Okay. Good. No questions. Okay. Let's decouple even more. So what I need is now to discuss with you. So my plan is the following. So I would like to give you some basic facts about index theorems and also particular two examples, CP1 and CP2. And then next two lectures, I would like to give you idea how to calculate determinants for S3, and then how to calculate determinants for S5 and introducing for your gauge theories. Again, my warning is that I will have many more fields and more complicated determinants. But now it's a tool. So it's just mathematical, you know, deviation from, but I need it basically. The whole idea is the following that I have to know for operators, how to, let me see what I wanted to say. And again, about index theorems, let me just make a warning. It's a big subject. And transverse elliptic operators, it's a very subtle subject. So majority of physicists don't know and it's hard to read idea thing. So I will try to do for you just examples to give you hints of this. Okay. So what's idea of index theorem? So I have my operator D, elliptic operator. And one of the problem is the following. So every operator, so elliptic I told you, so far we're on compact space, compact space M. So as I told you, D is thread hole, thread hole. So then I can define, this is what's called index. So I can take index of D, this is a dimensionality of kernel of D minus dimensionality of kernel of D. Okay. And this is a number. So this is because of ellipticity, it's a number. You just have to calculate. So if in principle you wanna calculate the thing analytically, so this thing is called analytical index, then it's pretty complicated. But in sixties what was realized is that this number actually controlled by topology. So on compact manifolds, I can write for you the formula which would tell me that this object is controlled by topology. So I can write basically some generalized Gauss-Bonne theorem. Okay. And for this we are using different characteristic classes. So let me write for you example idea bought theorem. So this is just example. Again, I'm basically since because of my applications I will use for you the Dalbo operator. Okay. And also quite often I need a Dalbo operator which is twisted by a vector bundle. It means that I need a covariant version. So I have a vector bundle, I have a connection. So actually I mean by this it means that I have a covariant version of this guy plus connection term. So analytically for example, this is written for one operator. So quite often what we need we need and this is what we have quite often we have. So this is elliptic operator. There is a notion of elliptic complex. So when we go, so example of elliptic complex keep in mind it's for example a DRAM. So the complex calls elliptic if I will calculate the symbols of these operators and they exact. So if I compose this symbol with this I mean this would be exact sequence. So quite often the number you would like to calculate here. So the index of this thing of del bar of E will be basically just minus one K would have a dimensionality of this cohomologist for this operator. I mean cohomologists and kernels are naturally related. Again it's a number. If I try to solve it analytically it's very complicated. So it will give me some finite number. So there is a prescription which is at your boat. So let me write for your formula and just to give you idea minus two pi i n. And so instead of solving PDEs I can write this integral of certain objects. So this is different characteristic classes. So this is for example I start to feel stupid. Raise the hands who knows characteristic classes. Okay, who knows, okay lower the hands. I will be done. Who knows if you are in characteristic classes. Okay, so I have zero minutes. But I started later. Now let me just tell you things. So actually Google it in the theorem. So I will give you tomorrow a current version of this and we will calculate examples and I will tell you how to calculate determinants. But I will give you example wise. Anyhow, so since most of you know what's characteristic class this is thought class of this bundle. This is churn character. It's certain classes. So typically characteristic classes are produced by putting curvatures in variant polynomials, et cetera. So this integral when you actually wanna calculate it depends on connection. But eventually the result is independent of connection. So this is purely topological thing. So I have to finish now but let me just give you idea. So again, look at this as a black box. If you don't understand there is some recipe. This is hard to calculate from false principles. You do some magic and you have to integrate this. It's basically generalization of Gauss-Bann theorem. Now, typically in old days people would calculate by doing some complicated things with characteristic classes, et cetera for concrete geometries. Now what we actually can do if the manifold admits you one action then we can actually upgrade this formula and we can introduce characteristic classes and we can use IT about formula. So eventually on IT about formula I can write this as a sum of a contribution from fixed points. And then the calculation becomes very, very nice and simple. So to calculate the index I just need to know certain data discretely at fixed points. And this I can do both for usually elliptic operators and transverse elliptic operators. So I will finish here and so tomorrow I will continue this index theorem and then right away I will give you application of index theorem for determinants in S3 and S5 case. Yeah, thank you.