 Thank you very much for invitation, for possibility to give lectures. First of all, let me apologize. I'm very jet-lucked. So I just came from Canada, and I woke up at 3 AM. So I can be slow, but maybe it will be good. It will be pedagogical. Also, I have no idea what you were told last week. I have no idea what people promised that I will tell you. But anyhow, my intention is that the following, that I will try to talk about localization. I will stress number of mathematical aspects. So I will tell you about notions like, I will talk about index theorems, I will talk about elliptic, transversely elliptic operators. So I will touch all these mathematical aspects, and eventually one of my main application, I will tell you about five-digit theories in this mathematical thing. So actually, I don't know if you were told, but I will start by deriving a very basic things, and I will stress mathematical aspects. I will actually start from deriving for your ITA-bot theorem. So first in toy model, and then I will progress, I will make it more complicated, and from finite dimension, I will go to infinite dimension. So imagine I have a smooth manifold, and I have a U1 action. Okay. If you don't know something, let me know, but I will basically I'm a rather mathematically oriented person. So I would like to associate for this manifold, what's called typically super manifold or graded manifold. So what I will do, I take a tangent bundle, and I shift degree by one. So locally, if I think I would have a coordinate x mu, and then here I will have my psi mu. So psi mu, it's a grass manian variable. So it's a grass manian variable. And it transforms, so psi mu transforms the same way as dx mu. So it means that, remember the differential transforms in this way. So if I change coordinates, right? So if I change coordinates, again, I'm assuming that my psi mu would transform exactly like this. But the only difference is that it's odd guy. Okay. So what is important, one of the important facts about the thing that if I'm looking at this super manifold, so fact that this integration is canonically, whatever I integrate, we will discuss it, but this is canonically defined. So I would like, so I will give you some hints, but as a first exercise, prove it. So the hint is the following. So if you don't know grass, so grass man integral, so I will give you a hint how to define. So the grass man integral is defined in the following way. So if I have a one variable, if I integrate constant, this is equal to zero. If I integrate one psi, this is equal to one. So this is very important fact when we will deal with localization. And that is also important when we do this for path integral. So because of this very strange property, it's important that, for example, if I change my variable, so if psi tilde is equal to some constant times psi, I still have to have this property. Then if you look at this thing, then the only way to have this property is actually to say that delta psi tilde transform S1 over A delta psi. So you see, this is opposite what you would expect from X. So just to compare, if you would have X tilde equal to AX, then DX tilde is equal to ADX. So I'm asking you to prove it. So it means that the following thing what's going on. So this part of the integral changes by determinant of this matrix. And then you have to work out that this guy changes by one over determinant. They cancel out, okay? So this is first exercise. And there are a lot of conventions when you define these guys is how to order, et cetera. Because I have more than one psi and the anti-commuting, et cetera, okay? Good, yeah. It's T, this is important thing. So now if you wanna do another exercise, if you would do T star, for example. So this is a big deal. If it would be T star, then my psi will transform as a derivative. And for example, this is not true. So it's not canonically defined. So you need a density to defined. So just, I mean, so this fact is very, very important. So now let me say the following things. And as I told you, I have U1 action. And let me assume that I will have a corresponding vector field. So I have a vector field for this guy, okay? So then let me write the following transformations here. Delta of X mu is equal to delta psi mu, delta psi mu goes to V mu of X. So I will call it supersymmetry. So let's discuss what it means exactly geometrically. So first of all, what I would like to do, I would like to integrate functions on this manifold. So my functions will be from X and psi, okay? So if I expand them, right? Because psi is gross money. I basically will have mu1 muK of all Ks psi mu1 psi muK. So of course, my functions here one to one correspondence with differential forms. So if you wanna write it in a fancy language, you can write like this, differential forms. And in fact, this gradient, so this is, I mean, I put here one because I put a degree one here. I like a degree one. So the degree of the form corresponds exactly, I mean, to degree of the space, okay? Then what about the supersymmetry? So if I act on alpha X psi, right? So it's very obvious what I will do. So if I'm just writing this, this would be is the following. So I will first, when I act on X, I will have d rho alpha mu1 muK psi rho psi muK plus, then you will have different terms, alpha mu1 muK v mu1 psi mu2 psi muK, et cetera. So I'm just doing these transformations on this side, okay? So what you can convince yourself very easily and there are some conventions involved which I'm not actually fixing. So quite often, canonical people, if you look at the textbooks in differential geometry, the most canonical way is to put here one over K factorial just conventions. So in principle, I have to have here one over K factorial and then here we'll have one over K factorial, et cetera. Anyhow, the upshot is that at the level of differential form, this guy corresponds to the following operation, d plus IV. So d is a DRAM differential and this is a contraction. Contraction is a vector field. So this guy I will denote in the future as a dv and this has a name. It's called a QRN differential. So one of the thing what you can see and again you can do this in local coordinates but there is the following thing that if I square my delta to, well, let me say dv to square, right? This is just, so this by itself is zero. This is by itself is zero and the only thing which just survives is this operation d IV plus IV d. This is just a linearity for v. So by the way, I assume that you know differential geometry. Okay, so I'm very fast with this because if you don't know, then goodbye, basically. Okay, so this is equal to linearity of acting on the forms. So again, I will introduce now different notions. So what quite often people do when they discuss differential, this equivalent differential is the following thing. So the problem with this guy is that this would raise your degree by one up and this will allow degree by one, right? So one of the properties is a d acts, it's p form to p plus one form while IV is from p form to p minus one form. Okay, now quite often it's to introduce formal parameters. So this is just a formal parameter. And then I introduced what I would say that I would say that it's degree equal to two. And then typically mathematicians when they discuss this differential, they write like this. And this object now acts on differential forms on M which you regard as a formal expansion in psi. So it's a forms and then you allow to have multiply psi, I mean in polynomial fashion. So basically it makes, it allows you to have the subject of the same degree. So you change the notion of degree. So it's not only differential forms, it's these guys. Now, if you look at guys which is invariant. So if you look at invariant forms. So invariant meaning the following, that my form will be annihilated by, so it depends on, it may depends on psi, it's equal to zero. So if I look only invariant forms, then these guys squares to zero because before it was, so then I have a complex. So this is invariant form on M and then I have a current differential. So in all forms, it's not a differential. It does not square to zero. But if I look at the variance form, it squares to zero. So it's cohomology. So the cohomology typically denoted. So in this case, I'm just considering U1. So it's called a current cohomology. And this guy in called Carton model. Carton model of a QR and cohomology. There are many other models, okay? So I'm going to use later on about, so I'm deriving at ya both formulas, but I'm also telling you the K word. So I'm going to use all these words, element of a QR and cohomology. So this is Carton model. There are alternative ways of building a QR and cohomology one is called while model, then there is B or C models, et cetera, so there are many. But they give rise to the same cohomology. So just idea, so idea that this object actually is the same as a cohomology of a quotient space. But the main thing is that M over U1, if it exists, if it exists. Because quite often my action may not be free. It may have fixed points. So this does not exist. So this object always is well defined, but this object is defined only if U1 acts freely. So it does not have fixed points. And this is actually a bit boring in some sense for us. Okay. Good, so let me see. Okay, good. So this is a setup. So we would like, so this is my supersymmetry written where it's written somewhere on my day erased already. So I have my supersymmetry written in local coordinates. So this is in my Suzy. And I would like to study these functions, which is killed by this. So this is what I will call a QR, a QVarrently Closed Form. Closed Form. So what is important that this condition, so for now I will assume like physicists do, so just assume that Xi is equal to one. So I'm not carrying this formal parameter. But I just wanted to tell you that this is very important for keeping track of grading, et cetera. And for example, it's very important for this statement. Actually, if you wanna make precise. So it's, I mean this parameter Xi is required. But I will do like a physicist, I will just put it to one. Sorry, side here. So one of the comment is the following that if I actually look at these conditions because of my degrees. So actually this condition, if I expand this guys, it will involve a, you know, loud degree form, et cetera. For example, let's say for definiteness that dimensions of M is even then this would involve top form here. So if I wanna solve this condition, being currently closed, it's necessary involves all forms. So my goal is to calculate, so goal to calculate the following integral. So provided that delta of alpha is equal to zero. So let me denote this at z to zero. So the trick is, which I will, again I'm sure you were told about this trick already, but I will basically, I would like to stress some features of this trick and also explain you some mass. I will make it more and more complicated as we go along. But the trick is the following that I can deform this action by the term minus delta of w. It's another function. Then you can prove very simple theorem. Again, for now I'm working for compact manifolds, MS compact, et cetera. So it means that I can use integration by parts that's without any troubles. But the thing is that if I assume that delta square of w is equal to zero, then it follows then d zeta dt is equal to zero. So this is done, sorry I forgot to put here t. So this is done very trivially. You differentiate over t, this goes down and then you have to differentiate by parts. And then when you hit delta on alpha it's zero and then it hits another time, this delta w. So the only term which is actually problematic is this. One term which contain this guy and I put it to zero. So if I choose, so if we choose delta w equal to zero, then we can do the following thing. So if I wanna calculate it zero since it's independent of t I can send t to infinity. So I just have to calculate whatever, I mean it will survive at t goes to infinity. So let's discuss what is going on here. Okay, so for this concrete example I can take my w to be g mu, g mu, so I knew, so where g is the metric. So first of all, statement number one, which please you do it yourself, it's pretty trivial statement. This is equal to zero if linearity for v of g is equal to zero. So v is a killing, so v is a killing. So for compact manifold this is not a problem, it's not actually a condition because if the group you want acts you can always have a metric by averaging. So this is not actually a condition. If it's not a group action, if it's non-compact then it can be rather non-trail condition. So now let's calculate exactly what we have here. So our integral, so if I calculate delta w this is equal to the following thing. So just remember, so this is v mu, g mu nu, v nu plus psi mu, psi nu and then here I will have basically g mu rho v rho derivative of nu. So in case if I'm too fast just do it yourself at the evening instead of drinking alcoholic beverages so you can repeat my calculations but okay, so that's what you get here. So that's what I have to put there. So if I put there then, so let me just, I will denote by the same letter v until so when I contract form. So let me not introduce another letter so you will see from contents if it has index downstairs or upstairs. But what we get, so what we have to do with you is t goes to infinity dn x dn psi alpha x psi e minus t then I have a norm of v minus t psi mu psi rho and then I will have basically dv mu rho. So I have to take t to infinity. So the term which dominates, so if this is positive guy, so the only term which would survive it's when v equal to zero. So only when norm is equal to zero will contribute meaning that v equal to zero. This is exactly fixed points. So if the action is free then this integral actually will go to collapse so they will be nothing interesting because if the action is free this is never zero. So it will just be suppressed, okay. All right, so of course on the manifold I may have many points. So let's concentrate, let's look at one fixed point. Let me assume that this point is zero. Just I can choose coordinates I'm looking at neighborhood to one point, okay. So then what I can do, I can of course take this expression and I can do the expansion. So I will have these guys and I will expand first of all this term. So I will have, let me write abstract H mu nu x mu x nu. There will be no linear terms. I hope it's obvious. Plus if I expand this then the next terms will be order of cube. So I will give you later on a rather complicated exercise to repeat it for some case. So now I also assume that this point is isolated. So isolated it's in fact exactly mean that my expression will start from quadratic term and what is important is that this metric is not degenerate, okay. So this is I'm just expanding this term. Then I have to expand this term. So what I will have, I will have s mu nu psi mu psi mu. So this is some matrices. I will tell you that actually I don't need to know them. But in principle, if you're strong person you can just actually take and do expansion of metric or field combine different terms such as right where explicitly. So this is concrete geometrical object but actually I don't care how they look like. And I will tell and you will see later on why. And then here I will start to have terms so the next term will be of order of psi square of x, et cetera. Right, that's the terms will appear. So then let me do the following change of variable. So let me say that my x tilde is equal square root of t x and psi tilde is equal to square root of t psi. So then I'm looking at this object, right? So I wanna calculate zeta of zero equal to limit of t to infinity. So now here what is very, very important and that's what I told you. So when I do this change of variables my measure transforms canonically and so in fact, I did erase this already statement but there will be no any Jacobian in the transformations for this guy. So this is actually will be dn of x tilde dn psi of tilde, okay? So this is important crucial thing and that's actually an infinite dimensional set up it's a big deal. So you do this. So when I rescale this t's this will transform as a power of t this will transform exactly opposite way. Okay, then next thing is that I will have alpha and then here I have to write this as x tilde square root of t psi tilde square root of t, right? Right. And then what I will have there I will have here the minus h mu nu x tilde mu x tilde nu. So you see I had exactly I have t I have quadratic term I rescaled but the next term will be order of one of a square root of t because it starts from x cube there and then next thing is I will have here minus s mu nu psi tilde mu psi tilde nu and then again I have just to analyze how this term goes and this is what we gain of order one over square root of t. So now I actually can take t to zero sorry to infinity and then what I get I get the following thing. I will get alpha zero of zero. So because anything which has psi will just disappear so I have a law guy which does not have psi it's at zero and then I have to have inside the following integral dn x tilde dn psi tilde on e minus h mu mu x tilde. dn mu x tilde nu minus s mu nu psi tilde mu psi tilde mu. Well then I will just have to do this integral and up to some pies which you can figure out yourself up to some pies. This is the following thing. So this is a bazonic integration. So I will have alpha zero zero. So this is just a standard determinant of H. So this is fermionic integration. So this goes up. So this you can write two ways. I mean, so this is a sub five fan of s. So just remember that for anti, so s by definition anti-symmetric. So this is symmetric, this is anti-symmetric. Sub five fan of s this is just a way of taking a square root of s. And this you can do only when this is anti-symmetric. This is not the result we are satisfied, but this is the result. And so here what is important before we analyze this further and eventually we'll do these similar things with differential operators. The H has to be positive definite. Hm? The H has to be positive definite. Yes, H has to be positive definite, but this is guaranteed by this, right? If it's not positive definite, it's not defined. So for example, one of the things, I mean the typical problem you will have if you try to mimic this on non-compact space. So this is typically metric maybe not definite, et cetera. So now everything definite defined. So what is important for this expression to make sense, this matrices should be non-degenerate. So remember I had this suzi. So I had a suzi written there. So I had these things for you. So now I would like to write my transformations in these variables with a scale t and actually take t to infinity. So if you do it, you will have linearized. So I will have actually delta x tilde mu is equal to psi tilde mu. And then I will have here delta psi tilde mu is equal to now I will have d rho v mu at zero x tilde rho. So all other terms just will die out if I do the thing. So this is linearized suzi. So now actually why I told you you don't need to know this matrices exactly. So you can actually do this calculation right away and see. But you also can look at this term. So this term I can call it delta w up to quadratic order, quadratic order. And then what I can check is that, so let me put here subscript l, so it's a linear guys. So what I can do, I can actually ask that this term if I look at the linear guys, this guy, this is equal to zero. So it's analog of the conditions I had here, but now I can do it up to linearized level. So if you do it, so again, it's exercise, you can do it very easily. So just take these guys and apply to this combination and require that this is zero. What you will get, you will get the following fact, you will get that h mu nu is equal to d mu v rho from zero s rho nu. Is there is some way to get this board down? Stick, I thought it's more advanced. This is logic. 24th century. So I would like that you stare at this thing up there. Right, so let's put everything together. So I told you that you calculate this, you apply linearized thing and what you eventually get. So we had alpha zero at zero is equal to fathian of s determinant of h. Then what I told you here, it's exactly that h is equal to some matrix composed of first derivative of v times s, right? So then it just cancel out. So you will get alpha zero is equal to one of the square root of determinant delta mu v rho from zero. That's what you have to calculate. So finally, let me make a statement, a theorem and then let's discuss it and let me give you exercise. So the theorem which is goes under the name of fathian bought formula. So if I have to calculate for you this integral is the conditions that delta alpha is equal to zero. Then this is equal to sum over fixed points. So now let me be honest so that you can restore the thing. So this is equal to pi dimensions of m over two. And then here you will have alpha zero at fixed points. So I have to sum them up and then here I will have a determinant of this matrix. So this is called at your bought. Okay. Zero of x. x which? Yeah, sorry. x, thank you. Okay, so I would like to stress for you the things of what I'm doing is triviality. But it's important to keep in mind that there is different steps of calculation because as we go along, I mean, in a while I will consider more complicated models still find a dimensional and then we will have more fields, more stuff, et cetera. There always will be a two set of calculations and tell us what people do in localization literature. So either you take a sec, I mean matrices which is quadratic and you do Gaussian integrations, you get something like there or because your term has extra symmetries, I mean, these things linearized and then you actually go here. So what is important that this term, so I mean, if I want actually to produce answer for me, it's much more important to know this, not full supersymmetry nonlinear but only linearized things around fixed points. Okay. So this is very important and people actually when we will do index theorems, et cetera, I will explain to you that there is this calculation, it's a valid calculation but then you can simplify it even one step further. So now let me give you exercise and a marginal statement and this is hard exercise but if you're brave, you can do it. So what I did here and the way I derived for you, I assume that my fixed points isolated but quite often I may have not isolated fixed locus. So it can be some sub manifold. So actually the general idea about formula is the following types. So if I would have DNF, so let me write as a differential form. So if I integrate my currently closed differential form, then actually I can reduce over invariant subsets. So my, this is a set of fixed points and the U1 action. And then I have to evaluate my alpha at this set of fixed points and then here will be a earlier class of a normal bundle for this guy. So I suggest for you to derive this formula but not, let me tell you how to derive it. Not because in most of the books, it's written as a two line statement because there is a lot of formal nonsense, push forward, pull backs, et cetera, you derive this formula. Do it like a real physicist. So what you do, you're supposed to do, you have to repeat this calculation. The thing is that when you'll expand over the things, so remember what you have to do, you have your manifold, you have a fixed locus. So you have to split coordinates along the fixed locus and transfer to fixed locus. And before you assume that your metric, I mean, you just need to know metric at one point. But in this case, you actually have to start to expand metric and that's why curvature will appear. So do it honestly. So then basically what you have to do, you still have to expand everything up to quadratic order but you have to differentiate. So what I'm saying to you, this is my fixed locus manifold. And then I would have, for example, x i's here and this would be x alphas. So the way I would expand, for example, things will differ and that's curvature turns your life here. And eventually if you do it very explicitly, you have to, for example, very convenient to use normal coordinates and then you can get this expression very explicitly and then you can identify with a formal nonsense. But I actually suggest to you once in your life to derive it explicitly. It's hard exercise, okay? You have to remember a lot of stuff but in principle, it's a straightforward exercise. Okay. Good. Any questions before I go to a more complicated setup? So everything is crystal clear, right? So next I would like to also derive for you some finite dimensional model which is a bit more complicated and also start to discuss some mathematical issues. So here you were integrating manifold. So here in this example, my supersymmetry was simple. So this example just to confront, I will have two examples. So here this was my super manifold and that was my Suzie. So next thing, which is also from this example I will jump right away to infinite dimensional setting. So assume that I have some vector bundle. So I have a vector bundle. So quite often it's, so my super manifold here would be the following. I will take a vector bundle, shift by degree one. So this is notation as a following. It means that my fiber coordinates is odd but my base guys are even. So this is just, so what I'm doing is I'm shifting. So this is a vector bundle. So I have my x mu still coordinate and then I have chi A in other coordinates. So this is odd and this is just some expanded in the basis of sections. So now my super manifold where I actually would like to solve the problem is one, this one. I don't need, well, this is. So I take a vector bundle, I shift degree one. I will write for you in local coordinate. Then I take a tangent bundle and I shift by degree one. Okay. So if you want I, then I'm writing everything. So my coordinates will be the following x mu. So I will still have part of the old problem I had before I will have psi mu. So this guy transforms S dx mu. Then I will have chi A. This is just a section transforms as a section of E. And then here h A and this is also odd. So this is even odd, odd, and this is even. And here where life becomes complicated but nothing you can do, it transforms a D of x A. Sexually it's a coordinate, right? I mean, it's a coordinate. H, h, h A. Ah, sorry, I'm just lucky. Thank you. D chi A. So now let me give you a big, so this is unfortunately half of the physicists don't understand it. Vector bundle over vector bundle is not a vector bundle. This is a mantra easy to remember but this is in a way very trivial thing is that if I have a vector bundle, for example E and I take another vector bundle over this, so I'm taking T of this guy, I actually cannot think of this as a vector bundle as of original space, okay? So what I'm telling you that I take a vector bundle over M, then I take it again bundle over this space as a total manifold. I cannot think of a whole as a vector bundle over M. And in local coordinates, it's actually mean the following. It mean that this transform as a section but this does not transform as a section, does not transform as a section. I mean, this is very simple exercise. So let me just give you some hints and then you can do. What do you put D here? Does this mean D over the initial manifold or over the vector bundle? Over vector bundle. I wanna have differential for this guy. So if you think, so what you are doing is the following. I'm thinking this as a, so I have my manifold, right? Which is E one. So it has a coordinates X, mu and chi A. So this is a section meaning that chi A tilde transforms so there is some transition functions GAB of X as a chi B. So I choose my bundle, whatever my bundle is, there is a transition function so you can define them and that's how it transforms. And then over this super manifold, I again would like to introduce, I mean DRAM would like to switch part, et cetera. So then of course my fibers formula, right? So it will be DX mu. So for this I would like to associate, well associate psi mu and then there will be D chi A to associate HA. So now the problem what I'm highlighting for you that this is, I mean, this is very well defined object geometrically in certain sense but this is not a section simply because if I take this formula, so D chi tilde A, this is GAB of X chi B plus GAB rho D X rho chi B. So if I, sorry here I forgot to write D chi. So what I'm saying that if I would identify this with a chi A tilde, this is my chi A, then it does not transform as a section. That's exactly this magical statement, okay? So it means typically there are two ways of resolving these things, either you introduce connection or you just don't care about this but then you have to be preoccupied with coherence. So it's all important thing is because why I'm telling you this stuff because I could easily lie to you but for example I cannot try it for you, I mean any object like this on the bundle. I mean this is bad. This is bad because if my H will be a section, then it's fine if I introduce metric but because of this property, okay? So it's a complication but there is nothing we can do actually because many formulas sometimes involve some non-canonical choices, et cetera. So one of the way of writing it, so this is one thing I wanted to highlight for your, so another thing that what I'm going to do, I'm going to discuss so my bundle E would be a QRN bundle. So when we will discuss, so it means the following that I will have action on M, right? Don't forget that I'm always would have my action on U1 on M. So this section also exists on E so I can lift it on E and this section is compatible with this projection map. So this is calling the QRN bundle. I will give you examples so when we'll do calculations for example for S5 I will give you very concrete examples. But this is quite natural that I have a bundle, for example if I have S2 it has a rotation of U1 and most of the bundles I can put I mean there the most interesting bundle they will be QRN so they will respect the action of U1. Okay, so then I would like to write for your supersymmetry here, so naively and again if I have action of vector I mean U1 here I'm saying it's lifted here so they will exist an alloc of literative here. Sometimes it looks complicated depends on the bundle I'm not going to write for you the formula. Anyhow in a moment I will give you a linearized version but let's write things here. So I had my supersymmetry this one so this what we discussed before. And then here I have to write so let me write first naïve thing. So this is operation whatever will corresponds to my infinitimal I mean it's infinitimal action of U1 which I lift on the vector bundle. Okay, so this is appropriately defined literative. Again it's a tricky thing how to define this literative because depending on the bundles for example for spin bundles defining literative requires certain imagination they can be defined but it's an important thing that it's a linear operation which is compatible with the action of U1 with all this maps. So what I wanted to tell you that this is not good thing so typically there is a way of co-orientizing. So quite often it is actually a coherent version of this guys plus some extra terms plus extra terms. So just let me say you that this is just some connection which I choose. So whole thing is here done that I'm introducing connection such a way that I can claim that this is a transformation section this transformation section then I can write expression like this. Sorry Ky, sorry Ky, sorry. Okay, so be aware of this facts and you know there is this extra complication. So for example when you derive I mean this is in fact a setup to derive earlier theorem through the curve which et cetera and the curve which actually appears exactly because of these terms. Otherwise you cannot write burst exact terms. So when I will write construct for you this I'm not going to do it so I will but basically I have to do this guys et cetera. It's important that I'm using this formulas not this guy. So this correct and this correct but squaring H I have to make sure that it transforms a section, okay. So this is one of the complications. So now I would like. Is that canonically defined or the coherent form? No, no, no, no, no. This is exactly a pain. It's not canonically defined. You have to make a choice. So these formulas I mean I have to write you know I have to make choice and that's it. Unfortunately nothing you can do. This is not canonically but you have to do it. If you don't do it you cannot actually write things like this, okay. But in a way this is not now this new guy is not a diram of chi formula. So it's a co-varnetized version. But what is important that because of its diram again there is a sort of the same statements as before. So if I write for you this guy so I can write this as a just to differentiate them, okay. So dn chi rank of r h. It's again canonical measure. So one of the important thing of localization if you believe in localization sexually gives you a canonical measure for doing in path integral, et cetera. Okay, so this is canonical and in fact you can switch to these guys and also since there is a linear shift transformations it's also will be canonical measure in this new coordinates. Okay, so let me skip the following step where basically I can do this same mantra. Okay, now let construct me w. Let's me take a form integrate over this, blah, blah, blah. Everything goes to fixed points, et cetera, et cetera. Sorry, in this case e is not compact, right? Yes, but life is getting complicated. That's exactly, yeah. I mean, you're absolutely correct. I mean, life, it's not compact. You have to think what to define, et cetera. So for example, one of the thing is that you cannot put any object and integrate. So for example, if you integrate things which decays fast enough, then it's okay. Okay, so this mantra I'm skipping but in principle I have more fields and now my, so I have collections of even a note field so I would like to repeat the same mantra but let me not tell you this idea about argument. It goes through completely. So let me actually try to analyze the situation at the level of linearized things and try to understand how final answer will look like. And this is in fact, much more close to the field theory. I have 15 minutes, so let me start this story. Okay, so what my goal is the following. I have these transformations, whatever the way to look. Yep. You actually have half an hour. You have half an hour. Half an hour, why I thought that. I have 15 minutes. Okay, I forgot, sorry. Good. Okay, so let me tell you what I want to do. So I'm skipping this part where in this more complicated model you can do stuff. So by the way, as exercise for you, you can look for Mathai equivalent representative, you can play with these formulas, et cetera. And I don't give you references, but for example, you can definitely, I mean, there are millions of places in this. There is summary of this construction if you look at Peston has a small paper in our big volume on localization. So for example, he discusses this issue there. So you can write, I mean, this is the level one, two pages. So this is a very nice model. I suggest you to play with this and get it. If I'm alive, I can tell you maybe, I mean, later during exercise classes, how it's done actually. But right now I would like to actually linearize this problem. So let's just look how it will look about. So this will go to psi mu, right? Then let me draw appendices because it will be obvious, right? So delta psi goes to I zero X. So this is exactly this formula. So I'm just saying is delta X goes to psi, delta psi goes to some matrix times X. So I'm just doing linearizing. So this is a linear operator now. And then delta chi goes to H. And delta H goes again to another linear operator, chi. So that's what I'm doing. So here, I'm still have to tell you how localization works here. So before I go to linearizing. So actually in this story, so I'm writing this. This I typically cover and ties are replaced by this. And then I have to write for you this term. And the most interesting thing would arise from the following AB HB minus SB of X. So where SB, so where S is a section of E. So by the way, of course, you can add the same term you added before, I mean, we had it before. So you see now it's important that my H is covalent and contains connection because otherwise one of the terms when I hit by delta I will get H paired with the metric H. So this is metric on fibers, metric on E. This is metric on M. So S is a section. So I told you there will be two conditions. So if I will start now to check this thing, then of course there will be one conditions coming from here that it should be killing guy. But then there also will be conditions for the section which let me tell you now in the fancy words but we will derive it at linearized level. So my section, this should be a QVARant section with respect, with respect. So again, remember that section if I have E over M. So my section is a map from M to E. So a QVARant means the following thing that because eventually when localized at zeros of this section, so I have U1, it moves points here. What is important then when I will put look at zeros of the section, my action should be such that it does not move points. So if I pick up arbitrary section, then my U1 action may move with zero. So QVARant, I mean, is exactly basically this condition. There is a purely mathematical definition which I don't wanna give you. I mean, I will derive for you at linearized level as a physicist, but this is two conditions. I mean, you have to keep in mind. And this is what we are writing here, okay? So why I'm discussing this section because for me would be important actually think about this section as something linearized, right? So I will have for you the following operation. So there will be S, B, rho at zero. So this would be this matrix, which I will call D. So let's forget all this geometry and let's just look things at linearized level and analyze them. Because this is what actually matters when we will do very concrete calculation. Sir, does it hold some constraint on metric on section? No, I mean, there is no, I mean, what you would assume only that it's invariant with respect to V. Yeah, that's what, yeah, I mean, this type of things. I mean, typically whatever you do should be invariant under V. But again, it's typically not a problem for compact manifolds, et cetera. Right, so let's write this W in linearized terms. So my W would be the following. So just look what I would like to do. So I'm now looking at everything abstractly. So imagine I have some, a scalar product. So psi, I zero X. So this term is this term. So I'm pulling W, I mean, I don't write delta first. So this is just this term up to linearized level. And then here I will have next thing is a chi, h minus i dx, I put a i just because I like to put i, okay? So it's just exactly these things, but everything written at linearized level, et cetera. And again, I start to use more abstract notations because I wanna give you infinite dimensional interpretation pretty soon, okay? So what I would like that my algebra is satisfied in one of the conditions, I would like that this is true. So actually the only thing we have to fulfill it's the following property between r d minus d r zero is equal to zero. So you can check it. I leave it for you as a exercise is to do it. So again, I try to use different operations. So this is my operations, linear operations which enters in the transformations, then this is some term related to zero section which I wanna localize, et cetera. So this is the third thing. Another thing I have to ask you, and again, this would be over over that with respect to products, I typically will have this property. I will tell you why I'm assuming this in a moment. So when I'm moving, so this would be both for zero and one, zero and one. So it meaning that r zero dagger is equal to minus r zero, r one dagger is equal to minus r one. So this is very obvious property which you know, this is typically, so r zero, it's like a derivative. And when you move the derivative in a scalar product assuming that metric is invariant, it's exactly gives you minus. So Lv is not a Hermitian operator but I Lv is a Hermitian operator. So that's why I'm assuming that these operations anti Hermitian. Of course, I can make them Hermitian by putting appropriate ties. But in a way, this is very important when you do this calculation, subtractly, so you write these operators, you have to figure out how to move the things around. But here, I'm not assuming anything. If I move here, I will just put explicitly d dagger. Good. So for example, if I write rho zero, rho zero dagger, this is the same as a minus rho square. So now let's work it out and let's write, let me hit by delta this. I will have delta, w is equal to, this is i zero x, i zero x, minus psi i zero psi plus h minus id x h minus, minus chi r one, chi plus e, d psi. So you can see that h centers only quadratically and there are no any operations on h. So if I put in the integral, we can integrate h. So after this, I will get the following. Delta w is equal to x minus r zero square plus one over four d dagger d x minus psi i zero psi minus chi r one, chi, minus chi i over two d psi plus psi i over two d dagger chi. I'm just copying, but again, it's just applying. So what I did here, because h just centers quadratically, I can integrate this in pass integral. I can do this Gaussian integration. So now you see that I have, this is basonic term. So this is an analog of this term we had before h mu nu x mu x nu. So this is an analog of this term. And then all remaining terms, it's a fermionic terms. It's exactly these terms s mu nu psi mu psi nu. And the story is exactly the same as before. That's why I'm not repeating for some details, but that's what we get the level of operators. So now it seems that, you know, we know the answer, whatever we observe while we integrate, it will be evaluated at zero, but then we have to think about determinants here. So in determinants, well, we know what will be up there. So if you calculate, then the answer will be looking the following. You will get here downstream determinant one over two minus r zero square plus one over four d dagger d. So this is a basonic integration. And up there, I will get a fermionic. So let me instead of Fafn write like this, so you can write Fafn, but this will be r zero, i over two d dagger minus one over two d r one. So this is my basically zeta. I mean, there can be some observable evaluated the points, but I mean, we are concentrating on determinants. This is something much less trivial. So psi and chi are like real speed numbers. So I'm talking about spinners. I'm not talking about spinners. I'm just saying, you know, it's one of the thing is that don't try to bring all words at once. I did not talk about spinners. They're not here at all. I said, I'm taking a vector bundle. I just reverse parity. Where is spinner? No, either you are a spinner or not. I mean, you know, it's, don't bring stuff which I'm not talking about. Maybe Guida talk last week. I'm not a super similar. I mean, I'm talking, I'm stressing these. There are no spinners here. It's a grass money and variables. The only thing you have to keep track, I will comment towards them about supersymmetry, but right now, you know, everything is pretty formal, but it's important. Okay. This is a vector bundle and I just reverse parity. So some guys are treated as a grass money and variable. Some as non grass money. I guess the question was also, is Psi dagger or just Psi? Is Psi dagger just Psi? You can ask, what is Psi? Psi is DX, is DX dagger is DX? Yeah. No, but you're asking questions which do not make sense. It's simply your problem is you know too much. I mean, you should not ask these questions. Otherwise, you get in a trouble. You start to discuss my run a condition and the place where they do not exist. Okay, so it's, I'm just, basically what I'm doing is I'm integrating a differential form on the vector bundle. The problem is that if you actually try, I mean, one of the thing is why you apply, I mean, use the supersymmetry, et cetera, it's very hard if I will start for you to write D chi, you get even more confused. So instead of writing differentially, you introduce more and more letters. And in infinite dimensional setup, you cannot do it without this, right? I mean, that's exactly the problem. But eventually what we are doing, we're just integrating differential forms on the vector bundle. And we have to construct things which is well-defined, et cetera. Any other questions? I have questions about the variation of W. Why the second term is minus? I don't know. I mean, just do calculate. Maybe I cover, I mean, I don't wanna do calculation. You just do, you know, you take the things you did. I act the data on W that you defined. The second term must be plus sign, right? But don't, I mean, delta, when it jumps over odd guy, it pickups minus, right? So here what is important? Okay, I mean, so this is even, this is odd. So this is odd, right? So when I go write for your delta psi on, I don't know, mu, v mu, then this is formula as I've fallen. This is delta psi mu v mu minus psi mu delta v mu. Did you use this property? So here, whatever odd stuff, jumps of odd stuff, you put minus. So this is the rule of the game. Okay, so because delta is odd, it jumps over this, you put minus. Otherwise, it will not be, have a correct property of super derivation. Any other questions? So let's massage the thing, et cetera. So first of all, what is important for these guys to make sense? So this operator should be well-defined. And for example, this operator should be non-degenerate because if this operator will have a big kernels in trouble, right? So if these guys have a kernel, which is big kernel, then, you know, this is not good. So typically, this is second order. So what I think I will not have a time today, but let me already mention some words. So basically, what would be important for this calculation to make sense? The word is non-degenerate positive, et cetera. You actually will have to require. So this is r zero square in the field theory. You'll have to be a second order, second order elliptic operator. So now about this operator, which stands there. So let, you can do the following exercise. So let me write for you these four moves. So this will be r zero, i over two, d-degener minus i over two, d r one. Then you write the same guy, but you put a dagger on the result, I don't know. You check it. I think I have i here, but I'm not sure. I mean, my handwriting is not very clear, but it's easy to establish. But I want to get here, so I will get minus r zero square plus one over two, one over four d-dagger d. Then here we'll have one over two, i zero d-dagger minus d-dagger r one, i over two d-i zero minus r one, d minus one, sorry, minus r one square plus one over four d-dagger. I mean, I hope I did not screw up, I mean, I was writing my notes four a.m. today the morning, so I hope I did it. But anyhow, it's very trivial, you can check everything. So I mean, conceptually I'm not lying in details, I may be lying. Now what is important here, if you remember original condition, so at some point I told you that this condition r one, d minus d-r zero equal to zero. So this terms are gone, so this is very important. So then if you now plug these things in this stuff and massage them a bit more, then actually what you will get, so what you will get eventually blah, blah, blah. So d-ta will be proportional to determinant one over four minus r one square plus one over four d-dagger, determinant one over four minus r zero square plus one over four d-dagger d. So you see why one over four appears because this is like a square of this operator times conjugate. So I have a determinant and this is determinant one over two. So I have to take another square and then I will have here some two copies so some things will cancel out. So it's one over four actually. And also remember that here the order is different because this is operators in different spaces. One thinks it's acts on this space and some it's act on this space. So the question is which I will start to make some comments now but then we'll discuss when does it make sense? Again, just to keep in mind, so let me just make two statements for you. So this is analog of this statement actually. So this is what we are calculating. It's a determinant of S over determinant of H. If I use these conditions, I actually can reduce this a bit further on but we will discuss it so actually it can formally go down to the following things. So it's a determinant of one over two of r one, determinant one over two r zero. So this is analogous to this thing. So this is second order things. This is first order. This I can see just from linearized transformations because linearized transformations have only this and this. However, in fact, if you think about it, so there will be huge cancellations. So in fact, these things are done. So to be more precise, if you massage the thing, so you have to play with this condition. So this condition is extremely important. So this is actually also equal to a determinant on the kernel of d dagger of minus r one square. And I don't need to put squares. I can put here one halves. I mean, I'm not discussing phases, d dagger d of minus r zero square. So if you want to derive this formula, you just have to use, so use r one d minus d r zero. So basically the thing is that outside of this kernel, there will be huge cancellations, okay? So please remember all these formulas because we need them quite a lot. So I have to finish, but let me say some words. What next I will discuss for you. So now the issue is the following. So when we can, so imagine we're doing this in infinite dimensional setting. So in finite dimensional setting, everything is fine. And the only conditions actually, you have to choose good vector field. You have to choose good section. You have to make sure that things are decaying and integral is well defined, et cetera. And then you calculate this determinants, which is again not that bad. In infinite dimensional setup, things becomes more complicated. So you have this linear operations. And the thing is I have to tell you exactly when I can calculate because despite the fact that it's written well, right? For example, nobody tells me that I will be able to calculate this. So there are conditions on the operators to be defined, et cetera. So for example, what I will discuss for you until also next time, I will give you some very concrete models. So we'll discuss what does mean operator to be elliptic. What does it mean to be transversely elliptic? And also I have to figure out which conditions I have. So for example, I mean, I start that this should be always second order elliptic operator. Then for example, for these things to make sense, I mean, when these things is actually well defined. So I will tell you that D should be, for example, transversely elliptic. It can be elliptic, but it can be relaxed to be transversely elliptic. So I don't wanna start with this, but next time I'm going to talk about elliptic, transversely elliptic operators. And also I will mention for you index theorems. So if by any chance you don't know anything about differential geometry, you have one day to read Nakahara. And I don't know yet. So I will assume that you have a good idea about characteristic classes, et cetera. So tomorrow, I mean, so remember this formula. So tomorrow I will write them again and I will go through some infinite dimensional examples which feed this framework completely. And I will try to explain you, it's still not 5D, eventually I will come to 5D, but I will do 2D, 3D some examples to tell you what's the difference, et cetera. And also to explain you the stages, history of localization, because in old days in 80s, I mean, Wheaton would write like this and it will be equal to one or minus one. That was the highest science. Now it's a bit more complicated. So this is in fact, if you put all things, it can be rather complicated and interesting functions. So this is what I will do tomorrow. So I will do this more math part, et cetera, index theorems. And yeah, I think that's it for today. Any questions you have on the stuff I told you? Sorry, why? Why is the first equality true? Why is the first equality true? Why this is true? Yes. Because of this. So one of the thing is what I suggest to you is that, I mean, the formulas which I told you, derive them yourself. And this has become the complication now that I always will do the same what I did. So the whole localization in life, it's at your boat, nothing else. But it's at your boat on infinite dimensional super manifolds, et cetera. You always will have this tangent bundles, et cetera. So always keep in mind, and I highlight for you, this is one step calculation, this is second step calculation. So if you want to arrive here, you use exactly the same thing. This is like, this is the type of conditions you are using. So this is equal to zero. So that's how you use it. But it's exactly this. So dv is zero is equal to zero. So a priority naively, nothing should depend on this dd dagger. Because d, it's a thing which enters through various exact term. But these are operations. It's the only operations which enter transformations. So I mean, derive yourself, but this is exactly step from here to here.