 Dobro. Čekaj. To je veliko prav, da sem tukaj tukaj. To je moja najboljstva izinstitut in je veliko prav, da se leče. Zato je to občatno. V štih lečesih, nekaj zelo, je, da vse občatno vse občatnje, da se svoje tudi obrženje, ki sem videli, v svoj časih. Beli dve dnešnega, in pričo smo način, v supersimetrih, kvantofil teori, tudi o neproturbativne metoder, da zelo pomembajimo neproturbativne komputationi v nekaj kvantori v supersimetrih teori, tako v delovih koreljators, vzvečenje vzvečenja vzvečenja, in zelo, da se vzvečenje potretiljstv, kako je vzvečenje vzvečenja, vsak vzvečenje in vzvečenje. Vzvečenje je zelo, do lokalizacijovih technikov. Nisem, tvoje večne technikov, da se vzvečenje izgledaj, vzvečenje, in zelo, da se vzvečenje for dimensions by Wolf-Ger. And these techniques are very powerful because they allow us to reduce an infinite dimensional integral, which is the patin integral, to something much simpler, to some finite dimensional integral, or to some counting problem, to some series, and so on. But in fact, localization has a very long history, both in the applications to field theory, so what I call supersymmetrical localization, but especially in the version in math that was originally applied to finite dimensional integrals. And so this dates back to theorans of Duistermann and Heckmann, Atja Bott, and Berlin Verne, that appeared in the beginning of the 80s. So this is between 82 and 84. And in fact, as we will see, supersymmetrical localization is somehow an infinite dimensional version of these localization theorems. And so I think it would be a good thing, since we have a lot of time in this call, to start briefly discussing these important results, at least sketch how these are obtained. And so this will be useful, because you will see all the main ideas that will be applied and will appear in the more complicated setup of quantum field theory, but essentially all the main ideas are here. So I think this will be useful. So I will start discussing what we can call, with modern perspective, bosonic equivarian localization. And so supposedly we want to compute some integral on some manifold m. And on this manifold we have some symmetry g, some isometry. So if we are in this situation, of course, a natural idea to perform this integral would be to first perform the integral on the orbits of g and then integrate over the orbits. So of course it depends what type of function we are integrating, but so in particular we could try to reduce the problem to m mod g by doing this integration in two steps. However, in general, m mod g is not a manifold, in particular if g has fixed points, m mod g is not a manifold, a smooth manifold. And so equivarian chromology is in fact a generalization of what would be the chromology of m mod g in the case in which this is not actually a smooth manifold. So if this is a smooth manifold, of course one can define the chromology of this space, but when it is not, equivarian chromology generalizes that concept to this situation. So just to discuss the main idea, so let me focus on the case in which g is just u1. So the simplest example. We are assuming g is compact always. Yes, yes, of course everything can be generalized, but this is a compact manifold. And g is some, yes, so I'm taking the example in which g is a compact. This equivarian localization exists for g compact or non-compact, I don't know. Yeah, I will discuss the simplest case in which my symmetry is u1, so it's compact. OK, so OK, we have in this situation, so we have our compact manifold m and we take a metric on it, so this is some Riemannian manifold. And in particular we take the case in which the dimension is even, so this will be some 2l. And well, since you have some u1 symmetry on it, there is a vector field that describes this symmetry. And so let's take v, that we can write in components as v mu d mu as some killing vector field. And so in particular the lead derivative along v of the metric is zero, which is equivalent in component to saying that this symmetrization of the covariant derivative of the vector field is zero, this is the killing equation. And yeah, as it was stressed, so we are assuming that the symmetry g is really compact, so this is really u1 and so there is a common period in the orbits of this u1 on the manifold m. I want to sketch some picture, this vector field v generates some u1 action on the manifold. Now, so we can consider forms on m and in particular it is useful to consider the space of polyforms, the space of polyforms, that we can indicate as hm, so these are just objects in which it's essentially a polyform as many components, this component is a form of different degree. The font is a formal sum of all possible degrees and we have all possible components. Each of these is a standard form. Now, this is useful because then on this space we can define v-equivarian differential, which we call dv, and this dv is defined as d, the standard exterior differential, minus the contraction with the vector field v. So, if you wish, d is the standard differential, so this maps an n form to an n plus one form, but the contraction with v maps an n form to an n minus one form. And so in particular, this object mixes forms of different degree because if you start acting on some polyform and all these m has one component of a single degree, you get two pieces of different degree. And so this is the reason why we need this space of polyforms to define it. Now, okay, this might look a little bit inconvenient, the fact that so we had a grading on the space of forms and then we are losing it because this object mixes forms of different degree. Now, this could be solved by introducing some parameter psi here, and this parameter in general should take values in the algebra of our group, g, and then we could assign to this parameter some degree, in particular degree two, in such a way that actually this differential preserves the degree, but in the case of the one is not gonna buy as much, so we'll not do that, but in general, this is some useful thing to do, okay? So we'll not do this. But cool. I'm not familiar with the terminology. The contraction is just an integral over v. No, no, no, no. So I'm not sure we get the coefficients correct, but essentially if you have a form, so if it is just a one form, the contraction, you just contract the vector field with the form in components, and if you have more indices, I'm not sure what, okay, it depends on the convention, which section number you have to put here, but essentially you contract the indices with the vector field, and in particular, this operation does not require, of course, the metric, because the vector field is already indexed up, it's already indexed down, so, okay, any other questions so far? So, and is the first index is contracted by the convention? Yes. I mean, of course, this is, I mean, this is anti-symmetric object. So, okay, so, so, okay, so what are the properties of this differential? Well, we can take the square, and of course, this square is zero, contraction with v is zero, this is obvious from this presentation in components, because if you contract another time, this object is a symmetric tensor, but this is contracted with an anti-symmetric tensor, and so you get zero, but you get the anti-commutator of these two guys, and in fact, this anti-commutator is the lead derivative along the vector field v, and so, and so in particular, we can restrict now the space of polyforms to the space of v-equivariant polyforms. This is just the space of polyforms, which are invariant under v. So, in particular, when you apply this lead derivative, you get zero. Here, so these are all the polyforms, such that the lead derivative. So, why do we do that? Because if we do that, it's on the restricted space, this is nilpotent, and so in particular, then we can define a homology of this operator. So, let's call it h star with v, and as usual, this is just the quotient, so if you want closed forms, modula exact forms, so this is the quotient of the kernel of dv, restricted to this space, mod the image of dv, restricted to this space. Now, in fact, the interesting thing, as I said at the beginning, is that if this g acts without fixed points on m, then m mod g is a manifold, and it turns out that this homology, which is called the v-equivarian homology, precisely reproduces, precisely equal to the homology of the quotient space. However, in the case in which m mod g is not smooth, and so in particular when g has fixed points, this is an interesting generalization of this homology. Now, okay, this was pretty obvious from here, but we extend the terminology that we use with forms to the equivalent case, so we say that the form is equivalent to closed, if it is closed under dv, not alpha, equivalent to closed is dv alpha is equal to zero, and it is equivalently exact if alpha is dv of beta for some equivalent form beta. Notice that when we impose this condition on an equivalent closed form, of course this condition mixes forms of different degree, because so you have to take each degree of alpha, apply dv, but this, as we saw, mixes the components. Ah, we have here, so it mixes two components, and in particular one can write this condition as a series of condition that relates the values components. So it looks like all components are mixed, however still if you look at components of even degree and odd degree, they do not talk to each other, right? Because if you apply this condition to some, let's say, to the even, to the degree, to the components of even degree, we get object that only involves odd degree parts, and vice versa, if you apply it on the odd degree parts, these conditions involve the even ones. So there is no mixing between even and odds, although within the events and within the odds there is mixing. So you're assuming there is no group action still here on these forms. That is not what? Group is not acting on the forms yet. Yeah, so I'm restricting to this space in which they are invariant on the action of g. So they are not equivalent with respect to g. Yes, yes. Yeah, I'm saying v equivalent, you could call it g equivalent. No, but then, so if it is v equivalent, but you have to impose the condition gq variance. Otherwise, I don't find it sensible, because. So the condition I impose is this. But this is not the gq variance, g in the compact, the group you're taking. So is it the gq variance condition also? Well, the condition is that they are invariant under g. Do you mean that v exponentiates to some action of u1 and then this is? Yeah, yeah, so I said at the beginning that the condition is that this u1 has common periods. So it's really u1 that acts on the manifold. So then, if u1 has, I don't know the notation common periods, but if it has fixed points, then how do you understand this? Because then this homology may be defined, the homology you define with the quotient space. This lambda subscript vm. So if g has, if the group u1 has fixed points, then how do you define the homology there? Yeah, so our dv is close, is nilpotent, and we define the homology in this way. It also comes with a g action. Yes. No, dv just represents a g action. There isn't a very difference. Yeah, I mean g acts by the killing vector v. So you can say that this is the g equivalent action. I mean you can say that under rotations the forms are invariant, and infinitemzimal version of that is that I act with the killing vector v. Okay, so now we want to define integration of these polyforms, and okay, so we simply define integration of polyform as the integral of a top form, which is the one that we can integrate, and in particular, notice that if we try to integrate a form which is equivalently exact, so let's integrate dv beta. Well, dv beta, if you look at the top component, which is the one that we integrate, can only get contribution from the next to top component. So this will be the integral of d of the component to l minus one of beta because the top degree is 12, so we cannot go at 12 plus one and they contract with v. And so in particular, well, this is zero in a compact m. And so if you wish to talk, theorem still applies to equivalent polyforms, okay? Okay, so in particular the integral of a polyform, in particular of a closed, equivalently closed polyform, only depends on the homologi class, does not depends on the representative. So another way to put is that if we integrate over m, alpha plus dv beta, well, this is equal to the integral of just alpha. So only the homologi class, the equivalent homologi class matters and not the particular representative. So what we are interested in, we are interested in computing now integrals on the manifold, in particular we are interested in computing integrals of equivalently closed forms and the equivalent localization theorems of Atja Bot and Berlin Wern, that I mentioned in the beginning. In fact, tell us that these integrals only get contributions not from the whole of the manifold, but in fact only from the fixed points of this action on the manifold. So essentially if we have this picture that we have before and there is this one action on the manifold and in general this action can have fixed points where the action look like the following. So only these points contribute to the integral of this form and not the totality of the manifold. And so of course this is a great simplification, because if there is a finite number of points and if we have a finite number of points, only we have to some of these, some finite number of contribution, we don't really have to perform the full integration over the manifold. So only the neighborhoods of the fixed points, so let me call this space mv, is the set of points such that the vector field vanishes. So only the neighborhood of this space really contributes. So let's see why this is the case. So I will give you two arguments. So let's see the first localization argument. Are there questions before going into that? Prej, I don't see how this definition of 50-vajda for homology solves the problem with the fixed points. Well, so far, sorry, okay, I'm not sure what problem you are. So one of the interesting thing is that as I said, if g does not act with fixed points, m mod g is a smooth manifold and so in particular we can define its homology, okay? But if this is not the case, if g has fixed points, m mod g is not a smooth manifold. So equivalent homology generalizes somehow the concept of the homology of the quotient space to the case in which the quotient space is not a smooth manifold because the equivalent homology is defined even when g has fixed points. Now, of course, this is just one aspect. The aspect that will be most important for us would be that, well, equivalent homology, which is a part of equivalent localization, will allow us to simplify the composition of these integrals. So for us, this will be the problem. We want to compute these integrals and the equivalent localization simplifies the problem. The task. Sorry, about the definition of this equivalent homology, so you had to define this dv external differential in some sense. Is it canonical or could you somehow? Yeah, I wrote it, what it is. So you take the external differential and you take the contraction with v. Of course, you need v. Okay, so yeah, there is no choice, no there is no choice. Well, I mean, there are many things that you can do in general, but this is, okay, this is what I will do. So of course in general there are infinite number of generalizations that you can do. To begin with, you can have an abelian group which has, what I mentioned, bigger than one. You can have non-abelian group and so on. So there are infinite number of generalizations. So I mean, all the homology theories you end up with are all in some sense, based on what is to be done? No, no, no. I mean, I consider this particular case, which is the basic example and, well, essentially we'll capture all the ideas and the features that we need and we will find the supersymmetric case. But of course you can do much, much more. Is there a generation for discrete groups? Right? For discrete groups, how do you define IV? For discrete groups, well, there is no v. Of course there is no vector field. So of course these are not applied to discrete groups. In fact, this applies to U1. I'm doing U1. But, yeah, I'm sure you can do something with discrete group actions. Okay, no, it doesn't come to my mind something you can do, but I'm sure you can do something with that. Where do you use it? It's a kidding. Where do I use it? Well, first of all, really what I want is that, so I want a U1 action on the manifold. Yes, but you don't need me for U1 action. You need three milt again, right? Yes. An importance comes, you don't need v to be killing to have an importance, because I said the importance only comes from, sorry, one question at a time, I cannot. Yeah, so it's not, so an importance is not an issue because an importance, you see, if I act twice with v, I will get up to factors v mu1, v mu2, omega mu1, mu2, and so on. And this is a symmetric tensor, this anti-symmetric, so an importance will come out with no problem. But I really want a U1 action because of what I'm gonna discuss here, okay? You don't need a metric, no, to define vector field, but it will come. So, I mean, I could have started with say, okay, there is no metric, and then at this point I say, okay, at some point I need the metric, okay, I started with the metric in the beginning. Okay, any other question? So, I think it's misleading to say there is no metric, it's implicitly, metric is implicit there. In my mind, it's misleading to say that. Where is, if I didn't define it, where was it? No, when you started the reman in many fold, you are defining vector field with respect to that metric, and when you are defining this, whatever the structure, this forms, et cetera, metric is implicit there. Okay, maybe it's not useful, maybe let's continue. No, but yeah, okay. But in any case, I can define v without a metric. Okay, if you find it misleading, I'm sorry, you can find, I mean there are a lot of reviews with, maybe I'm better, but you don't need a metric to define the vector field, as you know. Okay, sorry, there was another question. Yeah, just a quick question, because I didn't really get your argument for seeing that in such an integration, only the fixed points of the action contribute. Yes, so I'm gonna give you two arguments, and the first one is coming, the second one become afterwards. Okay, any other question? Okay, so first localization argument. So this first argument uses a version of the point karelema that applies in this case, and in particular what you can prove is that if you have a v-equivariant closed polyform on m, then it turns out that this form in fact is exact, equivariantely exact, but not on the full of m, of course, but at least on m minus the fixed points of v. Okay, so if you remove these fixed points, in fact the form is automatically exact. So how do we see this? So how do we see this? Essentially by construction. So, first of all, we can construct a one form, which is due to the vector field, and at this point I do need the metric. So this one form is just you lower the index with the metric. So if you want this form, is, okay, you can define it as using the metric, but in component is just v mu, g mu nu, v x nu. So what are the properties of this form? So first of all, this form is equivariant, in the sense that the literarity of this eta is zero, and this, where does it come from? Well, when the literarity acts on v, you would get the commutator of the two fields, that is the very same field, so the commutator of v with v is zero, and when the literarity acts on g, where you use that, in fact the metric was an isometry with respect to this vector field. Now we can compute the differential of this eta, and so of course this has two pieces. So one piece is d, and the other piece is contraction with v, but contraction with v will give you v squared, the modulus of v. Now it turns out that this differential is invertible on m mod v, sorry, not mod m minus the fixed point, so on the manifold where you would remove the fixed points, and essentially, so you take this, you put it in the denominator, you take out v minus v, so you can write it, so you can think that this is formal, but I will be more precise in a minute. So it's here, essentially it just took minus v squared out, and then you can use the Taylor expansion of this object, but of course since we are dealing with form, this Taylor expansion is not gonna be an infinite series, but it's gonna stop, because when, so this is a two form, and when we reach the dimension of the manifold, it's gonna stop, and so we get a finite expression. So here you have forms, this is a poly form, you go up, up to the degree 12, so which is the maximum degree, and okay, so if you don't like this formal manipulations, essentially the meaning of this expression is just that, so of course this is well defined on m minus the fixed point, just because v is always different from zero, and then meaning of this inverse is just that if you compute dv eta minus one, taking these, if you wish as a definition, and we wedge with dv eta, this give you one. So it's a simple computation, you can just do it, and this is what you get. So this object here that we can call in the inverse, it's well defined on this excised part of the manifold. You can also check that it is equivalently closed, and in order to do that, so essentially you just act with dv on this expression, here you get zero, when you have dv, so here you have dv squared, which is zero, so you have this, and so what you get is that dv or this expression wedge, this is equal to zero, this is almost what you get, it's not yet because you are wedging with something, but again you use that you have an inverse, so you can multiply by the inverse and remove this factor here, and you can get your equation. So it's just simple algebra. So now this fact allows us to define another polyform, we can call theta v, and this is the nice property that dv theta v is equal to one. And so finally, if we go on m minus mv, we can write alpha as dv theta v alpha. And so well, when dv acts on alpha, it gives you zero because this was closed, and when dv acts on theta v, it gives you one, so this gives you what you want. And so this explicitly shows that alpha is dv or something, which is well defined on this manifold. And so in particular, since alpha is exact, it means that when we integrate on this manifold, we can reduce to a boundary term. And so we don't get any contribution from the manifold, we only get contribution from the boundary of this space. So if you want the integral over m of alpha gets contributions from the boundary of m minus v, from the boundary of m minus v, but in fact, this is precisely a neighborhood of the fixed points of v. So, okay, so this is not yet showing us what is the result of this integration, but at least it's showing us that, in fact, this integral only gets contribution from special points on the manifold. We don't have to care about the whole integral, only about the special points. Is there any question on this? Why do we want the dimension of the manifold to be even? Let's see. Maybe for the inverse. I mean, it will be important for us later on, because we want, so I will discuss, so this is just a simple example, as I said, this is not a complete theory of equivarian form, homology, equivarian integration. And in particular, I will show the simplest example in which there are just fixed points. And if you want to have just fixed points, you need the manifold to be even dimensional. Otherwise, in general, you have manifolds, I mean sub manifolds, which are the fixed points, which, of course, you can do, but then it's more complicated. So, as I said, I mean, this is not the most general thing we can do. I just want to present a simple example, okay? Any other question? Okay, so this argument gives us this this localization result, but it doesn't tell you yet what the integral is, so now let's see a second argument, which will actually give us the result. Usually, what we want is we have an integral, which we want then to localize it. Yes. So, is it clear that I can always write it in this point, that I can always find this form alpha? No, I mean, this is our starting point. So, as I said, we are interested in computing integrals of equivariantly closed alpha. I mean, you are given such an integral and somebody ask you, can you compute for me this integral? So, you are given the alpha and then what we want to argue is, first of all, that this integral, this particular integral, it has to be closed, otherwise, of course, this argument doesn't apply. Then it localizes to fixed points. So, first of all, it has to be equivariant, so it must be invariant under the action. Otherwise, I mean, if the thing that you integrate is not invariant under the action, you don't expect that the action plays a bigger role. But is that the only requirement that is what I don't understand? So, this alpha is that you are integrating, all it has to be is equivalent to... Yes, so far, I give you an argument why only the fixed point, the neighborhood of fixed point should contribute. And what I use was that alpha is equivalent to closed. Because it looks like a very specific alpha that you found by... So, it's not clear to me that it's sufficient that it's just equivariantly closed. Well, I mean, we can go through the steps. I mean, I showed you that it has to be so, right? I mean, of course, the condition that is equivariantly closed is a constraint. But assuming that constraint, I mean, I just showed it. Just one more quick question. Like, when you write the dv eta in versus 1, what's the one on the right? And is it the top form? Is it just the number one? Oh, it's a bottom form. It's a, I mean, a zero form. Zero form, just one. Yes, yes. I mean, it's very simple to check. Okay, I have another question, sorry. In a different way, what alpha is dv of theta v times alpha, again? When you write alpha in equal dv of theta v alpha? Yes. So alpha is again in its own. So alpha is equivariantly closed form. And what I want to show that if you restrict to m minus the fixed points of v, then it is also exact. Is the d of something. How do I prove it? Well, I tell you what is the d of what. And you can just take the dv, right? dv of theta give you one. So you get the alpha. And the v of alpha is zero because we took it closed. So this is a proof by construction. Okay. Any other questions? Okay, so let's go to the second localization argument. Yes, by the way, I think that next week you will have a much more formal and maybe rigorous treatment of the subject by there's a zine or, I don't know, Nikita. Go much more in the details. Well, it will be probably more formal with less details but more formal and rigorous. So combining the two, you should have a good picture. Okay, so second argument and evaluation. So now we really want to do this integral. So since we already discussed the fact that when we integrate, so once again, we have the covariantly closed form alpha. So if you want dv alpha is equal to zero, is our starting point. So we already discussed that when we compute these integrals, only the homology class matters. So we can deform this alpha. We can take another representative of the same homology class and the integral on a compact manifold is gonna be the same. And so in particular, let's deform it by some equivalent exact piece. So let's see, instead of studying alpha, we study alpha t, where t is a parameter, t is a number, which is alpha wedge e to the t dv beta. So as I said, t is a number, this exponential, the meaning of this exponential is just to expand, you do the Taylor expansion and since these are forms, this is gonna, well, actually this is a poly form, so this is probably gonna give you an infinite number of terms, but okay, this is the definition. And, but we insist that beta is a equivalent poly form, of course, because then, well, because then this is also close, right? If you act with dv on this, here we get dv squared, which is zero only if this condition is satisfied. Okay, so in particular, okay, this should be obvious from here, this is a equivalent exact deformation, but if you are not convinced, we can just compute the derivative with respect to t of this alpha. So the dependence is here, and so what we get, so let me suppress this wedge in the following. Okay, I'm not writing wedge, but it is implicit. So we bring down this factor here, but since everything is closed, this is dv of something. Okay, so this deformation by t is exact, and so integrals are not affected. Okay, so in particular, instead of integrating alpha, we can integrate alpha t, we are gonna get the same result, and we can choose any t, this is always gonna give us the same number, the same result. And so in particular, what will be useful to do, so, okay, let me say the expertise. So if we take t equal to zero, we have the original integral, but we can take any other t, and in particular consider the limit in which t goes to infinity, plus or minus infinity, now we will see, is useful. Okay, so we will evaluate the integral in this limit, because anyway, it does not depend on it. Okay, so what do we do? Well, in fact, we can choose a particular beta, but there is a natural choice, which is the eta that we define above. So we can make any choice, but it's particularly useful to choose as our beta, the very eta. And so the statement now is that the integral of alpha that we want to compute is equal to the limit, as t goes to plus infinity, integral over m, alpha e to the t, and so now here we have dv eta, now dv is made of two pieces, there is the differential, and there is the contraction, so the differential give us this, and the contraction gives us v squared. Okay, because now we are contracting v with the dual of v, so. So this is the statement. And now here we notice something interesting. So first of all, this is still a form. So in particular we can expand, so this is now a standard, so this is just a one form, so this is just a two form, it's not a poly form, so we can expand this definition, this time we truly find a finite number of terms in this expansion, and in particular we get a polynomial. Okay, so this is a polynomial in t of degree l. But on the other hand, so this v squared is just a function on the manifold, and so if we want a zero form, so this is a true exponential, okay? And so in particular, when you take t to plus infinity, this gives us an exponential suppression of the integrand, and this exponential suppression cannot be compensated by the fact that maybe this piece is becoming large because this is just a polynomial, okay? So this wins, so this is a true exponential, and in particular for any point where this v squared is not zero, you get in the limit an infinite exponential suppression, and so these points do not contribute to the integral, only the points where this is zero and the neighborhood of them where this is infinitary is minus small can contribute to this integral. And so once again, we have found this localization argument that only the fixed points of v where this is zero can contribute to this integral, okay? Okay, so once again, we find the result that integral localizes to fixed points of v. If m is the infinite dimensional, as it might want to use for quantum fuzer applications, then this argument doesn't apply. Well, let's see. I mean, of course you have infinities, so I mean, it's always more difficult to deal with infinities, but the same argument applies because at least up to infinities because you will still have an exponential suppression performed by some bosonic term, and this will still be, essentially this will be fermions, so this will still be some polynomial. So this in the quantum fuzer case, let's see, this will play, this piece will be given by fermions. So in particular, you might have some number of fermions zero modes, but it's a finite number of fermions zero modes. But of course, I mean, in quantum fuzer there are infinities, I mean, the very beginning of your, I mean, you start with the path integral, yes, you can go to Euclidean, still you have to deal with an infinite dimensional integral. So, of course, one has to be careful at least. Okay. So, okay, so we find, again, are there other questions? Okay, so we rediscover these results, but in fact we can go on from this expression and we can use it to actually evaluate the integral. And so, let me assume for simplicity once again, I just want to start the simplest example and we will discuss the more general case in the context of quantum field theory, but let's assume that v has only isolated fixed points. Okay, of course, this is not the most general case, because I say in general v can have isolated, well, so the set of fixed points can be, can have some dimension, I mean, doesn't have to be zero dimensional. If you want the most trivial example in which, is in which the U1 action does not act at all, is a trivial action, and then the vector field is zero, I mean, formally everything that I said go through, but then the fixed points are the all manifold, okay? I mean, still what I said is correct, but of course there are intermediate situations. Okay, so let me, for simplicity consider this case here. So now, since the integral localizes to the fixed points, we can zoom on the fixed points and we can perform, essentially, we can consider the expression only in a neighborhood of the fixed points. So, and so if you zoom on a fixed point, well, the metric is essentially flat. So all points are smooth. So a neighborhood of the point is essentially R2L. So let's zoom at p, some fixed point, and then the metric up to corrections will be the flat metric, and okay, we are in even dimension. So we separate into copies of, if you want copies of the complex plane of R2, and in each of these R2 we take radial coordinates. So let me consider, let me write the coordinates in this way. Okay, so each of these is a copy of R2 in radial coordinates, L, if you wish. And then, and so the vector field takes, so and we choose this coordinate in such a way that the vector field takes a simplified form. So, since we are assuming that these are only fixed points, essentially this V, what it does is a rotation in each of these planes. And so this will be the eigenvalues of the rotation. Okay, so in particular, so I've chosen this coordinate in such a way that vector field takes this simplified form. And then we can also write what the dual form is. This is also simplified. And finally we can compute the equivalent differential of this. And so now that we have all the ingredients, we take those ingredients and we plug them into the integral. Again, take into account that we are taking this limit in t goes to infinity. And so, and so what we get, let's see. So we get an integral only on the neighborhood of p. So this is the object. This is precisely the same object that it was decomposed here using dV. And so if we plug in the value space is what we get. So let me write it and then comment. So what I've done here, so I've decomposed these into the two pieces. And we said before, the first piece where we have d eta, if we expand the exponential, this is a polynomial in t, so that I'm only taking the leading contribution, the one which has the largest degree in t, which is degree l. And so these are the terms. Essentially from this piece here. Why here I've done nothing. However, now notice that so this leading piece in t, in fact give you the maximal degree that you can have on the manifold. And so out of this alpha, the only term that contributes is alpha zero. Is the bottom component, right? Because here you have all the other terms. And so you have alpha zero. So this alpha zero is integrated over the neighborhood, but there is this Gaussian factor. So we are still taking the limit in which t goes to infinity. So this Gaussian factor is strongly peaked around the origin. And so essentially around the origin in the limit we can take that this alpha is constant, because this is a smooth form and around the point where it is peaked. This is only the value at zero meters. So we can take this out. And then, once we take these, well these are just numbers, alpha has been taken out and this is just a Gaussian integral, so we can do it on each copy of r2. And what we get is this. And so what we see nicely is that, in fact, powers of t cancel out. So this was, so we saw at the beginning, that the integral does not depend on t. So we better find something that does not depend on t. Of course we are in the limit that t is large. So here there are corrections, but this correction go to zero. But we want to see that at least the leading term does not depend on t. And in fact, powers of t precisely cancel. Also notice that if we were going to keep some of the subleading terms in the polynomial, those would have had less powers of t here. And so those contribution are indeed suppressed as you take t to be large. And so, okay, so this simplifies. So this is alpha zero of p and then to pi to the l divided by the product from one to l omega pi. So what we have completed is the contribution of this integral from a neighborhood of one of the fixed points. And in fact, it's very simple. Now, so we have this product of these factors here, but in fact, these are the eigen values of the u1 action on the tangent space around the point, because around the point this u1 action is a rotation, these are the eigen values. And so we can give a more geometric or invariant definition of this object here as, well, this is a product of the eigen values, so this is essentially the determinant, well, not precise, this is the pfafjan, because only contains l terms instead of 2l. And so this is the pfafjan of the rotation, well, the u1 action on the tangent space at p, which is a rotation. So this is your u1 action on tpm. And of course there are, I mean, on the manifold in general you have many fixed points. And so, collecting the various results, we get a final nice formula, which is, in fact, the content of this Atia-Bott Berlin-Vern localization formula, which is the integral of alpha under our assumptions, is given by a sum over the fixed points of v. And what you collect are these local contributions. So the value of the bottom component of alpha is weighted by the pfafjan of the u1 action at p. OK, so as you see, this is quite a nice simplification of our original integral. Any question? So at some point we kind of thought, OK, this integral over alpha depends only on the top component right over and on tpm. And now this bottom component pops up, so to me it seems a bit mysterious. No, this is a good observation, because, well, it's not. Yeah, so we wanted to integrate, so the integral of alpha is really the integral of the top component. This is the integral we want to compute, and this is really the integral we are computing. And it is true that this integral localizes to the fixed point, however, what appears here is the bottom component. Of course, as we said, these two objects, so if these two objects were unrelated, this formula could not be correct. But in fact, these two objects are related because one of the requirements for these two works is that alpha is equivalently closed. So this has to be closed. And as I said, at the beginning, this condition mixes up the various components of alpha, because if you look at the top component of this equation, this is telling you essentially that minus IV alpha top is equal to d of alpha. So maybe this is not minus. This is equal to d of alpha 2l minus 2. But then there is another equation that is telling you that ev alpha 2l minus 2 is equal to d alpha 2l minus 4, and so on. And so by this, so this equation is a constraint on the various components. And in the end of the day, it relates the bottom component to the top component. It has to be so. Of course, if you just start from a standard form and you're just interested in integrating a form and you want to apply this form, there is some work that you have to do. You have to, first of all, make into an equivariant polyform, which is closed, and then you can use the theorem. But still, this formula is just. And the question is probably related to this. So this procedure here is a very formal way of separating polar and angular components. Like this 2 pi to the l is just integration over angular components. And then you still have an integral over r. And somehow, presumably, this is going to be this alpha 0. And probably is related to the original integral. So let's say we can. Oh, you see it from here, right? So here, what we did was, essentially, the point was that in the form that we obtain over there, the crucial point was that there is this exponential factor. And if you imagine the manifold, so let's say that this is our manifold, this is exponential factor in the limit in which t becomes very large, is picked. But I don't think this is related to what I'm asking. Because the pedestrian way of doing this you will find a coordinate system in which your integrant is only dependent on r. And therefore, angular integration will just give you, this is simple, this is trivial, because nothing depends on r. But that depends on finding that particular coordinate system. And then you just have r integrals, which are somehow presumably related to this alpha 0 at p. Because these are still integrals that you have to do by hand. I mean, there's no way to simplify it. You still have an L dimensional integral, which you have to do, which you cannot. Well, as we saw in the beginning, sorry, the beginning in the first argument, which is here, essentially in the bulk of the manifold, the form is exact. So it reduces to a boundary term. So these integrals are that you're doing, you will find that what you integrate is just a total derivative. So you don't really have to do the integral, you just have to evaluate this object that the integral is total derivative of at the extreme, or the points, which are the boundary of this space. Now, in my assumption, which is that really I'm removing just points, you see the boundary of these are small spheres around these points. But because of smoothness, on these spheres, everything is constant, the form alpha is constant. Because when this sphere is very, very small, since I'm assuming that everything is smooth here, then alpha is constant. And this is why I can, instead, you see, this alpha becomes just alpha zero. I mean, I can take this just constant, and this is probably a realization of what you're saying. OK. And why can we assume that such coordinates can be found? Yeah, so this is just local around the point. So you have a point, and the U1 action does not, I mean, it is only a fixed point. And so this U1 action is, if you want, you diagonalize a matrix. This is U1. This is U1. I think so, probably you don't need to use the metric. I mean, the metric simplifies your life because you can construct this eta and so on. But, well, in fact, what you need is to take any, essentially, here what I need is to construct this eta. And so I need just to construct a metric, if you wish, which is invariant under the U1 action, which is probably can be done, always, I'm not sure. Yeah, so this was a convenience because, yeah, I wanted to show things in an explicit way. Any other question? So the fat union is always positive, that vector? Fat union is always positive? No. OK, never mind. I was going to ask whether the only non-trivial signs that come up in alpha zero. No, no, you do have signs, right? This is, I mean, you have to multiply the eigenvalues of this rote. I mean, you have a rotation matrix. This rotation matrix is just a rotation in each plane, and you can have positive or negative eigenvalues. So I cannot make out my point a little bit more crisp. So the way you presented it, it looks like it was completely done away with all integrals. But if somebody gives you a two L-dimensional real integral, then you just have alpha to L. And then you have to relate, you have to find all these alphas, in particular alpha zero. But to do that, you have to do integrals, because you relate the lower rank alpha to the higher one by doing an integral. No, you have to, yeah, you have to solve this. You'll not really have to integrate over the whole manifold, you just, but you have to find to solve these integrals. So there's still an L-dimensional integral, basically, that you have to do if somebody gave you. You have to solve these differential equations. Yeah, wow, this is how it is. OK, I just wanted to make a point that there is still some integrals you need to do, right? You don't have to integrate over the whole manifold, you just have to solve these equations, but yes. But instead of a two L-dimensional integral, you have to do L integrals. Yeah, there is still some work to do. But this simplifies a lot, the task. If you sit down and try to do one example. So in fact, OK, again, if you want to familiarize with these, I can just suggest one simple example that, I don't know, maybe it might be, might look too easy for some of you, but maybe for some other ones, it's interesting, I don't know. So as a simple example, you could try to, so suppose that you want to compute integral on S2 e to the i, some constant C cos theta v all S2. So suppose you have some around S2, you want to compute this integral. Of course, this is elementary integral, you can just do the integral. But if you wish to clarify your ideas, you can try to solve it with the equivalent localization. So you need to, so this will be the top component, then you need to construct what, the equivalently closed polyform, such that this is the top component, and then you can apply the localization theorem, OK? It's a simple example. OK, so I'm going to change topic, so if you have more questions, isn't there time? Just in the derivation of the localization formula before, first you choose very specific coordinates, locally, near the point, and then when you integrate it for each pair, you integrate it on the entire real plane. Well, I can do that because I have this dumping factor, so let's see what is it. So I have this Gaussian factor, so this is going to be picked at the point. So everything else is not going to contribute, and so in the limit, this is correct. For that finite t, you might say this is an approximation because, as you said, I mean this should be on the neighborhood of the point, and here there are all these, so there are corrections everywhere. The crucial point is that since this formula is, so we can take any value of t, and for any value of t we're going to get the same result, we can take t as large as we want, and so in particular we can make the corrections as small as we want, and because of that this is the final result, so this is not an approximation. It's the exact answer. Also the question about the local coordinate, that is truth, so you set up your coordinate system like that. V is the entire neighborhood, has component omega pi, omega pi is the constant in the entire neighborhood. Yes. Or just only at the point. Well, I mean this is, so if I go close to the point, this V is just a rotation around the point. So, I mean this is, this is not equal to that, there are corrections, but the leading term up to corrections is just a rotation around the point, and so this is just the eigen value of the rotation. So that's how you can simplify the equation there. Yes. In that p of this omega pi. Yeah, this is just numbers. Yeah, this notation, so this is the eigen value of the an action at p in, okay, this is the eigen value. Yeah, but if you say it's kind of thing, then the entire neighborhood is... Yes, that's true. I mean, a neighborhood is infinitesimal, right? Yeah. Yeah. Okay. Okay, so I guess I have 11 minutes. Yeah. Okay, so we'll introduce, yeah, the next topic. So now that we understood, so as I said, here we saw the basic ideas of localization, but really what we want to do is to apply this to quantum field theories. So let me say just in some introductory words about Euclidean alpha integrals, that results. So I already Guido said a few words about what is the general strategy or the general goal that one wants to achieve. And, okay, in this school probably you will hear the same things in various lectures, because of course it's a very focused set of lectures. And so one of the ideas, at least one of the ideas is the following, that so if you're given a quantum field theory in principle, at least formally, of the information about this quantum field theory can be encoded in the Euclidean party integral. So if you can compute these objects, so these are integrals over an infinite dimensional space of field configurations. And these field configurations are weighted by the classical action, which is a functional over these field configurations. These, of course, include both bosons and fermions, weighted by h bar. So in principle, all the information is contained in these objects up to the fact that, okay, we should introduce sources if we want to compute expectation value of operators. And then, of course, we have to do some weak rotation to Lorentz and signature. But it is formally everything is containing here. But unfortunately, these objects is too hard to compute. This is an integral of an infinite dimensional space. We don't know how to do that in general. Of course, we know the standard approximation scheme or the standard paradigm that we can try to apply. We expand this as a perturbative expansion and then we try to compute perturbative corrections. And, of course, this works very well if you are at weak coupling, but it does not work if you are, if the couplings, if you wish, have order one, what we call strong coupling. And, of course, this is for a couple of reasons. First of all, because there is an infinite number of corrections, so, first of all, you should compute all these corrections. But in general, even if you are able to compute all these corrections and resum them, at least in general, you are still missing the non-perturbative contributions. In general, you have some asymptotic series. And so, at least in general, the non-perturbative contribution has to be computed. So, what do we do? So, one possible approach to address this problem is to study some special theories for which at least some of these pat integrals can actually be computed. This is not the only strategy, but this is one possible strategy. However, before this localization revolution, we can call it that way, that started with nekrosov and then with pastum, it was thought that only specific, very special and maybe peculiar or exotic theories were amenable to such a treatment, particular some homological theories or some topological theories. But in fact, this big development started from the understanding that in fact is not true. So, there is a very large class of quantum field theories and of observables that can be exactly computed and addressed with various techniques and in particular supersymmetrical localization. Now, okay, this might be obvious, but let me just stress from the very beginning that even though we will be able to compute some pat integrals with localization, we will not be able to compute every pat integrals. So, if I take a supersymmetric theory, I will not be able to compute any possible pat integrals in that theory. So, that will mean that I have solved the theory, I compute any observable. This will be awesome, but it's not the case. Because quantum field theory is still very complicated. We will be able to compute some pat integrals, but as we will see, still this sum will be a pretty large class of observables. In fact, we don't have a complete classification. So, it's still an open set, if you want. And these observables that we can compute, even though they are not all, they contain a lot of very interesting physical information. But, of course, I mean, localization by no means is the only non perturbative tool that we have at our disposal. For other classes of theories, we have other tools, for instance, for conformal theories, we have the conformal bootstrap, for integrable theories, we have integrability, and many other methods. So, of course, this is just one strategy, but it's very interesting. Okay, so, if you want the objective of these lectures, we'll be to compute euclidean pat integrals of this form. But, more specifically, and this is a point that already Guido stressed or mentioned, we will be interested in euclidean pat integrals. And, moreover, we will put the theories on compact manifolds. So, we'll choose some... So, the spacetime will be some compact euclidean manifold M. And what we want to compute is the pat integral of these theories. Let me set h bar to one. So, this will be... So, the action is a functional of the field configurations, but it's also a function of various parameters. And we call this C, which is not to create confusion with what t was before. So, this will be a function of various parameters, and so, the partition function will not just be a number, it will be a function of these parameters. So, what are these parameters? Well, they can be couplings in the theory, of course, but they could also be parameters of the manifold, where we put the theory. And, moreover, they could also be parameters that control the background, the supersymmetric background that we use on these manifolds. So, this is precisely the thing that Guido will describe, how to preserve supersymmetry on chord manifolds, and he will show you that, in fact, there are parameters. When you do that, there is not just one way to do that, there are parameters, and so, in general, these parameters become parameters of the partition function. And so, our objective will be to compute these objects, and we will discuss how to compute these objects with the localization techniques. And so, well, in fact, you might ask, why should we do that? I mean, we are interested in the Lorentzian theory on flat space. I mean, yeah, it's true that space time is curved, but, I mean, if you are in a lab, it's essentially flat, up to gravitational pulling. So, why should we care about studying supersymmetric theory on Euclidean and compact, very weird manifold? What's the point with that? Besides the fact that it's theoretically interesting if we can do that. And in fact, there are, if you want more, phenomenological reasons to do that. So, it turns out that it is a profitable exercise to study supersymmetric quantum field theories on compact, they don't have to be compact, but let me say compact on, well, the point is that on generic, at least as much as supersymmetry allows us to do, are compact manifolds and backgrounds. And I just want to mention two reasons why this is a good exercise. So, the first reason is that, so, we will see that the localization will allows us to reduce this complicated problem to a simpler problem. But in fact, how much simpler this problem becomes depends on which particular manifold we are and which particular backgrounds. So, on some manifolds and backgrounds, this becomes much, much simpler and we can actually solve this in a very explicit way. In some other cases, it becomes simpler, but it might still be some non-trivial problem that involves non-trivial mathematical problem. So, if you wish, some manifolds, m and backgrounds, are easier than others in the sense of computing partition functions. The second reason which is more important is that in fact, different manifolds and different backgrounds grant us access to different sectors of observables in the theory. So, different m manifolds and backgrounds give access to different observables. And in particular, this set of observables that we can reach by vaning the manifold and the backgrounds can be quite rich. So, we can have access to correlation functions of chiral operators, and this has been known for a long time, but in fact, we can have much more. For instance, we can have correlation function of olomorphic with anti-olomorphic operators. We can have correlation function of conserved currents, which are not olomorphic. We can have access to various counting problems of counting states or counting operators, various classes of operators. So, we really have access to much larger sets of observables than was thought in the past. Okay, and I think that my time is over, so I will stop here.