 Then the literature is really vast, and I'm going to just mention a few things that I found useful for better understanding. So there's Richard Sabo has a book. The book is called Equivariant Co-Homology and Localization of Path Integrals. And this is from 2000, Springer, okay? So one of the inventors himself has a lecture that is called an introduction to Equivariant Co-Homology, and this is actually some LeSouche lecture from 1999. And it's quite nice because, of course, it's addressed to physicists. So it is very rigorous, but it has everything in a language that we like, in a language of differential forms. And the last note that you can look online is Mavé Libin, and this is 07-09-3615. This is really written for a math audience, but then it has, as a mathematician, it has the oracle, it has the full truth, and it's also fun to read, okay? So this is what I'm going to be talking about in the next hour. So first, let me tell you the class of problems that we want to discuss, okay? So I guess I did not say anything about the beauties of supersymmetric localization. You're already here that proves that you're interested. So I don't have to motivate it a lot. But what I'm going to discuss is a simpler setup where things can be done very, very rigorously, find a dimensional setup, okay? So this is, let me now spend a few minutes talking about the setup. So I'm interested. So let me take some manifold M, maybe with a metric. So I'm going to, of course, at the end, I will tell you how to relax a number of conditions, but I'm going to start assuming everything that I need. So I have some manifold M. I want to call it 2N dimensional, okay? So I'm going to also put some local coordinates x, and I'm going to consider some function f. And the kind of integrals that we're interested, I'm going to write. So they depend on some parameter t. And they are going to be integrals over this manifold of dx squared root of g. And then I have something like this, oops, tf of x, okay? So this is the kind of integrals that we all know and love. You think that this is some simpler version, some finer dimensional version of a path integral. So what we usually understand, even at the level of quantum mechanics, is the following statement. In the t goes to infinity limit, okay? So the phase changes very rapidly, so you have some sort of interference. Okay, I'm just reminding you a thing from quantum mechanics. And in that case, what you're really concerned about, or the main contribution, come from the stationary points of this function f. So in fact, one of the main formulas that we like to write, I'm going to write approximate so that I don't have to worry about 2 pi's, but I will do the 2 pi's later. But right now I just want to make sure that we roughly assume or realize that in this case, so xk are the fixed points or points such that this differential is 0. And essentially what we are going to have is some sum, and here I'm going to write something like this, very, very determinant of Haitian. So I'm going to write it like this. So that's the kind of formula that we have that I'm interested. So this type of integral is going to be my object. Of course, in the context of quantum mechanics, this t is related to 1 over h bar, and this f is related to the action. So now we know what we want to understand. I want to understand these kind of integrals. I want to understand exactly when such an approximation, so this actually will have correction t to the, let me write the corrections, minus l minus 1, at least again the dimension of the manifold. So I'm interested, my question, now I can formulate it a little bit more precise. My question is, so here's my main question, is this when are such integrals exact? And I will, of course, I have to define what exact mean, but this is my question. So I want to understand the mathematical machinery that will guarantee for me that when I compute something using the fixed points, that would be it. That is my full answer, and there are no other corrections. That's the context. So now let me do this one example in this, that if I had an extra blackboard, which I done, I would just write it and leave it there, because this is the example that would be recurrent in this whole lecture. So let me, let me give you, again, I'm still in the formulation of the problem. I want to very clearly state for you what is it that I want to solve. So my example is the following. So we want to consider the round S2 with metric, canonical metric, dT squared plus sine squared of Qeta d phi squared. And I'm going to consider my function F. In principle, it can be a function of all the coordinates, but I'm going to take it to be cosine of theta. And you will see that this is a miraculous choice. And then I will, at the end of the lecture, you will see why this is a particular important choice. So in that case, my integral, I can write it here. My integral can be written 0 to pi d phi, 0 to pi d theta. The square root of the determinant of the metric is sine of theta. And then I have it cosine of theta. Okay, because I don't have a very big backboard, this design, I'm going to write as d cosine minus sine here. And now you see that this is a very simple integral, dx e to the something x. And I evaluate this and my answer is going to be 2 pi, 2 pi comes from here, i divided by t. And then I have minus e to the i t plus e to the minus i t. So this is the answer. And of course, 4 pi sine of t divided by t. So this example, of course, is the canonical example that I want to develop. But because I'm still in the business of defining my question very carefully, I'm going to restate the question again. So let me make a couple of remarks. So the integral is a sum of terms that correspond to the stationary points of this function, derivative of cosine is sine, so 0 and pi. And if the sum of two terms, and these two terms come from fixed points of the circle action. So I haven't said it, let me say, the circle action, of course, I'm talking about is 2. Let me draw it like here. And I'm considering the action which is phi goes to phi plus unconstant. And of course, it has two fixed points. These are the points that essentially give me these two terms. Okay? So in this sense, you can say, this is a very baby, it's a baby case of the baby case of finite dimensional. But in this case, you can see that the integral localizes and it gets contribution only from the stationary points of the action, which in this case is just translation by phi. So that is more general case. So that's essentially what allows me to... Sorry? Yes, in this case, it's a hessie. I evaluate it, I mean, there's two points, minus one and one, and I evaluate it at the two points. Thanks. Okay, so what is the problem with this kind of integral? So let me now say again, the question again is computing or if you wish understanding integrals over M with some symmetry, some symmetry group G. So that's really what I'm interested, right? So that's the kind of problem that I have. Now, the naive or the first impulse that you might have is the one thing that comes from physicists and from gaze transformation, et cetera. So you think, well, one possibility is, of course, to identify all points that are connected by a transformation. That would be a natural thing for you to do to say, well, I'm interested in manifolds with some symmetry. Let me question by the symmetry. And then that's what you do, let's say, in electrodynamics, et cetera. But you see the problem with that question. So this is S2, if I divide by S1, the space that I get, so now I wake up the students, what is the space that I get when I do that question? An interval, right? So I get something like this, okay? I'm from your kid that got the cohomology, you know that the cohomology of something that is contractible in this case, is empty. So cohomology is the precise way to talk about integrals. So all the integrals that I would like to compute if I assume that what I need to do is just to question the space are going to be trivial. I wouldn't have anything interesting like what I have here. So it tells me that the naive, the most natural thing to do, is going to lead us to a problem and, of course, the problem comes from fixed points. If I didn't have any fixed points here, then, of course, this will work. If I didn't have any fixed points, my construction doesn't work. My construction essentially doesn't give me what I'm interested in, okay? So now that set up, set the stage to define equivariant cohomology. That's what I want to talk about now. So I have to now tell you what is the construction that will take care of integrals over spaces with the action of a group, but that doesn't go this route because this route is dead-end for us. Okay. So again, in my same category, so two-dimensional manifolds, maybe with even with a metric, and now let's consider this group G to be a billion. Let's do the simpler case. So in this case, of course, what I mean is that I have some vector. I'm going to write it like this in some appropriate coordinates. This vector has to be a Keeling vector of the metric, and this is essentially what you saw here. This is again, this example is really very, very nice. So again, the Keeling derivative of the metric, which is d mu v mu plus d mu. This is zero. So this is the derivative, so now I will introduce a little bit of notation. Again, my goal now is here. I want to define equivariant cohomology, so I need, as I said, I need to introduce a number of objects. So the definition, so let me define, this is the notation of cremonacy, but sometimes people write like this, so omega of n. This is going to be the space, the definition of the space of forms, poly forms indeed, so from n zero to 12, or forms of the differential forms of degree n, so basically this is space. Alpha n is again lambda n of n. So what I do is instead of considering some particular form, I'm going to some, this is a formal sum of all forms of degree from zero to 12. It's not very easy, because maybe I can project it like this. Okay, so, okay, so first definition is the space of poly forms, is some sum of forms from zero, differential form of degrees zero to 12. The other definition is the v-equivariant differential. Okay, so the d-equivariant differential, I'm going to write it like this, and it's going to be the standard stereo differential, but then I, this is a contraction. If it is, okay, let me now remind you what this is. So these 10, it takes an m-form and it sends it to m plus one. IV, the contraction, it sends an m-form to an m minus one. I guess that's more or less what you need to do. If you forgot, let me remind you, let me remind you if I have some form alpha, which is one over k factorial, alpha mu one, mu k, dx one, which dx k, then IV of alpha is going to be essentially one over k minus one factorial, v mu one, then I have alpha mu one, you see this contraction, then mu two, mu k, dx mu two, which dx mu k. So now, I'm very close, I have my differential, I now need to show you, remember this part is about defining the equivariant cohomology, I define the equivariant differential, you more or less know where I'm going, so equivariant differential, if I have a, with one condition of new potency, I can show you, I can now define cohomology, so let's do that. It's clear? Any questions so far? Okay, so, so dv squared is essentially, I wrote it, this squared is going to be this squared minus d IV minus IV d plus IV squared, okay? So this is zero, sort of from kindergarten, this might not be obvious, but it acts on antisymmetric tensor, right? So if I, this is going to be symmetric, if I do mu one, mu two, I'm acting on antisymmetric, so it gives me zero as well, so this is also zero. And this, if you remember your differential geometry, this is, I'll let somebody from, of the students to tell me what is the answer here? Leader derivative, right? So minus leader derivative. And that's essentially, now we're in business because it tells me on, on the space of v-equivarian forms or polyforms, lambda vm, so this is the space of polyforms, I keep mixing my notation, sorry, such that the leader derivative acting on them is zero. On that space, I do have the importance, so I'm ready to define my cohomology, I'm going to write it quickly right here. So the n-equivarian cohomology of my space m is going to be kernel of my operator, my equivarian stereo differential, this let me remind you that this acts, I can regulate this on m, modulo, oops, the image of that operator, which acts of course on v m-1. That's my definition. Okay, so now we introduce a little bit of machinery, but again our ultimate goal was integrals, right? So I have to get back to integrals, but of course integrals, I'm going to relate integrals to cohomology classes very quickly. So that's, that's what I want to do. Next, so let's do that. So remark. So if a polyform alpha of highest degree, highest differential degree n is equivariately closed, then alpha n, and again it's, remember I did my decomposition, sorry that I erased it, but n is sort of the top degree form, then alpha n and alpha m-1 are closed forms. Okay, so let's check that. That is immediate. And again, many of the statements here are proven very easily. So this is my statement. Some form is equivariately closed of highest differential degree, sorry, sorry, sorry. So again, remember my polyform is a formal sum of degree n from n to 0 to the top, to l. Okay, so here's my condition, and I want to now prove that these two, let's say top n degree and m-1 degree are also closed. So the proof again is immediate because all I need to do is to write this on the note. So remember the definition of the equivariant differential was d-iv. So what I need to do is to say, okay, I will have d alpha n plus d alpha m-1 plus d alpha blah blah blah et cetera, right? And then I have minus i-v at n alpha n minus i-v at n alpha m-1 et cetera. Yes, I wrote it. Now, I'm saying that this is 0. And now, look at the degree of this form. This form has degree m plus 1, right? There is nothing else in this sum that can have that degree because this will have degree n and this one will have degree a minus 1. So this has to be 0 by itself. So that's my statement. d alpha n is close. Okay? The top degree form of an equivariantly close form is close. And similarly, you can chase that basically this is degree n. The next one would be a minus 1. This is a minus 1, so this also has to be close. There's no other form of degree n that can contribute to this being 0. So this is also 0. That's my statement. Okay, so and similarly, but I will not even write it. Similarly, I can show that if some polyphone beta of highest degree n is equivariantly exact, then beta n and beta m minus 1 have to be equivariantly, it has to be exact in the normal sense. Okay? So now, bear with me for a minute because again, our goal was to go to integrals. And I have diverged a little bit, but now we're very close because why? Because I have now this statement is very similar to what you do in ordinary cohomology. So let's say ordinary, I mean the RAM. So let's see why this statement is powerful for us. Because one more definition. The integral over the manifold m of a polyphone alpha is by definition equals to the integral of everything else is 0. That's the definition. Okay? And you see now I'm talking now I'm connecting. So what happens now is that I can do this. So I can integrate some form. And that in email deny the RAM cohomology this is defined up to a cohomology class. Right? So I'm going to let me do it like this so I can add an equivariantly exact piece and I might so my my statement is that cohomology classes right? So this integral is independent of the cohomology class that I, the representative that I use. And then this is very easy because my second statement is the integral of an exact form means that the top form is exact right? So maybe I'll write one formula. So the contribution from here would be something like this. We will have the top degree. But now I use the stocks and this tells me that if my space is compact, this is crucial. Question? Okay. I take it a definition because you can try to define intersection. You can try to define more complicated. Yeah, yeah. But you have to say what the integral means for the equivariant form. Okay, so here I use the stocks and now if my boundary was empty, this would be zero. Okay? So essentially, and now we are closer to what we want. So now we know that I can pick any representative in the same cohomology class, equivariant cohomology class, and the integral will not change. So now you should be, you should now realize that we are closer to one of the main elements of localization. And I guess that's what I'm going to talk about now. So now I want to my next ten minutes is to prove that equivariant integrals localize. Let me have that slogan. So let's see that. So again, remember your original example. So in that original example, the answer came from the fixed line of the U1 action. Okay, so my statement more generally is that the which I want to prove right now is that the integrals localize to the following locals. So the set of points that belong to M on which I have to use a little bit of notation, sorry, the norm I'm still talking about a billion, right? So I introduce a vector on points on which the norm of the vector is zero. So that's the main statement that I want to prove. I'm pretty close. So of course one possibility, so let me do, I'm going to now do two ways. One way is of course sleek, but not very obvious or not very... So one way is of course easy to show, but this is not what we do in localization. So I also want to emphasize the connection with supersymmetric localization. So one way is of course to construct a form no T here, no I here, sorry. So if I construct this form okay, my claim is that this form is in the same cohomology class as alpha and this you can see more or less by saying this formal exponential is T dv of beta plus dot dot dot which is alpha plus alpha let me have this T here dv of beta plus dot dot dot but alpha is closed, so I can put alpha inside and similarly in other terms so these two forms alpha T and alpha differ by some exact form, complicated but exact form. So they are in the same cohomology and therefore the integral is independent of T. So that's argument number one which is again it was very obvious, but I want to propose the argument that it nicely connects with what we do in localization, so let's let me remind you or introduce so the integral is independent of T, so therefore I haven't proven that it localizes yet, but therefore I can take T equals 0, T equals infinity and compute this. This is just that statement for computing the integral, sorry beta is some poly form, some poly form there's no alpha 2, I'm sorry alpha T, so I'm defining a deformation of alpha alpha is a poly form too, yes everything is poly form, sorry yes, yes, yes and this alpha is equivalently closed of course that's what I can put it in here alpha is equivalently closed because it defines a cohomology, right so I need equivalently close okay so this argument again this argument is not my favorite argument the favorite argument of course is to define some z that depends on T which is going to be no T is still some parameter, some real parameter, so now of course the best object for us to define and this is what connects nicely with localization is to let me compute this integral so now you can forget about argument one, now this is your argument I wanted to compute this integral, I know that at T equals zero this is the integral that I'm interested but now I deform it and of course a way to see how it depends on T is by differentiating with respect to T by differentiating with respect to T what I would get is alpha is here, dV of beta is going to be here and I still have T dV of beta okay I can do integration by parts so this is dV of the whole thing and then sign that I would not be very precise so I write there's a sign here that I have to be more careful but then I'll have oops beta is out dV of alpha e to the T dV of beta alpha is close it defines a cohomology class for us so essentially this dV can act on this exponential and what I would get is essentially let me know no worry about signs m, beta, alpha dV square of beta e to the T dV of beta but dV square we already showed that it was the derivative from beta and because if beta is b this is zero so that integral is independent of T that's another, this is the more direct the more the definition align with what we do in localization okay so now because it's independent of T I can compute it for T equals zero which is what I want for T goes to infinity I still have to show that it localizes I haven't I'm just so far I have just shown you that it is independent of T now let me let me kill the argument or finish the argument with showing that it does localize so it leaves only in a very specific neighborhood the one that I claim questions? let me show you again now that it localizes so I have shown you that it's independent of T but I haven't shown you that it leaves completely determined in this neighborhood that's what I need to show you now so let's do that so what I need to do is to construct so let me say differently construct your dV of beta explicitly because I have a vector so I have a manifold with metric I said I wouldn't assume everything that I want then at the end I would tell you how things are relaxed so because I have a manifold with metric I also have a vector so of course I can use the metric to construct the dual form natural dual form so my dual form I'm going to construct it as gV a little bit formal but of course what this means in coordinates is that I do this okay? this is I will show it to you now not yet I want to show you now that not only localizes but it localizes precisely on the logos of course so I have a natural one form this is the one form that I'm going to use so IV of eta let me wake up again the students what is IV of beta of eta sorry v squared the norm exactly right? you can see here the contraction substitutes that by b so I get v what did I write here v squared or the norm of v so now we go back to our expression to our z of t if you wish and we had so I want to compute this I can compute it for whatever t I like so let me take t goes to infinity and now my integral over the whole manifold of alpha e to the t d eta d eta is not zero but then I have let me write it ostensibly like this maybe without the word okay? so now when t goes to infinity the only part this is kind of like a delta function right? so it's a Gaussian so the only contribution can come from points where v is non zero sorry when v is zero so this proves that a billion equivalent integrals localize on the logos where the vector has zero points or zero norm which is precisely again in the case I don't have time to do explicitly to go back to do explicitly the sample but in that example you see that the norm is sin square of theta when zero and pi is essentially when the norm is zero so now we understand why that example had to work okay? yes z of t is independent of t let me compute it when t goes to infinity so beta is eta of course because I took a specific eta in this case right? so let me let me sorry let me maybe one formula there so d eta minus i v of eta but i v of eta the smartest students in the back told me it was exactly the norm so I see that this integral can only get contribution when v is zero v is zero so that essentially finishes this part so equivalent integrals of course there are many more things to do I did the abelian case I can go to non abelian I will tell you at the end the full extent of this but that have shown to you that equivalent integrals localized to the locus where the norm of the u1 action vanishes because I'm using only one vector right if I have I use only one vector here otherwise I have to consider extending this space and then the lead derivative will have anti-commitator I can do it but I'm just one vector in my space and this is my answer of course if I have many vectors what's going to happen is that it will localized to the intersection of all the points where the norm vanishes now the master formula or the big theorem in this in all this story is atia bot and then I have to confess my cultural ignorance so there is berlin and vernier I don't know if I'm pronouncing correctly but essentially what I want to show you now is that theorem so let me write the name first so that I don't have to worry much about it so atia bot so berlin verge I guess so this is okay this is 84 this is a little later this is 82 there's another famous theorem that I will talk about at the end if I have time which is deuster mack heckman and this is also 82 at least it was published in 82 so the story is essentially that this guy is founded in a very specific context localization for some Hamiltonian actions this guy founded in a more general context and of course the super generalization is atia bot then I G and E sorry thank you for the French French speaking colleagues here okay good so now let's try to prove this theorem we almost have all the ingredients so it should be easy as soon as I find another piece of chalk so so again the theorem you can say is going to be a explicit formula for the integral of equivalently of equivalent forms okay so so let's assume just to make seems simpler assume the localization logos of the killing vector V is and V which is essentially a set of points and I'm going to assume that of course the points are isolated I'm going for the easy case the points are isolated isolated points okay so if I have some manifold and I'm interested in one point that is separate from other complicated points of course I go to I use coordinates that are appropriate to that point so I go locally now to that point and close to in a neighborhood of that point of course my metric is going to be flat that's what a manifold means so from I from 1 to L this is 12 dimensional the Xi squared plus I squared and I also going to write it in polar coordinates because it makes some computation more obvious the R I squared plus R squared I again I is like you see I have some um the composition in planes okay so now what is my vector in this in this particular coordinates my vector is going to be again in that neighborhood I have some weights y I d d Xi plus Xi d dy I which I of course can also write from L 1 to I 1 to L of omega p d d phi I okay so you can see that this is more or less there is a side that we did before so the circle action um so the circle action I just split it into planes going to write it like this where this R is going to be what you expect cosine p I sine omega p I minus sine omega p I and cosine omega p I so wait let me let me tell you where I am um this is a fight here sorry so let me let me let me tell you where I am I'm going to I'm making my computation in a very clean uh coordinate system that is particular to that point okay but as we always do in differential geometry you go to some nice coordinates you do the computation and then if you recognize that your answer is tensorially then you're done then this is true for any coordinate and and I have lost nothing right so let me let me that's that's where we are so that's what I have the action of the generator of that vector it would act on each sub space it acts with this with this rotation and I have my vector of course when I have my vector I automatically have my one form which I already use for localization my one form is going to roughly be some from I one to L omega p I and this time I'm going to just write in the vector because lack of time oops no it's just this one form some of this one form and of course now let's see so the only computation that we need to do because we know that we localized with this eta so let me write for you let me write what I have already done so I can be very quick so dv of eta is going to be twice the sum from I from one omega p I dxi which dyi sorry that I keep mixing okay doesn't matter omega p I square xi square plus yi okay so remember this is a contraction contraction means wherever I see this I substitute by um sorry in my form wherever I see this I substitute by the appropriate coefficient so that's what I have so this is my one form and now remember my z shit my z that I wanted to compute is essentially of t was the limit as t went to infinity of alpha e to the t dvn okay so t is a large parameter here I'm computing in that in the limit so here's the only non trivial step here when I take t to infinity so this I have to expand right and I expand formally one plus this plus I have this square etc but you see I'm looking for the term that contributes the most when t goes to infinity so that would be the term with the most t obviously and I can have my my space is two dimensional so I can have at most this is a two form the end of it I said to form so I can have lt so that's the only term that I worry about all the other terms are so leading in t and that's the the that's the gist of this computation that I now you see because I constructed my tool form strictly from this from this field so from alpha what can alpha be what can alpha give me it cannot give me any form because it will be then lower degree now so the only alpha that can contribute has to be I'm going from t goes to infinity of alpha zero the only thing that I can take from alpha is function the function that was there so alpha zero at my point p and the integrals that I have to compute are going to be from one to l of t omega p i integral from zero to Dubai of d phi i and then I have integral from zero to infinity of d r square e to the e to the minus t omega p i square r i square okay that's my integral and this is again this is a simple integral just exponential but the beauty here is that you see that t will cancel and the answer is essentially so I'm going to already write the full answer because we have limited time the integral over this locus of alpha e to the t dv eta is essentially alpha zero on p to pi to the l to pi come from this integral I have it out times to pi to the l divided by product from i l omega p i that's the integral in the t goes to infinity but again it's independent of t so now this is the product that I was talking about I use a very specific representation of the action of my killing vectors but I got the product of the eigenvalues so that of course smells like a determinant in a second but let me make um let me make the two comments so this is still not the I mean this is already maybe I'll let me just write the formula co-variantly let me write the formula co-variantly this is the main result of a co-homology so the integral of alpha was 2 pi to the l sum of the fixed points of alpha 0 evaluated at those fixed points divided by the fafien of the action of the vector field at that point okay so this is the main localization formula this is again due to all the people that I have already mentioned the formula indeed the formula is more general I assume it just to be quick to be able to compute I agree with you I can use I can find the dual without necessarily doing that but this is the main localization formula again it gives you an integral of this poly form exactly at the fixed point of this metric so our construction also shows you that the only points that can contribute the only points so it also tells you precisely that the width of this Gaussian is precisely proportional to D so you can take it take it very small so in the last 5 minutes I will make one comment about this termangetman that came before of this story but this termangetman is a very specific case so in that case you have a 2L implicated manifold V is the vector it's my vector so L is the action of the vector I wrote R but I want to write it infinitesimally so this is the ok so what happens there again there are many many examples in this class that you know and love classical phase space maybe Kalawiyao, Kotanje bundle so there are many examples that fall in this category but the key thing for us is that you have this implicated form this form by definition is close ok and also if you have a Hamiltonian action so some people might recognize this Hamiltonian action also as moment map but essentially what happens is that Hamiltonian action generated but again some vector V so this quantity the contraction of the of the V form with this vector is D of a function in this case this function is the Hamiltonian function or moment map but you see if I rewrite I didn't do it it's sleek enough but I can write this I can write I can write it like this and another way of saying this is that the equi-variant differential of H plus omega is 0 so contraction of a function is 0 and D of H is this and when this act D of so the random differential of omega is 0 and the contraction is this so this condition tells me that I can now localize using this form and this is the formula that Mr. Manhekman wrote so let me ok I wouldn't write the formula because it's essentially the same formula that I wrote before but ok so we have 5 minutes so let me kill the original example with this gun that we now have so the metric I already wrote the vector was D d phi again back to example back to the example so now we can see why we were lucky that we chose cosine and then we would be in trouble so in that case and I also going to write it in the language of ok this is now a question of language but so my Hamiltonian which is also sometimes called in the literature the height function it's kind of a most function as well was cosine of theta I v of omega in that case well omega what is this inflective form on the two sphere is sine of theta d theta d phi which I'm going to write as d phi which d cosine of theta ok so this object is going to be what is it so here it is so whatever I see differential I substitute by v so essentially this is going to be d cosine of theta which is precisely h ok so that ok I have to talk a little bit more so the of the action of that vector in the north pole one of the fixed point is plus one and the of the action of the vector in the south pole is minus one and these are the two fixed point that we wrote in our original example so this is roughly the story of equivariant localization this is the main theorem is here it tells you when integrals yes sorry ok so essentially this is the story as I said it's an appetizer it's an introduction to the introduction to localization it's a very clean example my hope is that it inspires you to see the roots of localization you see here there was no there were no fermions it was a it's all the structure that is behind but the answer is that the exact integral is not approximated this is the exact answer also for those who want to to pursue this story a little further so I'll tell you what I did not say so most of my discussion was of course about the the whole space but what I created that I was going to talk about was equivariant cohomology so I need to tell you about all cohomology classes so so I want to mention that the proper definition that I only talk about the top cohomology class the proper definition of this subject is something that is of course there within algebraic topology so you need to find some contractible space on which the group acts without fixed point all our problem came from fixed points so if I construct a space on which the group without fixed points and then I multiply that by my original space I take the question let me write the formula because so the formula which requires a little bit of algebraic topology is this so I take some generalization so on a space on which the group acts freely I multiply now if I divide this is going to act freely and this defines my cohomology my equivariant cohomology without a problem then of course it's subtle right so that's what I forth reference so he will do this very carefully if you are interested you have to also prove that it's independent of this extension it's subtle but all the ingredients are there you can of course I didn't do it here but in the book of Sabo and also in Matvey Levin you can see more general like top class or chain class all this is defining in this context the non-abillion generalization there that somebody asked me so it's not it's not very complicated so I want to finish again saying that this is sort of like a very very simple skeleton but very rigorous and I invite you to as you listen to the introduction to localizations lecture find the analog the analogs find in that context what is the V what is this integral that you're trying to compute et cetera make that part of your homework so thank you very much