 So the first lecture is Francesco Benini, he will continue to introduce you to localization. Hello, good morning. So yesterday we discussed how to construct supersymmetric theories on chord manifolds. And let me just stress the last point that I said maybe quickly towards the end of the lecture. So our goal yesterday was to construct a Lagrangian for the preserved supersymmetry on a chord manifold, and these supersymmetry variations delta, because this depends on the parameters epsilon. And so how did we succeed to constructing them? So I said this quickly towards the end. So we discussed the supergravity. So we discussed how to find the solutions to the generalized scaling spinor equation. And then once we have done that, what we do, we take the solution for the background fields and for epsilon and plug them in the supergravity Lagrangian, or the action, and supersymmetry variations in supergravity and substituting them, we do find the action and the form algebra for the field theory. I said this towards the end, but let me stress it once again. So this was our goal, and with supergravity we do achieve this goal. Are there any other questions on what we did yesterday? So now that we have understood how to construct actions and supersymmetry algebra on chord spacetimes, now I want to see how to compute pat integrals. So we discussed the localization argument. That will be very similar to what Leo showed us in the first lecture. So our goal will be to compute objects like the following partition function, Euclidean partition function on some compact manifold M, a possible expansion of some parameters, and become them raw. So this is our pat integral, here there are all the fields in the theory, e to the minus s of phi in raw. And here this M is some manifold, some D dimensional manifold, where supersymmetric quantum field theory in dimension D lives. So our theory is supersymmetric. So assume that we have some fermionic symmetry, we call them q, and the action is invariant under them. Now since these symmetries are fermionic, the square is either zero, but more generally can be some bosonic symmetry in the theory. So q squared will be some delta b, some bosonic symmetry. If you want this q is what I call delta before, so let me call it q, I might use interchangeably the two symbols. So what are these bosonic symmetries? These are either some translations or rotations on the manifold, or it can be some internal symmetry, like some r symmetry or some other flavor symmetry in the theory. Now instead of studying directly this integral, which is the one that we are interested in, so in the following let me suppress this parameter rho, but of course it is meant that there are various sorts of parameters. As you said this parameter might either control the manifold or the background, because as we said still the background is ambiguous, so we have choices, whether they might control couplings or sources in here. So it will suppress rho, but still we want to construct one parameter family of partition functions. So let me call this new parameter t, so what we do is we deform the action by something which is q exact. So it is the q variation of some other functional, as I call v, and we put a parameter t in front of it. So t is a number, but v is a functional of the fields, and we choose this v, such that the bosonic variation of this p is zero. In other words, this assures us that this term is actually supersymmetric. If you apply q to this, it is zero. Ok, so we have generalized our pat integral, now it depends on one more parameter, we have one parameter family of pat integrals, so we can ask what is the actual dependence on t, and this is simple, and we just bring down this factor here. However, since both these factors are closed under supersymmetry, we can actually write this in the following way. And so what we see is that, in fact this is a sort of total derivative, so it is a variation, a supersymmetric variation of something. And so in particular, by the standard argument, this means that by a fear definition we can see that this object is zero. More precisely, so let me say this more precisely, so this is a, so if this was a bosonic variation, it would just say, ok, we can write the pat integral, then we can write the same object, but with a fear definition when we act with the bosonic symmetry, and we ask, what is the difference of the two when the variation is infinitesimal, of course the difference is given by the action of the bosonic symmetry, but since it was just a fear definition, the action of the bosonic symmetry is zero. No, otherwise this is kind of foolish. So this is not zero, this is the q variation of something, of some object, this is q exact, this call. Yeah, so if this was a bosonic symmetry, so what we would say, we would say something like, ok, if you compute the integral in the x of some function of x, so let me do a fear definition, this is the same as the integral in some dx prime of xx prime, because it's just a fear definition. However, if I take the infinitesimal version, the difference of the two is that the action of the bosonic symmetry, and so this object will be zero. Now this is fermionic, so maybe we have to be a little bit more careful, but in fact it's exactly the same thing, because a fermionic symmetry acts like a derivative, so this is essentially a derivative, and so this is an integral of a total derivative, and it is zero. So you know that in with Gauss-Mann variables, so the action of the variables acts like a derivative, and so this would be a total derivative in Gauss-Mann variables. Now, of course, we have to be a little bit careful, because the integral of the derivative is not automatically zero, so this would be a total derivative not on the manifold m, it would be a total derivative in field space, because this acts on the fields, so this is a total derivative in the space of fields, and it's not in spacetime. In the space of fields it's not compact, and so the integral of the total derivative is not automatically zero, in general it's unboundary terms. So one might get some boundary terms and one has to be careful, where in general there are boundary terms. However, if these functionals give you a suppression, some exponential suppression for large values of the fields, then you can make sure that there are no boundary terms at infinity in field space, and so this is in fact zero. But there is something to be careful about. This is not always the case. There are various explicit examples that if you ask me after what I can tell, where this is not so, and there are boundary terms. But if these objects are large enough in field space at infinity, then this gives you a exponential suppression and the boundary terms are killed. Oh, yeah, sorry. We also have to assume that there are no anomalies. So we also have to assume that the measure, thank you for this remark, that the measure is invariant, so no anomalies. And in particular, since our supersymmetry azobar contains a bosonic symmetry, we know that bosonic symmetry can be anomalous. So for sure if there is an anomaly here, there is the danger that the measure is not invariant, because if this is not invariant, q cannot be invariant. The measure for q cannot be invariant under q, if it's not invariant under delta b. So under these assumptions that the measure is invariant and there are no boundary terms, this is zero. So this is interesting because so we have deformed our pat integral and we thought that we got a one parameter family of pat integrals, so this is not dependent on t. So we can compute this pat integral for any value of t that we like, we are gonna get always the same result. So in particular we can draw some number of conclusions. So first of all, suppose that in fact this was not some deformation that we are doing, but this was some term in the action. In other words, suppose that in the action there is some term that we can write as q of something else. So there is some q exact term in the action and t would just be the coupling in front of this term in the action. Then the argument just shows us that this pat integral does not depend on those coupling. So the pat integral does not depend on couplings, so z not dependent on couplings in front of q exact terms. So this already simplifies our life because it's telling us that even though here we might have many couplings it doesn't really depend on all of them. It can be simpler. Then of course this argument is still valid if we insert some operator in this pat integral. So we might make some insertion. This operator might be local or non-local. It could be a widths online. It could be something else. So let's look at order operators. So standard functions of the fundamental fits in the theory. And in particular let's consider supersymmetric operators. Then it is clear that the very same argument goes through, so once again that there is respect to t or can be put inside this parenthesis once again because it is annihilated by q. And so we learn that when we insert operators in our pat integral only the q-comology class of this operator matters. Not the representative of this of this comology class. So let me first write it and then say it again. So only dependence on q- comology class of operators. In other word if you compute an expectation value of operator o this is going to be the same as the expectation value of o plus q of something else. So this is some o tilde. So in particular if you want to make a parallel in particular with the Laos lecture so q is like a differential and so in particular we can construct q-comology classes in the same way as with the differential one comology classes so you take an operator and you shift it by something which is q-exact expectation value so the observables that we can actually compute are not going to change. And so only the class matters. In particular if an operator is q-exact all these in session are going to give us zero. Very good. And then the last point which is so important z is not modified by our deformation so not dependent on our deformation qv and we want to use this last fact at our advantage to compute this integral. Ok, so first before going there which let me say once again it will be an infinite dimensional version of what you already saw I am repeating somehow something that you already saw but hopefully this is useful So first of all let me make a remark about the fact that we are going in the Euclidean signature ok, because then everything is better defined and we have better properties of convergence however when we go to the Euclidean signature we have to complexify the fields as we mentioned towards the end of the lecture yesterday in the Euclidean signature it becomes a complex field and if you have a complex field the field and what would be its complex conjugate becomes two independent fields Ok, so in particular we might use a symbol like phi instead of using phi dagger we might use phi tilde to indicate the field that would be the complex conjugate in Lorentzian signature but then becomes an independent field so all the fields get doubled however of course we are not interesting in now doing a pat integral on a double number of variables because there will be an object which is not related to the Lorentzian theory that we started with and wanted to understand of course we want to understand the analytic continuation of the Lorentzian theory and so in particular we have to choose a contour in this space of fields a middle dimensional contour so in particular we should find a contour such that this integral is convergent in particular when we go to infinity in field space in such a way that these are well defined objects and in particular why we should also choose now that we have this deformation we should choose a contour such that this object is convergent for any value of t if you want to play with this t so if you want to choose contour in field space such that z is convergent ok so now let's go back for integral so of course at t equal to zero is the object that we want to compute but now since it does not depend on t let's play with t and in particular suppose since we can choose this b at our wheel so suppose that we chose where we can find some v such that the bosonic part of qv is semi positive definite is non negative along the contour in general we cannot you cannot obtain this in general because this will be some complex object so it's not the convergence and if I choose a different contour then they are sub supersymmetry ah yeah so you want that the contour is invariant on the super it's closed if you want this middle dimensional subspace it's closed on the supersymmetry well you know the convergence is a problem with the bosons it's not a problem with the fermions I mean with the fermions you have you know since if you expand you are going to get a polynomial in the fermions while here you are exponential so you don't have to worry about convergence in the fermionic sector it's a bosonic sector which is trouble ok so suppose that you can find some v that satisfies these properties so this deformation is non negative along the contour then we can take the limit t plus infinity and of course we say from here that any field configuration for which this function is not zero is going to be infinitely suppressed as we take t larger and larger and so only the configurations for which this function is zero are going to survive in this pat integral yes and in particular we can take t as larger as we want because anyway we just show dependence on t so we are not going to change the result doing this operation and so essentially what happens is that we started with our big space of field configurations but in this limit the pat integral localizes on some sub manifold where that we can call phi zero so some special field configuration so to zero ok so what happens here we should be careful to see what happens close to this sub manifold so let's parameterize the fields around this phi zero in terms of some small deformation or oscillation in particular if you want to understand so of course one can always do some theory definition here and rescale the fields in some way but this is traded by an opposite change of normalization in the measure so if you want to keep the measure if you want to order one we should use canonically normalize fields and our term is that when we take t to infinity we have a large deformation here and this dominates over the action so if you want to keep using normalize fields we will see we have to rescale the fields by some power of t and this power of t will turn out to be this so let me put this power now and then we will see that this is the correct power that we have to use but apart from that we are just parameterizing oscillations around this special locus and so when t is large we tell or expand this action around phi zero so what about the first term so we want to evaluate s plus tqv s gets a contribution from phi zero but then all other higher terms are suppressed by powers of t at least by t one over square root of t what about this other term at zero order in the variation of course is zero it was our condition that on these special configurations are zeros moreover, since it is non-negative these are also minima this configuration also minima of the potential so also the first variation is zero so we have jumped directly to the second variation and in the second variation you see that this power of t cancel with this power of t and so here we get the quadratic expansion of qv around here we should probably put so this is the quadratic expansion of qv around phi zero which is a function of phi hat and then once again if the higher orders are suppressed by powers of t and in fact as I promised we should make sure that here the powers of t so we chose this power of t precisely in such a way that the first leading order does not depend on t the leads are not canonical normalized and then it means that there is actually some power of t in the measure so we want to avoid them and then this is why we put this specific power here so is this point clear? yeah, fermionic term doesn't have to be zero yes, in general you will have something in the fermions no, in the sense that if you want in these variations you have all the fermions and in any way the fermions if you expand are polynomials somehow you can always solve the pent integral in the fermions I mean the fermions are not a problem because it's just some polynomial so you always know how to do the grasman integration of the fermions yes, exactly so that is not gonna mess things because the exponential suppression wins so let's say that I am at a configuration in which I have some fermionic contribution if the bosonic part is not zero this is gonna suppress that configuration even though it's a polynomial in the fermions but you have exponential suppression in the bosons so the only configurations that matters are the ones in which the bosonic suppression is not there well in the fermions you can have more than quadratic if you have more than... yeah, ok, that case, yes yeah, sure if you have more than one fermion you can have for ferm interactions in general you can have a polynomial you should just do saddle point evaluation yeah yes this is precise what it is ok, I hope that turns result yeah 5x squared well, this would be a quadratic so this is the quadratic expansion so you take this, so this in general is some maybe complicated object you do the quadratic expansion at around this point and so this would be a quadratic function of the oscillations so this is in fact a quadratic function of phi hat, but ok but in general it would be some maybe complicated function of phi zero but in general we have 1 over that is part of t as well yes that is precisely because you want the chronically normalized ones yes I should ask him well, ok, maybe ok so ok, so essentially here the message is that it is true that we localize to this the manifold phi zero but in fact the first quadratic contribution is still important so we localize but still small quadratic fluctuations around this some manifolds are important they are captured by this term here and this term here is not suppressed when you take t to infinity because of this canonical normalization and but that's ok I mean this is well, ok this is what we have and so in particular since in phi hat is a quadratic function we know how to do exactly the part integral of phi hat and so we can reduce this integral to this part integral over an integral only over the configurations phi zero that in principle is arbitrary yes well up to this property that we ask that q v because it is definite but in principle we can have two different choices of v so I hope we will see one example tomorrow of that but of v that I made yes, that's correct but I don't have only one partition function of the theory of what, yes this is the object we want to compute that depends on v yes so what this is a very good observation and in fact it is true that so far we do have dependence on v of course we know that the final result must not depend on our source of v because it is just the original part integral but at this stage we might get different expressions and as I said I hope that tomorrow we will see one example of this but in fact this is a procedure that can give us different looking different expressions for the same part integral and for instance one of them might be easier than the other one but yeah this is definitely a feature that happens and I hope we will be able to see it and it is very important very good observation any other question yes so okay so we restrict our part integral to a special class of configurations 50 and this BPS we come back to this so maybe in quotation marks so far these are zeros of this functional which as you see depend on the functional and then we have e to the minus s of phi zero and then we can exactly do the this quadratic integral so it is a Gaussian integral and what we are going to get is determinant so let me write it as super determinant so let me first write so this is object here well of course here we have bosons and fermions and so in the denominator we are going to get the determinant of the bosonic operator and then the numerator we are going to get the determinant of the fermionic operator let me put this one half of course if you are dealing with complex fields and there is no one half but if they are real there is a one half and this is going to be the determinant of course of the quadratic expansion so this is in fact our operator and of course this operator itself depend on the point where we compute it so it depends on this phi zero so specific field configuration where we compute this determinant ok oh there is also prime and again this is a standard thing in quantum field theory so we might have zero modes ok so if we have a bosonic zero mode it means that the determinant in the denominator there is a zero so what we should do we should remove the zeros from the determinant so this is a determinant prime we do not put the zeros but then we will have an integral over the zero modes but in fact what are the zero modes well the bosonic zero modes are precisely this configuration phi zero and this is precisely what this integral is if we have a fermionic zero mode then that would imply that there is a zero in the numerator in that case once again we remove the zero but then we need to expand this action in such a way to absorb the fermionic zero modes because of course in grasman variables if we integrate deep psi we have zero and it is only integral of deep psi times psi which is equal to one then this means that we should actually expand this object at some order in the fermion in such a way that there are no zeros anymore so this is standard I don't think we will have time to see one example of this maybe Samir will have examples of this ok so ok good so this is our main formula and in fact you recognize that it is an infinite dimensional version but it looks the same as what Leo brought yesterday this object is called one loop determinant because in fact it is computed at one loop while this object is called the classical action because of course it is the classical action evaluated on these special configurations ok so so what do we do with this so what about these configurations well it turns out that in fact one can in fact one can almost always so this is not a theorem but essentially these configurations and this is why I call them BPS because almost always or at least often these are supersymmetric configurations or at least one can arrange so we can make a choice of V in such a way that these are BPS configurations so in fact one can give an area ok so we will see why in a second but one can also give an heuristic argument why one should expect that in fact one localizes two BPS configurations which is so the argument is the following so here we have an integral on field space this is a super manifold configuration and now we have symmetries in general when we have symmetries we can divide the space into orbits and perform the integration orbit by orbit ok now if you have a bosonic symmetry there might be some rotation symmetry in our space so we can do this integration shell by shell now if you have a fermionic symmetry then there is also a fermionic orbit and we might do the integration over the fermionic orbit now the statement of supersymmetry is that in fact this integrand is invariant under supersymmetry so in particular it does not depend on the Grassmann coordinate along the fermionic orbit and so this tells us that along the fermionic orbits we have precisely something like this so here psi will be some of the components some of the modes of fermionic field but the integrand will not depend on psi because it is it is invariant under the fermionic symmetry and so all the integrations along fermionic orbits are going to give us zero then they do not contribute the only exception is that when we have actually a fixed point of the fermionic symmetry and then we don't really have a fermionic orbit we have a single point so this argument has that only those points should contribute but in fact those points are in configurations because they are fixed under the fermionic symmetry this is a bit heuristic but in fact in this framework we can make it more precise because we can make some canonical choice of V which is the following we take a sum over all fermions of cube psi double dagger and what this double dagger is as you see we are in Euclidean so we have complexified the fields so the complex conjugate of a field is really an independent field so we can just define some anti-linear operator in our space and we want to define to find some anti-linear operator such that qV but the bosonic part of qV is positive definite so we try to see whether we can define this object such that qV is positive definite and this is not always possible so when this is not possible one has to find V in some other way it's not obvious that such a V exists but when this is possible we can make this choice and then you see that if you compute qV now there are two terms one term is obtained when qV would be equal to a sum of q of cube psi double dagger psi plus cube psi double dagger cube psi and so this term is fermionic because you see there is at least one fermion so we don't care about it because it is fermionic piece and in fact you see that this is precisely if this turns out to be positive definite this precisely is zero when cube psi is equal to zero so if you want the bosonic part so qV bosonic equal to zero I can probably write if and only if cube psi is equal to zero configurations are BPS configurations but we should probably put like this I mean ok we have if and only if but this is along the contour so this is restricted along the contour ok and so in fact what we reduce to is an integral of BPS configurations yes no so the point is that so we are in Euclidean space we have complexified fields and we might look for a complexified BPS configurations and there are a lot of them but what we care about here is the ones which are BPS and along the contour so it depends on the contour that we chose and is only with a contour that this can be positive definite otherwise this is not positive definite and so these if you want these are not all possible BPS configurations only the ones that are along the contour no it's not always possible because you need to see whether there is an anti-unitary operator sorry and also and you also need of course the Q squared you also have to ensure this so as far as I know one example is for dimension n equal 1 on S4 this doesn't work so this is why people say that localization is more an art than systematic procedure because you need to find this which is not algorithm to find it so ok ok so this is essentially our our formula now let me just make a comment that at this stage we have to compute this super determinant now this might be, might look exceedingly hard because essentially completed this determinant amounts to finding the spectrum of a laplace operator or a Dirac operator which in general we don't know how to do in some simple situation we might be able to do it if we have enough symmetry but not in general but in fact one can one can use some powerful techniques related to index theorems to in fact reduce this problem to a simple problem and the reason is that since this is a super determinant so is the ratio of two determinants in general there are many cancellations and we don't need to compute cancel between numerator and denominator we only care about the terms that do not cancel and for them one can reduce the simple problem I will not have time to do that maybe some of you will do that because you cannot find a V that satisfies these properties I have not tried myself so I cannot tell you well you try to find one and well there is no one well I don't know if there is no one yes because this was the so this is the Taylor expansion that we did all the other higher terms are suppressed by T so in the limit that T goes to infinity only the quadratic term this is second order in the fields not in the derivatives so the fact that we will have two derivative operators comes from the fact that we will start with theories which are standard theories with two derivatives you could study higher derivative theories and then this will contain more derivatives but these two refers not to the number of derivatives refers to the number of fields in principle at this level you could, I mean you could start with more of course if we make if you make this choice of V you are in QV you are only gonna get two derivatives but you might say okay I want to make some other choice and okay this is not something that I explored but it might be useful for something yeah okay I write it as a one half so so what happens you it depends on the example so for instance if you have complex fields then you don't have the one half if you have real fields you have to make sure that there is no sign problem and in the example that I'm gonna show it will be just gauge fields in that case there are no problems but of course there are theories where you might have a sign problem and then it's more complicated you might have framing anomalies or stuff like that so yeah so here I just wrote a one half but of course define in the square root might be not completely obvious okay so this was very abstract and so what I would like to do next is to study a specific example so we see in some details what sort of computations and the results that one can obtain after all localization is all about computing and so so I will consider a particularly simple case so in particular we study two dimensional theories yes so in fact I didn't fully understood this comment but maybe you can explain to me I think there is though I mean is this you know it's the old minimal supergravity so you do have an option formulation the problem is with the complexity so the action of q squared give you a bosonic action which sense you into the outside of the contour so there is a problem in defining the contour which is closed under delta B again I didn't do it myself but this is what I have been told it should be fine then yes yes yes if you don't have such a I think generator of q squared is not compact yeah yeah I'm not sure if it's compact or not but I know that it goes into the complex plane it's not compatible with the contour I don't remember if it's compact or not ok so let's try to go to the example ok ok so I want to consider two-dimensional theories with 2.2 supersymmetry which means four supercharges this is just one example but it's simple enough that we will be able to do the computation and get to the end let me also say this is dimensional reduction of 4d n equal to 1 ok ok so what can we say on this supersymmetry algebra so if we go in Lorentzian signature there are two complex supercharges and they have opposite chirality as this notation suggests and the biggest possible r symmetry is given by u1 times u1 so when u1 acts on left moving supercharge and the other one on the right moving supercharge and sometimes we can rearrange this into a diagonal part that we call the vector like and the non-diagonal part that we call the axial part but of course this r symmetry as we know is an outer automorphism of the algebra is not part of the algebra so we don't have so a t supersymmetric theory does not have to have this r symmetry in general however we want to focus on theories in which the vector like r symmetry is there so we will assume that these are symmetries there why we will not say anything about this in fact in this algebra now we can have a complex central charge and this complex central charge comes precisely because we are allowing a breaking of this so the complex central charge in general breaks the UNA if for instance we have a super conformal theory then we have both factors but then we don't have central charges and if we had two central charges then we would not have this r symmetry so if you go now in Euclidean then this algebra looks like the following so it's very similar to the four-dimensional one since it is a dimensional reduction so we take the anti-commutator of the supercharges and now I'm using this tilde to mean what in Lorentz's signature would be the complex conjugation but now in Euclidean it's no longer complex conjugation it's an independent quantity and this is going to be momentum of course but then there are also these two central charges sorry there is this complex central charge and ok, let me use the by-spin or notation in particular this projector are chirality projectors positive chirality and negative chirality well the other ones are zero q, q and q tilde, q tilde are zero so let's call this complex central charge while these are the z and z that is the complex central charge ok, so now we want to study how to put this theories on chord manifolds so we have to go through the machinery of of yesterday now since we are focusing on theories with these are symmetry so as we said yesterday in general so we need to find this s what is a supercurrent multiplet which corresponds to the stress tensor or the graviton multiplet of some official supergravity in general we have to do the s multiplet which is rather long however since we have some are symmetry we can reduce it we can consider a smaller one which is called the r-multiplet in fact this is the reduction from four dimensions of the r-multiplet in four dimensions and so what is contained in this r-multiplet so I will not use super space notation here because otherwise I make a lot of confusion and mistakes so I will just use coordinates so r-multiplet we have the following components so of course there is the stress tensor and of course there is the supersymmetry current but then besides there are three gauge fields where there is a if you want a real gauge field and then a complex or a gauge fields for a complex for a complex central charge and so essentially this is you remember this is paired and now the r-multiplet is precisely what corresponds to the gravity-multiplet that Stefan presented yesterday in the new minimal supergravity in four dimensions if you reduce it to two we get this but we understand why these guys are there because we have an r-symmetry and we have a complex central charge so we have gauge fields a couple to them in fact correspondingly there is some off-shell supergravity some 2D off-shell so it corresponds to this supercurrent-multiplet and this is the dimensional reduction of four-dimensional new minimal new minimal supergravity so in particular in the in the gravity-multiplet what do we have? We have the metric we have the gravitino and then we have so these are conserved currents and so what corresponds to them are gauge fields and we can call them v mu and c mu tilde and how do these fields appear? Well they appear in covariant derivatives so in general covariant derivative will contain a term that depends on the r-charge this couples to v and then there will be terms that couples to these central charges and they couple to c for some reason there are some the details here, the factors and the signs are not important the important thing is that it corresponds to this vector v and the central charge couples to c now it turns out that in fact these also the field strength of this c appear in the theory now since we are in two dimensions it's convenient to dualize the field strength into a scalar so it is convenient to define some scalars can call h which is just what the odd dual to the field strength and similarly there is an object still them ok and now what we need is the generalized gravitino sorry the generalized killing speed or equation I think we call it which is nothing else the variation of the gravitino in this supergravity theory and so it looks like the following so we have so let me restrict to the case in which what the fields do not have central charges sorry actually I think that this is general and there is a similar thing for psi t so this is the gravitino variation so there is as always this covariant derivative of the parameter epsilon and then there was this matrix of the fields that was acting on epsilon that I wrote schematically yesterday so here we see that these the field strength for these central charge fields appear and then here I'm suppressing some terms which are a higher order in the fermions and we don't care them because we set the fermions to zero so I don't need to consider those so when we solve this equation we set all the fermions to zero so in general here there are some terms but they are not important for finding supersymmetry on core manifold and moreover what is this matrix but here I'm just using the spinor notation so this is a two component spinor and here I'm writing a matrix in spinor notation in which if you want gamma 3 is 1 minus 1 so this is I've diagonalized the chirality here so I could write in terms of this projector but then becomes long inconvenient and here there is a similar equation ok let me just ok let me just write it so what happens here is the epsilon tilde which has an opposite R charge and here there is a similar thing so we have to solve these equations both for the background fields so both for the matrix that appears here this gauge fields V this field strength H tilde not that C does not appear only in the field strength and of course for epsilon and epsilon tilde we can solve this we can find supersymmetry on a core manifold and so I guess this is what we will study tomorrow