 Okay, so let me begin with two clarifications from yesterday's lecture. There were two confusions. One of them was public, one of them was not, but I think it's worth writing it out. So the first confusion was my stupidity. I just use two different, I use the same notation for two different things. Psi mu always in all the textbooks in all our presentation is the gravitino, there are two of them, psi mu i equal to 1, 2. So those are the gravitini and epsilon i I've always used as scaling spinner. But when I started to do this method of bilinear and forming these scaling vectors and scalars and so on, I used the notion psi and there was a little bit of confusion just to say what I did was essentially psi is just epsilon, there are two of these things, epsilon 1 and epsilon 2. Okay, so I just use epsilon 1 plus i epsilon 2 equal to psi, maybe a better notation would have been epsilon. But in the paper I used with Rajesh Gupta, I used psi and then I took that in my notes and so on. Okay, that's all it is and you see that, so these are symplectic maran, the two of them, the reason we combine it like that is then you get a Dirac, you can just easily write bars instead of always writing ij and epsilon ij. Okay, so it's a very silly confusion. I say this because today also there'll be many psi's and epsilon's, so okay. The other one was a slightly more serious one, it's just that I didn't write enough yesterday because I was out of town, so let me, out of time, so let me clarify it. So the action, so there are three parts of the action, the bulk action which is the usual one which is given to you, this can be, I wrote it down two lectures ago, the Wilson line and the bound reaction. The bulk action and the Wilson line I spent enough time on and then when you, so these are local, this is a local gauge invariant expression, this whole thing is gauge invariant. When you evaluate it on this localization manifold, you get these expressions which are wrote down on the board and then the same thing is true of the S boundary. S boundary, this is the part I didn't write, has a local gauge invariant expression, okay, here it is and you can check that this thing is supersymmetric, A plus B plus C is supersymmetric and then when you evaluate it on M localization, this is the expression I wrote yesterday, okay. I just wrote this directly as if it's just something proportional to R naught, it actually comes from a gauge invariant local term and the whole thing is finite, okay. So these are the two clarifications I wanted to make. So today, so now let me recap, I'm going to use a combination of computer and blackboard. So first let me just recap yesterday. So there were four steps in the localization, the 0th step was the first step here, there are more people here, so I'll start here. So the first one is just you find the formalism that we discussed a lot in the school. The second one is you have to find all solutions of these localization equations with the ADS 2 times S 2 boundary conditions and the answer was that in a certain gauge, the vial multiplet has no other solution except the classical full BPS ADS 2 times S 2 and in each vector multiplet there is a solution where all the vector fields, so the gauge fields take the classical values, the attractive values and the scalar fields xi, complex scalars, the real part of the complex scalar is excited of shell. So there's a shape like this which I wrote down yesterday and this shape is dictated by supersymmetry. It's not on shell and it's supported by one auxiliary field, yij which is not zero. The size is arbitrary, the size was called phi i and the problem then reduced to integrating over these sizes, this n plus 1, nv plus 1 parameters, real parameters phi i. Then you have to take the full super gravity action, the one over there, evaluate them on the solution, that's just an exercise and I remind you that this includes all kinds of terms, all kinds of local terms that you can add and then there was a D terms don't contribute statement and when you do all this you get this expression here is the sum of these three things and then you have to compute the determinant and measure for which there's a lot of work and today I'm going to talk about this. At the end of the day you get this very nice formula that for n equal to 2, 4 dimensional n equal to 2 super gravity, on gauge super gravity coupled to nv vector multiplets. If you take a black hole carrying charges p i and q i then this ads 2 functional integral which is exponential of the quantum entropy was equal to this n dimensional integral n plus 1 dimensional integral. So, it's the integral of exponential of s renormalized which itself is a Legendre transform of imaginary f evaluated on this localizing solution. So, the answer is very nice and pretty times a z 1 loop f here is the holomorphic pre potential of super gravity. So, this was the summary any questions so far? Can you just shout out here this one this plus this goes to that that's the boundary term. So, that's a b and c are the questions? Yeah, so just a word of so I've been mostly careful with my formula in all the talks. So, if you've been taking notes but I might be off with signs and eyes and stuff don't trust that completely but I give you references and you can look at them. So, today my goal is to want to talk about this term is one loop determinant. So, let me close that. So, today I want to show you an expression but so let me first announce the result the result is the following. The result is this z 1 loop. So, suppose you have a theory of while and then there are vectors and hypers we didn't treat hypers but suppose there are hypers then the answer is this is exponential of minus k evaluated at 5 plus i p i times 2 minus chi over 24 where chi is 2 times n v minus n 8 plus 1 and for those of you learned any string theory you will immediately recognize this as the Euler number the Calabiao in the type 2. If you think of this n equal to 2 super gravity as coming out of type 2 on a certain Calabiao this combination is precisely the Euler number and it's that that appears okay but from the low energy point of view we just see it as the number of fields n v and n h sorry thank you okay. So, that's the answer and what I am going to do I won't derive this in full detail but what I would like to do since there were many discussions but not too many discussions of this determinant I want to go through the method of computing the determinants in this off shell theory okay. So, there are two methods so one loop z 1 loop so that's so you can either do an explicit computation this is this is I think Francesco did this in his second or third lecture one of the lectures he at least gets some kind of an explicit computation what does it mean explicit computation well you know what this is what you have to compute is you have some q v action I wrote it out it's psi bar cube psi summed over out size and q of that so it's some action you have to truncate it to quadratic terms and you have to ask what is the one loop what is the quadratic fluctuation determinant that's the task it's a very explicit action you can try to do it but it's it's a it becomes harder and harder as the theory becomes more and more complicated and you have to do it off shell you have to evaluate this at a generic point on this localization manifold this off shell manifold this should be a function of phi i as you see the answer okay. The other method is more sophisticated so it involves some learning some mathematics but it's actually extremely useful and that's what I want to talk about today so these are methods of index theory and in a sense it's a sort of localization within the localization so the whole we've already localized the integral to this to this finite dimensional integral so it started with something infinite dimensional functional integral you went to finite and then I'm just saying there's another method you can there is a method to compute this that one loop which also uses some kind of localization okay so that's optional if you want but it's very nice this is very close to what Leopoldo showed in the first lecture so it relies on this fixed point formula of a T unbought which I just want to I want to show you how that happens okay. So what you need is an off shell algebra of the type q square equals h all right so that's your starting point that's your main point you have some theory and you have some in fact for the purposes of exposition let me assume it's just a quantum field theory so you have some rigid symmetry q okay which squares to H and then later I want to make comments about super graph okay so just assume q some rigid symmetry of the theory have some fields or some variations okay so this is generated by q is generated by some killing spin or epsilon so the idea is the following what you do is you the first step is you organize all fields in representations of q okay it's off shell you have an off shell theory so q acts on all the fields so you want to organize the fields and what we will call it like this we'll say there's a bunch of fields which I'll call Phi A and then their super partners q Phi A these are bosonic and then there's a bunch of fields Psi Alpha and q Psi Alpha these are for me on it so so that's bosonic and that's bosonic and this one is for me on it and for me on okay let me just tell you an example instead of telling you sort of formally what this means I just work this out an example it'll become immediately clear what I'm doing so this thing is called the co homological basis co homology because you have some equivalent co homology of this q as Leopold talked about let me just do an example so take n equal to 1 4 dimensions chiral multiple okay so if you've done any supersymmetry this is the first example that you see so this has complex scalar Phi 2 fermions let me take them to be vile but it's not important I mean one one fermion Psi and f okay so this is complex this is complex and so this is 4 plus so this is 2 2 and 4 so let's write the Susie variations Delta Phi is epsilon bar Psi okay so you have some supersymmetry parameter epsilon not so you have some Susie let me just call it epsilon epsilon alpha it's a spinner but it's one spinner okay I'm just taking one supercharge generated by some particular epsilon alpha okay so that's Delta Phi then there's a Delta Phi bar then Delta Psi Alpha is epsilon bar Alpha dot gamma mu Alpha dot d mu Phi plus epsilon Alpha f okay I shouldn't look at the notes for this clearly nobody should but I will epsilon bar dot alpha dot let me not put the indices okay indices okay okay so that's like basic supersymmetry and now what this means is that huh yeah so epsilon is really just a spinner you just write down the killing spinner that's okay so yeah I'm not worried I'm not going to keep track of the anti-commuting and this is just everything these are just fields classical fields if you want now the idea of this is that you choose this epsilon to make projections so so Phi so here is my let me organize it so this we know what it is so let me organize it for this example so here Phi a is Phi and Phi bar okay so by the way these things are called sometimes people call them elementary bosons and fermions elementary fields elementary bosons elementary fermions Phi is this now space I think it's what I'm doing is clear so I'll use this so Phi is Phi Phi bar Q Phi is this particular combination epsilon bar psi it's one combination okay and it's complex okay so there are two here and there are two here so I'm just using epsilon so those are precisely the Q super partners given an epsilon so the Q super partners of Phi and Phi bar if epsilon changes that's going to change as well but it's some particular combination and the ones that appear here are the orthogonal combinations okay so orthogonal in this case means you can use gamma matrix algebra okay to do that and Q psi this contains F NF bar I mean here I'm being a little bit loose in this last thing because it might be some combination which it's not so important okay so that's all I've done so this I think everybody should be completely comfortable with I've just used the super symmetry to organize my fields and now it's very simple I have these fields and I just so these are scalars these are also scalars and these are also scalars and those are also scalars so I've organized everything into just reps where each of which has a scaler scaler scaler scaler okay so sometimes people call this twisting it's it's the same all this is the same thing okay now once that's done so apologies to the experts I know that many people who have worked on this but I know some people asked me about this so I thought I'll do it so the idea now is to organize oops it feels like this so Q is offshore and I have explicitly made a pairing so when I go and computer determinant any whenever there's a mode here there's a mode here okay so there's an automatic cancellation of this and that's ensured by the offshore algebraic nature of Q okay and similarly for psi and Q psi okay so the horizontal direction is automatically taken care of by super symmetry algebra now when you computer determinant notice that now this is a formula is a boson as well so the only thing I have to check is what are the unpaired modes in the determinant of the ratio of the determinants of these two okay and that's the idea so what you want to do is sort of algebraically write down what is this operator okay and then just ask what are the modes that are not paired between this operator okay if I do that I'll show you how if you do that then then you fully finish the determinant because you've made maximal use of super symmetry all this cancellation is manifest and this one you have to compute okay that's the idea any questions about this one question people ask me often is what is this operator in order to get this operator you have to ask what is it you want to compute depends on your problem and what we want to compute is something like this you have psi I Q psi I where I runs over all the fermions of the theory and what we want is so our task is to calculate Z of Q V so that means so let me describe determinant of Q V that means that it's firstly it's a super determinant let me assume for today there's no issue with zero modes and so on people have already talked about it super determined means you this is a bosonic object Q is fermionic sorry V is fermionic Q V is bosonic that means that you'll have two types of term either have something like a boson so it's a roughly Q V we'll have two types of terms something like this or something like that okay these are just actions and what the task is to compute the determinant of this and divide that by square root of determinant okay that's what we want to do so that's V and that's so you write Q V and what now what you want to do is to take that Q V and reorganize it in terms of these fields okay that's an algebraic problem once you've done this classification of what these fields are that's an algebraic problem yeah no I just said that there is it contains so if you really do Q psi so here Q phi is literally this if you do Q psi you might get F plus some other combination of this but the variables which weren't there are four variables right so two of them went there the other two must belong here but maybe there are linear combinations of F and something else that's all I wanted to say okay now as you will see that's not important that's why I was not very careful the only as I already hinted at the only thing that's important are what are the elementary fields because everything there is taken care of by Q alright okay so the next one is to write this okay so I'll do this now I wanted one full blackboard so let me start here let me just do it and erase so let me write this whatever it is so I'm not I'm not going to do it for any example let me just show you what the linear algebra is so it's some you have fermions and fermions and psi will be now a combination of these fields psi and Q psi Q phi those are the fermions of the theory okay so let me write it in some kind of block diagonal form like this so Q phi psi okay that just means I'm writing a vector of all those things and there's some 2 cross 2 matrix and here I have phi times Q psi okay so this whole thing is fermionic so it must be fermion boson operator and this operator is typically linear right and these operators I'm going to call D 0 0 D 1 1 D 0 1 D 1 0 so I hope this procedure is clear so what I've done is taken the Lagrangian that I have re expressed it in terms of these co homological basis but this is this looks like a each of these D of course is a matrix because each phi and psi is is a vector okay and now you see that this operator is just D 1 0 operator so this one actually goes like this and this one was like okay you can just track what I mean is that in this Lagrangian if you just so this is Q phi D 0 0 phi so this thing is called D 0 0 right then let me do this one I want to phi which is here what do I want phi and psi this one is D 1 0 so let me see how that works D 1 yeah so phi so the second line would be D 1 0 times phi and then that multiplies by psi so that's indeed D 1 0 so this one is D 1 1 and this one is D 0 okay so this is linear okay so that's so please keep this in mind I'm going to erase this but I need the space right so then you compute Q V and it's once I have so this is the non-trivial algebraic step takes quite some time to do it but you can do it once you've done it the next step is completely trivial I just act on this by Q but now I use the algebra Q square equal to H okay so Q V looks like this so there'll be a Bosonic part and a fermionic part Bosonic part now we'll have phi Q psi times something times the same thing okay because now you want bosons it's either boson boson or fermion fermion okay and what comes here you can just compute this is an exercise it's just that okay it's almost it's too many it's a two line calculation plus now I have the fermionic term phi Q phi psi Q phi psi Q phi and here is the other way around the sorry in fact that's not right I'm going to organize it like this it's Q phi psi okay good so that's you have to take my word for it but if it's I assure you it's not if you if you spend more than five minutes on this from here please come and see me I'll tell you how to do it okay so now you can just write down what Z 1 loop is so what you want is determinant of Q V fermionic divided by determinant of Q V bosonic to the power half just here I'm using some complex notation okay so bosons always to the power half fermions depends on how you write it let me just write it like that but now you see I can just read it off these are the operators right just write it as a block diagonal so it's the determinant of this operator sorry determinant of this operator divided by determinant of that operator okay now as determinant determinant of two matrices is determinant of a times the terminal B so that's going to cancel all the complication is going to cancel all your left with is determinant of this divided by determinant of that okay here you just have H acting on psi right you just have the bottom component of this is psi H psi here is just phi H phi so you left with something very easy determinant of psi sorry H on psi okay that's all it is okay now this is an enormous simplification because what is H remember H was Q square Q square typically some some translation and some rigid rigid action inside your gauge there and so on so that means that all you need to know is what all you need to compute is take that operator the linear operator and compute the determinants of that operator so much simpler problem than computing this complicated kinetic term that you have okay now there's a further simplification for this and that's where this beautiful index theorem comes in okay so firstly notice that this can be written as hope everyone can see this it's so it's the same ratio of determinants but now I'm going to make a write it and then be more or less obvious to you right what I've said here is kernel of D 1 0 and co kernel of D 1 0 sorry for maybe I'll write it again this is important so where can I write it here I don't want to take up another so this okay is this visible determinant of H so this equals it's just kernel of D 1 0 and co kernel of D 1 0 to the power half all I'm saying is that there's a further simplification so H is some kind of H roughly looks like some V mu D mu or something like this okay if you act on on all the fields in phi and psi by H you see that there can be a further cancellation that cancellation is the one we've been trying to track if there is a you can check now that H commutes with this okay and therefore if two there are nonzero modes paired with D 1 0 then they won't contribute to that ratio right because H commutes with this so whenever you have a mode here with some value of H there's always a mode here nonzero value of H you have a mode here with a nonzero value of H okay so there is a pairing here but this pairing is not algebraic this pairing is not sort of off shell and I don't want to say that all I want to say that this pairing can have zeros there might be a kernel and co kernel whereas here there was nothing of the sort okay we've taken gotten rid of that okay so the only thing you need to keep track of is the unpaired modes in this pairing so the unpaired modes are the modes which go to zero or the modes here which don't come from anything that's called the kernel and co kernel so hold on hold on so my assumption in the beginning was you can always organize your fields in this in this manner okay so that's the next step I'm going to do so the example I gave you for the was the chiral multiplet but I'm saying suppose you can do this kind of organization with respect to Q okay what you're saying is exactly the next complication but holy horses just a second all right so if you could so basically if you can organize your fields in this co homology co homology complex then that's what it boils down to now what I wanted to show was that this thing why it's so nice because this this ratio okay is actually determined by something called the index of the operator D1 0 okay so the index of D1 0 it's a function of T okay so this is a it's a power series which I run a series which are defined like this trace over kernel of D1 0 e to the minus iht minus trace of e to the minus iht of co kernel over co kernel of D1 0 suppose so this is just a definition the statement is that if I know this object I can compute this determinant okay that's a moment start will show you how that works so suppose this index it has the following form a of n e to the minus i lambda n t suppose the eigenvalues of h are lambda n so there are essentially these are bosonic and fermionic as you see from here right bosonic and fermionic eigenvalues so they can be plus or minus sign okay so write this as some infinite Laurent series okay so all you have to do is to ask what is the current so suppose I know what the kernel of D1 0 is and the co kernel I just evaluate this so I have not finished the same so I get this suppose I get I have this already suppose someone gives this to you then the determinant that I want is so this determinant co kernel h over determinant of kernel h it's just product over n of lambda n to the minus a m because those are the eigenvalues okay this is just some formal trick right so these what you want here is the eigenvalues of h okay in the co kernel and kernel I'm saying reorganize that so start from here if you want suppose this thing looks like this it's a product over again values first you make this series okay for the index okay now we are in business because I'm going to waste this because there's a very pretty formula for the index okay so there's a nice formula I'm going to present in a second for the index once I have that formula I can read off the eigenvalues plug it into this so the formula is this it's by so maybe I move here so it's roughly half an hour no okay yeah good so the formula is the following so this is a T m 1 fixed point formula for this index it says that okay there are technical conditions about about this operator D1 0 that we have been talking about I'm not going to discuss that please look up the papers so it has to be so it's a differential operator and there's some conditions on it it's called transfer so ellipticity and so on and so suppose you have some manifold so this is your space time manifold okay so this is let me just call it space time of course it's Euclidean so maybe I shouldn't say it manifold but this will be our ADS 2 times S2 or S4 or whatever you have and on these there are fields okay those fields in this in this theorem will be called sections of some bundles I think most of you know this better than I do essentially there's some space time with some fields and H has an action on space time okay like this call this X tilde and then the index of D1 0 of T so function of T is the following it's just so remember the index actually is a function of the operator defined on the whole manifold right this it's a very complicated thing still a priori there's some differential operator which is local okay and the statement is that that index reduces to the fixed points of H on the space time manifold so says that look at all X says that X tilde equal to H X and then you just look at trace of E to the minus IHT kernel minus co kernel and I'm going to write this in physics notation essentially that's just minus onto the F okay but now only at that point okay divided by determinant of 1 minus DX tilde there's some geometric it's some kind of this comes from some Jacobian factor as I'm sure all of you have seen and this trace remember goes should run over the kernel and co kernel of D1 0 but remember D1 0 just took Phi to psi okay that was what D1 0 was okay so I don't actually lose any information by just saying that this is a trace over Phi and I can keep the massive modes as well they'll automatically cancel okay and that's your final formula and it's a very beautiful formula and you'll see how powerful it is in a second yeah H has to be compact yeah so what has happened is that the whole index calculation has reduced to some over fixed points on the space time manifold let's assume these are discrete so it's just a quantum mechanics problem so what you have to do is to find write down your so the algorithm is following start enough start with an off shell q q square is some H it's some some d phi or something like this ask what are the fixed points of that operator on the manifold just take your fields organize them into this co homology basis and then just ask all you need to know is what are the charges of the fields at that point so essentially how do the fields rotate near the fixed points okay so it's a very simple quantum mechanical problem the whole thing reduced all the complication is to actually find goes into constructing this complex okay once you've done that now it's a machine this trace from yeah yeah okay so thank you so maybe it's at that point phi of x psi of x so that's what I said this is really a quantum mechanics problem so yeah thank you I should have said it's not a trace over the field space but just a quantum mechanical modes of those fields at that point at the fixed point okay thank you okay so now I'm going to do apply this to our problem so now case so now case we have some ADS 2 times an S2 ADS 2 and remember Q square was L0 minus J0 okay L0 was this so fixed points of a lot means the center of ADS 2 are the R equal to 1 in my notation okay there's only one six point and you also need a fixed point the J0 so that means that it's these two points okay so there are two fixed points J0 went like this the north pole and south pole of the sphere that sits at the origin of ADS 2 okay there is two points and the task is now to compute the charges of all the fields and this determinant so let me do it extremely quickly so first let's do the determinant so let me write ADS 2 times S2 in this complex coordinates so that's the stereographic projection dz dz bar over 1 plus dz bar square that's the same stereography it's the same construction for ADS 2 therefore there's a minus okay L square here is the size of the ADS 2 but now I'm going to remember in some core so let me write L square is the physical L square is just e to the minus k of 5 that's the physical size of your manifold remember so this was if you want square root of G in the physical metric k was the killer potential k e to the minus k is minus i x i f i bar minus x i bar sorry sorry 5 plus ipi thank you so I can speak up this yes then phi is on what yes sorry I'm not I don't understand what you're saying can just pick up I just can't hear you actually I have a killer potential there it is yes yes it's x times yeah okay phi is my coordinate on the x is sorry I know you're saying something which is important but yes yes yes look all I'm saying here is that there is a there's a certain size to the ADS metric right it's like in our paper what we call square root of G maybe does that help if you write that yes okay can you say what you think should you are you saying this is incorrect or something I just don't understand your statement it shouldn't what should this be okay then can we just take it offline can we discuss this afterwards so can we discuss this okay I can give you a reference for this and we can discuss this later I really don't understand exactly what you're saying so the references are I mean you know the references so there's one paper that I wrote with valentine rice who's skipped and another that imtac jon who also skipped wrote with two of the collaborators imtac is there maybe imtac can explain to you no this is this is a e to the minus k turns out to be real you see this is the it's a real thing okay let me let me carry on and we'll come back to this it's okay so what are the fixed points the fixed points are w equal to 0 and z equal to 0 and z equal to so combined with z equal to 0 and 1 over z equal to 0 okay those are the two fixed points okay w equal to 0 is here z equal to 0 is here 1 over z equal to 0 is here okay so then you can easily compute the determinant of 1 this thing is 1 minus q square times 1 minus q inverse square the q is e to the minus it over there all right so that's the determinant that's very easy to compute and then you have that you want to compute this trace the small trace of e to the minus it at the fixed points over phi and psi so this equals it turns out to be equal so there are four for this now I have to say what it is so so far everything was very general and now I'm going to do this for the hypermultiplet okay for which there is actually an off shell formulation if you take so this is for hypermultiplet if you take one super charge okay and that's just minus 2q 1 minus q inverse square okay there are eight modes in the hypermultiplet therefore bosons and for fermions and those are the eight modes minus so it's minus 1 minus 1 minus 1 sorry plus 1 plus 1 minus 1 0 0 0 0 0 those are the charges okay just have to compute okay so I put this all together and what you get is well sorry now what you get is z 1 loop this product of n greater than equal to 1 minus n over LADS those are the eigenvalues should write this continue here so this implies that the index is minus 2q over 1 minus q inverse square divided by that such 1 minus q square such sum n greater than equal to 1 n q to the n for for so this is at the north pole so this is at the north pole and there's a similar computation at the south pole so you get 2 plus 2 which is 4 and you put this okay for for okay all right so from this then you can just read off what is the the determinant it's just that to the power 4 okay those are the eigenvalues n over L LADS is the is the eigenvalue and the degeneracy is 4n so I just write that this was for the hyper okay so you get this and so now you see there's some infinite product and you have to do some zeta function regularization the numerator is just some number which I don't care about what I want to release how the determinant scales as a function of this LADS so you can see this is just equal e to the minus k times so minus 2k time actually it's yeah minus k times sum over n n greater than equal to 1 times 2 you can just check that and so that becomes just e to the minus that's minus 1 over plus k plus k okay so that's minus 1 over 24 and by zeta function and you get this okay so it was a little bit sketchy but that's the main point okay so this is how you compute the terminates so if you have n h hyper multiplets and that's going to be multiplied by n h so remember this formula I showed you e to the k or 12 times n v minus that's how you get this I might have made sin or a 7 a little bit of a hurry okay so this I just wanted to give you an idea of how the determinant works okay now I have about 15 minutes left roughly 13 13 minutes left let me what I want to do now I want to use the projector so this was what I just showed you this is right you have these fixed points you have to do this thing and you do it for for the hyper multiplet and you get this okay now now the next question is you want to do this for the next level of complication is you want to do this for the vector multiple okay why is that a complication and this is what I want to spend some time on why is that a complication so the complication comes in the very first step the starting point is you want to organize your fields in representations of off-shell super symmetry okay so let's look at the fields so the vector n equal to 2 vector multiplied looks like this there's an a mu I'm going to use this so if you can't see then just either move or yeah there's a vector field a complex scalar which I write as x 1 x 2 y i j which is this auxiliary field and these are the gauge you need I used to call this omega until now but now it's lambda okay but now you count degrees of freedom so here there are 8 degrees of freedom in the fermions in the bosons there are 3 4 5 and then there's a gauge field so if you put the gauge invariance it's 3 so it's also 8 so it should be 8 plus 8 as Stefan said but here we want to do everything off-shell you want to do everything off-shell and you don't want to so there are two options you can do you can either say if I want to do off-shell I must gauge fix to get 8 plus 8 degrees of freedom that's one way of doing it okay but if I gauge fix a supersymmetry variation sorry if I gauge fix a gauge multiplied you'll see the supersymmetry is not consistent with that start with some gauge for a mu take the Lorentz gauge for a mu act on it by q act on it by q back and you're out of the Lorentz gauge okay and you'll see that in a second actually so this is a very small calculate if you've not done this you should do this so here are the variations q a mu is some function of lambda q lambda is is some function of f and d mu x and y ij and so on and you'll see so if you just naively compute q square you'll get q square is some v mu d mu d mu v mu is this fermion bilinear that's the keeling vector I have d mu is partial d mu plus something inside the in this case there is nothing else actually it's just partial d mu plus a gauge transformation with parameter being x so field dependent gauge transformation this is not something that teach you in high school but anyone who has actually done this has run into this problem okay this problem becomes more serious in supergravity but already at just u1 gauge Maxwell gauge multiplied you have this problem it's not a problem it's just that q square equal to h is not it's not really true it's this equal to h plus some gauge variation of course in physics you don't care about it because you say it's it's still a symmetry here I care about it because I don't want to I want to do a functional integral with gauge fixing and so on and gauge symmetry is no longer a symmetry right I want to use this off-shelf formulation right let me emphasize that I want to do a functional integral a la fa je popo so I put a gauge fixing term once I put a gauge fixing term there is gauge symmetry is no longer a symmetry right I only have BRST or something okay so that's one issue and here you can also see that if I start with a certain gauge I'll get out of that gauge because of this variation okay so one thing people do is to actually gauge fix by hand and you can modify the super symmetry transformation according to the gauge so that you stay within the gauge that's a little bit ugly you can do it you have to do it case by case but there's another nice way to do it co-variantly it's much more elegant but there if you want to do it co-variantly you have to ask a fight this problem that the number of degrees of freedom are not the same there's nine and eight but co-variantly means that you put ghosts and there are two fermionic ghosts so BC and capital B okay that's the Lagrange multiplier so that's one boson and two fermions and indeed miraculously you have 10 plus 10 degrees of freedom okay so there's some hope to realize the offshore algebra on all these 10 plus 10 fields there are still problems the problem is what do you do with this okay that's one problem the second problem is what is Q on the ghosts so I introduce new fields but I haven't told you how super symmetry acts on these fields okay in fact no one tells you how super symmetry they're not supposed to be in super symmetry multiples okay so what you do this problem is an old one and people have solved it long ago use the fact that there is a BRST charge which acts on this BC so I mean an abelian theory so Q of C is 0 Q of small b is big b so that's the sometimes this is called the ghost anti-ghost and Lagrange multiplier so then you make this combination called Q plus Q BRST called Q equivariant and let's look at this algebra so Q equivalent square is Q square plus Q BRST square plus the anticommunist Q square is H plus delta gauge Q BRST square is 0 and this commuted anticommunist is what it is what you want is you want Q square Q equivalent square equal to H I want to construct a charge which is really equal to H and now it's the two problems cancel each other remember these things acts on all the fields so I just demand that these two are cancelled and a it gets rid of this gauge variation if and it also tells you how Q acts in the ghost so there's a consistent way that Q acts in the ghost by just solving this plus this equal to 0 okay and that's the answer so that's how you do it and now you have Q equivalent square equal to H genuinely offshore once I have this I have this 10 plus 10 degrees of freedom I have to repeat this whole exercise write down how this Q equivalent acts on it organizes into these boxes run the whole program and what you'll get is exactly e to the minus k over 12 for one vector okay I still have maybe three four minutes I have more have seven eight minutes actually yeah I good good good yeah question yes yes so what's going to happen is here so now this is going to be Q is going to be replaced by Q equivalent so it's just Q plus QB so let's take the let's take the boson Q plus QB Q is this what is QB of a mu QB of a mu it's a gauge transformation with gauge parameter replaced by ghost so QB of this is that plus D mu of C here nothing happens and so on okay with a completely algorithmic way to write so it's slightly more complicated but it's nice so once you do this you get yeah okay so here are your 10 fields and you want to organize this into boxes and as I said one nice way to do it is write these sort of make these epsilon projections so these are typically this is what is called twisting so take epsilon in the current multiplied I just did epsilon psi here is the same thing that's the scalar then I can form a vector and a two form symmetric two forms so that's one plus four plus three those are the eight degrees of freedom I just reorganized my eight degrees of freedom according to epsilon okay I've just done epsilon projections and because of the twist I just get bosons and when you do this organization you'll see that what lives in the elementary fields is a mu and x2 and that's one part of the complex scalar and in the elementary fermions you have so qb of a mu is precisely lambda mu that has to be there's only one vector and so on q me of qb sorry q equivalent of a mu is lambda mu plus d mu c that's that's and here these three fields sit here and b and c sit here okay so that's this takes some time to do what you can do okay you run this and as I said you get you do the same thing you get so in general you get something like this the z1 loop is e to the minus ak a hypers this a vector is that if you put this into the original formula that I started with today and I ended with yesterday you're fine and then start do some saddle point evaluation you'll see that the black hole entropy goes as area over four plus this number a naught times log of a h okay so the leading log correction is actually just now just a sum of this thing which was this two minus chi over 12 that I showed you this so after having gathered all the technology if you if you're just interested in some leading log which people were interested for a long time it becomes a tool and calculation after all this technology but you have of course the reason to do this is not to only calculate a 0 but to calculate the whole function but of course it's consistent pretty good now now in the last five minutes I want to talk about one more thing okay thank you very much I'll just go slowly and I'll finish I have a few more slides actually one question for Jean are you going to talk about this Bessel function and so on at all then I won't show it you have a slide with the numbers and so on good then I won't show it no that's fine but okay so then I want to talk about a formal thing so the simplest example was just a chiral multiplied when there was no issue of gauge invariance the next example was a gauge multiply regular a mu gauge Maxwell field now I'm going to do this for super gravity so this was a thorny problem it stayed for a long time what is the problem in fact there are two problems not just a technical problem of computing determinants okay there was that problem which was around for a long time but in fact this whole story of localization has to be revisited because the starting point of localization is that I have a path integral so in the context of a gauge theory it's some for the upper part of times type gauge fixed path integral and I need a rigid symmetry I need a rigid supersymmetric q which squares to a rigid symmetry college that was how the whole localization argument went on okay but okay sorry this I want to skip this was maybe I'll show this later if at all but otherwise yeah so or maybe I'll just show you exactly no it'll waste time so I want to keep the momentum there's a slogan which you might have heard that there is no global symmetry in gravity you might have heard it in different ways I sometimes you must you heard there's no global symmetry in gravity other times you'll say well it takes super gravity super gravity is a theory of gauge symmetries including supersymmetry supersymmetry is also gauge symmetry in super gravity unlike in quantum field theory and if you ever but gauge symmetry is not a symmetry of the action because I've gauge fixed another slogan you've heard is gauge symmetry is a redundancy of description it's not really a symmetry all these things you have to start worrying about now what the hell am I doing at step zero so a lot of people actually worried about this already you know like at the point where I was two lectures ago and I said let's start with super gravity and do the variations and many people said well but what does this even mean and we said well I gave you this kind of thing which sounded nice that you solve for all geometries which admit some supersymmetry something like this but now I can do it in a better way so this is true well this is kind of true but it's we know that this is not true if there is a boundary okay like take ads CFT if I have ads CFT the symmetries the global symmetries of the bulk theory is also a symmetry by the statement of ads CFT it must be a global symmetry of the bulk theory if the if the boundary theory has a rotation it's also a rotation in the in the bulk theory now again this is little bit of confusion you might have had if you're actually a little more advanced you're like well global symmetries in the bulk is realized as gate symmetries in the boundary etc etc but there's always a global part of it which is not local you can't write a local current in the bulk but there is a global symmetry okay so the way to do this is the following so think of the path integral you fix a background let's just fix ads 2 times s2 the fully supersymmetric background and then the idea is to fluctuate integrate over all possible fluctuations okay not just small fluctuations now this prescription of course loses some very high level background independence okay but I didn't expect anything more than that because ads CFT asked me to fix the boundary conditions anyway so I'm taking the fixing the boundary conditions pulling that into the interior with some fixed solution background solution and then fluctuating over all integrating over all fluctuations the question is how to do this okay so you have to implement this you want some super global supersymmetry which squares to some rigid surface near the boundary the boundary conditions that I said admit such a thing the boundary condition is just ads 2 times s2 there I know there's a q square equal to h and so if I just declare the background to be ads 2 times s2 there's a q square equal to h everywhere but now I want to ask how does that act on all the fluctuations of the graviton and then I want to ask how should I gauge fix that okay that's the technical problem okay so that's a little bit difficult and what this suggests I mean the words I'm saying reminds you of the background field method in gauge theory that you fix some background there's some symmetry of that and then there's some fluctuations on which there are now there's a gauge symmetry of the fluctuations and the background symmetry and you want to integrate out the gauge degrees of freedom that's the background field method unfortunately the background field method has not been developed for supergravity unless except in very exceptional cases where there's some super field formalism okay certainly not for this now that's the first question okay so first sort of deep question conceptual question the second question is relates to this determinant calculation that I suppose I want to do this like I did then you would have said well now you did a mu so first I did chiral multipliers you say what about gauge symmetry I showed you how to do for a mu so now what about supergravity and what I need to do is to tell you what these twisted variables are that's my first step this co homological variables and then again you run into this problem because super symmetry doesn't respect choice gauge of choice of gauge and so on so there are two methods like I said so you either change q or introduce goes and do this thing this was the method we saw so far but in supergravity there is only q b r s t there is no q there is no rigid symmetry to begin with you the only rigid symmetry is q b r s t but q b r s t squares to zero it's an important of it okay so everything seems to be a problem so at best you have something like that but you want some equivalent algebra and you want to twist okay so I'm effectively going to leave you with these problems I'll just tell you how these are solved and if you're not going to be present next week please come talk to me because there'll be two talks next week one by bernard do it and one by Imtac John sitting there was going to essentially answer question one and question two okay so let me sketch what the answer is the answer for question one is that so I should tell you very quickly what the problem the problem technical problem of why you can't do background field quantization for supergravity is that the the gauge algebra is not a lee algebra if you have a lee algebra you just take out peskin shredder and just write it down all right it's not a lee algebra the structure constants are not constants and they're field dependent so if you don't know this you should read this book which Stefan brought by Friedman and from front brilliant they explained very nicely but it's also many other review talks review papers so that's really the problem but this is already no not for your gauge theory is just so you want of course it's very trivial but even if you take non-abillion it's a lee algebra but the gauge term you add yeah but supersymmetry itself is rigid so you don't even run into this problem that's so you can just fix that by the fix that I just gave you this this thing I mean not I it goes past to bullio singer and so this is the thing that was used in festoon and how that's completely fixable here you're you're you know you're compelled to face this problem of solving this thing okay so what we do is we develop oops the brst quantization first we just do brst quantization okay but doing a background split so we tell you how to do a background split in theories for generic gauge algebra including the soft gauge algebras namely when the field structure constant or field dependent that's a that's a new thing and I think that's very very interesting um the next thing we do is once you have such a thing we actually we show that there's a consistent deformation of this algebra which gives you an aquarium charge okay this deformation depends on the background okay as it should in ADS CFT and you just get this but this is now a rigid symmetry so we actually construct such a rigid symmetry so it's nice for many points of view firstly it solves this problem I'm talking about but very generally it tells you how to do in a sense at least formally quantization of supergravity not the uv problem but these formal problems and once you have localization you can do it and then you can sort of for mathematicians you can present the answer as a definition of these class of supersymmetric integrals so that's the first resolution the second resolution is now you have a q rigid thing you just organize all the fields okay this is a now the number of fields in fact I forgot how many fields so you started with 24 plus 24 of super here for the gauge multiplied you started with 8 plus 8 which became 10 plus 10 for the supergravity multiplier start with 24 plus 24 but help me I think it becomes 96 something like 96 plus 96 because there's an enormous amount of gauge symmetries the whole gauge algebra of supergravity has to be now gauge fixed so you give goes for all of them of course it's reducible and you make this big complex but okay it's a it's an algebraic exercise we spend pages doing it and at the end of the day you have a small table that's all okay and and that essentially computes this this pairing huh black hole I define bh in my very first lecture black hole if you if you've taken notes you can check this so yeah so we call this the black hole cohomology it's the it's the cohomology of the supersymmetric black hole that's the title the title of the paper is also something like that I forgot what it is BPS something like this okay okay so I think I'm going to stop here so this paper so there's a so both these things resolutions please look for these talks Bernard I think Val will talk about something else but Bernard will talk about this for sure and in touch we'll talk about this all right thank you very much