 So it's nice to be back in Trieste as always and I'm grateful to the organizers that they have invited me to speak I'm going to be talking about the supersymmetrical localization with dynamical gravitons And most of you will know what I mean by that, but it will probably become clear in just a moment Yeah, it's not Can you hear him? No, not good enough Yeah, maybe this should move up Is it better now? No, I think it's better now Okay, thank you. In supergravity, that's what I'm going to talk about applications in supergravity as another of engaged theories, PRST quantization is a very good framework for having to be able to consistently integrate over dynamic modes in a gauge invariant way and so on and The problem is that this when there are backgrounds or when there are boundaries then in this in the Supergravity case you have a problem because the boundary is sort of, you know, it's gravitational and also the bulk is gravitational so there are two issues there that is you want to know how the isometries that's of course a Rigid symmetry of the boundary how they act on the the quantum modes on the fluctuating modes and You wonder whether there's a charge associated with the boundary and Somehow you have to do this to set this up in a way that one thing doesn't interfere the other Although they are coming from the same source but the question is to how to consistently deal with these two different but very related symmetries and that plays a role in the applications of localization in supergravity Which usually does not appear when you have Separate symmetries, you know when you for instance take you have rigid supersymmetry and You have local gauge symmetry then you can use these two symmetries and you can take the BLST charge You can combine it with a convenient supersymmetry and you can get an equilibrium in cosmology But when you consider full supergravity Then you have to insist that also the gravitational modes all of them play a role in the localization And so then it becomes slightly more complicated so The above observation is important because there are applications where you actually do want to integrate over all the gravitational modes and all actually super gravitational modes and the mean to so that you cannot freeze this space-time metric and That means that the all the dynamic degrees have to be summed over in a functional into And until recently this was not entirely clear how to carry this out Although it was believed that it could be done And the application that it was that we were motivated by is the application of black holes Where the geometry the near-the-horizon geometry is ad s2 cross s2 and What you can do it normally Sort of simple approximations is you know that there is supersymmetry enhancement at the near-horizon region So that everything and the near-horizon is determined by the charges of the black hole And then you can just either use baking Hawking Biggestine Hawking or you can use the world entropy formula and you it can extract the entropy and That was done in several cases also with higher derivative theories But obviously this approach does not take into account that you have to Indicate over all the dynamical fluctuations of the quantum fluctuations of the underlying theory Now that can be done by using sense and to be function. There is a precise prescription. Are you should do that? And that means that you have to evaluate the path integral over all the fields that live in the space time with a boundary because there is the horizon as a boundary and There you have to integrate over all the super gravitational degrees of freedom So far that calculation hasn't been done Although there is now one paper where it has been done and that will be next speaker but we were just Startling with the conceptual problem of how to set this up in such a way that you have full control and you know What you're doing and it makes sense Now in recent years, we have seen several calculations of quantum entropy functions But as I said already, you know, they there was no integration of the gravitational modes There was only an integration over the over the gauge mode and over the matter fields So here I will report on recent progress on this issue and explain how you can apply localization Spaces with a boundary and at the same time integrating over all the fluctuations And as I also said, you know, there is now paper out where this formalism was used and that will be discussed by Intech in the next talk So the contact of my the content of my talk is as follows Since it's a gauge theory the first thing that you have to do in any case is you have to quantize the theory and for that You use BRST. That's an obvious statement So we have to I will briefly introduce just to introduce the notation say say a few things about the RST homology then you have to struggle with the incorporation of the boundary details and The only way to sort of make the boundary treated independently of the bulk is to use a background field splitting Usually the background field splitting is used for simple technical things, but here you're really using it for physical purposes And you have to do that in such a way that you can apply to a theory which has a soft gauge algebra and In the soft gauge algebra that is in gauge algebra where the extract constants are not constants, but they depend on the fields Then for localization one needs an equi variant homology So you can maybe I'm going to introduce a functional integral, which is BRST completely Okay, but you have to figure out what this relation is with the equi variant homology Now you how you generate it and it should not change too much because otherwise you're destroying the previous Contestation that you have already introduced And then finally for localization you have to introduce a deformation So that you can move to a convenient phase in the field configuration space And You have to set it up in the formalism and that's maybe the hardest part So let me first go to the BRST homology that will be familiar to all of you But at least that you will understand my notation. So we write the gauge transformations that these are just generic fields as a function are of the fields and this is the parameter and This function are can also contain a derivative So every gauge transformation can be written in this form. Of course, we use the instant Standard convention where we some of repeated indices which involve also space time So there are no integrals or whatever and these transformations close on the commutations That means that the usual thing if you apply two infinitesimal transformations Then you find a new infinitesimal transformation back and the new one is written in terms of the structure constant times these two parameters But the structure constants are not seems to be constants They may depend on the fields and that means that the algebra soft So this statement applies and this is going to be a rigorous requirement First of all closure that the transformations close without using equations of motion or anything else And this is the place where you define the structure constants and then correspondingly the Jacobi identity Which is a very special part here because the structure constant depends on the fields So that means it's not a trivial Jacobi identity, but it has an extra term so in this introduction I just you know discuss only Commuting symmetries commuting transformation parameters to make keep it light But of course eventually I'm interested in mixed transformations both Anticommuting and commuting parameters So the corresponding BST transformation which follows straightforward from the transformation rules Just an application of the algebra value chronology they They read like this and this is the font for the Postfields and there is this object lambda which is there just to make sure that the relative Signs with the commuting and anti-commuting parents are eventually. Okay. Now if people only use Anticommuting goes then you can put this lambda everywhere when when you have both of them present You have to write it and precisely this way. Otherwise, it doesn't match the symmetry structure of the structure constant And if you work this out, then you can use the Jacobi identities also in the soft case And you'll find then that the BST transformations are no potent and Both on the ordinary fields and on the ghost fields The next step is to introduce the gauge fixing and for that you write something. That's the BST transformation I take out the parameter lambda here That's left derivative of this parameter lambda of the BST transfer is also sometimes called the gauge formula That is the high contains the anti-ghost and the gauge fixing term and the gauge fixing term in this case Well, the anti-ghost in this case goes to a Lagrange multiplier field and it goes to the anti-ghost and these are the corresponding Transformations which are also no potent. So there is no potency everywhere The fields as I said already there are Lagrange multipliers So if you use them then you get the condition that this is equal to zero and of obviously F has to be such that it is a Meaningful condition because otherwise it's not going to work That means that it really fixes the fields or one of the fields depending on what the another one associated with alpha doesn't matter which one and be alpha as the ghost the anti-ghost And the statistics of the ghost is and the anti-ghost is always opposite to the statistics of the corresponding parameter and gauge fields so now let me go through the which is for sort of straightforward you make the The background field split so you write the field phi as a background field and As a quantum field and in this case we are thinking in terms of a boundary So phi dot here is then supposed to be the value at the boundary And this one is supposed to be the fluctuating field about which you are going to integrate in the functional integral Now the boundary fields are fixed at the boundary and that depends on you what they are that depends on the physics in principle And you can also continue them into the ballad because you have to sort of give the deviation of the fields in the bulk with respect to the boundary field and The precise continuation is not very important And you will have to integrate over the fight till to the quantum field And then you split the transformation into background transformation where a background field only transforms into a background Into under a background transformation in background fields again And the quantum field transforms in a complicated way because it has to take up the remainder and The remainder in this case is written like that It's all very simple and it's you know, if you work it out you're a little bit familiar with it. You'll see that this is Everything is off shell because I already declared the closure of the symmetry so everything every field that I need for being off shell at them I'm Insisting on complete off shell off shell closure the back the background is the backgrounds I mean, it's just the field so every field has a background field and it has a quantum field So this is unrelated to that. I mean at the moment. I have just doubled the set of fields So then here have the quantum transformations and they look like this and that is very close to what they Originally are and of course the background fields do not transform under the quantum fields So it's important to them determine the algebra of these background and these these doubled transformations the background and the Quantum transformation and here they are these are the old transfer the commutators that you can find on the background fields and of course you see that with the Quantum transformations, you know, it always gives zero and here it just gives a new background transformation That's completely consistent, but here things are becoming more complicated Of course the background transformations on the back on the background or sorry on the quantum transformation on the quantum fields They just close nicely. This one is also sort of standard But this one is different and the main difference is here now that there is a term which is there because of the soft algebra that means that you know sometimes the background refers to the background field and The and the other one. So this is the comp. This is the sum of the background field and the quantum field And this is just only the background field And we thought that that would be a major problem. It certainly has a structure for the algebra You see two background transformations will not close into background transformation because of this term But the system that is that you usually see in a gauge theory is lost here That's and we we worried about that and we tried to do it in another way actually this whole Exercise was a little bit like crossing the rivers jumping from one stone to another and then turning around because it was not I know finally when we were at the other side, we realized that everything was alright So now that you have these transformation rules now you can just write down the BST transformations and there is a weird thing because there are background ghosts something that is hardly ever mentioned in the literature I think in fine moves within what Wineberg yeah in Wineberg's book is mentioned, but that's in the linear case. So he doesn't and he doesn't do anything with it Staff from mathematical reasons, I guess So I give the transformation here and he indicated what is now really the crucial points here First of all, you see there is this extra addition which was necessary in order to make sure that this thing You know that you could split the transformations correctly. This is not this is generic This is always there, but here you see in the first rule you see that there is this difference coming up And you can rearrange these things in this way which is convenient And then you stuck here with something that refers to the background ghost and the background structure constant Now if the algebra were not solved, of course, it would still work This would just be the same structure constant So it doesn't do any harm but the point is to Explain that this phenomenon exists and it's not going to do any harm So now you go back to the action and now you have doubled all the fields and you have doubled the ghosts and the action isn't just like the classical one where the of course the classical Fields are the sum of the two different kinds of fields there is in The gauge fixing the gauge fixing has to be such of course that you can check you that you can fix the phytildes So find not would not be a gauge fixing. You're not going to integrate over find not find notice You are free to choose what it is and Then for now for a change. I added all the statistical factors So you hear you have the anti-ghost here and then there are different derivative of respect to the background field and the quantum field and this follows precisely from In from the transfer from the transformations and then you have a full BRST algebra, which is nilpotent with double set of fields and which is Which is all this nilpotent and the action is invariant So nothing major over it looked a little bit scary here and there So now you have to define a functional integral and obviously the functional integral You want to integrate over phytilded. That's the quantum field and the ghost the quantum ghost and the anti-ghost and the Lagrange multiplier and then you write down this whole action and Then you can prove that if you and this thing I already indicated depends on find not and you may wonder why it doesn't depend on C not, but it's just a matter of Ghost conservation conservation. I mean there is ghost number and so there is no C not on the left-hand side. It thinks Depends exclusively on phi not and if you then just do on this whole expression you do your BST BRST transformation It's just the derivative of this with respect to phi not times the plate the background transformation on phi not That means that if you take the background invariant or if you do something with the background It's exactly BRST invariant And I think that's a statically pleasing and that's what you would expect as long as you have still Background fields outside and they are not invariant. They will pick up something from the BRST transformation Now the functional is independent of the gauge condition. You can prove this There are many proofs, but one of the standard proof is to just set up the water entities and then Fix it and show that it is the same and of course if you would add an interaction Which would be BRST exact like this. It would not affect the functional integral So this is there's nothing Disasters happened at this moment Another implication of the water entities at the Lagrange multiplier fields at the quantum level must have a vanishing expectation value that is true because Otherwise the BRST would only be spontaneously broken and you want to make sure that it is exactly broken You know that it's manifest symmetry Now everywhere I say it again that I have assumed and I will assume that the BST transformations close off shell I'm really starting from a theory where BRST is nilpotent off shell If that's not the theory you have you will have to be smart and do something We may make a comment on that later on Now towards a covariant comology. You see what I defined now I in some sense this must be the answer because I have quantized the theory I cannot doubly quantize or do something else So I somehow have to understand how I get the covariant to comology by going through this situation The first comment that we made is that You know that the background field for practical purposes you have to take an invariant background field That means you can restrict it to an isometry group Which is smaller than the maximal isometry for the simple reason that they can also choose to choose some of the C Nots equal to zero some of the background ghosts I can just say that there is a finite subgroup of the Potentially the full the maximal isometry group and the put the other ghost to zero If I do that I still have this term in the transformation rules But this would be consistent with the algebra. So that means that if I Impose a restriction that some of the C knots are equal to zero namely the ones that do not be deep depend on the isometry You then it doesn't do any harm because this thing will then automatically also be in the isometry group So this is a stick system and you have the freedom at this point to restrict the number of ghosts depending on the background Situation that you have chosen at this point the BST is in fact, it hasn't been changed I have just chosen my external parameters in a certain way But and the functional integral is still BST invariant Now subsequently I'm going to make a deformation Which is of course closely related to BST because BST is sort of the initial framework that I have to respect And the deformation is that I'm going to assume that all the background fields are invariant I already assumed that so that's fine But the ghost all the ghost remain invariant and the Background ghost and the background ghost did have a transformation on the BST So that is going to be a real deformation and on the next transparency. You're going to see what that is So here I give you the new transformation, which as it turns out will define the equivalent chronology And I've indicated the changes So this one was already zero because I chose the background in a certain way This one was not zero But I just put it equal to zero Now when I put this thing equal to zero that something goes wrong in the algebra as you would appreciate But actually it's not too serious if I look at these fields The only thing that happens is that the square I have no longer no potency, but they get a background transformation back for it So I really have an equivariant map here And then the only thing you have to do is you have to look at the B and the Lagrange multiplier fields and the anti-ghosts Who are quantum fields? So you have the freedom of manipulating them a little bit and you find that there is a new transformation here Which has the following form and note that all these forms this this garbage here depends on the background fields And it's just linear in the field B itself This one is the same one as it was before Now the other thing is that this psi naught takes values in the isometry algebra It gives the expression in a moment, I think so it is really a background transformation And so the background goes now will only play an ancillary role because they are just the parameters of the background transformation of the isometry group and the isometry transformation act both on the background and on the quantum fields and that was the Obstacle that I mentioned at the beginning of this talk So you see whenever there is a C naught here, it's a background transformation And it's this background transformations. Well, one of them is associated with this one here So all the quantum fields transform under this background transformation And they know of that and everything is internally consistent. So here you have this delta naught this psi naught this speciality The equivariant map How it acts on the quantum fields and you see it really looks like well It looks almost like you know if you forget that there is phi dot plus phi tilde here This is really a rigid transformation It's like a rotation on phi and it's sort of a slightly more complete rotation on C and B and B Capital B is the same so it acts very system systematically as you expect a background transformation to transform And it doesn't interfere with the quantum fields because they're still there the equivariant transformations are still there Now this is an important feature and note that these variation vanish at the boundary if you move Phi tilde to zero which means that you're at the boundary then the variation is also equal to zero So the bonus that you get here is I didn't say anything about boundary conditions But the theory knows about boundary condition by the way I set it up if the quantum field goes back to zero So that I am at the boundary then the variation is also is also zero. That means that it's consistent with the boundary The funny thing was that we didn't realize that as first and then we suddenly noticed and then so ah So it's really working because if this would not work it would make no sense so having established this I can go back to the Functional integral so here's the equivariant integral and now I have only changed made the change here I just wrote Equivariant here and note remember that I changed the transformation also on the equivariant transformation It's not a BRST transformation And if you now go through close inspection of this expression here This expression is the same integral that I had before So the BRST quantized functional integral is the same one that you would intuitively want to write Down when you believe in equivariant Chormology, it's just the same Now you want to check now and that is critical by for the corresponding functional integral is also invariant under the equivariant Variations and that is not trivial in this case because the equivariant map the equivariant Variations they square to the background field and everything transforms under the background So what you get there is the following thing you derive that the equivariant variation of the action here is equal to the background transformation on the integral the spacetime integral over the Over the gauge formula, but we use that the boundary was invariant on the delta So I moved it out of the spacetime integral I say spacetime, but you know, it doesn't have to be spacetime. It can also be a pudding And the right-hand side will in principle contribute when if I in principle when you evaluate the equivariant variation on the functional integral However, you can suppress that by realizing that if the background symmetry is compact then it has to be zero in a compact manifold so you can Use maze these ansatz and then effectively you get the bonus that you have from the RST You can you also use that here and that's why it keeps working but the functional integral is You know term by term the same one as the B and C one of course under certain conditions because I had background conditions And so on I have to take the same condition and the compactness of the manifold is also a requirement of the spacetime Now on the basis of this assumption it follows that the functional integral must be independent of the gauge condition because that still applies because it's the same one and like what's when it can introduce an Delta exact delta equivariant exact affirmation in such a way that it doesn't affect anything in the integral You know the integral will give the same result as before And this is the statement that you need in order to make To take advantage of or to apply localization. This is sort of the most crucial statement there So let me now go back to the integral and having this machinery behind me. Let me now see how this is going to work So I'm going to add a deformation which is Exact under the equivariant transformation Remember because this will keep coming back and it has to satisfy this which means that we will are going to use Make use of this compactness as before So I have this then this is And I'm sorry, and now I assume then that it doesn't it's not going to contribute to the lambda variation and what I do here is that I write this thing down and The functional integral when I take the derivative of perspective lambda takes this form I can move the Equivariant derivative all to the beginning and I can use I can assume that I can use in this situation The sober stokes theorem that means that I did this in differential operator that works consistently on the configurations way So in other words, I can integrate by parts and then it follows that this functional integral is independent of lambda This is nothing to do with what we do that is a basic assumption that you always have to make Otherwise you cannot go on at this point So in that case You are you can add this at this deformation You can move yourself to a very strange point in configuration space where life is maybe very simple But where you don't understand any physics, but you can calculate and you can calculate the functional integral there That's the hallmark of localization So more qualitatively when this is done correctly and you take this thing through extreme point You will take the integral you will discover you will get the value of the integrant on the localization manifold The localization manifold is the manifold. Well, this thing is satisfied When this is yeah here when the deformation is Equivariant, you know vanishes under the very in variation That it's not a point. It can be a point. It can be the localization manifold can be many things and for us It's not that's not really important So more qualitatively when you do this correctly the results given in the value give the value of the locomotion or the localization And if all and then you can extend the flow you can take into account of fluctuations about the manifold the localization manifold and you can do a semi classical approximation and if that semi classical Approximation is in the limit of lambda goes to infinity that means that it is exact and then you have an exact result so the typical Affirmation that you choose is that you Write down an interaction Where you have you take all the fermions of the theory now normally this would be no subtle You just take all the fermions and you take the equivariant variation the fermion and then you take multiplied by an alpha fermion and in the the variation in the the Equivariant variation of this thing is the actual is the complete deformation that you're going to use so that's well at the end be Masonic But here it's slightly more subtle because we had already a constraints on one of the fermions because we have fermion engaged conditions So you have to make sure that you treat you know the gauge condition differently from these fluctuating fields No, there's a slight subtlety. You will see that how that goes in the moment In the lambda to infinity limit. I already said is the critical points of the deformation Of this one here. I worked out the Delta Now the Delta equivalent on this thing here and then they get here This is a bosonic It's something that is quadratic in the bosons because these things for fermions and the equivalent one is the equivariant Supersymmetry variation and this one here is the fermion stuff And I assume that I the manifold the localization manifold So the manifold of all the points that satisfy this condition is going to be bosonic Therefore this one is irrelevant because this is a fermion But this one since it's the square tells you that actually this psi has to be equal to zero So I'm assuming here that the bar on the frame is defined something that is positive definite So that I can make that assumption that they can say that this has to be equal to zero So it means that Delta on psi has to be equal to zero And furthermore as I said already I'm assuming that the locus of this station at point of these points is is bosonic So the localization manifold in this case now is a complicated beast It first of all is the contains all the points which are invariant on the equivalent transformations And these are only the fermions the bar on top of the I means that I'm talking about a fermion But I have to pay attention to the fact that I don't mix them up with the ghosts with the gauge fixing the Fermion a gauge fixing which has to be imposed independently and That is then a manifold and that's a complicated manifold that will contain both the propagate the Quantum fields fight tilde and the ghosts associated with the fermion transformations, and this is something that you know in most cases you don't see So you have a complicated manifold with bosonic ghosts the commuting ghosts and the commuting fields because the equation Delta Q on Delta Delta on Psi gives you a constraint on the bosons into which this psi I bar transforms So the localization many falls involves as I just said the bosonic quantum fields of the original super death D and the bosonic ghost associated with the fermion a gauge transformations and the bosonic multiplier fields and the end of fields will eventually mix with this But at the moment we will not consider them They will come up when you really go to the end of the story So the relevant Lagrangian that you take now the relevant action depends on this lambda this is the classical action and then you here you have the The part that is from the gauge fixing and then here you have this deformation And when you take the limit to lambda then this thing goes to infinity dominates the position of the integral and then you have to make small fluctuations Because the lambda also applies on the interaction and only say the the propagator that you induce and they cancel each other So this is a balanced set of quantum fields with an equal number of fermions and bosons I could have shown you already that at the beginning Something that I did in the BRST the way I dealt with that The difference is the fact that I chose Lagrange multipliers was critical for making sure that you had the same number of fermions and bosons Princess you would not like to take quadratic gauge fixing terms then it would not work So now in order to make clear what happens in this limit you rescale the deformation so you have The fight till this and you localize them You know somewhere on the localization manifold and this T tells you where you are on the localization manifold And then there is a small fluctuation which you take what proportion to one over square root of lambda And you do the same thing with the ghosts Like this and then the anti-ghost and the Lagrange multiplier field you also scale so that they all get the same value Now there is one extremely nice feature, which is one of these features that tells you that you're on the right way You see normally you would say Somebody could already have objected, you know, you you know you have the measure So you rescale the field so you get a factor is that not important? No, it's not important because there is a balance between fermions and bosons So you can if you rescale uniformly then you don't have any problem here. It's completely exact But the second contribution originates from the integral over the fluctuations as I said already so here This is actually not what I did so forget this sentence Since there is a balance between fermions and bosonic fields. We reskill the fluctuation fields So we reskill the fluctuation fields and here there is the one loop determinant coming the one loop the Correction coming from the semi-classical corrections and here is the value of this the classical Lagrangian and Then because this is an embedded surface in the in the full space of quantum fields There is principle a measure here, which is associated with that embedding and which is in principle Calculable so now you can work out the classical contribution of the path integral. I mean the first term was of course very simple that That just contained the classical action and On the on the on the localization manifold and here this is the semi-classical determinant You're only integrated over five till the but about Or yeah, there's the prime here means that it's the actual fluctuation. So there is the fight tilting fluctuation There is the one of the ghost field there are the Lagrange multiplier here and there is the anti-ghost here and they're completely balanced and here you have the Combination of the Deformation the lambda has now completely disappeared and here you have the Standard say for they have pop-up for the quantization and this is now becoming one thing And you have to expand this in these quantum fields only in first order because in first order There's no dependence and the next order fields are down by orders of one of square root of lambda So in the limit that lambda goes to infinity everything is exact and You only well only It's not so simple, but in principle you have to only calculate, you know this expression here The first term was simple because that's just the classical action this thing here where this term here when you expand This is quadratic in the bosonic fluctuations. Of course, there's still background fields all over the place and This thing is quadratic in the fermionic fluctuations So you get a very nice Gaussian integral a super determinant because there are fermions and bosons But because the background in this case is bosonic So you get just the ratio over bosonic and a fermionic determinant Of all these fluctuations and then you have to mix them with these that is a principle That's that is a trivial step, but it is a complication, but it is Nothing wrong with it Now remember that I introduced here the square root of G naught, which I didn't specify where it is But in the semi-classical, you know, it's very kind in the semi-classical world. It cancels anyway I could have written any strange pie there or something and I would still cancel So this is the end of the story. This is the thing that you have to calculate and I Believe and we believe that this is a completely Consistent prescription for doing localization for spaces with a boundary and especially for super gravity So let me come to my conclusions There is enough time So the conclusion is that we think that we have presented a general framework for applying super symmetric localization in this It's super gravity and the first practical application will be presented in the next all talked by in time You know and one condition Let me stress again is that it has to be satisfied that has to be satisfied is that the supersymmetry also must close off shall That means without the need of field equations This ensures the nil-potent beer stick symmetry and therefore a consistent a covariant Coromology. Oh not cosmology Yeah, that's an interesting thing. That's maybe a next subject of study So there may be ways of circumventing this I mean you can have higher ghost interactions as well But things will we get complicated, but you may be lucky because you know that may be Special things to the background ghosts in such a way that certain things that would generically be there would not show up And it's still an organisable something So all the other conditions that I mentioned have to do with localization itself I mean there is no law of nature that every system is Meanable to localization you have to have to create special criteria in order to be able to do this And of course these conditions have to be met separately Okay. Thank you very much