 I'm very happy to be here, I thank the organizers for the invitation. So I will be talking about taking the localization results and trying to learn something from holography about these results. So this is work I've been learning with my friends and collaborator, Jeremías, which is a PhD student who has finished, Diego Correa, who is here, Alberto Faraghi-Leo Vimal, who is a student of Leo and Diego Trancanelli. Okay, so what I will be talking about is taking these localization results for these vessel functions that currently appears in all the talks about Wilson loops. This is a gauge theory result. This is something that you may know nothing about holography and compute. And then Malda comes, tells you that there is a conjecture, and if this exists, then you would like to reproduce this result, a strong coupling from string theory computations. So this is what I will be talking about and telling you about the issue. So the two sides of the, this is, I will be talking most, I will be talking completely in N equal 4, but this is something that can be done in N equal 4 or in N equal 6 where the two sides of the correspondence are precisely defined. So in this way we will see the intricacies that appear when trying to get a number and make it match. So I will be summarizing just a transparency on the Suzy localization, then I will make a summary of the different Wilson loops that appears in the context of these gauge theories. Then Malda comes and tells you if you have something in the gauge theory, what is the relation to string theory? So you have to know the parameters, then expand, and then see what we should do in the string theory side and then what is the outcome of all this that we've been working on. Okay, so exact results in any field theory are rare things. We usually learn perturbative computations, so it's not common to have an exact result in all the parameters of our theory. This is unusual, but if you have enough symmetry, you may be able to compute quantities in an exact form and this is what happens in equal 4 and in equal 6, and mainly there have been two techniques to obtain exact results. These are integrability, which has already been mentioned, and localization. Integrability typically works at the planar level, but there have been recent work by the Derobiere collaborators working beyond the planar. These are like two situations where integrability you can compute for any operator, but you have to be at the leading order in N, but localization allows you to compute for any N, but only for sushi objects, so depending on what you want to do, what kind of exact results you want, you may use one or the other, or maybe combine them, I don't know. But we will be mainly discussing localization with relates to supersymmetry and supersymmetric objects that you can compute by the techniques that you may know because you are experts and have been lectured last week in very nice talks here. So I will start talking about the observables we want to compute. So this is gauge theory, you know nothing about holography. You are given a Lagrangian, you learn a technique to compute the exact things for this Lagrangian, and there are particular observables that can also be computed exactly. Well, in particular, for the N equal 4, there is this Wilso loop called Maldacena Wilso loop. It's the Wilso loop we are used to, that Diego was talking this morning, in the sushi context we have to add some coupling to the scalars in order for this to preserve some supersymmetry. So when you compute the vacuum expectation value of this object is a quantity that depends on say two curves, the curves in the space time where you, the curve of the loop you are drawing in the space time, and also due to the coupling to the scalars you have to tell me along the curve, how does the scalar couples, how does the curve couple to the scalars in the N equal 4? I think we all know, but here you have six scalars. This X prime is here to preserve diphomorphous invariance. So C is the curve on the space time. It can be time-like, space-like, null. You may need to add some eyes here. For example, for a null line, this is supersymmetric without adding these. N is what is called internal space, is how the curve couples to the scalars along the curve. And this can be computed in many representations. I will be talking mostly about the fundamental representation, which is for which this object has a representation in terms of a string as Diego was telling this morning. So, okay, here these are all on the adjoint, but being in the adjoint, they are expanded in generator. The generator can be put in any representation. So this is also a quantity that depends on the representation. Super symmetry demands that then this N, which in principle is something in R6, should be on the five-sphere, and there is a particular relation between the curve and the internal space vector that has to be satisfied in order for this to be supersymmetric. This has been worked on by a lot of people. Nadab has made a lot of work along these lines. Maldacena, Sujon Ray, sorry that I forgot to add him. Well, these guys are here. Okay, so now what is the simplest Wilson loop you can look for, straight line, okay? So you put a straight line and you began doing computations. This was nicely explained last week. You can do perturbative computation. There is a combined propagator. Nadab was talking about this combined propagator this morning for the straight line, this propagator vanishes, so the straight Wilson line has zero corrections and the vacuum expectation value is just one. This is independent of coupling constant, rank of the gauge group or representation. The straight line preserves one half supersymmetry, this is jargon. This means that preserves 16 of the 32 supersymmetry then equal four has. Okay, so perturbative is a rise in the cancellation between gauge and scalar propagators. As Diego Trancanelli was telling us last week, when we pick the circle and the vector coupling to the scalar is a constant, this is related to what Nadab was saying that he put phi 1 only in the coupling, in the autonomy of the Wilson loop he was computing. The Wilson loops is still one half VPS preserves 16, but now the answer is non-trivial and gives you this Laguerre polynomial. This was conjectured originally by Erikson Semenos and Sarembro doing perturbative computations as Diego was talking us last week. The vertices cancel, so they conjectured that the vacuum expect, this operator should be given by a matrix model, Nadab and Gross, they worked at the planar level, Nadab and Gross generalized this to the non-planar for any n, conjectured, obtained this Laguerre for the Gaussian matrix model and then Pestum with the localization technique proved that what Nadab and Gross were conjecturing and Semenos was purissimo. Okay, so once you have this, this is a very nice result because this is exact in lambda and in n, so now you can expand it. Since this is an exact result and now we are having the Maldacena conjecture in our minds, we can expand it at large lambda and we can expand it in large n and try to see what happens. If we expand these Laguerres, we obtain a series, we expand these Laguerres in one over n, we obtain this series, this would be the leading piece, this would be subleading in n, this will be given, say, by the team Pest. Okay, this is one over n, then when I explain how this is related to string wall sheets, oops, was it me? Maybe it's because of the power. Okay, I will stay here in case. No, maybe it has a turn off after two minutes of this. Well, okay, so this will be related to a disk topology, a disk with a handle, a disk with two handles, et cetera. We will be mainly discussing this, that will be the disk topology, which is the leading contribution in the string topology, in the string topology's expansion, so this can be expanded in this, in different topologies and then we can expand this at large n and we obtain this. What we would be aiming to reproduce is these numbers, from a string theory point of view. Okay, so here you have all these. Well, Diego said this last week, this result arises from, and it's similar to what Diego Correa was saying, that the summation of ladder diagrams gives you these expressions. This is a one-half and now you can put an additional parameter on to the Wilson loops and these are these one-quarter VPS, eight suces, and this corresponds to couple, non-trivial, to the scalars around the curve in internal space. So when you put theta zero equal to zero, everything vanishes and this is a fixed point. When you take a non-trivial theta zero, this is a curve inside the S5. This is a generalized Wilson loop, preserved some supersymmetries, another conjecture in 2006, I think, that the result for this Wilson loop with this particular coupling was just the same as before this, but replacing lambda by lambda cos square theta. This was checked by perturbative computations. Then what Nada was saying this morning, Pestum did some localization also for this Wilson loop, but piece was not completely, as far as I understand, not completely fixed, but if you suppose that this determinant, instant on determinant, were one, everything seemed consistent with this conjecture. So we have the line, the circle one-half, and the circle one-quarter. In particular, there's a particular limit for also for this Wilson loop that amounts to take theta zero to the equator, and if you take theta zero to the equator, you may see that this is going to give you one, and this turns out to be also an interesting limit. Okay, that was gauge theory. Then Malda comes and tells you, okay, if you have n equal to four super young mills, and you have, this should be dual to type 2b, in n equal to 4 you have gauge coupling, range of the gauge group, the natural parameter is the tuft coupling. Why? Because if you make this change of parameters, and you re-express everything in terms of lambda and one over lambda, this naturally decomposes all the quantities you want to compute in this gauge theory in terms of planar and non-planar diagrams, and this naturally maps to the topological expansion of the string. And on the string side, one writes the Polyaco action, one has the Diego was saying this morning, one over alpha prime, you also have the target space metric, the target space metric depends, has an R square because it's the size of the, this comes out of the action, so you get an effective tension being L square over alpha prime, and this is the relation. These are the gauge theory parameters, and they relate to the string theory parameters in this way. So now that I have an exact expression in gauge theory, the natural thing to do is to re-express it in terms of string parameters, see what is the perturbative limit I can take in strings and expand in this appropriate limit and try too much. So then you go to the string theory side, you, the string theory side, as I was saying, you have the target space, this string tension, R square comes out, so the natural thing to do in the string side is to make a semi-classical expansion. The role of H bar is going to be played by this effective tension, so H bar going to zero, this, this is very big, so I have to take large lambda, large lambda will allow to make classical plus fluctuations over classical, over classical worksheet, and this is what we are going to do now. We are going to keep n equal infinity, that amounts to what we are going to say, not to consider handles on the on the on the worksheet, and the remaining piece I have to tell you is what is a Wilson loop from the string theory point of view. The Wilson loop has a curve, as Diego was saying, the worksheet has to end on this curve. So this is the problem you have to solve, and depending on the curve you choose to parallel lines, a cask, a circle, a circle, you have different worksheets in the bulk. I stolen this from many different papers. This is from Valentina Forini, this is from Nadab, this is from the internet, this is from Sarembro, and this is from Eduardo Vesco Vitesis. So in the end what you have to do is compute, in technical words, the discamplet of the type 2 super string with Ramon-Ramon flexing with a precise contour attached to the boundary. So now you have all the pieces. So now take the localization results for the one half, for the one quarter, and if you write them down you will see different pieces. This is the leading, now you're supposed to expand that large lambda because in the large lambda you're expect to be able to reproduce this using string theory computations. You find different terms. This is the leading, sub-leading, sub-sub-leading, sub-sub-sub-leading. If this is h-bar, what is the meaning of all these terms? This is the classical thing. This logarithm we know. We go to Coleman Instantons and he says, when you see a logarithm of h-bar this has to do with zero modes. So this kind of expected to be related to some zero modes and these numbers should be the value of the determinants from a quantum theory boundary view. So now we would like to see how this kind of, we would like to reproduce this minus three halves and this one half log of two pi by computing determinants. Well okay this is what he said. Here the logarithm and this three here of, this precisely, this one half logarithm of square root of lambda, this is precisely amounts, if you go to all the numerology, this amounts to having three zero modes for some determinant. Nadab and Drucker suggested that these three was coming from zeros of the Fadeye Popov gauge fixing determinant in the string, in the string diff of gauge fixing. Okay you go to chapter two of Polchenki where he talks about this. So in order not to deal with this what, okay this work I will be talking about this work I've been done with Leo, the collaborators and another group of Italian guys working also on computing, on trying to compute this. In order to forget about the log of lambda and Sarene as well you compute this quotient. If you compute these quotients if this is topological it's not going to be there so you don't care. So if you compute the quotient of these two things and you trust on the fact that this is going to be topological this is the natural thing you would like to reproduce. Okay so you will have a leading, this will be the quotient of the say classical surfaces and this should come from the determinants of the fluctuating of the fluctuating fields of the string action around a classical solution. You mean this? The argument is that the h-bar is outside the action so the logarithms of h-bar the cost theta is coming from something that has to do with the worksheet that if you if you believe in this because the argument was that the Faye Popo, Riemann, Rochtheorem have three for the disk it's h-bar. No it showed up in the when you when you see the expression it came out from there but it's it has to come from the determinant. You will see when we compute now the determinants you will see that it shows up by computing the determinants. It's not solved the problem I will go on to that. Okay so I think that now it I well okay okay so what is the string partition function? This is a string partition function typically you will have a normalizing factor that you don't know much what is. If you expand to second order you will have the classical action, the determinants, the Faye Popo. This is all we have. Oops just for not to vanish. Okay so this is typical quantum field theory perturbation theory. This is all the stuff you have inside. Nadab's argument with with Gros was that this Faye Popo that you go and see what is the form of this operator it's a p-p dagger related to the fixing of the D-Pho plus by then this that they were saying Riemann-Roch disk the disk has three conformal key vectors. These three conformal key vectors of that operator correspond to the three zero mode correspond to the three halves of the logarithm of square root of lambda that I was referring to and this is topological. If I question this with a with a similar topological string I forget. Okay so now I will go to the details. Any question? Because I was asking this the argument it's a fancy argument because when you go to the type 2a in abjm context it doesn't show up to have the same kind of three halves although you have a disk topology. So it's not clear that the argument is true or not. No no he was saying the three halves of the log if I if I trust the three halves of the log why don't you I say that. Okay so now we go to the solution we want to find a solution which is a circle on the S5 a circle on the configuration space this is at the boundary of ADS we write ADS5 process 5 if you are lucky you choose a nice parametrization of ADS we did this this is embedding so the embedding corresponds to setting a place in the string at the u equal to zero surface so this is collapsed u equal to zero this is one this is not there and it's embedding all these hyperbolic slides on ADS5 and then in for the five sphere we put a non-trivial profile on this angle this angle this three sphere is fibered over this angle you plug everything into the machine you find the classical solution this is the solution okay so rho is like a sigma coordinate the angles are tau this is wrapping all the surface and it's also this tau at the same time you are wrapping here the angle on the on the on the ADS you are moving on this on this angle here on the five sphere note that we have put a fixed and we sit at a fixed point inside this three sphere okay due to the homogeneity of the s3 we expect these these angles the point at which we are sitting on the three spheres nothing should depend on that so now we have the string solution this can be shown to to be supersymmetric looking at the supersymmetries of the type to be green schwarz action so now what we have to do is to compute the on shell action ah okay before this well or no at the same time okay so if you have a surface you look at the induced metric on the surface by when you you put it inside the ADS what is the induced metric for this embedding the induced metrics are asymptotic to ADS to this is in correspondence with the symmetries we expect that they will salute should have this can be analyzed also in a in terms of symmetries I won't talk about that we did that in in our papers but the important fact I want to stress is that the induced metric can be written as an ADS to with a conformal factor and this conformal factor as you go to the boundary goes to one so it's like having ADS to the boundary and then a deformation inside a smooth conformal factor which deforms ADS to inside so this is the geometry it's a disc if it's a disc it's it has three conformal green vectors no matter what is the theta zero this theta zero I'm placing here it's just a boundary condition which is this theta zero is the theta zero which is supposed to be related to the gauge theory theta zero which was parameterizing the the circle inside the five sphere we can compute the on shell action put this over here of course if the surface extends all the way to to the boundary it will have infinite area this is very well understood how to regularize this you essentially drop away this number and you end up with this negative area after regularization now you go here e to the action on shell minus minus gives you a plus and you find that the leading contribution matches this thing that you have here okay in some way we have found a worksheet that has the same the same supersymmetries than the gauge theory side you can match the supers you can you can make a matching between the gauge theory supersymmetries and the bulk supersymmetries and you can check both present the same supersymmetries it has the same area so now we would like to do the sub leading computation this is one quarter the the generic ones louder no they if you take a tita set equal to zero no they are measured with this what do you mean the half bps in this in this drawing means a point here above it's a point and it doesn't move so it's a mu and five six just a single one without moving it's here you put zero theta is constant all the way in row doesn't change tita zero since since i am comparing between the okay okay okay just just to have an intuition on on these geometries as samir was saying the one half bps corresponds to tt igual a zero the geometry is precisely a ds2 when it's precisely a ds2 a ds2 it has sl2r symmetry the three conformal in this particular case are three killin vectors okay if you go to a particular coordinate system you can you can draw them these are the three killin vectors one this is the d tita the angular and these are the other two okay these notice that the these these killin vectors they are killin vectors which their action on the on the worksheet leaves the boundary invariant but the boundary moves okay it points on the boundary to points on the boundary and then the one quarter the particular i was saying the equator the equatorial one quarter when you put this angle to the equator of ds5 well it has this particular thing now only one killin vector survives which is dd5 two conformal killings but something interesting happened in this case and that's why i was mentioning because in this case these classical parameters that you are expected to give from a boundary point of view become zero modes the reason for these parameters to become zero mode is that when you go to the equator when you place the boundary conditions the boundary conditions at the boundary for the for the worksheet on the on the internal sphere on the on the internal space you see that if theta is equal to pi over two these cos cancels so you don't have to give the boundary conditions for this is similar to the king in one plus one you put going to the two to the two back you are and suddenly the solution a typical solution in a second-order differential equation is i give you two boundary conditions the solution is unique okay except when you have zero modes the king sees you put the two on the recondition and there remains the position of the king this is a similar thing so we expect to have zero modes on the bosonic related to the existence of this feature for tt-walla pi over two and this is a serenbo-quatorial loops let's go to the fluctuations so now we want to expand around this worksheet compute the operators representing the expansion and then compute their determinants okay this is there's a very nice formalism to do these fluctuations are properly defined on tangent space the original the original formalism for this was also developed by nadab, druker and sightling and now you might have to do gauge fixing since we are at second order it can it is kind of irrelevant whether you do an ambugotto or you do poliaco if you do an ambugotto you go to static gauge you look for the faie popo this gives you a trivial faie popo determinant and you end up with eight physical bosons and fermions operators this is oh sorry i should have added here also the italiens i put uh no here i put them okay the italiens the italian group or you can go to a poliaco go to a conformal gauge if you go to a conformal gauge you will find out the faie popo but the faie popo is going to cancel uh this is also uh an assumption which has been worked out you looked at the faie popo determinant and you look at the longitudinal fluctuations and they are exactly the same operator so you say they cancel they cancel but the boundary conditions not being the same doesn't mean that they should cancel so assuming that they cancel you end up with the eight and eight and the operators we found were similar were the same as these italian guys found and then the kappa symmetry fixing gives a trivial contribution so we don't care okay so now we want to compute a functional determinants if you want to compute a functional determinant you need a differential operator boundary conditions and an inner product the inner product is going to say to you which which is the precise operator because once you have fluctuated you you write down this this is ambiguous if i put an m here an arbitrary function here and here so it's not clear what is the operator i am diagonalizing uh until i give you what is the precise measure on which i will uh auto analyze the ahem functions good so there's an important recurring theorem that appears every time in two dimensions which is this what happens if i change them how these two determinants related by a by rescaling how do they relate okay there is a formula short 79 which is nothing but hand-wavingly the vile anomaly if you interpret parts you may see that this is the typical eluville mode so this is important because um as you may guess in the end we will make this kind of vile conformal transformations and they will vanish because the string is supersymmetric and conformally variant uh okay so this is one piece of a ingredient you need to to know in order to compute the determinants because now i'm going to compute the determinants so you go you plug the formalism of fluctuating strings and you end up with a bunch of operators fermionic and bosonic ones fermionic this is the one quarter so it's the deformed ads2 so you you you can write them in this way in this particular induced metric i am assuming that i use i am using as a measure the induced metric on the worksheet so this is the the operators one finds they are kind of la la plus operators with gauge fields on a deformed ad asymptotic ads2 space okay this is the details it is important to say that the gauge field is smooth it collapses smooth at the center of the disk it vanishes in the vps limit these are okay originally the the first guy who nadab i think in in in their paper with gross and sitling computed this operator then martin kruczynski and tirsiu try to evaluate this for the one half vps and the italian sanas were computed this operator for the one quarter and we try to go on and try to compute the determinants another important thing to say that the fermions have chiral and non-chiral masses and the this non-chiral vanishes in the one half vps limit so these operators in the one half vps limit are just a free fermion and ads2 with a chiral mass and la plus operator and ads2 good okay so we would like to reproduce this term by computing the determinants these determinants we can do two things we can scale the induced metric to flat space because in two dimensions every metric is conformally flat so we can write write it as a these operators as if they were in flat space or we can use a set of functions we have done both both tricks this is kruczynski tirsiu the italian group us and this is sarembo recently this year drucker gross cycling buhbinder cycling or any cycling beskovi okay i don't know what this is ah this is us because it has two surnames so i put the second surname in lower case sorry okay so we have let's let's do the first idea to scale the metric to flat space okay so this is induced metric take everything out and write it as a flat and forget about the conformal factor you cannot do this in principle not because the conformal factor i've already showed a formula of shorts that shows that there there should be an additional piece okay but when you when you see the contribution of bosons and fermions everything cancels so yeah i can do it okay bosons and fermions the different contributions coming from this conformal factor goes away how do i compute the determinants when i write them as if they were in flat space well there's a well-known technique coloman did everything in the 70s because if you go to appendix of the uses of instanton this is already said they are how to compute determinants in terms of he doesn't call him call it gelfand yaglom but it's this trick of looking for an homogeneous solution of the operator you want to compute the determinant with particular boundary conditions this is for 1d determinants we are in a 2d situation we cannot use this yes we can if we define the operator as a sum over the Fourier modes okay this is a lot of delicate issues in order to see whether this is convergent or not okay this is okay this is what the italians and us did in in 2016 so we we we cope with this with these things and we ended up with the result we want to obtain plus an additional reminder okay it's important to say that the potential potentially divergence in the Fourier modes cancels and this is what we did look into the sucy multiplet so this is correct so we are having a thing that is worrisome in the end this will turn out to be related to the fact that we have changed topology from this to cylinder this is what recently sarembo realized okay so let's go to the second perspective to do a set of function computation this has a long story Camporese Wucchi in the 90s began studying this set of function on on ads maybe Samir knows better than this because he used it every day then it was used in this bunch of papers with and without agreement in different situations for example in Buchbinder-Seitling and in in this Fourier-Seitling best copy when computing using the heat kernel set of functions since the space is homogeneous you have an integration over heat kernel kk and you have to integrate over x sorry heat kernel k xx and you have an integration over x since the space homogeneous doesn't depend on x so this is the volume the volume typically is put to minus 2 pi although it's infinity the argument is that if you take the the volume traca traca traca traca you find the divergence the divergence this 2 pi was appearing this morning also in in adab's in one of nadab's formula so we decided to work out this in the more difficult or more involved problem of a conformally related geometry to ads with no change of topology so typically we have this kind of look notice this when i wrote them down i this is the the one quarter the one quarter we have already written them in terms of some conformal on ads2 this is pure ads2 with a conformal factor here so these are the typical operators we have where these are just ads connections so our idea was to write to say that the determinants were the determinant in the scaled in the vile scaled operator plot the anomaly now the anomaly is not going to cancel and it's going to be our contribution since we are not going to flat space to set a function we do the same idea logarithm of the 2d written as sum over Fourier modes this is a tricky because now the sum over l is divergent this problem was studied by ketten and duni in 2006 for flat space we follow their their their their ideas they didn't consider gauge field we introduce gauge field typically the set of function i am getting faster because he's saying i have no time so the set of function is determinant as usual is expressed in terms of the derivative of the seta there might be an ambiguity related to the normalization scale of course this is going to cancel you can show by adding all the all the pieces of the contributions typical typical seta function typical seta function expressions for for writing down a determinant they involve a function with simple pose at the location of the singularities euclidean rotation end up with this what is called a just function which is nothing but the phase shift of of of the problem note this resemblance with the gelfanieblon in fact the gelfanieblon can be deduced from these from these expressions we end up with a bunch of expression with with a finite finally a finite expression for the bosons and fermion determinants add get the three halves but maldito reminder and this reminder i told you that forini sightling and bescovie had done a heat kernel computation that was matching to the localisational result in fact our reminder is to an order that forini sightling and bescovie didn't compute it so this is an uneasy situation because we found agreement with what has been done here but we perform these two all orders in tita and tita zero and this disagrees so we don't understand what this is doing there well this is what's embodied but i have no more time so let's finish with a loop boundary conditions link all the function matches fluctuations with different techniques give the same result so it's not a problem of the technique because you can you can see it matching i didn't have time to talk to speak about sarembo but my name is not sarembo and there are a lot of things that are not clear which are because originally we started this project trying to to learn about supersymmetric boundary conditions on eds and this kind of stuff but this didn't work the role of zero modes in equatorial will salute with an index the three quarters and this doesn't match with the type 2a why is the mismatch okay and this is a student of sarembo which ended up with some result saying okay we have this happened last month it would be to have a better understanding thank you