 Okay, so I think it's time to start so we'll have one lecture and one discussion this afternoon So stay more tuned on more on two more days So then we'll be over. Okay. All right. So this afternoon the first the lecture is Diego It's a third part of the person know. Okay. Thanks Okay, so let me Remind you the results the main the main result from the first two lectures. So we consider a Wilson loop in the Fundamental representation for the line and we discovered that this was one and then the Wilson loop Always for the mental representation, but for the circle Was some Laguerre polynomial if you have done the exercise but when n goes to infinity is Vessel function that can also be expanded when lambda goes to infinity as Two square root of pi lambda minus three quarters e to the square root of lambda Okay, and then I've seen some people Doing exercises diligently and then if you have done it you you know What these guys are So the symmetric anti-symmetric representation, okay, I'm not going to write down the result, but We are going to recover it from supergravity Okay, so this is What we had so far so we were equipped with a set of exact results obtained using localization and now the idea is to Recover these results from Holography, so I'm going to start part two of my lectures which is holographic Wilson loops and Okay, so we are in n equal to four super young males and And the dual theory is type to be Strings on edges five process five that string coupling is lambda over four pi n and The tension is Remember this okay, so so this is the dictionary between the parameters and okay So in the end going to infinity limits We have non-interacting strings and on top of that we can take the lambda going to infinity limit in which we have a Supergravity limit Okay, good. So 2.1 is minimal Surfaces, okay, so the question is what is the holographic dual of a Wilson loop and starting from the case In which the representation is the fundamental one. So this is the question we want to start with Okay, so let's remind so what so this is n equal to force of your mills in a is a theory with a joint field so all the Fields the scalars that we consider the gauge field and the fermions that we didn't consider but they're only the joint representation of the gauge group but we We we probe this theory with particles that can be in any representation you want so typically you as I told you you think about Particles in the fundamental representation, but you don't need to do that you can be more generic and You can consider higher representations different representation of these particles coupled to the gauge field and so essentially what we're We are testing we are probing and adjoin theory with other particles in different representations but in This theory we don't really have particles in we don't have fields in the phone for example in the fundamental representation so we have we have to we have to In we have to understand what would be a Heavy source in the fundamental representation in this holographic setup so n equal to force to bring a mills all fields in the adjoint Okay But so we can introduce fundamental representation particles Using the Higgs mechanism. So you use a Higgs thing of membranes think about starting with a stack of and D3 brains or if you want and one n plus one D3 brains Okay, and then you are going to have all possible strings right attached to this to this D3 brains, but now let's do a Higgs thing so let's Separate one of them From the others so we have now and D3 brains and one single D3 brain Far away and so here I still have strings In the stack, but then I'm also going to have macroscopic strings stretching between the two the two different Settings, okay, so This would be like breaking as you on as you and plus one into as you and Times you one Or if you think about this thing as a matrix you are going to have something like this n times n block a One times one block so this would be n plus one times n plus one and then you have two of the diagonal blocks that Transforming the fundamental and anti fundamental representation of this big block Which is sent by n. Okay, so we can call them Like w bosons with some abuse of language, okay So the endpoints of the string are going to be a like fundamental and anti fundamental particles and then And these are going to be heavy because these are Largely separated. Okay, so this is endpoints of this macroscopic string Is going to be heavy Particle in the fundamental should I should write bigger? Okay, not whiter white white is is wet enough Okay, so now so what is the prescription then if I want to compute this holographic with some loops is to Compute a disc well a string amplitude so prescription compute a string amplitude for a string Landing on the with some loop contour that we called see okay, so now now you see that this guy is So this is an heuristic argument of course But I mean you see that this guy is a fundamental Particle now you go through the in your horizon limit you created yes five process five so this is going to be the boundary and now you are going to have this guy propagate attached to the boundary and so it's an heavy particle in the fundamental representation of the boundary and That contour of this particle is going to give you the with some loop Okay, so now we want to compute a string amplitude for a string which has an end point landing on the contour so The expectation value then is going to be something like this so we are going to have An integration of the metric the embeddings the fermions of a string action as a function of the metric embedding and Worship ferments, okay Well I don't know if it is very specific so in fact so that that coupling to the ferments is going to sorry to their skin This is going to appear in a moment. I'm going to go back to this point But so I mean I mean of course it's a string Partition functions, so I have to integrate overall possibilities a priori But of course, I don't know how to do that and then I'm going to take a limit now But okay, so this is a priori what you would expect from this picture what you would expect you have to do from this picture, right? but of course about Bear with me for a moment. Okay, so this is very difficult but We can we can take a limit So we can take lambda going to infinity because now we have a this becomes a subtle point So this other points is going to dominate and this other point is just the a Minimal surface So this becomes the minimum the area of the minimal surface of The string attached to the Wilson loop Okay, so this is a classical string classical string action Which is a minimal? surface ending on the contour Okay, so now the contour Now we so I only brought C But actually we know that it's not just C C and theta i so theta i is an integral part of the contour So so far this is again. It is very very Sketchy, and it's not very precise. I mean what what I mean which minimal surface We see that we have to be more more careful, and this is not precisely the right thing to compute Okay, but anyway, so this is this is the idea so So now you let's think about This is the boundary of the DS This was the circle C, and then you have some minimal surface Which extends in there in the back so because of the gravitational pool the minimal surface is not Lying on the boundary right there is some it's like a catenary a question It goes into into the back But when you have this minimal surface has to be minimal because it's subtle point It has to land on this loop, but okay, so let's try To take this as a working Definition of what I have to do. Okay. Yeah Well after the new rise on limit this essentially disappears you so okay good so You you would think that this this string goes deeply in the infrared now if you take a close contour Somehow this this thing in order to be minimal. It has to close. Yeah, it is It's a little bit realistic. Yes. Yeah, so grab it. So it yes as some gravitational pull toward the center and Just because of the curvature Yes, so this guy doesn't want to stay on the on the boundary the minimal surface You actually want to avoid as much as possible this region because of this is the original high curvature So you want to go you want to go Good so now And actually I should say that so young we seem to be odd. Yeah where is that is Wrote one of the one of the two original papers about this this so one is so young Ray and yee And the other one is my descent Okay, so let me take one correct coordinates and so so the boundary is at z equal to zero and So let me take worship coordinate what I call tau and sigma and okay So this would be a string But somehow that the notion of the contour as to enter in this computation and it enters Through a boundary condition. So x mu at zero and sigma is Little x mu of sigma so the contour C was Parametrized if you remember by x mu and then we had this theta I then Z of zero and sigma is equal to zero and Let me take big capital theta of zero and sigma Just to be Okay, so the contour enters through some some Boundary condition Okay, so now we we understand This contour in this internal space gets interpreted. So theta I of sigma gets now interpreted as a contour On the S5 so in particular Teta I equal to one zero zero zero zero zero is Is a north pole of the S5? So we just have a string which is sitting still at the north pole of the S5 good. So now Okay, so now we can we can essentially Compute this minimal this minimal surface. For example, we can use the namu goto action. So we have detail the sigma square root of a determinant of gamma AB Where gamma AB is the induced matrix. So it's da x mu db x mu Plus da z db z Over z square you could use polio cove action doesn't matter Very good, but now if you if you try to go ahead like this you immediately discover That there are large volume so from the gravity point of view or infrared divergences Coming from regions of small z right because you are computing a minimal surface of Something that lands on the boundary of the space where the curvature diverges Okay, so We need to regularize and also we need to understand what is what is the meaning of this of this divergence And so are we really computing the right object or not? So but anyway, so for the moment, let's regularize. So let's cut The surface at z equal to epsilon So let me cut it here So I got z equal to epsilon and then I'm just going to integrate Everything only up to up to epsilon then I have to do something and then in the end I can read I can take epsilon to two to zero so You can understand the origin of this. I mean you can understand the structure of these divergences in a simple way you just Do an expansion that it's it's in my notes So structure of Divergence essentially, okay, so You can use for example the poliac of action Then you write down the equations of motion plus the Viras Oro Constraint then you expand these fields Close To the boundary So the details are in my notes, but I I'm going to So they're only I'm going to go a little bit faster than in the notes because I want to emphasize another point, which is not in the notes Okay, and then Okay, so you you expand this and you find for example that x mu tau and sigma is equal to x mu of sigma So this is the this is the boundary condition plus tau squared over 2 x dot square sigma Del sigma x dot mu of sigma over x tau cube and so on Z of tau and sigma starts by zero plus tau x dot sigma plus orders Type tau cube and some okay Okay, then you plug this into the action and then you discover that the divergence Goes like the perimeter of the curve the length of the perimeter of the curve Divided by epsilon. Okay, so it's a linear divergence Proportion to the perimeter Okay, then if you want you can You can say okay very good. So let's let's redefine what I mean by minimal surface So the area of the minimal surface You can define it as the limit epsilon that goes to zero of the area Only Computed up to epsilon minus the length of the curve over epsilon. So you can you can give this proposal okay But okay, so let's try to understand this this point a little bit better because essentially we we have been sloppy About boundary conditions So this works and in practice is going to work for what I'm going to to say today, but yes You can think of it as so so you are you are inserting an infinitely heavy object in there in the In your contour so you can think of it as Renormalizing this or subtracting this infinite object, but okay, so the thing is that so I Think this is something that maybe Yanis is going to talk about At least it was in his outline I don't know if he's going to have time to do it, but so the legion transform so we yes So you start from zero you don't have the linear term because essentially you want to avoid the region of high curvature as much As possible, so you go straight up tau squared Plus all plus orders tau cube This is zero the boundary condition plus tau plus order tau cube now there is not there is no No, maybe there is a tau square here. Sorry. I mean, of course, there are two there probably yeah You cannot determine fully the the expansion of course because you have two Integration constants, but it could be tau square Good Okay, so the thing is that we are the boundary So these opens up the possibility of introducing boundary terms and boundary terms are important and Of course, they do not change the equation of motion. So After you find a minimal surface, even if you include a boundary term is going to be the same Equation of motion is going to be the same solution But of course once you have to evaluate the action on shell This is going to change the result of the action on shell. So they change The on shell action and the main point is the following. So a string Landing on a Wilson loop obeys complementary boundary conditions Then a string On a brain. So on a D3 brain we have of course for nine on boundary condition and six Dirichlet boundary condition in the transverse direction for the string landing on the Wilson loop we have six Neumann boundary condition and four Dirichlet boundary conditions so let me give you just a An heuristic way to understand how this switching of boundary Condition comes from so let's start Let's start from it like a D9 brain with a curve in in ten dimensions, so you have a You don't have any brain. You have a space filling brain You have a curve in ten dimension and then the Wilson loop is going to land that this Sorry, the string end point is going to land on this ten dimensional curve. So it's going to be Ten Dirichlet boundary conditions, right instead of having because it has to land on the on this curve So it is ten Dirichlet boundary conditions Without the curve would be ten Neumann boundary conditions, right? So now Do 60 dualities so you end up with a D3 brain and You end up with with with four Dirichlet and six Neumann, which is What I what I wrote here so the six Neumann so the four Dirichlet are going to be the directions parallel to the ideal to the ADS boundary and The six Neumann is going to be the radial direction Plus DS5, okay So essentially what we were doing so far was wrong because we were using a Polyakov action or an ambiguity action which is appropriate For Dirichlet boundary condition, but we have something which obeys different point boundary conditions So we have to do a legion transform to compensate for for for this so the Nambu Gato action It is appropriate For so it depends on the on the on the embedding coordinates not on the momenta right for D boundary conditions So we need to replace Coordinates with momenta for the directions That switched boundary condition for the for the direction that changed boundary conditions, okay Yes, yes Right, right. Yeah, it has yeah Yes, it is in between yes if the boundary direction is Yes, so you don't Yeah, so you can use Nambu Gato for for both their for both embedding functions in both directions But there is one embedding function, which is would be the radial direction that for which you have to change the boundary condition I'm going to write right right now Okay, so So the string was sitting At a point on S5 So we don't we don't have we don't have to we don't have to do anything about this as the S5 direction so the only coordinate to be replaced By its momentum is the radial one So Z okay, so I need to add a term. I need to define PZ some radial some some normal derivative to the boundary and then I need to Define a new action, which is was the old action With with a with a legion transform, okay And of course now if you if you do that if you do the variation You are going to see that while this other while this guy was a function of Z This guy is going to be a function of PZ. Okay, okay, so after we do that the action the the action is finite and And actually this Yeah, so I'm I'm okay. So yeah, good. I'm So this will be Z PZ right the legion transform But then I it's a boundary term At epsilon that's where I where I cut my surface. I add this and now the is going to be finite Now the second term is going to be divergent, but it's going to be is going to cancel the divergence here in the in the divergence of the action Possibly give you a finite contribution sometimes Okay, but now if you if you consider this legend, so the proposal is then the dual Object the holographic dual to a with a loop is not the minimal area is the legion transform of the minimal area Okay Yanis will give you more detail or not. Okay But so you don't need to add the legion transform is Okay, but it's analogous I guess Yes, so this is this will be Z PZ right legion transform is Z PZ evaluated in this Regulated surface so Z becomes epsilon if you compute so this is of course is D. I mean this is going to introduce He's going to introduce someone over epsilon squared whatever Okay, so the important thing is that so if you if you really expand, so let me actually expand this Say something more. So If you expand this guy, you're going to find that this is there is a divergence one over 2 pi epsilon then the integral ds along the loop and This integral is proportional to x squared minus square root of t theta i theta i Plus finite pieces So you see that in general actually this is this is this doesn't doesn't solve the problem It solves the problem for the supersymmetric loops that I was Introducing in the first So if you remember in the first lecture, we had that a condition for supersymmetry is what the theta i theta i Would be equal to one So when when when when the loop obeys a supersymmetric condition, this is zero that a virgin piece is zero So answering just this question. So you see that there is some this is a more defined definition Which knows about the supersymmetry of the loop? Okay, good, but anyway, so let's let's given this this digression, let's just Take this prescription that I'm subtracting the length of the curve over epsom. Yes Well, I know it from the gauge theory that in the gauge theory the Wilson loop is in general So if you own if you don't have both zones, it has a linear divergence When the when the propagator school line Yeah, it's due to this fine-tuning of the coupling of the fields Okay, good. So now Yes If I add this what a spin Right And Okay, sometimes so yeah, you get this you mean this non-super symmetric with on loops with zeta equal to zero Well, okay, you could do this this this constructions, but so let's let me table that question for Okay, so now So the one half bps Circle and so what we want to do we want to match this thing with the localization result and so let me so let me take So this is the circle then you have this minimal surface on top of it So this is equal to zero. This is the center of the circle. This is growing Z and now let me take Let me call it so I apologize for the first Switching the orthodox the canonical names, but okay Now in Sigma, so let me call it throwing Sigma So then by by by symmetry well my my picture is not very symmetric, but by symmetry you you you know that these sections of the surface are going to be circles, so you can You can write down R cosine of Sigma R sine of Sigma Zero zero and Z R of Sigma is just a function of R So this is this is your answers Okay, you can compute Let let me just let me just Skip I mean this is in the notes. You can compute the number go to action the induced metric the equations of motion And okay, so then you are going to have some equation of motion that you have to that you have to solve with boundary condition that Z of This is R. So Z of one has to be equal to zero So when the radius so this is the radius R so when the radius is equal to 1 of course I mean this this is the radius goes this coordinate R goes like this right So when the radius is one is like maximum possible you are on the boundary So Z of one has to be has to be zero So this is some equation of motion with this boundary condition Okay, so this is a way of doing it you just solve an equation but there is another way of doing it which is a little bit more instructive which is Start from a line and Do an inversion So if you have a solution for the line, you are going to find the solution for the circle very simple very simply so So I got an inversion So this is the operation that I want to do and And so at the boundary we have that x mu of sigma is equal to cosine of sigma sign of sigma and Also, okay, so let me let me remove the circle from the origin. I don't like to have it at the origin So let me translate it up plus one So now if I if I invert This this is going to be cosine of sigma to one plus Sign of sigma one-half Okay, so this inversion that happens at the boundary can be extended to an AdS isometry So in the bike I've got I AdS Which now depends on the big z big x and z and so now this isometry is going to be x squared x squared plus z squared z x squared plus z squared, okay, so this is This is the the inversion of the boundary extends as an AdS isometry Which I can do the same inversion but involving more coordinates Okay, so now what do we have to do we just have to So we take a line So the the the solute the minimal surface for the line is trivial. It's just an infinite sheet Going into AdS so this is the sigma direction, this is the z direction and so x mu of tau and sigma is going to be sigma one-half and Z of tau and sigma is going to be tau Good, so the inversion gives you Big x mu of tau and sigma for sigma one plus four Sigma square plus tau squared one minus four Very good. So actually what I did here. I also translated back So I was removing the circle from the origin and now I do the inversion I move it I move it I had I had the shift in the in the line so the line was not at sigma zero was sigma one-half and I move I move back Just to remove just avoid the origin Okay, so this is a hemisphere so we We We find them is fierce as you would imagine Now note that x square plus z squared is equal to one Then you find that z squared is Equaled to one minus big x squared Which is One minus r squared, okay So this holds the equation of motion that I didn't write down But this this would solve that equation of motion that is in the notes So now we need to we need to compute the on shell action So the on shell action is the integral from 0 to 2 pi the sigma integral from 0 to r at the cutoff of epsilon d r r over One minus r squared three-halves So our epsilon is such that when z equal to epsilon So our epsilon has to be equal to one minus epsilon squared From this equation, right? So then the the minimal Is the limit epsilon going to zero of this a Minus 2 pi over epsilon. So this is the length of the circle divided by epsilon I'm regularizing but doing the lesion transform will give you the same answer So this is the limit Epsilon going to zero to pi Integral between zero one minus epsilon square root Rdr One minus r squared three-halves minus one over epsilon You plug it into what you do this integral and you find minus 2 pi good So now you find that this is So in this limit of lambda going to infinity This is e to minus square root of lambda to pi This was the pre-factor in front of a number go to I mean in the number go to action in the Times minus 2 pi. So you find e to the square root of lambda and these matches Precisely the leading behavior of the Bessel function that I wrote down in the beginning of the lecture okay, okay, so remember so that this we had so the leading behavior was Was like this was square root of 2 over pi Lambda minus three-quarters e to the square root of lambda So we managed very easily to reproduce this e to the square root of lambda Okay, so now of course You would like also to understand the other the other pieces in this in this expansion of the Bessel function, right? and this is going to be the subject of Maybe a couple of talks next next week In which okay, so this is this is this was quite easy to understand now they're producing this thing is it's actually turned out to be a long-standing problem and You know there's progress recently, but we still don't have a really very very, you know, solid understanding of So you can interpret this in terms of zero modes of some residual SL to our symmetry that's why so I didn't tell you but so this this the string as a Worship with induced metric which which is an ADS to So there is some residual SL to our symmetry which are some Generators every generator contributes to some power of lambda or h bar in a sense and okay So you can you can argue Where this comes from, but you know getting getting this number is actually very very difficult anyway, so you're going to learn more about this this Next time okay, so just let also let me give you a comment About the matrix model So if you remember the maximum eigenvalue that we had in the matrix model was lambda over 4 pi squared so so we had a vigner semicircle distribution and This this endpoint here was square root of lambda over 4 pi squared and So you see immediately that So you have This 2 pi so this this is square root of lambda over 2 pi You multiply with this 2 pi and you get square root of lambda So the contribution of the matrix model is the contribution from this point essentially because essentially this distribution is exponentially picked To the to the to the right, okay? So this is this is the point this point is what what gives you this okay good so in the in the next 10 minutes, let me start Exploring something more interesting than the string Well, this is very interesting especially understanding. This is is very very interesting but Let me talk about so-called giant Wilson loops Okay, so I guess I repeated many times that We can take different representations of the Wilson loop because taking different representation gives us new Parameters and access to different directions in which we can explore Holography, so if the exercise you were asked to compute the matrix model in this case in which I've got a Symmetric representation of rank K and anti-symmetric representation as well of rank K So this is a K and SK and now we can take K To be order n Okay So this is still sublady compared To the Vigna distribution or if you want to n equal to force of a to the action of n equal to for superior Mills so we can just Evaluate the integral of the Vigna distribution with this particular insertions and in the exercises you were asked to Introduce some generating function of these representations and and evaluate this this thing But now if we take this So there so we discover that when R is equal to to a single box So we had one string So we are led to think that every box in the young tableau represents one string. So when I have This row or this column I'm taking combinations of K strings Okay, and when I take combinations of the case things interesting things things can happen Okay, so in particular the string that we are considering can grow extra cycles, so this is a known effect in in In a biography where you have Ramon Ramon background forms. This is called Myers effect is a is like a Polarization of single direction. So you have a string that grows extra cycles So let's start to understand what what can happen from symmetry So take the gauge to operator This can be either the line or the circle is going to be the same analysis I mean, it's going to be the same upshot except that the the symmetries are going to be realized differently in the two cases But so we have as you to two slash four Symmetry of the vacuum, but now you introduce a line. So let's think about the line Which is simpler to to to visualize. So you have a line in the vacuum. So this line is going to break as You to two slash four down to a cell to R Times so three Times S of five. So now let's let's understand So this is just as a geometric object. This is what what is going to happen. So you have so these two guys are the remnants of SO 2 comma 4 So you have a line so you are going to have Dilatation on the line Translation on the line and special conformal transformations on the line So which is three generators that produces L2R and then at every point of the around every point of the line You haven't you have a sphere and S2 which is going to be the S3. So this is that The conformal group of the vacuum gets broken down to this by the presence of this geometric object the line and then The SO 6 of the of the our symmetry gets broken down to SO 5 Because if you remember we took this North pole of the sphere. So I'm taking My string to see that the north pole of the sphere. So I I select one point of the sphere and this breaks SO 6 down to SO 5 okay, so in terms of Isometries, this is ADS 2 times S2 times S4 okay And of course, I mean you can do the same analysis for the fermionic generators And you are going to see that actually it's a half EPS operator and it preserves 16 supercharges which are precisely the supercharges in OSP 4 star 4, okay So this is a bosonic subgroup of a supergroup which has the right Super symmetries and all the right bosonic generators as as as that you want Good so now we saw that the fundamental string. So the string which would be one box of a young of a young tableau so the string had ADS 2 Warshets in dios metric So they're having a string means having an SL2R means having a ADS 2 factor but you have an S2 and s4 that you could that you could wrap okay So you immediately see just from symmetries that I mean what else can it be? the string can So the fundamental string corresponding to the fundamental representation Can either grow to a D3 brain? So this was wrapping and ADS 2 This is ADS 2 Times s2 or it could it could grow into a d5 brain wrapping and ADS 2 Times s4 So it can only be d3 or d5. There's no other option, right? Of course now the question is okay. What is this? So is this a k or sk? And then is this a k or sk or maybe I don't know some other representation with k blocks with k k box I don't know what it is So in order to really understand this and it took some time I mean of course there was some intuition in the early days of this analysis But it took some time to really sort things out and you have to do something more More precise for example, you have to consider brain intersections in which have the ND3 brains plus a probe D3 or D5 you write down Really the defect conformal field theories on these intersections when you integrate out fields and you discover Okay, this is equivalent to inserting what in the party integral and then you discover after you do this analysis that Actually, we can answer this question and this is the symmetric representation sk And this is the anti-symmetric representation a k Okay, I cannot really go through the details of how you discovered this But I can give you a piece of evidence that this guy This from the from the holographic point of view I can give you a piece of evidence that this is that this is correct And so then by exclusion I guess you can take the other one to be sk Okay Okay, so we really haven't haven't said much about Okay, so I've got three minutes So just let me tell you Okay, how do you how do you see the string charge in this brain? So right because now you you think about having many strings So let's say think about having Eurystically again think about having this fundamental string landing on the line on the on the circle Now now you add K of them one on top of each other Right, so you know you have yes K strings but of course they can they can interact in very complicated ways and it turns out that The interaction between these things is going to be captured by by by this D3 and if I Brains, right, so why this really happens. I think nobody really Understands so because you have an effective description So this could be something very very complicated. Of course, imagine you have coincident warships with all possible, you know branch cuts or four points of intersection or handles whatever but it turns out that a DBI action captures this picture Okay Okay, so now let me let me just tell you how to see so how to capture there in the last minute or so how to see the Now now they don't really so they are really relevant for the so the multi winds Multi-wound string is for again This was something that took some time to sort out You have to look really carefully about all the subtle points in the in the matrix model to understand What is the difference between the multi wind and symmetric? But anyway, so now is this is just to try to understand how you had this effective description in terms of a deed of a D-brain action of DBI action for something which is a priority is something different It's just multiple strings which are combined tension in order to give you some particular representation of a gauge group Sorry Many people. Okay, so let's see So I am going to tell you the name Yamaguchi, Drucker, Fjall, Gomez, Paserini There's lots of people that work. So in particular associating what what what is what in a in a nice way It's Gomez Paserini, but finding solutions for this thing is Yamaguchi, Drucker, Fjall, and many people have worked on it Okay, so so this this case so we have this parameter k in the in the representation that it has to have a counterpart in holography so now what is k in holography and k is going to be k units of String flux across the relevant Sphere that is going to appear in the in the in the solution Okay, and so in order to understand precisely what I have to do. So let me write the DBI action, which is so the tension of a brain Dp plus 1 sigma determinant of G mu nu plus B mu nu Plus one over the tension of a fundamental string F mu nu So this is a volume gauge field. This is the B field So it's a pullback of the B field and the metric Okay, then expand this Expand this to linear order in the B field and so this thing contains a term, which is the fundamental string D P plus 1 sigma B mu nu pi mu nu where pi mu nu is Dbi Del Dbi Del F mu nu so we need to compare this with a string action Which which has a coupling To the B field which is TF TF 1 then you have I in Euclidean signature B AB Epsilon AB so you want to compare these two things So you you see that the the the the D brain will carry the right amount of string flux if you take The integral over as P minus 1 D P minus 1 sigma Pi AB equal I k over 2 epsilon AB Okay, so this tells you essentially that okay, so in the case of a D3 brain You have to integrate across the S2 for the D5 brains to integrate across the S4 this this Momentum conjugate to the volume Gauge field and and you put it equal to K So this is where K enters holographically K is the charge Associated to the strings that have been dissolved into this D brain Okay, so you understand that P P mu nu or if you want F mu nu has to have only components so P mu nu as components only Along the string directions So there the directions of a worship and okay, so this is where K enters Okay, so now what you have to do essentially is to Just repeat what we have done for the string in this more complicated setup So you have to compute the DBI action There's a vessel mean or term that you have to include there are all these Legend transformations that you have to do and you do them you compute the action of the of the brain on shell Is going to be finite after you include this this I mean actually is going to be fine it by itself the DBI action as a divergence that cancels against the divergence of a Minoterm, but anyway after you include all the legion transforms you recover the result that should have been Your answer from the matrix model point of view and I'm going to do it tomorrow Okay, so tomorrow I'm going to do this I'm going to sketch how you do these computations for the D3 a D5 brain and then I then we are going to see what happens when you Take even larger representations when you instead of having n boxes You have n square boxes. No now you have a back reaction on the geometry and then we are going to do and a nice we're going to talk about nice connections between Seemingly different operators that you could consider like deformations of the line or of the circle brimstrahlung functions with some loops with casps or Latitude with some loops and they are all connected in a very nice way and you you can also use localization in those examples, okay