Zato sem nekaj ne možem poživati, da je tvoj dimensičnji kaj, tvoj bljekstrinji kaj. Vse možete vse informacije v zelo, da imam tvoje zelo. Zelo se, da vzelo je vzelo, in tvoje vzelo je bolj vzelo, danes vse. To ne vzelo? Perfect. The last two lecture are, well, you will see, is there are a lot of cancellations, especially the fourth one, but the fourth one will not be covered in this lecture. So, if you want more information, well, you can find in papers and maybe one day I will put the lecture in tech, in latech. Ok, so, most of what I'm going to say today is containing this paper to have a reference. In the notes you will find references to other relevant papers on the subject. So, let's go back to the black hole. Black hole in ages four times seven. Now we have discussed the gravitational picture, so what are these black holes. Remember that there is a family of black hole with three independent magnetic charges and three independent electric charges. Yesterday, we also, and the day before, we also discussed the localization in field theory. So, the partition function corresponding to the index count in the microstate of this black hole is precisely what we discussed yesterday. And now we will apply to the dual of this theory. The dual of this theory is the famous ABJM theory, Arony Bergman, Jeff Ferris, Maldasena, and is a three-dimensional chairman's theory. I will use n equal to two notation. The theory is actually n equal to eight, or n equal to six in general, and in particular, the choice of transimus is n equal to eight. And a very convenient way of encoding the matter information is this quiver picture, where essentially for each node you assign a gauge group. So, the gauge group is u n times u n. And the arrow correspond to matter field, and arrow going from here to there, a i correspond to a field in the fundamental representation, say, of the first group and anti-fundamental of the second. There are two guys here, so i goes from one to two. There are also b fields that goes in the opposite direction, meaning that are anti-fundamental of the first group and fundamental of the second, and there are, again, two of them. There are transimus term that I can put, I put a label here, meaning that the first gauge group has transimus level k, transimus level, and the second group has transimus level minus k. So, if you want a review about ABJM also from the point of view of localization, there are many things that you can do, the three-sphere partition function, I can recommend this nice lecture by Marinho, that contains basics about ABJM, basics about localization, computation of localization, and many interesting things that you may need to know. So, let me tell you briefly, I'm interested in the lography, in the lography in the legend limit. So, the things that are known about this theory, the relevant for us, are the following. So, it has been known from lography since the 90s that the free energy of this theory on S3, essentially you evaluate the classical action in ADS-4, the Euclidean ADS-4, and you discover mapping the parameter nk to the supergravity value that the free energy on S3 should scale like n to the 3 over 2. Actually, in more recent time with localization, we are actually able to compute this, and you will find in the Marinho lectures the computation and the result, and the result of localization on the three-sphere is a different manifold, it's not the one that I'm considering, it's a different type of localization, it's this one even with coefficient, scale like n to the 3 over 2, which is a funny behavior. The theory in general has n equal to 6 supersymmetry. In the special case where k equal 1 or 2, supersymmetry is announced non perturbatively to n equal to 8, and then there is nso8 are symmetry. This k equal 1 is the case that we will consider, so the maximally symmetric case. And in, well, one can be more precise, but roughly speaking, there is a regime where n is much larger than k, then you have a weakly coupled supergravity background, and the supergravity background is just 8 s4 times s7, and k just enter as a quotient of s7. I will consider k equal 1, so that I really am in m theory. It's not that k different from 1 makes a big difference, the computation goes pretty much in the same way, but then I'm not completely sure about black hole, you have quotient, so probably there are quantization conditions to be taken in account more carefully, and we actually never did it. So let's focus on k equal 1, so localization doesn't matter too much. Then for global properties, analysis that should be done. So in k equal 1, everything is perfectly fine, there are no quotient to take and so on. So I wrote this in n equal to two notations, and for the game I'm going to play, I need an asymmetry, because I'm doing a topological test. The full asymmetry is so 8R, we can easily find cartons of algebra that you want to the four that I discussed the first day, the very simple one is the following. Oh sorry, I forgot a crucial, let me, something very crucial, there is a super potential, there is a super potential, which is important, the super potential is quartic, and this very simple is this one, a1, sorry, a2 b1. So there is a super potential that constrain the interaction of the fields. So what's the role of the super potential in this localization business? I was telling you yesterday, it's just to fix the global symmetry, there are symmetries and so on, you will not enter explicitly in the computation, and if I select, when I can select the basis of our symmetry in the following way, the only b1 and b2, I can choose four of them, let me call it r1, r2, r3, r4, so I need to give charge to the chiral field, the constraint is that the r charge of the super potential should be 2, in the standard normalization, it's not normalization, but in the standard normalization, you just want a dark charge of the super potential, it's 2 because the super potential is integrated over the square theta and theta is charge 1. So then it's easy to construct the basis of our symmetries. Let's see that I assign charge 1 to, sorry, charge 2 to a1, 0, 0, 0 to the other, or the other possible charges are just this one. And this makes a sort of democratic basis in the set of all our symmetries. For the twist, for the topological twist, we said yesterday we need to choose a large symmetry, there are many, so you can choose whatever you like and you get a different result, and I will show you how different. Now, so let me write here the index with the ingredients that we introduced yesterday. There is a sum of our magnetic fluxes of the groups. In this case, the lattice of the magnetic fluxes, the gauge group is just z to the n, so you have a collection of vector, I call it m and n tilde, and this vector has components that goes from 1 to n, and are integers. You need for symmetry to divide by the value group. Value group is just a permutation of the entries of the diagonal matrix in u n, so you get a factor and factorial square, and then the structure is an integral of the classical plus one loop contribution that I am going to write in a moment, and there is a slight complication when you have genus greater than 1, you need to take this guy, to take the log, take the determinant of the derivatives of these guys with respect to the variable u and the variable n, and this is to the power of g. So, for example, for the sphere, we are g equal to zero, this is absent. And what is this z classical plus one loop using the rules that we learned yesterday, the only classical contribution comes from the Chang-Simons. I have two groups, so I associate variables in the carton of the gauge group to the two groups, u and u tilde, and the u long is used yesterday in the notation that whenever there is a u, I will spoil it in the variable x. So, I will have two set of variables, x and x tilde, or u and u tilde, as you prefer you write in terms of x and x tilde. And the contribution of the Chang-Simons is just as I wrote yesterday the variable x to the Chang-Simons is k multiplied by the corresponding magnetic flux. So, this is for the first group. The second group is similar, there is x i tilde, the Chang-Simons is opposite, so you put minus k, and you put n tilde. This is the classical contribution. So, this is classical. Then there is the one loop determinant for the gauge field, and the one loop determinant is just a Vandermonde determinant. You saw it in Diego Lecce. So, there is a Vandermonde for the first gauge group. There is a Vandermonde for the second gauge group to some power that depends on the genus of the Riemann surface. And this is the vector contribution. Finally, you have the chiral fields. For each field you have a particular rational function. So, you have four contributions that we are splitting two sets, the A's and the B's that have different gauge representation. And the result is that you just sum over all elements of the representation. A and B are matrices, n times n matrices. So, for each entry of this matrix you put index i and j and you get an expression like this. So, for the, say, this is the A's, the expression is always rational. The A's are fundamental of this group. So, I put x at numerator are anti-fundamental of this group. So, they have opposite charge and I put x still the j at denominator. This take in account the charge and I put a fugacity that counts the global symmetry. Now, we come back to the symmetry in a moment. So, for a moment, just let me write this in this way and I put this guy to some exponent. The exponent yesterday was written rho of m plus some external flux plus 3 minus 1 alpha minus 1. Now, what is rho of m? You need to take in account the representation of your guy. So, if this is a bifundamental guy so xi divided x still the j here you will put mi with a plus is on top minus mj tilde. So, the meaning of putting this rho is just count the weight of the guy. And I'll come back to this point in a moment. So, it's convenient to parameterize these guys with integer.  It will play the role of the magnetic charge of the black hole. We come back in a second. These are the a's, there are two and then there are the b's. So, there are so again let me introduce an index a that goes from 1 to 4. 1 and 2 correspond to the a field. 3 and 4 correspond to the fields b but the structure is pretty much the same. There is this rational function and let me write. So, there is the corresponding fugacity. This time the guy is in the fundamental of the second gauge group so I put x tilde on the top and in the fundamental anti-fundamental of the first group so I put this guy on bottom and here I do something that you can read immediately so this is on the top, I put m tilde i minus mj and I put minus pa the black of this over and again plus 1 minus j the structure is pretty much the same. And there he goes from 1 to 4 over the four guy. So, you see and then you have to integrate over the integration measure is just the xi over 2 pi i xi the i tilde 2 pi i xi tilde xi tilde OK So, what's the point here in order to introduce external parameters the y's and the p I choose to use a sort of democratic point of view to each of the field I assign something so a1 is associated to y1 p1 a2 is associated to y2 p2 b1 is associated to y3 p3 and so on b2 is associated to y4 p4 so this guy looks like four chemical potential for fugacity magnetic charges for the symmetry that I have here but remember that actually the magnetic charges are not totally independent you need to choose to fix a reference one and with that cancel the spin connection so there should be a linear combination of this p that is fixed by supersymmetry and also the y's at this point looks like fugacity so it's like assign symmetry to a1 global symmetry symmetry to a2 b1 and b2 but the index depends on global symmetry not on r symmetry the fugacity for global symmetry and there is a superpotential superpotential I cancel unfortunately is this one so the global symmetry should respect this this superpotential which means that the rotation in a1 rotation in b1, a2 and b2 the sum of all this rotation should be the identity that this you can enforce by saying if I assign a fugacity to this guy to this and this and this the product of this guy is actually constrained to b1 in order that the superpotential is invariant why in the notation that we introduced yesterday is again an exponential precisely as x is the exponential of u y is the same guy for a global symmetry meaning that is e to the i some guy is a Wilson line for the flavor gauge symmetry plus i beta and the real scalar in the flavor symmetry multiplet for sake of simplicity I will call this guy for now on and there are a's here simply delta a so I will just use a letter where delta is a chemical potential but I think that is complacify everything is complacify in this game and what is the meaning of this guy this is really a chemical potential we see yesterday this is the Wilson line of the flavor symmetry but this can be interpreted as trace minus 1 to the f e to the minus beta h e to the i delta f j so delta f enters precisely as a chemical potential in the index it's counting the states according to the electric charge of the states this is really the real part of this guy let me ask what is sigma sigma was mentioning yesterday sigma has always appeared in the Lagrangian multiplied by 5 square psi psi so it's a mass term it's a real mass term for all the fields so what's the role of this guy well it will modify your Hamiltonian so if you start with a quantum mechanic that is as many massless states you're just introducing masses real masses for all fermions and bosons a little bit the Hamiltonian this is an important point because this will give you a regularization if you look at the old paper on Witten index the first basic thing that Witten says take a quantum mechanics with a discrete set of states and the gap between the vacuum and the first one whenever you have a continuous states in the Witten index you need to be careful usually it doesn't work most of the time it doesn't work so the actual effort of this sigma is to regularize my theory give mass to everything so the result is really a Witten index and so what I'm computing is what is typically called a equivalent Witten index so you regularize in this way at the end of the computation we'll send the sigma to zero and I still will get an integer so taking account this I'm regularizing my supposedly existing super conformal quantum mechanics in this particular way which is a natural ways dictated by homomorphicity of the problem ok so I will call this guy just delta for simplicity for the you will see how it enters the fact that these are complex so this for the global symmetries if the product of the guy is one then the sum of the delta i is zero or better and this way you will we will see that in this game there are many multivalued functions logarithms so it's better to keep factor of two pi because otherwise you can get the wrong result because if you just forgot the monodrome of your theory so the constraint is that the sum should be an integer power an integer multiple of two pi this for the oh no the superpotential I could avoid to write it again ok so what's about the the asymmetry well I choose a particular parametrization here so this is what I yesterday call the exponent of each chiral field depends on g-genus it depends on the choice of our symmetry and the point is that in this theory you have many different choice of our symmetry so you can do many different choice many different twist the only constraint is that this guy here let me call it this A because I call A from one to four parametra is my fields the only constraint is that since this A is running over in B1, A2, B2 and they are symmetry and they are charge of the superpotential should be two should be two you can choose whatever you like here with the constraint that the sum over these guys is two in the table there were four possible choices you can also take a linear combination such that the sum is two so I decided to reparametriz writing this which is an integer in this language 1-g-pa where also p are integers and can we do that yes well you just need to solve this question and call pa the guy that appears here since the sum of the r i is equal to 2 this guy will be also constraint how well you can sum over A here you can sum over A here and see what you get so when you sum the array you get factor of two when you sum minus one you have four factor so you get minus four here you will get four times 1-j minus the sum of the pa which means that sum over the pa is equal to 1-j ok in my parametrization that's the constraint that you need to impose on the on the magnetic charges this will be the filter ideal of what we wrote couple of days ago remember that we wrote that in gravity the sum of the p lambda is fixed to something k was one minus one or zero g unfortunately is not the genus notation is a different object but and also obviously this guy is not exactly this is this up to some multiplicative factor ok we will see exactly which one soon so ok so this is for the notation now I have three magnetic charges independent this will correspond to the three independent magnetic charges of the black hole and I can introduce three independent chemical potential for three global symmetry of the theory but remember that the black hole we say couple of days ago depends only on three electric charges not four so we are lucky so we have enough ingredients to match the entropy of the black hole now what you have to do now ok I cancel this which is just notations so the tricky to solve this this model in the large and limit is the following as I was mentioning yesterday the if I focus for example on the magnetic fluxes of the of one of the gauge group you see that they always enter at exponent but linearly so there is always a structure inside the loop integral which is what we wrote yesterday you can write as m i multiplied by something this something as a physical interpretation you can rewrite as the derivative of a twisted super potential of an effective two dimensional theory but it's not really necessary for the moment you can explicit check that you can write in this form for a function w that I'm going to write in a second and the logic yesterday was here there is a complicated contour this Jeffrey Kirwan but luckily enough as an example that I show yesterday that Jeffrey Kirwan choose in this lattice actually half of the lattice say the positive m or the negative m the positive m tilde, the negative m tilde so the trick is the following you use the geometric series to resend this guy here and you will get an integrand that has a complicated numerator and at the dominator you have these guys here the derivative of i the w respect the ui the same for the derivative with respect w tilde is the tilde multiply from 1 to n so the structure of your of your integral when you resend the magnetic flux is simple you have to do a contour integral and the logic is that Jeffrey Kirwan essentially is there to tell you that just to take the poles that you see at the dominator so you need to essentially solve this equation here equal 1 I write only for one gauge group but there is another one for utility and which which has a physical interpretation in terms of beta v aqua of the two-dimensional theory which are equivalent to derivative of this guy equal 2 pi n i where n i is a certain integer again these 2 pi's are overall they are seemingly relevant so if you set it to zero and you choose the wrong determination for your multivalent function you get nothing so it's crucial to keep everything including factor of 2 pi's so the set of the equation that you need to solve is this one and there is the same for utility the variable appear in a democratic way the two groups appear in a democratic way what is w? as I gave you yesterday which is general formula for the two-dimensional supervisor or compute it's not difficult and you can check easily that the form of w and u tilde is the following sum over 1 to then one half I'm setting k equal 1 otherwise k would appear here this is the chairman-simon's contribution the chairman-simon contribution is quadratic in the fields and each of the matter fields contributes this function that I already wrote yesterday which is a polylog and is this one u i minus u tilde j plus delta a and then there is another guy which is minus sum over i j from 1 to n le 2 e to the i u tilde i minus u j plus delta a and here is sorry the sum is a equal this was 3 and 4 this is 1 and 2 and this is a le 2 it's pretty much what I wrote yesterday essentially this is the chairman-simon's contribution each matter field brings le 2 because le 2 is essentially the object that you get in an effective Lagrangian by integrating out a set of kaluzakline model in a circle so each matter field brings le 2 this corresponds to the gauge representation of your guys remember that x is e to the i u i and here there is a minus sign we can also write a formula with a plus because there are nice identities with the le 2 so the le 2 of e to the i u is essentially equal to the le 2 of e to the minus i u plus some polynomial terms so you can use some identity to write in a nice symmetric form and the reason of this minus 1 is that the second gauge group has transimus opposite to the first one so this is the function you need to solve this and the point is that your index here at the end will be a sum overall the possible solution of these equations and once you solve the equation you plug it in this mess you get something you plug there you compute the residue theorem so you need to be careful if you have a denominator it depends on the denominator you need also to take the normalization of the denominator so you do all of this and you will get some expression that is essentially the residue theorem applied to this guy sum over all the solution of this set of equations now we need to these are solutions between you cannot solve it algebraically but there are solutions of this set of equations in principle depends what you call solution in principle solution is a solution of this set of equations theorem you can solve it numerically yes so there is a set of solutions that you cannot find analytically still solutions suppose that you can find analytically you do this but I don't want to do this the logic is the following I want to study the problem in the large and limit so what is the expectation here you have a sum say I want something that scale like the entropy should go like n to the 3 over 2 so very big so you will get a bunch of guys most likely here that goes like e to the n to the 3 over 2 and as usual you use the saddle point someone that will dominate over all the others in the large and limit so what you expect is that here that dominates the sum is the saddle point approximation so you need to find essentially the solution the saddle point of this set of equation in the large and limit so now the point is to find the set of solution of this object well this is a typical matrix model problem suppose that interpret ui and ui tilde as the set of eigenvalue of a matrix is pretty much similar to what Diego did for the Gaussian matrix model you get a set of equation for the eigenvalue then you need to solve in the large and limit in the large and limit there are many techniques to solve this kind of guys the one that you saw with Diego the traditional one is to find resolvent and use analyticity and so on I would like to be able to do that I don't know what the resolvent what the resolvent associated to this set of equation if someone of you find it please tell me because you can do many other things obviously all the tricks that you typically use doesn't work in particular the trick that you Diego use here is not obviously working essentially because in that trick there was a competition between a potential and a Coulomb repulsion here essentially the Coulomb repulsion is not there in the game obviously this guy is not depending on him typically the repulsion come from the van der Mond there is no am here so this is just the contribution of the matter field it's like having a matrix model with only the potential then what you do you need to solve the equation not to use trick like using resolvent so if you find a way of solving this in a more clean way tell me because it could be a very interesting result but luckily enough there are other ways of studying this problem in the large limit and curiously enough this way of solving the problem has already appeared in other model if you look at the marigno leccia you will see that for the matrix model on S3 on the contrary there is a nice resolvent well people were lucky so the matrix model on S3 for ABJM can be mapped to a problem that has been solved in the past where you know precisely the resolvent everything and you can use this trick but in the leccia you will also find other papers in the large and limit by Kleybanov, Pufug, Ferris and collaborators where instead of using the resolvent they use another trick that has been used for three sphere five sphere so it is used quite a lot in this business of localization and the same tricks I'll tell you what it is or let me maybe write one of these equations there are two, let me write one just to see how it looks like when you take the derivative and exponentiate essentially I was telling you yesterday the derivative of a polylog 2 is a polylog 1 polylog 1 is the log is log of 1 minus x so you can transform everything to a rational function so if for j that goes from 1 to n what you have to solve is sorry, no sorry for e that goes from 1 to n what you have to solve is this set of equation after some repackaging you can write this rational set of equation and you need to solve it in equal 1 and you need to solve it in the large and limit this is the set of equation that you need to solve now so what's the trick well essentially what you can do is you take a laptop you write the equation of motion and you run some numerics and you see how the eigenvalues are distributed and what you discover running numerics is that the eigenvalue are distributed in a which is one of the peculiarity of this kind of matrix model the eigenvalues are such that they say the imaginary part of u grows with n so if you look at the solution of the Gaussian matrix model that Diego shows couple of days ago eigenvalue are not scaling with n they are bounded in this case instead i goes and then the trick is the usual one in the large and limit this guy become a continuous variable you have so many so you can introduce a continuous variable a continuous function t such that these guys here are t evaluated in i over n you can do the same for the real part you introduce a continuous function v such that evaluated in i over n is vi v tild is i over n and you can introduce a density of eigenvalues this precisely as in in Diego set of Leccia which you can roughly speaking right in this way with the density is normalized such that the integral over the distribution of eigenvalue is one in this way you typically can write transform the discretize function w in a continuous functional of this variable here t has become a continuous variable t that I used to parameterize some distribution of eigenvalue and all the other guys become functions functional of the density and of the imaginary parts of the fields now the interesting part of the story this is different from other matrix model is that due to the growing of eigenvalues with n this guy here is a local function of the variable actually can be written quite explicitly the channel simons contribution gives s t rho t so one discover that the only relevant thing is v tilde of t which is the difference sorry delta v of t which is the difference between v tilde and v okay essentially you can see using in variance that this is the only relevant object and indeed these guys actually only depend on this delta v which is the difference and there is another term that is a sum over the matter field so there is this sum over a from 1 to 4 of some function let me call it g is very simple we write in a second of the delta plus delta a where plus you need to use plus when a is equal 3 and 4 and minus when a is equal 1 and 2 what is this g it function is a cubic function not I will not do all the computation let me write just to so this is pi 2 u over 3 so it's a cubic function so you see that everything is local and polynomial in all the fields the chemical potential centers here contribution why is that okay I don't have time to do the the logic is the following in the leaf function you have expression like this so this is the argument of the leaf function le2 le2 okay here for sure I will do mistakes so you will correct me so the sum of x to the n divided n to the k is that correct let's see no sorry k sorry this is k I cannot get it right anytime so le n should be this let's check the log le1 is sum over x to the k divided by k and the other are powers so you see that for small value of x this is a power series and it goes to 0 and the point is that you have this guy here with difference between ui and uj but there is a line here so typically if i is greater than j say you have n to the alpha times something positive and you have e to the minus n to the alpha which goes to 0 exponentially so typically when say i is greater than j let me see or j is greater than i in this notation i think j is greater than i this is very small so it's oppressed when on the other hand when i is greater than j you see it's okay to say okay well I'm in trouble the guy blows up and you need to study how the le2 behaves at infinity but you discover I was mentioning before that there are identities on the le2 for example le2 of e to the iu plus le2 of e to the minus iu so it's just a polynomial it's a polynomial I don't need to write quadratic polynomial very simple so that when this guy blows up you can replace with something that goes to 0 plus a polynomial so you just get polynomials if you evenly plug it this behavior in the w that I wrote before you will discover that it soon seems to grows too fast with n but when you do this computer you discover that the nasty part that goes with bigger power than cancel and if you also choose in a clever way this angular ambiguity that I was mentioning before you can see that all the nasty stuff goes away and this is the final result which is local because essentially only the contribution where i is almost j matter this is the computation that you can find in the paper the full computation is not a simple one but it's not particularly complicated mathematically but you just need to be careful about all things, integers this is the final result polynomial and you see that here there are different powers of n so that the point that you choose this alpha in this way you want some competition between the two terms in order to find a solution so it's better that this guy is equal to this one and in this way you find that the correct scaling this you can check numerically is alpha equal one half such that this is n to the 3 over 2 this is n to the 3 over 2 and remember n to the 3 over 2 is precisely the scaling limit that you expect for theory of m2 brains so once you have this this this expression here you can solve it now is a yes so this dependence on the difference between the minus v so from gauge invariance oh gauge invariance because because u and util essentially this is it's getting contribution from i equal j take the u1 associated to i and j v transform with positive charge v tilde transform with negative charge and gauge transformation so whatever is not the sum can be gaged away so this good question since once I did I should know the answer since I did few months ago I should really know the answer there is some subtlety but it cancel you can do the for example the first few terms and you see that even if you keep it the equation of motion will tell you that you have to set them to zero so once you have this expression here you solve it this is set of polynomials so you get algebraic equation you get delta v the raw and look that here there is a factor of t the variable so the result will not a constant will be a function of t and the kind of result that you find is the following as we draw pictures so the behavior of the density is something like this is a piecewise linear function and delta v is slightly more complicated it's not really linear but it goes like 1 over t and it has a behavior like this this is delta v now so what is this what is this well this is what you see from the numerics at a certain point the the delta v becomes constant in the larger limit ok why is that this constant these constants appear precisely when these combinations here are zero or two pi ok this is essentially the remnant of combination like this that appears in this guy here and the point is that I wrote so for example take the polylog 1 that enters in the equation of motion the polylog 1 is something like a log of e to the i util the minus uj plus delta something like that essentially these values here correspond to the branch cuts of this function so the point is that you need to be careful about the termination your computation with multi-valued function you choose one determination and numerics tells you that when you are way far from the cuts everything is smooth and you find a smooth solution when they get value and approach the cut instead of going to the other determination, they freeze and you can check that even analytically that this is true the point is that this particular quantity here near the plateau is suppressing in powers of n to be precise what you discover by solving the analytic equations this is actually non-trivial first time I see something like that in matrix model is that the effect in the large and limit is that you phrase things think by analogy in time conversation certain point the picture that you see in statistical mechanics is aligned for the chemical potential obviously at finite n is not aligned everything is smooth so this is not actually zero but it's very small you might expect that it's 1 over n 1 over n squared and so on at the very beginning we say it's 1 over n 1 over n squared who cares let's throw this stuff away no it's not like that the next order in the expansion and goes exponentially to zero so that you say ok who cares it's exponentially suppressed well in doing this computation to be very careful when you plug this solution back here you have many terms including terms that come from for example here when i is equal to j when you exponentiate those terms you will get something like this minus yi xi over x tilde j delta v is precisely the difference of the arguments of these two guys here and this combination here is parameterizing precisely this guy here so what happens on the plateau is that this guy goes to 1 not to zero logarithm of zero is blowing up so you immediately see that you have pieces in your matrix model that are blowing up on this plateau that's also the reason why there are plateau so typically terms like this you say who cares when you do matrix model you have matrices you have sum over i and j and the typical first thing that you say ok whenever you have a sum over i and j you get something that goes like n square when you have sum over a single variable i you get things that goes like n so typically these are suppressed you typically throw away in any in any matrix model computation but this is tricky because if this difference was going like 1 over n that was true you will get n log n but this dumb guy goes like e to the minus square root of n so this is e to the minus square root of n something when you take the log instead of getting n you will get n from the sum and you will get square root of n from the log and n times square root of n is n to the 3 over 2 so this plateau is quite important in the computation you need to be taking account you need to compute also and once you have everything you go here you plug it back you do some computation you need to be careful otherwise the result is not correct and finally closely enough everything simplifies everything simplifies and the final result is pretty simple the final result is pretty simple so if you take the the solutions of the of this better equation and you plug it back in this functional w you will get something very simple that depends on the chemical potential I have four of them and this is the result very simple function and when you compute when you take the solution you plug it back you compute the determinant also from the determinant you get contribution you do everything carefully log of z also become very simple can be written as a derivative of this function here multiplied by the corresponding magnetic fluxes and sum from 1 to 4 now in the secondary in the last 5 minutes I will compare with the gravity result let me tell you that I cheat a little bit because again there are logs polylogs all multivalued functions so in order to do this computation you need to choose branch cuts and this kind of things actually since the computation you can do with all possible choices of branch cuts but democratic choice of branch cuts where these angles go from 0 to 2 pi the point is that the correct result instead of being this one is this one where this guy is essentially delta A mod 2 pi again there are periodicities so you can restrict to a choice of determination and you have this constraint remember that the sum of delta was should be an integer modulo 2 pi clearly if you choose the guy in between 0 and 2 pi cannot be 0 can be only if the deltas are 0 so this condition here become essentially a condition that this integer part should be either 0 but it's the trivial case where all are 0 can be 2 pi since you have 4 of them you have a couple of choices for this to satisfy this condition that is just the condition that the superpotential is invariant and when you do explicitly the computation you should do the computation all the possible cases to see if there is a solution this is essentially trivial this is trivial here you find no solution here you find a solution delta is 2 pi it also exists here but you will discover that it is not an independent solution because there is a discrete symmetry of the matrix model that maps one into the other so the correct result should be written in this way let me finish comparing the comparing gravity and and filter let me write a table filter on one side gravity on the other so localization localization remember logic is you compute the partition function then 2 pi as we discussed in the very first day is the Legend transform with respect to the electric charges that I call qa of the logarithm of z ok so this you can write explicitly as using the formula that just erased delta 1 delta 2 delta 3 delta 4 pa over delta a ok 10 turns in this way minus i qa delta a and there is a sum again a equal 1 4 so this is the just to find the entropy you need to do the Legend transform so you need to take the derivative with respect to delta set them to zero and plug it back delta ok that's the meaning of Legend transform remember this and you will get this complicated function of p and q I show you an example first day ok but this is the compact expression and this is valid for sum over delta a equal to real part of delta a in zero to pi ok so these are actually the guy I should better call this guy the mod 2 version of delta but let me complicate too much equations with that gravity and it is slightly bigger part of the black board ok maybe you remember this formula I wrote it couple of days ago is the tractor mechanism where g are the fire heliopolis and for our black hole as we discuss all the electric one are constant equal to a particular parameter this fire heliopolis is not the genus sorry for the notation f lambda is the derivative of pre-potential with reference to the sections pre-potential was written 2 days ago is this one then these are the charges that you typically find in supergravity literature if you correctly quantize them using dirac quantization condition you get these expressions here couple of days ago I gave you the expression but essentially couple of days ago I chose a particular normalization for the volume of sigma without telling you now I'm general without choosing this and the correct quantization condition is the following depends on the charges and fire heliopolis p lambda and there is a similar one for the electric charges depends on the volume for g n g lambda q lambda equal 2 pi q lambda square where now p lambda hat and q lambda hat are integer these are the normalization then you also for example open marine lectures and you go after the correct supergravity normalization of the Newton constant and fire heliopolis term fire heliopolis is just a coupling so you will find this information in marine and you discover that is the result and then you plug all the ingredients back here you plug all the ingredients back here and something that you can do this is zero something that you can do very easily because this is homogenous of degree one and this is homogenous of degree one let's define this new variable except which are 2 pi x lambda divided sum over x lambda which have the property that the sum over x lambda is equal to pi definition so you can reabsorb this guy here because the numerator is linear and and if you plug all this information you will find the following lambda from 0 to 3 and to the 3 over 2 divided by 3 2 x 0 x 1 x 3 p at lambda divided x at lambda minus i q lambda x at lambda where the p hat and q hat are the correctly normalized guy now you compare the two side of the black board this one and this and you see immediately that if you send the field theory charges that are integers into the integer supergravity charges and you send a delta a into the sections there are four of them which the two things are manifestly equal so there is no need of computing really the entropy the field theory result is the localization result can I still two minutes to finish so the address are just comments one comment is the following here I write from 1 to 4 these are the letter charges but remember that there are only independent guys so I can find the black philanthropy depending on 3 electric charges but as I told you indeed the black hole only depends on 3 electric charges so the point is that with this field theory computation you cannot tell anything about the fourth electric charge there is in principle a fourth electric charge that you can associate to their symmetry what is the value of the charge for the black hole from field theory I cannot predict from gravity I can well in principle yes in the gravity limit you know that there is precisely one choice of this charge that makes a macroscopic black hole comparing the entropy you don't need this information gravity choose also the other charge so the expectation is that maybe there is another field theory computation not this one, not the topological twist something else with insertion of currents that also compute the value of the fourth charge that you can compare with gravity this is the first comment I am quite below schedule so I cannot comment much about the other result in this set of topic let me just mention few of them and you find more references in the notes so this field theory computation you can generalize to any two dimensional field theory so there are many theories for M2 brain if you want to know what is the corresponding topologically twisted index you use the same trick goes over and you discover a set of interesting things for example this function here that appear in this way which was the value of delta of the twisted superpotential on shell curiously enough is also something that is total looks completely unrelated if you look at the literature people compute in the large and limit the value of the free energy on a street and as a function of r charges of the three dimensional theory and the expression that you find is precisely the same there are reasons maybe we will see some next week in I don't know will let or close at talks there is a unfortunately there are not so many black hole associated with this quivers there is a class of black hole which is a massive type 2a which is as simple as the one in a BJM massive type 2a you have a chairman's theory with UN gauge theory and three a joint and the matching has been explicitly done by Kirill there Osseni, Passe, Fassias is there Francesco Milan what's the name of the no no Catrice Cian it was done last year and matches perfectly went beyond the larger limit there are interesting logarithmic correction to the entropy here is a log of n they were able to extract it numerically from the matrix model and compute the same in gravity there are many other result related to the fact that people see the log of z and start asking the log of z is a free energy I should be able to find not as an entropy but as a on-shell value of the gravitational action after holographic renormalization and indeed you can there are papers by Almagi again Leo Papadimitrio and Kavo Bizet is also here somewhere so this you can check and finally there is a big game also with localization in supergravity you saw the asymptotically flat version in samir lecture localization in for radius black hole has been started by Kirill Val Rice is not here and Ivan Odato they first they computed at least for the moment the classical part of the action and I think they are computing the one loop we wait for this result this would be a quantum gravity result that you in principle you should correspond to the full matrix model not only in the larger limit and finally and with this I stop this game works very well in 4D if you try to do this similar kind of computation in for five dimensional black hole the electrically charged black hole in the first lecture you will discover that the problem is that you compute an index trace of minus 1 to the f is not strictly speaking the entropy and for this kind of black hole minus 1 to the f leads to big cancellation so you get something of order 1 nevertheless it is quite interesting I am not trying to discuss it obviously but you can repackage the entropy of this five dimensional seven dimensional black hole in terms of very simple things so if you ask are they the general transform of something very simple like this function here the answer is yes and what you get is something that looks pretty much coefficient in some anomaly polynomial of the gauge theory so it is not trivial because the entropy is really a mess when it is written in terms of electric charges and rotations but still there is this observation that is a sort of smoking gun that you can rewrite the entropy as the general transform of something that it has a field theory meaning the computation with localization is still missing and clearly if someone of you can find a way of computing from field theory would be a quite nice result sorry for being late and I stop here