 Thank you very much. Thank you very much to the organizers. It's a very nice privilege to speak here. I will keep the first lecture quite elementary. In particular, the first hour will be mostly things about Gaussian ensembles and motivations, and we will get slightly more technical about universality in the last half an hour today. So my plan for these lectures is today to talk about some motivations and the so-called local law. Tomorrow we will introduce Dyson-Brahme motion and let's call it just dynamics because we will also look at a bit of eigenvectors. And the last lecture will be about how you may want to combine information about eigenvalues and eigenvectors in terms of universality to prove things about models which are not Minfield. So there has been a lot of nice progress, a lot under the impulse of Erdos and Yao, about universality for Minfield random matrices. In particular, Balint talked about the Gaussian ensembles, but you may wonder whether the local statistics at the edge or in the bulk of the spectrum are the same if your matrix is not made of Gaussians but of Bernoulli, for example. And this is now very well understood. I think what will be more and more challenging in the future of the theory is really to go beyond Minfield. Namely, the models of interest for physicists are when the amount of randomness is much smaller and can you create random matrix statistics with much smaller amount of randomness. So in particular, we will talk about it in the last lecture. So today let's talk a little bit about Gaussian ensembles first. So Balint introduced a Gaussian orthogonal ensemble, GOE, which I will scale in the following manner. The entries H, I, J are distributed like standard Gaussian scale by a factor of square root n, because this is going to be n by n symmetric. And you have a slightly different normalization on the diagonal with a square root of 2 over n n01. And because of this slightly different normalization, this ensemble of random matrices is invariant as Balint told you by any orthogonal conjugacy. So for any fixed O, which is in the orthogonal group of size n, O star HO has exactly the same distribution as H. And this invariance allows you to calculate some racobians in some contexts, and that's why, for example, we will see that the eigenvalue distribution is explicit. Now you may wonder, are there many such examples of random matrices which are invariant by orthogonal conjugacy? So here is a statement. If you take n by n symmetric matrices with the following requirement, all entries are independent up to the symmetric condition, of course. And you have invariance by any orthogonal conjugacy, and you want to avoid some degenerate situations like identity. Then only GOE up to the scaling is going to satisfy these conditions. So this is why this is such a natural ensemble. So what do we know about GOE? In particular, why did I scale it with this square root n? So here is a scaling ID. So as I told you, I will just do some heuristics and simple things first here. If you calculate, for example, the trace of the square of your matrix, this gives you some information about the typical amplitude of eigenvalues. And if we want the limiting spectrum measure to be on a compact set of other one, then it happens that this is a good scaling. So in particular, you can calculate 1 over n times the sum of your lambda i square, trace of h square. You expand your trace, you get the sum of h ij square. Each one of them is going to be of size 1 over n because of my scaling. You have n square of them, but they have an extra factor n here. So altogether, this is another one. So this is another one quantity. So this is a good scaling. Now, because of this orthogonal conjugacy invariance, what else do we know about this GOE? We can actually compute the joint eigenvalue distribution. So my notation for the eigenvalues lambda i here in doing these lectures is always going to be unordering in this way. And the distribution, so the law of my vector lambda is exactly the following one. You have a repulsive part product of lambda i minus lambda j times exponential minus n over 4 sum of lambda i square. And I will not pretend it's just an exercise, but I will not prove it. And this can be seen in any good book on random matrix theory. Okay. So why is this important? Because from these types of expressions, you can actually calculate things like the correlation functions of these points, the global joint eigenvalue density. You can have some phenomenology as well. For example, here one thing that is important and recurrent in our lectures is that it's very unlikely to get lambda i and lambda j close together. And we want to understand from variety of points of view why this is true. Okay. So Balin told you that you have this Wigner semicircle though, namely the rescaled eigenvalue empirical spectral measure converges to the projection of the uniform measure on the unit disk on the line. So 1 over pi square root of this 4 minus x square, okay, which is a well-known thing. Typically, if you make an histogram of the eigenvalues of a very large, actually just go by, just make a simulation of 20 by 20 or something, and you will see that it converges to this type of shape. So Wigner's contribution goes way beyond that. He actually proved that this convergence holds even if you don't have Gaussian entries. Any standard normalized random variable with high enough moments works here. And he did it by the method of moments. So as Balin told you, the moments characterize the limiting spectral measure. And in particular, you can take the traces of high powers of h. You end up with a counting problem like counting some paths and you have some Gaussian moments involved and so on. And you will find out that it converges to the moments of this measure, which are the Catalan numbers. Okay. So again, not an exercise, but not something we will prove either. If you want to understand from a different point of view why this is true, you may forget about the matrix and just look at this measure. And if you write it in a Hamiltonian form, exponential minus h, you will find out that the minimizer of h is exactly in the large n limit, the semicircular law, okay, by some kind of remand sum approximation. So this is one thing to know about this Gaussian ensemble. One object which will be of recurrent interest for us is the still just transform. In other words, the trace of the resolvent. So I will quite often denote g of z to be 1 over h minus z, where z is in the upper half complex plane. And if you take the trace of this, 1 over n times the trace of g of z, denoted as s for stitches, this gives you information about the spectrum. If you know jointly in many z's information about this object. So it's known that for the semicircle distribution, the still just transform as a following expression. So this 1 over n trace of g of z converges as n goes to plus infinity to just what you expect. If I give a name to this semicircle distribution, this is just this number here, okay, and we will not really care about what this is, even though you can just calculate it by some residue theorem argument, okay. Yes, that's my definition for a still just transform. Thank you. Okay, so this is my Gaussian ensemble. This s of z converges to this quantity, which we will call m of z. And I don't care about what m of z here, but I care about the fact that it satisfies one quadratic equation, which is the following one m of z plus its inverse plus z equals zero. And it's not clear at all from what I wrote on the blackboard so far. Okay, so why? So we will just see that in a minute, okay. Now we want to understand the fact that we have repulsion between eigenvalues, even though we don't, imagine you don't know this explicit formula, do you have a heuristics physics type of argument for repulsion between eigenvalues of this ensemble, okay. And this is actually one of the, one of the origins after the statistical aspects of random matrix theory. From a computer science perspective, von Neumann was interested in random matrices at some point. And he just mentioned some qualitative aspect of the, of this repulsion in the following way. So just take a two-by-two matrix. I tell you it's going to be elementary, okay. So the story about universality, all of these quantities we are looking at at some point will become universal. And this was this vision of Eugene Wigner, who wanted to give a model for the energy, the stable energy level of heavy nuclei. And he observed that this energy level has a small discrepancy. They look kind of regularly spaced, but not exactly. So for example the Poisson point process will not be a good model for that. So he just decided to model it by the eigenvalues of a random matrix because this energy level are actually eigenvalues of some operator in describing the quantum system. So, but he started with a two-by-two matrix, really. And what he found as a, as a, as a distribution ends up to be very close to the truth of the large n random matrix, random matrix here. So if you just start with, with your matrix A, for example, imagine you, you have such a matrix here. Okay, my symmetric matrix, what was the eigenvalues are? So here, here we are, this is one half of the trace minus plus or minus one half of this square, okay. So these are my, both of my eigenvalues. Let's call them lambda, or this is bad notation, sorry, lambda one, lambda two. And from there what you can observe first is if you want to create an eigen, a matrix with these two eigenvalues coinciding, what kind of choice do you have? If you, if you want lambda one to be equal to lambda two, well, obviously you need to, to have the sum of these two squares, which is zero. So A needs to be C and B needs to be zero. Okay, very simple observation. But what this tells you is that in these three parameters family, it's those matrices with multiplicity in the spectrum of co-dimension two, not co-dimension one. Okay, so this gives some idea of a repulsion, which appears also in the large n limit. Okay, it's very unlikely too. So it happens that this co-dimension two is true in any dimension. So let's try to understand why. So imagine you have h n, the set of symmetric matrices of size n. And let's call it h n tilde. Those who have some multiplicity in the spectrum. So in particular, lambda one equal lambda two. So let's try to understand as a manifold the dimension of h n tilde. But I'm going to start even easier. Let's try to understand the dimension of h n. So for h n, obviously you have you can choose your entries. It's an n by n matrix. So it's n plus n minus one. So for h n, dimension is going to be this guy. Now imagine I want to choose my matrix with multiple spectrum. So this n, the first n here can be understood as choosing your eigenvalues. So but if I have a multiplicity in multiplicity, I just have n minus one choice. So n minus one and this one is for the spectrum choice. Okay, then this n minus one here could be understood as a dimension for the first eigenvector because it's on a sphere of dimension n minus one. Okay, so this is first eigenvector. But here still it's n minus one. This one is for the eigenvector for my first eigenvalue, say lambda n. Then I need to choose any eigenvector in the orthogonal complement. I have n minus two as a dimension for the next second value and so on. And where do we stop? So we end up with lambda three, for example. So what about the eigenvector for lambda three? For lambda three, I just have a dimension two. But at that point, I have no choice at all for my eigenvector for lambda two. And the reason is no matter which choice I will make, I will end up with the same matrix. Because lambda two equals lambda one, if you take a two by two or two by one matrix, O star times your diagonal times O is going to be identity no matter what. Okay, so it's not two plus one, it's two plus zero plus zero. And this zero plus zero is because of you see for any two by two orthogonal matrix, you have your O star lambda one, lambda two. Okay, because these two are chosen to be the same. Sorry. Yes, it was two plus one plus zero. This plus one needs to be read as plus one plus zero. Yes. That's true. That's true. All right. So now exercise. Do the same for the so-called Hermitian ensemble, not the symmetric one. So you define now the GUE. Instead of my definition of invariance by orthogonal conjugacy, I take a definition of invariance by unitary conjugacy. And I enlarge my space of matrices to Hermitian. So it happens that the entries are Gaussian with some normalization which is chosen so that you have the good invariance property and so on. Okay, same thing. But now what is the co-dimension? You will find that the co-dimension now is actually not two but three. So then the co-dimension of let's call it h n, but for the complex, for the Hermitian ensemble in h n, this is actually three. Exercise. You will start the same way, but something more happened by the end. Now. Is the real co-dimension? Yes. So all of this is quite qualitative. So what did what else did Wigner do? So Wigner's work went beyond that. Okay. He, for example, as the first step beyond that, he said, well now let's take n a and c to be Gaussian. And what is exactly the distribution of this object? Okay. So this is Gaussian plus Gaussian. You take Gaussian squares to the one half. So in particular, this tells you something about the tail which is Gaussian. Okay, so the gap between lambda one and lambda two, which is just this number, is going to be having a Gaussian tail. So if a and b, a, b, and c are Gaussian, lambda one or lambda two minus lambda one has Gaussian tail. And it happens also that you can see that the density of lambda two minus lambda one vanishes at zero because you, you need this being zero. You just see that the density vanishes at zero. So in particular, the density of lambda two minus lambda one is going to look something like this. And this is called the Wigner's harmonies. Gaussian tail and vanishing density at zero. And actually it vanishes linearly here. Okay. If I, if I was doing the same thing for the Hermitian ensemble, because I have more repulsion because of my co-dimension argument, instead of having a linear vanishing, I would have a quadratic one. How accurate was this? So let's go back to the large n dimension and let me state the results about what happens in the microscopic limit for large n dimensions. So remember I have my semicircle distribution. If you consider your eigenvalues which are lambda one, lambda two, so I had my ordering there, up to lambda n, you know thanks to the previous lectures that lambda n is actually very close to two. Okay. And that it should live actually on a scale n to the minus two-thirds. How can you understand this scale n to the minus two-thirds? Because of the quadratic vanishing of the circle here. So let's, let's just talk a bit about the scale. So the scale for lambda n. Assume that you have your lambda i's which are always close to that typical location. So let me define the typical locations as the quantiles of the semicircle distribution. So, so you define it as the integral of your semicircle distribution up to gamma k which is equal say to, so I'm not going to say exactly k over n, but k minus one half over n. The reason being that I want my lambda one and lambda n to have the same rule. Okay. So if you define it this way, then what you really expect is that lambda n is going to be two plus or minus n to the minus two-thirds. Just a calculation with this quadratic vanishing, the derivative and so on. So the good heuristics is the following that if lambda n minus gamma n, if this distance behaves typically like gamma n minus one minus gamma n, then we expect this random variable here xn which has typical size of order n to the minus two-thirds. And it's really something like this. If you look at this density here, this lambda i is actually going to oscillate, be able to oscillate up to its neighbor. But it's not an easy thing to justify this here. Okay. Now let me go a bit to the description of my microscopic limits. So to describe my microscopic limit, I'm going to introduce a generalization of my measure on n points up there with a different inverse temperature. I will introduce an extra parameter. So let mu n beta to be my measure one over z n beta. Okay. So as you can see, I changed the strength of the interaction between my lambda i's. And the reason I did it is because this parameter beta is natural from two perspectives. The first perspective is if I don't consider my Gaussian autogonal ensemble, but my Gaussian unitary ensemble, then my the eigenvalues distribution is going to be exactly this one, but with a new beta equal to, as I will just write. And also because this is a, you can think about it as the equilibrium measure for particles in Coulomb interaction in dimension two, but restricted to a line. But the Coulomb interaction in dimension two is log and restricted to a line. So there are reasons why this is natural. Now, so for beta equal to, this is a spectrum distribution of GOE. So what is known is that for any beta, both in the edge scaling limit or in the bulk scaling limit, there are limiting random variables. Balint will tell much more about it. I will not focus on the description of these random variables. I'm just stating you that they exist. So this is a theorem to which many people participated that for this limit. So this is as n goes to infinity. First you have the bulk gap distribution, which you need to rescale by a factor n in the bulk. So that's lambda j plus 1 minus lambda j, the gap between two successive eigenvalues. And you rescale it by the density of the semicircle distribution. Yes, at lambda i. So that this is a quantity of order one typically of expectation one. And this converges to a limit which will not depend on j in the bulk, which I'm going to call here go down with parameter beta, but this is not historically really correct. For beta equal 1, 2, and 4, these are things which have been found thanks to grand verdict works of Godin, Meta, and Eisen. And the extension to general beta comes back to what Balint tells in in in these lectures. So what is the range for so so first i is j. Thank you. And second, so you take j between some epsilon n and 1 minus epsilon n. This is for any such j. This is what we mean by in the bulk for epsilon fixed. It's for any such j, which is just one j. This is a random variable depending on j. And n goes to infinity. And no matter which deterministic j you have chosen in this range, this you have this convergence. Okay, so this is for the bulk and for the edge. This your lambda j, your lambda n minus 2, n to the 2 3rd converges to this T w, which some people were worried about before. That's T w beta tracing with them with general parameter beta. And I will not describe it just one distribution. Okay, so that's one thing. And this is about the spectrum. And the goal of these lectures will be to prove that for general random matrices, the same types of distribution with beta equal 1 and 2 occur. And I will also talk a little bit about eigenvectors in the lecture. And for eigenvectors, things look much simpler in some sense because, you know, what is universality? This is a very high dimensional system and goes to infinity to n by n matrix. So you need to identify universal laws by restricting to very small dimension things. This is a small dimension one, for example. And for eigenvectors, this is still a high dimensional object. If you want to talk about universality, you can, for example, look at the projection of the coordinates of a eigenvector. And for GOE, this is particularly simple because this is how measure. So any eigenvector is uniformly distributed on the sphere and you know this result which can be attributed to Boral or Levy, that when you project on any deterministic direction, uniform point on the sphere, up to a scaling, it becomes a Gaussian. So for eigenvectors, universality statements are much more simple and you can just derive them. But we will also prove them by the end of the lectures. Now, once I have stated how these distributions occur, I can give a bit of motivation about fields which are completely unrelated to random matrix theory and where they are found as well. And this is really just a side part of my talk and it will be just 15 minutes or something. So historically, one of the main examples is that the Goudin distribution with parameter 2 appears in analytic number theory. So let's call this Mongomer. So it's actually very easy to state what Mongomer's conjecture is. You know that it remains a function which can be defined in two manners. This is a function of importance for describing primes, the set of primes here and there's a sum for n greater or equal to 1. This definition makes sense for any real part of s greater than 1, but there is an analytic extension everywhere to the complex plane except one pole at one. And of fundamental importance is the study of the zeroes of the Riemann zeta function and I assume the Riemann hypothesis here in my statement of Mongomer's conjecture. So let's assume the Riemann hypothesis. So we are here. Here is my critical axis and we assume that the non-trivial zeroes of zeta are aligned on this dotted line here. So here they are. They come in pairs by an obvious symmetry with respect to the relaxes and let me call them as a consequence one half plus or minus i whole n for ordered whole one which are positive. So this is assuming here. So there are things we know very well in particular how many of them are there up to some level. So the number of i such that rho i is between zero and some level t grows like t log t which is a normalization 1 over 2 pi here. This is something that you can see in any good book about the Riemann zeta function. Now what this tells you in particular is that they get more and more packed so if I want to have a statement relating to random matrix theory I want to rescale them to have an average spacing one and I rescale them in the following way as a consequence. Define omega n to be rho n so 1 over 2 pi rho n log rho n. So now the average spacing between these guys becomes 1 and it happens that in the large n limit when you do the histogram it coincides with this random matrix thing but that's a conjecture. What is this for? So the conjecture is as n goes to infinity if you take the average of omega i plus 1 minus omega i so this is a distribution and this converges weakly to your go down with parameter 2 and of course this is related to this Hilbert-Pollier idea that there may be some spectral interpretation of the zeros of zeta but this is very speculative. But the numerics are extremely convincing. This is actually one of the most remarkable parts here. Okay so but this may be seen a bit as a non-universal thing but it's supposed to hold to a very much wider class of L functions and also it happens that these types of discrete analogs, discrete sums converging to go down occurs in much more universal context. For example if you look at the Laplacian on a compact manifold. So this is the so-called Borigas Giannone and Schmitt. So here I'm really trying to exhibit deterministic settings for which these distributions are supposed to occur. So imagine you have a manifold, a compact. Let's take dimension 2 to simplify in my notation. That's m. If you do I endow it with some Riemannian metric and let's call mu the measure associated to my Riemannian metric. And I consider delta which is the Laplace-Baltramie for the Riemannian metric. And you can look at the Helmholtz equation. So you look for diagonalizing the Laplace-Baltramie. Okay and I order my lambda i's again. Maybe I'm sorry I should choose another notation instead of lambda i here. Let's call these guys okay. So you mean closed without boundary? I take it without boundary actually it's our analogs with boundaries. Borigas Giannone and Schmitt conjecture was originally for Beats with Dirichlet boundary condition but I just want to simplify things here. Psyche are just my eigenstates. So now you how many mu k's do you have up to some levels? So these mu k's are ordered. And the Weyl law tells you that the number of mu k up to some level l is equivalent as l goes to plus infinity to some constant depending on the manifold times l. It just grows linearly. The mean one is non-zero the first one. The first eigenvalue. I mean I just okay so both of them are correct. So we now rescale to have an average gap equal to one and what happens is now if I look at one over let's say so I look up to some level l and I look at my gaps between mu k plus one minus mu k. I rescale by cm but I never get up to one and this is for all the sum of all new k's more than l. This converges as l goes to plus infinity to now the Godin distribution but with parameter one. And I must say that this this is not a clean cut conjecture because it depends on the manifold but it's true for typical manifolds. Let's say for most manifolds even though it doesn't mean much. So in other words what I mean by most manifolds is that the classical dynamics just the geodesic type of dynamics are mixing enough. In particular if the classical dynamics and the manifolds are not integrable you cannot predict where the trajectory is going to be in the long term. Then this is supposed to be true. What it means is that any time you have a lot of mixing underlying randomness and so on you supposedly observe the random matrix statistics. This is completely out of reach to the best of my knowledge. Numerics are very common thing so I saw many numerics for for example in the Beard context in the context if you take a cardioid it really fits well if you take a so dozens of them but whether I can tell you it is true no if yeah for stadiums yes the numerics work. All right are there any questions about this this introduction? If eta is not equal to one for two do we have the main question is the representation of normal variables under coefficients? So matrix representations of this beta ensemble type of measure is going to be I'm sure I mentioned by valent in this week so the answer is yes. Now what else do we want to do here? Okay so let me state the results of universality. So what I will prove this week is the following fact I changed my matrix I still keep independent entries but I consider the variance that may change and the fact that they are not Gaussian. So this is a so-called generalized Wigner matrix so I'm still going to denote them H so let's assume that they are centered I'm going to denote the variance of H ij S ij and I will always assume that it's of order one over n between a small and a large constant over n. The reason is that I we only can deal for these results for in mean field models okay I wish I could impose some geometry here on the randomness but this is not the topic in the next two days. And I to have the limiting semicircle distribution I will also assume that for any i the sum of S ij is one we will see that under these assumptions you always end up with a semicircle distribution. That's my variance it's defined just a line above okay so now the theorem is here and this is a theorem for which ideas I like a lot are due to Erdos line and Jao and but many people participated and it would be hard to mention all of them but I will try in doing the proof to to to mention some some key ideas yes yes thank you yes yeah I'm coming to that. So under the assumption of moments that for that the supremum over ij of that for any p if I take my square root n H ij and I assume that it has a finite p moment and as this assumption then you have university so this is easily said in the following manner oh I erased it so so if you take n times a semicircle distribution at lambda j times lambda j minus lambda j this one converges to my godin with parameter one because I'm looking at symmetric random matrices here but the same thing holds for Hermitian ones so that's the first point and the second point is whatever you remember about tracy-wedam is true so that means that your n to the two-third times lambda the greatest eigenvalue lambda n converges in fact in distribution to this tracy-wedam with parameter with parameter one okay so this is the main goal here and if you want to start proving such thing it's really not clear what's the first step right there is no formula you can start with but maybe for example for the last line here when you want to prove convergence in distribution the first step is to prove its right scale okay so so let's talk about scales and let's try to to prove that these things are on the good scale so to talk about the scales I need to mention the local laws and to talk about the local laws let me talk let me tell a little bit about the green function so probably Simone last week told a lot about green functions or maybe I'm completely wrong but I think she did so let's introduce this well let me redefine this h of z g of z is one over h minus z now my h is not g oe anymore it's for example your favorite generalized beginner matrix you can think about plus or minus one entries for example okay the expectation is missing that's true thank you so let's try things to this object to understand the semicircle distribution and let's see why by pushing the methods you can understand that you have the good scales so imagine you want to prove the semicircle distribution um an idea that goes back to past tour I think uh is the following one I have my big matrix if I remove one line and one column what remains is basically the same in the large dimension limit so it looks just should look the same but I know that I have a coherence relation by the short complement formula between both objects so um in such cases either the formula tells you something on trivial or it would be completely useless so but it happens that it gives something highly non trivial okay um so actually let's just let's just do it okay so I have my h11 entry here h1 h1 star and let's call the remaining the remainder of h1 okay and this is just a one size one and n minus one what the short complement tells you is um that your if I define g tilde g1 of z to be one over h1 minus z um the entry one one of g of z can also be written in the following form okay so um go back to your um undergrad classes short complement linear algebra and this is just okay um now um there are obviously many ways to to prove it so let's talk about the size of these objects here I I have my my semicircle distribution like this and I see it also as a domain of the complex plane and I have my z in the complex plane okay so um my circle is not in the complex plane but anyways um so h11 remember is this is a random variable divided by square root n so compared to this basically nothing um my h1 star g1 h1 this is a quadratic form uh with my um entries independent here if you take the expectation for example you will see that the diagonal the diagonal terms in the quadratic form should be prominent okay so but if the diagonal terms are prominent then this is something like one over n times the trace of g1 so you you may think it's it we should be close to one over n times the trace of g1 this we don't care and this was proved for g11 but of course the same should hold for any distinguished entry I chose first so if I take the average over all of these entries one that I chose first I get something some coherent solution for the trace okay which is nontrivial so hence we expect the following fact that one over n the sum of gi i n is not too far from one over minus e minus the trace of g1 one problem is here I have the trace of g1 but here this is the trace of g so if you believe in this past tour self similarity id both of them should be the same in the large n limit if you want to argue about it uh you can for example use interlacing of the eigenvalues okay by interlacing you have some constraints and you are good so we assume it's correct we assume that this is actually not too far by interlacing of uh not too far from one over n trace of g um so now the quadratic equation you obtain is exactly the same as the one that was characterizing the still just transform of the g o e or du e that I mentioned before right so um hence we have the following fact which this which is that this s of z one over n trace of g of z g1 or g doesn't matter um so you can rewrite your equation as s of z plus one over s of z plus z not too far from zero it happens that you have some stability estimates for this quality equation first of course there are two solutions how do you choose you have some constraints that the imaginary part of s of z needs to be positive for example with my choice of of definition so this gives only one possible definition and once you are dealing with this solution having a good estimate for the error of this term implies a good estimate on s of z so as a consequence what by stability s of z is not too far from what I was calling m of z the still just transform of the semicircle now many things are unclear from a mathematical rigor point of view um in particular this is not too hard to be made rigorous when z has a fixed imaginary part but if the imaginary part of z decreases with n um the types of estimates for this is keeping only the trace they deteriorate a lot so it's not clear you can make this work um so it's a theorem by Erdösch, Jaou and Yen that you can you can get optimal error estimates on any mesoscopic scale and by um running all of this you get the following local laws that I'm going to write now so um how far can you get yes please go ahead what are the end names? I think I introduced it as being the still just transform of the semicircle distribution so that's my uh or maybe I forgot and my apologies if I did that's my definition my question is how far can you get with just you know with some very difficult tricks using this? How far can you get using difficult tricks you said? Not difficult okay well it's difficult for me to judge about the difficulty of what others did that's the first thing but um but the second thing is um that there is one very important trick that was used for getting to uh the optimal rigidity estimate so actually I will comment on on this just in a minute after stating the result so here is the the optimal local law so um here is the definition if I have a sequence of two random variables Xn and Yn I'm going to write Xn dominated in this way by Yn um when the following holds for any small epsilon n big uh d the probability that Xn is greater than n to the epsilon Yn that's more than n to the minus d for large enough n so in particular Yn is really dominated by Xn is really dominated by Yn up to some n to the epsilon factor okay um as n goes to infinity um so after this very short notation here is a the the optimal um local law so it's um it's barely two statements um the first one is here I told you about the trace but we will also be interested in the individual entries and what you have is a gii but not only gii also gjj minus um point z this is bounded by something but this something needs to depend on z because my estimates need to deteriorate when I get to the microscopic scale okay so I need to define the domain for z first so I defined d to be um the set of z which would always be denoted e plus i eta such that um let's say e smaller than 10 and eta is between one and n to the minus one plus some small epsilon okay so I choose a very small epsilon I define this d and for any z in d this is um of order at most one over n eta plus something which deteriorates at the edge and is a actually a one over square root of n eta and the imaginary part of m of z okay so so let's let's just um identify whether this really defines the leading order m of z uh imagine you're in the bulk m of z that the imaginary part of your still just transform at one over n times the sum um that's your limiting still just transform it's of order one m of z when you get to the to the axis actually converges to the density so this is order one and what this tells you is that this is on the diagonal this object of order one and of diagonal something which is close to zero and indeed for eta greater than n to the minus one plus epsilon this these errors are much smaller than order one okay this you can check so this tells you that the as a matrix g the resolvent converges to the diagonal matrix with m on the diagonal but something more happens when you look at the trace you get a bit um you get a better error or estimate because here you see um actually if you're in the bulk the main term is this one because you have a one over square root of an eta which is worse than one over an eta and um but if you use this it will not be sufficient for our estimates later on this is for individual entries but when you make an average you gain something so in particular s of z minus m of z is always bounded by one over an eta okay so the average works better as expected um now to answer your question if you push the method um alapastour trying to go to small scales and so on um getting these estimates for gi i is something very much untriggered you need to do a multi-scale analysis but does not require uh additional incredible tricks to quantify the fact that the average you have an extra cancellation is actually very hard and for this you don't need to do just one short complement you need to look at large moments of this quantity and do quite a few short complements together but this is way too technical for what i want to say here so this is a result that implies the rigidity of the again values and the fact that they are on the good scale this gives the rigidity this would not okay so we we just have a flavor about how to prove these things and i assume these holds i just want to derive one simple consequence two a couple simple consequences before the end here so here is a corollary one so this result i probably mentioned but i'm not sure in the optimal form here was derived by erdos yaw and yin the first corollary is eigenvector delocalization so there is this meta conjecture that when you have random matrix statistics for the spectrum it means that the eigenstates are delocalized namely if you look at them as a as a measure by the l2 norm the this measure gives mass a bit everywhere and the counterpart of this meta conjecture is that when you have a Poisson type of statistics for the eigenvalues you should have localization of the eigenvectors and here what we just confirmed is that indeed we have eigenvector delocalization for these generalized victim matrices in the following sense so if i name so remember that my spectrum is lambda 1 up to lambda n and let's call u1 u n the associated eigenvectors and they are normalized in l2 so then what you have is that the subnorm of any of this uk no matter which one is of order at most 1 over square root n so this is obviously the best you can hope for in terms of polynomial scales right because the sum of the squares sums to 1 okay so each entry is of is at most 1 over square root n is the best you can hope for there the true conjecture that was maximum should be involves some logarithmic factors but in terms of polynomial scales that's that's what it is and how do you derive this this estimate from the local law is very easy to just look at one entry of the resolvent and it tells you something about the eigenvector so is everything clear here about my notations and so on so you you can take the super work case that's also correct the reason it's correct is you know in my definition of domination here my error is so small and descent to the minus d any union bounce we might will do the job okay so um so imagine you're interested in lambda k in uk you want to prove that the subnorm of uk is not too big so you know that your lambda k is somewhere and you're going to choose one point just above lambda k okay at distance n to the minus one as epsilon i assume that my lambda k is in the bulk of the spectrum the proof at the edge is basically the same just adapting the scales and let's look at the resolvent but i want to bound for example the entry number i of uk so i will look at the resolvent the coordinate i i so of course for the resolvent you have this formula when you diagonalize your h which is clear so if you have some bound on g it can tell you something about the eigenvectors but for this you need to choose your z in an appropriate way so let me repeat that i'm choosing my z to be exactly this point so you may say how can you replace the local law with this z which is random because it depends on lambda k because i want to say that my g of z the coordinate i i is of order at most one and so strictly speaking my i stated the local law for fixed z's but again because the the error estimates are so strong you can prove it simultaneously for a grid which is extremely dense and then use as use a standard leap sheets argument for g to prove that it's true uniformly for the in the domain in other words i can yes but what i mean is like i could i could write the soup over z here uh and have bound by one okay that's that's an easy thing to do okay um so in particular i'm allowed to take this z so g of z one one is of order at most one okay so in particular the imaginary part um of the i i entry is of order at most one but what is the imaginary part so that's my sum one over n sum um so i have my um uh so it's um a minus m dot j square plus eta square eta and here i have u um k of i k write what it is that's what you get um so we know that this is of order at most one so in particular which i can value am i interested in i'm i'm interested in the case one so i take j equal lambda k this is a sum of positive numbers so the j equal k factor is also dominated by this so you just take j equal k for j equal k uh my um my e my as lambda k is at zero i chose my e accordingly and my eta over eta square but that's a one over eta so i have one over n eta uh so this is bounded by my n eta so um so i forget my one over n and i get u um k of i square which is bounded by n eta and by my choice of eta this is an n to the epsilon so um so this is true for any epsilon and you're done so you understand the localization properties from the resultant that's basically the only message here okay so the second application i want to mention is this rigidity of eigen values and it will be um quite useful um in the following for us let's go right um okay so what this says is uh if you define let's say k hat to be the minimum between k and n plus one minus k so this k hat is in terms of indices the distance to the edge okay um then your lambda k is going to be a distance from from gamma k remember gamma k is the quantile that's that's where you expect it after after ordering so that's going to be of order n to the minus two-third k hat to the minus one-third so this is uh something quantifying exactly um in term is optimal in an optimal way in terms of polynomial scales where you expect these eigen values okay um so let's let's check um the at least the bulk and the edge scaling limits so if k is in the bulk this is of order n and i have an n to the minus one oscillation meaning that my lambda k is going to be very rigid around this typical location it's basically only allowed to go visit its first neighbors and at the edge you will really get n to the minus two-third all right so how do you prove some such things from from the steady response form you know it's it's a well known fact that the steady response form convergence implies convergence of the measure uh but here we want to quantify so we need to just you know do the job um so one way to do that that's for those five minutes for the proof so you um you want to have you want to count your eigenvalues how many do you have up to some level and have optimal fluctuations and so on so you will find a function which approximates a step function for counting and you would like this step function or approximate step function to be possibly represented thanks to the steady response form by some push integral formula so you cannot do it exactly of course if it's a step function but if you choose this function here let's take let's take your f of so this is a function that looks like this and then you can rescale it here you are between minus two and two you are now considering an energy level e and you want to count how many eigenvalues do i have up to e and if fluctuations are small this implies this guy um so g of z so you introduce this function okay for for your rescaling so this is a function which will be here and oscillating on a scale n to the minus one plus epsilon and it's basically constant except in this window so window size plus epsilon and my e is supposed here to be in the bulk let me just prove it in the bulk the edges but adapting the scales sorry so i want to prove this for fixed k that's right and here i have a fixed e so now my lambda k is not related to e at all anymore i just choose a fixed e if for any fixed e you have a good estimate about how many before and how many after with small fluctuations this implies this guy so so now i need to be careful with my with my scales so what you can write is that your average of so you want to count how many eigenvalues so the sum of g of lambda i and what you hope um is that this is going to be very close to n times the integral of g let's go back to semicircle but now the cushy formula tells you because this is analytic there are some poles but we will talk about them after and so i'm going to write it this way on some contour of your g of z times your still just once form minus m of z okay not too hard to find this just take this is of course true without the sum this can be understood as a sum and uh and then it's true but for s of the minus m of z that's exactly the object we have a good bound on and yeah okay so you say your g of lambda so it's just about one over uh so my notation for that's what i do and you submit and you're done and um now this is not true for any control because there are some poles okay but this is true if i choose my control properly and this tells you why you needed the scales because the poles where are they so remember you have your semicircle between minus two and two and your poles are exactly here at these points which are of the following type e plus or minus and to the epsilon over n and i 2k plus one if you have a point of this type then it's a pole so your contour needs to go below that point so you are choosing a contour which will be this way i take 45 degrees here for the slope and i go far and um and remember that this s of z minus m of z is one over n eta i multiply by n it's one over eta but i have no n dependence anymore it's just the integral of one over eta along this contour one over the distance over this contour but one over the distance you just get a log factor when you get close so that's fine you're just with a log factor you don't care if you had the local law so coming back to your question if you had the local law only with one over square root of n eta instead of one over n eta you would just lose too much by this process well one over n eta is really necessary okay if you have the local law with one over square root n eta that corresponds to the same accuracy as a Poisson point process basically the one over n eta makes a difference so i will stop here you can you can finish this proof easily by just checking the whole thing and uh tomorrow we we've completely forgot about this first and we do the dynamics of Dyson Ramosh okay so what does it mean it means um i mean are you wondering about the why is this wording okay so you know that uh if you look at uk of iso squares it defines a measure okay so the sum of these guys is one it could be that you have a just a spike and uh with the whole math there but what this tells you is that it cannot be true it needs to be that all of the entries have a have a size order one over square root n so there is massive error that's why we call it delocalization just nothing complicated