 the random matrix part. So again, the idea is to study, to present you another example of extreme statistics for strongly correlated system, which as we will see partly today or maybe a little bit later on Monday, has actually many very nice applications, which is essentially the problem of studying the largest eigenvalue of random matrices. So the largest real eigenvalues is that the matrices have some special symmetry. More generically, the study of the eigenvalues of highest, a lot of largest modulus, if you deal with non-remission, for instance, random matrices. And I started with a simple example to just give you some motivation and show you how this question of largest eigenvalue naturally arises in some simple model. And I chose to present this model introduced by May, which is at the end boils down to this sort of dynamical system. So you have x i's, basically, measures. So you suppose that you have a species of certain density, y, and you say that, OK, I mean, in the absence of any interactions, they actually form a stable system. And that means that their density is typically given by a rho i star, some equilibrium density. And then you want to measure the fluctuations or the distance from this initial density when you slightly perturb the system. Now you make the assumption that in the absence of interactions, essentially, your system is pretty much stable. And that means that all the species, all the densities, will slowly, I mean, not slowly, in fact, exponentially relax to their equilibrium value. Another question that one may ask is what happens if you switch on interaction? So again, we model this interaction by this random matrix because we estimate that the system is quite complex and also quite big enough. And you ask, again, this question, which is what's the probability that the system remains stable as a function of alpha? So we know that if alpha is 0, this probability is just 1. So that's basically this case here. So I plot it as a function of 1 over alpha. And then what I may observe is that if you increase the interaction and when n is large, then suddenly what you see is that there is a rather sudden transition between a stable phase and an unstable phase as you cross the sum critical value Wc. So that means that if alpha is large enough, then with probability 1, your system will be unstable or probability 0, the system will be stable. And this, of course, happens only in the limit when n goes to infinity. And you see that this has the feature of some phase transition. And what you would like to understand is what is this transition, what kind of transition is that? Now, of course, well, to study that, it's not of course, but what I showed you is that there is in fact a very nice connection between this probability of stability and the cumulative probability of the largest eigenvalue of some set of random matrices. So, namely, if you consider this case for this situation, if you consider the case of Gaussian orthogonal matrices, that means basically you just take real symmetric matrices and you fill it with random numbers which are taken from Gaussian distributions. They are IAD, except that they need to, of course, respect this constraint of symmetric matrices, as we discussed yesterday. So, basically, the problem boils down to study the distribution of the largest eigenvalue of this set of random matrices. Now, these GOE type of matrices have been widely studied. I mentioned yesterday in various contexts, starting from the realm of nuclear physics. I will not enter too much in these historical details. That's not my purpose here. But I just first showed you that if you look at the distribution of this matrix, you'll think that now you want to put some probability measure on the set of your matrices, and you can do that easily by just setting some measures on these GIs. Then eventually you get this kind of Gaussian distribution, so that's a sort of Gaussian measure generalized to the case of matrices. I mean, you have this trace of J squared. You have this quadratic term. And it turns out that you can actually write explicitly the joint law of the eigenvalues of this matrix. So I remind you that these matrix being symmetric, real symmetric, its eigenvalues are, of course, real, all real. And you see that you have this explicit expression. So the case that we need to study here, in principle, is the case B corresponds to beta equal to 1. So you have an explicit expression. So you have this product of Gaussians, which essentially comes from this term here. And you have an additional term, which is this Vandermonde term, which actually corresponds to, again, this change of variable when you initially have the set of matrix elements. And when you switch on to eigenvalues and eigenvectors, there will be a Jacobian when you compute the probability measure. And this Jacobian gives rise to this non-trivial term. Now I put a beta here, because the case that we are studying is actually beta equal 1. But there is, in principle, one can study these models for any beta. And in particular, there are three values of beta, which are singular in this context. Beta equal 1, I already discussed it. Beta equal 2 is actually the case of, instead of having real symmetric matrix, you take a complex Hermitian. So you put, again, a matrix which is Hermitian. And you put just real and complex parts as IID Gaussian random variables. You need to respect the constraints that your matrix has to be Hermitian. And then again, because it's Hermitian, you would have real eigenvalues, and that's what you get. You can have other values of beta. I mentioned beta equal 4, which corresponds to these Gaussian symplectic ensemble, which I will not discuss here. Now it turns out that one can also give meaning to any values of beta. But this is also a little bit more technical. And I will not enter too much into the details. But still, I want to focus on this part here. And again, it tells you that you immediately see that because of this term here, you have actually a strong correlations between these lambda i's. So that means that if I had only these product of Gaussians, then obviously there would be IID. But because of this term, you see that they were interacting. And in fact, they were interacting in such a way that because of this term, you see that two eigenvalues don't want to sit very close together. Because you see that if lambda i is very close to lambda j, then this term will be very small. And the probability of such configuration will be extremely small. So this is called level repulsion. So that means that the eigenvalues of such random matrices, they really don't like to sit very much close by. Now, and this is why this is an interacting set of random variables. Now, it turns out that although it, in principle, here comes out from a rather mathematical way, that if you think about this joint law, it has actually a very nice physical interpretation. And it has actually a physical interpretation in terms of what I would like to call here a Coulomb gas. So what is it about? Well, the idea is relatively simple. And basically, this idea is due to Dyson, the same as the Schringer Dyson. So what you can realize is that you see this product here. So let's write it in a slightly different way. So we'll just use this stupid identity, which is that x to the power beta is just exponential of beta log x. I'm sure you will agree with this identity. And so we'll apply it to that. So let's write it this way. So I just have this. So I have a product of such terms here, which with each of them I will write it in this way. Now, the product of exponential is just the exponential of the sum. So I will just write it like this. And I will write it as exponential of beta sum from i less than j log of lambda i minus lambda j minus n over 2 beta sum over i lambda i squared. That fine? So for convenience, I prefer to symmetrize this a little bit. And I will just rewrite it as the sum over all ij's. But then I need to have a factor of 2 here. What I want to emphasize is that I can interpret this joint law as Boltzmann weight associated to a set of particles, which are sitting in a quadratic potential and interacting via this logarithmic potential, repulsive logarithmic potential. So that means that I will write this as exponential of minus beta times some function of the lambda i's. It's because you see here, the standard convention is to have i strictly smaller than j. But I prefer to have the full sum over i and j unrestricted. But then I will count i less than j and i bigger than j. And since they are the same, I will just count it with a factor of half. Why is it so? Why they are the same? i because the absolute value here is symmetric. This guy, I could also write it as lambda j minus lambda i. Is it OK? If you don't like it, I mean, you can just say with that I will not. It's just because that's the standard way I work with that. And so in my notes, it's like this. And because, OK, usually I do something else with that, then it's much better to have a symmetric expression. Today, I will not use that. So if you don't like it, don't use it. OK, so now what is this E? So again, it has the form of a Boltzmann weight. So this beta now, you see, I mean, it plays the role of an inverse temperature. That's why I called it beta. I mean, it's not me who called it beta, by the way. And so what is E? So E, so there are two components, two parts. There is the first part which has to do with the interaction. And then I have a confining potential. I have n over 2. So maybe just to finish to answer your question, I mean, when you really want to do computations with that, it's nice to have a half here because I have a half there. This is just the idea. Just to simplify a bit the calculations. OK, so now you see that I have two parts again. So that means that I can interpret this in the eyes as the positions of some particles. So these are my eigenvalues, principle purely mathematical objects. But in fact, I want to interpret them as the position of some particles. Because of this term here, that means that they live actually in a confining potential, quadratic potential. So they are confined. So that's basically this term here. So it's n over 2 lambda squared. And so for that reason, of course, they would prefer to stay all of them in the center of the trap. But it turns out that because of this level repulsion that I mentioned before, so they don't like to sit close by, there is this logarithmic interaction term. So these particles actually are interacting via repulsive interaction. So it's a logarithmic repulsive interaction. So at the end, this problem of random matrices, which in fact, in principle, is a purely mathematical problem, now becomes a problem of statistical mechanics, which is a problem of studying particles confined in a trap and interacting via long range interactions. Logarithmic repulsive interaction. So that's actually the, and that's how certainly many physicists then have contributed to this field, is that you see that this principle purely mathematical problem have a very nice and direct interpretation in terms of statistical mechanics problem. So now, why Coulomb gas? Well, gas is clear. Just we have a gas of particles. Now, why Coulomb? Why Coulomb? Because this logarithmic potential corresponds to the Coulomb interaction in two dimensions. If you solve the Poisson equation, if you look at the nature of the Coulomb interactions, then you would find, of course, that in 3D, I mean that the interaction is like 1 over r, then if you look at, for the potential is 1 over r, in 2D, this is logarithmic, and in 1D, this is linear. Now here, we have the logarithmic potential. So this logarithmic potential is like the Coulomb interaction. So in other words, that's why it's called Coulomb gas. And now, the more concrete physical realization is that, OK, you take these particles, which in principle would like to live in the plane, and they would then, these are charges. They have all the same sign. You only have plus or only minus charges here. So it's a one component plus, minus, in some sense. And they are repulsively interacting, and they are confined to stay on the line and fill this confining potential. So that's a very nice interpretation, and which then opens the way to use the standard tools of statistical mechanics. Is that fine? Yeah? It's completely classical. Yeah, it's completely classical. So it's a classical system, yeah. Absolutely, so it's a classical picture. Turns out that you can give completely different interpretation of this joint law in terms of a quantum system. In fact, in terms of free fermions, very similar to what Maurizio has been discussing in the early set of these office lectures, but I will not enter too much into that. What I'm saying is that this p-joint can be interpreted as the modular square of some wave function of a multi-particle system. But this is yet something else, which I could discuss if you want, but which is not really the topic of these lectures. So I would prefer. We can discuss in private, but so let's stick to this picture now. Classical picture, indeed. And let's try to understand what is happening. So first, you see that if you think a little bit about these two terms, there is one that really wants to confine all the particles in the center. So if there were only this term here, all the particle in the ground state, or at equilibrium somehow, the state with lowest energy would correspond to having all the particles just sitting at zero in the center of the trap. But on the other hand, because of these interactions, they cannot do that. So there is a competition between these two terms. And as a matter of fact, so maybe one could also consider the other case. That means, suppose that you don't have any confining. If I have only the repulsive interactions, then the particle would like to be very far from each other, and the density will be spread out along all the real lines. So you have a competition between these terms who would like to have all the particles at zero in the middle of the trap. This one would like to have all the particles spread over the whole real line. And as a result of competition between these two terms, what will happen is that the density actually has a finite support. So that means that what we will see is that the competition between, so there is a competition between repulsion and trapping potential between repulsion and the trapping potential essentially will lead to the fact that if you look at the density, so we will see that in a minute. If I look at the density of particles, I will be a little bit more precise in a minute. But if you look at the density of particles as a function of lambda, what you will see is that it's actually confined. So it's symmetric because you see that the Boltzmann Wain is completely symmetric when you change lambda in minus lambda. So the density itself will be symmetric, but there will be a finite support. And in fact, the values I can already tell you, this is square root of 2 and minus square root of 2. And this is related to this critical value that I mentioned yesterday for the Maze model. I told you that the WC is 1 over square root of 2. Well, this square root of 2 is here. We will see why. So this number doesn't really matter. I just want to insist on the fact that the competition of these two things, to these two types of energy, will lead to confinement. So before giving you the explicit form of this, which is quite simple, what we would like to do, so since I'm saying that here there is indeed some confinement, I was already telling you here on this that the typical value is over the 1. I want to show you that. So I want to estimate the typical value of the lambda i. So that means where, I mean, if I look at, so suppose that I have this kind of interactions, then if I look at one of these charge, what is this typical value as a function of n? Is it something that goes with n? Is it something that decreases with n? And if it goes or decreases with which power or how does it behave with n? So that's what I want to estimate. Yeah? Yeah, essentially, yes. Yeah, exactly. So that's what happens, precisely, yeah. So the density is higher there. And if you go there, really, I mean, the density is very small. And it will be also much more sensitive to fluctuations. Yes, that's what happens. So let's try to estimate the typical value of lambda. So to estimate it, I will do the standard heuristic argument. Of course, it's just an estimate. And I will see that at equilibrium, as I already anticipated, these two energies, interactions energy, confining potential energies should be of the same order. So that's how I will basically estimate this term here. So let's try to estimate the potential energy, which is n over 2 sum from 1 to n of lambda i square. Then this, OK, I mean, if I am saying that there is a typical scale for lambda i, then that should be of the order of n over 2 sum from 1 i equal to n. So I will say that this lambda i square is typically lambda typical square. And this is where we have n terms here. So n times n times n, so it means n square. So that means that I have something which scales like n square over 2 times lambda square typical. So that's the first estimate. Now let's do the same for this term here, for the interaction energy. Well, I'm saying that most of them, at least the macroscopic part of it, have the same scale, yes. Of course, not all the spectrum actually have this, not all of them. This is the typical one. So that means that that's what I mean by typical. So most of them, the macroscopic fraction of this eigenvalues have the same scale with them. So again, that means that, OK, if you want to do it more carefully, you will write that lambda i is basically some scale here times mu i, if you want. Now this guy is of order 1. And lambda typical might be whatever you are after. Again, it's a heuristic argument. There is a fully rigorous way to do that. But let's go for a heuristic argument to try to estimate things. You can compute this exactly, if you want. I mean, there are many tools. But let's try to just get the scale. Again, that's how a physicist should try to look at these models first. So if I look at this, then you see that it's almost a dimensional analysis. But this log here, when you have a typical scale here, this log here will be typically order 1. The log is just a number. It doesn't have any scale with n. And so that means that to estimate the scale of it, you need to count the number of terms that you have here. Now the number of terms that you have here is just n minus 1 over 2. And for large n, this is just n squared, or minus n squared over 4. So I can do that because I have a log, right? Suppose that if I had something like this, absolute value, for instance, in one dimension, which would be the Coulomb interaction in one dimension, then, of course, I would say that this typically is over the lambda typical. But because of this log here, this is just simply over the 1. Over the 1, or something which has some logarithmic correction in lambda typical, which at this level don't really matter. So if you balance these two terms here, well, you see that, so now if you say that the potential energy is like the interaction energy, if you assume that this minimization profile results from this competition, which indeed happens, otherwise you would have, as I said, this is the only way to have a non-trivial physics, by the way, otherwise either they are very strongly interacting, and you don't see them, and they are completely widespread. Otherwise they are not interacting at all, and the physics is also a bit trivial. But if you do that, then you see that you need to have n squared over 2, lambda typical square, should be over the n squared over 4. So if you forget about the factors, because of course this will not allow you to get the correct pre-factors here, but what it tells you is that lambda typical is over the 1. So a bit related to what we were discussing last time, when I was mentioning that to get this measure here, you need to have a variance which is proportional to 1 over n. Then you see, of course, that if I suppose that I take my random matrices which have Gaussian with a variance over the 1, then here you would have something over the 1 here. And then of course the scaling will be different, and the scaling will be square root of n. It's clear that this estimate needs to be redone at each time that you are really fixing the scales. They depend pretty much on how the coefficients here, either the confining potential, either the strengths of your interaction do scale with that. But you see that it's fairly standard, and actually it's this heuristic argument. I mean, I don't know any example where this heuristic argument does not predict correctly the scale of the lambda. So that's extremely powerful. So that tells you indeed that this lambda is actually typically over the 1. Now what is the density? Well, the density can be computed. There are many ways to do that. Various methods. One would use this Coulomb gas approach associated coupled to a kind of saddle point calculation. Other methods would involve, for instance, orthogonal polynomials, which are actually extremely useful for random matrices. I will not do that here. I will just show you the result. So the density, let's keep this in mind. So if I want to compute the average density, so what is the average density? So for rho n, so I would define rho n, which would be this quantity here. So I take the average and draw that. So that's the definition of my density, the average density. So now the average that I take here is with respect to this Boltzmann weight. So that's with respect to 1 over z times n. I really compute as I would do in standard statistical mechanics. Now you can almost forget that if you don't like matrices, then you can just think about it in terms of Coulomb gas picture. So you have this computation to do. It's not a completely obvious computation to do, but what happens is that when you do the limit of large n, again, I'm doing the things very qualitatively, not qualitatively, but at least I don't have really time to enter into the details. I'm not doing the course on RNT. I just want to go rather fast to the largest eigenvalues. So what is the limiting form? Well, the density is known under the Wigner semicircle. So it converges to some function, which is independent of n. So that is expected because we already know that the typical scale will be of order 1. So one should expect that this simply goes to something which is well defined as a function of n. And this rho of w, I'm sure you have already seen it, has this form. So remarkably, it has a finite support. And it's given by this semicircle. So this is called the Wigner semicircle. So this is, of course, true for lambda less than square root of 2. And this is the Wigner semicircle. So if you think a little bit about it, it's not completely trivial that it does really a finite support. Because, of course, for finite n, there will be particles everywhere. And only in the large n limit, it becomes completely finite. But one has to, maybe I should comment on that. If you look at how the, sorry. So this is the large n limit. So that gives you the density of particles. But, of course, if you run a simulation, for instance, and if you have, I don't know, 1,000 or 10,000 particles for n large but finite, it's a remark. But the density actually will be more like that. So you have your limit limiting form. So that's the Wigner semicircle. And in fact, what will happen is that, so here I have square root of 2. Here I have minus square root of 2. And if you look at the density, well, here it will be very close to what it is. But then, of course, here it will be rounded up, smeared out. So you will have, so that's the finite n results. Of course, I'm sort of exaggerating a little bit because, in fact, it's really tiny. That's the finite n results. So finite n, what I'm saying is that, of course, there will be some particles which are far away from square root of 2. But the probability that this happens actually goes to 0 when n goes to infinity. But nevertheless, there will be some. So now what I would like to study is the, now I come to the main point of what I want to study is the largest eigenvalue. So that's more or less what you need to know about this RNT, just to understand a bit or appreciate a bit the extreme statistics of this largest eigenvalue. So let's go. And let's try, again, to apply what we know to this rather complicated problem. So what do I mean by that? That OK? So now I come to the largest eigenvalue. Because now that I know the density, there is something that one must probably have to appreciate is that it's the meaning of the density. So the density, you see, OK, I define it as the average density of particles. Now it has also a slightly different meaning, which is suppose that you are looking at these eigenvalues. And if you ask what's the probability to find a particle in Ceylanda, then the PDF of a single eigenvalue is precisely the density. So that's something that one should appreciate is that I will come back to that. So I want to look at this as this guy. Now the remark that I did is that we have seen, I think this is clear. I mean the way it's defined, I mean this is fine. It's the limit for large n of this object. So that's the density. But I can also view it as equivalent, but it's not always very clear to everyone that this is just the PDF of lambda i, of any lambda i. Of course they play an identical role. I mean if you look at the joint law, joint law of the lambda i's is completely symmetric under any permutation of this lambda i's. So they all behave in the same way, statistically. But now if I look at this, well, PDF of any eigenvalue. Let's do it this way. Let's write it in this way. It's a one point sometimes. People would call one point function. So now we know quite a few things. We know that the density here has this kind of variables. So let's, for a moment, forget about we know that they are correlated these variables. Let's try to see what we can get if we forget about the fact that they are correlated. And then just let's try to apply what we know from the theory of IID random variables. So let's, for a while, and let's see how far we can go. And at some point we will see that it fails. But at first glance, the main observations actually can be more or less the main behavior can be obtained using the theory of IID random variable. So let's see how it goes. So there is a first remark that probably should be relatively intuitive, is that if you look at the behavior of this random ax, so of course it will be a random variable, it will fluctuate from one realization to the other of your random matrix. But when n is very large, since the support here is bounded, well, you would not naively expect that this random ax would be very close to square root of 2. So that means that if you take a huge matrix, then typically, random ax will be as close as you want from square root of 2. I mean, as n is large. So that means that random ax, almost surely, will go to square root of 2 when n goes to infinity. That means that with probability 1, when n goes to infinity, if you look at where random ax is, that will be in square root of 2. So that's something that we know for sure. If the eigenvalues here were completely uncorrelated, I mean, this is something that I gave you results at least for that. But we turned out that you can show that. So it's a theorem, and it's not something that is easy to show, but that's a fact. And this fact is, at the end, quite intuitive. Yeah. Yeah, OK. I do not assume that they are ordered. I could do that, but that would lead to something equivalent. I divide by n such that it is normalized to 1. So it's really a PDF. Yeah, that's a good point. Well, I mean, if I don't do that, that will give me n times the PDF of the eigenvalues. Without the sum. Yeah, yeah, yeah, that's true. I could do that. Yeah, absolutely. Yeah. It's just that you're right. I could just do that. Then you are absolutely right. Sorry. Well, the total weight is 1. The integral of this is 1. OK, so yeah. By definition of my density, initially, d lambda of O w of lambda. Yes, you are right. This is 1. This is 1 even for any. Of course, you can check that this is well normalized. Now concerning your question, you are right. The reason why we usually write it in this way is that when you really want to do the computations, it's much better to have an object which is completely symmetric under the different lambda i's. But you are right. I mean, that's, of course, so what he suggests is that this is just equal to delta of lambda minus lambda i for any i. That's basically that's what I wrote here, average. Is that clear? That's probably a good way to see it, indeed. OK, so that's now the first thing. Now, suppose that you would like to know more. And we have seen that in the IID case, knowing just the limiting value. So that would be the equivalent of an, if you want, in our language of standard. So that would be, that means that an is basically equal to square root of 2. So in other words, for finite n, one expects, I mean, this is just one expects. I mean, if not, there will be a leading term, which is an, square root of 2, say. And then there will be some bn times some random variable. I'm not assuming anything here, but it's just that I take the notations that we had before for IID. So now we have seen that this guy is just an. Again, in our previous language. So let's try to say something about this full thing if one assumes that these lambda i's were IID random variables. So you see that if the lambda i's were IID, we are in a case where here we have something which looks like variable. So we would end up in the variable case. Because we have a finite support. And on top of that, the density vanishes as a square root here. So if the lambda i's were IID, essentially chi would be distributed according to some variable type of distribution, which turns out to be completely one. But still, one can still try to say something about the scale bn. So if the lambda i's were IID, so again, that would be that could correspond into the variable case, let's write it. How do I determine bn? Well, bn, remember, I mean, we had a formula for bn, which is the following. So bn is related in this case to basically the typical value. So what is the formula for the typical value? So here I'm bounded that square root of 2. And then bn is such that there is essentially one eigenvalues in this that should be equal to 1 over n. OK, so that would be the formula that you get from IID. That's the variable case. And we have seen that there is a nice formula for it. Is that fine? Now, of course, bn will be, one expects that bn will be very small, so that interval of integration is really very close to square root of 2. So in this region here, I can just replace rho of w. OK, I can do the explicit integral if you want, but I can also just replace this expression here by its asymptotic behavior close to square root of 2. Now close to square root of 2, you see that I want to estimate bn. Again, I know that I am a bit off the realm of IID, but OK, let's try something. I mean, this is the only thing that we can say in principle without going to very heavy calculations. So let's do that. So rho w of lambda, I mean, you see how it behaves here. It's basically, OK, there is a constant. OK, you can still write this like this if you want. So it's just 1 over square root of pi, square root of 2 minus lambda, times square root of 2 plus lambda. So if you look at how it behaves when lambda goes to square root of 2, then this is 2 square root of 2, so you have square root of 2, square root of 2, divided by pi, and times this I don't care. But what is important is this square root behavior. So when you really are close to that, I mean, you have something which vanishes as a square root. So now you can replace this behavior in this integral because we expect bn to be close, I mean, to be very close to 0. So within this interval here, I can replace this guy by its asymptotic expansion. And what you get is the following. Let's just write it explicitly. So that tells you that you have this integral here, square root of 2 minus bn, times square root of 2, and you have d lambda. So you have some constant here. Let's call it, let's call this a. So I have a a, which doesn't matter too much here. And then you have square root of square root of 2 minus lambda, and this is equal to 1 over n. That's this. So now when you do this integral, what you will get is that, OK, there will be some constant that I don't want to compute. But as a function of bn, you immediately see that when you integrate this square root, you will get something to the power 3 by 2. And that gives you bn to the power 3 by 2 is equal to 1 over n. A prime is some number that I don't care of. And immediately, you get that bn is of order n to the power minus 2 sorry. So that's what you would get, and this exponent is quite non-trivial. I mean, people have looked at this exponent for years. It turns out that you can get it very simply here. This 2 third here is related to the kpz 2 third exponent that we have seen yesterday. We will not explain why, but this is the same exponent. So this 2 third here, of course, we obtained it by assuming that the lambda i's are iad. And we use this assumption. But as all the good assumptions or approximations, they hold far beyond the domain of applicability. And it turns out that this, although in principle, it's not very clear as to why I could really apply this kind of formula. It turns out this actually gives me the good answer. That's a very easy, and there is a reason. The reason is that the correlations are not strong enough to break the validity of this. And in fact, in all the matrix models that we know, this is, in fact, in all the models, even of that's probably interesting to emphasize that. But in all the models that we know, even of strongly correlated variables, this formula that allows you to extract the typical value of the maximum always works. I don't know any example. There is no proof of that. But I don't know any examples where this estimate of the typical fluctuations actually fails. So that's also why I started to present this by presenting this formula when I started my lectures on iad random variables. This is probably the most robust approximation that you can get out of this iad random variables. So that means that we know actually now this bn, which is this n to the power 2 third. And now, basically, that's where the iad approximation stops, because if I want to go one step further, step further would be to say, OK, I am. I would be in the viable class. So that would be essentially the viable distribution. And this already is completely one. So that's, we are also now to mention that, that this iad statement or this iad approximation correctly predicts the scale of the fluctuations, but does not predict properly the detailed structure of this random variable. So that's kind of rough approximation, which is not completely off, because it gives you already the good exponent. But if you want to know more, then you have to work more. So the iad approximation correctly predicts, maybe I can just write this here, correctly predicts. So the typical scale of fluctuations, so here this is bn, but in general, this is correctly predicts the scale fluctuations, but not the distribution itself, but not the chi, but not chi. So we don't know what is chi, unfortunately, with that. One reason why we cannot get chi with that is that, of course, we are doing a very strong assumption here. We are saying that the density is bounded by square root of 2. So we are considering the fact where the lambda i's actually cannot exceed the value square root of 2. But I've told you that for finite n, of course, there will be some probability to escape far away from this big notion. So remember, I did this picture here. So this is the n goes to infinity limit. And that's what I played with. I replaced here, I really replaced its value n goes to infinity, but still having a 1 over n on that side, so it's a little bit inconsistent in some way. And plus, the point is that for finite n, we know that, as I said, there will be some tails there, some kind of leaf sheet tails, if you want to say it like that. Some kind of tails. So of course, lambda max, in fact, have the possibility to be larger than square root of 2. While if you stick to this IID case and say that they have all this kind of viable distribution, this will not just be possible. So that's mainly the reason why this approximation fails. And in fact, the full distribution, or say the fluctuations of lambda max, have been properly computed 20 years ago, or almost in 1994, by Troisi and Riedem. And I will just tell you now what the results are. So the question is, what is this chi random variable? So again, we know some piece of this. We know A and Bn. Now what is chi? So again, chi is given by another name of the Troisi-Riedem distribution. We have heard about it yesterday in this KPZ talk. And OK, there were basically two seminal papers, 1994 and 1996. Usually in this random matrix case, the Gaussian unitary ensemble is the simplest one. So beta equal 2 is usually quite simple. And it was studied first. And beta equal 1 was studied later, because it's much, much harder. And in fact, it was studied with the case 1 equal to beta equal 4 also, because 1 and 4, I mean, there are some similarities at the technical level. Basically, in this case, since we are dealing a bit with algebra, we expect, OK, that's not a big surprise. But here, the central objects are determinants. So basically, you have to manipulate complicated determinants. Unfortunately, in this case, you have to deal with some more complicated objects, which are called Phafian. And OK, you can still do things with Phafian. I mean, at least these guys were able to do very nice things. But it's more complicated. So I will probably discuss mainly the case beta equal 2, although I will also. I mean, just to give you a bit of the flavor. So what do we know? And then I will try to show you or to give you some interpretation of how to compute this distribution. So what we want to compute is really this quantity. We would like to compute basically this guy. So this is the probability that lambda max is smaller than w. So this is, again, I mean, I will come back to this later. But you remember that in the case of this maze model, this ecological particles models, this is just the stability property of this. So maybe already here we can already comment on something and make a link with this before going to the, yeah, maybe that's a good point, maybe a good way to do. Before really going to show into you the traceridum distribution, let's make a small contact with the stability that May will study. So just to make contact in May's problem, the P stable of alpha is just fn of 1 over alpha. It was just 1w equal to 1 over alpha. Remember that? So the probability that the system is stable is just the probability that lambda max is smaller than w. So we can already understand to some extent this phase transition that May was observing. Because you see that fn of w is a cumulative distribution. So it's basically 0, it's in between 0 and 1. So if I plot this guy, fn of w as a function of w, well, we have seen that you see that it will have typically this shape, right? We understand now that it will have this shape, because obviously it's a cumulative distribution. So that's fairly easy to understand. So this is 1, this is 0. So if w is very small, then the probability that lambda max is very small is also very small. This probability is also very small. On the other hand, if w is very large, the probability that lambda max is smaller than a very large number, this probability will be almost 1. Now what we have seen here is that in the limit when n goes to infinity, lambda max is essentially equal to square root of 2. So it asymptotically, I mean, almost surely it will be square root of 2. So in other words, that means that this value here, so if I really look at what happens at n goes to infinity, actually this cumulative distribution will be a very sharp step function. And the value that you have here is precisely square root of 2. Yes, because here, so at the moment we cannot say more about the nature of the transition. I mean, there is a transition, but we don't say anything about the nature of this transition. So in principle, to go further, so I realized that what I wrote yesterday probably was a bit, I wrote w is equal to 1 over square root of 2, actually. I was 1, actually. It's alpha which is equal to 1 over square root of 2. I mixed up alpha and w. That's the correct picture. So now if you want to say something about the nature of the transition, yeah, you need to go further. I mean, you need to say more about essentially about the free energies in the two phases. Now the free energies of what is not yet clear, but maybe I will just write it in a moment. So that's basically the free energy associated to this model. So that's the main transition. So now we understand at least why you observe something. So this is the limit n goes to infinity. And again, at least we have already something about the main transition. Very nice. Now let's write maybe and try to write why this fn of w is. And maybe we will begin to understand that this phase transition is really associated to some thermodynamic phase transition. Because if I write fn of w, so this is the probability, again, that all the eigenvalues are less than w. But so that means that I want all the eigenvalues. So I want to have lambda 1 smaller than w, lambda 2 smaller than w, lambda n smaller than w. What is that? Well, I can write it explicitly, actually, because I know the joint distribution of these lambda i's. And so I just need to integrate this joint flow from minus infinity to w. So that's just formally that should be written like this. So I would have w of the joint law lambda 1, lambda n. Is that OK? And this is, again, let me write it. Now we have a sort of statistical physics interpretation of this. Because this joint law is actually a Boltzmann weight associated to Macurangas. So you remember that this is just the Boltzmann weight associated to particles which are sitting on the real line confined in a quadratic potential and interacting repulsively via the logarithmic potential. But now there is a new ingredient. This new ingredient is that these particles actually have to stay between these. They cannot live on the older real line, but now they are bounded to stay between minus infinity and w. So in other words, I have now my, one has to think about it in this way. So I have this collection of particles. They live in this quadratic potential. So they are there. They are interacting. But now I am imposing something else. I am imposing that they cannot be on the right side of this place. So I would like to think of this model as a Coulomb gas with a wall. So I have an impenetrable wall here. And my particles actually cannot go to the right of it. So my particles are bounded here to stay on this semi-axis here. So as you can imagine, I mean if I push them, that will sort of, there will be some crowding effects close to the wall. While if you are far away, you don't feel too much to the wall. But what you are seeing on this model is that, again, this cumulative distribution, which is something which is, in principle, a complete mathematical object, again, is just the partition function. So that's just the partition function. OK, forget about this. So this is the ratio of two partition functions. But this is just the partition function of the Coulomb gas in presence of a wall. And now you see what is happening. So your control parameter is W. So it's like the temperature if you want in the Ising model. I mean you have an external parameter. Here this is W. It's a kind of pressure. It's the pressure that you exert on your gas here. And what you see is that this is fn of W. So fn of W, we know it's just the partition function here. So we already anticipate that the better object would be to look at the log of this, or minus log of this, OK? Because fn being a partition function, minus log of fn will be the free energy. And that's usually the good example, I mean the good object to look at. And from the minus log of fn of W, we would be able to see or to identify the nature of the transition. OK, that's the idea. So again, what I like in this setup here, and I think which is quite interesting, is that you see that you have initially a problem which is the principle of purely probabilistic nature, but you have eventually translated it in a quite physical language, OK? So yes, so now what do I do with that? So I just maybe, that's fine, let's do like this. So now I want to tell you what we know about it and what happens in the large n limit, OK? So there is, as we said, as I said, there is something happening close to square root of 2 here. And I want now to comment and be more specific on what these guys, Tressien, with them have shown. So what these gentlemen have shown is that you can write explicitly this distribution. So I will be brief because, OK, it's a little bit technical and mathematical, but still it's nice to see it, I think. So again, what we know is that lambda max, we have seen it, is square root of 2 plus n to the power of minus 2 third times some distribution here. Now this distribution that we call Tw because of these guys Tressien and we don't, they are mathematicians. And usually to define it, OK, it's better to have, I will let me introduce a square root of 2 here. It's just a matter of definition. Now it turns out that this, the distribution of Tw, it has an explicit expression. And maybe I can just, yeah, let me just give some, maybe some details, not details, but let's show it. So for beta equal 1 to 4, at least, it's possible to compute explicitly the probability distribution of chi of Tw. So I can compute the probability that chi of Tw is smaller than x. And it has a fairly explicit expression and, OK. There is a first way, so if I look at beta equal 1, at 2, for instance, beta equal 2, which is the simpler case. In principle, this distribution can be written in what is known under the name of Fredholm determinants. For 1 and 4, in terms of Fredholm-Phaffian, so I will not show you, I don't want to frighten you, so I will not show you these formulas. But it turns out that in many cases, so in these three cases, they have an alternative expression, and that's actually what was uncovered by Tressien-Reeder, is that they can be expressed in terms of the solution of some equations. And, OK, if you remember a little bit of, I mean, or if you have studied that, maybe I'll just tell you this. When you classified the second order differential equations, ordinary differential equations, there was a classification, which is due to Parlevé, which was a French mathematician. He was also a politician, actually, but quite a politician, but a good mathematician. So he actually classified the second order differential equations with respect to the singularity that they may exhibit. And among them, they have numbers. And with the second one, which is now quite well known because of this, so you take a function Q of s, which is basically a solution of this equation. So it's quite nonlinear. So there is no explicit expression, if you want, for this solution. It's quite explicit. So this is your equation. Now, to specify it, you only need to specify its behavior for largesse. And for largesse, it has to behave right. If you know a bit of special function, this is nice. Otherwise, it doesn't matter. This is just to show you. OK, so for largesse, basically, this term goes to 0. And you need to solve this equation. Q double prime is equal to sqs. And this is actually the solution of this equation. It's just what is known under the name of the airy function. Now, once you know this equation, OK, no matter. Once you know this equation, then basically, you can compute for beta equal 1 to 4, you can get explicitly this function here, this traceridum distribution, as an integral of this thing. Just to give you the flavor, I'll just give you. Because it's as simple as for beta equal 2. And then I just stop with this Paneuver type of things, which I like very much, but OK, which is not. OK, I don't know. Maybe there are some of you who have worked a bit on integrable systems. These Paneuver equations actually are very important in the realm of integrable systems. So f2 of x, as I said, is just exponential. It's just to show you that there is some explicit expression. ds x minus s q square less. OK, so you first need to solve this equation. I mean, suppose that you want to do it numerically. You can solve this equation numerically. It's pretty simple. And then you inject it, and you get whatever you want. So OK, it's a bit formal, but what is interesting is that you can get the asymptotic behaviors. And it turns out that this function, so that one can actually look at the distribution of f beta and to see how it looks a little bit. How is it close to something like a Gaussian that we know, or something that's a. So let's look at the asymptotic behavior of this quantity. And let me just tell you what they are. So if you look at very negative x, you just want to show you that it is quite an asymmetric function when x goes to minus infinity. So it behaves like, so it's quite non-Gaussian on the left. So it behaves like exponential minus beta over 12, I don't know, 24. Beta over 24, for x goes to minus infinity. And if you look at how it behaves, so x going to plus infinity, it's, on top of that, it's also quite asymmetric. And it behaves like minus 2 beta over 3 times x to the power 3 by 2. So I just will not probably comment too much on this, except to say that they have quite asymmetric tails. And it's clearly non-Gaussians. OK, so it's highly non-Gaussians, non-Gaussian tails. This guy? So this is a function f. OK, sorry. So I'm saying, yeah, maybe I was a bit fast. So OK, so the statement is as follows. So you have this lambda max. So in the large and limit, this goes to a deterministic value, square root of 2. And then there are some small fluctuations, which are of the order n to the power minus 2, sir. And they are given by this guy here, chi. Now what is this chi? So chi is a random variable, which does not depend on n. And the cumulative distribution of chi is given by f beta. That means that it depends on the parameter beta that we have seen before. This beta, beta equal 1 for GOE, beta equal 2 for GOE, et cetera. So OK, I just wanted to show you an example and tell you the simplest case, which is already a bit complicated, but still explicit. So if you look at beta equal 2, in fact, 1, 2, and 4, you can express this f beta as a specific, a particular solution of an ordinary differential equation, which is called here, perlovay equation. It's a perlovay 2 equation. Now for beta equal 2, so this is this function. For beta equal 2, it can be written as this integral. This integral of this, so you need to know the function q of s, which is the solution of that horrible equation. And now I'm looking at f prime beta, which is the PDF. So I'm doing f prime beta because f beta, you see, is the cumulative distribution. And OK, I usually prefer to look at the PDF, because otherwise it's a bit. Is that OK? So now f prime beta, you can show. I mean, it's more than an exercise, actually. But you can show that it has these tails. I just want to show you this because, of course, when you see this formula here, probably it doesn't tell you anything. But on the other hand, that's why it's always very useful to know the asymptotic behaviors of a PDF. Because if you have an explicit expression like that, it's very useful to evaluate it numerically or to plot it. But at some point, you want to see how it looks like. Since I cannot really show you the plot of it, I prefer to be a bit more quantitative and show you the asymptotic behavior of this distribution, which is actually very precise already. So they describe quite well the tails. So it behaves like exponential minus mod x cubed for x negative and large. And it has this exponential minus x to the power of 3 by 2 for large positive x. So it's pretty non-garesian. It's also quite asymmetric. So garesian would be just x square. So none of the tails actually have that form. And on top of that, they are quite asymmetric. This is how they look like. So now, I guess, that I want to, I mean, OK, that's something that I want to show you. So this was this traceridum distribution. And this has to deal with what I want to name or to call, have to do with the typical fluctuations. Now, somewhat unfortunately, if you want to understand a little bit more this may transition, you need to go beyond these typical fluctuations. And what you would like to understand is what happens when you have large deviations of lambda max. That means when lambda max is very far away from square root of 2. So by this, I mean the following. Oh, by the way, I mean, if you want to think already about it, this cubing here is the indication of this third order phase transition. It's still a bit hidden, but this is what it is. So now, I want to comment on, as I said, to understand better this part of the, I mean, to understand better this may transitions, one needs to go further and to study the large deviations of lambda max. So this is something that we have not studied too much up to now, so I just want to comment on that. So let me give the motivation for it. And then, so OK. So I like the following picture. You look at your eigenvalues. I have here the density, minus square root of 2 plus square root of 2. And I was actually telling you this, Tracy and William, what they tell us, and this is also what we got with our argument, that if you look at lambda max, actually the scale of the distribution that we have studied is very narrow around square root of 2, right? Because the scale is of order n to the power minus 2, sir. So we are looking at, we are describing the fluctuations on a very small scale around square root of 2. So typically, I will plot, so this is the density of eigenvalue. On the same plot, I'm plotting the density of eigenvalue and the PDF of lambda max. So this is, this is it. So this has something like this. This is basically this distribution that we have studied here. So that's choicey rhythm. But you see, I mean, choicey rhythm is very nice. But unfortunately, it's also restricted to a very narrow scale. And in particular, it doesn't tell you anything on how the things behave here. Lambda max, you see, I mean, lambda max in principle can be anywhere. It can be 0, for instance. So how would you describe this scale here? Very far away from square root of 2. Even down to minus square root of 2 if I want to. Minus infinity, in principle. So this regime here are called, so you have to see that, for instance, a natural question could be, what's the probability that all the eigenvalues are negative? It's a natural question, I mean. That means what's the probability that lambda max is smaller than 0? So you see that you are very far away from square root of 2, typically over the 1. So it turns out that this regime here of large deviations, that's what I call large deviations. So this is the large deviations of lambda max. So this is the left, left from some obvious reason. And it turns out that this large, left large deviations are not described by traceridum. They are described by something else, by some large deviation regime, some large deviation functions that we would need to compute to say something on that regime. But this traceridum doesn't tell you anything about it. OK, I mean, they are crossing here, but it's just a purely incidental. I mean, it's not that point. Yeah, yes. Well, here I didn't show you really the computation of it. So they are, of course, you are assuming, that when you do this computation, so essentially what you show is that I want to compute this fn, this fn. So I wrote it as square root of 2 plus, as I said, basically 1 over square root of 2. I mean, that's the way you would do the computational. And to the power minus 2 third times s. So that's my variable chi. And this actually goes, when n goes to infinity, this converges in the sense of distribution, blah, blah, blah. This converges to fb times s. OK, so that's somehow the approximation that I did it. Well, it's not an approximation. This is the statement. But that really tells you that you see that you are only looking at the fluctuations which are of the order n to the power minus 2 third. And obviously what you will see is that, that is true, is that when we look at the left large deviations and the right large deviations, that means that if I take another scale for s, so suppose that I take s equal to 1 here, I mean, over the 1, then formally it will go to 0. But 0 doesn't tell you so much. I mean, you would like to see really how small it is. And you enter a so-called large deviation regime, and which has a specific form which I will describe in a minute. Well, here that means that basically, so that means that in minus infinity means that you are touching here this regime here. So you are approaching this regime. But still over the n to the power minus 2 third. So that means that s is large, but still you have to remind that you are in a regime of order n to the power minus 2 third. And the same, you could ask also what's the probability that the distribution is very far on the right. That's also legitimate question. Again, so that would be the right large deviation. And unfortunately, I mean, fortunately, I don't know if it's unfortunate, but there is more than three series in this problem. And this is the right large deviation. So you would like to know more about this one. So I don't know how much familiar you are with these large deviations, but I thought maybe it would be nice to show you some before illustrating on the traceridum, which is a bit involved. Let's maybe look at some large deviation problems on very simple system, except if every one of you is very familiar to large deviations, otherwise I will just immediately go and continue on that. But how many of you are quite familiar with large deviations? OK. So let's do it. I'm sorry for you guys, but I will just try to illustrate it on a very simple question. No. On a simple problem, the simplest example, which is a coin tossing example. I mean, it's quite simple, but just illustrate what large deviations is, because otherwise you will really get bored with what I will say, and that would just mean anything, not mean anything for you. It's just a very reminder on large deviations. So in fact, it's the large deviations for the sum of IID random variables. So let's consider the simple game, just a coin tossing problem. So with probability half, you will have a head. With probability half, that will be h. And with probability half, you will have tail. So just do some coin tossing experiments, and you do it n times. So you have n trials. And after n trials, you count the number of heads that you have. So why is it related to the sum of IID? Well, I mean, obviously, this will be, so if you do n trials, let me introduce an indicator function, sigma i, is essentially like a spin variable, which is 0 or 1, depending on whether you have a. So it's 1 if you have a head, and it's 0 if you have tail. So nh is obviously just the sum of random variables. So what you want to do, obviously, so it will be a random number. And you want to compute, say, the probability that nh is equal to, say, a given value, say, m. So this is something that you can compute easily. This is just a combinatorial factor divided by the total number of possibilities, which is 1 over 2 to the n. So this is just n choose n, and you divide by n by this. That should be OK. Of course, it's very simple. Now, once you have this, you can compute various statistics of nh. For instance, you can compute the average value. The average value is very simple. The average value, obviously, you don't need any computation. It will be just n by 2. And you can also compute the variance, which requires a bit of computation. But you can compute sigma h square, which will be nh square minus nh square. And it's a small computation to show that this is n over 4. We know from the central limit theorem that the typical situations of nh will be given by Gaussian distribution around its mean. So let's forget our friends here for a while, or probably until Monday. So the typical fluctuations are given by, so that's nice to have in mind, that these are the typical fluctuations. I will come this, you will see in a minute what I mean. So we know that they will be given by the central limit theorem. So that means that that's what I mean by the typical fluctuations are given by the central limit theorem, CLT. So what does it mean? It means that this probability nh of m, in the large n limit, it will reach a Gaussian form around the mean value n by 2 and with some variance sigma square. You don't like this? It should be, what, sorry? Why do you want to have it n over 4 n square? No, because is there a sigma h? Sigma h will be square root of n. So that's the central limit theorem. So you would expect, because you see I mean this is the variance, so I have subtracted the leading term. The leading term of this will be n square by 4, but it will be cancelled by this. Is that OK? So what is that? So again, this will be a Gaussian. So let's write it explicitly. This will be 1 over square root of 2 over pi n exponential of minus 2 over n times m minus n over 2 square. That's right now. And we know that it describes very well the typical fluctuations around n by 2. Now let's look at a rare event. What do I mean by fluctuations? You see that the fluctuations here are typically over the square root of n, because of this. So typical fluctuations, maybe I just will do the same drawing that I did before. I have this, so m is positive. So this is this quantity as a function of n. And so what I'm saying is that here it's very nice. I have this explicit expression, which is that. This is a Gaussian. It should be perfectly symmetric. And this is over the square root of n. So these are the typical fluctuations. I'm claiming that these are the typical fluctuations. Why so? Why to see that? Let's go to the very left to n over 2. Let's consider, for instance, the probability that m is equal to 0. This will be some event, possible event, right? What's the probability for that? Well, very simple. Sorry, that nh is equal to 0. This is what? This is 1 over 2 to the power n, right? Because that means that I obtain tails every time. Now, if you evaluate this formula for m equal to 0, what you see here is that if you evaluate it for m at m equal to 0, it gives you, OK, forget about this. OK, you can put it if you like, but I won't need it. And then I will get what? I will get exponential of minus. So m is 0. So I have n squared by 4. So that's minus n by 2. Now here, obviously, this is exponential of minus n log 2. So obviously, this formula cannot hold up to that point. So in other words, that means that this regime here is only, so again, this is different from the exact result, 1 over 2 to the n. And so that tells you that this regime here can only has a limited domain of validity, which we expect, which is over the Scrabble domain. And if you go to now, if m instead is over the n, if you really are out of distance over the n from n by 2, then you will enter a different regime. So we have another regime here and another regime there. Now, question is what is this regime? So that's a large deviation. That's what is called large deviation. So when you are very far away from the typical value, you will face another regime, which is not described by the Gaussian. The question is, I mean, what is the good approximation or what is the form of the distribution? Now, it turns out that in this case, so now if you start from this formula, it turns out that the large deviation regime is in fact, this is easy to get actually from the, so let's look at the case where, basically, I want to have, it turns out that the large deviation regime, I can just have it. So I'm just writing that m, I just write it of this form. I want to have m to be over the n, and that means m will be, c, sorry, will be in between 0 and 1. Now, if you do that in this limit, and if you play with a sterling formula, it's a fairly simple exercise to show that in the large n limit, this will be of that exponential of minus n times some function of c, phi of c, from that formula. I mean, you just need to take the large n limit of this combinatorial factor, and this phi of c, OK. I'm sure you have seen several times. It has a relatively simple expression. You can compute it here explicitly here. So it's just c log c plus 1 minus c. It's kind of entropy. And you have an addition here because you have it. Now, this form is very nice. This form is very nice because, for instance, if you compute phi of 0, phi of 0, you really recover the exact result. If you compute this in particular, which is this exact result. Why is the function of this theorem saving in the limit of, say, a and b? Yeah, because? The system that I'm explaining is not working, but then it's minus n of 5. Now, what it means is that this Gaussian form, it only describes pretty well the fluctuations around n over 2 on a scale of order square root of n. But if you go beyond that, that means if you ask for some probability of m far from n over 2, which are over the n, say, when you put m equals 0, then the distance is over the n, n by 2, much larger than square root of n, then this regime is not correct anymore. This, in the central limit theorem, you are assuming that you are looking at the fluctuations around square root of n. So that's the same as before. So if you really, that means that central limit theorem tells you, basically, that if you do p of n by 2 plus, say, x square root of n, then this will go to the Gaussian when n goes to infinity, a bit like in the triceuridum case that I mentioned. But obviously, if you don't scale the things like that, something different happens. Of course, and I will show you it in a minute, and this is also nice in this case. The large deviation regime actually encodes also the Gaussian part. So that means that if you look at this distribution around the minimum, it has a quadratic behavior which precisely encodes this guy here. So if you look at, so now that comes what I wanted to say, is that if you look at phi of c as a function of c, so it's on 0, 1. It's symmetric around half for some obvious reason. And it goes like that, right? So phi of 0 is log 2. Phi of 1, for the same reason, phi of 1 is also log 2. So you go to log 2, go to log 2. So you have something non-trivial here. And around this point here, around this point, is the central limit theorem, OK? So that's the Gaussian behavior. What do I mean by that? If you look at this function close to, it behaves like c minus half square. And there is a pre-factor here. Maybe it's good to give it if I want to stick, I don't know. Yes, let's do it properly. It's 2. So now let's, so what I'm saying is that this form here is very nice because, so in other words, if I would, maybe you have to remember that c is, so this is the last, so this has typically this kind of thing, OK? So this is really of that form, OK? So now let's see that explicitly that this large deviation regime actually also encodes the Gaussian. So it has much more because it has the large deviation form, but it encodes also the Gaussian peak. So let's see it. And I guess if I have time, OK, I have a few minutes, but that's what I need. I mean, a few minutes, I see I have one minute, but I would need maybe two or three minutes. So what I'm saying is that this form here encodes more and more. So the quadratic form will be something like that, right? So the escalating parabola, if you want, probably something like this, OK? So that's the parabola. So now let's look at what we had before. That means that suppose that from that formula, let's look at it under this form. I will look at it like this, so you remember that, right. So you remember what? You remember that NH was N by 2, and we have sigma H, which was square root of N by 4. So that's what I showed you. So now let's look at this distribution if I set probability that NH is precisely of this form. It's basically N over 2 plus sigma H times some number X. OK, so maybe let's do it this way. So it's basically what I said before, roughly speaking. So this will be of that form, right? This will be of the exponential of minus N phi of what? Phi of M divided by N. So that means N over 2 plus square root of N by 4 times X divided by N. So I'm just taking for granted this large deviation form there. And when I have inserted M, which is N by 2 plus whatever. So now you see that, OK, it's nice. It's of that form, exponential phi. So I have half plus what? Plus 1 over square root of 4N X. OK? Oh, excuse me, there is N here. Is that OK? And now I just use the fact that I want to make a Taylor expansion of this function around 1 half, right, because this term is small. So I can just use this behavior, OK? So let's do it. So if I do it, you may get to see that you will get exponential of minus N times what? Times C minus half square. So C minus half is just this term here. So that's 1 over square root of 4N times X square. And I have a factor of 2 here, 2, OK? Now if you combine everything, what you find is that, of course, the N dependence goes away, N divided by square root of N square. So the N dependence goes away. That's what you want. And here you get 2 divided by 4. So this is half. So you get exactly your Gaussian, exponential minus X square over 2. So this large deviation regime here encodes also captures the Gaussian fluctuations up to a prefactor. You will notice that I don't get here the 1 over square root of 2 pi sigma, because, of course, I mean, I can only get the things. I did this computation at the level of the exponential. I mean, these are just logarithmic equivalents, OK? So I don't get the prefactors here. But at least you get this quadratic behavior, and that's the Gaussian behavior. And that's the Gaussian that you get from the CLT up to the prefactor. OK, so yeah, I wanted to show you this example. I find it fairly simple, but it's also rather instructive. I think it explains relatively well what the large deviations are. They usually pay also quite an important role in extreme statistics. And we will see later on, OK, next time I will show you. I mean, of course, I will not show you all the details, but I will show you a bit what are the large deviations for the largest eigenvalue of random matrices, and what kind of information one can extract regarding these main models, these ecological models that I started with. OK, there you have the question. Yes, you can view it like this. You mean for general problem of statistical mechanics? Well, in general, OK, these functions are quite non-universal. So for instance, if I could ask the same question, so here I was looking at the sum of random variables, you could have asked me, OK, let's take, for instance, the sum of IID random variables with some initial distribution p of x. Then usually, I mean, this large deviation function will be quite non-universal, will depend on p of x. Only its behavior here will be locally, I mean, the quadratic behavior near the minimum will be the same. This is just the result of the central limit theorem. But the terms here will depend on many things. Now, usually, you can obtain them via a saddle point calculation. And this is basically some legendary transform of some generating function that you can compute. So basically, in that case, this is typically the this is the legendary transform of the generating function of the rate function that enters in the generating function. That's usually the way you can get it. So again, I mean, here I didn't want to make a full course on large deviations. I just wanted to give you the flavor of what it is, how important it is, already on simple models. And I will then, later, so next time, I will show you how it works for the largest eigenvalues and see you how one can uncover the order of this transition and find the third order of this transition. OK, so with this, I will leave you and see you on Monday then. Yeah? I think you can think of the more higher momentum. Sorry? That's not so. In this case, for the large deviation case, it is nothing. No.