 lectures about beta ensembles, and the main theme of the today's lecture is going to be stochastic area operator, but I decided I'm not going to leave you in the lurch completely about this lemma that we used, which is the representation of beta ensembles in terms of tridiagonal matrices. So you already know that you can represent the GOE as a tridiagonal matrix, and you're probably not extremely surprised that you can do the same thing for the GOE as well. And if you think about how the process went, then the only difference is going to be a factor of two in the parameters of the chi's. So that's just going to be beta equals two. And all this thing can be proved and more general things can be proved. So in fact, as one of you mentioned last time, you can diagonalize the invariant ensembles, tridiagonalize the invariant ensembles, even for general beta, by the following theorem. We left off with last time. So what did the theorem say? So if you had the tridiagonal matrix, which is random, and you had a joint distribution of the entries, which was proportional to, so if the density of a's and b's is proportional to x of minus trace v of j, so v's, you know, is perfectly, everything is okay if it's even polynomial with positive coefficient, positive coefficient, coefficient, right? But this can be generally any function as long as this thing is integrable, normalizable. All of these things can be made sense, except maybe this density is going to be kind of messy, but so it can always make sense out of it. So plus, there was a sum i equals 1 to n minus 1, and there was a beta, right, is it beta log bi times i, maybe i may beta minus 1, I think, I hope I did it properly, like that, this was bn minus i. So if the density of the ab is this, then the eigenvalues have joint density, again proportional to exponential of minus sum of v over lambda j, right? So they like to be independent with this density whose log is v of j, that's what they like to do without any interaction, but then there is repulsion, which is just a random one, let me write it like this to the power beta, and the qi, the spectral weights are independent of the eigenvalues and have densities, have Dirichlet distribution, and this beta comes into Dirichlet distribution, okay? So for example, when beta equals 2, then you just have Dirichlet 111, which is just the same as if you took the unit interval and dropped the independent points to divide it up into pieces, and took the gaps to be those qs. So beta equals 2 is nice. Okay, so I'm going to tell you, you know, I already started discussing what this means. So first of all, when vj is just x squared, or I think what you usually take it to be is x squared over 4, or maybe beta x squared over 4, doesn't matter if it's some constant times x squared, then this is just the sum of the squares of the entries of the matrix, right? So this density becomes a product, so the entries become independent. So that tells you, and in that case, these things will be the densities of the chi betas, so that will be the general edamon limitry matrix, which looks like this. Okay, so I think, I hope I did it right. I think, yes, that's right, so n minus 1 here. So what am I doing? Yeah, this normal 0, 2 comes up in here. Okay, so you'll have the same normals in the diagonal, but the chi's will be slightly different, and you sometimes like to divide this by root beta, or some constant times root beta, so that the main asymptotics of this matrix is the same independent of what beta is. And then this has the Gaussian beta density for eigenvalues. This is the corollary, right? Now, here is a question to you. Let's say that v is x to the fourth, then you look at this process of A, I, and B, I, so the I is greater than or equal to 1. This is, maybe it's a too hard question. So when beta is, when v is 2, then this is just independent, right? So what happens when beta v is 4? What kind of stochastic process is that? It's a little exercise. Any guesses? Well, you think of this, you know, the AIs and BIs have a joint density. I specified them this way. Trace of vj. Okay, so what is it, what is trace of vj, right? So you look at v to the fourth power. So you look at paths of length 4, all loops of length 4, and multiply the element entries over, over those loops of length 4, cycles of length 4, and add them together. That's what, that's what you'll get in here. So you get this kind of thing that only consecutive things interact. Okay, because you have things of length 4. Okay, so let me give you the answer. The answer is that this is a Markov chain. Okay, and if you think about it a little bit, it's not time homogeneous Markov chain. So it's a time dependent Markov chain. That's because this, this coefficient here depends on time. Okay, but it's still a Markov chain. And knowing that it's a Markov chain helps you, helps you understand how this thing works. And in fact, this is true even no matter how, what polynomial you take for v. If you take a higher degree polynomial, then there are more interactions. So you have to record more of these, maybe have to record them off group of d over 2, and it will still be a Markov chain. So this is a nice tool to analyze these invariant ensembles. And in fact, we have a proof with Brian Ryder and Manju Krishnipur that for these invariant ensembles, the top eigenvalue has the right asymptotics, and the whole operator limit, which I will tell you about, works the same way as for v equals x squared. And the whole proof goes through analyzing these Markov chains. Well, the problem is that the matrices for the invariant ensembles are not particularly nice. So I never tried putting it through the householder algorithm because of their reason. But maybe you can. I'm not saying it's impossible. But you don't really need to, because it just tells you it works. What this thing tells you is that there is essentially no difference between various values of beta as long as you use our analysis. So the first two lectures, they work for all beta ensembles. So it's a kind of thing that's nice for beta ensembles. I mean, not the invariant ensembles, but the hermit bits beta ensembles, they work perfectly. You don't have to change anything. Just the same proofs, Wigner semicircle law, all those things. Good. So when I first worked in this area, our goal was to find a beta version of the traceridum distribution. So as you know, there are nice closed formulas for beta equals one, two and four for the traceridum distribution. But at the time, nobody knew how to define it for general beta. And nobody knew that actually the kind of the limit that you need to take for this to be existed. So what I want to talk about today is how to do that. And the idea is basically to look at this matrix. I'm just going to go back to beta equals one because everything is the same. So you want to look at this matrix. And you want to understand the top eigenvalue. But again, your goal is not to write formulas and all the things for it, but just somehow use the geometry of this picture to understand the top eigenvalue. So you want to take a limit of the top eigenvalue. You don't just forget everything and take a limit, but you want to take the whole matrix with it. And then you have more information than all the structure that's in here will remain. So that's the goal. And how can you do that? Well, first of all, if you want to understand the top eigenvalue, you'd better look at what the top eigenvector looks like. So you do some simulations and see what the top eigenvector of this matrix looks like. If you do it for a large n, what you find is that the top few eigenvectors, they all are going to live in this top, in the beginning of your discrete line. So it's in the first coordinates. Not first finite, it's growing with n, but you can certainly observe, as far as you can do simulations, that it's growing slower than n. The support is growing slower than n. So it's actually little over n. So what that suggests is that the information required for the top eigenvalue, for this matrix, for this particular matrices, should be somehow contained in the top corner of this whole thing. The next observation is that it's something that we've already seen before. If you take a limit of this matrix, the one that leaves you the semicircle law, you get something like this. You get this matrix if you divide by root n. So now what is this matrix? Well, this matrix is not especially interesting. It just has ones on the off diagonal. But let's say if I subtract from this twice the identity, like that, then you know what it is, right? This is the discrete second derivative. And what does that mean that this is the discrete second derivative? So let's try to make some meaning out of it. So let's say that I'm going to take a function, so this is an n by n matrix. But I don't really care so much about it. I'm going to take a function that lives on 1 to m, according to that's 1 to m, where m can be now a little low of n. And I told you why we do that. You already know it. We're looking at the top eigenvalue. And it's probably going to live somewhere slower. So what do I do? So I take a function f, which goes from, let's say, 0. Let me just take a compact interval. But generally, it could be longer as well for simplicity to take a complex interval to the reals, right? And I look at, I make a vector of f0, f1 over m, right? And take f2 over m, or sorry, fm over m. So we go all the way to m and so on. I don't care how it continues. Okay? And I apply this matrix to that vector. And then I get some discretization of f prime prime, right? As long as it had a second derivative. And in fact, to do that, I have to multiply things by m squared, right? So if I take this function, I discretize it, apply this matrix, I get a discretization, and multiply by m squared, I get a discretization of f prime prime. Okay? So in this sense, this function is a discrete second derivative. So that's not a bad first approximation of this object. But we, you know, we really would like to take something more refined. We already taken this where m was just the finite window, right? We'd like to have m to be a growing window. And then we have to do some slightly more precise asymptotics. Okay? So for this, let's understand this chi's, right? So we know that chi of n minus k, what does that look like? All right? So chi of n minus k is roughly root n minus k plus a normal, 0, 1 half, right? And we are in the regime when this m is going to be much smaller than m. So we have to, so you want to do a Taylor expansion of this, right? So we want to do this as root n minus k over 2n, right? Did I do this correctly? I hope so. Excuse me? 2 square root of n, thanks. So I think now we captured everything I wanted. So what are we going to do? So we're going to take this matrix J. Let's see. You're going to write J, let's see how should we do it? Yeah, we're going to take 2 root n minus J times the identity minus J. Okay? And we want to multiply it by some power of, let me see, m to the gamma, okay? And we hope that this is going to be asymptotics to some nice thing. So what do we get here? Well, first of all, we're going to get m to the gamma. Excuse me? So it's a multiplication of m to the gamma. I take 2 root n minus J, okay? This is to put those two on the diagonals, okay? And 2 root n times the identity, sorry. We often just drop the identity, so it's a bad habit. So what is this? Okay, so first of all, the main asymptotics is there is a square root of n and m to the gamma times this discrete derivative, which is minus this discrete second derivative. Okay, so that's one thing. And then there is a term plus, right? There is this matrix, which looks like 1, 2, 3, 4, the dot. And this one is multiplied by, this one is multiplied by, let me see, m to the gamma and divided by 2 root n. I guess these are off diagonal, okay? That's from the kites. So you see what I'm doing? I'm taking this, taking sort of the top corner of this matrix apart into pieces. Let's look at the last one. So then there is some matrix which has basically IID normals coming from the kites. Well, they're not IID. One of them, the off diagonals are different. But it's a tri-diagonal matrix of normals of order 1, right? And there is no, in there there is only the m to the gamma multiplication. Okay? So here is a challenge. Yes? There is no square root of n because I put that in here. In here, you see? Because you subtract, because this is the Taylor expansion of root. So, j has a minus sign here. That's why it's increasing. And the Taylor expansion of root n minus k has this k over 2 root n. Okay. So remember I told you something. I told you that lambda 1 of GOE is approximately 2 root n plus n to the minus 1, 6 trace of eta. Okay? So what should we write here so that we get a nice interesting limit? What we should be getting here is n to the 1, 6. Right? So that the top eigenvalue of this matrix will converge to something like a constant. Right? The top eigenvalue of this matrix is like 2 root n plus this. If I subtract from 2 root n, there will be an order n to the 1, 6. You have to multiply it back by n to the 1, 6 to get something nice. And now, this is kind of the fun part, you can get this n to the 1, 6 just by looking at this and trying to make sense out of everything. Okay? So let's look at, let's try, let's write some equations. Okay. So first of all, I want to write that n, m is going to be n to the, n to the, what should we use, alpha. Okay? That's fair enough. I mean, it should be some power of n. That's reasonable, right? At least not crazy. So what is the equation? So this you should, you want it to converge to the second derivative, minus second derivative. So what is the equation here? Well, we're going to take it on an interval of length m. Right? Maybe you should be writing it the other way. Maybe it's easier. Maybe it's easier. So m is going to be, how should I say, n is going to be m to the alpha. Okay, let's do it like this. So this thing, what you write here has to be a square of that. Right? That's what we just saw. For this to actually look like a second derivative. So what does this tell you? What is the power here? Let's measure it in powers of m. It's gamma plus alpha over 2. Right? That should be equal to 2. Does this make sense? Right? This number should be the square of that number. Hm? One of m. One in m. M, yeah. We're going to put the grid only on the top of the matrix. Mary. Mary is grid spacing. Okay, so that gives you an equation. Now let's look at this. You would like this to go to multiplication by t. Or what should be, what should be a variable, x, multiplication by x. Right? These things are on the off diagonal, but that really doesn't matter too much. You could imagine them, putting them on the diagonal. It only causes something of little o of everything. Okay, so like this is the function x. It's just growing. So when will this be, when will this go to multiplication by x? Well, I guess this thing then should be of order m. Right? So what you write in here is going to be x. It's going to be, you know, like at m over 2, you should have a number. The return is just of a constant size. So what does this tell you? So you have a gamma minus 1. So what is n? It's alpha. Okay, so minus alpha over 2 is equal to 1. Okay? Yes. Yes, minus 1. Thank you. Okay. We have two equations now. And the third one is harder to think about because it's multiplication by noise, but you can do the third one. And I swear that it's going to give you the same answer. Okay? So it's not going to contradict the first thing. But doing, you know, how to scale multiplication by noise in real time, I think that's unfair. So we won't do that. No, let's solve things. Can anybody solve this by, you know, in their head? What's the solution? I know the answer. So there is only one solution. These are not linearly dependent. And the solution should be alpha equals 3. And gamma equals, if alpha is 3, gamma is minus 1.5 plus 1.5. Yes. Okay. So what does this tell you? Well, it tells you that I just have to look at the part of the matrix which is n to the one-third, which is m. And I have to scale by, where is it? m to the gamma, which is n to the one-sixth. Okay? So we're happy. We got the right scaling. And trust me that this is the right scaling for this as well. What did you, gamma equals 1.5? Yeah, this should be a half, right? It's Gaussian stuff. Okay. And so what is the conclusion? Okay. So I write it like this. So 2 root n minus j n times n to the one-sixth. It's not really the interval, right? So you basically let this to be the whole R. I mean, you scale by, the steps are going to be 1 over m. And the tail of the matrix will just run out to infinity. We don't care. Nothing's happening there. And this here on the scale 1 over n to the one-third looks like some differential operator. Okay? And what's the differential differential operator? There is a minus-second derivative. Okay? Plus multiplication by x. Okay? Plus multiplication by noise. I think the right thing is 2 over root beta. And let me write it b prime. Okay? So this is just the derivative of Brownian motion. So this guy is called the stochastic area operator. And there is actually a theorem that this happens. Okay? The proof, the ideas that's happened is due to Edelman and Sutton. And the proof is due to Jose Ramirez and Ryan Ryder and myself. And I won't give you the whole proof, but I'm going to give you some ideas. At least, at least, at least I give you a precise statement. Get it? That's helpful. Okay? So let's try to make the precise statement. Okay? So one problem is that, of course, these things live in completely different worlds. This is like a discrete thing. This is continuous. But so how do you solve this problem? So the way you solve this problem is you embed r to the n.