 Last time we discussed that we would like to put some structure into random matrices, some geometric structure, and then right to understand this geometric structure to prove certain things. And as you know, we looked at the GOE, and we saw that the following pulling fact, just to recall, and actually recalling I'm going to tell you a little bit more. So the GOE had a model, or was conjugate to a Jacobi matrix in a way that the spectral measure was preserved at the first vertex, and the Jacobi matrix looked like this. So you had this normal zero two, and you had here a chi of n minus one, chi one, okay, and this is, I think it's very independent. So this is actually, this was a result of Trotter, and we have used this to deduce the semicircle law two ways, right, and also used it to prove the Komlo-Achiradi result, about the top, again, we are sticking to the semicircle. And we started talking about the Baig-Bender-Respachet result. Okay. So let me just tell you one thing that this is also useful, although I'm not going to give you the proofs in this series of lectures, although these are very nice proofs. And that's of course the result of Dmitry U Edaman. And that result is the following. So if I actually change the distribution slightly, okay, so if I take these parameters of the chi's and put here a beta, it's nice to write here this one over root beta just for normalization. This is the normalization that makes these chi's of the same order independent of beta. And then actually, the eigenvalues of this are those of the Gaussian beta ensemble, okay? So the beta ensemble looks like the joint distribution of eigenvalues. So this has a joint distribution eigenvalues, which is e to the minus beta over 4 sum of lambda i squared, okay, times the Venderman lambda i minus lambda j to the beta, and there is one over z beta. Okay. So this is a beautiful result to Dmitry U Edaman. These kind of beta ensembles are often called Dyson's beta ensembles. They're introduced by Dyson in the 60s. And for beta equals 2, this corresponds to GUE, so the Gaussian unitary ensemble. But for general beta, this is the first time that you actually, that there was a nice random matrix model for these things. I mean, you can always, you know, you can put these lambdas in the diagonals, a matrix, so that would be a matrix model, but actually something that has some meaning and has some structure, this is the first time that that that that was found. And in fact, you know, the more general theorem, which I would say due to Dmitry U Edaman, Brian Ryder, I'm not writing in the right order, Krishna for myself, is something slightly more general than this, which is the following. So if you look at the Jacobi matrix, remember it has these a1 up to an, b1, bn minus 1. So if you look at this matrix and put a density on the entries, which looks like this, so it's 1 over z exponential minus trace of v of j, okay, times a product of k equals 1 to n minus 1, or b to the n minus, v of n minus k of k beta minus 1. Okay, so this is some kind of descending power of the b's. So you put this thing on the matrix and this, of course, if this is most interesting and v is some polynomial, okay, better be an even degree with positive coefficients so that this is integrable, so that's a condition, because then it's just some, you know, this is just some polynomial of the entries, the same degree as v. So it's very explicit. When v is x squared, then you just get something like this, some scaling of that. So then, you know, the eigenvalues have density, again, something very similar, so 1 over z beta is exponential minus sum of v of lambda i, because it's the same v as before, and then you have a Venderman like that. So this, you know, what this shows you is that this general, these are called invariant ensembles, for beta equals 1 or 2, okay, because the density that you put on the ensembles is invariant under orthogonal or unitary transformations. They also have a trigonal representation. They're not as simple. The entries are not independent anymore, okay, when v is something more complicated, but there's still something that you may be able to study. So I think we're just going to focus on the beta-hermit case, which is when this is just x squared, but even in this most general case, I just want to give you one remark, which is which is, let's look at the case when v is, for example, x to the fourth. So what do you have? You get some kind of exponential, okay, which has polynomials, which has monomials in these a's and b's, but they come from this trace of the fourth power, so there are going to be sums over closed loops, cross-close-likeals of length four of what's written on those cycles in this matrix, right? So they have this interesting property, okay, so if you think, if you fix a i and b i for some fixed i, then the indices of a i, the variables with lower indices and the variable of higher indices will be independent, just because of the way this density looks like, okay, because all those monomials, the other monomials will be independent of each other, that's the only thing that connects to two, the a i and b i. So what is this property called, and anybody, so if you fix a i and b i for some i, and then look at, so a 10 and b 10, okay, and then you look, look a 1 up to a 9, b 1 up to b 9, and a 11 up to, up further and b 11 up further, then these two things are conditionally independent, right? So, so what is that property? Okay, so, so, so, come on, come on, so you look at this pair, okay, what is this process? So you look at the process of pairs, so again I'm setting v equals x squared, x to the fourth. Well, this is the Markov chain property, okay, the past and the future are conditionally independent, given the present, you may have heard of this thing, so these a i's and b i's are going to be a Markov chain, okay, they're not going to be time homogeneous, okay, because there is this time dependent thing over here, okay, but it's still going to be a Markov chain, so it's something that we like, okay, so it's possible to analyze it, and in fact this can be done, and even if there is some larger polynomial, you know, like degree 10, it's still going to be a Markov chain, you have to take more than just one a i's and b i's, maybe you have to take five of them or something in the case of 10, but that, that thing will still be a Markov chain, and you can still analyze it, and when you do the analysis of this Markov chain, you can actually say a lot about this matrix, and so this was actually even the first proof of edge universality for beta ensembles, was for this particular setup, where we some even polynomial, and we analyzed the structure of this chain, so you know this, the kind of things that we're doing now are just you know, the GOE and the simpler models, but in fact you can use this for more general things. I have also had questions whether you can use it for Wigner matrices, and the answer is no, so I don't think in that role this works well, so it somehow, it does a different thing than Wigner universality, but it sort of identifies the limiting objects, and shows how invariant ensembles connect to that. Okay, so the first remark of course, so I'm going to put this aside, just a side remark, but one thing that you can see from here right away is that all the proofs were for Komno-Schiridi and Wigner-Sambiso-Korol and everything works in this case, exactly the same proofs, we only changed some parameters in the kite. Okay, so this was, so we already got some results that are otherwise not at all easy to prove. All right, yes, so one more question, so how do you, one more thing, how do you prove this kind of thing? So actually the proof in this case is not hard at all, it's just a computation, and the reason is because when you look at these a's and b's, right, what they do is they parameterize your spectral measure, okay, the spectral measure has parameters lambda one, lambda n, and q one, which are the spectral weights, so the weights on those q n, but this sum to one, right, so this is also, these guys also live on two and minus one dimensions, okay, so if you want to understand how something like this is distributed and how something like that is distributed, well you just do a change of variables, okay, there is, you don't have to integrate anything out, it's just simply a change of variables, and by the, it's not that simple to, to get from the matrix entries to the eigenvalues, there is the most classical trick to solve this problem, what is the most classical trick, how did Wigner prove his semicircular law? Well he proved it by the method of moments, right, so what you do in fact to prove it is you write down the moments m one up to m two n minus one, okay, and these moments are very simple functions of both of these, they're actually just polynomials, right, here you just, here moments of the spectral measure is what I'm talking about, so they're simple polynomials, here you're just counting paths, and you put you know the weights over the path, and then here you're just summing squares and third powers and so on, so they're extremely simple functions, and it's also not hard to prove that everything here is an analytic bijection, okay, so you, so this is an analytic bijection and that also, so what you do, you just compute the Jacobian from here to there and from there to there, put them together and you get the change of variables formulae, yes? Both, both of us that the Wigner semicircular law could be seen from, I mean the, the potential of the V here, if you change the V, you don't have Wigner semicircular law, and so how, how the proof that you gave me today adapts in this case? So if you change the V, you'll get a different proof, a different, a different distribution in general, and but, but one thing that is, which is still going to be true in that proof is that you're still going to get a Z with a random weight on it, okay, except that distribution on the weight on the, on the, on the edges is going to be different, okay, so you're still going to get some law which is a multiplicative convolution of, of arc sign and something, so that's one thing you can do, almost just, but just by the fact that this thing sort of changes continuously what's written on the, okay, and, and that's actually, that, that is true, and, and also, and I think about, here is that there really isn't anything special about beta, less than one or greater than one here, so, so most of the proofs just go, go, go the same way, okay, and, oh, and one more thing, okay, so, now that we have introduced this, so you may want to take beta go to infinity here, so what happens in that case? Well, what happens, so, oh, you, you got, you're just going to get this matrix, right, which is, which has what, zero here, square root of n, square root of n minus one, and so on, right, zero, zero, that's the limit of the matrix, and so, so what is, what is this thing? Well, what it is, is actually the recursion matrix for Hermit polynomials, so, so, so you just get, so, you know, the Hermit polynomials in this case are going to be the analogs of the characteristic polynomials of the corresponding GUEs, okay, and the zeros of the Hermit polynomials are going to be the analogs of the, of the eigenvalues, so the eigenvalues of this are actually exactly the zeros of the Hermit polynomial, so the same proof gives you semicircular law for, for, for, for that process, so it's, it's much more classical and unknown, but, but, but, but, it's the same story goes through here, so, so you can, and, and, you know, basically the proof doesn't use anything except this, I mean, the randomness here in all of this proof just didn't matter so far, all right, okay, so that's, that's the remark for general betas, a good exercise is to compute the Jacobian, a simple one actually, is to compute the Jacobian from the matrix entries to M to the moment, because that matrix is actually upper triangular, so it's either easy to compute the Jacobian, so the other one is a little bit more massive, but it's, it's not, it's not that hard in any case, okay, so what I'd like to do is, is continue where we, where we left off last time, right, which is the VBP transition, and I'm not gonna keep this beta come along with me, I'm just gonna, you know, say that everything that I do works for general beta, okay, and, and this is what we, this is the theorem that we wanted to prove, which is the GOE with mean, okay, so, so, so you look at the GOE, you look at the top eigenvalue of that, and look at GOE of M, plus you add a mean, so a simple space just to put an all one matrix, an all one matrix to it, and the right scaling here is A over square root of N, and then you want to divide by root N, right, because typically it's the order root N, and the theorem is that this converges to pi over A, which is, which is one, or sorry, two, if A is less than or equal to one, and A plus one over A, if A is greater than or equal to one, okay, so this tiny change in the mean of the normals is actually, actually doing something in the top eigenvalue that's noticeable, so here is first of all an interesting, a nice, an interesting question, doesn't even really know what the Wigner semicircle law will look like for this thing, is there, is there gonna be a change in the distribution by adding this mean, so no, why not? Exactly, so this is a rank one matrix, and there is a very nice theorem that says that if you have a matrix with some eigenvalues, symmetric matrix, and you do a rank one additive perturbation, let's say that it's positive, okay, then what will happen to the eigenvalues is that they all move up, but they can't overcome the next one, okay, so there is interlacing, so the next eigenvalues have to be, have to be distributed something like this, okay, so the top one of course can go as far as it wants, but it's just one eigenvalue, in the semicircle law you don't see any of it, okay, and it's not hard to prove that if you have, you know, something going to the semicircle, then anything that interlaces with that will also go to the semicircle, that's just generally true, so there is, so you don't see this in, in this level, but in the top eigenvalue you see, okay, and the motivation that I try to give for this was, was the following, so if you look at lambda one in the sense of real equations, right, so over the following, so you look at z plus, plus a loop at a, loop weight a at zero, okay, so it's this, this graph, okay, so this is equal to two, and a plus one over a, two if a is less than or equal to one, and a plus one over a, this is far away, okay, and, although we're not going to prove this, this is, this is going to be a nice exercise for, for you guys, one thing that you can see right away is that if you look at this vector, which is one, a inverse, a minus two to the minus three dot, dot, dot, okay, then this satisfies the eigenvalue equation for that, okay, so what does the eigenvalue equation say, well with, with eigenvalue lambda equals a plus one over a, so the eigenvalue equation says that when I write this function here, one a to the minus three to the minus three and so on, then the sum of your neighbors is just a, a plus one over a times, times what you, what's written there, okay, and that's kind of obvious, the only thing you have to check that it's satisfied here, because that's not obvious, but it's also satisfied there, okay, so this is actually true for any, any a, also a less than one, but when a is less than one or a is one, then this is not, you know, two, so it's not really an eigenvector or anything, otherwise it is, you know, two, so in fact, if you look at the spectrum of this guy, it's also true that this is pure point, this is absolutely continuous, all the way until a is, a gets greater than one, at which it will get an isolated eigenvalue of a plus one over a, okay, so let's look at the proof, actually we, I need a, I need a lemma for the proof of this VBB transition, and the lemma is very simple, so, so, so it's about approximate eigenfunctions, okay, so here's, here's what it is, and it will be an exercise to prove, okay, so, so a is a symmetric matrix, and, and let's say that you have some V, which is almost an eigenfunction, okay, so we're almost an, almost an eigenvector in the sense that A V minus lambda V, okay, is less than or equal to, in norm is less than or equal to epsilon, okay, so this tells you that V, and let's also assume that V has, the V, the norm of V is greater than or equal to one, of course, otherwise this can happen just with V having small norm, okay, so, so then, then A has an eigenvalue epsilon close to lambda, and you just prove that by taking the inverse of this, okay, and looking at the norm, okay, so this is, this is what we need, okay, so let's do a proof of the BBB, BBB transition, okay, so let's look at the A less than or equal to one case, so first of all, because you have the same Ziegner semicircle law, okay, you already know that the limit of this lambda one has to be, has to be at least two, because even if I don't do a perturbation, this is at least two, so limb, limb soup, whatever in, in probabilities is greater than or equal to two, but, but then actually, I just have to show that it's not more than two, but for this actually, we can use this, this inequality, remember that each year the Jacobian matrix, we have this inequality that this is less than or equal to the max of Bi plus Bi minus one plus Ai over I, with the convention that B0 is Bn, which don't exist, so we just set them to zero, okay, okay, so I'm a little bit running ahead of me, so myself, so let's, let's, let's backtrack a little bit, so we're gonna use this, but how? Well, the trick is the following, so you see that GOE over there, right, and I'm adding, adding, adding to it this rank one, rank one matrix, because the GOE is invariant under orthogonal transformation, I can orthogonally transform that whole thing in the parenthesis, including the perturbation, and I don't change the top eigenvalue, so why don't I just transform it so that that one one vector goes to, again, a force coordinate vector times a constant, so when that happens, then the first coordinate vector will, will be, so, so will be changed, and then I can apply the trigonalization the same way that I did before, okay, or you could say that the rank one perturbation on the first coordinate vector doesn't change, it commutes with the trigonalization, that's also true, so, so the model that you get is, is exactly this, except you get here an a times root n, okay, and the times root n and, and over root n is just, comes from the fact that those vectors are not of length one, whereas the first coordinate vector is of length one, okay, so, so that's it, that's the, that's the difference, okay, so this a1 has changed, but remember that these things were of order root n typically, at least the top ones, the ones that are important in this thing, and this thing was of order root n, and this is where, this is how we got the upper bound to root n, but for the first coordinate, we don't have one of these root n's, because bi minus one is just zero, so we have an extra, we have some room for another root n there, okay, so we can still put in this ai, so even if we set a0 equal root n, or a1 equal root n plus this normal, this band works the same way, okay, so this takes care of the case when a less than or equal to one, now you may try to use it for a greater than or equal to one, to get an upper bound, but what do you get as an, as an upper bound, well, you know, basically if a is greater than or equal to one, then the upper bound in this max is going to be dominated by the first term, right, and so you're going to get, you're going to get, so the first term is b1 plus square root of, plus a1, right, that's what the first term is, because b0 doesn't exist, so this is of order square root of n, and you know, this is like a times square root of n, right, so this gives you an upper bound of a plus one, okay, but that's not what we want, we want a plus one over a, so it's, this upper bound is a little bit weak, so what do you do, well, here's what you do, you take this vector, v, okay, and you cut it down after k steps, and actually you're just going to chop it off, even to be an n-dimensional, n-dimensional, so you're going to want a inverse a to the minus k, and then you just put zero, zero, zero, and you're going to make it lengthen, it should really be v and k, so what happens when I plug in, and you set lambda equal to a plus one over a, so this vector is an eigenvector for this, this operator, right, for z plus plus a loop of weight a at zero, which is almost the same as this operator on the top corner, right, that's, that's how we prove this Rigner semicircle law, the top corner just looks like a z plus here, and this noise, noise doesn't really matter, okay, so that means that when I plug in, so when I plug in jv, maybe I divide by n, so one over root n, one over root n j is what looks like that, one over root n jv minus lambda v, okay, so look at the norm of this, so v is going to be almost an eigenvector for j, there are two kinds of noises, two kinds of problems, right, there's some, some problem coming from the noise, which is, which is just saying that those are not exactly, these are not exactly ones that are written there, so that's actually of order, of order one over root n, okay, and there is another problem coming from the boundary condition, because this we stopped here, so everywhere it's almost satisfied, but here there's going to be some error, so, so, so that's, so that's c times a to the minus k, vk, yes, okay, so, so we, what we did is we looked at this limiting operator here, we looked at its eigenvector and we tested the eigenvector on the finite matrix, makes sense, basically, we cut down the eigenvector and tested on the finite matrix, okay, so, so this means that, so there is, so that j, one over root n j has an eigenvector a plus one over a plus or minus, you know, this thing, o one over root n plus c times a to the minus k, okay, eigenvalue, sorry, now this is when a is greater than one and this is outside the semicircular law, right, so we just know that we have an eigenvalue, all we have to argue is that this is the top eigenvalue, because now this is close enough, right, just setting this thing small, of course we'll get what we want, but why is it the top eigenvalue, because of interlacing, right, or the other eigenvalues, we know, we'll have to be below the eigenvalues of the unperturbed operator and that's going to be separated from here, okay, so, so lambda two of j, j, well, let's say this is j a, okay, we know that it's less than or equal to lambda one of just the j, without the, let me put square root of n, without the perturbation and we know that this is, this converges to root, this converges to two, which is less than a plus one over a, okay, so already the second eigenvalue is too small to be this, so it just has to be the first one, okay, so this, let's call this lambda star and that completes the proof, right, so if we have seen proofs of this thing, it's probably the shortest one, but of course, there is much more you can say about this rank one perturbation, but the basic idea you can get from here, any questions? So you can do the same thing with the bottom of the spectrum, you just do a, you do a change, sign change and then you have to change signs of certain normals, oh right, right, so, okay, so the bottom of the spectrum corresponds to a negative, but a negative is not interesting because every, all the eigenvalues can only can go down and because for the top of the spectrum, you know, there is no way to go, nowhere to go, they're all, so there is nothing interesting happening, yeah, in that sense. Can you add up the proof if you have a rank two perturbation? No, not this proof, yeah, so what you can do then, and this has been done in papers, is that you look at a five diagonal matrix, and that's the setup in which you can study rank two perturbations because then there are two things that are fixed, right, and then anything that you do rank two, there works, so then you can do it, but actually, we haven't done it in this setting, so we've done it in the more precise setting, so when, you know, you can then ask how to tune A that the tracovidum distribution changes, and in that setting, this has been done for rank two as well, so what's next? Yeah, so I got some questions last time about whether you can do the tracovidum distribution this way, okay, because we did something about the top eigenvalue, we did BBP, can you do the tracovidum distribution? And well, of course, you know, you can try, and the first thing you would do, right, is you want to understand the top eigenvalue of this matrix, and if you want to understand the top eigenvalue and you want to understand the structure as well, you'd better look at the top eigenvectors, so what do you do? You put it in your computer, you do some simulations, okay, and you see what the top eigenvectors of this look like, okay, and what you find for a very large N is the top eigenvector, you know, it looks, you know, something like this, and it lives in a small fraction in the corner of this matrix, okay, that's, you just look and see, there's no, you don't have to do any measurements or anything, it's just obvious right away, so that somehow suggests that the information about the top eigenvector is contained in some small parts over here, so you start with that and try to make some progress, okay, and what I'm talking about now is a result of Edelman and Sutton, non-rigorous, so basically that, because I'm not going to be very rigorous, and we made this rigorous with Jose Ramirez and Brian Ryder, so whether we start with, so you want to understand the top corner of this matrix, which I just erased, okay, so what does this, what does this matrix look like? Well, let's, okay, so let's, let's look at what the matrix itself looks like, so it would be nice to do some expansion, the expansion that you want to use, so we said beta equals one, you're going to write chi of n minus k, and now we're interested in the k's where k is smaller than n, right, because we're just looking at the top corner, is, well of course you know this is square root of n minus k, plus a normal, plus, well, okay, and maybe we write this, we expand this as well, so it's like square root of n minus one over two root n, okay, right, just the expansion of the, what's this normal zero one half, plus the log, okay, so again there's some little terms, okay, so let's look at this in the matrix form, okay, so I'm going to look at the following, let's take two root n times the identity, okay, subtract j from that, okay, and I'm going to scale it by m to the gamma, I'm going to, and I'm going to look at the top m by m corner, okay, so m is going to be, m is going to be less than m, and I'm going to scale it by some factor m to the gamma, and I'm going to look at the top m by n corner, okay, so let me just write what you get, so this is, so first of all you get, you get this j which was mainly, you know, just z plus, right, the first term was just z plus, so you get a two minus a z plus, so what is that, that's just, that's just discrete second derivative, actually minus, okay, so that matrix and it's scaled by m to the gamma, square root of m, so that's the first term, the second term m to the gamma over two root n of, okay, so zero is on the undiagonal and you just have this k's, okay, these terms here, okay, and then you have the third term which I'm going to write here, so this is just m to the gamma, and this thing, right, so it's n1, n2, these normals from the diagonal and then you also have normals coming from the k's, until the one, until the two, they're not the same variance, but they're not, not worrying about that now, okay, so you have this expansion and then you realize as Edelman and Sutton did that these things, well first of all this thing is the most telling one, right, because this is a discrete second derivative, so what you hope to do of course is when n goes to infinity it should converge to a continuous second derivative in some sense, if you scale things the right way, okay, and in what sense does this mean, does this, in what sense, so let's try to state that somewhat more precisely, so you take some, you know, take some fantastic function f, okay, it's completely supported and smooth and everything you want, and you sample this f, right, so you sample it at 1 over m and 2 over m, f is gonna be, m is gonna be your scale, 2 over m, f3 over m, okay, so this, let's call this f sub m, okay, this vector, you go all the way, this is an n vector, and this is a discrete second derivative in the sense that if I plug in this vector, plug in this discretization of f, then what I get after applying the matrix is roughly the discretization of the second derivative, that's exactly the sense that this is a second derivative, and because it's a second derivative, you know, you have to have, you have to also scale it properly, okay, and what has to be the scale, well the scale has to be, it has to be a following, you have to take this matrix and multiply it by the inverse spacing squared, that's how you take second derivative, okay, so if, so this thing will converge to a discrete second derivative, if m to the gamma is equal, m to the gamma times square root of n, okay, is equal to m squared, okay, then this guy will converge to the second derivative, right, so this scaling factor is gonna be, has to be the right thing, so in the end we'll have to figure out this gamma, what it is, okay, that's our goal, okay, so this, we'll write it down, so what is this gonna be, okay, what is this matrix gonna go to? Well, you know, this is just by multiplication by k essentially, it's of the diagonal, but in fact you can move these things on the diagonal for free, it's just a little error, so once you move them to the diagonal, let's do it here, not during the diagonal, here is the two because there's two of them that I moved there, this is gonna be multiplication by x in the limit, right, but you know actually what you want is that when you're about m into this matrix then you get about the constant number there, okay, so what this gives you is that, what do I want, so what does it give you, so m to the gamma over root n right, times m equal to 1, okay, that's saying that's about roughly, about m into this thing you get roughly an order of one multiplier, so that's the next thing, and the last thing is this, so well this is gonna be some kind of multiplication by white noise, okay, we're multiplying by these independent things, and in fact when you multiply by white noise you don't really get functions, so there's gonna be some side problem here, but you at least want to get a distribution that makes sense, and it makes sense actually, you know, if this multiplier is exactly square root of n, it's just a standard thing from Brownian motion, so m to the gamma is equal to root n, okay, so we got these three equations, and actually, you know, it makes sense to say m equals n to the alpha, right, so you're gonna say what should be the scaling of this, so this is just a new parameter here, so what do you get, so here you get alpha gamma plus one half is two, right, this is what you get here, here you get alpha gamma minus one half plus one is zero, okay, so alpha times gamma, and here you get gamma, this is the best one, gamma equals one half, so is that correct, it should be, and what's the solution for, so there's three equations actually, and two unknowns, it's a problem, unless they're consistent, are they consistent? No, they're not, all is free. Ah, yeah, thank you, yes, yes, yes, like that, and it's a two alpha in the left equation. In the right answer. Yeah, and the gamma should be negative one, I think, yeah, gamma is minus, yeah, this should be minus one half, okay, let's see, so let's see if I get everything right, I'm sorry, and this is two alpha, okay, I was, I was worried, you know, that after lunch everybody's gonna be too tired, but I wasn't thinking of myself, I was thinking of you guys, anyway, so yeah, actually, right, so maybe this should be, now I got a bit confused, but I don't know, that's, that's correct, right, so, so yeah, so gamma should be minus one half, not one half, I'm sorry for the last one, and, and alpha should be one third, right, so now you tell me, is that, is that working? Come make was minus one half and alpha one third, yeah, that's good, I think it's roughly good, there's a mistake tell me later, okay, so, so, so what does this mean, okay, so it means that if I look at actually a tiny portion of my matrix, which is one third on the corner, on the top corner, then that's why, that's why these eigenvectors live there, right, because, because, because this thing tells you that this will converge to some operator, okay, and, and indeed this is true, okay, so, so if you look at this guy here, so you'll call this h n, okay, and then h n converges to the following operator, differential operator, and in a second derivative, plus multiplication by x, but plus, plus, actually there is a two over root beta, so I write the general beta version, but, but you know, in this, in our case it would just be two, or b prime, okay, which is multiplication by y, no, it, um, okay, so this is a theorem, so, yes, so I'm gonna tell you about that, so, so there, we have a lot of things to do, okay, what sends in this converges actually further down the line, first I have to tell you what the hell this is, okay, so this is actually acting on l 2, l 2 of, of r plus, okay, it's a positive real line because there's a beginning there, and, and also you, you, you want functions which, which started at zero, so Dirichlet boundary conditions, um, and the problem is you take a function which is smooth here, right, you multiply it by this white noise, you get a mess, it's not gonna be a function, and multiply it by x, that's fine, so in fact it turns out this operator, the only way it works, it can work nicely, so you can start with a nice function and get it, get a nice function, but the only way you do it is by the second derivative canceling out the nastiness of the b, b prime, and that's really exciting how that happens, but we don't wanna, we don't wanna worry about it, so we'll, we'll have to do a different approach, um, so let me first tell you how the convergence is, okay, so the convergence, okay, so, so you embed r to the n in l 2, okay, uh, you know, with functions, okay, this one is linear, okay, so e 1, so the first coordinate vector goes, or e k goes to, you know, the, the indicator of k over m, k minus 1 over m, k over m, interval, okay, and you, you have to multiply this by maybe root n, so that, so that this is actually a vector of length 1, so, so this, this way, you know, this is exactly what I was doing here before, just, just representing the functions by the, by their piecewise linear versions, okay, so now, at least there is a common space, okay, so these two things live in the common space, but, but then there is still this huge r n is the, so r n is where h a, in which h n, so j n acts, right, so that, that the, the r n is the vector space on which the, on which j n acts, and it, it, you just embed it into piecewise linear functions, okay, so of, of, of, of, of, of R plus, that's what you do, okay, so, so, so this actually puts those functions, you know, at least there is some, some meaning of, of this now, but there is still a problem that, you know, here there is a second derivative, so this is not going to work for piecewise linear functions, so basically, you know, the, the places, the domains, if you like, of these operators are completely disjoint, but it turns out that this is not a problem, okay, the reason it's not a problem is because you can take the inverses of these operators, or resolvents, but, you know, let's just say inverses, we don't have to be so fancy, okay, and let's call that operator the stochastic carry operator, okay, and these inverses are actually compact operators, okay, so, so this kind of makes sense, if you think about h n, right, we took a limit at the top eigenvalue and we subtracted it from the identity, so actually the eigenvalues are h n, are going to start around zero and then they're going to go up, they're going to be larger and larger, in the limit you expect the eigenvalues to be, to be going all the way to infinity, in fact we will prove that, but when you take the inverse then all of those things get inverted, so the eigenvalues will converge to zero, everything's going to be nice, okay, so, so h n inverse and, and this inverse actually have compact extensions, so, so they can, it can be, they can be defined everywhere on L2, okay, so you can look at their difference, you can couple the noise here and there, look at their difference and you can show that this goes to zero, this norm, okay, this is the operator norm, so this is just the operator norm goes to zero, okay, so this thing is called norm resolve and convergence, okay, because the norm of the inverse is convergent and, and it implies everything you want, okay, so it implies the eigenvalues converge, it implies that the eigenvectors converge in norm, so the eigenvectors of this guy have to be close to the eigenvectors of that guy, so really everything you want is implied, so that's, that's, that's, that's the, that's the notion, I say your beta is that, so, which I haven't quite defined to you how this works, okay, but I'll, I'll, I'll define to you in a second, I won't tell you exactly but I'll tell you how we work with it, okay, so how, how can you define this actually, so one way to do it is just to define the specific functions, domain of it on which, which is self-adjoint, okay, so that's the abstract theory and you can do that and we won't do that now, because it's messy and it's also not especially useful, but instead what you'd like to do is define, define a bilinear form, okay, so just, just we're going to define, if you can define using, eigenvalues using a rarely quotient characterization, so we just have to define a bilinear form that corresponds to this SAO beta, okay, and then we can define what the eigenvalues are and so on, okay, so, so let's, so that's what I'm going to do now, so bilinear form, so just some motivation first, so SAO beta is stochastic array operator, okay, so stochastic array operator and array is written here and why is it array, well it's array because if you erase this, okay, then you get this operator, which is just the array operator, okay, so, so let's first look at the array operator, what it does, okay, so let's say, so, so you look at this thing, so this is the array operator, so let's say, let's say you want to solve this differential equation, right, so let's forget about the boundary condition, sorry, so you just want to solve it, so let's, sorry, right, so let's say first lambda equals zero, okay, so, sorry, lambda equals zero, so what do you do? You look in Wikipedia, okay, and you find two functions, ai of x and bi of x, the array ai functions and the array bi function, what does these functions look like? Well, they look like this, this is the ai, okay, I think it grows maybe, and the bi, bi just blows up if you go this way exponentially, okay, so, so if you want to find eigenfunctions of this, they have to be now too, so this guy we can forget about, the other thing you notice that if you have a general lambda, then you can move it into the x and put x minus lambda there, so the solution for zero is the same as the solution for lambda is just shifted, so, so there is a, you know, it's a second order, so there is a, there is a two-dimensional space of solutions always, we're not allowed to use this one because that blows up, so all we're, all we left with is this ai and shifts of that, okay, so when I, when do you, where are the eigenvalues of this, of this array differential equation or array operator? Well, we have, if we set this, if we set this boundary condition, okay, then it just happens, if we happen to shift the array function in a way that the zero gets to zero, okay, so the eigenvalues are lambda i equal to minus zi, where zi are the zeros of the array function, okay, and the eigenfunctions are just the shifts of this single function, we just shift them, so it's actually very simple and beautiful, okay, and what do you expect? And there is classical asymptotics, so actually you know that lambda i is asymptotic to some constant times, times i to the two-thirds, okay, so this, this eigenvalues get denser and denser here, okay, so, so this is the array operator, and now you add some noise to it, so you expect somehow the eigenvectors to behave similarly and everything to behave similarly, and actually that is the case, and I'm going to prove a few senses in which this is the case, yes. Why do you want f of zero equals zero? So, so f of zero equals zero, this is a typical thing when you do a discrete approximation of, of some differential equation, so, so you know there is some strange condition at the, at the end of the matrix which forces any solution to, to, to, to be small there, and, and it becomes in the limit, it's not part of the differential equation, it becomes a boundary condition, it's, it's not special to hear, this always happens, so for matrices you don't need boundary conditions, basically, and, but in the continuous case, yes, okay, any other questions? Okay, and actually let me, let me just say one more remark, so, so what you can do is you can change, put the rank one perturbation on the right scale to this matrix, and you can change the boundary condition with the rank one perturbation, so if you do the right rank one perturbation you can, you can get a Neumann or, or anything in between, okay, so let's say that we define the L star norm of a function and simply f af for the area operator, okay, so I take the af and take the inner product of f, so what is this? This is the integral of xf squared plus, well, there is the second derivative, you do change of variables, you start at zero, so, so you're going to get integral of f prime squared, okay, so, so that's what it is, it's dx, okay, so, so then you can look at the function of L star, the set of functions in L star which are just f such that their f star norm is finite, right, and so, so we basically start with the area operator in this sense, we just look at the functions for which this makes sense in the area operator, so now what we'll do is you can define the bilinear form for this stochastic area operator for all these functions, so what is it going to be? Well, you want to define, right, refinal star, what would you like to define? Well, you know, the first part is okay, I mean, the area, stochastic area operator is just the area operator process multiplication by white noise, so just, I just have to define multiplication by white noise for this, so what I'll need to define is integral f squared v prime dx, okay, so you could, okay, I'm going to call this, you could call this, well, anyway, that's, that's, that's fine, so this is what we would like to define, if you can define it then we're good, so the first attempt, of course, is you do integration by parts, so f is differentiable, so, so you're good, actually the zeroth attempt is to try to use ito, ito theory, but the problem with ito theory is that you really like to prove this for every function almost surely, okay, so some kind of L2 overall is not good enough, okay, so we need the stronger, stronger sense than ito sense, but it's fine, you can, you can try to do it like this, f prime f dx, so, well, this is, looks, looks actually pretty nice, now this is completely a fine function, okay, this thing also is a fine function, because f is differentiable, so the only thing is we don't know that it converges, okay, so it may not converge, so, so for this you do, you have to do something slightly differently to fix this, and here's what you do, so you write b, the Brownian motion, as some average, so b of x, so we call this, this is some average, we call this b bar, plus, plus some remaining term, b tilde, so why is this good? Actually, it's good because it's very good for this integral, if you, if you look at it, because this term here is going to be smooth, okay, and it's differentiable because you have averaged it, and this term here is going to be small, because it's just the difference of Brownian motion and an average on an interval of length 1, okay, so in fact that's the way you will define this, so you're going to do the first part b tilde, let me see if I'm doing it right, yeah, b tilde prime dx, sorry, b tilde dx, so this, this is, this is the one you're going to just do with integration by parts, plus there is the, the other part, you're just going to keep it as it is, so f squared b bar prime dx, okay, so this is, this you can take as a definition, no boundary term, because that has zero at zero, so now let's see, does this make sense? Well, we can make sense out of it, and here is first of all a lemma, if you look at b tilde, b tilde prime squared, okay, so both of these are less than or equal to s functions in absolute value, so I'm a bit constant times log 2 plus x, so let's see why this is the case, well what is b tilde? Well, what is b bar? Well, b bar is just the, this, sorry, b bar prime, yeah, so what is b bar prime? Well, b bar prime is b bar prime, b bar prime is just b x plus 1 minus b x, right, so this is actually a stationary process, okay, not only a stationary process, it becomes independent as long as you get more than a distance away from it, so this basically behaves like an independent sequence of normal variables, okay, and so what is, how is that gonna grow? It's gonna grow that root log, okay, just tender thing about normal variables, so you square it and you have this, and b tilde the same story, b tilde is also a stationary process, actually this is a, this is a good exercise, so you already see that this term is well defined, why is it well defined? Because this b tilde is bounded by log, and the log is bounded by x, so, and this guy is well defined, so, so this, this integral is perfectly fine, this integral you have to schwarz it, okay, so you're gonna, gonna get the term of f prime squared, that's fine, okay, and you're gonna, gonna get the, gonna, gonna get the term of f squared b squared, b tilde squared, when you schwarz it, okay, and that's also fine, because that's again just growing like log, all right, so, so what, what's the conclusion? I'm gonna write, let me write a conclusion which is, which is very easy to show if you play with the constants, well, okay, okay, so the conclusion is this, okay, so for every epsilon there exists a random constant c, so that if you look at this integral, which, which I defined this way, so integral f squared b prime dx, okay, again, defined that way in absolute value is less than or equal to, so there is the L star norm of f times epsilon plus c epsilon times the L2 norm of f, so what does this say? It says that, yeah, okay, so for every f in L star this is true, so this is basically just what I said, okay, it's a little bit more, it's a little bit more that I said that basically all these terms can be bounded by L star norms, but in fact, if you play around with the constants properly you can make the L star norm part as little as possible for an expense of getting an L2 norm, which is larger, so let me have 10 more minutes and I want to get to a very specific point, which is the large numbers for the stochastic area operator, and it's just gonna be enough for that, okay, so you probably have seen the Courant-Vichier characterization, okay, and this is gonna be the first time that I precisely defined to you what the case eigenvalue of this operator is, okay, and now I'm actually defining it and there's no hand waving, okay, so the Courant-Vichier characterization is a generalization of the Rayleigh-Quachin formula, right, so the, so, so we have scaled everything so that the eigenvalues are, there's a minimal eigenvalue and then there are ones that are larger, so lambda 1 is not gonna be the minimal eigenvalue, so lambda 1 is what, is the inf over all b, which is subspace of L star with dimension k, right, the soup over f in b of norm 1 of, you know, this actually works for any operator, so let's call this anyway, but let's write it like that, that's it, and this is lambda k, okay, so let's see what, what is it when lambda 1, when k equals to 1, right, so when k equals to 1, then, then you just want to look at the one-dimensional subspace, so there is basically one vector in there, or two of length 1, so you're just looking over all one-dimensional, over all vectors of length 1, the inf over that, so that's the lowest eigenvalue, you have seen that, but it also works for general k with this trick, and, and of course you can generally define this for any operator over L star, and notice that, so the eigenvalues behave nicely with operator domination, okay, so if you have an operator a, that's never equal to b, what does this mean, this is operator monotonicity, okay, what this means is that for all f, a f, f is less than or equal to b f, f, okay, that's, that's, that's what this means, for all f in L star, say for, for, for now, okay, so why would, what, so, so, so let's see, so what does this imply first of all, notice that if a is less than or equal to b, then lambda k of a is less than or equal to lambda k of b, why, well simply just put this in here, okay, both the sup and the inf are monotone-increasing functions, so, so, so everything is fine, this is obvious actually, so now we're gonna do, and what have we proved for the stochastic core area operator, right, we showed that multiplication by b, right, is, is less than or equal to b prime, multiplication by b prime is less than or equal to in the operator sense by epsilon times the area operator plus c epsilon times the identity, this is what we showed there, right, for every f, this inner product here is less than or equal to that, that's exactly what this thing means, don't even need the absolute values there for that, okay, so what does this tell you, well you can, you can rearrange things, right, so you, so the stochastic area operator is just the area operator plus this multiplication by b prime, so you have the following thing, so the stochastic area operator is less than or equal to 1 plus epsilon times the area operator plus c epsilon times multiplication by the identity, just, just by adding the area operator to both sides of this and do, and the same thing you can get this way, right, so 1 minus epsilon times the area operator minus c epsilon times multiplication by the identity, okay, so this is the operator sense, okay, so, so if you look at the eigenvalue of this lambda k of beta then this is less than or equal to 1 plus epsilon times lambda k of the area plus c epsilon and it's greater than or equal to c epsilon minus c epsilon plus 1 minus epsilon lambda k of the area, okay, and the quantifier is for r or epsilon there exists a c epsilon, okay, so we've given upper bounds with some random constants, upper and lower bounds for these eigenvalues, so what is the conclusion? Since we know that these lambda k's, remember we said that lambda k is approximately k to the two-thirds and some constant, this implies that lambda k the beta is also approximately k to the two-thirds times some constant, the same constant, okay, so this is the law of large numbers for eigenvalues because you know it's the first order behavior, okay, k goes to infinity in that sense, it's not the fluctuations or it should be a clt as well, okay, good, so I got tired, okay, so let me recap what we did, okay, so what do we do? We looked at the top corner of this matrix and we guessed that it should look like some differential operator and I stated the theorem that it converges in a very strong sense to actually this differential operator, I didn't tell you what this differential operator exactly means but I did tell you what it means exactly as a bilinear form, okay, so if you believe me that the eigenvalues mean the same thing are actually the same in both definitions then we should be fine, okay, so believe me that and from that everything, from there, from these two things which I didn't show you, everything else is rigorous, okay, so I have proved now that this guy actually has eigenvalues defined this way and I have proved that those eigenvalues actually behave like this, okay, so these are perfectly fine proofs, so, you know, the punch line is of course that the test trace if we don't beta distribution which was of course defined for beta equals 1, 2 and 4 but first defined through this paper for general beta, you just define it as minus lambda 1 of the stochastic area operator, okay, and the minus is just because this is a kind of random Schrodinger operator and in there the tradition is to, you know, to have a lowest to make it almost positive definite, so there is a lowest eigenvalue but it's bounded from below, so this is, this actually is a definition for beta not equals to 1, 2 and 4 and it's a theorem for beta equals 1, 2 and 4 and so, so I'm thinking about what to do next time because I wanted to do a little bit more, completely unrealistically, so there are two things I can do, one thing I wanted to show you is that this representation as a random operator is actually extremely useful, so, so you can very quickly get tail bounds for example for the trace if we don't distribution and you can get various asymptotics that you might be interested in, so we already had one, right, a non-trivial one, these things, if you want to try to prove this in other methods it's actually quite difficult even for beta equals 2, especially if you want a little bit more precise bounds and it's more difficult, so that's one option and the other option is that I'm going to go and talk about the bulk limit which is a whole different story, so anyway I stop now and you can tell me what to do if you have an opinion later, Paul, right, okay And are there questions on this lecture? So Sina, you have this, so Sina, the title of the talk was operator limits and depending on what statistic you want to understand they're different operator limits, right, so we already saw two of them, one of these Benjamini-Strand limit, the other one was the rooted limit and this is the third one which is a differential operator kind of limit and all of these capture different things, so if you want to look at the bulk you again get an operator but it's very different from this, it's a dirac operator as opposed to a Schrodinger, so SAO is actually considered as a random Schrodinger operator it's one dimension, yes, yes, complete, it's X and R, it has the strong localization of this Yeah, I mean I don't know what you mean by strong localization, yeah, so they're actually even stronger, so the K of the L2 mass of the eigenfunctions is like e to the minus X to the three halves, so faster than exponential No, no, this is due to the X, yes, so in fact a lot of things don't change from the ordinary area, as you say, yeah, what about some models, some models with potential that you introduced in the beginning of signature? Yes, so you can prove the same things for the models with those potential, same limit in the case that V is, so the conditions we had, we had to be uniformly convex polynomial, so if you have some exotic things then they don't fit into this but you can try to deduce other limits and their works in progress for that What about the top eigenvalue of the V? It converges there Well, as the limit of the top eigenvalue distribution, but for that you have to prove that there exists a limit and this is the first proof that there exists a limit So you do prove the limit? Yeah, so this convergence, this implies the convergence of the top eigenvalue, it implies everything you want, it's extremely strong, so in particular the distribution of the top eigenvalue converges to the distribution of the, I guess your bottom eigenvalues