 I would like to thank the organizers for inviting me. It has been quite educational so far. Hopefully, this talk will be interesting to some of you. This is joint work with Anish Malik, who is currently in Chile, is okay. So the question, so the entire talk is about two random matrices. So random toplits and random Hankel matrices. So let me define them. So this is the random toplits matrix. And also symmetric. So the random toplits matrix looks like this. You have x1, x2, xn minus one. And it's also symmetric. So the same. And it's a toplit. So it is the same x0 on the diagonal, x1 here, et cetera. So it's constant on these diagonals. And the random Hankel matrix is the matrix which is constant on this anti-diagonals. So you have, let's say x1 here. x2 and x2. x3, x3, x3, and so on up to xn, xn plus one up to x2n minus one. And it's constant on these diagonals. Where xi's are iid, normal zero one random variables. So that will be the assumption. So the entire talk is about these two random matrices. So these were first introduced by Bai, Zidong Bai in 1999. He introduced these as more restricted matrices compared to Wigner and other things and asked various questions about it. In particular, about the limiting spectral distribution and stuff like that. So some of it was proved. So in 2003, I think, 2005 and six, so let's also let me write LTN for the empirical spectral distribution of TN. And similarly, and I'll write LTN bar for the expected empirical measure. So this is the measure which puts mass one over and at the eigenvalues of TN. And this is the expected measure. So these are, so in paper of Brick, Dembo and Xiang. This is from 2006. There's also a paper of Hammond and Miller in 2005. They showed that the limiting spectral distribution exists. So these things have a limit. So they showed that LTN and LTN bar, they converge to some measure mu sub T. And the same is true for Hankel matrices. So the Hammond and Miller did it for the top plates and Brick, Dembo, Xiang did it for both. So what are these muti and muh? So these are not random, of course. So they are deterministic measures. And this does not require the normality assumption. So under, so you only need to assume that the excise have mean zero and variance one, but they showed that these limits don't matter. I mean, it's the same for all such distributions. And yeah, and they are not random. So that's the result. So their approach was by the method of moments. So the proof was just, I mean, of course, the method of moments, the combinatorics is much more complicated here because the same excise are there in many places. And you have to keep track of many things. So, but it can be carried through and they found some descriptions for the moments. So they showed that the moments of these quantities converge and they determine a unique measure which they call mu sub t and mu sub h. But the moments are highly inexplicit. They're complicated. So it's enough to get some properties. Moments are good to get certain properties of a measure. For instance, from this one, it's firstly, it uniquely determines a measure that is one. Second, you can say something about the tails of the distribution. In particular, both of these are unbounded support. So mu sub t and mu sub h have unbounded support. All that can be got from the moments. But if you draw a picture, it's a very simple to do a simulation. So the picture of mu sub t, if you draw random topletes matrix, take its eigenvalues, draw the histogram. This is what mu sub t looks like. And mu sub h looks like this. They are both symmetric about the origin. This looks unimodal, that looks bimodal. So some things that moments don't tell us, so these are the questions. So moments are good for certain things, but certain properties are hard to say from the moments. So for example, the questions are, do they have densities? And if they do have densities, can we say whether the densities are smooth and smoothness and properties that you see in the picture, such as unimodality or bimodality? So these are questions which are not apparent. So in the very basic case of Wigner random matrix, of course it was done by the method of moments, but the moments are just Catalan numbers. And even if Wigner did not know it before, it is not hard to compute the generating function and see that the measure is the semi-circle law. But here every moment is complicated. To calculate say the k-th moment, odd moments are zero and even moment. To calculate the k-th moment, there are something like two to the k integrals. And the integrals can have dimension going from zero to k or so on the unit cube. So they are highly inexplicit. Even numerically, I was never able to compute more than up to the 10th moment of these things. Maybe if somebody is more savvy, they can compute. And the point is if you can just guess the measure, of course, from the moments, then well and good, but nobody has ventured a guess on what exactly this mu sub t and mu sub h are. So what I'm going to say may have some suggestion for that. So this question, there is not much work and the only thing I know about this is that this answer, this was answered as yes for mu sub t. This is a paper of Arunabh Sain and Balint Vevar in 2011. So they showed that mu sub t is absolutely continuous and has a bounded density. So what we have proved is the same for also the Hankel case. And our proof works for both the topolites and Hankel. And of course, mu sub t is a repetition. This is proof of the theorem of Sain and Vevar, but our bounds are better. So we have bound something like the bounds are like, it's not important and I'll not talk about that here, but bounds are actually reasonable looking numbers like one by square root of pi and one by square root of e. So they are off, if you compare with the numerical simulation, this one by square root pi is off by a factor of something like 1.4, a little more than that, 1.45. This is off by a factor of something like 1.85 also. So it's close to, yeah, it's, so that is the theorem of which I want to explain the proof. And well, I'll not talk about other random matrices, but our methods work for some more general class of matrices. So for instance, one can, so which again, I'll not talk about this. So what is generalization here? So if you have a countable group, you can form a topolites matrix. So let's say we have XG, G in the ID normal zero one. So you can form a topolites matrix, which is just XG inverse H indexed by the elements of the group or a Hankel matrix, which is the G, H entry depends only on G times H. So these are topolites and Hankel matrices associated to a group. What you can do is you can take finite subsets of the group going to the full group and take the corresponding matrices and ask about their limiting distribution. So our methods work for the case when if you take G to be ZD and you take going boxes, then we can show all the things go through and we can show absolute continuity and stuff like that. So in the more general case, I'm quite confused about this question, whether there's a good answer or not, I don't know. Please ask me if there are any questions at any time. What I want to do is to give an idea of the proof of this theory. So the proof has really two steps. No, no, they are both unbounded. So support is unbounded, yes, both of them. Because the moments go faster than exponentially, one can see that they cannot be bounded. So there are many interesting questions about these things, like the largest eigenvalue and various questions about this, which I'll not be talking about. I have nothing to say. So there are really two parts in the proof. Step one, actually it's quite interesting and if some, I have, it's a, okay, I'll, at this level, let me just say it like this. So we can, the main idea is that we can write this toplit matrix as a sum of two random matrices. So we write Tn as a sum of two random matrices where an and bn are independent. So this is independent random matrices. And if you look at the LSD of this, both of them converge to normal distribution. So you have two independent random matrices. They're limiting spectral distributions are actually normal. And so, and similarly in the case of Hankel, we write it as an plus bn, okay, not the same an plus bn. Again, they are independent and they're limiting distributions converge to this symmetric early distribution. So this, so it's just the symmetrized version of the early distribution. Mod Xe power minus X square by two is the density. Now anybody with experience in random matrix should think at this point, the problem is done. You have two independent random matrices and they have very nice limiting distributions and you're adding them. So how can it not be absolutely continuous? So indeed, it sort of explains the picture too. You have your, there is some kind of convolution. So the limiting distribution of this mu sub T has to be some kind of a convolution of normal with itself and mu sub H has to be some kind of convolution of this early distribution with itself, which sort of seems to explain why this picture should look like this. It's clearly this density looks like this. So you can, if you take a normal convolution of this, you will see a picture a bit like this. But the point is the operation here is neither the ordinary convolution or free convolution. It's a different convolution which we sort of understand at the level of moments but we don't really understand well. Maybe if one understands it, maybe one can even tell exactly what is this mu sub T and mu sub H. And so a step is needed. So from here it's not immediate. So at first I thought maybe it is free convolution but it is not at some 10th moment they differ actually. So the 10th moment, one of them is 42, the other is 44 or something. Okay, it's not free that much is clear. So there is a second step which is a spectral averaging. I'll explain it later. So spectral averaging is a technique that is well known to many people in random operators and so on. So here there is a spectral averaging. There is a different spectral averaging we have to do in the topolitz case and in the Hankel case. So these are the two main steps and from that we can get the absolute continuity of mu sub T and mu sub H, okay? So what I would like to do in the remaining time is explain these two steps. Now, so I'll explain the first step. So the first step in the topolitz case. So this is something curious to me. I think this must be, there is an observation about topolitz matrices, nothing random which probably is known but I could not find anywhere and so first don't, let's think of just a general topolitz matrix with not even Hermitian. It can be a general topolitz matrix and so this is not random, not symmetric, okay? So there is a presentation that we found for general topolitz like this. So to explain that let me say the main character in all this is this matrix L which is zero one, one's on the super diagonal and zero elsewhere. So clearly everything, the topolitz matrix can be written in terms of this as X naught times identity plus X K times L raised to the power K. L raised to the power K will have once on the Kth diagonal here and X minus K, so if you want it on the other side you just transpose it L star raised to the power K. So this any topolitz matrix can be written like this. Now a lot of things in topolitz matrices, I mean for instance also the same in Virak paper, the general idea is it looks almost like a circulant matrix, okay? Which would be much better but it is not circulant, it's a in fact a slightly problematic matrix, it's not normal. So the Kajia step here is to actually write this zero as one plus minus one. So we'll write it as zero one, one, one and put a one at the N comma one entry and everything else is zero plus you consider a similar matrix but put a minus one here, okay? So this matrix I call C and this matrix is D, okay? It's actually two times L, sorry. Two L is two times this and that is C plus D, okay? So I just put a one here and a minus one here. So this is very, this representation, what is the point of it? So here are some basic points. C and D are unitary, they are unitary matrices. It's quite clear, C is a circulant matrix. So this one I'll just call it twisted circulant. So what is C doing? If you think of a discrete cycle, C is just a shift operator on the discrete cycle and D is also a shift except that at one location it flips the sign, okay? So C and D are both unitary. In fact, C to the power N is identity and D to the power N is minus identity. If you apply D N times, you just get minus identity, okay? And most importantly, not only is L is half C plus half D, but also if I take L to any power P, it is just half of C to the power P plus half of D to the power P for, okay? Because if you take L square, L square is the one with ones here on the next diagonal and C square has ones on the next diagonal as well as two ones here and D square will have two minus ones here, et cetera. So this works. So for all P less than equal to N minus one and if you take the transpose, you just get that C star is same as C inverse. This is unitary, C to the minus P plus half of D raise to the power minus P. This is true for all P up to N minus one. So the point is this L is written in terms of these two unitaries. So the entire topletes matrix can be written also in terms of C and D, okay? So what does that lead to about the matrix itself? So this implies from this we get that T is, well, it is some polynomial of C and a polynomial of D because you write each L power K as and L power minus K in terms of C and D. So what we get is, let me write it as A plus B where A is summation Aj C to the power J, okay? So here I'll make an assumption. Let me take N to be odd, okay? This is just convenient. There is a slightly different representation when N is even but let me take N to be odd. Then you can write and this is of the form BJ D raise to the power J, J equals minus M plus one to M minus one. So where you can write down exactly what Aj and BJ are. Aj is something like XJ plus, okay? So you have this matrix. So these two matrices as you see A is just a polynomial of C. So we know it's full spectral decomposition. So C, it has the same eigenvectors as C. Only the eigenvalues change. So the eigenvalues of C are in fact, okay, we can, I did not write it here. Maybe I should say what are the eigenvalues and eigenvectors of C? So let's write V theta for the vector one e to the i theta, e to the i N minus one theta. Then C eigenvectors are just C V theta is e to the i theta V theta for any Nth root of unity. And for D, the eigenvalues are it's similar, i T V tau where e to the i N tau is minus one. So the eigenvalues of C are roots of Nth roots of one, eigenvalues of D are Nth roots of minus one. And eigenvectors are just these things. So it's just very explicit. So thus of course A will have exactly the same eigenvectors as this except the eigenvalues change. And B will have the same eigenvalues as, eigenvectors as D, eigenvectors as D, eigenvalues change, that's all there is. So in particular, so this representation, so it may be of some use in general, I don't know, but if you, for example, a basic theorem, this is just a small digression, if you think of the Ziegolimit theorem, which is a theorem about determinants of topolitz matrices, the Ziegolimit theorem is completely obvious separately for this A and for B. Because if you know exactly what the eigenvalues are, if you start with a topolitz matrix with a certain symbol, and you write down this A, the eigenvalues of A will be the values of the symbol at the roots of unity. And the eigenvalues of B are the values of the symbol at roots of minus one. And so the determinants are just these products which if you exponentiate right as e to the log, you get the Ziegolimit theorem. I don't know if for instance one can prove the Ziegolimit theorem for T from this, this I don't know. But maybe there is, so it's an elementary looking observation, so it must be somewhere, but if somebody knows, I would like to know it afterwards, okay? So in particular, what happens when you apply it to the random matrix? So in the random case, xj and x minus j are the same, and that is some normally distributed random variable. Therefore, aj and bj are independent, so it can work out everything. So this is the claim. So you have broken that matrix T into some of two independent random matrices. And okay, what are the eigenvalues of A? They are of the form, well, they will look like, so you take any nth root of one, and then you form this polynomial, so that is an eigenvalue. So you get this, you have this eigenvalues of A, which are very explicit, and they are all linear combination of the same edges, so they are all normal. Actually the eigenvalues of this big matrix A and B are actually normal, and they are almost independent, in fact. So I'll just write that A is, so I'll just write the conclusion, because I want to tell a little bit about the second step. So let me write what is the conclusion. So this A and B we can exactly solve using this. So what happens is, so A looks like this, lambda zero times P zero plus lambda one P one plus lambda m minus one P m minus one, where these are projections. These are rank two. When you work out, you see that one eigenvalue occurs with multiplicity one, others occur with multiplicity two, because of the n odd case. If n is even, then there are two eigenvalues of multiplicity one, et cetera, not so important. But these are rank two projections, and this is rank one. And their P's are not random at all. They are completely coming from the eigenvectors of C. And lambda J are almost IID normal, zero half random variables. So this is, I write almost, one can work out the correlation of these things. It's very small. So for the talk, I'll just pretend that they are actually IID. And the representation for B is quite similar. There are some, and mu J are almost IID normal, zero half. And Q, these are rank one, and these are rank two. So P J and Q J's are not random, that's the point. So this is the representation. This is step one, where you have written the topolitz matrix as a sum of two matrices, which are completely understandable. So from this, it's obvious that the limiting spectral distribution of A and B are actually just normal zero one. Sorry, normal zero half. So the Hankel case is similar, except that maybe I don't write it now in the interest of time. In the Hankel case, we can do the same. Yeah, what is the difference? So maybe I finish the topolitz case and in the end, I'll mention the Hankel case. So this is what works in the topolitz matrix. So from this, how do we see the absolute continuity of the, of T, for example? So we have written T as A plus B. And we want to, okay, so the main way in which we will show that the absolute continuity of this is to show that the still just transform of the expected empirical measure is bounded. Uniformly, there's an N. So we will consider GN of Z, which is the expectation trace, ZI minus TN inverse, one over N. And we show that Z in the upper half plane, that is imaginary Z positive. So there is a uniform bound for the still just transforms of the expected empirical distributions of all TNs. So from that, it is immediate that the limiting spectral distribution will have exactly the same bound. And it is a well-known fact that if you get a bound on the still just transform, that means the density has exactly the same bound. Because when you approach the real line from the upper half plane, the imaginary part of the still just transform converges to the density at the point. So you get exactly the same bound maybe with a one over pi factor for the density. That's what we have to show. So how we do that is, so we have this. So let me write what the spectral averaging is. So let me write the spectral averaging lemma. So this is the general nature of spectral averaging lemma. Incidentally, in the work of Arnab Sain and Balint Virag, where they proved this, they also use spectral averaging, but it's a much more complicated one. Ours is a very simple one-dimensional spectral averaging. So let's consider M. So we have a, so this is a real symmetric matrix. And then we add, so this is a projection and lambda is a parameter. So we think of lambda as random distributed according to some density. So this is a probability density function on the real line. So think of it as a random operator where you have a fixed one plus lambda times a projection. This will be relevant here since we know how to, the spectral decomposition of A and S, what we are going to do is just fix one of those lambda case, a condition on all the rest, then we have a matrix in this form. So if we condition on everything except say lambda one, we have lambda one P one plus the entire rest of A and B together will be M naught. So that will be the idea. So then the fact is that if you take trace of, P times Z i minus M lambda inverse P expectation, well expectation is with respect to phi. If we can, if we average this over the density phi, then we just get pi times, okay. Let me put absolute value pi times the sup norm of phi times the dimension of the projection. So this is the spectral averaging lemma. No, I mean, so of course, if pi is not a bounded density, I don't get anything, but assumptions on which, no M is much higher dimensional, it's okay, no problem. Yeah, but here we are not actually studying the full steel just transform. It's only on the projection in this P direction that is important, yeah. So for instance, how does that help here? So when we come to the random situation, so how is this useful? So let me indicate how this is useful. If you apply that with some say, you fix the kth one and so on. So we get that the expectation of trace P k, zi minus T inverse P sub k. From this, we get that it is bounded by pi times here, okay, so phi is the normal zero. So here I'm fixing say lambda one, so some kth one, then we have the normal density. So let me just write it as some constant times two, because rank of the projection there is two or one, depending on which one you take. So you just get a constant for each k up to, yeah. So I wrote almost IID. What it is is that when you condition on the rest of them, the variance of this is still very close to half, okay. It is always more than one fourth, but if with n going large, it is in fact, you can make it as close to half as possible. So what we do is we condition on the rest, then we still have a normal random variable with that much variance, and that is what is used in the spectral averaging. And then you take a further expectation over all the variables to get this. And then when you sum it over k, you get the sum of all these with all the projections will give you the still just transform. So you just get that the still just transform, which is the average of these things will give you, that's the, that's all. So this is the key point in the proof. Now, those pks, they sum to identity, right? This is the spectral, yeah. That's the spectral decomposition of A. So the sum of those projections is identity. When you add them, you just get the trace of Zi minus T inverse. No, no, but we have to average. G of Z has a one over n also. So it is the number of summands is also n and it's exactly canceling that. Any other questions? So this is how it works. Now, yeah, okay. How many minutes do I have? Two, three, four minutes. Okay, so yeah, so I'm thinking whether to indicate a proof of this or to just tell what happens in the Hankel case, maybe I just mentioned what happens in the Hankel case. So in the Hankel case is actually similar but a bit more complicated. So in the Hankel case, the first step is actually quite similar. Maybe I just say it in words. So here what happened was we take any toplit matrix. We could write it as a sum of two matrices. One is a polynomial of C and one is a polynomial of D. In the Hankel case, we can, you take any Hankel case without randomness. We can write it as a sum of two matrices, a and b. Now, the point is a is, everything is slightly more complicated. So in the toplit case, a can be reduced to diagonal form. I mean, the eigenvectors don't depend on the entries of D. That was the point. But here it depends, but you can bring it to block diagonal form where these are two curves to blocks. So this matrix A can be brought to two curves to blocks by a unitary transformation conjugation which does not depend on the entries of the matrix. In the toplit case, the same could be done up to diagonal but here it is only up to two curves to block form. And the similar thing for B, we can bring it to two curves to block form, et cetera. So the spectral averaging also got a bit more complicated because well, there is a spectral averaging glamour where this quantity that we do spectral averaging over, this is not a positive definite matrix. So at least when I asked a few spectral averaging experts, they said it only works when you have a positive definite perturbation. But in our case, it's a very small rank perturbation. It's only two. Therefore, we could handle it directly. And we have an analogous spectral averaging glamour from which we can get the conclusion. So the key point is the proof of this but okay, I don't have time for that. So I'll stop here.