 Ortonal Polinomiel, Gina Brin ensemble with finite point charge. OK, hello. Yeah, thank you. First, I want to apologize. My voice today is very bad, so I'm going to use this and talk a very silent, small voice. First, I want to thank the organizer for giving me the chance to speak here. First time visiting Triste. And I want to speak about Plainer Othagonal Polinomiel. And this is the ongoing project with Menyang. So let me first describe the model. So this is Coulomb gas. The difference between what Professor Guionne taught us is that it's on the planer. It's on the full complex plane. So they repel with each other with beta equal 1 or 2, depending on the convention, and with some confining potential q. This is the probability distribution for n point particles at the complex plane. And I'm going to use Zn for the cap normalization constant called partition function. And so partition function is one interest. Another thing of interest is the characteristic polynomial, which is just degree n polynomial whose zeros are at the Coulomb gas. So it's a random polynomial, because Coulomb gas is randomly distributed according to this distribution. And if you take this random polynomial average, then I get this deterministic polynomial. And these are objects of interest. There's two objects, partition function and this polynomial, Pn. In the large n limit, that's the interest of the talk. OK, so I will use the specific potential, not general potential. Potential is Gaussian. If I have only this quadratic term, then it will be a Gini-Brown symbol, where the particle is on the unit disk, uniformly distributed. But I will have L point fixed charge at the location a1, a2 to al, with charge strength c1 to cl. So for example, in this picture, I fix 3. I can even make it move. So these are the blue dots, our Coulomb gas. And I fix 3 point charge with different strength. So this one I choose charge 4, this one charge 2, and this one charge 0.2 or something. And all the other are charge 1. So you can see there's a little more distance from all the other charge because it has more repelling force. And actually charge can be slightly attractive also, like charge can be any number bigger than minus 1. And for this model, we want to calculate the partition function, which is now explicitly written in this form. So for all the fixed charge, you also have the interaction term. And first, we will look at the polynomial, the average characteristic polynomial. What is known is this average characteristic polynomial is also orthogonal polynomial, which satisfies this planar orthogonality with respect to the measure given by the same queue that defines the probability distribution. So the same queue, this one. And the orthogonal polynomial more explicitly can be written in this form. So that's, yes? How did I make this animation? I just add Brownian noise. So this is the cheating because I don't know how much more noise I have to add to make the beta equal 2. So this is for some unknown beta. So each particle is moved by the force, deterministic force plus Brownian noise. Deterministic force is from the Gaussian and the force from all the other particles. But this is a cheating, not beta equal 2. C is not an integer. C can be any number larger than minus 1. So here is what is known so far is the asymptotics of the orthogonal polynomial. The message here is pretty less is known. For very simple potential, they could obtain the location of the zeros in the large and limited. Location of the zeros are usually concentrated on some one-dimensional structure even though the particles are supported on two-dimensional support. So there are few examples where this shaded part is where the Coulomb gas will stay, but the zeros of this average characteristic polynomial will stay on a one-dimensional, called skeleton of this domain. And most of the result is used using Riemann-Hilbert technique developed by DK MVC. Only M is here in this crowd, like Laughlin. But the goal is whether we can use this tool which was so successful for other complex orthogonal polynomial, whether we can apply the same technique to planar orthogonal polynomial, and that's so far almost what we have so far. Okay, and also there's a general result given by Heidenbaum and his student. This is the general asymptotics of the planar orthogonal polynomial for broad class of Q. So here Q can be any smooth potential, but there's also some subtle assumption about its shape of the equilibrium measure. But with subtle assumption, they obtain the asymptotics of this orthogonal polynomial. But here the asymptotics is only in the domain which is outside the support of the Coulomb gas. So outside, in our case, it's outside the unit disk. And all the zeros are some subset of some one-dimensional structure of the unit disk. So this doesn't tell you where the zeros are. The location of the zeros is still an unknown problem. Okay, so let me describe the answer for my system. So first I will define what is the Segel curve. The Segel curve is the curve defined by this equation, a very simple equation, which is this closed curve between yellow and dark region. So this closed curve is the Segel curve and that's actually the limiting zero locus of this truncated and little scaled exponential function. So exponential function doesn't have zeros, but if you truncate the exponential function, Taylor series truncation, then it will have zeros and zero converges to the Segel curve in the scaling limit. Okay, and I will describe the result when L equal one. So if you have just single fixed point, if you just have single fixed point, then the zeros of the characteristic polynomial goes to this Segel curve. That's already proven. But it's not just the Segel curve. It can be some generalized version of Segel curve, which I will describe soon. So depending on whether this fixed charge is inside the disk or at the boundary of the disk or outside the disk, you will have different shape of the Segel curve. Okay, so to describe the Segel curve for general case, I will describe the Segel curve in a different language. So again, the same equation for the Segel curve and this Segel curve can be understood as the following. So first I draw the graph of log of the absolute value as the yellow cone. That's the shape of the graph of log of C. And I draw the graph of the other parts, real part of A bar Z, and that gives, of course, the plane. And where this two graph meets is where this equation is satisfied. So I draw this yellow cone plus this blue plane and where it meets and those solution sets is where the Segel curve. So here, this white line is the Segel curve. And the right-hand side is, if I take only, so at every point, if I take the maximal value of the two graph, then I obtain this right-hand side. So that's the maximal between these two yellow part and the blue plane. And actually, that's exactly the log of the polynomial in the limit. That's the result from when there's only one fixed point. Okay, now I describe when there's a finite number of fixed points. So here I will describe multiple Segel curves. So multiple Segel curve is the same. We consider log of C, the previous cone-shaped yellow curve. And then we consider many planes and for each plane coming from all new addition of the fixed point. So fixed point A1 gives the plane, which is directed in the A1 direction, another plane A2. So there's L planes if you add L fixed points. And you draw all the planes and all the cones and you look at the graph of the maximum. Then you see this kind of graph, which is for three fixed points. So here the red dots are the location of the fixed points. I call the maximal function, which is represented by this graph, capital phi. And I just define each region where the jth plane is dominating, is omega j, for example, here. And I also note that each plane can move up or down by the level which I denote by L. Lj, if you increase Lj, then plane goes up, Lj down, place goes down. And you have to match this height such that the plane is exactly intersecting this log of the other plane, other graph, at the location of this aj, the point. So once you match the height, such that the point, the fixed location of the fixed point is at the boundary, exactly at the boundary, then that determines the height Lj. And it's not very trivial, but you can show that even aj such multiple sego curve is unique. So all the Lj's, the height, is chosen uniquely. Here I try to show how this lemma can be proven. First, you choose one plane until a1 is at the boundary, and then move the second plane until a2 is at the boundary. Actually, this point can be at the boundary between the two planes also. So whenever two planes meet, the boundary is the straight line. So you see many straight lines, and aj's can be in the straight line. So such curve is unique, and the theorem goes as the same as in the one fixed point case. So this maximal function that I showed in this graph here gives the asymptotics of the polynomial. And actually Riemann-Hilbert problem is always overkill. So this is the demonstration of the result. So here you cannot actually see this line, but there's a white line between all the intersection of the planar graphs. And the location of the zeros is exactly matching on the intersecting lines. So it shows that the result is actually very good. These red dots are the location of the fixed points. Actually, one can put the red dots also outside. Then, yes, question. Computing the zeros, I have to go through the Riemann-Hilbert problem. And I need a lot of mathematics. No, away from the zero locus. But as Riemann-Hilbert problem always does, it gives stronger result everywhere. Also, as common in Riemann-Hilbert problem, you need special asymptotics. So you have different form of asymptotics outside, inside, or on the locus of the zeros. And you need special asymptotics called local parametrics near each fixed point, each fixed charge. And if you zoom in those fixed charge, then it's given by also truncated exponential function, but it's the other way. It's subtracting the finite truncation. And the other part gives the local behavior of the zeros. Actually, this is only for integer C, but if it is not integer, then it's replaced by some integral. So I will explain how I obtain this in a very crude way. I don't want to go to the detail. So I will define the multiple orthogonal polynomial as exactly defined by Silva. So multiple orthogonal polynomial is, instead of just one measure, you have multiple measures to define orthogonal polynomials. So P, multiple orthogonal polynomials defined by index N1, N2, and L, not just one degree, so that their sum becomes the degree of the polynomial. And it requires L measure. So first measure, mu1, gives N1 orthogonality. The second measure gives N2 orthogonality. So in total, they give exactly N orthogonality that can define the N degree N polynomial uniquely. And this is the measure of this multiple orthogonality, which is given by some crazy integral. So I said L different measures, so difference is coming from this blue index. So if, oh no, ignore the blue index. Blue index ignore, just red index. So if k different, then one of the power in the integral changes. So sometimes you subtract one of the power, then that gives L different measures. So they are very, quite close to each other. And all in, so L different measure is in my problem, it's all living in the same contour. The contour is actually here. Contour is from going around each fixed point, each fixed charge once, while rotating the origin once. You can always arrange the contour in such way. So that's my definition of the orthogonal polynomial. And what we proved is that our planar orthogonal polynomial is actually same as the multiple orthogonal polynomial for such choice of index. So if I give a degree 100 and 10 fixed charge, then I divide the 100 into most uniform way into 10 decomposition, so 10 plus 10 plus 10 plus 10. So for such index, you define the multiple orthogonal polynomial with the previous measure, then it gives exactly the planar orthogonal polynomial. So having such multiple orthogonal polynomial, then we have immediately the Riemann-Hilbert problem, which is the matrix Riemann-Hilbert problem of size L plus 1 by L plus 1. So if you have additional charge, the size of the Riemann-Hilbert problem increases. And this is the corresponding jump condition and boundary condition. And the rest is the same steepest descent analysis developed by DKMBC. So I will just describe very quickly. So we want approximate solutions to the exact solution, which is unknown. The exact solution is unknown, but we want the approximate solution. To find the approximate solution, instead of looking at trying to find the solution, we look at the Riemann-Hilbert problem, exact Riemann-Hilbert problem, and compare it with the approximate Riemann-Hilbert problem that can be easily obtained from the approximate solution. And the theorem is the exact solution, the distance from the exact solution to the approximate solution is given by the distance of the problem, distance between the Riemann-Hilbert problem, exact Riemann-Hilbert problem to approximate Riemann-Hilbert problem. So if we can show that these two Riemann-Hilbert problems is close, then we can prove that these approximate solutions are also close. That's the main strategy of using Riemann-Hilbert problem. But there's other benefit. First, Riemann-Hilbert problem can give you a lax pair and it also can give a recurrence relation. And that's how I obtain actually the numerical calculation, the zeros of the polynomial. Because finding zeros by Gram-Schmidt is, I believe it's impossible, because there are many moments with exponentially small contribution, which actually the zeros are very sensitive to exponentially small quantity. And without this recurrence relation, it's impossible to plot the zeros. And the second thing is the partition function, which I will explain now. So partition function is related to the moment determinant of this multiple orthogonal polynomial system. Okay, so this is more thing about Riemann-Hilbert problem side. So I said the multiple orthogonal polynomial is supported on the contour that goes through all the A1 to A2, A3, A, all the fixed point. But some of the fixed point is actually inside the multiple sega curve. So first you fix the contour to go around the most outer boundary of the multiple sega curve. But whenever the fixed point is inside, you have to come back and go out following this branch cut, because all this charge strength Cj is not equal to integer. Then there's a nasty branch cut and that actually gave me so much headache. By the way, this talk has been similar talk has been given a few months ago and there's some people in the audience apologizing. So at the time, it's a few months ago, but non-integer C was still not completely proven, but I thought it's a small hurdle, but it took much more than that. So now non-integer C can also be handled. So that's how you draw the contour. So you have the contour that goes around the branch cut coming from the non-integer C. So some proof that I really worked and you have to write many notation with the general size matrix because it's for arbitrary size L plus 1 by L plus 1. So you have to use some crazy notation. Okay, now partition function. We are interested in the partition function because we are actually interested in the moments of the characteristic polynomial, which has been appeared in other talks. So I remind you the characteristic polynomial is chi is the polynomial that has zeros at the Coulomb gas. And if I take the absolute value and take the power by C1, then it becomes the moment and I can add multiple moments of the characteristic polynomial. So we want to calculate this expectation value of the moments of the characteristic polynomial. So why do we look at this quantity? So physically that's the following. So if we have L fixed charge with certain charge, then of course each charge feels some force, electric force coming from the external source and also between each other. But the question is what if you throw in this background Coulomb gas? What will be the force? How the force of this fixed charge change? And the answer to that is after this background charge is added, they don't feel the force from each other. So the background electrons will completely screen the electric field between each other so they don't feel any electric field. That's exactly the statement of this conjecture by Weben Wang. And there's more exciting recent result. So Weben Wang actually when they make this conjecture, they prove this formula for single charge, electrical one. And further, so this is the case when the fixed charge is inside the bulk but when the fixed charge is near the boundary of this unit disk, then there's a special function, Pagnet V4 appears. I think basically given by this paper but for some slightly different model which is actually equivalent. And then less than a month ago, there's another few weeks ago, there's another paper that proved this conjecture for L equal 2. They used, okay, so their method is different so they can only do this strength to be integer. So that's good because I can still contribute. So they can do strength integer actually after they can also make one of them non-integer. So C1 integer then C2 non-integer, then they can prove this fact for 2 fixed point. So this is the plot of actually the correlation, two fixed points. So when we have two fixed point, the moment can be plotted for different number of background gas and it converges to this, the top line is the theoretical line and this color line is converging to the top line. And of course the top line diverges when the two charges meet but what they did is when they are very close they can see more fine structure of the Coulomb gas and it's regularized by Pangneve 5. So this regularization is described in their paper. So I will describe how we can also study these moments of the characteristic polynomial and a little bit of result because this is ongoing work. So first, the solvability of the Riemann-Hilbert problem because Riemann-Hilbert problem is related to the multiple orthogonal polynomial, it's related to the unique existence of the multiple orthogonal polynomial. Whenever there's an orthogonal polynomial, the existence is related to the non-vanishing determinant of its moment. So moment determinant, non-vanishing then some orthogonal polynomial exists. So in this case also there's a moment determinant corresponding to the multiple orthogonal polynomial. So since there's many, many measure, for each measure you write down the moment matrix, rectangular matrix of some size and if you combine it for each measure then you will get the moment matrix of N by N, the degree of the multiple orthogonal polynomial. And if that determinant of that matrix is non-vanishing, then it's equivalent to having unique multiple orthogonal polynomial and it's also equivalent to the solvability of the Riemann-Hilbert problem. Also in another context, solvability of Riemann-Hilbert problem is known to be related to so-called tau function and tau function is given any Riemann-Hilbert problem or given matrix ODE system like here. Then there's an associated tau function. Tau function is defined by one of the tau called Jinbo Miwa Ueno tau function is defined this way. Also there's other definition. So the property of the tau function is the non-vanishing tau function means the solvability of the corresponding Riemann-Hilbert problem. So in our case we have the Riemann-Hilbert problem for YN and we can make the lax pair system or linear ODE by multiplying some deterministic matrix B which I show for example here. And once you do that you can find all the quantities to define the tau function. Actually this is variation of the tau function. Tau function is defined up to constant with respect to the variation of the parameters in the problem. So you can find the tau variance variation if you know the solutions to the Riemann-Hilbert problem. So if you have the asymptotic solution then you have the asymptotic behavior of the tau function. And what is known is this tau function is equivalent to the moment determinant because both, so the right hand side tau function is telling you the solvability of Riemann-Hilbert problem. The left hand side is showing you the unique existence of the orthogonal polynomial so they must be related but they are actually the same. And that's shown for example by Bertola and here's the main idea how to show them. So whenever, okay, so instead of showing that so it's shown by induction by looking at the tau quotient. So tau n divided by tau one length is equivalent to this ratio between the moment determinant and you can easily show the tau at n equals zero is equivalent to the moment determinant at n equals zero and that shows the equivalence. And therefore now we have the tool to calculate the moment determinant. And moment determinant is actually equivalent, not equivalent but related to the partition function by some crazy relation which come out of nowhere. I don't have good explanation. And from that we can calculate the variation of the partition function which can be obtained with a quick calculation using the asymptotics we have to be this and once we integrate then we have the antiderivative up to constant the partition function is behaving as the conjecture says for arbitrary number of L and for arbitrary set of Cj's. Okay. Yeah, so my work is too little to fill the whole 45 minutes so it's finished but since I have too many too many times I will say one more possible future work. That worked on but just so if we have two fixed point and the fixed point was have the same strength similar order strengths as all the other Coulomb gas but if we make this strength much bigger than the other Coulomb particle then you can actually change the droplet, the support of the Coulomb gas can drastically change so it's no longer you need disk and in such case such domain becomes some special domain sometimes called quadrature domain and the location of the zero is then conjecture to be the mother body of such domains. So those can also be studied using the formulation in terms of the remandable problem in the future and that's all. Thank you.