 So I'd like to first thank the organizers for allowing me to speak in this place and thank Ken for introducing discrete orthogonal polynomials because it will be the beginning of my topic. So I will consider discrete orthogonal polynomial ensemble. So my state space will be either n or z or z plus one half. And I take omega to be a measure with port e. So yes, I'll start with general things and then move to the examples I'm interested in. So with, well, the support will have to be infinite, it's a little bit different. And with finite moment of all orders of x to the n omega dx is finite. So I have the corresponding sequence of orthogonal polynomials. And just to fix notations, I will write pn of x as cn xn plus our terms and their squared norm is just hn. To this I can associate, well, a probability measure, pn e to the n just given by pn of x1, pn is well, a normalization constant times the square of under month and the product of the weight. And, well, as it was say many times here, so you can see it does not charge n tuples with two equal coordinates, so it can be seen as a point process on e. And this is a determinant all point process. So that is, we have the probability that subset h fixed is included in my random configuration x equals the determinant of kn ai a little n. And kn is the Christoffel-Darbu kernel, so it will be expressed as pn of xn of y divided by hn and I include the norm here in the, the weight sorry in the kernel. And it also has a nicer expression for me, for what follows. So again, x omega of y times cn divided by hn cn plus 1, x of y, pn of y, x divided by x minus y. Okay, so now I want to talk about Christoffel deformations of these kinds of things. So I take k points, real points, which lie outside the support of omega. And I modify, and I do what is also called Christoffel deformation of the measure. So that is, I replace omega by omega k, omega k of x, and it will be the product i equals 1 to k, x minus ui, where omega of x. Well, so I have this new measure. So I have a new orthogonal polynomial ensembles, p and k, with kernel k and k. And so why to do this? Just informally speaking, this is related to averages of characteristic polynomials. So product of characteristic polynomials because if I have, let's say, so if this were the density probability of eigenvalue of a random matrix, the average of, under this new measure, will be the average of product of characteristic polynomials as ui with respect to the oldest measure, yes? And also a question that arises, if I know something about pn as n tends to infinity, what can I say about p and k as n tends to infinity? And can I perform such deformations for more general determinant point processes? Okay, and the answer is yes in three examples. So you are not in the, it's a note set, so as this new measure is not degenerate in a sense. Okay, so the first example I will talk about is the Scharlie ensemble and the discrete Bessel process. So the Scharlie ensemble is the orthogonal polynomial ensemble with weight, which weight which is Poisson distribution. So it has a parameter and I take omega a of x to be exponential minus a, a to the x over x factorial. So the corresponding orthogonal polynomials are the Scharlie polynomials and it is known from Johansson from year 2000 I think that in some regime I will now describe this ensemble approaches the discrete Bessel process, the one, so that is the Poisson-Eis-Pamelschrell measure Sacha was talking about this morning. And well, what happens for omega a k? So let me state the theorem. I take alpha to be positive, a will be alpha over n, I have to scale this parameter and I take ui to be some fixed ui tilde plus n in this definition. And so the theorem is that the new Christopher Darbu kernel, k and k has the limits in this regime as n tends to infinity, so that is to say. I have this constant factor, I have another constant factor here, here I have some interaction between my particles and the ui tilde and I still have the integrable form of the kernel in the limits and it is described with a discrete run scan of Bessel functions, I will write formula later. So let me write the formula here, yes, so gp of x, I said it's a discrete function of Bessel functions, here I have, no, it's for p equals 0 or 1, so this is a 2k plus 1 times 2k plus 1 determinant, that's 2k. So I write them for all my ui's, then I have to add the derivatives with respect to the index of the Bessel function, so I don't know if there's a usual notation, I note them l, luk and at the last line I have my Bessel function at point x here and I also write what is ck. So ck is the upper left corner of this big determinant, so it's 2k times 2k determinant, okay, so this means in a sense that such deformation can be defined for the discrete Bessel process, yes? So here I introduced, so I fixed point ui, but here my points ui have to move with n and u tilde is just what is fixed here. I have a similar result for z measures on partition, so let me state its result before I, before I move to the proof, so what are the z measures on partitions? So partitions means young diagrams, so these are measures on the set of early on diagrams, which depends on three parameters, into a partition lambda, it assigns a weight, 1 minus, so this is the normalizing constant, I will write later what, if there is no mistake. So here is the generalized Pochamer symbol, so it means a subscript lambda is the product of all boxes with coordinate ij in lambda of a minus i plus j. But well, there are, first of all, the constant c is in zero one, and there are three cases where this defines the probability measure, namely when it is positive, so the first one is, well, when z and z prime are complex conjugates, the other one is when z and z prime both lie in the same interval with integer coordinates, and the last one is when, let's say, z is an integer, and z prime is bigger than, than minus one. Okay, so what I want to do now is to perform, can I do Christopher deformations of this ensemble, and can I say that it is a determinantal point process, and in what sense? Yes? Yes? An integral representation? I would love to have one. Well, by a simple limit, I can write it now, if you want. It's just, with the way that is here, scale, times the orthonormal charlie polynomials with index n plus l in x plus n. So this stands as n tends to infinity to the Bessel functions with index x minus l to square root of five. Using an explicit expression of k for k and k, I have, I have, I directly obtain this limit. Okay. So let, let me talk about the z measure before. So it is known since the work of Borodin Alamschanski, who introduced these measures that they, so to a partition, you associate a point process, and this is known as this determinantal point process, and in the third case, this is an orthogonal polynomial ensemble with, with a miceno weight. So this is a miceno ensemble. So let me say in, let's say this is one, this is two, this is three, case three. So first in case three, what I have, if you look at well, I have that, the length of my partition. So this is the, I have a partition. Its length is, well, its length. The last part of my particle with, which is not zero of my partition, which is not zero. So in case three, the length is fixed. So it is convenient to associate to it a point process on the positive integers. So, and under the map, lambda associate, lambda i minus i plus big n. So for i equals zero to n, mz z prime xi is the miceno ensemble. And well, so it is fine because here I have, well from i equals one to n, I always have the, so this is an orthogonal polynomial ensemble. I can perform Christopher deformation of this, on this ensemble, but for the other cases, this is not obvious that I can do this because I'm, here I have a fixed deterministic number of particles on the integer, on the integers. But for the other cases, the, the length of the partitions are random, or even infinite, if I consider infinite point processes, I will say this. So, and the theorem is, so let me just give a definition. So the Christopher deformation of mz z prime xi is defined by, so I introduce the product from i equals one to l of lambda. Product from j equals one to k of lambda i minus i plus z minus, what it is, plus g minus ui. And I take, no, I have to break the symmetry somehow. Here I said j, yes, z prime minus some other things, vj, times the original measure. And well, if I choose well the uj's and vj's, in the, in the third case, in, in, in this case, I will, I will have the, the, the, the Christopher deformation of the Meichner ensemble in the sense I explained in, at the beginning, yes. And the theorem is that with suitable choices of the uj's and vj's, this new measure also, also gives a determinantal point process with an explicit kernel and given by, which has the same form as this one, but with discrete Vronk's kernel of Gauss hyper geometric functions. So, the ui to be real and not in z plus one half. So this will be the convenient set of, for the, the convenient configure phase space for my particle to live for the case one and two. And I take the i to be ui plus z prime minus z, plus one half, minus one half. Then case one and two under, now my new configuration of points is this one under i minus i plus one half. And this is an infinite configuration. It's for all i. So at some point I will, I will have just particles on the left. And mz, z prime k, say, is a determinantal point process with a kernel I, I will just say described by discrete Vronk's of Gauss hyper geometric functions. Two f ones, yes. Okay. And, well, a reason why I wanted these deformations to be determinantal point processes is because I wanted to have deformations of the process with the gamma kernel, which is a limit without any scaling of the z measures, but only in the cases one and two. For the Meitzner ensemble, we don't have the convergence to the process with the gamma kernel. And so I wanted these deformations to be determinantal processes with nice formulas to obtain deformations of the process with the gamma kernel. And this is the theorem of this talk. But unfortunately, I can obtain it just for k equals one. So let me denote, I don't know, kz z prime xz for the k here for my new kernel. With k equals one, k1, and in cases one and two, k1 z z prime c of x and y. And here I have to add something, one minus xz converges as xz tends to one to a kernel described by the gamma function. So the formula is no more on the blackboard, but when xz tends to one, it gives more and more chance for big young diagrams. So what I have at the end is not a probability measure on the set of all young diagrams, but I have a point process on the lattice, which does not correspond to a young diagram. So now let me move to the proof of all of this theorem, or at least some ideas. This one is too long to write down. So this one is not that long, so maybe I think I will write it at some point. So it is as long as this one, but the one with the gamma function is hard to write and with quite heavy formulas. Okay, so the proof. So this is a general formula for the modification of my Christopher Dabau kernel. I have the square root of the original weight. Yeah, well, because I changed my weight, I must have something like this. And well, there is a constant factor which depends on the normalization. So and here I have, so I write what is d and delta. So we remember cn and hn where the leading coefficients are and square norm for the original orthogonal polynomials. And dn of x is again a determinant. It's pn of x, no, pn of u1, sorry, pn plus 2k 1, pn of uk, n plus 2k k. Then I have the derivatives, pn prime of u1, n plus 2k prime of u1, et cetera, pn prime of uk. And at the last line, I have pn of x, n plus 2k of x here. And delta n, okay, so what I write like this is the upper left corner. Okay, so now I have everything to finish the proof for the deformation of Bessel because as Marco asked me, I have the following lemma. And I have an analog lemma for the derivatives of the Scharlie polynomials which converge to, so minus something, they converge to the derivative of the Bessel functions with respect to the argument. And well, as you can see here, we have some homogeneity in this formula. Namely, I have here products of two determinants of size 2k plus 1s. And here I have a determinant of size 2k which is squared. So I can multiply both sides of this ratio by the corresponding weight I'm interested in. I already have the weight here which, which play a role for the plus 1 in the, because these determinants have size 2k plus 1. And then, perhaps using the lemma, passing through the limit, so I can, everything is explicit, so I also, I also can control this quantity. I obtain precisely this result of the first theorem. Now, for the Z measure, it's more, it's more delicate because there is no, there's no limit procedure, but the argument is an analytic continuation argument. So just for notation, I will write sigma of lambda will be lambda i minus i plus 1 half. And so this argument is due to Borodin and Olszanski and because here I have to break the symmetry to, to make the, for the definition to make sense. This argument works, but we have to do some more efforts to make it work. So the argument is as follows. So you know that when z equals n and z prime is bigger than z minus 1, you have an orthogonal polynomial ensemble. So it is a determinant of the process with an explicit count. Even for my deformed case. So the second fact is that, so this is Borodin Olszanski, mz z prime psi that a1, a n are included in sigma of lambda. So for any a1, a1, a n in z plus 1 half, this quantity has an analytic expansion and expansion psi with polynomials, with coefficients which are polynomials in z, z prime. Nice. So, and you know that at least for the case when z is, so you know that for the case when z is an integer, that this is a determinant of some, of some kernel. If you also have that, so if this kernel would make sense for other value of z and z prime, which is the case and which satisfies the same property, that is it has an expansion in psi with coefficients which are polynomials in z, z prime. So you can say that these two series in psi should have the same coefficients. These coefficients are polynomials in z, z prime. And well, this set in C2 is a set of uniqueness for polynomials. So the same formula should hold, so the same determinant formula should hold for the other cases of z and z prime. And well, this is precisely how it works. Yes. So in notation with parameter z, z prime and psi to be the maximum polynomials, psi a point x tilde divided by its norm l2 omega, well, which depends on z and z prime times the square root of the weight. So I have some shift playing a role here. So my new variables are expressed through the other one. Where is it? In my notes. Yes. Five minutes. Okay. And yes. With x tilde being x plus n minus 1 half and a is n minus a minus 1 half. At this point, z equals n and z prime is n plus, well, n plus some beta minus 1 where beta is defined through z prime and z. And so this holds for these choices of parameters. And there was an explicit expression for this function say psi a through Gauss hypergiometric function to f1, which I will not write since I only have three minutes and I will spend like four minutes to write the formula. But still, there is a formula for these functions, which depends only on z, z prime and x and not on n, n little n specifically, if you want. Yes. And the trick is, so in your Christopher Darbu kernel for the deformed process, you can replace by homogeneity arguments, the same as before. You can replace your Mike's polynomials by the function psi a. And then you want to transform your function psi a so that they satisfy the same property as the z measure. Namely, they have an analytic expansion in psi with coefficient, which are polynomials in z and z prime. And, well, let's do this at least some words. So if I define f z z prime of x to be gamma of x plus z plus 1 half divided by square root of gamma of x plus z plus 1 half gamma x plus z prime plus 1 half proposition, not mine, it's Borodin Olszanski. Psi a of x z z prime psi equals f z z prime of x divided by f z z prime minus a times some integral, which depends on a on z on z prime and on x a and in x. And this integral has an analytic expansion in psi with polynomial coefficients in z z prime. So in here, you can you can remark that I have broken the symmetry in the function. So I did not say it, but the function psi a is r psi r symmetric with respect to the changing z z prime. But here it is it is not of use at all because, well, this one is not symmetric. This one is well, it did not write it, but it's not symmetric as well, but it should be symmetric because psi a is. So for example, for this factor, I can use both expression one with z z prime in this order, the other one with z with z with z for z prime and z. And I will get rid of this factor f z z prime minus a in the formula because, well, there is a product structure in this determinant. And as you can remark, f z z prime is just the inverse of f z prime z. So this factor will disappear. And the same thing happens for the upper side of the ratio. You have boundary terms because the determinants don't have the same indexes, but still it works. It has an analytic expansion and we in psi with where is polynomial coefficients in z z prime. And this concludes the proof of this theory. So thank you.