OPSF, Random Matrices and Riemann-Hilbert problems Walter Van Assche School on Orthogonal Polynomials in Approximation Theory and Mathematical Physics, ICMAT 23{27 October, 2017 Walter Van Assche OPSF, Random Matrices and Riemann-Hilbert problems Plan of the course lecture 1: Orthogonal Polynomials and Random Matrices lecture 2: The Riemann-Hilbert problem for orthogonal polynomials lecture 3: Logarithmic potentials and equilibrium measures lecture 4: Asymptotics and universality Walter Van Assche OPSF, Random Matrices and Riemann-Hilbert problems Orthogonal polynomials We only use absolutely continuous measures: dµ(x) = w(x) dx The monic orthogonal polynomial Pn(x) for the weight w satisfies Z k Pn(x)x w(x) dx = 0; k = 0; 1;:::; n − 1; Z n 1 Pn(x)x w(x) dx = 2 6= 0: γn The Riemann-Hilbert1 problem for orthogonal polynomials gives a very different characterization in terms of a boundary value problem. It was first formulated by Fokas, Its and Kitaev in 1992 to investigate matrix models in 2D quantum gravity. 1Bernhard Riemann (1826{1866), David Hilbert (1862{1943) Walter Van Assche OPSF, Random Matrices and Riemann-Hilbert problems A scalar Riemann-Hilbert problem First we deal with a scalar Riemann-Hilbert problem, which will be needed later. Find a function f : C ! C for which the following properties hold 1 f is analytic in C n R. 2 On R the boundary values f±(x) = lim!0+ f (x ± i) exist, and they are connected by f+(x) = f−(x) + w(x); x 2 R: 3 The asymptotic behavior is 1 f (z) = O( ); z ! 1: z Does such a function f exist? Is such a function unique? Walter Van Assche OPSF, Random Matrices and Riemann-Hilbert problems A scalar Riemann-Hilbert problem Theorem (Sokhotsky-Plemelj) Suppose w is H¨oldercontinuous on R and integrable, then −1 Z 1 w(t) f (z) = dt; 2πi −∞ z − t and this is the only solution. −1 Z w(t) −1 Z x − t − i f (x + i) = dt = w(t) dt 2πi x + i − t 2πi (x − t)2 + 2 −1 Z w(t) −1 Z x − t + i f (x − i) = dt = w(t) dt 2πi x − i − t 2πi (x − t)2 + 2 Subtract to find 2i Z w(t) f+(x) − f−(x) = lim dt !0 2πi (x − t)2 + 2 1Julian Sokhotski (1842{1927), Josip Plemelj (1873{1967) Walter Van Assche OPSF, Random Matrices and Riemann-Hilbert problems A scalar Riemann-Hilbert problem Change variable t = x + s to find Z w(t) 1 Z w(x + s) lim dt = lim ds !0 π (x − t)2 + 2 !0 π s2 + 1 1 Z 1 = w(x) ds = w(x): π 1 + s2 Hence the boundary conditions hold. The asymptotic behavior is also true and the function is analytic in C n R. Unicity: suppose g is another solution and consider f − g f − g is analytic in C n R, it has no jump on R, hence analytic everywhere in C. It is bounded, hence by Liouville's theorem it is a constant, and by the asymptotic behavior g(z) = f (z). Walter Van Assche OPSF, Random Matrices and Riemann-Hilbert problems Riemann-Hilbert problem for orthogonal polynomials 2×2 Find a matrix function Y : C ! C such that 1 Y is analytic in C n R. 2 On R the boundary values lim!0+ Y (x ± i) = Y±(x) exist, and they are related by 1 w(x) Y (x) = Y (x) ; x 2 : + − 0 1 R 3 Asymptotic behaviour 1 zn 0 Y (z) = + O( ) ; z ! 1: I z 0 z−n Walter Van Assche OPSF, Random Matrices and Riemann-Hilbert problems Riemann-Hilbert problem for orthogonal polynomials Suppose w is H¨oldercontinuous, then Z 0 1 Pn(x)w(x) 1 Pn(z) dx Y (z) = B 2πi x − z C @ Z P (x)w(x) A −2πiγ2 P (z) −γ2 n−1 dx n−1 n−1 n−1 x − z where Pn and Pn−1 are the monic orthogonal polynomials for w. Walter Van Assche OPSF, Random Matrices and Riemann-Hilbert problems Unicity of the solution Property For every solution of this RHP one has det Y = 1. Let Yb be a solution of the RHP, and consider YYb −1. −1 YYb is analytic in C n R. YYb −1 has no jump over the real line, hence YYb −1 is analytic in C. −1 1 −1 YYb = I + O( z ), hence by Liouville's theorem YYb = I, and thus Yb = Y . Walter Van Assche OPSF, Random Matrices and Riemann-Hilbert problems Recurrence relation Let Yn denote the solution with Pn in the upper left corner. −1 Consider R = Yn+1Yn , then 1 R is analytic in C n R. 2 R has no jump on the real line, hence R is analytic in C. 3 As z ! 1 z + O(1) O(1) 1 R(z) = + O( ) O(1) 0 z Hence Liouville's theorem gives that z − b c R(z) = n n ; dn 0 Conclusion z − bn cn Yn+1(z) = Yn(z) dn 0 2 Pn+1(z) = (z − bn)Pn(z) − 2πiγn−1cnPn−1(z) Walter Van Assche OPSF, Random Matrices and Riemann-Hilbert problems Liouville-Ostrogradski formula det Y (z) = 1 gives Z P (x)w(x) Z P (x)w(x) γ2 P (z) n−1 dx − P (z) n dx = 1 n−1 n z − x n−1 z − x Z p (x)w(x) In terms of p (z) = γ P (z) and q (z) = n dx n n n n z − x an pn(z)qn−1(z) − pn−1(z)qn(z) = 1: This is known as the Liouville-Ostrogradski formula: Wronskian [Casorati determinant] of two solutions of the recurrence relation. Walter Van Assche OPSF, Random Matrices and Riemann-Hilbert problems Christoffel-Darboux kernel Recall n−1 X Pn(x)Pn−1(y) − Pn−1(x)Pn(y) K (x; y) = γ2P (x)P (y) = γ2 : n k k k n−1 x − y k=0 Pn(x) ∗ −1 ∗ ∗ Yn(x) = 2 ; Yn (y) = 2 −2πiγn−1Pn−1(x) ∗ 2πiγn−1Pn−1(y) Pn(y) 1 0 1 Y −1(y)Y (x) = 2πiγ2 P (x)P (y)−P (x)P (y) n n 0 n−1 n n−1 n−1 n Property 1 1 K (x; y) = 0 1 Y −1(y)Y (x) : n 2πi(x − y) n n 0 Walter Van Assche OPSF, Random Matrices and Riemann-Hilbert problems Motivation for doing asymptotics In random matrix theory one wants to investigate the asymptotic distribution of the eigenvalues for large matrices (n ! 1). Recall that Z ρ1(x) dx = E(Nn(A)); Nn(A) = number of eigenvalues in A: A ρ1(x) = Kn(x; x)w(x): Hence Z 1 Nn(A) lim Kn(x; x)w(x) dx = E lim : n!1 n A n!1 n But more is true Z Z Nn(A) 1 lim = lim Kn(x; x)w(x) dx = v(x) dx n!1 n n!1 n A A and v is called the asymptotic density of the eigenvalues. Walter Van Assche OPSF, Random Matrices and Riemann-Hilbert problems Motivation for doing asymptotics One needs to know the asymptotic behavior of Kn(x; x) as n ! 1. But this only gives the global distribution of the eigenvalues. One really wants to understand the local behavior of the eigenvalues: fix a point in the spectrum, how do the eigenvalues behave near that point? In particular one wants to understand the spacing between eigenvalues. This immediately gives rise to investigating gap probabilities pA(0) = P(no eigenvalues in A): Walter Van Assche OPSF, Random Matrices and Riemann-Hilbert problems Gap probabilities Recall that for disjoint sets k Z Z Z Y ··· ρk (x1;:::; xk ) dx1 ::: dxk = E Nn(Aj ) A1 A2 Ak j=1 k ρk (x1;:::; xk ) = det Kn(xi ; xj ) : i;j=1 If all the sets are the same then k−1 Z Y ρk (x1;:::; xk ) dx1 ::: dxk = E (Nn(A) − j) k A j=0 is the average number of points of the n-point process for which k components are in A. Z 1 1 Nn(A) X j ρk (x1;:::; xk ) dx1 ::: dxk = E = pA(j); k! k k k A j=k where pA(j) = P(j eigenvalues in A). Walter Van Assche OPSF, Random Matrices and Riemann-Hilbert problems Gap probabilities Consider the infinite series 1 k Z 1 1 X (−1) X k X j ρk (x1;:::; xk ) dx1 ::: dxk = (−1) pA(j) k! k k k=0 A k=0 j=k interchange sums 1 j X X j = p (j) (−1)k A k j=0 k=0 Now use j X j (−1)k = δ k 0;j k=0 to find 1 X (−1)k Z ρk (x1;:::; xk ) dx1 ::: dxk = pA(0) = P(Nn(A) = 0): k! k k=0 A Walter Van Assche OPSF, Random Matrices and Riemann-Hilbert problems Gap probabilities 1 X (−1)k Z ρk (x1;:::; xk ) dx1 ::: dxk = pA(0) = P(Nn(A) = 0): k! k k=0 A The sum is the known as the Fredholm2 determinant of the operator Kn;A 1 X (−1)k Z k det(1 − Kn;A) = det Kn(xi ; xj ) dx1 ::: dxk k! k i;j=1 k=0 A with Z (Kn;Af )(x) = Kn(x; y)f (y) dy: A 2Erik Ivar Fredholm, Sweden (1866{1927) Walter Van Assche OPSF, Random Matrices and Riemann-Hilbert problems Gap probabilities For local analysis around a point x∗ in the spectrum, we need 1 ∗ u ∗ v lim Kn x + ; x + = K(u; v) n!1 nc nγ nγ and the gap probablity will be the Fredholm determinant of KA with Z (KAf )(u) = K(u; v)f (v) dv: A If x∗ is in the bulk of the spectrum, then we have the sine kernel sin π(u − v) K(u; v) = : π(u − v) If x∗ is at the end of the spectrum, then we have the Airy kernel Ai(u)Ai0(v) − Ai0(u)Ai(v) Z 1 K(u; v) = = Ai(u+s)Ai(v+s) ds: u − v 0 Walter Van Assche OPSF, Random Matrices and Riemann-Hilbert problems References P.