OPSF, Random Matrices and Riemann-Hilbert problems

Transcription

1 OPSF, Random Matrices and Riemann-Hilbert problems School on Orthogonal Polynomials in Approximation Theory and Mathematical Physics, ICMAT October, 2017

2 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

3 Orthogonal polynomials We only use absolutely continuous measures: dµ(x) = w(x) dx The monic orthogonal polynomial P n (x) for the weight w satisfies P n (x)x k w(x) dx = 0, k = 0, 1,..., n 1, P n (x)x n w(x) dx = 1 γn 2 0. The Riemann-Hilbert 1 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. 1 Bernhard Riemann ( ), David Hilbert ( )

4 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 \ R. 2 On R the boundary values f ± (x) = lim ɛ 0+ f (x ± iɛ) exist, and they are connected by 3 The asymptotic behavior is f + (x) = f (x) + w(x), x R. f (z) = O( 1 ), z. z Does such a function f exist? Is such a function unique?

5 A scalar Riemann-Hilbert problem Theorem (Sokhotsky-Plemelj) Suppose w is Hölder continuous on R and integrable, then f (z) = 1 2πi and this is the only solution. f (x + iɛ) = 1 2πi f (x iɛ) = 1 2πi Subtract to find w(t) z t dt, w(t) 1 dt = x + iɛ t 2πi w(t) 1 dt = x iɛ t 2πi 2iɛ f + (x) f (x) = lim ɛ 0 2πi x t iɛ (x t) 2 w(t) dt + ɛ2 x t + iɛ (x t) 2 w(t) dt + ɛ2 w(t) (x t) 2 + ɛ 2 dt 1 Julian Sokhotski ( ), Josip Plemelj ( )

6 A scalar Riemann-Hilbert problem Change variable t = x + ɛs to find ɛ w(t) 1 lim ɛ 0 π (x t) 2 dt = lim + ɛ2 ɛ 0 π = 1 π w(x) Hence the boundary conditions hold. w(x + ɛs) s 2 ds ds = w(x). 1 + s2 The asymptotic behavior is also true and the function is analytic in C \ R. Unicity: suppose g is another solution and consider f g f g is analytic in C \ 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).

7 Riemann-Hilbert problem for orthogonal polynomials Find a matrix function Y : C C 2 2 such that 1 Y is analytic in C \ R. 2 On R the boundary values lim ɛ 0+ Y (x ± iɛ) = Y ± (x) exist, and they are related by ( ) 1 w(x) Y + (x) = Y (x), x R Asymptotic behaviour ( Y (z) = I + O( 1 ) ( z ) z n ) 0 0 z n, z.

8 Riemann-Hilbert problem for orthogonal polynomials Suppose w is Hölder continuous, then P n (z) Y (z) = 2πiγn 1 2 P n 1(z) 1 2πi γ 2 n 1 Pn (x)w(x) dx x z Pn 1 (x)w(x) dx x z where P n and P n 1 are the monic orthogonal polynomials for w.

9 Unicity of the solution Property For every solution of this RHP one has det Y = 1. Let Ŷ be a solution of the RHP, and consider Ŷ Y 1. Ŷ Y 1 is analytic in C \ R. Ŷ Y 1 has no jump over the real line, hence Ŷ Y 1 is analytic in C. Ŷ Y 1 = I + O( 1 z ), hence by Liouville s theorem Ŷ Y 1 = I, and thus Ŷ = Y.

10 Recurrence relation Let Y n denote the solution with P n in the upper left corner. Consider R = Y n+1 Yn 1, then 1 R is analytic in C \ R. 2 R has no jump on the real line, hence R is analytic in C. 3 As z ( ) z + O(1) O(1) R(z) = + O( 1 O(1) 0 z ) Hence Liouville s theorem gives that Conclusion R(z) = Y n+1 (z) = ( z bn c n d n 0 ), ( ) z bn c n Y d n 0 n (z) P n+1 (z) = (z b n )P n (z) 2πiγ 2 n 1c n P n 1 (z)

11 Liouville-Ostrogradski formula det Y (z) = 1 gives ( ) γn 1 2 Pn 1 (x)w(x) Pn (x)w(x) P n (z) dx P n 1 (z) dx = 1 z x z x pn (x)w(x) In terms of p n (z) = γ n P n (z) and q n (z) = dx z x ) a n (p n (z)q n 1 (z) p n 1 (z)q n (z) = 1. This is known as the Liouville-Ostrogradski formula: Wronskian [Casorati determinant] of two solutions of the recurrence relation.

12 Christoffel-Darboux kernel Recall n 1 K n (x, y) = γk 2 P k(x)p k (y) = γn 1 2 P n (x)p n 1 (y) P n 1 (x)p n (y). x y k=0 ( P Y n (x) = n (x) 2πiγn 1 2 P n 1(x) ) (, Y 1 n (y) = 2πiγn 1 2 P n 1(y) ) P n (y) ( ) ( ) 0 1 Y 1 1 ( ) n (y)y n (x) = 2πiγn 1 2 P 0 n (x)p n 1 (y) P n 1 (x)p n (y) Property K n (x, y) = 1 2πi(x y) ( ) 0 1 Y 1 n (y)y n (x) ( ) 1. 0

13 Motivation for doing asymptotics In random matrix theory one wants to investigate the asymptotic distribution of the eigenvalues for large matrices (n ). Recall that ρ 1 (x) dx = E(N n (A)), N n (A) = number of eigenvalues in A. A Hence But more is true ρ 1 (x) = K n (x, x)w(x). 1 ( lim K n (x, x)w(x) dx = E lim n n A n N n (A) 1 lim = lim n n n n A K n (x, x)w(x) dx = N n (A) ). n and v is called the asymptotic density of the eigenvalues. A v(x) dx

14 Motivation for doing asymptotics One needs to know the asymptotic behavior of K n (x, x) as n. 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 p A (0) = P(no eigenvalues in A).

15 Gap probabilities Recall that for disjoint sets ( k ) ρ k (x 1,..., x k ) dx 1... dx k = E N n (A j ) A 1 A 2 A k j=1 ( ) k ρ k (x 1,..., x k ) = det K n (x i, x j ). i,j=1 If all the sets are the same then (k 1 ) ρ k (x 1,..., x k ) dx 1... dx k = E (N n (A) j) A k is the average number of points of the n-point process for which k components are in A. ( ) 1 Nn (A) ( ) j ρ k (x 1,..., x k ) dx 1... dx k = E = p A (j), k! A k k k where p A (j) = P(j eigenvalues in A). j=0 j=k

16 Gap probabilities Consider the infinite series ( 1) k ρ k (x 1,..., x k ) dx 1... dx k = k! A k k=0 interchange sums Now use to find ( 1) k k=0 k! = p A (j) j=0 ( 1) k k=0 j=k j ( ) j ( 1) k k k=0 j ( ) j ( 1) k = δ 0,j k k=0 ( ) j p A (j) k A k ρ k (x 1,..., x k ) dx 1... dx k = p A (0) = P(N n (A) = 0).

17 Gap probabilities ( 1) k k=0 k! A k ρ k (x 1,..., x k ) dx 1... dx k = p A (0) = P(N n (A) = 0). The sum is the known as the Fredholm 2 determinant of the operator K n,a det(1 K n,a ) = ( 1) k ( det K n (x i, x j ) k! A k k=0 ) k i,j=1 dx 1... dx k with (K n,a f )(x) = A K n (x, y)f (y) dy. 2 Erik Ivar Fredholm, Sweden ( )

18 Gap probabilities For local analysis around a point x in the spectrum, we need 1 lim n n c K ( n x + u n γ, x + v ) = K(u, v) n γ and the gap probablity will be the Fredholm determinant of K A with (K A f )(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 K(u, v) = Ai(u)Ai (v) Ai (u)ai(v) = Ai(u+s)Ai(v +s) ds. u v 0

19 References P. Deift, Orthogonal Polynomials and Random Matrices: A Riemann-Hilbert Approach, Courant Lecture Notes in Mathematics 3, Courant Institute, New York, NY, and Amer. Math. Soc., Providence, RI., M.E.H. Ismail, Classical and Quantum Orthogonal Polynomials in One Variable, Encyclopedia of Mathematics and its Applications 98, Cambridge University Press, 2005 (paperback edition 2009) M.L. Mehta, Random Matrices, revised and enlarged second edition, Academic Press, San Diego, CA, E.B. Saff, V. Totik, Logarithmic Potentials with External Fields, Grundlehren der mathematischen Wissenschaften 316, Springer-Verlag, Berlin, G. Szegő, Orthogonal Polynomials, Amer. Math. Soc. Colloq. Publ. 23, Providence, RI, 1939; fourth edition 1975.