Riemann-Hilbert techniques in the theory of orthogonal matrix polynomials
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Riemann-Hilbert techniques in the theory of orthogonal matrix polynomials 1 Manuel Dom´ınguez de la Iglesia Courant Institute of Mathematical Sciences, New York University Departamento de An´alisis Matem´atico, Universidad de Sevilla XI Encuentros de An´alisis Real y Complejo Chinch´on, May 9, 2009 1joint work with F. A. Gr¨unbaum and A. Mart´ınez Finkelshtein
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials Outline 1What is a Riemann-Hilbert problem? 2The Riemann-Hilbert problem for orthogonal matrix polynomials Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials Outline 1What is a Riemann-Hilbert problem? 2The Riemann-Hilbert problem for orthogonal matrix polynomials Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials Hilbert’s 21st problem Riemann: The problem that an analytic function could be completely defined by its singularities and monodromy properties. Hilbert: A N×Nlinear system of differential equations dΨ(x) dx =A(x)Ψ(x) is called Fuchsian if A(x) is a rational function with simple poles. If Sis the punctured (at the poles) Riemann sphere C∪ {∞} then Ψ : Π1(S)→GL(N,C) gives a representation, called monodromy group. Question: Does always exist a Fuchsian system with given poles and a given monodromy group? Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials Hilbert’s 21st problem Riemann: The problem that an analytic function could be completely defined by its singularities and monodromy properties. Hilbert: A N×Nlinear system of differential equations dΨ(x) dx =A(x)Ψ(x) is called Fuchsian if A(x) is a rational function with simple poles. If Sis the punctured (at the poles) Riemann sphere C∪ {∞} then Ψ : Π1(S)→GL(N,C) gives a representation, called monodromy group. Question: Does always exist a Fuchsian system with given poles and a given monodromy group? Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials Hilbert’s 21st problem Riemann: The problem that an analytic function could be completely defined by its singularities and monodromy properties. Hilbert: A N×Nlinear system of differential equations dΨ(x) dx =A(x)Ψ(x) is called Fuchsian if A(x) is a rational function with simple poles. If Sis the punctured (at the poles) Riemann sphere C∪ {∞} then Ψ : Π1(S)→GL(N,C) gives a representation, called monodromy group. Question: Does always exist a Fuchsian system with given poles and a given monodromy group? Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials Hilbert’s 21st problem Riemann: The problem that an analytic function could be completely defined by its singularities and monodromy properties. Hilbert: A N×Nlinear system of differential equations dΨ(x) dx =A(x)Ψ(x) is called Fuchsian if A(x) is a rational function with simple poles. If Sis the punctured (at the poles) Riemann sphere C∪ {∞} then Ψ : Π1(S)→GL(N,C) gives a representation, called monodromy group. Question: Does always exist a Fuchsian system with given poles and a given monodromy group? Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials Some history J. Plemelj (1909) published a solution relating the problem into the context of analytic factorization of matrix-valued functions and methods of singular integral equations. A. Bolibruch (1989) gave a counterexample of size N= 3. For N= 2 Plemelj’s claim is true (Dekkers, 1978). There is now a modern version (D-module and derived category), the Riemann-Hilbert correspondence. The Riemann-Hilbert method consists of reconstructing an analytic function from jump conditions or the analytic factorization of a given matrix-valued function defined on a curve. This is what we will call Riemann-Hilbert problems (RHP). Applications: Integrable systems, Orthogonal polynomials and Random matrices, Combinatorial probability. Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials Some history J. Plemelj (1909) published a solution relating the problem into the context of analytic factorization of matrix-valued functions and methods of singular integral equations. A. Bolibruch (1989) gave a counterexample of size N= 3. For N= 2 Plemelj’s claim is true (Dekkers, 1978). There is now a modern version (D-module and derived category), the Riemann-Hilbert correspondence. The Riemann-Hilbert method consists of reconstructing an analytic function from jump conditions or the analytic factorization of a given matrix-valued function defined on a curve. This is what we will call Riemann-Hilbert problems (RHP). Applications: Integrable systems, Orthogonal polynomials and Random matrices, Combinatorial probability. Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials Riemann-Hilbert factorization problem Let G: Σ →GL(N,C) be a matrix-valued function. The RHP determined by (Σ,G) consists of finding an N×Nmatrix-valued function Y(z) s.t. 1Y(z) is analytic in C\Σ 2Y+(z) = Y−(z)G(z) when z∈Σ Y±(z) = limz′→z,±side Y(z) 3Y(z) = Ias z→ ∞ + − + − + −+ − + − + − + − Σ Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials Scalar and additive RHP (N= 1) Let ω:R→Rbe a L1(R) and H¨older continous function. The Riemann-Hilbert problem determined by (R, ω) consists of finding a function f:C→Csuch that 1f(z) is analytic in C\R 2f+(x) = f−(x) + ω(x) when x∈R 3f(z) = O(1/z) as z→ ∞ The unique solution of the RH problem above is the Stieltjes or Cauchy transform of ω, i.e. f(z) = C(ω)(z). =1 2πiZR ω(t) t−z Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials Scalar and additive RHP (N= 1) Let ω:R→Rbe a L1(R) and H¨older continous function. The Riemann-Hilbert problem determined by (R, ω) consists of finding a function f:C→Csuch that 1f(z) is analytic in C\R 2f+(x) = f−(x) + ω(x) when x∈R 3f(z) = O(1/z) as z→ ∞ The unique solution of the RH problem above is the Stieltjes or Cauchy transform of ω, i.e. f(z) = C(ω)(z). =1 2πiZR ω(t) t−z Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials Scalar and additive RHP (N= 1) Let ω:R→Rbe a L1(R) and H¨older continous function. The Riemann-Hilbert problem determined by (R, ω) consists of finding a function f:C→Csuch that 1f(z) is analytic in C\R 2f+(x) = f−(x) + ω(x) when x∈R 3f(z) = O(1/z) as z→ ∞ The unique solution of the RH problem above is the Stieltjes or Cauchy transform of ω, i.e. f(z) = C(ω)(z). =1 2πiZR ω(t) t−z Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials Scalar and additive RHP (N= 1) Let ω:R→Rbe a L1(R) and H¨older continous function. The Riemann-Hilbert problem determined by (R, ω) consists of finding a function f:C→Csuch that 1f(z) is analytic in C\R 2f+(x) = f−(x) + ω(x) when x∈R 3f(z) = O(1/z) as z→ ∞ The unique solution of the RH problem above is the Stieltjes or Cauchy transform of ω, i.e. f(z) = C(ω)(z). =1 2πiZR ω(t) t−z Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials Scalar and additive RHP (N= 1) Let ω:R→Rbe a L1(R) and H¨older continous function. The Riemann-Hilbert problem determined by (R, ω) consists of finding a function f:C→Csuch that 1f(z) is analytic in C\R 2f+(x) = f−(x) + ω(x) when x∈R 3f(z) = O(1/z) as z→ ∞ The unique solution of the RH problem above is the Stieltjes or Cauchy transform of ω, i.e. f(z) = C(ω)(z). =1 2πiZR ω(t) t−z Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials RHP for orthogonal polynomials (N= 2) Let dµbe a positive Borel measure. We will assume dµ(x) = ω(x)dx,ω≥0 and xiω, xjω′∈L1(R). We can then construct a family of orthonormal polynomials (pn)ns.t. (pn,pm)ω=ZR pn(x)pm(x)ω(x)dx =δn,m,n,m≥0 pn(x) = γn(xn+an,n−1xn−1+···) = γnb pn(x) We try to find a 2 ×2 matrix-valued function Yn:C→C2×2such that 1Ynis analytic in C\R 2Yn +(x) = Yn −(x)1ω(x) 0 1 when x∈R 3Yn(z) = (I+O(1/z)) zn0 0z−nas z→ ∞ Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials RHP for orthogonal polynomials (N= 2) Let dµbe a positive Borel measure. We will assume dµ(x) = ω(x)dx,ω≥0 and xiω, xjω′∈L1(R). We can then construct a family of orthonormal polynomials (pn)ns.t. (pn,pm)ω=ZR pn(x)pm(x)ω(x)dx =δn,m,n,m≥0 pn(x) = γn(xn+an,n−1xn−1+···) = γnb pn(x) We try to find a 2 ×2 matrix-valued function Yn:C→C2×2such that 1Ynis analytic in C\R 2Yn +(x) = Yn −(x)1ω(x) 0 1 when x∈R 3Yn(z) = (I+O(1/z)) zn0 0z−nas z→ ∞ Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials RHP for orthogonal polynomials (N= 2) Let dµbe a positive Borel measure. We will assume dµ(x) = ω(x)dx,ω≥0 and xiω, xjω′∈L1(R). We can then construct a family of orthonormal polynomials (pn)ns.t. (pn,pm)ω=ZR pn(x)pm(x)ω(x)dx =δn,m,n,m≥0 pn(x) = γn(xn+an,n−1xn−1+···) = γnb pn(x) We try to find a 2 ×2 matrix-valued function Yn:C→C2×2such that 1Ynis analytic in C\R 2Yn +(x) = Yn −(x)1ω(x) 0 1 when x∈R 3Yn(z) = (I+O(1/z)) zn0 0z−nas z→ ∞ Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials RHP for orthogonal polynomials (N= 2) Let dµbe a positive Borel measure. We will assume dµ(x) = ω(x)dx,ω≥0 and xiω, xjω′∈L1(R). We can then construct a family of orthonormal polynomials (pn)ns.t. (pn,pm)ω=ZR pn(x)pm(x)ω(x)dx =δn,m,n,m≥0 pn(x) = γn(xn+an,n−1xn−1+···) = γnb pn(x) We try to find a 2 ×2 matrix-valued function Yn:C→C2×2such that 1Ynis analytic in C\R 2Yn +(x) = Yn −(x)1ω(x) 0 1 when x∈R 3Yn(z) = (I+O(1/z)) zn0 0z−nas z→ ∞ Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials A recurrence relation If we call R=Yn+1(Yn)−1and denoting Yn(z) = I+1 zdnen fngn+On(1/z2)zn0 0z−n,z→ ∞ we have that 1Ris analytic in C\R 2R+(x) = R−(x) for all x∈R 3R(z) = z+dn+1 −dn−en fn+1 0as z→ ∞ Therefore Yn+1(z) = z+dn+1 −dn−en fn+1 0Yn(z) Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials A recurrence relation If we call R=Yn+1(Yn)−1and denoting Yn(z) = I+1 zdnen fngn+On(1/z2)zn0 0z−n,z→ ∞ we have that 1Ris analytic in C\R 2R+(x) = R−(x) for all x∈R 3R(z) = z+dn+1 −dn−en fn+1 0as z→ ∞ Therefore Yn+1(z) = z+dn+1 −dn−en fn+1 0Yn(z) Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials A recurrence relation If we call R=Yn+1(Yn)−1and denoting Yn(z) = I+1 zdnen fngn+On(1/z2)zn0 0z−n,z→ ∞ we have that 1Ris analytic in C\R 2R+(x) = R−(x) for all x∈R 3R(z) = z+dn+1 −dn−en fn+1 0as z→ ∞ Therefore Yn+1(z) = z+dn+1 −dn−en fn+1 0Yn(z) Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials A recurrence relation If we call R=Yn+1(Yn)−1and denoting Yn(z) = I+1 zdnen fngn+On(1/z2)zn0 0z−n,z→ ∞ we have that 1Ris analytic in C\R 2R+(x) = R−(x) for all x∈R 3R(z) = z+dn+1 −dn−en fn+1 0as z→ ∞ Therefore Yn+1(z) = z+dn+1 −dn−en fn+1 0Yn(z) Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials A recurrence relation II If we put αn=dn−dn+1,and βn=enfn entry (1,1) gives the three-term recurrence formula zhn(z) = hn+1(z) + αnhn(z) + βnhn−1(z) The entry (2,1) gives fn=cn=−2πiγ2 n−1 From the explicit expression of αnand βnwe have dn=an,n−1,and c−1 n+1 =en Also we have the following fn+1en= 1 dn+gn= 0 Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials A recurrence relation II If we put αn=dn−dn+1,and βn=enfn entry (1,1) gives the three-term recurrence formula zhn(z) = hn+1(z) + αnhn(z) + βnhn−1(z) The entry (2,1) gives fn=cn=−2πiγ2 n−1 From the explicit expression of αnand βnwe have dn=an,n−1,and c−1 n+1 =en Also we have the following fn+1en= 1 dn+gn= 0 Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials A recurrence relation II If we put αn=dn−dn+1,and βn=enfn entry (1,1) gives the three-term recurrence formula zhn(z) = hn+1(z) + αnhn(z) + βnhn−1(z) The entry (2,1) gives fn=cn=−2πiγ2 n−1 From the explicit expression of αnand βnwe have dn=an,n−1,and c−1 n+1 =en Also we have the following fn+1en= 1 dn+gn= 0 Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials A recurrence relation II If we put αn=dn−dn+1,and βn=enfn entry (1,1) gives the three-term recurrence formula zhn(z) = hn+1(z) + αnhn(z) + βnhn−1(z) The entry (2,1) gives fn=cn=−2πiγ2 n−1 From the explicit expression of αnand βnwe have dn=an,n−1,and c−1 n+1 =en Also we have the following fn+1en= 1 dn+gn= 0 Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials A differential equation Consider the transformation Xn(z) = Yn(z) e−z2/20 0ez2/2! We observe that Xnis invertible and that Xn +(x) = Yn +(x) e−x2/20 0ex2/2!=Yn −(x)1e−x2 0 1 e−x2/20 0ex2/2! =Yn −(x) e−x2/20 0ex2/2! ex2/20 0e−x2/2!1e−x2 0 1 e−x2/20 0ex2/2! =Xn −(x)1 1 0 1 That means that Xnhas a constant jump. Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials A differential equation Consider the transformation Xn(z) = Yn(z) e−z2/20 0ez2/2! We observe that Xnis invertible and that Xn +(x) = Yn +(x) e−x2/20 0ex2/2!=Yn −(x)1e−x2 0 1 e−x2/20 0ex2/2! =Yn −(x) e−x2/20 0ex2/2! ex2/20 0e−x2/2!1e−x2 0 1 e−x2/20 0ex2/2! =Xn −(x)1 1 0 1 That means that Xnhas a constant jump. Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials The Lax pair and compatibility conditions Xnsatisfies the following Lax pair Xn+1(z) = z−αn−c−1 n+1 cn+1 0 | {z } En(z) Xn(z),d dz Xn(z) = −z2c−1 n+1 −2cnz | {z } Fn(z) Xn(z) Combining them we get E′ n(z) + En(z)Fn(z) = Fn+1(z)En(z) In this case we get (using βn=cn/cn+1) αn= 0, βn+1 −βn=1 2⇒βn=n 2 Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials The Lax pair and compatibility conditions Xnsatisfies the following Lax pair Xn+1(z) = z−αn−c−1 n+1 cn+1 0 | {z } En(z) Xn(z),d dz Xn(z) = −z2c−1 n+1 −2cnz | {z } Fn(z) Xn(z) Combining them we get E′ n(z) + En(z)Fn(z) = Fn+1(z)En(z) In this case we get (using βn=cn/cn+1) αn= 0, βn+1 −βn=1 2⇒βn=n 2 Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials The Lax pair and compatibility conditions Xnsatisfies the following Lax pair Xn+1(z) = z−αn−c−1 n+1 cn+1 0 | {z } En(z) Xn(z),d dz Xn(z) = −z2c−1 n+1 −2cnz | {z } Fn(z) Xn(z) Combining them we get E′ n(z) + En(z)Fn(z) = Fn+1(z)En(z) In this case we get (using βn=cn/cn+1) αn= 0, βn+1 −βn=1 2⇒βn=n 2 Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials The ladder operators and second order differential equation From entries (1,1) and (2,1) of d dz Xn(z) = −z2c−1 n+1 −2cnzXn(z) we get the ladder operators h′ n(z) = nhn−1(z),h′ n−1(z)−2zhn−1(z) = −2hn(z) Combining them we get the second order differential equation h′′ n(z)−2zh′ n(z) + 2nhn(z) = 0 Conclusion: advantages 1Algebraic properties: three term recurrence relation, ladder operators, second order differential equation 2Uniform asymptotics:steepest descent analysis for RHP (Deift-Zhou,1993). The idea is to transform Yninto another RHP Rnwhere Rn(z) = I+O(1/z) as z→ ∞ and R+(z) = R−(z)(I+O(1/n)). Therefore Rn(z) = I+O(1/n) as n→ ∞ uniformly for all z∈C. Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials The ladder operators and second order differential equation From entries (1,1) and (2,1) of d dz Xn(z) = −z2c−1 n+1 −2cnzXn(z) we get the ladder operators h′ n(z) = nhn−1(z),h′ n−1(z)−2zhn−1(z) = −2hn(z) Combining them we get the second order differential equation h′′ n(z)−2zh′ n(z) + 2nhn(z) = 0 Conclusion: advantages 1Algebraic properties: three term recurrence relation, ladder operators, second order differential equation 2Uniform asymptotics:steepest descent analysis for RHP (Deift-Zhou,1993). The idea is to transform Yninto another RHP Rnwhere Rn(z) = I+O(1/z) as z→ ∞ and R+(z) = R−(z)(I+O(1/n)). Therefore Rn(z) = I+O(1/n) as n→ ∞ uniformly for all z∈C. Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials The ladder operators and second order differential equation From entries (1,1) and (2,1) of d dz Xn(z) = −z2c−1 n+1 −2cnzXn(z) we get the ladder operators h′ n(z) = nhn−1(z),h′ n−1(z)−2zhn−1(z) = −2hn(z) Combining them we get the second order differential equation h′′ n(z)−2zh′ n(z) + 2nhn(z) = 0 Conclusion: advantages 1Algebraic properties: three term recurrence relation, ladder operators, second order differential equation 2Uniform asymptotics:steepest descent analysis for RHP (Deift-Zhou,1993). The idea is to transform Yninto another RHP Rnwhere Rn(z) = I+O(1/z) as z→ ∞ and R+(z) = R−(z)(I+O(1/n)). Therefore Rn(z) = I+O(1/n) as n→ ∞ uniformly for all z∈C. Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials The ladder operators and second order differential equation From entries (1,1) and (2,1) of d dz Xn(z) = −z2c−1 n+1 −2cnzXn(z) we get the ladder operators h′ n(z) = nhn−1(z),h′ n−1(z)−2zhn−1(z) = −2hn(z) Combining them we get the second order differential equation h′′ n(z)−2zh′ n(z) + 2nhn(z) = 0 Conclusion: advantages 1Algebraic properties: three term recurrence relation, ladder operators, second order differential equation 2Uniform asymptotics:steepest descent analysis for RHP (Deift-Zhou,1993). The idea is to transform Yninto another RHP Rnwhere Rn(z) = I+O(1/z) as z→ ∞ and R+(z) = R−(z)(I+O(1/n)). Therefore Rn(z) = I+O(1/n) as n→ ∞ uniformly for all z∈C. Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials Outline 1What is a Riemann-Hilbert problem? 2The Riemann-Hilbert problem for orthogonal matrix polynomials Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials Orthogonal matrix polynomials The theory of orthogonal matrix polynomials (OMP) was introduced by Krein in 1949. AN×Nmatrix polynomial on the real line is P(x) = Anxn+An−1xn−1+···+A0,x∈RAi∈CN×N Let Wbe a N×Na matrix of measures or weight matrix. We will assume dW (x) = W(x)dx and RxnW(x)dx RxmW′(x)dx exist. We can construct a family of OMP with respect to the inner product (P,Q)W=ZR P(x)W(x)Q∗(x)dx ∈CN×N such that (Pn,Pm)W=ZR Pn(x)W(x)P∗ m(x)dx =δn,mI,n,m≥0 Pn(x) = γn(xn+an,n−1xn−1+· · · ) = γnb Pn(x) Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials Orthogonal matrix polynomials The theory of orthogonal matrix polynomials (OMP) was introduced by Krein in 1949. AN×Nmatrix polynomial on the real line is P(x) = Anxn+An−1xn−1+···+A0,x∈RAi∈CN×N Let Wbe a N×Na matrix of measures or weight matrix. We will assume dW (x) = W(x)dx and RxnW(x)dx RxmW′(x)dx exist. We can construct a family of OMP with respect to the inner product (P,Q)W=ZR P(x)W(x)Q∗(x)dx ∈CN×N such that (Pn,Pm)W=ZR Pn(x)W(x)P∗ m(x)dx =δn,mI,n,m≥0 Pn(x) = γn(xn+an,n−1xn−1+· · · ) = γnb Pn(x) Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials Again, we have that detYn(z) = 1, so we find a solution of the inverse (Yn)−1= C(Wb P∗ n−1)(z)cn−C(Wb P∗ n)(z) −b P∗ n−1(¯z)cnb P∗ n(¯z)! The Liouville-Ostrogradski formula (Dur´an, 1996) Qn(z)P∗ n−1(¯z)−Pn(z)Q∗ n−1(¯z) = γnγ−1 n−1 The Hermitian property (Dur´an, 1996) Qn(z)P∗ n(¯z) = Pn(z)Q∗ n(¯z) Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials A special case Let us consider W(x) = e−x2eAx eA∗x,x∈R for any A∈CN×N. The solution Yn(z) of the RHP 1Ynis analytic in C\R 2Yn +(x) = Yn −(x)I e−x2eAx eA∗x 0Iwhen x∈R 3Yn(z) = (I+O(1/z)) znI0 0z−nIas z→ ∞ is given by Yn(z) = b Pn(z)C(b Pne−t2eAt eA∗t)(z) cnb Pn−1(z)cnC(b Pn−1e−t2eAt eA∗t)(z)! where (b Pn)nis the family of monic polynomials with respect to W. Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials A special case Let us consider W(x) = e−x2eAx eA∗x,x∈R for any A∈CN×N. The solution Yn(z) of the RHP 1Ynis analytic in C\R 2Yn +(x) = Yn −(x)I e−x2eAx eA∗x 0Iwhen x∈R 3Yn(z) = (I+O(1/z)) znI0 0z−nIas z→ ∞ is given by Yn(z) = b Pn(z)C(b Pne−t2eAt eA∗t)(z) cnb Pn−1(z)cnC(b Pn−1e−t2eAt eA∗t)(z)! where (b Pn)nis the family of monic polynomials with respect to W. Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials A recurrence relation If we call R=Yn+1(Yn)−1and denoting Yn(z) = I+1 zYn 1+On(1/z2)znI0 0z−nI,z→ ∞ we have that 1Ris analytic in C\R 2R+(x) = R−(x) for all x∈R 3R(z) = zI +Yn+1 111 −Yn 111 −Yn 112 Yn+1 121 0as z→ ∞ Therefore Yn+1(z) = zI +Yn+1 111 −Yn 111 −Yn 112 Yn+1 121 0Yn(z) Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials A recurrence relation If we call R=Yn+1(Yn)−1and denoting Yn(z) = I+1 zYn 1+On(1/z2)znI0 0z−nI,z→ ∞ we have that 1Ris analytic in C\R 2R+(x) = R−(x) for all x∈R 3R(z) = zI +Yn+1 111 −Yn 111 −Yn 112 Yn+1 121 0as z→ ∞ Therefore Yn+1(z) = zI +Yn+1 111 −Yn 111 −Yn 112 Yn+1 121 0Yn(z) Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials A recurrence relation If we call R=Yn+1(Yn)−1and denoting Yn(z) = I+1 zYn 1+On(1/z2)znI0 0z−nI,z→ ∞ we have that 1Ris analytic in C\R 2R+(x) = R−(x) for all x∈R 3R(z) = zI +Yn+1 111 −Yn 111 −Yn 112 Yn+1 121 0as z→ ∞ Therefore Yn+1(z) = zI +Yn+1 111 −Yn 111 −Yn 112 Yn+1 121 0Yn(z) Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials A recurrence relation If we call R=Yn+1(Yn)−1and denoting Yn(z) = I+1 zYn 1+On(1/z2)znI0 0z−nI,z→ ∞ we have that 1Ris analytic in C\R 2R+(x) = R−(x) for all x∈R 3R(z) = zI +Yn+1 111 −Yn 111 −Yn 112 Yn+1 121 0as z→ ∞ Therefore Yn+1(z) = zI +Yn+1 111 −Yn 111 −Yn 112 Yn+1 121 0Yn(z) Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials A recurrence relation If we call R=Yn+1(Yn)−1and denoting Yn(z) = I+1 zYn 1+On(1/z2)znI0 0z−nI,z→ ∞ we have that 1Ris analytic in C\R 2R+(x) = R−(x) for all x∈R 3R(z) = zI +Yn+1 111 −Yn 111 −Yn 112 Yn+1 121 0as z→ ∞ Therefore Yn+1(z) = zI +Yn+1 111 −Yn 111 −Yn 112 Yn+1 121 0Yn(z) Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials A recurrence relation II If we put αn=Yn 111 −Yn+1 111, βn=Yn 112Yn 121 block entry (1,1) gives the three-term recurrence formula zb Pn(z) = b Pn+1(z) + αnb Pn(z) + βnb Pn−1(z) The block entry (2,1) gives cn=Yn 121 From the explicit expression of αnand βnwe have Yn+1 111 =an,n−1,and c−1 n+1 =Yn−1 112 Also we have the following Yn+1 121Yn 112 =Yn 112Yn+1 121 =I,Yn 111 +Yn 1∗ 22 = 0 Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials A recurrence relation II If we put αn=Yn 111 −Yn+1 111, βn=Yn 112Yn 121 block entry (1,1) gives the three-term recurrence formula zb Pn(z) = b Pn+1(z) + αnb Pn(z) + βnb Pn−1(z) The block entry (2,1) gives cn=Yn 121 From the explicit expression of αnand βnwe have Yn+1 111 =an,n−1,and c−1 n+1 =Yn−1 112 Also we have the following Yn+1 121Yn 112 =Yn 112Yn+1 121 =I,Yn 111 +Yn 1∗ 22 = 0 Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials A differential equation II Consider R=d dz Xn(Xn)−1. We have that 1Ris analytic in C\R 2R+(x) = R−(x) for all x∈R 3 R(z) = −zI +A0 0zI −A∗+Yn 1−I0 0I−−I0 0IYn 1 =−zI +A2Yn 112 −2Yn+1 121 zI −A∗ as z→ ∞ Therefore d dz Xn(z) = −zI +A2c−1 n+1 −2cnzI −A∗Xn(z) Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials A differential equation II Consider R=d dz Xn(Xn)−1. We have that 1Ris analytic in C\R 2R+(x) = R−(x) for all x∈R 3 R(z) = −zI +A0 0zI −A∗+Yn 1−I0 0I−−I0 0IYn 1 =−zI +A2Yn 112 −2Yn+1 121 zI −A∗ as z→ ∞ Therefore d dz Xn(z) = −zI +A2c−1 n+1 −2cnzI −A∗Xn(z) Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials A differential equation II Consider R=d dz Xn(Xn)−1. We have that 1Ris analytic in C\R 2R+(x) = R−(x) for all x∈R 3 R(z) = −zI +A0 0zI −A∗+Yn 1−I0 0I−−I0 0IYn 1 =−zI +A2Yn 112 −2Yn+1 121 zI −A∗ as z→ ∞ Therefore d dz Xn(z) = −zI +A2c−1 n+1 −2cnzI −A∗Xn(z) Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials A differential equation II Consider R=d dz Xn(Xn)−1. We have that 1Ris analytic in C\R 2R+(x) = R−(x) for all x∈R 3 R(z) = −zI +A0 0zI −A∗+Yn 1−I0 0I−−I0 0IYn 1 =−zI +A2Yn 112 −2Yn+1 121 zI −A∗ as z→ ∞ Therefore d dz Xn(z) = −zI +A2c−1 n+1 −2cnzI −A∗Xn(z) Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials The Lax pair and compatibility conditions Xnsatisfies the following Lax pair Xn+1(z) = zI −αn−c−1 n+1 cn+1 0 | {z } En(z) Xn(z),d dz Xn(z) = −zI +A2c−1 n+1 −2cnzI −A∗ | {z } Fn(z) Xn(z) Combining them we get E′ n(z) + En(z)Fn(z) = Fn+1(z)En(z) In this case we get (using βn=cnc−1 n+1) Compatibility conditions 2(βn+1 −βn) = Aαn−αnA+I αn=1 2(A+c−1 n+1A∗cn+1) Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials The Lax pair and compatibility conditions Xnsatisfies the following Lax pair Xn+1(z) = zI −αn−c−1 n+1 cn+1 0 | {z } En(z) Xn(z),d dz Xn(z) = −zI +A2c−1 n+1 −2cnzI −A∗ | {z } Fn(z) Xn(z) Combining them we get E′ n(z) + En(z)Fn(z) = Fn+1(z)En(z) In this case we get (using βn=cnc−1 n+1) Compatibility conditions 2(βn+1 −βn) = Aαn−αnA+I αn=1 2(A+c−1 n+1A∗cn+1) Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials The Lax pair and compatibility conditions Xnsatisfies the following Lax pair Xn+1(z) = zI −αn−c−1 n+1 cn+1 0 | {z } En(z) Xn(z),d dz Xn(z) = −zI +A2c−1 n+1 −2cnzI −A∗ | {z } Fn(z) Xn(z) Combining them we get E′ n(z) + En(z)Fn(z) = Fn+1(z)En(z) In this case we get (using βn=cnc−1 n+1) Compatibility conditions 2(βn+1 −βn) = Aαn−αnA+I αn=1 2(A+c−1 n+1A∗cn+1) Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials The ladder operators and second order differential equation From block entries (1,1) and (2,1) of d dz Xn(z) = −zI +A2c−1 n+1 −2cnzI −A∗Xn(z) we get the ladder operators Ladder operators b P′ n(z) + b Pn(z)A−Ab Pn(z) = 2βnb Pn−1(z) −b P′ n(z) + 2(z−αn)b Pn(z) + Ab Pn(z)−b Pn(z)A= 2b Pn+1(z) Combining them we get the second order differential equation Second order differential equation b P′′ n(z) + 2b P′ n(z)(A−zI) + b Pn(z)A2−A2b Pn(z) + 4βnb Pn(z) = −2z(b Pn(z)A−Ab Pn(z)) + 2(αn−A)(b P′ n(z) + b Pn(z)A−Ab Pn(z)) Manuel Dom´ınguez de la Iglesia The RH problem for OMP
What is a Riemann-Hilbert problem? The Riemann-Hilbert problem for orthogonal matrix polynomials The ladder operators and second order differential equation From block entries (1,1) and (2,1) of d dz Xn(z) = −zI +A2c−1 n+1 −2cnzI −A∗Xn(z) we get the ladder operators Ladder operators b P′ n(z) + b Pn(z)A−Ab Pn(z) = 2βnb Pn−1(z) −b P′ n(z) + 2(z−αn)b Pn(z) + Ab Pn(z)−b Pn(z)A= 2b Pn+1(z) Combining them we get the second order differential equation Second order differential equation b P′′ n(z) + 2b P′ n(z)(A−zI) + b Pn(z)A2−A2b Pn(z) + 4βnb Pn(z) = −2z(b Pn(z)A−Ab Pn(z)) + 2(αn−A)(b P′ n(z) + b Pn(z)A−Ab Pn(z)) Manuel Dom´ınguez de la Iglesia The RH problem for OMP