R andom M atrices , I ntefaces and H ydrodynamics S ingularities P. - PowerPoint PPT Presentation

R andom M atrices , I ntefaces and H ydrodynamics S ingularities P. Wiegmann review of works with friends: Anton Zabrodin, Eldad Bettelheim, Razvan Teodorescu, Seun Yeop Lee June 26, 2015 1 / 38

List of Objects • Random Matrix Models: Equilibrium Measure; • Geometrical Growth Models; • Orthogonal Polynomials: Distribution of zeros; • Hydrodynamics Singularities; 2 / 38

Normal Random Matrices Normal matrix M ⇔ [ M , M † ] = 0 ⇔ diagonalizable by a unitary transform. M = U − 1 diag ( z 1 , . . . , z N ) U , z i − complex The eigenvalues of N × N normal matrices with the probability distribution Prob ( M ) d M = 1 Z e − 1 h Tr Q ( M ) d M � distributes by the probability density   n N P ( z 1 , ..., z N ) = 1 � 2  − 1 � � �  , ( z j − z k ) Q ( z j ) exp � � Z � � h � j < k j = 1 Q1 . What is the distribution of eigenvalues for h → 0, N → ∞ , t = � h N = fixed? � The answer depends on the potential Q . 3 / 38

2D Dyson’s Di ff usion Brownian motion of a Normal Matrix M = M † + V ′ ( M ) + Brownian Motion ˙ Eqenvalues (complex) perform 2D Dyson di ff usion � h � z i + V ′ ( z i ) + ˙ � ξ i ( t ) ¯ ξ j ( t ′ ) � = 4 δ ij ( t − t ′ ) . z i = ˙ + ¯ ξ i , z i − ¯ ¯ z j i � j Z e − 1 h Tr Q ( M ) is the Gibbs distribution of Dyson’s di ff usion. Probability 1 � Depending on V ′ there may or not may be Gibbs distribution. 4 / 38

Ginibre Ensemple and its deformations   n N P ( z 1 , ..., z N ) = 1 � 2  − 1 � � �  , ( z j − z k ) exp Q ( z j ) � � Z � � h � j < k j = 1 A choice of Q ( z ) - Gaussian plus harmonic function when V is holomorphic. Q ( z ) = | z | 2 , Ginibre ensemble: Q ( z ) = | z | 2 + V ( z ) + V ( z ) , Deformed Ginibre ensemble: ∆ Q = 4. 5 / 38

Ginibre Ensemble Q = | z | 2 Support is the disk of the area π � h N 6 / 38

Equilibrium measure Continuum limit: N ρ ( z ) = 1 � δ ( z − z j ) N j = 1 ∆ Q 1 � ρ � = 4 Area = on the support of ρ . Area What is support of density? It depends on the deformation holomorphic function V ( z ) 7 / 38

The eigenvalues are 2D Coulomb interacting electrons : N 1 h E ( z 1 , ... z N ) := 1 1 � � e − 1 h E ( z 1 ,..., z N ) , Q ( z j ) − 2 log | z j − z k | . � Z n h � � j = 1 j < k � N Continuum limit: Defining ρ ( z ) = 1 j = 1 δ ( z − z j ) , we have N � � � � Q ( z ′ ) ρ ( z ′ ) d 2 z ′ − � C 2 ρ ( z ) ρ ( z ′ ) log | z − z ′ | d 2 z d 2 z ′ E ( z 1 , ..., z n ) = � h N h N . C the condition for the optimal configuration is obtained when � log | z − z ′ | ρ ( z ′ ) d 2 z ′ 0 = Q ( z ) − � h N on the support of ρ . D Applying Laplace operator 1 1 ρ ( z ) = h N = on the support of ρ . Area π � 8 / 38

Bratwurst Take V ( z ) = − c log ( z − a ) such that Q ( z ) = | z | 2 − 2 c log | z − a | ( c > 0). 9 / 38

Growth Change the size of the matrix N → N + n Area of Equilibrium measure changes t → t + δ t , δ t = π � h n Q: What is the velocity? 10 / 38

Growth process Area t := π N � h is identified with time. Define the Newtonian potential U ( z ) by � log | z − w | d 2 w U ( z ) = t D Equilibrium condition: π Q ( z ) = U ( z ) , inside D , ¯ z = ∂ z U , inside D , � d d � dt ¯ z = velocity = ∂ z dtU ( z ) , on the boundary d dtU ( z ) is a harmonic function outside D , d dtU ( z ) = log | z | + O ( 1 ) , z → ∞ , d dtU ( z ) = 0 on ∂ D , Velocity of the boundary = d dt U ( z ) is the Harmonic Measure of D 11 / 38

Harmonic measure: Brownian excursion with a free boundary A probability for BM to arrive on an element of the boundary is a harmonic measure of the boundary: � � d f � � Probability to land on ds : � = | ∇ n G ( z , ∞ ) | ds , z ∈ ∂ D � � d z � − ∆ G ( z , z ′ ) = δ ( z − z ′ ) , G | z ∈ ∂ D = 0 f ( z ) is a univalent map from the exterior of the domain to the exterior of the unit circle 12 / 38

Geometrical (Laplacian) Growth Hele-Shaw Problem 13 / 38

Physical setup 1898 • Navier-Stokes Equation: v + ( v · ∇ ) v = ρ − 1 ∇ p + µ∆ v ˙ • Small Reynolds number - no inertia 0 = ρ − 1 ∇ p + µ∆ v • incompresibility: ρ = const, ∇ · v = 0; • 2D Geometry - Poiseuille’s law: d 2 ⇒ v = − d 2 v ∆ v ≈ ∂ 2 z v ≈ 12 ν p ; • no viscosity on the boundary: ⇒ p = 0 on the boundary. Darcy Law: v = − ∇ p , ∆ p = 0; p | ∂ D = 0; p | ∞ = − log | z | 14 / 38

Experiment: Hele-Shaw cell, Fingering instability F igure : Viscous incompressible fluid pushed out by inviscid incompressible fluid Blow hard, otherwise the surface tension will take over. 15 / 38

Fingering Instability F igure : Flame (no convection), Serenga river (Russia), Lung vessels 16 / 38

Cusp-Singularities F igure : Cusp: end of a smooth growth 17 / 38

Cusp-Singularities: Growing Deltoid   n N P ( z 1 , ..., z N ) = 1 � 2  − 1 � � �  , ( z j − z k ) exp Q ( z j ) � � Z � � h � j < k j = 1 Q ( z ) = | z | 2 + t 3 z 3 + t 3 z 3 Deformed Ginibre ensemble: 18 / 38

Cusp-Singularities Q ( z ) = | z | 2 + V ( z ) + V ( z ) Deformed Ginibre ensemble: Almost any deformation leads to a cusp singularity: y p ∼ x q The most generic is (2,3)- singularity y 2 ∼ x 3 19 / 38

Di ff usion limited aggregation (DLA) Fractal pattern with (numerically computed) dimension D H = 1.71004... Structure of this pattern is the main problem one the subject Zeros of Complexified Orthogonal Polynomials 20 / 38

Unstable Di ff usion � V = t 3 z 3 - an example when the integral e − 1 h tr Q d M diverges, there is no � Gibbs distribution: � h � z i + V ′ ( z i ) + ˙ � ξ i ( t ) ¯ ξ j ( t ′ ) � = 4 δ ij ( t − t ′ ) . z i = ˙ + ¯ ξ i , ¯ z i − ¯ z j i � j Particle escape. One keeps to pump particles to compensate escaping particles. 21 / 38

Bi-orthogonal polynomials and growth process The measure for the subset of the eigenvalues, z 1 , ..., z k , ( k � n ) , is given by   n N P ( z 1 , ..., z N ) = 1 � 2  − 1 � � �  , ( z j − z k ) exp Q ( z j ) � � Z � � h � j = 1 j < k Bi-orthogonal polynomials p j = z j + ... � p i ( z ) p j ( z ) e − 1 h Q ( z ) d 2 z . h j δ ij = � C Polynomial � � � ( z − z j ) P ( z 1 , . . . , z N ) d 2 z 1 . . . d 2 z N p n ( z ) = � ( z − z j ) � = j j Q : What is the asymptotic distribution of the roots of p n ( z ) for n → ∞ , h → 0? � 22 / 38

Christo ff el - Darboux formula Density � ρ N ( z ) = 1 � P ( z ; z 2 , . . . , z N ) d 2 z 2 . . . d 2 z N N � δ ( z − z j ) � = j Christo ff el - Darboux formula ρ N + 1 − ρ N ( z ) = | Ψ N ( z ) | 2 where h ( − 1 1 2 | z | 2 + V ( z ) ) p n ( z ) Ψ n ( z ) = h − 1 / 2 e � n are weighted orthogonal polynomials � Ψ n ( z ) Ψ m ( z ) d 2 z δ nm = | Ψ n | 2 can be seen as a velocity of growth. 23 / 38

Asymptotes of Orthogonal Polynomials solve the growth problem solve Important result: At a properly defined n → ∞ | Ψ n ( z ) | 2 is localized on ∂ D and proportional to the width of the infinitesimal strip: | Ψ n ( z ) | 2 | dz | ∼ | f ′ ( z ) dz | ≈ Harmonic measure z ∈ ∂ D : 24 / 38

The simplest example: Circle When V ( z ) = 0 the orthogonal polynomials are simply Ψ n ( z ) ∝ z n e − 1 h | z | 2 � The di ff erence between the consecutive kernels | Ψ n ( z ) | 2 is localized on ∂ D and proportional to the width of the infinitesimal strip. 25 / 38

Another example: Bratwurst Take V ( z ) = − c log ( z − a ) such that Q ( z ) = | z | 2 − 2 c log | z − a | ( c > 0). The plots of p n ( z ) p n ( z ) e − NQ ( z ) for various times. 26 / 38

Zeros of Orthogonal Polynomials • Szego theorem: Zeros of Orthogonal Polynomials with real coe ffi cients defined on R are distributed on R . • Zeros of Orthogonal Polynomials with real coe ffi cients defined on C are distributed on C . F igure : Deltoid: Q ( z ) = | z | 2 + t 3 z 3 + t 3 z 3 27 / 38

Balayage A minimal body (an open curve) which produces the same Newton potential as a domain D - mother body - Γ � � � log | z − w | d 2 w = log | z − w | σ ( w ) | d w | D Γ z ∈ Γ : S ( z ) dz = σ ( z ) | dz | A graph Γ : � z S ( z ′ ) d z ′ Ω = Level lines of Ω : Re Ω ( z ) | Γ = 0, Re Ω ( z ) | z → Γ > 0; are branch cuts drawn such that jump of S ( z ) is imaginary. Balayage reduces the domain to a curve Γ 28 / 38

Zeros of Orthogonal Polynomials Important result: A locus of zeros of Orthogonal Polynomials is identical to balayage � ( Stokes coe ffi cients ) k e − 1 Ψ ∼ f ′ ( z ) h Ω k ( z ) � all branches of Ω A graph of zeros is identical to level lines of Ω Re Ω ( z ) | Γ = 0, Re Ω ( z ) | z → Γ > 0; 29 / 38

Boutroux Curves Definition: ( ¯ z , S ( z )) : Real Riemann surface d Ω = S ( z ) dz � Re d Ω = 0 − all periods are imaginary B − cycles number of conditions - number of parameters = g - there is no general proof that these curves exist. Important result: Zeros of Orthogonal Polynomials are distributed along levels of Boutroux curves A graph Γ : Re Ω ( z ) | Γ = 0, Re Ω ( z ) | z → Γ > 0; 30 / 38