On the interface between Hermitian and normal random matrices Yacin - PowerPoint PPT Presentation

On the interface between Hermitian and normal random matrices Yacin Ameur. Centre for Mathematical Sciences Lund University, Sweden Yacin.Ameur@maths.lth.se Random Matrices and Related Topics, KIAS, May 7, 2019 1 / 41

External potential Let Q : C → R ∪ { + ∞} . C ω -smooth where finite. 2 / 41

External potential Let Q : C → R ∪ { + ∞} . C ω -smooth where finite. µ p.m. on C �� 1 � I Q [ µ ] = log | ζ − η | d µ ( η ) d µ ( ζ ) + Q d µ. The equilibrium measure σ minimizes I Q . 2 / 41

External potential Let Q : C → R ∪ { + ∞} . C ω -smooth where finite. µ p.m. on C �� 1 � I Q [ µ ] = log | ζ − η | d µ ( η ) d µ ( ζ ) + Q d µ. The equilibrium measure σ minimizes I Q . The droplet S = supp σ . We assume that S ⊂ Int { Q < + ∞} . Then d σ = ∆ Q · 1 S · dA . 2 / 41

External potential Let Q : C → R ∪ { + ∞} . C ω -smooth where finite. µ p.m. on C �� 1 � I Q [ µ ] = log | ζ − η | d µ ( η ) d µ ( ζ ) + Q d µ. The equilibrium measure σ minimizes I Q . The droplet S = supp σ . We assume that S ⊂ Int { Q < + ∞} . Then d σ = ∆ Q · 1 S · dA . Note that ∆ Q ≥ 0 on S . We assume that ∆ Q > 0 on the boundary ∂ S. 2 / 41

Sakai’s theorem The boundary ∂ S is regular with possible cusps/double points. Figure: The figure on the left shows a boundary with two singular points: one double point and one cusp. Not all cusps are possible: 3 2 , 7 2 ,... are excluded. 3 / 41

RNM-model Particles/eigenvalues { ζ j } n 1 in external field nQ . Energy: n 1 � � H n = log | ζ j − ζ k | + n Q ( ζ j ) . 1 j � = k 4 / 41

RNM-model Particles/eigenvalues { ζ j } n 1 in external field nQ . Energy: n 1 � � H n = log | ζ j − ζ k | + n Q ( ζ j ) . 1 j � = k Probability law: � dP n ( ζ ) = e − H n ( ζ ) dA n ( ζ ) / C n e − H n dA n . ( dA n Lebesgue measure). 4 / 41

RNM-model Particles/eigenvalues { ζ j } n 1 in external field nQ . Energy: n 1 � � H n = log | ζ j − ζ k | + n Q ( ζ j ) . 1 j � = k Probability law: � dP n ( ζ ) = e − H n ( ζ ) dA n ( ζ ) / C n e − H n dA n . ( dA n Lebesgue measure). Coulomb gas at β = 1 / ( k B T ): replace H n ↔ β H n . 4 / 41

Ginibre ensemble � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � Figure: A sample from the standard Ginibre ensemble. 5 / 41

Correlation kernel We have a determinantal process, k -point functions R n , k ( ζ 1 , . . . , ζ k ) = det( K n ( ζ i , ζ j )) k × k . Here R n ( ζ ) := R n , 1 ( ζ ) is expected number of particles per unit area. The kernel K n is reproducing kernel for the subspace of L 2 of weighted polynomials p ( ζ ) e − nQ ( ζ ) / 2 where degree( p ) ≤ n . 6 / 41

Rescaling � Put r n = 1 / n ∆ Q ( p n ). Rescaled system about p n = 0: z j = r − 1 { z j } n j =1 , n ζ j , j = 1 , . . . , n . δ n δ n p n p n We fix T > 0, take δ n = Tr n , p n closest point to the cusp with boundary distance δ n . 7 / 41

Rescaled density function R n ( z ) Rescaled droplet as n → ∞ is strip − T ≤ Re z ≤ T . Im(z) D ( z , ǫ ) -T T Re(z) 8 / 41

Rescaled density function R n ( z ) Rescaled droplet as n → ∞ is strip − T ≤ Re z ≤ T . Im(z) D ( z , ǫ ) -T T Re(z) Random sample { z j } n 1 . R n ( z ) = expected number of points in D ( z , ǫ ) , ǫ 2 ( ǫ → 0). 8 / 41

Structure lemma (normal families) Lemma Subsequential limits R ( z ) := lim k →∞ R n k ( z ) exist. Each such limit determines a unique determinantal point field { z j } ∞ 1 . Each limit point field is determined by a Hermitian-entire function L ( z , w ) via R ( z ) = L ( z , z ) e −| z | 2 . L is the Bergman kernel of a contractively embedded subspace of a ( e −| z | 2 ). Fock space L 2 w is Bargmann-Fock kernel. Infinite Ginibre ensemble: L ( z , w ) = e z ¯ 9 / 41

Ward’s identity Let ψ any test-function and { ζ j } n 1 random sample. Random variable W n [ ψ ] = 1 ψ ( ζ j ) − ψ ( ζ k ) � 2 ζ j − ζ k j � = k � � + n ∂ Q ( ζ j ) ψ ( ζ j ) + ∂ψ ( ζ j ) . Lemma E n W n [ ψ ] = 0 . Proofs: Reparametrization invariance of Z n /integration by parts. Gives an exact relation between 1- and 2-point functions R n , 1 and R n , 2 . 10 / 41

Ward’s equation, zero-one law, ... Theorem (AKMW) 1 Zero-one law: either R = 0 identically or R > 0 everywhere. 2 If the width T is large enough then R > 0 everywhere. 3 If R > 0 then ¯ ∂ C = R − 1 − ∆ log R where B ( z , w ) = | K ( z , w ) | 2 / K ( z , z ) and � B ( z , w ) C ( z ) = z − w dA ( w ) . 4 If z = x + iy then R ( z ) ≤ Ce − 2( | x |− T ) 2 . 11 / 41

Limiting correlation kernel With K ( z , w ) = L ( z , w ) e −| z | 2 / 2 −| w | 2 / 2 the k -point intensities are given by R k ( z 1 , . . . , z k ) = det( K ( z i , z j )) k i , j =1 . Often easier to write L ( z , w ) = e z ¯ w Ψ( z , w ) and look for Ψ. Then R ( z ) = Ψ( z , z ). 12 / 41

Limiting correlation kernel With K ( z , w ) = L ( z , w ) e −| z | 2 / 2 −| w | 2 / 2 the k -point intensities are given by R k ( z 1 , . . . , z k ) = det( K ( z i , z j )) k i , j =1 . Often easier to write L ( z , w ) = e z ¯ w Ψ( z , w ) and look for Ψ. Then R ( z ) = Ψ( z , z ). Infinite Ginibre: Ψ = 1. Regular boundary point: Ψ( z , w ) = 1 2 erfc( z + ¯ w √ 2 ). 12 / 41

Translation invariant solutions Theorem Suppose 1 T is large enough that R > 0 , 2 Translation invariance: R ( z ) depends only on x = Re ( z ) . Then there is an interval [ A , B ] ⊂ [ − 2 T , 2 T ] such that � B 1 z − t ) 2 / 2 dt , e − ( z +¯ R ( z ) = Ψ( z + ¯ z ) := √ 2 π A Can be written 1 e − t 2 / 2 . Ψ( z ) = γ ∗ 1 [ A , B ] ( z ) , γ ( t ) = √ 2 π 13 / 41

Conjecture of AKMW We believe that R is nontrivial and translation invariant for every T > 0 and that [ A , B ] = [ − 2 T , 2 T ]. 1.0 1.0 1.0 0.8 0.8 0.8 0.6 0.6 0.6 0.4 0.4 0.4 0.2 0.2 0.2 - 4 - 2 0 2 4 - 4 - 2 0 2 4 - 4 - 2 0 2 4 Figure: The graph of R ( x ) := γ ∗ 1 [ − 2 T , 2 T ] (2 x ) for T = 1 / 2, T = 3 / 2, and T = 5 / 2. 14 / 41

The hard edge strip: natural candidates The limit R = lim R n for the strip satisfies Ward’s equation ¯ ∂ C = R − 1 − ∆ log R . Natural solutions (AKMW): 1.1 1.1 1.1 1.0 1.0 1.0 0.9 0.9 0.9 0.8 0.8 0.8 - 3 - 2 - 1 0 1 2 3 - 3 - 2 - 1 0 1 2 3 - 3 - 2 - 1 0 1 2 3 15 / 41

Fyodorov-Khoruzhenko-Sommers setting Potential x 2 y 2 1 1 − τ 2 ( | ζ | 2 − τ Im ( ζ 2 )) = Q ( ζ ) = 1 − τ + 1 + τ . FKS introduced weakly skew-Hermitian regime τ = τ n = 1 − ( πα ) 2 . 2 n The droplet is a narrow ellipse about the y -axis: ( α 2 / n ) 2 + y 2 x 2 C 2 2 ≤ 1 + o (1) . 16 / 41

Rescaling The droplet has width ∼ α 2 / n and area ∼ α 2 / n . 2 -2 1 random sample. Particle density is ∼ n 2 so it is natural Let { ζ j } n to rescale by 1 z = cn ζ, i.e. R n ( z ) = c 2 n 2 R n ( ζ ) . FKS choice is c = 1 /π . 17 / 41

Fyodorov-Khoruzhenko-Sommers theorem Theorem (FKS) R n → R where � π � 2 y + α 2 t 1 � e − 2 √ α 2 R ( z ) = 2 dt . 2 πα − π We make the transformation R n ,α ( z ) := α 2 R n ( i α z ) . Corollary As n → ∞ , R n ,α converges to � 2 T 1 T = απ z − t ) 2 / 2 dt , e − ( z +¯ R α ( z ) := √ 2 . 2 π − 2 T Note: Corollary is equivalent to theorem and is well-suited for our approach. 18 / 41

FKS densities √ √ Figure: Density profiles R α ( y ), α = 1 / 2 , 1 , 2. FKS noted that α → 0 K α ( x , y ) = K sin ( x , y ) = sin( π ( x − y )) ( x , y ) ∈ R 2 lim , π ( x − y ) and w −| z | 2 / 2 −| w | 2 / 2 . α →∞ K α ( z , w ) = K Ginibre ( z , w ) = e z ¯ lim 19 / 41

Model case: thin annular ensembles Put Q n ( ζ ) = 1 ( | ζ | 2 − 2 c n log | ζ | ) . a n For suitable choices of a n , c n the droplet is a thin annulus S n . r 2 r 1 2 T / n 2 T / √ n S n r 2 r 1 S n Fix T > 0 and choose: 1 � S n ∆ Q n dA = 1 , i.e. a n = Area( S n ) = 1 / ∆ Q n , 2 r 2 − r 1 ∼ 2 T / √ n ∆ Q n . 20 / 41