Asymptotic Analysis of Random Matrices and Orthogonal Polynomials - PowerPoint PPT Presentation

Asymptotic Analysis of Random Matrices and Orthogonal Polynomials Arno Kuijlaars University of Leuven, Belgium Les Houches, 5-9 March 2012

Point processes A configuration X is a subset of R with #( X ∩ [ a , b ]) < + ∞ for every bounded interval [ a , b ] ⊂ R . A (locally finite) point process on R is a probability measure on the space of all configurations. A point process P is an n -point process if P (# X = n ) = 1 . If P ( x 1 , . . . , x n ) is a probability density function on R n which is invariant under permutation of coordinates, P ( x σ (1) , . . . , x σ ( n ) ) = P ( x 1 , . . . , x n ) then P defines an n -point process.

Correlation functions The 1 -point correlation function ρ 1 ( x ) of X satisfies � ρ 1 ( x ) dx = E [#( X ∩ A )] A ρ 1 ( x ) is the particle density. The 2 -point correlation function ρ 2 ( x , y ) is such that for disjoint sets A and B � � � � #( x , y ) ∈ X 2 | x ∈ A , y ∈ B ρ 2 ( x , y ) dxdy = E , A B for any set A � � � � #( x , y ) ∈ X 2 | x ∈ A , y ∈ A , x < y ρ 2 ( x , y ) dxdy = E . A A

Higher order correlation functions The k -point correlation function ρ k (if it exists) has for disjoint sets A j � � · · · ρ k ( x 1 , . . . , x k ) dx 1 · · · dx k A 1 A k � � #( x 1 , . . . , x k ) ∈ X k | x j ∈ A j = E , for a single set A � � · · · ρ k ( x 1 , . . . , x k ) dx 1 · · · dx k A A � � #( x 1 , . . . , x k ) ∈ ( X ∩ A ) k | x 1 < · · · < x k = E .

Marginal densities For an invariant pdf P ( x 1 , . . . , x n ) on R n the n -point process has correlation functions � � n ! ρ k ( x 1 , . . . , x k ) = · · · P ( x 1 , . . . , x n ) dx k +1 · · · dx n ( n − k )! � �� � n − k times

Determinantal point process A point process with correlation functions ρ k is determinantal (fermionic) if there exists a kernel K ( x , y ) such that ρ k ( x 1 , . . . , x k ) = det [ K ( x i , x j )] k i , j =1 for every k and every x 1 , . . . , x k . K is called the correlation kernel.

Biorthogonal ensembles An n -point process is a biorthogonal ensemble if there exist two sequences of functions f 1 , . . . , f n and g 1 , . . . , g n P ( x 1 , x 2 , . . . , x n ) = 1 det[ f i ( x j )] n i , j =1 · det[ g i ( x j )] n i , j =1 . Z n This is a determinantal point process with correlation kernel n n � � � M − 1 � K n ( x , y ) = f i ( x ) g j ( y ) j , i i =1 j =1 where M is the matrix � M = ( M i , j ) , M i , j = f i ( x ) g j ( x ) dx

Biorthogonal functions We may find φ j ∈ span { f 1 , . . . , f j } , ψ k ∈ span { g 1 , . . . , g k } , such that � ∞ φ j ( x ) ψ k ( x ) dx = δ jk . −∞ Then � n K n ( x , y ) = φ j ( x ) ψ j ( y ) j =1 and P ( x 1 , . . . , x n ) = 1 n ! det[ K n ( x i , x j )] n i , j =1 . � w ( x ) x j − 1 An OP ensemble has f j ( x ) = g j ( x ) = Other examples come from non-intersecting paths.

The Karlin-McGregor theorem (1959) Let p t ( a ; x ) be the transition probability density of a one-dimensional strong Markov process with continuous sample paths. Consider n independent copies X 1 ( t ) , . . . , X n ( t ) conditioned so that X j (0) = a j where a 1 < a 2 < · · · < a n are given values. Let E 1 , . . . , E n be Borel sets so that sup E j < inf E j +1 for j = 1 , . . . , n − 1 . Then � � det [ p t ( a i , x j )] n · · · i , j =1 dx 1 · · · dx n E 1 E n is equal to the probability that X j ( t ) ∈ E j for j = 1 , . . . , n in such a way that the paths have not intersected in the time interval [0 , t ]

Proof of the Karlin-McGregor theorem, step 1 Write � p t ( a i , E j ) = p t ( a i , x j ) dx j E j so that we have the determinant det [ p t ( a i , E j )] n i , j =1 . Expand the determinant � � n det [ p t ( a i , E j )] n i , j =1 = sgn( σ ) p t ( a j , E σ ( j ) ) σ j =1 � = sgn( σ ) P ( A σ ) , σ where for a permutation σ , we use A σ to denote the event that X j ( t ) ∈ E σ ( j ) for every j = 1 , . . . , n .

Proof of the Karlin-McGregor theorem, step 2 We decompose A σ = B σ ∪ C σ where B σ is the event that X j ( t ) ∈ E σ ( j ) for j = 1 , . . . , n and the paths have not intersected in the time interval [0 , t ] , and C σ = A σ \ B σ . If σ � = id then P ( B σ ) = 0 (because of continuous sample paths). Hence � det [ p t ( a i , E j )] n i , j =1 = P ( B id ) + sgn( σ ) P ( C σ ) . σ � It remains to show that sgn( σ ) P ( C σ ) = 0 . σ

Proof of the Karlin-McGregor theorem, step 3 For a transposition τ = ( i , i ′ ) , we use C σ,τ to denote the event (1) X j ( t ) ∈ E σ ( j ) for every j = 1 , . . . , n , and (2) there is s ∈ (0 , t ] so that the paths do not intersect in the time interval (0 , s ) , 1 X i ( s ) = X i ′ ( s ) , and 2 if X j ( s ) = X j ′ ( s ) , for some 1 ≤ j < j ′ ≤ n , then i ≤ j , 3 and if i = j , then i ′ ≤ j ′ . We have a disjoint union C σ = � τ C σ,τ so that � P ( C σ ) = P ( C σ,τ ) . τ Crucial observation P ( C σ,τ ) = P ( C σ ◦ τ,τ ) . This follows from the strong Markov property.

Proof of the Karlin-McGregor theorem, step 4 Now we have � � � sgn( σ ) P ( C σ ) = sgn( σ ) P ( C σ,τ ) σ σ τ � � = sgn( σ ) P ( C σ ◦ τ,τ ) τ σ Make a “change of variables” σ �→ σ ◦ τ − 1 � � � sgn( σ ◦ τ − 1 ) P ( C σ,τ ) sgn( σ ) P ( C σ ) = σ τ σ � � = − sgn( σ ) P ( C σ,τ ) σ τ � = − sgn( σ ) P ( C σ ) σ Thus � σ sgn( σ ) P ( C σ ) = 0 , which completes the proof.

Consequences In the situation of the Karlin-McGregor theorem, if we condition on the event that the paths have not intersected in [0 , t ] , then the positions of the paths 1 det [ p t ( a i , x j )] n at time t have joint pdf i , j =1 Z n This is NOT a determinantal point process. (We need two determinants). Also condition at a later time T > t . Starting positions a 1 < a 2 < · · · < a n at time 0 End positions b 1 < b 2 < · · · < b n at time T Non intersecting paths in full time interval [0 , T ] Then the positions at time t ∈ (0 , T ) have joint pdf 1 det [ p t ( a i , x j )] n i , j =1 det [ p T − t ( x i , b j )] n i , j =1 Z n Biorthogonal ensemble with f j ( x ) = p t ( a j , x ) , g j ( x ) = p T − t ( x , b j ) .

Non-intersecting path ensembles Let p t ( a ; x ) be the transition probability density of a one-dimensional strong Markov process with continuous sample paths. Consider n independent copies X 1 ( t ) , . . . , X n ( t ) conditioned so that X j (0) = a j , X j ( T ) = b j where a 1 < · · · < a n , b 1 < · · · < b n are given values, The paths do not intersect in time interval (0 , T ) . Then the joint p.d.f. for the positions of the paths at time t ∈ (0 , T ) is equal to 1 det [ p t ( a i , x j )] n i , j =1 · det [ p T − t ( x j , b i )] n i , j =1 Z n This is a determinantal point process.

Confluent case Take Brownian motion in the limit a j → a , b j → b . This leads to same p.d.f. (after centering and scaling) as for the eigenvalues of GUE. 1.5 t=0.4 1 0.5 0 −0.5 −1 −1.5 0 0.2 0.4 0.6 0.8 1

Two different endpoints 1.5 1.5 1 1 0.5 0.5 0 0 −0.5 −0.5 −1 −1 −1.5 −1.5 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 This is not an OP ensemble! Still Sine kernel in the bulk and Airy kernel at the edges Pearcey kernels at the cusp point (double scaling limit) Bleher-Kuijlaars (2007)

Non-intersecting squared Bessel paths 2.5 2 1.5 x 1 0.5 0 0 0.2 0.4 0.6 0.8 1 t Squared Bessel paths are always positive. Sine kernel in the bulk, Airy kernel at soft edges, and Bessel kernel at the hard edge New kernel at critical time Kuijlaars-Mart´ ınez Finkelshtein-Wielonsky (2011)

Matrix Riemann-Hilbert problem for OPs Given weight w = e − V on R and n ∈ N , find 2 × 2 matrix valued function Y ( z ) such that RH-Y1 Y : C \ R → C 2 × 2 is analytic. RH-Y2 Y has boundary values for x ∈ R , denoted by Y ± ( x ) , and � � e − V ( x ) 1 Y + ( x ) = Y − ( x ) , x ∈ R . 0 1 RH-Y3 As z → ∞ , � � 1 �� � z n � 0 Y ( z ) = I + O z − n 0 z

Fokas, Its, Kitaev RH problem for OP Theorem (Fokas, Its, Kitaev (1992)) The Riemann-Hilbert problem has the unique solution   � p n ( s ) w ( s ) 1 γ − 1 2 π i γ − 1 n p n ( z ) ds n s − z   R Y ( z ) = �   p n − 1 ( s ) w ( s ) − 2 π i γ n − 1 p n − 1 ( z ) − γ n − 1 ds s − z R p n is the orthonormal polynomial w.r.t. e − V ( x ) dx γ n is the leading coefficient of p n .

OP kernel in terms of the RH problem OP kernel is √ e − V ( x ) √ � � e − V ( y ) Y − 1 K n ( x , y ) = + ( y ) Y + ( x ) 2 , 1 2 π i ( x − y ) √ e − V ( x ) √ � 1 � � � e − V ( y ) Y − 1 = 0 1 + ( y ) Y + ( x ) . 0 2 π i ( x − y )

Airy function y ′′ ( z ) = zy ( z ) The Airy equation has the special solution � 1 e − 1 3 t 3 + zt dt Ai( z ) = 2 π i C where C is a contour in the complex t -plane that starts at infinity at angle arg t = − 2 π/ 3 and ends at angle arg t = 2 π/ 3 . This solution is characterized by its asymptotics as z → ∞ in the sector − π < arg z < π , 3 z 3 / 2 � � 1 2 √ π z 1 / 4 e − 2 1 + O ( z − 3 / 2 ) Ai( z ) = , Ai( z ) = − z 1 / 4 3 z 3 / 2 � � ′ 2 √ π e − 2 1 + O ( z − 3 / 2 ) .