Asymptotic Analysis of Random Matrices and Orthogonal Polynomials Arno Kuijlaars University of Leuven, Belgium Les Houches, 5-9 March 2012
Multiple orthogonal polynomials Given weight functions w 1 , . . . , w r on the real line n = ( n 1 , . . . , n r ) ∈ N r . Notation | � and � n | = n 1 + · · · + n r The type II multiple orthogonal polynomial (MOP) is a monic polynomial P � n of degree | � n | such that � n ( x ) x k w 1 ( x ) dx = 0 , k = 0 , 1 , . . . , n 1 − 1 , P � � n ( x ) x k w 2 ( x ) dx = 0 , P � k = 0 , 1 , . . . , n 2 − 1 , . . . . . . , � n ( x ) x k w r ( x ) dx = 0 , P � k = 0 , 1 , . . . , n r − 1 , These are | � n | conditions for the | � n | free coefficients of P � n . In typical cases there is existence and uniqueness, but not always.
Type I multiple orthogonality Type I multiple orthogonal polynomials are r polynomials A (1) n , A (2) n , · · · A ( r ) n , of degrees � � � deg A ( j ) n ≤ n j − 1 , j = 1 , . . . , r � They are such that the linear form n ( x ) = A (1) n ( x ) w 1 ( x ) + · · · + A ( r ) n ( x ) w r ( x ) Q � � � satisfies � x k Q � n ( x ) dx = 0 , k = 0 , 1 , . . . , | � n | − 2 , � x k Q � n ( x ) dx = 1 , k = | � n | − 1 .
Block Hankel matrix � Moments µ ( i ) x j w i ( x ) dx = j n × m Hankel matrix for i th weight � � µ ( i ) H ( i ) n , m = j + k − 2 j =1 ,..., n , k =1 ,..., m Block Hankel matrix � � H (1) H ( r ) n = , n = | � n | H � · · · n , n 1 n , n r Conditions for type I MOPs give linear system with matrix H � n . Conditions for type II MOP give linear system with matrix H T n . � Both type of MOPs exist if and only if det H � n � = 0 .
Riemann-Hilbert problem (case r = 2) In the RH problem we look for a 3 × 3 matrix valued function Y ( z ) satisfying RH-Y1 Y : C \ R → C 3 × 3 is analytic. RH-Y2 Y has boundary values for x ∈ R , denoted by Y ± ( x ) , and 1 w 1 ( x ) w 2 ( x ) , Y + ( x ) = Y − ( x ) 0 1 0 for x ∈ R . 0 0 1 RH-Y3 As z → ∞ , �� z n 1 + n 2 0 0 � � 1 z − n 1 Y ( z ) = I + O 0 0 z z − n 2 0 0
Solution in terms of type II MOPs Theorem ( Van Assche, Geronimo, K (2001)) RH problem has a unique solution if and only if the type II MOP P � n uniquely exists. In that case the first row of Y is given by � ∞ � ∞ 1 P � n ( s ) w 1 ( s ) 1 P � n ( s ) w 2 ( s ) P � n ( z ) ds ds 2 π i s − z 2 π i s − z −∞ −∞ ∗ ∗ ∗ ∗ ∗ ∗ Other rows are filled using P � e 1 and P � e 2 (if they n − � n − � exist).
Inverse of Y The type I MOPs are in the inverse of Y . � ∞ Q � n ( s ) − s − z ds ∗ ∗ −∞ Y − 1 ( z ) = 2 π iA (1) n ( z ) ∗ ∗ � 2 π iA (2) n ( z ) ∗ ∗ � n = A (1) n w 1 + A (2) where Q � n w 2 � � Other columns contain type I MOPs with multi-indices � n + � e 1 and � n + � e 2 .
Biorthogonal ensembles Probability density function on R n of the form 1 det [ f i ( x j )] n i , j =1 · det [ g i ( x j )] n i , j =1 , Z n Normalization constant � R n det [ f i ( x j )] n i , j =1 · det [ g i ( x j )] n Z n = i , j =1 dx 1 · · · dx n � = 0 By Andr´ eief (1883) identity �� ∞ � n Z n = n ! det M n , M n = f i ( x ) g j ( x ) dx −∞ i , j =1 Corollary: det M n � = 0
Correlation kernel Biorthogonal ensemble is a determinantal point process with correlation kernel n n � � � � M − 1 K n ( x , y ) = ji f i ( x ) g j ( y ) . n i =1 j =1 Representation as determinant f 1 ( x ) . 1 M n . . K n ( x , y ) = − det det M n f n ( x ) 0 g 1 ( y ) ··· g n ( y ) Perform elementary row and column transformations to transform M n to the identity matrix I n
Correlation kernel (cont.) After transformation M n �→ I n φ 1 ( x ) . I n . . K n ( x , y ) = − det φ n ( x ) 0 ψ 1 ( y ) ··· ψ n ( y ) with functions φ j and ψ j satisfying � ∞ φ i ( x ) ψ j ( x ) dx = δ i , j (biorthogonality) −∞ n � Also single sum K n ( x , y ) = φ j ( x ) ψ j ( y ) j =1
Correlation kernel (cont.) Characterization: K n is the kernel of the projection operator onto the linear span of f 1 , . . . , f n , whose kernel is the orthogonal complement of the linear span of g 1 , . . . , g n . Operator � K n : h �→ K n h , K n h ( x ) = K n ( x , y ) h ( y ) dy Characterization K n h = h if h = α 1 f 1 + α 2 f 2 + · · · + α n f n , � K n h = 0 if h ( x ) g j ( x ) dx = 0 for j = 1 , . . . , n .
MOP ensembles Definition A multiple orthogonal polynomial (MOP) ensemble is a biorthogonal ensemble with functions f i ( x ) = x i − 1 , for i = 1 , . . . , n , g i ( y ) = y i − 1 w 1 ( y ) , for i = 1 , . . . , n 1 , g n 1 + i ( y ) = y i − 1 w 2 ( y ) , for i = 1 , . . . , n 2 , . . . g n 1 + ··· + n r − 1 + i ( y ) = y i − 1 w r ( y ) , for i = 1 , . . . , n r . Here w 1 , . . . , w r are given functions, and n 1 , . . . , n r are non-negative integers such that n = n 1 + · · · + n r .
Block Hankel matrix In a MOP ensemble the matrix M n is the block Hankel matrix � � H (1) H ( r ) M n = H � n = , n = | � n | · · · n , n 1 n , n r det H � n � = 0 and so the MOPs exist. The RH problem has a unique solution.
Christoffel Darboux formula Theorem (Bleher-K (2004) for r = 2, Daems-K (2004)) The correlation kernel K n for the MOP ensemble is given by 1 K n ( x , y ) = 2 π i ( x − y ) × 1 0 � � Y − 1 0 w 1 ( y ) · · · w r ( y ) + ( y ) Y + ( x ) . . . 0
Proof for case r = 2 Assume r = 2 . Let L n ( x , y ) be the right-hand side 1 1 � � Y − 1 L n ( x , y ) = 0 w 1 ( y ) w 2 ( y ) + ( y ) Y + ( x ) 0 2 π i ( x − y ) 0 We show (a) L n h = h if h is a polynomial of degree ≤ n − 1 , � h ( y ) y j − 1 w i ( y ) dy = 0 for j = 1 , . . . , n i , and (b) L n h = 0 if i = 1 , 2 .
Proof of (a) Let h be a polynomial of degree ≤ n − 1 . 1 h ( y ) � � Y − 1 L n ( x , y ) h ( y ) = 0 w 1 ( y ) w 2 ( y ) + ( y ) Y + ( x ) 0 2 π i ( x − y ) 0 1 = h ( y ) − h ( x ) � � Y − 1 0 w 1 ( y ) w 2 ( y ) 0 + ( y ) Y + ( x ) 2 π i ( x − y ) 0 1 h ( x ) � � Y − 1 + 0 w 1 ( y ) w 2 ( y ) + ( y ) Y + ( x ) 0 2 π i ( x − y ) 0 � L n ( x , y ) h ( y ) dy splits into two integrals.
Proof of (a), first integral First integral has 1 h ( y ) − h ( x ) � � Y − 1 0 w 1 ( y ) w 2 ( y ) + ( y ) Y + ( x ) 0 2 π i ( x − y ) � �� � 0 � �� � vector with linear forms polynomial in y of type I MOPs of degree ≤ n − 2 Integral with respect to y is 0 for every x because of type I multiple orthogonality.
Proof of (a), second integral Second integral is 1 � ∞ h ( x ) dy � � Y − 1 0 w 1 ( y ) w 2 ( y ) + ( y ) Y + ( x ) 0 2 π i x − y −∞ 0 From jump condition in RH problem � � � � � � Y − 1 Y − 1 − ( y ) − Y − 1 0 w 1 ( y ) w 2 ( y ) 1 0 0 + ( y ) = + ( y ) It remains to prove � ∞ � Y − 1 � − ( y ) − Y − 1 1 + ( y ) Y + ( x ) dy = 1 . 2 π i x − y −∞ 1 , 1
Proof of (a), second integral (cont.) � ∞ � Y − 1 � − ( y ) − Y − 1 1 + ( y ) Y + ( x ) dy = 1 . 2 π i x − y −∞ 1 , 1 Replace x ∈ R by z with Im z > 0 . � � Y − 1 ( y ) y �→ z − y Y ( z ) 1 , 1 is analytic in lower half plane and is O ( y − n − 1 ) as y → ∞ . By Cauchy’s theorem � ∞ � Y − 1 � 1 − ( y ) z − y Y ( z ) dy = 0 2 π i −∞ 1 , 1 � � Y − 1 ( y ) y �→ z − y Y ( z ) 1 , 1 has pole in upper half plane and same behavior at infinity. By residue calculation � ∞ � Y − 1 � 1 + ( y ) z − y Y ( z ) dy = − 1 2 π i −∞ 1 , 1 Subtract the two results and then let z → x ∈ R .
Proof of (b) � h ( y ) y j − 1 w i ( y ) dy = 0 for j = 1 , . . . , n j , Assume i = 1 , 2 . We have to prove L n h ( x ) = 0 We have that L n h ( x ) = 1 � ∞ � Y − 1 + ( y ) − Y − 1 1 + ( x ) � dy h ( y ) 0 w 1 ( y ) w 2 ( y ) Y + ( x ) 0 2 π i x − y −∞ 0 1 � ∞ � Y − 1 + 1 + ( x ) � dy . h ( y ) 0 w 1 ( y ) w 2 ( y ) x − y Y + ( x ) 0 2 π i −∞ 0 Second integral is obviously zero. 1 In first integral we can take out Y + ( x ) 0 . 0
Proof of (b), (cont.) We are left to evaluate � ∞ � Y − 1 + ( y ) − Y − 1 1 + ( x ) � h ( y ) 0 w 1 ( y ) w 2 ( y ) dy 2 π i x − y −∞ Second row of Y − 1 has polynomials of degree ≤ n 1 Third row of Y − 1 has polynomials of degree ≤ n 2 Hence for every x , the entries of � Y − 1 + ( y ) − Y − 1 + ( x ) � 0 w 1 ( y ) w 2 ( y ) x − y take the form w 1 ( y )( poly of deg ≤ n 1 − 1)+ w 2 ( y )( poly of deg ≤ n 2 − 1) This is in the linear span of g 1 , . . . , g n and the integral is zero.
Examples of MOP ensembles Non-intersecting Brownian motions Non-intersecting squared Bessel paths Random matrix model with external source Two matrix model
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