Tilings of a hexagon and non-hermitian orthogonality on a contour Arno Kuijlaars (KU Leuven) joint work with Christophe Charlier, Maurice Duits, and Jonatan Lenells (KTH Stockholm) Integrability and Randomness in Mathematical Physics and Geometry Luminy, France, 11 April 2019
Outline 1. Hexagon tilings 2. Non-intersecting paths 3. Tile probabilities 4. Saddle points 5. Equilibrium measure 6. Riemann-Hilbert problem 7. Deformation of contours
1. Hexagon tilings
Lozenge tiling of a hexagon three types of lozenges
Large random tiling Arctic circle phenomenon
Affine change of coordinates Hexagon with corner points at (0 , 0) , ( N , 0) , (2 N , N ) , (2 N , 2 N ) , ( N , 2 N ) , and (0 , N ) .
Non uniform model W ( T ) Probability of tiling: P ( T ) = � T ′ W ( T ′ ) � Weight on a tiling: W ( T ) = w ( � ) � ∈T Weight of � depends on its position: α, if � is in odd numbered column w ( � ) = 1 , if � is in even numbered column α = 1 is the usual uniform model α < 1 means punishment if � is in an odd column
Case α = 0 Only ground state in case α = 0
Small α > 0 Liquid region consists of two ellipses if α > 0 is small Special frozen region with two tiles in the middle
Larger α > 0 Liquid region is bounded by more complicated curve Special frozen region is broken. It no longer goes all the way from left to right.
2. Non-intersecting paths
Non-intersecting paths Tiling is equivalent to N non-intersecting paths starting at (0 , 0) , . . . , (0 , N − 1) and ending at (2 N , N ) , . . . , (2 N , 2 N − 1)
Paths fit on a directed graph 12 11 10 9 8 7 6 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 101112
Weighted graph N � � Weight of a path system w ( P 1 , . . . , P N ) = w ( e ) j =1 e ∈ P j Weight of an edge if e is a horizontal edge α, w ( e ) = in an odd numbered column 1 , otherwise Interacting particle system is determinantal [Lindstrom Gessel Viennot]
LGV formula Probability for particle configuration ( x ( m ) ) j , m is j 2 N 1 � � �� x ( m − 1) , x ( m ) � det T m i j Z N i , j =0 ,..., N − 1 m =1 with transition matrices α if y = x and m is odd 1 if y = x and m is even T m ( x , y ) = 1 if y = x + 1 0 otherwise Sum formula for correlation kernel [Eynard Mehta]
Transition matrices are Toeplitz Observation: T m is an infinite Toeplitz matrix with symbol z + α, if m is odd a m ( z ) = z + 1 , if m is even
Theorem [Duits-K]; (scalar version) Correlation kernel is x 1 K ( x 1 , y 1 ; x 2 , y 2 ) = − χ x 1 > x 2 � � a m ( z ) · z y 2 − y 1 − 1 dz 2 π i γ m = x 2 +1 N x 1 a m ( z ) · w y 2 1 � dz � dw � � + a m ( w ) · R N ( w , z ) · (2 π i ) 2 w 2 N z z y 1 γ γ m = x 2 +1 m =1 R N is the reproducing kernel for orthogonal polynomials on γ with weight 2 N 1 a m ( z ) = ( z + 1) N ( z + α ) N � W ( z ) = z 2 N z 2 N m =1
Orthogonal polynomials p n ( z ) z k ( z + 1) N ( z + α ) N 1 � dz = κ n δ k , n , k = 0 , . . . , n − 1 2 π i z 2 N γ with reproducing kernel N − 1 p n ( w ) p n ( z ) � R N ( w , z ) = κ n n =0 p N ( z ) p N − 1 ( w ) − p N − 1 ( z ) p N ( w ) = κ − 1 N z − w Non-hermitian orthogonality! Existence of OP is not automatic but can be proved for degrees n ≤ 2 N OP is Jacobi polynomials P ( − 2 N , 2 N ) in case α = 1 n
3. Tile probabilities
Probabilities for lozenges (one-point functions) � � = 1 − K ( x , y ; x , y ) P ( x , y ) R N ( w , z )( w + 1) N ( w + α ) N 1 � � = 1 − (2 π i ) 2 w 2 N γ γ × ( z + 1) ⌊ x 2 ⌋ ( z + α ) ⌊ x +1 2 ⌋ w y dwdz . ( w + 1) ⌊ x 2 ⌋ ( w + α ) ⌊ x +1 z y z 2 ⌋ with similar double contour integral formulas for � � and P P ( x , y ) ( x , y )
Large N limit Suppose x and y vary with N such that x y lim N = 1 + ξ, lim N = 1 + η N →∞ N →∞ ( ξ, η ) are coordinates for the hexagon H Double contour integral (0 , 1) (1 , 1) has relevant saddle point s ( ξ, η ) ( − 1 , 0) (1 , 0) H Liquid region is character- ized by ( − 1 , − 1) (0 , − 1) Im s ( ξ, η ) > 0
Main result on limiting tile probabilities s ( ξ, η ) s ( ξ, η ) ψ 2 φ 2 ψ 1 φ 1 ψ 3 φ 3 0 0 − 1 − α � � = φ 3 π = ψ 3 π = 1 − 1 lim π arg s ( ξ, η ) , N →∞ P ( x , y ) φ 1 π , x odd, = N →∞ P lim ψ 1 ( x , y ) π , x even, φ 2 � � π , x odd, lim = N →∞ P ψ 2 ( x , y ) π , x even,
4. Saddle points
Saddle point equation Asymptotic analysis of the orthogonal polynomials. If p N ( z ) ≈ e g ( z ) N then (very roughly) R N ( w , z ) ≈ e ( g ( w )+ g ( z )) N and the integrand of the double integral is ≈ e g ( z ) N ( z + 1) 1+ ξ 1+ ξ 2 N ( z + α ) 2 N z − (1+ η ) N × e g ( w ) N ( w + 1) 2 N ( w + α ) 1 − ξ 1 − ξ 2 N w − (1 − η ) N Saddle point equations 1 + ξ 2( z + α ) − 1 + η 1 + ξ g ′ ( z ) + 2( z + 1) + = 0 z 1 − ξ 2( w + α ) − 1 − η 1 − ξ g ′ ( w ) + 2( w + 1) + = 0 2
g -function g function typically takes the form � g ( z ) = log( z − s ) d µ 0 ( s ) where µ 0 is the weak limit of the normalized zero counting measures of the orthogonal polynomials 1 ∗ � δ z → µ 0 N p N ( z )=0 Where are the zeros of the orthogonal polynomials?
Zeros of orthogonal polynomials: α = 1 Zeros of P ( − 2 N , 2 N ) cluster as N → ∞ to an arc on the N unit circle. [Mart´ ınez-Finkelshtein Orive] [Driver Duren]
Zeros of orthogonal polynomials: 1 / 9 < α < 1 Zeros of P N cluster as N → ∞ to an arc on the circle of radius √ α .
Zeros of orthogonal polynomials: α = 1 / 9 The circular arc closes at α = 1 / 9 The density of zeros vanishes quadratically at − 1 / 3 Local behavior in terms of Lax pair solutions for Hastings-McLeod solution of Painlev´ e II
5. Equilibrium measure
Equilibrium conditions Take V ( z ) = 2 log z − log( z + 1) − log( z + α ) µ 0 should be probability measure on contour γ 0 going � around 0 such that g ( z ) = log( z − s ) d µ 0 ( s ) satisfies � = 0 , for z ∈ supp( µ 0 ) , Re [ g + ( z ) + g − ( z ) − V ( z ) + ℓ ] ≤ 0 , for z ∈ γ 0 \ supp( µ 0 ) , Im [ g + ( z ) + g − ( z ) − V ( z )] is constant on each connected component of supp( µ 0 ) , µ 0 is equilibrium measure of γ 0 in external field Re V γ 0 is a contour with the S -property [Stahl]
Rational function Q α − = V ′ on the support Since g ′ + + g ′ �� d µ 0 ( s ) � 2 z − s − V ′ ( z ) = Q α ( z ) is a rational function 2
Rational function Q α − = V ′ on the support Since g ′ + + g ′ �� d µ 0 ( s ) � 2 z − s − V ′ ( z ) = Q α ( z ) is a rational function 2 If α ≥ 1 / 9 then Q α ( z ) = ( z + √ α ) 2 ( z − z + ( α ))( z − z − ( α )) z 2 ( z + 1) 2 ( z + α ) 2 with z ± ( α ) = √ α e ± i θ α for some 2 π 3 ≤ θ α ≤ π If α < 1 / 9 then Q α ( z ) = ( z − z + ( α )) 2 ( z − z − ( α )) 2 z 2 ( z + 1) 2 ( z + α ) 2 with real − 1 < z − ( α ) < −√ α < z + ( α ) < − α
Liquid/frozen regions Saddle point equation � 2 � ξ 2( z + α ) + η ξ Q α ( z ) = − 2( z + 1) − z becomes a degree four polynomial equation in z . It has four solutions (four saddles). Lemma If ( ξ, η ) belongs to the hexagon, then at least two saddles in ( − 1 , − α ) . Hence at most one saddle in C + . Liquid region L α : there is a saddle z = s ( ξ, η ) in C + . Otherwise frozen region: all saddles are real.
Liquid region for α < 1 9 C 1 B 1 L + D 1 α A 2 A 1 D 2 L − α B 2 C 2
Liquid region for α > 1 9 C 1 B 1 L + D 1 α A 2 A 1 D 2 L − α B 2 C 2 Transition at α = 1 9 : tangent ellipses and tacnode...
6. Riemann-Hilbert problem
RH problem for orthogonal polynomials RH problem [Fokas Its Kitaev] ( z +1) N ( z + α ) N � � 1 Y + ( z ) = Y − ( z ) z 2 N on γ 0 0 1 � � z N � 0 I + O ( z − 1 ) � Y ( z ) = as z → ∞ z − N 0 Reproducing kernel in terms of solution of RH problem 1 � 1 � Y − 1 ( w ) Y ( z ) � � R N ( w , z ) = 0 1 0 z − w
Transformation First transformation in RH analysis � � � � e N ℓ e − N ( g ( z )+ ℓ 2 ) 0 0 2 T ( z ) = Y ( z ) e − N ℓ e N ( g ( z )+ ℓ 2 ) 0 0 2 Steepest descent analysis as in [Deift Kriecherbauer McLaughlin Venakides Zhou] Main outcome T ( z ) and T − 1 ( z ) remain bounded as N → ∞ , uniformly for z away from the branch points.
7. Deformation of contours
Possible contours for ( ξ, η ) ∈ L − α , ξ < 0 s − 1 − α 0 s Blue: Level lines Re Φ( z ) = Re Φ( s ) of 1 + ξ 2( z + α ) − 1 + η 1 + ξ Φ( z ) = g ( z ) + 2( z + 1) + z Figure is for α = 1 8 .
More contours γ 0 = γ w γ z γ z is in region where Re Φ < Re Φ( s )
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