Two-periodic Aztec diamond Arno Kuijlaars (KU Leuven) joint work - - PowerPoint PPT Presentation

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Two-periodic Aztec diamond Arno Kuijlaars (KU Leuven) joint work - - PowerPoint PPT Presentation

Two-periodic Aztec diamond Arno Kuijlaars (KU Leuven) joint work with Maurice Duits (KTH Stockholm) Optimal and Random Point Configurations ICERM, Providence, RI, U.S.A., 27 February 2018 Outline 1. Aztec diamond 2. The model and main result


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Two-periodic Aztec diamond

Arno Kuijlaars (KU Leuven) joint work with Maurice Duits (KTH Stockholm) Optimal and Random Point Configurations ICERM, Providence, RI, U.S.A., 27 February 2018

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Outline

  • 1. Aztec diamond
  • 2. The model and main result
  • 3. Non-intersecting paths
  • 4. Matrix Valued Orthogonal Polynomials (MVOP)
  • 5. Analysis of RH problem
  • 6. Saddle point analysis
  • 7. Periodic tilings of a hexagon
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  • 1. Aztec diamond
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Aztec diamond

West North South East

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Tiling of an Aztec diamond

West North South East Tiling with 2 × 1 and 1 × 2 rectangles (dominos) Four types of dominos

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Large random tiling

Deterministic pattern near corners Solid region

  • r

Frozen region Disorder in the middle Liquid region Boundary curve Arctic circle

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Recent development

Two-periodic weighting Chhita, Johansson (2016) Beffara, Chhita, Johansson (2018 to appear)

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Two-periodic weights

A new phase within the liquid region: gas region

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Phase diagram solid solid solid solid gas liquid

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  • 2. The model and main result
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Two periodic weights

b b a a a a a a a a b b b b a a b b b b Aztec diamond of size 2N Weight w(T)

  • f a tiling T

is the product

  • f

the weights

  • f

dominos Partition function ZN =

  • T

w(T) Probability for T Prob(T) = w(T) ZN

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Equivalent weights

β α α α α β β α β β α = a2 and β = b2 North and East dominos have weight 1 Without loss

  • f

generality αβ = 1 and α ≥ 1 Since North dominos have weight 1, we can transfer the weights to non-intersecting paths.

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Particles in West and South dominos

Particles along diagonal lines are interlacing Positions

  • f

particles are random in the two-periodic Aztec diamond. Structure

  • f

determinantal point process We found explicit formula for kernel KN using matrix valued orthogonal polynomials (MVOP).

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Coordinates

m runs from 0 to 2N n runs from 0 to 2N − 1

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Formula for correlation kernel

THEOREM 1 Assume N is even and m + n and m′ + n′ are even.

  • KN(m, n; m′, n′)

KN(m, n + 1; m′, n′) KN(m, n; m′, n′ + 1) KN(m, n + 1; m′, n′ + 1)

  • = −χm>m′

2πi

  • γ0,1

Am−m′(z)z

m′−m+n′−n 2

dz z + 1 (2πi)2

  • γ0,1

dz z

  • γ1

dw z − w z

N−m−n 2

(z − 1)N w

N−m′−n′ 2

(w − 1)N AN−m′(w)F(w)A−N+m(z) where A(z) = 1 z − 1

  • 2αz

α(z + 1) βz(z + 1) 2βz

  • F(z) = 1

2I2 + 1 2

  • z(z + α2)(z + β2)

(α − β)z α(z + 1) βz(z + 1) −(α − β)z

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  • 3. Non-intersecting paths
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Non-intersecting paths

Line segments on West, East and South dominos North West East South

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Double Aztec diamond

2N particles along each diagonal line

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Non-intersecting paths on a graph

Paths are transformed to fit on a graph 0 0.5 1 1.5 2 2.5 3 3.5 4

  • 2
  • 1

1 2 3 4 5 6

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SLIDE 20

Weights on the graph

0 0.5 1 1.5 2 2.5 3 3.5 4

  • 2
  • 1

1 2 3 4 5 6 α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β β

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Weights on non-intersecting paths

Any tiling of double Aztec diamond is equivalent to system (P0, . . . , P2N−1) of 2N non-intersecting paths Pj is path on the graph from (0, j) to (2N, j), Pi is vertex disjoint from Pj if i = j.

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Transitions and LGV theorem

There are 2N + 1 levels, 0, 1, . . . , 2N. Transition from level m to level m′ > m Tm,m′(x, y) =

  • P:(m,x)→(m′,y)

w(P), x, y ∈ Z

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Transitions and LGV theorem

There are 2N + 1 levels, 0, 1, . . . , 2N. Transition from level m to level m′ > m Tm,m′(x, y) =

  • P:(m,x)→(m′,y)

w(P), x, y ∈ Z Lindstr¨

  • m-Gessel-Viennot theorem

Probability that paths at level m are at positions x(m) < x(m)

1

< · · · < x(m)

2N−1:

1 ZN det

  • T0,m(i, x(m)

k

) 2N−1

i,k=0 · det

  • Tm,2N(x(m)

k

, j) 2N−1

k,j=0

Lindstr¨

  • m (1973)

Gessel-Viennot (1985)

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Determinantal point process

Corollary: The positions at level m are determinantal with kernel KN,m(x, y) =

2N−1

  • i,j=0

T0,m(i, x)

  • G −t

i,j Tm,2N(y, j)

where G = [T0,2N(i, j)]2N−1

i,j=0

Multi-level extension is known as Eynard-Mehta theorem.

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Block Toeplitz matrices

In our case: Transition matrices are 2 periodic T(x + 2, y + 2) = T(x, y) Block Toeplitz matrices, infinite in both directions, with block symbol A(z) =

  • j=−∞

Bjzj if T =           ... ... ... ... B0 B1 ... ... B−1 B0 B1 ... ... B−1 B0 ... ... ... ...          

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Double contour integral formula

THEOREM 2: Suppose transition matrices are 2-periodic. Then

  • KN,m(2x, 2y)

KN,m(2x + 1, 2y) KN,m(2x, 2y + 1) KN,m(2x + 1, 2y + 1)

  • =

1 (2πi)2

  • γ
  • γ

Am,2N(w)RN(w, z)A0,m(z) w y zx+1w N dzdw Am,2N and A0,m are block symbols for the transition matrices Tm,2N and T0,m. RN(w, z) is a reproducing kernel for matrix valued polynomials.

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  • 4. Matrix Valued Orthogonal

Polynomials (MVOP)

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MVOP

Matrix valued polynomial of degree j, Pj(z) =

j

  • i=0

Cizi each Ci is d × d matrix, det Cj = 0 W (z) is d × d matrix valued weight Orthogonality 1 2πi

  • γ

Pj(z)W (z)Pt

k(z) dz = Hjδj,k

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Reproducing kernel

RN(w, z) =

N−1

  • j=0

Pt

j (w)H−1 j

Pj(z) is reproducing kernel for matrix polynomials of degree ≤ N − 1 If Q has degree ≤ N − 1, then 1 2πi

  • γ

Q(w)W (w)RN(w, z)dw = Q(z) There is a Christoffel-Darboux formula for RN and a Riemann Hilbert problem

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Riemann-Hilbert problem

Y : C \ γ → C2d×2d satisfies Y is analytic, Y+ = Y− Id W 0d Id

  • n γ,

Y (z) = (I2d + O(z−1)) zNId 0d 0d z−NId

  • as z → ∞.

Gr¨ unbaum, de la Iglesia, Mart´ ınez-Finkelshtein (2011)

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Solution of RH problem

Unique solution (provided PN uniquely exists) is Y (z) =     PN(z) 1 2πi

  • γ

PN(s)W (s) s − z ds QN−1(z) 1 2πi

  • γ

QN−1(s)W (s) s − z ds     where PN is monic MVOP of degree N and QN−1 = −H−1

N−1PN−1 has degree N − 1

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Solution of RH problem

Unique solution (provided PN uniquely exists) is Y (z) =     PN(z) 1 2πi

  • γ

PN(s)W (s) s − z ds QN−1(z) 1 2πi

  • γ

QN−1(s)W (s) s − z ds     where PN is monic MVOP of degree N and QN−1 = −H−1

N−1PN−1 has degree N − 1

Christoffel Darboux formula RN(w, z) = 1 z − w

  • 0d

Id

  • Y −1(w)Y (z)

Id 0d

  • Delvaux (2010)
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Our case of interest

Weight matrix in special case of two periodic Aztec diamond is W N(z), with W (z) = 1 (z − 1)2 (z + 1)2 + 4α2z 2α(α + β)(z + 1) 2β(α + β)z(z + 1) (z + 1)2 + 4β2z

  • No symmetry in W . Existence and uniqueness of

MVOP are not immediate.

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Our case of interest

Weight matrix in special case of two periodic Aztec diamond is W N(z), with W (z) = 1 (z − 1)2 (z + 1)2 + 4α2z 2α(α + β)(z + 1) 2β(α + β)z(z + 1) (z + 1)2 + 4β2z

  • No symmetry in W . Existence and uniqueness of

MVOP are not immediate. Scalar valued analogue Weight z+1

z−1

N on circle around z = 1 and OPs are Jacobi polynomials P(−N,N)

j

(z) with nonstandard parameters

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  • 5. Analysis of RH problem
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Surprise

Steepest descent analysis of RH problem leads to explicit formula RH problem is solved in terms of contour integrals. For example: MVOP is PN(z) = (z − 1)NW N/2

∞ W −N/2(z),

if N is even.

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Surprise

Steepest descent analysis of RH problem leads to explicit formula RH problem is solved in terms of contour integrals. For example: MVOP is PN(z) = (z − 1)NW N/2

∞ W −N/2(z),

if N is even. It leads to proof of THEOREM 1

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  • 6. Saddle point analysis
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Asymptotic analysis

Saddle point analysis on the double contour integral 1 (2πi)2

  • γ0,1

dz z

  • γ1

dw z − w z

N−2x 2 (z − 1)N

w

N−2y 2 (w − 1)N AN−m(w)F(w)A−N+m(z)

when N → ∞ m, x, y scale with N in such a way that m ≈ (1 + ξ1)N, x, y ≈ (1 + ξ1+ξ2

2

)N Saddle points are critical points of 2 log(z − 1) − (1 + ξ2) log z + ξ1 log λ(z) where λ(z) is an eigenvalue of W (z) = A2(z) z .

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Saddle point analysis

Let −1 < ξ1, ξ2 < 1. There are always four saddle points, depending on ξ1, ξ2, and they lie on the Riemann surface for y 2 = z(z + α2)(z + β2) (genus one) with branch points −α2 < −β2 < 0 and infinity. At least two saddles are in z ∈ [−α2, −β2].

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Classification of phases

Location of other two saddles determines the phase. Two saddles are in [0, ∞): solid phase Two saddles are in C \ ([−α2, −β2] ∪ [0, ∞)): liquid phase All four saddles are in [−α2, −β2]: gas phase Transitions between phases occur when saddles coalesce.

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Phase diagram solid solid solid solid gas liquid liquid liquid liquid

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  • 7. Periodic tilings of a hexagon
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Tiling of a hexagon

Lozenge tiling of a regular hexagon Also admits a non-intersecting path formulation

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Large random tiling

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Two periodic tiling of a hexagon

Ongoing work with Charlier, Duits, and Lenells

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Thanks Thank you for your attention