Statistical mechanics via Answers: GUE asymptotics of symmetric - - PowerPoint PPT Presentation

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

Statistical mechanics via asymptotics of symmetric functions

Greta Panova (University of Pennsylvania) based on: V.Gorin, G.Panova, Asymptotics of symmetric polynomials with applications to statistical mechanics and representation theory, Annals of Probability arXiv:1301.0634

• G. Panova, Lozenge tilings with free boundaries, arXiv:1408.0417.

April 2015

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Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

Overview

Normalized Schur functions: Sλ(x1, . . . , xk; N) = sλ(x1, . . . , xk, 1N−k) sλ(1N) Lozenge tilings: Dense loop model:

x y ζ1 ζ2 L

Alternating Sign Matrices (ASM)/ 6-Vertex model:     1 1 −1 1 1 −1 1 1     Characters of U(∞), boundary

• f the Gelfand-Tsetlin graph

1 1 1 2 2 . . . 2 2 3 . . . . . .

2

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

Lozenge tilings

Tilings of a domain Ω (on a triangular lattice) with elementary rhombi of 3 types (“lozenges”).

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Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

The many faces of lozenge tilings

5 4 4 4 3 2 5 3 3 2 2 1 4 3 2 2 1 3 2 2 1 2 1 1 1 1 1 5 4 4 4 3 2 5 3 3 2 2 1 4 3 2 2 1 3 2 2 1 2 1 1 1 1 1

4

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

The many faces of lozenge tilings

5 4 4 4 3 2 5 3 3 2 2 1 4 3 2 2 1 3 2 2 1 2 1 1 1 1 1 5 4 4 4 3 2 5 3 3 2 2 1 4 3 2 2 1 3 2 2 1 2 1 1 1 1 1

4

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

The many faces of lozenge tilings

5 4 4 4 3 2 5 3 3 2 2 1 4 3 2 2 1 3 2 2 1 2 1 1 1 1 1 5 4 4 4 3 2 5 3 3 2 2 1 4 3 2 2 1 3 2 2 1 2 1 1 1 1 1

5

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

The many faces of lozenge tilings

5 4 4 4 3 2 5 3 3 2 2 1 4 3 2 2 1 3 2 2 1 2 1 1 1 1 1 5 4 4 4 3 2 5 3 3 2 2 1 4 3 2 2 1 3 2 2 1 2 1 1 1 1 1

5

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

The many faces of lozenge tilings

5 4 4 4 3 2 5 3 3 2 2 1 4 3 2 2 1 3 2 2 1 2 1 1 1 1 1 5 4 4 4 3 2 5 3 3 2 2 1 4 3 2 2 1 3 2 2 1 2 1 1 1 1 1

5

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

The many faces of lozenge tilings

5 4 4 4 3 2 5 3 3 2 2 1 4 3 2 2 1 3 2 2 1 2 1 1 1 1 1 5 4 4 4 3 2 5 3 3 2 2 1 4 3 2 2 1 3 2 2 1 2 1 1 1 1 1

5

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

The many faces of lozenge tilings

5 4 4 4 3 2 5 3 3 2 2 1 4 3 2 2 1 3 2 2 1 2 1 1 1 1 1 5 4 4 4 3 2 5 3 3 2 2 1 4 3 2 2 1 3 2 2 1 2 1 1 1 1 1

5

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

The many faces of lozenge tilings

5 4 4 4 3 2 5 3 3 2 2 1 4 3 2 2 1 3 2 2 1 2 1 1 1 1 1 5 4 4 4 3 2 5 3 3 2 2 1 4 3 2 2 1 3 2 2 1 2 1 1 1 1 1

5

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

The many faces of lozenge tilings

5 4 4 4 3 2 5 3 3 2 2 1 4 3 2 2 1 3 2 2 1 2 1 1 1 1 1 5 4 4 4 3 2 5 3 3 2 2 1 4 3 2 2 1 3 2 2 1 2 1 1 1 1 1

5

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

The many faces of lozenge tilings

5 4 4 4 3 2 5 3 3 2 2 1 4 3 2 2 1 3 2 2 1 2 1 1 1 1 1 5 4 4 4 3 2 5 3 3 2 2 1 4 3 2 2 1 3 2 2 1 2 1 1 1 1 1

5

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

The many faces of lozenge tilings

5 4 4 4 3 2 5 3 3 2 2 1 4 3 2 2 1 3 2 2 1 2 1 1 1 1 1 5 4 4 4 3 2 5 3 3 2 2 1 4 3 2 2 1 3 2 2 1 2 1 1 1 1 1

5

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

The many faces of lozenge tilings

5 4 4 4 3 2 5 3 3 2 2 1 4 3 2 2 1 3 2 2 1 2 1 1 1 1 1 5 4 4 4 3 2 5 3 3 2 2 1 4 3 2 2 1 3 2 2 1 2 1 1 1 1 1

5

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

The many faces of lozenge tilings

5 4 4 4 3 2 5 3 3 2 2 1 4 3 2 2 1 3 2 2 1 2 1 1 1 1 1 5 4 4 4 3 2 5 3 3 2 2 1 4 3 2 2 1 3 2 2 1 2 1 1 1 1 1

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Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

Limit behavior

Question: Fix Ω in the plane and let grid size → 0, what are the properties of uniformly random tilings of Ω? [Cohn–Larsen–Propp, 1998] Hexagonal domain: Tiling is asymptotically frozen

• utside inscribed ellipse.

[Kenyon–Okounkov, 2005] Polygonal domain: Tiling is asymptotically frozen

• utside inscribed algebraic curve.

2.jpg

[Cohn–Kenyon–Propp, 2001; Kenyon-Okounkov-Sheffield, 2006] There exists a “limit shape” for the surface of the height function (plane partition).

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Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

Behavior near the boundary, interlacing particles

x x1

1

x2

2

x2

1

x3

3

x3

2

x3

1

N

Horizontal lozenges near a flat boundary: interlacing particle configuration ↔ Gelfand-Tsetlin patterns. x1

1

x2

2

x2

1

≤ ≤ x3

3

x3

2

x3

1

≤ ≤ ≤ ≤ Question: What is the joint distribution

• f {xi

j }k i=1 as N → ∞ (scale = 1 N )?

1with an explanation what the answer should be. 2Subsequent results: [Gorin-P,2012], [Novak, 2014],[Mkrtchyan, 2013, periodic weights, unbounded

region]

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Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

Behavior near the boundary, interlacing particles

x x1

1

x2

2

x2

1

x3

3

x3

2

x3

1

N

Horizontal lozenges near a flat boundary: interlacing particle configuration ↔ Gelfand-Tsetlin patterns. x1

1

x2

2

x2

1

≤ ≤ x3

3

x3

2

x3

1

≤ ≤ ≤ ≤ Question: What is the joint distribution

• f {xi

j }k i=1 as N → ∞ (scale = 1 N )?

Conjecture [Okounkov–Reshetikhin, 20061]: The joint distribution converges to a GUE-corners (aka GUE-minors) process: eigenvalues of GUE matrices. Proven for the hexagon by Johansson- Nordenstam (2006). 2

1with an explanation what the answer should be. 2Subsequent results: [Gorin-P,2012], [Novak, 2014],[Mkrtchyan, 2013, periodic weights, unbounded

region]

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Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

GUE in tilings: our setup

Domain Ωλ(N): width N and the positions λ(N)1 + N − 1 > λ(N)2 + N − 2 > · · · > λ(N)N

• f its N horizontal lozenges at the right boundary.

3+2

+1

5 1 2 +3 4

+4

λ(5) = (4, 3, 3, 0, 0) Note:

1 N Ωλ(N) is

not necessarily a finite polygon as N → ∞ , e.g. λ(N) = (N, N − 1, . . . , 2, 1)

0 + 0 0 + 1 0 + 2 0 + 3 0 + 4 0 + 5 8 + 6 8 + 7 8 + 8 8 + 9 8 + 10

E.g. λ = (a, . . . , a

c

, 0, . . . , 0

• b

) ↔ the hexagon with side lengths (a, b, c, a, b, c).

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Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

Plane partitions/Gelfand-Tsetlin patterns

λ1 + N − 1 λ2 + N − 2 λN x3

1

x3

2

x3

3

Line j = 3

5 4 4 4 3 3 3 3 2 1 2 1 N λ = (5, 4, 3, 1, 0) x3 = (4, 3, 0) 4 3 3 3 3 3 3 1 2 λ = (4, 3, 3, 0, 0) Question: What is the joint distribution of the positions

• xi

j

• (under correct

rescaling) near the boundary as N → ∞ (i.e. lattice scale = 1

N )?

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Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

Answers: the Gaussian Unitary Ensemble (GUE)

Gaussian Unitary Ensemble: matrices [Xij]i,j: X = X T ReXij, ImXij – i.i.d. ∼ N(0, 1/2) for i = j and Xii – i.i.d. ∼ N(0, 1)     a11 a12 a13 a14 a21 a22 a23 a24 a31 a32 a33 a34 a41 a42 a43 a44     (xk

1 ≥ xk 2 ≥ · · · ≥ xk k )

– eigenvalues of [Xi,j]k

i,j=1 (top

k × k). Interlacing condition: xj

i−1 ≤ xj−1 i−1 ≤ xj i

x4

1

x4

2

x4

3

x4

4

x3

1

x3

2

x3

3

≤ x2

1

x2

2

≤ x1

1

The joint distribution of {xj

i }1≤i≤j≤k is known as

GUE–corners (also, GUE-minors) process, =: GUEk .

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Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

GUE in tilings: our results

Limit profile f (t) of λ(N) as N → ∞: λ(N)i N → f i N

• f (t)

N λ(N) Ωλ(N) domain:

λ1 + N − 1 λ2 + N − 2 λN x3

1

x3

2

x3

3

Line j = 3

Theorem (Gorin-P (2012), Novak (2014))

Let λ(N) = (λ1(N) ≥ . . . ≥ λN(N)), N = 1, 2, . . . be a partition. If ∃ a piecewise-differentiable weakly decreasing function f (t) (limit profile of λ(N)) s.t.

N

• i=1
• λi(N)

N − f i N

• = o(

√ N) as N → ∞ and also supi,N |λi(N)/N| < ∞. Let Υk

λ(N) = {xj i } be the collection of the

positions of the horizontal lozenges on lines j = 1, . . . , k. Then for every k as N → ∞ Υk

λ(N) − NE(f )

• NS(f )

→ GUEk (GUE-corners proc. rank k) in the sense of weak convergence, where E(f ) = 1 f (t)dt, S(f ) = 1 f (t)2dt−E(f )2+ 1 f (t)(1−2t)dt.

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Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

Towards the proof: Schur functions

...or more generally – symmetric functions, Lie group characters. Irreducible (rational) representations Vλ of GL(N) (or U(N)) are indexed by dominant weights (signatures/Young diagrams/integer partitions) λ: λ1 ≥ λ2 ≥ · · · ≥ λN, where λi ∈ Z, e.g. λ = (4, 3, 1) , Schur functions: sλ(x1, . . . , xN) – characters of Vλ. Weyl’s determinantal formula: sλ(x1, . . . , xN) = det

• x

λj +N−j i

N

ij=1

• i<j(xi − xj)

Semi-Standard Young tableaux(⇔ Gelfand-Tsetlin patterns) of shape λ : s(2,2)(x1, x2, x3) = s (x1, x2, x3) = x2

1 x2 2 1 1 2 2

+x2

1 x2 3 1 1 3 3

+x2

2 x2 3 2 2 3 3

+x2

1 x2x3 1 1 2 3

+x1x2

2 x3 1 2 2 3

+x1x2x2

3 1 2 3 3

.

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Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

Main object: Normalized Schur functions

Sλ(N)(x1, . . . , xk) := sλ(N)(x1, . . . , xk,

N−k

• 1, . . . , 1)

sλ(N)(1, . . . , 1

• N

)

• r more generally normalized Lie group characters:

Xγ(N)(x1, . . . , xk) := χγ(N)(x1, . . . , xk, 1N−k) χγ(N)(1N) Meaning also via the Harish-Chandra/Itzykson–Zuber integral: sλ(ea1, . . . , ean) sλ(1, . . . , 1)

• bj =λj +N−j =
• i<j

ai − aj eai − eaj

• U(N)

exp(Trace(AUBU−1))dU

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Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

Integral formula, k = 1 asymptotics

Theorem (Gorin-P)

For any partition λ and any x ∈ C \ {0, 1} we have Sλ(x; N, 1) = (N − 1)! (x − 1)N−1 1 2πi

• C

xz N

i=1(z − (λi + N − i))

dz, where the contour C includes all the poles of the integrand. Similar formulas hold for the other normalized Lie group characters.

Theorem (Gorin-P)

If λ(N)

N

→ f

• i

N

• [under certain convergence conditions], for all fixed y = 0:

lim

N→∞

1 N ln Sλ(N)(ey; N, 1) = yw0 − F(w0) − 1 − ln(ey − 1), where F(w; f ) = 1

0 ln(w − f (t) − 1 + t)dt, w0 – root of ∂ ∂w F(w; f ) = y. If λ(N) N

→ f

• i

N

• [”other” conv. cond.], for any fixed h ∈ R:

Sλ(N)(eh/

√ N; N, 1) = exp

√ NE(f )h + 1 2 S(f )h2 + o(1)

• ,

where E(f ) = 1 f (t)dt, S(f ) = 1 f (t)2dt − E(f )2 + 1 f (t)(1 − 2t)dt.

Remark 1. Similar statements hold for a larger class of functions, e.g symplectic characters, Jacobi...+ q–analogues. Remark 2. Integral formula appears also in [Colomo,Pronko,Zinn-Justin], random matrix interpretation in [Guionnet–Maida], new analysis in [Novak]...

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Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

From k = 1 asymptotics to general k, multiplicativity

Theorem (Gorin-P3 )

Let Di,1 = xi

∂ ∂xi , ∆– Vandermonde det. Then ∀ λ, k ≤ N, we have

Sλ(x1, . . . , xk; N) = sλ(x1, . . . , xk,

N−k

• 1, . . . , 1)

sλ(1, . . . , 1

• N

) =

k

• i=1

(N − i)! (N − 1)!(xi − 1)N−k × det

• Dj−1

i,1

k

i,j=1

∆(x1, . . . , xk)

k

• j=1

Sλ(xj; N, 1)(xj − 1)N−1.

Corollary (Gorin-P)

Suppose that the sequence λ(N) is such that, as N → ∞, ln

• Sλ(N)(x; N, 1)
• N

→ Ψ(x) uniformly on a compact M ⊂ C. Then for any k lim

N→∞

ln

• Sλ(N)(x1, . . . , xk; N, 1)
• N

= Ψ(x1) + · · · + Ψ(xk) uniformly on Mk. More informally, under various regimes of convergence for λ(N) we have Sλ(N)(x1, . . . , xk) ∼ Sλ(N)(x1) · · · Sλ(N)(xk).

3Similar for symplectic characters, Jacobi; and q-analogues. Note: appears in [de Gier, Nienhuis,

Ponsaing] for symplectic characters.

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Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

GUE in tilings I: combinatorics

3+2

+1

5 1 2 +3 4

+4

x 2 3 x3

3

x3

2

x3

1

3 3 4 3 3 Tilings of domain Ωλ(N) ⇔ Gelfand-Tsetlin schemes with bottom row λ(N) 2 3 1 3 3 3 3 3 4 ⇔ Semi-Standard Young Tableaux of shape λ(N) T= 1 1 2 5 3 4 4 5 5 5 Positions of the horizontal lozenges

• n line j:

xj –shape of the subtableaux of T of the entries 1, . . . , j.

Note: actual positions on the 2d projection are at (xk + (k/2 − 1, k/2 − 2, . . . , 1 − k/2, −k/2)) √ 3 2 16

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

GUE in tilings I: combinatorics

3+2

+1

5 1 2 +3 4

+4

x 2 3 x3

3

x3

2

x3

1

3 3 4 3 3 Tilings of domain Ωλ(N) ⇔ Gelfand-Tsetlin schemes with bottom row λ(N) 2 3 1 3 3 3 3 3 4 ⇔ Semi-Standard Young Tableaux of shape λ(N) T= 1 1 2 5 3 4 4 5 5 5 x3 = (3, 1, 0). Positions of the horizontal lozenges

• n line j:

xj –shape of the subtableaux of T of the entries 1, . . . , j.

Note: actual positions on the 2d projection are at (xk + (k/2 − 1, k/2 − 2, . . . , 1 − k/2, −k/2)) √ 3 2 16

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

GUE in tilings II: moment generating functions

Proposition

In a uniformly random tiling of Ωλ the distribution of the positions of the horizontal lozenges on the kth line xk(λ) is given by: Prob{xk(λ) = η} = sη(1k)sλ/η(1N−k) sλ(1N) , where sλ/η is the skew Schur polynomial. Proof: combinatorial definition of Schur functions as sums over SSYTs.

Proposition

For any variables y1, . . . , yk, the following moment generating function of xk (as above) is given by E      sxk (y1, . . . , yk) sxk (1, . . . , 1

• k

)      = sλ(y1, . . . , yk,

N−k

• 1, . . . , 1)

sλ(1, . . . , 1

• N

) = Sλ(y1, . . . , yk).

17

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

GUE in tilings III: MGF asymptotics

Proposition

EBk(y; GUEk) = exp 1 2 (y2

1 + · · · + y2 k )

• ,

where Bk(y; ν) = sν−δk (y1, . . . , yk) sν−δk (1, . . . , 1

• k

) when ν — strict partition.

18

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

GUE in tilings III: MGF asymptotics

Proposition

EBk(y; GUEk) = exp 1 2 (y2

1 + · · · + y2 k )

• ,

where Bk(y; ν) = sν−δk (y1, . . . , yk) sν−δk (1, . . . , 1

• k

) when ν — strict partition. Compare: Sλ(y1, . . . , yk) = Etiling      sxk (y1, . . . , yk) sxk (1, . . . , 1

• k

)      = EtilingBk(y; xk + δk)

Proposition (Gorin-P)

For any k real numbers h1, . . . , hk and λ(N)/N → f (as earlier) we have: lim

N→∞ Sλ(N)

• e

h1

√

NS(f ) , . . . , e hk

√

NS(f )

• e
• −

E(f )

√

NS(f )

k

i=1 hi

• = exp
• 1

2

k

• i=1

h2

i

• .

18

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

GUE in tilings III: MGF asymptotics

Proposition

EBk(y; GUEk) = exp 1 2 (y2

1 + · · · + y2 k )

• ,

where Bk(y; ν) = sν−δk (y1, . . . , yk) sν−δk (1, . . . , 1

• k

) when ν — strict partition. Compare: Sλ(y1, . . . , yk) = Etiling      sxk (y1, . . . , yk) sxk (1, . . . , 1

• k

)      = EtilingBk(y; xk + δk)

Proposition (Gorin-P)

For any k real numbers h1, . . . , hk and λ(N)/N → f (as earlier) we have: lim

N→∞ Sλ(N)

• e

h1

√

NS(f ) , . . . , e hk

√

NS(f )

• e
• −

E(f )

√

NS(f )

k

i=1 hi

• = exp
• 1

2

k

• i=1

h2

i

• .
• Theorem. Let Υk

λ(N) = {xk, xk−1, . . .} –collection of positions of the

horizontal lozenges on lines k, k − 1, . . . , 1 of tiling from Ωλ(N), then Υk

λ(N) − NE(f )

• NS(f )

→ GUEk (GUE-corners process of rank k).

18

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

Free boundary domains

N M

Tf (N, M) :=

• λ :

ℓ(λ)=N,λ1≤M

tilings of Ωλ, – set of all tilings in an N ×M ×N trape- zoid with N free (unrestricted) horizontal rhombi on the right border. ⇔ Symmetric tilings of the N ×M ×N × N × M × N hexagon/ symmetric boxed Plane Partitions.

M N N

Questions:

• 1. Existence of a “limit shape”

(surface)?

Equation derived in [Di Francesco – Reshetikhin, 2009].

2. Behavior near boundary (GUE?).

19

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

Lozenge tilings with free boundaries: GUE

λ1 + N − 1 λ2 + N − 2 λN m n y3

1

y3

2

y3

3

Line k = 3

Theorem (P)

Let Y k

n,m = (yk 1 , . . . , yk k ) denote the

positions of the horizontal lozenges on the kth vertical line. As n, m → ∞ with m/n → a for 0 < a < ∞, we have Y k

n,m − m/2

• n(a2 + 2a)/8

→ GUEk weakly as RVs. Moreover, the collection

• Y j

n,m−m/2

√

n(a2+2a)/8

k

j=1

weakly converge as RVs to the GUE–corners process Proof method: Generating function for “free boundary tilings” is an SO2n+1 character:

• λ∈(mn)

sλ(x1, . . . , xn) = det[xm+2n−i

j

− xi−1

j

]1≤i,j≤n det[x2n−i

j

− xi−1

j

]1≤i,j≤n = γ(mn)(x), Apply the same asymptotic techniques to this MGF as for Schur functions → GUE eigenvalues MGF as N → ∞.

20

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

Lozenge tilings with free boundaries: limit shape (limit surface)

Theorem (P)

Let n, m ∈ Z, such that m/n → a as n → ∞, where a ∈ (0, +∞). Let H(u, v) be the height function (plane partition) of the uniformly random lozenge tiling in Tf (n, m),i.e. H(u, v) = 1 n y⌊nu⌋

⌊nv⌋ − v.

For all u ∈ (0, 1), u ≥ v ≥ 0, as n → ∞ we have that H(u, v) converges uniformly in probability to a deterministic function L(u, v), referred to as “the limit shape”. Moreover, for any fixed u ∈ (0, 1), the function L(u, v) is the distribution function of the limit measure m whose moments are given by

• R

xrm(dx) =

r

• ℓ=0

r ℓ

• 1

(ℓ + 1)! ∂ℓ ∂tℓ Φa(t)

• t=1

, where Φa(ey) = y a

2 + 2φ(y; a) − 2 and...

h(y) = 1 4

• (ey + 1) +
• (ey + 1)2 + 4(a2 + a) (ey − 1)2
• φ(y; a) = (

a 2 + 1) ln

• h(y) − (

a 2 + 1)(ey − 1)

• − (

a 2 + 1 2 ) ln

• h(y) − (

a 2 + 1 2 )(ey − 1)

• +

a 2 ln

• h(y) +

a 2 (ey − 1)

• − (

a 2 − 1 2 ) ln

• h(y) + (

a 2 − 1 2 )(ey − 1)

• 21

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

Lozenge tilings with free boundaries: limit shape (limit surface)

Theorem (P)

Let n, m ∈ Z, such that m/n → a as n → ∞, where a ∈ (0, +∞). Let H(u, v) be the height function (plane partition) of the uniformly random lozenge tiling in Tf (n, m),i.e. H(u, v) = 1 n y⌊nu⌋

⌊nv⌋ − v.

For all u ∈ (0, 1), u ≥ v ≥ 0, as n → ∞ we have that H(u, v) converges uniformly in probability to a deterministic function L(u, v), referred to as “the limit shape”. Proof, idea:4 Horizontal lozenges at line x = un at positions µ = yun distributed ∼ ρ (unif. on all tilings), giving a sequence of random measures m[µ] = 1 n

• i

δ µi n

• .

Sρ(u1, . . . , uN) :=

• µ:ℓ(µ)=N

ρ(µ) sµ(u1, . . . , uN) sµ(1N) . (and observe Sρ = γmn (x1,...,xN)

γ(mn)(1N)

– our MGF) Using this MGF, its asymptotics and ∇ operators, obtain a concentration phenomenon: m[µ] → m – a deterministic measure, which is the limit shape L.

4after [Borodin-Bufetov-Olshanski, Bufetov-Gorin] 21

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

Lozenge tilings with free boundaries: limit shape (limit surface)

Theorem (P)

Let n, m ∈ Z, such that m/n → a as n → ∞, where a ∈ (0, +∞). Let H(u, v) be the height function (plane partition) of the uniformly random lozenge tiling in Tf (n, m),i.e. H(u, v) = 1 n y⌊nu⌋

⌊nv⌋ − v.

For all u ∈ (0, 1), u ≥ v ≥ 0, as n → ∞ we have that H(u, v) converges uniformly in probability to a deterministic function L(u, v), referred to as “the limit shape”.

Corollary (P)

The height function of the uniformly random lozenge tilings of a half-hexagon with free right boundary converges (in probability) to a unique limit shape (surface) H(x, y), which coincides over the half-hexagon with the limit shape for the tilings of the full hexagon (fixed boundary). The shifted by m/2 and rescaled by

• n(a2 + 2a)/8 positions of the horizontal lozenges on the k-th

vertical line have the same joint distributions as n, m → ∞, which is GUEk.

21

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

6 Vertex model / ASM

Six vertex types:

O O O H H H H O O H O H H H H H H

a a

H

b b c c 1 −1

Alternating Sign Matrix:

     1 1 −1 1 1 1 1     

A 6 vertex model configuration:

O O O O O H H H H H H O O O O O H H H H H H O O O O O H H H H H H O O O O O H H H H H H O O O O O H H H H H H H H H H H H H H H H H H H H H H H H H H

22

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

Definitions and background on ASMs

Definition: A is an Alternating Sign Matrix ASM of size n if: A ∈ {0, +1, −1}n×n,

n

• i=1

Ai,j = 1,

n

• j=1

Ai,j = 1 and (Ai,k, i = 1 . . . n s.t. Ai,k = 0) = (1, −1, 1, −1, . . . , −1, 1) A monotone triangle is a Gelfand-Tsetlin pattern with strictly increasing rows. 6 Vertex model ↔ ASM ↔ monotone triangles. Uniform measure on ASMs ↔ vertices in 6V model have same weights (”ice”). ASM:      1 1 −1 1 1 1 1      positions of 1s ⇐ ⇒ in sum of first k rows Monotone triangle: 4 ≤ 2 5 ≤ 2 3 5 1 2 3 5 1 2 3 4 5 < Question: As n → ∞: Uniformly random ASM. What is the distribution of the positions of the 1s and −1s near the boundary ⇔ Distribution of the top k rows of the monotone triangle? Known results:

• Limit behavior(conj): Behrend,

Colomo, Pronko, Zinn-Justin, Di Francesco.

• Free fermions point

(weight(1,-1)=2) ↔ domino tilings, Aztec diamond.

• Exact gen. fs. for certain statistics

(e.g. positions of 1s on boundary).

23

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

ASMs/6Vertex: new results

ASM A: Ψk(A) :=

• j=1:n , Akj =1

j −

• j=1:n , Akj =−1

j Monotone triangle M = [mi

j]j≤i:

Ψk(M) =

k

• j=1

mk

j − k−1

• j=1

mk−1

j

k :      1 1 −1 1 1 1 1      4 2 5 2 3 5 1 2 3 5 1 2 3 4 5 2 Ψ2 = 2 + 5 − 4 = 3 Ψ2 = (2 + 5) − (4) = 3 3 Ψ3 = 3 Ψ3 = (2 + 3 + 5) − (2 + 5)

24

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

ASMs/6Vertex: new results

ASM A: Ψk(A) :=

• j=1:n , Akj =1

j −

• j=1:n , Akj =−1

j Monotone triangle M = [mi

j]j≤i:

Ψk(M) =

k

• j=1

mk

j − k−1

• j=1

mk−1

j

k :      1 1 −1 1 1 1 1      4 2 5 2 3 5 1 2 3 5 1 2 3 4 5 2 Ψ2 = 2 + 5 − 4 = 3 Ψ2 = (2 + 5) − (4) = 3 3 Ψ3 = 3 Ψ3 = (2 + 3 + 5) − (2 + 5)

Theorem (Gorin-P)

If A ∼ unif. rand. n × n ASM, then Ψk (A)−n/2

√n

, k = 1, 2, . . . converge as n → ∞ to the collection of i.i.d. Gaussian random variables, N(0,

• 3/8).

24

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

ASMs/6Vertex: new results

ASM A: Ψk(A) :=

• j=1:n , Akj =1

j −

• j=1:n , Akj =−1

j Monotone triangle M = [mi

j]j≤i:

Ψk(M) =

k

• j=1

mk

j − k−1

• j=1

mk−1

j

k :      1 1 −1 1 1 1 1      4 2 5 2 3 5 1 2 3 5 1 2 3 4 5 2 Ψ2 = 2 + 5 − 4 = 3 Ψ2 = (2 + 5) − (4) = 3 3 Ψ3 = 3 Ψ3 = (2 + 3 + 5) − (2 + 5)

Theorem (Gorin-P)

If A ∼ unif. rand. n × n ASM, then Ψk (A)−n/2

√n

, k = 1, 2, . . . converge as n → ∞ to the collection of i.i.d. Gaussian random variables, N(0,

• 3/8).

Using this Theorem on Ψk(n) and the Gibbs property:

Theorem (G, 2013; Conjecture in [Gorin-P] )

Fix any k. The [centered,rescaled] positions of 1s on the first k rows (top k rows of the monotone triangle M) tend to the GUE-corners process:

• 8

3n

• [M]i=1:k − n

2

• → GUEk.

24

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

6Vertex/ASMs: proofs

Vertex weights at (i, j): type a : q−1u2

i − qv2 j ,

type b : q−1v2

j − qu2 i ,

type c : (q−1 − q)uivj (v1, . . . , vN, u1, . . . , uN – parameters, q = exp(πi/3) ) Weight W (ϑ) of a configuration ϑ is =

• vtx∈ϑ

weight(vtx). Set λ(N) := (N − 1, N − 1, N − 2, N − 2, . . . , 1, 1, 0, 0) ∈ GT2N.

Proposition (Okada;Stroganov)

LetגN be the set of all 6-Vertex configurations on an N × N grid.

• ϑ∈גN

W (ϑ) = (−1)N(N−1)/2(q−1−q)N

N

• i=1

(viui)−1sλ(N)(u2

1, . . . , u2 N, v2 1 , . . . , v2 N).

25

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

6Vertex/ASMs: proofs

Vertex weights at (i, j): type a : q−1u2

i − qv2 j ,

type b : q−1v2

j − qu2 i ,

type c : (q−1 − q)uivj (v1, . . . , vN, u1, . . . , uN – parameters, q = exp(πi/3) ) Weight W (ϑ) of a configuration ϑ is =

• vtx∈ϑ

weight(vtx). Set λ(N) := (N − 1, N − 1, N − 2, N − 2, . . . , 1, 1, 0, 0) ∈ GT2N.

Proposition

Let xi =number of vertices of type x on row i, then ∀ collection of rows i1, . . . , im EN

m

• ℓ=1

 

• q−1 − qv2

ℓ

q−1 − q

• aiℓ

q−1v2

ℓ − q

q−1 − q

• biℓ

(vℓ)

cjℓ

  = n

• ℓ=1

v−1

ℓ

• sλ(N)(v1, . . . , vm, 12N−m)

sλ(N)(12N) = n

• ℓ=1

v−1

ℓ

• Sλ(N)(v1, . . . , vm)

25

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

6Vertex/ASMs: proofs

Vertex weights at (i, j): type a : q−1u2

i − qv2 j ,

type b : q−1v2

j − qu2 i ,

type c : (q−1 − q)uivj (v1, . . . , vN, u1, . . . , uN – parameters, q = exp(πi/3) ) Weight W (ϑ) of a configuration ϑ is =

• vtx∈ϑ

weight(vtx). Set λ(N) := (N − 1, N − 1, N − 2, N − 2, . . . , 1, 1, 0, 0) ∈ GT2N.

Proposition

Let xi =number of vertices of type x on row i, then ∀ collection of rows i1, . . . , im EN

m

• ℓ=1

 

• q−1 − qv2

ℓ

q−1 − q

• aiℓ

q−1v2

ℓ − q

q−1 − q

• biℓ

(vℓ)

cjℓ

  = n

• ℓ=1

v−1

ℓ

• sλ(N)(v1, . . . , vm, 12N−m)

sλ(N)(12N) = n

• ℓ=1

v−1

ℓ

• Sλ(N)(v1, . . . , vm)

Proof of Theorem: Proposition → moment generating function for Ψk = ak as a normalized Schur function. Using ck ≤ 2k − 1 (bounded), approximate the weights to get the MGF for

• ak. Asymptotics:

Sλ(N)(ey1/√n, . . . , eyk /√n) =

k

• i=1

exp √nyi + 5 12 y2

i + o(1)

• 25

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

The dense loop model

Finite grid (here: vertical strip of width L and height → ∞) tiled with squares, boundary – triangales:

x y ζ1 ζ2 L

The mean total current between points x and y: F x,y – the average number

• f paths connecting the 2 boundaries and passing between x and y.

Similar observables in the critical percolation model [Smirnov, 2009].

26

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

Dense loop model: the mean current

Let λL = (⌊ L−1

2 ⌋, ⌊ L−2 2 ⌋, . . . , 1, 0, 0)

Define: uL(ζ1, ζ2; z1, . . . , zL) = (−1)Lı √ 3 2 ln

• χλL+1(ζ2

1, z2 1, . . . , z2 L)χλL+1(ζ2 2, z2 1, . . . , z2 L)

χλL(z2

1, . . . , z2 L)χλL+2(ζ2 1, ζ2 2, z2 1, . . . , z2 L)

• where χν is the character for the irreducible representation of highest weight ν
• f the symplectic group Sp(C).

X (j)

L

= zj ∂ ∂zj uL(ζ1, ζ2; z1, . . . , zL) YL = w ∂ ∂w uL+2(ζ1, ζ2; z1, . . . , zL, vq−1, w)|v=w,

Proposition (De Gier, Nienhuis, Ponsaing)

Under certain assumptions the mean total current between two horizontally adjacent points is X (j)

L

= F (j,i),(j+1,i), and Y is the mean total current between two vertically adjacent points in the strip of width L: Y (j)

L

= F (j,i),(j,i+1).

27

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

Dense loop model: asymptotics of the mean current

Theorem

As L → ∞ we have X (j)

L

• zj =z; zi =1, i=j = i

√ 3 4L (z3 − z−3) + o 1 L

• and

YL

• zi =1, i=1,...,L = i

√ 3 4L (w3 − w−3) + o 1 L

• Remark 1. When z = 1, F (j,i),(j+1,i) is (trivially) identically zero.

Remark 2. The fully homogeneous case when w = exp−iπ/6, q = e2πi/3, then YL = √ 3 2L + o 1 L

• .

Proof: same type of asymptotic methods and results hold for symplectic characters + some tricks with the multivariate formula.

28

Asymptotics of symmetric functions Greta Panova Lozenge tilings The objects Probabilistic questions Answers: GUE Probability via Schur functions Schur functions asymptotics GUE proof Tilings with free boundaries ASMs and GUE Dense loop model

Thank you

29