Statistical mechanics via asymptotics of symmetric functions

symmetric Statistical mechanics via of symmetric (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:0.064 G. Panova, with free, arxiv:408.047. April 05

symmetric Overview Characters of U( ), boundary of the Gelfand-Tsetlin graph......... Normalized : S λ (x,..., x k ; N) = s λ(x,..., x k, N k ) s λ ( N ) Alternating Sign Matrices (ASM)/ 6-Vertex model: 0 0 0 0 0 0 0 0 : : ζ y ζ x L

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

symmetric The many faces of lozenge tilings 5 4 4 4 5 4 5 4 4 5 4 4 4

symmetric The many faces of lozenge tilings 5 4 4 4 5 4 4 5 5 4 4 4 4

symmetric The many faces of lozenge tilings 5 4 4 4 5 4 5

symmetric 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, 998] Hexagonal domain: Tiling is asymptotically frozen outside inscribed ellipse. [Kenyon Okounkov, 005] Polygonal domain: Tiling is asymptotically frozen outside inscribed algebraic curve..jpg [Cohn Kenyon Propp, 00; Kenyon-Okounkov-Sheffield, 006] There exists a limit shape for the surface of the height function (plane partition). 6

symmetric Behavior near the boundary, interlacing particles x x x x x x x N Horizontal lozenges near a flat boundary: interlacing particle configuration Gelfand-Tsetlin patterns. Question: What is the joint distribution of {x i j }k i= as N (scale = N )? x x x x x x with an explanation what the answer should be. Subsequent results: [Gorin-P,0], [Novak, 04],[Mkrtchyan, 0, periodic weights, unbounded 7

symmetric Behavior near the boundary, interlacing particles x x x x x x x N Horizontal lozenges near a flat boundary: interlacing particle configuration Gelfand-Tsetlin patterns. x x x x x x Question: What is the joint distribution of {x i j }k i= as N (scale = N )? Conjecture [Okounkov Reshetikhin, 006 ]: The joint distribution converges to a GUE-corners (aka GUE-minors) process: eigenvalues of GUE matrices. Proven for the hexagon by Johansson- Nordenstam (006). with an explanation what the answer should be. Subsequent results: [Gorin-P,0], [Novak, 04],[Mkrtchyan, 0, periodic weights, unbounded 7

symmetric GUE in tilings: our setup Domain Ω λ(n) : width N and the positions λ(n) + N > λ(n) + N > > λ(n) N of its N horizontal lozenges at the right boundary. +4 + + 8 + 0 8 + 9 8 + 8 8 + 7 8 + 6 4 5 + λ(5) = (4,,, 0, 0) Note: N Ω λ(n) is not necessarily a finite polygon as N, e.g. λ(n) = (N, N,...,, ) 0 + 5 0 + 4 0 + 0 + 0 + 0 + 0 E.g. λ = (a,..., a, 0,..., 0) }{{}}{{} c b the hexagon with side lengths (a, b, c, a, b, c). 8

symmetric Plane partitions/gelfand-tsetlin patterns λ + N x x x Line j = λ + N λn N 5 4 4 4 0 0 0 λ = (5, 4,,, 0) x = (4,, 0) 4 0 0 0 0 0 0 λ = (4,,, 0, 0) Question: What is the joint distribution of the positions rescaling) near the boundary as N (i.e. lattice scale = N )? { } xj i (under correct 9

symmetric Answers: the Gaussian Unitary Ensemble (GUE) Gaussian Unitary Ensemble: matrices [X ij ] i,j : X = X T ReX ij, ImX ij i.i.d. N (0, /) for i j and X ii i.i.d. N (0, ) a a a a 4 a a a a 4 a a a a 4 a 4 a 4 a 4 a 44 (x k xk xk k ) eigenvalues of [X i,j ] k i,j= (top k k). Interlacing condition: x j i xj i xj i x 4 x 4 x 4 x4 4 x x x x x The joint distribution of {x j i } i j k is known as GUE corners (also, GUE-minors) process, =: GUE k. x 0

symmetric GUE in tilings: our results Limit profile f (t) of λ(n) as N ( : ) λ(n) i i N f N N λ(n) f (t) Ω λ(n) domain: Theorem (Gorin-P (0), Novak (04)) Let λ(n) = (λ (N)... λ N (N)), N =,,... be a partition. If a piecewise-differentiable weakly decreasing function f (t) (limit profile of λ(n)) s.t. N ( ) λ i (N) i N f = o( N) as N N i= and also sup i,n λ i (N)/N <. Let Υ k λ(n) = {xj i } be the collection of the positions of the horizontal lozenges on lines j =,..., k. Then for every k as N x λ + N λ + N Υ k λ(n) NE(f ) NS(f ) GUE k (GUE-corners proc. rank k) x in the sense of weak convergence, where x E(f ) = f (t)dt, 0 Line j = λn S(f ) = f (t) dt E(f ) + f (t)( t)dt. 0 0

symmetric Towards the proof:...or more generally symmetric, Lie group characters. Irreducible (rational) representations V λ of GL(N) (or U(N)) are indexed by dominant weights (signatures/young diagrams/integer partitions) λ: where λ i Z, e.g. λ = (4,, ), λ λ λ N, : s λ (x,..., x N ) characters of V λ. Weyl s determinantal formula: s λ (x,..., x N ) = [ det x λ j +N j i ] N ij= i<j (x i x j ) Semi-Standard Young tableaux( Gelfand-Tsetlin patterns) of shape λ : s (,) (x, x, x ) = s (x, x, x ) = x x +x x +x x +x x x +x x x +x x x.

symmetric Main object: Normalized N k {}}{ S λ(n) (x,..., x k ) := s λ(n)(x,..., x k,,..., ) s λ(n) (,..., ) }{{} N or more generally normalized Lie group characters: X γ(n) (x,..., x k ) := χ γ(n)(x,..., x k, N k ) χ γ(n) ( N ) Meaning also via the Harish-Chandra/Itzykson Zuber integral: s λ (e a,..., e an ) s λ (,..., ) = a i a j bj =λ j +N j e a i e a j i<j exp(trace(aubu ))du U(N)

symmetric Integral formula, k = Theorem (Gorin-P) For any partition λ and any x C \ {0, } we have (N )! x z S λ (x; N, ) = (x ) N πi N C i= (z (λ i + 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 [under certain convergence conditions], for all fixed y 0: N lim N N ln S λ(n)(e y ; N, ) = yw 0 F(w 0) ln(e y ), where F(w; f = 0 ln(w f (t) + t)dt, w0 root of F(w; f ) = y. If ( ) w λ(n) N f i [ other conv. cond.], for any fixed h R: N ( ) NE(f S λ(n) (e h/ N ; N, ) = exp )h + S(f )h + o(), where E(f ) = f (t)dt, S(f ) = f (t) dt E(f ) + f (t)( t)dt. 0 0 0 Remark. Similar statements hold for a larger class of, e.g symplectic characters, Jacobi...+ q analogues. Remark. Integral formula appears also in [Colomo,Pronko,Zinn-Justin], random matrix interpretation in [Guionnet Maida], new analysis in [Novak]... 4

symmetric From k = to general k, multiplicativity Theorem (Gorin-P ) Let D i, = x i, Vandermonde det. Then λ, k N, we have x i N k {}}{ S λ (x,..., x k ; N) = s λ(x,..., x k,,..., ) s λ (,..., ) }{{} N [ ] k det D j k (N i)! = (N )!(x i= i ) N k i, k i,j= S λ (x j ; N, )(x j ) N. (x,..., x k ) j= Corollary (Gorin-P) Suppose that the sequence λ(n) is such that, as N, ln ( S λ(n) (x; N, ) ) Ψ(x) uniformly on a compact M C. Then for any k N ln ( S λ(n) (x,..., x k ; N, ) ) lim = Ψ(x ) + + Ψ(x k ) N N uniformly on M k. More informally, under various regimes of convergence for λ(n) we have S λ(n) (x,..., x k ) S λ(n) (x ) S λ(n) (x k ). Similar for symplectic characters, Jacobi; and q-analogues. Note: appears in [de Gier, Nienhuis, 5

symmetric GUE in tilings I: combinatorics x 0 x x x 0 0 4 0 0 + + 4 5 +4 + Tilings of domain Ω λ(n) Gelfand-Tsetlin schemes with bottom row λ(n) 0 0 0 0 0 0 4 Semi-Standard Young Tableaux of shape λ(n) T = 5 4 4 5 5 5 Positions of the horizontal lozenges on line j: x j shape of the subtableaux of T of the entries,..., j. Note: actual positions on the d projection are at (x k + (k/, k/,..., k/, k/)) 6

symmetric GUE in tilings I: combinatorics x 0 x x x 0 0 4 0 0 + + 4 5 +4 + Tilings of domain Ω λ(n) Gelfand-Tsetlin schemes with bottom row λ(n) 0 0 0 0 0 0 4 Semi-Standard Young Tableaux of shape λ(n) T = 5 4 4 5 5 5 x = (,, 0). Positions of the horizontal lozenges on line j: x j shape of the subtableaux of T of the entries,..., j. Note: actual positions on the d projection are at (x k + (k/, k/,..., k/, k/)) 6

symmetric GUE in tilings II: moment generating Proposition In a uniformly random tiling of Ω λ the distribution of the positions of the horizontal lozenges on the kth line x k (λ) is given by: Prob{x k (λ) = η} = sη(k )s λ/η ( N k ) s λ ( N, ) where s λ/η is the skew Schur polynomial. Proof: combinatorial definition of as sums over SSYTs. Proposition For any variables y,..., y k, the following moment generating function of x k (as above) is given by N k {}}{ s E x k (y,..., y k ) s x k (,..., ) = s λ(y,..., y k,,..., ) = S λ (y,..., y k ). s λ (,..., ) }{{}}{{} k N 7

symmetric GUE in tilings III: MGF Proposition ( ) EB k (y; GUE k ) = exp (y + + y k ), where B k (y; ν) = s ν δ k (y,..., y k ) when ν strict partition. s ν δk (,..., ) }{{} k 8

symmetric GUE in tilings III: MGF Proposition ( ) EB k (y; GUE k ) = exp (y + + y k ), where B k (y; ν) = s ν δ k (y,..., y k ) when ν strict partition. s ν δk (,..., ) }{{} k Compare: s S λ (y,..., y k ) = E tiling x k (y,..., y k ) s x k (,..., ) = E tiling B k (y; x k + δ k ) }{{} k Proposition (Gorin-P) For any k real numbers h,..., h k and λ(n)/n f (as earlier) we have: ( h lim S NS(f λ(n) e ),..., e N ) h NS(f k ) e ( ) E(f ) ( ki= h NS(f ) i = exp ) k hi. i= 8

symmetric Free boundary domains N M N T f (N, M) := λ : l(λ)=n,λ M tilings of Ω λ, set of all tilings in an N M N trapezoid 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 Questions:. Existence of a limit shape (surface)? Equation derived in [Di Francesco Reshetikhin, 009].. Behavior near boundary (GUE?). 9

symmetric with free : GUE m n y y y Line k = λ + N λ + N λ N Theorem (P) Let Yn,m k = (y k,..., y k 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/ n(a + a)/8 GUE k weakly as RVs. Moreover, the collection { } Y j k n,m m/ weakly converge as n(a +a)/8 j= RVs to the GUE corners process Proof method: Generating function for free boundary tilings is an SO n+ character: λ (m n ) s λ (x,..., x n) = det[xm+n i j det[x n i j x i j ] i,j n x i j ] i,j n = γ (m n )(x), Apply the same asymptotic techniques to this MGF as for GUE eigenvalues MGF as N. 0

symmetric with free : 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 T f (n, m),i.e. H(u, v) = n y nu nv v. For all u (0, ), 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, ), the function L(u, v) is the distribution function of the limit measure m whose moments are given by r ( r x r m(dx) = R l) l (l + )! t l Φa(t), l=0 t= where Φ a(e y ) = y a + φ(y; a) and... h(y) = 4 ( (e y ) + ) + (e y + ) + 4(a + a) (e y ) φ(y; a) = ( a ( + ) ln h(y) ( a + )(e y ) ) ( a + ( ) ln h(y) ( a + )(e y ) ) + a ( ln h(y) + a (e y ) ) ( a ( ) ln h(y) + ( a )(e y ) )

symmetric with free : 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 T f (n, m),i.e. H(u, v) = n y nu nv v. For all u (0, ), 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 = y un distributed ρ (unif. on all tilings), giving a sequence of random measures m[µ] = ( µi ) δ. n n i S ρ(u,..., u N ) := µ:l(µ)=n ρ(µ) sµ(u,..., u N ) s µ( N. ) (and observe S ρ = γ m n (x,...,x N ) our MGF) γ (m n ) ( N ) Using this MGF, its and operators, obtain a concentration phenomenon: m[µ] m a deterministic measure, which is the limit shape L. 4 after [Borodin-Bufetov-Olshanski, Bufetov-Gorin]

symmetric with free : 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 T f (n, m),i.e. H(u, v) = n y nu nv v. For all u (0, ), 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/ and rescaled by n(a + a)/8 positions of the horizontal lozenges on the k-th vertical line have the same joint distributions as n, m, which is GUE k.

symmetric 6 Vertex model / ASM Six vertex types: H H O O H O H H O H O H O H H a a b b c c 0 0 0 0 Alternating Sign Matrix: H H H A 6 vertex model configuration: 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 H H H O H O H O H O H O H H H H H O H O H O H O H O H H H H H O H O H O H O H O H H H H H H H H H O H O H O H O H O H H H H H H H O H O H O H O H O H

symmetric Definitions and background on ASMs Definition: A is an Alternating Sign Matrix ASM of size n if: n n A {0, +, } n n, A i,j =, A i,j = i= j= and (A i,k, i =... n s.t. A i,k 0) = (,,,,...,, ) 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 ). 0 0 0 0 0 0 ASM: 0 0 0 0 0 0 0 0 0 0 0 0 positions of s Monotone triangle: 4 5 in sum of first k rows 5 5 4 5 < Question: As n : Uniformly random ASM. What is the distribution of the positions of the s and s 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(,-)=) domino tilings, Aztec diamond. Exact gen. fs. for certain statistics (e.g. positions of s on boundary).

symmetric ASMs/6Vertex: new results k : ASM A: Ψ k (A) := j=:n, A kj = j Monotone triangle M = [m i j ] j i : Ψ k (M) = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 j=:n, A kj = k j= j k mj k m k j j= 4 5 5 5 4 5 Ψ = + 5 4 = Ψ = ( + 5) (4) = Ψ = Ψ = ( + + 5) ( + 5) Theorem (Gorin-P) If A unif. rand. n n ASM, then Ψ k (A) n/ n, k =,,... converge as n to the collection of i.i.d. Gaussian random variables, N(0, /8). Using this Theorem on Ψ k (n) and the Gibbs property: Theorem (G, 0; Conjecture in [Gorin-P] ) Fix any k. The [centered,rescaled] positions of s on the first k rows (top k rows of the monotone triangle M) tend to the GUE-corners process: 8 ( [M] i=:k n ) GUE k. n 4

symmetric 6Vertex/ASMs: proofs Vertex weights at (i, j): type a : q ui qvj, type b : q vj qui, type c : (q q)u i v j (v,..., v N, u,..., u N parameters, q = exp(πi/) ) Weight W (ϑ) of a configuration ϑ is = weight(vtx). vtx ϑ Set λ(n) := (N, N, N, N,...,,, 0, 0) GT N. Proposition Let x i =number of vertices of type x on row i, then collection of rows i,..., i m m q E N ( qv )âil ( l q vl q ) bil q q q (v l q )ĉjl l= ( n ) ( = v sλ(n) (v,..., v m, N m ) n ) l s λ(n) ( N = v l S λ(n) (v,..., v m) ) l= l= Proof of Theorem: Proposition moment generating function for Ψ k = â k as a normalized Schur function. Using c k k (bounded), approximate the weights to get the MGF for â k. Asymptotics: S λ(n) (e y / n,..., e y k / n ) = k i= [ nyi exp + 5 ] y i + o() 5

symmetric The dense loop model Finite grid (here: vertical strip of width L and height ) tiled with squares, boundary triangales: y ζ ζ x The mean total current between points x and y: F x,y the average number of paths connecting the and passing between x and y. Similar observables in the critical percolation model [Smirnov, 009]. L 6

symmetric : the mean current Let λ L = ( L L,,...,, 0, 0) Define: [ u L (ζ, ζ ; z,..., z L ) = ( ) L χλ ı ln L+ (ζ, z,..., z L )χ λ L+ (ζ, z,..., z L ) ] χ λ L (z,..., z L )χ λ L+ (ζ, ζ, z,..., z L ) where χ ν is the character for the irreducible representation of highest weight ν of the symplectic group Sp(C). X (j) L = z j z j u L (ζ, ζ ; z,..., z L ) Y L = w w u L+(ζ, ζ ; z,..., z L, vq, 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+,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+). 7

symmetric : of the mean current Theorem As L we have and X (j) L zj =z; z i =, i j = i 4L (z z ) + o ( ) L Y L = i ( ) zi =, i=,...,l 4L (w w ) + o L Remark. When z =, F (j,i),(j+,i) is (trivially) identically zero. Remark. The fully homogeneous case when w = exp iπ/6, q = e πi/, then ( ) Y L = L + o. L Proof: same type of asymptotic methods and results hold for symplectic characters + some tricks with the multivariate formula. 8

symmetric Thank you 9