Statistical Mechanics and Combinatorics : Lecture IV

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1 Statistical Mechanics and Combinatorics : Lecture IV Dimer Model Local Statistics We ve already discussed the partition function Z for dimer coverings in a graph G which can be computed by Kasteleyn matrix K Theorem (Kasteleyn, (temporally Fisher)) Z = det K Theorem 2 Given a subset X = {e = w b,, e k = w k b k } E of edges, the probability of all edges of X in a dimer cover M is Pr(e,, e k M) = det(k (w i, b j )) i,j k k K(b i, w i ) Proof To count the number of dimer coverings with all fixed edges in X is equivalent to remove the black and white vertices b,, b k, w, w k from G and count the dimer coverings in the remaining graph One can check that the remaining graph still have the Kasteleyn condition Let Z be the partition function of dimer coverings with all edges e i X erased Let K be a submatrix of the Kasteleyn matrix K deleting rows b,, b k and columns w,, w k The probability of all edges in X in M is i= Pr(e,, e k ) = Z Z = det K det K k ν(e i ) i= where ν is the weight function The proof is done by applying Jacobi Lemma: For any square matrix M, let MB A the submatrix obtained by keeping only rows indexed by A and columns indexed by B Then det(m ) A Ac det MB B = c det M ( ) k l= i l+j l Remark 3 The probability of a set of k edges is the determinant of a k k matrix, which is independent of the size of G

2 2 Partition Function on torus 2 Kasteleyn weighting It is known that any finite planar graph has a Kasteleyn weighting However, a graph G on tours might not have a Kasteleyn weighting In this section, we will discuss how to construct a local Kasteleyn matrix on torus providing the number of dimer covers in G Let H n be a honeycomb lattice on a torus such that the number of black and the number of white vertices are both n 2 and the number of edges is 3n 2 The edge weights a, b, c are determined by direction: a the horizontal, b the NW-SE and c the NE-SW edges Let ˆx, ŷ be the direction indicated in the picture below which show H 3 Figure : Fundamental domain for the honeycomb graph on a torus Let K be the weighted adjacency matrix of H n defined as: a if an edge joining b i, w j in the horizontal direction b if an edge joining b K = i, w j in the direction NW-SE c if an edge joining b i, w j in the direction NE-SW 0 otherwise Let {T i,j } be translation operators shifting i and j units in x, y directions respectively As the spanning tree model, K commutes with translation operator {T i,j } i,j Zn Z n, so they are simultaneously diagonalizable Since (T,0 ) n = I, (T,0 ) n = I Simultaneous eigenfunctions of T,0 and T 0, are z, ω respectively such that z n = I, ω n = We have the following graph restricting in a small neighborhood: zx 2 w x b c a So, K ( x x 2 ) ( = c wx 2 x b x 2 z x 0 a + bz + cw a + b z + c w 0 ) ( x On this block, the correspoding eigenvalue for K : R B R W = R B is a + bz + cw Moreover, all eigenvalues of K belongs to {a + bz + cw : z n = w n = } Thus, det K = a + bz + cw z n =,w n = x 2 ) 2

3 22 The homology class of a dimer cover Let G = (V, E) be a connected planar graph embedded on a surface Let Λ 0 = Λ 0 (G) be the space of functions on V Let Λ be the space of -form on G that is Define a linear operator d : Λ 0 Λ by Λ = {ω : E R ω(e) = ω( e)} dg(v v 2 ) = g(v 2 ) g(v ) A dimer cover M on a graph G provides a function ω M : E {0, } defined by { if e M ω M (e) = 0 otherwise Define [M] = e E ω M(e)[e] as a - chain where [e] is the edge e oriented from black to white Since M is a dimer covers d[m] = v V sgn(v)[v] where { if v is black sgn(v) = if v is white If M, M 2 are two dimer covers, [M ] [M 2 ] is a -cycle so defines a homology class [M ] [M 2 ] We fix a dimer covering M 0 such that M 0 is composed of all horizontal edges in the graph H n Thus, [M] [M 0 ] is an element of the homology class Figure 2: dimer covering in homology class (4, 2) 23 Partition function Definition 2 Define the function P n m (z, w) = a + bζ + cξ ζ n =z,ξ m =w 3

4 Example 22 P 2 2 (z, w) = (a + b z + c w)(a b z + c w)(a + b z c w)(a b z c w) Remark 23 = a 4 + b 4 z 2 + c 4 w 2 2a 2 b 2 z 2a 2 c 2 w 2b 2 c 2 zw P n n (, ) = det K n n P n n (z, w) = ζ n =z,ξ n =w P (ζ, ξ) Let B and W be the set of black and white vertices respectively in H n Then for each n, P n n (z, w) = det K B W = det K n 2 n 2, Theorem 24 P n n (z, w) = c ij z i w j i,j where is the triangle with vertices (0, 0), (0, n), (n, 0) and c ij = ( ) ij weights of dimer configurations with homology class (i, j), Idea of Proof By the remark above, P n n (z, w) = det K N N where N = n 2 Recall that det K N N = σ S N sgn(σ)k(b, w σ() ) K(b N, b σ(n) ) Similar as the rectangle case, each nonzero term corresponds to a dimer covering The modulus of this term is the product of edge weights We are left to check the signs Let M be a dimer covering of H n For each edge in M with weight a, b, c, we can draw a corresponding 60 rhombi respectively as the following example: This provides a lozenge tiling 4

5 Figure 3: A dimer covering on the honeycomb graph and the corresponding lozenges tiling Figure 4: A hexagon flip Figure 5: A hexagon flip Claim: If two dimer coverings have the same homology class, then all the corresponding terms in the expansions of P n n (z, w) have the same sign For a honeycomb graph, if two dimer coverings differ only on a single face, then they can be obtained from one another by a hexagon flip as following: The hexagon flip changes σ by a 3-cycle which is an even permutation If the homology class is not one of the cases: (k, 0), (0, k) or (k, n k), then two dimer coverings with same homology classes can be obtained from one another by a sequence of hexagon flips This corresponds to add or subtract a cube from the stepped surface in the lozenge tiling picture Note that the homology class is determined by the number of a, b, c s A dimer covering in homology class (i, j) has weight a (n2 ni nj) b ni c nj Thus, the distribution of homology class is a function of the weights a, b, c Corollary 25 When n is odd, the partition function Z is Z = (P (, ) + P (, ) + P (, ) P (, )) 2 5

6 3 Free energy Consider the free energy F = lim n n log Z 2 Theorem 3 F is given by the following: lim n n log Z = log P 2 (2πi) 2 (z, w) dz S S z = (2πi) 2 where Log denotes the principal branch dw w Log(a + bz + cw) dz S S z Proof Idea: log P (z, w) is not a continuous function in general It has two singularities and the Riemann sums log P n 2 n n (±, ±) might be very small near either one of the singularities However, four terms cannot vanish at the same time Since each term converges to the integral and max Z α0 α Z 2 max Z α0 α α 0,α α 0,α so n 2 log Z converges to the integral as well dw w Lemma 32 log(a + bz) dz { log(a) if a > b 2πi S z = log(b) if b > a By the above lemma, F = (2πi) 2 = 2πi a+cw > b Log(a + bz + cw) dz S S z log(a + cw) dw w + 2πi dw w Therefore, F is a piecewise linear function of a, b, c given by log(a) if a > b + c F = log(b) if b > a + c log(c) if c > a + b a+cw < b log(b) dw w Consider the parameter space of a, b, c We are left to compute F inside the inner triangle b > a + c c > a + b a > b + c 6

7 Suppose that a, b, c satisfy the triangle inequality Let θ a, θ b, θ c be the angles opposite sides a, b, c in a triangle in R 2 b θ a c θ b a The probability of an edge of type a in a dimer covering of the honeycomb graph is θ c Prob(edge of type a) = a a F a dz dw = (2πi) 2 S S a + bz + cw z w a dw = [ Res w=0 ( a+bz < a + bz + cw w ) + Res a w= ( a bz c a + bz + cw c a dw + Res w=0 ( a+bz > a + bz + w w )dz z c = a dz 2πi a + bz z a+bz >c = 2πi log( bz (π θ c) a + bz ) ei e i(π θc) = θ a π Apply the same argument for the probability of an edge of type b and c in a dimer cover which gives dw w )]dz z Prob(edge of type b) = θ b b, Prob(edge of type c) = θ c c To compute F of a honeycomb graph Let s first think about a -dimensional example Consider a path on integers of length N Let a, b be the weight for stepping left and right respectively Let p, q be the probability for left and right steps respectively p = a a + b, q = b a + b Let m be the number of left steps in the path Then the total weight of the path is ( N m) a m b N m For a typical path, m = pn + o(n), N m = qn + o(n) and Z = (a + b) N Then F = N log Z = ( ) N N log + p log a + q log b pn = p log p ( p) log( p) 7

8 For the honeycomb graph F = π (L(θ a) + L(θ b ) + L(θ c )) + θ a }{{} π log a + θ b π log b + θ c π log c }{{} entropy mean energy where L is the Lobachevsky function given by θ L(θ) = log 2 sin tdt 0 which is a 2-dimensional generalization of log 8