Statistical Mechanics and Combinatorics : Lecture III
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1 Statstcal Mechancs and Combnatorcs : Lecture III Dmer Model Dmer defntons Defnton A dmer coverng (perfect matchng) of a fnte graph s a set of edges whch covers every vertex exactly once, e every vertex s the endpont of exactly one edge Defnton 2 A graph s bpartte f the vertces can be colored black and whte and each edge connects vertces of dfferent colors Let G be a fnte bpartte planar graph such that each edge has weght ω e = e βee Let M be the set of all dmer coverngs of G For a dmer coverng m M, we the assocated probablty measure µ on dmer covers s defned by e m µ(m) = ω e Z where the partton functon Z s gven by 2 Kasteleyn matrx Z = m M ω(e) Two weght functons ω, ω are equvalent, ω ω f one can be obtaned from the other by multplyng edge weghts of all edges ncdent to a sngle vertex v by a constant λ For example, two weght functons ω and ω gven by the followng fgure are equvalent e m a b λa v v c λc λb When we multply λ to edge weghts of all ncdent edges to v, Z s also multpled by λ, and the measure µ does not change Thus, ω ω µ ω = µ ω
2 Q: How do we tell that ω ω? The answer s gven by the alternatng products: Take an even-length cycle wth edges e, e 2,, e 2k n cyclc order ω ω ω(e ) ω(e 2k ) ω(e 2 ) ω(e 2k ) = ω (e ) ω (e 2k ) ω (e 2 ) ω (e 2k ) Q: How to compute the partton functon Z? Kasteleyn(65) show that Z can be computed as the Pfaffan of a Kasteleyn matrx for G planar Defnton 2 (Pfaffan) Let M = (M j ) be a 2n 2n skew-symmetrc matrx The Pfaffan of M s defned by pf(m) = n sgn(σ) M σ(2 ),σ(2) σ S 2n = Example 22 M = 0 a b c a 0 d e b d 0 f c e f 0 pf(m) = M 2 M 34 M 3 M 24 + M 4 M 23 = af be + cd Defnton 23 A Kasteleyn weghtng of a planar bpartte graph s an assgnment of sgns to edges so that each face wth 0 mod 4 edges has an odd number of sgns and each face wth 2 mod 4 edges has an even number of sgns For example, quadrlaterals have odd number of sgns, hexagons have even number of sgns and octagons have odd number of sgns Defnton 24 A Kasteleyn matrx K s a weghted, sgned adjacency matrx from black to whte vertces of G Let B, W be the set of black and whte vertces respectvely Let ν be weght functons Defne K as a B W matrx such that { ±νb w K(b, w j ) = j f b w j 0 otherwse where the sgns are gven by the Kasteleyn weghtng Example 25 The Kasteleyn matrx K for the followng graph wth Kasteleyn weghtng w b 2 a b c w 2 w 3 b b 3 2
3 s K = w w 2 w 3 b a 0 b 2 b b 3 0 c Lemma 26 If γ s a smple cycle of length 2k enclosng l ponts, the alternatng product of sgns around the cycle s ( ) +k+l Proof Idea: We apply nducton on the number of edges of γ or the number of vertces enclosed n the graph G startng from a spannng tree Theorem 27 For a bpartte planar graph G, Z = det K Proof If K s not square or det K = 0, there are no dmer coverngs If K s a n n matrx, det K = σ S n sgn(σ)k(b, w σ() )K(b 2, w σ(2) ) K(b n, w σ(n) ) Each nonzero term corresponds a dmer coverng The modulus of ths term s the product of ts edge weghts We are left to check sgns Gven two dmer coverngs, draw them on G smultaneously whch form a set of loops To convert one dmer coverng nto the other, we can take a loop and move dmer from the frst coverng cyclcally around by one edge so that the resultng dmers match the dmers from the second coverng The number of the sgnature changed n the operaton s equal to the number of sgn changng from the Kasteleyn weghtng Snce the operaton on a loop wth length 2k changes the permutaton σ by a k-cycle Accordng to lemma 26, the sgn change of the edge weghts dependng on 2k s 2 or 0 mod 4 whch agree wth the Kasteleyn weghtng Thus, two dmer coverngs have the same sgn Remark 28 Any two dmer coverngs can be obtaned from one another by a sequence of hexagon flps Fgure : A hexagon flp If G s a planar but not necessarly bpartte, then G has a Kasteleyn orentaton, e an orentaton of the edges so that every face has an odd number of clockwse orented edges Let K = (K j ) be a weghted adjacency matrx wth entres of G ν j f the edge jonng v to v j orented from v to v j K j = ν j f the edge jonng v to v j orented from v j to v 0 otherwse 3
4 where ν j s the edge weghts Theorem 29 For a planar graph G, Z = pf(k) = det K 3 Partton functon Let G be an m n grd such that mn s even One can use complex numbers of modulus for Kasteleyn sgns under the condton that each face of length l, the alternatng product where z, =,, 4 s gven by z z 2 z 3 z 4 = ( ) l/2+ z 3 z 4 z 2 z Example 3 The followng pcture gves an example of Kast-Perc sgnng of the grd Let s consder -dmensonal lne segment 2 3 n We compute the egenvalue for the matrx K = Let z = e πj/(n+) The functon f j (x) = z x z x = 2 sn( πjx n + ) s an egenvector of K The correspondng egenvalue λ j can be computed by the equaton λ j f j (x) = f j (x + ) + f j (x ), j {,, n} 4
5 so that λ = 2 cos( πj ) Thus, n+ Z = det K /2 = ( n j= 2 cos( πj n + ))/2 For the m n grd G (mn even), ( the correspondng ) Kasteleyn matrx K has sze mn/2 mn/2 Consder the matrx K 0 K = K t Let z = e 0 πj/(m+), w = e πk/(n+) The functon f j,k (x, y) = (z x z x )(w y w y ) = 4 sn( πjx πky ) sn( m + n + ) s an egenvector of K The correspondng egenvalue λ j,k can be computed by that s λ j,k f j,k (x, y) = f(x +, y) + f(x, y) + f(x, y + ) + f(x, y ), πj πk λ j,k = 2 cos( ) + 2 cos( m + m + ), for j {,, m}, k {,, n} Therefore, For large m, n we have Z m,n = det m = ( K /2 j= k= n πj 2 cos( m + Z = e mn( G π +o()) where G s the Catalan s constant G = ) + 2 cos( πk n + ))/2 If we wrte P (z, w) = z + z + w + w and G s the Catalan s constant Then G π = lm m,n mn log Z m,n = log P (z, w) dz 2π 2 S S z dw w 5
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