Statistical Mechanics and Combinatorics : Lecture II

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1 Statistical Mechanics and Combinatorics : Lecture II Kircho s matrix-tree theorem Deinition (Spanning tree) A subgraph T o a graph G(V, E) is a spanning tree i it is a tree that contains every vertex in V We deine a map rom unctions on V to unctions on directed edges by d(xy) = (y) (x) This map is a -orm, that is, satisies d( e) = d(e) where e represents the edge with the reversed orientation Once we choose a preerred direction or each edge, we can think o d as a unction on (undirected) edges and thus represent d as a m n matrix Deinition 2 (Incidence matrix) The incidence matrix o a directed graph G is a m n matrix (d ij ), where n = V and m = E, such that i the edge e j enters vertex v i d ij = i the edge e j leaves vertex v j 0 otherwise Example 3 Let G be a directed graph as ollowing The incidence matrix d o G is d = 2 3 a 0 b 0 c 0 Let E e be an energy on each edge and C : E R 0 (the conductance) be deined by c e = e βee

2 Deinition 4 (The Laplacian) Assign a positive real weight (a conductance) c e to each edge The Laplacian on R V is the linear operator : R V R V deined by (v) = w v c e ((v) (w)) Lemma 5 The Laplacian matrix satisies where C is the set o conductances Example 6 For the example above, = d T Cd = 2 3 b + c c b 2 c a + c a 3 b a a + b Theorem 7 (Kircho 847) Let c e, be deined as above Let be the matrix obtained rom by removing row and column i rom Denote the determinant o by det Then we have the equation c e ) = det Idea o Proo: sptree T ( e T Write A = d T and B = Cd, so by lemma 5, = AB Then = A B where A is the matrix given by deleting a row rom A and B is the matrix by deleting a column rom B 2 Recall Cauchy-Binet ormula: Let [m] = {,, m} I A is an n m matrix, B is an m n matrix, then det(ab) = det(a S ) det(b S ) S [m], S =n where A S is the n n matrix whose columns are the columns o A indices rom S and B S is the n n matrix whose rows are the rows o B at indices S So, det(a B ) = det(a S ) det(b S ) S [m ], S =n 3 I S is disconnected, then B S is singular since it contains a row o zeros So nonsingular matrix corresponds to spanning trees 4 I S is a spanning tree, then det d S = ± We have 5 det ( ) = S [m ], S =n det B S = det(c S d S ) = ± e T det(a S ) det(b S ) = c e sptreet e T c e 2

3 2 Cayley s Theorem Theorem 2 (Cayley s theorem) Let k n be the directed complete graph on n vertices Let κ(k n ) be the number o spanning trees o k n without weights κ(k n ) = n n 2 Proo We apply the Matrix-Tree theorem so that κ(k n ) = det The Laplacian matrix n n = = ni M n where M is the n n matrix o s Since M has rank so M has eigenvalues n, 0,, 0 with 0 having multiplicity n This implies that has eigenvalues 0, n,, n with n having multiplicity n It means that det = n n 2 Thereore, 3 Spanning Tree Model κ(k n ) = n n 2 Let the coniguration space be the set o all spanning trees o graph G(V, E) Let Z be the partition unction deined as Z = det = c e sptreet e T The Boltzmann measure µ o a spanning tree T is µ(t ) = c e Z The probability o an edge e in a random spanning tree T is Write P r(e) = c e Z c e Z e T Z = Z 0 + Z c e = c ln Z e c e where Z 0 are the spanning trees without edge c e and Z c e are the weighted trees containing edge c e Let x, y be the vertices contained in edge e Write ab or the matrix obtained rom by deleting column a and row b with a, b {x, y} Then, P r(e) = c e d dc e det det = c e det( xx) + det( yy) det( xy) det( yx) det = c e (G xx + G yy G xy G yx ), 3

4 where G ab = ( ) ab or a, b {x, y} What about the probability o a subset o edges in uniorm spanning trees? Theorem 3 (Transer-Impedance Theorem) Let G(V, E) be any inite connected graph Let T be a weighted uniorm random spanning tree o G For any e,, e k G, Pr(e,, e k T ) = det(t ei e j ) k C ei where T : R E R E is the transer impedance matrix deined by Note that T 2 = T T = Cd ( ) d i= How do we take the limit or larger and larger determinant? Example 32 Let the graph G be a n n grid in Z 2, that is, Z 2 /nz 2 with conductance C e =, e E Since commutes with all translates with {T x,y } x,y Zn Z n, and T x,y are simultaneously diagonalizable (T,0 ) n = I The eigenvalues o T,0 are nth root o unity z = e 2πi n The eigenspace o z is c b a az n az Similarly, (T 0, ) n = I The eigenvalues o T 0, are nth root o unity ω = e 2πj n The eigenspace o ω is eω eω e g 4

5 Intersection the eigenspace o z with the eigenspace o ω Denote the -dimensional subspace by W aω aω a az The Laplacian : W (z, ω) W (z, ω) is given by az az = (4 z z ω ω ) aω a aω az The eigenvalues o on W (z, ω) are 4 z ω where z, ω ranges over the nth z ω root o unity The Laplacian determinant is det = z n =,ω n = The number o spanning tress in (Z n ) 2 is The ree energy F is det = F = lim n n log Z 2 n = z n =,ω n = (z,ω) (,) 4 z z ω ω (4 z z ω ω ) log(4 z S S z ω ω ) dz dω 2πiz 2πiω Example 33 Let, g be unctions on V The connections φ vv : V v V v ollowing graph are given by the b b b 2 b 2 a a 2 a (T 2,0 ) n 2 = I The eigenvalues o T 2,0 are n 4πj th root o unity z = e n 2 (T 0, ) n = I The eigenvalues o T 0, are nth root o unity ω = e 2πi n 5

6 Let W (z, ω) be the intersection o eigenspaces o z and ω respectively ω ω g z The Laplacian : W (z, ω) W (z, ω) is given by ( ) = g ( (a + a 2 + 2b ) ga gza 2 b ω b ω (a + a 2 + 2b 2 )g a a 2 ω b 2gω b 2 ω ) Let P (z, w) = det( W (z,ω) ) Then det Z 2 /nz 2 = z n 2 =,ω n = (z,ω) (,) P (z, ω)tr( W (,) ) Theorem 34 (Pemantle 90) For any sequence o graphs in Z 2 converging to Z 2, the spanning tree measure converges to a measure independent o approximation sequence 4 Spanning tress and random walks The problem we discuss in this section is that o sampling a uniorm spanning tree T Loop-erased random walk Let G be a graph and γ is some inite path o length n on G The ith vertex visited by γ is denoted by γ(i) beginning at γ(0) The loop erasure LE(γ) o γ is a new simple path obtained by erasing all the loops o γ in chronological order Intuitively, i γ is a sel-avioding path (ie the vertices γ(i) are distinct), then LE(γ) = γ I γ(i) = γ(j), i < j and j is minimal satisying the equation, delete all the vertices γ(k) with i < k j rom the sequence {γ(k)} I the step until the result is sel-avoiding Theorem 4 (Pemantle) For any a, b V in a uniorm spanning tree, the unique path connecting a to b is distributed as a LERW rom a to b David Wilson Algorithm Wilson provides an algorithm or sampling a uniorm spanning tree: Do LERW rom a vertex a to a root r 6

7 2 Pick any other vertex b in G and do LERW until hitting the already constructed path 3 Continue until the tree spans all the vertices Theorem 42 Wilson s algorithm generates a uniorm spanning tree Proo Associate with vertex r an empty stack and associate with each vertex u r an ininite stack S u The ininite stack is constructed by choosing a neighbor o u uniormly at random rom all neighbors, pushing it onto the stack and repeat ininitely All the items in all the stacks are mutually independent I S u,i is the ith element in the stack, we have P r(s u,i = v) = P r(randomsuccessor(u) = v) The tops o the stacks deine a directed graph G S = (V, E S ) where E S = {(u, top(u) : u r)} I there is a directed cycle in G S, we pop the top item o the stack o each vertex in the cycle by replacing the current stack {S u,i } with {S u,i+ } Apply the cycle popping algorithm while G has a cycle Choose a cycle C in G S at random and pop it return tree let on stacks I this process terminates, the resulting tree T will be a directed spanning tree with root r By the act that the order o popped-cycle is irrelevant, T is uniquely determined by the stacks Wilson s algorithm is an implementation o the cycle popping algorithm Consider the Prob(T ) Let C be the set o cycles Since the cycle weights ω(c) is independent o T, we have P rob(t ) = ω(c)ω(t ) = c deg(v) v V or a ixed constant c 7