Perfect Matchings and Pfaffian Orientation

Jeanette Nguyen

Bachelor Thesis

Supervisor: dr. Dion Gijswijt

KdV Instituut voor wiskunde

Faculteit der Natuurwetenschappen, Wiskunde en Informatica

Universiteit van Amsterdam

AbstractA graph can be covered with tiles consisting of two vertices andone edge connecting the vertices. If it is possible to cover a graphby those tiles such that each site of the graph is occupied byexactly one tile, the obtained configuration is called a perfectmatching. Not all graphs have perfect matchings, but when aperfect matching exists on a graph, the question is how manysuch configurations does this graph have? For a certain class ofgraphs, a method for finding the number of perfect matchings isused involving so-called Pfaffians. Some graphs can be oriented ina way that, when representing the graph by its adjacency matrix,the Pfaffian of the matrix enumerates the perfect matchings ofthe graph. The orientation is called a Pfaffian orientation. Thephysicist Kasteleyn has shown that every planar graph has a Pfaf-fian orientation. The method of constructing such orientation isapplied to a rectangular lattice of dimension m× n and a similarapproach has been used to find the number of perfect matchingof a rectangular lattice with periodic boundary conditions. Cov-erings of lattice graphs have applications in the field of condensedmatter physics and statistical physics.

InformationTitle: Perfect Matchings and Pfaffian OrientationAuthor: Jeanette Nguyen, ttnguyen@science.uva.nl, 0321877Supervisor: dr. Dion GijswijtSecond Assesor: dr. Jasper StokmanEnd Date: May 15, 2008

Korteweg de Vries Instituut voor WiskundeUniversiteit van AmsterdamPlantage Muidergracht 24, 1018 TV Amsterdamhttp://www.science.uva.nl/math

Contents

1 Introduction 5

2 Graphs and Matchings 72.1 What is a graph? . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Matchings and Perfect Matchings . . . . . . . . . . . . . . . . 9

3 Pfaffians 133.1 The adjacency Matrix and the Pfaffian . . . . . . . . . . . . . 133.2 Pfaffian Orientation . . . . . . . . . . . . . . . . . . . . . . . . 17

4 The Rectangular Lattice 244.1 The Planar Rectangular Lattice . . . . . . . . . . . . . . . . . 244.2 The Rectangular Lattice on a Torus . . . . . . . . . . . . . . . 31

5 Discussion and Conclusion 37

2

3

Preface

My search for a subject for my bachelor thesis started at the Institute forTheoretical Physics, since I was very interested in physics that time and mywish was to study a mathematical subject with applications in the physicsworld. However, string theory and quantum field theory, maybe a morecommon or fashionable combination of mathematics and physics, did notappeal to me. Luckily, the coordinator for Mathematical Physics mentionedthe Dimer Problem and advised me to visit the department of StatisticalPhysics. Here, I came across the dimer problem and it turned out to bea graph theoretical problem. Since I haven’t taken any course on discretemathematics and graph theory I needed to start from the very beginning andmust become familiar with graphs and matchings.My supervisor provided me with the article of Kasteleyn on dimers andPfaffians. Under his guidance I studied the theory of perfect matchingsand Pfaffian orientations. And now, after a long time and the usual writer’sblock, the result is this thesis. Even more, my interest in graph theory hasbecome larger with time.I would like to express my gratitude to Marianne Hoogeveen and Timo Kluck,whom I forced to read (parts of) this thesis, their suggestions and help havebeen useful and sometimes even motivated me, when I just did not feel likewriting anymore. They assured me that this is usual behaviour for studentswho are writing reports.I would like to thank my supervisor, Dion Gijswijt, for his guidance andinstructions to improve my understanding in the theory and my style ofwriting. His enthusiastic and patient way of explaining has been most helpful.I do not think this will be the last time that I encounter Graph Theory.

Jeanette NguyenAmsterdam, May 15, 2008

4

Chapter 1

Introduction

The problem of enumerating perfect matchings has applications in bothphysics and chemistry. Perfect matchings in physics are known as dimerconfigurations and in chemistry the term Kekule structures is used. Thecomponents of matter, the molecules and atoms, are arranged on a crystal,a regular lattice. The arrangement of the particles on a lattice, in partic-ular the number of ways the particles can be arranged, is of great interestin physics, prompting physicists such as Kasteleyn to explore the theory ofgraphs and matchings.A dimer is a molecule consisting of two atoms linked by a bond. A dimercovering of a lattice is a collection of dimers that covers all the vertices ofthe lattice exactly once.The most occurring regular lattices are the rectangular lattice, the triangu-lar lattice and the hexagonal lattice, also known as the honeycomb lattice.Kasteleyn [4][5] and Fisher[2] have found the number of perfect matchingson a planar rectangular lattice, using tools from linear algebra such as deter-minants and Pfaffians of a matrix. They have also determined the behaviourof the number of perfect matchings when the dimensions of the lattice go toinfinity.In this thesis the results of Kasteleyn [4] of perfect matchings on some graphs,especially a rectangular lattice graph, are treated.First, in Chapter 2 the concept of a graph is introduced and examples ofsome kinds of graphs are given. Then matchings and perfect matchings of agraph are defined.In Chapter 3 the tools to enumerate perfect matchings in graphs will be ex-plored. A graph will be represented by its adjacency matrix and the Pfaffianof a matrix is defined. We find out that some graphs have a Pfaffian orienta-tion. When such an orientation is found, enumerating the perfect matchingscan be done by computing the determinant.

5

Chapter 4 treats perfect matchings on a planar rectangular lattice. Themethod of the previous chapter will be used to find an expression for thenumber of perfect matchings of an m× n lattice graph. Also, the behaviourof this expression will be examined when letting the dimension of the graphgo to infinity. Then periodic boundary conditions are introduced. It appearsthat a slightly different approach is required for counting the matchings.Finally, the triangular lattice and the hexagonal lattice are briefly discussedand the known results [3],[10] of these lattices are compared with the resultsof the rectangular lattice.

6

Chapter 2

Graphs and Matchings

To treat the theory of perfect matchings on various lattices, an introductioninto graph theory is required. In this chapter a definition of a graph isgiven and examples of various graphs are shown. Furthermore, matchingsand especially perfect matchings are defined. Hall’s Matching Theorem andTutte’s 1-factor Theorem are briefly discussed.

2.1 What is a graph?

Definition 2.1. A graph G consists of a set of vertices V and a set of edgesE connecting pairs of vertices.

The order of a graph, n(G), is the number of vertices of a graph G. The sizeof a graph, e(G), is the number of edges of a graph G.Two edges are adjacent if they have exactly one common vertex.

There are different types of graphs and they can be categorised according totheir properties. Here, some kinds of graphs are briefly treated.

Figure 2.1: The complete graph K5, a bipartite graph, a planar graph and atree.

7

Example 2.2. A complete graph Kn is a graph whose vertices are pairwiseconnected by an edge.

Example 2.3. A tree is a connected graph without cycles.

Definition 2.4. A graph G is bipartite if V (G) is the union of two disjointsets of G. Thus, a vertex belonging to one of the sets can only have edgesconnecting to vertices of the other set of vertices.

Definition 2.5. A graph G is planar if it can be drawn without crossingsof edges on a plane surface. Such drawing of a graph G is called a planarembedding of G. A plane graph is defined by a mapping of the planar graphfrom every vertex into a position on R2 and from every edge into a curve withtwo extreme points, coinciding with the end vertices. All curves are disjointexcept on common endpoints of the graph.

The edges of a plane graph divides the plane into regions called faces of theplane graph. A finite plane graph has one unbounded face, called the infiniteface.The complete graphs K3,3 and K5 are examples of non-planar graphs.For plane graphs the formula of Euler holds.

Theorem 2.6 (Euler). If G is a connected plane graph with e the numberof edges of G, n the number of vertices and f the number of faces, then thefollowing equation holds:

n− e + f = 2. (2.1)

Proof. For the most simple plane graph consisting of just one vertex theequation holds, since e = 0, n = 1 and f = 1 (the infinite face).

u

v u

Figure 2.2: A graph with a vertex with degree 1, a connected graph G withouta vertex of degree 1 and edge u, graph G without edge u.

Consider now a connected plane graph G, an edge u and use induction onu. When subtracting edge u from G a plane graph G′ is obtained with n′

vertices, e′ edges and f ′ faces. Two cases can be distinguished: Either usaturates a vertex v with degree 1 or u does not saturate a vertex with

8

degree 1. In the latter case, u is an edge contained in a cycle C.The first case when removing u and vertex v will give n′ − e′ + f ′ = n− 1−(e − 1) + f = n − e + f = 2, since both sides of edge u is the same face asillustrated in Figure 2.2.The second case when removing u yields n′− e′+f ′ = n− (e−1)+(f −1) =n− e + f = 2. Hence, Euler’s formula is proved to be true.

Figure 2.3: A lattice graph and a directed graph

Example 2.7. A lattice graph is a graph where the points of a lattice arethe vertices and the edges of the graph are the line segments of the lattice.

The previous graphs are all examples of graphs consisting of unordered pairsof vertices. However, an orientation can be adopted on a graph. This meansthat the edges of the graph have a direction assigned to them. A graph withan orientation is called a directed graph and consists of a set of vertices V ,and a set E of ordered pairs of these vertices, called edges.

2.2 Matchings and Perfect Matchings

Definition 2.8. A matching M in a graph G is a set of non-loop edges ofthe graph such that each vertex of the graph is contained in at most oneedge. The vertices incident to the edges in M are called saturated by M . Aperfect matching in a graph is a matching saturating all vertices.

Definition 2.9. A maximum matching is a matching of maximum sizeamong all matchings in the graph.A maximal matching is a matching that cannot be enlarged by adding anedge.

A perfect matching is a largest possible matching in a graph. While not allgraphs have a perfect matching, all graphs do have maximal matchings. It

9

1

2

3

4

5

6

1

2

3

4

5

6

Figure 2.4: A graph G and a perfect matching M of G.

is obvious that a graph G of odd order does not have a perfect matching.

An M -alternating path is a path that alternates between the edges in amatching M and edges not in M .An M -augmenting path is an M -alternating path that begins and ends witha vertex not saturated by matching M .

Figure 2.5: An M -alternating path and an M -augmenting path.

Recall the symmetric difference of two sets M4N . This can also be writtenas M4N = (M −N) ∪ (N −M). When applying the symmetric differenceto graphs the following definition is in order.

Definition 2.10. The symmetric difference G4H of the graphs G and His the subgraph of G ∪H whose edges are the edges of G ∪H appearing inexactly one of G and H.

Figure 2.6: For example, the symmetric difference of matchings M and N ,where the black lines are the edges of matching M and the green lines arethe edges belonging to matching N , consists of a path of length 3 and a cycleof length 6. The common edge e is not included in the symmetric difference.

Lemma 2.11. The components of the graph induced by the symmetric dif-ference of two matchings are either paths or even cycles.

10

Proof. Consider matchings M and N of a graph G with vertex set V andtheir symmetric difference M4N . For a vertex there are four cases: eitherthe vertex has no incident edge or an incident edge from the matching Mor N or both, since M and N are matchings. A vertex in V (M4N) has atmost two incident edges. Thus, every component of (V (M4N), M4N) arepaths and cycles with edges belonging alternating to M − N and N − M ,which means that the cycles are even.

The symmetric difference can be a useful tool for investigating matchings ina graph.

Example 2.12. A tree T (V, E) has at most one perfect matching.

Proof. Consider a tree with even order and assume the tree has two perfectmatchings M and N . Since perfect matchings M and N saturate all verticesof the graph, (V, M4N) consists of even cycles and isolated vertices. How-ever, a tree does not contain a cycle, therefore neither does (V, M4N). Thismeans that M4N = ∅, which means M = N .

Definition 2.13. A k-factor of a graph G is a spanning k-regular subgraphof G. A subgraph H of graph G is a spanning subgraph if V (H) = V (G).

For instance, a 1-factor of a graph is a subgraph consisting of disjoint edges.This is similar to a perfect matching, the only difference being, that a perfectmatching is a set of edges, whereas a 1-factor is a subgraph.An odd component of a graph G is a component of odd order. The numberof odd components of G is denoted by o(G).

Figure 2.7: The square is a subset of the graph G called U , the circles arethe odd components of G−U and the triangles are de the even componentsof G− U .

There is a way to check whether a graph contains a perfect matching or a1-factor. Suppose a graph G with vertex set V has a perfect matching andlet U ⊆ V . Having a perfect matching implies that every odd componentof G − U must have a matching edge connecting to U (Figure 2.7). On the

11

other side if this is the case, every vertex in U is connected to at most one ofthese edges. Therefore, G−U can have at most |U | odd components. Tutteproved that also the converse holds.

Theorem 2.14 (Tutte’s 1-factor Theorem). A graph G has a perfectmatching if and only if o(G− U) ≤ |U | for every U ⊆ V .

A proof of Tutte’s 1-factor Theorem is provided by Tutte[1947] and Lo-vasz[1975].Now consider a bipartite graph G with bipartition V = X∪Y . If a matchingsaturates X, then for every S ⊆ X, there must be at least |S| vertices in Ythat have a neighbour in S. This is necessary since the neighbours of thevertices in S must be chosen from Y . Define N(S) to be the set of verticesin Y that have a neighbour in S. Hall’s Matching Theorem can be formallystated as:

Theorem 2.15 (Hall’s Matching Theorem). A bipartite graph G withbipartition X, Y has a matching that saturates X if and only if |N(S)| ≥ |S|for all S ⊆ X.

X

Y

Figure 2.8: A bipartite graph with bipartition |X| = |Y | without a perfectmatching.

The proof for Hall’s Matching Theorem is omitted.When the sets of bipartition have the same size, |X| = |Y |, Hall’s MatchingTheorem is also known as the Marriage Theorem. Back in the old days whenconsidering a group of men and an equal number of women, one wanted toknow if there is a possibility to match the men with the women with regardto compatibility. If, for some k ∈ N, every man is compatible with exactlyk women, and every woman is compatible with k men, a perfect matchingmust exist.It was Frobenius[1917] who originally proved the Marriage Theorem.

12

Chapter 3

Pfaffians

One way to count the perfect matchings of a graph is to manually countall the possible configurations. However, this task is complicated if the sizeof the graph becomes larger. In fact, determining the number of perfectmatchings of a graph is an NP-hard problem [9]. Hence, there is probablyno efficient algorithm to compute the number of perfect matchings in generalgraphs.However, for planar graphs an efficient algorithm involving so-called Pfaf-fians can be used to enumerate the perfect matchings of the graph. Firstan orientation is adopted on the graph and the graph is represented by itsadjacency matrix. Then the Pfaffian of a matrix is introduced. A graph isdirected in such a way to insure that the Pfaffian of the adjacency matrixcounts the perfect matchings of the graph. If the latter is the case, the ori-entation is called Pfaffian. When a Pfaffian orientation is known, countingthe perfect matchings can be reduced to just computing the determinant ofthe adjacency matrix. Thus a polynomial algorithm is found to enumeratethe perfect matchings of a special class of graphs.

3.1 The adjacency Matrix and the Pfaffian

A graph G with vertex set V = {v1, · · · , vn} can be represented by a matrixin a way that the entry aij of the matrix is the number of edges in G withendpoints {vi, vj}. The matrix is called the adjacency matrix and is denotedby A(G). Note that the adjacency matrix is symmetric since aij = aji. Fora graph H without multiple edges between pairs of vertices the adjacencymatrix A(H) has the form:

A(H)ij =

{1, if vivj ∈ E(H)0, if vi and vj are not-adjacent

. (3.1)

13

1

2

3

4 5

6

Figure 3.1: A graph G with labelled vertices.

Example 3.1. Let graph G be the graph as illustrated in Figure 3.1. Usingthe labelling of the vertices the adjacency matrix has the form:

A(G) =

0 0 0 0 1 10 0 1 0 1 00 1 0 1 1 00 0 1 0 1 01 1 1 1 0 11 0 0 0 1 0

. (3.2)

A perfect matching will have exactly one entry 1 in each row and column.

Consider a graph G with vertex set V (G) = {v1, . . . , v2n}. Now assign toeach edge e a variable xe and let an edge have a direction, i.e. apply anorientation to the graph. Define the matrix B of the oriented graph G asfollows:

B(x) = (bij)2n×2n,

where bij =

xe, if e = (vi, vj)

−xe, if e = (vj, vi)0, if vi and vj are not adjacent

.(3.3)

Matrix B is antisymmetric:

B(x) =

0 b1,2 · · · b1,2n

−b1,2 0. . .

......

. . . . . . b2n−1,2n

−b1,2n . . . −b2n−1,2n 0

. (3.4)

From now on we will only consider graphs with an even number of vertices,since an odd number of vertices excludes the existence of a perfect matching.The vertices of the graph are labelled by 1, 2, · · · , 2n. A partition of the set

14

{1, 2, . . . , 2n} into unordered pairs has the form {{i1, j1}, · · · , {in, jn}}.Define Π to be the set of all those partitions. Thus, a perfect matching ofthe graph is an element of Π.The Pfaffian Pf(B) of matrix B(x) is defined by:

Pf(B) =∑π∈Π

sgn(π)∏e∈Π

be, (3.5)

where be = bij if e = {i, j} with i < j and the sign sgn(π) of a partitionπ = {{i1, j1}, . . . , {in, jn}} is defined by:

sgn(π) := sgn

(1 2 3 . . . 2ni1 j1 i2 . . . jn

) n∏k=1

σ(ik, jk), (3.6)

where σ(i, j) =

1, if i < j

−1, if i > j0 otherwise

.

The sign of partition π is well-defined. The order of two numbers forming apair is arbitrary, as is the order of the pairs. Hence, when interchanging twopartners, say i3j3 → j3i3, the sign of the permutation will flip but one matrixelement will also gain an opposite sign. Nett, the sign of the perfect matchingwill not change. Switching two blocks of partners will leave the sign of thepermutation unchanged since the only consequence is the interchange of twomatrix elements.

Define the sign of a perfect matching to be the sign corresponding to theterm in the Pfaffian.

Lemma 3.2. If B is an antisymmetric matrix, then det(B) = (Pf(B))2

Proof. An antisymmetric matrix B of dimension n satisfies B = −BT , whereBT is the transposed matrix of B. Taking the determinant of the matrix leadsto:

det(B) = det(BT ) = det(−B) = (−1)n det(B). (3.7)

If B is a matrix of odd order, it says det(B) = − det(B). Hence, the deter-minant vanishes.Consider the case where B is of order 2n:Use the definition of the determinant of a matrix:

det(B) ≡∑

π∈S2n

sgn(π)2n∏i=1

bi,π(i) · σ(i, π(i)), (3.8)

15

where bi,π(i) is the variable assigned to edge {i, j} and σ(i, j) =

1, if i < j

−1, if i > j0, otherwise

.

The Pfaffian squared has the form:

(Pf(B))2 =

(∑M∈Π

sgn(M)∏e∈M

be

)(∑N∈Π

sgn(N)∏f∈N

bf

)=

∑M,N∈Π

sgn(M) · sgn(N)∏e∈M

be

∏f∈N

bf .

(3.9)

Now, to prove the lemma one must show that a term in the Pfaffian squaredhas a one to one correspondence with a term in the determinant. The re-maining terms of the determinant will have to cancel each other.Consider the expression for the determinant of B. The permutations π ∈ S2n

can be decomposed into even and odd cycles. Suppose the decomposition ofa term π contains an odd cycle (i1i2 · · · i2k+1), where k ≥ 1. Consider the oddcycle in the decomposition containing the smallest element. For each suchterm there is another term in the determinant containing this cycle in op-posite direction (i2k+1 · · · i2i1) and the rest of the decomposition remains thesame, say π′. Then

∏2ni=1 bi,π(i) = −

∏2ni=1 bi,π′(i) and since π and π′ consists of

cycles of the same length sgn(π) = sgn(π′). Hence, all terms that contain atleast one odd cycle in the decomposition cancel each other in pairs.Now one must show that the remaining terms of det(B), the permutationswhose decomposition consists only of even cycles, have a one to one corre-spondence with the terms of (Pf(B))2. Each term of the Pfaffian squared hasthe form:

sgn(M) · sgn(N)∏e∈M

be

∏f∈N

bf , (3.10)

where M = {{i1j1} · · · {injn}} and N = {{i′1j′1} · · · {i′nj′n}}. Choose i1 to beany number in the set {1, 2, · · · , 2n}, now j1 is fixed since it must be thepartner of i1. Let this line be an element of matching M . The number j1

occurs also in N , so choose j1 = i′1. This time j′1 is fixed, it can be either i1again or another number and setting this number j′1 = i2. This procedurecan be carried on until j′r is reached which is equal to i1.An even cycle is obtained with edges alternating between matching M and Nas illustrated in Figure 3.2 So, M ∪N corresponds to evenly oriented cycles,these, in their turn, correspond to permutations with only even cycles.

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ji1 1

Figure 3.2: Even cycles where the black lines are the edges of matching Mand the green lines are the edges of N .

The sign of M is given by:

sgn(M) = sgn

(1 2 · · · 2ni1 j1 · · · jn

)︸ ︷︷ ︸

ρM

n∏k=1

σ(ik, jk), (3.11)

and the sign of N is:

sgn(N) = sgn

(1 2 · · · 2nj1 j′1 · · · j′n

)︸ ︷︷ ︸

ρN

n∏l=1

σ(jl, j′l). (3.12)

The claim is that the permutation π in the term of the determinant is equalto ρπ = ρNρ−1

M .

ρNρ−1M =

(1 2 · · · 2nj1 j′1 · · · j′n

)(i1 j1 · · · jn

1 2 · · · 2n

)=

(i1 j1 · · · jn

j1 j′1 · · · j′n

)= ρπ.

(3.13)Since sgn(ρM) = sgn(ρ−1

M ) and σ(i, j) · σ(j, j′) = σ(j, j′), because i → j → j′,is sgn(π) = sgn(M) · sgn(N). Thus, proving the lemma.

3.2 Pfaffian Orientation

The Pfaffian of a matrix can be used to count the number of perfect match-ings of a graph. Let G be a graph with 2n vertices and given an arbitraryorientation ~G of G. The adjacency matrix of G is defined as follows:

As(~G) = (aij)2n×2n,

where aij =

1, if (ui, uj) ∈ E(~G),

−1, if (uj, ui) ∈ E(~G),0, if otherwise.

(3.14)

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This matrix is antisymmetric. Now recall the definition of the Pfaffian ofa matrix. The terms of the Pfaffian of this adjacency matrix are equal to−1, 1 or 0. Note that the nonzero terms of the Pfaffian have a one to onecorrespondence with the perfect matchings of the graph. To insure the Pfaf-fian of the matrix enumerates all perfect matchings with the right sign, anorientation must be adopted to the graph such that the corresponding termsof the Pfaffian are counted with the same sign. When this is the case, theorientation is called a Pfaffian orientation.The question now is: when does a graph have a Pfaffian orientation and howdoes one construct such orientation? The remainder of this section is dedi-cated to Kasteleyn’s result that every planar graph has a Pfaffian orientation.

Consider a directed graph G and let C be any even undirected cycle in G.For any given routing around C, if C has an even number of edges followingthe routing then C has also an even number of edges with an orientationopposite of the routing. Now the following can be defined:

Definition 3.3. If C is an even undirected cycle in G, then C is evenlyoriented if it has an even number of edges following the routing, otherwiseC is oddly oriented.

For a perfect matching M of a graph G the sign of M , sgn(M) is defined as:

sgn(M) := sgn

(1 2 . . . 2ni1 j1 . . . jn

) n∏k=1

σ(ik, jk), (3.15)

for M = {{i1, j1}, . . . , {in, jn}} where σ(i, j) =

{1, if i → j

−1, if j → i.

Lemma 3.4. Let ~G be an arbitrary orientation of a graph G and let Mand N be any two perfect matchings of G. Let k denote the number of evenlyoriented alternating cycles formed in M∪N . Then sgn(M)·sgn(N) = (−1)k.

Proof. First we note that if the lemma holds for an orientation, then thelemma holds for all orientations. To see this consider an edge e. This edge ecan be situated in:1. e 6∈ (M ∪N)2. e ∈ (M ∩N)3. e ∈ E(M)⊕ E(N), e is an edge in an alternating cycle.When reversing the direction of the edge e the following will occur:In the case of (1), e is not part of either M nor of N . There is no effect onthe equation of the lemma.

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For an edge e to be in both M and N , the sign of each matching will change,but the value of k remains the same.If, as in case (3), e is a component of an alternating cycle, then the sign ofmatching M will change, when reversing the direction of e, but k changes byone.Since an orientation can be chosen at will, orient the graph as follows: fora line e in the case of (1) and (2): orient the line arbitrarily. If the line isa component of an alternating cycle, orient e such that the cycle becomesdirected, evenly or oddly either way.Relabelling the vertices of G affects its adjacency matrix A(~G) such thatthe rows and the columns of the matrix are permuted by a permutation π.In the expression of the Pfaffian of the matrix, the perfect matchings aremultiplied by π. Therefore, this lemma remains invariant under relabellingof the vertices of the graph G.Now label the vertices as follows:For an edge e ∈ (M ∩N) label such that the head of the line equals the tailplus one. For an edge e ∈ E(M)⊕E(N), choose any line belonging to M andlabel it with the next unused consecutive numbers with its head equals itstail plus one. Continue this process until all alternating cycles are labelled.Now, the term of the Pfaffian corresponding to the perfect matching Mcontains the sign of the identity permutation, which is +1 and each termamn in the product is also positive. The Pfaffian of the perfect matching Ncontains only positive amn, due to the chosen labelling. Therefore, the signof N is the sign of the corresponding permutation ρ.Since each alternating even cycle of G correspond to an even cycle of thepermutation ρ and the cycles are disjoint, it follows that sgn(N) = (−1)k,thus proving the lemma.

Definition 3.5. A cycle C in a graph G is nice if the subgraph G − Ccontains a perfect matching.

Theorem 3.6. Let G be an even graph and ~G an orientation of G. Thenthe following are equivalent:1. ~G is a Pfaffian orientation.2. Every perfect matching of G has the same sign relative to ~G.3. Every nice cycle in G is oddly oriented relative to ~G.4. If G has a perfect matching, then for some perfect matching M , everyM-alternating cycle is oddly oriented relative to ~G.

Proof. By definition statement (1) implies (2).

To see that statement (2) implies (3) assume that ~G is a Pfaffian orientationand let C be a nice cycle of G. Let F be a perfect matching of G − C and

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C1 and C2 perfect matchings of C. Define two perfect matchings M and Nof G as follows: M := F ∪ C1 and N := F ∪ C2. By statement (2) M andN have the same sign and C is a M, N -alternating cycle in M ∪N . Thus byLemma 3.4 C is oddly oriented.Now assume that every nice cycle in G is oddly oriented as is stated in (3)and let M be a perfect matching of G. Let C be a M -alternating cycle. ThenM − C is a perfect matching of G− C, thus arriving at statement (4): C isnice and is oddly oriented.Now let M be a perfect matching of G and assume that every M -alternatingcycle is oddly oriented. Let N be another perfect matching of G. EveryM, N -alternating cycle is oddly oriented. Hence, by Lemma 3.4 M and Nhave the same sign. Therefore, ~G is Pfaffian.

Figure 3.3: The complete bipartite graph K3,3.

Example 3.7. Consider the complete graph K3,3 illustrated in Figure 3.2.As can be seen the graph has 6 perfect matchings. However, the graph doesnot have a Pfaffian orientation. To see this start with an orientation of K3,3

as illustrated in Figure 3.7: all lines are directed downwards.

1 2 3

4 5 6

Figure 3.4: An orientation of the complete bipartite graph K3,3

All nice cycles consist of 4 or 6 edges. With the chosen direction the cyclesare evenly oriented. For the cycles to be oddly oriented the direction of anodd number of edges of each cycle must be reversed.There are 9 cycles of length 4. Each edge of K3,3 appears 4 times in these 9cycles. If F ⊆ E, where E is the set of edges of K3,3, are the reversed edges,

20

then 4 · |F | = 9 (mod2) to have an odd number of clockwise oriented edges.This is not possible.

From now on when considering a planar graph the chosen routing will beclockwise.

Lemma 3.8. If ~G is a connected plane directed graph such that every bound-ary face, except for the infinite face, has an odd number of edges orientedclockwise, then in every cycle the number of edges oriented clockwise is ofopposite parity to the number of points of ~G inside the cycle. Therefore, ~Gis Pfaffian.

Proof. Let C be any cycle in ~G and define the following:f is the number of faces inside C, f1, f2, . . . are the faces of Cci is the number of edges on the boundary of face fi oriented clockwise.e is the number of edges inside Cv is the number of vertices inside Ck is the number of edges on C, this is equal to the number of vertices on C.c is the number of edges on C oriented clockwise.Now we want to show that c is of opposite parity to v. Let G′ be the graphobtained by deleting all vertices in the exterior of C. Recall Euler’s formulafor a connected plane graph:

number of vertices− number of edges + number of face = 2. (3.16)

That means that when applying Euler’s formula to G′:

(v + k)− (e + k) + (f + 1) = 2,

v − e + f = 1.(3.17)

If ci is the number of edges oriented clockwise on the boundary of face fi, thenit is by hypothesis odd, since faces do not have vertices inside its interior.Thus, ci ≡ 1(mod2) and f ≡

∑fi=1 ci(mod2). Because any edge inside of C

is counted as clockwise exactly once∑f

i=1 ci = c + e. Implementing Euler’sformula leads to:

f = c + e = c− 1 + f + v (mod2),

c− 1 + v = 0 (mod2).(3.18)

This means that c + v = 1 (mod2), thus proving the lemma.In particular, if C is a M ,N alternating with M and N two perfect matchings

21

of G, then the number of vertices inside C must be even, otherwise it willcontradict planarity, therefore, the number of edges on C oriented clockwiseis odd. Thus, ~G is Pfaffian.

Theorem 3.9 (Kasteleyn). Every planar graph G has a Pfaffian orienta-tion.

Proof. We may assume that G is connected. If G is a tree, then any orien-tation on this graph is Pfaffian.Let e be an edge in a cycle C in the border of the infinite face. The graphG − e has a Pfaffian orientation by induction. Adding the edge e to G − eand orient it in a way that the face had an odd number of clockwise orientededges, results in a Pfaffian orientation of G by Lemma 3.8. Every planargraph has a Pfaffian orientation by induction.

Lemma 3.8 can be used to construct a Pfaffian orientation on a planar graph:Consider a planar graph G and remove a line e belonging to a cycle borderingthe infinite face.

Figure 3.5: To construct a Pfaffian orientation on a graph G, first remove aline bordering the infinite face.

Now apply an orientation to the remaining lines in a way such that all bound-ary faces have an odd number of clockwise oriented edges. Now reinstall linee and give it an orientation such that the faces remain oddly clockwise ori-ented.The graph G now has a Pfaffian orientation.When labelling the vertices as illustrated in Figure 3.6, the adjacency matrix

22

1

3

4

5

7

2

8

6

Figure 3.6: The orientation of the graph is Pfaffian.

corresponding to this directed graph is given by:

A(G) =

0 −1 0 1 −1 0 0 01 0 −1 0 0 −1 0 00 1 0 1 0 0 1 0−1 0 −1 0 0 0 0 −11 0 0 0 0 1 0 −10 1 0 0 −1 0 −1 00 0 −1 0 0 1 0 10 0 0 1 1 0 −1 0

. (3.19)

The determinant of this matrix is 81 and using Lemma 3.2, this implies thatthe number of perfect matchings is 9. Of course, the graph is of such formthat it can be easily checked that the number of perfect matchings is indeed9 by constructing the configurations.

23

Chapter 4

The Rectangular Lattice

An application of matchings in physics is the dimer problem. A dimer is amolecule consisting of two atoms linked by a bond. The dimers are arrangedon lattices and one wants to find the number of possibilities of the dimer cov-erings on the lattice. In this chapter a rectangular lattice graph of dimensionm × n is treated. Using the method of the previous chapter, an expressionis derived for the number of configurations of the dimers on the lattice, i.e.the number of perfect matchings of the lattice graph.By winding the lattice on a torus, the lattice does not have a Pfaffian orien-tation anymore. However, counting the perfect matchings on a lattice graphon a torus does involve Pfaffians. It appears that there is a relation betweenthe genus of the surface on which the lattice graph is embedded and thenumber of Pfaffians needed to enumerate the perfect matchings of the graph.

4.1 The Planar Rectangular Lattice

Consider a rectangular lattice graph L(m, n) of dimension m × n, where mand n are even. A Pfaffian orientation on the graph can be constructed asillustrated in Figure 4.1.

Considering only the horizontal bonds each row of the graph can be repre-sented by an m×m matrix A:

A =

0 1 0 · · · 0

−1 0 1. . .

...

0 −1. . . . . . 0

.... . . . . . . . .

...0 · · · 0 −1 0

. (4.1)

24

Figure 4.1: A rectangular lattice graph with Pfaffian orientation.

Label the vertices of the first row by 1, 2, . . . ,m and the vertices of the nextrow by m + 1, m + 2, . . . , 2m and proceed in this way until all vertices arelabelled. The adjacency matrix As(~G) belonging to the orientation is:

As(~G) =

A I 0 · · · 0

−I −A I. . .

...

0 −I A. . . 0

.... . . . . . . . . I

0 · · · 0 −I (−1)n−1A

, (4.2)

where I is the identity matrix representing the vertical bonds connectingeach row.When multiplying the first row of the adjacency matrix by −1, then the thirdand fourth row, the fourth and fifth column, then the seventh and eighth rowand so on, the absolute value of the matrix is not changed and the followingmatrix M is obtained:

M =

−A I 0 · · · 0

I −A I. . .

...

0 I −A. . . 0

.... . . . . . . . . I

0 · · · 0 I −A

. (4.3)

25

If B is the n× n matrix given by:

B =

0 1 0 · · · 0

1 0 1. . .

...

0 1. . . . . . 0

.... . . . . . . . . 1

0 · · · 0 1 0

, (4.4)

then M can be written as a sum of two tensor products:

M = −I ⊗ A + B ⊗ I. (4.5)

The number of perfect matchings of this lattice can be found by means ofthe obtained result det(M) = (Pf(M))2. The determinant of M can bedetermined by its eigenvalues. Let A have the eigenvalues λ1, . . . , λm witheigenvectors ~a1, . . . ,~am and let B have the eigenvalues µ1, . . . , µn with eigen-vectors ~b1, . . . ,~bn. Because

M(~bi ⊗ ~aj) = (−I ⊗ A + B ⊗ I)(~bi ⊗ ~aj)

= (−I ⊗ A)(~bi ⊗ ~aj) + (B ⊗ I)(~bi ⊗ ~aj)

= (−I~bi ⊗ A~aj) + (B~bi ⊗ I~aj)

= −λj(~bi ⊗ ~aj) + µi(~bi ⊗ ~aj)

= (µi − λj)(~bi ⊗ ~aj)

(4.6)

matrix M has eigenvalues µi − λj with eigenvectors ~bi ⊗ ~aj.To find the eigenvalues of matrix B we have the characteristic polynomialpn(µ):

26

pn(µ) =

∣∣∣∣∣∣∣∣∣∣∣∣

−µ 1 0 . . . 0

1 −µ 1. . .

...

0 1. . . . . . 0

.... . . . . . . . . 1

0 . . . 0 1 −µ

∣∣∣∣∣∣∣∣∣∣∣∣︸ ︷︷ ︸n

= −µ

∣∣∣∣∣∣∣∣∣∣∣∣

−µ 1 0 . . . 0

1 −µ 1. . .

...

0 1. . . . . . 0

.... . . . . . . . . 1

0 . . . 0 1 −µ

∣∣∣∣∣∣∣∣∣∣∣∣︸ ︷︷ ︸n−1

−

∣∣∣∣∣∣∣∣∣∣∣∣

−µ 1 0 . . . 0

1 −µ 1. . .

...

0 1. . . . . . 0

.... . . . . . . . . 1

0 . . . 0 1 −µ

∣∣∣∣∣∣∣∣∣∣∣∣︸ ︷︷ ︸n−2

. (4.7)

So, pn(µ) = −µpn−1(µ)− pn−2. This is valid for n ≥ 1 when setting p0(µ) =−1 and p−1(µ) = 0. We look for a solution of pn of the form pn = sn. ThenS must satisfy:

sn = −µsn−1 − sn−2,

and hences2 + µs + 1 = 0.(4.8)

The roots of the obtained expression are s =−µ+

√µ2−4

2and t =

−µ−√

µ2−4

2,

then:

pn = c1sn+1 + c2t

n+1, for some c1, c2 ∈ C. (4.9)

Taking n = 0 and n = −1, we find for c1 and c2:

0 = c1 + c2,

1 = c1s + c2t.(4.10)

Thus,

c1 =1√

µ2 − 4, c2 =

−1√µ2 − 4

, (4.11)

27

and:

pn(µ) =1√

µ2 − 4

(sn+1 − tn+1

). (4.12)

Now, to find the eigenvalues µi set this equation equal to zero. Then

sn+1 = tn+1,

s = ε2t, with ε = ekπin+1 , 0 ≤ k ≤ n.

(4.13)

Solving s = ε2t for µ:

µ = ±(

ε +1

ε

)= ±2 cos

(kπ

n + 1

). (4.14)

Since − cos(

kn+1

)= cos

(n+1−k

n+1

), the minus sign can be omitted and hence,

the eigenvalues of B are 2 cos((πk)/(n + 1)) where k = 1, . . . , n.The eigenvalues of A can be found in a similar way and are 2i cos((πk)/(m+1)), where k = 1, . . . ,m. Therefore the eigenvalues of M are 2

(cos( πk

m+1)− i cos( πl

n+1))

with k = 1, . . . ,m and l = 1, . . . , n. The determinant of matrix M is theproduct of these numbers. The absolute value of the determinant is givenby:

| det M | = 2mn

m∏k=1

n∏l=1

(cos2

(πk

m + 1

)+ cos2

(πl

n + 1

)) 12

. (4.15)

Hence, the number of perfect matchings is:

nPM = PfM = (det M)12

= 2mn/2

m∏k=1

n∏l=1

(cos2

(πk

m + 1

)+ cos2

(πl

n + 1

)) 14

.(4.16)

Example 4.1. Consider a chessboard. Its dimensions are 8×8 and a dominotile covers 2 sites. Using the obtained equation(4.16) of the number of perfectmatchings, it is found that the domino tiles can cover the chess board in12,988,816 ways.

To determine the behaviour of expression (4.16) in the limit as m, n →∞ take the logarithm of the equation and transform the summations into

28

integrals [7][8]:

1

mnln(nPM) =

1

2ln 2+

1

4

1

mn

m∑k=1

n∑l=1

ln

(cos2

(πk

m + 1

)+ cos2

(πl

n + 1

))→ 1

2ln 2 +

1

4π2

∫ π

0

∫ π

0

ln

{cos2

(πk

m + 1

)+ cos2

(πl

n + 1

)}(4.17)

The integral can be written as:

1

4π2

∫ π

0

∫ π

0

ln{cos2 x + cos2 y

}dxdy

=1

4π2

∫ π

0

∫ π

0

ln

{1

2+

1

2+

1

2cos 2x +

1

2cos 2y

}dxdy.

(4.18)

To evaluate the integral consider the following:

F (X,Y ) =1

π2

∫ π

0

∫ π

0

ln {X + Y + X cos u + Y cos v} dudv. (4.19)

Taking the X-derivative of F leads to:

∂F

∂X=

1

π2

∫ π

0

∫ π

0

(1 + cos u)dudv

X + Y + X cos u + Y cos v. (4.20)

Evaluating the integral over v in the expression for ∂F∂X

gives:

∂F

∂X=

1

π

∫ π

0

(1 + cos u)du

{(X + Y + X cos u)2 − Y 2}12

. (4.21)

When substituting A = 1 + cos u and automatically dAdu

= − sin u, sin2 u =1− cos2 u = (1 + cos u)(1− cos u), 1− cos u = −A + 2, the expression can bewritten like:

∂F

∂X=

1

π

∫ 2

0

dA

{X(2− A)(XA + 2Y )}12

. (4.22)

Using the identity:

d

dxtan−1 p =

p′

1 + p2, (4.23)

29

the following can be found:

{X(2− A)(XA + 2Y )}−12 = − 2

X

d

dAtan−1

{X(2− A)

XA + 2Y

} 12

. (4.24)

With the expansion of the arctangent it follows that:

∂F

∂X=

2

πXtan−1

(X

Y

) 12

=2

πX

{(X

Y

) 12

− 1

3

(X

Y

) 32

+1

5

(X

Y

) 52

− . . .

}.

(4.25)

Integrating each term we obtain:

F (X, Y ) =1

2ln(X) +

4

π

{(X

Y

) 12

− 1

32

(X

Y

) 32

+1

52

(X

Y

) 52

− . . .

}. (4.26)

Now, returning to equation (4.17), evaluating F (X, Y ) where X = Y = 12,

the result is:

1

mnln(nPM) → 1

2ln 2 +

1

4

1

π2

∫ π

0

∫ π

0

ln

{cos2

(πk

m + 1

)+ cos2

(πl

n + 1

)}=

1

2ln 2 +

1

4F (

1

2,1

2)

=1

π

{1 +

1

32− 1

52+ . . .

}=

1

πC ≈ 0.29.

(4.27)

where C = (1− 132 + 1

52 − 172 + · · · ), is the Catalan constant.

Thus, the number of perfect matchings of a planar rectangular lattice isasymptotically:

nPM ∼ emnC

π . (4.28)

30

4.2 The Rectangular Lattice on a Torus

Consider now a rectangular lattice on a torus. One might expect to find thenumber of perfect matchings in the previous way of constructing a Pfaffianorientation. However, it appears that wrapped around a torus a rectangularlattice does not have a Pfaffian orientation.

Theorem 4.2. A rectangular lattice L with even dimensions on a torus doesnot have a Pfaffian orientation.

Proof. Consider a lattice with even dimensions on a torus.

Figure 4.2: A directed rectangular lattice graph of dimension 4×4 on a torus

If the same orientation is applied as on the planar rectangular lattice withthe extra directions of the boundaries to be given by:

(m, j; 1, j) = (−1)j−1,

(i, n; i, 1) = −1.(4.29)

where i, 1 ≤ i ≤ m, denotes the vertex in the horizontal direction and j,1 ≤ j ≤ n, the vertex in the vertical direction, the cycles formed by goingstraight in the horizontal and in the vertical direction are evenly orientednice cycles as can be seen in Figure 4.2. For an orientation to be Pfaffian,every nice cycle must have an odd orientation.

The chosen routing for the squares is clockwise and for cycles looping in thehorizontal direction the routing is to the left.Suppose the graph has a Pfaffian orientation, thus the vertical and the hor-izontal bonds are oddly oriented and the unit cells, given by the squaresurrounded by the dashed lines in Figure 4.3, as well. Consider now two unitcells bordering each other as illustrated in Figure 4.3. Define the following:H is the cycle looping in the horizontal direction.

31

e

c ca a

b

21

21b

H

P Q

Figure 4.3: Two bordering unit cells.

P and Q are the left and right unit cells.a1 is the number of oddly oriented edges of P ∩H.a2 is the number of oddly oriented edges of Q ∩H.e is the common edge of P and Q.b1 is the number of oddly oriented edges of P − a1 − e.b2 is the number of oddly oriented edges of Q− A2 − e.The number of oddly oriented edges of H is given by c + a1 + a2. The bonde is counted as oddly oriented for one cell and evenly oriented for the othercell. Hence, the cycle formed by these cells is of odd parity. Choose e to bedirected upwards. By hypothesis c + a1 + a2 ≡ 1 (mod2). We want to showthat the cycle denoted by the green line in Figure 4.4 is evenly oriented. So,c + b1 + b2 ≡ 0 (mod2).

e

c ca a

b

21

21b

Figure 4.4: The cycle denoted by green is an evenly oriented nice cycle.

Consider the cycle formed by c+ b2 +a1 + e. The parity of this cycle is equalto (1−a1)+b2 ≡ 1 ( mod 2) since a cycle formed by two oddly oriented cycleswith one common border is also oddly oriented. The cycle c + e + b1 + a2 isequal to (1− a1)+ b1 +1 ≡ 1 (mod2). This all leads to b1 + b2 = a1 + a2 +1.So:

c + b1 + b2 = c + a1 + a2 + 1 ≡ 0 (mod2). (4.30)

This obtained evenly oriented cycle G is also nice; a perfect matching of L−Gcan be found by pairing the ’left over’ vertex with the vertex situated aboveit. On this horizontal cycle all vertices can be paired with its horizontalneighbour except for one vertex. This one, in its turn, can be matched to

32

the vertex above it and continue in this way until a situation like illustratedin Figure 4.5 is reached.

Figure 4.5: A perfect matching of the graph L−G.

Thus, even if graph L is oriented such that the horizontal, vertical and unitcells are oddly oriented nice cycles, there is always within this configurationanother cycle, denoted by green in Figure 4.4 that is evenly oriented andnice. Therefore, a rectangular lattice on a torus does not have a Pfaffianorientation.

Since a rectangular lattice on a torus does not have a Pfaffian orientation thenumber of perfect matchings of this lattice must be obtained in an other waythan used in the previous section. Recall the planar rectangular lattice withits Pfaffian orientation as can be seen in Figure 4.6. The green lines denotethe perfect matching M .

Figure 4.6: The standard configuration of a rectangular lattice with the greenlines the perfect matching M .

This will be the standard configuration. By winding this lattice on a torus,periodical boundary conditions are introduced. There can be four classesof configurations derived from the standard configuration as illustrated inFigure 4.7, where the standard configuration M is denoted by green andblack denotes the other configuration N :

33

• (e, e): M ∪N forming even numbers of loops in the horizontal as wellas in the vertical direction.

• (e, o): M ∪N forming even numbers of loops in the horizontal directionand odd numbers of loops in the vertical direction.

• (o, e): M ∪N forming odd numbers of loops in the horizontal directionand even numbers of loops in the vertical direction.

• (o, o): M ∪ N forming odd numbers of loops in the horizontal as wellas in the vertical direction.

(e,e) configuration (e,o) configuration

(o,e) configuration (o,o) configuration

Figure 4.7: The four configurations for a rectangular lattice graph derivedfrom the standard configuration.

The added bonds to wrap the lattice on a torus do not have an orientationyet. Define for 1 ≤ i ≤ m, where i denotes the site in the horizontal direction,and for 1 ≤ j ≤ n, where j is the site in the vertical direction:

A(m, j; 1, j) = 1, A(i, n; i, 1) = (−1)i. (4.31)

The matrix A1 is defined corresponding to this orientation. According toLemma 3.4, sgn(M) · sgn(N) = (−1)k, where M is the matching in thestandard configuration and N another perfect matching and k is the numberof evenly oriented cycles with alternating edges belonging to M and N . Withthis information one can see that the Pfaffian of matrix A1 counts only the(e, e) configurations correctly since there are even numbers of loops in thehorizontal and vertical directions. The loops are even oriented; non-loopingpolygons in this configuration are oddly oriented. The other configurationswill be counted with a negative sign.

34

To count the other configurations correctly, three additional matrices aredefined:

A2 with A(m, j; 1, j) = 1; A(i, n; i, 1) = (−1)i+1, (4.32)

A3 with A(m, j; 1, j) = −1; A(i, n; i, 1) = (−1)i, (4.33)

A4 with A(m, j; 1, j) = −1; A(i, n; i, 1) = (−1)i+1. (4.34)

Again, with aid of Lemma 3.4 one can see that the matrices do not count allconfigurations correctly. The counting rules are summarised as follows:

Class of configurations A1 A2 A3 A4

(e,e) + + + +(e,o) - + - +(o,e) - - + +(o,o) - + + -

Table 4.1: Summary of the counting rules of the Pfaffians.

Thus, the number of perfect matchings of a lattice on a torus can be foundby means of four Pfaffians and is given by:

nPM =1

2(−Pf(A1) + Pf(A2) + Pf(A3) + Pf(A4)) . (4.35)

In comparison with the planar rectangular lattice, the occurrence of matrixA is replaced by matrices P and Q:

P =

0 1 0 . . . 0 −1

−1 0 1. . . . . . 0

0 −1. . . . . . . . .

......

. . . . . . . . . . . . 0

0. . . . . . . . . . . . 1

1 0 . . . 0 −1 0

, Q =

0 1 0 . . . 0 1

−1 0 1. . . . . . 0

0 −1. . . . . . . . .

......

. . . . . . . . . . . . 0

0. . . . . . . . . . . . 1

−1 0 . . . 0 −1 0

(4.36)

due to the periodic boundary conditions. The four Pfaffians can be found

35

analogously as in the planar case and this leads to:

nPM(torus) = −1

2

12m∏

k=1

n∏l=1

2

{sin2

(2kπ

m

)+ sin2

(2lπ

n

)} 12

+1

2

12m∏

k=1

n∏l=1

2

{sin2

(2kπ

m

)+ sin2

((2l − 1)π

n

)} 12

+1

2

12m∏

k=1

n∏l=1

2

{sin2

((2k − 1)π

m

)+ sin2

((2lπ

n

)} 12

+1

2

12m∏

k=1

n∏l=1

2

{sin2

((2k − 1)π

m

)+ sin2

((2l − 1)π

n

)} 12

.

(4.37)

When m and n are pushed to infinity, the logarithm of the expression turnsout to be the same as for the planar case (4.17):

1

mnln(nPM(torus)) →

G

π. (4.38)

Example 4.3. Consider a very peculiar chessboard, namely a chessboardwrapped around a torus. So the dimension is 8×8, then the number of waysto cover the chessboard with domino tiles is 311, 853, 312.

We have seen that although a rectangular lattice on a torus does not permit aPfaffian orientation, the perfect matchings of this graph can be enumeratedwith a linear combination of four Pfaffians corresponding to four matricesthat are slightly different from the standard configuration. This can be gen-eralised to graphs that can be embedded in a surface of genus g. Then theenumeration of perfect matchings of the graph requires a linear combina-tion of 22g Pfaffians corresponding to a slightly modification of the adjacencymatrix(Kasteleyn [6]).

36

Chapter 5

Discussion and Conclusion

The previous chapter treats the enumeration of perfect matchings of rectan-gular lattices. However, there are more types of regular lattices, for instancethe triangular lattice and the hexagonal lattice. In this chapter we will givea review of some results of these lattices. What is the expression for thenumber of perfect matchings and what is the behaviour of this expression inthe limit m, n →∞, where m and n are the dimensions of the lattices.A Pfaffian orientation on the triangular lattice and the hexagonal lattice isillustrated in Figure 5.1.

Figure 5.1: The triangular lattice and the hexagonal lattice, both with Pfaf-fian orientation.

A calculation using Pfaffians of the number of perfect matchings on thetriangular lattice is given by Fendley, Moessner and Sondhi [1]. Taking thelogarithm of the expression for the number of perfect matchings in the limitas m, n →∞, the result is STriangular = 0.4286. For the hexagonal lattice thenumber of perfect matchings depends on the boundary conditions. For thelattices with dimensions m and n as illustrated in Figure 5.2 the number of

37

configurations of tilings is given by Hosoya [3]:

nP =

(m + n

n

)=

(m + n)!

m!n!,

nH =2m−1∏k=m

(n + k

n

)/

m−1∏k=1

(n + k

n

).

(5.1)

1 2 ... m

...

n

1 2 ... n

...

m

Figure 5.2: The hexagonal lattices P and H with different boundaries.

In the limit as the dimensions of the hexagonal lattice are infinitely large,the logarithm of the number of perfect matchings is SHexagonal = 0.1615. Theresults for the three types of lattices are summarised in Table 5.1:

Type of lattice SLattice type

Rectangular 0.29Triangular 0.43Hexagonal 0.16

Table 5.1: Summary of three types of regular lattices.

For a review on other lattices and the enumeration of coverings of the latticesthe reader is referred to Wu [10].

We have seen that an algorithm exists for counting the perfect matchings ofplanar lattices and this has been applied explicitly to a rectangular latticeand some results are given for several other regular lattices. However, whenconsidering a three dimensional regular lattice, an exact result has not yetbeen found.

38

Popular Summary

Imagine a long and rainy Sunday afternoon and you are locked up in a roomcontaining just a chess board. A nice game of chess would be a pleasantoccupation, even if the opponent is yourself. But alas, the chess pieces arenowhere to be found. Luckily, hidden in a small cabinet you find 32 dominotiles. In what way can these two games be combined, you ask yourself. Oneway to spend the dreary afternoon is to place the tiles on the squares of theboard such that each tile covers two squares without overlapping each otherand count the ways in which this can be accomplished.

Figure 5.3: A plain chessboard

Looking at the chess board and its 64 squares, you know for sure that thisactivity will cost definitely this afternoon, maybe even all the afternoonsfrom now on until some time next month! However, full with patience youstart with the most simple configurations and vary a little.After a while you are fed up with feeling like a robot in a huge factory oran accountant doing the same thing over and over again without a prospectof ever finishing in the near future. So, you decide to try a more thoughtfulapproach, you turn to mathematics.

39

Figure 5.4: Two possible coverings of domino tiles on the chess board.

Let’s transform the chess board into a mathematical object, say a graph.A graph consists of a collection of dots and a collection of lines pairwiseconnecting the dots. The chess board in graph form looks like Figure 5.5where the dots are the points in the middle of the squares and the lines thepossible positions of the domino tiles. When assigning to each dot a label,this graph, in its turn, can be transformed into a matrix. The matrix consistsof rows and columns with zero’s and ones.

Figure 5.5: A chess board transformed into a graph

This is a good thing, you can do calculations with matrices, for instance,the number of ways of covering a chess board with domino tiles! There are12988816 ways to arrange the domino tiles on the board. And this numbercan be found in less than an afternoon. Feeling quite pleased with yourself,you find out that you are not the first one to discover this ingenious way.The physicist Kasteleyn has beaten you in this point.Maybe a little bit disappointed you throw the board and the tiles away andgrumble a while. But wait! What if the chess board is three dimensional?

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Bibliography

[1] P. Fendley, R. Moessner, and S. L. Sondhi. Classical dimers on thetriangular lattice. Phys. Rev. B, 66(21):214513, Dec 2002.

[2] Michael E. Fisher. Statistical mechanics of dimers on a plane lattice.Phys. Rev. (2), 124:1664–1672, 1961.

[3] Haruo Hosoya. Matching and symmetry of graphs. Comput. Math. Appl.Part B, 12(1-2):271–290, 1986. Symmetry: unifying human understand-ing, I.

[4] P. W. Kasteleyn. The statistics of dimers on a lattice, i. the numberof dimer arrangements on a quadratic lattice. Physica, 27:1209–1225,1961.

[5] P. W. Kasteleyn. Dimer statistics and phase transitions. J. MathematicalPhys., 4:287–293, 1963.

[6] P.W. Kasteleyn. Graph theory and crystal physics. Graph Theory andTheoretical Physics. Academic Press IN (London) LTD., London, 1967.

[7] L. Lovasz and M. D. Plummer. Matching theory, volume 121 of North-Holland Mathematics Studies. North-Holland Publishing Co., Amster-dam, 1986. Annals of Discrete Mathematics, 29.

[8] E.W. Montroll. Lattice Statistics. Applied Combinatorial Mathematics.John Wiley ans Sons, Inc., New York, 1964. Edited by E. F. Beckenbach.

[9] Leslie G. Valiant. The complexity of enumeration and reliability prob-lems. SIAM J. Comput., 8(3):410–421, 1979.

[10] F. Y. Wu. Dimers on two-dimensional lattices. Internat. J. ModernPhys. B, 20(32):5357–5371, 2006.

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