aa r X i v : . [ m a t h . C O ] F e b A NEW SOLUTION FOR THE TWO DIMENSIONAL DIMER PROBLEM
MIHAI CIUCU
Abstract.
The classical 1961 solution to the problem of determining the number of perfectmatchings (or dimer coverings) of a rectangular grid graph — due independently to Kasteleynand to Temperley and Fisher — consists of changing the sign of some of the entries in theadjacency matrix so that the Pfaffian of the new matrix gives the number of perfect matchings,and then evaluating this Pfaffian. Another classical method is to use the Lindstr¨om-Gessel-Viennot theorem on non-intersecting lattice paths to express the number of perfect matchingsas a determinant, and then evaluate this determinant. In this paper we present a new methodfor solving the two dimensional dimer problem, which relies on the Cauchy-Binet theorem. Itonly involves facts that were known in the mid 1930’s when the dimer problem was phrased,so it could have been discovered while the dimer problem was still open.We provide explicit product formulas for both the square and the hexagonal lattice. Oneadvantage of our formula for the square lattice compared to the original formula of Kasteleyn,Temperley and Fisher is that ours has a linear number of factors, while the number of factorsin the former is quadratic. Our result for the hexagonal lattice yields a formula for the numberof periodic stepped surfaces that fit in an infinite tube of given cross-section, which can beregarded as a counterpart of MacMahon’s boxed plane partition theorem. Introduction
In 1961 Temperley and Fisher [17, 6], and Kasteleyn [10], independently determined thenumber of perfect matchings of a rectangular grid graph, thus solving the dimer problem posedin 1937 by Fowler and Rushbrooke [7] in the limiting case when the dimers completely fill thelattice (the so called close-packed or high density limit); we refer to the formula they discovered— included in this paper as equation (7.6) — as the TFK formula. Both solutions consist ofchanging the sign of some of the entries in the adjacency matrix of the grid graph so that thePfaffian of the new matrix gives the number of perfect matchings, and then evaluating thisPfaffian. The resulting formula was then analyzed asymptotically to calculate the so calledfree energy per site — the limit of the ratio between the logarithm of the number of perfectmatchings and the number of vertices in the rectangular graph.Even though this was not realized until later, MacMahon [12] solved an instance of the dimerproblem on the hexagonal lattice (in the equivalent language of plane partitions; see [5]), byproviding a simple product formula for the number of perfect matchings of honeycomb graphs(centrally symmetric hexagonal portions of the hexagonal lattice).A modern way of proving MacMahon’s result is to use the Lindstr¨om-Gessel-Viennot theorem(see [11, 9, 16]) on non-intersecting lattice paths to express the number of perfect matchings asa determinant, and then evaluate this determinant.The Pfaffian method and the non-intersecting lattice paths method are two classical toolsfor determining the number of perfect matchings of a planar graph. In this paper we presenta new method, based on the Cauchy-Binet theorem, and use it to give a new solution to the
Research supported in part by NSF grant DMS-1501052 and Simons Foundation Collaboration Grant 710477.
Figure 1.
Two strands.
Figure 2.
Connecting the two strands in Figure 1.dimer model on the hexagonal and on the square lattice. This solution only involves facts thatwere known in the mid 1930’s when the dimer problem was phrased, so it could have beendiscovered while the dimer problem was still open.The particular form of the involved graphs has cylindrical boundary conditions, and the exactproduct formulas we obtain seem not to have appeared previously in the literature.This paper is organized as follows. In Section 2 we present the new method, and state thegeneral result as Theorem 2.1. In Section 3 we apply the general result to honeycomb graphsembedded in a cylinder, and find an explicit product formula for the number of their perfectmatchings (see Theorem 3.1). In Section 4 we connect these results to the enumeration ofcylindric partitions of a certain double staircase shape. Theorem 4.1 gives a formula for thenumber of periodic stepped surfaces that fit in an infinite tube of given cross-section. It can beregarded as a counterpart of MacMahon’s boxed plane parition theorem [12] (see Remark 5).In Section 5 we turn to applications involving the square lattice. We show how Theorem 2.1can be used to find an explicit product formula for the number of perfect matchings of squarecylinder graphs of even girth (see Theorem 5.1).Square cylinder graphs of odd girth are dealt with in Section 7. We use there an extension ofour factorization theorem from [3] (which we present in Section 6; see Theorem 6.1) to obtain acommon formula for the number of perfect matchings of square cylinders of even or odd girth.Section 8 contains some concluding remarks.
NEW SOLUTION FOR THE TWO DIMENSIONAL DIMER PROBLEM 3 The new method A strand is a path with pending edges (see Figure 1 for an illustrative example); the edgesin a strand are allowed to carry arbitrary weights (in particular, these weights are allowed toequal zero). Thus, strands are weighted bipartite graphs. We always draw strands so that theirvertices are lined up along two horizontal lines, with white vertices on the bottom and blackvertices on top.Our new method for enumerating dimer coverings applies to a special kind of planar graphsbuilt from strands, which we call fabric graphs . There are two versions of them. Both consistof horizontal strands joined together by vertical edges.A rectangular fabric graph is a weighted graph obtained from horizontal strands S , . . . , S m as follows. Draw the strands in the plane successively one above the other, with S on thebottom. Denote the number of white (bottom) vertices in S i by k i , and the number of black(top) vertices by l i , i = 1 , . . . , m . Assume l i = k i +1 , for i = 1 , . . . , m −
1, and join the topvertices of the strand S i to the bottom vertices of the strand S i +1 by vertical edges of weight 1(see Figure 2; for easy distinguishing, vertical edges between the strands are indicated as wavylines). The resulting graph is called a rectangular fabric graph.The second family consists of cylindrical fabric graphs . Start as above with strands S , . . . , S m ,but assume now that l i = k i +1 , for i = 1 , . . . , m , where k m +1 := k . Then, in addition to joiningtogether strands S i and S i +1 by vertical edges for i = 1 , . . . , m −
1, as in the previous paragraph,join together also the top vertices of the strand S m with the bottom vertices of the strand S ,by vertical edges (which can be thought of as curving down behind a horizontal cylinder). Fur-thermore, more generally than for the case of their rectangular counterparts, weight the verticaledges of cylindrical fabric graphs as follows: weight the vertical edges connecting strand S i tostrand S i +1 by x i , i = 1 , . . . , m (where S m +1 := S ).The bi-adjacency matrix of strand S i is the matrix A i defined as follows. The rows of A i are indexed by the bottom vertices of S i (listed from left to right), and the columns of A i areindexed by the top vertices of S i (also listed from left to right). Then if u is a bottom vertexof S i and v is a top vertex of S i , define the ( u, v )-entry of the bi-adjacency matrix A i to be theweight of the edge { u, v } , if u and v are connected by an edge in S i , and zero otherwise.In general, given a weighted graph G , we denote by M( G ) the sum of the weights of all perfectmatchings of G . However, if G is a cylindrical fabric graph, we often write M( G ; x , . . . , x m )(or M( G ; x ), if x = · · · = x m = x ) for the sum of the weights of its perfect matchings, in orderto spell out the weights of the vertical edges.As a consequence of their definition, both rectangular fabric graphs and cylindrical fabricgraphs are bipartite. A necessary condition for a bipartite graph to admit a perfect matchingis to be balanced (i.e., to have the same number of vertices in the two bipartition classes). Thedefinition of cylindrical fabric graphs implies that they are balanced. A rectangular fabric graphhas k + · · · + k m white and l + · · · + l m black vertices, and since l i = k i +1 , i = 1 , . . . , m −
1, itis balanced if and only if k = l m . A perfect matching µ of a graph G is a collection of edges of G with the property that every vertex of G isincident to precisely one edge in µ . The weight of µ is defined to be the product of the weights of the edges in µ . If G is not weighted (equivalently, all edges of G have weight 1), then M( G ) is simply the number of perfectmatchings of G . MIHAI CIUCU
Therefore, without loss of generality, we may assume that for any fabric graph (be it rectan-gular or cylindrical) with m strands, there exist non-negative integers l , . . . , l m so that the i thstrand has l i − bottom vertices and l i top vertices, i = 1 , . . . , m (with l = l m ). Theorem 2.1. (a) . If G is a balanced rectangular fabric graph with m strands, then M( G ) = det( A A · · · A m ) . (2.1)(b) . If G is a cylindrical fabric graph with m strands, strand S i having l i − bottom vertices and l i top vertices, i = 1 , . . . , m , we have M( G ; x , . . . , x m ) = x l − l m · · · x l m − − l m m − det(( x · · · x m ) I + A A · · · A m ) , (2.2) where I is the identity matrix of order l m . Our proof of Theorem 2.1 is based on three preliminary lemmas. Recall that the permanentper A of an n × n matrix A = ( a ij ) is defined to beper A = X π ∈S n a ,π (1) a ,π (2) · · · a n,π ( n ) , (2.3)where S n is the set of permutations of order n . The following lemma is well known. Lemma 2.2.
Let G be a weighted bipartite graph with the same number of white and blackvertices. Let A be the bi-adjacency matrix of G ( i.e., the rows of A are indexed by the whitevertices, the columns by the black vertices, and the ( u, v ) entry is the weight of the edge { u, v } ,or if there is no edge between u and v ) . Then M( G ) = per A. (2.4) Proof.
Each non-zero term in the expansion of per A (see the right hand side of (2.3)) corre-sponds to a collection of n edges of G that share no white vertex (because no two a i,π ( i ) ’s arein the same row) and no black vertex (because no two a i,π ( i ) ’s are in the same column), i.e. toa perfect matching of G . (cid:3) Lemma 2.3.
Let S be a strand ( see definition at the beginning of this section ) , and let A be thebi-adjacency matrix of S , with the rows of A indexed by the bottom vertices of S listed from leftto right, and the columns of A indexed by the top vertices of S listed from left to right. Thenfor any square submatrix B of A we have per B = det B. (2.5) Proof.
One readily sees that for the indicated ordering of the vertices of the strand, if the entriesof the bi-adjacency matrix A are thought of as residing at the centers of the unit squares of an n × n chessboard, the support of A is contained in a zigzag strip — the path of a rook that isallowed to move only down or to the right. This implies that in the expansiondet B = X σ ∈S k ( − sgn( σ ) b ,σ (1) · · · b k,σ ( k ) (2.6)of the determinant of any k × k submatrix B = ( b i,j ) ≤ i,j ≤ k of A , all the non-zero termscorrespond to permutations σ with no inversions. (cid:3) NEW SOLUTION FOR THE TWO DIMENSIONAL DIMER PROBLEM 5
Lemma 2.4.
Let A be an n × n matrix. Then X J ⊂ [ n ] x n −| J | det A JJ = det( xI n + A ) , (2.7) where I n is the identity matrix of order n .Proof. Regard the j th column of the matrix xI n + A as the sum of the j th column of xI n and the j th column of A , for j = 1 , . . . , n . Using the linearity of the determinant in columns,det( xI n + A ) becomes a sum of 2 n determinants. Each term in this sum is the determinant ofa matrix obtained from A by picking some subset J ⊂ [ n ], keeping the entries of A in the rowsand columns indexed by J , and replacing them by the entries of xI n in the rows and columnswith indices outside J . But this determinant is clearly equal to x n −| J | det A JJ . (cid:3) Proof of Theorem . . (a). Perfect matchings of a rectangular fabric graph can be thought ofas being built in two stages: (1) specify which vertical edges participate in each level — this isequivalent to specifying subsets J i ⊂ [ l i ], for i = 1 , . . . , m −
1; and (2) then choosing the perfectmatching internally within each horizontal strand, independently.Denote by S i ( I, J ) the subgraph of the strand S i obtained by deleting bottom vertices withlabels in I ⊂ [ k i ] and top vertices with labels in J ⊂ [ l i ]. Viewing perfect matchings asdescribed in the previous paragraph we obtainM( G ) = X J ⊂ [ l ] ,...,J m − ⊂ [ l m − ] M( S ( ∅ , J )) M( S ( J , J )) · · · M( S ( J m − , J m − )) M( S ( J m − , ∅ ))= X J ⊂ [ l ] ,...,J m − ⊂ [ l m − ] per( A ) [ l ] \ J [ l m ] per( A ) [ l ] \ J [ l ] \ J · · · per( A m − ) [ l m − ] \ J m − [ l m − ] \ J m − per( A m ) [ l m ][ l m − ] \ J m − , = X J ⊂ [ l ] ,...,J m − ⊂ [ l m − ] det( A ) [ l ] \ J [ l m ] det( A ) [ l ] \ J [ l ] \ J · · · det( A m − ) [ l m − ] \ J m − [ l m − ] \ J m − det( A m ) [ l m ][ l m − ] \ J m − , (2.8)where at the second and third equalities we used Lemmas 2.2 and 2.3, respectively, and in thelast two summations all the involved matrices are required to be square (i.e., | J | = l − l m , | J i | − | J i − | = l i − l i − , i = 2 , . . . , m − | J m − | = l m − − l m , which in turn is equivalent to | J i | = l i − l m , i = 1 , . . . , m − § Following customary notation, we set [ n ] := { , . . . , n } . As usual, A JI denotes the submatrix of A obtainedby choosing its elements in rows with indices in the set I and columns with indices in the set J . The bottom vertices in a strand are labeled from left to right by consecutive integers starting with 1; similarlythe top vertices.
MIHAI CIUCU X J ⊂ [ l ,...,Jm − ⊂ [ lm − | J | = l − lm,..., | Jm − | = lm − − lm det( A ) [ l ] \ J [ l ] \ J · · · det( A m ) [ l m ][ l m − ] \ J m − X J ⊂ [ l | J | = l − lm det( A ) [ l ] \ J [ l m ] det( A ) [ l ] \ J [ l ] \ J = X J ⊂ [ l ,...,Jm − ⊂ [ lm − | J | = l − lm,..., | Jm − | = lm − − lm det( A A ) [ l ] \ J [ l m ] det( A ) [ l ] \ J [ l ] \ J · · · det( A m ) [ l m ][ l m − ] \ J m − . (2.9)Repeating the argument that gave (2.9) m − G is a cylindrical fabric graph, the same reasoning that gave (2.8) leads toM( G ; x , . . . , x m ) = X J ⊂ [ l ,...,Jm ⊂ [ lm ] | Ji |−| Ji − | = li − li − , i =1 ,...,m x | J | · · · x | J m | m det( A ) [ l ] \ J [ l m ] \ J m det( A ) [ l ] \ J [ l ] \ J · · · det( A m ) [ l m ] \ J m [ l m − ] \ J m − . (2.10)Using repeatedly the Cauchy-Binet theorem as in the proof of part (a), and the fact that theconditions on the second row under the summation in (2.10) are equivalent to | J i | = | J m | + l i − l m ,we obtain from (2.10) thatM( G ; x , . . . , x m ) = x l − l m · · · x l m − − l m m − X J m ⊂ [ l m ] ( x · · · x m ) | J m | det( A A · · · A m ) [ l m ] \ J m [ l m ] \ J m = x l − l m · · · x l m − − l m m − X J ⊂ [ l m ] ( x · · · x m ) l m −| J | det( A A · · · A m ) JJ . (2.11)However, by Lemma 2.4, the right hand side above is the same as the right hand side ofequation (2.2). (cid:3) Honeycomb cylinder graphs
In this section we consider the special case when the strands are paths of the same length.As we will see, in this case the determinant in Theorem 2.1 can be evaluated explicitly.Define the honeycomb cylinder graph H m,n to be the cylindrical fabric graph with m strands,each of which is a path with n vertices (see Figure 3). Note that the bottom and top strands of H m,n fit together “seamlessly” only when m is even, in which case H m,n has m/ m is even throughout this section.The main result of this section is the following. Theorem 3.1.
For m even, the number of perfect matchings of the cylindrical honeycomb graph H m,n is given by M( H m,n ; x , . . . , x m ) = Q ⌊ n/ ⌋ k =1 h x · · · x m + (cid:0) kπn +1 (cid:1) m/ i , n even ,x x · · · x m − Q ⌊ n/ ⌋ k =1 h x · · · x m + (cid:0) kπn +1 (cid:1) m/ i , n odd . (3.1) NEW SOLUTION FOR THE TWO DIMENSIONAL DIMER PROBLEM 7
Figure 3.
The graph H m,n for m = 6, n = 8 (left) and m = 6, n = 9 (right). Remark . The floor can be avoided at the upper limit of the product by writing ⌊ n/ ⌋ Y k =1 " x · · · x m + (cid:18) kπn + 1 (cid:19) m/ = vuut n Y k =1 " x · · · x m + (cid:18) kπn + 1 (cid:19) m/ . (3.2)In our proof we will employ the following lemma. Lemma 3.2. (a) . The eigenvalues of the n × n matrix · · · · · · · · · · · · · · · · · ·· · · · · · · · ·· · · · · · · · · · · · · · · · · · (3.3) are kπn + 1 , k = 1 , . . . , n. (3.4) MIHAI CIUCU (b) . The eigenvalues of the n × n matrix · · · · · · · · · · · · · · · · · ·· · · · · · · · ·· · · · · · · · · · · · · · · · · · − (3.5) are kπ n + 1 , k = 1 , . . . , n. (3.6) Proof.
Part (a) is a classical result. See for instance [4, Section 2.6, q n ( λ ) the characteristic polynomial of the matrix (3.5). Regarding the last columnof the defining determinant for q n ( λ ) as a sum of two columns and using the linearity of thedeterminant, one readily obtains that q n ( λ ) = p n ( λ ) + p n − ( λ ) , (3.7)where p n ( λ ) is the characteristic polynomial of the matrix (3.3). However, the latter is justthe characteristic polynomial of a path of length n . By [4, Section 2.6, p n ( λ ) = U n ( λ/ U n ( λ ) is the Chebyshev polynomial of the second kind, given by U n ( x ) = sin[( n + 1) arccos x ]sin(arccos x ) . (3.8)The zeros of q n ( λ ) are then the zeros of U n ( λ/
2) + U n − ( λ/ U n ( x ) + U n − ( x ) = sin[( n + 1) arccos x ]sin(arccos x ) + sin[ n arccos x ]sin(arccos x )= 2 sin (cid:2)(cid:0) n + (cid:1) arccos x (cid:3) cos (cid:0) arccos x (cid:1) sin(arccos x ) . (3.9)Thus the zeros of U n ( x ) + U n − ( x ) are cos kπ n +1 , k = 1 , . . . , n , and those of q n ( λ ) = U n ( λ/
2) + U n − ( λ/
2) are 2 cos kπ n +1 , k = 1 , . . . , n . This proves part (b). (cid:3) Proof of Theorem . . The details of the proof depend on the parity of n . We treat first thecase when n is even. Then the cylindrical hexagonal graph H m,n looks like in the picture on NEW SOLUTION FOR THE TWO DIMENSIONAL DIMER PROBLEM 9 the left in Figure 3. Let A n be the n × n matrix A n = · · · · · · · · · · · · · · · · · ·· · · · · · · · ·· · · · · · · · · · · · · · · · · · . (3.10)Then if we denote the strands of H m,n by S , . . . , S m starting from the bottom, the bi-adjacencymatrix of strand S i is A n/ for odd i , and its transpose A Tn/ for even i . Therefore, as each strandhas n/ H m,n ; x , . . . , x m ) = det (cid:0) x · · · x m I n/ + ( A n/ A Tn/ ) m/ (cid:1) . (3.11)We have A s A Ts = · · · · · · · · · · · · · · · · · ·· · · · · · · · ·· · · · · · · · · · · · · · · · · · . (3.12)Therefore, by Lemma 3.2(b) we see that the eigenvalues of A s A Ts are 2 + 2 cos kπ s +1 , k = 1 , . . . , s .Since these are distinct, A s A Ts is diagonalizable. It follows that the eigenvalues of the matrix( A s A Ts ) m/ are (cid:0) kπ s +1 (cid:1) m/ , k = 1 , . . . , s .For an s × s matrix A with characteristic polynomial p A and eigenvalues x , . . . , x s , we havedet( xI s + A ) = ( − s det( − xI s − A )= ( − s p A ( − x )= ( − s ( − x − x ) · · · ( − x − x s )= ( x + x ) · · · ( x + x s ) . (3.13)Then (3.1) follows from (3.11), (3.13) and the above identification of the eigenvalues of A s A Ts . Consider now the case when n is odd. Then the cylindrical hexagonal graph H m,n looks likein the picture on the right in Figure 3. Let B n be the n × ( n + 1) matrix B n = · · · · · · · · · · · · · · · · · ·· · · · · · · · ·· · · · · · · · · · · · · · · . (3.14)Then the bi-adjacency matrix of strand S i is B ( n − / for odd i , and its transpose B T ( n − / foreven i . Therefore, since odd-index strands have ( n + 1) / n − / H m,n ; x , . . . , x m ) = x x · · · x m − det (cid:0) x · · · x m I ( n +1) / + ( B ( n − / B T ( n − / ) m/ (cid:1) . (3.15)The form of the matrix raised to the power m/ B s B Ts = · · · · · · · · · · · · · · · · · ·· · · · · · · · ·· · · · · · · · · · · · · · · · · · . (3.16)Thus, in this case, by Lemma 3.2(a) we see that the eigenvalues of B s B Ts are 2 + 2 cos kπs +1 , k = 1 , . . . , s . These are again distinct, so the eigenvalues of the matrix ( B s B Ts ) m/ are (cid:0) kπs +1 (cid:1) m/ , k = 1 , . . . , s . Proceeding as in part (a) we are again led to formula (3.1). (cid:3) Remark . The formula provided by Theorem 3.1 is a TFK-style formula for a family ofhoneycomb graphs. This is an unusual situation, as virtually all explicit product formulas inthe literature for honeycomb style graphs are “round formulas,” in the sense that the size ofthe factors is linear in the parameters.
Remark . The free energy per site for the family H m,n of honeycomb cylinder graphs turns outto be the same as for toroidal honeycombs, which is known to be maximal. This is in contrastwith the family of centrally symmetric honeycombs whose perfect matching enumeration isequivalent to MacMahon’s boxed plane partition theorem [12]. Remark . Our graph H m,n has the structure of a nanotube with “armchair boundary” (seee.g. [13]). The other natural type of boundary, zig-zag boundary, is not so interesting from thepoint of view of perfect matching enumeration, as it simply yields a honeycomb cylinder whosenumber of perfect matchings is 2 to the number of strands. NEW SOLUTION FOR THE TWO DIMENSIONAL DIMER PROBLEM 11
Figure 4. An m -periodic cliff of height n and horizontal displacement s for m = 3, n = 3, s = 4. Figure 5.
The corresponding periodic lozenge tiling; southern boundary istranslation of northern boundary 2 n + s units in the polar direction − π/ Figure 6.
Paths that determine the periodic lozenge tiling; each traverses s horizontal lozenges.
13 2 1 13 2 2 12 2
Figure 7.
The cliff heights on the horizontal lozenges traversed by the pathsdetermine the paths (left); the corresponding cylindric plane partition (right).4.
Cylindric plane partitions and periodic cliffs
We define a cliff to be a stepped surface whose projection on the triangular lattice is boundedby two parallel infinite zigzags (see Figure 4 for an example). We say that the cliff is m -periodic if it is invariant under translation by m v , where v is the shortest non-zero vector such thattranslation by it leaves the zigzags invariant. The height of a cliff is the difference betweenthe heights of the horizontal planes containing its bounding zigzags. A cliff can be viewed as Figure 8.
The lowermost (left) and uppermost (right) periodic cliff of height n and horizontal displacement s , for n = 3, s = 4.consisting of slices composed of unit cubes cut out by equidistant parallel planes one unit apart;the horizontal displacement of a cliff is the number of horizontal faces in each such slice. Thecliff pictured in Figure 4 is 3-periodic, has height 3 and horizontal displacement 4.Cylindric plane partitions were introduced by Gessel and Krattenthaler in [8]. For Youngdiagrams µ ⊂ λ and a positive integer d , a cylindric plane partition of shape λ/µ/d is a fillingof the skew Young diagram λ/µ with non-negative integers that weakly decrease along rowsand columns, with the additional property that when the bottom row is copied above the toprow and translated d units to the right, the resulting extended array is still weakly decreasingalong columns (see Figure 7 for an example).The main result of this section is the following. Theorem 4.1.
There are as many m -periodic cliffs of height n and horizontal displacement s as cylindric partitions of shape ( s + m − , s + m − , . . . , s ) / ( m − , m − , . . . , m ) /m withentries less or equal than n . The common number is [ x ⌊ s/ ⌋ ] n + ⌊ s/ ⌋ Y k =1 (cid:20) x + (cid:18) kπ n + s + 1 (cid:19) m (cid:21) , (4.1) where [ x k ] p ( x ) denotes the coefficient of x k in p ( x ) .Remark . The result in Theorem 4.1 seems to be the first explicit product formula in theliterature for the number of cylindric partitions of a given shape and a given bound on the sizeof its entries.It can be interpreted as giving the number of m -periodic stepped surfaces that fit in theinfinite tube enclosed by the lowermost and uppermost surfaces shown in Figure 8. From thispoint of view, it is a counterpart of MacMahon’s boxed plane partition theorem [12] (and as faras the container shape is concerned, it resembles even more Proctor’s theorem on the numberof plane partitions that fit in a right prism with staircase shape base [14]).The proof will follow from Theorem 3.1 and Proposition 4.2 below. Define the zigzag strip Z k to be the region on the triangular lattice between two infinite horizontal zigzags, the lowerbeing the translation of the upper k units in the polar direction − π/ − π/ T of Z k and a “/”-edge e on its top boundary, start at e and followlozenges containing a “/”-edge in their boundary until a “/”-edge is reached on the bottomboundary of Z k . The resulting sequence of lozenges is called a path of lozenges . Drawn so that one family of lattice lines is vertical.
NEW SOLUTION FOR THE TWO DIMENSIONAL DIMER PROBLEM 13
Proposition 4.2.
The following are equinumerous: (1) m -periodic cliffs of height n and horizontal displacement s (2) m -periodic lozenge tilings of the zigzag strip Z n + s with s horizontal lozenges along eachpath of lozenges (3) perfect matchings of the honeycomb cylinder H m, n + s with ms vertical edges (4) m -periodic families of non-intersecting paths of lozenges in Z n + s with s horizontallozenges on each path (5) cylindric partitions of shape ( s + m − , s + m − , . . . , s ) / ( m − , m − , . . . , m ) /m withentries less or equal than n Proof.
Consider a rectangular system of coordinates in which the faces of the stepped surfaceare parallel to the coordinate planes, and so that when viewed along the line x = y = z , thefaces are seen as congruent rhombi with angles of 60 ◦ and 120 ◦ . Projecting the stepped surfaceon a plane perpendicular to the line x = y = z shows that the sets (1) and (2) are in one-to-onecorrespondence (see also [5]).Any lozenge tiling of the zigzag strip Z n + s naturally defines an infinite family of paths oflozenges, obtained by starting at the “/”-segments of the upper boundary, following alonglozenges in the tiling, and ending at the “/”-segments of the lower boundary. Since these pathscome from a tiling, they are non-intersecting. This implies that paths starting at consecutivesegments of the upper boundary end at consecutive segments of the lower boundary. Thisin turn implies that all these paths contain the same number of horizontal lozenges. Givenan m -periodic lozenge tiling of Z n + s in which this common number is s , associate to it thepreviously described family of paths of lozenges. This is a bijection between sets (2) and (4).For a bijection between sets (2) and (3), note that m -periodic tilings of Z n + s can be identifiedwith tilings of the quotient of Z n + s under the action of the horizontal translation that leavesthem invariant. The dual of the quotient region is precisely the honeycomb cylinder H m, n + s .Furthermore, if in the tiling each of the described paths of lozenges has s horizontal lozenges,the total number of horizontal lozenges in the quotient region (which correspond to verticaledges in H m, n + s ) is ms .The bijection between (4) and (5) is indicated in Figure 7. Simply note that the whole m -periodic family of paths of lozenges is determined by m consecutive paths, which in turnare determined by the sequences of heights of the horizontal lozenges in them. These form m weakly decreasing sequences of non-negative integers. Arranging them in an array from bottomto top, with each successive row one unit further to the right and omitting the 0’s, one obtainsa cyclic partition of shape ( s + m − , s + m − , . . . , s ) / ( m − , m − , . . . , m ) /m . (cid:3) Proof of Theorem . . By Proposition 4.2, both the number of m -periodic cliffs of height n andhorizontal displacement s , and the number of cylindric partitions of shape ( s + m − , s + m − , . . . , s ) / ( m − , m − , . . . , m ) /m with entries less or equal than n , is equal to the number ofperfect matchings of the cylindrical honeycomb H m, n + s with ms vertical edges.Denote by M ( H m,n ; x ) the sum of the weights of the perfect matchings of H m,n when allvertical edges are weighted by x , and all other edges by 1. If s is even, Theorem 3.1 gives M ( H m, n + s ; x ) = n + s/ Y k =1 (cid:20) x m + (cid:18) kπ n + s + 1 (cid:19) m (cid:21) . (4.2)The number of perfect matchings of H m, n + s with ms vertical edges is then the coefficient of x ms = ( x m ) s/ in the product (4.2), which agrees with the right hand side of (4.1). Figure 9.
The square cylinder graph C m,n for m = 4, n = 8 (left). A fabricgraph with the same number of perfect matchings (right). In both pictures, edgesextending up from the top vertices connect down to the corresponding bottomvertices.If on the other hand s is odd, Theorem 3.1 gives M ( H m, n + s ; x ) = x m n +( s − / Y k =1 (cid:20) x m + (cid:18) kπ n + s + 1 (cid:19) m (cid:21) . (4.3)The number of perfect matchings H m, n + s with ms vertical edges is now the coefficient of x ms − m = ( x m ) ( s − / in the product on the right hand side of (4.3), which again agrees withthe right hand side of (4.1). (cid:3) Square cylinder graphs of even girth
Define the square cylinder graph C m,n to be the graph obtained from the m × n rectangulargrid graph with vertices { ( i, j ) : i, j ∈ Z , ≤ i ≤ m − , ≤ j ≤ n − } by adding an edgeconnecting vertex ( i,
0) to vertex ( i, n −
1) for each i = 1 , . . . , m (see Figure 9 for an example).The girth of the square cylinder graph C m,n is defined to be equal to m .The main result of this section is the following. Theorem 5.1. If m is even, we have M( C m,n ) = vuut n Y k =1 " cos kπn + 1 + r kπn + 1 ! m cos kπn + 1 − r kπn + 1 ! m = 2 n − ⌊ n/ ⌋ ⌊ n/ ⌋ Y k =1 " cos kπn + 1 + r kπn + 1 ! m cos kπn + 1 − r kπn + 1 ! m . (5.1)Our proof of Theorem 5.1 employs the following result. NEW SOLUTION FOR THE TWO DIMENSIONAL DIMER PROBLEM 15
Lemma 5.2.
Let A = ( a ij ) i,j ≥ be the infinite matrix having support A = · · · , (5.2) and define A n to be its restriction to the first n rows and first n columns. Then the eigenvaluesof A n A Tn are cos kπn + 1 + r kπn + 1 ! , k = 1 , , . . . , ⌈ n/ ⌉ cos kπn + 1 − r kπn + 1 ! , k = 1 , , . . . , ⌈ n/ ⌉ , (5.3) with the convention that for n odd and k = ⌈ n/ ⌉ , the two expressions above — both of whichequal — supply the eigenvalue with a total multiplicity of one.Proof. The matrix A n A Tn has order n , and the pattern of its non-zero entries looks slightlydifferent depending on whether the index n is even or odd. For even indices we have A n A T n = · · · , (5.4)(where only the pattern followed by the non-zero entries is indicated). For odd indices thetruncation at the bottom right corner is like in the matrix obtained from the one above bydeleting the last row and column.However, the proof we present works for both even and odd indices. The reason is that allthe recurrences we work with are obtained by expanding the resulting determinants along theirfirst rows or columns, and the pattern of the non-zero entries of A n A Tn is the same around thetop left corner for both n even and n odd. We give here the details for the even case. Denote by q n ( λ ) the characteristic polynomial of A n A T n : q n ( λ ) = det λ − − − − λ − − − − λ − − − − λ − − − − λ − · · · λ − − − − λ − − − − λ − − − λ − . (5.5)Regarding the first column as ( λ − , − , − , , . . . ) T + (1 , , , . . . , T and using the linearityof the determinant in columns, we obtain q n ( λ ) = p n ( λ ) + r n ( λ ) , (5.6)with p n ( λ ) = det λ − − − − λ − − − − λ − − − − λ − − − − λ − · · · λ − − − − λ − − − − λ − − − λ − (5.7)(where the matrix has 2 n rows) and r n ( λ ) = det λ − − − λ − − − − λ − − − − λ − · · · λ − − − − λ − − − − λ − − − λ − (5.8)(where the matrix has 2 n − λ − r n ( λ ). The second NEW SOLUTION FOR THE TWO DIMENSIONAL DIMER PROBLEM 17 is readily seen (by expanding along its first column) to equal − p n − ( λ ) − r n − ( λ ). Similarly,expanding along the first column in the third term, one sees that it is equal to − λr n − ( λ ). Thisgives p n ( λ ) = ( λ − r n ( λ ) − p n − ( λ ) − ( λ + 1) r n − ( λ ) . (5.9)On the other hand, expanding along the first column in the determinant for r n ( λ ), we get r n ( λ ) = ( λ − p n − ( λ ) − r n − ( λ ) . (5.10)The latter gives p n − = 1 λ − r n + r n − ) , (5.11)which when substituted into (5.9) yields r n +1 − ( λ − λ + 1) r n + λ r n − = 0 . (5.12)This recurrence holds for n ≥ r = 0. The solutions of the characteristic equationof recurrence (5.12) are θ , = λ − λ + 1 ± ( λ − √ λ − λ + 12 . (5.13)Therefore, r n can be expressed as r n = c θ n + c θ n , n ≥ , (5.14)where the coefficients c and c can be determined from the initial conditions r = 0, r = λ − r n into (5.11), and then using the resulting expressionfor p n and the expression (5.14) for r n in equation (5.6), we obtain( λ − q n ( λ ) = c ( λ + θ ) θ n + c ( λ + θ ) θ n . (5.15)Using the initial conditions r = 0, r = λ −
1, one readily gets from (5.14) that c = 1 √ λ − λ + 1 , c = − √ λ − λ + 1 . (5.16)Our goal is to find the eigenvalues of A n A T n , which are the zeros of q n ( λ ). By (5.15) and(5.16), the zeros of ( λ − q n ( λ ) are those values of λ for which1 √ λ − λ + 1 [( λ + θ ) θ n − ( λ + θ ) θ n ] = 0 . (5.17)Using (5.13), one sees after some manipulation that ( λ + θ ) θ n = ( λ + θ ) θ n if and only if( λ − θ n +11 − λθ n ) = 0 . (5.18)Since θ θ = λ , if λ = 1 this amounts to θ n +11 = λ n +1 , which in turn means that θ = ελ, ε = e kπi n +1 , k = 0 , , . . . , n. (5.19)Expressing θ by formula (5.13), (5.19) reduces to a quadratic equation in lambda, with solutions12 ε (cid:16) ε + ε ± √ ε + 14 ε + 8 ε + ε (cid:17) . (5.20) Using ε + ε − = 2 cos kπ n +1 and then the formula cos 2 α = 2 cos α −
1, the expression (5.20) canbe transformed into cos kπ n + 1 ± r kπ n + 1 ! . (5.21)By the paragraph before (5.17), we obtain that the zeros of q n ( λ ) are among cos kπ n + 1 ± r kπ n + 1 ! , k = 0 , , . . . , n. (5.22)Since cos( π − x ) = − cos x , all the distinct values provided by (5.22) are obtained if k runs over0 , , . . . , n . Furthermore, for k = 0, these become 3 ± √
2, which are the values that makethe denominator in (5.17) equal to zero. It is not hard to see that the remaining 2 n values(obtained by taking k = 1 , , . . . , n in (5.22)) are all distinct. Therefore they are the eigenvaluesof q n ( λ ), as claimed by the statement of the lemma.For the case of odd index, the above arguments applied to the matrix A n − A T n − lead tothe conclusion that the zeros of its characteristic polynomial multiplied by λ − cos kπ n ± r kπ n ! , k = 1 , . . . , n. (5.23)These are 2 n − k = n , when both choices of thesign lead to the value 1. Thus the eigenvalues of A n − A T n − are indeed the ones described inthe statement of the lemma. (cid:3) Proof of Theorem . . Since m is even, using the vertex splitting lemma of [3] (see Lemma 1.3there) one can readily construct a fabric graph G m,n with the same number of perfect matchingsas the square cylinder graph C m,n — the fabric graph corresponding to the square cylinder onthe left in Figure 9 is shown on the right in the same figure . The bi-adjacency matrix ofodd-indexed strands in G m,n is the matrix A n from Lemma 5.2, and the bi-adjacency matrixof even-indexed strands in G m,n is its transpose A Tn . Therefore, by Theorem 2.1(b) we haveM( C m,n ) = M( G m,n ) = det (cid:0) I n + ( A n A Tn ) m/ (cid:1) . (5.24)Then formula (5.1) follows from (3.13) and Lemma 5.2. (cid:3) Remark . Note that despite the great similarity between our square cylinder graphs and therectangular grid graphs, our formula from Theorem 5.1 has a quite different form compared tothe TFK formula (see (7.6)). A partial connection between the two is established in Section 7.6.
A factorization theorem for perfect matchings of symmetricturn-bipartite graphs
In this section we give an extension of our factorization theorem of [3] that applies to certainplanar non-bipartite graphs, which we call turn-bipartite.For a planar graph G embedded in a region R with holes, we say that G is turn-bipartite ifthe length of any cycle C of G has the same parity as the number of holes of R that are in the The indicated construction does not work if m is odd, as there is a periodicity involving every two consecutivestrands in the picture on the right in Figure 9. NEW SOLUTION FOR THE TWO DIMENSIONAL DIMER PROBLEM 19 interior of C . The picture on the left in Figure 10 shows a turn-bipartite graph embedded inan annulus.Following the terminology introduced in [3], we say that a plane graph G is symmetric if itis invariant under the reflection across some straight line. Clearly, a symmetric graph has noperfect matching unless the axis of symmetry contains an even number of vertices (otherwise,the total number of vertices is odd); we will assume this throughout this section.A weighted symmetric graph is a symmetric graph with a weight function on the edges thatis constant on the orbits of the reflection. The width of a symmetric graph G , denoted w( G ),is defined to be half the number of vertices of G lying on the symmetry axis.Let G be a weighted symmetric graph with symmetry axis ℓ , which we consider to be hor-izontal. Let a , b , a , b , . . . , a w( G ) , b w( G ) be the vertices lying on ℓ , as they occur from left toright. A reduced subgraph of G is a graph obtained from G by deleting at each vertex a i eitherall incident edges above ℓ (we refer to this operation for short as “cutting above a i ”) or allincident edges below ℓ (“cutting below a i ,” for short). We recall the following result provedin [3] (see Lemma 1.1 there). Lemma 6.1 ([17]) . All w( G ) reduced subgraphs of a weighted symmetric graph G have the sameweighted count of perfect matchings. Let G be a turn-bipartite weighted symmetric graph with symmetry axis ℓ . Assume that allthe holes in the region R in which G is embedded are along ℓ . Then the subgraph G ≥ inducedby the vertices of G on or above ℓ is bipartite.Let us color the vertices in the two bipartition classes of G ≥ black and white. For definiteness,choose a to be white. We define a subgraph of G as follows. Perform cutting operations aboveall white a i ’s and black b i ’s, and below all black a i ’s and white b i ’s. Note that this procedureyields cuts of the same kind at the endpoints of each edge lying on ℓ . Reduce the weight ofeach such edge by half; leave all other weights unchanged. Denote by G ′ the resulting graph.The main result of this section is the following. Theorem 6.2.
Let G be a weighted, turn-bipartite symmetric graph embedded in a region R sothat all the holes of R are along the symmetry axis. Then M( G ) = 2 w( G ) M( G ′ ) . (6.1)Note that in the special case when the region R is simply connected, G is bipartite, and theabove result becomes the factorization theorem [3, Theorem 1.2].Our proof will follow from Lemma 6.1 and the following two additional lemmas. A doublyreduced subgraph of G is a graph obtained from G by cutting either above or below each a i and b i , i = 1 , . . . , w( G ). Lemma 6.3.
Let G be a symmetric turn-bipartite graph. (a) . If G is not bipartite, then G must have a symmetric odd cycle. (b) . All doubly reduced subgraphs of G are bipartite.Proof. (a). As G is not bipartite, it has an odd cycle C . Since C is odd and G is turn-bipartite, C goes around an odd number of holes of R , so in particular C must cross the symmetry axis ℓ .Let C ′ be the mirror image of C across ℓ . Let ¯ C be the cycle consisting of the edges of theunbounded face of the graph C ∪ C ′ . Then ¯ C is symmetric, and goes around precisely thoseholes of the region R that C goes around. Since G is turn-bipartite and C is odd, ¯ C is alsoodd. (b). Suppose C is an odd cycle of G . We claim that C must have at least one vertex on ℓ .Indeed, consider the symmetric cycle ¯ C from the proof of part (a). By construction, ¯ C and C have the same set of vertices on ℓ . But if ¯ C does not have any vertex on ℓ , then all vertices on itcome in symmetric pairs, which implies that ¯ C has an even number of vertices, a contradiction.Since the cutting operations involved in the definition of each doubly reduced subgraph H guarantee that all such cycles C of G are interrupted, it follows that there is no odd cyclein H . (cid:3) Lemma 6.4.
Let G be a connected, symmetric turn-bipartite graph which is not bipartite.Set k = w( G ) . Assume that { a , . . . , b k } is an independent set, and that each vertex in it hasprecisely one incident edge from above and one from below. (a) . Let u and u ′ be vertices of G that are mirror images across ℓ . Then for any doublyreduced subgraph H of G , u and u ′ have opposite colors in H . (b) . Let x be a fixed reference vertex in G \ { a , . . . , b k } , and let H be a doubly reducedsubgraph of G . Color the vertices in the bipartition classes of H black and white so that x iswhite. Then: ( i ) for any vertex v ∈ G \ { a , . . . , b k } , the color of v is uniquely determined, and isindependent of H . ( ii ) the color of each a i and b i is determined by whether the cut above or the cut belowthem was made to obtain H .Proof. Note that G connected implies that G ≥ is connected. Indeed, let u and v be two verticesof G on or above ℓ . Since G is connected, there is a path in G that connects u to v . Reflectingany portion of P that is below ℓ across the symmetry axis yields a path in G ≥ connecting u to v .Note also that this in turn implies that the subgraph graph G > induced by the vertices of G above ℓ is connected. Indeed, the path P in G ≥ between two vertices above ℓ cannot visit avertex on ℓ , otherwise the unique neighbor in G ≥ of that vertex would be visited twice by P .(a). By Lemma 6.3(a), G has a symmetric odd cycle C . Let c and c ′ be mirror image verticeson C , with c above ℓ . Since G > is connected, there is a path P in it from u to c . Follow P from u to c , then follow C to c ′ choosing the odd arc (the sum of the lengths of the two arcs of C between c and c ′ is odd, so one of them is odd). Travel from c ′ to u ′ along the mirror image P ′ of P . This is a walk from u to u ′ with an odd number of edges, and since it does not containany vertex on ℓ ( P and P ′ clearly don’t, and the odd arc of C doesn’t either, otherwise bysymmetry it would have even length), this walk is contained in all doubly reduced subgraphsof G .(b). ( i ). Suppose first that v is above ℓ . As G > is connected, we can choose a path P in itconnecting x to v . Clearly, P is included in H , and is the same for all doubly reduced subgraphs H . Since the parity of its length determines whether v and x have the same or opposite colorin H , this proves the statement in this case.If v is below ℓ , use the same argument, but with x replaced by its mirror image x ′ . By part (a),the color of x ′ in H is determined (namely, it must be black), and the proof is complete.( ii ). In H , each a i is incident to precisely one vertex v / ∈ ℓ . Since by ( i ) the color of v isdetermined, so is the color of a i . Since, by part (a), v and its mirror image v ′ have oppositecolors, the color of a i is reversed if the cut at a i is reversed. The same argument works forthe b i ’s. (cid:3) NEW SOLUTION FOR THE TWO DIMENSIONAL DIMER PROBLEM 21
Proof of Theorem . . Without loss of generality we may assume that G is connected (otherwisejust apply the statement for each of its connected components, and multiply).Set k = w( G ). We prove first the special case when { a , . . . , b k } is an independent set, andeach vertex in it has precisely one incident edge from above, and one from below.We claim that if H is a doubly reduced subgraph of G , x is the number of its vertices a i thathave the same color as a , and y is the number of its vertices b i that have color opposite to thecolor of a , then H is balanced if and only if x = y .Indeed, suppose for definiteness that a is white, and let α and β be the number of whiteand black vertices of H above ℓ , respectively. By Lemma 6.4(a), the number of white verticesof H below ℓ is β , and the number of its black vertices below ℓ is α . Thus, the total numberof white vertices of H is α + β + x + ( k − y ), while the total number of its black vertices is β + α + ( k − x ) + y . These are equal precisely if x = y , as claimed.By Lemma 6.4(b)( ii ), there are exactly two doubly reduced subgraphs of G in which all a i ’shave the same color, and all b i ’s have the opposite color. Out of these two, let H be the onein which the cut above a was chosen.Let G , . . . , G k be the reduced subgraphs of G (recall that reduced subgraphs are obtainedfrom G by cutting only at the a i ’s), G being the one in which the cuts at the a i ’s are the sameas in H .We clearly have M( G ) = k X i =1 M( G i ) . (6.2)Furthermore, by Lemma 6.1, the subgraphs G , . . . , G k have the same weighted count of perfectmatchings. Therefore (6.2) implies M( G ) = 2 k M( G ) . (6.3)Let H , . . . , H k be the subgraphs of G obtained by cutting above or below b i , i = 1 , . . . , k , inall possible ways (the doubly reduced subgraph H defined above is clearly one of them). ThenM( G ) = k X i =1 M( H i ) . (6.4)However, by the third paragraph in this proof, in order for H i to be balanced, it needs to havethe same number of a i ’s of the same color as a as b i ’s of opposite color to a . Our definitionsof G and H , together with Lemma 6.4(b)( ii ), imply that all a i ’s have the same color ineach H i . It follows that all b i ’s must have the opposite color in order for H i to be balanced.Lemma 6.4(b)( ii ) implies that this happens in precisely one H i , namely in H . As all the othersummands in (6.3) are zero , equation (6.1) follows from (6.3), (6.4), and the readily checkedfact (which follows from Lemma 6.4) that the cuts made in H are precisely the ones made inthe definition of G ′ (which implies that G ′ = H in this case)The general case follows using the vertex splitting lemma [3, Lemma 1.3], as in the proof ofTheorem 1.2 in [3] . (cid:3) I.e., has the same number of vertices in its two bipartition classes. Since a bipartite graph has no perfect matching unless it is balanced. Square cylinder graphs of odd girth
As pointed out in the previous section, for odd m the approach presented there for findingM( C m,n ) does not work. However, there is an alternative approach that works for odd m (butnot when m is even!), based on the TFK formula and the factorization theorem of [3]. Theresult we obtain this way for odd m can be put together with the formula we obtained inTheorem 5.1 for even m , obtaining the following. Theorem 7.1.
For all non-negative integers m and n we have M( C m,n ) = vuut n Y k =1 " cos kπn + 1 + r kπn + 1 ! m cos kπn + 1 − r kπn + 1 ! m × vuut n Y k =1 " − cos kπn + 1 + r kπn + 1 ! m − cos kπn + 1 − r kπn + 1 ! m . (7.1)In fact, the alternative approach mentioned above leads to a formula (see (7.2) below) thatlooks quite different from (7.1). We prove this formula first, and then show that it agreeswith (7.1). Theorem 7.2.
For non-negative integers m and n we have M( C m +1 , n ) = n Y k =1 Y j ∈{ , ,..., m +1 } (cid:18) jπ m + 2 + 4 cos kπ n + 1 (cid:19) . (7.2) Proof.
Embed the rectangular cylinder graph C m +1 , n in an annulus as indicated on the leftin Figure 10. Then C m +1 , n is clearly symmetric and turn-bipartite. Apply Theorem 6.2 for G = C m +1 , n . One readily sees that the resulting graph G ′ is the (2 m + 1) × n rectangular gridgraph ˙ R m +1 , n , with the 2 n − /
2, and all others weightedby 1. Then Theorem 6.2 gives M( C m +1 , n ) = 2 n M( ˙ R m +1 , n ) . (7.3)In turn, M( ˙ R m +1 , n ) can be expressed in terms of numbers of matchings of unweighted rectan-gular grid graphs as follows. Consider the rectangular grid graph R m +1 , n , and apply to it theoriginal factorization theorem [3, Theorem 1.2] (this is illustrated in the picture on the right inFigure 10). This yields M( R m +1 , n ) = 2 n M( R m, n ) M( ˙ R m +1 , n ) . (7.4) For even m this becomes precisely Theorem 5.1. Suppose therefore that m is odd. Then the factors inthe first three products are always non-negative, and the ones in the fourth are always less or equal than zero.Therefore for even n , the overall product under the fourth root is non-negative. If n is odd, then — as we areassuming m odd — C m,n has an odd number of vertices, and thus no perfect matchings. This agrees with thegiven formula, as in that case the factor in the fourth product corresponding to k = ( n + 1) / NEW SOLUTION FOR THE TWO DIMENSIONAL DIMER PROBLEM 23
Figure 10.
Embedding the rectangular cylinder grid C , in the plane (left).Applying the factorization theorem to the rectangular grid R , (right).Combining (7.3) and (7.4) leads toM( C m +1 , n ) = M( R m +1 , n )M( R m +1 , n ) . (7.5)Using the TFK formula (see [3, 10])M( R m,n ) = ⌈ m/ ⌉ Y j =1 ⌈ n/ ⌉ Y k =1 (cid:18) jπm + 1 + 4 cos kπn + 1 (cid:19) (7.6)in (7.5), all factors at the denominator cancel out, and one obtains formula (7.2). (cid:3) Lemma 7.3.
For z an indeterminate we have Y k ∈{ , ,..., m +1 } (cid:18) z + 4 cos kπ m + 2 (cid:19) =2 z s − (cid:20) − (cid:16) z + √ z (cid:17) m +2 (cid:21) (cid:20) − (cid:16) z − √ z (cid:17) m +2 (cid:21) . (7.7) Proof.
The n th Chebyshev polynomial of the first kind, T n ( x ), is defined by T n (cos θ ) = cos( nθ ),and thus satisfies the recurrence T n +1 ( x ) = 2 xT n ( x ) − T n − ( x ). Together with the initialvalues T ( x ) = 1 and T ( x ) = x , this implies that T n ( x ) is a polynomial of degree n , and thatfor n ≥ n − . By its defining equation, the roots of T n ( x ) are thencos (2 k +1) π n , k = 0 , . . . , n −
1. Thus we have T m +1 ( x ) = 2 m x Y k ∈{ , ,..., m +1 } (cid:18) x − cos kπ m + 2 (cid:19) (cid:18) x + cos kπ m + 2 (cid:19) , (7.8) where the factor x in the denominator accounts for the fact that the root x = 0 appears twicein the product above. Using this, we obtain Y k ∈{ , ,..., m +1 } (cid:18) z + 4 cos kπ m + 2 (cid:19) = Y k ∈{ , ,..., m +1 } (cid:18) z − i cos kπ m + 2 (cid:19) (cid:18) z + 2 i cos kπ m + 2 (cid:19) = ( − i ) m +2 iz m T m +1 ( iz ) = ( − m +1 iz T m +1 ( iz ) . (7.9)On the other hand, we have (cid:20) − (cid:16) z + √ z (cid:17) m +2 (cid:21) (cid:20) − (cid:16) z − √ z (cid:17) m +2 (cid:21) = 1 − (cid:16) z + √ z (cid:17) m +2 − (cid:16) z − √ z (cid:17) m +2 + 1= − (cid:20)(cid:16) z + √ z (cid:17) m +1 + (cid:16) z − √ z (cid:17) m +1 (cid:21) , (7.10)so the right hand side of (7.7) becomes2 z (cid:20)(cid:16) z + √ z (cid:17) m +1 + (cid:16) z − √ z (cid:17) m +1 (cid:21) . (7.11)Therefore, to complete the proof it suffices to show that the right hand side of (7.9) agrees with(7.11), which in turn amounts to T m +1 ( iz ) = 12 ( z + √ z ) m +1 + ( z − √ z ) m +1 ( − m +1 i . (7.12)Writing ( − m +1 i = ( − i ) m +1 , this readily follows from the well-known fact (see e.g. [1]) thatthe Chebyshev polynomial of the first kind T n ( x ) can be written as T n ( x ) = ( x + √ x − n + ( x − √ x − n . (cid:3) (7.13) Proof of Theorem . . For even m , the statement follows directly from Theorem 5.1. Supposetherefore that m is odd. Then, if n is also odd, the expression on the right hand side of (7.1)is zero, and (7.1) holds, as C m,n has then an odd number of vertices, and thus no perfectmatchings. NEW SOLUTION FOR THE TWO DIMENSIONAL DIMER PROBLEM 25
For the remaining case, write the cylinder graph as C m +1 , n , to spell out the parities of itsparameters. One readily sees that (7.1) is then equivalent toM( C m +1 , n ) = vuuut ( − n n Y k =1 cos kπ n + 1 + r kπ n + 1 ! m +1 cos kπ n + 1 − r kπ n + 1 ! m +1 × vuuut n Y k =1 − cos kπ n + 1 + r kπ n + 1 ! m +1 − cos kπ n + 1 − r kπ n + 1 ! m +1 . (7.14)Apply Lemma 7.3 to each inner product in equation (7.2), once for each z = cos kπ n +1 , k =1 , . . . , n . Since Q nk =1 cos kπ n +1 = n , this yields equation (7.14). (cid:3) Remark . Formula (7.1) looks quite different from formula (7.2), but as we saw above the twoare the same when m is odd in the former. In the proof of Lemma 7.3, the fact that the m -parameter (denoted there by 2 m + 1) is odd was essential in the proof (both for equations (7.10)and (7.12)). It doesn’t seem that for even m one can rewrite formula (7.1) in the style of theTFK formula. 8. Concluding remarks
In this paper we presented a new method for solving the two dimensional dimer problem. Ourgeneral result (see Theorem 2.1) expresses the number of dimer coverings of graphs built froma certain type of linear building blocks (called strands) as the determinant of a matrix whichis the product of the bi-adjacency matrices of its strands, or the latter plus a multiple of theidentity matrix. For the case of the hexagonal and square lattices, we evaluated the resultingdeterminants and obtained explicit product formulas for the number of dimer coverings ofhexagonal and square cylinder graphs (see Theorems 3.1 and 7.1). An interesting (and puzzling)feature of our formulas is that they look quite different from the classical TFK formula (seeequation (7.6)), which enumerates the dimer coverings of the closely related rectangular gridgraphs. For one thing, our formulas have a linear number of factors, while the number of factorsin the TFK formula is quadratic. A partial connection between these two types of formulas wasgiven in Section 7. It would be interesting to elucidate further the relationship between them.
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