Resonance Graphs and Perfect Matchings of Graphs on Surfaces
aa r X i v : . [ m a t h . C O ] O c t Resonance Graphs and Perfect Matchings of Graphs on Surfaces
Niko Tratnik ∗ and Dong Ye † May 21, 2018
Abstract
Let G be a graph embedded in a surface and let F be a set of even faces of G (faces boundedby a cycle of even length). The resonance graph of G with respect to F , denoted by R ( G ; F ), is agraph such that its vertex set is the set of all perfect matchings of G and two vertices M and M areadjacent to each other if and only if the symmetric difference M ⊕ M is a cycle bounding some facein F . It has been shown that if G is a matching-covered plane bipartite graph, the resonance graphof G with respect to the set of all inner faces is isomorphic to the covering graph of a distributivelattice. It is evident that the resonance graph of a plane graph G with respect to an even-face set F may not be the covering graph of a distributive lattice. In this paper, we show the resonance graphof a graph G on a surface with respect to a given even-face set F can always be embedded into ahypercube as an induced subgraph. Furthermore, we show that the Clar covering polynomial of G with respect to F is equal to the cube polynomial of the resonance graph R ( G ; F ), which generalizesprevious results on some subfamilies of plane graphs. Keywords:
Perfect Matching, Resonance graph, Cube polynomial, Clar covering polynomial
Unless stated otherwise, the graphs considered in this paper are simple and finite. Let G be a graph withvertex set V ( G ) and edge set E ( G ). A perfect matching of G is a set of independent edges of G whichcovers all vertices of G . In other words, the edge induced subgraph of a perfect matching is a spanning1-regular subgraph, also called . Denote the set of all perfect matchings of a graph G by M ( G ).A surface in this paper always means a closed surface which is a compact and connected 2-dimensionalmanifold without boundary. An embedding of a graph G in a surface Σ is an injective mapping whichmaps G into the surface Σ such that a vertex of G is mapped to a point and an edge is mapped to a simplepath connecting two points corresponding to two end-vertices of the edge. Let G be a graph embeddedin a surface Σ. For convenience, a face of G is defined as the closure of a connected component of Σ \ G .The boundary of a face f is denoted by ∂f and the set of edges on the boundary of f is denoted by E ( f ).An embedding of G is a or cellular embedding if every face is homomorphic to a close disc. Note ∗ Faculty of Natural Sciences and Mathematics, University of Maribor, Slovenia. Email: [email protected]. Supportedby the Slovenian Research Agency. † Department of Mathematical Sciences and Center for Computational Sciences, Middle Tennessee State University,Murfreesboro, TN 37132, USA. Email: [email protected]. Partially supported by a grant from the Simons Foundation(No. 359516). G onΣ is a strong embedding (or closed 2-cell embedding) if the boundary of every face is a cycle. Denote theset of all faces of a graph G on a surface by F ( G ). A face f of G is even if it is bounded by an even cycleand a set of even faces is also called an even-face set . A cycle is a facial cycle of G if it is the boundary ofa face. If there is no confusion, a graph G on a surface always means an embedding of G in the surface,and a face sometime means its boundary cycle.Let G be a graph embedded in a surface Σ with a perfect matching M . A cycle C of G is M -alternating if the edges of C appear alternately between M and E ( G ) \ M . Moreover, a face f of G is M -alternating if ∂f is an M -alternating cycle. For a given set of even faces F ⊆ F ( G ), the resonancegraph of G with respect to F (or Z-transformation graph of G [27]), denoted by R ( G ; F ), is a graph withvertex set M ( G ) such that two vertices M and M are adjacent if and only if the symmetric difference M ⊕ M = ( M ∪ M ) \ ( M ∩ M ) is the boundary of a face f ∈ F , i.e. E ( f ) = M ⊕ M .Resonance graph was first introduced for hexagonal systems (also called benzenoid systems) whichare plane bipartite graphs with only hexagons as inner faces in [13], and reintroduced by Zhang, Guo andChen [27] in the name Z-transformation graph, and has been extensively studied for hexagonal systems[15, 22, 23, 28]. Later, the concept was extended to all plane bipartite graphs by Lam and Zhang [16, 26]and fullerenes [9, 25].A graph G is elementary if the edges of G contained in a perfect matching induce a connected subgraph(cf. [17]). An elementary bipartite graph is also matching-covered (or 1-extendable), i.e., every edge iscontained in a perfect matching. It is known that a matching-covered graph is always 2-connected [20]. Sois an elementary bipartite graph. It has been shown in [31] that a plane bipartite graph G is elementaryif and only if each face boundary of G is an M -alternating cycle for some perfect matching M of G . Theorem 1.1 (Lam and Zhang, [16]) . Let G be a plane elementary bipartite graph and F be the set ofall inner faces of G . Then R ( G ; F ) is the covering graph of a distributive lattice. The above result was also obtained independently by Propp [21]. Similar operations on other combi-natorial structures such as, spanning trees, orientations and flows have been discovered to have similarproperties [11, 21]. Moreover, such distributive lattice structure is also established on the set of all perfectmatchings of open-ended carbon nanotubes [24]. However, if G is not plane bipartite graph, the reso-nance graph of G with respect to some set of even faces may not be the covering graph of a distributivelattice (cf. [25]). It was conjectured in [25] that every connected component of the resonance graph of afullerene is a median graph and these graphs are a family of well-studied graphs (see [7, 8, 14]), containingthe covering graphs of distributive lattices. Median graphs are a subfamily of cubical graphs [1, 12, 18],which are defined as subgraphs of hypercubes, and have important applications in coding theory, datatransmission, and linguistics (cf. [12, 18]).In this paper, we consider the resonance graphs of graphs embedded in a surface in a very generalmanner. The following is one of the main results. Theorem 1.2.
Let G be a graph embedded in a surface and let F = F ( G ) be an even-face set. Thenevery connected component of the resonance graph R ( G ; F ) is an induced cubical graph. A of a graph G embedded in a surface is a spanning subgraph consisting of independentedges and cycles. A Clar cover of G is a 2-matching in which every cycle is a facial cycle. It has been2vident that the enumeration of Clar covers of a molecular graph has physical meaning in chemistry andstatistical physics. The Clar covering polynomial or Zhang-Zhang polynomial of graph G embedded in asurface is a polynomial used to enumerate all Clar covers of G . Zhang-Zhang polynomial was introducedin [30] for hexagonal systems. A definition of Zhang-Zhang polynomial will be given in the next section.Zhang et al. [29] demonstrate the equivalence between the Clar covering polynomial of a hexagonal systemand the cube polynomial of its resonance graph, which is further generalized to spherical hexagonalsystems by Berliˇc et al. [4] and to fullerenes [25]. In this paper, we show the equivalence between theZhang-Zhang polynomial of a graph G embedded in a surface and the cube polynomial of its resonancegraph as follows. Theorem 1.3.
Let G be a graph embedded in a surface and let F = F ( G ) be a set of even faces. Thenthe Zhang-Zhang polynomial of G with respect to F is equal to the cube polynomial of the resonance graph R ( G ; F ) . The paper is organized as follows: some detailed definitions are given in Section 2, the proofs ofTheorem 1.2 and Theorem 1.3 are given in Section 3 and Section 4, respectively. We conclude the paperwith some problems as Section 5.
Let G be a graph and let u, v be two vertices of G . The distance between u and v , denoted by d G ( u, v )(or d ( u, v ) if there is no confusion) is the length of a shortest path joining u and v . A median of a tripleof vertices { u, v, w } of G is a vertex x that lies on a shortest ( u, v )-path, on a shortest ( u, w )-path andon a shortest ( v, w )-path. Note that x could be one vertex from { u, v, w } . A graph is a median graph ifevery triple of vertices has a unique median. Median graphs were first introduced by Avann [2]. Mediangraphs arise naturally in the study of ordered sets and distributive lattices. A lattice is a poset such thatany two elements have a greatest lower bound and a least upper bound. The covering graph of a lattice L is a graph whose vertex set consists of all elements in L and two vertices x and y are adjacent if andonly if either x covers y or y covers x . A distributive lattice is a lattice in which the operations of thejoin and meet distribute over each other. It is known that the covering graph of a distributive lattice isa median graph but not vice versa [10].The n - dimensional hypercube Q n with n ≥
1, is the graph whose vertices are all binary strings oflength n and two vertices are adjacent if and only if their strings differ exactly in one position. Forconvenience, define Q to be the one-vertex graph. The cube polynomial of a graph G is defined asfollows, C ( G, x ) = X i ≥ α i ( G ) x i , where α i ( G ) denotes the number of induced subgraphs of G that are isomorphic to the i -dimensionalhypercube. The cube polynomials of median graphs have been studied by Breˇsar, Klavˇzar and ˇSkrekovski[7, 8].Let H and G be two graphs. A function ℓ : V ( H ) → V ( G ) is called an embedding of H into G if ℓ is injective and, for any two vertices x, y ∈ V ( H ), ℓ ( x ) ℓ ( y ) ∈ E ( G ) if xy ∈ E ( H ). If such a function ℓ exists, we say that H can be embedded in G . In other words, H is a subgraph of G . Moreover, if ℓ is an3mbedding such that for any two vertices x, y ∈ V ( H ), ℓ ( x ) ℓ ( y ) ∈ E ( G ) if and only if xy ∈ E ( H ), then H can be embedded in G as an induced subgraph. An embedding ℓ of graph H into graph G is calledan isometric embedding if for any two vertices x, y ∈ V ( H ) it holds d H ( x, y ) = d G ( ℓ ( x ) , ℓ ( y )). A graph H is ( induced ) cubical if H can be embedded into Q n for some integer n ≥ H is called a partial cube if H can be isometrically embedded into Q n for some integer n ≥
1. Formore information and properties on cubical graphs, readers may refer to [1, 5, 6, 12]. It holds that amedian graph is a partial cube (in fact, even stronger result is true, i.e. median graphs are retracts ofhypercubes, see [3]). Therefore, we have the nested relations for these interesting families of graphs: { covering graphs of distributive lattices } ( { median graphs } ( { partial cubes } (( { induced cubical graphs } ( { cubical graphs } . In the following, let G be a graph embedded in a surface Σ and let f be a face bounded by a cycleof G . If G has two perfect matchings M and M such that the symmetric difference M ⊕ M is acycle which bounds face f , then we say that M can be obtained from M by rotating the edges of f .Therefore, two perfect matchings M and M of G are adjacent in the resonance graph R ( G ; F ) if andonly if M can be obtained from M by rotating the edges of some face f ∈ F . We sometimes also saythat edge M M corresponds to face f or face f corresponds to edge M M .A Clar cover of G is a spanning subgraph S of G such that every component of S is either theboundary of an even face or an edge. Let F ⊆ F ( G ) be an even-face set. The Zhang-Zhang polynomialof G with respect to F (also called the Clar covering polynomial, see [30]) is defined as follows, ZZ F ( G, x ) = X k ≥ z k ( G, F ) x k , where z k ( G, F ) is the number of Clar covers of G with exact k faces and all the k faces belong to F . Notethat z ( G, F ) equals the number of of perfect matchings of G , i.e., the number of vertices of the resonancegraph R ( G ; F ). Let G be a graph embedded in a surface and let F be a set of even faces of G such that F = F ( G ). Inthis section, we investigate the resonance graph R ( G ; F ) and show that every connected component of R ( G ; F ) is an induced cubical graph. Lemma 3.1.
Let G be a graph embedded in a surface and let F = F ( G ) be an even-face set. Assume that C = M M . . . M t − M is a cycle of the resonance graph R ( G ; F ) . Let f i be the face of G correspondingto the edge M i M i +1 for i ∈ { , , .., t − } where subscripts take modulo t . Then every face of G appearsan even number of times in the face sequence ( f , f , ..., f t − ) .Proof. Let f be a face of G , and let δ ( f ) be the number of times f appears in the face sequence( f , f , . . . , f t − ). It suffices to show that δ ( f ) ≡ C = M M . . . M t − M is a cy-cle of R ( G ; F ) and f i is the corresponding face of the edge M i M i +1 , it follows that M i ⊕ M i +1 = E ( f i )for i ∈ { , , ..., t − } . So E ( f ) ⊕ E ( f ) ⊕ . . . ⊕ E ( f t − ) = ⊕ t − i =0 ( M i ⊕ M i +1 ) = ∅ (1)4here all subscripts take modulo t .Let f and g be two faces of G such that E ( f ) ∩ E ( g ) = ∅ , and let e ∈ E ( f ) ∩ E ( g ). Since e is containedby only f and g , and the total number of faces in the sequence ( f , f , . . . , f t − ) containing e is even by(1), it follows that δ ( f ) + δ ( g ) ≡ δ ( f ) ≡ δ ( g ) (mod 2). Therefore, all faces f of G havethe same parity for δ ( f ).Note that F = F ( G ). So G has a face g / ∈ F . Hence g does not appear in the face sequence. It followsthat δ ( g ) = 0. Hence δ ( f ) ≡ δ ( g ) ≡ f of G . This completes the proof.The following proposition follows immediately from Lemma 3.1. Proposition 3.2.
Let G be a graph embedded in a surface, and let F = F ( G ) be a set of even faces.Then the resonance graph R ( G ; F ) is bipartite.Proof. Let C = M M . . . M t − M be a cycle of R ( G ; F ) and let f i be the face corresponding to theedge M i M i +1 for i ∈ { , ..., t − } (subscripts take modulo t ). By Lemma 3.1, every face f of G appearsan even number of times in the sequence ( f , f , . . . , f t − ). So C is a cycle of even length. Therefore, R ( G ; F ) is a bipartite graph. f f f G M M M f f f R ( G ; F ( G ))Figure 1: A non-bipartite resonance graph where the double edges of G form M . The above proposition shows that the resonance graph R ( G ; F ) is bipartite if F = F ( G ). However,if F = F ( G ), then R ( G ; F ) may not be bipartite. For example, the graph G on the left in Figure 1 isa plane graph with three faces f , f and f . If F = { f , f , f } , then its resonance graph R ( G ; F ) is atriangle as shown on the right in Figure 1.It is known that a resonance graph R ( G ; F ) may not be connected [25]. In the following, we focus ona connected component H of R ( G ; F ), and always assume { f , . . . , f k } to be the set of all the faces thatcorrespond to the edges of H , which is a subset of F . Denote the set of all the edges of H that correspondto the face f i by E i for i ∈ { , . . . , k } . In the rest of this section, H \ E i denotes the graph obtained from H by deleting all the edges from E i . Lemma 3.3.
Let R ( G ; F ) be the resonance graph of a graph G on a surface with respect to a set of evenfaces F = F ( G ) , and let H be a connected component of R ( G ; F ) . Assume that M M ∈ E i where E i is the set of all the edges of H corresponding to some face f i ∈ F . Then M and M belong to differentcomponents of H \ E i .Proof. Suppose to the contrary that M and M belong to the same component of H \ E i . Then H \ E i hasa path P joining M and M . In other words, H has a path P joining M and M such that E ( P ) ∩ E i = ∅ .5et C = P ∪ { M M } be a cycle of H . Then f appears exactly once in the face sequence correspondingto the edges in the cycle C , which contradicts Lemma 3.1. The contradiction implies that M and M belong to different components of H \ E i .By Lemma 3.3, the graph H \ E i is disconnected for any face f i of G which corresponds to the edges in E i . Define the quotient graph H i of H with respect to f i to be a graph obtained from H by contracting alledges in E ( H ) \ E i and replacing any set of parallel edges by a single edge. So a vertex of H i correspondsto a connected component of H \ E i . Lemma 3.4.
Let R ( G ; F ) be the resonance graph of a graph G on a surface with respect to an even-faceset F = F ( G ) . Moreover, let f i be a face of G that corresponds to some edge of a connected component H of R ( G ; F ) . Then the quotient graph H i with respect to f i is bipartite.Proof. Suppose to the contrary that H i has an odd cycle. Then H has a cycle C which contains an oddnumber of edges corresponding to the face f i . In other words, the face f i appears an odd number oftimes in the face sequence corresponding to edges of C , which contradicts Lemma 3.1. Therefore, H i isbipartite.Recall that { f , . . . , f k } is the set of all the faces that correspond to the edges of a connected component H of R ( G ; F ), and H i is the quotient graph of H with respect to the face f i for i ∈ { , . . . , k } . ByLemma 3.4, let ( A i , B i ) be the bipartition of H i , and let M A i and M B i be the sets of perfect matchingsof G which are vertices of connected components of H \ E i corresponding to vertices of H i in A i and B i ,respectively. Define a function ℓ i : V ( H ) → { , } as follows, for any M ∈ V ( H ), ℓ i ( M ) = M ∈ M A i M ∈ M B i . Further, define a function ℓ : V ( H ) → { , } k such that, for any M ∈ V ( H ), ℓ ( M ) = ( ℓ ( M ) , . . . , ℓ k ( M )) . (2) Theorem 3.5.
Let G be a graph embedded in a surface, and let H be a connected component of theresonance graph R ( G ; F ) of G with respect to an even-face set F = F ( G ) . Then the function ℓ : V ( H ) →{ , } k defines an embedding of H into a k -dimensional hypercube as an induced subgraph.Proof. If G has no perfect matching, the result holds trivially. So, in the following, assume that G has aperfect matching.First, we show that the function ℓ : V ( H ) → { , } k defined above is injective, i.e., for any M , M ∈ V ( H ), it holds that ℓ ( M ) = ℓ ( M ) if M = M . Let P = M X . . . X t − M be a shortest path of H between M and M . Moreover, let g , . . . , g t be the faces corresponding to the edges of P such that g j corresponds to X j − X j for j ∈ { , ..., t } (where X = M and X t = M ). Note that, some faces g i and g j may be the same face for different i, j ∈ { , ..., t } . If every face of G appears an even number oftimes in the sequence ( g , . . . , g s ), then M = M ⊕ E ( g ) ⊕ E ( g ) ⊕ . . . ⊕ E ( g s ) = M , contradictingthat M = M . Therefore, there exists a face appearing an odd number of times in the face sequence( g , . . . , g s ). Without loss of generality, assume the face is f i . By Lemma 3.3, two end-vertices of an edgein E i belong to different connected components of H \ E i . Since the face f i appears an odd number of6imes in the face sequence ( g , . . . , g s ), it follows that if we contract all edges of P not in E i , the resultingwalk P ′ of H i joining the two vertices corresponding to the two components containing M and M hasan odd number of edges. Note that H i is bipartite by Lemma 3.4. So one of M and M belongs to M A i and the other belongs to M B i . So ℓ i ( M ) = ℓ i ( M ). Therefore, ℓ ( M ) = ℓ ( M ).Next, we show that ℓ defines an embedding of H into a k -dimensional hypercube. It suffices toshow that for any edge M M ∈ E ( H ), it holds that ℓ ( M ) and ℓ ( M ) differ in exactly one position.Assume that M M corresponds to a face f i ∈ F . In other words, the symmetric difference of twoperfect matchings M and M is the boundary of the face f i . For any j ∈ { , . . . , k } and j = i , theedge M M ∈ E ( H \ E j ) because M M ∈ E i and E i ∩ E j = ∅ . Therefore M and M belong to thesame connected component of H \ E j . Hence ℓ j ( M ) = ℓ j ( M ) for any j ∈ { , ..., k } , j = i . Since ℓ ( M ) = ℓ ( M ), it follows that ℓ ( M ) and ℓ ( M ) differ in exactly one position, the i -th position. Hence, ℓ defines an embedding of H into a k -dimensional hypercube.Finally, we are going to show that ℓ embeds H in a k -dimensional hypercube as an induced subgraph.It suffices to show that, for any M , M ∈ V ( H ), M M ∈ E ( H ) if ℓ ( M ) and ℓ ( M ) differ in exactlyone position. Without loss of generality, assume that ℓ i ( M ) = 0 and ℓ i ( M ) = 1 but ℓ j ( M ) = ℓ j ( M )for any j ∈ { , ..., k }\{ i } . By the definition of the function ℓ , we have M ∈ M A i and M ∈ M B i . Let P be a path of H joining M and M . Then contract all edges of P not in E i and the resulting walk P ′ of H i joins two vertices from different partitions of H i . So P ′ has an odd number of edges. In other words, | P ∩ E i | is odd. But, for any j ∈ { , ..., k } and j = i , contract all edges of P not in E j and the resultingwalk P ′′ joins two vertices from the same partition of H j . So | P ∩ E j | ≡ e ∈ E ( G ), the edge e is rotated an odd number of times along path P if e ∈ E ( f i ), but an even numberof times if e / ∈ E ( f i ). It follows that M ⊕ M = E ( f i ), which implies M M ∈ E ( G ). This completesthe proof.Our main result, Theorem 1.2, follows directly from Theorem 3.5. In this section, we show that the Zhang-Zhang polynomial (or Clar covering polynomial) of a graph G on asurface with respect to an even-face set F = F ( G ) is equal to the cube polynomial of the resonance graph R ( G ; F ), which generalizes the main results from papers [29, 4, 25] on benzenoid systems, nanotubes(also called tubulenes), and fullerenes. The proof of the equivalence of two polynomials, our main resultTheorem 1.3, combines ideas from [29, 4, 25] and [22]. However, this general setting of our result requiressome new ideas and additional insights into the role and structure of the resonance graph. The followingessential lemma generalizes a result of [29] originally proved for benzenoid systems. Lemma 4.1.
Let G be a graph embedded in a surface. If the resonance graph R ( G ; F ) of G with respectto an even-face set F = F ( G ) contains a 4-cycle M M M M M , then M ⊕ M = M ⊕ M and M ⊕ M = M ⊕ M . Further, the two faces bounded by M ⊕ M and M ⊕ M are disjoint.Proof. Since M M M M M is a 4-cycle in the resonance graph R ( G ; F ), let f i be the face of G suchthat E ( f i ) = M i ⊕ M i +1 where i ∈ { , , , } and subscripts take modulo 4. Note that E ( f ) ⊕ E ( f ) ⊕ E ( f ) ⊕ E ( f ) = ( M ⊕ M ) ⊕ ( M ⊕ M ) ⊕ ( M ⊕ M ) ⊕ ( M ⊕ M ) = ∅ . (3)7ince M i = M i +2 , it follows that f i = f i +1 , where i ∈ { , , , } and subscripts take modulo 4. So f i = f i +1 and f i = f i − . By (3), every edge on f i appears on another face f j with j = i . It follows that E ( f i ) ⊆ ∪ j = i E ( f j ) for i, j ∈ { , , , } . If f i is distinct from f j for any j = i , then all these faces togetherform a closed surface, which means that F ( G ) = F , contradicting F = F ( G ). Therefore, f i = f i +2 for i ∈ { , , , } . So it follows that f = f and f = f . In other words, M ⊕ M = M ⊕ M and M ⊕ M = M ⊕ M .To finish the proof, we need to show that the faces f and f are disjoint. Suppose to the contrarythat ∂f ∩ ∂f = ∅ . Note that f = f . So every component of ∂f ∩ ∂f is a path on at least twovertices. Let v be an end vertex of some component of ∂f ∩ ∂f . Therefore, v is incident with threeedges e , e and e such that e , e ∈ E ( f ) but e , e ∈ E ( f ). Since both f and f are M -alternating,it follows that e ∈ M . Note that M = M ⊕ E ( f ). So e / ∈ M . Since f = f , both f and f = f are M -alternating. Hence e ∈ M , contradicting e / ∈ M . This completes the proof. Remark.
Lemma 4.1 does not hold if F = F ( G ). For example, the resonance graph R ( G ; F ( G )) ofthe plane graph in Figure 2 (left) has a 4-cycle M M M M M which does not satisfy the property ofLemma 4.1, where F ( G ) = { f , ..., f } . f f f f G M M M M f f f f f f R ( G ; F ( G ))Figure 2: The resonance graph where the double edges of G form M . Now, we are going to prove Theorem 1.3.
Proof of Theorem 1.3.
Let G be a graph on a surface and let R ( G ; F ) be the resonance graph of G with respect to F . If G has no perfect matching, then the result holds trivially. So, in the following, wealways assume that G has a perfect matching.For an nonnegative integer k , let Z k ( G, F ) be the set of all Clar covers of G with exactly k faces suchthat all these faces are included in F , and let Q k ( R ( G ; F )) be the set of all labeled subgraphs of R ( G ; F )that are isomorphic to the k -dimensional hypercube. For a Clar cover S ∈ Z k ( G ; F ), let M , M , . . . , M t be all the perfect matchings of G such that all faces in S are M i -alternating and all isolated edges of S belong to M i for all i ∈ { , ..., t } . Define a mapping m k : Z k ( G, F ) −→ Q k ( R ( G ; F ))such that m k ( S ) is the subgraph of R ( G ; F ) induced by the vertex set { M , M , . . . , M t } . Since thesubgraph induced by { M , . . . , M t } is unique, the mapping m k is well-defined, which follows from thefollowing claim. 8 laim 1. For each Clar cover S ∈ Z k ( G ; F ) , the image m k ( S ) ∈ Q k ( R ( G ; F )) . Proof of Claim 1.
It is sufficient to show that m k ( S ) is isomorphic to the k -dimensional hypercube Q k .Let f , f , . . . , f k be the faces in the Clar cover S . Then { f , . . . , f k } ⊆ F . So each f i with i ∈ { , . . . , k } is even and hence has two perfect matchings labelled by “0” and “1” respectively. For any vertex M of m k ( S ), let b ( M ) = ( b , b , . . . , b k ), where b i = α if M ∩ E ( f i ) is the perfect matching of ∂f i withlabel α ∈ { , } for each i ∈ { , , . . . , k } . It is obvious that b : V ( m k ( S )) → V ( Q k ) is a bijection. For M ′ ∈ V ( m k ( S )), let b ( M ′ ) = ( b ′ , b ′ , . . . , b ′ k ). If M and M ′ are adjacent in m k ( S ) then M ⊕ M ′ = E ( f i )for some i ∈ { , , . . . , k } . Therefore, b j = b ′ j for each j = i and b i = b ′ i , which implies ( b , b , . . . , b k ) and( b ′ , b ′ , . . . , b ′ k ) are adjacent in Q k . Conversely, if ( b , b , . . . , b k ) and ( b ′ , b ′ , . . . , b ′ k ) are adjacent in Q k , itfollows that M and M ′ are adjacent in m k ( S ). Hence b is an isomorphism between m k ( S ) and Q k . Thiscompletes the proof of Claim 1.In order to show ZZ F ( G, x ) = C ( R ( G ; F ) , x ), it suffices to show that mapping m k is bijective for any k . Note that in the case of k = 0, a Clar cover S is a perfect matching of G and hence m k ( S ) is a vertexof R ( G ; F ). So the mapping m k is obviously bijective for k = 0. In the following, assume that k is apositive integer.First, we show that m k is injective. Let S and S ′ be two different Clar covers from Z k ( G ; F ). If S ∩ F = S ′ ∩ F , then the isolated edges of S and S ′ are different. So a perfect matching of S is differentfrom a perfect matching of S ′ . Therefore, the vertex sets of m k ( S ) and m k ( S ′ ) are disjoint. Hence m k ( S ) and m k ( S ′ ) are disjoint induced subgraphs of R ( G ; F ). So m k ( S ) = m k ( S ′ ). Now suppose that S ∩ F = S ′ ∩ F . Note that | S ∩ F | = | S ′ ∩ F | = k . So S ∩ F has a face f / ∈ S ′ ∩ F . Note that the facesadjacent to f do not all belong to S ′ since the faces in S ′ are independent. Hence the face f containsat least one edge e which does not belong to S ′ . From the definition of the function m k , the edge e does not belong to those perfect matchings of G that correspond to the vertices of m k ( S ′ ). For anyperfect matching M corresponding to a vertex of m k ( S ), the face f is M -alternating. Hence either M or M ′ = M ⊕ E ( f ) contains e . Without loss of generality, assume that e ∈ M ′ . So M ′ is not a vertexof M k ( S ′ ). Since both M and M ′ are perfect matchings of S , both M and M ′ are vertices of m k ( S ). So m k ( S ) = m k ( S ′ ). This shows that m k is injective.In the following, we are going to show that m k is surjective. Let Q ∈ Q k ( R ( G ; F )), isomorphic to a k -dimensional hypercube. Then every vertex u of Q can be represented by a binary string ( u , u , . . . , u k )such that two vertices of Q are adjacent in Q if and only if their binary strings differ in precisely oneposition. Label the vertices of Q by M = (0 , , , . . . , M = (1 , , , . . . , M = (0 , , , . . . , M k = (0 , , , . . . , M M i is an edge of R ( G ; F ) for every i ∈ { , . . . , k } . By definition of R ( G ; F ),the symmetric difference of perfect matchings M and M i is the boundary of an even face in F , denotedby f i . Then we have a set of faces { f , . . . , f k } ⊆ F . Note that f i = f j for i, j ∈ { , . . . , k } and i = j since M i = M j . Hence, all faces in { f , . . . f k } are distinct. In order to show that m k is surjective, it issufficient to show that G has a Clar cover S such that S ∩ F = { f , . . . , f k } . Claim 2.
All faces in { f , . . . , f k } are pairwise disjoint. Proof of Claim 2:
Let f i , f j ∈ { f , . . . , f k } with i = j and let W be a vertex of Q having exactly two 1’swhich are in the i -th and j -th position. Then M M i W M j M is a 4-cycle such that E ( f i ) = M ⊕ M i and E ( f j ) = M ⊕ M j . Then by Lemma 4.1, it follows that f i and f j are disjoint.9y Claim 2, we only need to show that G − ∪ ki =1 V ( f i ) has a perfect matching M so that S = M ∪ { f , . . . , f k } is a Clar cover of G . Consider the perfect matching M corresponding to the vertex of Q labelled by the string with k zero’s. Recall that E ( f i ) = M ⊕ M i and hence every f i is M -alternatingfor any i ∈ { , , . . . , k } . Therefore, M := M \ ( ∪ ki =1 E ( f i )) is a perfect matching of G − ∪ ki =1 V ( f i ). So S = M ∪ { f , . . . , f k } is a Clar cover of G such that m k ( S ) = Q . This completes the proof of that m k ( G )is surjective.From the above, m k is a bijection between the set of all Clar covers of G with k facial cycles and theset of all labeled subgraphs isomorphic to the k -dimensional hypercube for any integer k . Therefore, wehave ZZ F ( G, x ) = C ( R ( G ; F ) , x ) and this completes the proof of Theorem 1.3. Let G be a graph embedded in a surface Σ and let F be an even-face set. Assume that H is a connectedcomponent of R ( G ; F ). Let G H be the subgraph of G induced by the faces corresponding to edges of H . Proposition 5.1.
Let G be a graph embedded in a surface Σ and let F be an even-face set. If a perfectmatching M is a vertex of a connected component H of R ( G ; F ) , then M ∩ E ( G H ) is a perfect matchingof G H .Proof. Let M be a perfect matching corresponding to a vertex of H . Suppose to the contrary that M ∩ E ( G H ) is not a perfect matching of G H . Then G H has a vertex v which is not covered by M ∩ E ( G H ).Let f ∈ F be a face containing v , which corresponds to an edge of H . Then f is M ′ -alternatingfor some perfect matching M ′ which is a vertex of H . Since H is connected, there is a path P of H joining M and M ′ . Assume that the faces corresponding to the edges of P are f , . . . , f k . Then M = M ′ ⊕ E ( f ) ⊕ . . . ⊕ E ( f k ). Note that E ( f i ) ⊆ E ( G H ) for all i ∈ { , . . . , k } . So v is incident with anedge in M ∩ E ( G H ), a contradiction. This completes the proof.For a connected component H of R ( G ; F ), if the union of all faces corresponding to edges of H ishomeomorphic to a closed disc, then G H with the embedding inherited from the embedding from G in Σis a plane elementary bipartite graph. By Theorem 1.1, we have the following proposition. Proposition 5.2.
Let G be a graph on a surface Σ and F be an even-face set. Assume that H is aconnected component of R ( G ; F ) . If the union of all faces corresponding to edges of H is homeomorphicto a closed disc, then H is the covering graph of a distributive lattice. It has been evident in [25] that, if the subgraph induced by faces in F is non-bipartite, a connectedcomponent of R ( G ; F ) may not be the covering graph of a distributive lattice. But the condition inProposition 5.2 is not a necessary condition. It has been shown in [24] that a connected component of anannulus graph (a plane graph excluding two faces) could be the covering graph of a distributive lattice. Itis natural to ask what is the necessary and sufficient condition for F so that every connected componentof R ( G ; F ) is the covering graph of a distributive lattice.But so far, in all examples we have, a connected component of R ( G ; F ) is always a median graph.Therefore, we risk the following conjecture. Conjecture 5.3.
Let G be a graph embedded in a surface and let F = F ( G ) be an even-face set. Thenevery connected component of the resonance graph R ( G ; F ) is a median graph.