Diagonal Transformations in Quadrangulations and Dehn Twists Preserving Cycle Parities

Journal of Combinatorial Theory, Series B  TB1730 journal of combinatorial theory, Series B 69, 125141 (1997) article no. TB971730

Diagonal Transformations in Quadrangulations and Dehn Twists Preserving Cycle Parities Atsuhiro Nakamoto* Department of Mathematics, Faculty of Education, Yokohama National University, 156 Tokiwadai, Hodogaya-ku, Yokohama 240, Japan

and Katsuhiro Ota Department of Mathematics, Faculty of Science and Technology, Keio University, 3-14-1 Hiyoshi, Kohoku-ku, Yokohama 223, Japan Received October 19, 1995

It will be shown that any two quadrangulations of an orientable closed surface with the same and sufficiently large number of vertices can be transformed into each other, up to isotopy, by a sequence of diagonal slides and diagonal rotations if they have the same homological structure.  1997 Academic Press

1. INTRODUCTION A quadrangulation G of a closed surface F 2 is a simple graph embedded in F 2 so that each face of G is quadrilateral. The diagonal slide and the diagonal rotation were defined in [4] as two transformations of quadrangulations. (See Fig. 1.) If the graph obtained by a diagonal slide is not a simple graph, then we do not apply it. Two quadrangulations G and G$ of F 2 are said to be equivalent up to homeomorphism and denoted by GtG$ if they can be transformed into each other by a sequence of diagonal slides and diagonal rotations, up to homeomorphism. If two quadrangulations G and G$ can be transformed into each other by a sequence of diagonal slides and diagonal rotations, up to isotopy, then they are said to be equivalent up to isotopy and are denoted by GrG$. There are two classes of quadrangulations, in which one is bipartite and the other is non-bipartite. Since both of the two transformations preserve the bipartiteness of quadrangulations, a bipartite quadrangulation and a non-bipartite one call never be * A research fellow of the Japan Society for the Promotion of Science.

125 0095-895697 25.00 Copyright  1997 by Academic Press All rights of reproduction in any form reserved.

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equivalent to each other. Observe that a diagonal slide preserves each partite set of bipartite quadrangulations while a diagonal rotation does not. Hence a diagonal rotation is actually needed to transform two bipartite quadrangulations with the same number of vertices but different size of partite sets. The following theorem was shown in [4]. Theorem 1 (Nakamoto [4]). For any closed surface F 2, there exists a positive integer N(F 2 ) such that any two bipartite quadrangulations G 1 and G 2 of F 2 are equivalent up to homeomorphism if |V(G 1 )| = |V(G 2 )| N(F 2 ). In general, the above theorem does not hold for non-bipartite quadrangulations and there exists a pair of inequivalent non-bipartite quadrangulations of the Klein bottle with the same and arbitrarily large number of vertices. (See [4] for details.) However, Negami and Nakamoto [9] have defined an invariant for quadrangulations which is preserved by diagonal slides and diagonal rotations, called the cycle parity. Let G be a graph 2-cell embedded in a closed surface F 2 so that each face of G is bounded by a cycle of even length. Any two cycles or closed walks of G on F 2 have the same length modulo 2 if they are homotopic to each other on F 2. Thus, we can assign ``0'' or ``1'' to each element of the fundamental group ? 1(F 2 ) of F 2, according to whether its length is even or odd. (For any # # ? 1(F 2 ), we can find a cycle (or a closed walk) in G homotopic to #, by the following procedures. First, draw # on F 2 where G is embedded. For each face f of G, consider a segment of # both of whose ends lie on the boundary cycle of f and whose middle lies in the face's interior. Using a local homotopy, we can push the middle of the segment to the boundary since the embedding is cellular. Doing this for all such segments, we can transform # to be a cycle or a closed walk homotopic to #.) Denote this assignment by \~ G : ?1(F 2 )  Z 2 . It is clear that \~ G is a homomorphism. By well known facts on homotopy and homology groups, \~ G factors through H 1(F 2 ; Z 2 ) uniquely, that is, there is a unique homomorphism \ G : H 1(F 2 ; Z 2 )  Z 2 such that \~ G =\ G } q, where q : ? 1(F 2 )  H 1(F 2 ; Z 2 ) is the composition of

FIG. 1.

The diagonal slide and the diagonal rotation.

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the abelianization and reduction modulo 2. This homomorphism \ G can be regarded as an element of the cohomology group H 1(F 2 ; Z 2 ) and is said to be the cycle parity of G. Note that for a quadrangulation G, \ G is trivial if and only if G is bipartite. Let G 1 and G 2 be two quadrangulations of a closed surface F 2. It is easy to see that both a diagonal slide and a diagonal rotation preserve the parity of length of cycles with the same homotopy type on F 2 before and after the transformations. Thus, if G 1 rG 2 , then \ G 1 =\ G 2 . Two cycle parities \ 1 and \ 2 in H 1(F 2 ; Z 2 ) are said to be congruent and are denoted by \ 1 #\ 2 if there is a homeomorphism h : F 2  F 2 which induces a homomorphism h* : H 1(F 2 ; Z 2 )  H 1(F 2 ; Z 2 ) with h*( \ 2 )=\ 1 . Thus, if G 1 tG 2 , then \ G 1 #\ G 2 . Using this invariant, we could show the following theorem which implies Theorem 1 [3]. As is mentioned above, a diagonal rotation is needed in the bipartite cases, but it is known that a diagonal rotation in non-bipartite quadrangulations can be obtained by a sequence of diagonal slides [9]. Theorem 2 (Nakamoto [3]). For any closed surface F 2, there exists a positive integer M(F 2 ) such that any two quadrangulations G 1 and G 2 of F 2 with |V(G 1 )| = |V(G 2 )| M(F 2 ) are equivalent up to homeomorphism if and only if \ G 1 #\ G 2 . For the sphere S 2, the projective plane P 2, the torus T 2, and the Klein bottle K 2, the values of M( } )'s in Theorem 2 have been determined as follows: M(S 2 )=4 [9];

M(P 2 )=4[9];

M(T 2 )=10 [6];

M(K 2 )=10 [5].

In proving Theorems 1 and 2, it is essential that for any closed surface, the number of irreducible quadrangulations is finite, up to homeomorphism. (The term ``irreducible'' means minimal with respect to the face-contraction for quadrangulations defined in [4]. It has been known that for any closed surfaces the number of vertices of irreducible quadrangulations is finite [7].) Since we showed the finiteness by focusing the combinatorial structure of irreducible quadrangulations, that is, we bound the number of vertices, we cannot distinguish the difference of two homeomorphic quadrangulations, and hence Theorems 1 and 2 include the condition ``up to homeomorphism''. We cannot omit the condition ``up to homeomorphism'' so easily since there exist infinitely many non-isotopic irreducible quadrangulations with the same combinatorial structure. In this paper, we will regard two homeomorphic but non-isotopic quadrangulations as two distinct ones and establish the theorem with the condition ``up to isotopy.'' The following theorem is our main theorem which is an extension of Theorem 2 for orientable closed surfaces.

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Theorem 3. For any orientable closed surface S g , there exists a positive integer m(Sg ) such that any two quadrangulations G 1 and G 2 of S g with |V(G 1 )| = |V(G 2 )| m(S g ) are equivalent up to isotopy if and only if \ G 1 =\ G 2 . Indeed, investigating the group consisting of the isotopy classes of homeomorphisms of closed surfaces, called the mapping class group, the isotopy version of Theorem 1 has already been established in [8]. Roughly speaking, a homeomorphism of an orientable closed surface S g can be represented as a composition of ``twists'' of S g . (The twist of S g is to cut S g along a simple closed curve, full-twist one boundary and identify again.) We could show the isotopy version of Theorem 1 by expressing each twist of a closed surface as a sequence of diagonal slides. This method depends on the fact that any homeomorphism preserves the bipartiteness of quadrangulations. However, a cycle parity is not always preserved by homeomorphisms in general, while each diagonal slide must preserve the cycle parity as mentioned above. Thus, no sequence of diagonal slides can realize a homeomorphism which changes the cycle parity. In this paper, breaking such a dilemma, we shall show Theorem 3. We will need a more topological argument to show our main theorem than the theorem in [8], but our method is quite combinatorial and requires little knowledge of topology.

2. DEHN TWISTS AND DIAGONAL SLIDES Through this paper, we denote the orientable closed surface of genus g by S g . Let F 2 be a closed surface and let 1(F 2 ) denote the group of all the homeomorphisms from F 2 onto itself. Then the set consisting of homeomorphisms isotopic to the identity map, denoted by 8(F 2 ), forms a normal subgroup in 1(F 2 ). The quotient group 1(F 2 )8(F 2 ) is called the mapping class group (or the homeotopy group) of F 2 and is denoted by 4(F 2 ). The mapping class group 4(F 2 ) is the set of isotopy classes of homeomorphisms of F 2 and contains at most countably many elements. Let # be a 2-sided simple closed curve on a closed surface F 2, that is, one with an annular neighborhood. Cut open F 2 along #, twist one of the boundary components 360 degree and identify the resulting two boundary components again. This deformation of F 2 naturally induces an autohomeomorphism of F 2. We call this homeomorphism (or its isotopy class) the Dehn twist along # and denote by {(#). We also call {(#) n a n-times Dehn twist along #. It is known that the mapping class group 4(S g ) of S g(g1) is generated by Dehn twists and one orientation-reversing homeomorphism, and that

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Dehn twists generate a subgroup of 4(S g ) of index 2 [2]. Humphries has determined generators of the subgroup as in the following lemma. Lemma 4 (Humphries [1]). Any orientation-preserving homeomorphism of S g(g2) is expressed by a sequence of Dehn twists along the 2g+1 simple closed curves x 1 , ..., x g , y 1 , y 2 , z 1 , ..., z g&1 shown in Fig. 2. Any orientationpreserving homeomorphism of the torus S 1 is expressed by a sequence of Dehn twists along x 1 and y 1 . Let [u 0 , ..., u n&1 ] and [v 0 , ..., v n&1 ] be the sets of the vertices of two cycles C n of length n, respectively. Then join u i and v i for each i to make the graph C n _K 2 . If this graph is embedded in an annulus so that the two cycles u 0 } } } u n&1 and v 0 } } } v n&1 coincide with the two boundaries of the annulus respectively and that each v i u i u i+1 v i+1 bounds a quadrilateral face (the subscripts are taken modulo n), then the graph embedded in the annulus is called the ladder strip L. A simple closed curve in L which runs across the middle points of u i v i 's for n=0, ..., n&1 in this order is called the center line of L. If n is even (resp., odd), then L is called even (resp., odd). A quadrangulation G is said to have or contain a ladder strip L if G contains L as an induced subgraph and each cycle v i u i u i+1 v i+1 of L bounds a face in G. Lemma 5. Let G be a quadrangulation of a closed surface containing a ladder strip L with its center line l. Then {(l) is obtained by a sequence of diagonal slides if L is even. If L is odd, we can obtain {(l) 2. Proof. Suppose that G contains a ladder strip L with its center line l. Apply a diagonal slide to replace u i v i with u i+1 v i&1 for i=0, 1, ..., n&1, in this order. Notice that each diagonal slide in the sequence does not break the simpleness of a graph since the ladder strip is induced. Then, by this operation, the resulting quadrangulation contains the edges u i v i&2 for i=0, 1, ..., n&1 in the ladder strip. Repeating this operation, we can obtain {(l) if n is even. However, if n is odd, then we can not obtain {(l) but can obtain {(l ) 2. K By Lemma 4, Dehn twists along the system of the simple closed curves x 1 , ..., x g , y 1 , y 2 , z 1 , ..., z g&1 shown in Fig. 2 generate the subgroup of 4(S g ) of index 2. Assign the positive and negative directions to each simple closed curve in the system since there are two directions to apply a Dehn twist there. For example, we denote a Dehn twist along as x i in the positive and negative directions by {(x i ) and {(x i ) &1, respectively. Let G be a quadrangulation of S g . The quadrangulation obtained from G by {(#) is denoted by G {(#) , where # denotes a simple closed curve on S g . It is clear that G {(#) {(#) &1 =G. Similarly, we denote the quadrangulation obtained from G

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FIG. 2.

A set of Dehn twist generators of mapping class group of S g .

by a composition | of Dehn twists by G | . For a simple closed curve # on a closed surface F 2 , the image of # by a homeomorphism | of F 2 is also denoted by # | . It is easy to see that the following lemma holds. Lemma 6. Let # be a simple closed curve on a closed surface F 2. Then, for any homeomorphism | of F 2, {(# | ) is expressed by | &1{(#) |.

3. TRANSITION OF CYCLE PARITIES Let G be a quadrangulation of S g and \ G its cycle parity. As is mentioned in Section 1, \ G : H 1(S g ; Z 2 )  Z 2 is an element of the cohomology group H 1(S g ; Z 2 ) of S g , that is, \ G # Z 22 g . Let # be a simple closed curve on S g . Generally, G and G {(#) do not have the same cycle parities, that is, if \G (#)=0, then \ G =\ G {(#) while if \ G (#)=1, then \ G {\ G {(#) . Note that even if \ G (#)=1, then \ G =\ G {(#)2 , which implies that \ G {(#) =\ G {(#) &1 . This fact corresponds to that we can obtain {(l ) along the center line l of an even ladder strip L by a sequence of diagonal slides, but we can not obtain {(l) when L is odd. (See Lemma 5). In this section, we shall consider the transition of cycle parities by Dehn twists. We define the transition diagram H g of cycle parities \ : H 1(S g ; Z 2 )  Z 2 on S g as follows: V(H g ) :=[ \ :=b 1 b 2 } } } b 2 g # H 1(S g ; Z 2 )| \(x i )=b 2i&1 # Z 2 , \( y i )=b 2i # Z 2 ], where x i and y i (i=1, 2, ..., g) are simple closed curves on S g shown in Fig. 3. For :, ; # V(H g ), there is a directed edge :; # E(H g ) labeled {(#), denoted by (:, ;) {(#) , if and only if a Dehn twist along the positive direction of # changes the cycle parity : to ;, where # # [x 1 , ..., x g , y 1 , y 2 ,

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FIG. 3.

131

A generator x i and y i of H 1(S g ; Z 2 ).

z 1 , ..., z g&1 ]. In this case, if :{;, there also exists a directed edge ( ;, :) {(#) since a quadrangulation G and G {(#) 2 have the same cycle parity. We denote the pair of symmetric edges (:, ;) {(#) and ( ;, :) {(#) by (:, ;) # . Observe that H g contains many loops, which corresponds to Dehn twists preserving cycle parities, and that the vertex 00 } } } 0 =0 2 g of H g is an isolated vertex, 2g

which means that any Dehn twist in a bipartite quadrangulation preserves the bipartiteness. It is clear that each vertex of H g has 2 g+1 outgoing directed edges and 2 g+1 incoming ones, and that H g with all the loops omitted is a symmetric digraph. Let H g*=H g &[0 2 g ]. Figure 4 shows H 3 with all the loops omitted, in which we can find H 1 and H 2 without loops. Let : and ; be two distinct vertices of H g joined by a pair of symmetric edges (:, ;) {(#) and ( ;, :) {(#) . Observe that there are two ways to move from : to ;. One is along (:, ;) {(#) , and the other is along the reverse of (;, :) {(#) . The former means {(#) while the latter means {(#) &1. Thus, any orientation-preserving homeomorphism h of S g can be represented as a sequence | of Dehn twists {(#) and {(#) &1, where # # [x 1 , ..., x g , y 1 , y 2 , z 1 , ..., z g&1 ]. When the homeomorphism h changes the cycle parity \ G :=: of a quadrangulation G of S g into \ h(G) :=;, the sequence | corresponds to a walk (which might traverse some edges in the reverse direction) from : to ; in H g . If :=; (i.e., h preserves the cycle parity), then it corresponds to a closed walk in H g starting from :. Let D be a digraph. A subgraph T of D is said to be a pseudo-tree if T satisfies the following conditions: (i) with x,

if for x # V(D), x # V(T ), then T contains all the loops incident

(ii) for any symmetric pair of directed edges e, e$ # E(D), if e # E(T ), then e$ # E(T), (iii) T contains no cycle of length more than 2. (By (i) and (ii), T contains cycles of length 1 and 2.)

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FIG. 4.

The transition diagrams H 1 , H 2 and H 3 .

Now we shall define the spanning pseudo-tree T g of H g* as follows; 1. T 1 :=H * 1 . 2. T 2 :=H *&E, where E=[(0011, 1001) z 1 , (1101, 1111) y 2 , (0101, 2 0111) y 2 , (1111, 1110) x 2 , (0111, 0110) x 2 , (0110, 1100) z 1 ]. 3. T g(g3): (a) For :, ; # Z 22 g&2 , $ # Z 22 and # # [x 1 , ..., x g&1 , y 1 , y 2 , z 1 , ..., z g&2 ], (:$, ;$) # # E(T g ) if and only if (:, ;) # # E(T g&1 ). (b) (0 2 g&210, 0 2 g&211) x g , (0 2 g&40001, 0 2 g&41011) zg&1 , (0 2 g&40011, 0 2 g&41001) z g&1 # E(T g ).

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(0

(c) (0 2 g&41010, 0 2 g&41011) x g , 0100, 0 2 g&41110) z g&1 # E(T g ).

2 g&4

133

(0 2 g&6011001, 0 2 g&6010011) z g&1 ,

The following lemma is an essential argument for our purpose, but the proof of the lemma will be tedious for the readers. Lemma 7. For any edge (a, b) {(#) # E(H g )(a{b, # # [x 1 , ..., x g , y 1 , y 2 , z 1 , ..., z g&1 ]), there exists a finite sequence |=h 1 h 2 } } } h s such that (i) each h i (i=1, ..., s) is either {(l ) or {(l ) &1 for l # [x 1 , ..., x g , y 1 , y 2 , z 1 , ..., z g&1 ], (ii) {(#) and | represents the same homeomorphism of S g , (iii)

| corresponds to an a-b walk in T g .

Proof. It suffices to check each edge (a, b) {(#) # E(H g*)&E(T g ). In case g=1, since T 1 =H 1* , we need to check no edge. In case g=2, we check each edge of E(H *)&E(T 2 2 ) independently. For example, v For (0011, 1001) {(z 1 ), put (0011, 0001, 1011, 1001) {( y 2) {(z 1 ) {( y 2 ) &1 . Since the two simple closed curves y 2 and z 1 on S 2 do not cross, {( y 2 ) and {(z 1 ) are commutative. Thus, we have {( y 2 ) {(z 1 ) {( y 2 ) &1 ={(z 1 ). v For (0111, 0110) {(x 2 ) , put (0111, 1111, 1011, 1010, 1110, 0110) {( y 1 ) {(x 1 ) {(x 2 ) {(x 1 ) &1 {( y1 ) &1 . Since {(x 2 ) and {(x 1 ) are commutative and so are {( y 1 ) and {(x 2 ), we have {( y 1 ) {(x 1 ) {(x 2 ) {(x 1 ) &1{( y 1 ) &1 ={(x 2 ). In the case when g3, we shall show inductively that the lemma holds under the condition that the lemma holds if the genus is equal to g&1 and g&2. For a directed edge of T g , there is a symmetric edge. We shall check one edge of the symmetric pair since the same argument repeats for the other edge. Then we consider the following four cases. Case 1. (:$, ;$) {(#) for :, ; # Z 22 g&2 , $ # Z 22 , # # [x 1 , ..., x g&1 , y 1 , y 2 , z 1 , ..., z g&2 ]. If either :=0 2 g&2 or ;=0 2 g&2, then the edge considered here is not contained in H * g and hence we need not check it. Otherwise, by the hypothesis of induction, such an edge is replaceable with a path in T g&1 . Case 2. (:10, :11) {(xg ) for : # Z 22 g&2 . If :=0 2 g&2, 0 2 g&410, then (:10, :11) {(x g ) # E(T g ). So suppose that :{0 2 g&2, 0 2 g&410. By the hypothesis of induction, there exists a homeomorphism | corresponding to a path (:, 0 2 g&410) | in T g&1 , where | is a homeomorphism containing neither z g&1 nor x g . Thus, we put (:10, :11) {(x g ) =(:10, 0 2 g&41010, 0 2 g&41011, :11) |{(x g ) | &1 . Since | and {(x g ) are commutative, we have |{(x g ) | &1 ={(x g ).

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Case 3. (:00, :10) {(z g&1 ) for : # Z 22 g&2 . Since there are two types of edges which are represented as (:00, :10) {(z g&1) , we consider the following two cases. Subcase (a). (:0100, :1110) {(z g&1 ) for : # Z 2g&4 . We show that for any 2 , (:0100, :1110) {(z g&1 ) is replaceable with a path in T g containing : # Z 2g&4 2 (0 2g&40100, 0 2g&41110) {(z g&1 ) # E(T g ). If :=0 2g&4, then we have (0 2g&40100, 0 2g&41110) {(z g&1 ) # E(T g ). So we may suppose that :{0 2g&4. By the hypothesis of induction for Tg&2 , there exists a homeomorphism | corresponding to a path (:, 0 2g&610) | in T g&2 . Thus, we have (:0100, 0 2g&6100100, 0 2g&6101100, 0 2g&6000100) |{( y g&1 ) {(z g&2 ) , (0 2g&6000100, 0 2g&6001110) {(z g&1 ) , (:1110, 0 2g&6101110, 0 2g&6000110, 0 2g&6001110) |{(z g&2 ) {( y g&1 ) , and hence we have (:0100, ..., 0 2g&40100, 0 2g&41110, ..., :1110) |{( y g&1 ) {(z g&2 ) {(z g&1 ) {( y g&1 ) &1{(z g&2 ) &1 |&1 . Note that {( y g&1 ) is obtained as a sequence of {(x 1 ), ..., {(x g&1 ), {( y 1 ), {( y 2 ), {(z 1 ), ..., {(z g&2 ) when g3. Since y g&1 , z g&1 and z g&2 don't cross on S g , {( y g&1 ), {(z g&1 ) and {(z g&2 ) are all commutative. And {(z g&1 ) and {(z g&2 ) are commutative. Since | is a composition of {(x 1 ), ..., {(x g&2 ), {( y 1 ), {( y 2 ), {(z 1 ), ..., {(z g&3 ), | and {(z g&1 ) are commutative. Thus, we have |{( y g&1 ) {(z g&2 ) {(z g&1 ) {( y g&1 ) &1{(z g&2 ) &1| &1 ={(z g&1 ). Subcase (b). (:1100, :0110) {(zg&1 ) for : # Z 2g&4 . We show that for any 2 , (:1100, :0110) {(zg&1 ) is replaceable with a path in T g containing : # Z 2g&4 2 (0 2g&40100, 0 2g&41110) {(zg&1 ) # E(T g ). If :=0 2g&4, then we have (0 2g&41100, 0 2g&40100, 0 2g&41110, 0 2g&40110) {( y g&1 ) {(z g&1 ) {( y g&1 ) &1 . Note that {( y g&1 ) is obtained as a sequence of {(x 1 ), ..., {(x g&1 ), {( y 1 ), {( y 2 ), {(z 1 ), ..., {(z g&2 ) when g3. Since {(z g&1 ) and {( y g&1 ) are commutative, we have {( y g&1 ) {(z g&1 ) {( y g&1 ) &1 ={(z g&1 ).

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So now suppose that :{0 2g&4. By the hypothesis of induction for T g&2 , there exists a homeomorphism | corresponding to a path (:, 0 2g&610) | in T g&2 . Thus, we have (:1100, 0 2g&6101100, 0 2g&40100) |{(z g&2 ) , (0 2g&40100, 0 2g&41110) {(z g&1 ) , (:0110, 0 2g&6100110, 0 2g&6101110, 0 2g&40110, 0 2g&41110) |{( y g&1 ) {(z g&2 ) {( y g&1 ) &1 and hence we have (:1100, ..., 0 2g&40100, 0 2g&41110, ..., :0110) |{(z g&2 ) {(z g&1 ) {( y g&1 ) {(z g&2 ) &1 {( y g&1 ) &1 |&1 . Since {(z g&2 ) and {( y g&1 ) are commutative and since {(z g&1 ) is commutative with {(z g&2 ) and |, we have |{(z g&2 ) {(z g&1 ) {( y g&1 ) {(z g&2 ) &1 {( y g&1 ) &1 | &1 ={(z g&1 ). Case 4.

(:01, :11) {(z g&1 ) for : # Z 2g&2 . 2

Subcase (a).

(:1001, :0011) {(z g&1 ) for : # Z 2g&4 . 2

Subcase (b).

(:0001, :1011) {(z g&1 ) for : # Z 2g&4 . 2

The edges considered in Case 4 can be checked by the same way as in the previous case. Thus, we shall omit the proof and leave it to the reader. Considering the four cases, we have checked all the edges of H * g . Thus, the lemma follows. K Lemma 8. Let G be a quadrangulation of S g with the cycle parity \ G . Then there exists a finite set 7 \ G =[* 1 , ..., * k , + k+1 , ..., + n ] of simple closed curves on S g such that \ G (* i )=0 and \ G (+ j )=1 for all i and j, and any orientation-preserving homeomorphism h of S g which preserves \ G can be represented as a finite sequence h 1 h 2 } } } h s , where each h l (l=1, ..., s) is either {(* i ), {(* i ) &1, {(+ j ) 2 or {(+ j ) &2 for some i and j. Proof. First, we define a finite set 7 \G of simple closed curves on S g . If G is bipartite, then 7 \ G =[* 1 , ..., * k ] :=[x 1 , ..., x g , y 1 , y 2 , z 1 , ..., z g&1 ]. Then, the lemma holds, by Lemma 4. Suppose that G is non-bipartite and \ G =a # V(H g*). Since T g is a pseudo-tree, for a vertex v # V(H g*), the path P a, v in T g (through no loop) from a to v along the positive direction of each edge is uniquely determined. Suppose that there are p loops incident with v, labeled {(l 1 ), ..., {(l p ), where [l 1 , ..., l p ]/[x 1 , ..., x g , y 1 , y 2 , z 1 , ..., z g&1 ]. Since P a, v can be regarded as a homeomorphism which changes the

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cycle parity a into v, the homeomorphism P a, v {(l i ) P &1 a, v can be represented as {((l i ) P a,&1v ), by Lemma 6. Let 0 v :=[(l 1 ) P a,&1v , ..., (l p ) P a,&1v ]. Now suppose that the length of P a, v is m. Let u 1 , ..., u q be the vertices of T g such that each u i is incident with v in T g and that the length of P a, u i in T g is m+1. Since each P v, u i is represented as a Dehn twist, say {(# i ), the 2 &1 homeomorphism P a, v P v, u i P u i, v P &1 a, v =P a, v {(# i ) P a, v can be represented as 2 {((# i ) P a,&1v ) , by Lemma 6. Let 9 v :=[(# 1 ) P a,&1v , ..., (# q ) P a,&1v ]. Let 7 \ G := . v # V(Hg )

0v _

.

9v

v # V(Hg )

:=[* 1 , ..., * k ] _ [+ k+1 , ..., + n ], where each of * i and + j is a simple closed curve on S g . Any orientation-preserving homeomorphism h can be represented as a sequence | of Dehn twists along x 1 , ..., x g , y 1 , y 2 , z 1 , ..., z g&1 , by Lemma 4, and hence h corresponds to a closed walk | in H * g , in which the reverse of directed edges might be also used in |. By Lemma 7, we can transform | into |$ so that |$ goes through the edges only in T g and that h and |$ represent the same homeomorphism of S g . If |$ contains a Dehn twist of negative direction, say {(#) &1, which is not a loop, then replace it with {(#) &2 {(#). Applying this operation to all such Dehn twists of negative direction, we suppose the closed walk |$ to be transformed into the closed walk |" in T g . Clearly, |" represents the same homeomorphism as h. Let \ G :=a and |" :=e 1 e 2 } } } e m and let W :=ae 1 v 1 e 2 v 2 } } } e m a be the closed walk corresponding to |" in T g , where for l=1, ..., m, v l # V(H *) g and e l is either {(* i ), {(* i ) &1, {(+ j ) 2 or {(+ j ) &2 for some i or j. If m=0, then |" represents the identity map. So, we suppose that m>0 and use induction on m. Since W is a closed walk in the pseudo-tree T g , that is, has no cycle of length more than 2, W must includes a closed walk C of length at most 2. Thus, the walk C in W can be represented as either v i e i+1 v i+1 with v i =v i+1 or v i&1 e i v i e i+1 v i+1 with v i&1 =v i+1 . Now choose C to be the cycle of length 1 or 2 which first arises in W. First, consider the case when C=v i e i+1 v i+1 with v i =v i+1 . Observe that a } } } v i is a directed path P a, vi (i.e., using only the positive direction of each edge) in T g by the definition of |" and the assumption of C. In this case,

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e i+1 is a loop, say {(#), incident with v i ( =v i+1 ). Since # # [x 1 , ..., x g , y 1 , y 2 , z 1 , ..., z g&1 ], we have that # P a,&1v # 7 \ G . Consider a Dehn twist {(# P a,&1v ) i i along # P a,&1v . By Lemma 6, i

&1 {(# P a,&1v )=P a, v i {(#) P &1 . a, v i =(e 1 e 2 } } } e i ) } e i+1 } (e 1 e 2 } } } e i ) i

Hence, we have that |"=e 1 e 2 } } } e m =(e 1 e 2 } } } e i ) } e i+1 } (e 1 e 2 } } } e i ) &1_e 1 e 2 } } } e i e i+2 } } } e m ={(# P a,&1v )_e 1 e 2 } } } e i e i+2 } } } e m . i

The second part e 1 } } } e i e i+2 } } } e m forms a closed walk, starting from a, in T g of length m&1 and hence, by the hypothesis of induction, it can be represented as a sequence of Dehn twists along the simple closed curves in 7 \ G . If e i+1 is labeled {(#) &1, then consider {(# P a,&1v ) &1 instead of {(# P a,&1v ). i i Now consider the case when C=v i&1 e i v i e i+1 v i+1 with v i&1 =v i+1 . Since C is a closed walk of length 2 in T g and since any pair of vertices of H* g is joined by at most one pair of symmetric edges, e i is supposed to be {(#) or {(#) &1 and so is e i+1 . If the composition of e i and e i+1 is the identity, then |" can be shortened and it can be represented as the sequence of Dehn twists along the simple closed curves in 7 \ G by the hypothesis of induction. If both e i and e i+1 are {(#), then similarly to the above case, we have, by Lemma 6, that {(# P a,&1v ) 2 =P a, v i {(#) 2 P &1 a, v i i

=e 1 e 2 } } } e i&1 } e i e i+1 } (e 1 e 2 } } } e i&1 ) &1. Thus, |"=e 1 e 2 } } } e m =(e 1 e 2 } } } e i&1 ) } e i e i+1 } (e 1 e 2 } } } e i&1 ) &1_e 1 e 2 } } } e i&1 e i+2 } } } e m ={(# P a,&1v ) 2_e 1 e 2 } } } e i&1 e i+2 } } } e m . i

By the hypothesis of induction, the second part can be represented as a sequence of Dehn twists along the simple closed curves in 7 \ G . If both e i&1 and e i+1 are {(#) &1, then consider {(# P a,&1v ) &2 instead of {(# P a,&1v ) 2. Thus, the i i lemma follows. K Remark. We know that a finite number of Dehn twists generate the subgroup of 4(S g ) of index 2, denoted by 4 +(S g ). Equivalently, any orientation-preserving homeomorphism of S g is expressed by a sequence of a finite number of Dehn twists. Lemma 8 claims that the isotopy classes of orientation-preserving homeomorphisms of S g which preserve the cycle

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parity form a subgroup of 4 +(S g ) with the finite number of generators, which are {(* 1 ), ..., {(* k ), {(+ k+1 ) 2, ..., {(+ n ) 2 as in the proof of Lemma 8. Lemma 9. Let G be a quadrangulation of a closed surface F 2 with the cycle parity \ and 7 :=[# 1 , ..., # n ] be a set of simple closed curves on F 2. Then there is a quadrangulation H of F 2 with \ H =\ which has the system of ladder strips L 1 _ } } } _ L n such that the center line of L i is isotopic to # i for i=1, ..., n. Proof. We shall construct the desired H from G by the following procedures. First, draw # 1 _ } } } _ # n on F 2 (where G is embedded) so that (i)

each # i crosses only edges of G transversely,

(ii)

more than two closed curves in 7 don't cross at one point,

(iii)

each intersection of two closed curves in 7 lies on no edge of G.

Secondly, put a vertex on each intersection of G _ # 1 _ } } } _ # n to make a new graph, say T, embedded in F 2. Observe that some faces of G are subdivided in T. Let f be one of such face of G and F the corresponding 2-cell region in T. If # i passes through the interior of f for some i, then # i crosses the boundary of f even times. Then, the boundary cycle of F has an even length in T. So, adding vertices on # i 's and adding edges suitably, we can quadrangulate F and hence obtain a quadrangulation T $ which contains a subdivision of G and # 1 _ } } } _ # n . Clearly, each # i (i=1, ..., n) is subdivided to form a cycle in T$, and we denote the cycle by #$i for i=1, ..., n. Finally, replace #$i with a ladder strip #$i _K 2 :=L i in T $ for i=1, ..., n simultaneously to obtain the graph H on F 2, where a crossing of two cycles #$i and #$j should be replaced with a quadrilateral face shared by L i and L j . Since each face of T $ and each face in ladder strips are quadrilateral, H is clearly a quadrangulation of F 2. Since each edge of G is subdivided by an even number of vertices in H, we can see that \ G =\ H . Now, check whether or not each ladder strip is induced. If H has an edge e such that e joins two vertices of a ladder strip and is not contained in the ladder strip, then subdivide e by two vertices and add edges to be quadrangular. Doing this for all such edges, we obtain the new H with the system of induced ladder strips L 1 _ } } } _ L n (in which each # i is the center line of L i ). Thus, H is a desired quadrangulation of F 2. K By Lemma 8, if we fix a cycle parity \, then the set 7 \ =[* 1 , ..., * k , + k+1 , ..., + n ] of simple closed curves on S g can uniquely be determined. Moreover, by Lemma 9, we can construct a quadrangulation H with \H =\ containing the system of ladder strips L 1 _ } } } _ L n such that each L i (resp., L j ) has * i (resp., + j ) as its center line.

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Notice that in order to show the following lemma, Lemma 8 is not sufficient. We know that any homeomorphism of Sg preserving the cycle parity can be represented as a sequence of {(* 1 ), ..., {(* k ), {(+ k+1 ) 2, ..., {(+ n ) 2 and their inverses, by Lemma 8, and that the quadrangulation H contains L 1 _ } } } _ L n . Moreover, we know that each Dehn twist in the sequence can be obtained by a sequence of diagonal slides, by Lemma 5. However, the quadrangulation obtained from H after some Dehn twists in the sequence does not always contain the system L1 _ } } } _ L n . For example, even if H contains L1 _ } } } _ L n , H {(*1 ) does not have it since {(* 1 ) deforms the ladder strips crossing L 1 . Thus, the following lemma is not so obvious. Lemma 10. Let H be the quadrangulation of S g as above. Then, for any orientation-preserving homeomorphism h of S g which preserves \, H and the image H h of H by h are equivalent, up to isotopy. Proof. By Lemma 5, a Dehn twist along the center line of a ladder strip is obtained by a sequence of diagonal slides. So, we shall show that for any orientation-preserving homeomorphism h : S g  S g which preserves \, H h can be transformed into H by a sequence of Dehn twists at the center line of ladder strips preserving the cycle parity. By Lemma 8, h can be represented as a sequence | of Dehn twists such that |=h 1 h 2 } } } h s , where each h l (l=1, ..., s) is either {(* i ), {(* i ) &1, {(+ j ) 2 or {(+ j ) &2 for some i or j. We use induction on the length s of | (denoted by ||| ). In the case when ||| =0, we have nothing to do. Assume that for any | with ||| s&1, H | rH. Now let |=h 1 |$, where ||$| =s&1. We suppose that h 1 ={(* i ) for some i since the same arguments follow when h 1 ={(* i ) &1, {(+ j ) 2, {(+ j ) &2. Since the first Dehn twist applied to H is {(* i ), L i itself is not changed by {(* i ), and hence H {(* i ) has a ladder strip L i . Thus, H | has a ladder strip L |$ . Now, apply the Dehn twist along L |$ of the negative direction. By Lemma 6, we obtain H {(*i )|$ _|$ &1{(*) &1 |$=H |$ . Thus, H | rH |$ . Since ||$| =s&1, H |$ rH, by the hypothesis of induction. Therefore, H | rH for any |. K

4. PROOF OF THEOREM 3 In this section, we shall prove Theorem 3. Before we do it, recall Theorem 2 which contains the condition ``up to homeomorphism''. We claim here that it is replaceable with ``up to orientation-preserving homeomorphism''. (It is easily realized as follows. In the proof of Theorem 2 for orientable closed surfaces, instead of the set of irreducible ones up to homeomorphism, prepare the set of irreducible ones up to orientationpreserving homeomorphism. That is, we have only to take, in addition to

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the set of irreducible ones up to homeomorphism, the set of their mirror images. See [3] for the detail.) Thus, the following proposition holds. Proposition 11. For any orientable closed surface S g , there exists a positive integer n(S g ) such that any two quadrangulations G 1 and G 2 with |V(G 1 )| = |V(G 2 )| n(S g ) are equivalent, up to orientation-preserving homeomorphism, if \ G 1 =\ G 2 . Recall the quadrangulation H of S g containing the system of ladder strips as in Lemma 10. For any S g and any \, there actually exists such H of Sg , by Lemma 9. Moreover, adding vertices of degree 2 into faces of H not to break the ladder strips, we can increase the number of vertices of H arbitrarily. Now we shall show Theorem 3. Proof of Theorem 3. Let m(S g ) :=max[n(S g ), |V(H)| ]. Let G 1 and G 2 be two quadrangulations of S g with |V(G 1 )| = |V(G 2 )| m(S g ) and \ G 1 = \ G 2 =\ H . We may suppose that |V(G 1 )| =|V(G 2 )| = |V(H)|. By Proposition 11, since |V(G 1 )| = |V(G 2 )| = |V(H)| n(S g ) and since \ G 1 =\ G 2 =\ H , we have that G 1 rh 1(H) and G 2 rh 2(H), where h i (i=1, 2) denotes some orientation-preserving homeomorphism of S g which preserves the cycle parity. Since h 1(H)rh 2(H) by Lemma 10, we obtain that G 1 rG 2 . K In the transition diagram H g of cycle parities on S g , the vertex 0 2g is 2g isolated and H *=H g g &[0 ] is connected. Hence H g has two components. By Theorem 2 and the definition of the transition diagram of cycle parities, we can say that the number of components in the transition diagram of cycle parities on a closed surface F 2 is equal to the number of congruence classes of those on F 2. For an orientable closed surface S g , since any two cycle parities of non-bipartite quadrangulations are congruent, the condition of congruent cycle parities in Theorem 2 is replaceable with the bipartiteness of quadrangulations. Therefore, the following theorem holds. Theorem 12. For any orientable closed surface S g , there exists a positive integer M(S g ) such that any two quadrangulations G 1 and G 2 with |V(G 1 )| = |V(G 2 )| M(S g ) are equivalent, up to homeomorphism, if and only if G 1 and G 2 are either both bipartite or both non-bipartite.

REFERENCES 1. S. P. Humphries, Generators for the mapping class group, in ``Topology and LowDimensional Manifolds,'' Lecture Notes in Mathematics, Vol. 772, pp. 4447, SpringerVerlag, New YorkBerlin, 1979.

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2. W. B. R. Lickorish, A finite set of generators for the homeotopy group of a 2-manifold, Proc. Cambridge Philos. Soc. 60 (1964), 769778. 3. A. Nakamoto, Diagonal transformations and cycle parities of quadrangulations on surfaces, J. Combin. Theory Ser. B 67 (1996), 202211. 4. A. Nakamoto, Diagonal transformations in quadrangulations of surfaces, J. Graph Theory 21 (1996), 289299. 5. A. Nakamoto, Irreducible quadrangulations of the Klein bottle, Yokohama Math. J. 43 (1996), 135149. 6. A. Nakamoto, Irreducible quadrangulations of the torus, J. Combin. Theory Ser. B 67 (1996), 183201. 7. A. Nakamoto and K. Ota, Note on irreducible triangulations of surfaces, J. Graph Theory 20 (1995), 227233. 8. A. Nakamoto and K. Ota, Diagonal transformations of graphs and Dehn twists on surfaces, J. Combin. Theory Ser. B, to appear. 9. S. Negami and A. Nakamoto, Diagonal transformations of graphs on closed surfaces, Sci. Rep. Yokohama Natl. Univ. Sec. I 40 (1993), 7197.

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