ars mathematica contemporanea Also available at http://amc-journal.eu ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 10 (2016) 281-290 Cube-contractions in 3-connected quadrangulations Yusuke Suzuki * Department of Mathematics, Niigata University, 8050 Ikarashi 2-no-cho, Nishi-ku, Niigata, 950-2181, Japan Received 4 October 2013, accepted 10 January 2016, published online 30 January 2016 Abstract A 3-connected quadrangulation of a closed surface is said to be K' -irreducible if no face- or cube-contraction preserves simplicity and 3-connectedness. In this paper, we prove that a K'-irreducible quadrangulation of a closed surface except the sphere and the projective plane is either (i) irreducible or (ii) obtained from an irreducible quadrangulation H by applying 4-cycle additions to F0 C F(H) where F(H) stands for the set of faces of H. We also determine K'-irreducible quadrangulations of the sphere and the projective plane. These results imply new generating theorems of 3-connected quadrangulations of closed surfaces. Keywords: Quadrangulation, closed surface, generating theorem. Math. Subj. Class.: 05C10 1 Introduction In this paper, we only consider simple graphs which have no loops and no multiple edges. We denote the vertex set and the edge set of a graph G by V(G) and E(G), respectively. We say that S c V(G) is a cut of G if G - S is disconnected. In particular, S is called a k-cut if S is a cut with |S| = k. A cycle C of G is said to be separating if V(C) is a cut. Similarly, a simple closed curve 7 on a closed surface F2 is said to be separating if F2 - 7 is disconnected. A quadrangulation G of a closed surface F2 is a simple graph cellularily embedded on the surface so that each face is quadrilateral; thus, a 2-path on the sphere is not a quadrangulation. We denote the set of faces of G by F(G) throughout the paper. For quadrangulations we consider applying three reductions, called a face-contraction, a 4-cycle removal *This work was supported by JSPS Grant-in-Aid for Young Scientists (B) No. 24740056. E-mail address: y-suzuki@math.sc.niigata-u.ac.jp (Yusuke Suzuki) ©® This work is licensed under http://creativecommons.org/licenses/by/3.0/ 282 Ars Math. Contemp. 10 (2016) 255-268 [voVi]!/ Avi face-contraction 4-cycle removal "t 3) into the sphere, put a vertex x on one side and a vertex y on Y. Suzuki: Cube-contractions in 3-connected quadrangulations 285 the other side and add edges xv and yu for i = 0,..., k —1. The resulting quadrangulation of the sphere with 2k + 2 vertices is said to be apseudo double wheel and denoted by W2k (see the left-hand side of Figure 3). The smallest pseudo double wheel is W6, which is isomorphic to a cube, when the graphs are assumed to be 3-connected. The cycle C of length 2k is called the rim of W2k. We call a quadrangulation of the sphere obtained from W6 by a single 4-cycle addition a double cube, which is isomorphic to C4 x P2. Secondly, embed a (2k -1)-cycle C = v0vi... v2k-2 (k > 2) into the projective plane so that the tubular neighborhood of C forms a Mobius band. Next, put a vertex x on the center of the unique face of the embedding and join x to v for all i so that the resulting graph is a quadrangulation. The resulting quadrangulation of the projective plane with 2k vertices is said to be a Mobius wheel and denoted by TW2fc-1 (see the right-hand side of Figure 3). 3 Lemmas to prove Theorem 1.4 The following lemma holds not only for quadrangulations but also for even embeddings of closed surfaces F2, that is, for graphs embedded on F2 with each face bounded by a cycle of even length. Taking a dual of an even embedding and using the odd point theorem, we can easily obtain this lemma. Lemma 3.1. An even embedding of a closed surface has no separating closed walk of odd length. Let G be a quadrangulation of a closed surface F2 and let f = v0v1 v2v3 be a face of G. Then a pair {vj, vi+2} is called a diagonal pair of f in G, where the subscripts are taken modulo 4. A closed curve 7 on F2 is said to be a diagonal k-curve for G if 7 passes only through distinct k faces f0,..., fk-1 and distinct k vertices x0,..., xk-1 of G such that for each i, f and fi+1 share xj, and that for each i, {xi-1, x4} forms a diagonal pair of f of G, where the subscripts are taken modulo k. Lemma 3.2. Let G be a quadrangulation of a closed surface F2 with a 2-cut {x, y}. Then there exists a separating diagonal 2-curve for G only through x and y. Proof. Observe that every quadrangulation of any closed surface F2 is 2-connected and admits no closed curve on F2 crossing G at most once. Thus there exists a surface separating simple closed curve 7 on F2 crossing only x and y, since {x, y} is a cut of G. We shall show that 7 is a diagonal 2-curve. Suppose that 7 passes through two faces f1 and f2 meeting at two vertices x and y. If 7 is not a diagonal 2-curve, then x and y are adjacent on df1 or df2. Since G has no multiple edges between x and y, and since {x, y} is a 2-cut of G, we may suppose that x and y are adjacent in df1, but not in df2. Here we can take a separating 3-cycle of G along 7. This contradicts Lemma 3.1. □ Lemma 3.3. Let G be a 3-connected quadrangulation of a closed surface F2, and let f = v0v1v2v3 be a face of G. If the face-contraction of f at {v0, v2} breaks 3-connectedness of the graph but preserves simplicity, then G has a separating diagonal 3-curve passing through v0,v2 and another vertex x G V(G) — {v0, v1, v2, v3}. Proof. Let G' be the quadrangulation of F2 obtained from G by the face-contraction of f at {v0, v2}. Since G' has connectivity 2, G' has a 2-cut. By Lemma 3.2, G' has a separating diagonal 2-curve 7' passing through two vertices of the 2-cut. Clearly, one of the two 286 Ars Math. Contemp. 10 (2016) 255-268 vertices must be [v0v2] of G', which is the image of v0 and v2 by the face-contraction of f. (Otherwise, G would not be 3-connected, a contradiction.) Let x be a vertex of G' on Y' other than [v0v2]. Note that x is not a neighbor of [v0v2] in G'. Now apply the vertex-splitting of [v0v2] to G' to recover G. Then a diagonal 3-curve for G passing through only v0,v2 and x arises from y' for G'. □ The next lemma plays an important role in a later argument. Lemma 3.4. Let G be a 3-connected quadrangulation on a closed surface F2. If G has a separating 4-cycle C = x0x1x2x3 and a face f of G such that (i) one of the diagonal pairs of f is {x^ xi+2} for some i, and (ii) f has a separating diagonal 3-curve y intersecting C only at xi and xi+2 transversely, then there exists a 3-contractible face in G. Proof. Suppose that G has a separating 4-cycle C = x0xix2x3 and a face f bounded by axicx3. Since C is separating, G has two subgraphs GR and GL such that GR U GL = G and Gr n Gl = C. Suppose that f is contained in GR. Furthermore, we assume that GR contains as few vertices of G as possible. Since C is separating, we have df = C. By (ii), f has a separating diagonal 3-curve Y through x1, x3 and some vertex x. Note that x G V (GL) - V (C) by the condition (ii) in the lemma. Now assume that f is not 3-contractible at {a, c}. Observe that y (or the 3-cut {x1, x, x3}) separates a from c. Further, G does not have both of edges ax and cx since df = C. Therefore, there is no path of G of length at most 2 joining a and c other than ax1c and ax3c. Moreover, if {a, c} n {x0, x2} = 0, then f has no separating diagonal 3-curve joining a and c. This contradicts our assumption by Lemma 3.3 and so we may suppose that a = x0 and c = x2, and f has a separating diagonal 3-curve, say y', through a (= x0 ) and c. Since y' separates x1 and x3 and since x2 is a common neighbors of x1 and x3, y' must pass through x2, and hence we can find a face f' of GR one of whose diagonal pair is {c, x2}. Let C' be the 4-cycle x1x2x3c of G. Since deg(c) > 3, we have df' = C', and hence C' is a separating 4-cycle in GR such that C' = C. Moreover, y' and C' cross transversely at x2 and c. Therefore, C' and f' are a 4-cycle and a face which satisfy the assumption of the lemma, and moreover, C' can cut a strictly smaller graph than GR from G. Therefore, this contradicts the choice of C. □ Lemma 3.5. Let G be a 3-connected quadrangulation of a closed surface F2. If G is K3-irreducible then G is K3-irreducible. Proof. Let G be a 3-connected quadrangulation of a closed surface. Assume that G is not K3-irreducible. Then, G has either a 3-contractible face or a contractible cube. If G has a 3-contractible face, then G is not K3-irreducible. Therefore, we suppose that G has no 3-contractible face but has a contractible cube Q with an inner 4-cycle C in the following argument. Now, we apply a 4-cycle removal of C to G and let G' be the resulting quadrangulation. Let f' = dQ be the new face of G' into which C was inserted. If G' is 3-connected, G is not K3-irreducible by the definition, and we are done. Therefore, we assume that G' is not 3-connected. By Lemma 3.2, there is a diagonal 2-curve y passing through f' and another Y. Suzuki: Cube-contractions in 3-connected quadrangulations 287 face f''; otherwise, G would have a 2-cut, contrary to our assumption. Note that f'' is also a face in G. Now dQ and f" satisfy the conditions of Lemma 3.4, and hence there exists a 3-contractible face in G. However, this contradicts the above assumption. Thus, the lemma follows. □ In the following argument, we denote the set of K3-irreducible (resp. K3-irreducible) quadrangulations of a closed surface F2 by K3I(F2) (resp. K31(F2)). Lemma 3.6. Let G be a 3-connected quadrangulation of F2. If G G K3I(F2) \K3I(F2), then G has an attached cube Q such that the graph obtained from G by applying a 4-cycle removal of Q is in K3I(F2). Proof. Let G be in K3I(F2) \ K3I(F2). By the definition, G has an attached cube Q with an inner 4-cycle C which is removable, but is not contractible. We apply a 4-cycle removal of C and let G- be the resulting quadrangulation. We denote the new face of G- by f-, where f- = dQ. First, we confirm that G- is 3-connected. Otherwise, G- has a 2-cut and has a separating diagonal 2-curve 7 on F2 by Lemma 3.2. If 7 does not pass through f- then 7 would also be a diagonal 2-curve in G, a contradiction. Let f0 be the other face passed by 7. Here, fo and dQ in G satisfy the conditions in Lemma 3.4 and there exists a 3-contractible face, contrary to G being K3-irreducible. By way of contradiction, assume that G is not in K3I(F2). That is, G has either (a) a 3-contractible face or (b) a contractible cube. First, we assume (a) and let f be a 3-contractible face in G-. If f- = f, the attached cube Q in G would be contractible, contrary to G being K3-irreducible. Thus, suppose f- = f. In this case, let G' be the resulting 3-connected quadrangulation after applying a face-contraction of f in G-. Since any 4-cycle addition doesn't break the 3-connectedness of a quadrangulation, the graph obtained from G' by a 4-cycle addition to f- is clearly 3-connected. This means that f is also 3-contractible in G, a contradiction. Next, suppose (b) and let Q' be such a contractible cube with dQ' = v0viv2v3. If Q' does not contain f- as one of its five faces, Q' is also contractible in G and G would not be K3-irreducible by the similar argument as above. Thus, we assume that Q' contains f-. Let C = m0m1m2m3 denotes the inner 4-cycle of Q' where WjVj G E(Q') for i = 0,1,2,3. We consider the following two cases up to symmetry; (b-1) f- = C and (b-2) f- = v0u0u1v1. At first, suppose (b-1). Here, we apply a face-contraction of f1 = v0u0u1v1 at {u0, v1} to G. If the above face-contraction breaks the 3-connectedness of G, there exists a face f2 = v1xv3y in the outside of Q' by Lemma 3.3; note that it clearly preserves the simplicity of the graph since v1 = v3. Now, a separating diagonal 3-curve passing through {v1, u0, v3} satisfies the conditions of Lemma 3.4 and hence G is not K3-irreducible, contrary to our assumption. In fact, an analogous proof is valid for (b-2) if we try to apply a face contraction at {v1, u2} to G. Therefore the lemma follows. □ Lemma 3.7. Let G be a 3-connected quadrangulation of a closed surface F2. If G G K3I (F2) \ K3I (F2), then G can be obtained from H G K3I (F2) by applying 4-cycle additions to F0 C F(H). Proof. Assume that G G K3I(F2) \ K3I(F2). By the previous lemma, there exists a sequence of K3-irreducible quadrangulations G = G0, G1,..., Gk such that Gi+1 is obtained from Gj by a single 4-cycle removal of Cj, where Gk G K3I(F2). (Since the 288 Ars Math. Contemp. 10 (2016) 255-268 number of vertices of G is finite, Gk € K3I(F2).) Let Qi denote an attached cube in Gi with an inner 4-cycle Ci. For a contradiction, we assume that there exists l € {0,..., k — 2} such that G; is obtained from G;+1 by a 4-cycle addition which is put on a face not of F(Gk); this l should be maximal. This implies that C; is put on a face of Q;+1 as one of its five faces. Then the same argument as the proof of Lemma 3.6 holds and hence G; would not be K'3-irreducible, contrary to our assumption. Thus for each i € {0,..., k — 1}, Gi is obtained from Gi+1 by a 4-cycle addition which is put on a face of F(Gk). □ Proof of Theorem 1.4. By Lemma 3.5, we have K3I (F2) C K3I(F2). Furthermore, by Theorem 1.3 and Lemma 3.7, we obtain (i) and (ii) in the statement. Thus, we have got a conclusion. □ 4 Spherical and projective-planar cases In this section, we discuss the spherical case and the projective-planar case. Proof of Theorem 1.5. Let G be a K3-irreducible quadrangulation of the sphere. We have K3I(S2) C K^I(S2) by Lemma 3.5, where S2 stands for the sphere. If G is K3 -irreducible, then G is isomorphic to a pseudo double wheel by Theorem 1.1. If G is in K3I(S2) \ K31(S2), G can be obtained from a pseudo double wheel W2k (k > 3) by some 4-cycle additions to faces of W2k by Lemma 3.7. However if k > 4, G has a 3-contractible face (or a contractible cube), as shown in the first operation in Figure 4. (For example, the entire Figure 4 presents a sequence of a face-contraction and a cube-contraction which deforms W8 with an attached cube Q into W6, preserving the 3-connectedness.) Figure 4: W8 with an attached cube Q deformed into W6. Therefore, we only consider the case of k = 3 in the following argument. Assume that G is obtained from W6 by at least two 4-cycle additions to faces of W6. Similarly to the above argument, G would have a 3-contractible face (or a contractible cube) , as in Figure 5, contrary to G being K3-irreducible; note that it suffices to discuss these two cases, up to symmetry. Therefore, we conclude that G is obtained from W6 by exactly one 4-cycle addition. This is nothing but a double cube; observe that a double cube has no 3-contractible face and no contractible cube. □ To conclude with, we prove the projective-planar case. Y. Suzuki: Cube-contractions in 3-connected quadrangulations 289 Proof of Theorem 1.6. In this case, we use Mobius wheels Wk (k > 3) and Qp as base graphs by Theorem 1.2. First we consider the former case. Similarly to the previous proof (and see Figure 6), we consider only a Mobius wheel W3 as a base to which we apply some 4-cycle additions. However, W3 (= Qp) is isomorphic to the complete graph with four vertices, and hence it is irreducible. This fact implies that every G obtained from W3 by applying at most three 4-cycle additions is K'3-irreducible since any face-contraction and any cube-contraction to G destroys the simplicity of the graph, or results in a vertex of degree 2. From this case, we obtain exactly three quadrangulations in K'3I(P2) \ K3I(P2), up to homeomorphism, where P2 stands for the projective plane. Figure 6: W5 with an attached cube Q deformed into W3. Similarly, as the latter case, we obtain the other ten quadrangulations in K31(P2) \ K3I(P2) from Qp; consider all the way to put attached cubes into faces of Qp, up to symmetry. As aresult, we have \K'3I(P2) \ K3I(P2)| = 13 in total. □ 290 Ars Math. Contemp. 10 (2016) 255-268 References [1] V. Batagelj, An inductive definition of the class of 3-connected quadrangulations of the plane, Discrete Math. 78 (1989), 45-53, doi:10.1016/0012-365X(89)90159-3. 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