Re-embedding structures of 4-connected projective-planar graphs

Re-Embedding Structuresof 4-ConnectedProjective-Planar Graphs

Yusuke Suzuki

DEPARTMENT OF GENERAL SCIENCETSURUOKA NATIONAL COLLEGE OF

TECHNOLOGY, TSURUOKAYAMAGATA 997–8511, JAPAN

E-mail: [email protected]

Received September 11, 2008; Revised July 29, 2010

Published online 12 November 2010 in Wiley Online Library (wileyonlinelibrary.com).DOI 10.1002/jgt.20553

Abstract: We identify the structures of 4-connected projective-planargraphs which generate their inequivalent embeddings on the projectiveplane, showing two series of graphs the number of whose inequivalentembeddings is held by O(n) with respect to the number of its vertices n.� 2010 Wiley Periodicals, Inc. J Graph Theory 68: 213–228, 2011

1. INTRODUCTION

A graph in this paper is a simple graph, that is, one with no loops and no multipleedges. An embedding of a graph G on a closed surface F2 is a drawing of G onF2 without edge crossings. However, we should regard it as an injective continuousmap from a one-dimensional topological space G to F2 when formulating delicateproperties of embeddings. A graph is said to be projective-planar if it is embeddablein the projective plane.

Two embeddings f1, f2 :G→F2 are said to be equivalent to each other if there is ahomeomorphism h :F2 →F2 such that hf1 = f2, and they are inequivalent otherwise. Ifthere is precisely only one equivalence class of embeddings of G in F2, we say that GJournal of Graph Theory� 2010 Wiley Periodicals, Inc.

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214 JOURNAL OF GRAPH THEORY

is uniquely embeddable in F2, up to equivalence. Under this equivalence, the completegraph K6 with six vertices has exactly 12 inequivalent embeddings into the projectiveplane. However, they can be presented with the same picture if we neglect the labelingof vertices. Two embeddings f1 and f2 are said to be weakly equivalent to each otherif there exists a homeomorphism h :F2 →F2 and an automorphism � :G→G such thathf1 =h2�. Now we can say that K6 is uniquely embeddable in the projective plane, upto weak equivalence. (To simplify our notation, we often consider that a first givenprojective-planar graph G is already mapped on the projective plane, and denote itsimage by G itself in the later arguments. In that case, we denote another embeddingimage by f (G).)

Then we have the following natural questions: (A) How many inequivalent embed-dings into a closed surface F2 does a graph G have? (B) What structure generatesinequivalent embeddings of G? (Such a structure is often called the re-embeddingstructure of G.)

In fact, the number of inequivalent embeddings of a graph into a fixed closedsurface is closely related with its connectivity and representativity. (The represen-tativity of an embedded graph G, denoted by r(G), is the minimum number ofintersecting points of G and �, where � ranges over all essential simple closedcurve on the surface. A graph G on F2 is said to be r-representative if r(G)≥r.)However, if the surface has high genus, the right parameter should be representativity,since the average degree is no more than 6 if the graph has reasonable number ofvertices. Especially, it was proved in [5, 13] that the graph on a closed surface exceptthe sphere with sufficiently large representativity is uniquely embeddable into thesurface.

On the other hand, if the surface has low genus, such as the sphere, torus,projective plane, the parameter of the connectivity plays an essential role. Inparticular, on the sphere S2 (the representativity is not defined on S2), Whitney’sresults [14, 15] are well-known and he had proved that any 3-connected graph isuniquely embeddable into S2; however it does not hold without 3-connectednessin general. Recently in [10], Robertson et al. proved that every 3-connected and4-representative toroidal graph is uniquely embeddable into the torus with a fewexceptions.

If the surface is fixed to the projective plane P2, Mohar et al. [6] showed re-embeddingstructures of planar graphs on P2. (Vitray [11, 12] discussed re-embeddings of graphson P2 with conditions of representativity, although the surface has low genus. In [4, 8],re-embeddings of triangular embeddings on P2 are discussed.) Kitakubo and Negamihad studied the re-embedding structure in the projective plane throughout [1–3, 7]and proved the following theorem for 5-connected projective-planar “nonplanar”graphs.

Theorem 1 (Kitakubo and Negami [3]). Every 5-connected projective-planarnonplanar graph admits exactly 1, 2,3,4,6,9 or 12 inequivalent embeddings into theprojective plane.

It is easy to see that there exists a series of 3-connected nonplanar graphs which haveexponentially many inequivalent embeddings in terms of the number of its verticesin the projective plane. (See the right-hand side of Fig. 1. However, we do notexplain it in details here.) In this paper, we consider the re-embedding structure of a

Journal of Graph Theory DOI 10.1002/jgt

4-CONNECTED PROJECTIVE-PLANAR GRAPHS 215

FIGURE 1. Two structures.

4-connected projective-planar nonplanar graph, denoted by 4PNG for a simple notation.The following is our main theorem:

Theorem 2. Let G be a 4PNG with n≥11 vertices and assume that G admits inequiv-alent embeddings into the projective plane other than itself. Then G contains a sub-division of the graph (with its re-embedding structure) shown in the left-hand side orthe center in Figure 1 and the number of embeddings is at most 5n−10.

In fact, the re-embedding structure of G in the above theorem depends on the factthat r(G)=2. Thus, we have the following corollary.

Corollary 3. Let G be a 4PNG with n≥11 vertices having representativity at least 3.Then G is uniquely embeddable into the projective plane, up to equivalence.

Our result builds on the previous work on the re-embedding of graphs into a closedsurface. Throughout the paper, we shall show the re-embedding structure of 4PNG’s,not only giving the upper bound in the above theorem. In the process, we shall showthat a series of graphs whose number of embeddings increase with the number of itsvertices is classified into two structures given by the left-hand side and the center ofFigure 1. In fact, the graphs shown on the left-hand side of the figure admits 5n−10its inequivalent embeddings. Every graph that can be represented in this way is calleda double wheel with axle. The point of our proof is to focus on a Mobius ladder, to beintroduced in the next section. We can recognize its explicit re-embedding structure andcan find its subdivision in any 4PNG except some cases as shown later. In Section 3,we extend the notion of a Mobius ladder. Remaining part is devoted to considering thepossibility of a series of graphs having a lot of inequivalent embeddings whose numbercannot be bounded by any constant.

2. MOBIUS LADDERS

The Mobius ladder is a typical projective-planar graph and consists of an even cyclev1 . . .v2k with chords vivk+i (i=1, . . . ,k). The cycle is called its rim and the chordsare called its rungs. We denote the Mobius ladder with rim of length 2k by MLk. Thelength of MLk itself is k. In particular, ML3 is isomorphic to K3,3.

The standard embedding of MLk on the projective plane is shown in the left-handside of Figure 2, where each antipodal pair of points on the boundary of the disk



FIGURE 2. Mobius ladders in the projective plane.

should be identified to a single point on the projective plane. The two vertical linesegments form the rim of MLk, which divides the projective plane into a 2-cell regionand a Mobius band. The left and right half-moon regions correspond to the 2-cellregion, while the central region between the two vertical line segments corresponds tothe Mobius band. All of the rungs are placed in the Mobius band in parallel in thispicture.

The right-hand side of Figure 2 presents another embedding of MLk, which isnot equivalent to the standard one. Clearly, we can construct k such embeddings,corresponding to the rung that jumps out from the Mobius band into the 2-cell region,which are weakly equivalent to each other.

In fact, MLk admits no other embeddings on the projective plane if k≥4. Thatis, it has precisely k+1 embeddings up to equivalence and two embeddings up toweak equivalence. This “re-embedding structure” of MLk is typical among those ofprojective-planar graphs and will play an important role with the following fact in ourlater arguments.

Fact 4. The rim of MLk with k≥4 bounds a 2-cell on the projective plane in anyembedding.

This allows us to present the rim of MLk(k≥4) in any embedding with two fixedvertical line segments in a disk, as mentioned above. The Mobius ladder ML4 of length4 is often called the Wagner graph and there have been several studies about sufficientconditions for a graph to contain a subdivision of ML4. For example, Robertson [9]has proved the following theorem. A graph G is said to be internally 4-connected if Gis 3-connected and if any cut of three vertices yields a trivial component.

Theorem 5 (Robertson [9]). Let G be an internally 4-connected graph. If G does notcontain a subdivision of ML4, then one of the following five cases holds:

(1) G is planar.(2) G−{x,y} is a cycle for some x, y∈V(G).(3) G−{w,x,y,z} has no edge for some w, x, y, z∈V(G).(4) G is isomorphic to the line graph L(K3,3) of K3,3.(5) |V(G)|≤7.

Lemma 6. Let G be a 4PNG with n≥10 vertices and G is not a double wheel withaxle. Then G contains a subdivision of ML4.



Proof. The nonplanarity of G excludes Case (1) in Theorem 5, and the condition ofthe number of vertices immediately does Case (4) and (5). In Case (3), each vertex otherthan x, y, z and w must adjacent to all of x, y, z and w; otherwise, such a vertex wouldhave degree 3. If there are at least four vertices other than x, y, z and w, G includesthe subdivision of K4,4. However, it is well known that K4,4 cannot be embedded inthe projective plane. Therefore, G has at most 7 vertices and hence this case is alsoexcluded. By 4-connectedness and the nonplanarity, the graph in Case (2) is nothingbut a double wheel with axle, which is the exception in the lemma. �

3. FIXED RIM AND ITS BRIDGES

Let G be a connected graph and C a cycle in G throughout this section. A bridge Bfor C in G (with body B) is the subgraph induced by a component B of G−V(C) withedges incident to B. We call each of the vertices of B lying on C a foot and each edgeof B incident to a foot a leg of B. Also, each chord of C is called a bridge for C, butis said to be singular. That is, a singular bridge consists of only one leg and two feetwithout its body. A bridge consists of only one vertex as its body and four legs andtwo pairs of adjacent feet on C is said to be a ribbon bridge. In fact, these singularand ribbon bridges play important roles in the later arguments.

Suppose that G is embedded on the projective plane P2 and that C bounds a 2-cellregion D on P2, possibly not a face of G. Let f :G→P2 be another embedding. Iff (C) also bounds a 2-cell region, we can assume that f fixes C pointwise since we candeform f isotopically on P2. Such an assumption will make it easier to describe thestructure of f as below. The cycle C divides the projective plane P2 into a 2-cell D anda Mobius band M. An arc joining two points of C across M is said to be essential if itcuts open M into a rectangle, and to be inessential otherwise. For example, each rungof MLn in the standard embedding is essential in M and an inessential arc cuts off adiagonal 2-cell from M.

In G, each bridge B for C is placed in either D or M. In the latter case, B is said tobe essential if B contains an essential arc joining two of its feet. Otherwise, B is planarsince it is contained in a digonal 2-cell region bounded by C and an inessential arc in M.Any bridge placed in D also is planar. Let B be a nonsingular essential bridge for C,placed in M. Then we can take two segments SL and SR on C so that they include allfeet of B and four end vertices of SL and SR are also B’s feet, and so that any path inB joining two vertices l∈SL and r∈SR is essential. We call those SL and SR are twofoot segments of the essential bridge B.

Assume that there is another embedding f :G→P2 such that f (C)=C. Then, f mightchange the location of bridges for C, but always fixes its feet on C. If f (B) cannot bedeformed into B isotopically, then B is said to be relocatable (by f for fixed C). Itis easy to show the following lemmas on relocatable bridges; they present the samephenomena as the behavior of rungs of a Mobius ladder.

Lemma 7. Let P1 and P2 be two disjoint paths in D joining vertices of C. Then P1and P2 cannot be relocated together into M so that they are essential.

Lemma 8. Let P1 and P2 be two disjoint essential paths in M joining vertices of C.Then P1 and P2 cannot be relocated together into D.



Lemma 9. Let B be a nonsingular essential bridge for C placed in M having footsegments SL and SR. If there is another embedding fixing C such that f (B) is placed inD, then one of the followings holds:

(a) one of two foot segments consists of only one vertex(b) B splits into two subgraphs B′ and B′′ which have only one common vertex so

that all feet of B′ are on SL and those of B′′ are on SR.

Lemma 10. Let G be a 4PNG with n≥11 vertices and G is not a double wheel withaxle. Then G contains a cycle C such that:

(1) The cycle C divides P2 into a 2-cell D and a Mobius band M.(2) The interior of M includes no vertex of G.(3) Each edge of G across M is essential and there are at least three such mutually

disjoint edges.(4) The cycle f (C) bounds a 2-cell on P2 for any embedding f :G→P2.

Proof. Let C be a cycle in G which satisfies Conditions (1) and (4) in the lemma.We can actually consider the cycle corresponding to the rim of ML4 as such a cycle Cby Fact 4 and Lemma 6, and the Mobius band M includes at least three disjoint pathsP1, . . . ,Pk so that H =C∪P1 ∪·· ·∪Pk forms a subdivision of a standard embedding ofMLk with k≥3. We take k to be maximal. Let v1, v2, . . . ,v2k be vertices lying on C inthis order such that Pi connects vi and vk+i. Furthermore, we assume that M containsas few faces of G as possible. To simplify our arguments, we can assume that anyembedding f :G→P2 fixes C pointwise.

The subdivision H of MLk divides the Mobius band M into k rectangular regionsAi(i=1, . . . ,k). The two paths Pi and Pi+1 form a pair of parallel sides of Ai and theother pair consists of two segments of C, say Li and Ri. We may assume that Pi, Pi+1,Li and Ri are placed at the top, bottom, left and right of Ai. Further, we assume thatLi ∩Pi =vi.

First, assume that one of P1 and P2, say P1, has a length at least 2. Let u be an innervertex of P1. Then, we may assume that there is a path Q running across A1 from u toanother vertex v not lying on P1. Otherwise, the two ends of P1 would form a 2-vertexcut of G, contrary to G being 4-connected. We have the following two cases for this Q.

Case (A). The vertex v is an inner vertex of L1 (or R1): Let A be a triangular regionbounded by L1, P1 and Q in A1. Let C′ be the cycle with E(C′)=E(C)E(�A), whereXY denotes the symmetric difference of X and Y , and �A stands for the boundary cycleof the region A. Then C′ bounds a 2-cell D′ =D∪A and a Mobius band M′ =M−A,and hence this new cycle C′ satisfies (1). It is easy to see that f (�A) always bounds a2-cell, in D or in A1. This implies that f (C′) also bounds a 2-cell, and hence (4) holdsfor C′. The Mobius band M′ contains fewer faces than M, contrary to the minimalityof M. Thus, this is not the case.

Case (B). The vertex v lies on P2: Assume that v=v2 if v is one of the ends of P2,without the loss of generality. Let A be a rectangular (or triangular) region bounded byL1, P1, Q and P2 in A1 and let C′ be the cycle with E(C′)=E(C)E(�A). Then, C′ splitsP2 into a 2-cell D′ =D∪A and a Mobius band M′ =M−A and hence C′ satisfies (1).For any embedding f , if f (Q) always stays in A1, then f (C′) bounds the 2-cell D′ andhence (4) holds for C′. However, there are some cases that f (Q) does not stay in A1.



For example, if v=v2 then an embedding f :G→P2 may map Q into D. In such acase, f (�A) bounds a 2-cell and so does f (C′), as well as in Case (A). Thus, it sufficesto discuss only the case that f (Q) lies in M.

If k≥4, then one of f (P3) to f (Pk) stays in M and it forces Q to stay in A1. Thisimplies that C′ satisfies (4). So, we suppose that k=3 and that f (Q) is placed in M−A1for some embedding f :G→P2. In this case, f (P3) must be mapped into D, and bothf (P1) and f (P2) stay in M so that they bound A1 together with L1 and R1. Recall thatG contains a subdivision of ML4. Since k=3, the Mobius ladder must be embeddedin a nonstandard way. That is, only three of its rungs are placed in the Mobius bandbounded by its rim. The three paths P1, P2 and P3 can be assumed to run along thoserungs and there is a bridge for C placed in D, say B, which contains the fourth rungof ML4. The relocation of P3 by f forces B to be mapped into A1. Thus, we can findtwo foot segments SL and SR of B lying on L1 and R1, respectively. Note that there isa path which works like Q in neither A2 nor A3. Because, such another path requires abridge like B, they clearly cannot be placed in D together. Under these conditions, weretake P1 and P2 (and Q) so that A1 includes as many faces as possible.

Now, let C denote the cycle such that E(C)=E(C)E(�A1) and let T be the bridgeincluding Q for C; note that C bounds a 2-cell A2 ∪A3 in the projective plane. Assumethat T has its body. Then, T is essential for C and hence has two foot segments S1 andS2 which lie on P1 and P2, respectively. Under the assumption, T is relocatable for Cand satisfies either (a) or (b) in Lemma 9.

First, assume the former and that a foot segment S2 corresponds to a single vertexv without loss of generality. Then, there must be a path Q′ from an inner vertex ofS1; otherwise, two end vertices of S1 and v would be 3-cut. If Q′ has its other endalso on P1, it is contrary to the maximality of the number of faces of A1. Thus, Q′should be running across A3 to the vertex not lying on P1; however, it is impossibleby considering the relocation of T for C.

Next, we suppose (b) in Lemma 9 for the bridge T . Then, T splits into two subgraphsT1 and T2 which have only one common vertex t; T1 and T2 have their feet on S1 andS2, respectively. If T1 (resp. T2) includes a vertex other than t and two end vertices of S1(resp. S2), there must be a path like above Q′ from an inner vertex of S1 (resp. S2) andthe same argument stands for. Therefore, we conclude that T is a ribbon bridge for C.

Since u has degree at least 4, there should be another path from u. By the abovearguments, we only have to consider that such a path is running across A1 and has avertex of P2 as its one end vertex. However, the path and the ribbon bridge cannot getout of A1 simultaneously; it disturbs the other embedding f , a contradiction. Therefore,we assume that the bridge including Q is singular. Assume that v is an inner vertexof P2. Then v requires other path from itself, so as not to be degree 3. However, theabove arguments don’t admit such a path except an edge vv1. By the same reason,there should be a uv5 but it disturbs the relocation of B to be in M, by Lemma 8 for C.Therefore, we have v=v2 and an edge uv5. By careful observation, P1, P2 and P3 donot have any inner vertex other than u (see (i) of Fig. 3).

Next, we consider the bridge B including P4. Similarly to the case of T , B shouldbe either a ribbon bridge or a singular. Suppose the former and let l, l′, r and r′ bethe four feet of B on L1 and R1. Here, l is nearer or corresponding to v1. If bothof l and l′ are ends of L1, G would not include the subdivision of fourth rung P4.Therefore, one of l and l′ should be an inner vertex of L1. First, assume that l is



FIGURE 3. Graph without fixed rim and planar singular bridge.

such an inner vertex. Then, there must exist a bridge which includes the path in Dfrom l to another vertex not lying on L1. (If there is not such a path, then l wouldbecome degree 3. Furthermore, if such a path places in M, it clearly causes Case (A).)The bridge does not have its foot on the segment between r′ and v6 (including R2)except one end vertex v6, because the condition would disturb the relocation of P3.However, l, u and v6 would form a 3-cut which cuts off the part of G containingv1, contrary to G being 4-connected. Therefore, we may assume that l correspondsto v1.

Thus, we may suppose that l′ is an inner vertex of L1. Similarly to the abovecase, there should be a bridge which includes a path from l′. If the bridge hasits body, we have almost the same conclusion and find a 3-cut. Therefore, wesuppose that the bridge is singular. In this case, we must connect l′ and v3 (thesymmetrical structure also occurs around R1) and have the graph presented by(ii) of Figure 3; observe that the configuration does not admit any other vertex.However, this graph has exactly 10 vertices and it is excluded in the lemma. (Weomit the case that P4 consists of a singular bridge, since it is more easier to reach acontradiction.)

Since both (A) and (B) imply contradictions, each of P1 to Pk must not be subdivided,that is, it consists of a single edge. We assume that the interior of A1 contains a vertexw of G. Since G is 4-connected, there are four inner-disjoint paths from w to fourdistinct vertices of �A1. This implies the following case.

Case (C). There is a path Q running across A1 and joining a vertex u of L1 to anothervertex v on L1: If one of u and v, say u, is an inner vertex of L1, then Q bounds adigonal region A together with L1 in A1. In this case, the similar argument as that ofCase (A) holds. If both u and v correspond to the two ends of L1, Q=uwv plays thesame role as that in the Case (B). Considering all the arguments in this lemma, wereach (iii) in Figure 3. However, the condition |V(G)|≥11 is sufficient to exclude thecase. Hence we conclude that there is no vertex in the interior of M, that is, (2) holdsfor C′. Finally, suppose that there is an inessential edge uv in M. Then, the edge uv iscontained in one of A1, . . . ,Ak, say A1. If at least one of u and v is an inner vertex of L1,the same argument as in Case (C) will work for this case, and it implies a contradictionto the minimality of M. Otherwise, uv joins the two ends of L1 and a similar argumentto Case (B) works; in this case, we cannot construct a counter example like (ii) and(iii) in the figure. Therefore, C′ satisfies (3). �



We call such a cycle C in Lemma 10 a fixed rim of G. Although a fixed rim isnot unique in general, it is very useful to analyze re-embeddings of projective-planargraphs since it only suffices to consider how the bridges for the fixed rim are mapped.If we especially do not mention the way to choose a fixed rim, we assume that Mincludes as many essential edges as possible, re-embedding G and reselecting C. Afterfinding the proper fixed rim C as above, we call the embedding a standard embeddingof G. In the standard embedding, we call the edges placed in M its rungs. Thus,we specially use the word “bridge” when we indicate a bridge in D in the standardembedding below. A bridge B (or rung) which can be relocated into M (or D) is saidto be a relocatable bridge (or a relocatable rung) by some embedding f . For anotherembedding f of G, if f (B) is in M and includes essential arc, then f (B) is said to beessential in the embedding. On the other hand, f (B) does not contain such an arc, f (B)is said to be planar. Two rungs e and e′ in M are said to be parallel if they haveno common end. The following lemmas give fundamental properties of bridges andrungs for the fixed rim C.

Lemma 11. Two parallel rungs for a fixed rim C cannot be relocated together into D.

Proof. It is clear by Lemma 8. �Lemma 12. Any relocatable nonsingular bridge for a fixed rim C is essential in Mby any relocation.

Proof. Let B be one of nonsingular bridges for C and suppose that B is relocatableby an embedding f :G→P2 with f (C)=C. Suppose that f (B) is planar in M. Then,f (B) is placed in a 2-cell region in M which is bounded by a segment S of C and aninessential arc in M. The segment S contains all feet of B and both the ends shouldbe feet of B. All rungs of C incident to inner vertices of S must be mapped into D byf and hence they have a common end vertex w; for otherwise, there are two parallelrungs, contrary to Lemma 11. In this case, the two end vertices of S and w form a3-cut, contrary to G being 4-connected. �

Next, we consider the behavior of a relocatable singular bridge B in the standardembedding of G.

Lemma 13. Let B be a singular bridge of a fixed rim C in D. If f (B) is planar in Mby some embedding f , then there does not exist another embedding f ′ such that f ′(B)is essential.

Proof. In this case, the fixed rim C cannot become a subdivision of rim ofMLk(k≥ 5) and M contains exactly three rungs as those of ML4 in f (G). Since a similarargument to the proof of Lemma 10 holds, we can exclude such a singular bridge. �

By the same reason, we have the following lemma.

Lemma 14. Let e be a rung of a fixed rim C in M. Then there does not exist anotherembedding f such that f (e) is planar in M.

Lemma 15. Let B be a singular bridge of a fixed rim C in D. If f (B) is planar inM by some embedding f , then B has the structure presented by (vi) of Figure 5 in thestandard embedding.



FIGURE 4. Inequivalent embeddings of a double wheel with axle.

Proof. Similarly to the proof of Lemma 12, we take a segment S of C so that Sand f (B) bound a 2-cell. If S includes at least 2 vertices as inner vertices, we havethe same conclusion as Lemma 12. Thus, S includes the unique inner vertex and it isincident to at least two rungs in M; for otherwise, the inner vertex would have degreeat most 3. �

Before classifying the re-embedding structures of 4PNG’s, we exhibit two series of4PNG’s the number of whose inequivalent embeddings in the projective plane increasein linear order with respect to the number of its vertices. The left-hand side and thecenter of Figure 1 present two 4PNG’s each of which has such a structure. The left-handshows a double wheel with axle with n vertices, which has an essential cycle v1 . . .vn−2of length n−2 (called a rim) and two central vertices a and b. By the following reason,this graph admits three inequivalent embeddings into the projective plane, up to weakequivalence, as shown in Figure 4.

The cycle v1 . . .vn−2 in the figure cannot bound a 2-cell region on the projectiveplane and hence we can fix it pointwise in any embedding. In the first configurationin the figure, the 4-cycle av1vn−2b bounds a 2-cell region (or face) R. We constructanother embedding so that the edge bv1 runs across R, and obtain the center of Figure 4.Further, we replace the edge bvn−2 and obtain the third one. In the second configuration,one might replace the edge av1 into the pentagonal region R′ bounded by v1v2bavn−2.However, the resulting embedding is equivalent to the left-hand side of Figure 4, upto weak equivalence. Similarly, in the second configuration, the relocation of av2 intoR′ yields the third configuration. By the above arguments, we can replace at most 2edges to make inequivalent embeddings, up to weak equivalence, and hence there isno drawing other than those three in Figure 4.

As noted in Figure 4, they have n−2, 2(n−2) and 2(n−2) inequivalent embeddings,respectively, up to equivalence; e.g. in the first one, we clearly have n−2 ways to takevivi+1 such that it does not run along boundaries of triangular faces. In the second andthird ones, we further have to change two central vertices a and b and have to multiply(n−2) by 2. Therefore in total, if this graph has n vertices, it admits exactly 5n−10inequivalent embeddings in the projective plane, up to equivalence.

Next, we explain the graph given by the center in Figure 1. The shaded polygon inthe graph should be a suitable graph so that the whole graph becomes 4-connected,and two vertical segments represent the fixed rim C explained in this section. Thereare two disjoint segment L and R of C and there is no bridge in D for these L and R.



On the other hand, many rungs lie in M between L and R, so that the ends of eachrung have degree 4. Now let e1, . . . ,em be such rungs and assume that ei and ei+1 havea common end for i=1, . . . ,m−1. Note that each ei is relocatable.

When we enumerate the number of inequivalent embeddings other than standardone, that is, each ei is placed in M, it suffices to enumerate the number of theserelocatable rungs (m in this case), and the number of the pairs of these rungs whichhave a common end (m−1 in this case), since two parallel rungs cannot be placedinto D together by Lemma 11. In total, it admits m+(m−1)+1=2m inequivalentembeddings; the last “1” means its standard embedding. Then, we can conclude thatthe number of embeddings of this graph with n vertices is not more than 2n since weclearly have n>m.

The above two re-embedding structures of 4PNG’s are very important in this paper.If a 4PNG G contains a subdivision of a subgraph of double wheel with axle, denotedby H, and G’s re-embeddings are realized as the behavior of H, we call the structureDWA-structure; G’s re-embeddings might be restricted by some conditions. On the otherhand, re-embeddings are realized as the latter case; we call it zigzag-structure hereafter.

4. DWA-STRUCTURE

Let G be a 4PNG in which we can find a fixed rim C. Two relocatable bridges B andB′ for C are said to be gearing if B is relocated into M if and only if so is B′; that is,there is not any embedding f :G→P2 such that one of f (B) and f (B′) is placed in Mand the other is in D. (We also use the definition for two relocatable rungs.) Otherwise,two bridges (or rungs) are said to be independent.

Let G be a 4PNG having a fixed rim C. Assume that D contains two paths Pi joiningvertices x and vi on C for i=1, 2 and that there are other embeddings f1 and f2 suchthat each of f1(P1) and f2(P2) is essential in M; f1 might coincide with f2. Note that P1and P2 are contained in either one bridge or two bridges separately. Furthermore, wesuppose that there is a rung e=uv in M such that v is an inner vertex of the segment Sbetween v1 and v2 and that u =x (see (i) in Fig. 5). In fact, such a subgraph of G (withthe above conditions) restrict the existence of other relocatable bridges and rungs. Wecall it umbrella-structure for C and call two vertices x and u central vertices of theumbrella-structure.

Lemma 16. Let G be a 4PNG having a fixed rim C. Assume that G admits umbrella-structure for C defined as above. If there is another relocatable bridge B (or rung)having its feet on C−S, then one of the followings holds:

(a) If B is planar in M, then B=yz is singular and the 2-path yuz lies along C (see(ii) in Fig. 5).

(b) If B is essential in M, one of two foot segments SL and SR of B consists of asingle vertex corresponding to either u or x.

Proof. First, assume that B is planar in M after a relocation. By Lemma 12, B issingular and we denote B=yz. Furthermore by Lemma 15, B has the structure givenby (vi) in Figure 5 and has rungs st1, . . . ,stj for j≥2 having a common end s; note thatysz forms a 2-path along C. Regardless of the position of vertices y and z, we can findtwo parallel rungs e and e′ =stk(1≤k≤ j), except the case of s=u (see (ii) in Fig. 5).



FIGURE 5. Around the umbrella-structure.

Assume that we have such parallel rungs. If the one end of e′ is included in thesegment between u and x as an inner vertex, we cannot relocate the bridge containingP1 by Lemma 11. In other cases, when relocating B and st1, . . . ,stj simultaneously, P1and P2 should be replaced into M. Furthermore under these conditions, e must jump outfrom M. However, it is also impossible by Lemma 11. Thus in this case, we concludethe case (a) in the lemma.

Secondly, we consider the case that B is essential in M. For a contradiction, weassume that B does not satisfy (b) in the lemma. That is, B includes an essential pathP3 in M each of whose two ends corresponds to neither x nor u. Now we classifythe cases depending on the locations of ends of P3, and have (iii), (iv) and (v) inFigure 5.

In (iii) and (iv), relocating B (including P3) forces e to jump out from M and henceP2 must be relocated into M. However, we apply Lemma 7 to P2 and P3 and obtaina contradiction. Also in (v), it disturbs the relocation of P1, similarly to the argumentin (a). Therefore in this case, we have (b) in the lemma. Since the similar argument asabove stands for, one end of any relocatable rung corresponds to either x or u. Then,we got a conclusion. �

Lemma 17. Let G be a 4PNG having a fixed rim C. If G has a nonsingular relocatablebridge B for C which is not a ribbon bridge, then G has DWA-structure.

Proof. Throughout the proof, we draw the fixed rim C on the projective plane bytwo line segments, say L and R. To simplify our arguments, all the letters presentingvertices on C (sometimes presents other partial structures) have subscripts either L orR; e.g., a vertex uL lies on L. Since B is not singular, f (B) is essential in M for anotherembedding f :G→P2 by Lemma 12. Hence, we can take two foot segments SL and SR



FIGURE 6. Similar re-embedding structure to that of the double wheel with axle.

of B on C defined in Section 3. Furthermore by Lemma 9, we have either that one ofSL and SR consists of one vertex or that B splits into two subgraphs BL and BR whichhave only one common vertex x.

First, we assume the former, and suppose that SR consists of a single vertex yRwithout loss of generality. Then, there exists a relocatable rung e=uLuR such that uLis an inner vertex of SL and uR =yR; otherwise, two end vertices of SL and yR wouldbecome a 3-cut. See (i) in Figure 6.

Under the conditions, the partial graph as noted above contains an umbrella-structurefor C. There might be many relocatable bridges and rungs for C and each of themsatisfies either (a) or (b) in Lemma 16. (However, we do not have to consider (a) sinceit is unique and only restricts the re-embeddings of the whole graph. Furthermore, weomit rungs whose one end is an inner vertex of SL other than e since they are gearingwith e and also restrict G’s re-embeddings.)

Assume that there is another nonsingular independent relocatable bridge B′; then S′R

consists of one vertex y′R and it requires e′ =u′

Lu′R as well as B. For the relocation of

them, B′ and e′ need either (A) yR =u′R and y′

R =uR or (B) yR =y′R and uR =u′

R, byLemma 16 (see (ii) and (iii) in Fig. 6, respectively). Similarly, we may consider manyrelocatable bridges like B′ having structure either (A) or (B). Further, there might berelocatable singular bridges and rungs between these nonsingular bridges, so that thewhole graph becomes 4-connected and that each of them has one end correspondingto either uR or yR by Lemma 16. Now, we relocate the bridges with (A) in (ii) into Msimultaneously, and obtain (iv) of the figure.

Re-embedding structure of the graph is almost same as that of (v). (The shadedpolygon in the graph should be a suitable graph so that the whole graph becomes4-connected.) Actually, (v) includes the subdivision of a double wheel with axle; wecan find its rim running through the shaded part, otherwise G would not be 4-connected.



Then, we can regard the re-embedding structure of this graph as a very restrictedversion of DWA-structure with central vertices uL and yR; the number of its inequivalentembeddings is clearly less than 5n−10 if the graph has n vertices.

Finally, we assume that B admits (b) in Lemma 9 and that neither SL nor SR consistsof a single vertex. Let yR and zR be two end vertices of SR (see (vi) in Fig. 6). Inthis case, we also find an umbrella-structure which has either {uR,yR} or {uR,zR} asa pair of central vertices. However, G only admits at most 4 relocatable rungs (orsingular bridges) each of which has uR as an end vertex (see (vi) again). (For example,further rungs into (vi) require a bridge B in D which does not satisfy the conditions ofLemma 16. Then B is not relocatable and it disturbs the relocation of B or some rungs.)

We regard each graph excluded throughout the proof, which has a smallnumber of inequivalent embeddings, as that with strongly restricted version of DWA-structure. �

Two vertices uR and yR in the previous proof are said to be central vertices of theDWA-structure, as well as a double wheel with axle.

Lemma 18. Let G be a 4PNG having a fixed rim C. Assume that G has at least threerelocatable singular bridges for C which have a common end. If each of such singularbridges is essential in M for another embedding, then G has DWA-structure.

Proof. Suppose that there are such singular bridges xv1, xv2, . . . ,xvk(k≥3) in thestandard embedding, and assume that v1, v2, . . . ,vk(,x) lie on C in this order. Then thereis a rung e=uv in M such that v is contained in the segment between v1 and vk asan inner vertex and we have u =x; for otherwise, {x,v1,vk} would become a 3-cut ofthis graph. Note that each of the singular bridges and rung is always essential in M byLemmas 13 and 14. Then, we can find an umbrella-structure in G. By Lemma 16 andthe arguments in the previous lemma, we got a conclusion. �

Lemma 19. Let G be a 4PNG having a fixed rim C. Assume that G has at least threerelocatable rungs xv1 . . . xvk(k≥3) for C which have a common end x, and let SL denotethe segment between v1 and vk. Further, assume that there is a singular bridge e=uvsuch that v is contained in SL as an inner vertex and u∈V(C)−V(SL)−{x}. If there isanother embedding f such that f (e) is essential in M, then G has DWA-structure.

Proof. By Lemmas 13 and 14 again, each of those rungs and the singular bridge e isalways essential in M. Similarly to the above lemma, they derive DWA-structure. �Lemma 20. Let G be a 4PNG having a fixed rim C and assume that G does not haveDWA-structure. Suppose that there is a singular bridge e=uw such that f (e) is planarfor some embedding f . Then the configuration around e is presented by the left-handside of Figure 7.

Proof. By Lemma 15, we may assume that uxw forms the 2-path along C andthere are rungs xv1, . . . ,xvk(k≥2). Suppose that the segment SR on C between v1 andvk includes at least one inner vertex. Then, there should be a bridge B including a pathP which joins an inner vertex v of SR and y∈V(C)−V(SR)−{x}; otherwise, {x,v1,vk}would form a 3-cut. If there is another embedding f ′ such that f ′(P) is essential in M,we find an umbrella-structure in G. By the arguments in the former lemmas, bridgesand rungs around it would form DWA-structure, contrary to our assumption.



FIGURE 7. Re-taking the fixed rim and many relocatable bridges in M′.

Therefore, B is planar in M and should be a singular bridge e′ =vy by Lemma 12.Then by Lemma 15 again, either vv1y or vvky forms a 2-path on C, now say vv1y,without loss of generality. Furthermore, there are rungs v1z1, . . . ,v1zl(l≥2) which havea common end v1, where zl =x. However when we relocate e into M by f , two parallelrungs v1z1 and xvk have to be replaced into D simultaneously, but it is impossible byLemma 11. Thus, we have k=2 and v1v2 ∈E(C), and hence the lemma follows. �

The proof in the next section is our goal in this paper. It shows that any series of4PNG’s, the number of whose inequivalent embeddings increases in linear order withrespect to the number of their vertices, can be classified into that with DWA-structureor zigzag-structure.

5. PROOF OF THE MAIN THEOREM

Proof of Theorem 2. Let G be a 4PNG with n≥11 vertices. By Lemma 10, wecan find a fixed rim C in G. Here, we consider the case that there are many relocatablebridges and rungs for C but we exclude the case when G has DWA-structure. Note thateach such bridge in D is singular or a ribbon bridge by Lemmas 12 and 17. First, weassume that there is a singular bridge which can be relocated into M so that it is planar.Recall that it has the structure shown in the left-hand side of Figure 7, by Lemma 20.

Now, we retake the trivial cycle C′ on the projective plane so that it goes along allthose singular bridges which are planar in M. By Lemma 13, this C′ bounds a 2-cellD′ in any embedding, and hence we may regard C′ as a new fixed rim, although theMobius band M′ includes vertices of degree 4. Each such vertex of degree 4 with itslegs and feet looks like a ribbon bridge in M′ (see the center of Fig. 7). Note that eachsingular bridge for C′ is always essential for any embedding, by the way to take thecycle.

At first, suppose that for C′ there are many ribbon bridges and rungs in M′; note thatat most two rungs can have a common end vertex by Lemma 19 (see the right-handside of Fig. 7). The re-embedding structure of this graph clearly resulted in zigzag-structure explained in Section 3. Observe that we cannot put any relocatable singularor ribbon bridge into D′ in the configuration, since it clearly disturbs the relocation ofeither itself or bridges (rungs) in M′. (We can also consider the case that D′ includesmany essential singular and ribbon bridges but does not have relocatable rungs in



M′; note that at most two singular bridges have a common end by Lemma 18. However,the re-embedding structure completely coincides with the above one and we omit it.)Therefore, we got the conclusion. �

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Re-embedding structures of 4-connected projective-planar graphs