On Polyhedral Embeddings of Cubic Graphs

On polyhedral embeddings of cubic graphs Bojan Mohar Andrej Vodopivec Department of Mathematics University of Ljubljana 1000 Ljubljana, Slovenia [email protected] Department of Mathematics IMFM 1000 Ljubljana, Slovenia [email protected] January 29, 2004 Abstract Polyhedral embeddings of cubic graphs by means of certain operations are studied. It is proved that some known families of snarks have no (orientable) polyhedral embeddings. This result supports a conjecture of Grünbaum that no snark admits an orientable polyhedral embedding. This conjecture is verified for all snarks having up to 30 vertices using computer. On the other hand, for every nonorientable surface S, there exists a non 3-edge-colorable graph, which polyhedrally embeds in S. Keywords: polyhedral embedding, cubic graph, snark, flower snark, Goldberg snark. 1 Introduction In this paper we study embeddings of cubic graphs in closed surfaces. We refer to [5] for basic terminology and properties of embeddings. Following the approach of [5], all embeddings are assumed to be 2-cell embeddings. Two embeddings of a graph are considered to be (combinatorially) equal, if they have the same set of facial walks. If S is a surface with Euler characteristic χ(S), then ǫ(S) := 2 − χ(S) is a non-negative integer, which is called the Euler genus of S. If an embedding of a graph G in a non-orientable surface is given by a rotation system and a signature λ : E(G) → {+1, −1} and H is an acyclic subgraph of G, then we can assume that the edges of H have positive signature, λ(e) = 1 for all e ∈ E(H). We shall assume this in several instances 1 without explicitly mentioning it. Instead of describing an embedding by specifying rotation system an signature, it suffices to list all facial walks. An embedding of a graph G is called polyhedral, if all facial walks are cycles and any two of them are either disjoint or their intersection is a vertex or an edge. If G is a cubic graph, then any two facial walks are either disjoint or they intersect in an edge. It is also easy to see (cf. [5]) that every facial cycle of a polyhedral embedding is induced and non-separating. There is another way of looking at polyhedral embeddings (cf., e.g. [5], [6]). Proposition 1.1 An embedding of a graph G is polyhedral if and only if G is 3-connected and the embedding has face-width at least 3. If 1967 Grünbaum proposed a far-reaching generalization of the four color theorem (which had not yet been proved at that time). It is well known that the four color theorem is equivalent to the statement that every 3-connected planar cubic graph is 3-edge-colorable. This is no longer true for 3-connected cubic graphs on the torus since the Petersen graph P embeds in this surface. However no embedding of P in the torus is polyhedral. The lack of orientable polyhedral embeddings of other non 3-edge-colorable cubic graphs known at that time led Grünbaum to the following Conjecture 1.2 (Grünbaum [2]) If a cubic graph admits a polyhedral embedding in an orientable surface, then it is 3-edge-colorable. This Conjecture has been checked for all cubic graphs having up to 30 vertices (see Section 6). Cubic graphs that do not have 3-edge-colorings are said to be of class 2 . By Proposition 1.1 it suffices to check this Conjecture for 3-connected cubic graphs. It is not difficult to see that we may also restrict our attention to cyclically 4-edge-connected cubic graphs (cf. Theorem 3.1). Cyclically 4edge-connected cubic graphs of class 2 and with girth at least 5 are commonly known as snarks. In Section 2 it is proved that short cycles are necessarily facial in polyhedral embeddings of cubic graphs. In Section 3 we study reductions of graphs with nontrivial k-edge-cuts for k ≤ 5. In Section 4 it is proved that Isaacs graphs [4], except for the smallest one J 3 (which is just a one-vertextruncation of the Petersen graph and has a polyhedral embedding in the projective plane), have polyhedral embeddings neither in orientable nor in non-orientable surfaces. In Section 5 Goldberg graphs are considered. They have no polyhedral embeddings in orientable surfaces, but all of them have 2 polyhedral embeddings in non-orientable surfaces. It is also proved that for every non-orientable surface S there exists a cubic graph of class 2, which polyhedrally embeds in S. 2 Short cycles If a cubic graph with a short cycle C has a polyhedral embedding, then C is very likely to be a facial cycle. This is established by the following results. Lemma 2.1 Let G be a cubic graph and C a 3-cycle of G. Then C is a facial cycle in every polyhedral embedding of G. Proof. Let C = v0 v1 v2 v0 be a 3-cycle of G. Denote the neighbour of v i not in C with vi′ , i = 0, 1, 2. A facial cycle in a polyhedral embedding of G ′ , i = 0, 1, 2, indices modulo cannot contain any of the paths vi′ vi vi+1 vi+2 vi+2 3, since it must be induced. This implies that we have three facial cycles ′ , i = 0, 1, 2, indices modulo 3. Then C is a at C, which contain vi′ vi vi+1 vi+1 facial cycle. Lemma 2.2 Let G be a cubic graph other than K 4 and let C be a 4-cycle of G. Then C is a facial cycle in every polyhedral embedding of G. Proof. If G has a polyhedral embedding and G is not K 4 , then every 4-cycle of G is induced, since G is 3-connected by Proposition 1.1. Let C = v0 v1 v2 v3 v0 be a 4-cycle of G and let vi′ be the neighbour of vi not in C, i = 0, 1, 2, 3. Suppose that all facial cycles, which intersect C, intersect C in one edge only. Then it is easy to see that C is a facial cycle. Otherwise there is at least one facial cycle C 1 6= C that intersects C in more than one edge. Facial cycles in polyhedral embeddings are induced. Hence we may assume that C1 contains the path v0′ v0 v1 v2 v2′ . The other facial cycle C2 , which contains the edge v0′ v0 , must contain the path v0′ v0 v3 v3′ in order not to intersect C1 at v2 . The third facial cycle through v0 then contains edges v0 v1 , v0 v3 and v3 v2 , which is a contradiction. Let a graph G be embedded in a surface S, let F be a facial cycle and let C be a cycle of G. We say that F is k-forwarding at C, if F and C intersect precisely in k consecutive edges on C. Lemma 2.3 Let G be a cubic graph and C an induced 5-cycle of G. If G has a polyhedral embedding in a surface S, then the following holds. 3 (a) If S is orientable, then C is a facial cycle. (b) If S is non-orientable, then either C is a facial cycle or all facial cycles that intersect C are 2-forwarding at C. Proof. Let C = v0 v1 v2 v3 v4 v0 be a 5-cycle of G. Suppose that no facial cycle (other than possibly C) intersects C in more than one consecutive edge on C. Then it is easy to see that C is a facial cycle. Now let F be a facial cycle that intersects C in at least two consecutive edges on C. Facial cycles in polyhedral embeddings are induced. Therefore F is either 3-forwarding or 2-forwarding at C. If F is 3-forwarding, we can assume that the path v 0′ v0 v1 v2 v3 v3′ is in F . Then the facial cycle, which contains the path v 0 v4 v3 , intersects twice with F . This contradiction implies that no facial cycle is 3-forwarding at C. We may assume that F contains the path v0′ v0 v1 v2 v2′ . The facial cycle, which contains the path v1′ v1 v2 , must contain the path v1′ v1 v2 v3 so it is 2forwarding. If we continue along the cycle C, we see that all facial cycles at C are 2-forwarding at C. To complete the proof, we will show that S is not orientable, if all facial cycles at C are 2-forwarding. Suppose that S is orientable and let C i be the facial cycle, which contains the path v i vi+1 vi+2 , i = 0, 1, 2, 3, 4, indices modulo 5. We can assume that in the orientation of C 0 , induced by the orientation of S, vertices v0 v1 v2 are in clockwise order. Then the vertices v3 v2 v1 are in this clockwise order on C1 . If we continue along C, we see that in C4 vertices v4 v0 v1 are in clockwise order. But then C0 and C4 induce the same orientation of the edge v0 v1 , which is a contradiction with the assumption that S is orientable. Corollary 2.4 If a cubic graph G contains two induced 5-cycles, whose intersection is nonempty and is not just a common edge, then G has no orientable polyhedral embeddings. Proof. Suppose we have an orientable polyhedral embedding of G. By Lemma 2.3 both 5-cycles are facial. This is a contradiction with the fact that their intersection contains more than just one edge. In the Petersen graph P every edge is contained in four induced 5-cycles. Lemma 2.3 therefore implies that P has no orientable polyhedral embeddings. However, P has a polyhedral embedding in the projective plane (see Figure 1). 4 Figure 1: The Petersen graph embedded in the projective plane. Lemma 2.3 and its Corollary 2.4 can be applied on many other snarks, for example the Szekeres snark that is shown in Figure 2. Theorem 2.5 The Szekeres snark has no polyhedral embeddings. Proof. Each of the five “parts” of the Szekeres snark (see Figure 2) contains a path v1 v2 . . . v9 on 9 vertices and a vertex v0 that is adjacent with v2 , v5 , v8 and further there are edges v1 v6 and v4 v9 . There are four induced 5-cycles C1 = v0 v2 v1 v6 v5 v0 , C2 = v0 v2 v3 v4 v5 v0 , C3 = v0 v8 v9 v4 v5 v0 and C4 = v0 v8 v7 v6 v5 v0 . Cycles C1 and C2 intersect at two edges adjacent to v0 . Therefore they are not both facial cycles. If none of C 1 , C2 is facial, then the 2-forwarding facial cycles at C1 and C2 , which contain their intersection C1 ∩ C2 , are distinct and intersect in two edges. So one of them is facial and the other is not. Similarly, one of the cycles C 3 , C4 is facial and the other one is not. Suppose the cycle C2 is facial. Then it is 1-forwarding at C 4 , so C4 is facial and C1 and C3 are not facial. This implies that there is a facial cycle that contains the path v1 v6 v5 v4 v9 and another facial cycle that contains the path v1 v2 v0 v8 v9 , which is a contradiction. Suppose now that C2 is not facial. Then C1 is facial and is 1-forwarding at C4 . So C4 is a facial cycle and C3 is not. This implies that there is a facial cycle that contains the path v3 v2 v0 v8 v7 and another facial cycle that contains the path v3 v4 v5 v6 v7 , which is a contradiction. Nonexistence of orientable polyhedral embeddings of the Szekeres snark has been proved earlier by Szekeres [7]. 5 v2 v3 v1 v9 v0 Figure 2: The Szekeres snark. 3 Small edge-cuts Let G1 and G2 be cubic graphs and v1 ∈ V (G1 ), v2 ∈ V (G2 ). Denote the three neighbours of v1 in G1 by z0 , z1 , z2 and the three neighbours of v2 in G2 by u0 , u1 , u2 . Let G = G1 ∗ G2 be the cubic graph obtained from graphs G1 and G2 by deleting vertices v1 and v2 and connecting vertices ui with zi for i = 0, 1, 2. We call G the star product of G1 and G2 . It is easy to see that the graph G is 3-edge-colorable if and only if both G 1 and G2 are 3-edge-colorable. Theorem 3.1 The star product G = G1 ∗ G2 has a polyhedral embedding in an (orientable) surface if and only if both G 1 and G2 have polyhedral embeddings in some (orientable) surfaces. Proof. Suppose we have polyhedral embeddings of G 1 and G2 . At vertex v1 we have three facial cycles Ci = zi v1 zi+1 Pi zi for i = 0, 1, 2, indices modulo 3. At vertex v2 we have three facial cycles Di = ui Ri ui+1 v2 ui for i = 0, 1, 2. Since the embeddings are polyhedral, paths P 0 , P1 , P2 and paths R0 , R1 , R2 are pairwise disjoint. In the embedding of the star product G = G 1 ∗ G2 we keep all facial cycles from embeddings of G 1 and G2 , which do not contain vertices v1 and v2 , and add three new facial cycles Fi = zi ui Ri ui+1 zi+1 Pi zi , i = 0, 1, 2, indices modulo 3. Facial cycles in G, which are facial cycles in 6 z0 z0 u0 R0 P0 z1 v 1 z1 R2 P2 R1 z2 G1 R0 P0 v2 u1 P1 u0 u1 R2 P2 R1 P1 z2 u2 G2 u2 G Figure 3: The star product G of graphs G1 and G2 . G1 or G2 , intersect pairwise at most once. A facial cycle F , which is also a facial cycle in G1 or G2 , intersects the facial cycle Fi , i = 0, 1, 2, only on the path Pi or only on the path Ri . So it intersects Fi at most once. Facial cycles Fi and Fi+1 intersect only in the edge ui+1 zi+1 , i = 0, 1, 2, indices modulo 3, since the paths P0 , P1 , P2 and R0 , R1 , R2 are pairwise disjoint. So the embedding of G is polyhedral. It is easy to see that the embedding of G is orientable if and only if the embeddings of G 1 and G2 are orientable. Suppose now that G has a polyhedral embedding. The three edges z i ui , i = 0, 1, 2, form a 3-cut in G. Since the embedding is polyhedral, we have three facial cycles Fi = ui Ri ui+1 zi+1 Pi zi ui , such that Fi and Fi+1 intersect in the edge zi+1 ui+1 , i = 0, 1, 2, indices modulo 3. We may assume that there are no negative signatures on edges zi ui , i = 0, 1, 2. In the embedding of G1 (and G2 ) we keep all facial cycles, which do not intersect G 2 (respectively G1 ), and add vertices v1 , v2 with such local rotations that we obtain new facial cycles Ci = zi v1 zi+1 Pi zi in G1 and Di = ui Ri ui+1 v2 ui in G2 , i = 0, 1, 2, induces modulo 3. Since we have no new intersections between facial cycles (intersections on zi ui become intersections on zi v1 and ui v2 ), the embeddings of G1 and G2 are polyhedral. It is also clear that both embeddings are in orientable surfaces if and only if the embedding of G is orientable, since we did not change local rotation at any vertex or change the signature of any edge. 7 If the embedding of G = G1 ∗ G2 in a surface S is constructed as in the proof of Theorem 3.1 from embeddings of G1 and G2 in surfaces S1 and S2 of Euler genus ǫ(S1 ) = k1 and ǫ(S2 ) = k2 , respectively, then the Euler genus of S is ǫ(S) = k1 + k2 . This is easily proved by using Euler’s formula for G, G1 and G2 . Let G1 and G2 be cubic graphs. Choose an edge e = xy in G1 and two nonadjacent edges f1 = u0 u1 and f2 = u2 u3 in G2 . Denote the neighbours of x in G1 by v0 , v1 , and the neighbours of y by v2 , v3 . Let G be the graph obtained from G1 and G2 by deleting vertices x, y in G1 and edges f1 , f2 in G2 and joining pairs vi ui , i = 0, 1, 2, 3. The graph G = G1 · G2 is called the dot product of G1 and G2 . If both G1 and G2 are snarks, then their dot product is also a snark. Theorem 3.2 Let G1 and G2 be cubic graphs. If G1 and G2 have polyhedral embeddings in (orientable) surfaces S 1 and S2 , such that the geometric dual of G2 is not a complete graph, then a dot product G = G 1 · G2 exists, which has a polyhedral embedding in an (orientable) surface S. If the Euler genus of surfaces S1 and S2 are ǫ(S1 ) = k1 and ǫ(S2 ) = k2 , then the Euler genus of S is ǫ(S) = k1 + k2 . Proof. Suppose that we have polyhedral embeddings as described. We claim that G2 contains facial cycles D0 , D1 , D2 , such that D1 intersects D0 and D2 but D0 and D2 do not intersect. To see this, consider the dual graph R. Since it is not a complete graph, it has two vertices c 0 and c2 that are at distance two in R. If c1 is their common neighbor, then we can take D 0 , D1 , D2 to be the facial cycles corresponding to c 0 , c1 and c2 , respectively. Let f1 = u0 u1 and f2 = u2 u3 be the intersections between D0 , D1 and D1 , D2 , respectively, and choose an arbitrary edge e = xy in G 1 . Denote the neighbours of x and y in G1 so that the facial cycles, which contain x or y, are C0 = v0 xv1 P0 v0 , C1 = v1 xyv2 P1 v1 , C2 = v2 yv3 P2 v2 , and C3 = v3 yxv0 P3 v3 . Since the embedding of G1 is polyhedral, paths P0 , P1 , P2 , P3 are pairwise disjoint, except that P0 and P2 may intersect. In G2 we will use the following notation for facial cycles: D0 = u0 R0 u1 u0 , D1 = u0 u1 R1 u2 u3 R3 u0 and D2 = u2 R2 u3 u2 . The paths R0 , R1 , R2 , R3 are pairwise disjoint. In the embedding of G we keep all local rotations at vertices of G 1 and G2 , which are not deleted (with added edges naturally replacing deleted edges), and all edge signatures. Instead of facial cycles C i , Di we get a facial cycle Fi = vi ui Ri ui+1 vi+1 Pi vi , i = 0, 1, 2, 3, indices modulo 4. Since the paths Pi , Ri are pairwise disjoint, except for the possible intersection between P0 and P2 , all intersections between facial cycles F i , i = 0, 1, 2, 3, are the 8 v0 u0 x P0 v1 R3 P3 P1 P2 v3 G1 u1 R0 R3 P1 u2 y v1 P3 R1 v2 u0 P0 R0 u1 v0 R1 v2 u2 v3 u3 P2 R2 u3 R2 G2 G Figure 4: The dot product G of graphs G1 and G2 . intersections of Fi and Fi+1 in edges vi+1 ui+1 , i = 0, 1, 2, 3, indices modulo 4, and possibly one more intersection between F 0 and F2 . It is clear that any facial cycle F that does not contain any of the vertices v i , ui intersects at most once with any Fi and that two such facial cycles intersect at most once. So the embedding of G is polyhedral. It is also clear that if the embeddings of G1 and G2 are in orientable surfaces, the embedding of G is also in an orientable surface. The Euler genus of S is obtained from Euler’s formula and equalities |V (G)| = |V (G1 )| + |V (G2 )| − 2 |E(G)| = |E(G1 )| + |E(G2 )| − 3 |F (G)| = |F (G1 )| + |F (G2 )| − 3 from which we conclude that ǫ(S) = k1 + k2 . Theorem 3.3 Let G be a cubic graph and S a 4-cut in G. If G admits a polyhedral embedding (in an orientable surface), then there exist graphs G 1 and G2 , such that G = G1 · G2 and G1 admits a polyhedral embedding (in an orientable surface). Proof. Suppose that the edges ui vi , i = 0, 1, 2, 3, form a 4-cut S in G. If a facial cycle contains more than two edges of S, the embedding of G can 9 not be polyhedral. So we have four distinct facial cycles F 0 , F1 , F2 , F3 that contain edges of S. Since S is a cut, every cycle F i , i = 0, 1, 2, 3, contains two edges of S. Since the embedding is polyhedral, each of the F i intersects two other Fj , Fk . In the dual a subgraph induced by the vertices corresponding to F i , i = 0, 1, 2, 3, is a simple graph on four vertices in which all vertices are of degree 2. It must be a 4-cycle. Therefore we can assume that faces F i and Fi+1 intersect in the edge vi+1 ui+1 , i = 0, 1, 2, 3, indices modulo 4. Each facial cycle Fi is then of the form Fi = vi ui Ri ui+1 vi+1 Pi vi . Since F0 and F2 intersect at most once, we can assume they do not intersect at the paths P 0 and P2 . Let G1 be the component of G − S, which contains paths P i . If we set rotations of all vertices in G2 as they are in G (and replace deleted edges naturally with added edges), we can set rotations around vertices x and y so that the facial cycles in G1 , which do not contain x or y, remain unchanged and we have four new facial cycles C0 = v0 xv1 P0 v0 , C1 = v1 xyv2 P1 v1 , C2 = v2 yv3 P2 v2 , and C3 = v3 yxv0 P3 v3 . Since we added no new intersections between facial cycles, which were already in G, and facial cycles C i , i = 0, 1, 2, 3 intersect pairwise only once, the embedding of G 1 is polyhedral. If the embedding of G is in an orientable surface, it is clear that the embedding of G1 is in an orientable surface. Suppose we have polyhedral embeddings of cubic graphs G 1 and G2 , at least one of which is in a non-orientable surface. Let us construct the embedding of the dot product G = G1 · G2 as in the proof of Theorem 3.2. If the embedding of G is in orientable surface, then we may assume that all signatures of edges are positive. Now we can construct embeddings of G 1 and G2 similarly as the embedding of G1 in the proof of Theorem 3.3, which are both in orientable surfaces and have the same set of facial cycles as the embeddings of G1 and G2 with which we started. Since at least one of these two is an embedding in a non-orientable surface, we have a contradiction. This shows Corollary 3.4 If we have polyhedral embeddings of G 1 and G2 at least one of which is non-orientable and construct a polyhedral embedding of G = G1 · G2 as in the proof of Theorem 3.2, then the embedding of G is nonorientable. Let G1 and G2 be cubic graphs. Choose a vertex v in G1 , an edge v3 v4 in G1 and a vertex z0 in G2 . Let the three neighbours of v be v0 , v1 , v2 and let z1 , z2 , u4 be the neighbours of z0 . Let the neighbours of z1 , z2 other than 10 u be u0 , u1 and u2 , u3 , respectively. If all these vertices are distinct, remove the vertex v from G1 , vertices z0 , z1 , z2 from G2 and the edge v3 v4 from G1 . If we join pairs vi ui , i = 0, 1, 2, 3, 4, we get a cubic graph G = G1 ♦G2 , which is called a square product of graphs G1 and G2 (see also Figure 5). The cut Q = {vi ui | i = 0, . . . , 4} in G is said to be the product cut. It is well known (cf., e.g. [3]) that if G1 and G2 are snarks, then their square product is also a snark. Theorem 3.5 Let G be a cubic graph with a matching Q, which is a 5cut of G. If G admits a polyhedral embedding (in an orientable surface), then there exist graphs G1 and G2 such that G = G1 ♦G2 and Q is the corresponding product cut and such that G2 admits a polyhedral embedding (in an orientable surface). Proof. Suppose that G has a polyhedral embedding. Since Q is a cut, every facial cycle contains an even number of edges in Q. It is easy to see that none of them contains four edges of Q (since the embedding is polyhedral). This implies that there are precisely 5 facial cycles F 0 , . . . , F4 that intersect Q and that the edges vi ui of Q, i = 0, . . . , 4, can be enumerated so that F i contains edges vi ui and vi+1 ui+1 , indices modulo 5, and v0 , . . . , v4 are in the same component of G − Q. The facial cycles Fi are of the form Fi = vi ui Ri ui+1 vi+1 Pi vi , i = 0, . . . , 4, indices modulo 5. Since the embedding is polyhedral, every one of the pairs of paths Pi , Pi+1 and Ri , Ri+1 is disjoint. Suppose that the facial cycles Fi and Fi+2 are disjoint for some i. Then both pairs Pi , Pi+2 and Ri , Ri+2 are disjoint. One of the pairs Pi+2 , Pi+4 and Ri+2 , Ri+4 , i = 0, . . . , 4, is disjoint. Because of the symmetry, we can assume that the pair Ri+2 , Ri+4 is disjoint. Suppose now that all pairs of cycles Fi , Fi+2 , i = 0, . . . , 4, intersect. In at least three out of five pairs, Fi and Fi+2 intersect on the same “side” (Pi and Pi+2 or Ri and Ri+2 ). By symmetry, we may assume that intersections are between Pi and Pi+2 . Since facial cycles Fi and Fi+2 intersect at most once, it follows that there exists an index j such that R j , Rj+2 , Rj+4 are pairwise disjoint. By above, we can assume that R4 , R1 , R3 are pairwise disjoint. Now we can add to G − Q new vertices v, z0 , z1 , z2 and edges v0 v, v1 v, v2 v, v3 v4 and u0 z1 , u1 z1 , u2 z2 ,u3 z2 , z1 z0 , z2 z0 , u4 z0 so that the graph G is a square product of G1 and G2 . In the embedding of G2 we keep all rotations and signatures of vertices and edges that were already in G and we naturally replace deleted edges with the added ones. Around vertices z 0 , z1 , z2 we 11 u0 v0 v0 u0 v1 u1 v2 u2 z1 P0 P1 P4 z0 v2 P2 u1 v1 v u2 R0 P0 R1 P1 R4 z2 v3 P3 u3 u4 R0 R1 R4 P4 R2 P2 R3 P3 v2 u3 v4 u4 R2 R3 v4 G1 G G2 Figure 5: The square product of G1 and G2 . can set rotations so that facial cycles in G 2 , which were not already in G, are D0 = u0 R0 u1 z1 , D1 = z0 z1 u1 R1 u2 z2 z0 , D2 = z2 u2 R2 u3 z2 , D3 = z0 z2 u3 R3 u4 z0 and D4 = z0 u4 R4 u0 z1 z0 . The only new intersections of facial cycles of G2 are between D4 and D1 and between D1 and D3 . Hence the embedding of G2 is polyhedral and if the embedding of G is in an orientable surface, so is the embedding of G2 . 4 Flower snarks Let Jk be the graph with vertices ai , bi , ci , di and edges ai ai+1 , ai bi , bi ci , bi di , ci di+1 , di ci+1 for i = 0, . . . , k − 1, indices modulo k. If k ≥ 5 is odd, then the graph Jk is a snark and is called the flower snark of order k [4]. The graph J5 is shown in Figure 6. Szekeres proved that flower snarks have no polyhedral embeddings in orientable surfaces [8]. The goal of this section is to prove the following Theorem 4.1 For k ≥ 4 the flower graph Jk has no polyhedral embeddings. 12 c1 d1 b1 c0 b0 d0 a1 a0 Figure 6: The Flower snark J5 . The rest of this section is devoted to the proof of Theorem 4.1. The subgraph Yj of Jk induced on vertices {aj , bj , cj , dj } is called the jth star or simply just a star in Jk . Suppose that we have a polyhedral embedding of J k . Let us look at how facial cycles can traverse Yj . If we walk along a facial cycle C, come to Y j from Yj−1 and then leave Yj going back to Yj−1 , we say that C is a backward face at j. Similarly we define a forward face at j, which is a facial cycle that enters Yj from Yj+1 and leaves it towards Yj+1 . If a cubic graph G has a polyhedral embedding, then at every vertex v ∈ V (G) with neighbours v1 , v2 , v3 , each path P = vi vvj , j 6= i, defines a unique facial cycle, which we will denote by F (P ). Lemma 4.2 If C is a facial cycle that contains at least two vertices of Y j , then the intersection of C with Yj is one of the three possible paths: aj bj cj , aj bj dj or cj bj dj . Proof. A cycle C can enter and exit Y j only through vertices aj , cj or dj . Suppose now that aj , cj ∈ V (C). The facial cycle C ′ = F (aj bj cj ) intersects C in two nonadjacent vertices aj and cj , so C = C ′ and C ′ contains the path aj bj cj . Similar conclusion holds if aj and dj are on C or if cj and dj are on C. Since all facial cycles are induced, the intersection C ∩ Y j can consists only of one of the three paths. 13 A facial cycle, which is neither forward nor backward at Y j , is called a cross face. It follows from Lemma 4.2 that each facial cycle, which intersects Yj , is either a backward, forward or a cross face. Lemma 4.3 At Yj there can be at most one backward (forward) face. If there is one backward face, then there is also one forward face and four distinct cross faces. The backward face at Y j is forward at Yj−1 and the forward face at Yj is backward at Yj+1 . Proof. Suppose we have two backward (forward) faces. By Lemma 4.2 they intersect at an edge adjacent to b j . If they intersect at bj aj , they also intersect at aj−1 aj , which is a contradiction. Similarly we get a contradiction, if they intersect at bj cj or bj dj . This shows that there is at most one backward (forward) face. Suppose now that C is a backward face. The edges between Y j and Yj+1 are traversed twice by C and four times by cross faces. The cross faces therefore traverse the edges between Y j and Yj+1 at most four times, hence there must be a forward face at Yj . If C contains the path aj bj cj , then {aj−1 , dj−1 } ⊆ C ∩ Yj−1 . By Lemma 4.2, C ∩ Yj−1 = aj−1 bj−1 dj−1 , so C is a forward face at Yj−1 . A similar conclusion holds if C ∩ Yj is either aj bj dj or cj bj dj . Similarly we also show that a forward face at Yj is backward at Yj+1 . Out of facial cycles F (aj bj cj ), F (aj bj dj ) and F (ci bj dj ) one is a backward face, one is a forward face and one is a cross face. Since the one that is a cross face is the only cross face, which contains more than one vertex of Y j , all cross faces are distinct. A backward face at j is called a bottom face if it contains the edge a j−1 aj and is called a top face if it does not contain a j−1 aj . A top face at Yj is of the form cj−1 bj−1 dj−1 cj bj dj cj−1 . So it is clear that we cannot have backward top faces at Yj and Yj+1 at the same time. The star Yj is of type 0 , if all facial cycles, which intersect it, are cross faces. It is of type 1 , if there is one forward and one backward face at Y j . Lemma 4.2 implies that if the graph Jk has a polyhedral embedding, then all stars are of type 0 or all stars are of type 1. Lemma 4.4 If Jk has a polyhedral embedding, then k ≤ 6 and all stars are of type 1. Proof. By Lemma 4.2 every polyhedral embedding of J k has at least four cross faces. For each j = 0, . . . , k − 1 we have at least one intersection 14 between four selected cross faces on edges from Y j to Yj+1 . Since we can have at most 6 such intersections, we have k ≤ 6. If all stars are of type 0, then Jk has precisely 6 facial cycles. The geometric dual of G on S has 6 vertices and 4k·3 2 = 6k edges. Since the dual is a simple graph, it has at most 15 edges, so 6k ≤ 15. This implies that k ≤ 2. Lemma 4.5 The graph J4 has no polyhedral embeddings. Proof. Suppose we have a polyhedral embedding of J 4 . All stars are of type 1, so there are precisely 4 cross faces. We have three 4-cycles C 1 = a0 a1 a2 a3 a0 , C2 = d0 c1 d2 c3 d0 , C3 = c0 d1 c2 c3 c0 in J4 , which are facial cycles by Lemma 2.2. These cycles are all cross faces. As in the proof of Lemma 4.4, we see that there are at least four intersections of cross faces. But since C1 , C2 , C3 are pairwise disjoint, this is not possible. Lemma 4.6 The flower snark J5 has no polyhedral embeddings. Proof. Suppose we have a polyhedral embedding of J 5 . Each star must be of type 1. If all backward faces are bottom faces, then the inner cross face a0 a1 a2 a3 a4 a0 does not intersect any other cross faces. So we have 5 intersections between three cross faces, which is not possible. Since we cannot have two consecutive top faces, we must have two consecutive bottom faces at stars j and j + 1 and a top face at star j + 2. We can assume j = 1. The facial cycle F (a0 a1 a2 ) contains the path a0 a1 a2 a3 a4 . If not, it would intersect twice with one of the bottom faces at stars 1 or 2. So it must be a0 . . . a4 a0 . The facial cycle, which contains b2 a2 and is different from the backward face at star 2, must contain the path b 2 a2 a3 b3 . This facial cycle intersects twice with the facial cycle d 2 b2 c2 d3 b3 c3 d2 , which is a contradiction. Lemma 4.7 The graph J6 has no polyhedral embeddings. Proof. All stars in J6 are of type 1. We have three 6-cycles C 1 = a0 a1 . . . a5 a0 , C2 = c0 d1 c2 . . . d5 c0 and C3 = c0 d1 c2 . . . d5 c0 . From previous proofs it follows that at each star Y j one of the four cross faces goes from one of C1 , C2 , C3 to another. We say that this cross face has made a transition at Yj . It is obvious that if a cross face makes at least one 15 transition, it makes more than one transition. So one cross face makes no transitions, since we can have at most 6 transitions. Let the four cross faces be F1 , F2 , F3 , F4 and let F1 be the one, which does not make any transition. Because of the symmetry, we can assume that F 1 = C1 . There are four cross faces and six intersections between them. This implies that they must all pairwise intersect and in particular, all cycles F 2 , F3 , F4 intersect F1 . All transitions of cross faces are transitions of F i to C1 and from C1 , i = 2, 3, 4. In particular, the cycle F2 makes a transition to the cycle C1 at some star Yj and a transitions from C1 at the star Yi+1 . But then F2 is not induced, which is a contradiction. This completes the proof of Theorem 4.1. 5 Goldberg snarks In 1981 Goldberg discovered another infinite family of snarks [1]. Let G k be a cubic graph with vertices ai , bi , ci , di , gi , hi , ei , fi for i = 0, . . . , k − 1, and edges ai ai+1 , ai bi , bi ci , bi di , ci ei , ci gi , di fi , di hi , ei fi , gi hi , ei fi+1 , hi gi+1 , for i = 0, . . . k − 1, where indices are modulo k. If k ≥ 5 is odd, then G k is known as the Goldberg snark . Accordingly, we refer to all graphs G k as Goldberg graphs. The graph G5 is shown in Figure 7. Theorem 5.1 No Goldberg graph has a polyhedral embedding in an orientable surface. On the other hand, every Goldberg graph G k , k ≥ 3, has a polyhedral embedding in the non-orientable surface of Euler genus k. Proof. Suppose that the graph Gk has a polyhedral embedding in an orientable surface. For every i = 0, . . . , k − 1 we have have two 5-cycles Bi = bi di hi gi ci bi and Ci = bi di fi ei ci bi . By Lemma 2.3 both are facial cycles. This is a contradiction, since B i and Ci intersect in two edges ci bi and bi di . An embedding in a non-orientable surface has the following facial cycles: (a) A = a0 a1 . . . ak−1 a0 and B = f0 e0 f1 e1 . . . fk−1 ek−1 f0 , (b) Ci = bi di fi ei ci bi , i = 0, . . . , k − 1, (c) Di = gi hi gi+1 hi+1 di+1 fi+1 ei ci gi , i = 0, . . . , k − 1, (d) Ei = ai ai+1 bi+1 ci+1 gi+1 hi di bi ai , i = 0, . . . , k − 1. 16 e0 c0 g0 f0 b0 d0 h0 a0 Figure 7: The Goldberg snark G5 . It is easy to see that this determines a non-orientable polyhedral embedding. The Euler genus of the underlying surface of the embedding is calculated from Euler’s formula 2 − ǫ(Gk ) = |V (Gk )| − |E(Gk )| + |F (Gk )| = 8k − 32 8k + 3k + 2 = 2 − k. Goldberg graphs have more than one polyhedral embedding, not all of the same genus. They can be described as follows. Consider the subgraph Ti induced on vertices ai , bi , ci , di , ei , fi , gi and hi . Let us look at how facial cycles can traverse it. There are (at least) two possibilities. There is a facial 5-cycle C i = bi di hi gi ci bi and there are facial cycles that contain paths P1i = ai−1 ai ai+1 , P2i = gi−1 hi gi hi+1 , P3i = gi−1 hi di fi ei fi−1 , P4i = hi+1 gi ci di ei fi ei+1 , P5i = ei+1 fi di bi ai ai+1 and P6i = fi−1 ei ci bi ai ai−1 , where P1i and P2i can possibly be part of the same facial cycle. In such case, we say that Ti is of type 1 . The second possibility is the following. There is a facial 5-cycle D i = bi ci ei fi di bi and there are facial cycles that contain paths R 1i = ai−1 ai ai+1 , R2i = fi−1 ei fi ei+1 , R3i = ai−1 ai bi di hi gi−1 , R4i = ai+1 ai bi ci gi hi+1 , R5i = fi−1 ei ci gi hi gi−1 and R6i = ei+1 fi di bi hi gi hi+1 , where R1i and R2i can possibly 17 be part of the same facial cycle. We say that T i is of type 2 . We now choose arbitrary the types of all subgraphs T i and join facial segments described above into facial cycles as follows. There is an automorphism of the graph Gk , which sends all cycles C i into cycles D i , so we can assume that the subgraph Ti is of type 1. If not, we join facial segments symmetrically according to this automorphism. If subgraphs Ti and Ti+1 are both of type 1, we join facial segments P 1i and P1i+1 , P2i and P2i+1 , P4i and P3i+1 and facial segments P5i and P6i+1 . If the subgraph Ti is of type 1 and Ti+1 of type 2, we join facial segments P1i , R3i+1 and P2i , facial segments R1i+1 , P5i and R2i+1 and facial segments P4i and R5i+1 . If all subgraphs Ti are of type 1 (or all are of type 2), then the embedding is the one described in the proof of Theorem 5.1. If there are two consecutive subgraphs Ti and Ti+1 of different types, we say that there is a transition at i. It is easy to see that the embedding is polyhedral if we have at least 6 transitions. It is also easy to see that the number of facial cycles of the embedding is 3k. In this manner we have obtained a large number of (combinatorially) different polyhedral embeddings of the graph G k in a surface of Euler genus k − 2. This shows that Goldberg snarks admit polyhedral embeddings in distinct non-orientable surfaces (of Euler genus k and k−2) and that they admit combinatorially different polyhedral embeddings in the same non-orientable surface (of Euler genus k − 2). This fact by itself is of certain interest, see [5, Section 5]. Corollary 5.2 For every positive integer k there exists a cubic graph of class 2, which polyhedrally embeds in the non-orientable surface N k of Euler genus k. For every k > 0 and k 6= 2 there exists a snark, which polyhedrally embeds in Nk . Proof. The Petersen graph P has a polyhedral embedding in N 1 . By Theorem 5.1 the Goldberg snark G2k+1 has a polyhedral embedding in N2k+1 for every k ≥ 1. The graph G3 is not a snark since it contains a 3-cycle C = a0 a1 a2 a0 . If we contract C to a vertex, we obtain a snark G ′3 , which polyhedrally embeds in N3 (cf. Theorem 3.1). For k > 1 we have a snark H2k+2 = G2k+1 · P , which polyhedrally embeds in N 2k+2 , and H4 = G′3 · P , which polyhedrally embeds in N4 (cf. Theorem 3.2). The dot product H 2 = P · J3 polyhedrally embeds in N2 . The graph H2 is not 3-edge-colorable, but is not a snark, since the girth of H 2 is 4. 18 There are two non-isomorphic dot products of two copies of the Petersen graph P , which are known as Blanuša’s snarks. But since the dual of P in the projective plane is K6 , we cannot use Theorem 3.2 to obtain polyhedral embeddings in the Klein bottle for either of them. Indeed, it can be shown that they do not have such embeddings. It is possible that no snark exists which polyhedrally embeds in the Klein bottle. Problem 5.3 Is there a snark that has a polyhedral embedding in the Klein bottle? 6 Computer search All snarks and all cyclically 4-edge-connected cubic graphs of class 2 having up to 30 vertices are known. They were generated by computer and are available at [9]. We have used this database to verify Conjecture 1.2 on them (using computer). Combined with Theorem 3.1 this computation shows: Theorem 6.1 Every cubic graph with at most 30 vertices that has a polyhedral embedding in an orientable surface is 3-edge-colorable. References [1] M. K. Goldberg, Construction of class 2 graphs with maximum vertex degree 3, J. Comb. Theory Ser. B 31 (1981), 282–291. [2] B. Grünbaum, Conjecture 6, in Recent Progress in Combinatorics, Ed. W. T. Tutte, Academic Press, New York, 1969, 343. [3] D. A. Holton, J. Sheehan, The Petersen Graph, Austral. Math. Soc. Lecture Series 7, Cambridge University Press, 1993. [4] R. Isaacs, Infinite families of nontrivial trivalent graphs which are not Tait colorable, Amer. Math. Monthly 82 (1975), 221–239. [5] B. Mohar, C. Thomassen, Graphs on Surfaces, The Johns Hopkins University Press, Baltimore and London, 2001. [6] N. Robertson, R. P. Vitray, Representativity of surface embeddings, in: Paths, Flows, and VLSI-Layout, B. Korte, L. Lovász, H. J. Prömel, and A. Schrijver Eds., Springer-Verlag, Berlin, 1990, 293–328. 19 [7] G. Szekeres, Polyhedral decompositions of cubic graphs, Bull. Austral. Math. Soc. 8 (1973), 367–387. [8] G. Szekeres, Non-colourable trivalent graphs, Combinatorial Mathematics (Proc. Third Austral. Conf., Univ. Queensland, St. Lucia, 1974), Lecture Notes in Math., Vol. 452, Springer, Berlin, 1975, 227–233. [9] http://www.cs.uwa.edu.au/˜gordon/remote/cubics/index.html 20