Classification of vertex-transitive cubic partial cubes

Classification of vertex-transitive cubic partial cubes Tilen Marc∗

arXiv:1509.04565v1 [cs.DM] 15 Sep 2015

Institute of Mathematics, Physics, and Mechanics, Jadranska 19, 1000 Ljubljana, Slovenia September 16, 2015

Abstract Partial cubes are graphs isometrically embeddable into hypercubes. In this paper it is proved that every cubic, vertex-transitive partial cube is isomorphic to one of the following graphs: K2 ƒ C2n , for some n ≥ 2, generalized Petersen graph G(10, 3), permutahedron, truncated cuboctahedron, or truncated icosidodecahedron. This complete classification of cubic, vertex-transitive partial cubes, a family that includes all cubic, distance-regular partial cubes (Weichsel, 1992), is a generalization of results of Brešar et. al. from 2004 on cubic mirror graphs, and a contribution to the classification of all cubic partial cubes.

Keywords: partial cubes; vertex-transitive graphs; cubic graphs; isometric cycles

1

Introduction

Hypercubes are considered one of the classic examples of graphs that posses many symmetries. It is a fundamental question to ask how this symmetries are preserved on their subgraphs. To our knowledge the first ones who addresses this question were Brouwer, Dejter and Thomassen in 1992 in [3]. They provided many surprising and diverse examples of vertex-transitive subgraphs of hypercubes, but failed to make a classification. Based on their results, the examples are too diverse and a classification seems too ambiguous. They suggested that one of the reasons for the latter is that the group of symmetries of a subgraph of a hypercube does not need to be induced by the group of symmetries of the hypercube. On the other hand, Weichsel in 1992 [18] considered distance-regular subgraphs of hypercubes. He derived certain properties of them, and noticed that all his examples are not just subgraphs, but isometric subgraphs of hypercubes. It was thus a natural decision to study only the symmetries of partial cubes, i.e. isometric subgraphs of hypercubes. He classified all distance-regular partial cubes based on their girth: hypercubes are the only ones with girth four, the six cycle and the middle level graphs are the only ones with girth six, and even cycles ∗

Electronic address: [email protected]

1

of length at least eight are the only ones with higher girth. Notice that all this graphs are vertex-transitive, therefore they are a subfamily of vertex-transitive subgraphs of hypercubes. Probably the most well-known and studied subfamily of partial cubes are median graphs. It is a well-known result from [15] that hypercubes are the only regular - and thous the only vertex-transitive - median graphs (for infinite vertex-transitive median graphs check [13]). Due to this result, an extensive study of regular partial cubes has been performed [1, 2, 4, 5, 10, 11]. It especially focuses on the cubic case, since the variety of these graphs is far more rich than in the case of median graphs. Connections with other geometric structures were established, for example with platonic surfaces [2] and simplicial arrangements [5]. From the point of view of vertex-transitive partial cubes, the most interesting is the study [2], where a new family of graphs called mirror graphs was introduced. Moreover, it was proved that mirror graphs are a subfamily of vertex-transitive partial cubes, and all mirror graphs that are obtained by cubic inflation (thus cubic graphs) were classified. In [12], we made an analysis of isometric cycles in a partial cube. In particular, the results imply that there are no cubic partial cubes with girth more than six. This suggests that, as in the case of Weichsel’s distance-regular partial cubes, also the case of cubic vertex-transitive partial cubes should be approached from the study of their girth. In addition, every automorphism of a partial cube G preserves the so-called Θ-classes of G, therefore every symmetry of G is induced by a symmetry of a hypercube. In the present paper we form a natural connection between the study of vertex-transitive subgraph of hypercubes and the study of cubic partial cubes: we classify all cubic vertextransitive partial cubes. The results can be seen as a generalization of results from [2] since all the mirror graphs are vertex-transitive partial cubes, and, in the cubic case, results from [18] since every distance-regular partial cube is vertex-transitive. Let K2 denotes the complete graph of order 2, Ck the cycle of length k, and G(n, k) the generalized Petersen graph with parameters 3 ≤ n, 1 ≤ k < n/2. The main result of this paper is the following: Theorem 1.1. If G is a finite, cubic, vertex-transitive partial cube, then G is isomorphic to one of the following: K2 ƒ C2n , for some n ≥ 2, G(10, 3), the permutahedron, the truncated cuboctahedron, or the truncated icosidodecahedron. To our surprise, the variety of the graphs from Theorem 1.1 (Figures 1, 2) is small, and all graphs are classical graphs that were studied in many (especially geometric) views. In connection with the above introduction, we have to point out that the permutahedron, the truncated cuboctahedron, and the truncated icosidodecahedron are cubic inflations of graphs of platonic surfaces [2], K2 ƒ C2n are the only cubic Cartesian products of (vertex-transitive) partial cubes (this includes also the hypercube Q 3 ∼ = K2 ƒ C4 ), while G(10, 3) is the only known non-planar cubic partial cube and isomorphic to the middle level graph of valence three [10].

2

Preliminaries

The paper is organized as follows. In the next section we shorty present all the definitions and results needed to prove Theorem 1.1. 2

(a) G(10, 3)

(b) Permutahedron

Figure 1: Cubic, vertex-transitive partial cubes

(a) Truncated cuboctahedron

(b) Truncated icosidodecahedron

Figure 2: Cubic, vertex-transitive partial cubes We will consider only simple (finite) graphs in this paper. The Cartesian product G ƒ H of graphs G and H is the graph with the vertex set V (G) × V (H) and the edge set consisting of all pairs {(g1 , h1 ), (g2 , h2 )} of vertices with {g1 , g2 } ∈ E(G) and h1 = h2 , or g1 = g2 and {h1 , h2 } ∈ E(H). Hypercubes or n-cubes are the Cartesian products of n-copies of K2 . We say a subgraph H of G is isometric if for every pair of vertices in H also some shortest path in G connecting them is in H. A partial cube is a graph that is isomorphic to an isometric subgraph of some hypercube. The middle level graph of valency n ≥ 1 is the induced subgraphs of hypercube of dimension 2n − 1 on vertices that have precisely n or n − 1 coordinates equal to 1. For a graph G, we define the relation Θ on the edges of G as follows: abΘx y if d(a, x) + d(b, y) 6= d(a, y) + d(b, x), where d is the shortest path distance function. In partial cubes Θ is an equivalence relation [19], and we write Fuv for the set of all edges that are in relation Θ with uv. We define Wuv as the subgraph induced by all vertices that are closer to vertex u than to v, that is Wuv = 〈{w : d(u, w) < d(v, w)}〉. In any partial cube G, the sets V (Wuv ) and V (Wvu ) partition V (G), with Fuv being the set of edges joining them. We define Uuv to be the 3

subgraph induced by the set of vertices in Wuv which have a neighbor in Wvu . For details and further results, see [9]. We shall need a few simple results about partial cubes. All partial cubes are bipartite, since hypercubes are. If u1 v1 Θu2 v2 with u2 ∈ Uu1 v1 , then d(u1 , u2 ) = d(v1 , v2 ). A path P of a partial cube is a shortest path or a geodesic if and only if it has all of its edges in pairwise different Θ classes. If C is a cycle and e an edge on C, then there is another edge on C in relation Θ with e. For the details, we again refer to [9]. For a graph G, we shall denote with g(G) the girth of G, i.e. the length of a shortest cycle in G. For a vertex v in a graph G, denote with N (v) the set of all vertices adjacent to v. A cover of G is a graph H with a surjective homomorphism f from H to G, such that f maps bijectively N (v) to N ( f (v)), for every v ∈ V (G). A 2-cover is a cover such that the preimage of every vertex of G has exactly two elements. A major part of this paper depends on results developed in [12]. The following definition was introduced to study isometric cycles in partial cubes: Definition 2.1. Let v1 u1 Θv2 u2 in a partial cube G, with v2 ∈ U v1 u1 . Let D1 , . . . , Dn be a sequence of isometric cycles such that v1 u1 lies only on D1 , v2 u2 lies only on Dn , and each pair Di and Di+1 , for i ∈ {1, . . . , n − 1}, intersects in exactly one edge from F v1 u1 , all the other pairs do not intersect. If the shortest path from v1 to v2 on the union of D1 , . . . , Dn is isometric in G, then we call D1 , . . . , Dn a traverse from v1 u1 to v2 u2 . If D1 , . . . , Dn is a traverse from v1 u1 to v2 u2 , then also the shortest path from u1 to u2 on the union of D1 , . . . , Dn is isometric in G. We will call this u1 , u2 -shortest path the u1 , u2 -side of the traverse and, similarly, the shortest v1 , v2 -path on the union of D1 , . . . , Dn the v1 , v2 -side of the traverse. The length of this two shortest paths is the length of the traverse. It is not difficult to prove the following useful result. Lemma 2.2 ([12]). Let v1 u1 Θv2 u2 in a partial cube G. Then there exists a traverse from v1 u1 to v2 u2 . We shall also need the following definition: Definition 2.3. Let D1 = (v0 v1 . . . vm vm+1 . . . v2m+2n1 −1 ) and D2 = (u0 u1 . . . um um+1 . . . u2m+2n2 −1 ) be isometric cycles with u0 = v0 , . . . , um = vm for m ≥ 2, and all other vertices pairwise different. We say that D1 and D2 intertwine and define i(D1 , D2 ) = n1 + n2 ≥ 0 as the residue of intertwining. We can calculate the residue of intertwining as i(D1 , D2 ) = (l1 + l2 − 4m)/2, where l1 is the length of D1 , l2 the length of D2 , and m the number of edges in the intersection. In [12], it was proved that if at least two isometric cycles in G share more than an edge, then there must be at least two isometric cycles that intertwine.

3

Proof of Theorem 1.1

We start by analyzing two simple cases that strongly determine the structure of a cubic vertextransitive partial cube. 4

Lemma 3.1. If a vertex of a cubic vertex-transitive partial cube G lies on two 4-cycles, then G∼ = K2 ƒ C2n , for some n ≥ 2. Proof. Let v0 lie on two 4-cycles. This two 4-cycles cannot share two edges since otherwise we would have K2,3 as a subgraph, which is impossible for a graph isomorphic to a subgraph of a hypercube. On the other hand, they must share one edge since the graph is cubic. Let v0 u0 be the edge they share, and let v−1 and v1 be the other two neighbors of v0 . Moreover, let u1 be the common neighbor of v1 and u0 and similarly u−1 be the common neighbor of v−1 and u0 . If v−1 , v0 , v1 lie on a common 4-cycle, then, by vertex-transitivity, every vertex lies in three 4-cycles, that pairwise intersect in an edge. It is not hard to see that then G ∼ = Q 3 . Thus assume v−1 , v0 , v1 do not lie in a common 4-cycle. By vertex-transitivity, also v1 lies in two 4-cycles that intersect in an edge. Since G is cubic, the only possibility is that v1 u1 is the shared edge and that there exist vertices v2 , u2 such that (v1 v2 u2 u1 ) is a 4-cycle. If v2 = v−1 , then by the maximum degree limitation u2 = u−1 , and thus G ∼ = C3 ƒ K2 , which is not a partial cube. Also, if v2 = u−1 , then u2 = v−1 , and G is not a partial cube. Thus, v2 and u2 are new vertices. We can use the same argument for v2 , u2 , as we did for v1 , u1 , and find vertices v3 , u3 in a 4-cycle (v2 v3 u3 u2 ). Again, we have multiple options: u3 = u−1 and v3 = v−1 , u3 = v−1 and v3 = u−1 , or vertices v3 , u3 are different from all before. In the first case G ∼ = K2 ƒ C4 , and in the second G is not a partial cube. If the third case occurs, we inductively continue: at some point the induction stops therefore G ∼ = K2 ƒ C2n , for some n ∈ N. Notice that, if we considered also infinite graphs, a slight modification of the proof would provide that the only cubic vertex-transitive partial cubes with a vertex that lies in two 4-cycles are K2 ƒ C2n , for n ≥ 2, and K2 ƒ P∞ . Lemma 3.2. If an isometric 4-cycle and an isometric 6-cycle of a partial cube share two edges, there is a vertex that lies in three 4-cycles. Proof. Let (v0 v1 . . . v5 ) be an isometric 6-cycle and assume vertices v0 and v2 have a common neighbor u1 , different from v1 . Then v0 u1 Θv1 v2 Θv5 v4 . Since v0 and v5 are adjacent, also u1 and v4 are. Thus, vertex u1 lies in three 4-cycles. The last lemma holds for an arbitrary partial cube, but in a cubic vertex-transitive partial cube G, together with Lemma 3.1, it implies that if an isometric 4-cycle and an isometric 6-cycle in G share two edges, then G ∼ = K2 ƒ C4 ∼ = Q3. In [12] it was proved that there exists no regular partial cube with degree at least three and girth more than six. For the analysis of partial cubes with girth six, a graph X was introduced as shown in Figure 3. It was proved, that every regular partial cube with the minimum degree at least three and girth six must have an isometric subgraph isomorphic to X . We shall analyze this case in the next lemma. Firstly, notice the following: If D1 = (v0 v1 . . . v5 ) is an isometric 6-cycle in a partial cube G, for some vertices v0 , . . . , v5 , and also D2 = (v0 v1 v2 u3 u4 u5 ) is an isometric 6-cycle in G, for some vertices u3 , u4 , u5 not in D1 , then u3 u4 Θv0 v1 Θv3 v4 . Since d(u3 , v3 ) = 2, it holds d(u4 , v4 ) = 2. Thus, there is a vertex x adjacent to u4 and v4 . Summing up, if two isometric 6-cycles intertwine with intersection in two edges, then we have X as a subgraph. Moreover, no two isometric 6-cycles can share three or more edges, as a direct consequence of the transitivity of relation Θ. 5

v1 v8 v7

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Figure 3: Graph X Lemma 3.3. If X is an isometric subgraph of a cubic vertex-transitive partial cube G, then G ∼ = G(10, 3). Proof. Let X be an isometric subgraph of G, and denote its vertices as in Figure 3. Firstly, assume that there is a 4-cycle in G. By vertex-transitivity, v2 must be incident with a 4-cycle. But then an isometric 6-cycle and a 4-cycle must share two edges. By the previous two lemmas, this implies G = C4 ƒ K2 , and G does not have isometric subgraph isomorphic to X . Thus there are no 4-cycles in G. Notice that v2 in X lies in three isometric 6-cycles. Since G is vertex-transitive, also c1 must lie in at least three isometric 6-cycles. Since X is an isometric subgraph in a cubic graph without 4-cycles, and no two 6-cycles intersect in three edges, there must be a path P 0 of length 4, connecting c1 with one of the vertices v1 , v3 , v5 , v7 . Moreover, P 0 can intersect X only in its endpoints. Without the loss of generality, assume P 0 connects c1 and v7 and denote the vertices of P 0 with c1 , s1 , u6 , u7 , v7 , respecting the order in P 0 . Consider the isometric cycle D1 = (c1 s1 u6 u7 v7 v8 ) and the isometric cycle D2 = (c1 v4 v5 v6 v7 v8 ). Cycles D1 and D2 intertwine in two edges, thus, by the notice before this lemma, there is a vertex u5 adjacent to v5 and u6 . By the isometry of X , u6 is distinct from all vertices of X . Call X 0 the induced graph in V (X ) and s1 , u5 , u6 , u7 . Since there is no 4-cycle in G and X is an isometric subgraph, X 0 is as represented in Figure 4(a). v1

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(b) Graph X 00

Figure 4: Induced subgraphs Vertex c2 must lie in at least three isometric 6-cycles. As before, the fact that X is isometric, maximal degree of G is three, and the absence of 4-cycles imply that there exists a path P 00 6

of length 4, connecting c2 and one of v1 , v3 , v5 , v7 . By symmetry we can limit ourselves to two possibilities. If P 00 connects c2 and v5 , then P 00 must be on c2 , s2 , u4 , u5 , v5 , where s2 and u4 are some two vertices in G different from the vertices in V (X 0 ) (since X 0 is an induced graph). Then the cycle D3 = (c2 s2 u4 u5 v5 v6 ) and the cycle D4 = (c2 v2 v3 v4 v5 v6 ) meet in edges v5 v6 and v6 c2 . By the above remark, there must exist a vertex u3 connecting v3 and u4 . Since X 0 is an induced subgraph, u3 is distinct from all the vertices of X 0 . Denote the graph induced on V (X 0 ) and s2 , u3 , u4 by X 00 . Again, since G has no 4-cycles and X is an isometric subgraph, X 00 is as in Figure 4(b). On the other hand, P 00 can connect c2 and v3 . For the same reasons as before, there must exist vertices s2 , u4 , u3 , different from the vertices of X 0 such that P 00 lies on c2 , s2 , u4 , u3 , v3 . Then the cycle D4 from above and the cycle D5 = (c2 s2 u4 u3 v3 v2 ) share edges c2 v2 and v2 v3 . The remark before the lemma implies that there must be a common neighbor of v5 and u4 . This can only be u5 , thus we again have X 00 as an induced subgraph. We continue in the same fashion. The cycle D3 and the cycle (v5 v6 v7 u7 u6 u5 ) share v6 v5 and v5 u5 , thus there must be a vertex u8 connecting s2 and u7 . Similarly, the cycle (c1 s1 u6 u5 v5 v4 ) and the cycle (v3 v4 v5 u5 u4 u3 ) share v4 v5 and v5 u5 thus there must be a vertex u2 connecting s1 and u3 . Let X 000 be the subgraph induced on vertices V (X 00 ) and u8 , u2 . Since X is an isometric subgraph and G is without 4-cycles, X 000 is as in Figure 5(a). v1

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(b) Graph G(10, 3)

Figure 5: Induced subgraphs Notice that the only vertices in X 000 that do not have degree 3 are v1 , u2 , u8 . Also, observe that v5 lies in six 6-cycles. The only option that v1 lies in six 6-cycles is that there exists u1 connected to v1 , u2 and u8 . It can be checked directly that the obtained graph (shown in Figure 5(b)) is isomorphic to G(10, 3). Lemma 3.3 implies that if G is a cubic, vertex-transitive partial cube, then G ∼ = G(10, 3) or has girth four. We will use this in the following analysis. It is a well-known fact, that the edge-connectivity of a vertex-transitive graph equals to the degree of its vertices [8]. In a cubic, vertex-transitive partial cubes this implies that |Fab | ≥ 3 for every edge ab ∈ E(G). Lemma 3.4. In a finite, cubic, vertex-transitive partial cube G, every pair of incident edges lies in an isometric cycle. 7

Proof. Firstly, assume two isometric cycles in G share more than an edge. Than at least two isometric cycles intertwine. Let D1 = (v0 v1 . . . vm vm+1 . . . v2m+2n1 −1 ) and D2 = (u0 u1 . . . um um+1 . . . u2m+2n2 −1 ) be isometric cycles with u0 = v0 , . . . , um = vm for m ≥ 2, and all other vertices are pairwise different. It holds v2m+n1 −1 v2m+n1 Θvm vm−1 and vm vm−1 Θu2m+n2 −1 u2m+n2 , thus, by Lemma 2.2, there exists a traverse T1 from v2m+n1 −1 v2m+n1 to u2m+n2 −1 , u2m+n2 . Similarly, v2m+n1 −1 v2m+n1 −2 Θu2m+n2 −1 , u2m+n2 −2 , and there exists a traverse T2 from v2m+n1 −1 v2m+n1 −2 to u2m+n2 −1 , u2m+n2 −2 . Let P1 be the v2m+n1 −1 , u2m+n2 −1 -side of T1 and P2 the v2m+n1 −1 , u2m+n2 −1 -side of T2 . Let v be the neighbor of v2m+n1 −1 , different from v2m+n1 and v2m+n1 −2 . Since G is cubic, v lies in P1 and in P2 . Thus, all the pairs of edges that are incident with v2m+n1 −1 lie in some isometric cycle. By transitivity, this holds for all the vertices. Secondly, assume no two isometric cycles share more than an edge. Graph G is then not isomorphic to G(10, 3), thus the girth of G is 4. Let D = (v0 v1 v2 v3 ) be a 4-cycle in G. Let ab be an edge in F v0 v1 , distinct from v0 v1 and v2 v3 . By Lemma 2.2, there exists a traverse from v0 v1 to a b. Let D0 be the first isometric cycle on this traverse. Without loss of generality assume D0 6= D (if otherwise take the second cycle on the traverse and exchange edge v0 v1 with v2 v3 ). Similarly, without loss of generality v1 v2 lies in an isometric cycle D00 , different from D. Since isometric cycles D, D0 and D00 pairwise share at most an edge, all the pairs of edges incident with v1 lie in some isometric cycle. By transitivity, this holds for all the vertices. Let u be an arbitrary vertex of a cubic, vertex-transitive partial cube G, and let u1 , u2 , u3 be its neighbors. Let g1 (G) be the size of a shortest isometric cycle on u1 , u, u2 , let g2 (G) be the size of the shortest isometric cycle on u2 , u, u3 , and let g3 (G) be the size of the shortest isometric cycle on u3 , u, u1 . Without loss of generality assume g1 (G) ≤ g2 (G) ≤ g3 (G). Clearly, for vertex-transitive partial cubes functions g1 , g2 , g3 are independent of the choice of vertex u. As noted, in cubic vertex-transitive partial cubes it holds g1 (G) ≤ 6. Moreover, by Lemma 3.3, G∼ = G(10, 3) if and only if g1 (G) = 6. For the next lemma we will need the following easy fact that holds in cubic, vertex-transitive partial cubes with girth 4. It was already shown that if two isometric 6-cycles meet in two edges, we must have a subgraph isomorphic to X . Clearly, they cannot meet in more than two edges or just in a vertex. On the other hand, if they meet in one edge, we have two possibilities: ether g1 (G) = 4, g2 (G) = 6, g3 (G) = 6, or a 4-cycle and a 6-cycle meet in two edges. The latter case has already been covered. On the other hand, if g1 (G) = 4, g2 (G) = 6, g3 (G) = 6 and a vertex is incident with three isometric 6-cycles, then one of them must intertwine with a 4-cycle or a 6-cycle, since G is cubic. Again, this cannot be. To sum things up, if a vertex in a cubic, vertex-transitive graph G ∼ 6 G(10, 3) is incident with at least two isometric 6-cycles, = then g1 (G) = 4, g2 (G) = 6, g3 (G) = 6, and every vertex is incident with exactly two isometric 6-cycles and one isometric 4-cycle. We cover the latter case in the next lemma and obtain another well known cubic vertextransitive partial cube [7]. Lemma 3.5. Let G be a cubic vertex-transitive partial cube with g1 (G) = 4, g2 (G) = 6, g3 (G) = 6. Then G is isomorphic to the permutahedron. Proof. We paste a disc on every isometric 6-cycle and every 4-cycle. By the discussion before the lemma, we obtain a closed surface with G embedded into it. Denote with f6 the number of isometric 6-cycles, with f4 the number of 4-cycles, with f the number of faces of the embedding 8

of G, with n and e the number of vertices and edges of G, and with χ the Euler characteristic of the surface. We have 3n = 2e,

f4 + f6 = f ,

4 f4 = 6 f6 /2 = n, and n − e + f = 2 − χ.

From the second and third equation we get n( 14 + 26 ) = f . If we use the latter combined with the first equation in the Euler formula we get:   n 1 2 1 + − = . 2−χ = n 4 6 2 12 Since the right side of the equation is positive, it holds χ < 2, i.e. χ = 0 or χ = 1. In both cases n ≤ 24. Cubic partial cubes up to 32 vertices are known, the only vertex-transitive on 24 vertices are the permutahedron and K2 ƒ C12 , while K2 ƒ C6 is the only one on 12 vertices. Thus G must be isomorphic to the permutahedron. For the sake of convenience, we shall call the graphs C2n ƒ K2 (for n ≥ 2), G(10, 3), and the permutahedron the basic cubic graphs. In the following part we will find all the other cubic vertex-transitive partial cubes. For this we shall need a bit technical lemma. Lemma 3.6. Let u0 v0 Θum vm with um ∈ Uu0 v0 in a partial cube G. If P = u0 u1 . . . um is a geodesic, then at least one of the following holds (Cases (i)-(iii) are illustrated in Figure 6). (i) There exist vertices w i1 , w i2 , . . . , w il ∈ / V (P), for some l ≥ 0 and 0 < i1 < i2 − 1, i2 < i3 −1, . . . , il−1 < il −1 < m−1, such that the path u0 u1 . . . ui1 −1 w i1 ui1 +1 . . . uil −1 w il uil +1 . . . um is the u0 , um -side of some traverse T from v0 u0 to vm um . (ii) There exist edges ui zi , u j u j+1 , and vertices w i1 , w i2 , . . . , w il ∈ / V (P), for some l ≥ 0, 0 ≤ i < i1 < i2 − 1, i2 < i3 − 1, . . . , il−1 < il − 1 < j − 1 ≤ m − 1, such that the path ui ui+1 . . . ui1 −1 w i1 ui1 +1 . . . uil −1 w il uil +1 . . . u j is the ui , u j -side of some traverse T from zi ui to u j+1 u j of length at least two. (iii) There exist a vertex w i adjacent to ui−1 and ui+1 , edges w i zi and u j u j+1 , and vertices w i1 , . . . , w i l ∈ / V (P), for some l ≥ 0 and 0 < i < i1 − 1, i1 < i2 − 1, . . . , il−1 < il − 1 < j − 1 ≤ m − 1, such that the path w i ui+1 . . . uil −1 w il uil +1 . . . u j is the w i , u j -side of some traverse T from zi w i to u j+1 u j . Proof. Assume the lemma does not hold and let v0 u0 , vm um be counterexample edges with geodesic P = u0 u1 . . . um that has length as small as possible. By Lemma 2.2, there exists a traverse from v0 u0 to vm um , and let P1 be the u0 , um -side of it. If P1 = P, then Case (i) in the lemma holds, a contradiction. Thus there exists a cycle D0 = (uk1 uk1 +1 . . . uk0 zk0 −1 . . . zk1 +1 ) for some 0 ≤ k1 < k0 − 1 ≤ m − 1, where the path uk1 zk1 +1 . . . zk0 +1 uk0 is a part of the u0 , um -side of a traverse from v0 u0 to vm um . If this cycle is of length 4, take the next one on P of the same form. If all of them are 4-cycles, we have Case (i), a contradiction. Therefore assume D0 is not a 4-cycle. Firstly, assume it is isometric. Then edges uk1 zk1 +1 and uk0 uk0 −1 are in relation Θ, and D0 is a traverse from uk1 zk1 +1 to uk0 uk0 −1 of length at least two. Thus we have Case (ii). 9

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(a) An example of Case (i). The thick edges are the edges of P, w i1 = w2 , w i2 = w6 , l = 2.

(b) An example of Case (ii). The thick edges are edges in P, u j = ui+6 , w i1 = w i+4 , and l = 1.

(c) An example of Case (iii). The thick edges are edges in P, u j = ui+6 , w i1 = w i+4 , and l = 1.

Figure 6 Now assume D00 is not isometric. There must be exactly two edges in D0 that are in Fuk zk +1 . 1 1 Let uk2 uk2 +1 ∈ Fuk zk +1 , for some k1 < k2 < k0 . Then the path P 0 = uk1 . . . uk2 is shorter than P, 1 1 thus the lemma holds for it. If Case (ii), resp. Case (iii), holds for P 0 , then Case (ii), resp. Case (iii), holds for P as well. If Case (i) holds for P 0 with edges uk1 zk1 +1 , uk2 uk2 +1 , and the traverse from uk1 zk1 +1 to uk2 uk2 +1 is of length at least two, then Case (ii) holds for P. Thus assume the traverse from uk1 zk1 +1 to uk2 uk2 +1 is of length 1, i.e. there is a 4-cycle 00 D = (uk1 uk1 +1 uk1 +2 zk1 +1 ). Since D0 is not isometric, it holds D00 6= D0 . Let zk1 +2 be the neighbor of zk1 +1 in D0 , different from uk1 , and let uk3 uk3 +1 , for k1 + 2 ≤ k3 < k0 , be the other edge in D0 that is in Fzk +1 zk +2 . 1 1 Again, the isometric path P 00 = zk1 +1 uk1 +2 . . . uk3 is shorter than P, thus the lemma holds for P 00 with edges zk1 +1 zk1 +2 and uk3 uk3 +1 . If there exists a 4-cycle (zk1 +1 w k1 +2 uk1 +3 uk1 +2 ) for some vertex w k1 +2 , then Case (iii) holds for edges zk1 +1 w k1 +2 and uk1 +2 uk1 +3 with the traverse being a 4-cycle. Thus assume there is no such 4-cycle. If Case (i) holds for P 00 with zk1 +1 zk1 +2 and uk3 uk3 +1 , then Case (iii) holds for P. If Case (ii) holds for P 00 with zk1 +1 zk1 +2 and uk3 uk3 +1 , then Case (ii) or Case (iii) holds for P. Finally, if Case (iii) holds for P 00 with zk1 +1 zk1 +2 and uk3 uk3 +1 , then Case (iii) holds for P since there is no 4-cycle of the form (zk1 +1 w k1 +2 uk1 +3 uk1 +2 ). As we can see in Figure 6, Lemma 3.6 provides a traverse T that is attached to the path P as in one of the Cases (i)-(iii) with (possibly) some 4-cycles in between T and P.

10

In Figure 7 we define graphs H1 , H2 and H3 . The next lemma will show how Lemma 3.6 and subgraphs H1 , H2 , and H3 are connected. v1

u1

w1

v2

u2

w2

v1

u1

w1

v1

u1

w1

v3

u3

w3

v2

u2

w2

v2

u2

w2

v4

u4

w4

w3

v3

u3

w3

v5

w5

v4

u4

w5

v6

v5

u5

u3

v3 v4

u4

u5 u6

(a) Graph H1

(b) Graph H2

u5

w5 u7

u6

w7

u8 (c) Graph H3

Figure 7 Lemma 3.7. Let G be a cubic vertex-transitive partial cube. Let there be a traverse T1 from u0 x 0 to um x m with P1 = u0 . . . um being the u0 , um -side of it. If we have a traverse T2 attached to P1 as in one of the Cases (i)-(iii) from Lemma 3.6, with additional assumption that the isometric cycles on the traverse T1 , isometric cycles on T2 and the isometric 4-cycles in between, do not pairwise share more than an edge, then G is isomorphic to one of the basic graphs or has an isometric subgraph isomorphic to H1 , H2 or H3 . Proof. Firstly, assume the traverse T2 is attached to P1 as in Case (ii) or (iii). We fix the notation as in Lemma 3.6, i.e. T2 is a traverse from ui zi to u j u j+1 in Case (ii), or from w i zi to u j u j+1 in Case (iii), for some 0 ≤ i < j − 1 < m − 1, and some vertices w i and zi . Let C 2 be the last isometric cycle of T2 , the one that includes u j u j+1 . Assume the neighbor of u j on a side of T2 is a vertex w j−1 that is not on P1 , or in other words, we have a 4-cycle (u j u j−1 u j−2 w j−1 ) in between T1 and T2 . Since C 2 and this 4-cycle share at most an edge, there must exist an edge w j−1 z j−1 in relation Θ with u j u j+1 , i.e. C 2 is a 4-cycle. There exist two incident 4-cycles, therefore, by Lemma 3.1, G is isomorphic to a basic graph. Moreover, we see that if we have Case (iii), then the traverse T2 must have length at least 2. By above, we can assume that T2 starts with an isometric cycle D2 that includes u j u j+1 and the vertex u j−1 in P1 . Let D1 be the isometric cycle on T1 that is incident with the edge u j u j−1 . Since D2 and D1 share at most an edge, there must be an edge u j x j ∈ V (D1 ) in relation Θ with edge u0 x 0 . The part of the traverse T1 from u j x j to u0 x 0 is a traverse from u j x j to u0 x 0 . We have one of the following situations (we change notation for the sake of simplicity): There is a traverse T10 from u00 x 00 to u0k x k0 (in the above notation from u j x j to u0 x 0 ) with P10 = u00 . . . u0j being the u00 , u0m -side of it, and we have an edge u00 v00 (above u j u j+1 ), such that one of the following situations holds (check Figure 8 for examples): (a) There exists a traverse T20 from u00 v00 to some edge u0l vl0 , for some 2 ≤ l ≤ k, of length at least 2, such that T10 and T20 have at most 4-cycles in between. 11

(b) There exists a traverse T20 from u00 v00 to some edge w l0 vl0 , with w l0 adjacent to u0l−1 and u0l+1 for some 2 ≤ l < k, of length at least 2, such that T10 and T20 have at most 4-cycles in between.

v00 v10 v20 v30 v40

x 00

v00

x 10

v10

x 20

v20

u03

x 30

v30

u04

x 40

u05

x 50

u00 u01 w20

u02

u00 u01 u02 w30

u03 u04

u05

(a) An example of Situation (a) with l = 4, k = 5.

x 00 x 10 x 20 x 30 x 40 x 50

(b) An example of Situation (b) with l = 3, k = 5.

Figure 8 At the beginning of the proof we have assumed the traverse T2 is attached to P1 as in Case (ii) or (iii). Now assume the traverse T2 is attached as in Case (i). If T1 is of length 1, then also traverse T2 is of length 1, thus we have two incident 4-cycles. By Lemma 3.1, G is isomorphic to a basic graph. Therefore, assume T2 is of length at least 2. Then we again have Situation (a) from the above, where T10 = T1 and T20 = T2 . Now, let D20 be the isometric cycle of T20 that includes v00 , u00 (the first isometric cycle of T20 ), and D10 the isometric cycle of T1 that includes x 00 , u00 (the first isometric cycle of the traverse T10 ). Since G is cubic, they both include vertex u01 . Firstly, assume that we have an isometric 4-cycle E1 = (u01 u02 u03 w20 ) in between T10 and T20 , for w20 on a side of T2 . Since this isometric cycle does not intersect in more than one edge with D10 or D20 , we must have edges w20 v20 ∈ Fu00 v00 and u02 x 20 ∈ Fu00 x 00 . Vertex v10 is incident with one 4-cycle and two 6-cycles. Thus G is isomorphic to a permutahedron, by Lemma 3.5. Therefore, we can assume there is no such 4-cycle. Since D10 and D20 do not intersect in more than an edge, there must exist an edge u01 x 10 ∈ Fu00 x 00 or an edge u01 v10 ∈ Fu00 v00 . Assume there is an edge u01 v10 ∈ Fu00 v00 . Since the length of T2 is at least 2, there must exist another isometric cycle D21 on v10 u01 u02 , that is a part of T2 . Moreover, there cannot exist u01 x 10 ∈ Fu00 x 00 since G is cubic. Again, we have multiple options. Now assume there is an isometric 4-cycle E2 = (u02 u03 u04 w30 ) in between T10 and T20 , for w30 on a side of T2 . Since this isometric cycle does not intersect in more than one edge with D21 or D10 , we must have edges u03 x 30 ∈ Fu00 x 00 and z30 v30 ∈ Fu00 v00 . Thus we have the subgraph H1 from Figure 7(a). It is isometric, since all the shortest paths in H1 have all its edges in different Θ classes in G (as it is easily checked). Thus assume there is no 4-cycle E2 . Since D10 and D21 do not intersect in more than an edge, there must exist an edge u02 x 20 ∈ Fu00 x 00 or an edge u02 v20 ∈ Fu00 v00 . If there exists the edge u02 v20 , 12

then vertex u01 is incident with two 4-cycles, namely D20 and D21 , which implies G is isomorphic to one of the known graphs. Thus, assume there exists the edge u02 x 20 ∈ Fu00 x 00 . Moreover, by vertex-transitivity, vertex u02 must be incident with a 4-cycle. The only possibility, for which we do not get a vertex incident with two 4-cycles, is that u03 , u02 , x 20 lie in a 4-cycle D12 , i.e. there exists an edge u03 x 30 ∈ Fu00 x 00 . We will continue with this case. One possibility is that u03 x 30 = u0k x k0 , i.e. the end of the traverse T10 . Then there exists the edge u03 v30 ∈ Fu00 v00 . This implies that u01 lies in one 4-cycle and two 6-cycle. Then G is isomorphic to the permutahedron. On the other hand, let D13 be the isometric cycle on u04 , u03 , x 30 . We continue similarly as before. If there is an isometric 4-cycle E3 = (u04 , u05 , u06 , w50 ) in between T10 and T20 , for w50 on the side of T2 , then we must have edges u05 x 50 ∈ Fu00 x 00 and w50 v50 ∈ Fu00 v00 . Thus we have the subgraph H3 . It is isometric since we can check that all the shortest paths in H3 have all its edges in different Θ classes in G. If there is no 4-cycle E3 , there must exist an edge u04 x 40 ∈ Fu00 x 00 or an edge u04 v40 ∈ Fu00 v00 . Again, the existence of the edge u04 x 40 would imply that vertex u03 was incident with two 4cycles, thus G was isomorphic to one of the basic graphs. Therefore, we can assume there is an edge u04 v40 ∈ Fu00 v00 . We have found a subgraph isomorphic to H2 . It is isometric since all the shortest paths in H2 have all its edges in different Θ classes in G. At the beginning of the proof we assumed that there is an edge u01 v10 ∈ Fu00 v00 . If we assume, that there is no such edge, we have an edge u01 x 10 ∈ Fu00 x 00 , and the situation is symmetric with one minor difference. Assume we have (beside the edge u01 x 10 ) edges u02 v20 , u03 v30 ∈ Fu00 v00 and no edge u03 x 30 ∈ Fu00 x 00 (thus G is not isomorphic to a perutahedron), and no edge u04 x 40 ∈ Fu00 x 00 (thus we do not have a subgraph isomorphic to H2 ). Then the assumptions do not guarantee the third isometric cycle D23 of the traverse from u00 v00 to u0l vl0 , since l could be equal to 4. But in this case, vertex u00 lies in an isometric 6-cycle, thus also u03 must lie in an isometric 6-cycle D0 . Moreover, vertex u04 must lie in a 4-cycle (u04 u05 u06 w50 ), for some vertices zw0 , u06 . Since G is cubic and any isometric 6-cycle cannot share more than an edge with any 4-cycle (by Lemma 3.2) or intersect with another isometric 6-cycle, D0 can only be of a form (v30 u03 u04 u05 v50 v40 ) or (v30 u03 u04 z50 v50 v40 ) for some vertices v60 , v50 . In both cases, D0 can be seen as D23 , thus we do have a symmetric situation. Proposition 3.8. If G is a cubic vertex-transitive partial cube with two isometric cycles that share more than one edge, then G is isomorphic to G(10, 3) or contains an isometric subgraph isomorphic to H1 , H2 , or H3 . Proof. Assume G is not isomorphic to G(10, 3). It can easily be checked that none of the other basic cubic graphs contains two isometric cycles that share more than an edge. To prove the existence of H1 , H2 , or H3 , it is enough to prove that the situation from Lemma 3.7 occurs in G. We know that if two isometric cycles share more than an edge, there exist two isometric cycles that intertwine. Let (v0 v1 . . . vm vm+1 . . . v2m+2n1 −1 ) and (u0 u1 . . . um um+1 . . . u2m+2n2 −1 ) be isometric cycles that intertwine where u0 = v0 , . . . , um = vm . Assume m ≥ 2, and that the residue of intertwining n1 + n2 is minimal. As in the proof of Lemma 3.4, it holds v2m+n1 v2m+n1 −1 Θu2m+n2 u2m+n2 −1 and v2m+n1 −1 v2m+n1 −2 Θu2m+n2 −1 u2m+n2 −2 .

13

By Lemma 2.2, we have a traverse T1 from v2m+n1 −1 v2m+n1 to u2m+n2 −1 u2m+n2 . Let P1 be the v2m+n1 −1 , u2m+n2 −1 -side of it, and P2 the v2m+n1 , u2m+n2 -side of it. We denote with D1 , . . . , Di the isometric cycles on it, and let vertices in P1 be denoted by P = z0 z1 . . . zk , where v2m+n1 −1 = z0 and u2m+n2 −1 = zk . Notice that v2m+n1 v2m+n1 +1 . . . v2m+2n1 −1 v0 u2m+2n2 −1 . . . u2m+n2 is a v2m+n1 , u2m+n2 -path of length n1 + n2 , thus the length of the traverse T1 is at most n1 + n2 . Consider isometric path P1 and edges v2m+n1 −1 v2m+n1 −2 and u2m+n2 −1 u2m+n2 −2 that are in relation Θ. By Lemma 3.6, we have a traverse T2 attached to P1 as in one of the Cases (i)-(iii). Let w i1 , w i2 , . . . , w il , for some l ≥ 0 and 0 < i1 < i2 − 1, i2 < i3 − 1, . . . , il−1 < il − 1 < k − 1, be vertices on a side of T2 , that are not on a side T1 , and are a part of 4-cycles in between both traverses, as in Lemma 3.6. To prove that the situation from Lemma 3.7 occurs in G, we have to prove that none of the isometric cycles of T1 , isometric cycles of T2 , and isometric 4-cycles in between pairwise intersect in more than an edge. Firstly, we prove that none of the 4-cycles D1 = (zi1 −1 w i1 zi1 +1 zi1 ), . . . , D l = (zil −1 w l1 zil +1 zil ) 0 shares more than an edge with any of the cycles D1 , . . . , Di . Assume that D j = (zi j0 −1 zi j0 zi j0 +1 w i j0 ) shares two edges with isometric cycle D˜j . Since D˜j is an isometric cycle on the traverse from v2m+n1 −1 v2m+n1 to u2m+n2 −1 u2m+n2 of length at most n1 + n2 , the length of it is at most 2(n1 + n2 ) + 2. We calculate the residue of intertwining: 0

i(D˜j , D j ) ≤ (2(n1 + n2 ) + 2 + 4 − 8)/2 = n1 + n2 − 1 < n1 + n2 . By the minimality condition, this cannot be. Let E1 , . . . , E˜i be isometric cycles that form the traverse T2 from v2m+n1 v2m+n1 −1 to u2m+n2 u2m+n2 −1 in Case (i), from zi1 yi1 to zi2 zi2 +1 in Case (ii), or from w i1 yi1 to zi2 zi2 +1 in Case (iii), for some i1 , i2 , w i1 , yi1 . We want to prove that no two cycles from D1 , . . . , Di , D1 , . . . , D l , E1 , . . . , E˜i intersect in more than an edge. We have seen that D1 , . . . , Di , D1 , . . . , D l do not pairwise intersects in more than an edge. By the same arguments, also E1 , . . . , E˜i , D1 , . . . , D l do not. For the sake of contradiction, assume E˜l and D˜k intersect in more than an edge. We have multiple options. Firstly, assume that E˜l intersects with exactly one of the 4-cycles 1 D , . . . , D l . Without loss of generality we can assume a side of El 0 is of a form w l 0 , zl 0 +1 , . . . , zl 00 , where w l 0 zl 0 +1 is an edge of the 4-cycle in between. Assume a side of Dk0 is of a form zk0 , . . . , zk00 . Since D˜k does not intersect in more than an edge with cycles D1 , . . . , D l , we have 1 ≤ l 0 ≤ k0 . Moreover, because El 0 and Dk0 intersect in at least two edges, it holds k00 − k0 ≥ 3, and thus k00 ≥ 3. Assume l 0 = k0 and k00 ≤ l 00 . We have i(E˜l , D˜k ) = (2(l 00 − l 0 ) + 2 + 2(k00 − k0 ) + 2 − 4(k00 − (k0 + 1)))/2 = l 00 + 4 − 2k00 ≤ l 00 − 2 < n1 + n2 . A contradiction. We get similar results in all other cases, that is, if l 0 = k0 , but k00 ≤ l 00 , and if l 0 < k0 and k00 ≤ l 00 or k00 ≥ l 00 . Now assume E˜l intersects with exactly two of the D1 , . . . , D l , i.e. a part of E˜l is of the form w l 0 , zl 0 +1 , . . . , zl 00 −1 , w l 00 , where w l 0 zl 0 +1 and zl 00 −1 w l 00 are edges of the 4-cycles in between. Then, since a side of D˜k is of the form zk0 , . . . , zk00 , we can, similarly as above, show that D˜k and E˜l intertwine with the residue of intertwining smaller than n1 + n2 . Since E˜l does not intersect in pore than an edge with cycles D1 , . . . , D j , it cannot be incident with more than two of them. Therefore, we can assume it does not intersects with any. In this 14

case a side of E˜l is of the form zl 0 , zl 0 +1 , . . . , zl 00 , while a side of D˜k is of the form zk0 , . . . , zk00 . Assume l 0 ≤ k0 < l 00 ≤ k00 . We have i(E˜l , D˜k ) = (2(l 00 −l 0 )+2+2(k00 −k0 )+2−4(l 00 −k0 ))/2 = k00 −l 0 −2(l 00 −k0 −2) ≤ k00 −l 0 ≤ n1 +n2 . Since the equality must hold, we have l 00 − k0 = 2. We know that G must have girth 4. Thus the vertex zk0 +1 must be incident with a 4-cycle D4 . Let zk0 x k0 be an edge in D˜k that is in relation Θ with v2m+n1 −1 v2m+n1 . If D4 lies on zk0 +1 , zk0 , x k0 or zk0 +1 , zk0 +2 , zk0 +3 , then cycles Dk0 and D4 intertwine with i(Dk0 , D4 ) < n1 + n2 . On the other hand, let zk0 +2 yk0 +2 be an edge in E˜l that is in relation Θ with v2m+n1 −1 v2m+n1 −2 . If D4 lies on zk0 +1 , zk0 +2 , yk0 +2 or zk0 +1 , zk0 , zk0 −1 , then cycles El 0 and D4 intertwine with i(dl 0 , D4 ) < n1 + n2 . Since G is cubic, these are the only options. A contradiction. We get similar outlines in all other combinations of l 0 , k0 , l 00 , k00 . In the last proposition we analyzed graphs in which at least two isometric cycles share more than an edge. We cover all the other graphs in the next result. Proposition 3.9. Let G be a finite, cubic vertex-transitive partial cube. If no two isometric cycles share more than an edge, and G is not isomorphic to one of the basic cubic graphs, then G contains an isometric subgraph isomorphic to H1 , H2 , or H3 . Proof. As before we want to prove that the situation from Lemma 3.7 occurs in G. Take two edges z0 y0 and zm ym that are in relation Θ, and assume the distance between them is maximal among such. Let P = z0 z1 . . . zm be the z0 , zm -side of some traverse T1 from z0 y0 to zm ym . Let x 1 be the neighbor of z0 , different from z1 and y0 . By Lemma 3.4, vertices y0 , z0 , x 1 lie in some isometric cycle D. Denote the vertices of D by D = (x 0 x 1 x 2 . . . , x 2k−1 ), where x 0 = z0 and x 2k−1 = y0 . We have z0 y0 Θx k−1 x k . By the maximality assumption, the path x k−1 x k−2 . . . x 1 z0 . . . zm is not isometric. Let x i x i+1 be such that x i x i+1 Θz j z j+1 for some j ∈ {0, . . . , m − 1}, i ∈ {0, . . . , k − 2}. Without loss of generality, assume i is minimal. Then the path P = z j z j−1 . . . x i is isometric. Consider P and edges z j z j+1 , x i x i+1 . By Lemma 3.6, we have a traverse T2 attached to P as in one of Cases (i)-(iii). Firstly, consider Case (ii) or Case (iii). If traverse T2 is attached to a subpath of P on z0 z1 . . . z j , then T2 is attached to the z0 , zm -side of T1 as in the assumptions of Lemma 3.7. Then H1 , H2 or H3 is a subgraph of G. Now, consider that T2 is not attached just to a subpath on z0 z1 . . . z j (this also covers Case (i)). The last edge on T2 (denoted with u j u j+1 in Cases (ii)-(iii)) is of the form x n x n+1 , for some 0 ≤ n ≤ i (in Case (i) this is just x i x i+1 ). If n = 0, then we directly have Situation (a) or Situation (b) from the proof of Lemma 3.7, thus the proposition holds. Assume n > 0, and let D0 be the last isometric cycle on T2 , incident with x n+1 x n . Since D0 and D do not share more than an edge, there must be a 4-cycle (x n x n−1 uw n−1 ) in between T2 and P, such that x n x n−1 u lies in P and x n w n−1 u lies on a side of T2 . Since this 4-cycle and D0 do not share more than an edge, there must be an edge w n−1 yn−1 in relation Θ with x n x n+1 , i.e. D0 is a 4-cycle. Then x n is incident with two 4-cycles, and, by Lemma 3.1, G is isomorphic to one of the basic graphs. The following lemma, combined with Propositions 3.8 and 3.9 proves Theorem 1.1.

15

Lemma 3.10. If H1 or H2 is an isometric subgraph of a cubic vertex-transitive partial cube G, then G is isomorphic to the truncated cuboctahedron. If H3 is an isometric subgraph, then G is isomorphic to the truncated icosidodecahedron. Proof. If H1 , H2 , or H3 is an isometric subgraph of G, then g3 (G) ≤ 10. Since no basic graph has such an isometric subgraph, the only options are g3 = 8 or g3 = 10, while g1 (G) = 4 and g2 (G) = 6. Moreover, every pair of two isometric 6-cycles or two 4-cycles must be disjoint, while every isometric 6-cycle shares at most an edge with any 4-cycle. Assume that an isometric 8- or 10-cycle D1 intertwines with an isometric 4- or 6-cycle D2 . Then the residue of intertwining r(D1 , D2 ) ≤ (10 + 6 − 4 ∗ 2)/2 = 4. The latter implies that the minimal residue of intertwining is at most 4. Take two isometric cycles with minimal intertwining m ≤ 4. As in the proof of Proposition 3.8, we can find a traverse T1 of length at most m, and a traverse T2 attached to T1 as in one of the Cases (i)-(iii) from Lemma 6. Moreover, no pair of the isometric cycles of T1 , T2 , and the 4-cycles in between intersects in more than an edge. Let us analyze the latter case. If there exists a 4-cycle (ui−1 ui ui+1 zi ) in between T1 and T2 , for some ui−1 , ui , ui+1 on a side of T1 , then ui must lie in two isometric cycles of T1 . Since T1 has length at most 4, ui must lie in two 4-cycles, or in one 4-cycle and two 6-cycles. In both cases G is isomorphic to a basic graph. Thus assume that there is no 4-cycle in between T1 and T2 . This immediately implies that T2 must be attached as in Case (i) or (ii). Assume that it is attached as in Case (i). Since the length of T1 is at most 4, and no pair of the isometric cycles of T1 and T2 intersects in more than an edge, a simple case analysis shows that there exists a vertex incident with two 4-cycles, or with one 4-cycle and two isometric 6-cycles. Now assume that we have Case (ii). Thus T2 is a traverse from edge ui vi+1 to u j u j+1 , for some vertex vi , and vertices ui , u j , u j+1 on a side of T1 . Let D be the isometric cycle on T2 , incident with u j u j+1 . Since there is no 4-cycle in between T1 and T2 , D is of the form (ui ui+1 . . . u j+1 v j v j−1 . . . vi+1 ), for some vertices ui , . . . , u j+1 on a side of T1 and some vertices v j v j−1 . . . vi+1 . Again, since the length of T1 is at most 4, and D does not intersect with any isometric cycle of T1 in more than an edge, a simple case analysis shows that there exists a vertex incident with two 4-cycles, or with one 4-cycle and two isometric 6-cycles. We have proved that no isometric 8- or 10-cycle intertwines with an isometric 4- or 6-cycle. Since G is cubic, the latter implies that no isometric 8- or 10-cycle intersect in more than an edge with an isometric 4- or 6-cycle. Let v be an arbitrary vertex in G, and D0 an isometric g3 (G)-cycle incident with it. Every vertex of D0 is incident with a unique isometric 6-cycle and a unique 4-cycle. Since G is cubic, and D0 does not share more than an edge with any isometric 4- or 6-cycle, D0 must be a unique isometric g3 (G)-cycle incident with v. Now we can use similar arguments as in Lemma 3.5. We paste on every isometric 4-, 6-, and g3 (G)-cycle a disc and obtain a closed surface. Denote with f6 the number of isometric 6-cycles, with f4 the number of 4-cycles, with f g3 (G) the number of isometric g3 (G) cycles, with f the number of faces of the embedding of G, with n, resp. e, the number of vertices, resp. edges, of G, and with χ the Euler characteristic of the surface. We have 3n = 2e,

f4 + f6 + f g3 (G) = f ,

4 f4 = 6 f6 = g3 (G) f g3 (G) = n, and n − e + f = 2 − χ.

16

As before we get: 2−χ = n



1 4

+

1 6

+

1 g3 (G)

−

1 2

 .

Since 8 ≤ g3 (G) ≤ 10, the right side of the equation is positive, thus it holds χ < 2. This implies G is planar or embedded into a projective plane, i.e. χ ∈ {0, 1}. In the case of χ = 0, we get n = 48 if g3 (G) = 8, and n = 120 if g3 (G) = 10. On the other hand, if χ = 1, then n = 24 or n = 60. Graphs up to 32 vertices are known. In the case n = 24, all vertex-transitive, cubic partial cubes are basic graphs. Every vertex-transitive graph embeddable into a projective plane has a planar, vertextransitive 2-cover, naturally obtained by the 2-cover of the projective plane. Thus G is a planar vertex-transitive graph, or a 2-quotient of it. By the theorem of Mani [14, Theorem 2.8.12] (G is 3-connected, since it is cubic, vertex-transitive), every cubic, vertex-transitive, planar graph is isomorphic to a graph of so called Archimedean solid. The only Archimedean solid on 48vertices is the truncated cuboctahedron. On the other hand, the truncated icosidodecahedron is the only one on 120 vertices. They are both partial cubes [2]. We have to consider if G is a 2-quotient of the truncated icosidodecahedron. In this case, the automorphism of the truncated icosidodecahedron obtained by interchanging fibers can only be an antipodal symmetry of the truncated icosidodecahedron (this fact is easily checkable on Figure 2b, taking in the account a well know fact that for every pair of vertices of the truncated icosidodecahedron there exists only one automorphism that maps one into the other). For every antipodal symmetry, there exists a 4-cycle (v0 v1 v2 v3 ), such that v0 , v1 are in the first fiber and v2 , v3 are in the second. Let ui be the antipodal vertex of vi and w i its image in G, for i ∈ {0, 1, 2, 3}. There is a traverse from v0 v1 to u2 u3 , that is completely in one fiber. If G is a partial cube, the latter implies that w0 w1 Θw2 w3 in G with w2 ∈ Uw0 w1 . This is impossible since (w0 w1 w2 w4 ) is a 4-cycle, and thus w2 ∈ Uw1 w0 should hold.

4

Concluding remarks

We have also tested Theorem 1.1 by a computer search for partial cubes on a census of cubic vertex-transitive graphs up to 1280 vertices [16]. For the search we transferred the basis in Sage environment [17] and used Eppsteins algorithm [6] for a recognition of partial cubes. With this article we have provided a classification of cubic vertex-transitive partial cubes. Since the variety of such graphs is rather small, it suggests that a similar classification can be done in graphs with higher valencies. The latter problem is widely open. The only known examples of vertex-transitive partial cubes are graphs provided in this paper, even cycles, middle level graphs, and the Cartesian products of the latter. We do not assume that this are the only one, but we would like to propose the following conjecture: Conjecture 1. The middle level graphs are the only vertex-transitive partial cubes with girth six.

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