Classification of vertex-transitive cubic partial cubes
CClassification of vertex-transitive cubic partial cubes
Tilen Marc ∗ Institute of Mathematics, Physics, and Mechanics, Jadranska 19, 1000 Ljubljana,Slovenia
October 13, 2018
Abstract
Partial cubes are graphs isometrically embeddable into hypercubes. In this paper itis proved that every cubic, vertex-transitive partial cube is isomorphic to one of the fol-lowing graphs: K (cid:131) C n , for some n ≥
2, the generalized Petersen graph G (
10, 3 ) , thecubic permutahedron, the truncated cuboctahedron, or the truncated icosidodecahedron.This classification is a generalization of results of Brešar et al. from 2004 on cubic mirrorgraphs, it includes all cubic, distance-regular partial cubes (Weichsel, 1992), and presentsa contribution to the classification of all cubic partial cubes. Keywords: partial cubes; vertex-transitive graphs; cubic graphs; convex cycles
Hypercubes are considered to be one of the classic examples of graphs that posses many sym-metries. It is a fundamental question to ask how those symmetries are preserved on theirsubgraphs. To our knowledge the first ones who addressed this question were Brouwer, Dejterand Thomassen in 1992 in [ ] . They provided many surprising and diverse examples of vertex-transitive subgraphs of hypercubes, but did not make a classification. Based on their results,examples are very diverse hence a classification seems too ambitious. They suggested that oneof the reasons for the latter is that the group of symmetries of a subgraph of a hypercube neednot be induced by the group of symmetries of the hypercube.On the other hand, Weichsel in 1992 [ ] considered distance-regular subgraphs of hyper-cubes. He derived certain properties of them, and noticed that all his examples are not justsubgraphs, but isometric subgraphs of hypercubes. It was thus a natural decision to focus onthe symmetries of partial cubes. He classified all distance-regular partial cubes based on theirgirth: hypercubes are the only ones with girth four, the six cycle and the middle level graphsare the only ones with girth six, and even cycles of length at least eight are the only ones with ∗ Electronic address: [email protected] a r X i v : . [ c s . D M ] J u l igher girths. Notice that all these graphs are vertex-transitive, therefore they are a subfamilyof 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 [ ] that hypercubes are the only regular – and thus the onlyvertex-transitive – median graphs (for infinite vertex-transitive median graphs check [ ] ).Due to this result, an extensive study of regular partial cubes has been performed [
2, 3, 5, 6, 11,12 ] . It especially focuses on the cubic case, since the variety of these graphs is far richer thanin the case of median graphs. Connections with other geometric structures are established, forexample with platonic surfaces [ ] and simplicial arrangements [ ] .From the point of view of vertex-transitive partial cubes, the most interesting one is thestudy [ ] , where a new family of graphs called mirror graphs was introduced. Moreover, itwas proved that mirror graphs are a subfamily of vertex-transitive partial cubes, and all mirrorgraphs that are obtained by cubic inflation (thus are cubic graphs) were classified. In [ ] ,we made an analysis of isometric cycles in a partial cube. In particular, the results imply thatthere are no cubic partial cubes with girth more than six. This suggests that, as in the case ofWeichsel’s distance-regular partial cubes, also cubic, vertex-transitive partial cubes should beapproached 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 symmetryof a hypercube.In the present paper we form a natural connection between the study of vertex-transitivesubgraphs of hypercubes and the study of cubic partial cubes: we classify all cubic, vertex-transitive partial cubes. The results can be seen as a generalization of results from [ ] , sinceall mirror graphs are vertex-transitive partial cubes, and in the cubic case results from [ ] ,since every distance-regular partial cube is vertex-transitive.Let K denote the complete graph of order 2, C k the cycle of length k , and G ( n , k ) thegeneralized Petersen graph with parameters 3 ≤ n , 1 ≤ k < n /
2. The main result of this paperis the following:
Theorem 1.1.
If G is a finite, cubic, vertex-transitive partial cube, then G is isomorphic to oneof the following: K (cid:131) C n , for some n ≥ , G (
10, 3 ) , the cubic permutahedron, the truncatedcuboctahedron, or the truncated icosidodecahedron. To our surprise, the variety of the graphs from Theorem 1.1 (cf. Figure 1) is small, andall graphs are classical graphs that were studied from many (especially geometric) views. Wepoint out that the cubic permutahedron, the truncated cuboctahedron, and the truncated icosi-dodecahedron are cubic inflations of graphs of platonic surfaces [ ] , K (cid:131) C n are the onlycubic Cartesian products of (vertex-transitive) partial cubes (this includes also the hypercube Q ∼ = K (cid:131) C ), while G (
10, 3 ) is the only known non-planar cubic partial cube and is isomor-phic to the middle level graph of valence three [ ] . The paper is organized as follows. In this section we briefly present the definitions and resultsneeded to prove Theorem 1.1, while in the next section we give a proof of it.2 a) G (
10, 3 ) (b) Cubic permutahedron(c) Truncated cuboctahedron (d) Truncated icosidodecahedron Figure 1: The four sporadic examples of cubic, vertex-transitive partial cubesWe will consider only simple (finite) graphs in this paper. The
Cartesian product G (cid:131) H of graphs G and H is the graph with the vertex set V ( G ) × V ( H ) and the edge set consistsof all pairs { ( g , h ) , ( g , h ) } of vertices with { g , g } ∈ E ( G ) and h = h , or g = g and { h , h } ∈ E ( H ) . Hypercubes or n-cubes are the Cartesian products of n -copies of K . We saya 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 . It is convex if for every pair of vertices in H all shortest path in G connecting them are in H . A partial cube is a graph that is isomorphic to an isometric subgraphof some hypercube. The middle level graph of valency n ≥ n − n or n − K on vertices {
0, 1 } .For a graph G , we define the relation Θ on the edges of G as follows: a b Θ x y if d ( a , x ) + d ( b , y ) (cid:54) = d ( a , y ) + d ( b , x ) , where d is the shortest path distance function. In partial cubes Θ is an equivalence relation [ ] , and we write F uv for the set of all edges that are in relation Θ with uv . We define W uv as the subgraph induced by all vertices that are closer to vertex u than to v , that is W uv = 〈{ w : d ( u , w ) < d ( v , w ) }〉 . In any partial cube G , the sets V ( W uv ) and3 ( W vu ) partition V ( G ) , with F uv being the set of edges joining them. We define U uv to be thesubgraph induced by the set of vertices in W uv which have a neighbor in W vu . For details andfurther results, see [ ] .We shall need a few simple results about partial cubes. All partial cubes are bipartite, sincehypercubes are. If u v Θ u v with u ∈ U u v , then d ( u , u ) = d ( v , v ) . A path P of a partialcube is a shortest path or a geodesic if and only if all of its edges belong to pairwise different Θ -classes. If C is a closed walk passing edge uv , then C passes an edges in F uv at least two times.By so called Convexity lemma [ ] , convex subgraphs in partial cubes can be characterized asinduced, connected subgraphs such that no edge with exactly one end in the subgraph is inrelation Θ with any edge in the subgraph. For the details, we again refer to [ ] .An automorphism of a graph G is a permutation of V ( G ) that preserves the adjacency ofvertices. Graph G is vertex-transitive if for every pair u , v ∈ V ( G ) there exists an automorphismof G that maps u to v . A special subfamily of vertex-transitive graphs comes from groups: Fora group A with generator set S , such that S − = S and id / ∈ S , the Cayley graph
Cay ( A , S ) is agraph with vertex set A , elements α , α being adjacent if and only if α α − ∈ S . The stabilizerof a vertex v in G is the subgroup of all the automorphisms of G that map v to v . By [ ] , ifthe stabilizers of a vertex-transitive graph G are trivial, then G is a Cayley graph.A major part of this paper depends on results developed in [ ] . The following definitionwas introduced to study isometric cycles in partial cubes: Definition 2.1.
Let v u Θ v u in a partial cube G, with v ∈ U v u . Let D , . . . , D n be a sequence ofisometric cycles such that v u lies only on D , v u lies only on D n , and each pair D i and D i + , fori ∈ {
1, . . . , n − } , intersects in exactly one edge from F v u , all the other pairs do not intersect. Ifthe shortest path from v to v on the union of D , . . . , D n is isometric in G, then we call D , . . . , D n a traverse from v u to v u . If all the cycles D , . . . , D n are convex, we call it a convex traverse . If D , . . . , D n is a traverse from v u to v u , then also the shortest path from u to u on theunion of D , . . . , D n is isometric in G . We will call this u , u -shortest path the u , u -side of thetraverse and, similarly, the shortest v , v -path on the union of D , . . . , D n the v , v -side of thetraverse . The length of these two shortest paths is the length of the traverse . It is not difficult toprove the following useful result. Lemma 2.2 ( [ ] ) . Let v u Θ v u in a partial cube G. Then there exists a convex traverse fromv u to v u . We shall also need the following definition:
Definition 2.3.
Let D = ( v v . . . v m v m + . . . v m + n − ) and D = ( u u . . . u m u m + . . . u m + n − ) be isometric cycles with u = v , . . . , u m = v m for m ≥ , and all other vertices pairwise differ-ent. We say that D and D intertwine and define i ( D , D ) = n + n ≥ as the residue ofintertwining . We can calculate the residue of intertwining as i ( D , D ) = ( l + l − m ) /
2, where l is thelength of D , l the length of D , and m the number of edges in the intersection. To finish thissection we show two simple, but useful properties of convex cycles. Claim 2.4.
If two convex cycles share more than an edge or a vertex, then they intertwine. roof. Suppose that two distinct convex cycles D , D share two non-adjacent vertices. Let v , v ∈ V ( D ) ∩ V ( D ) with maximal distance between them. Since D , D are convex anddistinct, there exists at most one shortest v , v -path and it must be in V ( D ) ∩ V ( D ) . If anyother vertex u is in V ( D ) ∩ V ( D ) , then also shortest u , v -, u , v -paths must be in V ( D ) ∩ V ( D ) ,contradicting the choice of v , v . Thus D and D share exactly the shortest v , v path; by thedefinition they intertwine.In [ ] a similar statement was proved: if at least two isometric cycles in a partial cubeshare more than an edge or a vertex, then there must be at least two isometric cycles thatintertwine. Claim 2.5.
Every 4-cycle in a partial cube is convex and can share at most an edge or a vertexwith any other convex cycle.Proof.
Since hypercubes are bipartite and have no induced K (the complete bipartite graphwith the bipartition into two and three vertices), the same holds for partial cubes. Thereforeevery 4-cycle is convex. Moreover, if a convex cycle D shares more than a vertex or an edgewith a 4-cycle, then by Claim 2.4 it must share exactly two incident edges, say uv and uv .This implies that there are two shortest v , v -paths in D , thus D is a 4-cycle. A contradiction,since K is not an induced subgraph of a partial cube.Note that in a cubic graph two cycles cannot share exactly a vertex. This fact will be usedthroughout the paper, sometimes possibly not explicitly pointed out. We start the proof of Theorem 1.1 by analyzing a simple case that strongly determines thestructure of a cubic, vertex-transitive partial cube.
Lemma 3.1.
If asomevertex of a cubic, vertex-transitive partial cube G lies on two 4-cycles, thenG ∼ = K (cid:131) C n , for some n ≥ .Proof. Let v lie on two 4-cycles. By Claim 2.5, they share at most an edge. On the other hand,they must share one edge since the graph is cubic. Let v u be the edge they share, and let v − and v be the other two neighbors of v . Moreover, let u be the common neighbor of v and u , and similarly u − be the common neighbor of v − and u . If v − , v , v lie on a common4-cycle, then, by vertex-transitivity, every vertex lies in three 4-cycles, that pairwise intersect inan edge. It is not hard to see that then G ∼ = Q . Thus assume v − , v , v do not lie in a common4-cycle. By vertex-transitivity, also v lies in two 4-cycles that intersect in an edge. Since G iscubic, the only possibility is that v u is the shared edge and that there exist vertices v , u suchthat ( v v u u ) is a 4-cycle. If v = v − , then by the maximum degree limitation u = u − , andthus G ∼ = C (cid:131) K , which is not a partial cube. Also, if v = u − , then u = v − , and G is not apartial cube. Thus, v and u are new vertices. We can use the same argument for v , u , as wedid for v , u , and find vertices v , u in a 4-cycle ( v v u u ) . Again, we have multiple options: u = u − and v = v − , u = v − and v = u − , or vertices v , u are different from all before. Inthe first case G ∼ = K (cid:131) C , and in the second G is not a partial cube. If the third case occurs,5e continue inductively: at some point the induction stops, therefore G ∼ = K (cid:131) C n , for some n ∈ (cid:78) .Notice that, if we considered also infinite graphs, a slight modification of the proof wouldshow that the only cubic, vertex-transitive partial cubes with a vertex that lies in two 4-cyclesare K (cid:131) C n , for n ≥
2, and K (cid:131) P ∞ , where P ∞ is the two-way infinite path.In [ ] it was proved that there exists no regular partial cube with degree at least three andgirth more than six. For the analysis of partial cubes with girth six, a graph X was introducedas shown in Figure 2. It was proved, that every regular partial cube with the minimum degreeat least three and girth six must have an isometric subgraph isomorphic to X . We shall analyzethis case in the next two lemmas. Lemma 3.2.
Let G be a cubic partial cube. If a 4-cycle and an isometric 6-cycle in G share twoedges, there is a vertex that lies in three 4-cycles. If two isometric 6-cycles in G share more thanan edge, then they are a part of an isometric subgraph X or G is a hypercube of dimension 3.Proof.
Let ( v v . . . v ) be an isometric 6-cycle and assume vertices v and v have a commonneighbor u , different from v . Then v u Θ v v Θ v v . Since v and v are adjacent, also u and v are, by the definition of relation Θ . Thus vertex u lies in three 4-cycles.Let isometric 6-cycles D , D share more than an edge. They cannot share three consecutiveedges, by the direct consequence of the transitivity of relation Θ . If they share two oppositeedges, the transitivity of relation Θ implies that G is a hypercube. Since G is cubic, the onlyremaining option is that they share two consecutive edges. Let D = ( v v . . . v ) and D =( v v v u u u ) . It holds that u u Θ v v Θ v v . Since d ( u , v ) =
2, it holds d ( u , v ) =
2, bydefinition of Θ . Thus, there is a vertex x adjacent to u and v . We have found an isometricsubgraph X in G . v v v v c v v v v c Figure 2: Graph X Lemma 3.3.
If the graph X is an isometric subgraph of a cubic, vertex-transitive partial cube G,then G ∼ = G (
10, 3 ) .Proof. Assume X is an isometric subgraph of G , and denote its vertices as in Figure 2. Firstly,suppose that there is a 4-cycle in G . By vertex-transitivity, v must be incident with a 4-cycle.But then an isometric 6-cycle and a 4-cycle must share two edges. Lemmas 3.1 and 3.2 implythat G = C (cid:131) K , but then G does not have an isometric subgraph isomorphic to X . Thus thereare no 4-cycles in G . 6otice that v in X lies in three isometric 6-cycles. Since G is vertex-transitive, also c must lie in at least three isometric 6-cycles. Since X is an isometric subgraph in a cubic graphwithout 4-cycles, and no two 6-cycles intersect in three edges, there must be a path P (cid:48) oflength 4, connecting c with one of the vertices v , v , v , v . Moreover, P (cid:48) can intersect X only in its endpoints. Without loss of generality, assume P (cid:48) connects c and v and denotethe vertices of P (cid:48) by c , s , u , u , v , respecting the order in P (cid:48) . Consider the isometric cycle D = ( c s u u v v ) and the isometric cycle D = ( c v v v v v ) . Cycles D and D intertwinein two edges, thus, by Lemma 3.2, there is a vertex u adjacent to v and u . By the isometryof X , u is distinct from all vertices of X . Call X (cid:48) the graph induced by V ( X ) and s , u , u , u .Since there is no 4-cycle in G and X is an isometric subgraph, X (cid:48) is as represented in Figure 3a. v v v v c v v v v c s u u u (a) Graph X (cid:48) v v v v c v v v v c s u u u s u u (b) Graph X (cid:48)(cid:48) Figure 3: Induced subgraphsVertex c must lie in at least three isometric 6-cycles. As before, the fact that X is isometric,that the maximal degree of G is three, and the absence of 4-cycles imply that there exists apath P (cid:48)(cid:48) of length 4, connecting c and one of v , v , v , v . By symmetry we can limit ourselvesto two possibilities.If P (cid:48)(cid:48) connects c and v , then P (cid:48)(cid:48) must be on c , s , u , u , v , where s and u are some twovertices in G different from the vertices in V ( X (cid:48) ) (since X (cid:48) is an induced graph). Then the cycle D = ( c s u u v v ) and the cycle D = ( c v v v v v ) meet in edges v v and v c . By Lemma3.2, there must exist a vertex u connecting v and u . Since X (cid:48) is an induced subgraph, u isdistinct from all the vertices of X (cid:48) . Denote the graph induced on V ( X (cid:48) ) and s , u , u by X (cid:48)(cid:48) .Again, since G has no 4-cycles and X is an isometric subgraph, X (cid:48)(cid:48) is as in Figure 3b.On the other hand, P (cid:48)(cid:48) can connect c and v . For the same reasons as before, there mustexist vertices s , u , u , different from the vertices of X (cid:48) such that P (cid:48)(cid:48) lies on c , s , u , u , v .Then the cycle D from above and the cycle D = ( c s u u v v ) share edges c v and v v .Lemma 3.2 implies that there must be a common neighbor of v and u . This can only be u ,thus we again have X (cid:48)(cid:48) as an induced subgraph.We continue in the same fashion. Cycle D and the cycle ( v v v u u u ) share v v and v u , thus there must be a vertex u connecting s and u . Similarly, the cycle ( c s u u v v ) and the cycle ( v v v u u u ) share v v and v u thus there must be a vertex u connecting s and u . Let X (cid:48)(cid:48)(cid:48) be the subgraph induced on vertices V ( X (cid:48)(cid:48) ) and u , u . Since X is an isometricsubgraph and G is without 4-cycles, X (cid:48)(cid:48)(cid:48) is as in Figure 4a.7 v v v c v v v v c s u u u s u u u u (a) Graph X (cid:48)(cid:48)(cid:48) v v v v c v v v v c s u u u s u u u u u (b) Graph G (
10, 3 ) Figure 4: Induced subgraphsNotice that the only vertices in X (cid:48)(cid:48)(cid:48) that do not have degree 3 are v , u , u . Also, observethat v lies in six 6-cycles. The only option that v lies in six 6-cycles is that there exists u connected to v , u and u . It can be checked directly that the obtained graph (shown in Figure4b) is isomorphic to G (
10, 3 ) .It is a well-known fact, that the edge-connectivity of a vertex-transitive graph equals thedegree of its vertices [ ] . In a cubic, vertex-transitive partial cubes this implies that | F ab | ≥ a b ∈ E ( G ) . Lemma 3.4.
In a cubic, vertex-transitive partial cube G, every pair of incident edges lies in aconvex cycle.Proof.
If the girth of G is more than 4, then G has an isometric subgraph isomorphic to X and,by Lemma 3.3, G is isomorphic to G (
10, 3 ) . The assertion holds in this graph.Now assume that the girth of G is 4. Let D = ( v v v v ) be a 4-cycle in G . Let a b be anedge in F v v , distinct from v v and v v . By Lemma 2.2, there exists a convex traverse from v v to a b . Let D (cid:48) be the first convex cycle on this traverse. Without loss of generality assume D (cid:48) (cid:54) = D (if otherwise take the second cycle on the traverse and exchange edge v v with v v ).Similarly, without loss of generality v v lies in a convex cycle D (cid:48)(cid:48) , different from D . Sinceconvex cycles D (cid:48) and D (cid:48)(cid:48) each share at most an edge with D , by Claim 2.5, all the pairs of edgesincident with v lie in some convex 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 u , u , u be its neighbors. Let g ( G ) be the length of a shortest convex cycle on u , u , u , let g ( G ) bethe length of a shortest convex cycle on u , u , u , and let g ( G ) be the length of a shortestconvex cycle on u , u , u . Without loss of generality assume g ( G ) ≤ g ( G ) ≤ g ( G ) . Clearly,for a vertex-transitive partial cube functions g , g , g are independent of the choice of vertex u . Lemmas 3.1 and 3.3, together with the fact that a cubic partial cube with girth at least 6includes an isometric subgraph X [ ] , immediately give the following corollary. Corollary 3.5.
Let G be a cubic, vertex-transitive partial cube. Then g ( G ) ≤ . If g ( G ) = ,then G ∼ = G (
10, 3 ) , while if g ( G ) = g ( G ) = , then G ∼ = K (cid:131) C n , for some n ≥ .
8e cover another important case in the next lemma and obtain a well known cubic, vertex-transitive partial cube [ ] . Lemma 3.6.
Let G be a cubic, vertex-transitive partial cube with g ( G ) = g ( G ) = g ( G ) = .Then G is isomorphic to the cubic permutahedron.Proof. We paste a disc on every convex 6-cycle and every 4-cycle. By Claim 2.5, Lemma 3.2and the fact that G is not isomorphic to G (
10, 3 ) or K (cid:131) C n , two 4-cycles, two convex 6-cycles,or a 4-cycle and a 6-cycle share at most an edge. Thus we obtain a closed surface.Denote by f the number of convex 6-cycles, by f the number of 4-cycles, by f the numberof faces of the embedding of G , by n and e the number of vertices and edges of G , and by χ the Euler characteristic of the surface. We have3 n = e , f + f = f , 4 f = f / = n , and n − e + f = χ .From the second and third equation we get n ( + ) = f . If we use the latter combined withthe first equation in the Euler formula, we get: χ = n (cid:18) − + + (cid:19) = n
12 .Since the right hand side of the equation is positive, it holds that χ >
0, i.e., χ = χ =
2. Inboth cases n ≤
24. Cubic partial cubes up to 32 vertices are known, the only vertex-transitiveon 24 vertices are the cubic permutahedron and K (cid:131) C , while K (cid:131) C is the only one on 12vertices. Thus G must be isomorphic to the cubic permutahedron.For the sake of convenience, we shall call the graphs C n (cid:131) K (for n ≥ G (
10, 3 ) , and thecubic permutahedron the basic cubic graphs . In what follows, we will find all non-basic, cubic,vertex-transitive partial cubes. For this we shall need a simple but technical lemma. Lemma 3.7.
Let u v Θ u m v m with u m ∈ U u v in a partial cube G. If P = u u . . . u m is a geodesic,then at least one of the following holds (Cases (i)-(iii) are illustrated in Figure 5):(i) There exist vertices w i , w i , . . . , w i l / ∈ V ( P ) , for some l ≥ and < i < i − i < i −
1, . . . , i l − < i l − < m − , such that the path u u . . . u i − w i u i + . . . u i l − w i l u i l + . . . u m is the u , u m -side of some convex traverse T from v u to v m u m .(ii) There exist edges u i z i , u j u j + , and vertices w i , w i , . . . , w i l / ∈ V ( P ) , for some l ≥ , ≤ i < i < i − i < i −
1, . . . , i l − < i l − < j − ≤ m − , such that the pathu i u i + . . . u i − w i u i + . . . u i l − w i l u i l + . . . u j is the u i , u j -side of some convex traverse T fromz i u i to u j + u j of length at least two.(iii) There exist a vertex w i adjacent to u i − and u i + , edges w i z i and u j u j + , and verticesw i , . . . , w i l / ∈ V ( P ) , for some l ≥ and < i < i − i < i −
1, . . . , i l − < i l − < j − ≤ m − , such that the path w i u i + . . . u i l − w i l u i l + . . . u j is the w i , u j -side of someconvex traverse T from z i w i to u j + u j . u u u u u u u u w w v v (a) An example of Case (i).The thick edges are the edgesof P , w i = w , w i = w , l = u i + u i + u i + u i + u i + u i + u i w i + u i + z i (b) An example of Case (ii).The thick edges are edges in P , u j = u i + , w i = w i + , and l = u i + u i + u i + u i + u i + u i + w i w i + u i + z i (c) An example of Case (iii).The thick edges are edges in P , u j = u i + , w i = w i + , and l = Figure 5
Proof.
Assume the lemma does not hold and let v u , v m u m be counterexample edges withgeodesic P = u u . . . u m that has length as small as possible. By Lemma 2.2, there exists atraverse from v u to v m u m , and let P be the u , u m -side of it. If P = P , then Case (i) in thelemma holds, a contradiction. Thus there exists a cycle D (cid:48) = ( u k u k + . . . u k (cid:48) z k (cid:48) − . . . z k + ) forsome 0 ≤ k < k (cid:48) − ≤ m −
1, where the path u k z k + . . . z k (cid:48) + u k (cid:48) is a part of the u , u m -side ofa traverse from v u to v m u m . If this cycle is of length 4, take the next one on P of the sameform. If all of them are 4-cycles, we have Case (i), a contradiction.Therefore assume D (cid:48) is not a 4-cycle. First assume it is convex. Then edges u k z k + and u k (cid:48) u k (cid:48) − are in relation Θ , and D (cid:48) is a convex traverse from u k z k + to u k (cid:48) u k (cid:48) − of length at leasttwo. Thus we have Case (ii).Now assume D (cid:48) is not convex. Both u k u k + . . . u k (cid:48) and u k z k + . . . z k (cid:48) − u k (cid:48) are shortest u k u k (cid:48) -paths. Two such paths in a hypercube cross the same Θ -classes, thus the same musthold in a partial cube. Let u k u k + , for some k < k < k (cid:48) , be the edge in u k u k + . . . u k (cid:48) that isthe same Θ -class as the edge u k z k + . Then the path P (cid:48) = u k . . . u k is shorter than P , thus thelemma holds for it. If Case (ii), resp. Case (iii), holds for P (cid:48) , then Case (ii), resp. Case (iii),holds for P as well. If Case (i) holds for P (cid:48) with edges u k z k + , u k u k + , and the traverse from u k z k + to u k u k + is of length at least two, then Case (ii) holds for P . Thus assume the traversefrom u k z k + to u k u k + is of length one, i.e., there is a 4-cycle D (cid:48)(cid:48) = ( u k u k + u k + z k + ) .Since D (cid:48) is not a 4-cycle, it holds that D (cid:48)(cid:48) (cid:54) = D (cid:48) . Let z k + be the neighbor of z k + in D (cid:48) , different from u k . For the same reasons as above, there must exist an edge u k u k + , for k + ≤ k < k (cid:48) , in u k u k + . . . u k (cid:48) that is the same Θ -class as the edge z k + z k + . Again, theisometric path P (cid:48)(cid:48) = z k + u k + . . . u k is shorter than P , thus the lemma holds for P (cid:48)(cid:48) with edges10 k + z k + and u k u k + .If there exists a 4-cycle ( z k + w k + u k + u k + ) for some vertex w k + , then Case (iii) holdsfor edges z k + w k + and u k + u k + with the traverse being a 4-cycle. Thus assume there is nosuch 4-cycle.If Case (i) holds for P (cid:48)(cid:48) with z k + z k + and u k u k + , then Case (iii) holds for P . If Case (ii)holds for P (cid:48)(cid:48) with z k + z k + and u k u k + , then Case (ii) or Case (iii) holds for P . Finally, ifCase (iii) holds for P (cid:48)(cid:48) with z k + z k + and u k u k + , then Case (iii) holds for P since there is no4-cycle of the form ( z k + w k + u k + u k + ) .As we can see in Figure 5, Lemma 3.7 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 . We willuse this fact in the following way. First we will show that if we can find in a non-basic, cubic,vertex-transitive partial cube G a convex traverse attached to a side of another convex traversein a nice way, then the high density of convex cycles in this part of G will imply that somecycles must be small. In particular, we will show that g ( G ) =
6. Second, we will find thissituation in G by considering two complementary types of graph: ether we allow convex cyclesto intertwine or we do not. Finally, we will conclude the proof by classifying cubic, vertextransitive partial cubes with g ( G ) = g ( G ) = g ( G ) >
6, by using certain results fromgroup theory.
Lemma 3.8.
Let G be a non-basic, cubic, vertex-transitive partial cube. Let there be a traverse T from u x to u m x m with P = u . . . u m being the u , u m -side of it. If we have a traverse T attachedto P as in one of the Cases (i)-(iii) from Lemma 3.7, with additional assumption that the convexcycles on T and convex cycles on T pairwise share at most an edge, then g ( G ) = .Proof. Graph G is non-basic, therefore g ( G ) = g ( G ) ≥
6. Notice that to prove g ( G ) = g ( G ) = T , T be as in the assertion of the lemma; we adapt the notation from the Lemma3.7. If the length of T (which is the shortest of the two traverses) is 2, then there exist twoincident 4-cycles on T , which cannot be by Lemma 3.1, or it includes a convex 6-cycle and weare done. If the length is 1, we have Case (i) or (iii) - in both cases there exist two incident4-cycles, a contradiction. Thus assume the length of both traverses is at least 3. Also noticethat if a 4-cycle of the form ( u k − u k u k + w k ) is in between T and T , then by Claim 2.5 theremust be edges u k x k and w k v k such that two convex cycles of T share u k x k , and two convexcycles of T share w k v k .First assume that T is attached as in Case (i). The traverse T starts in the edge u x , T starts in the edge u v and they both share the edge u u since G is cubic. The first cycle on T and the first cycle on T cannot be both 4-cycles since such cycles are not incident, by Lemma3.1. If one of them starts with a 6-cycle, we are done. Without loss of generality assume that T starts with a ( l + ) -cycle for 2 l + ≥
8. By the last statement of the previous paragraph,there cannot be a 4-cycle in between T and T of the form ( u k (cid:48) − u k (cid:48) u k (cid:48) + w k (cid:48) ) for 0 ≤ k (cid:48) < l .Thus path u u . . . u l − must be the beginning of the sides of T and T . Since any two cycles of T and T share at most an edge and no two 4-cycles are incident, the only option is that l = T starts with a 4-cycle, and there is a 4-cycle ( u u u w ) in between T and T . Then the11econd cycle of T must end in some edge w v , thus it must be a convex 6-cycle (see Figure6a). v u x u u v w u x u (a) Subgraph in Case (i) u j + u j x j u j − u j − v j − w j − u j − x j − u j − (b) Subgraph in Cases (ii), (iii) Figure 6Second assume we have Case (ii) or (iii) and let u j u j + be the ending edge of T . Let D bethe last convex cycle of T , the one that includes u j u j + . Assume the neighbor of u j on a side of T is a vertex w j − that is not on P , or in other words, we have a 4-cycle ( u j u j − u j − w j − ) inbetween T and T . Since D and this 4-cycle share at most an edge, there must exist an edge w j − z j − in relation Θ with u j u j + , i.e., D is a 4-cycle. Then there exist two incident 4-cycleswhich cannot be.By the above, we can assume that T ends with a convex cycle D that includes u j − u j u j + in P . Let D be the convex cycle on T that is incident with the edge u j u j − . Since D and D share at most an edge, and u j − u j u j + is a part of a side of T , the cycle D must end in an edge u j x j ∈ V ( D ) in relation Θ with edge u x . The part of the traverse T from u x to u j x j is alsotraverse, say T (cid:48) . Now starting from the end of traverses T (cid:48) and T we have a similar situationas before: incident edges u j x j , u j u j + in which traverses end (before they started), and an edge u j u j − on the sides of T (cid:48) and T . Similar arguments as before lead us to a convex 6-cycle (seeFigure 6b).We will now consider two kinds of partial cubes: the ones that have intertwining convexcycles and the ones that do not. In both cases we will use Lemma 3.8 to show that g ( G ) = Proposition 3.9.
Let G be a non-basic, cubic, vertex-transitive partial cube. If no two convexcycles share more than an edge, then g ( G ) = .Proof. Let G be as in the assertion of the lemma. We want to prove that the situation fromLemma 3.8 occurs in G . Take two edges z y and z m y m that are in relation Θ , and assumethe distance between them is maximal among all such pairs. Let T be a convex traversefrom z y to z m y m , provided by Lemma 2.2. Let x be the neighbor of z m , different from itstwo neighbors on the traverse. By Lemma 3.4, vertices z m , y m , x lie in some convex cycle D .12enote the vertices of D by D = ( x x x . . . x k − ) , where x = z m and x k − = y m . We have z y Θ z m y m Θ x k − x k .By the maximality assumption, the sequence of cycles on T together with cycle D is nota traverse from z y to x k − x k . On the other hand, by Lemma 2.2, there is a convex traverse T from z y to x k − x k . Denote by P (cid:48) , P (cid:48)(cid:48) the z , z m -, y , y m -side of T and by P (cid:48) , P (cid:48)(cid:48) be the z , x k − -, y , x k -side of T , respectively. The edge z m x lies in D but does not lie in T . Assumeit lies in T , say on the convex cycle D (cid:48) of T . As noted in the preliminaries, none of theedges with exactly one end in D (cid:48) is in relation Θ with any edge of D (cid:48) since D (cid:48) is convex. Bydefinition, D (cid:48) has an edge in relation Θ with z m y m . Thus z m y m has both its ends in D (cid:48) , i.e., itlies in D (cid:48) . Since convex cycles share at most an edge, D (cid:48) = D and thus T includes D . Let D (cid:48)(cid:48) = ( z m z m − . . . z m (cid:48) y m (cid:48) y m (cid:48) + . . . y m ) be the last cycle on T . Since G is cubic, the traverse T must include edge z m z m − . Then also the cycle D (cid:48)(cid:48) must be in T since convex cycles share atmost an edge. Inductively we can show that T ⊂ T . A contradiction since D ∪ T is not atraverse, thus z m x do not lie on T .On the other hand, z m x lies on a closed walk starting in z passing a side of D , P (cid:48) and P (cid:48) .Since in partial cubes every such closed walk must pass the Θ -class of z m x at least twice, theremust be another edge on it in relation Θ with z m x . Since D is convex, it must be on P (cid:48) or P (cid:48) .If there is an edge in relation Θ with z m x on P (cid:48) , then Lemma 3.7, provides a convextraverse T attached to P (cid:48) as in Cases (i)-(iii). By Lemma 3.8 and the fact that convex cyclesin G share at most an edge, we have g ( G ) =
6. If there is an edge in relation Θ with z m x on P (cid:48) , then there also exists an edge in relation Θ with z m x on P (cid:48)(cid:48) . The latter holds since paths P (cid:48) and P (cid:48)(cid:48) cross the same Θ -classes. Considering this edge on P (cid:48)(cid:48) and the edge x k x k + , with x k x k + Θ z m x , Lemma 3.7 gives a convex traverse T attached to P (cid:48)(cid:48) as in Cases (i)-(iii). ByLemma 3.8, it holds that g ( G ) =
6. This concludes the proof.
Proposition 3.10.
If G is a non-basic, cubic, vertex-transitive partial cube with two convex cyclesthat share more than one edge, then g ( G ) = .Proof. Assume that two convex cycles in G share more than an edge. By Claim 2.4,the two convex cycles intertwine. Denote the vertices of the intertwining cycles by ( v v . . . v m v m + . . . v m + n − ) and ( u u . . . u m u m + . . . u m + n − ) where u = v , . . . , u m = v m .By definition m ≥
2; moreover assume that the residue of intertwining n + n is minimalamong all pairs of intertwining convex cycles.It holds that v m + n v m + n − Θ v m − v m Θ u m + n u m + n − and that v m + n − v m + n − Θ v m − v m − Θ u m + n − u m + n − . By Lemma 2.2, we have a traverse T from v m + n − v m + n to u m + n − u m + n . Let P be the v m + n − , u m + n − -side of it. We denote by D , . . . , D i the convex cycles on it, and let vertices in P be denoted by P = z z . . . z k , where v m + n − = z and u m + n − = z k . Notice that v m + n v m + n + . . . v m + n − v u m + n − . . . u m + n isa v m + n , u m + n -path of length n + n , thus the length of the traverse T is at most n + n .Consider the isometric path P and edges v m + n − v m + n − and u m + n − u m + n − that are inrelation Θ . By Lemma 3.7, we have a traverse T attached to P as in one of the Cases (i)-(iii).To prove that the situation from Lemma 3.8 occurs in G and thus that g ( G ) =
6, we have toprove that the convex cycles of T and the convex cycles of T pairwise intersect in at most anedge.Let E , . . . , E ˜ i be convex cycles that form the traverse T from v m + n − v m + n − to u m + n − u m + n − in Case (i), from z i y i to z i z i + in Case (ii), or from w i y i to z i z i + in13ase (iii), for some 0 ≤ i < i < k , w i , y i . We want to prove that no two cycles from D , . . . , D i , E , . . . , E ˜ i intersect in more than an edge. For the sake of contradiction, assume E ˜ l and D ˜ k share at least two edges. By Claim 2.4 they intertwine. We will get to a contradictionby showing that E ˜ l and D ˜ k have the residue of intertwining smaller than n + n .We have multiple options. Firstly, assume that E ˜ l intersects with exactly one of the 4-cyclesin between the traverses, say with ( z l (cid:48) − z l (cid:48) z l (cid:48) + w l (cid:48) ) for some 0 < l (cid:48) < k , w l (cid:48) . Since G is cubic z l (cid:48) − (cid:54) = z = v m + n − , thus l (cid:48) ≥
2. By Claim 2.5, E ˜ l shares exactly an edge with the 4-cycle.Without loss of generality we can assume a side of E ˜ l is of a form w l (cid:48) z l (cid:48) + . . . z l (cid:48)(cid:48) , for some2 ≤ l (cid:48) < l (cid:48)(cid:48) ≤ n + n , where w l (cid:48) z l (cid:48) + is an edge of the 4-cycle in between T and T . A side of D ˜ k is of a form z k (cid:48) z k (cid:48) + . . . z k (cid:48)(cid:48) , for some 0 ≤ k (cid:48) < k (cid:48)(cid:48) ≤ n + n . Since D ˜ k does not intersect inmore than an edge with the 4-cycle ( z l (cid:48) − z l (cid:48) z l (cid:48) + w l (cid:48) ) , we have 2 ≤ l (cid:48) ≤ k (cid:48) . Moreover, because E l (cid:48) and D k (cid:48) intersect in at least two edges, it holds that k (cid:48)(cid:48) − l (cid:48) ≥
3, and thus k (cid:48)(cid:48) ≥ l (cid:48) = k (cid:48) and k (cid:48)(cid:48) ≤ l (cid:48)(cid:48) . Recall that the residue of intertwining is calculated as i ( E ˜ l , D ˜ k ) = ( d + d − d ) /
2, where d is the length of E ˜ l , d the length of D ˜ k and d thenumber of edges they share. In particular, for E ˜ l and D ˜ k we get: i ( E ˜ l , D ˜ k ) = ( ( l (cid:48)(cid:48) − l (cid:48) ) + + ( k (cid:48)(cid:48) − k (cid:48) ) + − ( k (cid:48)(cid:48) − ( k (cid:48) + ))) / = l (cid:48)(cid:48) − k (cid:48)(cid:48) − l (cid:48) + k (cid:48) + ≤ ( n + n ) − + < n + n .A contradiction. We get similar results in all the other cases, that is, if l (cid:48) = k (cid:48) , but l (cid:48)(cid:48) ≤ k (cid:48)(cid:48) , andif l (cid:48) < k (cid:48) and k (cid:48)(cid:48) ≤ l (cid:48)(cid:48) or k (cid:48)(cid:48) ≥ l (cid:48)(cid:48) .Now assume E ˜ l intersects with exactly two of the 4-cycles in between the traverses, i.e., apart of E ˜ l is of the form w l (cid:48) z l (cid:48) + . . . z l (cid:48)(cid:48) − , w l (cid:48)(cid:48) , where w l (cid:48) z l (cid:48) + and z l (cid:48)(cid:48) − w l (cid:48)(cid:48) are edges of the 4-cycles in between the traverses. Then, since a side of D ˜ k is of the form z k (cid:48) z k (cid:48) + . . . z k (cid:48)(cid:48) , we can,similarly as above, show that D ˜ k and E ˜ l intertwine with the residue of intertwining smallerthan n + n .Since E ˜ l does not intersect in more than an edge with 4-cycles in between the traverses, itcannot be incident with more than two of them. Therefore, we can assume it does not intersectwith any of them. In this case a side of E ˜ l is of the form z l (cid:48) z l (cid:48) + . . . z l (cid:48)(cid:48) , while a side of D ˜ k is ofthe form z k (cid:48) z k (cid:48) + . . . z k (cid:48)(cid:48) . Assume l (cid:48) ≤ k (cid:48) < l (cid:48)(cid:48) ≤ k (cid:48)(cid:48) , i.e., they share the path between z k (cid:48) and z l (cid:48)(cid:48) (with l (cid:48)(cid:48) − k (cid:48) ≥ i ( E ˜ l , D ˜ k ) = ( ( l (cid:48)(cid:48) − l (cid:48) )+ + ( k (cid:48)(cid:48) − k (cid:48) )+ − ( l (cid:48)(cid:48) − k (cid:48) )) / = k (cid:48)(cid:48) − l (cid:48) − ( l (cid:48)(cid:48) − k (cid:48) − ) ≤ k (cid:48)(cid:48) − l (cid:48) ≤ n + n .Since the equality must hold, we have l (cid:48)(cid:48) − k (cid:48) =
2. We know that G must have girth 4. Thusthe vertex z k (cid:48) + must be incident with a 4-cycle D . Since G is cubic, this 4 cycle must sharetwo consecutive edges with D ˜ k or E ˜ l , a contradiction with Claim 2.5. We get similar outcomeswith the other positions of l (cid:48) , k (cid:48) , l (cid:48)(cid:48) , k (cid:48)(cid:48) . This proves that cycles of T and T pairwise share atmost an edge, concluding the proof.The following proposition, combined with Propositions 3.9, 3.10, Corollary 3.5, and Lemma3.6 proves Theorem 1.1. To prove it we shall need a result from group theory. A group of theform 〈 α , . . . , α n | ( α i α k ) k i j = 〉 , where k ii = k i j = k ji > Coxeter group . The finiteCoxeter groups are classified with their Cayley graphs being partial cubes due to their connec-tion with oriented matroids and reflection arrangements [
1, Chapter 2.3 ] . Limiting ourselvesto those Coxeter groups whose Cayley graphs are cubic, we are left with the following graphs:cubic permutahedron, the truncated cuboctahedron, and the truncated icosidodecahedron.14 roposition 3.11. If G is a non-basic, cubic, vertex-transitive partial cube with g ( G ) = , thenG is isomorphic to the truncated cuboctahedron or the truncated icosidodecahedron.Proof. Let G be a non-basic, cubic, vertex-transitive partial cube, with g ( G ) =
6. Then g ( G ) = g ( G ) = k >
6, by Corollary 3.5 and Lemma 3.6. It follows that the vertex-stabilizers ofthe automorphism group of G are trivial. Thus G is a Cayley graph. Notice that by Claim 2.5each 4-cycle shares at most one edge with any convex 6- and k -cycle. Moreover, if a convex6-cycle D shares more than an edge with a convex k -cycle D k , then they intertwine by Claim2.4. Since both cycles are convex, they can share at most two consecutive edges, say v v and v v . If this is the case, let D be the convex 4-cycle incident with v . Then D must share twoedges with D or D k since G is cubic. A contradiction, thus also convex 6-cycles and convex k -cycles share at most an edge.Color the edges that simultaneously lie in a 4-cycle and a convex 6-cycle green, theedges in a 4-cycle and a convex k -cycle red, and the edges in a convex 6-cycle and a con-vex k -cycle blue. By the above discussion the colors are well defined. We get the relations 〈 r , g , b | r , g , b , ( r g ) , ( g b ) , ( br ) k / 〉 . Then this is a Coxeter group. By the classification offinite Coxeter groups, and the fact that k >
6, the only possibility is that k equals 8 or 10.Moreover, in the first case G is isomorphic to the truncated cuboctahedron, while in the secondit is isomorphic to the truncated icosidodecahedron. 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 bedone for graphs with higher valencies. The latter problem is wide open. The regular graphs inthe subcubic cases can be seen as the beginnings of greater families of vertex-transitive partialcubes: G (
10, 3 ) as a middle level graph; the cubic permutahedron, the truncated cuboctahe-dron, the truncated icosidodecahedron, even cycles, and K as Cayley graphs of finite Coxetergroups; and even prisms as the Cartesian products of the latter graphs. To our knowledge thesefamilies are the only known examples of vertex-transitive partial cubes. We do not assume thatthese are the only ones, 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.
This paper was motivated by a computer search for partial cubes on a census of cubic,vertex-transitive graphs up to 1280 vertices [ ] . For the search we transferred the basis inSage environment [ ] and used Eppstein’s algorithm [ ] for the recognition of partial cubes.We are thankful to the authors of the census and the algorithms. We would also like to thankSandi Klavžar and the reviewers for useful comments on the text. References [ ] A. Björner, M. Las Vergnas, B. Sturmfels, N. White, and G. Ziegler,
Oriented matroids , 2nd ed., Cambridge:Cambridge University Press, 1999. ] C. P. Bonnington, S. Klavžar, and A. Lipovec,
On cubic and edge-critical isometric subgraphs of hypercubes ,Australas. J. Combin. (2003), 217–224. [ ] B. Brešar, S. Klavžar, A. Lipovec, and B. Mohar,
Cubic inflation, mirror graphs, regular maps, and partialcubes , European J. Combin (2004), 55–64. [ ] A. E. Brouwer, I. J. Dejter, and C. Thomassen,
Highly symmetric subgraphs of hypercubes , J. Algebraic Combin. (1993), 25–29. [ ] M. Deza, V. Grishukhin, and M. Shtogrin,
Scale-Isometric Polytopal Graphs in Hypercubes and Cubic Lattices:Polytopes in Hypercubes and z n , Imperial College Press, 2004. [ ] D. Eppstein,
Cubic partial cubes from simplicial arrangements , Electron. J. Combin (2006), 1–14. [ ] , Recognizing partial cubes in quadratic time. , J. Graph Algorithms Appl. (2011), 269–293. [ ] E. Gedeonová,
Constructions of S-lattices , Order (1990), 249–266. [ ] C. D. Godsil and G. Royle,
Algebraic Graph Theory , Vol. 207, Springer New York, 2001. [ ] R. Hammack, W. Imrich, and S. Klavžar,
Handbook of Product Graphs , 2nd ed., Discrete Math. Appl. (BocaRaton), CRC Press, 2011. [ ] S. Klavžar and A. Lipovec,
Partial cubes as subdivision graphs and as generalized Petersen graphs , DiscreteMath. (2003), 157–165. [ ] S. Klavžar and S. Shpectorov,
Tribes of cubic partial cubes , Discrete Math. Theor. Comput. Sci. (2007),273–291. [ ] T. Marc,
Vertex-transitive median graphs of non-exponential growth , Discrete Math. (2015), 191–198. [ ] , There are no finite partial cubes of girth more than 6 and minimum degree at least 3 , European J.Combin. (2016). [ ] H. M. Mulder, n-cubes and median graphs , J. Graph Theory (1980), 107–110. [ ] P. Potoˇcnik, P. Spiga, and G. Verret,
Cubic vertex-transitive graphs on up to 1280 vertices , J. Symbolic Comput. (2013), 465–477. [ ] G. Sabidussi,
On a class of fixed-point-free graphs , Proc. Amer. Math. Soc. (1958), 800–804. [ ] W. A. Stein et al.,
Sage Mathematics Software (Version 6.1.1) , The Sage Development Team, 2014. . [ ] P. M. Weichsel,
Distance regular subgraphs of a cube , Discrete Math. (1992), 297–306. [ ] P. M. Winkler,
Isometric embedding in products of complete graphs , Discrete Appl. Math. (1984), 221–225.(1984), 221–225.