Upper Embeddability of Graphs and Products of Transpositions Associated with Edges

Given a graph, we associate each edge with the transposition which exchanges the endvertices. Fixing a linear order on the edge set, we obtain a permutation of the vertices. D\'enes proved that the permutation is a full cyclic permutation for any linear order if and only if the graph is a tree. In this article, we characterize graphs having a linear order such that the associated permutation is a full cyclic permutation in terms of graph embeddings. Moreover, we give a counter example for Eden's question about an edge ordering whose associated permutation is the identity.


Introduction
In this article, a graph G stands for a connected multigraph G = (V G , E G , r G ), where • E G is a finite set of edges, • r G is a map from E G to V G 2 , the collection of subsets of V G consisting of 2 elements.Note that r G (e) represents the endvertices of the edge e.Also note that loops are not allowed.
Let n be a positive integer and suppose that V G = [n] := {1, 2, . . ., n}.When an edge e ∈ E G satisfies r G (e) = {u, v}, we associate the transposition τ e := (u v) ∈ S n with the edge e, where S n denotes the symmetric group of degree n.
An edge ordering of a graph G is a linear order ≤ ω on E G , denoted as a sequence ω = (e 1 , . . ., e m ) in which e i < ω e j if and only if i < j.Given an edge ordering ω = (e 1 , . . ., e m ), we associate the product π ω := τ em • • • τ e 1 ∈ S n .Definition 1.1.A permutation σ ∈ S n is called a full cyclic permutation if σ is a cyclic permutation of length n.An edge ordering ω = (e 1 , . . ., e m ) of a graph G is a full cyclic permutation ordering if the corresponding permutation π ω = τ em • • • τ e 1 is a full cyclic permutation.
Dénes proved the following theorem to state a connection between labeled trees and factorization of a full cyclic permutation into transpositions.Theorem 1.2 (Dénes [1].See also [7,Section 2] and [11,Lemma 2.1]).Given a graph G, the following are equivalent.
(i) Any edge ordering of G is a full cyclic permutation ordering.
(ii) G is a tree.
Note that Theorem 1.2 plays an important role in the studies of the chromatic symmetric functions and the chromatic operator for trees [3,11].Also, note that recently the second author [16] studied an analogue of Theorem 1.2 for signed graphs and the hyperoctahedral group.
It is a natural question to ask what graphs admit a full cyclic permutation ordering.For example, let G be the butterfly graph pictured in Figure 1  Therefore ω is a full cyclic permutation ordering of the butterfly graph G. Let denote the Betti number (also called the circuit rank) of the connected graph G.When a graph G has a full cyclic permutation ordering, considering the signature of a full cyclic permutation ordering, one can show that the Betti number β(G) is even.However, the converse is false.For example, the dumbbell graph (Figure 2) has no full cyclic permutation orderings although its Betti number is 2.
We regard a graph as a topological space by identifying each edge with the unit interval [0, 1] and gluing them at vertices.Then β(G) coincides with the Betti number of G as a topological space.In this article, we will show that having a full cyclic permutation ordering is a topological property as follows.
Let Σ be an orientable closed surface and ι : G → Σ an embedding.We call a connected component of the complement of the image of ι a face.An embedding ι is a 2-cell embedding if every face is homeomorphic to an open disk.For any 2-cell embedding ι : G → Σ, where f ι denotes the number of faces of the 2-cell embedding ι and g Σ the genus of Σ.Then Therefore maximizing the genus g Σ is equivalent to minimizing the number of faces f ι .Hence the maximum genus γ max (G), the maximum of genus g Σ such that there exists a 2-cell embedding G → Σ, satisfies where ⌊ ⌋ denotes the floor function.If the equality holds, then we say that G is upper embeddable.
Upper embeddable graphs are well-studied objects by many researchers [4, 5, 8-10, 12-15, 17, 18].Jungerman and Xuong gave a combinatorial characterization of upper embeddability independently.A connected graph G with even (odd) Betti number is upper embeddable if and only if there exists a spanning tree T of G such that all (all but one) connected components of G \ T consists of an even number of edges.Here is the main theorem of this article.
Theorem 1.4.Given a graph G, the following are equivalent.
(1) G has a full cyclic permutation ordering.
(3) The Betti number β(G) is even and (4) There exists a spanning tree T of G such that every connected components of G \ T consists of an even number of edges.
Note that the conditions ( 2), (3), and ( 4) are equivalent by the definition of upper embeddability and Theorem 1.3.Figure 3 shows how the butterfly graph can be embedded into a torus with exactly one face.
The organization of this article as follows.In Section 2, we will prove that (3) implies (1) and give an example of constructing a full cyclic permutation ordering.In Section 3, we will review the relation of 2-cell embeddings and rotation systems and we will prove that (1) implies (2).Combining the proofs in Section 2 and Section 3, we will complete the proof of Theorem 1.4.
In Section 4, we will study another extreme condition, that is, edge orderings ω such that π ω is the identity permutation, which Eden [2] studied.Eden gave necessary conditions for such orderings and asked whether the condition is also sufficient.We will give a counterexample for this question.
Lemma 2.1.Let π be a full cyclic permutation in S n .If the distinct numbers u, v, w appear in π in this cyclic order, then the product (u v)(v w)π is a full cyclic permutation.
Proof.By the assumption, we can write which is a full cyclic permutation.
We say that two edges e and e ′ are adjacent if r G (e) ∩ r G (e ′ ) = ∅, that is, they have a common endvertex.Note that the case r G (e) = r G (e ′ ) is allowed.The following lemma is required.Lemma 3]).Suppose that G is upper embeddable and β(G) is even.If G is not a tree, then there exist two adjacent edges e and e ′ such that G \ {e, e ′ } is upper embeddable.
Proof that ( 3) implies (1) in Theorem 1.4.We will show that G has a full cyclic permutation ordering by induction on the Betti number β(G).When β(G) = 0, G is a tree and has a full cyclic permutation ordering by Dénes' theorem (Theorem 1.2).
Assume β(G) > 0. By Lemma 2.3, there exist two adjacent edges e and e ′ such that G ′ := G \ {e, e ′ } is upper embeddable.By the induction hypothesis, G ′ has a full cyclic permutation ordering.By Lemma 2.2, G has a full cyclic permutation ordering.
Example 2.4.Consider the wheel graph W 5 pictured in Figure 4.The Betti number of W 5 is 4. Let T be the spanning tree of W 5 consisting of the edges 12, 23, 34, 45.Then W 5 \ T is connected and consisting of 4 edges.Therefore W 5 satisfies the condition (4) in Theorem 1.4 and hence has a full cyclic permutation ordering.We will construct a full cyclic permutation ordering following by the proof of Lemma 2.2.
Define D G by We call an element of D G a dart.When r G (e) = {u, v}, the dart (e, u) shows an orientation of the edge e from the source u to the target v.Define the involution α on D G by α(e, u) := (e, v).Given a rotation system ρ = (ρ v ) v∈V G , we will define bijections σ and φ from D G to itself.Suppose that ρ v = [e 1 , . . ., e s ].Define σ by σ(e i , v) := (e i+1 , v), where we consider e s+1 = e 1 .Let φ := σ • α.
For every dart d, the target of d coincides with the source of φ(d).Therefore each orbit in D G / φ determines a closed walk on G and we can make a polygon whose sides are formed by the darts in the orbit.Gluing the sides of the polygons obtained from the orbits in D G / φ by the involution α, we obtain an embedding of G on a closed surface.One can show that this surface is actually orientable (See [6, Subsection 3.2]) and hence this embedding is the desired 2-cell embedding.

2.4]). Given a graph G, there exists a one-to-one correspondence between rotation systems of G and 2-cell embeddings of G on oriented closed surfaces up to orientation-preserving homeomorphism.
Note that, from the construction, the number of the faces of the embedding corresponding to a rotation system is equal to the number of the orbits in D G / φ .
Let ω be an edge ordering of G.For each v ∈ V G , let ω v denote the linear order on I G (v) induced by ω.Moreover, let ρ ω,v be the cyclic order on I G (v) determined by ω v .Thus we obtain the rotation system ρ ω := (ρ ω,v ) v∈V G from an edge ordering ω and hence the corresponding bijections σ ω and φ ω = σ ω • α.Lemma 3.2.Let ω = (e 1 , . . ., e m ) be an edge ordering of a graph G.For each v ∈ V G , let f v denote the minimal edge in I G (v) with respect to ω. Define a map Ψ : where the brackets denote equivalence classes.Then Ψ is a bijection.
Proof.First, we will show that the map Ψ is well-defined.It is sufficient to show that The edge ordering ω = (e 1 , . . ., e m ) defines the set T v as follows.
where we agree with Then (e j 1 , . . ., e js ) is a trail from v to π ω (v).Let v 0 := v and for k ∈ {1, . . ., s} define v k recursively as the endvertex of e j k other than v k−1 .Note that From the definition of T v , for each k ∈ {1, . . ., s − 1}, e j k+1 covers e j k in I G (v k ) with respect to ω. Therefore φ ω (e j k , v k−1 ) = (e j k+1 , v k ).Since e js is the maximal element in I G (v s ) with respect to the order ω, we have φ ω (e js , v s−1 ) = (f vs , v s ) = (f πω(v) , π ω (v)).Thus Ψ is well-defined.
Next, to prove the surjectivity, take an orbit W ∈ D G / φ ω .Let f be the minimal element in Finally, we prove the injectivity.Let u, v ∈ V G and suppose that Ψ( for some s ∈ Z. Without loss of generality, we can assume that s > 0. Recall the edges of G are ordered by ω = (e 1 , . . ., e m ).We can write the edge f u as f u = e j 0 with some j 0 ∈ {1, . . ., m}.Moreover we can obtain the walk (e j 0 , e j 1 , . . ., e js ) by (e j k , v k ) := φ ω (e j k−1 , v k−1 ) for k ∈ {1, . . ., s}, where v 0 := u.Note that (e j 0 , v 0 ) = (f u , u) and (e js , v s ) = (f v , v).Suppose that with and hence f is injective.Now we are ready to prove that (1) implies (2).
Proof that ( 1) implies ( 2) in Theorem 1.4.Let ω be a full cyclic permutation ordering of G. Then the number of faces of the 2-cell embedding corresponding to the rotation system ρ ω is equal to

Identity permutation ordering
In this section, a graph is not necessarily connected.We say that an edge ordering ω of a graph G is an identity permutation ordering if π ω = ε, where ε denotes the identity permutation.Every edgeless graph vacuously has an identity permutation ordering.The minimal example of a non-trivial graph having an identity permutation ordering is the 2-cycle C 2 .Eden [2] studied simple graphs that have an identity permutation ordering and mentioned the complete graph K 4 is the minimal example.Figure 5 shows identity permutation orderings of C 2 and K 4 .
Eden gave necessary conditions (without proof) for simple connected graphs having an identity permutation ordering as follows.(1) m is even.
(2) There exist a set C consisting of closed trails and a map ψ : V G → C such that the following conditions hold.
(ii) Every v ∈ V G belongs to the closed trail ψ(v).
(iii) The sum of the number of edges of closed trails in C is 2m.
(iv) Each edge of G belongs to exactly two closed trails in C. Proof.Suppose that ω = (e 1 , . . ., e m ) be an identity permutation ordering.Since the identity permutation is an even permutation, we have m is even.Let Ψ : For each v ∈ V G , there exists no dart d ∈ D G such that both d and α(d) belong to Ψ(v) since π ω is the identity.Thus, forgetting the direction of each dart in Ψ(v), we obtain the closed trail ψ We will prove ψ is injective.Assume that there exist distinct vertices u, v such that ψ(u) = ψ(v).Let ω ′ = (f 1 , . . ., f r ) be the induced order of ω on ψ(u).Since π ω ′ (u) = u, the edges f 1 and f r are incident to u.Also, f 1 and f r are incident to v by the same reason.Hence f 1 and f r are parallel edges between u and v.This contradicts that G is simple.Therefore ψ is injective and hence bijective.
By the definition of maps Ψ and ψ, every v ∈ V G belongs to ψ(v).Moreover, For any e ∈ E G , the two darts on e belongs distinct orbits Ψ(u) and Ψ(v).Then e belongs to ψ(u) and ψ(v) and the other trails do not contain e.Thus the map ψ : V G → C has the desired properties.Eden asked whether the necessary conditions in Proposition 4.1 are also sufficient.We will give a counterexample for this question.Let G be the graph on 12 vertices with 20 edges pictured in Figure 6.Define a map ψ by

Figure 3 :
Figure 3: A 2-cell embedding of the butterfly graph into a torus with exactly one face.

Figure 4 : 3 =
Figure 4: Wheel graph W 5 and its spanning tree T

2 Figure 5 :
Figure 5: The 2-cycle C 2 and the complete graph K 4 have an identity permutation ordering.

Proposition 4 . 1 (
Eden [2, P. 130]).Let G be a simple connected graph on n vertices with m edges.If G has an identity permutation ordering, then the following conditions hold.

Problem 4 . 3 .
Characterize graphs that have an identity permutation ordering.