Signed Graphs and Signed Cycles of Hyperoctahedral Groups

For a graph with edge ordering, a linear order on the edge set, we obtain a permutation of vertices by considering the edges as transpositions of endvertices. It is known from D\'enes' results that the permutation of a tree is a full cyclic for any edge ordering. As a corollary, D\'enes counted up the number of representations of a full cyclic permutation by means of product of the minimal number of transpositions. Moreover, a graph with an edge ordering which the permutation is a full cyclic is characterized by graph embedding. In this article, we consider an analogy of these results for signed graphs and hyperoctahedral groups. We give a necessary and sufficient condition for a signed graph to have an edge ordering such that the permutation is an even (or odd) full cyclic. We show that the edge ordering of the signed tree with some loops always gives an even (or odd) full cyclic permutation and count up the number of representations of an odd full cyclic permutation by means of product of the minimal number of transpositions.


Background
Let n be a positive integer.Suppose that G = (V G , E G ) is a graph and V G = [n] := {1, . . ., n}.We associate an edge e = {i, j} ∈ E G with the transposition τ e := (i j) ∈ S n , where S n denotes the symmetric group of degree n.
An edge ordering of a graph G ≤ ω is a linear order on E G , denoted as a sequence ω = (e 1 , . . ., e m ) in which e i ≤ ω e j if i ≤ j.For an edge ordering ω = (e 1 , . . ., e m ), we give the product π ω := τ em • • • τ e 1 ∈ S n .Definition 1.1.An edge ordering ω is a full cyclic permutation ordering if the product π ω ∈ S n is a cyclic permutation of length n.
We can characterize a graph with a full cyclic permutation ordering by studying the orbit and the sign of π ω .For graphs with full cyclic permutation ordering, the following theorem by Dénes is known.
(i) Any edge ordering of G is a full cyclic permutation ordering.
(ii) G is a tree.
Dénes gives this theorem to count up the number of representations of a cyclic permutation of length n by means of a product of a minimal number of transpositions, and obtains the following corollary.
Corollary 1.4 (Dénes [4,§2,Corollary]).The number of representations of a cyclic permutation of length n by means of a product of n − 1 transpositions is n n−2 .
The author and Tsujie [10] also characterize graphs having a full cyclic permutation ordering in terms of graph embedding.

The hyperoctahedral group
In this article, we discuss these arguments in the case of the hyperoctahedral group H n , the Weyl group of type B n .It is a natural idea since S n is known as the Weyl group of type A n−1 .
Let I n := {−n, . . ., −1, 1, . . ., n}.The hyperoctahedral group H n is a subgroup of the symmetric group S In of I n , defined by An element η ∈ H n , called a signed permutation, is expressed as since η is determined by values at integers in [n].In particular, given a permutation in S n , the signed permutation in H n is naturally determined.Thus we can consider S n to be a subgroup of H n .
There are three types of signed transpositions in H n .For i, j be distinct numbers in [n] and ϵ ∈ {+, −}, which we consider as a multiplicative group of order two in the natural way, we say that is a positive transposition if ϵ = + and a negative transposition if ϵ = −.Note that we can consider a positive transposition (i +j) as a transposition in S n and abbreviate it as (i j).For i ∈ [n], we say that The finite group H n is generalized by these signed transpositions, but only some positive transpositions and one inversion transposition are sufficient for the generator.Also, The group H n is isomorphic to the semidirect product S n 2 ⋊ S n (the wreath product S 2 ≀ S n ).The group S n 2 is regarded as the group generalized by inversion transpositions.Let η ∈ H n , using the relation for signed transpositions given below, we can find σ 1 ∈ S n 2 and σ 2 ∈ S n such that η = σ 1 σ 2 (more on this later Example 2.3).Lemma 1.5.Let i, j, k ∈ [n] be distinct integers and ϵ ∈ {+, −}.Then the following three claims hold: A signed permutation is decomposed uniquely into a product of commuting signed cycles.Let i 1 , . . ., i l be distinct numbers in [n] and ϵ 1 , . . ., ϵ l ∈ {+, −}, given such a signed cycle of length l we say that σ is an even l In the case of l = n, the signed cycle σ is called an even (resp.odd) full cyclic permutation.Note that a cyclic permutation in S n is an even cyclic permutation in H n .In addition, formula (1) is written as where ε i := ϵ 1 • • • ϵ i .Also, we obtain the representation of an even cycle as the product of transpositions as follows: We know the following relation for signed cycles and inversion transpositions.
Lemma 1.6.Suppose that i 1 , . . ., i l ∈ I n have different absolute values and ε ∈ {+, −}.Then the following three claims hold: In other words, multiplying a signed cycle by one appropriate inversion transposition changes the even-oddness.
The representation of a signed permutation as the product of signed cycles determines the conjugacy class in H n .The conjugacy class of H n are parameterized by pair of two integer partition (λ, µ) such that |λ|+|µ| = n, and those containing even and odd n-cycle are represented by ((n), 0) and (0, (n)), respectively.

Signed graph
We introduce the signed graph to consider an analogy of Dénes' results for H n .Here, define a signed graph as a quadruple is called a positive (resp.negative) edge, and an element in T ∩ E − T = L T = ∅ and the unsigned graph T is a tree.For every edge e ∈ E G , define a signed transposition τ e ∈ H n by Let ω = (e 1 , . . ., e m ) be an edge ordering of a signed graph G (a linear order on E G ).We give a product Definition 1.7.An edge ordering of G ω is an even (resp.odd) full cyclic permutation ordering if π ω is an even (resp.odd) full cyclic permutation.

Main results
The main results are as follows: Theorem 1.8.Given a signed graph G, the following are equivalent.
(i) G has an even (resp.odd) full cyclic permutation ordering.
(ii) Ḡ has a full cyclic permutation ordering and |L G | is even (resp.odd).
Theorem 1.9.Given a signed graph G, the following are equivalent.
(i) Any edge ordering of G is an even (resp.odd) full cyclic permutation ordering.
(ii) Ḡ is a tree and |L G | is even (resp.odd), that is, G is a signed tree with even (resp odd) loops.
(A) The number of representations of an even n-cycle by means of a product of n−1 transpositions is n n−2 .
(B) The number of representations of an odd n-cycle by means of a product of n transpositions is n n .
More general results are known about Corollary 1.10 .Let W be a well-generalized complex reflection group of rank n with Coxeter number h.Then the number of factorizations of a fixed Coxeter element c ∈ W into a product of n reflections is given by the following formula (see [3, §1, Formula (1.1)]): For the hyperoctahedral group H n , since |H n | = 2 n n! and h = 2n, the value of the above formula is n n .In this paper, we prove Corollary 1.10 by counting the number of sequences consisting of n − 1 (resp.n) transpositions such that the product is an even (resp.odd) n-cycle.
Note that Chapuy and Stump [3, §1, Theorem 1.1] give the formula for the exponential generating function of factorizations of a fixed Coxeter element into a product of reflections.
The organization of this paper is as follows.In Section 2, we discuss the representation of π ω and prove Theorem 1.8.In Section 3, we prove Theorem 1.9 and Corollary 1.10.

The representation of the product obtained from the edge ordering
In this subsection, we prove a key lemma about the representation of the permutation π ω .Let G = (V G , E + G , E − G , L G ) be a signed graph.Definition 2.1.For an edge ordering ω of G, define φ(ω) as an edge ordering of Ḡ obtained by excluding loops from ω.For an edge ordering ω of Ḡ, define φ −1 (ω) as the set of edge orderings ω of G such that φ(ω) = ω.

Define the projection
Let ω be an edge ordering of G. Since where i, j are distinct integers in [n] and 1 is the identity element of S n , the permutation π φ(ω) is equal to Φ(π ω ).
Step 2. Change all negative transpositions contained in π ω to the product of two inversion transpositions and one positive transposition by (T-1).
Step 3. Move all inversion transpositions resulting from Step 2 to the left by (T-2) or (T-3).
Therefore, we obtain

The condition to have a full cyclic permutation ordering
Now, we prove one of the main theorems, the condition for a signed graph to have a full cyclic permutation ordering.We devise this theorem in two propositions.Proposition 2.5.Let G be a signed graph.If ω is an even (resp.odd) full cyclic permutation ordering of G, then φ(ω) is a full cyclic permutation ordering of Ḡ and |L G | is even (resp.odd).
Proposition 2.6.Let G be a signed graph.If ω is a full cyclic permutation ordering of Ḡ and |L G | is even (resp.odd), then any edge ordering in φ −1 (ω) is an even (resp.odd) full cyclic permutation ordering of G.
Proof.Assume that |L G | is even (resp.odd) and let ω be a full cyclic permutation ordering of Ḡ.
Since an unsigned graph with a full cyclic permutation ordering is connected, we get the following corollary: Corollary 2.7.If a signed graph G has an odd full cyclic permutation ordering and |E G | = n, then Ḡ is a tree and |L G | = 1, that is, G is a signed tree with one loop.

Signed trees and full cyclic permutations
Now, we prove the remaining main theorems.
3.1.The proof of Theorem 1.9 Theorem 3.1 (Restatement of Theorem 1.9).Given a signed graph G, the following are equivalent.
(i) Any edge ordering of G is an even (resp.odd) full cyclic permutation ordering.
(ii) Ḡ is a tree and |L G | is even (resp.odd), that is, G is a signed tree with even (resp odd) loops.
First we show (i) ⇒ (ii).We know that if a signed graph G has an even (resp.odd) full cyclic permutation ordering, then |L G | is even (resp.odd) by Proposition 2.5.Therefore we need to show the following.Proposition 3.2.If any edge ordering of G is an even (resp.odd) full cyclic permutation ordering, then Ḡ is a tree.
Proof.We prove the contraposition.Suppose that Ḡ is not a tree.Since a signed tree with a full cyclic permutation ordering is connected, we can assume that Ḡ has a cycle.Let C = (V C , E C ) denote the minimal cycle (minimal number of vertices) that Ḡ has, where . Let I Ḡ\C (v l ) denote the set of edges of Ḡ that are incident to the vertex v l but not contained in E C .Then we see that there exists no edge in where Then E G = {b 1 , . . ., b s , d 1 , . . ., d t , c 1 , . . ., c l }.We obtain Hence ω 0 is not a full cyclic permutation ordering.Therefore edge orderings in φ −1 (ω 0 ) are neither even nor odd full cyclic permutation orderings of G.
Now, we show (ii) ⇒ (i) by induction on the number of vertices n.We can suppose that n ≥ 2. Suppose that Ḡ is a tree and |L G | is even (resp.odd).Let ω = (e 1 , . . ., e m ) be an edge ordering of G. Assume that e 1 is a loop.Since G is a connected graph with at least two vertices, there exists an edge of G which is not a loop.Let ω ′ = (e k , . . ., e m , e 1 , . . ., e k−1 ) denote an edge ordering of G such that e k is not a loop.Since that is, π ω and π ω ′ are conjugate, ω is a full cyclic permutation ordering if and only if ω ′ is a full cyclic permutation ordering.Hence we can assume e 1 is not a loop.
Let e 1 = {i 1 , j 1 } and G\e 1 denote the graph excluding the edge e 1 from G. The graph G\e 1 has two connected components T 1 and T 2 which are signed trees with some loops.We assume that T 1 includes i 1 and T 2 includes j 1 .Let ω 1 = (c 1 , . . ., c p ) and ω 2 = (d 1 , . . ., d q ) denote the edge orderings of T 1 and T 2 which are obtained by restricting ω to E T 1 and E T 2 , where p + q = m − 1.
Since ω 1 and ω 2 are full cyclic permutation orderings by induction hypothesis, we can write π ω 1 and π ω 2 as where ν 1 , . . ., ν s , µ 1 , . . ., µ t are distinct inversion transpositions, i 1 , . . ., i l , j 1 , . . ., j k are distinct integers in Since the signed transposition corresponding to an edge of T 1 and the signed transposition corresponding to an edge of T 2 are commutative, we obtain , where ν ′ 1 , . . ., ν ′ r are inversion transpositions and r ≡ |L G | (mod 2).Thus ω is an even (resp.odd) full cyclic permutation ordering if |L G | is even (resp.odd).Now, Theorem 1.9 is completely proven.We can obtain (A) as the corollary of Corollary 1.4.Now, we prove (B).Let x be the number of representations of an odd full cyclic permutation by means of a product of n transpositions.Suppose that X := (τ n , . . ., τ 1 ) τ i is a signed transposition, the product τ n • • • τ 1 is an odd full cyclic permutation .
Define the set consisting of signed graphs by

Proposition 1 . 2 .
If a graph G has a full cyclic permutation ordering, then G is connected and the Betti number β(G) := |E G | − |V G | + 1 is even.

Theorem 2 . 4 (
Restatement of Theorem 1.8).Given a signed graph G, the following are equivalent.(i)G has an even (resp.odd) full cyclic permutation ordering.(ii)Ḡ has a full cyclic permutation ordering and |L G | is even (resp.odd).
then Ḡ has a cycle of length 2 consisting of b 0 and c l .We have l = 2 by minimality of C. Thus there exist three edges between v 1 and v l .It contradicts to the construction of Ḡ. Define an edge ordering of Ḡ ω 0 by ω 0 := (c 1 , . . ., c l−1 , d 1 , . . ., d t , c l , b 1 , . . ., b s ),

3. 2 .
The proof of Corollary 1.10 Corollary 3.3 (Restatement of Corollary 1.10).(A) The number of representations of an even n-cycle by means of a product of n−1 transpositions is n n−2 .(B) The number of representations of an odd n-cycle by means of a product of n transpositions is n n .
G has an odd full cyclic permutation ordering and define the set consisting of pairs of graphs and edge orderings of them by Z := (G, ω) G ∈ Y, ω is an odd full cyclic permutation ordering of G .Since we haveY = G = (V G , E + G , E − G , L G ) V G = [n], |L G | = 1, Ḡ is a treeby Corollary 2.7, we obtainZ = (G, ω) G ∈ Y, ω is an edge ordering of G .Since |Y | = n n−2 • n • 2 n−1 (Cayley's formula), we get |Z| = |Y | • n! = n n−1 • 2 n−1 • n!.Also, since there exists a bijection between X and Z, we see |X| = |Z|.Hence x = n n .