Partial-dual genus polynomials and signed intersection graphs
PPartial-dual genus polynomials and signed intersectiongraphs
Qi Yan
School of MathematicsChina University of Mining and TechnologyP. R. ChinaXian’an Jin School of Mathematical SciencesXiamen UniversityP. R. China
Email:[email protected]; [email protected]
Abstract
Recently, Gross, Mansour and Tucker introduced the partial-dual genus poly-nomial of a ribbon graph as a generating function that enumerates the partialduals of the ribbon graph by genus. It is analogous to the extensively-studiedpolynomial in topological graph theory that enumerates by genus all embed-dings of a given graph. To investigate the partial-dual genus polynomial oneonly needs to focus on bouquets, i.e. ribbon graphs with only one vertex. Inthis paper, we shall further show that the partial-dual genus polynomial of abouquet essentially depends on the signed intersection graph of the bouquetrather than on the bouquet itself. That is to say the bouquets with the samesigned intersection graph will have the same partial-dual genus polynomial.We then prove that the partial-dual genus polynomial of a bouquet containsnon-zero constant term if and only if its signed intersection graph is positiveand bipartite. Finally we consider a conjecture posed by Gross, Mansourand Tucker. that there is no orientable ribbon graph whose partial-dualgenus polynomial has only one non-constant term, we give a characterizationof non-empty bouquets whose partial-dual genus polynomials have only oneterm by consider non-orientable case and orientable case separately.
Keywords:
Ribbon graph, partial-dual genus polynomial, bouquet, signed Corresponding author.
Preprint submitted to Elsevier February 4, 2021 a r X i v : . [ m a t h . C O ] F e b ntersection graph, bipartite, complete.
1. Introduction
The concept of partial duality was introduced in [4] by Chmutov, andit, together with other partial twualities, has received ever-increasing at-tention, and their applications span topological graph theory, knot theory,matroids/delta matroids, and physics. We assume that the readers are fa-miliar with the basic knowledge of topological graph theory, see for example[15, 17] and in particular the ribbon graphs and partial duals, see for example[1, 4, 7, 9, 16]. Let G be a ribbon graph and A ⊆ E ( G ). We denote by G A the partial dual of G with respect to A .Similar to the extensively-studied polynomial in topological graph theorythat enumerates by genus all embeddings of a given graph. In [13], Gross,Mansour and Tucker introduced the partial-dual orientable genus polynomi-als for orientable ribbon graphs and the partial-dual Euler genus polynomialsfor arbitrary ribbon graphs. Definition 1. [13] The partial-dual Euler genus polynomial of any ribbongraph G is the generating function ∂ ε G ( z ) = (cid:88) A ⊆ E ( G ) z ε ( G A ) that enumerates all partial duals of G by Euler genus. The partial-dualorientable genus polynomial of an orientable ribbon graph G is the generatingfunction ∂ Γ G ( z ) = (cid:88) A ⊆ E ( G ) z γ ( G A ) that enumerates all partial duals of G by orientable genus.Clearly, if G is an orientable ribbon graph, then ∂ Γ G ( z ) = ∂ ε G ( z ). Either ∂ ε G ( z ) or ∂ Γ G ( z ) may be referred to as a partial-dual genus polynomial. A bouquet is a ribbon graph with only one vertex. It is known that the partial-dual genus polynomial of any connected ribbon graph is equal to that of abouquet, which is one of its partial duals. Hence it is natural to focus onbouquets. 2n [18], we introduced the notion of signed interlace sequences of bou-quets and proved that two bouquets with the same signed interlace sequencehave the same partial-dual Euler genus polynomial if the number of edgesof the bouquets is less than 4 and two orientable bouquets with the samesigned interlace sequence have the same partial-dual orientable genus poly-nomial if the number of edges of the bouquets is less than 5. As we observedin [18], there are bouquets with the same signed interlace sequence but dif-ferent partial-dual genus polynomials. The first purpose of this paper is tostrengthen the notion of signed interlace sequences such that it can determinethe partial-dual genus polynomial completely.Intersection graphs (also called circle graphs) appear and are very usefulin both graph theory and combinatorial knot theory [10]. For example, acharacterization of those graphs that can be realized as intersection graphsis given by an elegant theorem of Bouchet [3]. Signed interlace sequencesof bouquets are exactly degree sequences of their signed intersection graphs.Based on a theorem of Chmutov and Lando [6], we shall prove that anytwo bouquets with the same signed intersection graph will have the samepartial-dual genus polynomial.Then we focus on signed intersection graphs, the intersection polynomialis introduced, a recursion of this polynomial is given and is used to computeintersection polynomials of paths and stars. We also prove that the inter-section polynomial contains non-zero constant term if and only if the signedintersection graph is positive and bipartite.In [13], Gross, Mansour and Tucker characterized connected ribbon graphswith constant polynomials, i.e., one of its partial dual is a tree. They alsofound examples of non-orientable ribbon graphs whose polynomials have onlyone (non-constant) term. The second purpose of this paper is to characterizebouquets whose partial-dual genus polynomials have only one term.A bouquet is prime if its intersection graph is connected. Then we willshow that the partial-dual Euler genus polynomial of a prime non-orientablebouquet has only one non-constant term if and only if its intersection graphis trivial. For orientable ribbon graphs, they posed the following conjecture. Conjecture 2. [13] There is no orientable ribbon graph having a non-constant partial-dual genus polynomial with only one non-zero coefficient.The conjecture is not true. In [18] we found an infinite family of coun-terexamples (see Proposition 20), whose intersection graphs are non-trivial3omplete graph of odd order. In this paper, we shall prove that Conjec-ture 2 is actually true for all prime orientable bouquets except the family ofcounterexamples.This paper is organized as follows. In Section 2, we consider the cyclicinterlace sequences and an example is given to show that it can not determinethe partial-dual genus polynomial. In Section 3, we recall the notion ofmutant chord diagrams and a theorem of Chmutov and Lando on mutantchord diagrams and intersection graphs. In Section 4, we prove that thesigned intersection graph can determine the partial-dual genus polynomial.In Section 5, we introduce the intersection polynomial and discuss its basicproperties. In Section 6, we characterize all the bouquets whose partial-dualgenus polynomials have only one term. In the final section, we pose severalproblems for further study.
2. Cyclic interlace sequences and signed intersection graphs
Let e be an edge of a ribbon graph G . If the vertex-disks at the endsof e are distinct, we say that e is proper . If e is a loop at the vertex disk v and e ∪ v is homeomorphic to a M¨obius band, then we call e a twisted loop .Otherwise it is said to be an untwisted loop .A signed rotation of a bouquet is a cyclic ordering of the half-edges atthe vertex and if the edge is an untwisted loop, then we give the same sign+ to the corresponding two half-edges, and give the different signs (one +,the other − ) otherwise. The sign + is always omitted. See Figure 1 for anexample. Sometimes we will use the signed rotation to represent the bouquetitself.Let B be a bouquet and let E ( B ) = { e , · · · , e n } and e ∈ E ( B ). The interlace number of e , denoted by α ( e ), is defined to be the number of edgeswhich are all interlaced with e . We say that β ( e ) is the signed interlacenumber of e , where β ( e ) = (cid:26) α ( e ) , if e is an untwisted loop, − α ( e ) , if e is a twisted loop.The signed interlace sequence [18] of the bouquet B , denoted by S ( B ) =( β ( e ) , · · · , β ( e n )), is obtained by sorting the signed interlace number fromsmall to large, where β ( e ) ≤ β ( e ) ≤ · · · ≤ β ( e n ). We first strengthen asigned interlace sequence by considering its cyclic rotation of a bouquet.4 igure 1: A bouquet with the signed rotation ( a, c, − a, d, b, d, c, − b ) and its signed inter-section graph. Definition 3.
The cyclic interlace sequence of a bouquet can be obtainedfrom its cyclic rotation by replacing each entry with the signed interlacenumber of the corresponding edge.
Example 4.
Let B be a bouquet with signed rotation( a, b, a, c, b, d, e, f, d, e, c, f )and let B be a bouquet with signed rotation( a, b, a, c, d, e, c, f, e, d, b, f ) . It is easy to check that both bouquets B and B have the same cyclicinterlace sequence (1 , , , , , , , , , , , ∂ Γ B ( z ) = 12 z + 44 z + 8 z and ∂ Γ B ( z ) = 2 + 18 z + 36 z + 8 z . The intersection graph I ( B ) of a bouquet B is the graph with vertex set E ( B ) and in which two vertices e and f of I ( B ) are adjacent if and onlyif their ends are met in the cyclic order e · · · f · · · e · · · f · · · when travelingaround the boundary of the unique vertex of B . Clearly, the degree sequenceof the intersection graph is exactly the interlace sequence. The signed inter-section graph SI ( B ) of a bouquet B consists of I ( B ) and a + or − sign at5ach vertex of I ( B ) where the vertex corresponding to the untwisted loop of B is signed + and the vertex corresponding to the twisted loop of B is signed − . See Figure 1 for an example. A signed intersection graph is said to be positive if each of its vertices is signed +. The following lemma is obvious. Lemma 5.
A bouquet B is orientable if and only if its signed intersectiongraph SI ( B ) is positive. The signed intersection graphs of the bouquets B and B in Example4 are shown in Figure 2. It shows that two bouquets with the same cyclicinterlace sequence may have different signed intersection graphs. Figure 2: SI ( a, b, a, c, b, d, e, f, d, e, c, f ) and SI ( a, b, a, c, d, e, c, f, e, d, b, f ). In the following, we shall prove that signed intersection graphs can deter-mine the partial-dual genus polynomial completely. In next section we willfirst recall mutants.
3. Mutants
In knot theory mutants are a pair of knots obtained from one to the otherby rotating a tangle. Mutants are usually very difficult to distinguish by knotpolynomials.A chord diagram refers to a set of chords with distinct endpoints on acircle. A combinatorial analog of the tangle in mutant knots is a share . A share [6] in a chord diagram is a union of two arcs of the outer circle andchords ending on them possessing the following property: each chord one ofwhose ends belongs to these arcs has both ends on these arcs. A mutation [6] of a chord diagram is another chord diagram obtained by a rotation of a6hare about one of the three axes. Note that the composition of rotationsabout two of the three axes will be exactly the rotation about the third axis.Two chord diagrams are said to be mutant [6] if they can be transformedinto one another by a sequence of mutations.
Theorem 6. [6] Two chord diagrams have the same intersection graph ifand only if they are mutant.
For the details we refer the reader to [6]. Mutant can be defined for bou-quets similarly. Suppose P = p p · · · p k is a string and P − = p k p k − · · · p is called the inverse of P . Definition 7.
Let B be a bouquet with signed rotation ( M P N Q ) whereboth labels of each edge must belong to
M N or both not. A mutation of B is another bouquet with signed rotation ( M − P N − Q ) or ( N P M Q ). Twobouquets are said to be mutant if they can be transformed into one anotherby a sequence of mutations.In Definition 7, either
M, N, P or Q can be empty. Corollary 8.
Two bouquets have the same signed intersection graph if andonly if they are mutant.Proof.
Obviously, mutations preserve the signed intersection graphs of bou-quets, hence if two bouquets are mutant, they have the same signed inter-section graph. Conversely, if two bouquets have the same signed intersectiongraph, by Theorem 6, they are related by a sequence of mutations.In the next section, we will show that the signed intersection graphs candetermine the partial-dual genus polynomial completely.
4. First Main Theorem
Now we state our first main theorem as follows.
Theorem 9.
If two bouquets B and B have the same signed intersectiongraph, then ∂ ε B ( z ) = ∂ ε B ( z ) . Recall that the contraction
G/e of the edge e in the ribbon graph G isdefined by the equation G/e := G e − e . We denote by G/A the ribbongraph obtained from G by contracting each edge of A ⊆ E ( G ) and then7 /A = G A − A. It is an important observation [9, 12] that the operationof the contraction does not change the number of boundary components.Let v ( G ) , e ( G ) and f ( G ) denote the number of vertices, edges and boundarycomponents of a ribbon graph G , respectively. To prove Theorem 9, we needthree lemmas. Lemma 10.
Let B be a bouquet. Then the Euler genus ε ( B ) is given by theequation: ε ( B ) = 1 + e ( B ) − f ( B ) . (1) Proof.
Recall that if G is a connected ribbon graph then 2 − ε ( G ) = v ( G ) − e ( G ) + f ( G ). The Lemma then follows from v ( B ) = 1. Lemma 11.
If two bouquets B and B have the same signed intersectiongraph, then ε ( B ) = ε ( B ) .Proof. By Corollary 8, we can assume that B can be transformed into B by a mutation. Let B = ( M P N Q ). Then B = ( M − P N − Q ) or B =( N P M Q ) as in Figure 3. Assume that B = ( M − P N − Q ) and B =( N P M Q ). By Lemma 10, it suffices to prove that f ( B ) = f ( B ) = f ( B ). Figure 3: The bouquets B , B and B . Suppose that G = ( M f P f N eQe ) , G = ( M − f P f N − eQe ) and G =( N f P f M eQe ) as in Figure 4.Since B i = G i − { e, f } = ( G i { e,f } ) { e,f } − { e, f } = G i { e,f } / { e, f } for i ∈ { , , } and contraction does not change the number of boundary com-ponents, it follows that f ( B i ) = f ( G i { e,f } ). For the ribbon graph G i { e,f } ,arbitrarily orient the boundary of e and place an arrow on each of the twoarcs where e meets vertices of G i { e,f } such that the directions of these arrowsfollow the orientation of the boundary of e and label the two arrows with e (cid:48) igure 4: The bouquets G , G and G . and e (cid:48)(cid:48) . The same operating can be drawn for f and label the two arrowswith f (cid:48) and f (cid:48)(cid:48) . Let B i (cid:48) denote the ribbon graph obtained from G i { e,f } bydeleting the vertices v P , v Q together with all the edges incident with v P , v Q ,but reserving the marking arrows e (cid:48)(cid:48) and f (cid:48)(cid:48) as in Figure 5. Since both labels Figure 5: The ribbon graphs G i { e,f } and B i (cid:48) for i ∈ { , , } and G (cid:48) . of each edge must belong to M N or both not, this results in a bouquet, withexactly two labelled arrows e (cid:48)(cid:48) and f (cid:48)(cid:48) on its boundary of the vertex and thesemarking arrows only indicate the positions and no other significance. Notethat if we ignore the two labelled arrows e (cid:48)(cid:48) and f (cid:48)(cid:48) , the bouquets B (cid:48) , B (cid:48) B (cid:48) are equivalent. Hence f ( B (cid:48) ) = f ( B (cid:48) ) = f ( B (cid:48) ) . Similarly, let G i (cid:48) denote the ribbon graph obtained from G i { e,f } by deleting the vertex v MN together with all the edges incident with v MN , but reserving the markingarrows e (cid:48) and f (cid:48) . This results in a ribbon graph, with exactly two labelledarrows e (cid:48) and f (cid:48) on the boundaries of v P and v Q as in Figure 5. Note that G (cid:48) = G (cid:48) = G (cid:48) . Obviously, we can recover the boundaries of G i { e,f } from G i (cid:48) and B i (cid:48) as follows: draw a line segment from the head of e (cid:48) to the tailof e (cid:48)(cid:48) , and a line segment from the head of e (cid:48)(cid:48) to the tail of e (cid:48) . The sameoperating is applied to f (cid:48) and f (cid:48)(cid:48) . We observe that( a ) If e (cid:48)(cid:48) and f (cid:48)(cid:48) are contained in different boundary components of B (cid:48) ,then e (cid:48)(cid:48) and f (cid:48)(cid:48) are also contained in different boundary components of B (cid:48) and B (cid:48) .( b ) If e (cid:48)(cid:48) and f (cid:48)(cid:48) are contained in the same boundary component of B (cid:48) ,then e (cid:48)(cid:48) and f (cid:48)(cid:48) are also contained in the same boundary component of B (cid:48) and B (cid:48) . The arrows e (cid:48)(cid:48) and f (cid:48)(cid:48) are called consistent (inconsistent) in B (cid:48) ifthese two arrows are consistent (inconsistent) on the boundary component.We can also observe that if e (cid:48)(cid:48) and f (cid:48)(cid:48) are consistent (inconsistent) in B (cid:48) ,then e (cid:48)(cid:48) and f (cid:48)(cid:48) are also consistent (inconsistent) in B (cid:48) and B (cid:48) .If e (cid:48) and f (cid:48) are contained in the same boundary component of G (cid:48) and e (cid:48) , f (cid:48) are consistent in G (cid:48) , then there are three cases as following.1. If e (cid:48)(cid:48) and f (cid:48)(cid:48) are contained in different boundary components of B (cid:48) ,then by (a) f ( G { e,f } ) = f ( G { e,f } ) = f ( G { e,f } ) = f ( G (cid:48) ) + f ( B (cid:48) ) − Figure 6: Case 1.
10. If e (cid:48)(cid:48) and f (cid:48)(cid:48) are contained in the same boundary component of B (cid:48) and e (cid:48)(cid:48) , f (cid:48)(cid:48) are consistent in B (cid:48) , then by (b) f ( G { e,f } ) = f ( G { e,f } ) = f ( G { e,f } ) = f ( G (cid:48) ) + f ( B (cid:48) )as in Figure 7. Figure 7: Case 2.
3. If e (cid:48)(cid:48) and f (cid:48)(cid:48) are contained in the same boundary component of B (cid:48) and e (cid:48)(cid:48) , f (cid:48)(cid:48) are inconsistent in B (cid:48) , then by (b) f ( G { e,f } ) = f ( G { e,f } ) = f ( G { e,f } ) = f ( G (cid:48) ) + f ( B (cid:48) ) − Figure 8: Case 3.
Similar arguments apply to the case e (cid:48) and f (cid:48) are contained in differentboundary components of G (cid:48) or e (cid:48) and f (cid:48) are contained in the same boundarycomponent of G (cid:48) and e (cid:48) , f (cid:48) are inconsistent in G (cid:48) . Lemma 12. [13] Let B be a bouquet, and let A ⊆ E ( B ) . Then ε ( B A ) = ε ( A ) + ε ( A c ) , (2) where A c = E ( B ) − A . roof of Theorem 9 For any subset A of edges of B , we denoted its corresponding ver-tex subset of SI ( B ) also by A . Let SI ( B )[ A ] denote the induced sub-graph of SI ( B ) by the vertex subset A . Since SI ( B ) = SI ( B ), thereis a corresponding subset A of vertices of SI ( B ) such that SI ( B )[ A ] = SI ( B )[ A ] and SI ( B )[ A c ] = SI ( B )[ A c ]. It follows that ε ( A ) = ε ( A ) and ε ( A c ) = ε ( A c ) by Lemma 11. Hence, ε ( B A ) = ε ( B A ) by Lemma 12. Thus ∂ ε B ( z ) = ∂ ε B ( z ). Remark . Two bouquets with different signed intersection graphs may havethe same partial-dual genus polynomial. For example, let B = (1 , , − , B = (1 , , − , − ∂ ε B ( z ) = ∂ ε B ( z ) = 2 z + 2 z , see also[18], but the signed intersection graphs of B and B are different. In fact, B = B e .
5. Intersection polynomials
A graph SG with a + or − sign at each vertex is said to be signedintersection graph if there exists a bouquet B such that SG = SI ( B ). The intersection polynomial , IP SG ( z ), of a signed intersection graph SG is definedby IP SG ( z ) := ∂ ε B ( z ), where B is a bouquet such that SG = SI ( B ). ByTheorem 9, it is well-defined. Theorem 14.
Let SG be a signed intersection graph and v , v ∈ V ( SG ) .If v , v are adjacent and the degree of v is 1 and the sign of v is positive,then IP SG ( z ) = IP SG − v ( z ) + (2 z ) IP SG − v − v ( z ) . Proof.
Let B be a bouquet satisfying SG = SI ( B ). We have IP SG ( z ) = ∂ ε B ( z ). Note that v , v correspond to two edges of B , we denote them by e and e , respectively. Since the degree of v is 1 and the sign of v is positive,it follows that e is an untwisted loop and for any e ∈ E ( B ) − e − e , theends of e are therefore on α and β , or γ and θ (otherwise it interlaces e ) asshown in Figure 9. Let A be any subset of E ( B ). We partition the possiblesubsets A into two types:1. τ : those for which one of e , e is in A and the other is in A c ;2. τ : those for which e , e are both in A or both in A c .12 igure 9: Two cases for the bouquet B in the proof of Theorem 14. Then ∂ ε B ( z ) = Σ A ∈ τ z ε ( B A ) + Σ A ∈ τ z ε ( B A ) . Let D ⊆ E ( B − e ). Then D c = B − e − D . If e ∈ D , take A = D and A c = D c ∪ e ; if e / ∈ D , take A = D ∪ e and A c = D c . A 1-1 correspondencebetween the set of subsets of E ( B − e ) and τ is established. Furthermore,it is not difficult to see that ε ( D ) = ε ( A ) and ε ( A c ) = ε ( D c ) for each case.By Lemma 12 we have Σ A ∈ τ z ε ( B A ) = ∂ ε B − e ( z ) . Let D ⊆ E ( B − e − e ). Then D c = B − e − e − D . Take A = D ∪{ e , e } and A c = D c . Clearly ε ( A c ) = ε ( D c ) and it is not difficult to see that f ( A ) = f ( D ), hence ε ( A ) = ε ( D ) + 2. Thus, we haveΣ A ∈ τ z ε ( B A ) = 2Σ { e ,e }⊆ A ∈ τ z ε ( B A ) = (2 z ) ∂ ε B − e − e ( z ) . Hence, ∂ ε B ( z ) = ∂ ε B − e ( z )+(2 z ) ∂ ε B − e − e ( z ) , i.e. IP SG ( z ) = IP SG − v ( z )+(2 z ) IP SG − v − v ( z ). Example 15.
Let S n be a positive star which is a complete bipartite graphwhose vertex set can be partitioned into two subsets X and Y so that everyedge has one end in X and the other end in Y with | X | = 1 and | Y | = n .13hen we have initial condition IP S ( z ) = 2 + 2 z , and by Theorem 14, therecursion IP S n +1 ( z ) = IP S n ( z ) + 2 n +1 z . Then it is easy to obtain that IP S n ( z ) = (2 n +1 − z + 2 . (3) Example 16.
Let P n be a positive path with n vertices. Then IP P ( z ) = 2; IP P ( z ) = 2 + 2 z ; IP P n +2 ( z ) = IP P n +1 ( z ) + 2 z IP P n ( z ) . Theorem 17.
Let SG be a signed intersection graph. Then IP SG ( z ) containsnon-zero constant term if and only if SG is positive and bipartite.Proof. Let B be a bouquet satisfying SG = SI ( B ). We know IP SG ( z ) = ∂ ε B ( z ). Since IP SG ( z ) contains non-zero constant term, it follows that B is a partial dual of a plane ribbon graph. According to the property thatpartial duality preserving orientability, we have B is orientable and hence SG is positive. Suppose that SG is not bipartite. Then SG contains an oddcycle C . We denote by D the edge subset of B corresponding to verticesof C . It is obvious that deleting edges can not increase Euler genus. Thenfor any subset A of E ( B ), we have ε ( A ∩ D ) (cid:54) ε ( A ), ε ( A c ∩ D ) (cid:54) ε ( A c ).Note that there are two loops e, f ∈ A ∩ D or e, f ∈ A c ∩ D such that theirends are met in the cyclic order e · · · f · · · e · · · f · · · when traveling round theboundary of the unique vertex of B . Then ε ( A ∩ D ) + ε ( A c ∩ D ) >
0. Thus ε ( B A ) = ε ( A ) + ε ( A c ) >
0, a contradiction.Conversely, if SG is bipartite and non-trivial, then its vertex set can bepartitioned into two subsets X and Y so that every edge of SG has one endin X and the other end in Y . For these two subsets X and Y of the vertexset of SG , we denoted these two corresponding edge subsets of B also by X and Y . Obviously, X ∪ Y = E ( B ) , X ∩ Y = ∅ and ε ( X ) = ε ( Y ) = 0. Thus ε ( B X ) = 0 by Lemma 12. Hence ∂ ε B ( z ) (hence, IP SG ( z )) contains non-zeroconstant term.
6. Second Main Theorem
Gross, Mansour and Tucker [13] discussed the simplest partial-dual genuspolynomial, i.e., a constant polynomial and found examples of non-orientable14ibbon graphs whose polynomials have only one non-constant term. Theyproved:
Proposition 18. [13] Let G be a connected ribbon graph. Then ∂ ε G ( z ) =2 e ( G ) if and only if there is a subset A ⊆ E ( G ) such that G A is a tree. and Proposition 19. [13] For any n > and any m ≥ n , there is a non-orientable ribbon graph G such that ∂ ε G ( z ) = 2 m z n . For orientable ribbon graphs, Gross, Mansour and Tucker posed Conjec-ture 2 and we found an infinite family of counterexamples in [18]. Let B t bea bouquet with the signed rotation (1 , , , · · · , t, , , , · · · , t ). Proposition 20. [18] Let t be a positive integer. Then ∂ ε B t ( z ) = (cid:26) t z t − , if t is odd, t − z t + 2 t − z t − , if t is even. Note that B , B , B , · · · is an infinite family of counterexamples to Con-jecture 2. The purpose of this section is to characterize bouquets whosepartial-dual genus polynomial has only one non-constant term. Let
P, Q be two ribbon graphs, we denote by P ∨ Q the ribbon-join of P and Q . Note that in general the ribbon-join is not unique. A ribbon graph iscalled empty if it has no edges. We say that G is prime , if there don’t existnon-empty ribbon subgraphs G , · · · , G k of G such that G = G ∨ · · · ∨ G k where k ≥
2. Clearly, we have
Lemma 21.
A bouquet B is prime if and only if its intersection graph I ( B ) is connected. Let B = (1 , −
1) be the non-orientable bouquet with only one edge. Let B = { B , B , B , B , · · · } . Now we are in a position to state our second maintheorem as follows. Theorem 22.
Let B be a non-empty bouquet. Then ∂ ε B ( z ) = 2 e ( B ) z b ⇐⇒ B = B t ∨ · · · ∨ B t k , where k ≥ and B t i ∈ B for (cid:54) i (cid:54) k . Furthermore, if the number of theprime factors B in B is k , then b = e ( B ) − k + k . B is a negative isolated vertexand the signed intersection graph of B i +1 is a positive complete graph oforder 2 i + 1. In fact B i +1 is the only bouquet whose signed intersectiongraph is a positive complete graph of order 2 i + 1. Restate Theorem 22 inthe language of signed intersection graphs and intersection polynomial, wehave Corollary 23.
Let SG be a signed intersection graph. Then IP SG ( z ) =2 v ( SG ) z b if and only if each component of SG is complete graph of odd orderand each vertex, except isolated vertex, has positive sign. It is easy to see that ∂ ε B ( z ) = 2 and ∂ ε B ( z ) = 2 z . To prove Theorem22, we shall use the following lemma. Lemma 24. [13] Let G = G ∨ G . Then ∂ ε G ( z ) = ∂ ε G ( z ) ∂ ε G ( z ) . (4)It suffices to show that among all prime non-orientable bouquets thereis only B whose partial-dual genus polynomial has one (non-constant) termand among all prime orientable bouquets there are only B , B , B , · · · whosepartial-dual genus polynomials have one term.Let G ∗ denote the (full) dual of a ribbon graph G . Corresponding toeach edge e of G there is an edge e ∗ of G ∗ . We view each ribbon as anoriented rectangle, then the opposing two sides lying on face-disks are called ribbon-sides [13]. We need the following lemma. Lemma 25. [13] Let G be a ribbon graph and e ∈ E ( G ) . Then ε ( G ) = ε ( G e ) if and only if e ∗ is proper in G ∗ , if e is an untwisted loop, e ∗ is an untwisted loop in G ∗ , if e is proper, e ∗ is a twisted loop in G ∗ , if e is a twisted loop.6.2. Non-orientable case Theorem 26.
Let B be a prime non-orientable bouquet. Then ∂ ε B ( z ) =2 e ( B ) z b if and only if B = B . roof. The sufficiency is easily verified by calculation. For necessity, since B is non-orientable, we may assume that e ( B ) (cid:62) Claim 1. B does not contain a bouquet with signed rotation ( e, f, − e, f ).Suppose that Claim 1 is not true. Then e ∗ is a twisted loop and f ∗ isproper in B ∗ by Lemma 25. Thus the two ribbon-sides of e lie on a sameboundary component of B , denoted by C , and if we assign two arrows tothe two ribbon-sides of e such that these two arrows are consistent on theedge boundary of e , then these two arrows are non-consistent on C andthe two ribbon-sides of f lie on different boundary components of B as inFigure 10. Delete the edge e and note that f ( B ) = f ( B − e ) and the two Figure 10: Proof of Theorem 26. ribbon-sides of f also lie on different boundary components of B − e . Hence, f ( B − { e, f } ) = f ( B − e ) −
1, that is, f ( B − { e, f } ) = f ( B ) −
1. Since ε ( e, f, − e, f ) = 2 and ε ( B − { e, f } ) = e ( B ) − − f ( B − { e, f } ) by Eulerformula, we have ε ( B { e,f } ) = ε ( e, f, − e, f ) + ε ( B − { e, f } ) = e ( B ) + 1 − f ( B − { e, f } )by Lemma 12. Since ε ( B ) = e ( B ) + 1 − f ( B ), it is easy to check that ε ( B ) (cid:54) = ε ( B { e,f } ), contrary to ∂ ε B ( z ) = 2 e ( B ) z b . The claim then follows. Claim 2. B does not contain a bouquet with signed rotation ( e, f, − e, − f ).Assume that Claim 2 is not true. It is easily seen that B e contains abouquet with signed rotation ( e, f, − e, f ). Since ∂ ε B e ( z ) = ∂ ε B ( z ) = 2 e ( B ) z b ,this contradicts Claim 1.Since B is a non-orientable bouquet, there exists a twisted loop. Let e be any twisted loop. As B is prime and e ( B ) (cid:62)
2, there exists a loop f such that the loops e and f alternate, this contradicts Claim 1 or 2. Hence e ( B ) = 1, that is, B = B . 17 .3. Orientable case Theorem 27.
Let B be a non-empty prime orientable bouquet. Then ∂ ε B ( z ) =2 e ( B ) z b if and only if B = B i +1 .Proof. The sufficiency is easily verified by Proposition 20. For necessity,the result is easily verified when e ( B ) ∈ { , } . Assume that e ( B ) (cid:62)
3. Let e, f, g ∈ E ( B ). Note that e ∗ , f ∗ and g ∗ are proper in B ∗ by Lemma 25. Hencethe two ribbon-sides of e (or f or g ) lie on different boundary componentsof B . We denote the two ribbon-sides of e (or f or g ) lying on the twoboundary components of B by C e and C e (or C f and C f or C g and C g ),respectively.The following facts about ribbon graphs are well known and readily seento be true. Deleting any edge e of an orientable ribbon graph G can changethe number of boundary components exactly one. Otherwise, G ∗ contains atwisted loop, which is contrary to the orientability of G . More specifically,( T1 ) The two ribbon-sides of e lie on different boundary components of G if and only if f ( G − e ) = f ( G ) − T2 ) The two ribbon-sides of e lie on the same boundary component of G if and only if f ( G − e ) = f ( G ) + 1.From (T1) it follows that f ( B − e ) = f ( B ) − . Obviously, ε ( B ) = e ( B ) + 1 − f ( B ) and ε ( B − { e, f } ) = e ( B ) − − f ( B − { e, f } ) by Eulerformula. There are two cases to consider: Case 1. If B ( { e, f } ) = ( e, f, e, f ), we have ε ( B { e,f } ) = ε ( e, f, e, f ) + ε ( B − { e, f } ) = e ( B ) + 1 − f ( B − { e, f } )by Lemma 12. Since ε ( B { e,f } ) = ε ( B ), it follows that f ( B − { e, f } ) = f ( B ) = f ( B − e ) + 1 . Then the two ribbon-sides of f must lie on the same boundary component of B − e from (T2). Hence, the two ribbon-sides of f must lie on C e and C e ,respectively, in B . Thus { C e , C e } = { C f , C f } . Case 2. If B ( { e, f } ) = ( e, e, f, f ), then ε ( B { e,f } ) = ε ( e, e, f, f ) + ε ( B − { e, f } ) = e ( B ) − − f ( B − { e, f } )by Lemma 12. As ε ( B { e,f } ) = ε ( B ), we have f ( B − { e, f } ) = f ( B ) − f ( B − e ) − . f lie on different boundary components of B − e from (T1). Hence at most one of the two ribbon-sides of f lie on C e and C e in B . Thus { C e , C e } ∩ { C f , C f } (cid:54) = { C e , C e } . Claim 3. B does not contain a bouquet with signed rotation ( e, f, g, e, g, f ).Assume that Claim 3 is not true. Since B ( { e, f } ) = ( e, f, e, f ) and B ( { e, g } ) = ( e, g, e, g ), it follows that { C e , C e } = { C f , C f } = { C g , C g } by Case 1. Thus { C f , C f } ∩ { C g , C g } = { C f , C f } . But B ( { f, g } ) = ( f, f, g, g ), this contradicts Case 2.Suppose that I ( B ) is not a complete graph. Note that I ( B ) is con-nected. Then there is a vertex set { v e , v f , v g } of I ( B ) such that the inducedsubgraph I ( B )( { v e , v f , v g } ) is a 2-path (see Exercise 2.2.11 [2]). We may as-sume without loss of generality that the degree of v e is 2 in I ( B )( { v e , v f , v g } )and v e , v f , v g are corresponding to the loops e, f, g of B , respectively. Thus B ( { e, f, g } ) = ( e, f, g, e, g, f ), this contradicts Claim 3. Hence I ( B ) is acomplete graph and B = B i +1 by Proposition 20.Theorem 27 tells us that Conjecture 2 is actually true for all prime ori-entable bouquets except the family of counterexamples as in Proposition 20.
7. Concluding remarks
As shown in Remark 13, there are different signed intersection graphs withthe same intersection polynomial. More examples could be obtained by usingTheorem 22. For example, let K +5 be the positive K and 4 K − ∪ K +1 be thedisjoint union of 4 negative isolated vertices and 1 positive isolated vertex,then IP K +5 ( z ) = IP K − ∪ K +1 ( z ) = 32 z . Similar to the chromatic polynomial[8] and the Tutte polynomial [11], we could call two signed intersection graphs IP-equivalent if they have the same intersection polynomial. It is interestingto find more examples of equivalent signed intersection graphs and eventuallyclarify the IP-equivalence from the viewpoint of the structures of graphs.Not every signed graph is a signed intersection graph. We define theintersection polynomial of a signed intersection graph SG to be the partial-dual Euler genus polynomial of the bouquet B with SG = SI ( B ). Could we19edefine the intersection polynomial for signed intersection graphs indepen-dent of the bouquets? The recursion in Theorem 14 is a try, but fail even forthe negative v . If the answer is negative, could we define a polynomial on amore larger set of signed graphs including all signed intersection graphs suchthat when we restrict ourself to a signed intersection graph it is exactly theintersection polynomial?We have characterized non-empty bouquets whose partial-dual genus poly-nomials have only one term. One can continue to try to characterize bouquetswhose partial-dual genus polynomials have exactly two terms.As we mentioned a little in the introduction, except the partial-dual(i.e. partial- ∗ ) Euler genus polynomial, there are partial- × , partial- ∗× ,partial- ×∗ and partial- ∗ × ∗ Euler genus polynomials [14]. For investiga-tion of the partial- • Euler genus polynomial, one can focus on bouquets if • ∈ {∗× , ×∗ , ∗ × ∗} and focus on quasi-trees (i.e. ribbon graphs with onlyone face) if • = × . Could we derive something from bouquets or quasi-treeswhich could determine the partial- • Euler genus polynomial completely?Finally we point that we find Theorem 27 is also obtained independentlyby Chumutov and Vignes-Tourneret in [5], an arXiv paper appeared aboutone week ago, but the proof is not completely the same.
Acknowledgements
This work is supported by NSFC (No. 11671336) and the FundamentalResearch Funds for the Central Universities (No. 20720190062).
References [1] B. Bollob´as and O. Riordan, A polynomial of graphs on surfaces,
Math.Ann. (2002) 81–96.[2] J. A. Bondy and U. S. R. Murty,
Graph theory , Springer New York ,2008.[3] A. Bouchet, Circle graph obstructions,
J. Combin. Theory Ser. B (1994) 107–144.[4] S. Chmutov, Generalized duality for graphs on surfaces and the signedBollob´as-Riordan polynomial,
J. Combin. Theory Ser. B (2009) 617–638. 205] S. Chumutov and F. Vignes-Tourneret, On a conjecture of Gross, Man-sour and Tucker, https://arxiv.org/pdf/2101.09319.pdf.[6] S. Chmutov and S. Lando, Mutant knots and intersection graphs, Algebr.Geom. Topol. (2007) 1579–1598.[7] Q. Deng, X. Jin and M. Metsidik, Characterizations of bipartite and Eu-lerian partial duals of ribbon graphs, Discrete Math. (2020) 111637.[8] F. M. Dong, K. M. Koh and K. L. Teo,
Chromatic polynomials andchromaticity of graphs , World Scientific, Singapore, 2005.[9] J. A. Ellis-Monaghan and I. Moffatt,
Graphs on Surfaces , Springer NewYork, 2013.[10] C. Godsil and G. Royle,
Algebraic Graph Theory , Springer, 2001.[11] H. Gong and M. Metsidik, Constructions of pairs of Tutte-equivalentgraphs,
Ars Combin. (2017) 223-234.[12] X. Guo, X. Jin and Q. Yan, Characterization of regular checkerboardcolourable twisted duals of ribbon graphs,
J. Combin. Theory Ser. A ,accepted.[13] J. L. Gross, T. Mansour and T. W. Tucker, Partial duality for ribbongraphs, I: Distributions,
European J. Combin. (2020) 103084.[14] J. L. Gross, T. Mansour and T. W. Tucker, Partial duality for ribbongraphs, II: Partial-twuality polynomials and monodromy computations.[15] J. L. Gross and T. W. Tucker, Topological Graph Theory , John Wiley &Sons, Inc. New York, 1987.[16] M. Metsidik and X. Jin, Eulerian partial duals of plane graphs,
J. Graph.Theory (2018) 509-515.[17] B. Mohar, C. Thomassen,
Graphs on Surfaces , Johns Hopkins UniversityPress, Baltimore, MD, 2001.[18] Q. Yan and X. Jin, Counterexamples to a conjecture by Gross, Man-sour and Tucker on partial-dual genus polynomials of ribbon graphs,
European J. Combin.93