Iain Moffatt- Partial Duals of Plane Graphs, Separability and the Graphs of Knots

8/3/2019 Iain Moffatt- Partial Duals of Plane Graphs, Separability and the Graphs of Knots

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PARTIAL DUALS OF PLANE GRAPHS, SEPARABILITY AND THE GRAPHS

OF KNOTS

IAIN MOFFATT

Abstract. The concept of the partial dual of an embedded graph was recently introduced byS. Chmutov. Partial duals of an embedded graph are obtained by forming the geometric dualwith respect to only a subset of edges of the graph. Partial duality is proving to have far-reachingconsequences. In this paper we provide a characterization of the set of embedded graphs that arepartial duals of plane graphs. We characterize partial duals of plane graphs in terms of the existenceof 1-sum decompositions of the embedded graph into two (not necessarily connected) plane graphs.This provides a connection between the genus of partial duals and separability.

The class of embedded graphs that are partial duals of plane graphs is important since thisis precisely the class of embedded graphs that describes knots and links. Our characterizationof partial duals of plane graphs provides a characterization of the class of embedded graphs thatrepresents link diagrams. We apply this characterization to find a move that relates all of the link

diagrams that are presented by the same ribbon graph.

1. Introduction

Here we are interested in the structure of those embedded graphs that are partial duals of planegraphs. Partial duality, introduced by Chmutov in [8], is an extension of the geometric or Euler-Poincare dual of an embedded graph. Roughly speaking, the partial duals of an embedded graph areobtained by forming the dual with respect to only a subset of edges of the graph (a formal definitionis given Subsection 2.3). Chmutov introduced partial duals in [8] order to unify the various recentrealizations of the Jones polynomial of a link as an evaluation of Bollobas and Riordan [3, 4] and LasVergnas [23] topological Tutte polynomial (see also [26]). These connections extend the well known

theorem of Thistlethwaite, from [29], relating the Jones and Tutte polynomials. The connectionsbetween graph theory and knot theory provide the main motivation for the results presented inthis paper.

Partial duality appears to be a fundamental operation on embedded graphs, and many results andapplications of duality extend to partial duals. For example, the classical relationship relating theTutte polynomials of a plane graph and its duals, T(G; x, y) = T(G; y, x) (and the correspondingresult for the topological Tutte polynomial [14, 25, 15]), extends to a partial duality relation forthe topological Tutte polynomial [8] (see also [27, 31]). The algebraic characterization of geometricduals due to Whitney for plane graphs [34], and Edmonds for higher genus embeddings [13] wasshown to extend to partially dual graphs in [28]. The classical connections between duals andmedial graphs were also shown in [16] to extend to analogous relations between partial duals (andmore generally twisted duals). In addition to the applications to knot theory discussed above andbelow, partial duality has been used in an essential way to construct a pseudo-tree expansion of thetopological Tutte polynomial [32]; and has also found applications in quantum field theory [22].

Date: July 27, 2010.2010 Mathematics Subject Classification. Primary: 05C10, 57M15, Secondary: 57M25, 05C75.Key words and phrases. Partial duality, checkerboard graph, dual, Tait graph, separability, 1-sum, embedded

graph, ribbon graph, knots and links, plane graphs. Department of Mathematics and Statistics, University of South Alabama, Mobile, AL 36688, USA;[email protected].

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In the formation of a partial dual, the geometric dual is only formed with respect to a subset ofedges. Graph theoretic and topological structures can therefore vary widely over the set of partialduals of an embedded graph. In particular, the genus of an embedded graph and one of its partialduals can be very different. (This contrasts with geometric duals where G and G have the samegenus.) In this paper we study embedded graphs that admit plane (i.e. genus 0) partial duals.It turns out that the question of whether or not an embedded graph has a plane partial dual is

intimately related to the separability of that graph.Embedded graphs that are partial duals of plane graphs are important since such graphs are

exactly the set of graphs that describe knot and link diagrams. It is this connection between knotand graph theory that motivates the study of partial duals of plane graphs presented here. We nowwill briefly outline this motivation.

There is a classical and well-known way to associate a (signed) plane graph, called the Tait graph(or checkerboard graphor 2-face graph), to the diagram of a link: start with a checkerboard colouringof a link diagram; place a vertex in each black face; add an edge between vertices correspondingto incident black faces; and sign the edges to record the crossing type. (See Subsection 2.2 andFigure 4.) The graph constructed in this way is a Tait graph. Tait graphs are a standard tool in aknot theorists toolkit, and have found numerous applications in knot theory and graph theory.

We start by considering the following well-known, basic properties of Tait graphs:

(T1) the two Tait graphs associated with a link diagram are geometric duals (where dualityswitches the sign of an edge);

(T2) every signed plane graph is the Tait graph of a link diagram;(T3) a Tait graph gives rise to a unique link diagram.

Recently, Dasbach, Futer, Kalfagianni, Lin and Stoltzfus extended the idea of a Tait graph in[10] by associating a set of (signed) embedded graphs to a link diagram. In this construction, eachembedded graph arises by assigning one of the two possible smoothings at each crossing of thelink. (See Subsection 2.2 for the construction of the embedded graph and and Figure 5 for anexample.) The Tait graphs of a link diagram appear in this set of embedded graphs. One of thekey advantages of using non-plane graphs to describe knots is that it provides a way to encode thecrossing structure of a link diagram in the topology of the embedded graph (rather than by usingsigns on the edges).

The construction of the embedded graphs of a link diagram is extremely natural and is provingto be very useful to knot theorists. There have been many recent applications of the embeddedgraph of a link diagram. These include applications to the Jones and HOMFLY-PT polynomials[6, 7, 8, 10, 26, 27, 32]; Khovanov homology [5, 10]; knot Floer homology [24]; Turaev genus[30, 24, 1]; Quasi-alternating links [35]; the coloured Jones polynomial [18]; the signature of a knot[12]; the determinant of a knot [10, 11]; and hyperbolic knot theory [17].

Given the number of applications of the embedded graphs of a link diagram to knot theory,understanding the structure of the set of embedded graphs of a knot is a fundamental and importantproblem. The work presented in this paper is motivated by this problem of understanding theembedded graphs of a link. In particular we answer the following three questions on the natural

extensions of Properties T1-T3 of Tait graphs to the embedded graphs of a link:(Q1) how are the ribbon graphs of a link diagram related to each other?(Q2) Which ribbon graphs arise as ribbon graphs of a link diagram?(Q3) What is the relation between links that have the same signed ribbon graph?

This paper is structured as follows. We begin with preliminary definitions and results on embed-ded graphs, Tait graphs, the embedded graphs of a link and partial duals in Section 2. This sectioncontains an answer to Question Q1 and reformulations of and discussions about Questions Q2 andQ3. Section 3 is concerned with separability, 1-sums and decompositions of embedded graphs.

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These results are central to answering Questions Q2 and Q3. In Section 4, we characterize partialduals of plane graphs in terms of the existence of a 1-decomposition of an embedded graph intotwo (not necessarily connected) plane graphs. This provides a new connection between the genusof partial duals and separability and answers Question Q2. Section 5 discusses how plane partialduals are related to each other, as well as how 1-decompositions of an embedded graph are relatedto each other. Finally, Section 6 provides the solution to Question Q3 by describing how link

diagrams that are presented by the same ribbon graph are related to each other.A reader who is interested only in the graph theoretical results, and not the applications to knot

theory, may opt to read only Sections 2.1, 2.3, 3, 4 and 5. This sequence of sections contains thecomplete graph theoretical results of this paper with only a few references to the knot theoreticalapplications.

I would like to thank Scott Carter, Sergei Chmutov and Martin Loebl for helpful conversations.Part of this work was undertaken while I was visiting the Department of Applied Mathematics(KAM) of Charles University, Prague. I would like to thank KAM for its hospitality.

2. The graphs of link diagrams

In this section we set up our notation and describe the knot theoretical and the graph theoretical

problems that we investigate here.

2.1. Embedded graphs. Although we will primarily work in the language of ribbon graphs, attimes we will find it convenient to use other representations graphs in surfaces. We will use the termembedded graph loosely to mean any of three equivalent representations of graphs in surfaces:

(1) a cellularly embedded graph, that is, a graph embedded in a surface such that every face isa 2-cell;

(2) a ribbon graph, also known as a band decomposition and a fatgraph;(3) an arrow presentation.

These three equivalent representations will arise naturally in different settings throughout the paper(for example, cellularly embedded graphs arise as Tait graphs, ribbon graphs arise as the embedded

graph of a link diagram, and arrow presentations arise as partial duals). Although we will need allthree of the above representations of an embedded graph, we will generally work in the languageor ribbon graphs which is the most convenient and natural setting for the majority of the resultsand constructions in this paper. Accordingly, the following exposition of embedded graphs will bebiased towards the language of ribbon graphs.

We assume the reader is familiar with cellular embeddings of graphs, but we briefly review ribbongraphs, arrow presentations, and the equivalences.

Definition 1. A ribbon graphG = (V(G), E(G)) is a (possibly non-orientable) surface with bound-ary represented as the union of two sets of topological discs: a set V(G) ofvertices, and a set E(G)of edges such that

(1) the vertices and edges intersect in disjoint line segments;

(2) each such line segment lies on the boundary of precisely one vertex and precisely one edge;(3) every edge contains exactly two such line segments.

Ribbon graphs are considered up to homeomorphisms of the surface that preserve the vertex-edge structure. (In particular, the embedding of a ribbon graph in three-space is irrelevant.) Wewill say that a ribbon graph is orientable if it is orientable as a surface. The genus of an orientableribbon graph is defined to be the genus of the ribbon graph as a punctured surface. We will saythat a ribbon graph is plane if it is orientable and of genus zero. This means that a plane ribbongraph is the neighbourhood of a plane graph.

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* * * *

.

Figure 1. Equivalence of ribbon graphs and 2-cell embeddings.

Two vertices and anedge disc

aligning the arrows identifying arcs

Figure 2. Constructing a ribbon graph from a arrow presentation.

Definition 2 (Chmutov [8]). An arrow presentationconsists of a set of circles, each with a collectionof disjoint, labelled arrows, called marking arrows, indicated along their perimeters. Each labelappears on precisely two arrows. Two arrow presentations are considered equivalent if one can beobtained from the other by reversing the direction of all of the marking arrows which belong to somesubset of labels, or by changing the labelling set. The circles are considered up to homeomorphism.

The equivalence of the three representations is given rigorously in, for example [19]; however itis intuitively clear. If G is a cellularly embedded graph, a ribbon graph representation results fromtaking a small neighbourhood of the embedded graph G. Neighbourhoods of vertices of the graphG form the vertices of a ribbon graph, and neighbourhoods of the edges of the embedded graph

form the edges of the ribbon graph.On the other hand, if G is a ribbon graph, we simply sew discs into each boundary component

of the ribbon graph to get the desired surface.At times we will consider ribbon graphs that are cellularly embedded in a surface. By the above,

cellularly embedded ribbon graphs are equivalent to ribbon graphs.A ribbon graph can be obtained from an arrow presentation by viewing each circle as the bound-

ary of a disc that becomes a vertex of the ribbon graph. In this paragraph we will call these vertexdiscs. Edges are then added to the vertex discs by taking a disc for each label of the markingarrows. In this paragraph we will call these the edge discs. Orient the edge discs arbitrarilyand choose two non-intersecting arcs on the boundary of each of the edge discs. Orient these arcsaccording to the orientation of the edge disc. Finally, identify these two arcs with two markingarrows, both with the same label, aligning the direction of each arc consistently with the orientation

of the marking arrow. This is illustrated in Figure 2.Conversely, every ribbon graph gives rise to an arrow presentation. To describe a ribbon graph G

as an arrow presentation, start by arbitrarily orienting and labelling each edge disc in E(G). Thisinduces an orientation on the boundary of each edge in E(G). Now, on the arc where an edge discintersects a vertex disc, place a marked arrow on the vertex disc, labelling the arrow with the labelof the edge it meets and directing it consistently with the orientation of the edge disc boundary.The boundaries of the vertex set marked with these labelled arrows give the arrow marked circlesof an arrow presentation. This is illustrated in Figure 3.

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2

3

1 1

2

1

2

3

.

Figure 3. Equivalence of arrow presentations and ribbon graphs.

We will say that a ribbon graph is plane is it is a surface of genus zero. Plane ribbon graphscorrespond to plane graphs. Because of this identification, we will often abuse notation and referto a (non-)plane ribbon graph simply as a (non-)plane graph. This abuse of notation shouldcause no confusion. Plane ribbon graphs are the central objects of study in this paper.

We are motivated by the embedded graphs of link diagrams. In general link diagrams aredescribed by signed embedded graphs rather than embedded graphs. A ribbon graph or a cellularly

embedded graph G is said to be signed if it is equipped with a mapping from its edge set E(G) to{+, } (so a sign + or is assigned to each edge of G). An arrow presentation is said to besigned if there is a mapping from the set of labels of the arrows to {+.}. The equivalence betweencellularly embedded graph, ribbon graphs and arrow presentation clearly extends to their signedcounterparts.

We will often want to consider embedded graphs that are obtained from signed embedded graphsby forgetting the signs. We will refer to such embedded graphs as being unsigned. At times we willalso use the term unsigned to emphasize that a ribbon graph is not equipped with signs. Thisdouble use of the term should cause no confusion.

2.2. The graphs of knots.

2.2.1. Tait graphs. We assume some familiarity with the basics of knot theory, referring the readerto [9] or any other of the many excellent introductory knot theory texts if he or she needs additionaldetails.

Recall that a link diagram is a 4-regular graph embedded on the sphere S2 with an under/overcrossing structure assigned to each vertex, together with a number of embedded simple closed curve.We will consider link diagrams up to isotopy of S2.

Let D S2 be a link diagram. By a checkerboard colouring of D, we mean the assignment of thecolour black or white to each region of D in such a way that adjacent regions have different colours.The Tait sign of a crossing in a checkerboard coloured link diagram is an element of {+, } whichis assigned to the crossing according to the following scheme:

+

.

A Tait graph T(D) is a signed plane graph constructed from D as follows: checkerboard colourthe link diagram, place a vertex in each black region and add an edge between two vertices wheneverthe corresponding regions ofD meet at a crossing. Weight each edge of the graph with the medialsign of the corresponding crossing. An example of a link diagram and its Tait graph is shown inFigure 4.

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+

-

-

-

+

++

Figure 4. Forming a Tait graph T(D) of a link diagram D.Since there are two possible checkerboard colourings of D, every diagram D has exactly two

associated (signed) Tait graphs. The following facts about Tait graphs are well known and readilyseen to be true.

(T1) the two Tait graphs associated with a link diagram are geometric duals;(T2) every signed plane graph is the Tait graph of a link diagram;(T3) a Tait graph gives rise to a unique link diagram.

For the first property, we use the convention that duality switches the sign of an edge, that is if G

is a signed embedded graph, e E(G) with sign then the edge in G

corresponding to e is of sign. For the third property, note that D is recovered from T(D) by equipping the medial graph ofT(D) with crossings according to the Tait signs.2.2.2. The ribbon graphs of a link diagram. In [10], Dasbach et. al. extended the concept of a Taitgraph by describing how a set G(D) of ribbon graphs can be associated to a link diagram D. Asmentioned in the introduction, the set G(D) of ribbon graphs of a link diagram has proved to bevery useful in knot theory. We will now describe the construction of the set G(D), however, wewill also sign the edges of our ribbon graphs using the Tait signs of the crossing.

Let D be a link diagram. Every crossing of D admits an A-splicing and a B-splicing as definedin the following figure:

A crossing in D. An A-splicing. A B-splicing.

The lines in the figure are called markers.A state ofD is an assignment of an A- or B-splicing to each crossing of D. A signof a splicing is

an assignment of + or to its marker. If the diagram D is equipped with a checkerboard colouringthen we may use the Tait sign of each crossing to assign a sign to each splicing according to thefollowing rules:

if the Tait sign is +, then the A-splicing has sign , and the B-splicing has sign +; if the Tait sign is , then the A-splicing has sign , and the B-splicing has sign +.

A signed state of D is a state of D where a sign is assigned to each marking using the above rules.

A signed ribbon graph Gs() is obtained from from a signed state of D by resolving all of thecrossings according to the signed states to obtain a set of non-intersecting closed plane curves withmarkers between the curves; assigning a distinct colour to each marker; and assigning colouredarrows to the plane curves according to the following scheme:

+/-

+/-

e

e

+/-

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-

+

+

+

-

-

-

+

-

-

--

++

A link diagram D A state s of D. The ribbon graph Gs(D).Figure 5. A link diagram D, a state s of D and the corresponding ribbon graph Gs(D).

(The directions of the arrows should be consistent with an orientation of the plane) This results inan arrow presentation for a signed ribbon graph. The corresponding ribbon graph is Gs().Definition 3 (Dasbach et. al. [10]). Let D be a checkerboard coloured link diagram. Then thesetGs(D) of signed ribbon graphs associated with D is defined to be

Gs(D) = {Gs() is a signed state of D}.IfG Gs

(D

)then we say that G is a signed ribbon graph of D, and we will also say that G presents

D.An example of the ribbon graph of a link is shown in Figure 5We note that the (un-signed) ribbon graphs constructed by Dasbach et. al in [ 10] are obtained by

forgetting the signs in the constructions above. We will let G(D) obtain this set of ribbon graphs,which is obtained from Gs(D) by forgetting the signs.

The following proposition states that Gs(D) is independent of the choice of checkerboard colour-ing used in its construction, hence we may associate Gs(D) with the link diagram itself rather thanthe checkerboard coloured link diagram. The proposition also shows that the ribbon graphs of alink diagram generalize the concept of the Tait graphs of a link diagram.

Proposition 1. LetD be a link diagram.

(1) Gs

(D

)is independent of the choice of checkerboard colouring used in its construction.

(2) Both of the Tait graphs of D are inGs(D).Proof. Item 1, is easily verified by noting that a choice ofA- or B-splicing results in the same signedstate at each crossing regardless of the checkerboard colouring at that crossing.

For Item 2, given a checkerboard colouring of the diagram, we can recover one of the Tait graphsby, at each crossing, choosing the splicing where the arcs follow the black faces of the diagram (sotake a B-splicing at each positive crossing, and an A-splicing at each negative crossing). The otherTait graph is recovered by following the white faces (so take an A-splicing at each positive crossing,and a B-splicing at each negative crossing).

Remark 1. For each diagram D, Gs(D) contains exactly two elements where all of the edges are ofa single sign, either + or . We denote these ribbon graphs by G+(D) and G(D) respectively. Theribbon graph G

(D

)is constructed from D by choosing the state that consists wholly ofA-splicings.

It then follows that, after forgetting signs, G is the ribbon graph associated with the all-A-splicingfrom [10]. Similarly, G+(D) corresponds to the all-B-splicing from [10]. As was observed in [26](using the language to be defined in Subsection 2.3) G and the ribbon graph associated with theall-A-splicing are obtained from a Tait graph of D by forming the partial dual with respect to theset of + edges; and G+ and the ribbon graph associated with the all-B-splicing are obtained froma Tait graph of D by forming the partial dual with respect to the set of edges. The two ribbongraphs G+(D) and G(D) associated with the all-A- and all-B-splicings play a key role in theapplications of the ribbon graphs of a link diagram to knot theory.

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2.2.3. Three motivating questions. We now come to the three fundamental questions on the ribbongraphs of link diagrams that motivate the study presented here. We saw in Proposition 1 thatthe ribbon graphs of a link generalize Tait graphs. An important problem is to understand howproperties of and results involving Tait graphs extend to ribbon graphs. Our motivation comesfrom the following three questions which ask how the three fundamental properties T1-T3 of Taitgraphs extend to ribbon graphs.

(Q1) How are the ribbon graphs of a link diagram related to each other?(Q2) Which ribbon graphs arise as ribbon graphs of a link diagram?(Q3) What is the relation between link diagrams that are presented by the same signed ribbon

graph?

Question Q1 has previously been answered by Chmutov [8]. It turns out that the ribbon graphs arepartial duals of each other. We will describe partial duality in Subsection 2.3 and we will explainthe relation between the ribbon graphs of links in the Subsection 2.4.

Turning Question Q2, the first thing to observe is that not every ribbon graph arises as theribbon graph of a link diagram. It is not hard to verify this claim: consider the set of all linkdiagrams with exactly three crossings and by construct the set of ribbon graphs that present them.We find that the ribbon graph G shown in the second example in Example 2 does not appear in

this set. Therefore G is not a ribbon graph of a link diagram. (As the ribbon graph has threeedges, it can only arise from a link diagram with three crossings.)

Observe that since the signs of the ribbon graph record only the under and over crossing structureof the corresponding link diagram, a signed ribbon graph arises from a link diagram if and onlyif its underlying unsigned ribbon graph arises from a link diagram. Thus it is enough to answerQuestion Q2 for unsigned ribbon graphs (i.e. we can work with G(D) rather then Gs(D)).

We will see in Section 4 that the ribbon graphs that present link diagrams can be characterizedin terms of separability. It is shown that a ribbon graph represents a link diagram if and only if itadmits a particular type of 1-sum decomposition into plane graphs, answering Question Q2 . (The1-sum decomposition is described in Section 3.)

As for Question Q3, it turns out that although a signed ribbon graph does not, in general,present a unique link diagram (unlike the case with Tait graphs), all of the link diagrams that aribbon graph can present are very closely related to each other. In Section 6 it is shown that all ofthe link diagrams presented by the same ribbon graph are related by a summand-flip move (seeDefinition 9).

The solutions to all three of the questions Q1-Q3 above are intimately related to partial duality ofribbon graphs. In the following subsection we describe partial duality. Equipped with this concept,in Subsection 2.4, we will reformulate questions Q2 and Q3 in the way that we will solve them.

Remark 2. The relation between the ribbon graphs of a link diagram was a primary motivationfor the introduction of partial duality by Chmutov in [8]. Similarly, Tait graphs and their relationto medial graphs were also the primary motivation for twisted duality, a generalization of dualityintroduced by Ellis-Monaghan and the author in [16]. IfG and H are embedded graphs and Gm andHm are their embedded medial graphs, then G and H must be Tait graphs of Gm and Hm. It thenfollows that Gm and Hm are equal as embedded graphs if and only if G and H are geometric duals.Thus we can take the point of view that the relation of equality as embedded graphs generatesthe relation of geometric duality. This point of view suggests that other concepts of equality ofembedded graphs will generate other concepts of duality. In [16] we show that equality as abstractgraphs generates twisted duality, and that equality as combinatorial maps generates partial duality.Also, [16] contains a way to construct the Tait graphs of non-checkerboard colourable 4-regularembedded graph.

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2.3. Partial duality. Partial duality, introduced by Chmutov in [8], is an extension of the conceptof geometric (or Euler-Poincare) duality of embedded graphs. Just as geometric duality related thetwo Tait graphs of a link diagram, partial duality relates the ribbon graphs of a link diagram.This provides an answer to Question Q1. Partial duality will also provide the framework we use toanswer Questions Q2 and Q3. As we would expect, many properties of geometric duality extendto partial duality [8, 16, 26, 27, 28, 31].

Roughly speaking, a partial dual of an embedded graph is obtained by forming the dual withrespect to a subset of edges of the graph.

Definition 4 (Chmutov [8]). Let G be a signed ribbon graph and A E(G). Arbitrarily orientand label each of the edges of G (the orientation need not extend to an orientation of the ribbongraph). The boundary components of the spanning ribbon sub-graph (V(G), A) of G meet theedges of G in disjoint arcs (where the spanning ribbon sub-graph is naturally embedded in G). Oneach of these arcs, place an arrow which points in the direction of the orientation of the edge andis labelled by the edge it meets. Associate a sign to each label in the following way: if e is a labelof an edge of G with sign , then the arrow labelled by e has sign if the edge is in A, and hassign otherwise. The resulting decorated boundary components of the spanning ribbon sub-graph

(V

(G

), A

)define an signed arrow presentation. The signed ribbon graph corresponding to this

signed arrow presentation is the partial dual GA of G.Example 1. The signed ribbon graph G equipped with and arbitrary labelling and orientation ofits edges is shown in Step 1. For this example we take A = {2, 3}. The marked spanning ribbonsubgraph (V(G), A) is shown in Step 2 (note the change of signs of the edges in A). The boundarycomponents of this spanning ribbon subgraph define a signed arrow presentation (shown in Step

3). The corresponding signed ribbon graph is shown in Step 4. This is the partial dual G{2,3} ofG.

1

3

+

+

Step 1. Step 2.

21 12

+

++

+

+

+

Step 3. Steps 4.

Additional examples of partially dual graphs can be found in [8, 27, 28, 31].For completeness we include the definition of the geometric dual of a ribbon graph. The geometricdual (or Euler-Poincare dual) G of a ribbon graph G is the ribbon graph obtained from G in thefollowing way: regard G = (V(G), E(G)) as a punctured surface. By filling in the punctures usinga set of discs denoted V(G), we obtain a surface without boundary . G is then the ribbongraph with vertex set V(G) and edge set E(G).

We will use the following basic properties of partial duality.

Proposition 2 (Chmutov [8]). Let G be a ribbon graph and A, B E(G). Then9


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(1) G = G;

(2) GE(G) = G, where G is the geometric dual of G;(3) (GA)B = GAB, where AB = (A B)(A B) is the symmetric difference of A and B;(4) G is orientable if and only if GA is orientable;(5) partial duality acts disjointly on connected components.

2.4. Partial duals and the graphs of links. We now have the language to answer question Q1,which asks how the (signed) ribbon graphs of link diagram are related to each other. The followingproposition, which appears implicitly in [8] and [26], tells us that all of the ribbon graphs in Gs(D)are partial duals. This should be compared with the property T1 which states that Tait graphs aregeometric duals.

Proposition 3 (Chmutov [8]). LetD be a link diagram. Then the signed ribbon graph G presentsD (i.e. G Gs(D)) if and only if G is a partial dual of a Tait graph of D.Proof. In terms of arrow presentations, taking the partial dual with respect to an edge e can bedescribed as the following change in the arrow presentation: suppose and are the two arrowslabelled e in the arrow presentation. Draw a line segment with an arrow from the the head of to the tail of , and a line segment with an arrow from the head of to the tail of . Label bothof these arrows e, and delete and with the arcs containing them. Switch the sign of the edge.This is illustrated in the following figure:

e e

e

e

.

This is readily seen to correspond to the change in the state of a link diagram cased by switchingbetween A- and B-splicings at the crossing corresponding to e.

Corollary 1. G G(D) if and only if G is a partial dual of an unsigned Tait graph of D.Corollary 2. Any two embedded graphs inG(D) (orGs(D)) are partial duals.Corollary 3. G is the (signed) ribbon graph of a link diagram if and only if it is a partial dual ofa (signed) plane graph.

Corollary 3 provides following graph theoretical formulation of question Q2:

(Q2) Which ribbon graphs are the partial duals of plane graphs?

We will answer this question, and hence question Q2, in Section 4. We will give a characterizationof the ribbon graphs that are partial duals of plane graphs in terms of separability. The contrastbetween property T2, which states that every plane graph is a Tait graph, and the more complicatedsolution to question Q2 should be noted.

Having found a workable reformulation of Question Q2, let us turn our attention to question Q3.In order to address this question we need to understand how, if we are given a signed ribbon graph

G, we can recover the link diagrams that have G as their signed ribbon graph. We will begin withthe special case where G is a plane ribbon graph.

Let G be a signed plane ribbon graph. Then we can construct a link diagram on G by drawingthe following configuration on each signed edge of G and connecting the configurations by followingthe boundaries of the vertices of G:

+

,-

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Since G is topologically a punctured sphere, by capping off the punctures ofG we can obtain a linkdiagram D(G) on the 2-sphere. We will call D(G) the link diagram associated with D.

Clearly, the two Tait graphs of D(G) are G and G, and D(T(D)) = D.Now suppose that G is a (not necessarily plane) signed ribbon graph and that G is the ribbon

graph of some link diagram. We want to find the set D(G) of link diagrams that have G as a signedribbon graph. We know that by the construction of

Gs(D), G represents a link diagram D if andonly if G is the partial dual of a Tait graph of D. Since every signed plane graph is a Tait graph

of a link diagram, it follows that

(1) D(G) = {D(GA) A E(G) and GA is plane}.Notice that although there is a unique link diagram associated with a Tait graph, in general a

ribbon graph can have many plane partial duals, and therefore D(G) can contain more than onedistinct link diagram. However, we will see in Section 6, that all of the link diagrams in D(G) notonly represent the same link, but are very closely related. This result will provide an answer toquestion Q3.

3. 1-sums of ribbon graphs

In this section we discuss separability and 1-sums of ribbon graphs. These operations are thenatural extension of the corresponding operations for graphs to t ribbon graphs. We will go onto define 1-decompositions of ribbon graphs into two ribbon graphs. We will see in the followingsection that 1-decompositions provide a characterization of ribbon graphs that have plane partialduals. This provides a connection between the genus of a partial dual and the structure of a ribbongraph.

3.1. Decompositions of embedded graphs. In this subsection we will discuss the separationof ribbon graphs and its use in obtaining decompositions of ribbon graphs.

Just as with graphs, a decomposition of a ribbon graph G is a family F of edge-disjoint ribbonsubgraphs of G such that FFE(F) = E(G).

A separation of a connected ribbon graph G is a decomposition of ribbon graph into two non-

empty connected ribbon subgraphs P and Q which have exactly one vertex in common. If v is thevertex common to the two subgraphs, we will say that the separation occurs at v, and that v is aseparating vertex.

We will say that a connected ribbon graph G is a 1-sum of ribbon graphs P and Q, writtenG = P Q, if there is a separation of G into ribbon graphs P and Q. If the separation occurs at avertex v, we will say that the 1-sum occurs at v.

While it would be more conventional to use the symbol 1 to denote the 1-sum operation,here we use the symbol . This is in order to reduce the number of subscripts in the expressionsthat to follow.

Observe that the definition of the 1-sum given above is consistent with the usual definition of1-sums of graphs from the literature since G = P Q if and only if P Q = K1 E(K1) = K1 andG = P Q E

(K1

)= P Q. Also note that a ribbon graph G = P Q if and only if this identity

holds for the underlying abstract graphs ofG, P and Q. It is perhaps also worthwhile emphasizingthat a 1-sum is not a join of ribbon graphs (see Section 5 for a definition of a join), although a joinis a 1-sum.

It is important to keep in mind that if we take two different separations of G into subgraphs P1and Q1, and into P2 and Q2, say, where the separation is at the same vertex (so P1 Q1 = P2 Q2),then the topological properties of P1 and Q1, and P2 and Q2 can be quite different. For example,the genus 2 ribbon graph with one vertex and four edges can be separated into two subgraph ofgenus 2, or two subgraphs of genus 0, or one subgraph of genus 2 and one of genus 0. Here will be

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particularly interested in the existence separations of a ribbon graph into plane ribbon subgraphs.Also note that, non-equivalent ribbon graphs can be separated into equal ribbon graphs. Forexample, there are two ribbon graphs with one vertex and two edges - one plane and the other ofgenus one - but each of these ribbon graphs admits exactly one separation into two ribbon graphswith one vertex and one edge.

Here we are interested in using separation to obtain a decomposition. In our decompositions, wewill insist that at most one separation occurs at any vertex of the ribbon graph.

Definition 5. Let G be a ribbon graph. We say that G has a 1-decomposition into two graphs if

(1) G has a decomposition into two (not necessarily connected and possibly empty) ribbonsubgraphs P and Q;

(2) each vertex that is incident to edges in both P and Q is a separating vertex of the connectedcomponent of G on which it lies.

In addition, if P and Q are plane, we say that the 1-decomposition is a 1-decomposition into twoplane graphs.

If A E(G), then we say that A defines a 1-decomposition into two graphs if and only if G hasa 1-decomposition into two graphs and A is the edge set of one of these graphs.

Example 2. The following two genus 1 ribbon graphs admit 1-decompositions into two graphs.Observe that that example on the left admits a 1-decomposition into two plane graphs, while theexample on the right does not.

G =

= P

=Q

G =

= P

= Q

Note that the 1-decompositions in the above examples were separations. We will now give twoexamples that illustrate more complicated 1-separations into two plane graphs. The example belowgives a 1-decomposition of the signed ribbon graph Gs(D) = G from Figure 5.

+

-

-

--

++

--

+

+

-

-

+

G=

= P

= Q

In the example below, a 1-decompositions of G into two plane graphs P and Q shown. Notehowever that G admits several 1-decompositions into plane graphs P and Q, where P and Q are

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not isomorphic to the graphs P and Q shown below.

G =

=

= Q

Additional examples of 1-decompositions of the above graph are given in Example 7.

We will now describe a few of the basic properties of 1-decompositions that we need later. Theseproperties are analogies of standard results about separable graphs.

Proposition 4. LetG be a ribbon graph that has a 1-decomposition into the two graphs P and Q.Then

(1) each cycle of G is contained in a connected component of P or Q;(2) G is orientable if and only if P and Q are orientable.

Proof. A cycle ofG does not contain a separating vertex and so it must be contained in a connectedcomponent of P or Q.

If either P or Q is non-orientable, then G contains a non-orientable subgraph and is thereforenon-orientable. Conversely, if G is non-orientable then G contains a cycle that is homeomorphic toa Mobius band. By the first property, this cycle is contained in one of the connected componentsof P or Q. Therefore one of P or Q is non-orientable.

Now suppose that G is a connected ribbon graph that has a 1-decomposition into two graphs Pand Q. We will associate a vertex-labelled graph T(G,P,Q) to the 1-decomposition. For the vertex

set ofT(G,P,Q), take one vertex for each connected component of P and of Q, and label the vertexby the corresponding connected component. For the edge set, assign one edge between two distinctvertices if and only if the connected components labelling the vertices meet at a single vertex in G.

Let T(G,P,Q,) denote the rooted graph obtained from T(G,P,Q) by distinguishing a vertex. Wethen define the height of a vertex of T(G,P,Q,) to be the number of edges in the shortest pathbetween the given vertex and the root (we will see below that T(G,P,Q) is connected and so thisdefinition is sound). This assigns a height to each connected component of P and Q since eachconnected component labels a unique vertex of T(G,P,Q,). Finally by the height of T(G,P,Q,), wemean the maximum of the set of heights of its vertices.

Example 3. This example provides an illustration of the construction of the rooted graph T(G,P,Q,)associated with the 1-decomposition given in the last example of Example 2. The root is taken to

be the P1 labelled component of P.P1

P2

P3

Q1

Q2

Q3

P4

P1

P2P3

Q1

Q2 Q3

P4

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Proposition 5. Suppose that G is a connected ribbon graph that has a 1-decomposition into twographs P and Q. LetT(G,P,Q) be the graph associated with this 1-decomposition and T(G,P,Q,) bethe height h rooted graph associated with this 1-decomposition. Then the following properties hold.

(1) T(G,P,Q) is a tree.(2) If the root is a connected component of P, then a vertex of T(G,P,Q,) is of odd height if

and only if its label is a component of Q; and is of even height if and only if its label is acomponent of P.

(3) If the root is a connected component of Q, then a vertex of T(G,P,Q,) is of odd height ifand only if its label is a component of P; and is of even height if and only if its label is acomponent of Q.

(4) If{Ri,j j = 1, . . . ki} denotes the set of connected components of P and Q of height i, thenG =

h

i=0

ki

j=1

Ri,j

and each 1-sum in this expression acts on a different vertex.

Proof. (1) A path in T(G,P,Q) corresponds to a path in G between the connected components of thesubgraphs of P and Q. It then follows that T(G,P,Q) is connected since since G is connected, andthat T(G,P,Q) is acyclic since, by Proposition 4 all of the cycles in G are contained in the connectedcomponents of P and Q.

(2) & (3) If Ri and Rj are connected components of P or Q that label two vertices in T(G,P,Q)that are connected by an edge, then the components Ri and Rj meet at a single vertex in G. Bythe first property of Definition 5, it follows that one of the components, Ri or Rj, must belong toP and the other to Q. So if two vertices in T(G,P,Q) are adjacent then one must be labelled by aconnected component of P and the other by a connected component of Q, the results follow.

(4) G can be recovered from the tree T(G,P,Q,) as follows: start with the height 0 componentR0,1, call this G0. Given Gi, construct Gi+1 by 1-summing the height i + 1 components to the Ri,jcomponents Gi. Each Ri+1,j is summed to a component of Ri,j in Gi, and since the 1-sums acton disjoint components, these 1-sums may be carried out in any order. This process ends with Gh

which is equal to G. Thus we have G = Gh =hi=0 kij=1 Ri,j as required. 4. A characterization of the partial duals of plane graphs

In this section we state and prove our main theorem. This theorem characterizes ribbon graphsthat admit plane partial duals in terms of the existence of a 1-decomposition into two plane graphs.

Theorem 1. LetG be an embedded graph and A E(G). Then GA is a plane graph if and only ifA defines a 1-decomposition of G into two plane graphs.

The proof of the theorem is given in Subsections 4.1 and 4.2. We first state a few corollaries.Our first corollary obviously follows immediately from the theorem.

Corollary 4. An embedded graph G is a partial dual of a plane graph if and only if there exists a1-decomposition of G into two plane graphs.

With these results we can answer Question Q2, which was one of our main motivational problems.

Corollary 5. A connected (signed) embedded graph G represents a link diagram if and only if thereexists a 1-decomposition of G into two plane graphs.

We will apply this characterization in Section 6 to answer the final motivating problem, Ques-tion Q3, by relating all link diagrams that are presented by the same ribbon graph.

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4.1. Proof of the if part of Theorem 1. We begin our proof of Theorem 1 by proving thefollowing lemma. This lemma will allow us to reduce the height of a 1-decomposition into planegraphs, by taking the partial duals of the components of maximum height. This will be used in theinductive proof of the if part of Theorem 1.

Lemma 1. Let G be a ribbon graph such that G = P Q, where Q is a plane ribbon graph. Then

the genus of GE(Q)

is equal to the genus of P.In proving this lemma, the language of cellularly embedded ribbon graphs is particularly conve-

nient. As with graphs, we will say that an embedding of a ribbon graph G in a closed surface iscellular if each component of G is a disc. Note that since a ribbon graph is a punctured surface,we can obtain a cellular embedding of any ribbon graph by capping off the punctures. Also notethat a ribbon graph is of genus g if and only if it admits a cellular embedding into a surface ofgenus g.

We will prove the lemma by starting with a cellular embedding of P in a surface and usingthis embedding to construct a cellular embedding of GE(Q) = (P Q)E(Q) in the same surface .

While reading the the proof of the lemma below, the reader may find it helpful to consult Figure 6where the idea behind the proof is illustrated by an example.

Proof. It is enough to prove the lemma in the case where G is connected, so without loss of generalityassume that G is connected.

Suppose that v is the separating vertex of G that partitions G into P and Q, and that vP and vQare the vertices in P and Q respectively corresponding to the vertex v. So G = P Q is obtainedfrom P and Q by identifying vP and vQ. Let vQ vP denote the identifying map, and let ,denote the restriction of to the boundary.

Now cellularly embed P into a surface , and embed Q into the 2-sphere S2 in such a way thatvQ consists of the Southern hemisphere of S

2 with the equator of S2 is the boundary of vQ. (Theconditions on the embedding involving vQ are merely for convenience.) (For an example of theseembeddings see the top row of Figure 6.)

The partial dual GE(Q) = (P Q)E(Q) can be formed from the embeddings of P and Q in thefollowing way.

(1) Form the dual Q S2 of Q S2.(2) Delete the vP P . Also delete the Southern hemisphere of S

2 to obtain an embeddingof Q in a disc D2. (Note that Q

lies entirely in the Northern hemisphere of S2.)(3) Identify the boundaries of D2 and vP using . (Note the boundaries of vP and vP

are equal, and the boundaries of D2 and vQ are equal)

(For an example of see the bottom row of Figure 6.) This results in a ribbon graph cellularly

embedded in , and the resulting ribbon graph is readily seen to be the partial dual GE(Q) =(P Q)E(Q) (since the process above forms the dual at exactly the edges of G that belong to Q).Finally, since P and GE(Q) can both be cellularly embedded in the surface , they must be of thesame genus.

The idea behind the if part of the proof is as follows. We are given a 1-decomposition ofG intoplane graphs P or Q. Suppose that the decomposition is of height h and Si is the set of edges of Gbelonging to all of the components of the decomposition of height at least i. Then, by Lemma 1, itfollows that GSh admits a 1-decomposition of height h 1 into two plane graphs. By repeating this,it follows that H = (((GSh)Sh1))S1 admits a height 0 1-decomposition into two plane graphs,i.e. H is a plane graph. It then follows by Propositions 2 and 5 that H = GE(P) or H = GE(Q),proving the result. This argument is formalized in the following inductive proof.

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v

P

vQ

G = P Q P Q

Q GE(Q) Redrawing GE(Q)

Figure 6. An example of the construction used in the proof of Lemma 1

Proof of the if part of Theorem 1. Firstly, since partial duality acts disjointly on disconnectedcomponents (by Proposition 2), it is enough to prove the result for connected graphs. So, withoutloss of generality, assume that G is connected.

We will prove the if part of the theorem by induction on the height h of a tree associated withthe 1-decomposition of G.Base case: Suppose that G has a 1-decomposition into two plane graphs P and Q, and that a treeassociated to 1-decomposition has height h = 0. Then either P or Q is an empty graph. Withoutloss of generality assume that Q is empty. Therefore G = P and since P is plane, G plane. ThusGE(Q) = G = G and GE(P) = GE(G) = G are both plane.

Inductive hypothesis: Assume that if G has a a 1-decompositions of height h 1 into two planegraphs P and Q, then GE(Q) and GE(P) are both plane graphs.Inductive step: Let G = P Q be a 1-decomposition of height h, where P and Q are plane graphs.Let T(G,P,Q,) be a rooted tree associated with this 1-decomposition.

Let {Ri,j j = 1, . . . ki} denote the set of connected components of P and Q of height i, so that,by Proposition 5,

G =h

i=0

ki

j=1

Ri,j .

For simplicity, we will assume, without loss of generality, that the rooted vertex R0,1 is a connectedcomponent of P. Then by Proposition 5 The edges of Q are the components of odd numberedheight, so

(2) GE(Q) =

h

i=0

ki

j=1

Ri,jB

where B is set of edges of all of the components of odd height:

B =

h+12

i=1

ki

j=1

E(R2i1,j) .16


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We will partition (and relabel) the components of height h according to the components of heighth 1 as follows: let Sj = {Sj,p p = 1, . . . lj} denote the set of components of height h that are1-summed to Rh1,j. In addition, let S= {E(Rh,j) j = 1, . . . kh} be the set of edges of componentsof height h is the 1-decomposition.

With this notation we have

GS = hi=0

ki

j=1

Ri,jS

= h2i=0

ki

j=1

Ri,jkh1

j=1

Rh1,jkh

j=1

Rh,j

=

h2

i=0

ki

j=1

Ri,j

kh1

j=1

Rh1,j

lj

p=1

Sj,pS

.

Since each Sj,p meets Rh1,j at a distinct vertex and meets no other vertex of G Sj,p, we canwrite the above expression as

h2

i=0

ki

j=1

Ri,j

kh1

j=1

Rh1,j

lj

p=1

Sj,p

E(Sj) ,

where, for simplicity, we abuse notation and use E(Sj) to denote ljp=1 E(Sj,p), the set of all edgesof the ribbon subgraphs in Sj. Now by setting

Rh1,j =

Rh1,j

lj

p=1

Sj,p

E(Sj)

,

we can write

(3) GS =h2

i=0

ki

j=1

Ri,jkh1

j=1

Rh1,j.

Each

Rh1,j is plane by Lemma 1. From this we deduce that Equation 3 defines a 1-decomposition

of GS into two plane graphs (where one plane graphs consist of all of the components of oddheight, and the other of all of the components of even height). (Note that this uses that fact thatEquation 2 defines a 1-decomposition of G.)

It then follows from the inductive hypothesis and Proposition 5 that (GS)B is plane, whereB =

h12

i=1 kij=1 E(R2i1,j) if h 1 is evenh22i=1 kij=1 E(R2i1,j)kjj=1 E(Rh1,j) if h 1 is odd

is the set of components of odd height in the 1-decomposition for GS.To complete the proof, we need to show that (GS)B = GB. To do this we first note that by

Proposition 2

(GS

)B = GSB

where is the symmetric difference. Then if h is odd

SB = {E(Rh,j) j = 1, . . . , kh}h12

i=1

ki

j=1

E(R2i1,j)=

h+12

i=1

ki

j=1

E(R2i1,j) = B.If h is even, then

SB = {E(Rh,j) j = 1, . . . kh}

h22

i=1

ki

j=1

E(R2i1,j)

kj

j=1

E(Rh1,j)

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since kjj=1 E(Rh1,j) = {E(Ri,j) i = h 1, h, and j = 1, . . . , ki}

SB =

h22

i=1

ki

j=1

E(R2i1,j)

kj

j=1

E(Rh1,j) = B

Thus GB= (G

S

)B

and so GE(Q)

is plane. Also (GE(Q)

)

= GE(Q)

)E(G)

= GE(Q)E(G)

= GE(P)

isplane. This completes the proof.

4.2. Proof of the only if part of Theorem 1. We will need to introduce a certain amountof terminology for the proof. We begin with a graph G which is cellularly embedded in the sphereS2, and a set of edges A E(G). An arrow presentation for the partial dual GA can be obtained byconstructing a set of cycles that follow the boundaries of a small neighbourhood of G Ac, whereAc = E(G) A, and adding coloured arrows as indicated:

e

e

e

e

e

An edge in G . If e A. If e A.

.

Notice that, in this construction, the set of cycles that form the vertices of GA follow the boundariesof the faces of the (not necessarily cellularly) embedded graph G Ac. (Here the embedding ofG Ac is obtained from the embedding ofG by deleting the edges in A.) We will call the embeddedcycles in this set the vertex cycles. It is clear from the construction that for each colour of themarking arrows, we can embed a straight line, called a marking line, between the two markingarrows of the same colour as shown:

e

e

e

e

If e A. If e A.

.

There is a natural correspondence between the marking lines and the edges of G and the edges ofGA. We will say that a marking line is in A (respectively Ac) if it corresponds to an edge in A(respectively Ac) . We will refer to the set of vertex cycles equipped with the marking lines as thecycle diagram for GA. The cycle diagram is embedded in the sphere, and the vertex cycles dividethe sphere into disjoint regions which we will refer to as the regions of the cycle diagram. Tworegions are adjacent if they share a common boundary. Note that each marking line is embedded ina region. An example of these definitions and an illustration of their use in the proof of Theorem 1is given below.

Example 4. In this example, we provide an example to illustrate the ideas behind the proof of theonly if part of Theorem 1. For this example, we consider the embedded graph G shown below.In the figure the edges in Ac are dotted, and the edges in A are solid. GA and the cycle diagramfor GA are also shown below. (The arrows in the cycle diagrams are omitted for clarity.)

G. The cycle diagram for GA. GA.18


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Observe that in the cycle diagram for GA, the marking lines in A and the marking lines in GA lie indifferent regions. Also if two marking lines line are in adjacent regions then one belongs to A andthe other to Ac. It then follows that a vertex cycle that meets lines in both A and Ac correspondsto a separating vertex of GA. (These observations will be shown to hold in the general case inLemma 2.) Thus all of the edges in A, say, define a 1-decomposition into two graphs. Moreover,the each component of the graph must be plane because of the embedding of the cycle diagram in

S2:

The 1-summands arising from A. The 1-summands arising from Ac.

Thus we have obtained a 1-decomposition into two plane graphs of the partial dual GA of the planegraph G. This completes the example and the illustration of the ideas behind the proof of the onlyif part of Theorem 1.

Lemma 2. LetG be a connected graph that is cellularly embedded in the sphere, and let A E

(G

).

(1) If two marking lines lie in the same region of the cycle diagram for GA then they both belongto A or to Ac.

(2) If two marking lines lie in adjacent regions of the cycle diagram for GA then one belongs toA and the other belongs to Ac.

(3) If C is a vertex cycle in the cycle diagram which meets marking lines that belong to both Aand to Ac, then the vertex in GA corresponding to C is a separating vertex.

Proof. (1)& (2) We will prove these properties simultaneously using induction on the number ofedges in A. If A is the empty set then a vertex cycle bounds each vertex of G. Each vertex cyclebounds two regions one region contains a vertex of G and no marking lines, while the other regionmeets every marking cycle and contains marking lines in Ac. The results are then seen to be true.

Now suppose that that results stated as items (1) and (2) are true for all subsets of edges of size

m. Let A = m + 1 and e A. Then, by the inductive hypothesis, the cycle diagram for GA{e}satisfies the conditions in the lemma. Moreover, the cycle diagram for GA is obtained from thecycle diagram for GA{e} by the following change:

Ac

Ac

A

A

Ac

Ac

A

.

(The vertices shown in the figure need not be distinct.) In the figure above, the regions containingthe marking lines in A and the marking lines in Ac in a neighbourhood of the edge e is shown.From the figure and the inductive hypothesis, we see that a region of the cycle diagram of GA onlycontains A or Ac marking lines and adjacent regions contain marking lines from different sets out

of A and Ac

.(3) Consider the cycle diagram for GA. Since each vertex cycle C is an embedding of a circle inthe plane it partitions the plane into two discs. Call one of these discs the interior of C and theother the exterior of C.

Now let C be a vertex cycle in the cycle diagram which meets marking lines that belong to bothA and and to Ac. First of all we show that the vertex of GA corresponding to C is a separatingvertex of G. To show this, let P denote C together with the set of marking lines and vertex cyclescontained in the exterior of C, and let Q denote C together with the set of marking lines and

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vertex cycles contained in the interior of C. The sets P and Q correspond to subgraphs P and Qrespectively of GA. We will show the subgraphs P and Q define a separation of GA.

First we show that P and Q have only one vertex in common. Since the vertex cycles do notintersect, a vertex cycle is either C, in the interior of C, or in the exterior of C, and since themarking lines are embedded, each marking line is contained in the interior of C or the exterior ofC. This means that there are no marking lines between a vertex cycle in the interior and a vertex

cycle in the exterior of C. Therefore, the subgraphs P and Q of GA can only meet at the vertexcorresponding to C.

We also need to show that P and Q are not empty graphs. To do this it is enough to show thatthat the sets P and Q each contain a marking line. To see why this is, note that C meets markinglines in A and in Ac and, by property (2) of the lemma, each type or marking line must lie ondifferent sides of C so there are marking lines in the interior of C and the exterior of C.

Proof of the only if part of Theorem 1. Since partial duality acts disjointly on connected com-ponents, it is enough to prove the result for connected embedded graphs.

Let G be a connected cellularly embedded graph in the sphere and let A E(G). Let P denotethe subgraph ofGA determined by the edges in A and their incident vertices, and let Q denote the

subgraph of GA corresponding to the edges in Ac and their incident vertices. We will show thatP and Q define a 1-decomposition of GA into the two plane graphs. We will do this by lookingat the structures in the cycle diagram for GA corresponding to P and Q and to their connectedcomponents.

First observe that P and Q clearly define a decomposition of GA. We will now show that this isa 1-decomposition.

Let v be a vertex in GA that is incident to edges in both P and Q. Then v corresponds to avertex cycle Cv in the cycle diagram for G

A that meets marking lines belonging to both A and Ac.Therefore, by Lemma 2, v is a separating vertex of G. It follows that A defines a 1-decompositionof GA.

Having just shown that A defines a 1-decomposition into two graphs P and Q. It remains toshow that P and Q are plane graphs. We will show that P is plane, the fact that Q is planefollows similarly. Let Pi be a connected component of P. Let Pi denote the sub-diagram of thecycle diagram of GA that corresponds to the vertices and edges in Pi. Since all edges in A mustlie on a single side of a cycle in the cycle diagram of GA, it follows that Pi must consist of a setof embedded vertex cycles which lie inside one vertex cycle together with embedded marking linesbetween the vertex cycles. This set of vertex cycles and marking lines is easily seen to correspondto a plane graph.

5. Joins and partial duals of plane graphs

In this section we will be examine 1-decompositions of plane graphs and also how different 1-decompositions of the same ribbon graph are related to each other. The central construction usedin this study is the join operation of ribbon graphs. The join operation is a simple, special case ofthe 1-sum operation. We will see that every 1-decomposition of a plane graph can be expressed interms of joins, rather than the more general 1-sums. We will also see that 1-decompositions of thesame ribbon graph can only differ by moving join-summands between the subgraphsP and Q.

5.1. Joins and join-decompositions of plane graphs. We begin with the definition of the joinof two ribbon graphs.

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Definition 6. Let H1 and H2 be ribbon graphs and, for i = 1, 2, let vi V(Hi), and i be an arcon the boundary of vi such that i E(Hi) = . Then the join) H1 H2 of H1 and H2 is a ribbongraph obtained by identifying the arcs 1 and 2.

H1 and H2 are called the join-summands of H1 H2.

Example 5. An example of the join of two ribbon graphs is given below.

1 2

H1 and H2 with arcs 1 and 2. The join H1 H2.

The join operation for ribbon graphs is also known as the one-point join and the connectedsum in the literature.

We will make a few elementary observations about the join operation. Firstly, note that whileevery join of ribbon graphs is a 1-sum, 1-sums need not be joins. Secondly, observe that there is acertain amount of ambiguity in the notation H1 H2. The suppresses the choice of the arcs 1and 2 used in the formation of the join. Different choices of orientation of the arcs (and of coursedifferent choices of position on the vertices) used in the formation of the join can result in non-isomorphic ribbon graphs. This ambiguity in the notation, however, will cause no problemsfor us here.

It is also worth emphasizing the following complication of the join of ribbon graphs that does notarise with join of abstract graphs. If a ribbon graph G can be written as (H1 H2) H3 where H1,H2 and H3 all meet at the same vertex of G, in general the order of the join can not be changed,that is we may not be able to write G as (H1 H3) H2. This is since it is possible that the joininvolving H3 identifies arcs on H2 and H3, rather than H1 and H3. Consequently, when we use theconvention that when we write ki=1 Ri, we mean (((R1 R2) R3) Rk1) Rk.5.2. Joins and join-decompositions. As mentioned above, joins are a special case of 1-sums.We will see that every 1-sum into two plane graphs must be a join. As we will see, this means that

1-decompositions of plane graphs have a particularly simple structure.The following result tells us that every 1-decomposition of a plane graph can be expressed in

terms of join of plane graphs.

Theorem 2. LetG be a plane ribbon graph with a 1-decomposition into two (plane) ribbon graphsP and Q. Then

G =k

i=1

Ri,

where each subgraph Ri of G is a distinct subgraph of P or of Q, and G = ki=1 Ri.

Proof. We will prove the result by induction on the number of separating vertices in the 1-decomposition.

Base case: First assume that there are no separating vertices. In this case one of the ribbonsubgraphs in the 1-decomposition is trivial and we set G = R1.Inductive hypothesis: Assume that the result holds for all ribbon graphs with fewer than n sepa-rating vertices.Inductive step: Let v be a separating vertex. Then v meets edges in P and edges in Q. Lets = (e1, . . . , ep) denote the cyclic order of the edges of G that are incident to v (with respect to anarbitrary orientation of v). Since G has a 1-decomposition, this sequence of edges must contain asubsequence s = (eq, . . . , er) with the property that all of the edges in s belong to P, and that

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eq1 and er+1 belong to Q. (The indices are taken modulo p.) Cut the vertex v into two verticescalled v (again) and v in such a way that v incident to the edges eq, . . . , er, and v is incident tothe remaining edges er+1, . . . , eq1. Since G is plane, v and v

lie in different connected componentsof the graph that has been created by the cutting the vertex. Let H1 be the component containingv and H be the component containing v. We have G = H1 H. Repeat this process until v isonly incident to edges in Q. Let H1, . . . , H s be the components created by this process. We have

G = H1 Hs. Then by the inductive hypothesis, Hj = ji=1 Ri where the Ri are subgraphs of Gbelonging either of P or of Q. Finally, G = H1 Hs = (1i=1 Ri) (si=1 Ri), and the resultfollows by reindexing the join-summands.

If a ribbon graph G has an expression of the form G = ki=1 Ri, we will say that G admits ajoin-decomposition. The expression ki=1 Ri being the join-decomposition.

Example 6. This example provides an illustration of a join-decomposition. The original ribbongraph is shown on the left and a join-decomposition for the ribbon graph is shown on the right.The arrows indicate the positions of the joins.

The following corollary of Lemma 2 provides some useful additional structure of the join-decompositions of a ribbon graph G that admits a 1-decomposition into two plane graphs.

Corollary 6. Let G is a connected plane graph that admits a 1-decomposition into subgraphs Pand Q. The G admits a join-decomposition of the form

G =h

i=0

ki

j=0

Ri,j ,

where

(1) if Ri,j is a subgraph of P (respectively Q) , then Ri+1,k is a subgraph of Q (respectively P),for 0 i < h.

(2) For each 1 l h, Rl,j is joined to l1i=0 kij=0 Ri,j by identifying an arc on Rl,j with an

arc on R(l1),k, for some k.

Proof. By Lemma 2, we have G = ki=1 Ri, where each subgraph Ri of G is a distinct subgraph ofP or ofQ, and G = ki=1 Ri. All joins will be formed with respect to the set of identifying arcs usedin this expression.

Set H = R1, and S = {R2, . . . , Rk}. Without loss of generality, assume that R1 belongs to P. Ifany ribbon graph in S that belongs to P can be joined to H, do the join, denoting the resultingribbon graph also by H, and delete the ribbon graph from S. Continue to do this for as long as

possible. This gives the expression H= 0i=0 kij=0 Ri,j.Suppose now that H = l1i=0 kij=0 Ri,j, and that R(l1),j belong to P (respectively Q). If any

ribbon graph in S belongs to P (respectively Q) can be joined to H, do the join, denoting theresulting ribbon graph also by H, and delete the ribbon graph from S. Continue to do this for as

long as possible. Doing so gives H= li=0 kij=0 Ri,j.Continue this process until S is empty. The resulting expression for H is the expression of joins

required by the theorem.

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5.3. Relating 1-decompositions of G. It is obvious that 1-decompositions into two plane graphsare not unique. However, it turns out that any two 1-decompositions of a ribbon graph into twoplane graphs are related in a very simple way. Roughly speaking, the only choices that one canmake in the construction of a 1-decompositions into two plane graphs is whether to place a joinsummand in P or Q. In this subsection we will make this statement precise and describe therelation between different 1-decompositions into two plane graphs that a ribbon graph G can have.

Definition 7. A ribbon graph is said to be prime if it can not be expressed as the join of twonon-trivial ribbon graphs.

As an illustration of this definition, the second and third ribbon graphs named G in Example 2are prime, while the first and fourth ribbon graphs named G in Example 2 are not. Also each ofthe join-summands in the join-decomposition shown in Example 6 is prime.

The following result is obvious and its proof is therefore omitted.

Proposition 6. Every ribbon graph admits a factorization into unique prime ribbon subgraphs,that is, for every ribbon graph G, we can write

G =

k

i=1

Ri

,

where each Ri is prime. Moreover, any other expression of G as joins of prime join-summands canonly differ from the above expression by the order of the join-summands in the expression.

The following result tells us that prime ribbon graphs admit only trivial 1-decompositions intotwo plane graphs.

Lemma 3. LetG be a prime, ribbon graph. Then G admits at most one 1-decomposition into twoplane graphs.

Proof. Suppose that G admits at most one 1-decomposition into two plane graphs. We will showthat at each vertex v on which the 1-decomposition acts, there is only one choice of decomposition.It then follows that the 1-decomposition into two plane graphs is unique.

Let v be a vertex of G which is incident to edges both in P and Q. Arbitrarily choose a cyclicorder of the half-edges incident to v. Partition these half edges into sets S1, . . . S k according tothe equivalence relation that two half-edges e and e are in Si if and only if there is a path in Gconnecting e and e that does not pass through the vertex v. During this proof will call the sets Siblocks.

We say that two blocks Si and Sj interlace each other if there are half-edges e, e Si and f, f

Sjsuch that we meet the edges in the order e , f , e, f when travelling round the vertex v.

Observe that

(1) every block Si interlaces at least one other set Sj, since G is prime. (If not Si is a joinfactor.)

(2) IfS is a set of blocks and S is the complementary set of blocks, then a block in S interlaces

with a block in S. (Otherwise S and S define a join.)(3) In the 1-decomposition, all of the half-edges in a block must belong to one of P or Q. (Since

P and Q define a 1-decomposition into two graphs.)(4) The half-edges in interlacing blocks must belong to different sets P and Q. (Since the

1-decomposition is into plane graphs.)

Now we will show that there is a unique partition of the half-edges incident to v into the set ofthose belong into P and the set of those belonging to Q (up to switching the names P and Q). Wecolour all of the half-edges as follows. Without loss of generality assume that the half-edges in S1

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belong to P. Set S = {S1}. Now for Si i nS, Si interlaces a block Sj in S. If the half-edges in Sjare in P (respectively Q), then the half-edges in Si must be in Q (respectively P). Add Si to S.

Continue until all of the blocks are in S. In doing this we have seen that every edge in a blockhas a unique assignment to P or to Q, as required.

We will now define an operation on a 1-decomposition that will allow us to relate all of the

1-decompositions into two plane graphs that are admitted by a graph G.Definition 8. Let G be a ribbon graph equipped with a 1-decomposition into two subgraphs Pand Q. In addition, let H be a join-summand ofG. By switching the parity of H we mean that wealter the 1-decomposition by switching the subgraph, P or Q, that each edge in H belongs to (andswitching the appropriate vertices).

Example 7. The following 1-decompositions are related by switching the parity of summands.

Theorem 3. LetG be a ribbon graph that admits a 1-decomposition into two plane graphs P andQ,and admitting a second1-decomposition into two plane graphs P and Q. Then the 1-decompositioninto P and Q can be obtained from the 1-decomposition into P and Q by switching the parity ofprime join-summands.

Proof. Let G be a ribbon graph that admits two 1-decompositions into two plane graphs P andQ, and P and Q. By Proposition 6, G admits a unique prime factorization: G =

ki=1 Ri

. The

two 1-decompositions each induce a partition of the edges in each prime join-summand Ri. ByLemma 3, the two induced partition in each Ri must be equal or related by switching the parity ofthe join-summand. The result then follows.

5.4. Joins and partial duality. In this section we will see that partial duality preserves the joinstructure of a ribbon graph. We go on to use this fact, in conjunction with Theorem 1, to deducea relation between all of the plane partial duals of a given ribbon graph. We will use this relationbetween plane partial duals in the following section to relate all link diagrams that are presentedby the same ribbon graphs.

Suppose G is a ribbon graph that can be written as a join G = H1 H2. This, and the notationthat will follow, is illustrated schematically in the figure below on the left. Also let A E(G).

H1 H2

a b

cd

Consider what happens in the formation of the partial dual GA. Regardless of the choice of A,there will be a boundary component C of the ribbon subgraph (V(G), A) that contains the arcs aband cd as shown in the figure above on the right. Moreover, when we travel around this boundarycomponent we will meet the points a, b, c and d in the order (a,b,c,d) or (c,d,b,a). Now notice

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that since G = H1 H2, all marking arrows corresponding to edges in H1 will lie on the arc da ofC,and all marking arrows corresponding to edges in H2 will lie on the arc bc of C. Thus if G is splitinto H1 and H2 by cutting the vertex v through the line as shown in the left-hand figure above,

then GA can be split into HAE(H1)1

and HAE(H2)2

by cutting the vertex in GA corresponding to

C ,through the line . Thus GA = (H1 H2)A = HAE(H1)1 HAE(H2)2 and the join acts naturallywith respect to partial duality.The conclusion of this discussion is summarized by the following lemma.

Lemma 4. Suppose G be a ribbon graph that can be expressed as a join, G = H1H2, andA E(G).Then

GA = (H1 H2)A = HA1 HA2 ,where A = A E(H1), A = A E(H2), and the joins act naturally with respect to partial duality.

We will now go on to define a move that will relate all of the plane partial duals of a ribbongraph. We will see in the next section that this move is particularly natural when considering theribbon graphs of link diagrams.

Let G be a ribbon graph such that G = H1 H2. We will say that the ribbon graph GE(H2) =

H1HE(H2)2

= H1H

2is obtained from G by taking the dual of a join-summand. This corresponding

move on the set of ribbon graphs is called the dual of a join-summand move. Furthermore, we definean equivalence relation on the set of ribbon graphs by setting G H if and only if there is asequence of dual of a join-summand moves taking G to H.

Lemma 5. LetG be a ribbon graph with a join-decomposition into G = ki=1 Ri. Then Gki=l

E(Ri) =

ki=1 Riki=lE(Ri) can be obtained from G by a sequence of dual of a join-summand moves.Proof. Let H1, . . . , H p be the connected components of G l1i=1 Ri. The graph G can then bewritten as

l1i=1

Ri H1 Hp,where each join involving an Hi is formed by identifying an arc on Hi with an arc on l1i=1 Ri.Then

Gki=l

E(Ri) = ki=1

Riki=l

E(Ri)

= l1i=1

Ri H1 Hppi=1

E(Hi)

= l1i=1

Ri H1 Hp ,where the third equality follows by Lemma 4. From this identity it follows that G

ki=l

E(Ri) can beobtained by dual of a join-summand moves which act on H1, . . . , H p.

Theorem 4. LetG and H be plane graphs. Then G and H are partial duals if and only if G H.

Proof. If G H then G and H are partial dual by the definition of.To prove the converse we need to show that any partial dual GA of G can be obtained from G

by dual of a join-summand moves.Let A E

(G

). Then by Theorem 1, A defines a 1-decomposition into two plane graphs P and

Q. By Corollary 6 we can then write

G =h

i=0

ki

j=0

Ri,j ,

such that if Ri,j is a subgraph of P (respectively Q) , then Ri+1,k is a subgraph of Q (respectivelyP), where 0 i < h. We also have

A = i odd

j

E(Ri,j) or A = i even

j

E(Ri,j).25


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Assume that A = i oddj E(Ri,j). Then

(4) GA =

h

i=0

ki

j=0

Ri,ji oddj E(Ri,j)

= hi=0

ki

j=0

Ri,ji2

j

E(Ri,j)

i3j E(Ri,j)

i=hj E(Ri,j)

where the second equality follows by the third item of Proposition 2. Observe now that the partial

dual hi=0 kij=0 Ri,ji2j E(Ri,j) can be obtained a sequence of dual of a join-summand movesby Lemma 5, and the partial dual has a join-decomposition by Lemma 4. Then it follows that

h

i=0

ki

j=0

Ri,ji2j E(Ri,j)

i3j E(Ri,j)

can be obtained a sequence of dual-summand moves and has a join-decomposition. Continuing inthis way, we see that GA can be obtained from G by a sequence of dual of a join-summand moves.The case when A = i evenj E(Ri,j) follows similarly.

6. Link diagrams with the same graph

In this section we answer Question Q3 which asked how link diagrams that are presented bythe same ribbon graphs are related to each other. Recall that this question was motivated byconsidering that fact that a Tait graph presents a unique link diagram (Property T3). Unlike Taitgraphs, in general a ribbon graph can present more that one distinct link diagram, however it turnsout that all of the link diagrams that are presented by the same ribbon graph are related to eachother in an extremely simple way. One particular consequence of this is that all of the link diagrams

presented by a given ribbon graph are diagrams of isotopic links.Recall that we consider all link diagrams up to planar isotopy. Note that the construction of

ribbon graphs of a link diagram and the construction of link diagrams from plane graphs are bothwell defined up to planar isotopy.

In Section 5 we saw the importance of the join operation on ribbon graphs when consideringpartial duals of plane graphs. The operation on link diagrams corresponding to a join is a connectedsum.

Let D1 S2 and D2 S

2 be link diagrams. The connected sum D1#D2 S2 of D1 and D2 is

the link diagram formed by, for i = 1, 2, choosing a disc Di on S2 that intersects Di in an arc i,

deleting the interior or each Di, and identifying the boundaries of S2

D1 and S2

D2 in such away that each end point of 1 is identified with a distinct endpoint of 2. This process is illustratedin the following figure.

D1 D2 D1 D2

D1 and D2 D1#D2

The following well-known (and obvious) result describes the connection between joins and con-nected sums. Its proof is omitted.

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Lemma 6. Let G be a plane graph such that G = H1 H2, then D(H1 H2) = D(H1)#D(H2).Moreover, there is a natural correspondence between the arcs on H1 and H2 used to form the join,and the arcs on D(H1) and D(H2) used to form the connected sum.

We will also need to make use of the following well-known result.

Lemma 7. LetG be a signed plane ribbon graphs. Then D(G) = D(G

)Proof. This follows since G and G are the two Tait graphs (formed with respect to the two possiblecheckerboard colourings) of D(G) and D(G). (Remembering that taking the dual changes thesign of an edge.)

We will see that the following simple move on link diagrams will relate all link diagrams that arerepresented by the same ribbon graph.

Definition 9. Let D1 and D2 be link diagrams. Then we say that D1 and D2 are related by asummand flip if and only if D2 can be obtained from D1 by the following process: orient S

2 andchoose an disc D in S2 that intersects D1 transversally in exactly two points a and b. Cut out D

and glue it back in in such a way that the orientations ofD

and S

2D

disagree and the pointsa on the boundaries ofD and S2 D are identified, and the points b on the boundaries ofD andS2 D are identified.

We will say that two link diagrams D1 and D2 are related by summand-flips, written D1 D2,if and only if there is a sequence of summand-flips taking D1 to D2

The summand-flip move is illustrated in the following diagram.

D D

D D D D

D1 =

D#D

cut, flip and glue D2

The main result of this section, which describes how link diagrams that are presented by thesame ribbon graphs relate to each other, will follow from the following lemma.

Lemma 8. LetG and G be signed plane graphs. Then D(G) D(G) if and only if G G .Proof. First observe that to prove the lemma it is enough to show that G and G are related bya single dual of a join-summand move if and only if D(G) and D(G) are related by a singlesummand-flip. This is what we will show. To show this we begin by examining how the embeddedgraphs G and G are related to each other.

Assume that G and G are related by a single dual of a join-summand move. Then we can writeG = H1 H2 and G

= H1 H

2= (H1 H2)E(H2). We can form the partial dual (H1 H2)E(H2) in

the following way: supposing H1 H2 = v, then(1) detach E(H1) (V(H1) v) from the vertex v by cutting along the intersections of H1

and v. Record the position of the edges by using coloured marking arrows in such a waythat H1 H2 can be recovered by identifying the arrows on H2 and E(H1) (V(H1) v)of the same colour.

(2) Form the dual H2 of H2 retaining the coloured arrows on the boundary.(3) Attach E(H1) (V(H1) v) to H2 by identifying the arrows of the same colour. The

resulting ribbon graph is (H1 H2)E(H2).27


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This process is illustrated in the following figure:

H1 H2 H1 H2b b H1

a

b

c

a

b

c

H

2H1 H

2

.

Now G = H1 H2 is a plane graph so it admits a unique embedding in the sphere S2. We

consider this construction of the partial dual (H1 H2)E(H2) acting on the embedded ribbon graphH1 H2 S

2. Start off with an embedding of H1 H2 in S2 where the vertex v forms the Southern

hemisphere. Detaching E(H1) (V(H1) v) gives an embedding of H2 in S2 with marking arrowsrecording how H1 was attached. Since v H2 S

2 is the Southern Hemisphere, the dual H S2 isembedded in the Northern hemisphere with the marking arrows sitting on the equator. Finally, toattach and embed H1 (in order to obtain an embedding of H1 H

2= (H1 H2)E(H2) in S2) we flip

over H1 and embed it in the Southern hemisphere. The resulting embedding of (H1 H2)E(H2)is obtained from the embedding of H1 H2 by a single dual of a join-summand move acting in H1.The argument is illustrated in the following figure.

v

H1 H2

v

H1

a b ab

H2

H1

ab

b a

H

2

H1

H

2

.

Now since the embeddings of G = H1 H2 and G = H1 H

2 in S2 are unique (as they are

obtained by filling in the punctures of the ribbon graph), it follows that the embeddings of G andG are related as in the above figures. The lemma then follows by considering the correspondinglink diagrams D(G) and D(G) as illustrated:

v

H1 H2

dual of a join-summand H1

H

2

D(H1

H2

)

T(D(H1

)#D(H2

))

D(H1

H 2

)

T(D(H1

)#D(H 2

)

)

D(H1) D(H2)

flipping a summand D(H1)

D(H2 )

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Since the link diagrams D(H2) and D(H2) are equal by Lemma 7, it is clear from the figure that

G and G are related by a single dual of a join-summand move if and only if D(G) and D(G) arerelated by a single summand-flip. This completes the proof of the lemma.

Remark 3. It is worthwhile noting that Lemma 8 does not say that the medial graphs of G and G

are equal. (See [16] for a discussion on the relation between medial graphs and partial duals.) To

see why this is, notice that in the proof of Lemma 8 the interior and the exterior of the vertex v,where the join acts, is switched. This switching may add half-twists the ribbons. Consequently,twists may be added to the medial graphs. For link diagrams this switching is not a problemsince links are one dimensional. In particular, this means that the lemma does not contradict thewell known result that the medial graphs of G and H are equivalent if and only if H= G or H= G.

The following theorem is the main result of this section. It describes how link diagrams arisingfrom the same ribbon graph are related to each other. Consequently, the theorem provides ananswer to our final motivational question Q3 completing the exploration of the ribbon graphs of alink diagram presented in this paper.

Theorem 5. Let G be a signed ribbon graph. Then D, D D(G) if and only if D D.Proof. By Equation 1 and Proposition 2, D, D

D

(G) if and only if D = D(H) and D

= D(HA

),where H is a partial dual of G, and H and HA are both plane graphs. By Theorem 4, this happensif and only if H HA. Finally, by Lemma 8, H HA if and only if D = D(H) D(HA) = D.

The following corollary is a straight forward consequence of Theorem 5.

Corollary 7. Let D, D D(G). Then the links corresponding to D and D are isotopic.References

[1] T. Abe, The Turaev genus of an adequate knot. Topology Appl. 156 (2009), no. 17, 27042712.[2]

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Iain Moffatt- Partial Duals of Plane Graphs, Separability and the Graphs of Knots