Branched covering surfaceknots with degree three have the simplifying numbers less than three
Abstract.
A branched covering surfaceknot is a surfaceknot in the form of a branched covering over a surfaceknot. For a branched covering surfaceknot, we have a numerical invariant called the simplifying number. We show that branched covering surfaceknots with degree three have the simplifying numbers less than three.
Key words and phrases:
surfaceknot; 2dimensional braid; chart; 1handle2010 Mathematics Subject Classification:
Primary 57Q45; Secondary 57Q351. Introduction
A chart [1, 4] is an oriented and labeled graph satisfying certain conditions, and an oriented surfaceknot in 4space is described by a chart on a disk. Let be a positive integer. It is known that for a chart on a disk of degree , is equivalent to a “ribbon chart”, which consists of several “free edges” and edges connected with no vertices [2]. It follows that the unknotting number of , denoted by , which is the minimum number of free edges necessary to deform to a union of free edges by the addition, satisfies [3]. In this paper, we show a similar result for the simplifying number of a branched covering surfaceknot. A branched covering surfaceknot is a surfaceknot presented by a chart on a diagram of a surfaceknot , denoted by . The simplifying number of , denoted by , is the minimum number of “1handles with chart loops” necessary to deform to a “simplified form” by the addition [7, 9]. Our main result is as follows.
Theorem 1.1.
Let be a positive integer. Let be a branched covering surfaceknot of degree such that is connected. If , then .
A branched covering surfaceknot is a surfaceknot in the form of a branched covering over a surfaceknot. A surfaceknot is a closed surface embedded smoothly into the Euclidean 4space . We assume that surfaceknots are oriented. Let be a 2disk for an interval . For a surfaceknot , let be a tubular neighborhood of in . A closed surface embedded in is called a branched covering surfaceknot over of degree if it satisfies the following two conditions.

The restriction is a branched covering map of degree , where is the natural projection.

The number of points consisting is or for any point .
Take a base point of . We say that two branched covering surfaceknots over of degree are equivalent if there is a fiberpreserving ambient isotopy of rel which carries one to the other. In this paper, we assume that the base surfaceknot is connected.
We consider a branched covering surfaceknot of degree 2, as the simplest nontrivial case. We explain the chart presentation. Consider the singular set of the image of by the projection to . Perturbing if necessary, we assume that consists of double point curves and isolated branch points. The image of by the projection to forms a finite graph on such that the degree of a vertex of is . A chart is a graph obtained from by labelling each edge by the number and assigning an orientation to each edge by a certain rule. See Section 2 for details.
We explain the simplifying operation. Let be a unit 2disk and let . A 1handle is a 3ball smoothly embedded in such that . The surfaceknot obtained from by a 1handle addition along is the surface
which is denoted by . We assume that is orientable, and we give the orientation induced from that of . For a 1handle , we call and the ends of . We take the oriented core (or simply the core) as an oriented path with the orientation of in such that for . When both ends of are on a 2disk in , we determine the core loop of as the oriented closed path in obtained from the oriented core by connecting the initial point and the terminal point by a simple arc in , and we assign the core loop the induced orientation. We say that a 1handle attached to a surfaceknot is trivial if there is a 3ball satisfying .
Let be a branched covering surfaceknot of degree 2. We call a connected segment of an edge an arc. We call an edge/arc of a chart a chart edge/arc, and we call a chart edge connected with no vertices a chart loop or simply a loop. A chart edge is called a free edge if its end points are a pair of vertices of degree 1. Now, the chart of degree 2 consists of several free edges and chart loops. We consider an equivalence deformation as the local exchange of two parallel horizontal chart arcs with opposite orientations into two parallel vertical chart arcs with induced orientations. This deformation is called a CIM2 move; see Section 2 for precise definition and other equivalent modifications. When we have a 1handle with a chart loop in a neighborhood of an arc of a chart loop, applying a CIM2 move and sliding an end of the 1handle, eliminates the chart loop, as illustrated in the first row of Fig. 1. When the orientation of the chart loop is opposite, then we turn around the 1handle to make the loop along the core loop have the opposite orientation, as illustrated in the second row of Fig. 1. Repeating this operation, by an addition of , of degree 2 deforms to
(1.1) 
where is a chart consisting of several (maybe no) free edges.
We call the form (1.1) a simplified form. Thus the simplifying number, denoted by , satisfies
for a branched covering surfaceknot of degree 2.
In this paper, we focus on branched covering surfaceknots of degree 3. A surface diagram of a surfaceknot is the image of in by a generic projection, equipped with over/under information on sheets along each double point curve. A finite graph on a surface diagram is called a chart of degree if it satisfies the following conditions.

The intersection of and the singular set of consists of a finite number of transverse intersection points of edges of and double point curves of , which form vertices of degree .

Every edge of is oriented and labeled by an element of .

Every vertex has degree , , or , and the edges connected to each vertex satisfy one of the following conditions.

The adjacent edges of a vertex of degree 2 have labels and orientations as in Fig. 3.
We regard edges connected with vertices of degree 2 as one edge. Similarly to the case of charts of degree 2, a free edge is a chart edge whose end points are a pair of black vertices, and a chart loop (or simply a loop) is a chart edge connected with no vertices or a chart edge with vertices of degree 2. A chart is said to be empty if it is an empty graph.
We consider a 1handle on which a chart is drawn, attached to a fixed 2disk , where there are no chart edges nor vertices except those of the attached 1handle. We denote by a 1handle with a chart loop along the core loop with the orientation of the core loop and whose arc in has the label . And we denote by a 1handle with an empty chart. We use the same notation for any 1handle, and further, we do not distinguish the frame of a 1handle.
For a branched covering surfaceknot and a finite number of 1handles with chart loops attached to a fixed 2disk , where there are no chart edges nor vertices except those of the attached 1handles, we denote the branched covering surfaceknot which is the result of the 1handle addition by . By [7], for the degree 3 case, we have the following result. See [9] (see also Remark 2.1) for the same notations and terminologies we use here.
Theorem 1.2 ([7, Theorem 1.6]).
Let be a branched covering surfaceknots of degree 3. By an addition of finitely many 1handles in the form , or , to appropriate places in , is deformed to
(1.2) 
where , and is a chart consisting of several (maybe no) free edges.
We remark that the presentation (1.2) is welldefined: Since is a disjoint union of free edges, (1.2) does not depend on the place where 1handles are attached. We call the form (1.2) a simplified form. The simplifying number of , denoted by , is the minimum number of 1handles in the form , or necessary to deform to a simplified form. We further remark that in [7], we gave two notions of the simplifying number: the weak simplifying number and the simplifying number , but they are the same notion for the degree 3 case, where we have no vertices of degree 4. See [7, 8, 9] for the investigations of simplifying operations and the upper estimates of the (weak) simplifying numbers.
Together with the above argument for the case of degree 2, Theorem 1.1 is reduced to Theorem 1.3 as follows.
Theorem 1.3.
Let be a branched covering surfaceknot of degree . Then . Further, if has a positive number of black vertices, then .
In order to show the latter part of Theorem 1.3, we show the following theorem. The case when is a 2sphere is shown by Kamada [2].
Theorem 1.4.
Let be a branched covering surfaceknot of degree such that has a positive number of black vertices. Then, is equivalent to a chart consisting of free edges and chart loops.
2. Chart presentations, Cmoves and Roseman moves
A branched covering surfaceknot was a surface we formerly called a 2dimensional braid over a surfaceknot [6, 7]; we changed the terminology in [7]. We introduced a branched covering surfaceknot as an extended notion of a 2dimensional braid or a surface braid over a 2disk [2, 4, 11]. A branched covering surfaceknot over a surfaceknot is presented by a chart on a surface diagram of [6] (see also [2, 4]). For two branched covering surfaceknots of the same degree, they are equivalent if their surface diagrams with charts are related by a finite sequence of ambient isotopies of , and local modifications called Cmoves [2, 4] and Roseman moves [6] (see also [10]). In this paper, we review these notations for the case of degree 3.
2.1. Chart presentation
Let be a branched covering surfaceknot of degree 3 over a surfaceknot . We explain how to obtain a chart of degree 3 on a 2disk in a surface diagram of which does not intersect with singularities of . We identify with a 2disk whose projected image is , and we denote by , where is the projection . We identify by for an interval . Consider the singular set of the image of by the projection to . Perturbing if necessary, we assume that consists of double point curves, isolated triple points, and isolated branch points. Thus the image of by the projection to forms a finite graph on such that the degree of a vertex of is either or , where we ignore the points in . An edge of presents to a double point curve, and a vertex of degree and degree present a branch point and a triple point, respectively.
For such a graph obtained from a branched covering surface of degree 3, we assign labels and orientations to all edges of by the following method. We consider a path in such that is a point of an edge of . Then is a braid with one crossing in the cylinder such that corresponds to the crossing of the braid. Let (,
) be the presentation. We assign the edge the label , and the orientation such that
the normal vector of is coherent (respectively, is not coherent) with the orientation of if (respectively, ), where the normal vector of is the vector such that for a tangent vector of at , is coherent with the orientation of the 2disk . The resulting is the chart of degree 3 presenting .
Conversely, when we have a chart on a surface diagram, we can construct the branched covering surfaceknot with the chart presentation. See [6] for the structure of a branched covering surfaceknot for neighborhoods of the singularities of the surface diagram or vertices of degree 2 in a chart.
Around a white vertex of a chart, there are six chart arcs. We call the arc which is the middle of the three adjacent arcs with the coherent orientation a middle arc, and we call an arc which is not a middle arc a nonmiddle arc. Around a white vertex, there are two middle arcs and four nonmiddle arcs. For edges/arcs around a white vertex, we call a pair of edges/arcs separated by two edges at each side diagonal edges/arcs. See Fig. 4.
2.2. Cmoves
Cmoves (chart moves) are local modifications of a chart, consisting of three types: CImoves, CIImoves, and CIIImoves. CIImoves are modifications for charts of degree , where we have vertices of degree 4 called crossings [4]. We review CImoves and CIIImoves for charts of degree 3. Let and be two charts of degree 3 on a surface diagram . We say that and are related by a CImove or CIIImove if there exists a 2disk in such that does not intersect with the singularities of , and the loop is in general position with respect to and and , and the following conditions are satisfied.
(CI) There are no black vertices in nor . We use a CIM1 move, a CIM2 move and a CIM3 move, as shown in Fig. 5.
(CIII) and are as in Fig. 5, where , and the black vertex is connected to a nonmiddle arc of a white vertex.
2.3. Roseman moves
Roseman moves for surface diagrams with charts of degree 3 are defined by the original Roseman moves (see [10]) and moves for local surface diagrams with nonempty charts as in Fig. 8 (see [6]), where we regard the diagrams for the original Roseman moves as equipped with empty charts.
For charts and of degree 3 on a surface diagram of a surfaceknot , their presenting branched covering surfaceknots are equivalent if the charts are related by a finite sequence of ambient isotopies of , Cmoves and Roseman moves [6] (see also [2, 4, 10]).
Remark 2.1.
We remark that the deformations we use for the simplifying operation, such as sliding one end of a 1handle, can be investigated by a similar method whether we have vertices of degree 2 or not. Hence, in the proofs of lemmas and theorems, we ignore the knottedness of and 1handles. We assume that there are no singularities in surface diagrams of and 1handles; in particular, we assume that 1handles are trivial. But, one lemma, Lemma 4.5, requires the concerning 1handle to be trivial. So we must be careful when we use Lemma 4.5; see Remark 4.7.
3. Simplifying branched covering surfaceknots
In this section, we show Theorem 1.2. We review lemmas used in [7], and we explain the simplifying operation.
We consider branched covering surfaceknots of degree 3.
Lemma 3.1.
Let be a chart loop with the label in a branched covering surfaceknot . If there is a 1handle near a neighborhood of an arc of , then is equivalent to , where denotes the chart obtained from by elimination of .
Proof.
Applying a CIM2 move between arcs of and sliding an end of as in Fig. 1, we eliminate , and the other parts of remains unchanged. Thus we have the required result. ∎
Proposition 3.2.
Let be a branched covering surfaceknot of degree 3 such that has no white vertices. By an addition of , deforms to a simplified form, hence .
Proof.
Since has no white vertices, consists of several free edges and chart loops. By an addition of , applying Lemma 3.1 to each chart loop from the loop whose neighborhood has the 1handles, we eliminate all chart loops. Thus we have a simplified form, and . ∎
Lemma 3.3.
If there is a 1handle near a neighborhood of a nonmiddle arc with the label of a white vertex , then, by equivalent deformation, collects as illustrated in Fig. 9.
Proof.
Apply a CIM2 move between the arc of and the nonmiddle arc of . Move on , and apply CIM2 moves twice, first between arcs with the label and secondly between arcs with the label , where ; see Fig. 10. Then is on such that there are two diagonal arcs along the core, and the other four arcs become two edges whose both endpoints are connected with , as in the rightmost figure of Fig. 9. ∎
Lemma 3.4.
Let be a 1handle on which is a white vertex such that two edges are at both endpoints connected with , as in the rightmost figure of Fig. 9. Let be the labels of edges. Let be one of the edges whose both endpoints are connected with , and let be the label of . If we add a 1handle in a neighborhood of , then, after collecting from to , the orientations of the two edges connected at both endpoints with are reversed from when they were on , where we identify the oriented diagonal edges along the core of with those of .
Proof.
Note that for the two arcs of connected with , there is only one arc which is a nonmiddle arc, along which we slide an end of and collect . See Fig. 11 for the case . For the other case , the form of after collecting is the same with the case , and we have the required result. ∎
Proof of Theorem 1.2.
If has no white vertices, then the result follows from Proposition 3.2. We assume that has a positive number of white vertices. We consider the case when has no black vertices. We add a 1handle in a neighborhood of a nonmiddle arc of a white vertex . Apply a CIM2 move, slide an end of , and by moves illustrated in Figs. 9 and 10, collect on . Then we slide the end of . If comes to a nonmiddle arc of another white vertex , then we collect on . If comes to a middle arc of another white vertex , then we add another 1handle near a nonmiddle arc of , apply a CIM2 move, slide an end of and collect on , and let the end of pass under . Repeat this process, until the end of comes back to a neighborhood of the other end of . The resulting has several white vertices. Repeat this process, until we have all white vertices on 1handles.
Take a 1handle on which there are several white vertices, and consider edges connected to one white vertex at both endpoints. If the orientations of these edges for one white vertex are opposite to those for another adjacent white vertex , then, as shown in Fig. 12, by a CIM3’ move and two CIM1 moves, we eliminate the pair of vertices . If the orientations of the edges of and are coherent, then, as shown in Fig. 11, add a 1handle in a neighborhood of , apply a CIM2 move, slide the ends of and collect and then and the other vertices on . By Lemma 3.4, on , the orientations of the edges whose endpoints are connected with are reversed. By the same process with the former case, by modifications as shown in Fig. 12, we eliminate . The resulting is a 1handle obtained from by eliminating . Repeat this process, until we eliminate all white vertices. The resulting chart consists of several chart loops and 1handles with chart loops. Adding () if necessary and applying Lemma 3.1, we eliminate chart loops except those consisting . Thus we have a simplified form.
We consider the case when has a positive number of black vertices. By the same process as in the case when has no black vertices, we collect white vertices on 1handles. The middle arcs of a white vertex on a 1handle are connected at both endpoints with ; see the rightmost figure of Fig. 9. Hence, if a black vertex is connected with a chart edge on , then, the edge is always a nonmiddle edge of a white vertex on . Thus, applying CIIImoves if necessary, we assume that there are no black vertices on , and the ends of are in the neighborhood of each other when we have collected white vertices. Thus, by an addition of 1handles with chart loops, we eliminate all white vertices and the resulting chart consists of several free edges, chart loops and 1handles with chart loops. Then, adding () if necessary and applying Lemma 3.1, we eliminate unnecessary chart loops and we have a simplified form. ∎
4. Proofs of Theorems 1.3 and 1.4
4.1. Key Lemma
Lemma 4.1.
Let be a branched covering surfaceknot of degree .
If is has no black vertices, then, for any white vertex in , there is a closed path with the base point near such that several white vertices including can be collected on a 1handle for some by applying a CIM2 move and sliding one end of along .
If has a black vertex, then is equivalent to a chart such that each black vertex is at an endpoint of a free edge.
4.2. Notations and Lemmas
We determine the cocore of a 1handle by the oriented closed path , with the orientation of . Further, we determine the base point of the core loop and the cocore of by their intersection point. We call the ends and the initial end and the terminal end, respectively. We call the arc used to form the core loop of a 1handle from the core the base arc. See Fig. 13.
Let be a 1handle whose ends are on a 2disk . For braids and which commute, we denote by a 1handle whose cocore with the given orientation and core loop with the reversed orientation present the braids and , respectively. We draw such a chart on , and this notation is unique [5]. We call a 1handle with a chart, or simply a 1handle. We denote by the standard generator of , the braid group of degree 3, and we denote by the trivial braid in . Note that this definition of coincides with that of in the previous sections.
We say that 1handles with chart loops are equivalent if their presenting surfaces are equivalent, and we use the notation “” to denote the equivalence relation. When we denote for braids , we assume that and commute.
We define the type of a white vertex as follows. For a white vertex such that the three arcs oriented toward consist of two arcs with the label and one arc with the label (respectively, two arcs with the label and one arc with the label ), we call a white vertex of type (respectively, of type ); see Fig. 14.
Lemma 4.2.
Let be a branched covering surfaceknots of degree 3. Let and be disks in such that . Let be the result of an addition of on . Then, and are equivalent. Thus, by equivalent deformation, moves anywhere.
Lemma 4.3.
In a chart with no black vertices, the number of white vertices of type is the same with that of white vertices of type .
Lemma 4.4.
Let be a chart loop with the label . If there is a 1handle near a neighborhood of an arc of , then we can eliminate using and a CIM2 move, where is a braid.
Lemma 4.5.
If is a trivial 1handle, then
When a 1handle has collected several white vertices, and the chart arc along the base arc of has the orientation coherent with the base arc and labeled by , has the presentation for a nonzero integer . The 1handle has white vertices such that the chart loops parallel to the cocore have the coherent orientations as illustrated in Fig. 15.
Lemma 4.6.
Let be a nonzero integer. Let .

The 1handles deforms to if , or if , by equivalent deformation.

The 1handles deforms to if , or if , by equivalent deformation.
4.3. Proofs of Theorems 1.3 and 1.4
Proof of Theorem 1.3.
First we show the result for the case when has black vertices. If has a positive number of black vertices, then, by Theorem 1.4, by equivalence, we deform to the form consisting of several free edges and chart loops. We take a free edge , and let be the label of . We add in a neighborhood of . By Lemma 3.1, eliminates a chart loop of the label which has an arc in a neighborhood of . Similarly, eliminates a chart loop of the label which has an arc in a neighborhood of , by a CIM2 move and an ambient isotopy as illustrated in Fig. 16. Hence, using or , we eliminate all chart loops from the one in a neighborhood of and . The result is free edges and , which is a simplified form. Thus the simplifying number satisfies , which is the required result.
We show the result for the case when has no black vertices. Let and . We add in a neighborhood of a nonmiddle arc of the label 1 of a white vertex of type . By Lemma 4.6 (1), there is a closed path with the base point near such that several white vertices including are collected on a 1handle by applying a CIM2 move and sliding one end of the 1handle along the path. We collect the white vertices on . Then we have for some integer . By Lemma 4.2, we move to a neighborhood of a nonmiddle arc of the label 1 of a white vertex of type of the resulting , and by the same process, we collect white vertices on . Note that since the number of white vertices of type is the same as that of white vertices of type by Lemma 4.3, if there are white vertices other than those on , then there exists a white vertex of type . We repeat this process until we collect all white vertices on . The result is for some integer and several chart loops with the label 1 or 2. By Lemmas 3.1 and 4.4, applying CIM2 moves to the chart loops and the arc of or , we eliminate the loops. Thus we have
(4.1) 
If is even, then, applying Lemma 4.6, we have , which is a simplified form. If is odd, then applying Lemma 4.5, we have
(4.2) 
see Remark 4.7. Then, applying Lemma 4.6, in this case also we have , which is a simplified form. Thus the simplifying number satisfies , which is the required result. ∎
Remark 4.7.
In order to consider branched covering surfaceknots when may be knotted, we must be careful to remark when 1handles are trivial, since Lemma 4.5 requires the condition that the concerning 1handle is a trivial 1handle. We first add trivial 1handles with chart loops, . When we slide ends of a 1handle and collect white vertices, may become knotted. However, since we don’t slide ends of , in (4.1) is a trivial 1handle, and hence we can apply Lemma 4.5 and we obtain (4.2).
5. Proofs of Lemmas 4.1–4.6
For a chart edge/arc, we call the endpoint with the orientation from it (respectively, toward it) the initial point (respectively, the terminal point).
5.1. Proofs of Lemmas 4.2–4.6
Before the proofs, we give two lemmas which are also used in [7, 8, 9]. In Section 1, we used ; see Fig. 1. We have a similar lemma.
Lemma 5.1.
For a braid ,
(5.1) 
Proof.
Let be a 2disk where the ends of are attached. Turning around and regarding the orientationreversed core loop as a new core loop, we have : thus ; see Fig. 17. ∎
Lemma 5.2.
For braids , and ,
(5.2)  
(5.3) 
Proof.
We show (5.2). Sliding the initial end of along the core of and and applying CIM2 moves as in the first row of Fig. 18, becomes .
We show (5.3). Moving the initial end of through the chart edges along the base arc of and applying CIM2 moves, becomes ; see the second row of Fig. 18. In Fig. 18, the crossing change is the deformation such that the crossing of the cores of the 1handles are changed. Since the 1handles are in 4space, this deformation is an equivalent deformation. See [7, Claim 4.1] and the proof of [7, Lemma 4.6]. ∎
Proof of Lemma 4.2.
Let be the label of a chart edge in whose neighborhood we have . Apply a CIM2 move between an arc of and an arc along the base arc of or , move the other 1handle under to the other side of , and by a CIM2 move again, let be on the other side of ; see Fig. 19. Thus moves through any chart edge with any label. Hence we have the required result. ∎
Proof of Lemma 4.3.
A white vertex of type (respectively, ) have two arcs (respectively, one arc) with the label 1 oriented toward the vertex, and one arc (respectively, two arcs) with the label 1 oriented from the vertex. Since we have no black vertices, counting the numbers of arcs with the label 1, we see that the numbers of white vertices of type is the same with that of white vertices of type . ∎
Proof of Lemma 4.4.
By the same argument as in the proof of Lemma 3.1, we have the required result. ∎
Proof of Lemma 4.5.
Proof of Lemma 4.6.
(1) We consider the case when . The case is shown similarly. Put . Apply a CIM2 move to the chart arc of and the chart arc of presenting the first of , and move the ends of and collect all white vertices on . See Fig. 21 for the deformation when we collect the first white vertex on . See also Lemma 3.4 and Fig. 11. On , the orientations of the edges parallel to the cocore and around the first white vertex are reversed, and those of the edges around the other white vertices are unchanged. Hence, deforms to
By CIM3’ and CIM1 moves as in Fig. 12, or, since and a 1handle is unique [5], we have
By Lemma 5.1, , and it follows from Lemma 5.2 (5.2) that we have
By Lemma 5.1 again, ; thus we have
which is the required result.
(2) We consider the case when . The case is shown similarly. Put . By Lemma 5.1, . Thus deforms to
By Lemma 5.1 again, we have
Applying CIM2 moves, we deform to the form such that is equipped with an empty chart and the ends are attached to the places in as in the last figure of Fig. 22. Let and be the first and the second white vertices associated with the first sequence of . Then, move and on and apply a CIM3” move to cancel the pair . Then, on , there are no white vertices.
We see the form of the resulting . We denote the initial points and the terminal points of arcs connected to the white vertices and by and as in Fig. 23. See the right figure of Fig. 23. Then, and are connected by a chart arc on . Further, we take on the initial end of . We focus on the edges with the label . We start from and move along the edges. Then we move as follows. We start from , we move along the core of to , along the arc parallel to the cocore of to , along the core of to , along the arc parallel to the cocore of to , along the core of to . Thus, the arcs/edges with the label 1 form a connected path . Move the terminal end of along the base arc of and . Then, considering a closed path parallel to the cocore of in a neighborhood of as a new end; see Fig. 24. Then becomes . The resulting 1handle has only edges with the label 2. The 1handle becomes by a CIM1 move and CIM2 moves as illustrated in Fig. 25. Thus we have the required result. ∎
5.2. Proof of Key Lemma 4.1
Proof.
(1) Let be a branched covering surfaceknot of degree 3 such that has no black vertices. We show that when we attach a 1handle to a certain place, we can slide an end of , collecting white vertices including , and the end comes back to a neighborhood of the other end.
We have a given white vertex . We assume that is of type . The other case when is of type is shown similarly. We attach a 1handle in a neighborhood of a nonmiddle arc of with the label 1. We apply a CIM2 move and slide an end of and collect on . Let be the chart edge which comes out from the sliding end of . Then we have two cases: (Case 1) The edge is connected at the terminal point with a white vertex as a nonmiddle arc, and (Case 2) The edge is connected at the terminal point with a white vertex as a middle arc.
(Case 1) Assume that is connected at the terminal point with a white vertex as a nonmiddle arc. Then, we collect on , and we regard the resulting edge coming out from (the diagonal edge of with respect to ) the new .
(Case 2) Assume that the chart edge is connected at the terminal point with a white vertex as a middle arc. Let be the label of . Then, is of type . By equivalence deformation as in Fig. 26, we can connect the end of with an arc of any edge with the label which admits a path from to intersecting with no chart edges nor vertices. Since is a middle arc with respect to , the two edges sandwiching are distinct edges; thus there exists at least one edge with the label which admits such a path.
Let be the set of edges with the label such that each element has a path from to itself, intersecting with no chart edges nor vertices. We assume that the elements of are not contained in the 1handle . By the above argument, contains edges other than .
Suppose that contains an edge which is connected at the terminal point with a white vertex of type . Then, applying the deformation illustrated in Fig. 26 to and , we connect and by an edge. The arcs with the label of a white vertex of type with the orientation toward the vertex are nonmiddle arcs. Hence, we slide the end of , and we collect on .
Suppose that each element of is an edge connected at the terminal point with a white vertex of type . Since comes out of the sliding end of the 1handle , by Lemma 5.3, we take inductively a sequence of edges where and is an edge with the label such that the closure of the union of the edges forms a connected path , and the edges are mutually distinct, until or goes into or comes out of the fixed end of for some . We give an orientation induced from that of . The edge going into or coming out of the fixed end of is , which is an edge with the label and oriented coherently with that of . By Lemma 5.3, is oriented coherently with that of , and is oriented oppositely to that of ; thus, cannot be and we see that and for some . We apply a CIM2 move between and . Then, the sliding end of is back to a neighborhood of the other end, which is the required result.
Repeating these processes, we have the required result.
(2) Let be a branched covering surfaceknot such that has a positive number of black vertices. We take an edge which is connected at the initial point with a black vertex. Remark that counting the numbers of edges connected with black vertices at the initial points and the terminal points, we see that half of the black vertices are connected at the initial points of edges; thus, if there are black vertices, then there exists a black vertex connected at the initial point of an edge. Then, by the same argument as in the proof of (1), instead of collecting white vertices, we apply CIIImoves and we move the black vertex, as follows. If is connected at the terminal point with another black vertex, then we have a free edge. If is connected at the terminal point with a white vertex as a nonmiddle arc, then, applying a CIIImove, we eliminate , and we regard the resulting edge connected with the black vertex at the initial point as the new . Suppose that is connected at the terminal point with a white vertex as a middle arc. Let be the label of . Then, is of type . Because is connected with a black vertex, we cannot take a sequence of edges including such that the closure of the union of the edges forms a closed path; hence, Lemma 5.3 implies that there exists an edge with the label which is connected at the terminal point with a white vertex of type or a black vertex. Applying a CIM2 move to and , we have one of the two cases: (Case 1) The black vertex and the white vertex of type are connected by an edge oriented toward and with the label , or (Case 2) The black vertex is connected with another black vertex by an edge with the label . In Case 1, we eliminate by a CIM3 move. In Case 2, we have a free edge. Repeating these processes, every black vertex becomes an endpoint of a free edge, and we have the required result. ∎
Lemma 5.3.
Let be a branched covering surfaceknot of degree 3. If necessary, we regard another chart as which is Cmove equivalent to and has a smaller number of white vertices. Let be an edge with the label connected at the terminal point with a white vertex of type . Let be the set of edges with the label such that each element admits a path from to itself, intersecting with no edges nor vertices. Assume that each element of is an edge connected at the terminal point with a white vertex of type . Then, we can take inductively a sequence of edges (until we have for some ) satisfying the following conditions.

The edge is an element of and is an edge with the label .

The closure of the union of form a connected path.

The pair of edges and (respectively, and ) are both connected with the same vertex (respectively, ) at the terminal points (respectively, the initial points).

The edges are mutually distinct.
Proof.
We take the sequence of edges as follows. First we suppose that each edge is connected with a white vertex at each endpoint. We show later that we can assume this situation.
Let be the white vertex connected with at the terminal point. Let be the adjacent edge of with respect to in the anticlockwise direction. Let be the white vertex at the other endpoint of . Let be the adjacent edge of with respect to in the anticlockwise direction. By the similar method, we take a sequence of edges satisfying the conditions (1) and (2), and a sequence of white vertices ; see Fig. 27, where we don’t know the orientations yet.
Now we see the orientations of edges. See Fig. 27. By assumption, the edge is with the label and oriented toward . Since is of type , is oriented toward . If is oriented toward , then is of type and is connected with the white vertex of type at the terminal point, which implies , hence a contradiction. Thus we see that is oriented toward . Similarly, if is oriented toward , then is of type , and is connected with the white vertex of type at the terminal point, hence a contradiction. Thus we see that is oriented toward . By the same argument, we see that is an edge with the label which is oriented from to , and is an edge with the label which is oriented from to . Thus the sequence satisfies the condition (3).
Since is connected with at the terminal point, by assumption, is of type . We see that is also of type : If for some is of type , then, a by CIM3’ move as in the top right figure of Fig. 6, we eliminate , and we regard the resulting chart as the new . Thus, we have