Subgroups generated by two Dehn twists on a nonorientable surface
SSUBGROUPS GENERATED BY TWO DEHN TWISTS ON ANONORIENTABLE SURFACE
MICHA(cid:32)L STUKOW
Abstract.
Let a and b be two simple closed curves on an orientable surface S such that their geometric intersection number is greater than 1. The groupgenerated by corresponding Dehn twists t a and t b is known to be isomorphicto the free group of rank 2. In this paper we extend this result to the case ofa nonorientable surface. Introduction
Let N be a smooth, nonorientable, compact surface. We will mainly focus onthe local properties of N , hence we allow N to have some boundary componentsand/or punctures. Let H ( N ) be the group of all diffeomorphisms h : N → N suchthat h is the identity on each boundary component and h fixes the set of punctures(setwise). By M ( N ) we denote the quotient group of H ( N ) by the subgroup thatcomprises the maps isotopic to the identity with an isotopy which fixes the boundarypointwise. M ( N ) is known as the mapping class group of N . The mapping classgroup M ( S ) of an orientable surface S is defined analogously, but we consider onlyorientation preserving maps. Usually, we will use the same letter to denote a mapand its isotopy class.Important elements of the mapping class group M ( S ) are Dehn twists . Dehntwists generate M ( S ), thus obtaining a good understanding of possible relationsbetween them is important. One of the basic results in this direction is the followingtheorem: Theorem 1.1 (Ishida [3]) . If a and b are simple closed curves on an orientablesurface S such that the geometric intersection number of a and b is greater than 1,then the group generated by Dehn twists t a and t b is free of rank 2. The main goal of this paper is to extend the above result to the case of a nonori-entable surface: see Theorem 13.2. Let us mention that Dan Margalit observedthat if we lift the statement of Theorem 13.2 to the oriented double cover S , thenwe obtain some special cases of the well-known conjecture [4] that two elements ofthe Torelli subgroup of S either commute or generate a free group. For more detailsabout this correspondence see [6].The paper is organized as follows. In Section 2, we establish some basic notation.Section 3 contains some examples that show how the nonorientable case differs fromthe orientable one. In Section 4 we recall some language introduced in [5], namely Mathematics Subject Classification.
Primary 57N05; Secondary 20F38, 57M99.
Key words and phrases.
Mapping class group, Nonorientable surface, Dehn twist, Free group.Supported by grants 2012/05/B/ST1/02171 and 2015/17/B/ST1/03235 of National ScienceCentre, Poland. a r X i v : . [ m a t h . G T ] A ug MICHA(cid:32)L STUKOW the notion of adjacent and joinable segments. Sections 5, 6 and 8 are devoted tothe study of properties of curves in the neighborhood of a ∪ b . The main theoremof the paper (Theorem 13.2) is proved in Section 13. This proof is based on fivepropositions (Propositions 7.1, 9.12, 10.4, 11.3 and 12.3) that are proved in Sections7 and 9 to 12. 2. Preliminaries
By a circle on N we mean an oriented simple closed curve that is disjoint fromthe boundary of N . Usually, we identify a circle with its image. If two circles a and b intersect, then we always assume that they intersect transversely. Accordingto whether a regular neighborhood of a circle is an annulus or a M¨obius strip, wecall the circle two-sided or one-sided , respectively.We say that a circle is generic if it bounds neither a disk with fewer than twopunctures nor a M¨obius strip without punctures. It is known (Corollary 4.5 of [5])that if N is not a closed Klein bottle, then the circle a is generic if and only if t a has infinite order in M ( N ).For any two circles a and b we define their geometric intersection number asfollows: I ( a, b ) = inf {| a (cid:48) ∩ b | : a (cid:48) is isotopic to a } . We say that circles a and b form a bigon if a disk exists whose boundary is theunion of an arc of a and an arc of b . The following proposition provides a usefultool for checking if two circles are in a minimal position (with respect to | a ∩ b | ). Proposition 2.1 (Epstein [1]) . Let a and b be generic circles on N . Then | a ∩ b | = I ( a, b ) if and only if a and b do not form a bigon. (cid:3) Disappointing examples
Let a and b be two circles in an oriented surface S such that I ( a, b ) ≥
2. Thekey observation that leads to the conclusion that Dehn twists t a and t b generate afree group is the following lemma: Lemma 3.1 (Lemma 2.3 of [3]) . Assume that circles a, b, c ⊂ S satisfy I ( a, b ) ≥ .Then for any nonzero integer kI ( c, a ) > I ( c, b ) = ⇒ I ( t ka ( c ) , a ) < I ( t ka ( c ) , b ) . (cid:3) The above lemma allows to apply the so-called ’ping-pong lemma’ (Lemma 13.1)and easily conclude that (cid:104) t a , t b (cid:105) is a free group.Relations between Dehn twists and geometric intersection numbers are knownto become more complicated if we allow the surface to be nonorientable. Someresults in this direction were obtained in [5], but they were too weak to prove anonorientable version of the above lemma. The main goal of this section is toshow that there is a reason for this condition, namely, Lemma 3.1 is not true onnonorientable surfaces. Moreover, finding general families of counterexamples ispossible. Hence, no easy fix seems to exist for this situation (for a nontrivial fix,see Propositions 7.1, 9.12, 10.4, 11.3, and 12.3). Example . Let a, b, c be two-sided circles indicated in Figure 1 (shaded disks arecrosscaps, that is, the interiors are to be removed and the boundary points are tobe identified by the antipodal map). In particular I ( a, b ) = 2 and I ( c, a ) = 8 > UBGROUPS GENERATED BY TWO DEHN TWISTS . . . 3
Figure 1.
Circles a, b, c – Example 3.2. I ( c, b ) = 4. However, checking the following is straightforward: I ( t a ( c ) , a ) = 8 > I ( t a ( c ) , b ) = 4 . The above example can be generalized in the obvious way (by changing c ) to theexample where I ( a, b ) = 2, I ( c, a ) = 2 n > I ( c, b ) = n and I ( t a ( c ) , a ) = 2 n >I ( t a ( c ) , b ) = n , where n ≥ Example . Let a, b, c be two-sided circles indicated in Figure 2. In particular
Figure 2.
Circles a, b, c – Example 3.3. I ( a, b ) = 8 and I ( c, a ) = 2 > I ( c, b ) = 1. However, checking the following isstraightforward: I ( t a ( c ) , a ) = 2 > I ( t a ( c ) , b ) = 1 . The above example can be generalized in the obvious way (by changing b ) to theexample where I ( a, b ) = 2 n , I ( c, a ) = 2 > I ( c, b ) = 1 and I ( t a ( c ) , a ) = 2 >I ( t a ( c ) , b ) = 1, where n ≥ I ( a, b ) and I ( c, a )]. Example . Let a, c be two-sided circles indicated in Figure 3. The action of t a on c is trivial because a bounds a M¨obius strip; this is the case even though I ( a, c ) = 8(or in general I ( a, c ) = 2 n , n ≥ S – if I ( a, c ) > c on S , then t a is automatically nontrivial. MICHA(cid:32)L STUKOW
Figure 3.
Circles a, c – Example 3.4.4.
Joinable segments of a and b For the rest of the paper assume that a and b are two generic two-sided circlesin a nonorientable surface N such that | a ∩ b | = I ( a, b ) ≥ segment of b (with respect to a ) we mean any unoriented arc p of b that satisfies a ∩ p = ∂p . Similarly, we define an oriented segment of b . If p isan oriented segment, then by − p , we denote the segment equal to p as an orientedsegment but with the opposite orientation.We call a segment p of b one-sided [ two-sided ] if the union of p and an arc of a connecting ∂p is a one-sided [two-sided] circle. An oriented segment is one-sided[two-sided] if the underlying unoriented segment is one-sided [two-sided].If P, Q ∈ a ∩ b are two intersection points of a ∩ b consecutive on b , then by P Q we denote an oriented segment of b with endpoints P and Q . Oriented segments P P (cid:48) and QQ (cid:48) of b are called adjacent if both are one-sided and an open disk ∆exists on N with the following properties(1) ∂ ∆ consists of the segments P P (cid:48) , QQ (cid:48) of b and the arcs P Q , P (cid:48) Q (cid:48) of a ;(2) ∆ is disjoint from a ∪ b (Figure 4). Figure 4.
Adjacent segments of b .Oriented segments p (cid:54) = q are called joinable if oriented segments p , . . . , p k existsuch that p = p , p k = q and p i is adjacent to p i +1 for i = 1 , . . . , k − a (with respect to b ) and theirproperties. Remark . The main reason for the importance of adjacent/joinable segments of b is that they provide natural reductions of the intersection points of t a ( b ) and b (Figure 5). In fact, as observed in [5], these segments are the only nontrivial sourceof such reductions. UBGROUPS GENERATED BY TWO DEHN TWISTS . . . 5
Figure 5.
Joinable segments of b and reduction of intersectionpoints of b and t a ( b ).Let us recall some basic properties of joinable segments. Proposition 4.2 (Lemmas 3.4, 3.7 and 3.8 of [5]) . (1) Initial [terminal] points of oriented joinable segments of b are on the sameside of a . (2) Let p and q be oriented segments such that q begins at the terminal point of p (this includes the case q = − p ). Then p and q are not joinable. (cid:3) Still following [5], by a double segment of b , we mean an unordered pair of twodifferent oriented segments of b that have the same initial point. Exactly I ( a, b )double segments exist, which correspond to intersection points of a and b .Two double segments are called joinable if an oriented segment p exists in thefirst double segment and q exists in the other such that p and q are joinable.Studying the action of a twist t a on a circle b is important to obtain someobstructions for possible reductions of intersection points between t a ( b ) and b . Thebasic result in this direction is the following proposition: Proposition 4.3 (Lemma 3.9 of [5]) . Suppose I ( a, b ) ≥ . Then, for each doublesegment P , a double segment Q (cid:54) = P exists, which is not joinable to P . (cid:3) Curves in the neighborhood of a ∪ b A regular neighborhood N a ∪ b of a ∪ b is fixed. Topologically, N a ∪ b is the unionof regular neighborhoods N a and N b of a and b respectively. By changing N a , N b and N a ∪ b into their closures we can assume that all these sets are closed. If wedefine N a \ b = N a \ N b , N b \ a = N b \ N a , N a ∩ b = N a ∩ N b , then N a ∪ b = N a ∪ N b = N a \ b ∪ N b \ a ∪ N a ∩ b , where each three sets on the right hand side consist of I ( a, b ) disks with disjointinteriors. These disks correspond to the intersection points of a and b (Figure 6).We consider these disks as rectangles with two opposite sides parallel to a andthe other two parallel to b . The rectangles in N a \ b and N b \ a have a one-to-onecorrespondence with the segments of a and b , respectively.If r is one of the rectangles that constitute N a ∪ b and c is a circle in N a ∪ b thatintersects a ∪ b transversally, then by an arc of r ∩ c , we mean a connected componentof r ∩ c .Let C be the family of generic circles on N that satisfy the following properties:(1) Each circle in C is contained in N a ∪ b and intersects a ∪ b transversally.(2) Each intersection point of c and a ∪ b is contained in N a ∩ b . MICHA(cid:32)L STUKOW
Figure 6.
Neighborhood of a ∪ b as the union of 3 · I ( a, b ) rectangles.(3) If c ∈ C and r is one of the rectangles in N a ∪ b , then each arc of c ∩ r hasendpoints on two different sides of r .The third condition simply means that c does not turn back when crossing a rec-tangle. Each generic circle contained in N a ∪ b is obviously isotopic to a circle in C . Let c ∈ C and let r be one of the rectangles in N a ∪ b . If an arc of c contained in r crosses both sides of r parallel to a , then we say that r contains an arc of c parallel to b .If every rectangle in N b contains an arc of c parallel to b , then we say that c winds around b . Clearly, the sufficient condition for a circle c ∈ C to wind around b is that each rectangle in N a ∩ b contains an arc of c parallel to b .If r is a rectangle in N b \ a and r contains an arc q of c parallel to b , then we saythat q is a segment of c . Moreover, if p is a segment of b that corresponds to r ,then we say that q is parallel to p . Similarly, we define segments of c parallel tosegments of a . Lemma 5.1.
Let c ∈ C such that c winds around b , and let a (cid:48) be one of thecomponents of ∂N a . If | c ∩ a | = I ( c, a ) and ∆ is a bigon formed by c and a (cid:48) , then ∆ ⊂ N a .Proof. Suppose that a bigon ∆ with sides p ⊂ a (cid:48) and q ⊂ c exists, which is notcontained in N a . If ∆ and N a are on the same side of p (Figure 7), then we canfind a smaller bigon ∆ (cid:48) ⊂ ∆ with sides p (cid:48) ⊂ ∂N a and q (cid:48) ⊂ c such that ∆ (cid:48) and N a are on different sides of p (cid:48) (because ∆ \ N a (cid:54) = ∅ ). Hence we can assume that ∆ and Figure 7.
Case of ∆ and N a being on the same side of a (cid:48) , Lemma 5.1. N a are on different sides of p .If c intersects the interior of p , then we can pass to a smaller bigon that is stillnot contained in N a . Hence, we can assume that p ∩ c consists of two points A and B (note that q may still intersect a (cid:48) ). UBGROUPS GENERATED BY TWO DEHN TWISTS . . . 7
Let r A and r B be rectangles of N a ∩ b that correspond to A and B , respectively.If r A = r B , then the segments of q that start at A and B terminate at the samerectangle of N a ∩ b (Figure 8). Hence, we can pass to a smaller bigon ∆ (cid:48) ⊂ ∆ byremoving these segments of q . The obtained bigon ∆ (cid:48) is still not contained in N a because this would imply that c is not in C ( c would need to turn back in one ofthe rectangles of N a ). Hence we can assume that r A (cid:54) = r B . Figure 8.
The case of r A = r B , Lemma 5.1.Let c A , c B ⊂ N a be arcs of c that start at A and B , respectively.Recall that we assumed that c winds around b . Hence, r A and r B contain arcsof c that are parallel to b . Therefore, c A either crosses a in r A or c A turns in r A in the direction of p , and after running parallel to p , c must turn and cross a in r B . In fact, c cannot turn towards p and it cannot cross r B because r B contains anarc of c parallel to b (Figure 9(i)). Similar analysis applied to c B shows that the Figure 9.
Configurations of arcs, Lemmas 5.1 and 5.2.arc of c ∩ N a that contains c B must intersect a ; it can do it in r B , or in r A afterrunning parallel to p . However, this implies that c and a form a bigon, which is acontradiction. (cid:3) Lemma 5.2.
Let A and B be the endpoints of an arc b (cid:48) contained in one of thecomponents of ∂N b ∩ N a . Let c ∈ C be such that c winds around a , and let q ⊂ c be an arc with endpoints A and B which starts and ends on the same side of thecomponent of ∂N b containing b (cid:48) . Then b (cid:48) and q do not form a bigon with interiordisjoint from b ∪ c .Proof. Suppose to the contrary that b (cid:48) and q form a bigon ∆ with interior disjointfrom b ∪ c and consider the arcs of q that start at A and B [Figure 9(ii)]. If thesearcs enter some rectangle r of N a ∪ b , then they must be parallel in r , that is, they MICHA(cid:32)L STUKOW are disjoint and intersect the same sides of r . Clearly, this condition is true forrectangles in N a ∪ b \ N a ∩ b (since c ∈ C ), and for rectangles in N a ∩ b , this followsfrom our assumptions that the interior of ∆ is disjoint from b ∪ c and that c windsaround a . However, this implies that the arcs of q which start A and B will nevermeet. (cid:3) Let c ∈ C and p be one of the arcs of c ∩ N a . Four different possible configurationsof p exist [Figure 10] and are referred to as types A–D. Figure 10.
Possible configurations of arcs of c ∩ N a . Remark . If c winds around b , then arcs of c ∩ N a of types B–D can pass throughonly one rectangle in N a \ b . Otherwise, c would intersect itself.6. Rigidity of circles in C Remark . As we mentioned in Remark 4.1, adjacency between segments of b isthe only nontrivial source of reductions of the intersection points between b and t a ( b ). However, if we consider the intersection points between b and t a ( c ) for c ∈ C ,then other kinds of reductions exist. For example, if ∆ is a component of N \ N a ∪ b ,which is a disk, then t a ( c ) and b may reduce along ∆ (Figure 11). As we will see Figure 11.
Exterior hexagon ∆ and the reduction of intersectionpoints between b and t a ( c ).later, this type of reduction is rather exceptional, but we need additional definitionsto control it.Suppose that ∆ is a component of N \ N a ∪ b , which is a disk. If the boundaryof ∆ intersects exactly n boundaries of rectangles r , r , . . . , r n in N a ∩ b , then wesay that ∆ is an exterior n -gon with vertices r , r , . . . , r n (Figure 11). Note that r , . . . , r n does not need to be pairwise distinct (∆ may intersect r i in each of itsfour corners).Let p be an arc of c ∈ C , which is parallel to b in a rectangle r of N a ∩ b . Fix someorientation of p and follow p to the rectangle r of N a ∩ b following r . We say that UBGROUPS GENERATED BY TWO DEHN TWISTS . . . 9 p is rigid in r with respect to b if p is parallel to b in r . Equivalently (from theperspective of p intersecting N a ), p is of type A in r and r .We say that c ∈ C winds strongly around b if for every rectangle r in N a ∩ b anoriented arc p that is parallel to b in r exists, such that both p and − p are rigid in r . Equivalently, for each three rectangles r , r , r of N a ∩ b , which are consecutivealong b , an arc of c ∩ N b exists, which is parallel to b in r , r , and r . In particular,if a circle c ∈ C winds strongly around b , then c winds around b .Let R be a double segment of b and let p (cid:54) = q be oriented segments of b that arenot contained in R and do not start at the same point of a ∩ b . Assume also that p and q start on different sides of a . If we fix an orientation of N a , then two possiblemutual positions of p, q , and R exist (Figure 12). We say that the triple { p, q, R } Figure 12.
Two possible orientations of { p, q, R } .is positively oriented if the configuration is as in Figure 12(i).As we will see later, some special configurations of generic two-sided circles a, b exist in N , which will require additional analysis (Sections 9–12). These specialconfigurations are defined by the following properties:(S1) I ( a, b ) ≥ p, q of b exist, which start on differentsides of a such that each double segment of b contains an oriented segmentjoinable to p or q (see Figure 38 in Section 9).(S2) I ( a, b ) ≥ p, q of b , and a double segment R of b exist, such that: a and b are not in the special position (S1), p and q starton different sides of a , p starts and terminates on different sides of a , eachdouble segment of b different from R contains an oriented segment joinableto p or q , { p, q, R } is positively oriented (see Figure 38 in Section 9).(S3) I ( a, b ) ≥ p, q of b and a double segment R of b such that: p starts and terminates on one side of a , q starts andterminates on the other side of a , each double segment of b different from R contains an oriented segment joinable to p or q , { p, q, R } is positivelyoriented (see Figure 42 in Section 10).If one of the ordered pairs of circles ( a, b ) or ( b, a ) satisfies one of the above condi-tions, then we say that the unordered pair { a, b } is special .Let X a be the set of isotopy classes of circles c in N , which satisfy the followingconditions:(1) c ∈ C ,(2) I ( c, a ) = | c ∩ a | , I ( c, b ) = | c ∩ b | ,(3) I ( c, a ) < I ( c, b ),(4) c winds strongly around a .Similarly, we define X b by requiring (1)–(2) above and additionally (3’) I ( c, b ) < I ( c, a ),(4’) c winds strongly around b .7. The case of I ( a, b ) ≥ and { a, b } not special The main goal of this section is to prove the following:
Proposition 7.1.
Let a and b be two generic two-sided circles in N such that I ( a, b ) ≥ and { a, b } is not special. Then for any integer k (cid:54) = 0 we have t ka ( X b ) ⊆ X a and t kb ( X a ) ⊆ X b . Proof.
Of course proving that t ka ( X b ) ⊆ X a is sufficient. No canonical choice existsfor the orientation of the neighborhood N a . However, in our figures, we will assumethat t ka twists to the right. Construction of t ka ( c ) . Fix a circle c ∈ X b . It is enough to prove that t ka ( c ) ∈ X a .We begin by constructing the circle d = t ka ( c ). Outside N a and on each arc of d ∩ N a of type D, d is equal to c . For each arc of c ∩ N a of types A–C, d circles | k | times around a . In particular d winds around a and I ( d, a ) = | d ∩ a | = | c ∩ a | = I ( c, a ) | d ∩ b | = I ( c, a ) · I ( a, b ) · | k | . Now the problem is that d may not be an element of C and d does not need to bein a minimal position with respect to b .Before we start to reduce d , observe that if an arc of d enters N a and turns tothe left, then after passing through one rectangle in N a \ b it must turn back or leave N a through the same side of N a as it entered N a (Figure 13). In fact, arcs of d Figure 13.
Arcs of d turning to the left in N a .turning to the left in N a came from arcs of c ∩ N a of types C and D. As we observedin Remark 5.3 such arcs can pass through only one rectangle in N a \ b . Reduction of type I.
Suppose that one of the rectangles r in N a ∪ b contains anarc p of d such that the endpoints of p are on the same side of r ( d turns back in r ). Clearly, this situation cannot happen for r being one of the rectangles in N b \ a (because in such a rectangle, d coincides with c ) or r being a rectangle in N a \ b (by construction d runs parallel to a in each such rectangle). Hence, r must be arectangle in N a ∩ b and p must intersect the b -side of r (otherwise c would not be anelement of C ). Hence, we have the situation illustrated in Figure 14, and we canreplace d with the circle d (cid:48) shown in the same figure. In such a case, we say thatwe reduced d by a reduction of type I .Reductions of type I correspond to arcs of c ∩ N a of type C. Hence, on each arcof d ∩ N a , at most one reduction of type I exists. UBGROUPS GENERATED BY TWO DEHN TWISTS . . . 11
Figure 14.
Reduction of type I.Let d be the circle obtained from d by performing all possible reductions of typeI. In particular, d ∈ C . Remark . The only arcs of d ∩ N a that turn to the left after entering N a arearcs that correspond to (in fact, equal to) arcs of c ∩ N a of type D.We now argue that d winds around a . In fact, if we fix a rectangle r in N a ∩ b and r (cid:48) is another rectangle in N a ∩ b , then (because c winds around b ) r (cid:48) containsan arc q of c parallel to b (Figure 15). Now t ka ( q ) is strictly monotone in N a with Figure 15.
Action of t a on an arc q in a rectangle r (cid:48) .respect to a . Hence, this arc does not admit a reduction of type I. In particular, t ka ( q ) gives an arc of d , which is parallel to a in r .For further reference, note the following observation: Lemma 7.3.
Let q ⊂ d be an arc with endpoints A, B ∈ ∂N a such that q ∩ b = ∅ .Assume also that q intersects both N a \ b and N b \ a . Then q is an arc of c .Proof. By construction, d coincides with c in every rectangle of N b \ a and if q enters N a in a rectangle r (Figure 16), then q must leave r through the b -side of r giventhat d winds around a . Afterwards, q goes through one rectangle of N a \ b , and Figure 16.
Configuration of arcs – Lemma 7.3.then it must turn and leave N a (because q cannot intersect b ). Hence, each arc of q ∩ N a is an arc of type D, which proves that q is an arc of c . (cid:3) Reduction of type II.
Suppose that there exist arcs p and q of b and d , respec-tively, such that • p and q form a bigon with interior disjoint from b ∪ d , • p \ N a ∩ b is a subarc of a two-sided segment of b .In such a case, we can remove the bigon formed by p and q (Figure 17), and we saythat we reduced d to d (cid:48) by a reduction of type II . Figure 17.
Reduction of II.Let us describe the possible reductions of type II in more detail. Let A and B bevertices of the bigon ∆ formed by p and q , and let r A , r B be the rectangles of N a ∩ b that contains A and B , respectively. By Lemma 5.2, the arc q A ⊂ q , which connects A with the boundary of N a , is either entirely contained in r A or it passes throughone rectangle of N a \ b and then leaves N a (it can pass at most one rectangle of N a \ b because otherwise it would intersect b ). In other words the situation illustrated inFigure 18 is not possible. The same is true for the arc q B ⊂ q , which connects B Figure 18.
Types II and III reductions – impossible configurationof arcs.with the boundary of N a . For the same reason, either both arcs q A , q B are entirelycontained in r A , r B , or each of them passes through one rectangle of N a \ b .Given that we assume that p corresponds to a two-sided segment of b , at oneendpoint of q , say B , q turns to the left as it enters N a . We claim that q consistsof q A , q B and a single segment of c parallel to b . In order to prove this statement,showing that q B is entirely contained in r B is sufficient. Suppose to the contrarythat q B passes through a rectangle of N a \ b (Figure 19). Given that q B turns to theleft after entering N a , by Remark 7.2, this arc must be an arc of type D. Hence,after crossing p in B it must follow an arc t of d , which turns left in r B and leaves N a (it must leave N a in r B by Remark 5.3). After leaving N a , t runs parallel to p in a rectangle of N b \ a . In particular, q B and t are arcs of c [given that ( q B ∪ t ) ∩ N a is an arc of type D]. Moreover, by Lemma 7.3, q \ ( q A ∪ q B ) is also an arc of c .Hence, ( q \ q A ) ∪ t ⊂ c and an arc of ∂N a form a bigon ∆ (cid:48) not contained in N a (Figure 20). This finding contradicts Lemma 5.1. UBGROUPS GENERATED BY TWO DEHN TWISTS . . . 13
Figure 19.
Reduction of type II – impossible configurations of arcs.
Figure 20.
Reduction of type II – extending ∆ to ∆ (cid:48) .Therefore, we proved that if d (cid:48) and d differ by a reduction of type II thatcorresponds to the bigon formed by arcs p and q of b and d , respectively, then p and q are in the same rectangle of N b \ a , and q is parallel to p . This conditionmeans that d and d (cid:48) intersect rectangles in N a ∪ b in exactly the same way. Hence, d (cid:48) ∈ C and d (cid:48) winds around a .Let d be the circle obtained from d by performing all possible reductions oftype II. As we observed, d ∈ C and d winds around a . Remark . In the following, we use Lemma 7.3 with d instead of d . Thisapproach is somewhat problematic, because some arcs of d that satisfy the as-sumptions of that lemma may admit reductions of type II. To solve this problem,we mimic reductions of type II on the level of c . To be more precise, if q is an arc of c ∩ N b \ a which as an arc of d admits a reduction of type II, then we push q so thatit coincides with the corresponding arc of d (Figure 21). Then we extend this push Figure 21.
Applying reductions of type II to c .to the isotopy of c . If c is a circle obtained from c by all possible reductions of typeII (in the sense described above), then c still winds around b . Hence, it satisfies the assumptions of Lemma 5.1. Moreover, we have the following replacement forLemma 7.3: Lemma 7.5.
Let q ⊂ d be an arc with endpoints A, B ∈ ∂N a such that q ∩ b = ∅ .Assume also that q intersects both N a \ b and N b \ a . Then q is an arc of c . (cid:3) Reduction of type III.
Suppose that there exist arcs p and q of b and d respec-tively such that • p and q form a bigon with interior disjoint from b ∪ d , • p \ N a ∩ b is a subarc of a one-sided segment of b .In such a case we can remove the bigon formed by p and q , see Figure 22. We say Figure 22.
Reduction of type III.that we reduced d to d (cid:48) by a reduction of type III .Let us attempt to understand reductions of type III in further detail.Let A and B be the vertices of the bigon ∆ formed by p and q , and let r A , r B bethe rectangles of N a ∩ b that contains A and B , respectively. By Lemma 5.2, the arc q A ⊂ q that connects A with the boundary of N a is either entirely contained in r A or it passes through one rectangle of N a \ b and then leave N a (it can pass throughat most one rectangle of N a \ b because otherwise it would intersect b ). In otherwords, the situation illustrated in Figure 18 is not possible. The same is true forthe arc q B ⊂ q , which connects B with the boundary of N a . For the same reason,either both arcs q A , q B are entirely contained in r A , r B , or each arc passes throughone rectangle of N a \ b .If q \ ( q A ∪ q B ) intersects N a , then by Lemma 7.5, this is an arc of c (see Remark7.4). In particular, all components of ( q \ ( q A ∪ q B )) ∩ N a are arcs of type D (onarcs of other types, d does not coincide with c ).Observe also that q turns to the right when it enters N a (because q is one-sided,it enters N a in the same way on both ends). In fact, q A and q B would be arcs oftype D otherwise (Remark 7.2), which are not involved in reductions of type II.Hence, q A and q B would be arcs of c (Figure 23). Therefore, the whole q would be Figure 23.
Reduction of type III – impossible configuration of arcs.an arc of c , and that would imply that c and b form a bigon. UBGROUPS GENERATED BY TWO DEHN TWISTS . . . 15
Remark . Observe that d (cid:48) ∈ C . In fact, if d (cid:48) turns back in one of the rectangles r of N a ∪ b , then r must be a rectangle that contains one of the vertices of the bigon,which leads to the reduction (in all other rectangles, we either did not changeanything, or d (cid:48) runs parallel to b in them). Since d (cid:48) enters r through the a -side,after turning back, it must leave r also through the a -side. Hence, we have the Figure 24.
Reduction of type III – impossible configuration of arcs.situation shown in the right-hand side of Figure 24. The left-hand side of the samefigure shows how the situation looked before the reduction.Let t be an arc of d following q past the intersection point p ∩ q ∩ r . Arc t turnsright in r and leaves r as an arc parallel to p : see Figure 25. The same figure shows Figure 25.
Reduction of type III – impossible configuration of arcs.how the reduction disk ∆ can be deformed to the closed disk ∆ (cid:48) with boundarycomposed of q, t , and an arc of ∂N a . Note that q ∩ ∆ (cid:48) is an arc of c by Lemma7.5 and t is an arc of c because d intersects r as an arc of type D. Therefore theexistence of ∆ (cid:48) contradicts Lemma 5.1. Hence, we proved that d (cid:48) ∈ C (in particular, d (cid:48) does not admit reductions of type I). Remark . Reduction of type III does not create any new arcs of d (cid:48) ∩ N a , whichturn to the left after entering N a . This reduction does not affect the segments of d (cid:48) ,which run parallel to two-sided segments of b . Hence, d (cid:48) does not admit reductionsof type II.Before we go further, we divide possible reductions of type III into 3 subtypes.Let p and q be arcs of b and d respectively which define a reduction of type III. • If p and q enter N a in the same rectangles of N a ∩ b , then we say that p and q define a reduction of type IIIa . • If p and q enter N a in different rectangles of N a ∩ b and q meets a singlerectangle of N b \ a , then we say that p and q define a reduction of type IIIc . • Otherwise we say that p and q define a reduction of type IIIb . Reduction of type IIIa.
Let d be the circle obtained from d by performingall possible reductions of type IIIa. As we already observed, d ∈ C and d doesnot admit reductions of types I, II and IIIa (Remarks 7.6 and 7.7). Moreover, d
36 MICHA(cid:32)L STUKOW intersects the rectangles of N a ∪ b in exactly the same way as d , hence d windsaround a . Remark . We now follow Remark 7.4 to obtain a version of Lemma 7.3 for d .As in the construction of c , we mimic the reductions of type IIIa on c (Figure26). However, we do not need to mimic all reductions of type IIIa. To be more Figure 26.
Applying reductions of type IIIa to c .precise, we apply to c only these reductions of type IIIa, which are determined byan arc q of c , which enters N a on one side as an arc of type D ( q cannot enter N a on both sides as an arc of type D: this condition would imply that c and b bounda bigon). If q is an arc of c that enters N a on both sides as an arc of types A–C,then the corresponding arc of d does not satisfy the assumptions of Lemma 7.3(on both ends, it intersects b ). Hence, we do not consider these arcs as they do notmimic the setup of Lemma 7.3.Let c be the arc obtained from c by reductions of type IIIa described above.Since c still winds around b , it satisfies the assumptions of Lemma 5.1. As areplacement for Lemma 7.3, we have the following lemma: Lemma 7.9.
Let q ⊂ d be an arc with endpoints A, B ∈ ∂N a such that q ∩ b = ∅ .Assume also that q intersects both N a \ b and N b \ a . Then, q is an arc of c . (cid:3) Reduction of type IIIb.
Let d (cid:48) be the circle obtained from d by a single reductionof type IIIb. As in the general definition of the reduction of type III, by A and B we denote vertices of the bigon formed by p and q , r A and r B are rectangles of N a ∩ b that contains A and B respectively, and q A , q B are arcs of q that connect A and B with the boundary of N a . Remark . By definition, q \ ( q A ∪ q B ) is not a single segment of d parallel to b . Hence, it intersects N a at least once. By Lemma 7.9, q \ ( q A ∪ q B ) is an arc of c , so it intersects N a only in arcs of type D (on arcs of other types, d does notcoincide with c ).The existence of a bigon with vertices A, B between d and b implies that q and p bound an exterior n -gon ∆ (see Section 6 and the right-hand side of Figure27). Moreover, if the reduction is not of type IIIc, then n ≥
6. In such a case, anoriented arc s of c which is of type A in a rectangle r of N a ∩ b (see the left-hand sideof Figure 27) and is rigid in r with respect to b , cannot yield an arc of d , whichallows a reduction of type IIIb (because if we follow s to the next rectangle r (cid:48) of N a ∩ b then s is of type A in r (cid:48) and not of type D). Remark . Our assumption that c winds strongly around b (see Section 6) impliesthat for each rectangle r of N a ∩ b , an arc s of c exists, which is of type A in r andsuch that s yields an arc of d , which does not allow reductions of type IIIb on UBGROUPS GENERATED BY TWO DEHN TWISTS . . . 17
Figure 27.
Exterior hexagon ∆ and the reduction of intersectionpoints between b and t a ( c ).either side of r : for such s , an arc of c that is rigid on both sides of r should beselected. Lemma 7.12. If s (cid:48) is an arc of c of type A in a rectangle r of N a ∩ b , then s (cid:48) canyield a reduction of type IIIb only on one side of r .Proof. Suppose to the contrary that s (cid:48) yields an arc s of d which allows reductionsof types IIIb on both sides of r and assume that r , r, r are rectangles of N a ∩ b thatare consecutive along b (Figure 28). By Remark 7.10, s coincides with s (cid:48) in r and Figure 28.
Configuration of arcs – Lemma 7.12. r and enters each of these rectangles as an arc of type D. However, this conditionimplies that each arc t of c that is of type A in r must enter either r or r as anarc of type D (because it must follow s (cid:48) along the boundary of one of the exterior n -gons leading to reductions of s ). This contradicts our assumption that c windsstrongly around b . (cid:3) Lemma 7.13.
The arcs of c ∩ N a that correspond to q A and q B are arcs of typesA or B.Proof. If for example q A came from an arc of c of type D, then by Lemma 7.9, q \ q B is an arc of c and if we follow this arc past the point A , we obtain an arcof c which, together with an arc of ∂N a , bounds a disk ∆ that is not contained in N a [Figure 29(i)]: this contradicts Lemma 5.1.As for arcs of type C, if for example q \ q B was formed from the arc q (cid:48) of c oftype C and r is the rectangle of N a ∩ b in which q A meets ∂N a , then all the arcs of c that are parallel to b in r are on one side of q (cid:48) in r and they are bounded alongthe boundary of ∆ by q (cid:48) [Figure 29(ii)]. Hence, all these arcs must lead to arcs of d , which allow a reduction of type IIIb across ∆. In particular, if we follow thesearcs along the boundary of ∆, then they enter N a as arcs of type D . However, this Figure 29.
Reduction of type IIIb – impossible configurations of arcs.contradicts our assumption that arcs that are rigid with respect to b exist on bothsides of r (because c winds strongly along b ).Hence we proved that q A and q B were formed from arcs of c which are of typesA or B in r A and r B . (cid:3) Note that the distinction between types B and C of arcs of c ∩ N a is a consequenceof our assumption that t a twists to the right in N a . Remark . As a consequence of Lemma 7.13, if q (cid:48) is an arc of d (cid:48) obtained from q by a reduction of type IIIb, then the arcs of d (cid:48) ∩ N a that follow q (cid:48) (on each end)are not arcs of type D (because they intersect a ). Hence, q (cid:48) is not involved in anyfurther reductions of d (cid:48) of type IIIb (Remark 7.10). In other words, no cascadereductions of type IIIb exist: each arc of d ∩ N b \ a is involved in at most one suchreduction. However, q (cid:48) may be further reduced by reductions of type IIIc. Wepostpone this problem to the analysis of reductions of type IIIc. Remark . If q (cid:48) is as in the previous remark (that is, q (cid:48) is an arc of d (cid:48) obtainedfrom q by a reduction of type IIIb), then q (cid:48) does not admit a reduction of type IIIa.This conclusion follows because the arcs of d (cid:48) ∩ N a that follow q (cid:48) (on each end) donot intersect b in rectangles of N a ∩ b in which they enter N a (Figure 22). The onlyarc of d (cid:48) ∩ N b \ a in which d (cid:48) differs from d is q (cid:48) . Thus, no new reductions of typeIIIa exist on other arcs of d (cid:48) ∩ N b \ a .Let d be the circle obtained from d by performing all possible reductions oftype IIIb. As we observed in the general analysis of reductions of type III, d ∈ C (Remark 7.6) and d does not admit any reductions of types I and II (Remarks 7.6and 7.7). We also proved that d does not admit reductions of type IIIa (Remark7.15). Remark . A segment q of d obtained by a reduction of type IIIb can not enter N a as an arc of type D (Lemma 7.13). We will show later (Lemma 7.20) that insuch a case arcs that follow/precede q in N a must intersect b . Hence Lemma 7.9remains valid with d replaced by d (arcs modified by a reduction of type IIIb cannot satisfy assumptions of that lemma). Reduction of type IIIc.
Let d (cid:48) be the circle obtained from d by a single reductionof type IIIc. As in the general definition of the reduction of type III, by A and B we denote vertices of the bigon formed by p and q , r A and r B are rectangles of N a ∩ b that contain A and B respectively, and q A , q B are arcs of q that connect A and B with the boundary of N a . UBGROUPS GENERATED BY TWO DEHN TWISTS . . . 19
Remark . Reduction of type IIIc may be considered a special case of a reductionof type IIIb, where the exterior disk is a rectangle. For this reason, these two typesof reductions have some common properties: • the arcs of c that correspond to q A and q B cannot be of type D (Lemma7.13); • if an arc q (cid:48) is an arc of d (cid:48) obtained by a reduction of type IIIc, then q (cid:48) isnot involved in reductions of type IIIa nor IIIb (Remarks 7.14 and 7.15); • d (cid:48) does not admit reductions of type IIIa nor IIIb.The proofs of the above properties can be copied verbatim from the analysis of thereduction of type IIIb. Hence, we skip the proofs.Let d be the circle obtained from d by performing all possible reductions oftype IIIc. As we observed in the general analysis of reductions of type III, d ∈ C (Remark 7.6) and d does not admit any reductions of types I and II (Remark 7.7).As we noted above (Remark 7.17), d also does not admit reductions of type IIIaand IIIb. Remark . As in Remark 7.16, a segment q of d obtained by a reduction of typeIIIc cannot enter N a as an arc of type D. We will show later (Lemma 7.20) that insuch a case, arcs that follow/precede q in N a must intersect b . Hence Lemma 7.9remains valid with d replaced by d (arcs modified by a reduction of type IIIc cannot satisfy assumptions of that lemma). Remark . An arc s of d ∩ N a can be involved in multiple reductions of typeIIIc (Figure 30), that is, the arcs that start at the end-points of s can be involved inseveral reductions of type IIIc. However, all these reductions can change the initial Figure 30.
Cascade reductions of type IIIc.and the terminal rectangles r , r of s only to the rectangles joinable to r and r ,respectively (see Section 4). Lemma 7.20.
Let s (cid:48) be an arc of c ∩ N a of type A, B or C, and let s be the arc of d ∩ N a which corresponds to s (cid:48) . Then s is parallel to a in at least one rectangle of N a ∩ b .Proof. Let s (cid:48)(cid:48) be the arc of d that corresponds to s (cid:48) (that is, s (cid:48)(cid:48) is obtained from t ka ( s (cid:48) ) by reductions of types I, II, IIIa), and let p and q be segments of b thatcorrespond to the segments of d that start at the endpoints of s .Assume first that s (cid:48) is an arc of type C. Arcs of this type do not allow reductionsof type IIIb (Lemma 7.13). Hence, the only type of reductions that could decrease the number of rectangles in which s (cid:48)(cid:48) is parallel to a is the reduction of type IIIc.If we assume that no rectangles of N a ∩ b exist in which s is parallel to a , then eachdouble segment of b must contain a segment joinable either to p or q (Remark7.19). However, this implies that a and b are in the special position (S1), whichcontradicts our assumption that { a, b } is not special.If s (cid:48) is an arc of type A or B, then the situation is completely analogous. If s (cid:48) isof type A, then s (cid:48)(cid:48) can admit on one side one reduction of type IIIb (Lemma 7.12),and if s (cid:48) is of type B, then s (cid:48)(cid:48) can admit reductions of type IIIb on both sides (oneon each side – Remark 7.14). Hence, the assumption that no rectangle of N a ∩ b exists in which s is parallel to a implies that a and b are in the special position(S1). (cid:3) Strong winding of d . We need to show that for each three rectangles r , r, r of N a ∩ b , which are consecutive along a , an arc of d exists, which is parallel to a in r , r, r . Without loss of generality, we can assume that the configuration ofrectangles is as in Figure 31, that is, r is on the right of r . Let p and q be segments Figure 31.
Intersection of t ka ( s (cid:48) ) with rectangles of N a ∩ b .of b such that p goes up from r and q goes down from r , and let R be the doublesegment of b that corresponds to r .Let r (cid:48) be any rectangle of N a ∩ b that is different from r , r and r (such rectanglesexist given that I ( a, b ) ≥ s (cid:48) be an arc of c ∩ N a ∩ b that is rigid on bothsides of r (cid:48) (it exists because c winds strongly around b ). If s is an arc of d thatcorresponds to s (cid:48) , then s does not allow reductions of type IIIb on either side of r (cid:48) . Hence, it can only be reduced by reductions of type IIIc. By Remark 7.19, ifwe assume that s can be reduced so that the corresponding arc of d is not parallelto a in r or r , then one of the segments of b that start in r (cid:48) must be joinable toeither p or q .From the above analysis, if we assume that each arc s (cid:48) of c ∩ N a ∩ b leads to anarc s of d ∩ N a that is not parallel to a in either r or r , then each double segmentof b different from R is joinable either to p or q . Moreover, the triple { p, q, R } ispositively oriented (see Section 6). Hence, we are in the special position (S2) or(S3), which is a contradiction. Bigons formed by d and b . Let us prove that d and b are in the minimalposition so they do not form any bigon. Suppose on the contrary that ∆ is a bigonwith vertices A and B bounded by arcs p and q of d and b , respectively. By takingthe inner most bigon, we can assume that the interior of ∆ is disjoint from d ∪ b .Since d winds around a , each rectangle of N a ∩ b contains an intersection pointof d and b . Hence, q is either an arc of b in a single rectangle r A,B in N a ∩ b , or q is a segment of b that connects two different rectangles r A and r B of N a ∩ b . In thesecond case, d would admit a reduction of type II or III (depending on whether q istwo-sided or not). Hence, we concentrate on the first possibility. If ∆ is contained UBGROUPS GENERATED BY TWO DEHN TWISTS . . . 21 in N a , then d admits a reduction of type I, which is not possible. Hence ∆ is notcontained in N a .Let p (cid:48) be the subarc of p , which is obtained from p by removing the arcs containedin N a that connect A and B with the boundary of N a (Figure 32). By Lemma 7.9 Figure 32.
Bigon ∆ between b and d .and Remark 7.18, the arc p (cid:48) of d is in fact an arc of c . Hence, the existence of ∆contradicts Lemma 5.1 Counting intersection points between d and b . To finish the proof, we needto show that I ( d , b ) > I ( d , a ). The idea is to show that associated intersectionpoints of d ∩ b exist for each intersection point of d ∩ a .These arcs of d ∩ N a that intersect a have a one-to-one correspondence to arcsof c ∩ N a that intersect a , hence to arcs of types A–C. Moreover, all reductions weperformed on d preserved this bijection because during the reductions we did notcreate any new arcs that intersect a , and we did not remove any of the existingones.By Lemma 7.20, each arc s of d ∩ N a that corresponds to an arc s (cid:48) of c ∩ N a oftype A–C is parallel to a in at least one rectangle of N a ∩ b . Hence, it intersects b atleast once. Moreover, because d winds strongly around a , some arcs s of d ∩ N a intersect b in at least 3 points. Therefore I ( d , b ) > I ( d , a ). (cid:3) Weak rigidity
The proof of Proposition 7.1 is based on the notion of strong winding: for eachthree rectangles r , r , r that are consecutive on a , an arc of c ∩ N a exists, whichleads to an arc of d ∩ N a parallel to a in r , r , and r . This assertion can fail inspecial cases (S1)–(S3). Example . Let a and b be two circles that are in the special position (S1), asshown in Figure 33(i). Segments p and p are adjacent, thus the arcs of d , which Figure 33.
Possible failure in the proof of Proposition 7.1 – Ex-amples 8.1 and 8.2. are obtained from arcs of c of type A in the rectangle r , may admit a reduction oftype IIIc. Hence, none of these arcs may be parallel to a in r . Similarly, all arcs of d that are obtained from arcs of c of type A in r may reduce by reductions of typeIIIc. Hence, none of these arcs may be parallel to a in r . Therefore, a possibilitythat no arc of d ∩ N a which is parallel to a in r and r exists (hence, d does notwind strongly around a ). Example . Let a and b be two circles that are in the special position (S3) asshown in Figure 33(ii). Segments p and p are adjacent. Thus, the arcs of c ∩ N a which are of type A in r or r give arcs of d that can be reduced so that theresulting arcs of d ∩ N a are not parallel to a in r . Similarly, arcs of c ∩ N a that areof type A in r or r yield arcs of d ∩ N a that are not parallel to a in r . Therefore,a possibility is that no arc of d ∩ N a that is parallel to a in r and r exists (hence, d does not wind strongly around a ).To overcome the abovementioned problems, we redefine the rigidity of arcsslightly.Let r , r , r , r , r , r be rectangles of N a ∩ b that are consecutive vertices of anexterior hexagon ∆ (consecutive here means consecutive along ∂ ∆). Suppose alsothat q is an arc of a circle c ∈ C as in Figure 35(i), that is, q is of type A in r and r , q is of type D between r and r , and the segment of b that connects r and r is one-sided. In such a case we say that q is a one-sided boundary 3-segment of ∆. Remark . Let ∆ be an exterior hexagon and let q be an arc of c , which leads toan arc q (cid:48) of d and allows a reduction of type IIIb across ∆ [Figure 35(i)]. Accordingto the definition of the reduction of type IIIb, if q enters N a as an arc of type A(on both ends), then q is a one-sided boundary 3-segment of ∆.Let q be an arc of c ∈ C that is parallel to b in a rectangle r of N a ∩ b . Someorientation of q is fixed, and q is followed to the rectangles r , r , r of N a ∩ b following r . We say that q is weakly rigid in r with respect to b if either • q is rigid with respect to b in r , or • q does not intersect a in r and r , q is parallel to b in r , and q is not anone-sided boundary 3-segment of an exterior hexagon (Figure 34). Figure 34.
Arc q of c which is weakly rigid in a rectangle r of N a ∩ b .If c winds around b , then equivalently (from the perspective of q intersecting N a ), q is of type A in r and then either q is of type A in r , or q is of type D between r and r , q is of type A in r , and q is not a one-sided boundary 3-segment of anexterior hexagon.The following lemma shows that from the point of view of the proof of Proposition7.1, weakly rigid arcs are as good as rigid arcs. UBGROUPS GENERATED BY TWO DEHN TWISTS . . . 23
Lemma 8.4.
Suppose that c ∈ C winds around b and let q be an oriented arc of c which is weakly rigid in a rectangle r of N a ∩ b . Then t ka ( q ) does not admit areduction of type IIIb.Proof. Suppose that q after leaving r goes around the boundary of an exterior n -gon ∆ and then enters N a in a rectangle r of N a ∩ b as an arc of type A (Figure35(i)). If q leads to an arc that admits a reduction of type IIIb, then q between Figure 35.
Reductions of types IIIb and IIIc. r and r must follow n − N a \ b . Hence, it must n − N a as an arc of type D. However, by the definition of weak rigidity, q can intersect N a as an arc of type D only once. Hence, n = 6 and by Remark 8.3, q is a one-sided boundary 3-segment of ∆. However, this contradicts the assumption that q is weakly rigid in r . (cid:3) The next lemma shows that the reductions of types IIIb and IIIc produce arcsthat are good candidates for weakly rigid arcs.Let c, d , d be as in the proof of Proposition 7.1, that is, c ∈ C winds around b , d is obtained from t ka ( c ) by all possible reductions of types I–IIIa, and d isobtained from d by all possible reductions of types IIIb and IIIc. Assume alsothat c is such that the statements of Lemma 7.13 and Remark 7.17 holds true. Lemma 8.5.
Suppose that each arc of c ∩ N a of types A–C gives an arc of d ∩ N a that is parallel to a in at least one rectangle of N a ∩ b . Let q be an oriented arc of d which starts in a rectangle t of N a ∩ b as an arc parallel to a , then it follows a to thenext rectangle of N a ∩ b and then it follows an arc obtained from an arc of d \ N a byreductions of types IIIb and/or IIIc [Figure 35(ii)]. If q is not a one-sided boundary3-segment of an exterior hexagon, then q is weakly rigid in t with respect to a .Proof. Let q (cid:48) be an arc of d , which can be reduced to the arc q of d by reductionsof types IIIb and/or IIIc. Assume that the endpoints of q (cid:48) are A, B ∈ d ∩ b andlet r A , r B be rectangles of N a ∩ b that contain A and B , respectively. By the generalproperties of reductions of type III (see Lemma 7.13 and Remark 7.17), both endsof q (cid:48) were formed from the arcs of c intersecting a (hence not arcs of type D).Therefore, by our assumption, both ends of q after entering N a must run parallelto a in at least one rectangle of N a ∩ b . Orient q from r A to r B and extend q onboth ends so that it starts and terminates in rectangles r (cid:48) A and r (cid:48) B of N a ∩ b , whichrespectively precede and follow r A and r B along q . From the point of view of q intersecting N b , this arc is of type A in r (cid:48) A , then it intersects N b as an arc of typeD, and then it is again an arc of type A (in r (cid:48) B ). This finding means that if thisarc is not a one-sided boundary 3-segment of an exterior hexagon, then q is weaklyrigid in r (cid:48) A . (cid:3) We say that c ∈ C is weakly rigid with respect to b if each arc p of c that isparallel to b in a rectangle r of N a ∩ b is weakly rigid with some choice of orientation. Remark . If c is weakly rigid with respect to b , then by Lemma 8.4, c satisfiesthe statement of Lemma 7.12. Hence, the assumption that c is weakly rigid withrespect to b serves as a replacement for Lemma 7.12.We say that c ∈ C winds weakly around b if for every rectangle r of N a ∩ b andeach choice of orientation for the collection P of arcs of c that are parallel to b in r a weakly rigid arc exists in P . Remark . In the proof of Proposition 7.1, the assumption that c winds stronglyaround b was used to conclude that in each rectangle r of N a ∩ b an arc s of c exists,which is of type A in r and which leads to an arc of d ∩ N a that does not allowa reduction of type IIIb on either side of r (Remark 7.11). By Lemma 8.4, weakwinding provides a slightly weaker conclusion: for each a -side of r , an arc s of c exists, which is of type A in r and which yields an arc of d that does not allow areduction of type IIIb on the chosen side of r (for such s , choose an arc of c that isweakly rigid on the chosen side of r ). Remark . Remark 8.7 implies that the proof of Lemma 7.13 remains valid if wereplace the notion of strong winding with the notion of weak winding. In fact, thatproof was based on the fact that if r is a rectangle r of N a ∩ b , then for each a -sideof r , an arc s of c exists, which is of type A in r and which yields an arc of d thatdoes not allow a reduction of type IIIb on the chosen side of r .Let (cid:98) X a be the set of isotopy classes of circles c in N , which satisfy the followingconditions(1) c ∈ C ,(2) I ( c, a ) = | c ∩ a | , I ( c, b ) = | c ∩ b | ,(3) I ( c, a ) < I ( c, b ),(4) c is weakly rigid with respect to a ,(5) c winds weakly around a .Similarly, we define (cid:98) X b by requiring (1)–(2) above and additionally(3’) I ( c, b ) < I ( c, a ),(4’) c is weakly rigid with respect to b ,(5’) c winds weakly around b . Lemma 8.9.
Let c ∈ (cid:98) X b and assume that a and b are not in the special position(S1), (S2) nor (S3). Let d be as in the proof of Proposition 7.1 and let s (cid:48) be anarc of c ∩ N a of type A, B or C. If s is the arc of d ∩ N a which corresponds to s (cid:48) ,then s is parallel to a in at least two rectangles of N a ∩ b .Proof. Let s (cid:48)(cid:48) be the arc of d that corresponds to s (cid:48) (that is s (cid:48)(cid:48) is obtained from t ka ( s (cid:48) ) by reductions of types I, II, IIIa), and let p and q be segments of b thatcorrespond to the segments of d that start at the endpoints of s .Assume first that s (cid:48) is an arc of type C. Arcs of this type do not allow reductionsof type IIIb (Lemma 7.13 and Remark 8.8). Hence, the only type of reduction thatcould decrease the number of rectangles in which s (cid:48)(cid:48) is parallel to a is the reductionof type IIIc. If we assume that no rectangles of N a ∩ b exist in which s is parallel to a , then each double segment of b must contain a segment joinable either to p or q UBGROUPS GENERATED BY TWO DEHN TWISTS . . . 25 (Remark 7.19). But this implies that a and b are in the special position (S1), whichcontradicts our assumption. Analogously, if we assume that s is parallel to a inonly one rectangle r of N a ∩ b and R is the double segment of b that corresponds to r ,then each double segment of b that is different from R contains an oriented segmentjoinable to p or q . Moreover, the triple { p, q, R } must be positively oriented. Hence, a and b are in the special position (S2) or (S3), which again is a contradiction.If s (cid:48) is an arc of type A or B, then the situation is completely analogous. If s (cid:48) isof type A, then s (cid:48)(cid:48) can admit on one side one reduction of type IIIb (Lemma 7.12,Remarks 8.6 and 7.14), and if s (cid:48) is of type B, then s (cid:48)(cid:48) can admit reductions of typeIIIb on both sides (one on each side – Remark 7.14). Hence, the assumption thatno rectangle of N a ∩ b exists in which s is parallel to a implies that a and b are in thespecial position (S1), and if we assume that s is parallel to a in only one rectangleof N a ∩ b , then a and b are in the special position (S2) or (S3). (cid:3) Lemma 8.10.
Let c ∈ (cid:98) X b and assume that a and b are not in the special position(S1), (S2) nor (S3). If d is as in the proof of Proposition 7.1, then d winds weaklyaround a .Proof. We need to show that for each rectangle r of N a ∩ b and for each orientationof arcs parallel to a in r , an arc of d exists, which is parallel to a in r and is weaklyrigid with respect to the chosen orientation of arcs in r .Fix r and let r be the rectangle of N a ∩ b following r along a with respect to thechosen orientation of arcs in r . Without loss of generality we can assume that theconfiguration of rectangles is as in Figure 36, that is r is on the right of r . Next, Figure 36.
Intersection of t ka ( s (cid:48) ) with rectangles of N a ∩ b .follow a , with the chosen orientation of arcs in r , to the first rectangle r of N a ∩ b such that the top oriented segments of b in r and r are not joinable ( r existsby Proposition 4.3). If r = r , then each double segment of b contains a segmentjoinable to the top segment of r or to the bottom segment of r . Hence, a and b are in the special position (S1). Therefore, r (cid:54) = r .Let r be the rectangle of N a ∩ b that follows r . If r = r then a and b are ina special position [if the top segment of r is joinable to the top segment of r orif the bottom segment of r is joinable with the bottom segment of r , then we arein the special position (S1); otherwise, we are in the special position (S2) or (S3)].Hence, r (cid:54) = r .Let s (cid:48) be an arc of c ∩ N a that is parallel to b in r and that is weakly rigidon the top side of r (it exists because c winds weakly around b ). Under thisassumption the segment s of t ka ( c ) that starts at the top endpoint of s (cid:48) does notallow a reduction of type IIIb (Lemma 8.4). It may allow reductions of type IIIc,but they cannot reach r , because r is not joinable to r . Now, we concentrate on the segment s of t ka ( c ) starting at the bottom endpointof s (cid:48) . Let p i and q i , for i = 1 , ,
3, be segments of b , which go up and down from r i ,respectively. Segment s may admit a reduction of type IIIb and some reductionsof type IIIc, but if these reductions reach r , then either all double segments of b contain a segment joinable to p or q (this case happens when s can be reducedby reductions of type IIIc), or only one double segment of b (corresponding to r )exists, which does not contain a segment joinable to p or q (this happens when s is reduced by a reduction of type IIIb and then by reductions of type IIIc).According to our assumption that a and b are not special, such a situation is notpossible. Hence, the reductions on s cannot reach r , and as a consequence, the arc s of d , that corresponds to s (cid:48) is parallel to a in r and r . (cid:3) The special cases (S1) and (S2)
The common feature of cases (S1) and (S2) is the existence of oriented segments p and q starting on different sides of a , such that each double segment (or eachdouble segment except one in case (S2)) contains a segment joinable to p or q . Incase (S1) this implies the possibility that some arcs of c ∩ N a of types A–C maylead to arcs of d ∩ N a , which does not intersect b (see the proof of Lemma 7.20).We will show below (Lemma 9.13) that such reductions are not possible.The second problem in cases (S1) and (S2) is that in these cases, d may notwind strongly around a . We deal with this problem by replacing the notion ofstrong winding with the notion of weak winding defined in the previous section. Lemma 9.1.
Let π : M → S be a bundle over S with fiber I = [0 , and which ishomeomorphic to a M¨obius strip. If a simple oriented arc c in M is monotone withrespect to the fixed orientation of S and has endpoints in ∂M , then c intersectsevery fiber in at most two points.Proof. If c intersects some fiber in at least 3 points, then c winds infinitely manytimes around the core of M [Figure 37(i)]. (cid:3) Figure 37.
Impossible configurations of arcs – Lemmas 9.1 and 9.2.
Lemma 9.2.
Let a and b be two generic two-sided circles in N such that I ( a, b ) ≥ ,and assume that oriented segments p and q of b exist, which start on different sidesof a , such that each double segment of b contains an oriented segment joinable to p or q (that is a and b are in the special position (S1)). Then p is joinable to − q . UBGROUPS GENERATED BY TWO DEHN TWISTS . . . 27
Proof.
Suppose first that p starts and terminates on the same side of a . In sucha case, by Proposition 4.2, segments that start at the terminal points of segmentsjoinable to p must be joinable to q , and vice versa. However, this implies that b isa circle on an annulus [that is the union of twisted rectangles given by adjacency:see Figure 37(ii)]. This contradicts the assumption that I ( a, b ) ≥ p starts and terminates on two different sides of a . But then, the segmentstarting at the terminal point of p cannot be joinable to q (because p and q beginon two different sides of a ), and it cannot be joinable to p (by Proposition 4.2).Hence, − p must be joinable to q . (cid:3) Remark . Let a, b, p be as in the above lemma and let p , p , . . . , p n be all orientedsegments of b joinable to p . The union of adjacency disks between p , p , . . . , p n provides a rectangle Γ. We can assume that p goes up from a and that p and p n are on the boundary of Γ, where p n leaves N a to the left of p . By Lemma 9.2, q isequal to one of the segments: − p , − p , . . . , − p n . Hence, without loss of generality,we can assume that q = − p . The other two boundary arcs of Γ are arcs a , a of a . Observe that by Proposition 4.2, a ∩ a = ∅ , hence a \ ( a ∪ a ) consists of twoarcs and the configuration of a and b is as in the left part of Figure 38. Figure 38.
Configuration of arcs in cases (S1) and (S2).
Lemma 9.4.
Let a and b be two generic two-sided circles in N such that I ( a, b ) ≥ ,and assume that oriented segments p, q of b and a double segment R of b exist,such that a and b are not in the special position (S1), p and q start on differentsides of a , p starts and terminates on different sides of a , each double segmentof b different from R contains an oriented segment joinable to p or q , { p, q, R } is positively oriented [that is a and b are in the special position (S2)]. Then p isjoinable to − q .Proof. Our first claim is that an oriented segment of b exists, which is joinable to p and such that it neither starts nor terminates in R (that is neither p nor − p isa segment of R ). Given that I ( a, b ) ≥
4, at least 3 intersection points of a ∩ b different from R exist. Hence, at least two arcs s and t are joinable to p or joinableto q , which do not start in R . If both these arcs are joinable to p , then at leastone of them cannot terminate in R (given that they terminate on the same sideof a ). Hence, our claim follows. Now assume that these two arcs are joinable to q . If q starts and terminates on different sides of a , then the roles of p and q aresymmetric, and we can prove our claim by relabeling p to q , and vice versa. Hence,assume that q starts and terminates on the same side of a . Let s (cid:48) and t (cid:48) be orientedsegments of b that follow s and t , respectively [Figure 39(i)]. By Proposition 4.2,none of − s, − t, s (cid:48) , t (cid:48) is joinable to q and by our assumptions − s and − t are not Figure 39.
Configuration of arcs, Lemmas 9.4 and 9.8.joinable to p (because p and q start on different sides of a ). Hence, at least oneof s (cid:48) , t (cid:48) is joinable to p . If both s (cid:48) and t (cid:48) are joinable to p , then our claim followsbecause only one of them can terminate in R . If only one of them, say s (cid:48) , is joinableto p , then R corresponds to the terminal point of t . Hence, s (cid:48) cannot terminate in R , which again proves our claim.Assume that s is a segment of b joinable to p , which does not start nor terminatein R . Then, the segment that starts at the terminal point of p cannot be joinableto q (because p and q begin on two different sides of a ), and it cannot be joinableto p (by Proposition 4.2). Hence, − p must be joinable to q . (cid:3) Remark . Following the lines of the analysis conducted in Remark 9.3, we con-clude that the configuration of a and b in case (S2) differs from that in (S1) only byone additional double segment R , which intersects a in one of the arcs of a \ ( a ∪ a )( a and a are defined as in Remark 9.3). Hence the configuration is as in Figure38. (Note that the mutual position of p, q , and R is determined by the assumptionthat { p, q, R } is positively oriented.)For the rest of this section, we will use the notation introduced in Remarks 9.3and 9.5, that is p = p , p n are segments of b which together with arcs a , a of a bound a rectangle Γ that contains all segments of b joinable to p , p goes up from a and leaves N a to the right of p n (Figure 38). Moreover, assume that p starts at A in a rectangle r A of N a ∩ b and it terminates in B in a rectangle r B . Let r and r be rectangles of N b \ a which precede and follow p , respectively. Lemma 9.6.
Suppose that a and b are in the special position (S1) and let u , . . . , u n and v , . . . , v n be the oriented segments of b which respectively go down/up from theinitial/terminal points of p , . . . , p n . Then u i is not joinable to u j for some i (cid:54) = j and v i is not joinable to v j for some i (cid:54) = j .Proof. If for example all the u i are mutually joinable, then a and b are circleson the annulus given by adjacency disks between p i and u i .This contradicts theassumption I ( a, b ) ≥ (cid:3) Lemma 9.7.
Suppose that a and b are in the special position (S2) and let u , . . . , u n and v , . . . , v n be the oriented segments of b which respectively go down/up fromthe initial/terminal points of p , . . . , p n . Then, the segments that constitute R arejoinable neither to u nor to v . Moreover, u i is not joinable to u j for some i (cid:54) = j or v i is not joinable to v j for some i (cid:54) = j . UBGROUPS GENERATED BY TWO DEHN TWISTS . . . 29
Proof.
If a segment of R is joinable to u or v then a and b are in the specialposition (S1), which is not possible. Next, suppose to the contrary that all the u i are mutually joinable and all the v i are mutually joinable. Given that I ( a, b ) ≥ v i = − u j for some i, j . Hence, we can assume that each v i is joinable to − u . Oneof the u i must terminate in a rectangle r that corresponds to R and also one of the v i must terminate in r . However, this situation contradicts Proposition 4.2 giventhat v i is joinable to − u i . (cid:3) Suppose that the component of N \ N a ∪ b , which is determined by p = p andan arc of a \ ( a ∪ a ), is an exterior n -gon ∆. Let t , t be these boundary sidesof r and r , respectively, which enter r A and r B on the right of b ∩ r A and b ∩ r B (Figure 38).Now we define two M¨obius strips associated with p . The first one M is theunion of the rectangle r p of N b \ a that contains p , rectangles r A , r B , and a singlerectangle r AB of N a \ b that connects r A and r B . The second one M is the unionof r A , r B , r AB and all the rectangles that connect r A and r B along the boundary of∆.Finally, for the rest of this section assume that c ∈ C and c winds around b . Lemma 9.8.
Let s be an arc of type C of c ∩ N a which connects the initial pointsof oriented segments p (cid:48) and q (cid:48) of c which run parallel to p and − p respectively.Then at least one of the oriented arcs of c ∩ N a following p (cid:48) and q (cid:48) is an arc of type D which turns to the left as it enters N a .Proof. Two possible configurations of arcs p (cid:48) and q (cid:48) exist (they can pass each otherin two different ways). However, these configurations lead to the same conclusions.Hence, assume that we have the configuration shown in Figure 39(ii). Consider thearc t of c ∩ N a following p (cid:48) . After entering N a , this arc must turn to the left and iseither of type C or D. If t was of type C, then c would wind at least 3 times aroundthe core of the M¨obius strip M , which contradicts Lemma 9.1. Hence, t is an arcof type D. (cid:3) Let t be a common boundary of a rectangle r in N a \ b ∪ N b \ a and an exterior n -gon ∆. We say that an arc q of c ∩ r is a bounding segment for ∆ (with respectto t ) if no arcs of c exist between t and q in r . Clearly, c ∩ r is either empty or itcontains exactly one bounding segment for ∆ with respect to t . Lemma 9.9.
Let r p be the rectangle in N b \ a that contains p . If s is an arc of c that passes through r , r A , r p and is of type A in r A , and s is an arc of c whichpasses through r , r B , r p and is of type A in r B , then either s ∩ r is not a boundingsegment for ∆ with respect to t or s ∩ r is not a bounding segment for ∆ withrespect to t .Proof. Two possible configurations of arcs s and s exist (they can pass each otherin two different ways). However, these configurations lead to the same conclusions.Hence, assume that we have the configuration shown in Figure 40(i).Consider the arc s of c ∩ N a following s . If s was an arc of type C, then c would wind at least 3 times along the core of the M¨obius strip M and this wouldcontradict Lemma 9.1. Hence, s is either of type A or D. In the former case, s isnot a bounding segment for ∆ with respect to t , and in the later case, s is not abounding segment for ∆ with respect to t . (cid:3) Figure 40.
Configuration of arcs, Lemmas 9.9, 9.10 and 9.13.If s is an arc of c ∈ C with endpoints in r A and r B that connects r A and r B in M \ r p , then we say that s is a long bounding segment for ∆. Lemma 9.10.
Let s , s be oriented arcs of c ∩ N a of type B such that the arc s of c that starts at the terminal point of s and terminates at the starting point of s is a long bounding segment for ∆ (Figure 40(ii)). Then the arcs of c which precede s and follow s are not long bounding segments for ∆ .Proof. If for example the arc that precedes s was a long bounding segment for ∆,then c would be a curve in the M¨obius strip M which winds at least 3 times alongthe core of M ; this contradicts Lemma 9.1. (cid:3) Lemma 9.11.
Assume that an arc s of c ∩ N a of type B exists, which starts in r A and terminates in r B . If s [or s ] is an arc of c that is of type A in r A [or r B ],then s [or s ] is not a bounding segment for ∆ with respect to t [or t ].Proof. Given that c cannot intersect itself, s and s must enter N a on the left of s [our point of view here is along s /s towards r A /r B ; see Figure 41(i)]. (cid:3) Figure 41.
Configuration of arcs, Lemma 9.11 and Proposition 9.12.Now we are ready to adopt the proof of Proposition 7.1 to the special cases (S1)and (S2). However, as we observed in Example 8.1, the notion of rigidity is toostrong in these cases. Hence, we replace it with the notion of weak rigidity (seeSection 8).
Proposition 9.12.
Let a and b be two generic two-sided circles in N such that I ( a, b ) ≥ and { a, b } is not in the special position (S3). Then, for any integer k (cid:54) = 0 , we have t ka ( (cid:98) X b ) ⊆ (cid:98) X a and t kb ( (cid:98) X a ) ⊆ (cid:98) X b . UBGROUPS GENERATED BY TWO DEHN TWISTS . . . 31
Proof.
As we observed in Remark 8.8 the assumption about weak winding of c around b suffices to prove the statement of Lemma 7.13. As a substitute for Remark7.11, we have slightly weaker Remark 8.7. Moreover, as observed in Remark 8.6,the assumption that c is weakly rigid with respect to b serves as a replacement forLemma 7.12. Hence, in most parts, we can copy verbatim the proof of Proposition7.1, yet there are some differences, which we study in detail below.Suppose first that a and b are not in the special position (S1) or (S2). In sucha case, by Lemmas 8.9 and 8.10, d is weakly rigid with respect to a and d windsweakly around a . Moreover, Lemma 8.9 implies that I ( d , b ) > I ( d , a ).Therefore, for the rest of this section, we can concentrate on circles a and b , whichare in the special position (S1) or (S2) and are hence in the situation described inLemmas 9.2, 9.4 and Remarks 9.3, 9.5. Counting intersection points between d and b . If we follow the proof of Proposi-tion 7.1, then the first place it can fail in cases (S1) and (S2) is the proof of Lemma7.20. However, as we will show below, even the stronger statement of Lemma 8.9holds true in cases (S1) and (S2). The possible failure of the argument in Lemma8.9 follows from the fact that an arc s (cid:48) of c ∩ N a of type A, B, or C may exist,which yields an arc s of d ∩ N a that is parallel to a in fewer than two rectangles of N a ∩ b . Moreover, the proof of that lemma provides a specific description of possibleconfigurations of p, q and s (cid:48) that may lead to such a failure. Three possibilitiesexist in the case (S1)/(S2): s (cid:48) may be an arc of type A in r A /r B , s (cid:48) may be an arcof type B that connects r A and r B , or s (cid:48) may be an arc of type C that connects r A and r B . The following lemma shows that in each of these cases, strong restrictionsare given for reductions of types IIIb/IIIc. Lemma 9.13.
Let c be as in the proof of Proposition 7.1. (1) Let s be an arc of d ∩ N a that corresponds to an arc s (cid:48) of c ∩ N a . If s (cid:48) isan arc of type C with endpoints in r A and r B , then s can admit a reductionof type IIIc only on one side of s (cid:48) . (2) Let s be an arc of c , which is the innermost long bounding segment for ∆ (that is, s is a bounding segment for ∆ in rectangles of N ( a ∪ b ) \ ( a ∩ b ) ) andwhich leads to a reduction of type IIIb across ∆ . Then, the arcs of c ∩ N a that precede/follow s are arcs of type B. (3) Suppose that a long bounding segment of c exists, which leads to a reductionof type IIIb across ∆ . Then, no arc of c which is of type A in r A or r B leads to a reduction of type IIIb across ∆ . (4) No arc of c ∩ N a that is of type A in r A or r B leads to a reduction of typeIIIb across ∆ . (5) If s is an arc of c of type B with endpoints in r A and r B , then s can leadto a reduction of type IIIb only on one side of s .Proof. (1) If s (cid:48) is an arc of type C, then by Lemma 9.8, at least one of the arcs thatprecede/follow s (cid:48) cannot lead to a reduction of type IIIc (because it turnsto the left as it enters N a ).(2) By Lemma 9.9, the arcs that precede/follow s cannot be both of type A,and by Lemma 9.11, they are not arcs of types A and B. Hence, both arcsmust be of type B. (3) If we choose the arc s of c that admits a reduction of type IIIb across ∆and is the innermost long bounding segment for ∆, then by the previouspoint, we know that the arcs s , s of c that precede/follow s are of type Bin r A and r B (Figure 40(ii)). But then, by Lemma 9.10, we know that thearcs that respectively precede s and follow s do not admit reductions oftype IIIb. Hence, they are the obstacles for the arcs of type A in r A , r B toadmit such a reduction.(4) The conclusion is a direct consequence of the previous point.(5) At this point, we know that if an arc s of c admits a reduction of typeIIIb across ∆, then this must be a long bounding segment that connectstwo arcs s , s of c ∩ N a of type B. But then, by Lemma 9.10, the arc thatprecedes s , which is different from ± s , does not admit reductions of typeIIIb. (cid:3) The above lemma implies that even in cases (S1)/(S2), none of the arcs s (cid:48) of c ∩ N a can be reduced so that the resulting arc s of d ∩ N a is parallel to a infewer than 2 rectangles of N a ∩ b . Hence, Lemma 8.9 remains valid in these cases.In particular, d is weakly rigid with respect to a and I ( d , b ) > I ( d , a ). Weak winding of d . Finally, we need to show that d winds weakly around a .Denote by r p = r A , r p , . . . , r p n and r q = r B , r q , . . . , r q n the rectangles of N a ∩ b ,which contain the starting points of p , p , . . . , p n and − p , . . . , − p n , respectively(Figure 38). In case (S2), by r R denote the rectangle of N a ∩ b that corresponds to R . Let s (cid:48) be an arc of c that is of type A in r p , and let s (cid:48)(cid:48) and s be the correspondingarcs of d and d respectively. By Lemma 9.13, the bottom part of s (cid:48)(cid:48) does notallow a reduction of type IIIb. The top part of s (cid:48)(cid:48) may admit reductions of typeIIIc which can reach at most r p n . Hence s is parallel to a in all the rectangles r q , . . . , r q n and r R in case (S2). Moreover, if the reductions on the top part s (cid:48)(cid:48) t of s (cid:48)(cid:48) reach r p n , then the corresponding arc s t of d is two-sided with respect to b , thatis, s t together with an arc of b that connects the intersection points of s t and b isa two-sided circle [Figure 41(ii)]. Hence, s t is not a one-sided boundary 3-segmentof an exterior hexagon and by Lemma 8.5, s is weakly rigid with respect to a onboth sides of r q n in case (S1) and on both sides of r R in case (S2). Moreover, s isrigid with respect to a on both sides of r q , . . . , r q n − .In exactly the same way, by considering the arc of c which is of type A in r q ,we prove that d is weakly rigid with respect to a on both sides of rectangles r p , . . . , r p n . Hence, to finish the proof, we need to show the existence of an arcof d ∩ N a that is weakly rigid on both sides of r q and an arc of d ∩ N a that isweakly rigid on both sides of r p .In case (S1), let s (cid:48) be an arc of c that is of type A in r p n and that is weaklyrigid on both sides of r p n . Let s (cid:48)(cid:48) and s be the corresponding arcs of d and d respectively. By Lemma 8.4, the bottom part of s (cid:48)(cid:48) does not allow a reduction oftype IIIb. It may admit reductions of type IIIc, but by Lemma 9.6, they can notreach r p . The top part of s (cid:48)(cid:48) does not admit a reduction of type IIIb (Lemma 8.4)and it does not allow a reduction of type IIIc (Proposition 4.2 and Remark 7.19).Hence, s is parallel to a in r p , r q , r q . Similarly, by taking an arc s (cid:48) of c that isweakly rigid on both sides of r q , we construct an arc s of d that is parallel to a in r p , r p , r q . UBGROUPS GENERATED BY TWO DEHN TWISTS . . . 33
In case (S2), by Lemma 9.7, we know that either the segments that go down fromthe starting points of p , . . . , p n or the segments that go up from the terminal pointsof p , . . . , p n are not mutually joinable. The argument in both cases is completelyanalogous. Hence, the latter case is assumed. Let s (cid:48) be an arc of c that is of type Ain r q n and is weakly rigid on both sides of r q n . Let s (cid:48)(cid:48) and s be the correspondingarcs of d and d respectively. By Lemma 8.4, the top part of s (cid:48)(cid:48) does not allow areduction of type IIIb. It may admit reductions of type IIIc, but by our assumptionthey cannot reach r q . The bottom part of s (cid:48)(cid:48) does not admit a reduction of typeIIIb. Hence, s is parallel to a in r p , r p , r q . Similarly, if s (cid:48)(cid:48) is an arc of c that is oftype A in r p n and s is the corresponding arc of d , then s is parallel to a in r q , r q ,and is either parallel to a in r p , or is weakly rigid on the r p side of r q (Lemma8.5). This completes the proof that d winds weakly around a . (cid:3) The special case (S3)
Lemma 10.1.
Let a and b be two generic two-sided circles in N such that I ( a, b ) ≥ , and assume that oriented segments p, q of b and a double segment R of b existsuch that p starts and terminates on one side of a , q starts and terminates on theother side of a , each double segment of b , which is different from R contains anoriented segment joinable to p or q , { p, q, R } is positively oriented [that is, a and b are in the special position (S3)]. Then, the oriented arcs that constitute R arejoinable neither to p nor to q .Proof. Suppose that in each double segment of b , an oriented segment joinableeither to p or q exists. By Proposition 4.2, all segments that start at endpointsof segments joinable to p and are not joinable to − p are joinable to q . And viceversa, all segments that start at endpoints of segments joinable to q and are notjoinable to − q are joinable to p . Hence, b is a circle on an annulus [which is anunion of adjacency disks between the segments joinable to p and q – see Figure37(ii)]. However, this contradicts the assumption that I ( a, b ) ≥ (cid:3) Remark . Let a and b be circles in the special position (S3), let p , p , . . . , p n beall oriented segments of b joinable to p , and let q , . . . , q m be all oriented segmentsjoinable to q . The union of adjacency disks between p , p , . . . , p n gives a rectangle∆ p and the union of adjacency disks between q , q , . . . , q m gives a rectangle ∆ q .By Lemma 10.1, we know that one of the segments p , . . . , p n , q , . . . , q m ends in R and is followed by an arc that is joinable neither to p nor q . Without lossof generality we can assume that p is above a , p n ends in R , p , p n are on theboundary of ∆ p and q , q m are on the boundary of ∆ q . Given that all orientedsegments p , . . . , p n − must be followed by segments joinable to q , and all segments q , . . . , q m must be followed by segments joinable to p , we can also assume that p , . . . , p n − are followed by respectively q , . . . , q m . Hence the configuration is asin Figure 42. (Note that the mutual position of p, q and R is determined by theassumption that { p, q, R } is positively oriented.)For the rest of this section we will use the notation introduced in the aboveremark. Moreover, by r, r p , r q we denote the rectangles of N a ∩ b that correspondrespectively to R , the initial point of p n , and the initial point of q (Figure 42). Lemma 10.3.
No component of N \ N a ∪ b is an exterior hexagon. Figure 42.
Configuration of arcs in case (S3).
Proof.
Given that the segment that connects the terminal point of p n with theinitial point of p is one-sided, checking that r can be a vertex of an exterior n -gononly for n = 4 , n -gons are rectangles. (cid:3) Let (cid:98) X a and (cid:98) X b be defined as in Section 8. Proposition 10.4.
Let a and b be two generic two-sided circles in N such that I ( a, b ) ≥ . Then for any integer k (cid:54) = 0 we have t ka ( (cid:98) X b ) ⊆ (cid:98) X a and t kb ( (cid:98) X a ) ⊆ (cid:98) X b . Proof.
Observe first that if a and b are not in the special position (S3), then theproposition coincides with Proposition 9.12. Therefore, we can concentrate on thecircles a and b , which are in the special position (S3) and are hence in the situationdescribed in Remark 10.2.As in special cases (S1) and (S2), we observe that weak winding is sufficientfor repeating most of the proof of Proposition 7.1. Hence, we follow the lines ofthe proof of Proposition 9.12 and we concentrate on places where that proof canfail. The first such place is the proof that d is weakly rigid with respect to a . Inthe proof of Proposition 9.12 we obtained weak rigidity of d as a consequence ofLemma 8.9, which may not be true in the case (S3). However, as a replacement wehave the following lemma: Lemma 10.5.
Let s (cid:48) be an arc of c ∩ N a of type A, B or C, and let s be the arc of d ∩ N a which corresponds to s (cid:48) . (1) If s (cid:48) is an arc of type A in r p or s (cid:48) is an arc of type C that connects r p and r q , then s is parallel to a in r . (2) If I ( a, b ) = 4 and s (cid:48) is an arc of type A in r or s (cid:48) is an arc of type C withthe top part in r , then s is parallel to a in r p . (3) If s (cid:48) is not as in previous points, then s is parallel to a in at least tworectangles of N a ∩ b .Proof. Observe first that the boundary component of N a ∪ b that contains p n and p cannot bound an exterior rectangle. Otherwise, a would bound a M¨obius strip.Hence the segments of c , which run parallel to p , do not lead to reductions of typesIIIb nor IIIc. Let s (cid:48)(cid:48) be an arc of d that corresponds to s (cid:48) .(1) If s (cid:48) is an arc of type A in r p , then the bottom part of s (cid:48)(cid:48) can admita reduction of IIIb and then both sides of the obtained arc may admitreductions of type IIIc. By Lemma 10.1, these reductions cannot reach R .Hence, s is parallel to a in r . UBGROUPS GENERATED BY TWO DEHN TWISTS . . . 35
The situation is completely analogous if s (cid:48) is an arc of type C thatconnects r p and r q (by Lemma 7.13, s (cid:48)(cid:48) does not admit reductions of typeIIIb in this case).(2) If s (cid:48) is an arc of type A in r , then the top part of s (cid:48)(cid:48) may admit somereductions of type IIIc, which can reach at most r q . The bottom part of s (cid:48)(cid:48) may admit a reduction of type IIIb, but then it does not admit any furtherreductions of type IIIc. Hence, s is parallel to a in r p .The situation is completely analogous if s (cid:48) is an arc of type C with thetop part in r (by Lemma 7.13, s (cid:48)(cid:48) does not admit reductions of type IIIb inthis case).(3) If s (cid:48) is not as in the previous points, then it is straightforward to check that s is parallel to a in either the rectangles of N a ∩ b that contain the initialpoints of p n − and p n , or in rectangles of N a ∩ b that contain the terminalpoints of p and p . (cid:3) As a consequence of the above lemma, we obtain that every arc s of d that isparallel to a in a rectangle t of N a ∩ b is weakly rigid on one side of t . In fact, if s is parallel to a in at least two rectangles of N a ∩ b , then s is rigid in t . If s isparallel to a in only one rectangle of N a ∩ b , then s was obtained from an arc of d by reductions of types IIIb and IIIc. In such a case, by Lemmas 8.5 and 10.3, s isweakly rigid on one side of t . In particular, d is weakly rigid with respect to a .Finally, we will show that d winds weakly around a . Let s (cid:48) and s (cid:48) be arcs of c of type A in r q and the rectangle r p of N a ∩ b that contain the initial point of p ,respectively. Assume also that s (cid:48) and s (cid:48) are weakly rigid below a , and let s (cid:48)(cid:48) , s (cid:48)(cid:48) and s , s be arcs of d and d that correspond to s (cid:48) , s (cid:48) , respectively. The arc s (cid:48)(cid:48) does not admit any reductions of types IIIb and IIIc. Hence, s is parallel to a in allrectangles of N a ∩ b except r p . The bottom part of s (cid:48)(cid:48) may admit some reductionsof type IIIc, but these reductions cannot reach r . Hence, s is parallel to a in r andin all the rectangles of N a ∩ b between r p and r p . This proves that d winds weaklyaround a . (cid:3) The case of I ( a, b ) = 3 with nonorientable N a ∪ b If I ( a, b ) = 3, then we still follow the proofs of Propositions 7.1 and 9.12. How-ever, as in the special case (S3), the problem is that in general, if I ( a, b ) = 3,then Lemma 8.9 is not true. Fortunately, this case is quite special because of thefollowing proposition: Proposition 11.1. If a and b are two generic two-sided circles in a surface N suchthat | a ∩ b | = I ( a, b ) = 3 and N a ∪ b is nonorientable, then the exterior n -gons for N a ∪ b can exist only if n = 10 or n = 12 .Proof. If all segments of b are two-sided, then N a ∪ b is orientable. Hence, one-sidedsegments of b exist, and two such segments p , p must exist. If we denote thetwo-sided segment of b as p , then we can assume that p , p , p are oriented sothat p follows p and p follows p .If all segments p , p , p start and terminate on different sides of a , then thearc of a that connects the initial point of p with the initial point of p startsand terminates on the same side of b . Hence we can interchange a with b , andwe can always assume that at least one segment of b starts and terminates on the same side of a . In such a case one of the remaining segments must also start andterminate on the same side of a , and the final segment must connect two differentsides of a . Hence we have the configuration of arcs as in Figure 43. We still Figure 43.
Configurations of segments of b , Proposition 11.1.have two possibilities: either p is a segment that connects two different sides of a [Figure 43(i)], or p starts and terminates on the same side of a [Figure 43(ii)].Checking that in the first case the boundary of N a ∪ b is connected and it is a 12-gonis straightforward. In the second case, the boundary of N a ∪ b has two components:a bigon and a 10-gon. (cid:3) The above proposition implies that if I ( a, b ) = 3, then no adjacent segments of b exist. Hence, no reductions of type IIIc exist. Moreover, the notion of weakly rigidarcs simplifies, given that no exterior hexagons exist.Let X a be the set of isotopy classes of circles c in N , which satisfy the followingconditions:(1) c ∈ C ,(2) I ( c, a ) = | c ∩ a | , I ( c, b ) = | c ∩ b | ,(3) I ( c, a ) < I ( c, b ),(4) c winds weakly around a .Similarly, we define X b by requiring (1)–(2) above and additionally(3’) I ( c, b ) < I ( c, a ),(4’) c winds weakly around b .The main difference between sets X b and (cid:98) X b is the lack of the assumption that c isweakly rigid with respect to b . As a replacement for this assumption we have thefollowing lemma: Lemma 11.2.
Let a and b be two generic two-sided circles in N such that I ( a, b ) =3 and c ∈ X b . Let s be an arc of c ∩ N a of type A and let s (cid:48) be the arc of d that corresponds to s . Then, s (cid:48) can admit a reduction of type IIIb with only oneorientation of s .Proof. Suppose to the contrary that s leads to reductions of type IIIb with bothorientations of s (that is, on both sides of a ) and let r be the rectangle of N a ∩ b that contains s . Observe first that both ends of s (cid:48) must be involved in two differentreductions of type IIIb; otherwise, s would intersect a only once, which wouldcontradict the assumption that c winds around b . By Proposition 11.1, at most oneexterior n -gon ∆ exists. Hence, both reductions on s (cid:48) must correspond to the same UBGROUPS GENERATED BY TWO DEHN TWISTS . . . 37 one-sided segment p of b ; this segment corresponds to one side of the bigon defininga reduction. Therefore, if we follow s in both directions until we obtain the arcs s , s of c ∩ N a which intersect a , then s and s must enter N a in r . Moreover,the arcs of c that connect s with s and s run through rectangles of N a ∪ b , whichtogether with r , constitute a M¨obius strip strip M . Hence, s and s enter r on thesame side of s [Figure 44(i)]. Therefore, by Lemma 7.13 and Remark 8.8, at least Figure 44.
Configurations of segments of c and d , Lemma 11.2and Proposition 11.3.one of the arcs s and s must be an arc of type A, which leads to a contradictionwith Lemma 9.1. (cid:3) Proposition 11.3.
Let a and b be two generic two-sided circles in N such that I ( a, b ) = 3 . Then for any integer k (cid:54) = 0 , we have t ka ( X b ) ⊆ X a and t kb ( X a ) ⊆ X b . Proof.
As in the special case (S3), the main problem that may lead to the failure ofthe proof of Proposition 9.12 is the fact that if I ( a, b ) = 3, then Lemma 8.9 may notbe true. However, as a replacement for that lemma we have the following slightlyweaker result: Lemma 11.4.
Let s (cid:48) be an arc of c ∩ N a of type A, B or C, and let s be the arc of d ∩ N a that corresponds to s (cid:48) . Then, s is parallel to a in at least one rectangle of N a ∩ b Proof.
Let s (cid:48)(cid:48) be an arc of d ∩ N a that corresponds to s (cid:48) . If s (cid:48) is an arc of typeA, then by Lemma 11.2, s (cid:48)(cid:48) may admit a reduction of type IIIb only on one side of a . Hence, s is parallel to a in at least one rectangle of N a ∩ b . If s (cid:48) is an arc of typeB, then s (cid:48)(cid:48) may admit reductions of type IIIb on both sides of a , but this leads tosame conclusion as above. Arcs of type C does not allow reductions of type IIIb(see Lemma 7.13 and Remark 8.8). Hence, in this case, s is parallel to a in exactlyone rectangle of N a ∩ b . (cid:3) As a consequence of the above lemma and Proposition 11.1 we have the followingsimplified version of Lemma 8.5:
Lemma 11.5.
Let q be an oriented arc of d that starts in a rectangle t of N a ∩ b as an arc parallel to a , then it follows a to the next rectangle of N a ∩ b , and then itfollows an arc s obtained from an arc of d \ N a by a reduction of type IIIb. Then, q is weakly rigid in t with respect to a . (cid:3) Weak rigidity of d . Fix a rectangle r in N a ∩ b and assume that the orientationof arcs parallel to a in r is such that these arcs point to the rectangle r , which ison the right of r [if the orientation is opposite we can rotate the whole picture by180 ◦ ; see Figure 44(ii)]. Let s (cid:48) be an arc of c ∩ N a of type A in a rectangle r of N a ∩ b different from r and r , and assume that s (cid:48) is weakly rigid in r with respectto b and with the orientation pointing down (we use the assumption that c windsweakly around b ). The arc s (cid:48)(cid:48) of d that corresponds to s (cid:48) is parallel to a in r and r . If this arc does not admit a reduction of type IIIb, then s (cid:48) is in fact an arc of d and this arc is rigid in r . If, on the other hand, s (cid:48)(cid:48) admit a reduction of type IIIb,then this reduction must be on the top part of s (cid:48)(cid:48) . Hence, by Lemma 11.5, the arc s of d that corresponds to s (cid:48)(cid:48) is weakly rigid in r . Counting intersection points between d and b . Lemma 11.4 guarantees that foreach intersection point of c and a , we have at least one intersection point of d and b . To prove that I ( d , b ) > I ( c, a ), we need to show that for some arcs of c ∩ N a the corresponding arcs of d ∩ N a intersect b at least twice.In fact, by Proposition 11.1, at most one exterior n -gon exists. Hence, all re-ductions of type IIIb must correspond to the same segment p of b ; this segmentcorresponds to one side of the bigon that defines a reduction. Let r A and r B bethe rectangles of N a ∩ b that contain the endpoints A and B of p . Let r and r be rectangles of N a ∩ b that precede r A and r B , respectively, with respect to thetwisting direction in N a , that is r is the right [left] neighbor of r A if p approaches a from above [from below], similarly for r (see Figure 45). All arcs of c ∩ N a of Figure 45.
Configuration of rectangles of N a ∩ b , Proposition 11.3.type A that lead to the reduction of type IIIb must be contained in either r or r . Hence, if we choose an arc s (cid:48) of c ∩ N a that is of type A in a rectangle r of N a ∩ b different from r and r , then the corresponding arc s (cid:48)(cid:48) of d does not admitreductions of type IIIb. Therefore, the arc s of d ∩ N a that corresponds to s (cid:48) isparallel to a in two rectangles of N a ∩ b . (cid:3) The case of I ( a, b ) = 2 with nonorientable N a ∪ b Checking that Propositions 7.1 and 10.4 are false if I ( a, b ) = 2 and N a ∪ b isnonorientable is not difficult. Hence, we need slightly more sophisticated analysisin that case.The case in question is special because of the following proposition: Proposition 12.1. If a and b are two generic two-sided circles in a surface N such that | a ∩ b | = I ( a, b ) = 2 and N a ∪ b is nonorientable, then no component of N \ ( a ∪ b ) is a disk.Proof. Observe that N a ∪ b is a Klein bottle with two boundary components (Figure46(i)). Hence if one of the components of N \ ( a ∪ b ) is a disk, then one of the circles UBGROUPS GENERATED BY TWO DEHN TWISTS . . . 39
Figure 46.
Klein bottle with two holes as a regular neighborhoodof a ∪ b . a or b bounds a M¨obius strip which is a contradiction. (cid:3) For a circle c ∈ C define J ( c, a ) = number of connected components of c \ N a J ( c, b ) = number of connected components of c \ N b . Proposition 12.2.
Let a and b be two generic two-sided circles in a surface N such that | a ∩ b | = I ( a, b ) = 2 and N a ∪ b is nonorientable. If c, c (cid:48) ∈ C such that c isisotopic to c (cid:48) , then J ( c, a ) = J ( c (cid:48) , a ) and J ( c, b ) = J ( c (cid:48) , b ) .Proof. Suppose first that | c ∩ c (cid:48) | > c and c (cid:48) . Bytaking the inner-most bigon, we can assume that the interior of ∆ is disjoint from c ∪ c (cid:48) . Given that the boundary of N a ∪ b is disjoint from c ∪ c (cid:48) , if a component of N \ N a ∪ b intersects ∆, then this component must be a disk. By Proposition 12.1,this case is not possible. Hence, ∆ is contained in N a ∪ b . Moreover, we can assumethat the vertices of ∆ are in the interior of rectangles of N a ∪ b .Fix a rectangle r in N a \ b ∪ N b \ a , and then let ∆ r be a connected component of∆ ∩ r . Given that ∆ ⊂ N a ∪ b and c, c (cid:48) do not turn back in any of the rectanglesof N a ∪ b , ∆ r must be either a rectangle, a triangle, or a bigon with two sides beingarcs of c and c (cid:48) [Figure 47]. In any case, if we remove the bigon ∆, that is, if we Figure 47.
Arcs of c and c (cid:48) in a rectangle r .replace c (cid:48) with the circle c (cid:48)(cid:48) isotopic to c (cid:48) which is obtained by pushing c (cid:48) across ∆(Figure 48(i)), then J ( c (cid:48)(cid:48) , a ) = J ( c (cid:48) , a ) and J ( c (cid:48)(cid:48) , b ) = J ( c (cid:48) , b ) . Therefore, we can assume that c and c (cid:48) are disjoint. This assumption means thatan annulus M in N with boundary curves c and c (cid:48) exists. Figure 48.
Arcs of c and c (cid:48) in a rectangle r .If a component of N \ N a ∪ b intersects M , then this component must be an annuluswith boundary curves isotopic to c and c (cid:48) . In such a case N is a nonorientablesurface of genus 4 [Figure 46(ii)] and J ( c, a ) = J ( c (cid:48) , a ) = 2 and J ( c, b ) = J ( c (cid:48) , b ) = 2 . Finally, assume that M is contained in N a ∪ b . As in the case of a bigon formedby c and c (cid:48) , if r is a rectangle in N a \ b ∪ N b \ a and M r is a connected componentof M ∩ r , then M r must be a rectangle with two sides being arcs of c and c (cid:48) that connect opposite sides of r [Figure 48(ii)]. This implies that M r gives in r exactly one arc of c and one arc of c (cid:48) . This result means that J ( c, a ) = J ( c (cid:48) , a ) and J ( c, b ) = J ( c (cid:48) , b ). (cid:3) For two generic two-sided circles a, b in N such that | a ∩ b | = I ( a, b ) = 2, wedefine (cid:101) X a as the set of isotopy classes of circles in N , which satisfy the followingconditions:(1) c ∈ C ,(2) J ( c, a ) < J ( c, b ),(3) c winds around a .Similarly, we define (cid:101) X b by requiring (1) above and additionally(2’) J ( c, b ) < J ( c, a ),(3’) c winds around b .As an analog of Proposition 7.1, we have Proposition 12.3.
Let a and b be two generic circles in N such that I ( a, b ) = 2 and N a ∪ b is nonorientable. Then for any integer k (cid:54) = 0 , we have t ka ( (cid:101) X b ) ⊆ (cid:101) X a and t kb ( (cid:101) X a ) ⊆ (cid:101) X b . Proof.
As in the proof of Proposition 7.1, we concentrate on the inclusion t ka ( (cid:101) X b ) ⊆ (cid:101) X a . We begin by constructing the circle d = t ka ( c ) and as before, we assume that t ka twists to the right in N a . We perform reductions of type I on d , and as a result,we obtain a circle d ∈ C which winds around a . Observe that by Proposition 12.2,showing that J ( d , b ) > J ( d , a ) is sufficient (we do not need to focus on reductionsof types II–III).Let n A , n B , n C , n D be numbers of arcs of c ∩ N a of types A, B, C, D, respectively(Figure 10). In particular, J ( d , a ) = n A + n B + n C + n D . To determine the number J ( d , b ), suppose first that | k | = 1. Each arc of c ∩ N a oftype A gives an arc of d which goes once around a and therefore gives I ( a, b ) = 2in J ( d , b ). An arc of c ∩ N a of type B gives I ( a, b ) + 1 = 3 in J ( d , b ), and an arc UBGROUPS GENERATED BY TWO DEHN TWISTS . . . 41 of type C gives I ( a, b ) − c ∩ N a of type D does not change afterthe twist and gives 1 in J ( d , b ). Finally, if | k | >
1, then for each arc of c ∩ N a oftypes A–C, we have additional ( | k | − · I ( a, b ) = 2( | k | −
1) arcs of d \ N b . Hence,we proved the following formula: J ( d , b ) = 2 n A + 3 n B + n C + n D + 2( | k | − · I ( a, c ) . Given that c winds around b , we have n A >
0. Hence, J ( d , b ) > J ( d , a ). (cid:3) Remark . The proof of Proposition 12.3 can be repeated with minimal changeswhen I ( a, b ) ≥ N \ N a ∪ b is a disk or an annulus (forexample if N = N a ∪ b ). However, if disks are present in the complement of N a ∪ b ,then Proposition 12.2 is not true and the situation becomes difficult.13. Twists generating a free group
Recall the so called ’Ping Pong Lemma’ (see for example Lemma 3.15 of [2]).
Lemma 13.1.
Suppose that a group G acts on a set Y , and Y , Y ⊆ Y arenonempty and disjoint. Let g , g ∈ G such that for every nonzero integer k , g k ( Y ) ⊆ Y and g k ( Y ) ⊆ Y . Then the group generated by g and g is a free group of rank 2. (cid:3) Theorem 13.2.
Let a and b be two generic two-sided circles in a nonorientablesurface N . If I ( a, b ) ≥ , then the group generated by t a and t b is isomorphic tothe free group of rank 2.Proof. If I ( a, b ) ∈ { , } and N a ∪ b is orientable, then we can repeat Ishida’s proof [3]without any changes.Let ( Y a , Y b ) be equal to ( (cid:98) X a , (cid:98) X b ) or ( X a , X b ) or else ( (cid:101) X a , (cid:101) X b ), where the sets (cid:98) X a , (cid:98) X b , X a , X b , (cid:101) X a , (cid:101) X b are defined in Sections 8, 11, and 12. Observe that Y a and Y b satisfy Y a ∩ Y b = ∅ , a ∈ Y a , b ∈ Y b . Hence, if I ( a, b ) ∈ { , } and N a ∪ b is nonorientable, or I ( a, b ) ≥
4, then the theoremfollows from Lemma 13.1 and Propositions 10.4, 11.3, and 12.3. (cid:3)
References [1] D. B. A. Epstein. Curves on 2–manifolds and isotopies.
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