Andy Wand- Factorizations of Diffeomorphisms of Compact Surfaces with Boundary

8/3/2019 Andy Wand- Factorizations of Diffeomorphisms of Compact Surfaces with Boundary

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arXiv:0910

.5691v3[math.GT

]5May2011

FACTORIZATIONS OF DIFFEOMORPHISMS OF COMPACTSURFACES WITH BOUNDARY

ANDY WAND

Abstract. We study diffeomorphisms of compact, oriented surfaces, devel-oping methods of distinguishing those which have positive factorizations into

Dehn twists from those which satisfy the weaker condition of right veering. Weuse these to construct open book decompositions of Stein-fillable 3-manifoldswhose monodromies have no positive factorization.

1. Introduction

Let be a compact, orientable surface with nonempty boundary. The (re-stricted) mapping class group of , denoted , is the group of isotopy classes oforientation preserving diffeomorphisms of which restrict to the identity on .The goal of this paper is to study the monoid Dehn+() of products of rightDehn twists. In particular, we are able to show:

Theorem 1.1. There exist open book decompositions which support Stein fillablecontact structures but whose monodromies cannot be factorized into positive Dehntwists.

This result follows from much more general results which we develop concerning

Dehn+

(). We develop necessary conditions on the images of properly embeddedarcs under these diffeomorphisms, and also necessary conditions for elements ofthe set of curves which can appear as twists in some positive factorization of agiven . The first of these can be thought of as a refinement of the rightveering condition, which dates at least as far back as Thurstons proof of the leftorderability of the braid group, and which was introduced into the study of contactstructures by Honda, Kazez, and Matic in [HKM]. In that paper, it is shown thatthe monoid V eer() of right-veering diffeomorphisms (see Section 2 for def-initions) strictly contains Dehn+(). Our methods allow one to easily distinguishcertain elements of V eer() \ Dehn+().

Central to much of current research in contact geometry is the relation betweenthe monodromy of an open book decomposition and geometric properties of thecontact structure, such as tightness and fillability. The starting point is the remark-

able theorem of Giroux [Gi], demonstrating a one-to-one correspondence betweenoriented contact structures on a 3-manifold M up to isotopy and open book de-compositions of M up to positive stabilization. It has been shown by Giroux [Gi],Loi and Peirgallini [LP], and Akbulut and Ozbagci [AO], that any open book withmonodromy which can be factorized into positive Dehn twists supports a Stein-fillable contact structure, that every Stein-fillable contact structure is supported bysome open book with monodromy which can be factored into positive twists, andin [HKM] that a contact structure is tight if and only if the monodromy of each

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supporting open book is right-veering. The question of whether each open bookwhich supports a Stein-fillable contact structure must be positive, answered in thenegative by our Theorem 1.1, has been a long-standing open question.

Section 2 introduces some conventions and definitions, and provides motivationfor what is to come. Section 3 introduces the concept of a right position for themonodromy image of a properly embedded arc in , and compatibility conditionsfor right positions on collections of arc images under a given monodromy. Moreimportantly, we show that a monodromy can have a positive factorization only ifthe images of any collection of nonintersecting properly embedded arcs in admitright positions which are pairwise compatible. We use this to construct simpleexamples of open books whose monodromies, though right veering (see Section 2for a precise definition), have no positive factorization.

In Section 4 we use these results to address the question of under what condi-tions various isotopy classes of curves on a surface can be shown not to appear asDehn twists in any positive factorization of a given monodromy. We obtain various

necessary conditions under certain assumptions on the monodromy.Finally, in Section 5, we construct explicit examples of open book decompositionswhich support Stein fillable contact structures yet whose monodromies have no pos-itive factorizations. For the construction we demonstrate a method of modifyinga certain mapping class group relation (the lantern relation) into an immersedconfiguration, which we then use to modify certain open books with positivelyfactored monodromies (which therefore support Stein fillable contact manifolds)into stabilization-equivalent open books (which support the same contact struc-tures) whose monodromies now have non-trivial negative twisting. We then applythe methods developed in the previous sections to show that in fact this negativetwisting is essential.

As this work was being written, Baker, Etnyre, and Van Horn-Morris [BEV]were able to construct similar examples of non-positive open books supporting

Stein fillable contact structures, all of which use the same surface, 2,1, involved inour construction. Their method for demonstrating non-positivity is on the one handsubstantially shorter, but on the other is quite restrictive, being entirely specificto this surface, and also requiring complete knowledge of the Stein fillings of thecontact manifold in question, which is known only for a small class of contactmanifolds.

We would like to thank Danny Calegari for many helpful comments on a previousversion of this paper, and Rob Kirby for his support and encouragement throughoutthe much extended period over which it was completed. A revision was done whileat the Max Planck Institute for Mathematics.

2. Preliminaries

Throughout, denotes a compact, orientable surface with nonempty boundary.The (restricted) mapping class group of , denoted , is the group of isotopyclasses of orientation preserving diffeomorphisms of which restrict to the identityon . In general we will not distinguish between a diffeomorphism and its isotopyclass.

Let SC C() be the set of simple closed curves on . Given SC C() wemay define a self-diffeomorphism D of which is supported near as follows.Let a neighborhood N of be identified by oriented coordinate charts with the


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annulus {a C | 1 ||a|| 2}. Then D is the map a ei2(||a||1)a on N, andthe identity on \ . We call D the positive Dehn twist about . The inverseoperation, D1

, is a negative Dehn twist. We denote the mapping class of D

by

. It can easily be seen that depends only on the isotopy class of .We call positive if it can be factored as a product of positive Dehn twists,

and denote the monoid of such mapping classes as Dehn+().An open book decomposition (, ), where , for a 3-manifold M, with

binding K, is a homeomorphism between (( [0, 1])/ , ( [0, 1])/ ) and(M, K). The equivalence relation is generated by (x, 0) ((x), 1) for x and (y, t) (y, t) for y .

Definition 2.1. For a surface and positive mapping class , we define the positiveextension p.e.() SC C() as the set of all SC C() such that appears insome positive factorization of .

A recurring theme of this paper is to use properly embedded arcs i

to understand restrictions on p.e.() which can be derived from the monodromyimages (i) in (relative to the arcs themselves). Our general method is asfollows: Suppose that P is some property of pairs ((), ) (which we abbreviateby referring to P as a property of the image ()) which holds for the case is theidentity, and is preserved by positive Dehn twists. Suppose then that p.e.();then there is a positive factorization of in which is a twist. Using the wellknown braid relation, we can assume is the final twist, and so the monodromygiven by 1 is also positive. It follows that P holds for (

1 )() as well.

As a motivating example, consider a pair of arcs , : [0, 1] which sharean endpoint (0) = (0) = x , isotoped to minimize intersection. Following[HKM], we say is to the right of , denoted , if either the pair is isotopic,or if the tangent vectors ((0), (0)) define the orientation of at x. The propertyof being to the right of (at x) is then a property of images () which satisfies

the conditions of the previous paragraph. We conclude that, if () is to the rightof , then p.e.() only if (1 )() is to the right of .

The main motivation for the use of properly embedded arcs to characterize prop-erties of mapping classes is the well-known result that a mapping class of a surfacewith boundary is completely determined by the images of a set of properly em-bedded disjoint arcs whose complement is a disc (an application of the Alexandermethod, see e.g. [FM]). The most well-known example of such a property is whatis now generally known as right veering:

Definition 2.2. [HKM] Let be a mapping class in , a properly em-bedded arc with endpoint x . Then is right veering if, for each such andx, the image () is to the right of at x.

We denote the set of isotopy classes of right veering diffeomorphisms as V eer()

.The right veering property can perhaps best be thought of as a necessary con-

dition of positivity, and (by [HKM]), using the Giroux correspondence, also a nec-essary condition for tightness of the supported contact structure. It was howevershown in that paper that right veering is far from a sufficient condition; indeed,that any open book decomposition may be stabilized to a right veering one.

The failure of right veering to capture either tightness or positivity is at leastpartially due to two obvious shortcomings. Firstly, we observe that right veering


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only sees one arc at a time. Referring back to the original motivation of theAlexander method, we would rather a property which can see arbitrary collectionsof (disjoint) arcs. Secondly, right veering is localized to the boundary; in particularnegative twisting in the interior of a surface may be hidden by positivity nearer tothe boundary. The goal of Section 3 is to introduce a substantial refinement ofright veering which takes into account each of these issues.

As an initial step, we consider what it should means for two arcs to have thesame images under . An obvious candidate is the following:

Definition 2.3. Let be a properly embedded arc in , V eer(). Define thestep-down C() as the embedded multi-curve on obtained by slicing () along, and re-attaching the endpoints as in Figure 1.

Figure 1. Construction of step-down

The stepdown thus allows us to relate distinct arcs to a common object, a mul-ticurve. We want to say that a pair of arcs has in some sense the same image ifthey have the same stepdown:

Definition 2.4. Let 1, 2 be properly embedded arcs in , and V eer(). Wesay the pair 1, 2 is parallel under if C(1) is isotopic to C(2).

As it stands, however, this condition is far too strict to be of any practicaluse. Our approach then is to localize these ideas by decomposing the monodromyimages of arcs into segments which we may then compare across distinct arc images,as will be made clear in the following section.

3. Right position

In this section, we develop the idea of a right position for the diffeomorphic imageof a properly embedded arc in , and develop consistency conditions which allowus to prove:

Theorem 3.1. Let be a compact surface with boundary, admit a pos-itive factorization into Dehn twists, and {i}

ni=1, n 1, be a collection of non-

intersecting, properly embedded arcs in. Then{(, i)}ni=1 admit consistent right

positions.

3.1. Definitions and examples.

Definition 3.2. Suppose is a properly embedded arc with boundary {c, c},and V eer(), with () isotoped to minimize (). Let A be a subsetof the positively oriented interior intersections in () (sign conventions areillustrated in Figure 2). Then the set {c, c} A is a right position P = P(, ) forthe pair (, )


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Note that we are thinking of the set I of positively oriented interior intersectionsin () as depending only on the data of and the isotopy class , and thusindependent of isotopy of and () as long as intersection minimality holds. Inparticular,the integer |I| clearly depends only on this data. We may label elementsof I with indices 1, . . . , |I| increasing along from distinguished endpoint c. Rightpositions for the pair (, ) are thus in 1-1 correspondence with subsets of the setof these indices.

Figure 2. p1 is a positive, p2 a negative, point of ()

Associated to a right position is the set H(P) := {[v, v] () | v, v P }. Wedenote these segments by hv,v , or, if only a single endpoint is required, simply usehv to denote a subarc starting from v and extending as long as the context requires.In this case direction of extension along () will be clear from context.

Right position can thus be thought of as a way of decomposing the image ()into horizontal segments H(P), separated by the points P (Figure 3). Note thatfor any right-veering , and any properly embedded arc , (, ) has a trivial rightposition consisting of the points (), and so there is a single horizontal segmenthc,c = (). Of course ifI is nonempty, there are 2|I|1 non-trivial right positions.Intuitively, the points in P play a role analogous to the initial (boundary) points

of the arc, allowing us to localize the global rightness of an arc image though thedecomposition into horizontal segments (this will be made more precise further onin the paper).

Figure 3. P = {c,v,c} is a right position of (, ). There arethree distinct horizontal segments: hc,v, hv,c, and hc,c

We are interested in using right position to compare pairs of arcs; the idea beingthat, if is positive, we expect to be able to view any pair of horizontal segmentsas either unrelated, or as belonging to the same horizontal segment.


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To get started, we need to be able to compare horizontal segments. To that end,we have:

Definition 3.3. Suppose i, i = 1, . . . , 4 are properly embedded arcs in a surface. A rectangular region in is the image of an immersion of the standard disc[0, 1] [0, 1], such that the image of each side is a subset of one of the i, andcorners map to intersections i j .

Definition 3.4. Let 1, 2 be distinct, properly embedded arcs in , V eer(),and Pi = Pi(, i) a right position for i = 1, 2. We say that hv H(P1) andhw H(P2) are initially parallel (along B) if there is a rectangular region B in bounded by 1, hw, 2, hv, such that, if each of hv and hw is oriented away fromits respective vertical point, then the orientations give an orientation of B (seeFigure 4). We refer to v and w as the endpoints of the region, and the remainingtwo corners midpoints.

Remark 3.5. It is helpful to think of the rectangular region B along which two

segments are initially parallel as the image under the covering map : ofan embedded disc in the universal cover. As such, we will draw these regions as em-bedded discs whenever doing so does not result in any loss of essential information.

Figure 4. (a) hv and hw are initially parallel along B. (b) hv andhw are not initially parallel.

So, to say that a pair of horizontal segments is unrelated is to say that theyare not initially parallel. It is left to clarify what we expect of initially parallelsegments hv and hw. The idea is to extend the notion of being parallel up to asecond pair of vertical points v w, and thus to the horizontal segments hv,v hw,w,such that the pair is completed; i.e. if we switch the orientations so that hv andhw are taken as extending in the opposite direction, then hv and hw are them-selves initially parallel, and the rectangular regions complete one another to forma singular annulus (Figure 5) (notice that there is no restriction on intersectionsof the regions beyond those given by the properties of the boundary arcs). To beprecise:

Definition 3.6. Let i, Pi, i = 1, 2 be as in the previous definition. Suppose hor-izontal segments hv and hw are initially parallel along B, with corners v, w, y1, y2,where yi i. Then the pair hv, hw are completed if there are points v P1, w P2, which we call the endpoints, such that the arcs 1, 2, hv,v, hw,w bound a sec-ond rectangular region B, with corners v, w, y1, y2 (Figure 6). We say the pairhv, hw are completed along B. Note that, turning the picture upside-down, thearcs hv,v and hw,w are initially parallel along B

, and completed along B.


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Figure 5. Horizontal segments hv,v, hw,w complete one another,forming the boundary of a singular annulus.

Figure 6. The horizontal segments hv, hw are completed bypoints v, w. Strands which terminate in like brackets are meant tobe identified along arcs which are not drawn; a similar conventionwill be used throughout the paper.

Remark 3.7. Throughout the paper, we will draw vertical points as solid dots,midpoints as circles.

As justification of the term singular annulus, note that if initially parallel hvand hw are completed then the resolution of the midpoints y1 and y2 as in Figure7 transforms the singular B B into an immersed annulus.

Figure 7. resolution of y1

We have come to the key definition of this section:

Definition 3.8. Let {i}ni=1, n 1, be a collection of non-intersecting, properlyembedded arcs in , V eer(), and Pi = Pi(, i) a right position for eachi. We say the Pi are consistent if for all j = k, each pair of horizontal segments


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Finally, consider the open book decomposition indicated in Figure 10, wherethe monodromy is 1

n11

n22 34) (we have drawn the case n1 = 1). Using

the shaded initially parallel region, observe that the pair (, 1) and (, 2) arenot a right pair; indeed, for any n1, n2 > 0 this will clearly remain true, as thepicture will only be modified by twists about the boundary component parallelto 2. In particular, has no positive factorization for any n1 and n2. On theother hand, it is a simple exercise to show that for any positive n1 and n2 eachboundary component is protected; i.e. each properly embedded arc with endpointon the boundary component under consideration maps to the right at that point (itsuffices to check for the either of the boundary components parallel to 3 or 4). Inparticular, each such monodromy is right-veering. See e.g. [L] for a more in-depthexamination of this example.

Figure 10. (a) Surface and curves for the open book decompo-sition (0,4, 1

n11

n22 34). (b) The pair (, 1) (in blue), and

(, 2) (in red) is not a right pair, but is right veering for allm, n 1.

3.2. The right position associated to a factorization. Suppose we have asurface , a mapping class given as a factorization of positive Dehntwists, and a set {i}ni=1 of pairwise non-intersecting, properly embedded arcs in. While each such arc admits at least one, and possibly many, right positions,the goal of this subsection is to give an algorithm which associates to a uniqueright position, denoted P(i) for each i, such that the set {P(i)}ni=1 is consistent(Definition 3.8). Note that existence of such an algorithm proves Theorem 3.1.

We begin with the case of a single arc , and construct P() by induction onthe number of twists m in .

Base step: m = 1 Suppose = , for some SC C(). We begin byisotoping so as to minimize . Label = {x1, . . . , xp}, with indicesincreasing along . We then define the right position P(, ) associated to thefactorization to be the points {c = v0, v1, . . . , vp1, c = vp}, where {c, c} = (),and, for 0 < i < p, vi lies in the connected component of \ support() betweenxi and xi+1 (Figure 11).


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Figure 11. The base case: the right position P(, ) associatedto a single Dehn twist.

As an aside, note that, in this case, if hvi,vj denotes the closed curve obtainedfrom the horizontal segment hvi,vj by adding the segment along from vi to vj ,

then hvi,vi+1 for 0 i < p. Compare this with the stepdown of Definition 2.3.Now, this construction has as input only the isotopy class of , so we have:

Lemma 3.10. The right positionP() defined in base step depends only on the

isotopy class of .3.2.1. Description of the inductive step. For the inductive step, we demonstrate analgorithm which has as input a surface , mapping class V eer(), properlyembedded arc , right position P(, ), and SC C(). The output thenwill be a unique right position P = P( , ). A key attribute of the algorithmwill be that, while is given as a particular representative of its isotopy class [],the resulting right position P will be independent of choice of representative.

To keep track of things in an isotopy-independent way, we consider triangularregions in :

Definition 3.11. Suppose and are properly embedded arcs in a surface such that () = (), isotoped to minimize intersection. Then for SC C(), atriangular region T (of the triple (, , )) is the image of an immersion f : ,

where is a 2-simplex with vertices t1, t2, t3 and edges t1t2, t2t3, t3t1, such thatf(t1t2) , f(t2t3) , and f(t3t1) .

Definition 3.12. A triangular region T for the ordered triple (, , ) is

essential if can be isotoped relative to T so as to intersect and

in a minimal number of points (Figure 15(a)). upward (downward) if bounded by , , in clockwise (counterclockwise)

order (Figure 15(b)).

We begin with a brief description of the algorithm, with an illustrative example inFigure 12. Let be a representative of [] which minimizes and (), so inparticular all triangular regions for the triple (, , ()) are essential. Furthermore,chose support(D) so as not to intersect any point of () (Figure 12(a)) (recallthat D refers to the Dehn twist about fixed [], while is its mapping class).

Consider then the image D(()), which differs from () only in support(D)(Figure 12(b)). Now, the only bigons bounded by the arcs D(()) and containvertices which were in upward triangles in the original configuration. In particular,there is an isotopy of D(()) over (possibly some subset of) these bigons whichminimizes (()) (the issue of exactly when a proper subset of the bigonssuffices is taken up below, in Section 3.2.2). There is thus an inclusion map i of thosepoints of P which are not in upward triangles into the set of positive intersectionsof (()), whose image is therefore a right position (Figure 12(c)).


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Figure 12. Top row: (a) The setup for the above discussion.Triangular region T1 is upward, while T2 is downward. The supportof the twist is shaded. (b) The result of . There is a singlebigon, in which v1 is a vertex. (c) The result of isotoping (())over this bigon so as to minimize intersection with . Bottom row:Same as the top, but with . Note that while the images (()),(()) are of course isotopic, the right positions of (c) and (f)differ. Finally, (g) indicates the isotopy-independent right position(i.e. the points v and v, along with ) which our algorithm ismeant to pick out.

Note however that, as described, this resulting right position is not independentof the choice of - indeed, ifT1 is any downward triangle, we may isotope the given over T1 (e.g. go from Figure 12(a) to Figure 12(d)) to obtain [] such that and () are also minimal. The algorithm of the previous paragraphthen determines a distinct right position for ( , (Figure 12(f)). Motivated

by this observation, we refine the algorithm by adding all points of (())which would be obtained by running the algorithm for any allowable isotopy of (Figure 12(g)). As we shall see, this simply involves adding a single point for eachdownward triangle in the original configuration.

3.2.2. Details of the inductive step. This subsection gives the technical details nec-essary to make the algorithm work as advertised, and extracts and proves thevarious properties we will require the algorithm and its result to have.


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We begin with some results concerning triangular regions. Let , be properlyembedded, non-isotopic arcs in , with () = (), isotoped to minimize inter-section. Our motivating example, of course, is the case = (), so we orient thearcs as in Figure 2. Let G denote their union. A vertex of G is thus an intersectionpoint . Let SC C() be isotoped so as not to intersect any vertex ofG. A bigon in this setup is an immersed disc B bounded by the pair (, ) or by(, ). Similarly, a bigon chain is a set {Bi}ni=1 of bigons bounded exclusively byone of the pairs (, ), (, ), for which the union

{ (Bi)}ni=1 is a connected

segment of (Figure 13(a)). Finally, we say points p, p G are bigon-relatedif they are vertices in a common bigon chain.

Suppose then that T is a triangular region in this setup, with vertex v .We define the bigon collection associated to T, denoted BT, as the set of points{p G | p is bigon-related to a vertex of T} (Figure 13(b)). Note that, if avertex of T is not a vertex in any bigon, the vertex is still included in BT, so that,for example, the bigon collection associated to an essential triangle T is just the

vertices of T.Now, the point v divides each of, into two segments, which we label +, , +,

in accordance with the orientation. For each bigon collection B associated to a tri-angle with vertex v, we define v(B) := (|B +|, |B|, |B+|, |B

|) (Z2)

4,where intersections numbers are taken mod 2 (Figure 13(b)).

Figure 13. (a) The shaded regions are distinct maximal bigonchains. (b) As each chain from (a) has a vertex in each ofthe triangular regions T1, T2, the bigon collection BT1 = BT2 in-cludes each intersection point of each chain. In this example,v(BT1) = ( 0, 1, 1, 0), so the collection is downward. Note thatT1 is an essential downward triangular region (Definition 3.12),while T2 is non-essential.

Finally, we label a bigon collection B as upward (with respect to v) if v(B) {(1, 0, 1, 0), (0, 1, 0, 1)}, downward if v(B) {(1, 0, 0, 1), (0, 1, 1, 0)}, non-essentialotherwise (Figure 13(b)). Compare with Definition 3.12,

Lemma 3.13. Let [] be such that and () are minimal, andv () . Then the number of essential triangles with vertexv in the triple(, (), ) depends only on the isotopy class [].

Proof. Let be an arbitrary representative of [] (in particular and ()are not necessarily minimal), and v a vertex. We define an equivalence relation onthe set of triangular regions with vertex v by T1 T2 BT1 = BT2. In particular,there is a 1-1 correspondence between {triangular regions with vertex v} andthe set of bigon collections associated to triangular regions with vertex v.


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Now, if is another representative of the isotopy class [], we may break theisotopy into a sequence = 1 n = , where each isotopy i i+1 iseither a bigon birth, a bigon death, or does not affect either of || and |()|.Note that, if B is upward or downward, an isotopy i i+1 which does not crossv does not affect v(B), while an isotopy i i+1 which crosses v changes v(B)by addition with (1, 1, 1, 1). In either case, the classification is preserved; i.e. wecan keep track of an essential upward (downward) bigon collection through eachisotopy. Furthermore, any two distinct bigon collections will have distinct imagesunder each such isotopy (a bigon birth/death cannot cause two maximal bigonchains to merge). We therefore have an integer a(v), defined as the number ofessential bigon collections B associated to triangular regions with vertex v, whichdepends only on []

Finally, if we take to be a representative of [] which intersects and ()minimally, then {triangular regions with vertex v} is by definition just the setof essential triangular regions with vertex v, and has size a(v).

Lemma 3.14. Let be a fixed representative of the isotopy class [] which hasminimal intersection with (), andv () a vertex of an upward triangleT. Then v is not a vertex of any downward triangular region T for .

Proof. Consider the neighborhood of v as labeled in Figure 14. As T is upward,it must be T1 or T3, while T must be T2 or T4. Without loss of generality thensuppose T = T1 is a triangular region. But then neither of T2, T4 can be triangularregions without creating a bigon, violating minimality.

Figure 14. (a) A neighborhood ofv. (b) The upward triangle T1.(c) As has no self-intersection, T2 cannot be a triangular regionwithout creating a bigon.

It follows from Lemmas 3.13 and 3.14 that we may unambiguously refer to a ver-tex of an essential triangular region as downward, upward, or neither, in accordancewith any essential triangular region of which it is a vertex. In particular, if hasminimal intersection with , (), then all triangular regions are essential. Hence-forth we will drop the adjective essential. Figure 15(b) and (c) give examples of

the various types of vertices.

Definition 3.15. Suppose T is a triangular region with vertex v ().Isotope over T to obtain another triangular region T of the same type (up-ward/downward) as T with vertex v, as in Figure 16. Note that if T containssub-regions with vertex v then the process involves isotoping the innermost regionfirst, and proceeding to T itself. We call such an isotopy a shift over v.

We require a final pair of definitions before we bring the pieces together:


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Figure 15. (a) If the shaded region is a maximal bigon chain,then T1 and T4 are essential triangular regions for (, ,

), T2and T3 are not. (b) Upward triangular regions and vertices. (c)Downward triangular regions and vertices.

Figure 16. A shift over a vertex v.

Definition 3.16. Let be a properly embedded arc in , . Suppose , (),and are isotoped to minimize intersection. Let SC C() be a fixed represen-tative of its isotopy class which has minimal intersection with and (). Choosesupport(D) to be disjoint from (). Then let i : (()) (D(()))be the obvious inclusion. Note that D(()) will not in general have minimal in-tersection with .

Definition 3.17. Let and be as in the previous definition, and v ()

a positively oriented intersection point. If there is a representative SC C() ofthe isotopy class [] such that the image D(()) which differs from () only ina support neighborhood of can be isotoped to minimally intersect while fixinga neighborhood of i(v), we say v and i(v) are fixable under the mapping class. Similarly, for a factorization = n 1 of , we say v () is fixableunder if v is fixable under each successive i . Various examples are gathered inFigure 17.

Observation 3.18. Using this terminology, we have the following characterization ofthe right position P() which the reader may or may not find helpful: The rightposition P() is a maximal subset of the positive intersections in () suchthat each point ofP() is fixable under , and is in fact the unique such positionup to a choice involved in the coarsening process (described in detail in the proofof Lemma 3.21).

As a simple example, Figure 18 illustrates distinct right positions of a pair (, )associated to distinct factorizations of a mapping class .

We are now ready to precisely describe the inductive step. Let P = P(, )be a right position, and let SC C(). We construct the right position P =P( , ) in two steps. Firstly, we coarsen P by removing all points which arenot fixable by . This new right position, being fixable under , defines a right


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Figure 17. (a) A downward point is fixable. (b) An upward pointis not fixable, unless (c) there is another upward triangle shar-ing the other two vertices. Note that, in this case, while eitherof the points v, v is individually fixable, the pair is not simulta-neously fixable. Finally, (d) illustrates the reason for demandingthat a neighborhood of the point be fixed - though the intersection-minimizing isotopy may be done without removing the intersectionpoint v, there is no fixable neighborhood, and so v is not fixable.

position for , (()). Secondly we refine this new position by adding points soas to have each new fixable point of (()) accounted for.

3.2.3. The coarsening P P. The coarsening process takes the given right posi-tion P and a curve SC C(), and returns a right position P = P(, ) P.The intuitive idea is that we will remove a minimal subset of P such that the re-maining points are fixable by , thus allowing an identification of P with a subset


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16 ANDY WAND

Figure 18. On the left are the curves of the well-known lanternrelation on 0,4 - setting 1 = 4321 , 2 = 321 , thelantern relation tells us that 1, 2 are factorizations of a common . Clearly, for the arc indicated, the intersection pointv () in the upper right figure is fixable under 1. However,as is clear from the lower sequence of figures, v is not fixable underthe factorization 2. Thus the right position P(1, ) contains thepoint v as well as the endpoints of , while P(2, ) contains onlythe endpoints of .

of (()). Of course we must do so in a way which depends only on theisotopy class of .

We begin by choosing a representative of the isotopy class [] which has mini-mal intersection with and (). Choose support(D) to be disjoint from P. Now,the only bigons bounded by the arcs D(()) and correspond to upward trian-gular regions of the original triple (, (), ). Thus, any isotopy which minimizes(()) must isotope D(()) over some subset of the upward triangular re-gions. The coarsening must therefore consist of a removal of a subset of the upwardpoints of P. We want to determine which of these points are fixable under .

Lemma 3.19. Suppose that v P is a vertex of an upward triangular region T.Then if v is fixable under , then there is another upward triangular region T

such that T, T share the vertices of T other than v (Figure 17(c)).

Proof. Let a be the vertex ofT at and p the remaining vertex (Figure 19(a)).As v is upward, there is a bigon in the image, corresponding to T, bounded by((()) and , with vertices v and a (Figure 19(b)). Therefore, if v is to be

fixed through an isotopy of (()) which minimizes (()) , then a mustbe a vertex of a second bigon also bounded by (()) and , with vertices aand v for some v (()) (Figure 19(c)). By minimality of intersectionsin the original configuration, this new bigon then corresponds to a second upwardtriangle T with vertices a, v, p in the original configuration (Figure 19(d)). Asv is fixable, the bigons must cancel, and we have p = p, as desired (Figure 17(c)indicates the configuration). Note that, while each of v, v is fixable, the pair is notsimultaneously fixable.


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Figure 19. Figures for Lemma 3.19.

We refer to such a twist as surgering out from () relative to . Also,Lemma 3.20. If triangular regions T andT share exactly two vertices, then thesevertices are the two which lie on the closed curve .

Proof. Let the triples (v1, p1, a1), (v2, p2, a2) be the vertices of upward triangles T1and T2, where vi (), pi (), and ai , and suppose T1 andT2 have exactly two vertices in common. Essential triangular regions can have noedges in common, so they can have two vertices in common only if the vertices liealong a closed curve given by the union of the two edges defined by the pair ofvertices. In particular, as is the only closed curve in the construction, the pointsin common must be p1 = p2 and a1 = a2.

Motivated by Lemmas 3.19 and 3.20, we introduce an equivalence relation onthe set of upward triangular regions, such that T T if and only if they shareexactly two vertices. Now, any such equivalence class comprises of m triangularregions, each of which has a unique vertex in (). These vertices (and thusthe elements of ) may be ordered along () from the distinguished endpoint c.We then define a new right position P for the pair (, ) by removing the last (inthe sense of the previous sentence) point of P from each equivalence class in whicheach triangular region has a vertex in P.

Lemma 3.21. Let P P be as defined in the previous paragraph. Then P

satisfies the following properties:

(1) Each point of P is fixable under (2) P is a maximal subset of P for which property (1) holds.(3) P depends only on the isotopy class of .

Proof. Let be an equivalence class of upward triangles. We distinguish twosubcases, depending on whether each element of has a vertex in P or not. Toemphasize the distinction, we call a point p () a vertical point if p is in P,non-vertical otherwise.

(1) (Each element of has a vertex in P) Let T . Using Lemma 3.20, theequivalence class of T is formed of triangles sharing the points along (atypical equivalence class of three triangles is shown in the left side of Figure20(a) - note that for either choice of , exactly two of the three regions are


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18 ANDY WAND

embedded). Suppose then that contains m triangular regions. As notedin the proof of Lemma 3.19, at most m 1 vertical points from the verticesof elements of are simultaneously fixable under

, and conversely, any

set of m 1 vertical points from these is fixable. In particular, if eachtriangle in the class has a vertex in P , the removal of a single verticalpoint is a necessary and sufficient refinement of the vertical points involvedin the equivalence class so as to satisfy properties (1) and (2). Clearly theresulting set P obtained by repeating this refinement on each equivalenceclass is independent (as an unordered set) of the actual choice of whichvertical point to remove, and so property (3) is satisfied.

(2) ( contains some element which does not have a vertex in P) If there is atriangle in the class which does not have a vertex in P, then all of the abovegoes through for the neighborhood of the non-vertical point p of () ;i.e., the twist , along with any isotopy of the image () necessaryto minimize intersections can be done relative to a neighborhood of each

vertical point, so that there is no coarsening necessary. Again, property (3)follows immediately.

Figure 20. (a) For the case that each element of an equivalenceclass has a vertical vertex, can be isotoped so as to fix all butany single vertical point - the result is independent of the choiceof which subset to fix. (b) An upward equivalence class of trian-gles whose set of vertices contain non-consecutive (along ), non-vertical points of () (here separated by the vertical pointv). There is therefore a choice of vertical point in the image. Our

convention will be to choose the lower figure.

3.2.4. The inclusion map i. Now, let be a specific representative of the isotopyclass [] which has minimal intersection with and (). We again consider theeffect of the Dehn twist D restricted to a supporting neighborhood of the twistchosen so as not to intersect any point of (). As D(()) will not (in thepresence of upward triangular regions) have minimal intersection with , the image


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of the inclusion i(P) ( D(())) (Definition 3.16) will not in general be aright position. Indeed, i is no longer well-defined as a map into (()) once thishas been isotoped to minimize intersections with .

One may get around this by fixing conventions: if is an equivalence class ofm upward triangles, let {x1, . . . , xm} be the collection of its vertices in (),indexed in order along () from c. By Lemma 3.21, the image of the {xj} under iwill be a collection ofm points in D(()), and exactly one of these intersectionsmust be removed by the intersection-minimizing isotopy. The resulting set ofm 1simultaneously fixable points in (()) is independent of this choice; we labelthese points {y1, . . . , ym1}, ordered along (()) from c. Then, if xj P foreach j, set i(xj) = yj for i < m. Otherwise, let k := max{j | xj P}, and seti(xj) = yj for j < k , and i(xj) = yj1 for j > k (Figure 20).

With these conventions:

Lemma 3.22. For any [], the image i(P) is a right position for ( , ).Moreover, i(P) is simultaneously fixable under .

Proof. We again consider the effect of the Dehn twist restricted to a supportingneighborhood of the twist chosen so as not to intersect any point of ().Observe then that in the image, the only bigons bounded by the curves andD(()) correspond to essential upward triangles in the original configuration.Each singleton equivalence class contributes a single, isolated bigon, while classeswith multiple elements contribute canceling pairs of bigons. In particular, (()) may be minimized by an isotopy of (()) over one bigon from eachclass. This isotopy fixes a neighborhood of each of the points P.

3.2.5. The refinement i(P) P. For the final step, we wish to refine a given

i(P) by adding points to obtain the desired right position P for ( , ) so

that P depends only on [].So, let be a given representative of the isotopy class [] such that the sets

, () are minimal. By Lemma 3.22 we have an identification of P asa subset i(P) of the positive intersections of (()) , and thus as a rightposition for ( , ).

As we are after an isotopy-independent result, we shall require that P includeeach point which is fixed by any representative of [] which satisfies the intersection-minimality conditions. So, let v P be a downward point. Then for each verticaldownward triangular region T with vertex v, let a be the vertex . We label thepositive intersection of with (()) corresponding to a by da,v ( )() (Figure 21), and call the set of all such points Vv. We refer to i(v) Vv P

as the refinement set corresponding to v. Note that, for distinct v, v, the setsVv, Vv are not necessarily disjoint, but as we only require that each such point

be accounted for, this does not affect the construction. We now define the rightposition P of ( , ) as the union of all the refinement sets of each point,P := i(P) {Vv|v P}.

Finally,

Lemma 3.23. The right position P of ( , ) given by the refinement processis exactly the set

[] i(P

). In particular, P depends only on the isotopy classof .


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20 ANDY WAND

Proof. We will show that P =[] i(P

) by showing that each of these sets

are equivalent to the set W := {v ( )() | v is fixable under with

preimage ( under i) v

P}. Now, by construction, for any [] or v

P,each element of i(P) or Vv is in W, so we only need to show that, for arbitrary, each v W is in both P and

[] i(P

).

So, let v W, with preimage v, and a fixed representative of []. There aretwo cases:

(1) (v is downward) Suppose v i(P). Then there is [], given by ashift of over v, such that v i(P). Then v is in the refinement set Vvin the right position P, and also in i(P)

[] i(P

).

(2) (v is not downward) Then v P, and so v i(P) for any .

Figure 21. Construction of the set Vv for downward v P.Figure (a) is the setup in P, (b) shows the result of the refinementfor P by vertical points Vv = {da,v, da,v} on (()).

Figure 22. P is independent of isotopy of in the refinementprocess. (a) is the initial setup, with p, p (), (b) indicates

distinct isotopies, and (c) is the resulting P

for each choice. Whilethe labeling is different, the right positions are equivalent.

3.2.6. The associated right position P.

Definition 3.24. Let be given as a positive factorization . The uniqueright position associated to the pair , by the algorithm detailed in this sectionis the right position associated to , denoted P().


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Finally, we need to extend the associated right position to each pair i, (i) foran arbitrary set {i} of nonintersecting properly embedded arcs. We have:

Lemma 3.25. Let {i} be a set of pairwise non-intersecting properly embeddedarcs in . The algorithm as given above extends to an algorithm which assigns aunique right position P(i) to each pair i, (i).

Proof. This follows directly from independence of the algorithm from the choice ofrepresentative of the isotopy class of, and the observation that we may always findsome representative of which intersects each ofi and (i) minimally (by e.g. aninductive sequence of removing bigons which contain no other bigons - alternately,but at the expense of loosing self-containment, of course one could simply chooseeverything to be geodesic for some hyperbolic metric).

3.3. Consistency of the associated right positions. The goal of this subsectionis to show that any two right positions P(1), P(2) associated to a positivefactorization are consistent (Definition 3.8). Theorem 3.1 will follow from this

result coupled with Lemma 3.25.Our method to show consistency will be to use the right positions to localize

the situation. Recall from the previous section (in particular Lemma 3.23) that, ifP P denotes the inductive step of the algorithm, then, using the notation ofthat section, given a point z P, there is a representative of [] such that z isin the image of the inclusion map i. In particular, this identification ofz as i(v),for some v P, will hold for any isotopy of which does not involve a shift over v.Thus, having fixed up to this constraint, we may consider v as having propertiesanalogous to the endpoints of ; i.e. it is fixed by , and cannot be isotoped soas to cross it. We refer to v as the preimage of z.

Suppose then that we have a pair of nonintersecting properly embedded arcs ona surface with a given monodromy, with right positions P1 and P2. Then, for aDehn twist , the inductive step of the algorithm gives a pair P1 , P

2 . Consider

a pair of points, one from each of Pi . We need to show that for each such pair,the horizontal segments originating at the pair of points are completed if initiallyparallel. We call a pair satisfying this condition consistent.

Now, if there is an isotopy of such that each point in a given pair of points is inthe image of the associated inclusion map i, we call the pair a simultaneous image.The idea is that the horizontal segments originating at such a pair have well-definedpreimages in P1 and P2, so we only have to understand the local picture consistingof the images under of each such pair.

In keeping with analogy of fixable points as boundary points, we generalize thenotion of to the right (as presented in [HKM]) to paths with common endpoint onthe interior of a properly embedded arc . We will say that an isotopy t, t [0, 1],of path respects path if, for all s [0, 1], the interior of path s contains neitherendpoint of .

Definition 3.26. Let be a properly embedded arc in , , : [0, 1] pathsin with common endpoint (0) = (0) = x , such that each intersects minimally, and then isotoped (relative to the endpoints and respecting each other)to minimize intersection with each other. Further suppose that each lies to thesame side of (i.e. if we fix a lift in the universal cover of , and lifts , with

endpoints x then , lie in the same connected component of \ ). We say is to the right of (at x), denoted , if either the pair is homotopic (relative


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22 ANDY WAND

to the endpoints), or if the tangent vectors ( (0), (0)) define the orientation of at x (Figure 23(a)). Similarly we say is to the left of , denoted .

We also need a slightly stronger notion of comparative rightness:

Definition 3.27. Let , , x be as in the previous definition, and suppose :[0, 1] is an embedded arc in with endpoints (0) = x, and (1) = y ,for another properly embedded arc in , disjoint from . Then if there is atriangular region for the triple (, , ) in which x and y are vertices, we say isparallel along . If ( ) and and are not parallel along , then wesay is strictly to the right (left) of (Figure 23(b)).

Figure 23. (a)Arc is to the right of at x. (b) Arc is parallelalong , while is strictly to the left of .

Observation 3.28. Note that, for a pair of horizontal segments hv and hw, theproperty of being initially parallel along a rectangular region B implies that eachis parallel along the unique (up to isotopy) diagonal arc with endpoints v andw and contained in B. Conversely, if hv and hw are each parallel along and to theright of an arc with endpoints v and w, there is a rectangular region along whichthe segments are initially parallel (given by the union of the triangular regions of

Definition 3.27).As for the utility of these definitions, let Pi Pi denote the inductive step

of the algorithm, and consider a simultaneously fixable pair v P1, w P2. Ifthe images (hv) and (hw) are initially parallel along rectangular region B withdiagonal , then we may compare with the preimages hv and hw. Clearly, foreither ofhv and hw, the property of being strictly to the right of is preserved by, and so incompatible with the images being initially parallel along B. Moreover,if one of the pair, say hv, is to the left of , then if hw is to the right of , as ithas no intersection with hv it must be strictly to the right of , which as abovegives a contradiction. We summarize:

Lemma 3.29. Letv P1(, 1) andw P2(, 2) be simultaneously fixable under. Fix [], and let u = i(v) P

1 and z = i(w) P

2 . Suppose that hu and

hz are initially parallel along B, with diagonal . Then:(1) If hv , then hv is parallel along .(2) hv hw (and so hv hw )

We now have everything we need to show that the algorithm preserves con-sistency. We begin with a check that, under certain conditions, initially parallelimages of initially parallel and completed segments are themselves completed. Tobe precise:


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Lemma 3.30. Let Pi be consistent right positions for (, i), i = 1, 2. Supposev P1 andw P2 are fixable under, and horizontal segments hv andhw initiallyparallel along a rectangular region B with diagonal, and completed along a regionB with diagonal to points v and w. Then,

(1) If neither ofv andw are downward (Definition3.12), andhi(v) andhi(w)are initially parallel along (), then hi(v) and hi(w) are completed.

(2) Suppose at least one of v and w is downward, and let Vv and Vw denotethe refinement classes from the refinement step of the algorithm. Then forany u Vv and z Vw, if the horizontal segments hu and hw are initiallyparallel along (a parallel translate of) (here a parallel translate of

just refers to an isotopy which allows the endpoints to move along the arcsi), then they are completed.

Proof. For claim (1), note that, for any allowable [] (i.e. has minimalintersection with the i and their images), intersects B in arcs as in Figure 24(a).

In particular, if neither initial point is downward, then any arc of B eitherconnects parallel edges, or is the edge of an upward triangle in either (1, (2), )(so y1 is a vertex), or (2, (1), ) (so y2 is a vertex).

Suppose firstly that neither yi is a vertex of an upward triangle. Then for anyallowable , all arcs B connect parallel edges, and so all four corners of Bare fixable. We thus may view these as the corners of the initially parallel regioncontaining () (observe that this is well-defined when its endpoints are fixable)in the image, and so need only check the completing region B. Observe that anyarc of B either connects parallel edges, or is the edge of a downward trianglein which v or w is a vertex (Figure 24(b)). It follows then that hi(v) and hi(w)are completed by points u Vv and z Vw .

Figure 24. (a) Dashed lines illustrate possible intersections B.(b) If the corners yi are not upward, the completing disc is pre-served.

We may then assume that any point of hv is a vertex of an upward triangle

with vertex y2 (and similarly for hw and y1). Suppose then that y2 is a vertex ofan upward triangular region of the triple (, 2, (1)). Following the argument ofLemma 3.21, the image (hv) can be initially parallel along () only if there area pair of triangular regions T1 and T2 in the triple (2, (1), ) with two verticesin common (Figure 25(a)); i.e. is a surgering out of relative to 2 (using theterminology of the coarsening step of the algorithm). It is then straightforward tocheck that consistency is preserved: If the surgery can be done so as to fix the edgesof B, this is of course trivial. Otherwise, either an is surgered out of each of the


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24 ANDY WAND

arc-images along the edges ofB, thus fixing endpoints and preserving the region, orB itself is contained in a neighborhood of, in which case the surgery must removea pair of regions with common endpoints, which again preserves consistency.

To see this explicitly, we may separate the claim into 4 cases as follows: if welet p denote the vertex at 2 (1) of T2 (Figure 25(a)), then we may distinguishwhether p is on the edge of B along (1) (call this edge e(1)), and then alsowhether it is on the edge of B along 2 (which we refer to as e2). We see thenthat, if p e2 , then there is an isotopy of such that fixes each edge of B

(Figure 25(b) shows the case p e(1), (c) shows p e(1)), while if p e(1)and p e2 , then surgers out a subarc of each edge ofB

(Figure 25(d)). For theremaining case (Figure 25(e)), ifp B , then B is contained in a neighborhood of. As hw is also assumed parallel along , we find that hv and hw are constrainedto be initially parallel along a further region (call this B1, which for consistency iscompleted by some B1) again contained in a neighborhood of . The surgery thenhas the effect of removing B and B1, and connecting the midpoints ofB with those

of B1 via canceling bigon pairs. We thus have B completed by B

1 in the image.

Figure 25. (a) Surgering from (1) relative to 2. The figureis in the universal cover; all lifts of a given object are given the samelabel as the object itself. (b) p e2 , p e(1). (c) p e2e(1).We have added the extra step of isotoping so that the twist moreclearly fixes all of the edges of B. (d) p e2 , p e(1). (e)

p e2 e(1). Again we have added an extra step, this timeextending the picture to see the region B1 (with midpoints y3 andy4) as described in the proof.


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For claim (2), in addition to the intersections of with B and B as in theprevious case, we must deal with the case that our initial points v and w aredownward. So, let p h

vbe a vertex in a downward triangle T with vertex v.

We first observe that, as hu is assumed parallel along , for each such T there is acomplementary region T of the triple (, 2, (1)), such that T and T

share thevertex (1) (Figure 26(a)).

Figure 26. (a) If v is downward, then to preserve parallelity of

hiv along , must extend from the downward region T to forma complementary region T. (b) For an arbitrary point u in Vvsuch that hu is parallel along , we may isotope to fix y2.

In the case that u = iv, we may then use these complementary regions toassume is shifted over y2 so that hv B is fixed by . In particular, y2 itselfis fixed, and so we only need check that the images are completed, which followsfrom the proof of case (1) above.

Finally, we are left with arbitrary u Vv. Note firstly that (by definition) sucha u corresponds to the corner x 1 of a downward triangle T of the triple(, 1, (1)). We may assume is isotoped such that the edge of T along 1 iscontained in B . We are assuming hu is parallel to , so as above we may furtherassume isotoped such that y2 is fixed (Figure 26(b)) The same arguments hold

for z Vw, so hu and hz are again initially parallel along a region with corners y1and y2, which is again completed by the argument of part (1).

We are now able to show that the algorithm preserves consistency. We split thestatement into two lemmas, beginning with the case of a simultaneous image pair:

Lemma 3.31. Let Pi, i = 1, 2 be consistent, and let u P1 and z P2 be a

simultaneous image. Then if hu and hz are initially parallel, they are completed.

Proof. Suppose hu and hz are initially parallel along rectangular region B withdiagonal . As usual, we label the remaining two corners of B by yi i. Let [] be fixed such that u and z are in the image of i. Denote the preimages by

v P1 and w P2, so i(v) = u and i(w) = z. Using Lemma 3.29, there are twocases:

(1) hv , hw . We again fix a neighborhood U() of , and consider theisotopy D(hv) hi(v) which minimizes intersections of (hv) with and . We let p be the initial (along hv from v) point of hv, and p

the first point of (()) hv along hv from p (Figure 27). Now, the onlybigons bounded by D(hv) and or correspond to triangular regions inthe preimage. We distinguish two subcases:


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26 ANDY WAND

Figure 27. Notation for Lemma 3.31. The shaded region is aneighborhood of .

(a) The point v is the vertex of a downward triangular region T in thetriple (, 1, hv) (Figure 28(a)). Then p

(considered now as a point ofD(hv)) is not in a bigon bounded by D(hv) and 1 (else 1 and hvbound a bigon), so our isotopy may be done so as to fix p. Thus, forthe image (hv) to lie along , we find that there is a triangular regionT of (2, , ) which intersects T in the vertex of T. Similarly,as hw , it follows that an initial segment of hw is within T, and

thus w is also downward. Using T and T, we may then assume that runs from 1 to 2 within a neighborhood of , intersecting exactlyonce (Figure 29). There are then u P1 , z

P2 corresponding tothe vertices i of the associated triangles. The segments hu and hzare compatible, initially parallel along with completing region alongthe remainder of and endpoints u and z.

Figure 28. Figures for Lemma 3.31

Figure 29. (a) If hv (at v), then the same is true of hw(at w), so the pair is initially parallel along . (b) The segmentshu = (hv) and hz = (hw) are compatible, with endpoints u

, z,given by the refinement procedure, corresponding to the points i


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(b) The points v and p are not vertices in a downward triangular regionof (, 1, hv) (Figure 30(a)). By part (a), w is also not in a downwardtriangular region (of (,

2, )). Now, h

uis assumed parallel along ,

so we consider the respective triangular region T in (()(1), , 2)(Figure 30(c)). As the point w is not downward, all arcs of Tconnect hu to or 2. In particular, hv =

1 (hu) is parallel along

1 (). The same holds for the complementary triangular region T

in (( )(2), , 1). Now, B = T T, so we find the preimages hvand hw initially parallel along a region with diagonal

1 (Figure 31

gives a typical situation). The result then follows from Lemma 3.30.

Figure 30. (a) The points v and p are not vertices in a downwardtriangular region. (b) The image under D. (c) The triangularregion T in (( )(1), , 2), and some possible components of T.

Figure 31. (a) Some possible components of B. (b) Segmentshv and hw are initially parallel along

1(B)

(2) hv , hw By Lemma 3.29, hv and hw are each parallel along . Thepair is thus initially parallel along the same region B, and so completed bypoints v and w. Again, the result follows from Lemma 3.30.

We must also consider the case of non-simultaneous image pairs:

Lemma 3.32. Let Pi, i = 1, 2 be consistent, and let u P

1

and z P2

not be asimultaneous image. Then if hu and hz are initially parallel, they are completed.

Proof. As discussed at the beginning of this subsection, we may assume, by shifting, that any given point in P2 is the image under i of some w P2. We thereforebegin by fixing [] such that z = i(w) (i.e. up to isotopies which do not involvea shift over w). By assumption, this does not fix u, so there is a point v P1such that u is in the refinement set Vv (Figure 32(a)). In particular, v is downward.Note that w is in the interior of the associated downward triangular region T (else


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28 ANDY WAND

we could use T to isotope so as to fix each of u and z simultaneously). The pointw is thus also downward.

We wish to adapt the proof of Lemma 3.31 to this situation. We again supposehu and hz are initially parallel along B with diagonal , and label the remainingtwo corners ofB by yi i. However, we no longer have a preimage of hu. We mayget around this difficulty as follows: Observe that, after possibly shifting over u,we may assume u is not downward in (y1, hu, ) (Figure 32(b)), so that hu has awell-defined preimage path with endpoint x 1 as in Figure 32(c). Intuitively, starts at the corner x = 1 of T, follows along the length of the edge of T,then coincides with hv up to its endpoint. Thus is a path with the property thatall self intersections and/or intersections with hw lie along its coincidence with .Let be the (unique up to isotopy) properly embedded arc obtained from bysliding its endpoint on 1 along 1 to v. We proceed to analyze cases as in Lemma3.31:

Figure 32. The construction of the path with basepoint x in 3.32.

(1) : Similarly to the argument of Lemma 3.29, we observe that hv < ,and so hw cannot be both parallel along and to the right of (at w) withoutintersecting hv. Thus hw (Figure 33(a)). The proof follows exactly asin Lemma 3.31, with v and hv replaced by x and .

(2) : as in the argument for Lemma 3.29, cannot be strictly to theright of , so it is parallel along , and so by construction both hv andhw parallel along (Figure 33(b),(c)). We are thus in case (2) of Lemma3.30.

Figure 33. Figures for Lemma 3.32. Although in the pointw will actually lie in the region T, we keep to our convention ofdrawing pictures in the universal cover.

Finally, we bring all of the above together to prove Theorem 3.1.


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Proof. (of Theorem 3.1) It follows from Lemmas 3.31 and 3.32 that the inductivestep of the algorithm preserves consistency of right positions. It is left to show thatthe base step, in which is a single twist

, gives consistent right positions to

each pair 1, 2.So, let Pi be the right position associated to (i, ), i = 1, 2 by the base step.

We index the points and associated horizontal segments in increasing order alongi from the basepoint ci. Suppose then that hvj H(P1) and hwk H(P2) areinitially parallel horizontal segments, along region B (Figure 34(b)). As noted inthe construction of the base step, each hvj ,vj+1 is just the image of a segment of1 which has a single intersection with , with endpoints corresponding to eachconnected component of the fixable points 1 \ support(), and similarly for eachhwj,wk+1 . Thus, if n is the number of points of P1 B , then hvj and gvk arecompleted by the points vj+n+1 and wk+n+1. Figure 34(b) illustrates the localpicture for the case n = 1.

Figure 34. (a) An example of the right position associated to asingle Dehn twist. (b) Segments hvj and hwk are initially parallelalong the shaded disk in the leftmost figure, and are completed tovj+2, wk+2 along the shaded disk in the rightmost figure

By induction, then, given a surface , and with positive factorization ,the associated right positions P(), P() are consistent for any nonintersectingproperly embedded arcs , .

4. Restrictions on p.e.()

In this section, we switch focus back to the entirety of pairs of distinct properlyembedded arcs 1, 2 in a surface , and the images of these arcs under right-veering . The motivating observation here is that, as the endpoints of each arc are


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30 ANDY WAND

by definition included in any right position, the property of the initial horizontalsegments of the images being initially parallel is independent of right position, andso is a property of the pair (

1), (

2). In particular, for positive , if (

1) and

(2) are initially parallel, they must admit consistent right positions Pi(, i) inwhich the initial segments are completed (see Example 3.9). We are interested inunderstanding what necessary conditions on p.e.() (Definition 2.1) can bederived from the information that (1) and (2) are initially parallel. These aresummarized in Lemma 4.4 and Theorem 4.13.

4.1. Restrictions from initial horizontal segments. Throughout the section,we will be considering rectangular regions R in , and classifying various curvesand arcs by their intersection with such regions. We need:

Definition 4.1. Let R be a rectangular region with distinguished oriented edge e1,which we call the base of R. We label the remaining edges e2, e3, e4 in order, usingthe orientation on e1 (Figure 35). We classify properly embedded (unoriented) arcs

on R by the indices of the edges corresponding to their boundary points, so thatan arc with endpoints on ei and ej is of type [i, j] (on R). We are only concernedwith arcs whose endpoints are not on a single edge. An arc on R is then

horizontal if of type [2, 4] vertical if of type [1, 3] upward if of type [1, 2] or [3, 4] downward if of type [1, 4] or [2, 3] non-diagonal if horizontal or vertical type 1 if not downward

Figure 35. Representatives of each of the 6 possible types of arcon R

Definition 4.2. Let R in be an rectangular region with distinguished base as inDefinition 4.1. We say SC C() is type 1 on R if each arc R is type 1 on R.

Following Definition 4.1, we use a given pair of properly embedded arcs to define

a rectangular region D of as follows: Suppose 1 and 2 are disjoint properlyembedded arcs, where i = {ci, ci}, and 1 is a given properly embedded arc withendpoints c1 and c2. Let 2 be a parallel copy of 1 with endpoints isotoped along1 and 2 to c1 and c2. We then define D as the rectangular region bounded by1, 2, 1, 2 (the endpoints are labeled so that c1 and c2 are diagonally opposite),and base 2, oriented away from c1 (see Figure 36). For the remainder of thissection, D will always refer to this construction. Of course, for a given pair of arcs,D is unique only up to the choice of 1. We also make use of the diagonal of


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Figure 36. disc construction

D, a representative of the unique isotopy class of arcs with boundary {c1, c2} andinterior within D. In particular, the diagonal determines 1, and vice versa.

Definition 4.3. We say the pair (1), (2) is initially parallel (on D) if thereexists D such that (1), (2), 1, 2 bound a rectangular region B D on whichc1 and c2 are vertices. Initially parallel images are flat ifD can be constructed suchthat (i) j = for i, j {1, 2}.

As expected, then, the pair of arcs (1), (2) is initially parallel exactly whenthe horizontal segments hc1 and gc2 originating from the boundary of each pair ofright positions are initially parallel. Using this, we immediately obtain:

Lemma 4.4. Let be positive, and the pair (1), (2) initially parallel on D.Then p.e.() only if is type 1 on D.

Proof. Suppose otherwise - then D has an arc of type [1, 4] or [2, 3] on D. Now,if p.e.(), there is some positive factorization of in which is the initialDehn twist. However, if D has an arc of type [1, 4], then (1) is strictly tothe right of (1) (in the sense of section 2), while if the arc is of type [2 , 3], then(2) is strictly to the right of (2), either of which contradicts (1) and (2)

being initially parallel.

Now, by Theorem 3.1, if is positive, there are consistent right positions P1 andP2 for any pair (1), (2). If the initial horizontal segments hc1 H(P1), hc2 H(P2) are initially parallel, then there are v P1 and w P2 such that v andw are endpoints of the completed segments. Letting B be the completing disk(Definition 3.6), we can define a new rectangular region by extending B so that itscorners lie on i, rather than i :

Definition 4.5. A bounded path in (, D , ) is a rectangular region P bounded byi and (i), i = 1, 2, with corners c1 and c2 in common with D (Figure 37).

Lemma 4.6. If(1) and(2) are initially parallel, for positive , then(, D , )

has a bounded path.

Proof. This is a restatement of the above discussion.

Let , D, and be given such that the pair (1), (2) is flat (and thus initiallyparallel) with respect to D. The remainder of this section is an exploration of whatnecessary conditions this places on elements of p.e.().

We start by observing that under these conditions (i.e. initially parallel and flat),the pair of trivial right positions whose vertical sets consist only of the arc boundary


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Figure 37. P is a bounded path for type 1 (1) and (2).

points are consistent. Also, the entirety of each image (i) is in the boundary ofthe bounded path (see Figure 38). In particular, such images are parallel in thesense of Definition 2.4. Now, for SC C(), it is clear that if all arcs D arehorizontal on D (Definition 4.1), then (D, 1 ) is again initially parallel and flat,and so admits consistent right positions. Our interest then lies in such that D

contains diagonal arcs.

Figure 38. The bounded path of flat (D, )

Lemma 4.7. Let the pair (1), (2) be flat and initially parallel. Then p.e.() only if:

(1): 1 = 2 = (2): (1) = (2) =

Proof. For statement (1), suppose is such that 1 = and 2 = .Referring to Figure 35, this means that arcs D cannot be of type [2, 3], [1, 2] or[2, 4] on D, while any arc of type [1, 4] on D would result in 1 not being rightveering. Similarly, arcs P cannot be of type [1, 2] or [1, 3] on P, while the rightveering conditions eliminates [3, 4] and [1, 4]. Putting these together, we find that may be isotoped such that all intersections P, D are along subarcs isotopic toone of the , , shown in the left side of Figure 40, and in particular that thereis at least one along .

Now, for such , the pair 1 ((1)), 1 ((2)) is initially parallel, and so, if(1 ) is positive, must contain a bounded path. Also, arcs along

, preservethe bounded path, so we may assume all intersections are along . (Figure 39). Itis immediate then that (D, 1 ) has no bounded path.

The argument for statement (2) is nearly identical: We suppose (1) = and (2) = , and find that all intersections P and D are isotopic toone of the arcs , , shown in the right side of Figure 40, and in particular thatthere is at least one along . Again, (D, 1 ) has no bounded path.


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Figure 39. Illustration of the terminology of Lemma 4.7. (a)A typical , with one arc along , and one along . (b) is theimage under 1 . Notice that, to recover a standard picture of theassociated D, we must normalize the picture as in (c), thereby

justifying the assertion that arcs along and may be ignored.It is clear that there is no bounded path.

Note that this is a slight generalization of the the second example of Example3.9.

Figure 40.

We have one final application of the bounded path construction. Note that, for and D as above, an orientation of gives a derived orientation on each arc in D .Thus, given a pair of upward arcs, we can ask whether their derived orientationsagree, independently of the actual orientation of .

Lemma 4.8. Let(1) and (2) be flat and initially parallel. Let be type 1 onD, such that D has exactly 2 diagonal (so necessarily upward) arcs s1 and s2.Then p.e.() only if the derived orientations on s1 and s2 agree.

Proof. Note firstly that, by Lemma 4.7, s1 and s2 must be as in Figure 41(a); i.e.

one cuts the upper left corner of D, the other the lower right corner. We show that(D, 1 ) has a bounded path if and only if the derived orientations on s1 and s2agree, which, by Lemma 4.6 gives the result.

Note that if D has a horizontal arc, it cannot have a vertical arc, and vice-versa. We distinguish the two cases:

(1) D has no vertical arcs. The boundary of bounded path P must includethe initial segments of each arc 1 ((i)) from ci, i = 1, 2. We can thenstart from either corner ci and follow the initial segment around . Figure


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34 ANDY WAND

41 illustrates this for c1. It is clear then that these initial segments willclose up to bound P if the orientations match (Figure 41(b)). If, on theother hand, the orientations do not match, the segments do not form theboundary of a bounded path (Figure 41(c)).

(2) D has no horizontal strands. If there are vertical strands, we mustadjust D such that the diagonal is given by 1 (). Again, as in case (1),there is a bounded path if and only if the derived orientations on s1 and s2agree.

Figure 41. (a) D has exactly 2 diagonal arcs s1 and s2. (b)Orientations agree. (c) Orientations disagree.

4.2. Restrictions from the full set of horizontal segments. To summarizethe previous subsection, for flat (D, ), p.e.() only if is type 1 on D, andsatisfies the intersection conditions of Lemma 4.7. Furthermore, if D has onlytwo diagonal arcs, it must satisfy the orientation conditions of Lemma 4.8. Wenow wish to utilize the results of Section 3 to extend these results to the sets ofhorizontal segments of consistent right positions of the arc/monodromy pairs, andthus to obtain stronger necessary conditions on arbitrary type 1 for inclusion in

p.e.().

4.2.1. Motivating examples. We begin with a pair of examples, meant to motivatethe various definitions to come.

Example 4.9. Consider a flat (D, ), where Dehn+(), with consistent rightpositions P1 and P2 for the edges 1 and 2 ofD (Figure 42(a)). The segments hc1and hc2 are of course initially parallel, so completed by some v P1 and w P2(it is easiest to keep in mind the simplest case in which v and w are the oppositeendpoints of1 and 2, as in the Figure). Let be a type 1 simple closed curve suchthat D has exactly two diagonal arcs, and satisfies the intersection conditions of

Lemma 4.7 and orientation conditions of Lemma 4.8. It is straightforward to verifythat any consistent positions Pi for ((

1 ), i), i = 1, 2, must contain uniquely

determined points v2 P1 and w2 P2 which complete the initially parallel initial

segments, and further that these points define initially parallel segments hv2 andhw2 which are then completed by the original completing points v and w, whichare preserved by 1 . Furthermore, the corners of the regions are each fixable(indicated by circles in the Figure). The regions and points involved are illustratedin Figure 42(b).


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Consider then a slightly more complicated configuration, in which has 4 diag-onal arcs, arranged as in Figure 42(c). Again, one may verify that any consistentright positions for ((1

),

i) must contain one of the two positions indicated

in Figure 42(e) and (f). Again, each of these determines a sequence of regions andpoints which ends with the original completions. The goal of the remainder of thissection is to show that for any type 1 , any pair of consistent right positions for((1 ), i) must contain a similar sequence, which then determines necessaryconditions on p.e.() which generalize those of Lemmas 4.7 and 4.8.

(a) (b)

(c) (d)

(e) (f)

1

1

Figure 42.

4.2.2. Notation and conventions. Given flat (D, ), and a type 1 curve , we labelthe upward sloping arcs si D as in Figure 43. A subarc si is thus defined onlyin a neighborhood of the point xi. Let a1, a2, . . . am, where a1 = 1, be the indices ofthe xi with order given by the order in which they are encountered traveling along to the right from x1 1. We associate to the list = (a1, a2, . . . am), defined

up to cyclic permutation. We decorate an entry with a bar, aj, if the derivedorientations of saj and s1 do not agree.

Given a pair xi 2 and xj 1 (so i is even, j odd), the pair splits into two disjoint arcs. If the orientations of si and sj agree, we label these arcs i,jand j,i (where for even i, i,j refers to the arc connecting xi and xj and initiallyexterior to D at each endpoint, so for odd j, a neighborhood of each endpoint ofj,i is contained in D). We refer to each such arc as a path of the pair (, D).We say paths and are parallel if, after sliding endpoints along the i so as to


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38 ANDY WAND

Figure 45. Above left, the various points of i and (i).Above right, the inverse image 1 ((i)), for with = 136542(strands terminating in brackets with like letters are meant to be

identified). Below, a detail of a point-neighborhood N4.

(a) (b) (c)

Figure 46. (a) The region Ba,b corresponding to the path a,b,and complementary Bb,a corresponding to b,a. The curve is the

dashed line. (b) Non-nested, and (c) nested regions.

Throughout the rest of this section, all notation will be as in Figures 45 and 46.We will further assume that m1 m2 1; i.e. that there are at least as manyupward intersections 1 as there are upward intersections 2.

4.2.3. The positions P1 and P2 . Our immediate goal is to show that, if (

1

, i), i = 1, 2 admit consistent right positions, then any such pair of positions ex-tends non-trivially along the entire length of each arc, as described above. Theorem4.13 will then follow from various properties of these positions.

Lemma 4.15. Let(D, ) be flat and parallel, p.e.(), and Pi, i = 1, 2 consis-

tent right positions of (

1

, i). Then there are {vj}

n

j=1 P1, {wj}

n

j=1 P2,and {ek}2nk=1 N such that

(1) Each of the sets {vj} and{wj} is a right position (i.e. v1 = c1 andvn = c1,and w1 = c2 and wn = c2), and the indices 1, , n increase along each ofi and 1 ((i).

(2) For each j < n, segments hvj,vj+1 andhwj,wj+1 are initially parallel along aregion Bj, and completed along a region Bj with endpoints vj+1 and wj+1.Furthermore, the set of midpoints for these regions are fixable, and the


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midpoints of the completing pair Bj, Bj are ye2j1 and ye2j (in the set {yk}

of fixable points indicated in Figure 45).

(See Figure 47 for a schematic illustration.)

Figure 47. A schematic overview of the positions P1 and P2

constructed in Lemma 4.15. The picture is of the universal coverof , along a lift of 1 ((1)). As usual, all lifts of an object aregiven the same label as the original object. The circled verticesare the fixable points {yk}.

Proof. By assumption, y1 and y2 (as in Figure 45) are midpoints of an initiallyparallel region B1, which for consistency must be completed by some region B1with endpoints v2 and w2. We need then to show that the horizontal segmentshv2,c1 and hw2,c2 are themselves initially parallel along a region with midpointsagain in the fixable point set {yk}. The result will then follow, as we may repeatthe construction until we hit one of the opposite endpoints.

Using the coordinate notation introduced above, recall that v2 can be uniquelydescribed by the strand slr on which it lies. This also determines w2 as lying on s

l1r+1

(Figure 48(a)). Observe that it cannot be the case that both r > 1 and l > 2; inparticular this would imply that the arcs [y2, y4], [y4, y6]

1 ((1) are parallel,

and thus that is not connected (Figure 48(b)). We distinguish 3 cases:

(1) Suppose firstly that r = 1, l = 2, so v2 lies in the intersection of the arc[y2, y4] with N1, and thus w2 lies in the intersection of the arc [y1, y3] withN2. It follows from Lemma 4.8 that v2 and w2 are completing points if andonly if 2,1 is a symmetric path of , D, in which case hv2 and hw2 will beinitially parallel along a region with midpoints y3 and y4 (this is the caseillustrated in Figure 44. For later reference, we note that B1 and B2 are(respectively) exactly the basic regions B2,1 and B1,2 defined earlier.

(2) We next consider the case r > 1. Then v2 in Nr and w2 in Nr+1 define

a completing region B1, which in turn implies existence of a symmetricmultipath of , D with (r + 1)/2 components 2,r, 4,r2, . . . , r+1,1 (thecase r = 5 is illustrated in Figure 49(a)). So l = 2, and hv2 and hw2are initially parallel along a region B2 as in the Figure. Observe that B1contains each of the basic regions B2,r, B4,r2, . . . , Br+1,1, arranged in arow between 1 and 2, while B2 contains each of the complementary basicregions Br,2, Br2,4, . . . , B1,r+1, arranged in a column between 1 ((1))and 1 ((2)).


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(a) (b)

Figure 48. (a) The general situation for the completing disc B1.If both r > 1 and l > 2, then there is a rectangular subregionof B1 with edges [y2, y4] 2, [y2, y4]

1 ((1)), [y4, y6]

2, [y4, y6] 1 ((1)) as in (b). Observe then that (the dashedlines) cannot be a connected curve.

(3) Finally, suppose l > 2. By case (2), this implies r = 1. The configurationis complementary to that of case (2), in that hv2 and hw2 are now initiallyparallel along a region B2 determined by a symmetric multipath with l/2

components, 1,l, 3,l2, . . . , l1,2. Again, Figure 49(b) illustrates the sit-uation (for the case l = 6). Again B1 contains each of the basic regionsB2,l1, B4,l3, . . . , Bl,1, but they are now arranged in a column, while thecomplementary basic regions of B2 are in a row.

(a) (b)

Figure 49. Figures for Lemma 4.15. (a) The case r = 5, sov N5 and w N6. The completing region B1 for hc1 and hc2 islightly shaded, while the initially parallel region B2 for hv2 and hw2is darkly shaded. B1 contains basic regions B2,5, B4,3, and B6,1,while B2 contains B5,2, B3,4, and B6,1. Though irrelevant to ourargument, for intuition it is perhaps worth noting that the pathstaken by the basic regions of B2, which are here represented withour usual bracket notation, will each cross B1 exactly r 3 times(compare with Figure 42(e)). (b) The complementary case l = 6.

Continuing in this fashion, then, we obtain a sequence of points {vj} and {wj}satisfying condition (2), which continues until we hit an endpoint; i.e. wn = c2(recall that we are assuming m1 m2 1). Condition (1) will then be satisfied ifm1 = m2 1, so that the corresponding vn is the other endpoint c1. This howeverfollows using the argument of Lemma 4.7; replacing (1) and (2) in that Lemma

with the segments hvn1,c

1 and hwn1,c

2, we see that these segments are initiallyparallel, and completed only if vn = c1. Figure 50 shows the relevant regions.

We have the following immediate corollary, giving us half of Theorem 4.13:

Corollary 4.16. Let (D, ) be flat and parallel. Then p.e.() only if isbalanced with respect to D.

Andy Wand- Factorizations of Diffeomorphisms of Compact Surfaces with Boundary

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