Immersed Surfaces, Dehn Surgery and Essential Laminations Citation Li, Tao (2000) Immersed Surfaces, Dehn Surgery and Essential Laminations. Dissertation (Ph.D.), California Institute of Technology. doi:10.7907/4kag-zt09. https://resolver.caltech.edu/CaltechTHESIS:11212019-175145827 Abstract Let M be an orientable and irreducible 3-manifold whose boundary is an incompressible torus. We are interested in immersed essential surfaces in closed 3-manifolds obtained from Dehn fillings on M . We show the following two things. In Chapter 2, we suppose that M does not contain closed non-peripheral incompressible surfaces. We show that the immersed surfaces in M with the 4-plane property can realize only finitely many slopes. Moreover, we show that only finitely many Dehn fillings on M can yield 3-manifolds that admit non-positive cubing. This gives the first examples of hyperbolic 3-manifolds that cannot admit non-positive cubing. In Chapter 3, we suppose M is hyperbolic. We show that all but finitely many Dehn fillings on M yield 3-manifolds that contain closed essential surfaces. Moreover, we give a bound on the number of exceptional Dehn fillings. Item Type: Thesis (Dissertation (Ph.D.)) Subject Keywords: Mathematics Degree Grantor: California Institute of Technology Division: Physics, Mathematics and Astronomy Major Option: Mathematics Thesis Availability: Public (worldwide access) Research Advisor(s): Gabai, David Thesis Committee: Gabai, David (chair) Bonahon, Francis Candel, Alberto Pandharipande, Rahul Defense Date: 22 May 2000 Record Number: CaltechTHESIS:11212019-175145827 Persistent URL: https://resolver.caltech.edu/CaltechTHESIS:11212019-175145827 DOI: 10.7907/4kag-zt09 Default Usage Policy: No commercial reproduction, distribution, display or performance rights in this work are provided. ID Code: 13595 Collection: CaltechTHESIS Deposited By: Mel Ray Deposited On: 22 Nov 2019 17:46 Last Modified: 16 Apr 2021 23:33 Thesis Files Preview PDF - Final Version See Usage Policy. 4MB Repository Staff Only: item control page

Immersed surfaces, Dehn surgery and essential laminations Thesis by Tao Li In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 2000 (Submitted May 22, 2000) Acknowledgements First and foremost, I would like to thank my advisor, Dave Gabai, who introduced this subject to me, for his constant support. His insight into this field has always guided my research, and I could never have accomplished this without it. I have had great benefit from our weekly meetings. Countless times, his knowledge and intuition have put my research on the right track. His support has allowed me to attend many conferences, during which I had the chance to discuss my research with many people and to expand my knowledge in other areas. I would also like to thank Francis Bonahon. who has always been available to me when I need his help, both mathematical and non-mathematical. I am very grateful to Yanglim Choi for a series of meetings about his thesis. I would also like to thank my former office mates Siddhartha Gadgil and Douglas Zare, from whom I have learned much mathematics. I also thank everyone in the math department of Caltech for providing me with such a nice atmosphere. Finally. I would like to thank my wife, Tianxin Chen, who helped me survive the darkest time of my graduate study. Abstract Let M be an orientable and irreducible 3-manifold whose boundary is an incompress­ ible torus. We are interested in immersed essential surfaces in closed 3-manifolds obtained from Dehn fillings on M. We show the following two things. In Chapter 2, we suppose that M does not contain closed non-peripheral in­ compressible surfaces. We show that the immersed surfaces in M with the 4-plane property can realize only finitely many slopes. Moreover, we show that only finitely many Dehn fillings on M can yield 3-manifolds that admit non-positive cubing. This gives the first examples of hyperbolic 3-manifolds that cannot admit non-positive cubing. In Chapter 3. we suppose M is hyperbolic. We show that all but finitely many Dehn fillings on M yield 3-manifolds that contain closed essential surfaces. More­ over, we give a bound on the number of exceptional Dehn fillings. iii Contents Acknowledgements i Abstract ii 1 Summary 1 2 Surfaces with the 4-plane property 4 2.1 Introductioo................................................................................................ 4 2. 2 Hatdier’s trick......................................................................................................... 6 2.3 Cross disks............................................................................................... 9 2.4 Liιmts of crosssdisks................................................................................ 13 2.5 Meaaured sutblaπunatioos....................................................................... 20 2.6 Bound∑ay curres................................................................................... . 25 3 Immersed surfaces in hyperbolic 3-manifolds 46 3 .1 IntΓoductίoo............................................................................................... 46 3.2 I-bundle région......................................................................................... 48 3.3 ^η^ηη^ΐοη oo the injective surfaaes..................................................... 55 Bibliography 62 1 Chapter 1 Summary A closed irreducible 3-manifold is called Haken if it contains a two-sided incompress­ ible surface. Waldhausen has proved topological rigi^^^y for Haken 3-manifolds [35], i.e., if two Haken 3-manifolds are homotopy equivalent, then they are homeomorphic. However, a theorem of Hatcher [19] implies that, in some sense, most 3-manifolds are not Haken. Immersed 7Γι-injective surfaces are a natural generalization of incom­ pressible surfaces, and, conjecturally, 3-manifolds that contain 7Tiinjective surfaces have the same topological and geometric properties as Haken 3-manifolds. Another related major conjecture in 3-manifold topology is that any 3-manifold with infinite fundamental group contains a πT-ιnjective surface. In this thesis, we investigate immersed essential surfaces in closed 3-manifolds. In particular, we are interested in closed 3-manifolds obtained from Dehn surgery. Dehn surgery on a 3-manifold is the operation that takes out a solid torus in the 3-manifold and glues it back using a different homeomorphism of the boundary torus. If we have a 3-manifold whose boundary is a torus, we can glue a solid torus to this 3-manifold along its boundary and get a closed 3-manifold. This operation is called Dehn filling. Dehn surgery is a useful way of constructing closed 3-manifolds. It has been known for a long time that any closed 3-manifold can be obtained from Dehn surgery on a link in S3. Let M be an irreducible 3-manifold whose boundary is an incompressible torus. Dehn filling on M has been used extensively to construct examples and counter­ examples of closed 3-manifolds with certain properties. Thurston found first exam­ ples of non-Haken 3-manifolds that are not small Seifert fiber spaces by doing Dehn surgery on the figure eight knot. Later, Hatcher showed that, if M does not contain closed non-peripheral incompressible surfaces, only finitely many Dehn filings on M yield Haken 3-manifolds. Along this line, Thurston has shown that, if M is hyper­ bolic, all but finitely many Dehn fillings on M yield closed hyperbolic 3-manifolds. This gives positive evidence for the hyperbolization conjecture. We denote the closed manifold after Dehn filling (along slope s) by M(s). Sup­ 2 pose M(s) contains an essential surface, then there are two cases. Either there is an injective surface in M whose boundary is a union of closed curves of slope s, or M contains a closed essential surface that remains essential after the surgery. In Chapter 2, we consider the boundary slopes of immersed surfaces with small complexity in M, namely surfaces with the 4-plane property (see Chapter 2 for the history of surfaces with the 4-plane property). The following theorem is a generalization of a theorem of Hatcher [19]. Theorem 1. Let M be an orientable and irreducible 3-manifold whose boundary is an incompressible torus, and let H be the set of injective surfaces that are embedded along their boundaries and satisfy the 4~plane property. Suppose that M does not contain non-peripheral closed incompressible surfaces. Then the surfaces in H can realize only finitely many slopes. As a corollary of the theorem above, we give the first examples of hyperbolic 3manifolds without non-positive cubing (see Chapter 2 for the history of non-positive cubing). Theorem 2. Let M be an orientable and irreducible 3-manifold whose boundary is an incompressible torus. Suppose that M does not contain closed non-peripheral incompressible surfaces. Then only finitely many Dehn fillings on M can yield 3- manifolds that admit non-positive cubing. In the proof of Theorem 1, we apply Hatcher's observation to immersed branched surfaces. The key part of the proof of Hatcher’s theorem is a result of Floyd and Oertel [12]. We also generalize this result to immersed surfaces with the 4-plane property. Theorem 3. Let M be a closed, irreducible and non-Haken 3-manifold. Then sur­ faces with the 4-plane property in M are carried by finitely many immersed branched surfaces. A similar result of Theorem 3 for surfaces with 3-plane and 1-line properties has been shown by Choi [5] using similar approaches. The proofs of the theorems above are much more technical than the case of embedded surfaces. In particular, 3 we used techniques of essential lamination that are not commonly used in the field of immersed surfaces. In Chapter 3, we assume that M is hyperbolic and we construct closed immersed surfaces by closing up embedded surfaces with boundary using long annuli that wind around dM many times. We show that these immersed surfaces remain πTimi<eitiee after most Dehn fillings. Theorem 4. Suppose A is a hyperbolic 3-manifold whose boundary is a single torus. Then all but finitely many Dehn fillings on M yield 3-manifolds that contain 7T[ -injective surfaces. Moreover, an explicit bound (depending on M) on the number of exceptional surgeries is also given in Chapter 3. Cooper and Long [8] have also proved Theorem 4 (earlier) using different methods, but no bound was given in [8]. 4 Chapter 2 Surfaces with the 4-plane property 2.1 Introduction Hass and Scott [18] have generalized Waldhausen’s theorem by proving topological rigidity for 3-manifolds that contain 7rι-inj'ective surfaces with the 4-plane and 1line properties. A surface in a 3-manifold is said to have n-plane property if its pre-image in the universal cover of the 3-manifold is a union of planes, and among any n planes, there is a disjoint pair. The n-plane property is a good way to measure combinatorially how complicated an immersed surface is. Incompressible surfaces are surfaces with the 2-plane property. It has been shown [33] that any immersed 7Γχ-inj^^1iive surface in a hyperbolic 3-manifold satisfies n-plane property for some n. In this paper, we will use immersed branched surfaces to study surfaces with the 4-plane property. Branched surfaces have been used effectively in the study of incompressible surfaces and laminations [12, 15]. Many results in 3-manifold topol­ ogy (e.g. Hatcher's theorem [19]) are based on the theory of branched surfaces. We define an immersed branched surface in a 3-manifold M to be a local embed­ ding to M from a branched surface that can be embedded in some 3-manifold (see definition 2.2.1). Using lamination techniques, we will show: Theorem 2.1. Let M be a closed, irreducible and non-Haken 3-manifold. Then surfaces with the 4-plane property in M are carried by finitely many immersed branched surfaces. This theorem generalizes a fundamental result of Floyd and Oertel [12] in the theory of embedded branched surfaces. One important application of the theorem of Floyd and Oertel is the proof of a theorem of Hatcher [19], which says that incompressible surfaces in an orientable and irreducible 3-manifold can realize only finitely many slopes. However, Hatcher's theorem is not true for immersed 7rjinjective surfaces in general, since there are many 3-manifolds [2, 31, 3, 27] that injective surfaces can realize infinitely many slopes, and in some cases, can realize 5 every slope. Using Theorem 2.1, we will show that surfaces with the 4-plπae property are, in a sense, like incompressible surfaces. Theorem 2.2. Let M be an orientable and irreducible 3-manifold whose boundary is an incompressible torus, and let H be the set of inj'ective surfaces that are embedded along their boundaries and satisfy the 4-plane property. Suppose that M does not contain non-peripheral closed incompressible surfaces. Then the surfaces in Ή can realize only finitely many slopes. Aitchison and Rubinstein have shown that if a 3-mαaifrld has a ara-prsitive cubing, then it contains a surface with the 4-plane and 1-line properties, and hence topological rigidity holds for such 3-manifolds. Non-positive cubing was introduced by Gromov [16] as an example of non-positive polyhedral metric. A 3-maaiaold is said to have a ara-prsitive cubing if it is obtained by gluing cubes together along their square faces under the following conditions: (1) For each edge, there are at least four cubes sharing this edge; (2) for each vertex, in its link sphere, any simple 1-cycle consisting of no more than three edges must consist of exactly three edges, and must bound a triangle. Mosher [29] has shown that if a 3-manifold has a ioi- positive cubing, then it satisfies the weak hyperbo^a^on conjecture, i.e., either it is negatively curved in the sense of Gromov or its fundamental group has a Z © Z subgroup. So, non-positively cubulated 3-manifoldk have very nice topological and geomet­ ric properties. A natural question, then, is how large the class of such 3-manifolds is. Aitchison and Rubinstein have constructed many examples of such 3-manifolds, and only trivial examples, such as manifolds with finite fundamental groups, were known without such cubing. In this paper, we will give the first non-trivial ex­ amples of 3-maniaoldk, in particular, frst examples of hyperbolic 3-maaifoldk, that cannot have non-positive cubing. In fact, Theorem 2.3 says that, in some sense, most 3-manifoldk do not have such cubing. Theorem 2.3. Let M be an orientable and irreducible 3-manifold whose boundary is an incompressible torus. Suppose that M does not contain closed non-peripheral incompressible surfaces. Then only finitely many Dehn fillings on M can yield 3manifolds that admit non-positive cubing. 6 (a) (b) Figure 2.1: 2.2 Hatcher’s trick A branched surface in a 3-manifold is a closed subset locally diffeomorphic to the model in Figure 2.1 (a). A branched surface is said to carry a surface (or lamination) S if. after homotopies, S lies in a fibered regular neighborhood of B (as shown in Figure 2.1 (b)), which we denote by N(B), and is transverse to the interval fibers of N{B). We say that S is fully carried by a branched surface B if it meets every interval fiber of N(B). A branched surface B is said to be incompressible if it satisfies the following conditions: (1) The horizontal boundary of N(B), which we denote by ∂hN(B), is incompressible in the complement of N∖B), and dhN(B) has no sphere component; (2) B does not contain a disk of contact: (3) there is no monogon (see [12] for details). Theorem 2.2.1 (Floyd-Oertel). Let M be a compact and irreducible 3-manifold with incompressible boundary. Then there are finitely many incompressible branched surfaces such that every incompressible and ∂-incompressible surface is fully carried by one of these branched surfaces. Moreover, any surface fully carried by an incom­ pressible surface is incompressible and d-incompressible. Using this theorem and a simple trick, Hatcher has shown [19] that given a com­ pact, irreducible and orientable 3-manifold M whose boundary is an incompressible torus, the boundary curves of incompressible and ^-incompressible surfaces in M can realize only finitely many slopes. An immediate consequence of Hatcher’s theo­ rem is that if M contains no closed non-peripheral incompressible surfaces, then all 7 but finitely many Dehn fillings on M yield irreducible and non-Haken 3-manifolds. To prove Hatcher’s theorem, we need the following lemma [19]. Lemma 2.2.1 (Hatcher). Let T be a torus and r be a train track in T that fully carries a union of disjoint and non-trivial simple closed curves. Suppose that r does not bound a monogon. Then r is transversely orientable. In Theorem 2.2.1, if ∂M is a torus, then by the definition of incompressible branched surfaces, the boundaries of those branched surfaces are train tracks that satisfy the hypotheses in Lemma 2.2.1. This lemma together with a trick of Hatcher prove the following. Theorem 2.2.2 (Hatcher). Let M be a compact, orientable and irreducible 3manifold whose boundary is an incompressible torus. Suppose that (B.∂B) C (M.∂M) is an incompressible branched surface. If Si and So are two embedded surfaces car­ ried by B, then ∂S[ and ∂So have the same slope. Moreover, the incompressible and ∂-incompressible surfaces can realize only finitely many slopes. Proof. Since M is orientable, the normal direction of ∂M and the transverse orien­ tation of ∂B uniquely determine an orientation for every curve carried by ∂B. Now every component of ∂S, (i = 1 or 2) with this induced orientation represents the same element in Hi(∂M). If ∂Si and ∂So have different slopes, they must have a non-zero intersection number. There are two possible con^^uτations for the induced orientations of ∂Si and ∂So at endpoints of an arc q of Si fl So, as shown in Fig­ ure 2.2. In either case, the two ends of a give points of ∂Si fl ∂So with opposite intersection numbers. Thus, the intersection number ∂Si ■ ∂So = 0. So, they must have the same slope. The last assertion of the theorem follows from the theorem of Floyd and Oertel. □ In order to apply the trick about intersection numbers, we do not need the surfaces Si and So to be embedded. In fact, if Si and So are immersed 7T-ιnjective surfaces that are embedded along their boundaries and transversely intersect the interval fibers of N(B), then dSi and ∂Si must have the same slope by the same argument. This is the starting point of this paper. Actually, even the branched surface B can be immersed. An obstruction to applying Hatcher’s trick is the 8 existence of a local picture as in Figure 2.3 in B. Note that a branched surface with a local picture as in Figure 2.3 can never be embedded in any 3-manifold. Next, we will give our definition of immersed branched surfaces so that we can apply Hatcher’s trick to immersed surfaces. Definition 2.2.1. Let B be a branched surface properly embedded in some com­ pact 3-manifold; i.e. the local picture of B in this manifold is as in Figure 2.1 (a). Let i : B —► MI (resp. i : M(B) —» M) be a map from B (resp. N(B)) to a 3manifold M. We call i(B) an immersed branched surface in M if the map i is a local embedding. An immersed surface j : S — M is said to be carried by i{B) if, after some homotopy in M, j = i o h, where h : S —* N(B) is an embedded surface that transversely intersects the interval fibers of N(B). The following proposition is an extension of Hatcher’s theorem, and its proof is simply an application of Hatcher’s trick to immersed branched surfaces. Proposition 2.2.2. Let M be a compact, orientable and irreducible 3-manifold whose boundary is an incompressible torus. Let S\ and So 6e immersed iti-injective 9 surfaces carried by an immersed branched surface B. Suppose that i∖gg is an em­ bedding and i(∂B) does not bound a monogon, where i : B —* M is the immersion. Then ∂Sχ and ∂So have the same slope. □ 2.3 Cross disks We have seen in section 2.2 that Hatcher’s trick can be applied to immersed branched surfaces. However, we also need finiteness on the number of branched surfaces, as in the theorem of Floyd and Oertel. to get interesting results. This is impossible in general because there are many examples of d-manifolds that immersed 7T-injective surfaces can realize infinitely many slopes. In this section, we will show that this can be done under certain assumptions. By the normal surface theory, it is very easy to get anitenekk (of the number of branched surfaces) in the case of embedded branched surfaces. For any triangulation of a 3-maaifrld, an incompressible surface can be put in Kneser-Haken normal form [26, 17]. The intersection of the surface with each tetrahedron is a union of normal disks in at most 7 normal disk types. By identifying all the normal disks of the same type to a branch sector, we can naturally construct a branched surface carrying this surface, and the fniteness follows from the compactness of the 3-maaifold (see [12] for details). However, in the case of immersed surfaces, we cannot do this, although immersed Xi-injective surfaces can also be put in normal form. Given two normal disks (of the same normal disk type) that intersect each other, if we simply put them in the same branch sector, we may get a picture like that in Figure 2.3 in some tetrahedron along the double curves of the immersed surface, which makes Hatcher’s argument fail. Suppose that S is a 7T-injective surface in a 3-mαaiaold M with a triangulation T. By the normal surface theory, we can assume that S is in normal form. Let M be the universal cover of M, jt t M — M be the covering map, S = 7r-1(S), and T be the induced triangulation of M. For any arc a in M (or M) whose interior does not intersect the one skeleton T^1∖ we define the length of π to be ∖int(a) fl T^|, where int(E) denotes the interior of E and |i?| denotes the number of connected 10 components of E. Moreover, we define the distance between points x and y, d(x,y), to be the minima! length of all such arcs connecting x to y. In this paper, we will always use the distance defined above unless specified. A normal (immersed) surface f : F → M is said to have least weight if ∣∕- 1(T1^)∣ is minimal in the homotopy class of ∕. Let g : F — MI be a TT-injective map and F' be the universal cover of F. Suppose that g : F' —* M is a lift of g o p : F' —» M to Ml, where p : F' —» F is the covering map. Then we call g(F') a component of F in MI, where F is the pre-image of g(F) in M. We say that a component of F has least weight if any disk in this component has least weight among all the disks in M with the same boundary. A normal homotopy is defined to be a smooth map ff : F x [0,1] > M so that for each t 6 [0,1], the surface Ft given by H∣F×{t} is a normal surface. Note that the weight of Ft is fixed in a normal homotopy. A Tι-injective immersed surface f : (F.dF) —* (M, dM) (F # S- or P~) is said to have n-plane property if every component of the pre-image of f(F) in M is embedded and in any collection of n different components, there is a disj'oint pair. From the FL-minimal surface theory [22], we know that for any τ--injective surface f : F — M, there is a normal surface fι : F — M of least weight in the homotopy class of / such that any component of the pre-image of fι(F) in M is embedded. Moreover, it follows from Theorem 5 of [22] or Theorem 3.4 of [14] that fχ can be chosen so that any component of the pre-image of f(F) in M has least weight. By Theorem 8 of [22] (or Theorem 6.3 of [14]), if / has n-plane property, so does ∕χ. In this paper, we will assume that our 3-manifolds are compact and irreducible, and our immersed surfaces, when restricted to the boundary, are embedded. We will also assume that our injective surfaces are normal and have least weight, and any component of their pre-images in the universal cover of the ^manifold has least weight. To simplify notation, we will not distinguish f : F — MI, F and f(F) unless necessary. Definition 2.3.1. Let f : F —► MI be an n^f^^^i^tive normal surface. Let Fχ and Fo be two components of the pre-image of f(F) in MI. Suppose that D\ and Di are two embedded sub-surfaces in F\ and F> respectively. We say that D\ and D-2 are parallel if there is a normal homotopy H : D x I — M such that H(D, 0) = D\, 11 H(D, 1) = Do and H fixes the 2-ske!eton, i.e., if H(x, y) E T^ then H(x, I) C (i — 1,2). We call D\ U Do a cross disk if D\ and Do are parallel disks, I\ ^ Io, and Fi n Fo 7^ 0. We call D, (i = 1,2) a component of the cross disk D\ U Do. Let H be the normal homotopy as above. We call H(p, 0) U H(p, 1) a pair of points (resp. arcs, disks) in the cross disk, for any point (resp. arc, disk) p in D. A cross disk DιU Do (or the disk D\) is said to have size at least R if there ex^^ts a point x ∈ Di such that length(a) > R for any normal arc q C D[ connecting x and ∂D∖ — ∂M, and we call the normal disk of T n D\ that contains x a center of the cross disk, where T is a tetrahedron in the triangulation. To simplify notation, we will also call it(Di u Do) a cross disk and call the image (under the map 7r) of a pair of points (resp. arcs, disks) in D\ U Do a pair of points (resp. arcs, disks) in the cross disk, where π : M — M is the covering map. We denote by F the set of TT-mjective and d-injective surfaces in M whose boundaries are embedded in ∂M. Let Fr = {F 6 I : there are no cross disks of size R in F}, where F is the pre-image of F in M. The following lemma is due to Choi [5]. Lemma 2.3.1. The surfaces in Fr are carried by finitely many immersed branched surfaces. Proof. Let T be a tetrahedron in the triangulation T of M and d, c F fl T be a normal disk (i = 12,3), where F ∈ Fr. Suppose that T is a lift of T in M, di is a lift of dt in T, and Fi is a component of F in M that contains d, (i = 1,2,3), where F is the pre-image of F in M. We call D^(di) = {ar 6 F;|d(x,p) < N, where p E di} a surface of radius A" with center ∂i. Note that, topologically, Dn-(<) may not be a disk under this discrete metric. Next, we will define an equivalence relation. We say that ∂i is equivalent to d< if DkR(∂i) is parallel to DiRdo) and Fi fl F2 = 0 (or Fi = Io), where k is fixed. We assume that k is so large that DkRld) contains a sub-disk of size R whose center is ∂i (i = 1,2). Note that, since M is compact and every component of F has least weight, k can be chosen to be independent from the choices of F E Fr and the normal disk ∂i c F, i.e., k depends only on R and the triangulation of M. Suppose that there are three normal disks ∂i, ∂i and d$ in Ffl T so that ∂i is 12 equivalent to do and do is equivalent to d$. Then DkR(di) is parallel to DkR(ds) by definition. If iq ^ F3 and F∖ n F3 ≠ 0, by the assumption on k, there is a cross disk of size R that consists of two disks from Fi and F3 respectively. This contradicts the hypothesis that F € Fr. Thus d∖ is equivalent to rf, and the equivalence relation is well-defined. Since M is compact, for any normal disk d in M, the number of non-parallel (embedded) normal surface of radius kR (with center d) is bounded by a constant C that depends only on R and the triangulation of M. As there are no cross disks of size R, there are at most C equivalence classes in the normal disks of F n T of the same type, and hence at most 1C equivalence classes in FflT (since there are 7 different types of normal disks). For any tetrahedron T, suppose there are Cp (Cτ < 1C) equivalence classes in F fl T. We put Cp products Di x I (i = 1,..., Ct) in T such that Di x {£} is a normal disk and the normal disks of FflT in the same equivalence class lie in the same product Di x I. Along Ti'2'l, we can glue these products Di x Fs together according to the equivalence classes, as in the construction of embedded branched surfaces in [12]. In fact, we can abstractly construct a branched surface B and a map f : N(B) — M such that, for any tetrahedron T. f(∂υN(B)) C T<2* and ∕(V(B) — p~i(L)) fl T is exactly the union of the products int(Di) x Fs in T, where L is the branch locus of B, p : N{B) — B is the map that collapses every interval fiber of N(B) to a point, and int(Di) denotes the interior of Di. By our construction, B does not contain a local picture like that in Figure 2.3, and hence it can be embedded in some 3-manifold [6]. Since the number of equivalence classes is bounded by a constant, there are only finitely many such immersed branched surfaces that carry surfaces in Fr. □ Corollary 2.3.2. Suppose M is a compact, orientable and irreducible 3-manifold whose boundary is an incompressible torus. Then the surfaces in Fr can realize only finitely many slopes. Proof. Suppose that Fι, Fo 6 Fr are carried by the same immersed branched surface f : B —> M. To simplify notation, we will also denote by f the correspondent map from N(B) to M. Since the surfaces in Fr are embedded along their boundaries, after some normal homotopy if necessary, we can assume that ∕∖qb is an embedding. 13 Since the surfaces in J-r are TT-mjective, the horizontal boundary of f(dB) does not contain any component that is a trivial circle. Because of Lemma 2.3.1 and Proposition 2.2.2, we only need to show that f{dB) does not bound a monogon. Since f∖aβ is an embedding, to simplify notation, we do not distinguish ∂B and f(∂B), and denote f(N(dB)) by N(∂B), where N{dB) is a fibered neighborhood of the train track dB. By our definition of immersed branched surface, we can assume that Fi C ∕(V(B)) and ∕~l(F[) is an embedded surface carried by N(B). Suppose that D C dM is a monogon, i.e., dD = a U 0, where a is a vertical arc of ∂vN(∂B) and 3 c 3⅛iV(3B). The vertical boundary component of f(dυN(B)) that contains α is a rectangle E whose boundary consists of two vertical arcs a. a' on dM and two arcs 7,7' in f(dυN{B) tndhN(B)). By our definition, ∕^1(Ft) is embedded in N(B). So, after some normal homotopy, we may assume that E is embedded, dtlN(∂B) C 3Ft, and 7 U 7' C Ft. Then £ = 3 U 7 U 7' is an embedded arc in Ft with dδ C dF∖ c dM, and 5 can be homotoped rel dδ into dM. Since Ft is 3-injective, S must be 3-parallel in Ft, i.e., there is an arc S C <3 Ft such that SU S bounds a disk A in Ft. Moreover, a' U S also bounds a disk D, in dM, since a' U S forms a homotopically trivial curve in M. Note that Δ may not be embedded in M. So, DuEuAuD' forms an immersed sphere in M. Since M is irreducible, i.e.. t(M} is trivial, we can homotope the sphere DuEuΔuD' (fixing E) into E. After this homotopy, we get an immersed surface in the same homotopy class as Ft with less weight. This contradicts our least weight assumption on the surface F{. So, dB does not bound any monogon. By Proposition 2.2.2, 3Ft and 3Fo must have the same boundary slope, and the corollary follows from Lemma 2.3.1. □ 2.4 Limits of cross disks Let H be the set of injective surfaces rnth the 4-plane property in M. If there is a K € R such that H 6 T^κ, by Corollary 2.3.2, the surfaces in H can realize only finitely many slopes. Suppose no such number K exists. Then there must be a sequence of surfaces Ft, F2,..., F„, - - - € H such that, in the pre-image of Fi in M (denoted by Ft), there is a cross disk Di = D'i U D" of size at least i, where i 6 N. 14 Since M is compact, after passing to a sub-sequence if necessary, we can assume that D[ is parallel to a sub-disk Aj of D'i+i such that d(∂Δt,-∂M, ∂Di+l-∂M) > 1. We also assume that ∂D'i lies in the 2-sk^^∣^^on. Now we consider the image of Di in M, i.e., π(D,), where ir : M → M is the covering map. Proposition 2.4.1. The intersection of ir(Di) with any tetrahedron does not con­ tain two quadrilateral normal disks of different types. Proof. We know that any two quadrilateral normal disks of different types must intersect each other. Suppose that the intersection of ir (Di) with a tetrahedron contains two different types of quadrilateral normal disks. Let T be a lift of this tetrahedron in M. Then, in each of the two quadrilateral disk types, there is a pair of normal disks in I ∩ T that belong to different components of a cross disk. So, by the definition of cross disk, the two components of F; that contain the two normal disks must intersect each other. The two different quadrilateral disk types give rise to 4 components of I intersecting each other. Note that, since each component of Ii is embedded, the 4 components above are different components of F,. This contradicts the 4-plane property. □ Thus, as in [12], we can construct an embedded branched surface Bi that carries ft(Di), i.e.. τr(Di) lies in N(B,) transversely intersecting every interval fiber of iV(Bi). In fact, for each normal disk type of π(Di)(~T, we construct a product δ×I, where T is a tetrahedron and 5 x {f} is a normal disk of this disk type (f 6 /). Then, we can glue these products along naturally to get a fibered neighborhood of a branched surface Bi, and τr(Di) can be isotoped into N{Bi) transversely intersecting every interval fiber of N(B) Note that Bi may have non-trivial boundary. After some isotopy, we can assume that ∂υN(Bi) fl = 0 and N(Bi) fl is a union of interval fibers of N(Bi). Note that by the definition of cross disk, we can assume that the image of every pair of points in the cross disk lies in the same /-fiber of N{Bi). Proposition 2.4.2. N(B) can be split into an I-bundle over a compact surface such that, after isotopies, every pair of points in the cross disk ir(Di) lies in the same I-fiber of this I-bun∂le. 15 Proof. By our construction above, N(BX n 7^2', when restricted to a 2-simplex in T^2K is a fibered neighborhood of a union of train tracks. Suppose that ∂υN(Bi) transversely intersects T^~K First, we split N(Bi) near N(B) n to eliminate ∂vN(Bi)nTW. Let A be a 2-simplex in 2∖ S be a component of ∂υN(Bi) HA and r be a component of N(Bi) n A that contains 5. We associate every component 5 of ∂υN(Bi) fl A with a direction (in A) that is orthogonal to 6 and points into the interior of N(βi) n A. Let V be the union of interval fibers of r that contain some components of ∂υN(Bi) n A. By some isotopies, we can assume that every interval fiber in V contains only one component of ∂vN(Bt)∩ A. We give every interval fiber in V a direction induced from the direction of ∂v.N(Bi) fl A. Now r — V is a union of rectangles with two horizontal edges from ∂hN(Bi) and two vertical edges from V. Every vertical edge of a rectangle has an induced direction. Case 1. For any rectangle of r — V, the direction of at most one vertical edge points inwards. In this case, there is no ambiguity about the splitting near the rectangle. We split r as shown in Figure 2.4, pushing a component of ∂υN(B) across an edge of A. During the splitting we also push some double curves of Fi across this edge. The effect of the splitting to TD) is just an isotopy. Thus, we can assume that any pair of points in the cross disk lies in the same interval fiber of the fibered. neighborhood of the branched surface after this splitting. Case 2. There is a rectangle in r — V such that the directions of both vertical edges 16 (b) (a) Figure 2.5: point inwards. The local picture of such a rectangle must be as in Figure 2.5 (a), and there are (locally) three different splittings as shown in Figure 2.5 (b). We denote the rectangle by R and the part of r as in Figure 2.5 (a) by tr. Then tr — R consists of 4 components, and we call them UL (upper left) end, LL (lower left) end, UR (upper right) end and LR (lower right) end, as shown in Figure 2.5 (a). The intersection of the image of the cross disk and tr, i.e., r(Di) fl tr, consists of arcs connecting the ends on the left side to the ends on the right side. An arc in n(Di) ∩ tr is called a diagonal arc if it connects an upper end to a lower end. Claim. i(Dl) n tr does not contain two diagonal arcs, say a and 3, such that a connects the UL end to the LR end, and /3 connects the LL end to the UR end. Proof of the claim. Suppose that it contains such arcs a and j3. Then there is an­ other arc a' (resp. 3t) such that alia' (resp. βUβ'} is a pair of arcs in the cross disk. So, a' (resp. β,) also connects the UL end to the LR end (resp. the LL end to the UR end). Note that a (or a') and 3 (or 3') must have non-trivial intersection in tr. Next we consider a lift of tr in M and still use the same notation. By the definition of cross disk, the 4 components of Fi that contain a, a', 3 and β' respectively must intersect each other in M. Since every component of Fi is embedded in M, each is a different component of F. This contradicts the assumption that Fi has the 4-plane property. □ 17 Now we split N(Bi) near tr as follows. If there are no diagonal arcs in ^∙(D,∙)∩7f⅞, we split N(B) in a small neighborhood of tr as the splitting (1) in Figure 2.5. If there are diagonal arcs, we split it as the splitting (2) or (3) in Figure 2.5 according to the type of the diagonal arcs. Note that by the claim, diagonal arcs of different types cannot appear in tr at the same time. As in case 1, we can assume that any pair of points of the cross disk lies in the same /-fiber after the splitting. To simplify the notation, we will also denote the branched surface after the splitting by Bi. Since Di is bounded, after finitely many such splittings, ∂υN(Bl) n Ti∙= 0. Now ∂υN(Bi) is contained in the interior of the 3-simplices, i.e., in a collection of disjoint open 3-balls. So. every component of ∂vN(Bi) bounds a disk of contact (or a half disk of contact near the boundary). After we cut N(Bi) along these (half) disks of contact, as in [12], ∂υN∖B) = 0 and N(Bi) becomes an /-bundle over a compact surface. As before, we can assume that, after isotopies if necessary, every pair of points in the cross disk lies in the same /-fiber. □ In the splittings above, we can preserve the intersection pattern of F,. For any arc 7 C F fl A. since every arc in I, n A is a normal arc in the triangle A, we can assume that if an arc (in Ii n A) does not intersect 7 before the splitting, it does not intersect 7 after the splitting. Moreover, since the intersection of F; with any tetrahedron is a union of normal disks, we can assume that cutting the (half) disks of contact as above does not destroy the 4-plane property. The effect of the splitting on Ii is just a homotopy pushing some double curves out of the cross disk. So, after the splitting, I still satisfies the 4-plane property and has least weight. Therefore, we can assume for each i. τr{Di) is carried by an /-bundle over a compact surface. We will still denote this /-bundle by N(Bi). After collapsing every /-fiber of N(B{) to a point, we get a piece of embedded normal surface, which we denote by Si, in M. Furthermore, D[ is parallel to a sub-surface of a component of Si, where S, is the pre-image of Si in M. There are only finitely many embedded normal surfaces (up to normal isotopy) in M that are images (under the covering map ir) of normal surfaces that are parallel to D[. So, after passing to a sub-sequence and doing some isotopies if necessary, we can assume that Si is a sub-surface of Sj+i. By our assumption ∂(∂Di-∂M, ∂Di+i— 18 dM) > 1, it is easy to see that the direct limit of the sequence {⅜} is a surface in M whose boundary lies in dM, and its closure is a lamination in M. We denote this lamination by A. Next we will show that A is an essential lamination. Before we proceed, we will prove a useful lemma. Lemma 2.4.3. Let Fo be an injective normal surface in a 3-manifold M and F be a component of the pre-image of Fq in M. Suppose that F has least weight and there are two disks D∖ and Do embedded in F and pamllel to each other. Suppose that there is another embedded disk D with dD = a U 3, where (3 = D D (F — D∖ U Do), 3 fl D∖ and 3 ∩ Do are two endpoints of a, and a is an arc lying in a 2-simplex. Then weight(D∖) < weight(D). Proof. Since F has least weight, we can assume that F is embedded in M. As D\. and Do are parallel, there is an embedded region D2 x [1,2] in M, where D2 x {£} is parallel to D↑, for any t 6 [1,2] and D2 x {t} = Di for i = 1,2. Moreover, by our hypothesis on π, we can assume that a = {p} x [1τ2], where p € ∂D2. We take a parallel copy of D, say D, which is close to D. Let D' = a! n 3' and a' = {p'} x [1,2], where p' ∈ dD'2. Then dD2 — pUp' consists of two arcs 7 and q. By choosing D' to be close to D, we can assume that q is the shorter one. The four arcs 3, 3' and q x {1,2} form a circle that bounds a disk S in F. We can assume that D' is so close to D that the weight of 5 is zero. Di U Do U 6 is a disk in F whose boundary is ldU,J'U (7 x {1,2}). The circle ∕3∣u∕3'u(7 x {1,2}) also bounds another disk D U D' U (7 x [1,2]) in M. Since F has least weight. weight(Di U Do U 6) = 2weight{D 1) < weight(D U D' U 7 x [1,2]) = 2weight(D) 4- weight- x [1,2]). By our assumption, weight— x [1,2J) = 0. Thus, weight(Di) < weight(D). □ We call the disk D (in the lemma above) a monogon. Lemma 2.4.4. The lamination A is an essential lamination. Proof. First we will show that every leaf of A is 7ι--njective. Otherwise, there is a compressing disk D embedded in M — A and dD lies in π leaf 1 where A is the pre­ image of A in the universal cover M. By our construction of A, there is a cross disk Dκ = D'κ U D'K of size at least K that is parallel to a sub-surface of L Since F% is 19 Figure 2.6: TΓ-injective and has least weight, and since dD is an essential curve in I, if K is large, D'k does not contain a closed curve that is parallel to dD. By choosing K sufficiently large, we may assume that D,χ winds around dD (in a small neighborhood of D) many times, as shown in Figure 2.6 (a). Let N(D) be an enlargement of D (i.e. an embedded disk in M that contains D in its interior), and F be the component of Fr that contains D'κ. Since F is embedded in M, the component of Fn N(D) that contains the spiral arc in Figure 2.6 (a) must form a monogon with a long ‘tail’ that consists of two parallel spiral arcs, winding around dD many times, as shown in Figure 2.6 (b). The weight of the monogon is at most weight(D). If K is large enough, the length of each arc in the 'tail’ of the monogon is very large and, in a neighborhood of the 'tail’, we can choose two pieces of normal surfaces that are parallel to each other and have weight greater than weight(D). This contradicts Lemma 2.4.3. Next, we will show that every leaf of A is <9-injective. Otherwise, there is a dcompressing disk Df whose boundary consists of two arcs a and /3, where a C dM and d is an essential arc in I. By our construction of λ, there is a cross disk Dn = Dtn U D'n of size at least n such that there are arcs an C dM and (3n C ir(D,n) (dan = d3n} that are parallel and close to a and respectively. The two arcs q„ and 3 bound a disk dn that is parallel and close to Dl. Since the surface Fn is 9-injective, there is an arc jn C dFn such that jn U dn bounds an immersed disk 20 An in In. Since (3 is an essential arc in I, by choosing n sufficiently large, we can assume weight (An) > weight(D') = weight{dn). Note that 7n U an bounds an immersed disk δn in ∂M and that 4l U An U ^l is an immersed 2-sphere in M. Since πo(MI) is trivial, we can homotope At, U δn to dn fixing dn and get another immersed surface F' that is homotopic to In. Moreover, weight(If)—weight(In) = weight(dn) — weight(An) < 0, which contradicts the assumption that In has least weight. It is easy to see from our construction that no leaf is a sphere. Also, if A is not end-incompressible, there must be a monogon with a long ‘tail’, which contradicts Lemma 2.4.3 by the same argument as above. Therefore, Ais an essential lamination. □ 2.5 Measured sub-laminations In this section, we will show that any minimal sub-lamination of A has a transverse measure. A minimal lamination is a lamination that does not contain any proper sub-lamination. Using this result, we will prove Theorem 2.1 that can be viewed as a generalization of a theorem of Floyd and Oertel [12]. Let μbe a lamination in M and i : I×l — M be an immersion that is transverse to p, where I = [0,1]. We will call {p} x / a vertical arc, for any p E I, and call i(Ix I) a transverse rectangle if i(I x {0,1}) C p and the singular set of i is a collection of sub-arcs of the vertical arcs. To simplify notation, we will not distinguish I x I and its image in M. Lemma 2.5.1. Let p be a minimal lamination. If p has non-trivial holonomy, then a there is a transverse rectangle R : I x I — M such that Fi({1} x I) C F({0} x I), O where I = (0,1). Proof. Since p has non-trivial holonomy, there must be a map g : S1 x I → MI, which is transverse to p, such that g(S1 x {0}) C L C p (L is a leaf) and g~1(p) consists of a collection of spirals and one circle Sl x {0} that is the limiting circle of these spirals. Moreover, for any spiral leaf I of g~i(p), there is an embedding i : [0,oo) x I → Sl x I such that i~l(l) = [0, oo) x {1∕2} (see the shaded region in 21 (a) (b) Figure 2.7: Figure 2.7 (a)). Since S1 x {0} is the limit circle of ∕, for any arc {p} x [0, ej C 5l x I, there exists a number N, such that i{{N} x ∕) C {p} x (0, e). Since p is a minimal lamination, every leaf is dense in p. Thus, there is a O path p : I — L such that p( 0) = g(p, 0), where p 6 S1 and p(l) eg o i({0} x 7). Moreover, if e is small enough, there is a transverse rectangle r : 7 x I —► M such that r∣i×{0} = P∙ r({0} x ∕) = g({p} x [0,e]), and r({l> x I) = go i({0} x [<l, Jol), where [di, d-] C I. The concatenation of the transverse rectangle r and poi([0. N] x [di, do]), i.e., R : 7x7 — M where P([0,1/2] x 7) = r(IxI) and 72([1∕2,1] xI) = poi^,x [dl, d?]), is a transverse rectangle that we want. □ Remarks. 1. The kind of construction in Lemma 2.5.1 was also used in [20]. 2. After connecting two copies of such transverse rectangles if necessary, we can O assume that Λ({1} x I) c P({0} x I) in Lemma 2.5.1 preserves the orientation of the 7-fibers, In other words, we may assume that there is a map ∕ : A —* M transverse to μ, where A = S1 x 7, and an embedding (except for the boundary) h : I x I — A, as shown in Figure 2.7 (b), such that R = f o h and f(A) lies in a small neighborhood of R(I x I). 3. Let ∕, h, and R be the maps above. Suppose that Lq and L∖ are leaves in p containing R(I x {0}) and R(I x {1}) respectively. Then f~l(Lo U L↑) contains two spirals of different directions whose limiting circles are meridian circles of A (see 22 Figure 2.7 (b)). Note that Lq and L i may be the same leaf and the two spirals may have the same limiting circle. 4. If μ is casriad by ayrancand seufaue B, we can ianume that h({p} p ∕) ⅛ a sub-arc of an interval fiber of N(B). Lemma 2.5.2. Let A be the lamination constructed in section 2.4 and μ be any minimal sub-lamination of A. Then p has trivial holonomy. Proof. Suppose that p has ara-trieial holonomy. Since p is a minimal lamination, by the remarks above, there is an annulus g : A = Sl x I — M such that g~i(p) contains two spiral leaves, one clockwise and one counterclockwise, as shown in Figure 2.7 (b). From our construction of A, there is a cross disk D∖ = D'v U D'N such that <7-l(7r(D'v)) (resp. g~l(ιr(D'χ))) contains two arcs parallel and close to the two spirals respectively. We denote these two arcs by aθ and a'l (resp. aθ and a"), as shown in Figure 2.8 (a). Now we consider g~i(F1v). Since Fv is compact, <7-l(Fv) is compact. Denote the component of ≤r-l(Fv) that contains π' (resp. a") by di (resp. di'), where i = 0,1. Since Ftv is a normal surface, by Remark 4 above, we can assume that g-l(Fv) is transverse to each vertical arc {p} x I in A. If ∂y n 5l x {0} = 0, then C1 is either a closed curve, as shown in Figure 2.8 (c), or an arc with both endpoints on Sl x {1}, as shown in Figure 2.8 (b). Since A is an essential lamination, g(Sl x {0}) must be an essential curve in M, and we have the following commutative diagram, where q is a covering map. Rx I ,l -½→ M 4 A = Sl×I ~g→ M The pictures of q~l(c'l) c g~1(Fv) are shown in Figure 2.9 (a) or (b) depending on whether cz1 is an arc with both endpoints on 51 x {1} or a closed curve. If N is so large that a'1 winds around A more than four times, there are four components of q-l(dl) intersecting each other, as shown in Figure 2.9 (a) and (b), which contradicts the akkumptira that Fv has the ^plane property. Thus, by the argument above, dl , d[, ∂q and cθ must be arcs with endpoints in different components of dA, as shown in Figure 2.8 (d). In this case, 7-1(cθ Uc^U 23 Figure 2.8: 24 (a) (b) (c) Figure 2.9: 25 CQ U d[) must contain 4 arcs dQ, CQ,dl, d" as shown in Figure 2.9 (c), where 5(d- Ud") is the union of two arcs in different components of a cross disk (i = 0,1). By the definition of cross disk, the 4 components of ivy that contain g(dQ), g(^o)i g{d[) and g(d'i) respectively must intersect each other, which contradicts the assumption that Fjy has the 4-plane property. □ The next theorem is a generalization of a theorem of Floyd and Oertel [12]. Theorem 2.1. Let M be a closed, irreducible and non-Haken 3-manifold. Then the surfaces with the 4-plane property in M are carried by finitely many immersed branched surfaces Proof. If the set of immersed surfaces with the 4-plane property is a subset of Tr for some number R (see section 2.3 for the definition of Tr), then by Lemma 2.3.1 the surfaces are carried by finitely many immersed branched surfaces. If there is no such number R, by section 2.4, there are a sequence of cross disks that give rise to an essential lamination A. Let p be a minimal sub-lamination of A. Since p is also an essential lamination, by [15], we can assume that p is carried by an embedded incompressible branched surface B. By Lemma 2.5.2, p has no holonomy. A theorem of Candel [4] says that if a lamination has no holonomy then it has a transverse measure. So, p has a transverse measure, and hence the system of branch equations of B (see [32]) has positive solutions. Since each branch equation is a linear homogeneous equation with integer coefficients, the system of branch equations of B must have positive integer solutions. Every positive integer solution corresponds to an embedded surface fully carried by B. But, by a theorem of Floyd and Oertel [12], any surface fully carried by an incompressible branched surface must be incompressible. This contradicts the hypothesis that M is non-Haken. 2.6 □ Boundary curves Let M be an irreducible 3-manifold whose boundary is an incompressible torus, A be the lamination constructed in section 2.4 and p be a minimal sub-lamination of A. Let {Di = Dri U Df be the sequence of cross disks used in the construction of 26 the lamination A in section 2.4 and let F; be the immersed surface that contains π(Di). We denote the pre-image of F in M by Ii. Suppose that M does not contain non-peripheral closed incompressible surfaces. Lemma 2.6.1. /z fl ∂M ≠ 0 Proof. Suppose that μ n ∂M = 0. Then μ is fully carried by an incompressible branched surface B and B(~∂M = 0. As in the proof of Theorem 2.1 (see section 2.5), the linear system of branch equations must have integer solutions that correspond to incompressible surfaces. Since B n ∂M = 0 and M does not contain non-peripheral closed incompressible surfaces, those incompressible surfaces correspondent to the integer solutions must be 5-parallel tori. Let N(B) be a fibered neighborhood of B. C be the component of M-N(B) that contains ∂M. and Tι,F>,... T be a collection of 5-parallel tori that correspond to a positive integer solution of the system of branch equations. After isotopies, we can assume that every F, is transverse to the interval fibers of N{B) and ∂hN(B) C u-L17). Let A be a component of ∂⅛A(β) that lies in the closure of C. Claim. The surface .4 must be a torus. Proof of the claim. We first show that .4 is not a disk. Suppose A is a disk. Let u be the component of ∂υN(B) that contains ∂A. Then ∂v — ∂A is a circle in the boundary of a component D of ∂^N(B). Since ∂^N(B) is incompressible and A is a disk, D must be a disk. So Aut^D is a 2-sphere. Since M is irreducible, AUt'UD must bound a 3-ball that contains U∙=1Γ,, which contradicts the assumption that Ti is incompressible. If ∂A — 0, since ∂⅛iV(,B) C A must be a torus. Suppose ∂A ≠ 0 and A C Ti. If there is a component of ∂A that is a trivial circle in Ti then , since A is not a disk, there must be a trivial circle in ∂A that bounds a disk in Ti — A. We can isotope this disk by fixing its boundary and pushing its interior into the interior of N(B) so that it is still transverse to the /-fibers of N{B). This gives us a disk of contact [12], which contradicts the assumption that B is an incompressible branched surface. So, every circle of ∂A must be an essential curve in Ti, and hence A must be an annulus. 27 Let c be a component of dA, i/ be a component of dvN(B) that contain c, and d = du' — c be the other boundary component of id Denote the component of dhN(B) coatainiag d by A,. By the argument above, A! must also be an annulus. If A and A' belong to different tori, then id is a vertical annulus in the product region T2 x I bounded by the two tori. This contradicts the assumptions that those tori are 5^π^1,1 and dM C C. Thus, A and Al must belong to the same torus 7χ. Then, id must be an annulus in the T2 * I region bounded by Τχ and dM, and did c T∖. So, the vertical arcs of d can be homotoped rel did into T∖. This contradicts the assumption that B is an incompressible branched surface [12]. Therefore, dA = 0 and A must be a torus. □ By the claim and our assumptions, C must be a product region T~ x I where Γ2 x {1} = dM and T2 x {0} = A C dhN(B). Since μ is fully carried by B, we can assume that A C y is a leaf. After choosing a sub cross disk if necessary, we can assume that there is a cross disk Dκ = D'κ U D'^ of size at least K such that κ(D,κ) lies in a small neighborhood of .4 that we denote by T"2 x J, where J = [—e, e] and .4 = T2 x {0}. By choosing e small enough, we can assume T~ x {t} is normal for any t € J. Let E be the component of Fκ n (T~ x J) that contains τ(D'κ) and E, be a component of the pre-image of E in M. Let F, be the component of Fκ that contain E,. So E, is embedded in a region R2 x J in M, dE' C R2 x {±e}. By choosing e small enough, we can assume that E' is transverse to the /-ffibers of R2 x J. If E' is a compact disk, then dE, must be a circle in R2 x {±e} and Dκ must be in the region bounded by dE' x J. So, if AT is large, the disk in R2 x {±e} bounded by dE' is large. However, if the disk bounded by dE, is large enough, gk(dE') (k = 0,1,2,3) must intersect each other, where g is an element in τ (dM) that acts on M and fixes R2 x J. This violates the 4-plane property, and hence E' cannot be a compact disk. Suppose that Fκ n (R2 x {±e}) contain circular components. Let e be an innermost such circle and Fe be the component of Fκ that contain e. Then e bounds a disk D in R2 x {±e} and bounds another disk D' in Fe. We can assume that D' fl τr~l(T2 x {±e}) = dD,^. otherwise, we can choose e to be a circle in 28 D'∩7∙-l(T,2 x {±e}) that is innermost in D'. So, DuD' bounds a 3-ball in M and π(D' — dD') fl (T2 x J) = 0. Then, we can homotope π(D') to 7r(D) fixing 7∙(e). We denote the surface after this homotopy by FR and denote by F½ the component of FR (the pre-image of FR in M) that contains e. Let e' be a component of 7r~l(7r(e)) and Fe> (resp. F',) be the component of Fr (resp. FR) that contains e'. Since D is innermost, if Fe ∩ Fe> = 0, then F^ ∩ F'e, — 0. Hence, FR is a surface homotopic to Fr and FR also has the 4-plane property. Note that since Fr has least weight and ∕z is the ‘limit;’ of least weight cross disks, both D and D' have least weight and weight(D) = weight(D'). Thus, FR also has least weight and FR D T~ x {±e} has fewer trivial circles after a small homotopy. So, we can assume that Fr∩R2 x {±e} contains no trivial circles. Note that since E, can never be a compact disk as above, the homotopy above will not push the entire E' out of R2 x J. Therefore, we can assume that E' is a non-compact and simply connected surface. If dF,'nR2 x {e} has more than one component, then since we have assumed that E' is transverse to the ../-fibers of R2 x J, dE' D R2 x {e} bounds a (non-compact) region Q in R2 x {e} and D'κ C Q x J· Moreover, it is easy to see that, for any element g 6 ni{dM) that acts on M fixing R2 x J, if Q £ g(Q) and Q∩g(Q) £ 0 in R2 x {e}, then E'∏g(E') £ 0. If FT is large, the distance between any two lines in dQ must be large. Thus, by assuming D'κ to be large, we can find a non-trivial element g in πι(dMI) such that gk(Q), and hence the 4 components gk(Er) (R = 0,1,2,3) intersect each other, which contradicts the 4-plane property. Therefore, 9F'flR2 x {e} must have only one component that is a line, and hence E must be an immersed annulus inT~×J with one boundary component in T2 x {e} and the other boundary component in T2 x {—e}. By our construction, weight(E) is large if AT is large. We can always find an immersed annulus Ae C T2 x J with ∂Ar = dE and weight(AE) relatively small. So, the surface (Fr — E) U Ae is homotopic to Fr and has less weight. This contradicts the assumption that Fr has least weight. So, fi fl dM cannot be empty. □ Lemma 2.6.2. μ∖am ^ a lamination by circles. Proof. Suppose μ is fully carried by an incompressible branched surface B. Since fi is a measured lamination and dM is a torus, fi∖dM is either a lamination by circles or 29 a lamination by lines with irrational slope. Let S be the solution space of the system of branch equations. Since the coefficients of branched equations are integers, there are finitely many integer solutions that generate S, i.e., any point in S can be written as a linear combination of these integer solutions. Every integer solution gives rise to an incompressible surface carried by B. By Hatcher’s theorem, these surfaces have the same slope. The boundary slope of any measured lamination μ carried by B is equal to the measure of a longitude of ∂M divided by the measure of a meridian. Hence, it can be expressed as a fraction with both nominator and denominator linear functions of the weights of the branch sectors. Since the solution in S that corresponds to y is a linear combination of those integer solutions, and since the slopes of those integer solutions (plugging into the fraction described above) are the same, ∂y must have the same slope as the slope of the compact surfaces carried by B. Therefore, any measured lamination carried by B, restricted to ∂M, is a lamination by circles with the same slope. □ Lemma 2.6.3. Let {Dt = D[ U D'} be the sequence of cross disks used in the construction of A, Ii be the immersed surface with the 4-plane property that contains r^(Di), and y be a minimal sub-lamination of A. Then there exists a number N such that ∂Ii and ∂y have the same slope if i > N. Proof. Let B be an incompressible branched surface that fully carries y. Since ∂y is a union of parallel circles, we can assume that ∂B is a union of circles. Let N(B) be a fibered neighborhood of B, B = π~l(B) and N(B) = π~l(N(B)). We denote D[ ∩N(B) by Ei. Note that, since y is a sub-lamination of A, Ei is only a sub-disk of D[. Nevertheless, by our construction of A, we can assume that the size of Ei is large if i is large. By the modification of the construction of A above, we can assume that, after isotopies, ir(Ei) n ∂M C π{Ei+i) H ∂M. Suppose that ∂Ik has a different slope from ∂y. Then π(Ek) is a piece of immersed surface in N(B) transverse to every /-fiber, and n{Ek) n ∂M is a union of spirals in N(B) n ∂M. We give each component of ∂B an orientation so that they represent the same element in Hi(∂M). This orientation of ∂B determines an orientation for each /-fiber of N(B) fl ∂M. As in the proof of Hatcher’s theorem, the orientation of the /-fibers and a normal direction of ∂M uniquely determine an 30 (b) (a) Figure 2.10: orientation for every curve in N(B) fl dM that is transverse to the interval fibers. Claim I. Each circle in dFk admits an orientation that agrees with the induced orientation of dFk (~)N(B). Proof of claim 1. Suppose there is a circle in dFk that does not admit such an orientation. Then there must be a sub-arc C (of the circle) outside N(B) fl dM connecting two spirals that are either in the same component of N(B) fl dM, as shown in Figure 2.10 (a), or in diferent components of N(B)ndM with incompatible induced orientations, as shown in Figure 2.10 (b). After assuming the size of the cross disk to be large, we can rule out the first possibility, i.e., Figure 2.10 (a), by Lemma 2.4.3. To eliminate the second possibility, i.e., Figure 2.10 (b), we use a certain triangulation of M as follows. By [23], there is a one-vertex triangulation of M and this vertex must be on dM. Since dM = T2, the induced triangulation of dM must consist of two triangles as shown in Figure 2.11 (a). Now we glue a product region T2 x I (I = [0,1]) to M such that T2 x {0} = dM. Then (T^ D dM) x I gives a dellulatira of T2 x I that consists of a pair of triangular prisms. Then we add a diagonal to each rectangular face of the prisms, which gives a triangulation of T x I. Figure 2.11 (b) 31 Figure 2.11: is a picture of the induced triangulation of a fundamental region in the universal cover of T2 x /. In this triangulation, there is only one vertex on T~ x {1}, and its link hemisphere H intersects the 1-skeletorι in 10 points of which 6 points lie on ∂H. Since M U (T2 x I) is homeomorphic to Λ∕, we can assume that M has a triangulation as that of M U (T2 x I) above. We denote this triangulation also by T. Note that fl ∂M is a single vertex u and that the intersection of its link hemisphere H and T(1 consists of 10 points of which 6 points lie on ∂H C ∂M. We assume that our immersed surfaces are normal and have least weight with respect to the triangulation above. Suppose the second case, i.e.. Figure 2.10 (b), occurs. Let A be the annulus of ∂M — N{B) that contains the arc C. Then we isotope Ik by pushing C along > to 'unwrap' the spirals in a small neighborhood of ∂M, as shown in Figure 2.12. If the vertex v is not in A, then after this isotopy, ∣∂Ik n F(1)| deceeases <md ∖(F. — ∂F)t ∩T^∣∣ not . This co∏raadicts the assumption that Ik has least weight. So v € A. If every edge of fl ∂M intersects ∂A non-trivially, then after C passes through the vertex v during the isotopy, ∖∂Ik fl T^∣ decreases by 6 and ∣(Ik — ∂Ik) n F<1∣ increases by 4. Hence, the total weight of Ik decreases, which also gives a contradiction. Therefore, there is an edge e of fl ∂M lying inside A, as shown in Figure 2.12 (a). Then by our construction of the triangulation, e forms a meridian circle of the annulus A and there is at most one such edge. After C passes through v in the isotopy above, ∣∂Ik n T(11 decreases by 4, ∣(Ik — ∂Ik) n T^j increases by 4, and the total weight does not change. Now, we wifi see exactly what happens in a tetrahedron. Let T be a tetrahedron 32 (a) Figure 2.12: 33 with a face A on dM. There is a normal arc S in C fl A that cuts off a sub-triangle (in A n A) that contains the vertex v. The normal disk of Fr fl T containing 5 is either a triangle or a quadrilateral. If we do the isotopy above by pushing C across v, then the effect of this isotopy on the normal disk that contains S is either as in Figure 2.13 (a), in which case the normal disk is a triangle, or as in Figure 2.13 (b), in which case the normal disk is a quadrilateral. In the first case, as shown in Figure 2.13 (a), the disk is no longer a normal disk after the isotopy. So. we can perform another homotopy to make Fr (after the first isotopy) a normal surface. This homotopy reduces IjFfc7T^| by at least 2 as we push the disk in Figure 2.13 (a) across the edge, which contradicts the assumption that Fr has least weight. Thus, every normal disk that contains such an arc (as S) is a quadrilateral. Since there are only two triangles on dM, and since the edge e lies inside A, there must be two arcs <ι and S in C that cut off two corners of the same triangle (in the induced triangulation of dM). By the argument above, the two normal disks that contain <1 and So (respectively) must be two quadrilaterals of different normal disk types in the same tetrahedron. Note that, during the isotopy, we push parts of ∂Fr from N(B) n dM to the annulus .4. Suppose that the weight of Fr is fixed during these isotopies. After some isotopies as above, we can assume that there is a pair of normal disks of the cross disk in each of the two quadrilateral normal disk types. Since any two quadrilateral normal disks of different types must intersect each other, those 4 quadrilaterals give rise to 4 components of Fr intersecting each other, which contradicts the hypothesis that Fr has the 4-plane property. So, if A is large enough, we can reduce the weight of Fr at a certain stage of the isotopy above. Therefore, Figure 2.10 (b) cannot occur and claim 1 holds. □ Case 1. ft is a compact orientable surface. Let T = fi x [ c M (I = [0,1]), and P be a component of the pre-image of T in M. Suppose k is large: by our construction of the lamination, there is always a large sub cross disk of Dr = DR U DR lying in P. To simplify notation, we assume that Dr C P; otherwise we use a large sub cross disk of Dr and the proof is the same. Let FR be the component of Fr that contains DR, Hr = FR flP, H = π∙((Hr, and 34 Figure 2.13: 35 Ho be the smallest normal sub-surface of Fk that contains H. Since we can give every component of dFk an orientation that agrees with the induced orientation of ∂Fk n Γ, the sign of every intersection point of dFk n dS is always the same, where S = fx x {£} (t £ I). Then, H cannot be transverse to every /-fiber of T, because otherwise, by the same argument as in the proof of Hatcher’s theorem, dFk and dS would have the same slope, which contradicts our assumptions. Figure 2.14 gives a local picture of H where it is not transverse to an /-fiber of T. In fact, it is not hard to see that, in some tetrahedron T, there must be two quadrilateral normal disks of diferent types in T fl S and T n Ho respectively. Otherwise, by an argument in [12], Ho and 5 lie in N{Bτ) and are transverse to the /-fibers of N(Bτ), where N(Bτ) is a fibered neighborhood of an embedded normal branched surface Bj-. Hence, by the argument in the proof of Hatcher’s theorem, Fk and S have the same boundary slope (although Fk is not embedded), which contradicts our assumption. Moreover, since all these surfaces are normal, after a small homotopy, we can assume that each /-fiber of Γ either transversely intersects H or lies in H, in which case the local picture of this fiber is as shown in Figure 2.14. We call such fibers puncturing fibers. Any arc of Fk n S must pass through a puncturing fiber. Let g be the genus of μ and b be the number of boundary components. Then there are g disjoint annuli A^. A2,...,Ag such that fi — uf^A- is a planar surface. Let Ag+i,...,Ag+6 be the collection of annuli that are regular neighborhoods of the boundary components of μ. Suppose Aid Aj = 0 if i ^ j^. Let cj...., cκ be a maximal collection of disjoint non-parallel and non 3-parailel arcs in y — ufi^Ai, and let E = (uf^iAi) U (Uj=lN(cj)}, where N(cj) is a small neighborhood of ¾. 36 For each component a* of <3Aj, a component of Ft n (ai x I) is either a closed curve, or an arc with endpoints on different boundary components of a* x ∕, or a <3-parallel arc, i.e., an arc with both endpoints on the same boundary component of a; x I. By pushing the 3-parallel arcs out of a* x I, we can assume that Fr ∩ (a* x I) does not contain <3-parallel arcs. Similarly, after some homotopy, we can assume that Fr n (At x dl) has no <3-parallel arc for any i. Note that this homotopy can be done without destroying the 4-plane property. Moreover, since such a 3-parallel arc cannot wind around the annulus more than 4 times as in Figure 2.8 (b), the length of every 3-parallel arc is short compared with k, and hence the parts of Fr that we push out of S x ∕ cannot be parts of )(D^r^^^) (if k is large). Using the homotopy in the proof of Lemma 2.6.1, we can assume that Fr ∩ (4; x dl) contains no trivial closed curves. Since both y and Fr are normal surfaces and every component of Fr is embedded in M, after some homotopies as above, we can assume that each component of Ft n (4, x I) that intersects both components of (3A,) x ∕ is either an annulus or a quadrilateral with two edges on (34;) x I and two edges on 4; x dl. Moreover, we can assume that every component of Fr n (.4, x I) that intersects only one component of (34,) x I is an annulus, since we have assumed that there are no 3-parallel arcs in (34;) x I. If k is large, since the slopes of ∂Fr and dμ are different, iτ(D'',^^^rdMI contains a long spiral, where F∣f∕0 is a sub-disk of DR of size [A∕2]. Suppose (7^(Z)∣f(2∣) fl dM)f)(y x {£}) £ 0 for some t € (0,1). By claim 1 and the proof of Hatcher’s theorem, any double arc of F^fl (y x {£}) must pass through a puncturing fiber. Let 7 be such a double arc with at least one endpoint in (T(DμR^,,j)∩∂MF f)(yx {£}). By enlarging the annuli 4;’s and isotoping Fr (or y x {£}) if necessary, we can assume that Fr n (y x {£}) c E x ∕ and each component of Ft fl (N(cj) x F is a small neighborhood of Ft n (cj x I) (though two components of Ft n (N(cj ) x I) may intersect each other). Let 7' be an arc in Fr ∩ (y x dl). Then every component of Ft n (4j x I) that contains a sub-arc of 7' cannot be an annulus. Therefore, after some isotopy, we can assume our 7 as above also has this property, i.e., every component of Ft ∩ (4; x I) that contains a sub-arc of 7 is not an annulus, for any i. Let ui be a simple closed curve in y and a be an arc in cj x I with its endpoints on different boundary components of cj x I. We call a a puncturing arc if there is a 37 point p ∈ u such that a (up to homotopy fixing ∂a) does not intersect {p} x I. Let D be a component of Ik n (A, x I) for some i. We say that D is a level 0 surface in Ai x I if there exists an essential simple closed curve u G Ai such that D fl (ω x I) contains a puncturing arc. Note that the curve w can be chosen to have length no larger than the diameter of a fundamental region in the universal cover of Ai. Notation. For any Ai, we denote by Ai a component of τr- 1(A1) and by gl a generator of trι(A,). Suppose < acts on M fixing Ai, where i = 1,... ,g+b. For any component D of Ik n (Ai x /). we denote by D a component of 7t~1(∩) n (Ai x I) and by Id the component of Ik that contains D. Claim 2. If D is a level 0 surface in Ai x I, then Fq ^gi(Io>) = 0. Proof of claim 2. Suppose Iq ∩ gi(Io) ^ 0. Since D has level 0, there exists a simple closed curve u in Ai such that D n (ω x /) contains a puncturing arc a. Moreover. n(Dk) ∩ (u x I) contains a pair of curves that are either closed essential curves in u x [ or spirals winding around ω x I many times (if A is large). In both cases, q intersects these two curves non-trivially. Therefore, there is a cross disk Dk = D'k U D'! such that Dk fl D and D£ n D are not empty. Furthermore, since D and gi(D) are not far away, if k is large, D,k ∩gi{D) and D£ ∩ gi{D) are not empty. Thus, if Io∩9iiFD) ^ 0, we have 4 components of Ik intersecting each other, which contradicts the assumption that Ik has the 4-plane property. □ Claim 3. Let D be a component of Ft∩(A, x I) whose intersection with (3Aj) x I is non-empty. Suppose Iq ngi(Io) = 0 or Iq = gi(Io). Then D must be embedded in Ai x I. Proof of claim 3. We first consider the case that D is a disk. Since every component of Ik is embedded in M, D is an embedded disk in Ai x I. Topologically, Ai x I is an infinite solid cylinder and by our assumptions on Ik fl (A* x I), D must be a meridian disk. If D n gi(D) = 0 then gi(D) n gf(D) = 0. Moreover, D and gf(D) lie in different components of Ai x I — gi(D), thus D n gf(D) = 0. Inductively, D∩gf{D) = 0 for any n, and hence D is embedded in Ai x I. Now we suppose that D is an annulus. Since Iq is embedded in M, it is clear that D must be embedded in Ai x I. □ 38 Let D be a component of Fk fl (A; x I), let c be a simple arc (defined before) connecting .4; to Aj, and let ca fl A, = x,c3 n Aj = y, where 1 < i,j < g — b and 1 ≤ s ≤ k. Suppose D fl ({x} x I) = uf=iXfl. We denote the components of Fk fl (cs x I) that contain Xi,...,Xp by al,...,ap respectively (Xh 6 ah). Let Yh be the other end point of q/, (h = 1, ...,p). Now, we inductively define level and extended level of a component in Ai x [ as follows. If D is a disk and ^uft=ι^ι)∩({'.y} x /") = 0 (i∙e∙ uh=ιYh C c„ x dl), we say that D has extended level (or simply e-level) 0 in Ai x I. Moreover, if Aj x / contains a disk Dq of level (resp. e-level) 0 and if D n Do = 0. then we also say that D has level (resp. e-level) 0. We say that D is a surface with level (resp. e-level) at most n (n > 1) in Aj x / if (u^=iyf)∩({y} x /) 7^ 0 and at least one component of Fk n (Aj x I) that has non-empty intersection with is a surface with level (resp. e-level) at most n - 1 in Aj x I. We define the level (resp. e-level) of D to be the minimum of such n with respect to all the Aj x /’s and all c, 's. Note that Aj and Aj may be the same annulus. Let D be a surface with level (resp. e-Ievel) n in Aj x I as above and D\ be a surface with level (resp. e-level) n - I (that contains Yh for some h) in Aj x / as above. We say that D\ is a level (resp. e-level) n — 1 surface attached to D. If Do is a level (resp. e-level) n — 2 surface attached to Di, we say that Do is a level (resp. e-level) n — 2 surface attached to D. Repeatedly, for any 0 ≤ t < n, there is at least one surface with level (resp. e-level) t attached to D. Suppose that D C Aj x I has level (or e-level) n. We say that D is regular if there is a sequence {Kt} (0 ≤ t < n) such that: (1) Kt is a component of Fff As x I (for some s) and Kt has level (or e-Ievel) f, (2) Kt_i is a surface attached to Kt (Kn = D), and (3) every Kt is a disk. Remarks. Suppose that Dc AjX/ has e-Ievel 0. We can enlarge Aj along c3 (using the notation above). Namely, let A? be π small neighborhood of Aj U ds where ds is a sub-arc of cs such that ds contains x and ds x I contains is a disk by definition, the component of Ft ∩ (Af x Then, since D that contains D must have a puncturing arc in Af x I, and hence D has level 0 in Af x I. In particular, D must puncture through a cross disk. Level and e-Ievel can be considered as measures of the distance between D and an arc in Fd that puncture through a cross disk. 39 By the remarks above, for any disk D of level (or e-level) 0 in Ai x I, D must contain a short arc that connects the two components of A* x d[. We also call such short arcs puncturing arcs. By our assumptions on the cross disk Dr — DR U DR, any puncturing arc cannot lie in ir(Dr). Claim 4· Let D and D' be two components of Fr H (Ai x I) with level (or e-level) n and n' respectively. Suppose k is very large and n, n' are relatively small. Then, D and D' are embedded. Moreover, if D is an annulus, D' must also be an annulus and D fl D' = 0. If D has level (or e-level) 0, then D' n D = 0, and hence D' also has level (or e-level) 0. Proof of claim 4∙ We first show that D is embedded. By claim 3, it suffices to show that Fd n gi(Fo) = 0 (or Fd = ^(Fd))- Suppose that Fq n gi(Fo) £ 0 and Fd £ 9i(FD}. Then, since k is large and n.n' are relatively small, there is a cross disk Dr = DR U DR such that DR n Fd, DR ∩ gi(Fo), DR ∩ Fq and DR ngi(Fo) are not empty, as in the proofs of claim 2. This gives 4 components of Fr intersecting each other, which contradicts the assumption that Fr has the 4-plane property. So, D and D' are embedded. Suppose D is an annulus, then π~l(D) fl (A; x I) has only one component, say D. Let D' be a component of κ~l(D')fl (Ai x I) (using the notation before claim refclaim2) and Fq> be the component of FR that contains D'. If D fl D' £ 0, then D fl g™(D') £ 0, and hence Fd n gRl(Fq') £ 0 for any m. Since n and n' are relatively small, we can find a points x E D and x, ∈ gfD') such that x and x are closed to some puncturing arcs in Fq and g™(Fq') respectively. By choosing an appropriate m, we can assume that the distance between x and X is short. Thus, if A is large, the two puncturing arcs (in Fq and 3fx(ir0j)) puncture through the same cross disk. Since Fq n gfl(Fq') £ 0 for any m, we have 4 planes intersecting each other, which gives a contradiction. By our assumptions on Fr n (Ai x I), Dn D' = 0 implies that D' must also be an annulus. If D has level (or e-Ievel) 0, D contains a puncturing arc. If DnD' £ 0, similarly to the argument above, we can find a cross disk and an m such that the cross disk intersects both D and g^{Fq∣) and D fl g™(Fq>) £ 0- This also contradicts the assumption that Fr has the 4-plane property. □ 40 Claim. 5. Let D and D' be two components of Ik fl (Ai x I) with level (or e-level) n and n' respectively. Suppose that D is regular. If f is large and n, n' are relatively small, then n' < n and D' is regular. Proof of claim 5. By claim 4, D and D' are embedded. We prove claim 5 by in­ duction on n. If n = 0, by claim 4, D fl D' = 0. Hence, n' = 0 by our definition. Now. we assume claim 5 holds for n — 1. Since D is regular, D is a disk and there is another disk component Kn-↑∣ of Ikn(Aj x I) with level (or e-level) n — I such that D and Kn-ι are connected by an arc in Ik n (cs x I). Since D is a disk, by claim 4, D' cannot be an annulus. So, D' must be a disk. Then either D' has e-level 0, or there is an arc in Ikn (cs x I) connecting D' to a component l in Ikn(Aj x I). By our induction, ^n~l has level (or e-level) at most n — I and Kf~l is regular. Therefore, Claim 5 holds for D'. □ We have assumed (before claim 2) that there is a double arc 7 of Ik n p x {i} such that 7 C ∈ x I and 7 has an end point lying in π(∩^,] ) fl ∂M. Moreover, we have assumed that any component of Ik fl (At x I) that contains a sub-arc of 7 is not an annulus. Claim 6. Let T be the union of the components of Ff∩(.^ι×∕),s and Ik^∣(N(ci) xl),s that contain sub-arcs of 7. Then every component of 7,∩(Aι x /) has level (or e-level) at most g + 6 — 1, for any i. Proof of claim 6. The proof of claim 6 is an easy application of the claims 4 and 5. Since 7 passes through a puncturing fiber, there is a component of T n (Aj x I), say Tq, contains a puncturing fiber, then it must have level 0. We can inductive define the level and e-level of each component of T ∩ (Aj x I) for every i. By our assumptions on 7, every component of Tn (Aj x /) is regular. Thus, claim 6 follows from repeated application of Claim 4 and 6. □ Claim 7. If f is large (compared with g and 6), then z £ 7r(∩f∕2j) for any 2 6 T, where T is as in claim 6. Proof of claim 7. By claim 6, the level (or e-level) of the surface that contains 2 is less than g + b. Thus, there is a short arc that contains z and connects both 41 components of μ x dl. However, if k is large, any short arcs must lie in w(Dk), which contradicts our assumption on Dk- □ Therefore, Claim 7 and our assumption that 7 has an end point lying in dM give a contradiction. If y is a non-orientable compact surface, we can take a double cover of y and the proof is the same. So, we have proved Lemma 2.6.3 in the case that y is a compact surface. Case 2. fi i^ not a compact surface. If μ is not a compact surface, then fi is carried by a normal incompressible branched surface B whose boundary is a union of circles. By blowing air into each leaf (i.e. replacing each leaf of y by an /-bundle over this leaf and then deleting the interior of the /-bundle), we can assume that y is nowhere dense. Since B carries a compact surface, Ff∙∣'ΙiV(B) cannot be transverse to every /-fiber of N(B). Same as the case that y is a compact surface, we can assume that any /-fiber of N(B) either transversely intersects Fk, or lies in Fk (we also call it a puncturing fiber). By assuming that the branch locus L of B lies in the 2-skeleton, we can view B as a union of normal disks. For each normal disk D C B — L, we denote p~l(D) by N(D), where p : N(B) — B is the map that collapses every /-fiber to a point. Suppose that iV(Z?) = D~ x I. N(D) ∩ y = D~ x C, where C is a nowhere dense closed set in ∕, and suppose N(D) contains some puncturing fiber K. For any x € K, there is a non-trivial simple closed curve lx on a leaf of y such that x € lx and lx∩N(D) has only one component. Since y has no holonomy, there is an embedding bx : Sl x [—1,1] — N(B) such that bx(Sl x {0}) = lx, 6χ({9} x [—1,1)) is a sub-arc of an /-fiber of N(B) for any q 6 S1, and b~l(y) is a union of parallel circles. Suppose that bx(Sl x [—1,1)) fl N(D) = a× Jx, where a is an arc of D2 (in N(D) = D2 x I) and Jx is a sub-arc of an /-fiber of N(D) = D2 x I. Let bx be a small fibered neighborhood of the union of bx(Sl x [—1,1]) and D2 x Jx. Note that 6^ is a product of an annulus and an interval I. We can assume that each /-fiber of b'x is a sub-arc of an interval fiber of M(B) and bx fl y is a union of parallel annuli. We call bx a thick band. By compactness, there are finitely many disjoint thick bands in N(B) such that the union of these bands contains N(D) fl y. Moreover, 42 there are finitely many disjoint thick bands Bi,Bo,■.. Bn in N(B) such that, for every puncturing fiber K, K ∩ p belongs to U∙Lj,Bj. Let Bi = Ai x I, where Ai is an annulus and {<} x I is a sub-arc of an interval fiber of N(B) for any q E Ai∙ Let Bn+i, ,..,Bn+b be a small neighborhood of p~l(∂B). Then Bj = Aj × I (n + 1 < j < n -f 6) where Aj is a small neighborhood of a boundary component of ∂B and {<} x I is an interval fiber of N(B) for any q E Aj. By our construction, we can assume that pD B{ = Ai x Ct. where C, is a closed infinite set in I for each i. Similarly, since every leaf is dense in p. there are finitely many disjoint embed­ dings Qj : E x I — ι((B) (1 < i < m and E = I x [0, e]), where I is an arc, such that: 1. a~i(p) = E x C[, where C[ C I is a closed and infinite set, 2. aj({q} x I) is a sub-arc of an interval fiber of N(B) for any q € E, 3. ai(((∂l) x [0, ej) x ∕) = at{E x /) n (U"^.4j x J) c U^(A∙4j) x ∕, 4. p - Uj^Aj x I — uJLtQ(E x ∕) is a union of disks, 5. The intersection of each /-fiber of ∂A x I with U^aAE x I) has at most one component. Let fl = (UΓ=1M1x∕χj(∪f=iat(Ex∕^)) andO, = (^=i6(^Ai)x/^^U^(Uj=l /)). Since p — Ω is a union of disks, there are finitely many disjoint embeddings 3i : D x ∕ — N(B) (D is a disk and 1 < i < q), such that: 1. 3~l(p) = D x C". where C" is a closed and infinite set in I for each i. 2. 3i({p} x I) is a sub-arc of an interval fiber of N(B) for any p E D, 3. 0i(D x I) f Ω = i3(∂D) x /) c &u, 4. μ-nc∪L1β(P×∕). Furthermore, we can assume that (Ui=1 A(E x ∕))u∣Ω matches perfectly to form an /-bundle over another branched surface B' and ∂υN(B') C flu- Otherwise, we can replace Ai x I by Ai x /' where ∕' is a closed sub-interval of I, or modify ai and 3i. So, p is fully carried by N(B'). Since p is a measured lamination, by out previous discussion, N(B') must fully carry a compact orientable surface, which we denote by S. It follows from our construction that 5 — fl is a union of disks. We denote by AÇ,..., A'n, the collection of annuli in S fl ((JJL1 A'i x /), and denote by 43 Li,..., Lm the collection of disks in 5 fl ((J£Li Qt(∙F, x -0). Let S x J be a small neighborhood of S in N(B'), where J is a small interval. Since p intersects each at(E x I) infinitely many times, we can assume that fi n (A( x J) = A' x C (for each i), where C is a closed and infinite set in J. Moreover, we can assume that fi fl (Lj x J) (for each j) contains Lj x C for some closed infinite set C in J. Note that fiC(Lj x J) may contain components that are not parallel to Lj x {£} (t £ J). In other words, some components of f n (Lj x J) may intersect Li x d.J. Now, we apply the same argument in the case that fi is a compact surface. We only need to replace fi in that case by S, replace the N(ct) x Fs in that argument by c∏i(E x ∕),s in this case, and replace the Ai x Fs in that argument by A< x Fs here. We can assume that the image of a cross disk, i.e., ir(Dr), lies in a tiny neighborhood of a leaf of fi. Since the intersection of every leaf of fi with each L, x J contains infinitely many components of the form Li x {£} (t £ J), after choosing a sub cross disk of Dr (with large size) if necessary, we can assume that k(Dr) does not contain a puncturing arc (defined as before) in any A,i x J. Thus, x(-D[jk∕2∣ ) must be far away’ from any puncturing arc, and hence the argument in the case that fi is a compact surface also works here. □ Theorem 2.2, which is a generalization of Hatcher’s theorem, now follows easily from Corollary 2.3.2 and Lemmas 2.6.1 and 2.6.3. Theorem 2.2. Let M be an orientable and irreducible 3-manifold whose boundary is an incompressible torus, and let Ή be the set of injective surfaces that are embedded along their boundaries and satisfy the 4-plane property. Suppose that M does not contain non-peripheral closed incompressible surfaces. Then the surfaces in H can realize only finitely many slopes. Proof Suppose that the surfaces can realize infinitely many slopes. Let {F^a} be a sequence of surfaces in H with different slopes. Since they have different slopes, by Corollary 2.3.2, the surfaces in {Fre} cannot be carried by finitely many immersed branched surfaces. Then, by the argument is section 2.4, there exists a sequence of cross disks from {Fn} whose ‘limit’ is an essential lamination. However, Lemma 2.6.1 44 and 2.6.3 imply that the surfaces in {Fra} cannot realize infinitely many slopes, which gives a contradiction. □ As an application of Theorem 2.2, we will show that, in some sense, most 3manifolds do not have non-positive cubing. Theorem 2.3 gives the first non-trivial examples of 3-manifoldk that do not have aoa-prsitive cubing. Before we proceed, we would like to prove a lemma. Lemma 2.6.4. Let M be a closed and irreducible 3-manifold, S be an immersed surface in M with the 4-plane property, and C be a homotopically non-trivial simple closed curve that intersects S non-trivially. Then S — C is a surface with the 4-plane property in M — C. Proof. Let M be the universal cover of M and C be the pre-image of C in M. Then M — C is a cover of M — C. and among any 4 components of S — C, there is a disjoint pair, where 5 is the pre-image of S in M. Since each component of S — C is embedded, among any 4 components of the pre-image of S—C in the universal cover of M — C (i.e. the universal cover of M — C), there is a disjoint pair. Therefore, S - C satisfies the 4-plane property in M - C. □ Theorem 2.3. Let M be an orientable and irreducible 3-manifold whose boundary is an incompressible torus. Suppose that M does not contain closed non-peripheral incompressible surfaces. Then only finitely many Dehn fillings on M can yield 3manifolds that admit non-positive cubing. Proof. Let M(s) be the closed 3-maniaold after doing Dehn filling along slope s, and C3 be the core of the solid torus glued to M. Then, except for finitely many slopes, C3 is a hrmotrpicαllt non-trivial curve in M(s). Suppose that M(s) admits a non­ positive cubing. For each cube in the cubing, there are 3 disks parallel to the square faces and that intersect the edges of the cube in their mid-points. These mid-disks ^om all the cubes in the cubing match up and yield a union of immersed surfaces. Moreover, the complement of these immersed surfaces in M(s) is a union of fi-balls. ATchison and Rubinstein have shown that these surfaces (and their double cover in M(s) if they are one-sided) satisfy the 4-plane property. Since C3 is nra-trivial and the complement of these kurf∏dek is a union of 3-0π11^ C3 must noa-triviallt intersect 45 at least one of these surfaces. Hence, by Lemma 2.6.4, there is an injective surface in M that satisfies the 4-plane property and has boundary slope s. By Theorem 2.2, there are only finitely many such slopes. Therefore, the theorem holds. □ 46 Chapter 3 Immersed surfaces in hyperbolic 3-manifolds 3.1 Introduction An important question in 3-manifold topology is whether a closed 3-manifold con­ tains 7∙i-injective surfaces. Embedded 7Ti-njective surfaces give a lot of information about 3-manifolds, e.g. [35]. But unfortunately, in some sense, most 3-mamfolds do not contain embedded surfaces [19]. The main goal of this paper is to prove the contrary to [19] for immersed surfaces, i.e. (in some sense) most 3-manifolds do contain a surface subgroup. Theorem 3.1. Suppose X is a hyperbolic 3-manifold whose boundary is a single torus. Then all but finitely many Dehn fillings on X produce 3-manifolds containing πι-injective surfaces. This theorem was also proved by Cooper and Long [8] earlier using different methods. The proof that we give here is topological, and an advantage of this approach is that it gives an explicit bound on the number of exceptional surgeries. Theorem 3.1 follows directly from Theorem 3.2 by the deep results in [11, 10]. See below for definitions of Xp) and Xp, s). Theorem 3.2. Suppose X is a hyperbolic 3-Manifold whose boundary is a single torus, and S is a two-sided, embedded, incompressible and d-incoMpressible surface with boundary slope s, and S is not a virtual fiber of X. Then there exists a num­ ber Γ such that Xp) contains πι-injective surfaces for any boundary slope p with Xp,F > Γ. Proof of Theorem 3.1 from Theorem 3.2. It follows from [11, 10] that X contains such incompressible surfaces with at least two distinct boundary slopes arising from the splitting of τ^ι(A) associated with the ideal points of certain algebraic curves. 47 Then by Proposition 1.2.7 of [10], the fundamental groups of the splitting sur­ faces cannot be normal subgroups of 7t(X). hence they are not virtual fibers and Theorem 3.2 implies Theorem 3.1. □ In this paper, we will mainly prove Theorem 3.2. Unlike [8], we do not actually use the hyperbolic structure. The only thing that we need is that tti(X) has no ∩o∩O peripheral Z © Z subgroups, which is equivalent to saying that X has a complete hyperbolic structure by Thurston [34]. Moreover, we will give an explicit bound to the number of exceptional surgeries. Theorem 3.3. In theorem 3.2, T can be chosen to be an explicit linear function of the genus and number of boundary components of S. The idea of the proof is to construct a closed surface from S by connecting pairs of the boundary components of 5 using long annuli that wind around ∂X. By some combinatorial arguments, we show that if both the number of times that the annuli wind around ∂X and the distance between the surgery slope and the slope of ∂S are large, then this closed surface is nq-injective. Notice that the immersed surface constructed has no triple points. The techniques in this paper have been used on embedded incompressible sur­ faces in various papers (e.g. [28, 10]). The simplicity of the immersion in our construction allows us to apply them to this case. The idea of closing up boundaries of surfaces using long annuli was introduced by B. Freedman and M. H. Freedman in [13], and extensively used in [9, 7, 8]. Notation. Let a, 0 be two slopes on the boundary torus of X. X(a) denotes the closed manifold by ∩ehn filling along a, i.e., by adding a two-handle to X along a simple closed curve with slope a and then capping off the resulting 2-sphere bound­ ary component with a 3-cell. A {a, (3) denotes the minimal geometric intersection number between two closed curves representing a and (3. N(E) denotes a small regular neighborhood of E, and |1? denotes the number of components of E. We O use both E and int(E) to denote the interior of E. 48 3.2 I-bundle regions Definition 3.2.1. Suppose that M is an irreducible 3-mπalfrld with boundary, and Al,..., -4* are disjoint annuli in dM such that dM — Ui=i A, is incompressible in M and the vertical arcs of each annulus cannot be homotoped rel boundary into dM — Ui_i A,. Let i : D = I x I — M be a proper map. We call i a product disk of (M, A), if i(dl x Z) is a pair of vertical arcs of A = Ui=i ^* und i(I x ∂I) is a pair of immersed arcs in dM — A which cannot be homotoped rel boundary into A. We call {p} x I a vertical arc of the product disk for any pel. By our definition, any vertical arc of a product disk cannot be homotoped rel boundary into M — .A. The following lemma is a simple case of the characteristic pairs in [24, 21]; see also [25]. For completeness, we give a proof here. Lemma 3.2.1. Let (M.A) be as above. Then there is a maximal I-bundle region J in M such that any product disk can be homotoped into J. Proof. First, we will show that there exists such a region J for embedded product disks. Given any two embedded product disks, by the standard cutting and pasting argument, we can assume after isotopy that their intersection is a union of vertical axes. So, in our proof, we always assume that the intersection of any two embedded product disks is a union of vertical arcs. We start with A and an embedded product disk D\. We thicken them a little to get a small neighborhood of the union of A and D\, which is clearly an /-bundle. We call it J\. Assume that we have constructed Jk, which is a neighborhood of the union of D},Do,...,D∣i and A. If there is still an embedded product disk -Dj+i that cannot be isotoped into j then we let Jk-i be a neighborhood of the union of Dι,..,,Dk+i and A. Since Dk-i cannot be isotoped into Jk such operations increase the Euler characteristic of the non-disk components of dM — Jk. Thus the operations must stop at a certain stage, and we get an /-bundle Jr such that any embedded product disk can be isotoped into J'. Furthermore, suppose some component of dM—J', say D, is a disk. Then by the definition of a product disk, each fber of Jf cannot be homotoped rel boundary into 49 dM — A, and hence D together with the fibers of J' incident to dD is a disk, denoted by D'. Since dM — A is incompressible, dD' bounds a disk D" in dM — A, and D' U D" bounds a 3-ball because M is irreducible. Therefore, by adding such 3-balls to J', we can enlarge J' to another /-bundle J with canonical fibration such that no component of dM — J is a disk. We call (dM — A) n J the horizontal boundary, which we denote by d^J, and dJ — dhJ — A the vertical boundary of J, which we denote by dv J. Notice that M — J does not contain any embedded product disks with respect to (M — J,dυJ). Now we show that any product disk can be homotoped into J. Let i : P = I x ∕ —- M be a product disk. Then i~l(dυJ) is a union of disjoint simple arcs and simple closed curves in P since dJ is embedded in MI. If there are simple closed curves in i~l(dυJ), we choose an innermost one which bounds a disk A in P. Then i(dA) is a homotopically trivial curve in the vertical boundary of J, and i(A) lies in J (or M — J). Since τ∙o(M) is trivial, we can homotope i(A) to a point on dυJ and move it out of J (or M — J) reducing the number of components of i~i(dυJ). Hence, we can assume that i~i(dυJ) does not contain any simple closed curves. Since i(dl x ∕) c A C J, each component of i~l(dυJ) is either a vertical arc of P or an arc with both endpoints on the same component of I x dl. In the latter case, we choose an outermost such arc, say a. a together with a subarc of I x dl, say ,d, bounds a disk 6 in P. Now i(a) is a d-parallel arc in dυJ and, since dM — A is incompressible and M is irreducible, i(3) is a 5-parallel arc in d^J (or dM — J). Since tpojAi) is trivial, we can homotope i(6) out of J (or M — J) reducing the number of components of i~i(dυJ). So we can assume that i~l(dυJ) consists of only disjoint vertical arcs in P. If P cannot be homotoped into J, then i~l(M — J) is a collection of rectangles of the form [a, 6] x I in P, where [a, 6] is a subinterval of ∕^. i restricted to each of these rectangles is a product disk of (M — J,dυJ). By doing some cutting and pasting to P and dvJ, we get an embedded product disk in M—J, which contradicts the assumption that J is maximal. Thus any product disk can be homotoped into J. □ 50 Notation. Let S be an orientable surface and R C S be a subsurface of S with OS C R. Let Rf = RU(disk components of S—R), and R = R' — (disk components of R'). We define an equivalence relation: Rι ~ Ro, f Ri and R2 are isotopic in S. Denote the set of surfaces equivalent to R by [/?]. Proposition 3.2.2. Suppose Ri,Ro ore subsurfaces of S, and∂S C RyClRo. Then there exist R'∣ ∈ [Ri] and Ro € [Ro] such that if a non-trivial curve can be homotoped into each of R^ and Ro, it can be homotoped into R∣ fl R⅛. Proof. If 5 is a disk or an annulus, then the proof is trivial. So, we can assume that S is a hyperbolic surface with geodesic boundary. For simplicity, we only consider the case that Rγ and Ro. are connected. By our definition, there are no disk components in S - Ri. We isotope Ri and Ro to be subsurfaces of S with quasi­ geodesic boundaries as follows. If S — Ri ( i = 1 or 2) contains annular components, we isotope Rl such that the annular components are e-neighborhood of geodesics for some small e. For other boundary components of Ri, we first isotope Ri so that these boundary components are geodesics, then enlarge Ri by adding a 2e-neighborhood of the geodesics to it. By choosing e small enough, we can assume that there is no overlapping of R, with itself, i = 1,2. For any nontrivial curve of S which can be homotoped into both Ri and Ro, we first homotope it to be a geodesic. It then lies either in both surfaces constructed above or in an annular component of S — Ri, for some i. In the later case we homotope the curve out of the e-neighborhood so that it still remains in the 2eneighborhood of the geodesics. By our construction, it lies in the intersection of the two surfaces. □ Let Ri, Ro,R,∣.R,o be the surfaces of Proposition 2.2. We denote [R^ fl by [i?i] fl [Ro]- To simplify our notation, we do not distinguish between [R] and a properly chosen element in [R]. Definition 3.2.2. Let X be an irreducible 3-mamfold whose boundary components axe tori, S be a two-sided, incompressible, 3-incompressibIe, embedded surface in 0 X, and M be the 3-marnfoId obtained by cutting X along 5, i.e., M = X — N(S) O (M may not be connected). Let A = ∂X — N{∂S) be a union of annuli in ∂M. We 51 call a map i : I x [0, n]→ Xan essential rectangle of length n, if i intersects S transversely and i|ixpt,,k—i] is a product disk of (M, A) for each k 6 {0,1,,.., n — 1}. The following lemma is important to our proof. The same result was proved in [7] foo the nan-skpp∏rting ccπe. Lemma 3.2.3. Let X be a hyperbolic 3-manifold whose boundary is a single torus and S be a two-sided, incompressible, d-incompressible surface in X. Suppose S is not a virtual fiber. Then there exists a number P(S) € N such that the length of any essential rectangle is less than P(S), where P(S) = 6g -) 46 — 6 and g, b are the genus and the number of boundary components of S. Proof. We assume that S is separating. If not. we can take S together with a paπallel copy of 5 (disconnected) to be our sari∏de· Let Mi and Mo be the closures of the two components of X — S. and let M be the disjoint union of Mi and Mo. Let .4, = Mi fl dX for i = 1,2. Then dMi — .4,ι = ∂Mo — Ao. Let Ji be the maximal /-bundle region of (Mi, Ai) constructed in Lemma 3.2A. Let Si = Ji D (dMi - Ai) be the horizontal boundary of J, for i = 1,2. Note that Si and dMi — Ai - Si have no disk components. We can also assume that Ai C Ji. Define rj to be an involution of Si such that r, : po n p\, where po and p\ are the endpoints of π fiber of Ji, i = 1 or 2. If Ji = Mi for both i. then both Mi and Mo are /-bundles. Hence ττχ(S) is a normal subgroup of 7-.(.Y) and 5 is a virtual fiber. So we can assume that Si 7^ ∂Mi — .4i. Let p : dMi — Ai — ∂Mo — Ao be the gluing map, then X = Mi UMo/x ~ y(x). For any S, E [S,]. we can isotope Ji so that Ji∩(dMi — At) = Sf and define 7j coherently. So we do not distinguish between [5,] and an element in the equivalent class and always use Ji for the coherent /-bundUe. Let Ti = [S1], Tk = Kfr] 1 o r,([S2i n [φ(Tt)])]), ∏nd Tm = (Tk} n T]. Claim. [Tk] ^ [Tk—i] for any k unless [Tc] = [dAi}, where [<9Ai] is a small neighbor­ hood of ∂Ai in ∂Mi — Ax. Proof of the claim. We have [Ϊ*] = <ρ[I½] 2 lS2]a∣≤’(μfc)I — ¥ lαΓ2([S’2]a[[μ((Tc)]) D = CTt] D [Γf)∩lΓ(] = [7i)+1]- Equalities hold if and only 52 if Mπ∙)] = [⅜] n M(Γ*)I, [<-1oγ2([5⅛]∩ Mrfc)])l = [st] n [v→ oτ⅛([s⅛] n [<^(r*)])], and [Τ'] = py. If [Tfc+i] = [Tfc] and [Tt] £ [9AJ for some k, then there exists a boundary component 7 of [T⅛] such that 7 is not parallel to dA\. Note that 7 is a non-trivial curve by our construction. Hence <(7) is a boundary component of [<p(Tr)] = [So] n M(TTk)], and 71 = rι o £_l o r2 o <(^∕) is a boundary component of [T£] = [Γr]. By our construction, S U 17(<) bound an annulus or Mobius band in Ji for any simple closed curve S in S;. Hence, if 7 is isotopic to 71, we can close up the two annuli or Mobius bands bounded by <(7) U r2 o i^(7) and £_l o r2 o £(7) u 7, to get a torus or Klein bottle in M. If 7 is not isotopic to 71 then 72 = 77 oip^ior⅞0£(7j is also a boundary component of [T,.]. In this way, we can define 7, for any L Since [T,.] has only finitely many boundary components, there exist i £ j 6 N such that 7* is isotopic to 7j. So we can always close up some annuli or Mobius bands to get an immersed torus or Klein bottle. We shall show that the immersed torus (or Klein bottle) is TTinnj^ctive. To simplify notation, we do not distinguish between the torus (or Klein bottle) and its image in X under the immersion and denote both by T. The 7js are parallel non-trivial curves in T and their images are non-trivial curves in 5. Since S is two­ sided and incompressible, the 7, 's are non-trivial in M. If the immersion is not TTinnj^cttive, then there is a non-trivial curve, say 7', in T which intersects each 71 non-trivially and is mapped to a trivial curve in M. By our construction of T, we can homotope 7' so that it consists of vertical arcs of J\ and J2. Now that 7' bounds a disk D, in M. The pull-back of the intersection of Dl with S is a collection of simple curves in D'. By using homotopies as before we can assume that there are no simple closed curves In D,. Wc then choose an outermost arc. This arc together with a subarc of 7' bounds a subdisk in D'. This subdisk is mapped to either Mi or M2, which means that thc subarc of 7' (i.e. a vertical arc of J,, by assumption) can bc homotoped rel boundary into dMi — A;. This contradicts our assumption on Ji. Hcncc thc immersion is TΓ-injectivc. Now wc show that 7* is not homotopic into dX. Note- that, since 5 is 7∙1mjectivc, any non-trivial and ∩o∩ <3-parallcl curve in ∂Mr — Ar (h = 1 or 2) is not homotopic into Ar in Mr. Suppose that 71 is homotopic into dX in X. Then thcrc 53 is an immersed annulus f : E = Sl x [0,1] — X such that ∕ is transverse to S, f~l(∂X) — Sl x {0}τ and f (Sl x {1}) C int{M-∣) is a curve parallel and close to 7*. Thus f~l(S) fl Sl x {1} = 0. Since S is incompressible and ^-incompressible, by some homotopies, we can get rid of trivial circles in f~l{S) and those arcs in f~l{S) with, both endpoints on S1 x {0}. Hence, we can assume that f~l{S) is a union of disj'oint meridian circles in E. Since 7* is non-trivial and non 3-parallel in S, the image of each component of f~l{S) is non-trivial and non 3-parallel in S. Let Eq be the component of E — f~l{S) that contains S1 x {0}. Then f∣g0 is an annulus connecting Ah (h = 1 or 2) to a non-trivial and non 3-parallel curve in ∂Mh — Ah, which gives a contradiction. Therefore, we get a Ti-injective and non-peripheral torus in X, which contradicts the hypothesis that X is hyperbolic. □ It is easy to see that if [Tf-+i] ^ [Tfc] then there exists a ∩o∩-triviαl simple closed curve o< in pTk+i∣ — [TfC∙ Moreover, we can choose the simple closed curves such that a, is not parallel to aj if i ^ j. By an Euler characteristic argument, there are at most 3g + 26 — 3 disjoint, essential and non-parallel simple closed curves in ∂Mi — Ap By our assumption that Si ≠ ∂M∣∣ — A1, there is at least one non-trivial simple closed curve in ∂M∣ — At — St. Hence if k > Zg + 26 — 3, [Ft] = [3Ai]. Let i : I x [0, nj — M be an essential rectangle of length n. Suppose i^∣∕x[o.t] is a productdisk of (Mi, At), (3.1) i∣∕x[i .2| is a productdisk of (Mo, A2), (3.2) i|/x[2,3l is a productdisk of (Mi,A)), (3.3) 54 By(1), i(Ix {1}) 6 [Ti], by (2), i(Ix {1})6[5J∩ b(Ti)], i(x {2}) € roj^ n [⅛TO)]), i(I x {2}) 6 [Sj] ∩ £_1 o ιo([^si] ∩ [⅛5(Γ1)]), by (3), ( x {3})∈[T[]∩[TJ = [Γ2j, Thus, if n > 2(3g + 26 - 4) + 1, i(I x {2(35 + 26 - 4) + 1}) 6 [T3g+26-3] = [MJ, which contradicts our definition of a product disk. Hence wc have n < 2(3g + 26 — 4). Similarly, if i(I x [0,1]) is a product disk of (Mio, Ao) a∩d Jo = Mo then wc get n < 2(3g + 26 — 4) + 1. So i∩ any case, we have n ≤ 6g + 46 — 7. □ Corollary 3.2.4. Let X be a hyperbolic 3-Manifold whose boundary is a single torus and i : (S.dS) +*∙ (X,dX) a πγ-injective surface. Suppose there is a constant C ∈ N such that the genus of S is less than C. Then there are only finitely Many possible slopes for the boundary circles of i(dS). Proof Let S' bc a∩ embedded, two-sided, incompressible, ^-incompressible surface in X a∩d suppose S' is ∩ot a virtual fiber. Let thc boundary slope of dS, be s a∩d the boundary slope of i(dS) bc μ. As in thc proof of Theorem 3.1 it suffices to show that A(∕ι, s) is bounded. The components of i~l(Sr} are disjoint simple arcs or simple closed curves in S because S' is embedded. Let g bc thc gc∩us of S and 6 bc thc number of boundary components of S. Then i~i(S') consists of at least |6A(/z,s) simple arcs i∩ S. By 55 an Euler characteristic argument, there are at most 6g + 36 — 6 disjoint nonparallel nontrivial simple arcs in S. Since g < C, if A(μ, s) is large, there are many parallel arcs in S and we get an essential rectangle with large length violating Lemma 3.2.3. So A(/i, s) cannot be too large. □ Remarks. 1. In [2] and [31], it was shown that for many 3-mamfolds with boundary a torus there are infinitely many boundary slopes realizing 7T-mjective surfaces. Corollary 2.4 says that as the boundary slope increases, the genus of the surface increases. 2. Corollary 2.4 is not a deep repute The knowing ekgem angument is due to ∩ave Gabai. Let X be a hyperbolic 3-mamfold with a single cusp and let S be a Tri-injective surface mapping cusps to cusp. Suppose S has the least area in its homotopy class. Then, by Gauss-Bonnet, Area (5) < — y(S) = 2g — 2 + 6. On the other hand, we let T be a CorospCrrical torus in X. Then S n T is a union of b closed curves (in T) of length at least I, where I depends on the slope of the closed curves. By hyperbolic geometry, the area of the cusps of S is at least kbl, where k is a constant and b is the number of cusps of S. Hence we have kbl < Area(S) < 2g—2+b. Since g is bounded, I cannot be too large and S can realize only finitely many slopes. 3.3 Construction of the injective surfaces Let X be a hyperbolic 3-mamfold whose boundary is a torus and S be a two-sided, incompressible, <-incompressible, embedded surface which is not a virtual fiber. As before, we assume that S is separating; otherwise we take S together with a parallel copy of 5 (disconnected) to be our surface. For simplicity we only consider the case that 5 has two boundary components. The proof is similar for the case that 5 has more than two boundary components. Let T2 x f be a product neighborhood of ∂X and S' be a parallel copy of S. We construct our immersed surface T by connecting the two circles of â(S'—T2 x I) using an annulus that winds (in T2 x I) around ∂X K times as shown in Figure 3.1 (a). Thus T n S is a collection of 2K --parallel disjoint simple closed curves. We call this annulus the long annulus. 56 We define a retraction map x : X — X — T2 x I by fixing points in X — T2 x / and mapping every interval {p} x I of T'2 x I to the point (p, 1), where p 6 T’2. Lemma 3-3-1- If K > P(S) + 1 then T is xi-injective in X. Proof. Suppose not, then there exists an immersed closed curve I in T such that I is contractible in X but not contractible in T. Let p be the number of arcs in the intersection of I with the long annulus. Notice that p >1 since S is incompressible and two-sided. We homotope I to minimize p and the number of points in its intersection with S. Since / is contractible in X, there is a map j : D — X such that j(βD ) = I, where D is a disk. We see that |/ n S| = 2Kp and j~i(S) is a collection of disjoint simple arcs in D since S is embedded. The two circle components of d(S' — T2 x I) divides I into 2p subarca, namely aι,3ι,t22,>, .... ap,3p, where Uf=i(Qi Ufi) = 'd, i(βi) is a subarc of I lying entirely in the long annulus and j(ai) is π subarc of I lying entirely in S,. Thus j~l(S) fl ∂d C Uf=i A and |j—(∙S,) fl #| = 2K for each i. We call the π√s a-nrcs, and the dfs d-arcs. These π-πrcs and J-arcs appear on ∂d alternately. Claim 1. There are no arcs in j~γ(S) whose endpoints are both in the same (3- Proof of claim 1. Suppose there are such arcs. We choose an outermost one, say 7; then 7 together with a subarc (3 of 3i bounds a bigon in D. Hence x o j(7) must be a Opan-allel arc in x(S). Since 7 is outermost and S is ^-incompressible, both endpoints of j(β) must lie in the same component of Tfl S and jfint(jj3)) n 5 = 0. So we can homotope I to have fewer points of intersection with S, which contradicts our assumption. This proves claim 1. We call an arc in j~l(S) a long arc if it cuts D into two components such that each of them contains at least two a-arcs. Claim 2. There exists a k 6 N such that the endpoints of no long arc lie in 0k- Proof of claim 2. Consider all the long arcs of j~l(S) in D and choose an outermost one. This together with a subarc of dD bounds a bigon that does not contain long arcs. Suppose the bigon contains arcs a⅛ and ∏t+i, then 0k is πs needed because j'^1(S) consists of disjoint simple arcs. This proves claim 2. Consider πll the arcs with an endpoint in 0k (0k ∏s in daim 2). By claim 1, the 57 (a) (b) Figure 3.1: other endpoint of such an arc must lie in either -f._i or .df-c-i- Since ∣ βk n j~l{S)∣ = 2K, we have at least K parallel arcs which are all parallel to at (or c*i+1), as shown in Figure 3.1 (b). Notice that 7ro j'(c∩c) is not a --parallel arc in tt(S'); otherwise we can homotope I to have fewer points of intersection with S. Thus the images of the K arcs which are parallel to a*· (or at÷i) are essential arcs in tt(5). Hence we get an essential rectangle of length K - 1 > P(S) with respect to the 3-marnfold xGX and surface <(S), which contradicts Lemma 3.2.3. □ Proof of Theorem 3.2 and Theorem 3.3. We will prove that the surface T constructed at the beginning of this section is li-injective in X{y) if both A^s) and K are large. Suppose not, then there exists a closed essential curve I in T contractible in X(μ). Hence for any i > 1 there is an immersion j : P V X, where P is a planar surface with k -11 boundary components, ∂P — U†=o Pi, jpo) = I and j(pi) is an immersed curve of slope fi in ∂X. We assume that I has been homotoped to have the fewest points of intersection with S and k (> 1) is the minimal number for till such planar surfaces. The case where k = 0 follows from Lemma 3.3.1. Now j~l(S) is a 58 u∩io∩ of disjoint simple axes or simple closed curves i∩ P because S is embedded. By the same argument as before wc ca∩ assume that there are ∩o trivial circles in P. Claim. There are ∩o 9-parallel arcs in P with both e∩dpoi∩ts o∩ the same pi for any i > 1. Proof of the claim. Suppose there are such arcs. Wc choosc an outermost one, say 7. which together with a subarc V of p, bounds a bigon i∩ P. Hence 7'(7) can be homotoped rel boundary into dX. Since S is 9-mcompressible, both e∩dpoi∩ts of j(7') must lie i∩ the same component of dS and jpnt(-γ,)) ∩ OS = 0 (because 7 is outermost). So we ca∩ homotopc p1 to get fewer points of intersection with dS, which contradicts our assumption. □ Let B be the subset of j~l(S) consisting of arcs with at least o∩c endpoint on pj for some i > 1. Since j(pi) is a curve of slope μ i∩ dX for i > 1, jpi) intersects S i∩ at least 2A(p, s) points (wc have assumed that S has two boundary components). Hence ∣B∣ > kXp.s). By a∩ Euler characteristic argument, the maximal number of ∩o∩-parallel arcs in P is 3k — 3 if k > 1. a∩d 1 if k — 1. So, if ∣B∣ > kAp.s) > 3kN, there are at least N + 1 arcs i∩ B which are parallel to each other. Let So∙4i, ... ,6^ be thc N 4- 1 parallel arcs. Case 1. The iV +1 parallel arcs have e∩dpoi∩ts o∩ pi and pj with both i.j > 1. Recall that by our construction (pi) ∩ dX if i > 1. Suppose j(6i) is a --parallel arc i∩ S for some i. Then wc ca∩ homotopc j(Si) to dX, then cut along dt to get a map of a planar surface with fewer boundary components, which contradicts our assumption. Therefore j(δi) is an essential arc for every i a∩d So. d1.....form an essential rectangle of length iV. By Lemma 3.2.3, if N > P(S). no such essential rectangle exists. Case 2. Each of thc N + 1 parallel arcs has o∩c e∩dpoi∩t o∩ po and the other endpoi∩t on pi for some i > 1. As i∩ the proof of Lcmma 3.3.1, wc divide po i∩to segments ai,βι,.. .,aq,βq, whcrc po = Uf=ι(α* u A) a∩d each j(ft) is a subarc of I lying entirely i∩ thc long a∩∩ulus, a∩d each jpi) is a subarc of I lying entirely i∩ the surface S'. Wc call thc α,-⅛ a-arcs and the ∕%s ∕3-arcs. 59 (b) (a) Figure 3.2: So U Jy together with a subarc a of po and a subarc p of p* form a quadrilateral Q in P (see Figure 3.2 (a)). By the claim, there are no arcs in Q with both endpoints on p. So the arcs of j~i(S) — B in Q are arcs with both endpoints on σ. By the claim 1 in Lemma 3.3.1, there are no arcs in Q with both endpoints on the same J-arc. As in the proof of Lemma 3.3.1, we can assume that there is no long arc in Q, i.e., no arc cutting of a bigon in Q which contains at least two a-arcs. Next we choose K and N such that 2K > 3P(S) +1 and iV-f 1 > 2K—2P(S) -11. Since each J-arc contains exactly 2K endpoints of arcs in j~l(S), there is at least one a-are in Q. Suppose there are at least two a-ares, say a* and πο.·+ι, in Q (by choosing N larger, we can always ensure that.). Then all the arcs with one endpoint on 0k are contained in Q. As in the proof of Lemma 3.3.1, there are at most P(S) arcs parallel to a⅛ or a/t+i; otherwise we have an essential rectangle of length P(S). So there are at least 2K — 2P(S) > P(S) + 1 arcs with one endpoint on 0k and the other endpoint on p, as shown in Figure 3.2 (b). If the images of these P(S) +1 arcs under the map 7 o j are not trivial in ιr(S), we get an essential rectangle of length P(S), which contradicts Lemma 3·2·3· Therefore we assume that they are trivial arcs in tTS). 60 Since iN+ 1 > 2K + 2P(S) +1, we have at least P{S) +1 parallel arcs in B with one endpoint on Pk+i (or Pk-i)- Again we assume the image of these parallel arcs under no j are trivial in n(S); otherwise it contradicts Lemma 3.2.3. Let (Jli = T fl 5, where each k is a simple closed curve. Then by our construction of T and assumption on I, the intersection of each βi with T ft S appears either in the order (with respect of a certain orientation of 1) /[,/2,..., I^κ or in the order Ijk-hiK-- — ∙ fi- a∩ arc that has one endpoint in the central portion of 3k or Pk+i (or Pk-i) must have the other endpoint on p, and hence is one of the N + 1 parallel arcs that we considered above (see Figure 3.2 (b)). The reason is that we cannot have too many arcs parallel to the two a-arcs adjacent to this P-arc; otherwise we will get a long essential rectangle. So it is easy to see that there must be an arc Α with Si fl 3k = a and an arc Sj with SjHpk±i = b such that j(a) and j(b) lie on the same simple closed curve component of T n S. Since by our assumption no j(6i) and no j(δj) are trivial arcs in tτ(S). we can homotope j(fii), moving j(a) along the simple closed curve component of T fl 5 to j(b) and closing up j(δi) and jδj), as shown in Figure 3.3. to get an immersed annulus in X. One boundary component, say o', of this immersed annulus is mapped into T and the other boundary component is mapped into ∂X with slope different from that of ∂S. Notice that there is exactly one a-arc between Si and Sj in Q. If < is mapped to a trivial curve in 7,. then the a-arc between Si and Sj must be a ∂- parallel arc in S', and we can homotope I to get fewer points of intersection with S. wihch conntadiccs ouu assumpttons. So σ' is a ηοη--ηνί^ curve in T. Now the simple closed curve component of T fl 5 containing j(a) and (b) together with a boundary component of ∂S bounds another annulus in 5, and the intersections of the two annuli are vertical arcs in both of them. Therefore we get two elements in ni(T) simultaneously homotopic to two curves in ∂X of different slopes. Since T is 7Tiιnjecrive in X and clearly T is not peripheral, Z©Z is a non-peripheral subgroup of τi(X). This contradicts the hypothesis that X is hyperbolic. So, as long as 2K > 3P(S) + 1 and N+l > 2K+2P(S) + l (i.e. N > 5P(S) + 1), T is 7T--njective m Xfμ). Recall that we have chosen ∩Δ(^c, s) > 3kN to get N -11 parallel arcs. Hence it suffices that A(/z, s) > 3N > 15P(S) -f- 3. If Q contains exactly one a-arc, then as in the proof of Lemma 3.3.1, there are 61 at most P(S) arcs parallel to this a-arc. Since iV > 2K + 2P(S), incident to each d-arc that is adjacent to this a-arc, there are at least 2P(S) arcs belonging to thc set of thc iV + 1 parallel arcs that we considered above. Now thc proof is as earlier. □ Remarks. If 5 has more than two boundary components, thc∩ ∣B∣ > s). I∩ this case, there are ∣ long an∩uli a∩d wc need j(a) and j(b) to bc on thc same long a∩∩ulus. He∩ce the factor 6 will bc canceled a∩d we get a bound for Aiμ,F that is a Unear function of g a∩d 6. 62 Bibliography [1] Aitchisoa, I. R.; Rubinstein. J. H. An introduction to polyhedral metrics of nonpositive curvature on 3-manifolds. 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