Essential Laminations in Seifert-fibered Spaces

Mark Brittenham

0. Introduction

Over the past few decades, the importance of the incompressible surface in the study of 3-manifold topology has become apparent. In fact, nearly all of the important outstanding conjectures in the field have been \underline{proved}, for 3-manifolds containing incompressible surfaces (see, e.g., [20],[22]). Faced with such success, it becomes important to know just what 3-manifolds could contain an incompressible surface.

Historically, the first 3-manifolds (with infinite fundamental group) which were shown to contain no incompressible surfaces were a certain collection of Seifertfibered spaces. Waldhausen [21], in the 1960’s, showed that an incompressible surface in a Seifert-fibered space is isotopic to one which is either vertical or horizontal. This added structure puts a severe restriction on the existence of an incompressible surface, and led to the discovery of these ‘small’ Seifert-fibered spaces.

Now in recent years the essential lamination, a recently-defined hybrid of the incompressible surface and the codimension-one foliation without Reeb components, has begun to show similar power in tackling problems in 3-manifold topology (see [7]). It also has the added advantage of being (seemingly) far more widespread than either of its ‘parents’; its more general nature makes it far easier to construct in a wide variety of 3-manifolds (see, e.g., [6]). In light of these facts, it would be interesting to know if there are any 3-manifolds which contain no essential laminations, and only natural to look in the same place that Waldhausen found his examples.

In this paper we carry out such a program. We show that an essential lamination in a Seifert-fibered space satisfies a structure theorem similar to the one given for surfaces by Waldhausen. Together with work of Eisenbud-Hirsch-Neumann on the existence of horizontal foliations, this structure theorem allows us to show that some of the ‘small’ Seifert-fibered spaces above cannot contain any essential laminations.

We also obtain, as a further application of the structure theorem, a result which states that any codimension-one foliation with no compact leaves in a ‘small’ Seifert-fibered space is isotopic to a horizontal foliation; this completes (in some sense) a group of results on isotoping foliations in Seifert-fibered spaces, which began with Thurston’s thesis.

The author wishes to thank Allen Hatcher for his many comments and suggestions during the course of this work, in his capacity as the author’s thesis advisor. We should also mention that Wilhelmina Claus has, using different methods, obtained similar results [3] in the direction of this paper.

1. The Main Results

For definitions and notations concerning essential laminations, see [7].

In this paper the word ‘lamination’ will mean a lamination which is carried by a branched surface; technically, therefore, a foliation F, for example, is not a ‘lamination’. One must first split F along a finite number of its leaves, as in [7]. Because we are largely interested in the existence of essential laminations, splitting will cause no difficulty; the splitting of an essential lamination is essential.

For definitions and basic concepts regarding Seifert-fibered spaces, see [8] or [16].

Generalizing [21], we say that a lamination L ⊆M is vertical, w.r.t. a Seifertfibering p:M→F if p^{−1}(p(L))=L, i.e., L contains every circle fiber of M that it meets; L is horizontal if it is transverse to the circle fibers of M at every point.

Now let M be a compact orientable Seifert-fibered space, with Seifert-fibering π : M → F.

Theorem 1: Every essential lamination L in M contains a sublamination L_{0} which is isotopic to a vertical or horizontal lamination.

The proof of this theorem comprises the bulk of this paper.

Corollary 2: If a Seifert-fibered 3-manifold M contains an essential lamination, then it contains a horizontal or vertical one.

Therefore, if we wish to show that a Seifert-fibered space contains no essential laminations, it suffices to show that it contains no horizontal or vertical ones.

It is well-known that M contains a vertical essential surface unless either F=S^{2} and M has ≤3 multiple fibers, or F=RP^{2} and M has ≤1 multiple fiber. Of these cases the only one of interest is F=S^{2} with 3 multiple fibers; in the remaining cases M is either reducible or has finite fundamental group [7], so cannot contain an essential lamination for well-understood reasons [7].

Proposition 3: There are no vertical essential laminations in a Seifert-fibered space M with base S^{2} and 3 multiple fibers.

Proof: Suppose L is a vertical essential lamination. After splitting along some leaves of L, we may assume that L misses the multiple fibers γ_{1}, γ_{2}, γ_{3} of M, and so can be thought of as a (vertical) lamination in M_{0}=M\N(γ_{1}∪γ_{2}∪γ_{3})=F×S^{1}, where F is a pair of pants S^{2}\3D^{2}. Because L is vertical (and M_{0} has no multiple fibers), λ=p(L)⊆F is a (1-dimensional) lamination in F. Further, because L is essential in M, it is easy to see that λ is incompressible in F; we can think of F⊆M_{0} (by choosing a section of the (trivial) fibering of M_{0}) and λ=L∩F, and then any compressing or end-compressing disk for a leaf of λ will be a compressing or end-compressing disk for L in M, because L is vertical. But an easy Euler-characteristic calculation like those in [2] or [7], using an incompressible train track carrying λ, shows that any incompressible lamination in the interior of a pair of pants must contain a (∂-parallel) compact loop γ. But then p^{−1}γ=T is a vertical torus in L⊆M, which bounds a solid torus (one of the N(γ_{i})), and hence is compressible, a contradiction.

Corollary 4: Every essential lamination L in a Seifert-fibered space M with base S^{2} and 3 multiple fibers contains a horizontal sublamination.

Now it is easy to see that any horizontal lamination L can be \underline{completed} to a transverse foliation of M; L cuts the circle fibers of M into arcs, so M split along L, M|L, is a collection of I-bundles, and these bundles can be foliated by surfaces transverse to the I-fibers, completing L to a foliation of M. Because the I-fibers are contained in the circle fibers of M, this foliation is everywhere transverse to the circle fibers of M.

In [4] and [11] such foliations were studied, and criteria based on the normal Seifert invariants of M were given for determining their existence. More precisely, suppose M is a Seifert-fibered space with normal Seifert invariant M= Σ(0, 0; k, a_{1}/b_{1}, a_{2}/b_{2}, a_{3}/b_{3}) and suppose either

(a) k≠-1, -2, or

(b) k=-1, and (possibly after a permutation of the a_{i}/b_{i}) a_{i}/b_{i} ≥ a′_{i}/b′_{i} > 0, for some rational numbers a′_{i}/b′_{i} satisfying

a′_{1}/b′_{1} = 1 − (a′_{2}/b′_{2} + a′_{3}/(b′_{2}(b′_{3} − 1)))

or (c) k=-2; then after replacing M= Σ(0, 0; −2, a_{1}/b_{1}, a_{2}/b_{2}, a_{3}/b_{3}) with M= Σ(0, 0; −1, (b_{1} − a_{1})/b_{1}, (b_{2} − a_{2})/b_{2}, (b_{3} − a_{3})/b_{3}) (by reversing the orientation of M), apply the criterion (b).

Then M does not admit a transverse foliation.

In particular, M contains no essential laminations. Since it is well known that Seifert-fibered spaces M as above with 1/b_{1} + 1/b_{2} + 1/b_{3} < 1 have universal cover R^{3} (see [16]), we have the following corollary.

Corollary 5: There exist Seifert-fibered spaces M with M̃ = R^{3} which contain no essential laminations.

We now turn our attention to foliations without Reeb components of a Seifertfibered space M.

Proposition 6: If an essential lamination L with no compact leaves in a closed, orientable Seifert-fibered space M contains a horizontal sublamination L_{0}, then L is isotopic to a horizontal lamination.

Proof: Since L_{0} is horizontal, M|L_{0} is a collection of I-bundles foliated by subarcs of the circle fibers of M. Let N be a component of M|L_{0}, an I-bundle over some non-compact surface E, π : N → E, and consider L_{1} =L∩N⊆N. Every leaf L of L_{1} is π_{1}-injective in N (since the composition π_{1}(L)→π_{1}(N)→π_{1}(M) is injective).

Now let \{C_{i}\} be an \underline{exhaustion} of E by compact, connected subsurfaces, i.e., ∪C_{i} =E, and let E_{i} = E \ int(C_{i}). Because the leaves of L_{1} limit on leaves of L_{0} (in fact their limit set is contained in L_{0}), which are horizontal, one can then see that for some i, every leaf of L_{1} is horizontal over E_{i}. So to show L can be made horizontal, it suffices to show that L∩π^{−1}(C_{i}) can be isotoped to be horizontal in N_{i} = π^{−1}(C_{i}), rel π^{−1}∂(C_{i}) =A. Note that N_{i} is a compact handlebody.

We proceed by induction on the genus of N_{i} (see Figure 1). If genus=0, then C_{i} is a disk, and N_{i} = C_{i} × I, with C_{i} × ∂I ⊆ L_{0}, and L_{1} meeting ∂C_{i} × I horizontally. Therefore L_{1} ∩ N_{i} is a collection of taut disks, which can be pulled horizontal.

Figure 1: horizontal laminations

If genus> 0, then choose an essential arc α in C_{i} and look at the disk Δ = π^{−1}α. ∂Δ can be separated into four arcs, two contained in L_{0} and two transverse to L_{1}. By an isotopy of L_{1} we can remove any trivial loops of intersection L_{1} ∩ Δ; then L_{1} meets Δ in compact arcs. None of these arcs can have both endpoints in the same arc of ∂Δ; the disk it cuts off together with a (vertical) half-infinite rectangle going off to infinity in N would give an end-compressing disk for L.

So all of the arcs run from one side of Δ to the other; in particular, these arc can be pulled taut w.r.t. the I-fibering of Δ from N. If we then split open N_{i} along Δ, we get an I-bundle of smaller genus, with L_{1} meeting the ∂I-bundle horizontally, and with horizontal complement in N. By induction, therefore, we can isotope L_{1} (rel L_{0}) to be horizontal in N. Doing this for all of the components of M|L_{0}, we can isotope L to be horizontal in M.

Corollary 7: Every essential foliation with no compact leaves F in a Seifertfibered space M with base S^{2} and 3 multiple fibers is (C^{(0)}-) isotopic to a transverse foliation.

Proof: We can split F along a finite collection of leaves to give an essential lamination L carried by a branched surface. By the corollary above, L contains a horizontal sublamination. By the proposition (since L has no compact leaves) L itself is a horizontal lamination. The I-bundles M|L then are fibered by arcs in the circle fibers; crushing each fiber to a point retrieves F in M, and it is now transverse to the fibers of M.

This result can be thought of as an extension and completion (in the C^{0}-case) of results of Thurston [19] and Levitt [12], Eisenbud-Hirsch-Neumann [4], and Matsumoto [13]. Taken together these papers show that a C^{2}-foliation with no compact leaves, in any (closed) Seifert-fibered space other than the ones in the corollary, can be C^{2}-isotoped to a transverse one. The corollary says that a C^{2}- foliation in M with base S^{2} and 3 multiple fibers can be C^{0}-isotoped to a transverse one; it leaves open the question of whether such a foliation can be C^{2}-isotoped (the argument above cannot be adapted; at the very beginning, the splitting of the foliation to obtain a branched surface destroys the transverse C^{2}-structure).

It is worth noting that an extension in the other direction is not possible; there exist C^{0}-foliations of Seifert-fibered spaces, with no compact leaves, which contain vertical sublaminations. Examples are easily constructed from vertical essential laminations in F×S^{1}, where F is a compact surface of genus greater than or equal to 2.

2. Proof of the Theorem: Preliminaries

Every orientable Seifert-fibered 3-manifold M is the union of a (finite) collection of solid tori (with disjoint interiors) which meet along their boundaries. This view can be obtained from the standard one. Consider the base surface F of the Seifertfibering; it is a compact surface. Choose a triangulation τ of F, in general position with respect to the collection of multiple points of the fibering, so that each 2simplex contains \underline{at} \underline{most} one multiple point. Then every 2-simplex Δ_{i}^{2} has inverse image π^{−1}(Δ_{i}^{2}) = M_{i} a solid torus (it is an (orientable) Seifert-fibered space with base D^{2} and at most one multiple fiber), and these solid tori meet along the inverse images of the 1-simplices of τ, which meet each solid torus in its boundary.

The inverse images of the points of τ^{(0)} =the 0-skeleton of τ form a finite collection S of regular fibers of M in the boundary of the solid tori (they in fact constitute the points where three or more of the solid tori meet). These fibers will be of central importance to us; we will call them the \underline{sentinel} \underline{fibers} of M.

If the lamination L is carried by a branched surface B, then possibly after splitting along a finite number of leaves, we may assume that ∂_{h}N(B)⊆L. Then N(B) split open along L, denoted N(B)|L, is a collection of I-bundles over compact and non-compact surfaces (possibly with boundary). If we split B along the bundles over compact surfaces (i.e., remove their interiors from N(B)), we obtain a (possibly new) branched surface B, carrying L, which now has no such bundles in N(B)|L. Such a branched surface will be called a branched surface having no compact bundles w.r.t. L. Every lamination (up to splitting) is carried by such a branched surface, except when it has a compact isolated leaf.

For a lamination L⊆M carried by a branched surface B having no compact bundles w.r.t. L, and γ a loop transverse to L (i.e., to B), we can define a number ϵ, called a \underline{monogon} \underline{number} for L w.r.t. γ, in terms of the branched surface B, as follows: N(B) meets γ in a collection of vertical fibers, and L∩γ is contained in these subarcs of γ; we let ϵ = 1/2 of the smallest distance (along γ) from one of these subarcs to another. It then follows that any two points of L∩γ which are within ϵ of one another are contained in the same vertical fiber of N(B).

We also need to know something about how L meets typical surfaces S in the M_{i} (that it meets transversely), i.e., (meridian) disks, annuli=π^{−1}(e_{j}^{1}) ⊆ ∂M_{i}, and tori=∂M_{i}. Because L is essential, this is easy to categorize.

λ=L∩S is a 1-dimensional lamination in the surface S. There can be no holonomy around a loop γ of λ which is trivial in S (see [15] for the notion of holonomy), because γ bounds a disk in L and there can be no holonomy around the boundary of a disk. It follows by Reeb stability [17] that the collection λ_{0} of trivial loops and ∂-parallel arcs in λ ⊆S form a sublamination open and closed in λ.

λ \ λ_{0} ⊆S can have no monogons; because L is transverse to S they would give an end-compressing disk for L. An Euler-characteristic argument like that in [2] implies (since χ(S)≥0) that λ \ λ_{0} can be completed to a \underline{foliation} of S.

If S=torus, facts from dynamical systems about foliations of the torus (see, e.g., [9]) imply that there can be only 3 kinds of behavior in λ \ λ_{0}: either

(a) λ \ λ_{0} contains no compact leaves; it then contains an irrational (measured) sublamination, and all other leaves are parallel to this sublamination. In particular, λ\λ_{0} can be isotoped to be transverse to the leaves of any foliation of S by compact loops (since this is true of the completed foliation of S); or

(b) λ\λ_{0} contains compact leaves. The collection of compact leaves then forms a (closed) sublamination of λ \ λ_{0}, and all other leaves lie in the annular regions between the compact leaves, and are of either (1)‘Reeb’ type or (2) ‘Kronecker’ type (see Figures 2ab).

Figure 2: laminations in the 2-torus

If S=disk, then λ \ λ_{0} = ∅ (you can’t foliate a disk), so all of the leaves of L∩S are trivial loops and ∂-parallel arcs. If S=annulus, then by doubling S and λ \ λ_{0} ⊆S, and applying the above, we can conclude that either λ \ λ_{0} consists of parallel essential compact arcs, or λ \ λ_{0} contains essential (vertical) loops, with (possibly half-) Reeb or Kronecker leaves in between them. a. Recognizing good laminations in a solid torus.

Let L be an essential lamination in the 3-manifold M, and let M_{0} be a solid torus in M. By a small isotopy of L we can arrange that L is transverse to M_{0} (this amounts to making L transverse to ∂M_{0}). Then L∩M_{0}=L_{0} is a lamination in M_{0}. This lamination is almost certain not to have π_{1}-injective leaves. However, this lack of π_{1}-injectivity, basically, lives in the boundary ∂L_{0}=L ∩ ∂M_{0}, as the following lemma shows:

Lemma 2.1: Let L, M_{0}, and L_{0} be as above, with M\int(M_{0}) irreducible. If every embedded loop γ_{0} in ∂L_{0} which is null-homotopic in M_{0} bounds a disk in L_{0}, then every leaf of L_{0} is π_{1}-injective in M_{0}.

Proof: Look at the collection τ of loops in ∂L_{0}=L_{0}∩∂M_{0} which are trivial in ∂M_{0}. These loops, by hypothesis are trivial in the leaves of L_{0} which contain them, and therefore bound (a collection T of) boundary-parallel disk leaves in L_{0}. By a Reeb Stability argument, this collection T forms an open and closed set in L_{0}, and so is a sublamination of L_{0}, and so τ is a sublamination of ∂L_{0}. τ therefore consists of a finite number of parallel families of trivial loops in ∂M_{0}, bounding parallel families of ∂-parallel disks in M_{0}. We can then by an isotopy of L (choosing an outermost family of disks (meaning an innermost family of loops) and working in) remove these families of disks from L_{0}. Since T is closed in L_{0}, nothing else is changed, so after the isotopy L_{0} has been altered to L_{0}\T , i.e., ∂L_{0} no longer contains any loops trivial in ∂M_{0}. We will prove the lemma for this altered lamination.

Let γ be a (singular) loop in a leaf L_{0} of L_{0}, which is null-homotopic in M_{0}. Then γ is also null-homotopic in M. Since L_{0} is contained in a leaf L of L, and this leaf is π_{1}-injective in M, it follows then that γ is null-homotopic in L. Let H : D^{2} → L, be a null-homotopy, and make it transverse to ∂M_{0}. Then Γ = H^{−1}(∂M_{0}) is a (finite) collection of circles in a disk D^{2}. Consider a circle γ_{0} of Γ innermost in D^{2}, and consider the leaf l_{0} of the lamination λ_{0} =L_{0}∩∂M_{0}which H maps it into. This leaf is homeomorphic to either S^{1} or R. If it is homeomorphic to R, then γ_{0} is null-homotopic in l_{0}, and so by redefining H on the disk Δ_{0} of D^{2} cut off by γ_{0} so that it maps into l_{0} (and then homotoping H off of l_{0} slightly), we get a new null-homotopy for γ_{0} with fewer circles of intersection in Γ. If l_{0} is a circle, then one of two things will be true. In the most usual case γ_{0} again maps into l_{0} null-homotopically, in which case we proceed as before, finishing the proof by induction. When γ_{0} is essential in l_{0}, we must use a different argument which avoids the induction.

Because γ_{0} is innermost, it bounds a disk Δ_{0} in D^{2} which misses Γ, so the image of Δ_{0} under H misses ∂M_{0}, and hence maps into M_{0} or M_{1} =M\int(M_{0}). So some non-trivial power of l_{0} is null-homotopic in M_{0} or M_{1}.

If γ_{0} is null-homotopic in M_{1}, this means that the torus ∂M_{1} is \underline{compressible} in M_{1}. Because M_{1} is irreducible by hypothesis, it follows that M_{1} is in fact a solid torus. This implies that our original 3-manifold M is a union of two solid tori glued along their boundary, and hence is a lens space. But this is impossible, since a lens space cannot contain an essential lamination (it does not have universal cover R^{3} [7]).

If γ_{0} is null-homotopic in M_{0}, then because π_{1}(M_{0}) is torsion-free (it’s Z), l_{0} is also null-homotopic in M_{0}, and therefore bounds a disk leaf of L_{0}, by hypothesis. By our additional hypothesis, this disk is not ∂-parallel in M_{0}, so it must be \underline{essential} in M_{0}; in particular, L_{0} contains a meridian disk leaf. Now consider the collection μ of meridian loops of ∂L_{0}. By hypothesis, these loops bound a collection M of meridian disk leaves of L_{0}. Again, Reeb Stability implies that this collection M is closed in L_{0}, so μ is closed in ∂L_{0}. But then the leaves of ∂L_{0} not in μ live in the annular regions between loops of μ; they cannot be compact (they would then be trivial or meridional), but they cannot be non-compact, because they would have to limit on μ, giving non-trivial holonomy around a loop which bounds a disk. Therefore, M =L_{0}, so every leaf of L_{0} is a meridional disk, which obviously π_{1}-injects. b. Making intersections taut: solid tori.

Because a Seifert-fibered space can be thought of as a union of solid tori, which meet along their boundaries, it will also be useful to have a general procedure to isotope an essential lamination L so that it meets a vertical solid torus M_{0} in a Seifert-fibered M in a lamination, L ∩ M_{0} = L_{0}, which has π_{1}-injective leaves. We will show later that such a lamination L_{0} in fact has a rather simple structure; this result will then be exploited to give our structure theorem for essential laminations in Seifert-fibered spaces.

Now there is in fact a very easy way to do this: just think of a solid torus M_{0} as a regular neighborhood of its core circle γ_{0}, make γ_{0} transverse to a branched surface carrying L, and then L ∩ M_{0} will be a collection of meridian disks in M_{0}, which certainly has π_{1}-injective leaves.

Unfortunately, this is a far too destructive process for our uses; it loses alot of the information that we will be gathering in the proof of our theorem. Instead we will construct an isotopy which is much more ‘conservative’ (and which, incidentally, allows much more interesting laminations L ∩ M_{0} to be created).

We have seen already that in order to make a lamination meet a (nice) solid torus M_{0} in a π_{1}-injective lamination L_{0}=L∩M_{0}, we need only arrange that any loop of ∂L_{0} which is null homotopic in M_{0} bounds a disk in L_{0}. What we will now do is to describe an isotopy process which, given an essential lamination, will arrange exactly that.

First we deal with trivial loops of λ_{0} = ∂L_{0}. If ∂L_{0} contains loops which are trivial in ∂M_{0}, the collection C of such loops in ∂M_{0} is open and closed in ∂L_{0}, and (by transversality) consists of a finite number of families of parallel loops in ∂L_{0}.

Now take an \underline{outermost} loop γ of an \underline{innermost} family of trivial loops. γ bounds a disk D in ∂M_{0}, and a disk D_{0} in the leaf of L containing it, and they are isotopic, rel γ (because M is irreducible). An (ambient) isotopy of L taking D_{0} to D and a bit beyond has the effect of \underline{removing} the family of loops containing γ from λ_{0} (and possibly more). To be more exact, such an isotopy must be done in stages, since it is not immediate that D_{0}∩D=γ; it could consist of (a finite number of) loops in D_{0} (one then argues from innermost out). Then by induction on the number of parallel families in λ_{0}, we can assume that L∩F contains no trivial loops.

Now if ∂L_{0} still contains loops which are null-homotopic in M_{0}, then these loops must be meridional, i.e., bound disks in M_{0} but not in ∂M_{0}. What we first must establish is that at least one of these loops in ∂L_{0} bounds a disk leaf of L_{0}.

Choose a meridional loop γ of ∂L_{0}. Because L is essential, this (embedded) loop bounds a disk D in L. Consider the intersection D ∩ ∂M_{0} ⊆ ∂L_{0}; this intersection consists of (a finite number of) closed loops. Choose an innermost such loop γ_{0} in D, bounding a disk Δ in D (possibly γ_{0} = γ).

Claim: Δ is contained in M_{0}.

If not, then Δ ⊆ M\int(M_{0}) (because γ_{0} is innermost). γ_{0} cannot be trivial in ∂M_{0} (there are no trivial loops in λ_{0}, so it is essential in ∂M_{0}=∂(M\int(M_{0})). So Δ represents a compressing disk for ∂(M\int(M_{0})). Therefore, M\int(M_{0}) is a solid torus, making M a lens space (the union of two solid tori), a contradiction (a lens space doesn’t have universal cover R^{3}, and M does [7]).

Therefore there is a disk Δ in L_{0} ⊆M_{0} with boundary a loop γ_{0} ⊆ ∂L_{0} ⊆ ∂M_{0}. Consider now the collection M of meridian disk leaves of L_{0}. Reeb Stability implies that this collection is open and closed in M_{0}, as before. Moreover, because M ≠ ∅, the lamination L_{1} = L_{0}\M must have ∂L_{1} consisting of meridional loops; it cannot contain any trivial loops, by construction, and any non-compact leaf of ∂L_{1} would have to limit on a meridional loop, implying non-trivial holonomy around a loop null-homotopic in a leaf of L, a contradiction.

Every leaf of L_{1} has more than one boundary component; if a leaf had only one and were compact, then it would contain a non-separating loop contained in the ball M_{0}\Δ, implying the leaf of L containing it was not π_{1}-injective in M. If the leaf were non-compact, then the limit set of an end (see [14] for a definition) would be a lamination which did not meet ∂M_{0}; it would then be contained in the interior of a ball, implying the existence of an essential lamination in a sphere, which is impossible.

This implies that, although M_{0}|M is a possibly infinite collection of balls, only finitely many of them can contain any leaves of L_{1}. To see this, look at a loop γ having intersection number 1 with each loop of the meridional lamination ∂L_{0}. If there are an infinite number of regions containing leaves of L_{1}, then there are an infinite number of (distinct) arcs of γ|M meeting these leaves. Such a collection of arcs must have their lengths tending to 0. If we look at the top endpoints (in some orientation of γ) of these arcs, we have an infinite sequence of (distinct) points in ∂M, which (because M is closed) must limit on some point of γ ∩ M. This point therefore lies in a meridional disk leaf D of L_{0}; therefore by Reeb Stability, all nearby leaves are also meridian disks. But the top endpoints of the arcs are limiting on this leaf, and the lengths of the arcs are tending to zero (so the bottom endpoints are limiting on D, too), implying that these non-disk leaves pass arbitrarily close to D, a contradiction.

Now look at a component N of M_{0}|M, and the leaves of L_{1} contained in it. N is a ball with two leaves of M in its boundary.

Every loop of ∂L_{1}∩N= λ bounds a disk D in the leaf of L containing it; thinking of L⊆N(B), the set of these disks which are parallel to D in N(B) have boundaries forming an open and closed set in λ. Consequently, they fall into \underline{finitely}-\underline{many} parallel families (in N(B)). For (choosing an arc β running from the top to the bottom of the ball) every point of λ ∩ β has an open neighborhood in β whose points are in loops bounding parallel disks in N(B); because λ ∩ β is compact in β, there is a finite subcover, giving the finite number of families.

Therefore, the loops of ∂L_{0}\∂M fall into a finite number of such parallel families.

It is possible to see a finite sequence of \underline{surgeries} of L in M_{0} which makes every loop in ∂M_{0} bound a disk in M_{0} (see Figure 3). These surgeries represent our ‘template’; what we wish to do now is use this surgery picture to find an isotopy of L which will do the same thing.

Figure 3: surgery in the solid torus

We have a finite number λ_{1}, …, λ_{n} of families of loops in ∂L_{0}\∂M which bound a collection D_{i} of disks in L parallel in N(B). Think of doing these surgeries family by family. Choose a collection D_{i}; note that every disk in the collection meets λ_{i} only in its boundary (a disk cannot be parallel in N(B) to a proper subdisk of itself - it would imply that the disk met an I-fiber of N(B) infinitely often).

Therefore the disk in D_{i} together with \underline{one} of the disks from the surgery form an embedded sphere in M (all of which are parallel to one another); because M is irreducible, they bound (nested) balls (see Figure 4). This ball, together with the ball that the two ‘outermost’ surgery disks bound, forms a ball which can be used to describe an \underline{isotopy} taking the disks in D_{i} to the (other) collection of disks in M_{0}, making the collection of loops λ_{i} bound disks in M_{0}. This isotopy may have removed leaves of M, as well as loops from some of the λ_{i}, but since it can be thought of as a replacement (surgering, and then throwing away the spheres created), it \underline{adds} \underline{nothing} to any intersection L has with any object outside of the

Figure 4: surgeries to isotopies

interior of M_{0}; in particular, it adds no new intersections with ∂M_{0}, and moves none of the disks which it didn’t erase. By a finite application of this process, then, we can arrange that every loop in L∩∂L bounds a disk leaf in M_{0}, completing our isotopy.

3. π_{1}-injective, end-incompressible laminations in a solid torus

We have seen how to isotope an essential lamination L to make it meet a solid torus in a π_{1}-injective lamination L_{0} with no ∂-parallel disk leaves. It is easy to see that L_{0} is end-injective (this is in fact true for any transverse intersection of an essential lamination with a codimension-0 submanifold); any end-compressing disk for L_{0} is an end-compressing disk for L. L_{0} is in general, however, not ∂-injective.

Such a lamination, however, still has a great deal of identifiable structure. Theorem 3.1: A lamination L_{0} as above is either a collection of meridian

disks, or there is a (model) Seifert-fibering of M_{0} so that (after isotopy) L_{0} contains a vertical sublamination L_{1} (whose leaves are annuli, with possibly one Möbius band); all leaves of L_{0}\L_{1} are non-compact, simply-connected, and horizontal.

The proof contains two essential ingredients; first one needs that the ∂-lam-ination ∂L_{0} contains compact loops (which determine the regular fiber of the Seifert-fibering), and then that every such compact loop is in the boundary of a compact \underline{leaf} of L_{0}. The union of these leaves is the vertical sublamination L_{1}. First, though, we need a small catalogue of basic facts, so that we can more easily recognize when these two things are happening. a. Some basic facts about laminations in a solid torus

Fact 1: L_{0}⊆ D^{2} × S^{1} must meet the boundary torus; ∂L_{0}≠ ∅.

This is true more generally; an essential lamination cannot live in the interior of a handlebody. To see this, take a meridian disk D (or, in the general case, a compressing disk for one of the handles), and make L_{0} transverse to it. If L_{0}∩D=λ contains any compact loops, we can isotope L_{0} to remove them. So we can assume λ contains no compact loops. λ is carried by the train track τ = B_{0} ∩ D (where L_{0} is carried by B_{0}), and contains only non-compact leaves; an Euler characteristic calculation implies that, if λ ≠ ∅, λ will contain a monogon, so L_{0} does, which is essential since L_{0} is transverse to D. So λ = ∅, so L_{0} misses D, implying that L_{0} is contained in a \underline{ball}, B (or, inductively, is contained in the interior of a handlebody of lower genus). It is π_{1}-injective there (same argument as before), and contains no spheres (L didn’t), and so all of its leaves are \underline{planes}. Capping this ball off with a ball, we get a lamination in S^{3}, which is essential (because monogons can be pushed off the capping ball), a contradiction.

Fact 2: Every leaf L of L_{0} meets T= ∂M_{0}.

Otherwise, the closure \overline{L} of L would give a π_{1}-injective lamination missing T. Applying the argument above to this sublamination gives the same conclusion, unless \overline{L}∩D contains a monogon; but then Euler-χ arguments will find a monogon for L_{0}∩D inside that one, which gives an end-compressing disk for L_{0}, because L_{0} is transverse to D.

Fact 3: If a leaf L of L_{0} has more than one compact ∂-component, then it is an annulus.

This is standard; the two loops γ_{1}, γ_{2} are parallel, otherwise one of them is trivial (making L a boundary-parallel disk). We can assume that they are oriented coherently, so that they represent the same free homotopy class in the boundary torus. Draw an arc α in the leaf joining the two components; then γ_{1} ∗ α ∗ \overline{γ_{2}} ∗ \overline{α} is (almost) an embedded loop in L null-homotopic in D^{2} × S^{1}, hence bounds a disk in L. It follows that L is a disk with two arcs in its boundary identified, i.e. an annulus.

Fact 4: An annulus A with ∂A vertical (in a model fibering of a solid torus) is vertical.

This is also standard; from the previous argument it is easy to see that A is ∂-parallel, and so isotopic to an (of necessity vertical) annulus in the boundary of the solid torus. Pushing it back into the solid torus slightly, we see that A is isotopic to a (properly embedded) vertical annulus.

Fact 5: A non-orientable surface L with π_{1}(L)=Z and a compact ∂-component γ is a Möbius band.

Proof: Let p:L_{0} →L be the orientable double cover of L. γ is orientationpreserving in L, so p^{−1}(γ) = γ_{1} ∪ γ_{2}, disjoint simple loops mapping homeomorphically down to γ under p. Being simple loops, they do not represent a proper power in π_{1}(L_{0}) =Z [2]. So both represent the \underline{generator} (up to reorienting the curves), hence are freely-homotopic. By [5], they are then \underline{isotopic}, and cobound an annulus A in L_{0}. Since γ_{1} and γ_{2} are ∂-components, this implies that L_{0} itself is an annulus, hence compact.

So p(L_{0})=L is compact; by the classification of surfaces, it is therefore a Möbius band.

Fact 6: A Möbius band L with ∂L vertical (in a model fibering of a solid torus) is vertical.

This follows from a result of [18], which says that one-sided incompressible surfaces in a solid torus with a single boundary curve are determined up to isotopy by the slope of that curve (π_{1}-injective surfaces are incompressible). With this result in hand it remains then only to show that ∂L represents 2x the generator of π_{1}(solid torus), because a vertical π_{1}-injective Möbius band with that boundary slope can easily be constructed.

But this in turn follows readily from some π_{1} considerations; let M=solid torus, and consider M_{0}=M\int(N(L)). It π_{1}-injects into M (since L is π_{1}-injective), is irreducible (since M is) and has boundary =(∂M\∂N(L))∪(∂N(L)∩int(M)= A_{1}∪A_{2} =annulus∪annulus=torus. So M_{0} is a solid torus, and M=M_{0}∪_{A_{2}}N(L).

Claim: The core of A_{2} represents a generator of π_{1}(M_{0}). For if the map π_{1}(A_{2})→π_{1}(M_{0}) sends 1 to n, then by Van Kampen’s theorem π_{1}(M) is equal to π_{1}(M_{0}) ∗_{π_{1}(A_{2})} π_{1}(N(L)), and since the core of A_{2} represents 2 in π_{1}(N(L)), this implies that π_{1}(M)= Z =(a,b : a^{2} = b^{n}) = G. But the subgroup generated by a^{2} is normal (a^{2} commutes with both a and b), and G modulo this subgroup is (a,b : a^{2} = 1, b^{n} = 1) = Z_{2} ∗ Z_{n}. But every quotient group of π_{1}(M)=Z is cyclic, implying n=1.

In particular, M_{0} deformation retracts to A_{2}, so M deformation retracts to the regular neighborhood N(L) of L. Since ∂L represents 2×generator in π_{1}(N(L)) (it’s parallel to the core of A_{2}), it therefore represents 2×generator in π_{1}(M).

Note also that there cannot be two disjoint such Möbius bands in a solid torus M, because any other L′ would be contained in the solid torus complement M_{0} of the other. The boundary of L′ is parallel to ∂L in M, but ∂L now \underline{generates} the fundamental group of M_{0} (this is easy to see when L is vertical), and so ∂L′ cannot in fact bound a Möbius band in M_{0} (the generator can’t be divisible by 2). b. Proof of the theorem

Lemma 3.2: Any π_{1}-injective, end-incompressible lamination L_{0} in a solid torus M_{0} contains a compact ∂-leaf.

Proof: Suppose not; we know from Fact 1 above that ∂L_{0} is non-empty. From the catalogue of ∂-laminations in section 2, ∂L_{0} contains an irrational lamination, and so can be isotoped so that it is everywhere transverse to the meridional foliation.

Pick a meridian disk D in M_{0}. By an isotopy of L_{0} (supported away from ∂L_{0}) we can make L_{0} transverse to D. By the usual argument, L_{0}∩D= λ_{0} consists of circles and arcs, and by an isotopy of L_{0} we can remove the circles of intersection, using the π_{1}-injectivity of L_{0}. Pick an outermost arc α of this intersection. It cuts D into two disks, one of which D_{0} meets L_{0} only in an arc of its boundary. The other arc of ∂D_{0} lies in ∂M_{i}|∂L_{0}, and and splits the component containing it into two half-infinite rectangles. Pick one rectangle R, then it is easy to see that D_{0}∪R is an end-compressing disk for L_{0}, because L_{0} is transverse to ∂M_{i}, contradicting the end-incompressibility of L_{0}.

Proposition 3.3: Every compact loop in ∂L_{0} is contained in a compact leaf of L_{0}.

Proof: We will proceed by exhaustion. For a different proof, arguing by contradiction, see [1].

If the loop is null-homotopic in M_{0}, then it bounds a meridional disk leaf of L_{0}. Therefore, we can assume that it is not meridional. If any leaf L containing a compact ∂-loop is non-orientable, then by Fact 5, it is a Möbius band, hence compact. If we split L_{0} along L, and then split M_{0} along L, we get a new lamination in a new solid torus, with all the same essentiality properties that the originals had. But this lamination now has no Möbius band leaves, by Fact 5 (L_{0} had at most one). In other words, after possibly splitting L_{0} and M_{0} , we can assume that L_{0} contains no non-orientable leaf with a compact ∂-component.

Now let Δ be a meridonal disk of M_{0}, which we may assume meets ∂L_{0} transversely, and meets each compact loop of L_{0} tautly. By an isotopy of L_{0} supported away from ∂M_{0}, we may make L_{0} transverse to Δ, and by a further isotopy we can remove any loops of L_{0} ∩Δ=λ. λ then consists of compact arcs, which fall into a finite number of parallel families.

It is easy to see by inspection that the collection of compact loops C of ∂L_{0} is closed in ∂L_{0} . But also the collection C_{0} of loops in C which are in the boundary of a compact leaf of L_{0} is open and closed in C; the leaf L must be an annulus, because π_{1}(L) injects in π_{1}(M_{0} )=Z. Call the boundary components of L γ_{1} and γ_{2}. It is easy to see that an arc α of λ emanating from γ_{1} has its other endpoint in γ_{2} (otherwise L contains an orientation-reversing loop), and L split along α is a disk, with boundary δ_{1} ∪ α ∪ δ_{2} ∪ ᾱ. This disk then lifts to a disk in any nearby leaf in the normal fence over L; in particular, its boundary lifts to a closed loop. This implies that if there is a compact loop γ lying close enough to γ_{1} (say), then δ_{1} lifts to a closed loop in the leaf containing γ, so α and ᾱ are mapped onto one another, so δ_{2} also lifts to a closed loop. Therefore the leaf of L_{0} containing γ has two compact ∂-components; by Fact 3, it is then an annulus, hence compact. This shows that the set of loops bounding compact leaves is open in C . Showing C_{0} is closed is easier; the set of compact leaves of a lamination L_{0} is always closed [15, Lemma 1.2], so its set of ∂-components is also closed .

Suppose now that C\C_{0} is not empty. It then follows from the above that there is an arc of λ emanating from an element γ of C\C_{0} whose other endpoint is in a non-compact leaf. Because C\C_{0} is closed, we can find an outermost such arc α (i.e., one cutting off a subdisk Δ_{0} of Δ which misses C\L_{0} \(γ ∩ α)). γ is isolated in C on the Δ_{0}-side, because C_{0} is closed; and the arcs of λ joining the loops of C_{0} to one another one the Δ_{0}-side fall into a finite number of parallel families, so there are a finite number of innermost such arcs (i.e., closest to α), contained in a finite number of annulus leaves of L_{0}. If we remove small neighborhoods of these annuli, we split M_{0}into a finite number of solid tori (with L_{0} meeting each solid torus in a π_{1}-injective and end-incompressible lamination), and in the component containing γ, (what is left of) Δ_{0} no longer meets any other compact loops.

Figure 5: Finding the other compact loop

Now look at the arc α and the leaf of ∂L_{0} it joins to γ. We must have a situation like one of those pictured in Figure 5. If the other endpoint is in a Kronecker leaf (Figure 5a), or in the ‘inner half’ of a Reeb leaf (Figure 5b), then it is easy to find an end-compressing disk for L_{0}, a contradiction. If it is on the outer half of a Reeb leaf ℓ_{0}, then we will iterate our chase, to find a contradiction.

Notice first that all of the arcs of λ in Δ_{0} must be parallel, otherwise we can find an end-compressing disk (Figure 6a). If we follow ℓ_{0} around, it will return to Δ_{0} again at a point x_{0} after travelling at net 0-times around ∂M_{0}vertically, and there is an arc α_{0} joining ℓ_{0} to the outer half of some other Reeb leaf ℓ_{1} (otherwise we can find an end-compressing disk (Figure 6b,c)). We can continue this construction, finding a sequence of arcs α_{i} of λ, which have a limit α_{∞} in λ. But it is easy to see (by lifting the picture to the universal cover of M_{0} (Figure 6d)) that the endpoints of α_{∞} are in the same leaf of ∂L_{0} , and split off an arc β which wraps around ∂M_{0} a net 0-times longitudinally. Therefore, α_{∞}∪β is a null-homotopic simple loop in a leaf L of L_{0}, so bounds a disk in L. But it is easy to see that our construction (the α_{i}∪ the arcs of the ℓ_{i}) string together to form a half-line spiralling in on α_{∞} ∪ β, implying non-trivial holonomy around the boundary of a disk, which is impossible.

Figure 6: various cases

Consequently, C_{0}=C , i.e., every compact loop of ∂L_{0} is contained in a compact leaf of L_{0}.

To complete the theorem, consider our ‘essential’ lamination L_{0}. By Lemma 3.2, ∂L_{0} contains a compact loop γ. Choose the Seifert-fibering of the solid torus M_{0} whose regular fiber in ∂M_{0} is isotopic to γ. Since every compact loop of ∂L_{0} is parallel to γ, we can, after an isotopy of of L_{0} supported near ∂M_{0}, assume that every compact loop of ∂L_{0} is a fiber of M. Now by the proposition every leaf of L_{0} which contains a compact ∂-loop is compact. They have vertical boundaries, and so by the facts above, each can be isotoped to be vertical in M. They can in fact be so isotoped simultaneously; the leaves fall into a finite collection of parallel families, and each family can be isotoped in turn, from the innermost out; think of isotoping the innermost leaf of the family to the boundary and then back in slightly; this is an ambient isotopy which makes the entire family vertical. Subsequent isotopies will be supported away from the ones which have already been straightened. This isotopy gives the vertical sublamination L_{1} of the theorem.

Now consider the leaves of L_{0} which are not in L_{1}. These leaves all have noncompact boundary (which we assume runs transverse to the foliation of ∂M_{0} by fiber circles), and so limit on leaves of L_{1}. From holonomy considerations, this limiting takes place in a very simple way; see Figure 7a.

Figure 7: making the other leaves horizontal

Thus in each component M_{1} of M_{0}|L_{1}, it is possible to arrange the leaves of L_{0}, by an isotopy supported away from ∂M_{0}, to meet a saturated neighborhood of the boundary of the component as in Figure 7b. It is easy then to see that M_{0} =L_{0}∩(M_{1}\int(N(L_{1})) is π_{1}-injective in M_{2} = M_{1}\int(N(L_{1})) (we have just removed half-infinite rectangular ‘tails’ from the leaves of L_{0}, and the solid torus M_{2} π_{1}-injects into M_{0}), and end-incompressible (a monogon for M_{0} is a monogon for L_{0}, since L_{0} is transverse to ∂M_{1}). Also, its ∂-lamination is transverse to the circle fibering of ∂M_{1} induced from M_{0}, so it has no trivial leaves. Consequently, by the proof above, it either consists of meridian disks, or it contains an annulus or Möbius band leaf L. If the latter occurs, then L has boundary transverse to the vertical fibering of ∂M_{2} induced from M_{0}, and so meets every fiber of ∂M_{2}. In particular, since ∂M_{2}∩∂M_{0}≠ ∅, ∂L meets ∂M_{0}.

Now, there is an arc α in L which together with an arc δ in ∂M_{2} bounds a disk D in M_{2} (if L is an annulus this is because it is ∂-parallel; if L is a Möbius band, look at the boundary of a regular neighborhood of L; it is a ∂-parallel annulus, which supports such a disk, and then project back). By an isotopy of D (leaving α in L and δ in ∂M_{2}) we can arrange that δ is contained in an annulus A of ∂M_{2}∩∂M_{0}, and so we can make it lie in a circle fiber of this annulus. We may also assume that D is transverse to L_{0}, meeting it in a collection of compact arcs.

Now consider in what leaves of ∂L_{0}⊆∂M_{0} the endpoints of δ are lying in. None of the circle loops of ∂L_{0} meet A, so these points are contained in (distinct; these leaves run transverse to the circle fibering of ∂M_{0}) non-compact leaves of ∂L_{0}. Therefore (see Figure 7c) δ together with a pair of half-infinite arcs in ∂L_{0} cut off a half-infinite rectangle in ∂M_{0}; this together with the disk D form a monogon for L_{0}; embedded in this is a end-compressing disk for M_{0}|L_{0} (essential because its ‘tail’ is in ∂M_{0}, which is transverse to L_{0}).

This gives us a contradiction, so M_{0} consists of meridian disks with boundary transverse to the circle fibering; an isotopy rel boundary makes this a collection of \underline{horizontal} disks. Doing this for all of the components of M_{0}\L_{1} gives an isotopy of L_{0} which makes every leaf of L_{0}\L_{1} horizontal, in our chosen Seifert-fibering of M_{0}.

Since the lamination in the saturated neighborhood is also clearly horizontal, this implies that the leaves of L_{0}\L_{1} can be isotoped, rel L_{1}, to be horizontal in M_{0}. By gluing back, we have then arranged that

(*) the leaves in the complement of the vertical sublamination of L_{0} found above can be isotoped (rel the vertical sublamination) so that they are \underline{horizontal}.

Since these leaves are just disks with half-infinite rectangles glued to them, they are also simply-connected. This completes our proof.

4. A special case: ∂M≠ ∅ and L∩∂M=∅

In this section we give a proof of the theorem in the case stated in the title. In the next section we give the general proof; this preliminary result will need to be used in that proof.

In this case in fact only one of the stated conclusions can occur:

Theorem 4.1: If L is an essential lamination in the compact, orientable Seifert-fibered space M, with ∂M≠ ∅ and L∩∂M=∅, then up to isotopy, L contains a vertical sublamination.

The idea of the proof (as in the general case) is to split M up into a collection of solid tori M_{i}, and then isotope L so that it meets each solid torus in a π_{1}-injective lamination L_{i}⊆M_{i} with no ∂-parallel disk leaves. In each solid torus it is therefore is an ‘essential’ lamination, and so our structure theorem of the previous section tells us what each looks like.

The proof here involves a somewhat different decomposition of M into solid tori than the one described in section 2. The base of the Seifert-fibering is a compact surface with boundary. It is well-known that such a surface can be split along proper arcs to give a disk; splitting along additional arcs, as necessary, we can split the surface into a collection of disks, each containing at most one multiple point(=image of a multiple fiber) of the Seifert-fibering. Then as before, the inverse images of these disks are solid tori; the difference here is that each of the solid tori of the decomposition meets ∂M in one or more annuli, and L does not meet these annuli (because it misses the boundary). Let A= the union of the inverse images of the splitting arcs; it is a finite union of annuli. By an isotopy of L we can make L transverse to A, and by the usual methods, we can remove any trivial circles of intersection from L ∩A=λ. λ is then incompressible in A, so any compact loop in λ is parallel to ∂A; by an isotopy of L we can make such loops vertical in A. Set L_{i}=L∩M_{i}.

Each L_{i} is π_{1}-injective in M_{i}, by Lemma 2.1, sine M\int(M_{i}) is Seifert-fibered with non-empty boundary, hence is irreducible (see, for example, [Ha]), and ∂L_{i} contains no meridonal loops (they would have to meet M_{i}∩∂M).

Therefore each L_{i} is an ‘essential’ lamination in the solid torus M_{i} which contains it. Now if L∩A=∅, then L_{i}∩∂M_{i} =∅ for all i. But then by Fact 1, L_{i}=∅ for all i, so L=∅. If L∩A≠ ∅, then for some i, ∂L_{i} contains vertical loops, and so some L_{i} contains a vertical sublamination. Now consider all of the vertical sublaminations in all of the L_{i}. They each meet ∂M_{i} in the (\underline{entire}) collection of vertical loops of ∂L_{i}, and so they glue together across the A_{j} to give a lamination in M, which is the vertical sublamination of L required by the theorem.

5. Proof in the general case

For convenience we will assume that M is closed; ∂M = ∅. The proof in the bounded case is entirely similar, although some of the isotopies must be constructed slightly differently.

We think of M as a union of (embedded) solid tori M_{i} = π^{−1}(Δ_{i}^{2}), i=1,...,r which meet one another in the annuli A_{j} in their boundaries. We set S=π^{−1}(F^{(0)}), the collection of sentinel fibers of the decomposition of M into solid tori. a. The isotopy process

The strategy of the proof is to set up an isotopy process, i.e., a sequence of isotopies I_{j} which will, one by one, isotope L to meet the i^{th} solid torus (j ≡ i(mod r)) only in horizontal disks, while at the same time controlling the intersection of I_{j}(L) with the sentinel fibers S. What we will see is that if at any stage of the process we are \underline{unable} to continue the isotopy process, we can use this information to find a vertical sublamination of L (after possibly splitting one of the leaves of L). Otherwise, we are able to continue the isotopy process indefinitely, and then we will be able to see that (larger and larger pieces of) L begin to \underline{limit} on (larger and larger pieces of) some lamination L_{0}, which, by its construction, is horizontal; as it turns out, L_{0} is in fact isotopic to a sublamination of L.

We have seen how to isotope a lamination so that it meets a (vertical) solid torus M_{i} in a lamination L_{i} with π_{1}-injective leaves (M\M_{i} is irreducible because it is Seifert-fibered with non-empty boundary (see [7])). Consider now how this isotopy affects L∩S, the intersection of L with the sentinel fibers S. This isotopy was achieved by doing surgery on L in the solid torus, and then throwing away any 2-spheres which are created. In terms of the sentinel fibers, this means that L∩S (after surgery) is \underline{contained} in L∩S (from before the surgery). This is what we mean by controlling the isotopies. We will call an isotopy which has this control \underline{conservative}.

Now after this (preliminary) isotopy, we have arranged that L∩M_{i}=L_{i} is π_{1}- injective in M_{i}. It is also end-incompressible, and contains no spheres or ∂-parallel disks (by construction), so it is ‘essential’. By the Theorem it then either consists of meridional disks, or contains a vertical sublamination w.r.t. some Seifert-fibering of M_{i} (not necessarily the one that it inherits from M).

Let us consider first the case that L_{i} consists of \underline{meridional} \underline{disks}. We wish to show that, by an isotopy of L which controls the intersection of L with S, we can make L meet M_{i} in a collection of \underline{taut} disks (meaning each disk meets each annulus of ∂M_{i}|S in essential arcs). To do this, consider λ=∂L_{i}⊆∂M_{i}, and its intersection with each annulus complement A_{j} of S in ∂M_{i}. This intersection consists of a finite number of parallel families of essential and trivial arcs in A_{j}.

Note that because λ is (assumed to be) carried by a train track τ = B∩∂M_{i}, there is an upper bound on the number of times a loop in λ can meet S (the loops fall into a finite number of loops parallel \underline{in} \underline{τ}; each loop in a family meets S the same number of times). Now, any collection of trivial arcs in an A_{j} can be removed by an isotopy of L supported in a neighborhood of the disk which the innermost arc of the family splits off from A_{j}. This reduces the number of times the loops of λ containing these arcs meets S. By an inductive use of this process, eventually every loop of λ must be taut. Note that this isotopy never adds points to L∩S, only removes them.

If, on the other hand, ∂L_{i} contains non-meridional loops, then L_{i} contains an annulus or Möbius band leaf. Look at the collection C of compact leaves of L_{i}; it is a (closed) sublamination of L_{i}. C∩∂M_{i} consists of a collection of parallel loops in ∂M_{i}; by a process similar to that just described, we can make these loops meet S tautly.

There are now two cases to consider: Case 1: ∂C⊆∂M_{i} runs parallel to S (i.e., C∩S= ∅), or C contains a Möbius band leaf. Then (see Fact 6 of section 3 for the Möbius band case) we can isotope C (in so doing isotoping L) so that C contains a circle fiber of M. Therefore, possibly after splitting L along the leaf containing the fiber, we may assume that L \underline{misses} a circle fiber γ of M, and therefore misses a small (fibered) neighborhood of γ, and so we can consider L⊆M\int(N(γ))=M_{0}. Now, thought of in M_{0}, L is still essential: π_{1}-injectivity of leaves follows from the injectivity of the composition π_{1}(L)→π_{1}(M_{0})→π_{1}(M), ∂-injectivity is vacuous (L misses ∂M_{0}), irreducibility of M_{0}|L follows because γ is essential in M|L, and end-incompressibility follows easily (because M_{0} is a codimension-0 submanifold of M).

Therefore by Theorem 4.1, L contains a vertical sublamination L_{0} in M_{0}, and hence contains a vertical sublamination in M. We had to split L open to find this sublamination; we need to show that L also contains a vertical sublamination \underline{before} splitting.

Consider the component N of M|L created by the splitting. It is a (possibly non-compact) I-bundle, and it has one or two boundary components which are leaves of L. It is easy to see that they are contained in the vertical sublamination of L (they are the first leaves that the vertical annuli would meet travelling away from ∂M_{0}, so the leaves contain vertical loops). Therefore N is saturated by circle fibers, so it is a Seifert-fibered I-bundle, with vertical ∂I-subbundle. It is easy to see that such a bundle has a vertical section L (since N is orientable, there are only 4 cases); but by collapsing N onto L, we reverse the splitting, retrieving our original lamination L with the vertical sublamination (L_{0} \∂N)∪L. Case 2: ∂C⊆∂M_{i} meets S, and C does not contain a Möbius band leaf. Then every leaf of C is a ∂-parallel annulus, and the loops of ∂C meet S non-trivially and tautly. The leaves of C again fall into a finite number of parallel families in M_{i}. Choose an innermost leaf L of an outermost family in C, and choose a ∂- compressing disk Δ for L, ∂Δ = α ∪ δ, with α ⊆L and δ contained in a loop of S. By the usual methods we can assume that L meets Δ transversely in a collection of arcs.

Figure 8: killing annuli

Then by doing a ∂-surgery on L using (a disk slightly larger than) Δ, we can split the annulus leaves in the same family as L into a collection of trivial disks (see Figure 8), which we can then isotope away using our previous methods. Note that this creates no \underline{new} families of annuli or Möbius bands; the effect of surgery on leaves near L is to cut off half-infinite rectangular tails from simply-connected leaves (each parallel family is open and closed in L_{i}), and cut them into trivial disks. So simply-connected leaves remain simply-connected. It also adds no new points of intersection to S.

After a finite number of such surgeries, we can kill off all of the annulus leaves of C; L∩M_{i} then must consist of meridional disks (because it is still π_{1}-injective and end-incompressible), which we treat as before.

The construction above forms the core of our isotopy process. Starting with L, either it contains a vertical sublamination or there is a conservative isotopy I_{1} so that I_{1}(L) meets M_{1} in a collection of taut disks. We now continue cyclically through our list of solid tori M_{1}, …, M_{r}, so that at stage j, we are adding to the previous isotopies, trying to make I_{j}(L) meet M_{i} in taut disks, where j ≡ i(mod r). By the above construction, either this isotopy can be built, or L contains a vertical sublamination.

If we therefore assume that L does not contain a vertical sublamination, then are able to construct an infinite sequence of isotopies I_{j} with the property that I_{j}(L) meets M_{i} in a collection of taut disks. If at any stage I_{j}(L) meets \underline{all} of the solid tori M_{1}, …, M_{r} in taut disks, then as in section 4 these disks can be ‘straightened’ out, completing the isotopy of L to a horizontal lamination. Thus L is itself a horizontal lamination.

Because each of the above two situations justify the theorem, we (can and) will assume from now on that neither of them hold; i.e. L does not contain a vertical sublamination, and is not itself isotopic to a horizontal lamination. We will therefore think of these isotopies as defining an \underline{infinite} isotopy process; we find ourselves forever pushing L around, and are ‘not quite’ able to make it all horizontal.

We will need a little more notation to continue. We have defined I_{j} as the composition of the first j isotopies of L, making L meet the solid tori M_{i} cyclically in taut disks. We will let I_{(j)} represent any \underline{stage} of the isotopy between I_{j−1} and I_{j}. We will also let I_{j,k} denote the composition I_{k} ∘ I_{j}^{−1} (i.e., the composition of the isotopies built \underline{between} the j^{th} and the k^{th} stages), so that I_{j,k} ∘ I_{j} = I_{k}. b. Finding stable arcs

Now we have an isotopy process, and we assume that it continues indefinitely. This means that at no stage does it succeed in pulling L horizontal, but for all j, the isotopy I_{j} succeeds in making L meet M_{i} in taut disks, where j≡i(mod r). Now for each j, the points I_{j}(L)∩S form a (closed) collection of points in S, the set of sentinel fibers of our Seifert-fibering. By the construction of the isotopy I_{j}, these points were \underline{never} \underline{moved} by any of the isotopies that went into the construction of I_{j}, i.e., they are \underline{stable} under these isotopies. In particular, for j≤k, I_{j}(L) ∩ S ⊇ I_{k}(L) ∩ S, i.e., these sets are \underline{nested}. They are also non-empty; if I_{j}(L) ∩ S = ∅, then L misses a fiber of M (i.e., any of those in S), and so, by Theorem 4.1, contains a vertical sublamination. But we have assumed L contains no such sublamination.

So we have a nested sequence of closed, non-empty subsets of the compact set S; their intersection ∩(I_{j}(L) ∩ S) = P_{0} is therefore non-empty. P_{0} in fact meets every component of S (for otherwise I_{j}(L) must have missed that component for some j, allowing us to find a vertical sublamination again). By construction, P_{0} consists of all of the points of L∩S which are never moved by any of the isotopies in our isotopy process, i.e., they represent the \underline{stable} \underline{points} of our isotopy process. What we will now show is that, as we watch the isotopies progress, these points become ‘islands of stability’ for the process; a stable (horizontal) lamination starts to ‘grow’ out of them.

Now, consider a 1-simplex e_{i} ∈ B^{(1)} and the annulus A_{i}=π^{−1}(e_{i}), ∂A_{i} ⊆S. Pick points x, x′ of P_{0}, one in each component of ∂A_{i}. What we wish to look at now are the \underline{arcs} of I_{j}(L)∩A_{i} containing x, x′ (call them, respectively, α, α′), and how they change under further isotopies. Because for each arc one of its endpoints is anchored down (x, x′ are stable), the only way these arcs can change is by ‘boundary compressions’ (see Figure 9). Our intent is to show that for some k≥j, each of these arcs I_{j,k}(α), I_{j,k}(α′) has \underline{both} of its endpoints in P_{0}. This arc would therefore be stable, i.e., I_{j,k}(α) (say) would be fixed under all further isotopies.

We proceed as follows. Given α, α′⊆ A_{i}, there exists an arc ω_{j} (for ‘winding number’) joining x to x′ and not meeting α, α′ except at their endpoints. This is because A_{i} split on α, α′ has 2 or 3 (if α, α′ are both trivial arcs in A_{i}) components, at least one of which contains both x and x′.

Lemma 5.1: If at some further stage I_{j′} of the isotopy process, one of the arcs emanating from x, x′ has non-zero winding number wrt. ω_{j} (meaning it is not isotopic rel endpoints to an arc meeting ω_{j} only at its endpoints), then at some stage of the isotopy process between I_{j} and I_{j′}, one of the arcs emanating from x or x′ was trivial, i.e., ∂-parallel in A_{i}.

Proof: Since α, α′, have zero winding number wrt. ω_{j}, and change only by ∂-compressions, there is a first ∂-compression after which one of the arcs has nonzero winding number. We claim that, at the time of this compression, one of the arcs is trivial.

For suppose not; note that since the stable ends of the arcs α, α′ are on opposite sides of A_{i}, the ∂-compression leaves one of the arcs, say α′, fixed. Since this is the first ∂-compression where the winding number changes, we have that the winding number of α′ is zero. Now if α′ is not trivial, then its other endpoint is on the same side as x (see Figure 9). Since α is not trivial, its other endpoint is on the x′-side of A_{i}, so the ∂-compression is taking place on that side. But because after the compression the arc emanating from x cannot meet α′ (because after the compression, L still meets A_{i} in a lamination, which can’t have leaves intersecting), which hasn’t been moved, only one of two things can have occurred: either (1) the new arc α_{new} is a trivial arc, in which case it is isotopic rel endpoints to an arc in ∂A_{i}, with x as an endpoint, so has zero winding number, or (2) α_{new} is an essential arc (which lies in A_{i}|α′, which is a disk), and so is isotopic rel x to α, by an boundary-preserving isotopy which does not meet x′; and therefore α_{new} also has winding number zero w.r.t. ω_{j}, since it must then have the same winding number that α has. Both of these situations, however, violate our hypothesis, giving the necessary contradiction.

Figure 9: winding numbers

In other words, if one of the arcs moves alot, then one of the arcs had to be trivial (at some time).

It then follows, by an inductive use of the lemma (since the arcs emanating from x, x′ in I_{nr+i}(L) ∩ A_{i} are non-trivial (they are contained in the boundary of taut disks in M_{i})), that one of two things will happen:

(1) one of the points x, x′, is the endpoint of a trivial arc in I_{(k)}(L)∩A_{i} infinitely often (i.e., for arbitrarily large values of k),

or

(2) eventually, neither point is ever contained in a trivial arc, and there exists j, and ω_{j} so that for k≥j, the arcs of I_{(k)}(L) ∩ A_{i} emanating for x, x′, never have non-zero winding number wrt. ω_{j}.

What we now show is that the first of these possibilities must necessarily lead to a contradiction, while the second leads to the eventual stability of the arcs emanating from x, x′ (in order to avoid a contradiction similar to the one encountered in the first case).

First case: x (say) is contained in a trivial arc α_{k} of I_{(k)}(L)∩A_{i} for arbitrarily large values of k.

What we will do now is watch the proliferation of the intersections of these trivial arcs with γ, a loop in A_{i} lying parallel to the component of ∂A_{i} containing x. Recall that our isotopies are conservative, so that the only points of the intersection of L with the sentinel fibers which move are those which disappear. Now the effect of a ∂-compression on the arc α_{k} is to cut off a short arc near its non-stable end, and splice it to another arc by an arc running in the annulus between γ and the loop of ∂A_{i} it runs next to. We may assume that such compressions do not remove points of intersection of α_{k} with γ. We can therefore assume that the points of α_{k}∩γ are \underline{fixed} under all further isotopies, i.e., α_{k}∩γ ⊆ α_{k′} ∩ γ whenever k′ ≥ k. Since the arc containing x periodically becomes essential (every time L ∩ M_{i} is pulled taut), it follows that this inclusion is usually proper, i.e., these trivial arcs continue to pick up more and more points of intersection with the neighbor loop as k gets larger and larger. It is the fact that these points must be piling up on one another in the neighbor loop that is going to give us our contradiction.

First we need some notation. Let ω be an essential arc in A_{i} whose endpoints in ∂A_{i} are not in L (in fact, since L∩∂A_{i} is closed, we may assume ϵ -neighborhoods (in ∂A_{i}) of the endpoints do not meet L, for sufficiently small ϵ). Orient ω with tail z on the component S of ∂A_{i} containing x. z and x separate S into two arcs, called the \underline{left} \underline{side} and the \underline{right} \underline{side} of ω. Orient the α_{k} with tail at x, and orient the neighbor loop γ. Using these orientations, we can assign local orientations to the points of α_{k}∩γ, and winding numbers to the arcs of α_{k} between x and a point of α_{k}∩γ. Note that because the isotopies are constant near the points of α_{k}∩γ, and (α_{k}∩γ) ⊆ (α_{k+1} ∩ γ), it follows that the local orientation assigned to a point is the same as the one assigned when thought of as living in every further arc α_{k}. Also, the winding numbers associated to a subarc of α_{k} is actually a function of its \underline{endpoint} t ∈α_{k}∩γ, because the arcs in α_{k} and in α_{k+1} between x and t are \underline{identical}.

Call the other endpoint of α_{k} (i.e. the one which isn’t x) x_{k}, and the intersection point of α_{k} with γ, which is adjacent to x_{k} along α_{k}, y_{k} (see Figure 10).

Figure 10: stabilization: first case

Now, the winding number of the arc β_{i} of α_{i} between x and y_{i} is always either -1, 0, or 1. This is because β_{i} differs from α_{i} only in the short arc between x_{i} and y_{i} (which doesn’t meet ω), and α_{i} has one of the above mentioned winding numbers because, being trivial, it is homotopic (in fact isotopic) rel endpoints to a subarc of S∈ ∂A_{i}, which meets the winding arc ω at most once. Therefore, the winding numbers assigned to the points y_{i} in α_{k} are either -1, 0, or 1.

Now lift the α_{i} to the universal cover π:R×I → A_{i} of A_{i}, sending x to (0,0)=x̃, and let ỹ_{i} be the resulting lifts of the points y_{i}, obtained by lifting the α_{i}. Let γ̃=π^{−1}(γ), so y_{i}∈γ̃. Because we could calculate the winding number of α_{i} w.r.t. ω by lifting α_{i} to α̃_{i} and count the winding number w.r.t. \underline{all} of the lifts of ω in R×I, and this amounts basically to calculating the integer part of the first coordinate of ỹ_{i}∈R×I, it follows that the points ỹ_{i} must lie in a compact piece [-2,2]×I of R×I.

So these points ỹ_{i} must be piling up on one another. In particular, for any ϵ > 0, there exist points ỹ_{i}, ỹ_{j}, j < i, which are within ϵ of one another along γ̃. β̃_{i} \ β̃_{j} is the arc of α̃_{i} between ỹ_{j} and ỹ_{i}, which together with the arc of γ̃ between these two points, forms a (null-homotopic; R×I is contractible) loop in Ã_{i}. This loop projects down in A_{i} to a loop consisting of the arc β=β_{i}\β_{j} in α_{i}, together with an arc of length < ϵ in γ, and this loop is null-homotopic.

Now, consider this short arc δ between y_{i} and y_{j} in γ. If β ∩ δ ⊆ ∂δ, then β ∪ δ is an embedded null-homotopic loop in A_{i}, hence bounds a disk D in A_{i} with ∂D= β ∪ δ, where β ⊆L, and δ is a short arc (of length < ϵ) transverse to L. But looking back across the isotopies carried out so far, this disk demonstrates a homotopy of a vertical arc in N(B), rel its boundary, into a leaf of L. This, however, contradicts [G-O, Theorem 1(d)], which says that such homotopies are impossible.

If β meets δ in the interior of δ, then since β ⊆α_{i}, it follows that α_{i} meets δ in interior points. Now α_{i} cuts off a disk Δ in A_{i}; think of it as being colored \underline{green}. Δ meets γ in subarcs of γ\α_{i}; think of these as being colored green as well.

Because α_{i} separates A_{i}, it follows that γ\α_{i} consists of an even number of arcs, which (travelling along γ) are alternately colored green and left uncolored ( locally, α_{i} is colored green on only one side).

Since α_{i} meets δ ⊆ γ in interior points, it follows that δ\α_{i}⊆ γ\ α_{i} contains a \underline{colored} \underline{subarc}, δ_{0}. δ_{0} is contained in Δ, properly embedded, and so it splits Δ into two disks, one of which, Δ_{0}, does not contain the arc η = ∂Δ ∩ ∂A_{i}. Therefore, ∂Δ_{0} = δ_{0} ∪ α_{0}, with α_{0} ⊆α_{i}⊆L, and δ_{0} ⊆ γ, transverse to L, with length < the length of δ < ϵ. This, however, once again contradicts [G-O, Theorem 1(d)].

Therefore, this first situation is impossible.

Second case: α_{i} and α′_{i} are always essential (for i > i_{0}), and there is some essential arc ω ⊆A_{i} joining x and x′ so that α_{i} and α′_{i} always have winding number zero w.r.t. ω.

We wish now to show that eventually α_{i} (say) becomes \underline{stable}, i.e., for some i, α_{k}=α_{i}, for all k ≥ i. This amounts to saying that x_{k} = x_{i}, for all k ≥ i, i.e., x_{i} ∈P_{0}.

So assume the contrary; assume that x_{k_{i}} ≠ x_{k_{i−1}}, for k_{i} > k_{i−1}, infinitely often (to save the reader’s eyesight, we will conveniently forget that this expression has a ‘k’ in it, and write x_{i} instead). We will then obtain a contradiction, in a manner similar to the first case (with some slight technical additions).

We get an arbitrarily large collection of distinct points y_{i}∈ γ, i= 1, 2, …, in the α_{i} which are near neighbors to the endpoints x_{i} of the α_{i}. Now, as before, we can lift the α_{i}, α′_{i} to R × I =Ã_{i}, with x̃ =(a,0), x̃′ =(b,1) fixed. Because the winding number of the lifts of α_{i} can be counted across the lifts of ω, it follows that the endpoints x̃_{i} of the lifts of the α_{i} based at x̃ all lie in the interval [b-1,b+1]×1 and so the points y_{i} are contained in a compact piece ([b-1,b+1]×I)∩γ̃⊆γ̃ of the neighbor line on the x̃′_{i}-side of R×I. So as before we have an arbitrarily large number of ỹ_{i} accumulating in a fixed compact piece of γ̃, so eventually we can find (adjacent) points of (some) α̃_{i}∩γ̃ which are within ϵ of one another. the arc of α̃_{i} joining these two points, together with the arc of γ̃ joining them, form an (embedded) loop in R×I, which descends to a (singular) null-homotopic loop in A_{i}.

Lemma 5.2: If we orient α_{i}, α′_{i} so that x, x′ are at their tails, and look at the normal orientations that this induces on the set T= (α_{i} ∩ γ) ∪ (α′_{i} ∩ γ) of (transverse) intersection points with γ, then seen from γ they occur with \underline{opposite} \underline{sign}.

Proof: α_{i} and α′_{i} together separate A_{i} (although each separately doesn’t) into two disks D_{1}, D_{2} (see Figure 11), with the orientations of α_{i}, α′_{i}, giving orientations two two arcs in each boundary, as shown. Any arc δ of γ between two adjacent points of T must lie in either D_{1} or D_{2} (D_{1}, say). If the endpoints of δ both lie on the same end of ∂D_{1} then measured along δ the normal orientations of its endpoints are opposite; if they lie on opposite ends of ∂D_{1}, then, because we chose the orientations of α_{i} and α′_{i} to complement one another as they do, measured along δ the normal orientations of its endpoints are again opposite.

Figure 11: normal orientations

Note that this lemma would not be true if we dealt with only one arc (α_{i}, say) at a time; this is because by itself α_{i}, say, does not separate A_{i} (see Figure 11). Note also that if we \underline{lift} α_{i} to α̃_{i} in R×I, with the lifted orientation, and look at the normal orientations with which α̃_{i} meets the neighbor line γ̃, as you travel along γ̃ these also alternate; this is because α̃_{i} now \underline{does} separate R×I, so the situation is just as in the first case of the lemma above.

Now, we have already found adjacent points of (some) α̃_{i}∩γ̃ which are within ϵ of one another along γ̃. By the note above, these two points inherit opposite normal orientations in γ̃ from α̃_{i}. Together with the arc of γ̃ between them, the arc of α̃_{i} joining them forms an embedded null-homotopic loop in Ã_{i}, which descends to a null-homotopic loop in A_{i}, consisting of an arc β of α_{i} between points y_{i_{o}} and y_{i_{1}} of α_{i}∩γ, together with the short arc δ of γ between them. If β ∩ δ = ∂δ, then, as before, β ∪ δ is an embedded null-homotopic loop; the disk it bounds gives a null-homotopy violating [G-O, Theorem 1(d)], a contradiction.

If β ∩ δ ≠ ∂δ, then in particular α_{i}∪α′_{i} meets δ in interior points. Now these points of intersection inherit normal orientations from α_{i} and α′_{i}, which when seen along δ occur with opposite sign. The endpoints of δ also have opposite sign (their lifts did in γ̃, and they remain the same when projected); it then follows that there are an \underline{even} number of points in C=(α_{i}∪α′_{i})∩δ. Since the endpoints of δ both belong to α_{i}, it then also follows that some pair of points of C, adjacent along δ, both belong to α_{i} or α′_{i} (say α_{i}), joined by a subarc δ_{0} of δ. Now α_{i} and α′_{i} together separate A_{i} into two disks D_{1} and D_{2}, and since δ_{0} doesn’t meet α_{i} or α′_{i} except at its endpoints, δ_{0} is contained in one of these disks, say D_{1}. δ_{0} separates this disk into two sub-disks; because both of the endpoints of δ_{0} are in α_{i}, one of these disks Δ_{0} does not meet ∂A_{i} (see Figure 12), so its boundary ∂Δ_{0} = δ_{0} ∪ β_{0}, where β_{0} is a subarc of α_{i}. This disk Δ_{0} would again give a homotopy violating [G-O, Theorem 1(d)], and so gives a contradiction.

Figure 12: stabilization: second case

So all other possibilities lead us to a violation of [G-O, Theorem 1(d)]; we must therefore conclude that, eventually, the arcs α_{i}, α′_{i}, for some i, emanating from the points x,x′∈P_{0} are stable: their other endpoints are also in P_{0}. c. Proof of the theorem

We are now in a position to complete the proof of the theorem.

Given a point x ∈P_{0} in the stable set of our isotopy process, and an annulus A_{i} containing it, in the boundary of a solid torus M_{i}, we have shown that for some j, the arc α_{j} of I_{j}(L)∩A_{i} which contains x is \underline{stable}; all further isotopies of L fix α_{j}. This is equivalent to saying that its other endpoint is also in P_{0}; since such an arc would only be changed by ∂-compressions, and both its endpoints are stable, this means that the arc cannot be moved by further isotopies.

Lemma 5.3: Given x ∈P_{0}, there is a neighborhood U of x in S and a j so that for any x′ ∈ U∩P_{0} and A_{i} ⊆∂M_{i} containing x′, x′ is contained in a stable arc of I_{j}(L)∩A_{i}.

Proof: Fix an annulus A_{i} containing x. By the above, there is a j so that x is contained in stable arc α of I_{j}(L)∩A_{i}, with other endpoint x′. Let U be a (closed) ϵ-neighborhood of x in the loop of S containing x, intersected with P_{0}, and consider the (taut) arcs of some I_{kn+i}(L)∩A_{i}, with kn + i ≥ j, emanating from these points (then set j = kn + i). P_{0} is closed, so P_{0}∩U is closed in U; there is therefore a highest and lowest point of P_{0} in U. By choosing a larger j, if necessary, we may additionally assume that the arcs of I_{j}(L)∩A_{i} emanating from these points are also stable. The collection I_{j}(L)∩A_{i} of arcs is a 1-dimensional lamination in A_{i}, which are all parallel to one another.

Now, suppose an arc β of I_{j}(L)∩A_{i} emanating from a point in P_{0} ∩ U moves under a further isotopy. Consider the \underline{first} time such a move occurs. Because the endpoint of the arc on the x-side is stable, the change occurs as a ∂-compression on the x′-side of A_{i}.

If the resulting arc is trivial, then because the points at either end of U are in stable (essential) arcs, the disk that it cuts off of A_{i} therefore meets ∂A_{i} in an arc of U\x (because x is contained in a stable arc, too), which therefore has length less than ϵ.

If the resulting arc is still essential, then the ∂-compression joined β to a trivial arc on the x′-side of A_{i}. But such a trivial arc (since \underline{all} of the arcs between the highest and lowest (essential) arcs emanating out of U were essential at stage j) had to be created by some x-side ∂-compression at some stage after k; this trivial arc (immediately after the compression) had to meet the neighbor loop γ on the x-side, and a subarc, together with a short arc of γ (of length < ϵ), bounds a disk in A_{i}.

In each case we therefore have a situation which gives a disk violating [G-O,Theorem 1(d)], a contradiction.

Repeating this argument for each of the annuli containing x, taking the maximum of the j’s generated and the intersection of the U’s generated, completes the proof.

Now we have that for each x in P_{0} there exists a pair (U_{x}, j_{x}) given by the lemma. The collection of U_{x}’s form an open cover of P_{0}, which, because it is compact (P_{0} is closed in S, which is compact), has a finite subcover, {U_{1}, …, U_{n}}. Set j=max{j_{1}, …, j_{n}}, then it follows that every arc of I_{j}(L)∩A_{i} emanating from any point of P_{0}, for any A_{i}, is stable; it has \underline{both} of its endpoints in P_{0}.

Now choose a point x∈P_{0}∩M_{i}, for any given M_{i}. For some r, 0 ≤ r < n, x is contained in a taut disk D of I_{j+r}(L)∩M_{i}. But by dragging ourselves around ∂D starting from x, we see inductively (using the above) that every point of ∂D∩S is in fact contained in P_{0}, i.e., the boundary of this disk is stable, and therefore the disk containing x is stable. It therefore follows that for every x∈P_{0}, and every M_{i} containing x, x is contained in a stable, taut, disk of I_{j+n}(L)∩M_{i}. Because P_{0}∩M_{i} is a closed set in ∂M_{i}, it follows that the collection of disks of I_{j+n}(L)∩M_{i} containing points of P_{0} is a (closed) sublamination of I_{j+n}(L)∩M_{i}; the union of these disks over all of the M_{i} then forms a sublamination L_{0} of I_{j+n}(L) (they meet correctly along the ∂M_{i}, in the (stable) arcs emanating from P_{0}), which meets each M_{i} in a collection of tauts disks. By a small further isotopy of I_{j+n}(L) (first supported in a neighborhood of the ∂M_{i} to make the boundaries of the taut disks transverse to the circle fibering of ∂M_{i}, then supported away from ∂M_{i} to make the entire disks transverse) we can make L_{0} into a lamination meeting each solid torus in a collection of \underline{transverse} disks, i.e., L_{0} is a horizontal lamination.

Therefore, L contains a sublamination I_{j+n}^{−1}(L_{0}) which is isotopic to a horizontal lamination.

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