COMPRESSION BRANCHED SURFACES AND TIGHT
ESSENTIAL LAMINATIONS
(PRELIMINARY VERSION)
TAO LI
Abstract. We prove that if a 3-manifold contains a transversely orientable essential lamination, then it must contain a tight essential lamination. This theorem together with a theorem of Calegari implies that if
an atoroidal 3-manifold contains a transversely orientable taut foliation
then it must contain a genuine lamination. Hence, laminar 3-manifolds
satisfy the weak hyperbolization conjecture and have finite mapping
class groups.
1. Introduction
An essential lamination is called a tight lamination if its leaf space is
Hausdorff (or the leaf space has no branching using the terminology in [4, 5]).
A foliation is tight if and only if it is R -covered. Tight laminations have
many remarkable properties. For example, if a 3-manifold contains a tight
lamination, then any closed curve in the 3-manifold can be homotped to a
curve that intersects the lamination efficiently; a tight lamination can be
put in Kneser-Haken normal form with respect to any triangulation or celldecomposition of the 3-manifold. If a hyperbolic 3-manifold contains a tight
lamination, then one has a certain volume estimate of the 3-manifold [1].
Theorem 1.1. Let M be a closed, orientable and atoroidal 3-manifold. If
M contains a transversely orientable essential lamination, then M must
contains a transversely orientable tight essential lamination.
In [4], Calegari proved that any transversely orientable R -covered foliation has a pair of stable and unstable laminations transverse to the foliation,
and the pair of laminations are both tight and genuine. An essential lamination is called a genuine lamination if it cannot be trivially extended to
a foliation. If the foliation is not R -covered, Calegari also showed [5] that
there is also a transverse lamination, but the lamination may not be a genuine lamination. Thus, Theorem 1.1 and Calegari’s theorem imply that if an
atoroidal 3-manifold contains a transversely orientable taut foliation, then
it contains a tight genuine lamination.
Corollary 1.2. If an atoroidal 3-manifold contains a transversely orientable
taut foliation, then it must contain a tight genuine lamination.
Partially supported by an NSF grant.
1
A theorem of Gabai and Kazez [9] says that if a 3-manifold contains a
genuine lamination, then it satisfies the weak hyperbolization conjecture,
i.e., its fundamental group either contains a Z ⊕ Z subgroup, or is wordhyperbolic. They also show that the mapping class groups for 3-manifolds
with genuine laminations are finite [10]. Hence, we have:
Corollary 1.3. If a 3-manifold M contains an essential lamination, then
it satisfies the weak hyperbolization conjecture and has finite mapping class
group.
The idea of the proof is to construct a compression branched surface
using the assumption that the lamination is not tight, and then get another
essential lamination with greater gut number by evacuating a taut sutured
manifold as in [7]. The case of transversely orientable lamination is easier to
deal with because the compression branched surface that we constructed is
embedded. The techniques and results of this paper can also be applied to
non transversely orientable laminations, but we will have to perform some
cutting and re-gluing on the lamination, using the construction similar to
[12].
Notation. Throughout this paper, we will denote the interior of X by
int(X) , and denote the number of components of X by |X| .
2. Some modifications on the lamination
The main purpose of this section is to prove Lemma 2.2. We only need this
lemma to deal with a subtle technical issue in section 7. So, the reader can
skip the proof of Lemma 2.2 and comes back until section 7. It is conceivable
that the technical issue can be resolved without using Lemma 2.2, but I do
not see any easy way to do it.
Suppose M is a closed and orientable 3-manifold and λ is an essential
lamination in M . We will first modify our lamination and branched surface
using some properties of laminar branched surface and some techniques from
section 2 of [12].
Suppose λ is a transversely orientable essential lamination and we can
assume that λ is fully carried by a transversely orientable branched surface
B . Since the 3-torus T 3 contains a tight essential lamination, we may
assume M is not a 3-torus, and by [12], B can be split into a laminar
branched surface. If M − int(N (B)) has a D2 × I region, then ∂h N (B)
has a disk component. As in the section 2 of [12], there must be a subdisk
δ of D , which corresponds to a disk branch sector of B (i.e. the closure
of a disk component of B − L ) and the branch directions of arcs in ∂δ
either all point into δ or all point out of δ . Since we have assume that
B contains no sink disk, the branch direction of ∂δ must point outwards.
By Lemma 2.1 of [12], one can remove the branch sector δ from B , as
shown in Figure 2.1 of [12], and the new branched surface B − int(δ) is also
a laminar branched surface. So, by [12], B − int(δ) also fully carries an
2
essential lamination. Since δ is a disk in the boundary of a D2 × I region
of M − B , removing δ from B reduces the number of D2 × I regions.
By repeating this operation, we can assume that M − int(N (B)) contains
no D 2 × I region. Note that the gut number never decreases during this
operation, since the branch sectors that we remove are from the boundary
of the D2 × I regions.
Asumption 2.1. By the argument above, we may assume the branched surface B contains no sink disk and M − int(N (B)) has no D2 × I region.
Let α be an arc with ∂α ⊂ λ and suppose α ∩ N (B) consists of two
subarcs of the I -fibers of N (B) that contain the two endpoints of α . Suppose there is a disk ∆ embedded in M such that ∆ ∩ N (B) is a union
of subarcs of the I -fibers of N (B) and ∂∆ = α ∪ β , where β lies in a
leaf of λ . Now, we collapsing every I -fiber of N (B) to a point and get
the branched surface B . We may consider N (B) and B as subsets of M
and extend the collapsing map π : N (B) → B to a deformation retract
π : M → M . Suppose π(∆) is still an embedded disk in M . Thus, π(∆)
is transverse to B , π(β) is an arc embedded in B , and π(α) ∩ B = ∂π(α) .
We will call such a disk ∆ or π(∆) a transverse bigon, and call α or π(α)
the base arc of the transverse bigon. Let v1 and v2 be the two components
of α ∩ N (B) . Then, since there is no monogon, ∆ ∩ λ must consist of circles
and compact arcs with one endpoint in v1 and the other endpoint in v2 .
Lemma 2.2. Suppose B is a transversely orientable branched surface that
satisfies Assumption 2.1 and fully carries a lamination λ . Let ∆ be a
transverse bigon and α be its base arc as above. Then, ∆ can be isotoped,
fixing B and the base arc α , to a transverse bigon ∆ such that ∆ ∩ λ
contains no circles.
Proof. We will simultaneously consider ∆ and π(∆) . Note that B ∩ π(∆)
is a train track in π(∆) and the one-dimensional lamination ∆ ∩ λ contains
a circular leaf if and only if the closure of some component of B − π(∆) is
a disk with smooth boundary. We will perform isotopies on π(∆) , fixing
π(α) to get π(∆ ) which is still transverse to B .
Since B is transversely orientable and π(∆) is transverse to B , π(∆)∩B
and each leaf of ∆∩λ has an induced transverse orientation in π(∆) and ∆ .
For each circular leaf c of ∆ ∩ λ , we use ∆c to denote the disk bounded by
c in ∆ . Since λ is essential, c also bounds a disk in λ , which we denoted
by Dc . We denote π(∆)∩ B by τ . We may consider ∆ ∩ N (B) as a fibered
neighborhood of τ , and we still use π : ∆ ∩ N (B) → τ to denote the map
that collapses each I -fiber to a point. Since τ is transversely orientable, for
each circular leaf c in ∆ ∩ λ , π(c) must be embedded in τ . Hence, π(∆c )
is also an embedded disk in π(∆) .
Claim 1. Let c be a smooth circular sub train track of τ and ∆c be the
disk bounded by c in π(∆) . ∆c ∩ B can also be considered as a sub train
track of π(∆) ∩ B . Suppose the induced transverse direction of c points
3
into ∆c . Then, there must be a circular leaf α of ∆ ∩ λ carried by the sub
train track ∆c ∩ B and the induced transverse direction of α points into
the disk ∆α bounded by α in ∆ .
Proof of the claim. Since there is no monogon, the closure of some component of ∆c − B must be a disk with smooth boundary. Hence, the sub train
track ∆c ∩ B must carry some circle of ∆ ∩ λ , and we need to prove that
∆c ∩ B carries a circle with the required transverse direction.
Without loss of generality, we may assume that c is innermost, i.e., ∆c ∩B
does not contain any other smooth circle with induced direction pointing into
the disk it bounds. If c itself is a connected component of the train track
∆c ∩ B , then ∆c corresponds to a disk whose boundary is a component of
∆ ∩ ∂h N (B) . As we can assume ∂h N (B) ⊂ λ , c (as a sub train track)
must carry a circle of ∆ ∩ λ and the claim follows. Now, we suppose c is
not a component of ∆c ∩ B . Since ∆ ∩ λ consists of circles and compact
arcs, there must be a smooth arc β ⊂ ∆c ∩ B properly embedded in ∆c .
As τ is transversely oriented, we can perform some cutting and pasting on
c and β to get a smooth circle in ∆c ∩ B with induced transverse direction
point into the disk it bounds, which contradicts our assumption that c is
innermost. Note that if ∆ ∩ λ has non compact leaves, then the claim is
false and a Reeb train track of the annulus is a counter-example.
Let E ⊂ ∆ be a planar surface in ∆ , and suppose ∂E consists of smooth
circles carried by ∆ ∩ N (B) with direction (induced from the transverse
direction of B ) pointing out of E . We call E a good planar surface in
∆ if there is an embedded surface F ⊂ N (B) carried by N (B) such that
∂E = ∂F , F is homotopic to E fixing ∂F and F ∩ ∆ consists of circles
in λ ∩ ∆ . Note that, although F itself is embedded, it may intersect ∆
nontrivially, and F may not be totally in a leaf of λ . We define the area of
E to be the number of components of ∆ − N (B) inside E .
Before we describe how to construct a good planar surface, we would like
to describe certain operations that we will use later.
Let E and F be as above. F has a normal direction induced from
the transverse orientation of B . For each point x ∈ F , let Ix be the I fiber of N (B) containing x . Then, the transverse orientation of B gives
a direction for Ix along Ix . We say that a point y ∈ Ix − x is on the
positive side of Ix if the direction of Ix points from x to y , otherwise we
say y is on the negative side of x . Let G be the collection of points x in
F such that Ix ∩ ∂v N (B) contains a component on the positive side of x .
G can be considered as a trivalent graph whose vertices are precisely those
points x where Ix ∩ ∂v N (B) has two components both on the positive
side of Ix . Since Ix ∩ ∂v N (B) has two components if and only if π(Ix )
is a double point of the branch locus LB , π(x) is a double point of the
branch locus LB for each vertex x ∈ G . Moreover, each non-vertex point
G has a normal direction in F induced from the branch direction of the
corresponding component of ∂v N (B) . Then, we can deform this trivalent
4
G
τG
Figure 2.1
graph G into a train tack τG as shown in Figure 2.1 (or Figure 2.3 of [12]).
We call this train track the characteristic train track of F with respect to
this transverse direction of F .
The train track τG is transversely oriented by the induced branch direction of the branch locus LB . Suppose the closure of a component of F −τG ,
say δ , is a disk with smooth boundary. If the branch direction for ∂δ points
out of δ , then δ corresponds to a disk component of ∂h N (B) and hence
M − int(N (B)) has a D 2 × I region, which contradicts our Assumption 2.1
on B . If the branch direction for ∂δ points into δ , then as in section 2 of
[12], by studying the characteristic train track of δ from the negative side
of δ , one can easily conclude (using an Euler characteristic argument) that
either M − int(N (B)) has a D2 × I region or a subdisk of δ is a sink disk,
which again contradicts our Assumption 2.1 on B . Thus, we can assume
that the closure of any component of F − τG is not a disk with smooth
boundary.
By pushing F towards the normal direction of F a little and fixing ∂F ,
as shown in the one dimension lower picture Figure 2.2, we can obtain a
surface F so that F ∩ N (B) consists of subarcs of the I -fibers of N (B) .
Since π|∂F is an embedding in a small neighborhood of ∂F and since B
is transversely orientable, we may assume F is still an embedded surface.
So, we can consider F as an embedded surface transverse to B . It is easy
to see from the local pictures of this isotopy that the train track B ∩ F is
the same as the characteristic train track τG above. Moreover, since ∆ is
transverse to B , we may assume that ∂F is a collection of circles parallel to
∂E = ∂F and ∂F is outside E because of the assumption on the induced
transverse direction of ∂E . Let E be the planar surface in ∆ bounded by
∂F . So, E ⊂ E and (∆ − E ) ∪ F is a disk transverse to B .
We call a circular leaf of ∆ ∩ λ a type I circle, if the induced direction of
this circle points out of the disk it bounds in ∆ , otherwise we call it a type
II circle.
By our hypotheses on ∆ ∩ λ , there are finitely many type I circles
c1 , . . . , cn such that every type I circle lies in the disks they bound in ∆ . Let
∆ci be the disk bounded by ci in ∆ . Since λ is essential, each ci bounds
a disk Dci in λ . Each Dci is homotopic to ∆ci fixing ci and Dci ∩ ∆ is
a union of circles in ∆ ∩ λ . So, the union of the ∆ci is a (disconnected)
good planar surface. Let E be a maximal good planar surface in ∆ that
5
isotopy
N (B)
F
∂F
Figure 2.2
contains all the type I circles ( E may not be connected), and let F the
surface carried by N (B) as in the definition of a good planar surface.
Claim 2. Let τG be the characteristic train track of F , and let Q be the
closure of a component of F − τG . Suppose ∂Q consists of smooth circles
in τG . Then, Q cannot be a disk and the induced branch direction of ∂Q
must point into Q
proof of the claim. If there is a boundary circle η of Q with the induced
direction pointing out of Q , then η corresponds to an annular component
of ∂v N (B) . However, since E is a subsurface of the disk ∆ , η must be
homotopically trivial in M . Hence, the corresponding annular component of
∂v N (B) bounds either a disk of contact or a D2 ×I region, which contradicts
our assumption on B . So, there is no such a boundary component η of Q .
Moreover, Q cannot be a disk in the interior of F , because otherwise, by
the argument above, the branch direction of ∂Q must point into Q . Then,
as before (or see section 2 of [12]), we can apply the same argument to the
characteristic train track of Q with respect to opposite transverse direction
and conclude that either a subdisk of π(Q) is a sink disk or there is a D2 ×I
region, which again gives a contradiction.
Let ∆1 = (∆ − E ) ∪ F . By the argument above, we may consider F ∩ B
and τB as the same train track. Hence, ∆1 ∩ B and τG ∪ (∆ ∩ B − E) can
be considered as the same train track. We claim that ∆1 ∩ λ contains no
type I circle in ∆1 . Suppose there is a type I circle γ which bounds a disk
δ in ∆1 . We may assume that γ is innermost, i.e., there is not type I
circle in int(δ) . If δ ⊂ int(F ) , then since γ is innermost, there must be
a smooth disk component Q of F − B , which contradicts Claim 2. So, δ
cannot totally lie in int(F ) . In other words, either γ ∩ ∂F = ∅ or γ is
homotopically nontrivial in F . Figure 2.3 is a simple example, where E is
a disk in ∆ . In Figure 2.3, after one pushes F outwards, one create a new
smooth circle, and the point is that the union of E and the bigon outside
E is a bigger good planar surface.
After some isotopy, we may assume that ∂h N (B) ⊂ λ . Since γ is inner
most and B is transversely oriented, as in the proof of Claim 1, we may
assume γ is a circle in ∆1 ∩ ∂h N (B) . As ∂h N (B) is incompressible, γ
must bound a disk δh in ∂h N (B) . Note that δh ⊂ λ and δh ∩ ∆ consist
6
F
δ
E
∆
B
Figure 2.3
of circles in λ ∩ ∆ . Since F is constructed by pushing F outwards along
the transverse orientation of B and since γ if a type I circle in ∆1 , we can
extend δh a litter near ∂δh to get a slightly larger disk δh ⊂ N (B) with
∂δh ⊂ (∆ − E) ∪ F and δh ⊂ δh . In fact, if we identify the train track
∆1 ∩ B with τG ∪ (∆ ∩ B − E) , π(∂δh ) and π(γ) are the same circle in
this train track. Moreover, δh ∩ ∆ = δh ∩ ∆ consists of circles in ∆ ∩ λ . So,
∂δh bounds a disk δ in F ∪ (∆ − E) , which corresponds to δ , and δh is
homotopic to δ fixing ∂δ .
Let E1 = E ∪ (δ − F ) . Since the branch direction of γ points out of
δ , F1 = (F − δ ) ∪ δh is a surface carried by N (B) , ∂F1 = ∂E1 , and F1
is homotopic to E1 fixing ∂F1 . If F1 is not embedded, then the singular
curves are the possible intersections δh ∩ F and hence are disjoint circles
bounding disks. Since B does not carry any 2-sphere, we can perform some
cutting and pasting along these circles and obtain an embedded surface
which we also denote by F1 to simplify notation. Moreover, since δh ∩ ∆ =
δh ∩ ∆ consists of circles in ∆ ∩ λ and F1 = (F − δ ) ∪ δh , F1 ∩ ∆ consists
of circular leaves in ∆ ∩ λ . Thus, E1 is a good planar surface. Since δ
cannot lie in int(F ) as above, E1 must have greater area than E , which
contradicts our assumption that E is maximal. Therfore, ∆1 ∩ λ contains
only type II circles. Although ∆1 may not be embedded, since F ∩ ∆
consists of circles in ∆∩λ and ∆−E contains only type II circles, and since
F is obtained by pushing F along the transverse direction, the possible
double curves of ∆1 (i.e. F ∩ (∆ − E ) ) must lie in the disks bounded by
the type II circles F ∩ (∆ − E) .
Now we reverse the transverse direction of B , and the ∆1 above contains
only circles of type I with respect the new transverse orientation. Similar
to the argument above, there are finitely many circles d1 , . . . , dm in ∆1 ∩ λ
7
bounding disjoint disks ∆d1 , . . . , ∆dm such that all the circles in ∆1 ∩ λ lie
in these disks ∆d1 , . . . , ∆dm . If ∆1 is not embedded, by the construction
above, the singular
set is a union of double curves lying inside the ∆di ’s. In
particular, ∆1 − m
i also bound a
i=1 ∆di is an embedded
m surface. Each dm
in
λ
,
and
D
∩∆
⊂
λ∩∆
⊂
∆
.
Thus,
(
disk
D
1
1 i=1 di
di
i=1 Ddi )∪(∆1 −
m di
m
∆
)
is
an
embedded
disk.
So,
∆
is
a
good
planar
surface in
i=1 di
i=1 di
m
∆1 . Let Ê be a maximal good planar surface in ∆1 containing i=1 ∆di .
Let F̂ be the surface carried by N (B) with ∂ Ê = ∂ F̂ . By the definition
of good planar surface, F̂ is embedded and F̂ ∩ ∆1 consists of circles of
∆1 ∩ λ . Hence, F̂ ∩ ∆1 ⊂ Ê and we may assume that (∆1 − Ê) ∪ F̂ is an
embedded disk.
We claim that Ê must be a union of disjoint disks in ∆1 . If a component
H of E is not a disk, then a boundary circle of H must bound a disk, say
P , in ∆1 − int(H) . By the definition of good planar surface, the induced
direction of ∂ Ê points out of Ê and hence the direction of ∂P points into
P . By Claim 1, there must be a circular leaf of ∆1 ∩λ carried by P ∩B with
the induced direction pointing into the disk it bounds, which contradicts the
assumption that ∆1 ∩ λ contains no type II circle with respect to the new
transverse orientation of B .
Similar to the argument above, we can perform an isotopy by pushing F̂
towards the new transverse direction and get a surface F̂ which is transverse
to B and with ∂ F̂ ⊂ ∆1 . Since F̂ is a union of disks, if F̂ ∩ λ contain
circles, by the argument in Claim 2, either M − int(N (B)) has a D2 × I
region or B contains a sink disk, which contradicts the assumptions on B .
So, F̂ ∩ λ contains no circle.
Let Ê be the union of disks bounded by ∂ F̂ in ∆1 and ∆2 = (∆1 −
Ê ) ∪ F̂ . If ∆2 ∩ λ contains circles, then the closure of some component of
∆2 − B , say δ , is a disk with smooth boundary. Since F̂ ∩ λ contains no
circles, δ ∩∂ Ê = ∅ and δ ∩ F̂ = ∅ . If the induced transverse direction of ∂δ
point out of δ , similar to the argument above and as shown in Figure 2.3,
we can construct a good planar surface with greater area using Ê ∪δ , which
contradicts the assumption that Ê is maximal. If the induced transverse
direction of ∂δ point into δ , as shown in Figure 2.4, the boundary of each
component of δ − int(Ê) must be a smooth circle in the train track B ∩
(∆1 − int(Ê)) with induced direction pointing into the disk it bounds. So,
each component of δ − int(Ê) is a disk with smooth boundary. Thus, by
the Claim 1, there must be a type II circle of ∆1 ∩ λ lying outside Ê , which
contradicts our assumptions. Therefore, ∆2 ∩ λ contains no circles and the
lemma follows.
Remark 2.3.
1. A version of this lemma for immersed disks was included
in the first version of the paper [12], but the author later decided not to
exclude that part because it is not quite relevant to the main theorem.
8
F̂
Ê
∆1
δ
B
Figure 2.4
2. In the statement of this lemma, ∆ ∩ λ must be a union of compact
arcs with endpoints in v1 and v2 , and π(∆ ) − B is a union of bigon,
since α ∩ B = ∂α and B contains no monogon.
3. Nested almost vertical disks
In this section, will assume that λ is a transversely orientable essential
lamination fully carried by a transversely orientable branched surface B .
Suppose the leaf space is not Hausdorff. The goal of this section is to
construct a sequence of “nested” vertical disks.
Definition 3.1. We say that a surface S ⊂ N (B) is vertical if S is a
union of subarcs of the I -fibers of N (B) . Let D = I × I ( I = [0, 1] ) be
a disk properly embedded in M − int(N (B)) . We call D a product disk if
I × {0, 1} ⊂ ∂h N (B) and {0, 1} × I is a pair of vertical arcs in ∂v N (B) .
Let S be a surface embedded in M . We say that S is almost vertical if
S ∩ N (B) is a union of vertical arcs and S − int(N (B)) consists of product
disks. Suppose S is almost vertical. We define the weight of S weight(S)
to be the number of components of S ∩ ∂v N (B) .
Definition 3.2. Let D1 and D2 be two embedded disks whose intersection
with N (B) consist of subarcs of the I -fibers of N (B) . We say that D1
and D2 are λB -parallel if there is a local embedding f : D × [1, 2] → M
such that
1. f (D × {i}) = Di ( i = 1, 2 );
2. each component of f −1 (λ) is of the form α×[1, 2] where α is an curve
embedded in D .
3. f (D×{x}) is an embedded disk whose intersection with N (B) consists
of subarcs of the I -fibers of N (B) .
9
4. each component of f −1 (∂v N (B)) is of the form α × [1, 2] where α is
a compact arc in D and f (α × {i}) is a vertical arc of ∂v N (B) .
We will first use the assumption that λ is not tight to construct a sequence
of nested almost vertical disks. The next two propositions show that, under
certain assumptions, one can extend an embedded arc in a leaf of λ to a
vertical (or an almost vertical) disk. Since B fully carries λ , after some
isotopy if necessary, we may assume ∂h N (B) ⊂ λ .
Proposition 3.3. Let α be an embedded arc in a leaf of λ . Suppose α0
lies in int(N (B)) and π(α0 ) is also embedded, where π : N (B) → B is the
map collapsing every I -fiber to a point. Then, on each side of α0 , there is a
maximal vertical disk α × [0, 1] ⊂ N (B) such that α × {0} = α0 , {x} × [0, 1]
is a subarc of an I -fiber of N (B) for each x ∈ α , and α×{1}∩∂h N (B) = ∅
Proof. The proof of this proposition is obvious. Since π(α0 ) is embedded, on
each side of α0 , there must be such a maximal vertical disk in π −1 (π(α0 )) .
Proposition 3.4. Let Mλ be the closure of M − λ under the path metric,
E be a component of Mλ , and P = E ∩ N (B) . Let γ = α0 ∪ ρ ∪ α1 be a
simple arc in E , where αi ⊂ P is an I -fiber of the I -bundle P for each
i = 0, 1 , and ρ ⊂ ∂E ⊂ λ is a simple arc connecting α0 and α1 . Suppose
γ can be homotoped into ∂E fixing ∂γ . Then, there is an embedded almost
vertical disk D = I × I properly embedded in E such that ∂I × I = α0 ∪ α1
and I × {0} = ρ .
Proof. Since γ can be homotoped into ∂E fixing ∂γ , there exists a map
f : I × I → Mλ ( I = [0, 1] ) such that f |{i}×I = αi ( i = 0, 1 ), f |I×{0} = ρ ,
f (I × ∂I) = ∂E ∩ f (I × I) . We call an arc of the form {p} × I a vertical arc
in I × I . Since ∂h N (B) ⊂ λ , after some standard homotopy if necessary,
we can assume that f −1 (∂v N (B)) is a union of finitely many vertical arcs
in I × I . Thus, f restricted to I × I − f −1 (M − int(N (B))) is a collection
of immersed product disks in M − int(N (B)) . By a standard argument
of characteristic manifolds (see [14]), we can assume the image of f lies
in the union of N (B) and a maximal I -bundle region of M − int(N (B)) .
Therefore, we can assume the image of f lies in a maximal I -bundle region
of Mλ with the induced bundle structure. Since λ is transversely orientable,
every I -bundle region in Mλ is a product and the lemma follows.
Let M̃ be the universal cover of M and λ̃ be the preimage of λ in
M̃ . By [11], the lamination λ̃ is equivalent to λ̃ = λ0 × R , where λ0 is a
one-dimensional lamination in R2 .
Suppose the leaf space of λ is non-Hausdorff. For any pair of nonseparable
points x and y in the leaf space, there is an embedding f : I × [0, 1) → R2
with the following properties.
1. for any leaf l of λ0 , f −1 (l) is either empty or of the form I × {p} for
some point p ∈ [0, 1) ;
10
Figure 3.1
homotopy
Figure 3.2
2. f cannot be continously extended to f : I × [0, 1] → R2 ;
3. there is an open set J ⊂ I containing ∂I , such that f can be
continuously extended to an embedding f¯ : D → R2 , where D =
(I × [0, 1)) ∪ (J × {1}) (see Figure 3.1 for a picture of f¯(D) );
4. f¯(∂I × {1}) corresponds to the pair of points x and y in the leaf
space.
Let p : M̃ → M be the covering map and g = p◦ f¯ : D → M . After some
perturbation, we can assume that there is an ∈ (0, 1) such that g|∂I×[,1]
is a pair of embedded and disjoint vertical arcs (i.e., subarcs of the I -fibers
of N (B) ), and g(I × {}) lies in a leaf of λ . We denote g({0} × [, 1])
and g({1} × [, 1]) by vx , vy respectively. The two points x1 = g(0, 1) and
y1 = g(1, 1) correspond to the pair of nonseparable points x and y in the
non-Hausdorff leaf space. For any point px ∈ (vx − x1 ) ∩ λ , there is a point
py ∈ (vy − y1 ) ∩ λ such that px and py are the two endpoints of an arc in
g−1 (λ) , and we call px ∪py a standard pair of points. Clearly, g(0, )∪g(1, )
is a standard pair. We denote the curve g(I × {}) by γ0 and call it the
base arc. By definition, γ0 is a curve in a leaf of λ with endpoints g(0, )
and g(1, ) . Note that every compact arc in a surface can be homotoped
(fixing endpoints) to an embedded arc as illustrated in Figure 3.2. So, after
some homotopy on g near γ0 (fixing vx ∪ vy ), we can assume γ0 ∪ vx ∪ vy
is a simple arc in M .
11
Let L0 be the leaf of λ that contains γ0 . By some splitting on N (B) and
choosing an closer to 1 if necessary, we can assume that γ0 ⊂ ∂h N (B)
and α and β are still subarcs of I -fibers of N (B) ( i = 0, 1 ). Note that
by the argument in section 5 of [12], we can make this splitting preserve Assumption 2.1 of section 2 on B . Since γ0 ⊂ ∂h N (B) and B is transversely
orientable, π(γ0 ) is an embedded arc in B . Hence, by Proposition 3.3,
there is a maximal embedded vertical rectangle D1 = γ × [0, 1] in N (B)
such that γ × {0} = γ0 , γ × {1} ∩ ∂h N (B) = ∅ , and ∂γ × [0, 1] is a pair
of subarcs in vx and vy respectively. We use γ1 to denote γ × {1} . Since
γ1 ∩ ∂h N (B) = ∅ and ∂h N (B) ⊂ λ , by Proposition 3.4, there is an almost
vertical disk E1 = γ × [1, 2] ⊂ Mλ with γ × {1} = γ1 . Moreover, we can
choose E1 on the side of γ1 so that D1 ∪ E1 is almost vertical. We denote
γ × {2} by γ2 .
If E1 ∪ D1 is not an embedded disk, then since B is transversely orientable, the singular points must be γ2 ∩ γ0 , and in particular γ0 and γ2
must lie in the horizontal boundary of the same component of Mλ . Note
that give two embedded compact arcs α and β in a surface with α lying
in the interior of the surface, similar to Figure 3.2, one can perform some
isotopy on β so that α ∩ β = ∅ . Since γ0 is a compact arc in the interior of ∂h N (B) , we can isotope the product disks in E1 along γ0 so that
γ2 ∩ γ0 = ∅ . Since the isotopy on E1 above happens in M − int(N (B)) , after this isotopy, E1 is still an almost vertical disk, though the product disks
are tilted during the isotopy on E1 . So, we may assume that D2 = D1 ∪ E1
is an embedded almost vertical disk. The vertical arcs ∂γ × [0, 2] must lie in
vx − x1 and vy − y1 , since x1 and y1 correspond to a pair of nonseparable
points in the leaf space.
Inductively, we can use γn ( n ∈ N ) to construct a vertical or an almost
vertical disk En = γ × [n, n + 1] with γ × {n} = γn , and denote γ × {n + 1}
and Dn ∪ En by γn+1 and Dn+1 respectively. Moreover, we can assume
each Dn is an embedded almost vertical disk and ∂γ × [0, n] must lie in
vx − x1 and vy − y1 . So, we can construct an embedding h : I × [0, 1) → M
k
]) = Dk . Since vx and vy are compact arcs, h|∂I×[0,1)
such that h(I ×[0, k+1
can be continously extended to ∂I × [0, 1] .
Remark 3.5. From our construction, it is possible that h(∂I × [0, 1]) vx ∪
vy . If this happens, though h can not be embeddedly extended to I × [0, 1] ,
h can be extend to a singular map h : I × [0, 1] → M and h(I × [0, 1]) is
homotopic to g(I × [0, t]) for some t ∈ (0, 1) .
Let h : I × [0, 1) → M be the map above. Note that by choosing different
product disks, we will end up with different embedding h . Moreover, we
may assume that h(∂I × [0, 1)) is a pair of maximal open intervals in vx ∪ vy
among all such almost vertical embedding h : I × [0, 1) → M . In particular,
we can make the following assumption.
12
Asumption 3.6. Let h : I × [0, 1) → M be the map above. We can assume
that there is no almost vertical disk ∆ = I × I such that I × ∂I ⊂ λ ,
I × {0} = h(I × {0}) and ∂I × [0, 1) = h(∂I × [0, 1)) .
Since x1 and y1 correspond to a pair of nonseparable points in the leaf
space, we may assume that h can not be extended to an embedded almost
vertical disk h(I × [0, 1]) . There exists a maximal open set J of I such
that ∂I ⊂ J and h|J×[0,1) can be continuously extended to an embedding
h : D → M , where D = (I × [0, 1)) ∪ (J × {1}) . We denote h(D) by H .
For each component η in J , we call h(η × {1}) a limit curve of H .
4. A modification of the construction of H
In this section, we will modify our construction of H mostly by choosing
different product disks during the construction of the almost vertical disks
Dn , so that H satisfies certain properties.
Definition 4.1. Let P = α × I be a product disk, where α × ∂I ⊂ ∂h N (B)
and ∂α×I is a pair of vertical arc in ∂v N (B) . We say P is a trivial product
disk if there is an embedded 3-ball D × I in M such that
1. int(D × I) ⊂ M − N (B) and D × ∂I = (D × I) ∩ ∂h N (B) ,
2. ∂D = α ∪ β , α × I = P , and β × I ⊂ ∂v N (B) ,
3. π(D × ∂I) ∩ LB = π(β × I) , where LB is the branch locus of B and
π : N (B) → B is the collapsing map.
Note that, by item 3 in the definition, N (B) ∪ (D × I) is equivalent to
N (B) as fibered neighborhoods of branched surfaces. Thus, by adding such
D×I to N (B) , we may assume that H does not contain any trivial product
disk.
Property 4.2. We can assume that H contains no trivial product disk.
Let h and H be as above. Let NH = H ∩ N (B) . Since H is almost
vertical, NH is a union of subarcs of I -fibers of N (B) . Let π : N (B) → B
be the map that collapses every I -fiber to a point. The map π restricted
to NH maps NH to a train track and {Di } to a sequence of nested bigons.
Let λH = λ ∩ H . Then, λH is a one-dimensional lamination in H . The
limit curves of H are leaves in λH which we call the limit leaves of H .
We call h(I × {0}) the base leaf or the base arc of λH , and call other
leaves of λH the interior leaves. We call the two arcs h(∂I × [0, 1]) the
two sides of H . Each interior leaf is a compact arc properly embedded
in H with endpoints in h(∂I × [0, 1]) . The two limit leaves containing
h(∂I × {1}) are homeomorphic to the interval [0, ∞) , and all other limit
leaves are homeomorphic to R , as shown in Figure 3.1.
We give every limit leaf in the lamination λH an outwards normal direction. Since λH is transversely orientable, every curve has an induced
transverse orientation, which also determines a transverse orientation for λ
and B in M .
13
Recall that, in our construction, we have a sequence of nested almost
vertical disks {Di } , and Dn+1 = Dn ∪ En , H = limi→∞ Di . Moreover,
each E2n is a vertical disk and each E2n+1 is an almost vertical disk in a
component of Mλ . In our construction, once D2k is constructed, E2k and
hence D2k+1 is fixed by the I -fibers of N (B) . However, even if D2k+1 is
fixed, one can have more than one choice for the product disks in E2k+1 ,
because it is possible to have two isotopic product disks that are not λB parallel. So, we can modify the construction of H by choosing different
product disks at each stage of E2k+1 .
Let ∆1 = I × I be a product disk in E2k+1 with I × ∂I ⊂ ∂h N (B) and
∂I × I ⊂ ∂v N (B) . Suppose I × {0} is the arc ∂E2k+1 ∩ ∂D2k+1 . Hence,
the transverse direction (defined above) of I × {0} points into ∆1 = I × I
and the transverse direction of I × {1} points out of ∆1 . We call I × {1}
the top arc of ∆1 . Let α1 ⊂ ∆1 be a properly embedded arc with both
endpoints in the top arc I × {1} . So, α1 and a subarc of I × {1} bound
a subdisk ∆1 of ∆1 . Let ∆2 be a product disk in E2m+1 with m ≤ k
and α2 be a properly embedded arc in ∆2 with both endpoints in the top
arc of ∆2 . So, α2 and a subarc of the top arc of ∆2 bound a disk ∆2 in
∆2 . Suppose α1 is λB -parallel to α2 . Then, we can reconstruct our E2k+1
(since k ≥ m ) by cutting off ∆1 and gluing back a disk that is λB -parallel
to ∆2 . We can do this operation to each product disk at each stage and still
able to construct a sequence of almost vertical disks {Di } as in section 3.
Thus, we assume H has the following property.
Property 4.3. For any pair of product disks ∆1 and ∆2 in H and any
pair of arcs α1 , α2 with endpoints in the top arcs of the product disks. If
α1 and α2 are λB -parallel, then ∆1 and ∆2 are λB -parallel, where ∆i is
a subdisk of ∆i bounded by αi and a subarc of the top arc ( i = 1, 2 ).
By similar cutting and pasting on H , we may also assume H has the
following property.
Property 4.4. Let c1 and c2 be two disjoint simple closed curve in H
that bound two disjoint disks d1 and d2 respectively. Suppose c1 and c2
are λB -parallel, then d1 and d2 are λB -parallel.
Definition 4.5. Let α ⊂ H be an arc transverse to λH and ∂α lying in
a leaf l of λH . Let α be the arc in l with ∂α = ∂α and suppose α ∪ α
bounds an embedded disk ∆ in H . We say that α is a good arc if the
direction at α points out of ∆ . Moreover, we call the disk ∆ a good disk
bounded by the good arc α . We call ∆ a simple good disk and α a simple
good arc if λH ∩ ∆ consist of arcs with endpoints in α and the arcs in
λH ∩ ∆ cut ∆ into a collection of pairwise nested good disks.
Proposition 4.6. Let α be a simple good arc in H with α ∩ λH = ∂α ,
and let l , α , ∆ be as in Definition 4.5. Suppose there is another arc β in
H that is λB -parallel to α and β ∩ α = ∅ . Then, there must be an arc β 14
in a leaf of λH such that β ∪ β bounds a disk ∆ and ∆ is λB -parallel
to ∆ .
Proof. Since α and β are λB -parallel and by the hypotheses on α , we can
suppose α ⊂ Em and β ⊂ En , where Em and En are two almost vertical
disks in the construction of H above. In particular, β ∩ λH = ∂β . Hence,
there must be an arc β in a leaf of λH such that β ∪ β bounds a disk ∆ .
Moreover, ∆ and ∆ consist of some subarcs of the I -fibers of N (B) and
parts of the product disks (if any) in Em and En .
First, since α and β are λB -parallel and λ is transversely orientable,
the transverse direction of ∂∆ − β must point out of ∆ . By the definition
of λB -parallel, ∆ ∩ N (B) must be λB -parallel to ∆ ∩ N (B) . So, the
proposition follows from Property 4.3.
Lemma 4.7. Let α be a simple good arc, and l , α , ∆ be as in Definition 4.5. Suppose α is λB -parallel to another arc β ⊂ int(H) . Then, there
must be an arc β in a leaf l of λH with ∂β = ∂β and β ∪ β bounds a
disk ∆ in H . Moreover, ∆ is λB -parallel to ∆ .
Proof. First of all, the case that ∆ ⊂ NH = H ∩ N (B) is obvious.
By the definition of simple good arc, all the disks bounded by λH ∩∆ and
α are nested, there is an inner most disk ∆0 with ∂∆0 = α0 ∪ δ0 , where
α0 ⊂ α and δ0 is a component of λH ∩ ∆ . Similar to the construction of
the En ’s, there are finitely many components δ0 , . . . , δm in λH ∩ E with
δm = α , such that
1. ∆0 ⊂, . . . , ⊂ ∆m = ∆ , where ∆i is the disk bounded by δi and a
subarc of α ;
2. ∆2k+1 ⊂ NH = H ∩ N (B) for each k ≥ 0 ;
3. λH ∩ int(∆2k − ∆2k−1 ) = ∅ .
Let αi = ∂∆i − int(δi ) be the subarc of α . Hence, α0 ⊂, . . . , ⊂ αm = α .
Since α is λB -parallel to β , each αi is λB -parallel to a subarc βi in β .
By Proposition 4.6, there is an arc δ0 in λH such that β0 ∪δ0 bounds a disk
∆0 that is λB -parallel to ∆0 . Then, we consider the arc δ0 ∪ (α1 − α0 ) .
The arc δ0 ∪ (α1 − α0 ) is λB -parallel to δ0 ∪ (β1 − β0 ) , and it bounds
the disk ∆1 − int(∆0 ) with δ1 . After some splitting on λH if necessary,
the arc δ0 ∪ (α1 − α0 ) can be perturbed slightly to be a good arc. Since
∆1 − int(∆0 ) ⊂ H ∩ N (B) , by the definition of λB -parallel for arcs, there
must be an arc δ1 that bounds a disk ∆1 with β1 and ∆1 − int(∆0 ) is
λB -parallel to ∆1 − int(∆0 ) . Thus, ∆1 is λB -parallel to ∆1 . We can
proceed in this manner, repeatedly using Proposition 4.6 and the definition
of λB -parallel for arcs, and eventually get such a disk ∆ = ∆m that is
λB -parallel to ∆ = ∆m .
Lemma 4.8. Let α be a good arc in int(H) and l , α , ∆ be as in Definition 4.5. Suppose α is λB -parallel to another arc β ⊂ int(H) . Then,
there must be an arc β in a leaf l of λH that, together with β , bounds a
disk ∆ in H , and ∆ is λB -parallel to ∆ .
15
Proof. Since α ⊂ int(H) , the leaf l is not a limit curve in λH , i.e., l is
a compact arc with endpoints in the interior of the base arc of H . Then,
l and a subarc of the base arc of H bound a disk ∆0 in H . We consider
all the leaves of λ ∩ ∆0 . We say two leaves γ0 and γ1 of λ ∩ ∆0 are of
the same type if they are isotopic in NH ∩ ∆0 , more precisely, there is an
embedded disk I × I ⊂ ∆0 such that I × ∂I = γ0 ∪ γ1 and each {p} × I
( p ∈ I ) is a subarc of an I -fiber of NH . Since ∆0 is a compact disk, there
are only finitely many types of leaves in λH ∩ ∆0 . Thus, there are finitely
many leaves l = l0 , . . . , lk such that any leaf in λH ∩ ∆0 is of the same type
as one of the li ’s. Moreover, li and a subarc of the base arc of H bound a
disk ∆i ⊂ ∆0 , and we may assume ∆0 ⊃ · · · ⊃ ∆k . We call k the index of
α and we prove this lemma by induction on k .
We now consider the case that k = 0 . Let α1 , . . . , αm be the arcs of
∆0 ∩ α . Each αi bounds a good disk Fi in ∆0 with a subarc of l = l0 , and
any two such
good disks Fi and Fj are either disjoint or nested. Moreover,
∆0 ∩ ∆ ⊂ m
i=1 Fi . Since k = 0 , each Fi is a simple good disk. Since
α is λB -parallel to β , each αi is λB -parallel to a subarc βi of β . By
bounds a good disk Fi in H that is λB -parallel to
Lemma 4.7, each βi Fi . Since ∆0∩ ∆ ⊂ m
i=1 Fi , each component of ∆0 ∩ ∆ is λB -parallel to
.
F
a subdisk of m
i=1 i
Let P
be the closure of a component of ∆ − ∆0 . Since α ⊂ ∆0 ∩ ∆ ,
∂P ⊂ ( i=1 ∂Fi ) ∪ α , and hence ∂P is λB -parallel to a simple closed curve
c in ( i=1 ∂Fi ) ∪ β . By Property 4.4, the disk bounded by c must be
λB -parallel to P . By taking the union of ∆0 ∩ ∆ and the components of
∆ − ∆0 above, we have that ∆ is λB -parallel to a disk ∆ and the lemma
holds in the case of k = 0 .
We assume the lemma is true for k ≤ n − 1 and we prove it for k = n .
By our assumption, ∆1 ⊂ ∆0 is bounded by l1 . We may also assume
that l1 is outermost among the leaves of the same type, in other words,
every leaf of λH ∩ (∆0 − ∆1 ) has the same type as l0 = l . Let Q be
the closure of the disk component of ∆ ∩ (∆0 − ∆1 ) that contains α . So,
∂Q consists of α and some subarcs of α and l1 . Let η1 , . . . , ηt be the
arcs in ∂Q ∩ l1 . Then, each ηi is an arc properly embedded in ∆ with
∂ηi ⊂ α and induced direction pointing into Q . Let α̂i be the subarc
ˆ i be the disk bounded by α̂i ∪ ηi . By our
in α with ∂ α̂i = ∂ηi . Let ∆
ˆ
assumption on Q , ∆i lies outside Q and hence the direction of ηi points
ˆ i is a good disk, and by our construction, α̂i is a good
ˆ i . So, ∆
out of ∆
arc of index less than n . Thus, by the induction hypothesis, there is an
ˆ , such that α̂i and
corresponding subarc β̂i ⊂ β , which bounds a disk ∆
i
ˆ respectively. Note that (t ∆
ˆ
ˆ i are λB -parallel to β̂i and ∆
∆
i
i=1 i )∪Q = ∆
t
ˆ
and ∂Q − int(α ) ⊂ ( i=1 ∂ ∆i ) ∪ α . Thus, there is a corresponding arc in
ˆ ) ∪ β that is λB -parallel to ∂Q − int(α ) . After some small
( ti=1 ∂ ∆
i
perturbation and splitting of λH if necessary, ∂Q − int(α ) becomes a good
arc with index 0 . Hence, Q is λB -parallel to a corresponding disk Q , and
16
ˆ i ) ∪ Q is λB -parallel to the disk ∆ = (t ∆
ˆ
∆ = ( ti=1 ∆
i=1 i ) ∪ Q , and the
lemma follows.
Lemma 4.9. Let α1 ⊂ α2 ⊂ . . . be a sequence of good arcs in int(H)
that bound a sequence of nested good disks ∆1 ⊂ ∆2 ⊂ . . . . Suppose α =
limk→∞ αk is a compact arc in H . Suppose β1 ⊂ β2 ⊂ . . . is another
sequence of arcs such that βk is λB -parallel to αk for each k . If β =
limk→∞ βk is a compact arc lying in the interior of H , then α must be a
good arc that is λB -parallel to β . In particular, if β ⊂ int(H) , the two
endpoints of α must lie in the same limit curve of H .
Proof. By the Lemma 4.8 above, there is a sequence of nested good disks
∆1 ⊂ ∆2 ⊂ . . . bounded by β1 , β2 . . . respectively. Since β = limk→∞ βk ⊂
int(H) and each βi is a good arc, ∂β must lie in a leaf of λH , and β must
bound a good disk ∆ that contains every ∆i and limi→∞ ∆i = ∆ . By
Lemma 4.8, α also be a good arc bound a good disk ∆ that is λB -parallel
to ∆ . The last assertion is true because the endpoints of a good arc lie in
the same leaf of λH by definition and certainly cannot lie in different limit
curves of H .
The next corollary follows from the lemmas above easily.
Corollary 4.10. Let α be an arc properly embedded in H . Suppose the
endpoints of α lie in different limit curves. Then, α cannot be λB -parallel
to any arc in the interior of H .
Proof. Since ∂α lie in different limit curves, there is a sequence of subarcs
α1 ⊂ α2 ⊂ . . . such that α = limk→∞ αk and the αk ’s bound a sequence of
nested good disks ∆k in int(H) . Moreover, the size of the good disks ∆k
tends to infinity as k tends to infinity. Now, if α is λB -parallel to an arc β
in the interior of H , then β bounds a good disk ∆ and by Lemma 4.9, each
∆k is λB -parallel to a subdisk of ∆ , which contradicts the conclusions on
the size the ∆k ’s.
In normal surface theory, if one has a normal surface or lamination, by
identify all the normal disks of the same type, one can construct a branched
surface fully carrying this surface of lamination [6, 2, 13]. We say two product disks in M − int(N (B)) are of the same type if they are λB -parallel.
Since we have assumed that H contains no trivial product disks in Property 4.2, there are only finitely many different types of product disks in H .
Now, similar to the construction of branched surfaces using normal disks
[6, 2, 13], by identifying the product disks in the same type and taking a
closure (since H is not closed), we get a “special” branched surface Ω .
Since H has boundary, strictly speaking, Ω is not a branched surface. Ω
may contain certain singular points as shown in Figure 4.1(a), which we
call interior singular points. Except for the interior singular points, Ω is a
branched surface with boundary. We call Ω a compression branched surface. Similar to the definition for ordinary branched surfaces, we define the
17
A
(a)
(b)
Figure 4.1
branch locus of Ω to be the set of points where Ω is not locally a manifold
(with boundary). We denote the branch locus of Ω by LΩ .
We call a point in Ω an interior point if a small neighborhood of this
point in Ω has one of the three pictures of an ordinary branched surface
(see Figure 1.1 of [12]). The boundary of Ω , which we denote by ∂Ω , is the
union of non interior points. Note that ∂Ω is a train track, and the interior
singular points are the endpoints of the train track ∂Ω .
Now, we describe the fibered neighborhood of a compression branched
surface Ω . We can construct a neighborhood of Ω in M so that, near
any ordinary point, it is the same as a fibered neighborhood of a branched
surface, and the picture near an interior singular point is as shown if Figure 4.1(b). We will denote such a neighborhood of Ω by N (Ω) . Similar to a
fibered neighborhood of an ordinary branched surface, N (Ω) is also a union
of I -fibers, and there is a map π : N (Ω) → Ω that collapse every I -fiber to
a point. Similarly, we can also define the horizontal and vertical boundary
∂h N (Ω) and ∂v N (Ω) of N (Ω) . Note that ∂h N (Ω) near an interior singular
point looks like Figure 4.1(b). If ∂Ω = ∅ , then ∂N (Ω) − ∂h N (Ω) − ∂v N (Ω)
can be considered as a fibered neighborhood of the boundary train track
∂Ω , which we denote by N (∂Ω) .
By our construction, we may assume that H ⊂ N (Ω) and H is transverse
to the I -fibers of N (Ω) . Since H is not closed, we cannot guarantee that
every I -fiber of N (∂Ω) intersects H . However, we may assume that each
I -fiber in N (∂Ω) either contains a point of ∂H , or contains a limit point
of H . We say that Ω fully carries H if the I -fibers of N (∂Ω) satisfy
the condition above and every I -fiber in N (Ω) − N (∂Ω) intersects H .
Moreover, we may assume that any I -fiber of N (Ω) lies either in M − λ
or totally in a leaf of λ . In other words, the intersection of each leaf of λ
with N (Ω) is a union of I -fibers of N (Ω) .
The branch locus of Ω is a union of (immersed) closed curves and compact
arcs. Since H is almost vertical with respect to B , after some isotopy if
necessary, we may assume that the branch locus LΩ is transverse to λ . Let
α be a compact arc in the branch locus of Ω . Then we have the following 3
cases: the first case is that both endpoints of α are interior singular points;
the second case is that one endpoint of α is an interior singular point and
18
the other endpoint is a switch of the boundary train track ∂Ω ; the third
case is that both endpoints of α are switches of the boundary train track.
Definition 4.11. Let D0 and D1 be two disks or arcs in H . We say that
D0 and D1 are Ω -parallel if there is a local embedding f : D × I ⊂ N (Ω)
such that f (D×{i}) = Di ( i = 0, 1 ) and f ({x}×I) is a subarc of an I -fiber
in N (Ω) for any x ∈ D . Note that by our construction and assumptions
above, if D0 and D1 are Ω -parallel, then D0 and D1 are λB -parallel.
Notation. Let γ be a component of the branch locus LΩ . In this paper, we
alway denote the corresponding component of ∂v N (Ω) by γ × I with each
{x} × I ( x ∈ γ ) a vertical arc of ∂v N (Ω) .
Definition 4.12. Let Ω be the compression branched surface fully carrying
H as above, and γ be a component of the branch locus LΩ . Let γ × I be
the corresponding component of ∂v N (B) , and γi = γ × {i} ( i = 0, 1 ). Let
β be an arc in π −1 (γ) transverse to the I -fibers. We say that β is on the
γi ( i = 0, 1 ) side if there is an embedded disk c × I ⊂ π −1 (γ) such that
c × {0} ⊂ γi , c × {1} = β , (c × I) ∩ ∂v N (Ω) = γi , and each {p} × I is a
subarc of an I -fiber of N (Ω) .
At this stage, the notion of Ω -parallel are basically the same as the notion
of λB -parallel. Later, we will perform some splitting on Ω and after any
splitting, Ω -parallel disks are still λB -parallel.
By Lemma 4.8, the good arcs of H ⊂ N (Ω) have the following property.
Property 4.13. Let α1 be a good arc that bounds a good disk ∆1 in H ,
and α2 be an arc in H that is Ω -parallel to α1 . Then, α2 must be a good
arc that bounding a good disk ∆2 that is Ω -parallel to ∆1 .
Definition 4.14. Let γ be a component of LΩ . Suppose there is a subarc
α in γ with α ∩ λ = ∂α and the transverse direction of λ (at ∂α ) pointing
out of α . By our construction, there must be a good arc α bounding a
good disk ∆ in H such that π(α ) = α ⊂ LΩ , where π : N (Ω) → Ω is the
map collapsing every I -fiber to a point. We call π(∆) a bad bigon if the
branch direction of α (induced from the branch direction of α ) points out
of ∆ . We say the Ω is compatible with H and λ if LΩ is transverse to λ
and there is no bad bigon in Ω .
Proposition 4.15. The compression branched surface Ω constructed above
is compatible with H .
Proof. Let α be an arc in γ ⊂ LΩ as in the Definition above. By our
construction, there must be a pair of disjoint good arc α and α of H ,
one on γ0 side and the other on γ1 side, such that π(α ) = π(α ) = α . Let
∆ and ∆ be the two good disks bounded by α and α in H respectively.
Since π(α ) = π(α ) , α and α are λB -parallel. Hence, ∆ and ∆ are
λB -parallel and π(∆ ) must be equal to π(∆ ) . This implies that the induce
branch directions of α and α must point into ∆ and ∆ respectively,
since α and α are on different sides.
19
5. Some splittings on N (Ω)
In this section, we would like to perform some splittings on N (Ω) that
preserve Property 4.13. First of all, removing a disk of contact clearly preserves Property 4.13. Now, we discuss some other useful splittings that
preserve Property 4.13. To simplify notation, we still use Ω and N (Ω) to
denote the compression branched surface after splitting Ω and N (Ω) .
5.1. Excellent disks.
Definition 5.1. Let γ be a component of LΩ and γ ×I be the corresponding component of ∂v N (Ω) as before. Let α = η × {i} be an arc in γ × I ,
where η ⊆ γ . We call a disk Γ ⊂ N (Ω) an excellent disk if Γ satisfies the
following conditions.
1. Γ is transverse to the I -fibers of N (Ω) and Γ ∩ H = ∅ ;
2. ∂Γ = α ∪ β with β ⊂ λ ;
3. int(Γ) ⊂ int(N (Ω)) ;
4. the induced direction of β (from the transverse direction of λ ) points
out of Γ ;
5. for each point x ∈ Γ − β , the intersection of H with any of the two
components of Ix − x is not empty, where Ix denotes the I -fiber of
N (Ω) containing x .
We call the arc α the base arc of Γ . We call Γ a simple excellent disk if in
addition λ ∩ Γ cuts Γ into a sequence of nested sub excellent disks, Γ itself
lies in the interior of some excellent disk and Γ lies between two λB -parallel
good disks.
Remark 5.2. Let Γ be a simple excellent disk. We can splitting N (Ω) along
Γ by deleting a small product neighborhood of Γ from N (Ω) . By Proposition 4.15 and the argument above, after this splitting, N (Ω) still has Property 4.13. Moreover, after some splitting and isotopy on λ , we may assume
that branch locus LΩ is still transverse to λ . In fact, by Lemma 5.3, after
this splitting N (Ω) still fully carries H , i.e. every I -fiber of N (Ω)−N (∂Ω)
intersects H .
Lemma 5.3. Let Γ1 , . . . , Γn be a collection of disjoint excellent disks and
α1 , . . . , αn be the corresponding base arcs. Suppose each vertical arc of
∂v N (B) meets at most one αi . Then, for any point x ∈ Γk , the intersection of H with any component of Ix − ∪ni=k Γk is not empty, where Ix
denotes the I -fiber of N (Ω) containing x .
Proof. Let βk = ∂Γk − int(αk ) . By the definition of excellent disk, each βk
lies ina leaf of λ . By compactness, the intersection of each I -fiber of N (Ω)
with nk=1 Γk contains finitely many points.
Suppose the lemma is nottrue. Then, there must be a subarc J of an
I -fiber of N (Ω) with ∂J ⊂ ni=1 Γi and int(J) ∩ H = ∅ . Suppose the two
endpoints of J lies in Γi and Γj . Since the induced direction of each βk
points out of Γk for each k , we can choose the arc J so that both endpoints
20
lie in the interior of Γi and Γj . Then, there is a product D × I ⊂ N (Ω)
such that D × ∂I consists of two neighborhoods of the two endpoints of J
in Γi and Γj respectively, x × I is a subarc of an I -fiber of N (Ω) , and
D × int(I) ∩ H = ∅ . By taking such a D × I to be maximal, we have the
following: for each p ∈ ∂D , either ({p} × I) ∩ ∂v N (Ω) = ∅ , or at least one
endpoint of {p} × I lies in ∂Γi ∪ ∂Γj , or both.
Suppose ({p} × I) ∩ ∂v N (Ω) = ∅ for some p ∈ ∂D . Let A and B be the
two endpoints of {p} × I . Since Γk ∩ ∂N (Ω) = αk ⊂ ∂v N (Ω) , one endpoint
of {p} × I must lie in αi or αj , say αi . Since each vertical arc of ∂v N (B)
meets at most one αk , by choosing p ∈ ∂D so that {p} × I ⊂ M − λ
and π({p} × I) is not a double point of the branch locus LΩ , we can
assume that the other endpoint of {p} × I , say A , lies in int(N (Ω)) and
A ∈ int(Γj ) . Hence, there is a point B in the interior of {p} × I lying in
∂v N (Ω)∩∂h N (Ω) . Now, we consider the subarc AB of {p}×I , and we have
AB − B ⊂ int(N (Ω) . Since A ∈ int(Γj ) and B ∈ ∂v N (Ω) ∩ ∂h N (Ω) , by
pushing the arc AB along Γj in the direction opposite to branch direction,
one can find a point x ∈ Γj close to A such that the closure of a component
of Ix − x is close and parallel to AB , where Ix is the I -fiber of N (Ω)
containing x . Since AB ∩ H = A , the component of Ix − x does not
intersect H , which contradicts our assumption.
So, we can assume ({p}×I)∩∂v N (Ω) = ∅ ( p ∈ ∂D ). Hence one endpoint
of {p} × I must lie in ∂Γi ∪ ∂Γj . So, we can assume that D × {0} is a
subdisk of Γi and ∂(D × {0}) ∩ ∂Γi = ∅ . As ({p} × I) ∩ ∂v N (Ω) = ∅ for any
p ∈ ∂D , ∂(D × {0}) ∩ ∂Γi must a union of subarcs of βi . Hence, there must
exist an arc δ in ∂D such that δ × {0} is properly embedded in Γi with
both endpoints in βi . Since D × I is maximal by our assumption, δ × {1}
must be a boundary arc in Γj and hence a subarc of βj . However, since each
βk lies in a leaf of λ , the leaf of λ containing βi and the leaf containing
βj would have to intersect transversely at the endpoints of δ × {1} to make
the phenomenon above happen, which is impossible as λ is an essential
lamination.
5.2. Splitting along certain arcs. Let α be an embedded arc in Ω that
is transverse to λ . Suppose α admits a direction along α that agree with
the transverse orientation of λ at each point of λ ∩ α . If we perform a
splitting (of Ω along H ) in a small neighborhood of α along the direction
above, as shown in Figure 5.1, then after this local splitting, N (Ω) still
has Property 4.13. To see this, suppose the splitting along a small arc α
destroys Property 4.13. In other words, suppose there is a pair of Ω -parallel
good arcs α1 and α2 , but the good disks ∆1 and ∆2 that they bound are
not Ω -parallel in the N (Ω) after splitting. Since ∂∆i − αi ⊂ λ ∩ H and the
transverse direction of ∂∆i − αi points out of ∆i , the splitting must “cut”
the leaf of λ that contains ∂∆i − αi , which contradicts the assumptions on
the direction of α and the direction of the splitting along α . We call α
permissible arc.
21
splitting
α
Figure 5.1
5.3. Characteristic sections.
Definition 5.4. Let γ be a component of the branch locus LΩ and γ × I
be the corresponding component of ∂v N (Ω) . We denote γ × {i} by γi for
each i ∈ I = [0, 1] . If γ does not have self-intersection, then there is an
embedded disk or annulus (depending on whether γ is an arc or a closed
curve) Sγ = α × [−, ] in a small neighborhood of γ such that α × {0} = γ
and Sγ is transverse to Ω . So, Sγ ∩ Ω is a train track containing γ and the
switches of the train track are the double points of LΩ in γ . The train track
Sγ ∩Ω is simply the union of γ and some “tails” on γ . Let Nγ = π −1 (Sγ ∩Ω)
and Nγ can be considered as a fibered neighborhood of the train track Sγ ∩Ω
containing γ × I . Then, Nγ − γ × int(I) has two components, and we use
Nγi ( i = 0, 1 ) to denote the component of Nγ − γ × int(I) that contains
γi = γ × {i} . If γ has self-intersection, then we can abstractly construct
a train track by adding some tails to γ , a fibered neighborhood Nγ of the
train track, and a local embedding f : Nγ → N (Ω) which maps I -fibers
to I -fibers. We can consider f (Nγ ) to be the union of π −1 (γ) and some
“tails”. Note that if γ has no self-intersection, then f is an embedding and
the image of f is the Nγ described above. We may regard γ × I ⊂ Nγ
and identify γ × I in ∂v N (B) with its image under f . Similarly, we use
Nγi ( i = 0, 1 ) to denote the two component of Nγ − γ × int(I) and f (Nγi )
contains γi ⊂ ∂h N (Ω) ∩ ∂v N (Ω) . We call Nγi a characteristic section of
γi ( i = 0, 1 ), and we also call f |−1
Nγ (H) a characteristic lamination for γi
i
(this is an abuse of notation, since H is not be a closed set and f −1 (H)
may not be a lamination). Let T be the union of all the I -fibers of Nγk
that do not belong to π −1 (γ) , and we call each component of T a tail of
Nγi .
Definition 5.5. Let γ be a component of the branch locus LΩ and γ × I
be the corresponding component of ∂v N (Ω) . We denote γ × {i} by γi for
each i ∈ I = [0, 1] . Let Nγi be the characteristic section of γi ( i = 0, 1 ) and
f : Nγi → N (Ω) be the corresponding local embedding as in Definition 5.4.
Let β ⊂ f (Nγi ) be a compact arc and β ∩ H = ∅ . We say that β is a bad
splitting arc on the side of γi if β is transverse to the I -fibers, one endpoint
of β lies in γi and the other endpoint of β lies in f (Nγi ) ∩ ∂v N (Ω) − γi ,
22
Figure 5.2
as shown in Figure 5.2 (In Figure 5.2, the dotted arc is a bad splitting arc
and the shaded area is γ × I ).
Let γ be a component of LΩ and γ × I be the corresponding component
of ∂v N (B) . Let γi = γ × {i} ( i = 0, 1 ) and Nγ be as in Definition 5.4.
Let Nγi be the characteristic section and fi : Nγi → N (Ω) be the local
embedding as in Definition 5.4. Let µi = fi−1 (λ) be the characteristic
lamination in Nγi . We call an arc α ⊂ µi a good arc if f (α) ⊂ π −1 (γ) is
a good arc in H . Let αi ⊂ µi ( i = 0, 1 ) be two good arcs. We say that
α0 and α1 are Ω -parallel if f (α0 ) and f (α1 ) are Ω -parallel in N (Ω) . To
simplify notation, we often identify Nγi with its image f (Nγi ) in π −1 (γ) .
If γ is an arc component of LΩ , since the induced transverse direction
of ∂γ points out of γ , γ × I must contain some excellent arcs. Moreover,
if α1 and α2 are both excellent arcs in γ × I , let q : γ × I → γ be the
projection map, then q(α1 ) and q(α2 ) are either disjoint or nested. Let σi
be the union of arcs in µi that have at least one endpoint in the tail of Nγi .
So, µi − σi consists of good arcs that are Ω -parallel to γi ( i = 0, 1 ). Let
π be the map that collapse every I -fiber of Nγi to a point. Then, there is
a bad splitting arc in Nγi if and only if int(γi ) − π(µi − σi ) = ∅ . Let α be
the closure of a component of π(µi − σi ) in γi . Then, there is a sequence of
nested good arcs {αi } such that the limit of π(αi ) is α . Further, after some
splitting along simple excellent disks, we may assume {αi } is a sequence of
nested simple good arcs. Thus, we may assume that Ω has the property
that each component of π(µi − σi ) in each Nγi is the limit of a sequence
{π(αi )} , where the αi ’s are nested simple good arcs.
Next, we perform two types of splitting to eliminate bad splitting arcs.
The first is the splitting along a simple excellent disk. One can view this
splitting as deleting a small fibered neighborhood of the simple excellent
disk from N (Ω) . By some small isotopy and splitting of λ if necessary,
we may always assume that the branch locus is still transverse to λ after
23
splitting
A
B
Figure 5.3
the splitting. The second splitting is the splitting along a permissible arc α
with an induced transverse direction as shown in Figure 5.1 and described
above.
After some splitting along simple excellent disks, we may assume each bad
splitting arc β admits a direction along β that agree with the transverse
direction of β ∩ λ . Then, we split N (Ω) along β from the endpoint (of
β ) near the tail of Nγi to the endpoint in γi . So, this splitting preserve
Property 4.13. Note that after the splitting along bad splitting arc, we create
two interior singular points, as shown in Figure 5.3.
Note that during the splittings along excellent disks and permissible arcs,
we created some bubbles in Nγ , but the top and bottom of such a bubble
can be extended to two λB -parallel arcs if one undo the splitting. Although
such splittings may create some double points in the branch locus, we do not
really count them. After the splitting along a bad splitting arc, we eliminate
one double points of the original branch locus. Eventually, we can eliminate
all the bad splitting arcs connecting the double points of the original branch
locus. Now, since the top and bottom arcs can be extended to a pair of
λB -parallel arcs, after some more splitting along simple excellent disks, we
can eliminate all bad splitting arcs.
6. Simplifying the compression branched surface
By section 5, we may assume our N (Ω) contains no bad splitting arcs
and N (Ω) still satisfies Property 4.13.
Lemma 6.1. Let γ be a component of LΩ and γ × I be the corresponding
component of ∂v N (Ω) as before. Suppose γ is a circle. Then, there are a
maximal collection of disjoint excellent arcs α1 , . . . , αm ⊂ γ × {1/2} that
bound disjoint excellent disks ∆1 , . . . , ∆m , such that there is a direction
along the circle γ × {1/2} that agrees with the direction (induced
from the
transverse direction of λ ) of each point in λ ∩ (γ × {1/2} − m
i=1 αi ) . In
particular, after splitting along these excellent disks ∆1 , . . . , ∆m , γ becomes
a circle that is transverse to λ and admits a direction compatible with the
transverse direction of λ .
Proof. We consider the characteristic neighborhood Nγk and the characteristic lamination µk of γk = γ × {k} ( k = 0, 1 ). First note that µk does
24
not contain any closed circle, since if there is a circle c in µk ⊂ H , then c
must bound a disk in H , and hence either γ × I bounds a disk of contact
or we get a contradiction to Property 4.4.
Since there is no bad splitting arcs and µk contains no circle, we have
two cases. The first case is that there is an infinite spiral curve that wraps
around γk infinitely many times, and the second case is that there is a
sequence of spiral arcs {βi } such that βi has exactly one endpoint in a tail
of Nγk , βi is Ω -parallel to a subarc of βi+1 and the limit of the length
of {βi } is infinity. Note that in the second case, one endpoint of each βi
lies in a tail and the other endpoint lies in a limit curve and H . Moreover,
as i goes to infinity, the number of times that βi wraps around γi goes to
infinity.
If γ does not admit a direction that agree with the transverse orientation
of λ , by Property 4.13, there must be a simple excellent arc in γ × {1/2} .
Let π be the projection map that collapsing every I -fiber of Nγ to a point.
For any good arc α in µ0 and β in µ1 that do not intersect the tails of
Nγ , then π(α) and π(β) are either disjoint or nested.
If there is a sequence of nested good arcs {ηi } in µ0 , since γ is a circle,
there must a maximal good arc η in µ0 such that each ηi is Ω -parallel to a
subarc of η . Let α1 , . . . , αm be such maximal good arcs in µ0 and suppose
that is
the π(αi ) ’s are disjoint arcs in γ . So, if γ dose not admit a direction
)) ,
π(α
compactible with the induced direction of each point of λ∩(γ − m
i
i=1
then there must be another good arc β in µ0 such that either π(αi ) ⊂ π(β)
for some i or π(β) ∩ π(αi ) = ∅ for each i . By choose m and each αi to be
maximal, one can conclude that no such β exist. Similarly, we can find such
a sequence of maximal good arcs α1 , . . . , αm in µ1 . By the assumption on
the αi ’s above, the π(αi ) ’s and the π(αi ) ’s must be the same collection of
arcs. Thus, we may assume that αi and αi are a pair of Ω -parallel good
arcs for each i . Hence αi and αi bound a pair of Ω -parallel good disks.
So, there is an excellent arc αi ⊂ γ × {1/2} bounding an excellent disk ∆i
lying between the two good disks bounded by αi and αi , and the lemma
follows.
Lemma 6.2. Let γ be a component of LΩ and γ × I be the corresponding component of ∂v N (Ω) . Let γk = γ × {k} ( k = 0, 1 ) and µk be the
characteristic lamination in Nγi as before. Suppose γ is a compact arc.
Then, there is a sequence of good arcs {αi } in µ0 that is Ω -parallel to a
sequence of nested excellent arcs {βi } in γ × {1/2} such that βi ⊂ βi+1
and limi→∞ π(βi ) = γ .
Proof. The proof of this lemma is basically the same as the previous one. In
the case that γ is a compact arc, if η1 and η2 are two good arcs in µ0 and
π(η1 ) ∩ π(η2 ) = ∅ , then since there is no bad splitting arc, there must be
another good arc η in µ0 such that π(η) contains both π(η1 ) and π(η2 ) .
Thus, since the induced transverse direction of ∂γ points out of γ , there
must be sequences of nested Ω -parallel good arcs {αi } and {αi } in µ0 and
25
µ1 respectively, and there is an excellent arc βi ⊂ γ × {1/2} lying in the
middle of αi and αi for each i .
Asumption 6.3. After eliminating all the bad splitting arcs and splitting
along finitely many excellent disks for all the circular components of LΩ ,
we can assume the following:
1. each circular component of LΩ admits a direction compatible with the
transverse orientation of λ ;
2. for each arc component of LΩ , there is a sequence of excellent disk as
in Lemma 6.2.
Lemma 6.4. Let α ⊂ λ be a compact arc with ∂α ⊂ ∂v N (Ω) , int(α) ⊂
int(N (Ω)) and α ∩ H = ∅ . If one endpoint of α lies in γ × I , where γ is
a compact arc, then both endpoints of α must lie in γ × I . Moreover, ∂α
bounds an excellent arc β in γ × I and α ∪ β bounds an excellent disk.
Proof. By Lemma 6.2, if γ is a compact arc, then there is a sequence of
excellent disks with base arcs in γ × I and the limit of the base arcs is the
whole γ × {1/2} . Since α lies in a leaf of λ , α cannot cut through the top
arcs of the excellent disks which also lie in leaves of λ by the definition of
excellent disk. Hence, both endpoints of α must lie in γ × I and α together
with an arc in γ × I bound an excellent disk.
Remark 6.5. We can split N (Ω) along the excellent disk in Lemma 6.4
and get another compression branched surface still fully carrying H . This
splitting in Lemma 6.4 can also be viewed as first drilling a tunnel along α
and then eliminating the disk of contact created by the drilling.
Lemma 6.6. Let α ⊂ λ be a compact arc with ∂α ⊂ ∂v N (Ω) , int(α) ⊂
int(N (Ω)) and α ∩ H = ∅ . Suppose the two endpoints of α lie in γ1 × I
and γ2 × I , where γ1 and γ2 are circular components of LΩ . Then,
1. if γ1 = γ2 , then γ1 ×I and γ2 ×I bound a splitting annulus A ⊂ N (Ω)
transverse to the I -fibers of N (Ω) and with ∂A in γ1 × I and γ2 × I .
2. if γ1 = γ2 , then there is a splitting Möbius band A transverse to the
I -fibers of N (Ω) and with ∂A in γ1 × I .
In both cases, we can split along A and get a compression branched surface
that still satisfies Property 4.13 and has fewer circles in the branch locus.
Proof. We only prove the first case and the second case is very similar. By
Lemma 6.1, each circle admits a direction compatible with the transverse
orientation of λ . So, there is an arc ηi in γi × I ( i = 1, 2 ) such that
η1 ∪ α ∪ η2 is a arc which lies between two Ω -parallel good arcs. Hence,
there is a disk ∆ with η1 ∪ α ∪ η2 ⊂ ∂∆ and ∆ is Ω -parallel to a good
disk. Furthermore, we can extend at least one of ηi infinitely so that the
size of ∆ is large and π|∆ is not injective. Since ∆ ∩ H = ∅ , similar to
the proof of Lemma 5.3, π(∆) cannot have any branching. Now, it is not
difficult to see that ∆ must wrap around an annulus between γ1 × I and
γ2 × I , since ∆ is Ω -parallel to a good disk. Thus, there must be an such
26
splitting
(a)
splitting
(b)
Figure 6.1
annulus A between γ1 ×I and γ2 ×I . Since A∩H = ∅ , it is clear that after
cutting N (Ω) along A the new compression branched surface still satisfies
Property 4.13.
Therefore, we can assume there is no splitting annulus or Möbius band in
N (Ω) as in Lemma 6.6.
Let γ ×I ( γ is a compact arc) be a component of ∂v N (Ω) . Let {βi } be a
sequence of nested excellent arcs in γ×{1/2} and {∆i } be the corresponding
sequence of excellent disks bounded by {βi } as in Lemma 6.2. We say γ × I
is of infinite type if limi→∞ weight(∆i ) is infinite, and say γ × I is of finite
type if limi→∞ weight(∆i ) is finite.
If γ×I is of finite type, there must be an excellent disk Γ with ∂Γ = α∪δ ,
α = γ × {1/2} , δ ⊂ λ and each ∆i is a subdisk of Γ . Then, as shown
in Figure 6.1, we can split N (Ω) to eliminate all finite type components of
∂v N (Ω) . In particular, by Lemma 4.9, if any endpoint of γ is an interior
singular point in Ω , then γ×I must be of finite type. Hence, we can perform
some splitting as shown in Figure 6.1(b) to remove a pair of interior singular
points. Therefore, after these splittings, we eliminate all the interior singular
points and Ω becomes an ordinary branched surface.
Now, we consider the circular components of LΩ . Let γ be a circular
component of LΩ and γ×I be the corresponding component of ∂v N (Ω) . By
Lemma 6.1, γ admits a direction compatible with the transverse orientation
of λ . If γ ∩ λ = ∅ , then since H is cut by λH into disks, we either have
a contradiction to Property 4.4 or γ × I bounds a disk of contact. Hence,
γ ∩ λ = ∅ . Let x be a point of γ × λ and Ix = π −1 (x) be the I -fiber of
N (Ω) corresponding to x , and by our previous assumptions on λ ∩ N (Ω) ,
we have Ix ⊂ λ . By our construction Ix ∩ int(H) = ∅ . Let A be the
27
γ
splitting
α
β
∂Ω
∂Ω
Figure 6.2
point Ix ∩ γ × {1/2} and A be a point in Ix ∩ int(H) . So, we can find a
small arc α ⊂ N (Ω) − H with α ∩ ∂N (Ω) = A such that α is Ω -parallel
to an arc α ⊂ λH and A ∈ ∂α . So, we can split N (Ω) along α . By
Lemma 6.6, ∂α − A can never be a point in ∂v N (Ω) , hence we can assume
the length of α to be arbitrarily large. By choosing an appropriate α , we
may assume that the endpoint point B = ∂α−A of α is close to N (∂Ω) and
there is a permissible arc β with one endpoint B and the other endpoint
in N (∂Ω) . As shown in Figure 6.2, we first split N (Ω) along α and then
split it along the permissible arc β , in other words, we drill a tunnel along
α ∪ β connecting γ × I to N (∂Ω) . After this splitting, γ becomes an arc
and we get a compression branched surface that still satisfies Property 4.13
and has fewer circular components.
Therefore, we may assume that LΩ contains no circular component and
each arc component of LΩ is of infinite type.
Next, we will split each infinite type component of ∂v N (Ω) into a standard form. We first assume the branch locus LΩ is transverse to λ . Let γ
be a component of LΩ . After some isotopy, we may assume that there are
two disjoint small subarcs η and η of γ containing the two endpoints of
γ , and η and η are also subarcs of the I -fibers of N (B) . Let γ × I be
the corresponding component of ∂v N (B) as before. By Lemma 6.2, there
is an excellent arc βk ⊂ γ × {1/2} and a corresponding excellent disk Γk
such that the two endpoints of βk lie in η × I ⊂ γ × I and η × I . Let δ
be the arc ∂Γk − int(βk ) . Hence, by the definition of excellent disk, δ lies
in a leaf of λ .
As before, we can split N (Ω) across the excellent disk Γk by deleting a
product neighborhood of Γ from N (Ω) . To simplify notation, we still use
Ω and N (Ω) to denote the branched surface after splitting, and still call the
corresponding component of ∂v N (B) after this splitting γ × I . Moreover,
we will still use η , η and δ to denote the corresponding part of γ after
the splitting. More precisely, we may assume that after the splitting, γ × I
consists of 3 parts: η × I , η × I , and δ × I , where η and η are two
disjoint arcs in γ containing the two endpoints of γ and δ is the closure of
γ − η − η . By the definition of an excellent disk, we may assume δ × I lies
in a leaf of λ . Thus, after splitting along an excellent disk, we may assume
28
γ = η ∪ δη , where δ lies in a leaf of λ , η and η are subarcs of the I -fibers
of N (B) and contain the two endpoints of γ .
We can perform the splitting above to each component of LΩ . Let
γ1 , . . . , γm be the arc components of LΩ . After splitting along finitely many
excellent disks, we may assume that each γk is a union of three arcs ηk ,
ηk and δk , where δk ⊂ λ , ηk and ηk contain the two endpoints of γk and
are two subarcs of N (B) . By our construction, the branch direction of each
δk ⊂ γk agrees with the transverse direction of λ .
Next, we will split Ω further so that LΩ has no double point. Let γk =
ηk ∪ δk ∪ ηk be as above. Since the δk ’s lie in λ ∩ Ω , after some small
perturbation if necessary, we can assume that each double point of LΩ is
an intersection point of δi and ηj (or ηj ) for some i and j . Let X be
an intersection point of δi and ηj , and Y be the endpoint of ηj lying in
the boundary train track ∂Ω . We may choose X so that the subarc XY
of ηj does not contain any other double point of LΩ . Since ηj is a subarc
of N (B) and Y is a boundary point of Ω , the induced tranverse direction
(from λ ) of the arc XY must point from X to Y . Moreover, since the
branch direction of each δi agrees with the transverse orientation of λ , the
branch direction of δi at X points towards Y . Thus, we may consider XY
as a permissible arc and we split δi from X to Y along the arc XY , as
shown in Figure 6.3. Since XY is a permissible arc, after the splitting N (Ω)
still satisfies Property 4.13. After this splitting, γi breaks into two arcs, and
we may assume each is the union of two subarcs of I -fibers of N (B) and an
arc from λ ∩ Ω . If any of the two arcs is a component of finite type, we can
perform a splitting to eliminate such an arc as above. After this splitting
in Figure 6.3, the number of double points of LΩ is reduced. Hence, after
finitely many steps, LΩ is a collection of embedded and mutually disjoint
arcs of infinite type in Ω .
Now, the closure of each component C of Ω − LΩ is a compact. Since H
is a union of bigons. We can extend λ ∩ C to a foliation with a singularity
on each arc LΩ ∩ C . Thus, by an index argument, it is easy to show that
the Euler characteristic of C is positive. Hence, each component of Ω − LΩ
is a disk. Moreover, for every component C of Ω − LΩ , there must be an
arc in LΩ with branch direction pointing into C and an arc in LΩ with
branch direction pointing out of C .
7. Evacuating taut sutured manifolds
As in the last section, LΩ has no circular component and no double
points, and Ω − LΩ is a collection of disks. So, we can treat Ω as a 2complex with vertices the endpoints of LΩ , 2-cells the disks of Ω − LΩ .
Note that each component of LΩ is a 1-cell shared by three 2-cells. The
boundary of the branched surface has two parts. The first part, which we
denote by γ , is the union of the base arc γ0 of H and the two vertical arcs
at the endpoints of γ0 , and the remaining part is a train track TΩ lying in
the leaves of λ . Then, we “blow air” into the leaves of λ that contain TΩ .
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splitting
δi
X
Y
Figure 6.3
More precisely, we first replace the leaves of λ that contain the train track
TΩ by I -bundles over these leaves (we require that the train track TΩ lies
in the interior of these I -bundles), and then we delete the interior of these
I -bundles from λ . Now, the train track TΩ lies in M − λ and N (Ω) still
carries H (but not fully carries H ). So, π(H) ⊂ Ω − TΩ . After a small
isotopy, we can also assume that the 1-skeleton of Ω (i.e. LΩ ) is transverse
to λ .
It is easy to construct a cell decomposition containing Ω as a subcomplex
of the 2-skeleton and λ is normal except at Ω . As in [3, 7], there is a
standard isotopy on λ pushing the leaves of λ across the 1-skeleton to locally
normalize the lamination. By choosing an appropriate cell decomposition
for M , we can assume that the only non-normal parts of λ are always at
Ω after the normalizing isotopy above.
By our construction, also see [7], λ cannot be a normalized using finite
number of the normalizing isotopies. Nevertheless, by [7], we can get a new
normal lamination by evacuating a taut sutured manifold (N, s) . If (N, s)
is not a product sutured manifold, then the gut number of the essential
lamination obtained by evacuating (N, s) increases. Since the maximal gut
number of essential laminations in M is bounded from above, if the (N, s) is
never a product sutured manifold, then after finitely many steps, we obtain
a tight essential lamination.
Next, we will show that under our assumption and construction of H ,
(N, s) cannot be a product sutured manifold.
Since the only non-normal parts of λ are at Ω , by [7], the taut sutured
manifold can be described as follows. After some isotopy, we may assume
the train track TΩ lies in the leaves of λ . N can be consider as a small
neighborhood of Ω , and ∂N has two parts. The first part R− is a small
neighborhood of the train track TΩ and hence R− ⊂ λ , and the second part
is the closure of ∂N − R− , which we denoted by R+ . R− ∩ R+ is a union of
simple closed curves. By thickening R− ∩ R+ into a union of annuli s , we
may regard N as a sutured manifold (N, s) , where s is the suture, R+ and
R− are the ± -sides of (N, s) . By [7], (N, s) is a taut sutured manifold.
Moreover, by our construction of the cell decomposition, after evacuating
the taut sutured manifold (N, s) (as in [7]), one obtains a new lamination
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which is normal with respect to the cell decomposition, and after eliminate
all possible Reeb components, one gets a normal essential lamination.
By [7], the union of (N, s) and all the Reeb components is also a taut
sutured manifold, which we denote by (N̂ , ŝ) , and the final normal essential
lamination can be regarded as the lamination obtained by evacuating (N̂ , ŝ) .
By [7], if (N̂ , ŝ) is not a product sutured manifold, then the gut number
of the normal essential lamination in the end increases (suppose there is no
generalized cylindrical component). Now, we consider the case that (N̂ , ŝ)
is a product sutured manifold. It is easy to see from our construction that
(N, s) is not a product sutured manifold, and hence there must be Reeb
components in the lamination obtained by evacuating (N, s) . For each solid
torus T bounding a Reeb component, by [7], we may assume ∂T ∩N ⊂ R+ .
Let γ be the base arc of H . By our construction, ∂γ are the two
endpoints of the train track TΩ . Hence, after a small isotopy, we may assume
γ is an arc properly embedded in the sutured manifold N with ∂γ ⊂ R− .
Furthermore, we may assume that Ω lies in N and Ω ∩ N ⊂ R− . Note
that γ is not knotted, since γ is the base arc of H and there is a sequence
of disks {Di } in H , as in section 3, with lim ∂Di = ∂H − int(γ) . As (N̂ , ŝ)
is a product sutured manifold, γ can be isotoped into R− , in other words,
there is an embedded disk ∆ such that ∂∆ = γ ∪ β and ∆ ∩ ∂N = β ⊂ R− .
We can view the disk ∆ as a disk embedded in M with β ⊂ λ , and hence
we may assume the int(D) is transverse to the branched surface B and we
can perturbed γ a little so that γ ∩ N (B) consists of two subarcs of the
I -fibers of N (B) containing ∂γ , and ∆ ∩ N (B) is a union of subarcs of
I -fibers of N (B) . However, by Lemma 2.2, one can isotope ∆ fixing γ so
that β is still in λ and ∆ becomes a transverse bigon (see section 2), which
contradicts assumption 3.6. So, under our initial assumption, (N̂ , ŝ) cannot
be a product sutured manifold. This finishes the proof of Theorem 1.1.
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Department of Mathematics, Oklahoma State University, Stillwater, OK
74078, USA
E-mail address: [email protected]
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