Tài liệu Đề tài " The topological classification of minimal surfaces in R3 " pptx

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684 CHARLES FROHMAN AND WILLIAM H. MEEKS III
in the ball of radius R centered at the origin, then the results in [2] imply that
lim
R→∞
A(R)/πR
2
is an integer with the same parity as the end m. Thus, the
parity of m could also be defined geometrically in terms of its area growth.
This discussion proves the next proposition.
Proposition 2.2. If F is a properly embedded minimal surface in R
3
,
then each middle end of F has a parity.
In [9] Frohman and Meeks proved that the closures of the complements
of a minimal surface with one end in R
3
are handlebodies; that is, they are
homeomorphic to the closed regular neighborhood of a properly embedded con-
nected 1-complex in R
3
. Motivated by this result and their ordering theorem,
Freedman [4] proved the following decomposition theorem for the closure of a
complement of F when F has possibly more than one end.
Theorem 2.3 (Freedman). Suppose H is the closure of a complement
of a properly embedded minimal surface in R
3
. Then there exists a proper
collection D of pairwise disjoint minimal disks (D
n
,∂D
n
) ⊂ (H, ∂H),n ∈ N,
such that the closed complements of D in H form a proper decomposition of H.
Furthermore, each component in this decomposition is a compact ball or is
homeomorphic to A × [0, 1), where A is an open annulus.
3. Construction of the family of planes P
In [9] we proved the Topological Classification Theorem for Minimal Sur-
faces in the case the minimal surface F has one end. Throughout this section,
we assume that F has at least two ends.
Lemma 3.1. Let F be a properly embedded minimal surface in R
3
with
one or two limit ends and horizontal limit tangent plane. Suppose H
1
,H
2
are
the two closed complements of F and D
1
and D
2
are the proper families of disks
for H
1
,H
2
, respectively, whose existence is described in Freedman’s Theorem.
Then there exist a properly embedded family P of smooth planes transverse to
F satisfying:
1. Each plane in P has an end representative which is an end of a horizontal
plane or catenoid which is disjoint from F ;
2. In the slab S between two successive planes in P, F has only a finite
number of ends;
3. Every middle end of F has a representative in one of the just described
slab regions S.
Proof. Since we are assuming that the surface F has one or two limit ends,
the collections D
1
and D
2
of disks are each infinite sets. The disks in D
1
can
be chosen to be disks of least area in H
1
relative to their boundaries. In fact
TOPOLOGICAL CLASSIFICATION OF MINIMAL SURFACES
685
the disks used by Freedman in the proof of his theorem have this property.
Assume that the disks in D
2
also have this least area property. Suppose W
is a closed component of H
1
−∪D
1
or H
2
−∪D
2
which is homeomorphic to
A × [0, 1).
Let γ(W ) be a piecewise smooth simple closed curve in ∂W that generates
the fundamental group of W . The curve γ(W ) bounds two noncompact annuli
in ∂W. (Imagine W is the closed outer complement of a catenoid and γ(W )
is the waist circle of the catenoid.) By choosing γ(W ) to intersect the interior
of one of the disks in D
1
or D
2
on the boundary of W , we can insure that
neither annulus in ∂W bounded by γ(W ) is smooth. Fix one of these annuli
and an exhaustion of it by compact annuli A
1
⊂ A
2
⊂ A
n
⊂ with
γ(W ) ⊂ ∂A
1
. By [14] the boundary of W is a good barrier for solving Plateau-
type problems in W . Let

A
n
denote a least area annulus in W with the same
boundary as A
n
which is embedded by [14]. The curve γ(W ) bounds a properly
embedded least area annulus A(W )inW , where A(W) is the limit of some
subsequence of {

A
n
}; the existence of A(W ) depends on local curvature and
local area estimates given in a similar construction in [9]. Since the interior
of the minimal annulus A(W) is smooth, the maximum principle implies that
A(W ) intersects ∂W only along γ(W ). The stable minimal annulus A(W ) has
finite total curvature [3] and so is asymptotic to the end of a plane or catenoid
in R
3
. By the maximum principle at infinity [13], the end of A(W ) is a positive
distance from ∂W. Hence, one can choose the representative end of a plane or
catenoid to which A(W ) is asymptotic to lie in the interior of W.
Let S denote the collection of ends of planes and catenoids defined above
which arises from the collection of nonsimply connected components W of
H
i
−∪D
i
. It follows from the proof of the Ordering Theorem in [8] that S
induces the ordering of E(F ). Since the middle ends of F are not limit ends,
when F has one limit end, then, after a possible reflection of F across the
xy-plane, we may assume that the limit end of F is its top end. Thus, S will
be naturally indexed by the nonnegative integers N if F has one limit end or by
Z if F has two limit ends with the ordering on the index set N or Z coinciding
with the natural ordering on S, and the subset of nonlimit ends in E(F ).
Suppose that F has one limit end and let S = {E
0
,E
1
, }. Let B
0
be
a ball of radius r
0
centered at the origin with ∂E
0
⊂ B
0
and such that ∂B
0
intersects E
0
transversely in a single simple closed curve γ
0
. The curve γ
0
bounds a disk D
0
⊂ ∂B
0
. Attach D
0
to E
0
− B
0
to make a plane P
0
. Next
let B
1
be a ball centered at the origin of radius r
1
,r
1
≥ r
0
+ 1, such that
∂E
1
⊂ B
1
and ∂B
1
intersects E
1
transversely in a single simple closed curve
γ
1
. Let D
1
be the disk in ∂B
1
disjoint from P
0
. Let P
1
be the plane obtained
by attaching D
1
to E
1
− B
1
. Continuing in this manner we produce planes
P
n
,n ∈ N, that satisfy properties 1, 2, 3 in the lemma. These planes can be
modified by a small C
0
-perturbation so that the resulting planes are smooth.
686 CHARLES FROHMAN AND WILLIAM H. MEEKS III
If F had two limit ends instead of one limit end, then a simple modification of
this argument also would give a collection of planes P satisfying properties 1,
2, 3 in the lemma.
Remark 3.2. Lemma 3.1 only addresses the case where the surface F has
an infinite number of ends. When there are a finite number of ends greater
than two, then the proof of the lemma goes through with minor modifications.
If F has two ends and is an annulus, then extra care must be taken to find the
single plane in P (see for example, the proof of the Ordering Theorem for this
argument).
Proposition 3.3. There exists a collection of planes P satisfying the
properties described in Lemma 3.1 and such that each plane in P intersects
F in a single simple closed curve. Furthermore, in the slab between two suc-
cessive planes in P, F has exactly one end.
Proof. Suppose the limit tangent plane to F is horizontal and that P is
finite. Let P
T
and P
B
be the top and bottom planes in the ordering on P.
Since the inclusion of the fundamental group of F into the fundamental group
of either complement is surjective [9], the proof of Haken’s lemma [10] implies
that P
T
can be moved by an ambient isotopy supported in a large ball so that
the resulting plane P

T
intersects F in a single simple closed curve. Let

P
B
be
the image of P
B
under this ambient isotopy. Consider the part F
B
of F that
lies in the half space below P

T
and note that the fundamental group of F
B
maps onto the fundamental group of each complement of F
B
in the half space.
The proof of Haken’s lemma applied to

P
B
in the half space produces a plane
P

B
isotopic to

P
B
that intersects F
B
in a simple closed curve. We may assume
that this is an isotopy which is the identity outside of a compact domain in F
B
.
Consider the slab bounded by P

T
and P

B
. The following assertion implies
that {P

T
,P

B
} can be expanded to a collection of planes P satisfying all of the
conditions of Proposition 3.3.
Assertion 3.4. Suppose S is a slab bounded by two planes in P where
P satisfies Lemma 3.1. Suppose each of these planes intersects F in a simple
closed curve. Then there exists a finite collection of smooth planes in S, each
intersecting F in a simple closed curve, which separate S into subslabs each of
which contains a single end of F . Furthermore the addition of these planes to
P gives a new collection satisfying Lemma 3.1.
Proof. Here is the idea of the proof of the assertion. If F has more than
one end in S, then there is a plane in S which is topologically parallel to the
boundary planes of S and which separates two ends of F ∩ S. The proof of
Haken’s lemma then applies to give another such plane with the same end
which intersects F in a simple closed curve. This new plane separates S into
TOPOLOGICAL CLASSIFICATION OF MINIMAL SURFACES
687
two slabs each containing fewer ends of F . Since the number of ends of F ∩ S
is finite, the existence of the required collection of planes follows by induction.
Assume now that the number of planes in P satisfying Lemma 3.1 is
infinite. We first check that P can be refined to satisfy the following additional
property: If W is a closed complement of either H
1
− ∪D
1
or H
2
− ∪D
2
, then
W intersects at most one plane in P. We will prove this in the case that F
has one limit end. In what follows we will assume our standard conventions:
F has a horizontal limit tangent plane at infinity and the limit end is maximal
in the ordering of ends. The proof of the case where F has two limit ends is
similar.
Let W be the set of closures of the components of H
1
−∪D
1
and H
2
−∪D
2
.
Given W ∈W, let P(W ) be the collection of planes in P that intersect W .If
W is a compact ball, then P(W ) is a finite set of planes since P is proper. If
W is homeomorphic to A × [0, 1), then P(W ) is also finite. To see this choose
a plane P ∈Pwhose end lies above the end of W; the existence of such a
plane is clear from the construction of P in the previous lemma. Note that
the closed half space above P intersects W in a compact subset. Hence, only
a finite number of the planes above P can intersect W. Since there are an
infinite number of planes in P above P , there exists a plane

P above P so that

P is disjoint from W and any plane in P above

P is also disjoint from W .
Since there are only a finite number of planes below

P , only a finite number
of planes in P can intersect W .
We now refine P. First recall that the end of P
0
is contained in a single
component of W. Hence, the plane P
0
intersects a finite number of components
in W and each of these components intersects a finite collection of planes in
P different from P
0
. Remove this collection from P and reindex to get a new
collection P = {P
0
,P
1
, ···}. Note that P
1
does not intersect any component
W ∈Wthat also intersects P
0
. Now remove from P all the planes different
from P
1
that intersect some component W ∈Wthat P
1
intersects. Continuing
inductively one eventually arrives at a refinement of P such that for each
W ∈W, P(W ) has at most one element. This refinement of P satisfies
the conditions of Lemma 3.1 and so, henceforth, we may assume that P(W )
contains at most one plane for every P ∈P.
The next step in the proof is to modify each P ∈P so that the resulting
plane P

intersects F in a simple closed curve. We will do several modifications
of P to obtain P

and the reader will notice that each modification yields a
new plane that is a subset of the union of the closed components of W that
intersect the original plane P . This is important to make sure that further
modifications can be carried out.
Suppose P ∈Pand the end of P is contained in H
1
. Let A
2
be the set of
components of W∩H
2
that are homeomorphic to A×[0, 1). For each W ∈A
2
,
688 CHARLES FROHMAN AND WILLIAM H. MEEKS III
let T(W ) be a properly embedded half plane in W , disjoint from ∪D
2
, such
that the geodesic closure of W − T (W) is homeomorphic to a closed half space
of R
3
. Assume that P intersects transversely the half planes of the form T(W)
and the disks in D
2
.
We first modify P so that there are no closed curve components in P ∩
(∪D
2
). If D ∈D
2
and P ∩ D has a closed curve component, then there is an
innermost one and it can be removed by a disk replacement. Since the end of P
is contained in H
1
, there are only a finite number of closed curve components in
∪D
2
and they can be removed by successive innermost disk replacements. In a
similar way we can remove the closed curve components in P ∩(∪T (W ))
W ∈A
2
.
We next remove compact arc intersections in P ∩ (∪D
2
) by sliding P over
an outermost disk bounded by an outermost arc and into H
1
. In a similar
way we can remove the finite number of compact arc intersections of P with
∪T (W )
W ∈A
2
. Notice that P already intersects the region that we are pushing
it into.
After the disk replacements and slides described above, we may assume
that P is disjoint from the disks in D
2
and the half planes in A
2
. Let W ∈W
be the component which contains the end of P and let P(∗) be the component
of P ∩ W which contains the end of P. Cut H
2
along the disks in D
2
and half
planes in A
2
. Since every closed component of the result is a compact ball or a
closed half space, the boundary curves of P (∗), considered as subsets of these
components, bound a collection of pairwise disjoint disks in H
2
. The union
of these disks with P (∗) is a plane P

with P

∩ W = P (∗). If P(∗)isan
annulus, then we are done. Otherwise, since the fundamental group of W is
Z, the loop theorem implies that one can do surgery in W on P (∗) ⊂ P

such
that after the surgery, the component with the end of P

has fewer boundary
components. After further surgeries in W we obtain an annulus P

(∗) with
the same end as P(∗) and with boundary curve being one of the boundary
curves of P(∗). By our previous modifications, ∂P

(∗) lies on the boundary
of the closure of one of the components of H
2
− ∪D
2
and bounds a disk D in
this component. We obtain the required modified plane P

= P

(∗) ∪ D which
intersects F in the curve ∂P

(∗).
The above modification of a plane P ∈Pcan be carried out independently
of the other planes since the modified plane is contained in the union of the
components of W that intersect P and when P intersects W ∈W, then no
other plane in P intersects W . Now perform these modifications on all of
the even indexed planes in P to form a new collection. Note that the odd
indexed planes of P give rise to a proper collection of slabs with exactly one
even indexed plane in each of these slabs. Next remove all of the odd indexed
planes from P and reindex the remaining ones by N in an order preserving
manner.
TOPOLOGICAL CLASSIFICATION OF MINIMAL SURFACES
689
Finally, applying the Assertion 3.4 allows one to subdivide the slabs be-
tween successive planes in P so that each slab contains at most one end of F .
This completes the construction of P and the proof of Proposition 3.3.
4. The structure of a minimal surface in a slab or half space
Let F be an orientable surface and let C be a proper collection of disjoint
simple closed curves in F ×{0}.IfH is a three-manifold that is obtained
by adding 2-handles to F × [0, 1] along C and then capping off the sphere
components with balls, then H is a compression body. Alternatively, if H is
an irreducible three-manifold and ∂
−
H is a closed proper subsurface of ∂H
and Γ is a properly embedded 1-dimensional CW-complex in H so that there
is a proper deformation retraction r : H → ∂
−
H ∪ Γ, then H is a compression
body. The surface ∂
+
H = ∂H − ∂
−
H is called the inner boundary component
of H.If∂
−
H = ∅, then we say H is a handlebody. The compression body H
is properly embedded in the three-manifold M, if its inclusion map is a proper
embedding in the topological sense and ∂
−
H = H ∩ ∂M.AHeegaard splitting
of a three-manifold M is a pair of compression bodies H
1
and H
2
properly
embedded in M so that M = H
1
∪ H
2
and the intersection of H
1
and H
2
is
exactly their inner boundary components. The surface ∂
+
H
1
= ∂
+
H
2
is called
a Heegaard surface.
The 1-dimensional CW-complex Γ in the definition of compression body
is called a spine of the compression body. There are many choices of spines
for a given compression body. For the sake of combinatorial clarity we will
only work with spines whose vertices are all monovalent or trivalent, and the
monovalent vertices coincide with Γ ∩ ∂
−
H. We can further assume that the
restriction of the deformation retraction r : H → ∂
−
H ∪ Γ restricted to ∂
+
H
has the property that the inverse image of any point that is in the interior
of an edge is a single circle, the inverse image of any monovalent vertex is a
circle and the inverse image of any trivalent vertex is a trivalent graph with
three edges and two vertices (a theta curve). This leads to a corresponding
decomposition of ∂
+
H into pairs of pants, annuli, and a copy of ∂
−
H with a
disk removed for each monovalent vertex of Γ. There is a pair of pants for each
trivalent vertex, and an annulus for each edge that contains no vertex, and the
rest of the surface runs parallel to ∂
−
H. We can reconstruct Γ up to isotopy
from this decomposition.
Aside from isotopy there are two moves that we will be using on Γ. They
are both variants of the Whitehead move. We alter the graph according to one
of the two local operations shown in Figure 1 and Figure 2.
Dually the Whitehead move involves two pairs of pants meeting along a
simple closed curve γ which is the inverse image of a point in the interior of
the edge to be replaced. If γ

is any simple closed curve lying on that union
690 CHARLES FROHMAN AND WILLIAM H. MEEKS III
Figure 1: Whitehead move
Figure 2: Half Whitehead move with lower vertices on ∂
−
H
of pants that intersects γ transversely in exactly two points, and separates the
boundary components of the two pairs of pants into two sets of two, then we
can perform the Whitehead move so that the two new pairs of pants meet
along γ

.
The half Whitehead move can occur at a trivalent vertex that is adjacent
to a monovalent vertex (lying on ∂
−
H). You can think of it as collapsing the
edge with one endpoint on the boundary and one endpoint at the vertex to the
point on ∂
−
H and then pulling the ends of the two remaining edges apart.
Suppose that H is a compression body and δ is a simple closed curve on
the inner boundary component of H. We can extend δ to a singular surface
whose boundary lies in Γ ∪ ∂
−
H. First isotope δ so that with respect to the
decomposition into annuli, pants and a punctured ∂
−
H, the part of δ that
lies in each component is essential. There is a singular surface with boundary
δ obtained by adding “fins” going down to Γ based on the models shown in
Figure 3, along with fins in the annuli and near ∂
−
H.
Figure 3: Extending the disk D to a singular surface.
On a pair of pants there are six isotopy classes of essential proper arcs.
For each choice of a pair of boundary components there is an isotopy class
of essential arcs joining them, and for each boundary component there is an
isotopy class of essential arcs joining that boundary component to itself. We
call an essential proper arc good if its endpoints lie on distinct boundary com-
ponents, and bad if its endpoints lie in the same boundary component. Two
TOPOLOGICAL CLASSIFICATION OF MINIMAL SURFACES
691
such arcs are paral lel if they are disjoint and have their endpoints in the same
boundary components.
Lemma 4.1. Suppose that H is a compression body and δ is a simple
closed curve on ∂
+
H. Either δ bounds a disk in H or there is a graph Γ so
that H is a regular neighborhood of Γ ∪ ∂
−
H such that δ has no bad arcs.
Proof. The argument will be by induction on a complexity for δ. Let s
be the number of bad arcs. Given a bad arc k, the arcs (or arc) of δ adjacent
to k lie in the same pair of pants or in the punctured copy of ∂
−
H. If the
two endpoints of the bad arc coincide with the two endpoints of another bad
arc, then let d(k) = 0. If both arcs lie in the punctured copy of ∂
−
H, then let
d(k) = 1. If both arcs lie in the same pair of pants P, then either the two arcs
are parallel or not parallel . If they are not parallel, then d(k) = 1. If they
are parallel, then follow them into the next surface. If the next surface is the
punctured copy of ∂
−
H, then d(k) = 2, if the next surface is a pair of pants
and the next arcs are not parallel, then d(k) = 2, otherwise follow them into
the next surface, and keep counting. Let m = min
k bad
d(k). The complexity
of δ is the pair (s, m).
∂
4
∂
3
∂
1
∂
2
γ

γ
Figure 4: Reducing m when it is greater than 1. On the right-hand side of the
figure, P
1
is on the left, P
2
is on the right and P
1
∩ P
2
= γ

.
If m>1, then we do the Whitehead move to reduce m as follows; see
Figure 4. Let k be a bad arc with d(k)=m. Let Q be the union of the pair
of pants containing k and the pair of pants that contains the adjacent pair of
arcs k
1
and k
2
. Let γ be the curve that the two pairs of pants meet along.
Let ∂
1
,∂
2
,∂
3
,∂
4
be the boundary components of Q labeled so that ∂
1
and ∂
4
belong to one pair of pants, ∂
2
and ∂
3
belong to the other pair of pants, and
both k
1
and k
2
have an endpoint in ∂
4
. Let a = k ∪k
1
∪k
2
. There is an arc b of
∂
4
so that a push off γ

of a ∪ b lies in Q, misses a and separates the boundary
components of Q into two sets of two, say one set is ∂
1
and ∂
2
and the other
is ∂
3
and ∂
4
. Perform the Whitehead move so that γ

is the intersection of
the two new pairs of pants. Denote the new pairs of pants, resulting from the
Whitehead move corresponding to γ

,byP
1
where ∂P
1
= ∂
1
∪ ∂
2
and by P
2
where ∂P
2
= ∂
3
∪∂
4
. Notice that a is a bad arc and d(a)=m−1. To conclude
692 CHARLES FROHMAN AND WILLIAM H. MEEKS III
Figure 5: The endpoints of the bad arc are near ∂
−
H.
Figure 6: Reducing the number of bad arcs when m = 1. In the figure on the
left, the waist is the curve γ and the arc b lies in the lower pair of pants.
that we have simplified the picture we need to see that we have not increased
the number of bad arcs. If l is a bad arc in P
1
∪ P
2
and it has its endpoints
in some ∂
i
, then it contains some bad arc of the original picture. If l has its
endpoints in γ

and lies in P
2
,asδ is embedded it is trapped in the annulus
between γ

and a and is hence inessential. If l has its endpoints in γ

and is
contained in P
1
, once again the arc is trapped by a and hence there must be
two arcs in P
2
having one endpoint each in common with l and the other in b;
but this means l is contained inside a bad arc from the original picture. Hence
we did not increase s. On the other hand we have decreased m by 1.
If m = 1, then there are two cases. The first is when an adjacent pair of
arcs lies in the part of the surface parallel to ∂
−
H. In this case we do a half
Whitehead move to reduce the number of bad arcs; see Figure 5.
The other case is when an adjacent pair of arcs is contained in an adjacent
pair of pants. Once again, a Whitehead move can be applied to reduce the
number of bad arcs; see Figure 6. Let Q be the union of the two pairs of pants
that contain k, and let k
1
and k
2
be the arcs of δ adjacent to k and lying in
the other pair of pants. Let γ be the circle that the pants intersect along.
Let b be an arc in the other pair of pants that contains the endpoints of k,
only intersects k
1
and k
2
in the endpoints they share with k, and is transverse
to the other components of δ ∩ Q. Let γ

be a push off of b ∪ k such that it
intersects k in a single point, is disjoint from k
1
and k
2
, and such that during
the push off, the related arcs b
t
stay transverse to δ and the related arcs k
t
are
disjoint from δ for t = 0. Notice that the arc k
1
∪ k ∪ k
2
gets separated into
two good arcs by γ

. Hence, if we have not created any new bad arcs, then we
TOPOLOGICAL CLASSIFICATION OF MINIMAL SURFACES
693
have reduced the total number of bad arcs. If a bad arc enters and leaves the
new picture through a boundary component of Q, then it is either contained in
or contains a bad arc of the old picture. Hence, we only need to worry about
bad arcs with their endpoints in γ

. Since δ is embedded, such an arc misses
k
1
∪ k ∪ k
2
. The result of cutting Q along the arc k
1
∪ k ∪ k
2
is a pair of pants
and γ

gives rise to an arc of this pair of pants that has both its endpoints
in the same boundary component of the pair of pants. The only proper arcs
that intersect the arc corresponding to γ

in an essential manner in two points
must have both their endpoints in the same boundary component of the pair
of pants. This implies that such a bad arc is contained inside a bad arc from
the original picture.
Finally, when m = 0, there are two arcs joined end to end, and the disk
inside the regular neighborhood of Γ is readily visible; see Figure 7.
Figure 7: The interior disk.
Suppose S is a flat 3-manifold in R
3
that is homeomorphic to R
2
× [0, 1].
Denote the components of ∂S by ∂
0
S and ∂
1
S. Assume further that there are
simple closed curves C
0
⊂ ∂
0
S and C
1
⊂ ∂
1
S so that ∂
i
S is a union of two
minimal surfaces sharing C
i
as their joint boundary. As ∂
i
S is a plane, one of
these surfaces is a disk D
i
and the other is a once-punctured disk A
i
. Finally,
assume that F is a properly embedded minimal surface in S having one end,
boundary C
0
∪ C
1
and such that in each closed complement of F in S, the
interior angles along ∂F are less than π. Up until the end of this section, we
will assume these properties hold for S and F .
Proposition 4.2. The surface F separates S into two compression bodies
H
1
and H
2
, having F as their inner boundary components. That is, F is a
Heegaard surface.
Proof. We outline the idea for the sake of completeness. First consider
the region H
1
and suppose that ∂H
1
has one end. In this case, by [9], H
1
is
a handlebody. Assume now that ∂H
1
has two ends. By Freedman’s theorem
applied to H
1
, there exists a proper family of compressing disks D
1
which
can be chosen to have their boundary components disjoint from ∂S. After
possibly restricting to a subcollection of D
1
, we see that the result of cutting