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\begin{document}
\title{Basic Ideas about Laminations}
\author{Daniel Mathews}%
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\tableofcontents
\section{Preface}
The story of laminations is quite amazing. It throws around a whole
lot of interesting ideas from geometry, topology, and analysis.
I read Casson and Bleiler's book ``Automorphisms of Surfaces after
Nielsen and Thurston". I liked it. All the story is there in their
theorems. But stories can also be told directly. That's the idea
here.
\section{Prerequisites}
I'm assuming you know about basic topology. I assume you know the
classification of surfaces. And I'm assuming you're familiar with
the basics of hyperbolic geometry: for instance, the properties of
the hyperbolic plane; the types of hyperbolic isometries; and so on.
\section{Introduction}
We are dealing with laminations on surfaces. So what is a
lamination? You might have heard something like:
\begin{quotation}
A laminate is a material constructed by uniting two or more layers
of material together. The process of creating a laminate is
lamination, which usually refers to sandwiching something between
layers of plastic and sealing them with heat and/or
pressure.\footnote{From the Wikipedia entry on ``Laminate",
http://en.wikipedia.org/wiki/Laminate.}
\end{quotation}
Well, forget everything you thought you knew about \emph{this} kind
of lamination! This is completely different. Actually, if you have
an interest in composite materials, you may want to remember this
more well-known version of lamination, but be aware that it's
entirely unrelated.
No, a lamination on a surface is something completely different.
\subsection{So what is the definition, then?}
Before we give the definition, however, let's be clear what page
we're on. Our surface is not just any surface. First, it's a
\emph{closed} surface, that is, it's compact and has no boundary. No
punctures, no cusps, not infinite genus, nothing like that. Second,
it's an \emph{oriented} surface. No mobius strips, no crosscaps, no
klein bottles or anything like that. And third, it's a
\emph{geometric} surface, in fact we require it to have a very
particular geometry, \emph{hyperbolic} geometry. This means it has a
Riemannian metric of constant curvature $-1$, and implies that the
surface has negative Euler characteristic, hence has genus at least
$2$.
Thus, the topology of the surface we're talking about is completely
specified by the genus, which can be any positive integer $2$ or
greater. (On a particular surface, however, there are many possible
hyperbolic metrics --- in fact, a $(6g-6)$-dimensional space of
them.)
Now, a lamination will just be a bunch of geodesics. First, these
will be \emph{complete}, in that they don't stop. They can be closed
loops of finite length; or if they don't close up, they will be of
infinite length. No endpoints! Second, no intersections will be
allowed. So the geodesics must be \emph{simple}, that is, have no
self-intersections. And they must be \emph{disjoint}: they can't
intersect each other. The only other requirement --- which really is
the one that induces all the structure --- is that together this
bunch of geodesics forms a \emph{closed set}. Every limit point of
the set is in the set.
That's the definition.
\begin{defn}
Let $S$ be a closed oriented hyperbolic surface. A \emph{lamination}
on $S$ is a closed subset $L$ of $S$ which is a union of disjoint
complete simple geodesics.
\end{defn}
\subsection{Some examples}
There are some easy examples of laminations. For instance, there is
the empty lamination. A single simple closed geodesic forms a closed
set, hence is a lamination. A finite set of disjoint simple closed
geodesics will be closed, and hence taken together form a
lamination. A bit more tricky is an example where we take a
non-closed simple geodesic which \emph{spirals} towards a simple
closed geodesic. The spiralling geodesic does not form a lamination,
because it's not a closed set --- its limit set includes the simple
closed geodesic it's spiralling towards. But if we take the
spiralling geodesic, \emph{and} the simple closed geodesic it's
spiralling towards, \emph{together}, then we have a lamination.
Can you think of any other examples?
Well, it turns out that the types of phenomena I just mentioned ---
simple closed curves, and non-closed curves which spiral towards
closed curves --- are actually quite unusual. Laminations may have
properties such as: no closed geodesics; every geodesic is dense; no
isolated geodesics. This is something we haven't defined, but it
probably has the definition you think it does.
These generic examples are difficult to draw a picture of,
unfortunately. But there's another way to think about them, which
can be quite useful. This way relies on the fact that \emph{the set
of all laminations on $S$ is closed}. What does ``closed" mean here?
What's the topology on this space? Where does it naturally lie? We
will answer these questions shortly. But this fact means that
examples of laminations can be given as \emph{limits} of laminations
we can draw. For instance, consider a sequence of laminations $L_n$,
each consisting of a single simple closed geodesic, but where these
geodesics get more and more complicated, and as a closed set
``approaches" some subset $L_\infty$ of $S$. Once we define what all
this means, our fact means that $L_\infty$ will be a lamination; and
that may not have been an obvious thing to see. And since it's a
limit of increasingly complicated laminations, $L_\infty$ may have
more exotic properties than its relatively uncomplicated
progenitors.
\subsection{A quibble, and unique decomposition}
But first: there's a question with the definition. It might sound
like a quibble, but it's important and exposes something interesting
about hyperbolic geometry.
We'd like to say that a lamination $L$ consists of geodesics
$\gamma_i$, where $i$ is in some indexing set, possibly infinite,
possibly uncountable. We'd like to say that the geodesics $\gamma_i$
are \emph{the leaves} of $L$. Note the use of the definite article
``the".
But given the lamination $L$, there might be different ways to
decompose this closed set into geodesics. If that were the case, we
would have to be careful about the different decompositions of $L$.
If you think about Euclidean surfaces, you'll see that this can
actually happen. For instance, the whole of the Euclidean plane
(i.e. $L = S = \R^2$) can be decomposed as a disjoint union of
vertical geodesics, or as a disjoint union of horizontal geodesics,
in fact lines with any single given slope. The same is true on the
standard Euclidean torus $L = S = \R^2/\Z^2$: it can be decomposed
as a disjoint union of geodesics of any given slope.
Thankfully, it turns out that we don't have to worry.
\begin{prop}
\label{unique_decomposition}
Let $S$ be a closed oriented hyperbolic surface and let $L$ be
a lamination on $S$. Then $L$ can be decomposed as a
union of disjoint complete simple geodesics in one and only one way.
\end{prop}
This proposition tells us something about the nature of hyperbolic
geometry. In some sense, you can't have geodesics too close
together. If you do, the negative curvature doesn't allow them to do
so within a compact surface.
The same principle seems to be at work in the following simple
picture. If you take a small open ball in the hyperbolic plane, you
can try to fill it up with geodesics. It's possible to do so, in
fact in many different ways. Think of one way. But now look at where
those geodesics go after they leave the ball and continue through
the hyperbolic plane. They spread out like crazy through at least
some regions --- they will fill up a region of infinite area. That's
going to be a problem on a compact surface. Well, maybe you might
think that it's okay, because you're going to quotient the
hyperbolic plane by some group of isometries to get the surface, and
maybe it will all work out. But it won't, and you might like to
ponder why. We'll get to the proof soon enough.
In any case, pending the proof of proposition
\ref{unique_decomposition}, we will make a
\begin{defn}
The geodesics $\gamma_i$ in a lamination $L$ are called \emph{the
leaves} of $L$. At each point $x \in L$, we write $\gamma_x$ for the
leaf of $L$ passing through $x$.
\end{defn}
\section{What do laminations look like?}
Since it's hard to draw pictures of laminations, we can calibrate
our intuition by deriving some properties about them. These
properties have some pictorial value, which we can bear in mind for
later reference.
\subsection{Laminations are not jerky...}
Laminations don't jump around too much. By which I mean, the
geodesics involved can't change direction too sharply. That is, if
you go from one point $x$ in $L$ to another which is close by ---
call it $y$ --- then the directions of the geodesics $\gamma_x$ and
$\gamma_y$ don't change too much. We can state this as a
proposition, and it's so simple that we might even prove it.
Actually, it's \emph{so} simple that you can prove it for yourself!
\begin{prop}
\label{direction_varies_continuously}
Let $S$ be a closed oriented hyperbolic surface.
Let $L$ is a lamination which is decomposed as a union of
disjoint complete simple geodesics. Let $\gamma_x$
denote the geodesic passing through $x$.
The direction of $\gamma_x$ varies
continuously with $x$.
\end{prop}
(Why did I not just say that $\gamma_x$ is the leaf of $L$ through
$x$? Because I'm proving something, and I haven't proved anything
about unique decomposition into geodesics yet! Hence the slightly
more complicated phrasing. This fact about direction varying
continuously is true for \emph{any} decomposition of a given
lamination $L$.)
We can make this rigorously precise, if we like, though the
intuitive picture does not depend on it. At each point $x \in S$,
the set of directions from $S$ is a real projective line $\RP^1$,
which is homeomorphic to a circle. If you put together these
projective lines at each point, we have the projectivized tangent
bundle of $S$, which we write $PT(S)$. At each $x \in S$ the fibre
$PT(S)_x$ is a copy of $\RP^1$.
This bundle clearly restricts to $L \subset S$, and we call the
restricted bundle $PT(S)|_L$. (It's different from $PT(L)$!) There
is still an $\RP^1$ fibre above each point of $L$. Let $\pi:
PT(S)|_L \to L$ be the projection associated with the bundle. The
function assigning to each point $x$ of $L$ the direction of
$\gamma_x$ is a section of this bundle, that is a map $s: L \To
PT(S)|_L$ which takes each $x \in L$ to a point above it, $s(x) \in
(PT(S)|_L)_x$; equivalently $\pi \circ s = 1_L$.
Proposition \ref{direction_varies_continuously} says that the
section $s$ is continuous.
Did that general nonsense help? Probably not. The picture is simple
to draw in any case.
Locally our surface is isometric to the hyperbolic plane, so we can
just draw pictures there. If the direction of $\gamma_x$ doesn't
vary continuously with $x$, then you can find two points which are
close but where their directions are not close. Draw the geodesics
in these directions. Then they meet, so the geodesics are not
disjoint, neither in the hyperbolic plane or in the quotient surface
$S$. If you want to put in some epsilons and some hyperbolic
trigonometry to completely convince yourself, go right ahead.
\subsection{... but not very smooth, either.}
On the other hand, laminations do jump around a lot. They have lots
of holes in them. As we argued rather unconvincingly earlier, a
lamination can't fill up much of a surface. In fact it can't even
fill up an arbitrarily small ball in the surface. Proving this fact,
we will have shown that geodesics are ``not too close together", in
some sense. We will then be able to prove proposition
\ref{unique_decomposition}, the uniqueness of decomposition into
geodesics.
But we will start with a simpler proposition and build up to a
climax: $L$ can't be the \emph{whole} manifold. This is an extremely
weak result, in the light of what we're about to prove, but still an
interesting application of the Poincar\'{e}-Hopf index theorem.
\begin{prop}
Let $S$ be a closed oriented hyperbolic surface and $L$ a lamination
on $S$. Then $L$ is a proper subset of $S$, i.e. $L \neq S$.
\end{prop}
(Note that hyperbolic is essential here! We have already seen
examples of laminations of the Euclidean plane or Euclidean torus
which fill the whole manifold.)
\begin{Proof}
Take a decomposition of $L$ into geodesics (we still don't know
anything about uniqueness) and for $x \in L$ let $\gamma_x$ be the
geodesic through $x$. If $L=S$ then we consider the direction of
$\gamma_x$ at $x$. From above, proposition
\ref{direction_varies_continuously}, this direction varies
continuously. These directions thus give a line field on $S$. If you
like, you can obtain from this a nowhere vanishing vector field on
$S$. But in any case, this is prohibited by the Poincar\'{e}-Hopf
index theorem on a surface of negative Euler characteristic. The
indices of singularities must sum to the Euler characteristic, which
is nonzero!
\end{Proof}
And now for something stronger --- a \emph{lot} stronger --- which
piggy-backs on top of this rather weak proposition.
\begin{prop}
\label{nowhere_dense}
Let $S$ be a closed oriented hyperbolic surface and $L$ a
lamination on it. Then $L$ is nowhere dense in $S$.
\end{prop}
\begin{Proof}
Take a decomposition of $L$ into geodesics. Suppose to the contrary
that there's an open ball $U$ in $L$. Take $x_0 \in U$ and take an
arc $\alpha$ transverse to the geodesic $\gamma_{x_0}$ through
$x_0$. Since $L$ contains the neighbourhood $U$ of $x_0$, if
$\alpha$ is sufficiently small then $\alpha \subset U$. By
proposition \ref{direction_varies_continuously} the direction of the
geodesics $\gamma_x$ varies continuously, by replacing $\alpha$ with
a smaller arc if necessary we can assume $\alpha$ is transverse to
the lamination $L$ at every point of $\alpha$.
Now we have $\alpha \subset L$ which is transverse to $L$. This
spells doom for $L$. Identify the universal cover $\tilde{S}$ with
$\hyp^2$, and take a lift $\tilde{\alpha} \subset \hyp^2$. Define a
function $\Phi: \alpha \times \R \To \hyp^2$ which takes $(y,t)$ to
the point on $\gamma_y$, at (signed) distance $t$ from $\alpha$.
Want to worry about the technicalities of defining this properly in
the universal cover? Then go right ahead; it works fine. We see that
$\Phi(\alpha \times \R)$ is a huge region of $\tilde{S} \cong
\hyp^2$ bounded by two geodesics --- namely, the geodesics in $L$
through the endpoints of $\alpha$. These must enclose a region of
infinite area. This massive region of $\tilde{S}$ projects to $S$
and all lies in $L$. You can fit arbitrarily large balls in this
region, including ones larger than the diameter of $S$; such a large
ball will project onto the whole of $S$. So $L = S$; but this is a
fatal blow because it contradicts the previous proposition.
\end{Proof}
This proves that geodesics really aren't very close together.
\subsection{And hence they have unique decomposition.}
What really mattered in the previous proof was the arc $\alpha
\subset L$ transverse to the decomposition of $L$ into geodesics.
This allowed us to take the transverse geodesics --- and this smooth
family of disjoint geodesics had to fill up way too much of the
hyperbolic plane, hence way too much (all) of $S$.
But unique decomposition follows is proved in exactly the same way.
For suppose we can decompose $L$ into a union of disjoint complete
simple geodesics two different ways. Take a point $x_0$ where the
two decompositions have different geodesics through $x_0$. Call the
first $\gamma_{x_0}$; call the second $\alpha$. By shortening
$\alpha$ sufficiently, we obtain an arc $\alpha \subset L$ which is
transverse to the first decomposition of $L$. This spells doom for
$L$ in the same way that it did four paragraphs ago. The same proof
applies, word for word.
Now we will use the phrase ``the leaves of $L$" with a clear
conscience. Being careful with it was getting a little annoying!
\section{Closed in the set of closed sets}
We'll now return to the rather exciting prospect mentioned earlier
when discussing examples of laminations. That prospect is that we
can define really complicated laminations without being able to draw
them: we define them as the \emph{limit} of simpler laminations. To
do this properly, however, requires the \emph{set of laminations} to
be \emph{closed}.
We will now answer the questions raised in this previous discussion.
As always, $S$ is a closed oriented hyperbolic surface. Let $\L$ be
the set of all laminations on $S$.
\subsection{Where does $\L$ naturally lie?}
Every lamination is a closed subset of $S$ by definition. Hence $\L$
is a subset of the \emph{set of all closed subsets of $S$}, which we
will denote $\P(S)$.
Turns out $\P(S)$ has a nice topology. In fact, if $S$ has a metric,
then so does $\P(S)$. This metric is sometimes called
\emph{Hausdorff distance}. Given two closed sets $A, B \subseteq S$,
we define their Hausdorff distance $d(A,B)$ by
\[
d(A,B) \leq \epsilon \quad \text{if and only if} \quad A
\subseteq N_\epsilon (B) \text{ and } B \subseteq
N_\epsilon(A),
\]
or equivalently, as
\[
d(A,B) = \inf \left\{ \epsilon \geq 0 \; : \; A
\subseteq N_\epsilon (B) \text{ and } B \subseteq
N_\epsilon(A) \right\}.
\]
Here $N_\epsilon(A)$ denotes an open $\epsilon$-neighbourhood of
$A$.
So the distance between closed sets $A,B$ is essentially the least
amount required to expand them so that either one engulfs the other.
Clearly there's nothing special about any hyperbolic or geometric
properties of $S$ here. In fact we can define Hausdorff distance on
$\P(X)$ for any metric space $X$ and get a metric. If $X$
--- like $S$ --- is compact, then any two metrics on $X$ are
equivalent; hence the topology on $\P(X)$ doesn't depend on the
metric. This is left as an exercise for the reader; it is standard
fare in a course on metric spaces.
So $\P(S)$ is a well-defined metric space, and while the metric
depends on the metric on $S$, the topology does not. The set $\P(S)$
has one more important property: \emph{it is compact}. (In fact,
$\P(X)$ is compact for any compact topological space $X$.) This is
also a standard fact in metric spaces and left to the reader. But it
is important. If you take any sequence of closed subsets of $S$,
they contain a convergent subsequence, in the Hausdorff metric. And
the intuitive picture of Hausdorff convergence is one where the
pictures of sets look more and more similar.
\subsection{$\L$ is closed, the world rejoices.}
Every lamination is a closed subset of $S$, so $\L \subset \P(S)$.
(Clearly, a strict subset, since every lamination is nowhere dense!)
We endow it with the subspace topology.
Now all our talk previously of limits of laminations requires that
$\L$ be \emph{closed}. That is, we want $\L$ to be a closed subset
of $\P(S)$. If so, then as a closed subset of a compact space, it
will be compact itself. Any Hausdorff-convergent sequence of
laminations will converge not just to any old closed set, but to a
lamination. And, any sequence of laminations at all will have a
subsequence converging to a lamination.
This satisfies our aesthetic senses, and also means that we can
define examples in the way discussed earlier.
\begin{prop}
Let $S$ be a closed oriented hyperbolic surface and let $\L$ be the
set of all laminations on $S$. Then $\L$ with Hausdorff distance is
closed in $\P(S)$; so $\L$ is compact.
\end{prop}
The idea of the proof is as follows. Take a sequence of laminations
$L_n$ and suppose that $L_n$ converges to a closed set $L \in
\P(S)$. We must show $L$ is actually a lamination. Take $x \in L$,
and $x_n \in L_n$ converging to $x$, with $x_n$ lying on the leaf
$\gamma_n$ of $L_n$. Now we want to show that the geodesics
$\gamma_n$ converge (locally, Hausdorff-wise) to a geodesic $\gamma$
through $x$. This is not so easy to do directly.
What is easier to do, although more technical, is to take the
projectivized tangent bundle (alas!) $PT(S)$, and for each
lamination $L_n$ take a lift $\tilde{L_n}$ in $PT(S)$, which is the
image of a section of $PT(S)|_{L_n}$. We consider the $\tilde{L_n}$
as closed subsets of $PT(S)$, i.e. in $\P(PT(S))$. Now $\P(PT(S))$
is compact just like $\P(S)$ is. To make this proof rigorous, you
have to figure out how this works and that, whether we consider
laminations in $\P(S)$ or their lifts in $\P(PT(S))$, we get the
same topology on $\L$. With this fearsome technological structure in
place we return to the original argument.
The geodesics $\gamma_n$, or rather their lifts $\tilde{\gamma}_n
\subset \tilde{L}_n$, converge to a geodesic $\tilde{\gamma}$
\emph{by definition}: the points $x_n$ on them converge to $x$, and
the directions converge because that's what convergence in $PT(S)$
means.
\section{Isolated and boundary leaves}
We now turn to two particular types of leaves in a lamination, which
are useful and important: isolated leaves, and boundary leaves.
\subsection{Derivation: Destroying isolated leaves}
Sounds like a rather ruthless strategy for victory in a battle with
a tree. But it's a very useful thing to do with laminations and has
some amazing properties.
As we have seen, laminations can't have leaves which are so close
together that they fill up a ball in the surface. But they can have
leaves which are relatively close together, as in the
spiralling-into-a-closed-geodesic example, for instance. They can
also have leaves which are far apart, for instance, if the
lamination is merely a finite collection of simple closed geodesics.
The bits that are relatively close together are the interesting
part, because they can be very very subtle and weird and
Cantor-like. We therefore want an operation to remove everything
else. That's what derivation is: it removes isolated leaves.
What's an isolated leaf? Think about what it looks like, write the
definition you think it has and you're probably correct. In any case
I'll write a definition too.
\begin{defn}
Let $S$ be a closed oriented hyperbolic surface and $L$ be a
lamination on $S$. A leaf $\gamma$ of $S$ is \emph{isolated} if
every point $x$ of $\gamma$ has a neighbourhood $U_x$ in $S$ such that
$L \cap U_x$ only contains one arc, namely that of $\gamma$.
Equivalently, every $x \in \gamma$ has a neighbourhood $U_x$
such that the pair $(U, U \cap L)$ is homeomorphic to (disk,
diameter).
\end{defn}
Note that in this definition, the neighbourhood $U_x$ is allowed to
have different sizes for different points on $\gamma$. There is not
a uniform bound so that you can take a tubular neighbourhood of
$\gamma$ with that width. If the leaf $\gamma$ is closed, there will
be such a tubular neighbourhood. But if $\gamma$ is not closed, i.e.
$\gamma$ is infinitely long, there need not be any such tubular
neighbourhood.
So, in the example of a finite union of disjoint simple closed
geodesics, all the leaves are isolated. In the example of a closed
leaf and a non-closed leaf spiralling towards it, the closed leaf is
not isolated. But the spiralling leaf \emph{is} --- and the sizes of
the neighbourhoods in the above definition approach zero as we
approach the closed leaf. Maybe this isn't quite a consequence we
first had in mind, but we will go with it.
A derived lamination is obtained, as the heading suggests, by
destroying the closed leaves.
\begin{defn}
Let $S$ be a closed oriented hyperbolic surface and $L$ a
lamination on $S$. The \emph{derived lamination} $L'$ of $L$
consists of the non-isolated leaves of $L$.
\end{defn}
Is the derived lamination actually a lamination? Thankfully it is,
else the language would have been most unfortunate. We just need to
see why it's closed. Points on non-isolated leaves can't converge to
isolated leaves. (Of course not, because they're isolated! This
definition of ``isolated" is good: it adds rhetorical force to
arguments that rely not on rigour but on persuasion...) So a
sequence in $L'$, if it converges, can only converge to a point in
$L'$; hence $L'$ is closed.
To give you a taste of what this operation can do, here are some
theorems about derivation.
As always, let $S$ be a closed oriented hyperbolic surface, and $L$
a lamination on $S$.
\begin{thm}
\label{three_derivations} $L''' = L''$.
\end{thm}
That is, derivation has a sort of stability property: once you do it
enough times, you get to a stable sort of lamination to which
derivation does nothing. A lamination of this type has a special
name.
\begin{defn}
A lamination is said to be \emph{perfect} if $L' = L$.
\end{defn}
So the theorem above says that $L''$ is always perfect. In fact,
laminations can become perfect even earlier, if there are no closed
leaves.
\begin{thm}
\label{two_derivations}
If $L$ has no closed leaves then $L'' = L'$.
\end{thm}
In this sense, and as we will see later, closed leaves are actually
something of an annoyance to a lamination. We originally may like
closed leaves, because they are the easiest ones to see! But in fact
they just make life difficult. We will come to prefer laminations
with no closed leaves. When we come to discuss the correspondence
between laminations and the mapping class group, a closed leaf will
indicate that a surface automorphism is reducible, and reducible is
no fun for anybody. Pseudo-anosov is fun, and that means no closed
leaves.
\subsection{Boundary leaves and why they are everywhere}
In addition to isolated leaves, there are boundary leaves. These are
called boundary for a reason!
Soon, we will be considering what happens when we cut up the surface
along the lamination. We will get a disconnected bunch of smaller
surfaces. Leaves that are ``close" to others. Now what is the
boundary of these smaller surfaces? You might say, they are the
leaves of the lamination! Well, each boundary component will indeed
be a leaf of the lamination. But not every leaf will be a boundary
component. Can you see why?
If you can't see why immediately, consider that any leaf which forms
a boundary component must have a certain property --- it must be
``isolated on one side". If we have a leaf $\gamma$ which is not
isolated on either side, then any chunk of surface near it has other
leaves closer to it than $\gamma$. So $\gamma$ will not be a
boundary component. In fact you should be able to see that this is
condition of being isolated on one side is equivalent to being a
boundary component of the dissected surface. Just like the
definition of ``isolated", you can probably write a definition of
this idea of ``boundary" or ``isolated on one side" for yourself.
But I'll do it as well, again, for the sake of completeness.
\begin{defn}
Let $S$ be a complete oriented hyperbolic surface and $L$ a
lamination on $S$. A leaf $\gamma$ of $L$ is a \emph{boundary} leaf
if at each point of $x$ of $\gamma$ there is a half-disk
neighbourhood $V_x$ of $x$, with diameter lying on $\gamma$, such
that $L \cap V_x$ contains only the diameter of $V_x$.
(Equivalently, for all $x \in \gamma$ there exists a disk
neighbourhood $U_x$ such that $U_x \cap L$ contains at least one
component of $U_x - \sigma$, where $\sigma$ is the arc of $\gamma$
in $U_x$ containing $x$.)
\end{defn}
Now, if you cut the surface $S$ along the lamination $L$, you would
think that the regions you obtain should consist of most of the
original surface! If the leaves of the lamination were densely
packed somewhere, so they covered an entire ball in the surface,
then that whole part of the surface would be missing from the
dissected surface. But we know that our lamination $L$ is nowhere
dense; so that can't happen. Hence we expect that the remaining
leaves can't count for much. The boundary leaves account for most of
what's going on --- we lose non-boundary leaves after decomposing
the surface, but they don't count for much, in some sense.
In fact what we are trying to get at here is that the boundary
leaves are \emph{dense} in the lamination. Everything is close to a
boundary leaf; boundary leaves are everywhere!
\begin{lem}
\label{dense} Let $S$ be a compact oriented closed surface and $L$ a
lamination on $S$. The union of the boundary leaves of $L$ is dense
in $L$.
\end{lem}
\begin{Proof}
If you are a leaf of $L$, but not a boundary leaf, you can't bound
anything. So there have to be other things close by on both side ---
you are not isolated on either side. Hence there are leaves
arbitrarily close on both sides. What we want to show is that there
are \emph{boundary leaves} arbitrarily close; this is not difficult.
So, take a point $x \in L$ on a leaf $\gamma$; we must show there is
a point of a boundary leaf arbitrarily close to $L$. If $\gamma$ is
a boundary leaf we are done. If $\gamma$ is not a boundary leaf,
then as we just mentioned, there are leaves arbitrarily close to
$\gamma$. (In fact on either side, but we don't care.) Now take a
point $y \in S - L$ close to $x$; this is possible since $L$ is
nowhere dense. Take an arc connecting $y$ to $x$; proceeding from
$y$, this arc hits a first point of $L$, say $z$. Now $z$ lies on a
leaf which is not $\gamma$, which is a boundary leaf (bounding the
region containing $y$), and which is arbitrarily close to $x$.
\end{Proof}
\section{Principal regions and What They Look Like}
``Principal region" is the name given to the components of the
dissected surface we have been discussion --- the ones whose
boundary components are the boundary leaves of $L$.
\begin{defn}
Let $S$ be a closed oriented hyperbolic surface and $L$ a lamination
on $S$. A \emph{principal region} for $L$ is a component of $S - L$.
\end{defn}
Now that we have them, we may ask what these principal regions look
like! Let $U$ be a principal region for $L$. Clearly $U$ is also an
oriented hyperbolic surface. But $U$ may not be compact. Its
boundary components are the boundary leaves of $L$. A closed
boundary leaf will correspond to a closed geodesic boundary. A
non-closed, hence infinite length, boundary leaf will correspond to
a non-compact boundary component; the ends of this leaf will be ends
of the principal region.
A hyperbolic surface also has a lift to $\tilde{S} \cong \hyp^2$. So
a good way to understand principal regions will be to consider them
and how they sit in the hyperbolic plane, relative to the rest of
the surface. This is good, because if we try to draw them in the
surface itself they may look hideously complicated, with boundary
leaves that are a complete mess. In the hyperbolic plane, however,
geodesics look straight (or at least, circular in the unit disk
model), and hence the principal regions can't look too bad.
To understand what principal regions look like, we will make great
use of their lifts to $\hyp^2$.
\subsection{They are very finite...}
It might seem at first that our lamination could have extremely many
extremely complicated leaves, many of them boundary; and hence,
there could be finitely many principal regions. Thankfully, this is
not the case. The reason: the Gauss--Bonnet theorem, which has the
oft-forgotten consequence that the area of a hyperbolic surface
(with geodesic boundary, even if it has punctures) is tightly
constrained --- in fact, an integer multiple of $\pi$
\begin{lem}
Let $S$ be a closed oriented hyperbolic surface and $L$ a lamination
on $S$. Then $L$ has only finitely many principal regions, each with
only finitely many boundary leaves. (Hence only finitely many
boundary leaves.)
\end{lem}
The above observation shows that there are finitely many principal
regions. And if a principal region has infinitely many boundary
leaves, its Euler characteristic is infinite, implying infinite area
by Gauss--Bonnet, a contradiction.
\subsection{... can be lifted to the hyperbolic plane ... }
Let's now take a look at a lift of a principal region $P$ of a
lamination in $\hyp^2$. It's going to bound be a region of the
hyperbolic plane bounded by geodesics (possibly infinitely many,
since its lift may consist of infintiely many copies). The geodesics
bounding the plane have to be disjoint, else they would project to
intersecting leaves of $L$ in $S$. So topologically it's a disk, and
contractible; in fact, hyperbolically convex.
\begin{lem}
Let $S$ be a closed oriented surface and $L$ a lamination on $S$.
Let $P$ be a principal region for $L$ and $\tilde{P}$ a component of
the preimage of $P$ in $\hyp^2$. Then $\tilde{P}$ is a contractible
hyperbolic surface with geodesic boundary.
\end{lem}
We can also see that a principal region can't bound a disk: a
geodesic in a hyperbolic surface can't bound a disk.
\begin{lem}
If $P$ is a principal region for $L$, then $\pi_1(P) \To \pi_1(S)$
is injective.
\end{lem}
\subsection{... look nice when there aren't any closed leaves...}
Now we come to the first really interesting result about the
structure of principal regions. It turns out that when \emph{there
are no closed leaves}, they look particularly simple.
What will the regions look like? If there are no closed leaves, the
principal regions will be surfaces having boundary components
homeomorphic to $\R$. (Only boundary leaves which were closed leaves
could give boundary components homeomorphic to $S^1$).
One possibility is that a principal region $P$ is a polygon with
vertices at infinity. This possibility gives $P$ no interesting
topology. Now $P$ can have topology, however; but without any $S^1$
boundary components, the geometry at the boundary must be a little
subtle. If you have some familiarity with the geometry of hyperbolic
surfaces, you will know that a non-compact hyperbolic surface can
have a \emph{compact core}; that is, a connected compact subsurface
(or, in a degenerate case, a single geodesic) with boundary
consisting of geodesics, and which contains all the interesting
topology of the surface; and such that it separates the non-compact
ends of the surface, which ``flare" out to infinity. In our case,
the compact core will separate ends which have one $S^1$ boundary
component and finitely many $\R$ components, and have zero genus.
Thus each end looks like a ``crown": namely, an annulus with
finitely many punctures on one of its boundary components; the
geometry near each puncture goes out to infinity.
It turns out these are the only possibilities. If you know enough
about the geometry of hyperbolic surfaces, this may be obvious, once
you consider that we start with a surface of finite area. In any
case we will say something about why it's true.
\begin{lem}
Let $S$ be a closed oriented hyperbolic surface, and let $L$ be a
lamination on $S$ without closed leaves. Let $U$ be a principal
region for $L$. Then $U$ is either:
\begin{enumerate}
\item
isometric to a finite sided polygon with vertices at infinity;
or
\item
there exists a unique compact subset of $U_0$ (either a
subsurface, or a geodesic) such that $U - U_0$ is isometric to a
finite disjoint union of crowns.
\end{enumerate}
\end{lem}
To see why it's true, we consider $\tilde{U} \subset \hyp^2$. The
boundary of this region consists of $\R$ components. Take a boundary
leaf $\beta_0$; since the region has finite area, there are adjacent
boundary leaves adjacent to each end of $\beta_0$. So we obtain a
sequence $\beta_n$. This sequence may cycle around, in which case
$\tilde{U}$ is a finite sided polygon with vertices at infinity ---
and hence, so is $U$ (it is the universal cover of something,
something orientable, but that something can't have an infinite
fundamental group). Or it may not, in which case the $\beta_n$ in
the disk model get smaller and smaller (in the Euclidean metric!)
and converge to two limit points $\beta_{-\infty}$ and
$\beta_\infty$. (Note although $U$ can only have finitely many
boundary leaves, its universal cover can have infinitely many; and
will, if $\pi_1 (U)$ is infinite.) Joining these we obtain a convex
region $\tilde{W} \subset \tilde{U}$; it is the convex hull of the
$\beta_n$.
Now $\tilde{W}$ covers $W$, which will be a crown. Well clearly none
of the boundary leaves $\beta_n$ project to circles, since we have
no closed leaves. But the deck transformation taking $\beta_n
\mapsto \beta_{n+1}$ will project the axis joining $\beta_{-
\infty}$ to $\beta_\infty$ to a circle. So this gives a crown.
Cutting all these out gives the compact core.
\subsection{... without any annoying closed leaves, are finite at infinity...}
One more property. We now see that our principal regions will lift
to regions of $\hyp^2$, with compact cores and crown sets and
polygons tessellating the plane. These are all bounded in $\hyp^2$
by geodesics with endpoints at infinity. We know that there are only
finitely many principal regions, each with only finitely many
boundary leaves; but in $\hyp^2$, a compact core can have infinitely
many boundary components in its universal cover, and a crown has
infinitely many boundary components in its universal cover. So it
looks like things are much more infinite in $\hyp^2$, perhaps as
expected.
But not \emph{so} infinite. Turns out each vertex at infinity is
very finite! Again, \emph{provided} there are no closed leaves.
\begin{lem}
Let $S$ be a closed oriented hyperbolic surface and let $L$ be a
lamination on $S$ without closed leaves. Any point on the circle at
infinity is the endpoint of only finitely many leaves of
$\tilde{L}$.
\end{lem}
Why is this? Suppose there were infinitely many leaves meeting at
the same point at infinity $x$. Since there are only finitely many
boundary leaves, by the pigeon hole principle there's a deck
translation $g$ fixing $x$. Now translating boundary leaves ending
at $x$ under $g$, they converge to the axis of $g$. As a lamination
is closed, the axis of $g$ is a leaf of $L$. But since $g$ is a deck
transformation, this projects to a closed leaf of $L$,
contradiction.
It's precisely the absence of closed leaves that makes this
impossible. If closed leaves are allowed, all manner of infinities
in the degrees of these vertices can and will arise. In fact, if you
think about taking the universal cover of any surface with some
closed leaves in a lamination, you'll see that there will generally
be infinite degree vertices at infinity; these arise whenever there
is spiralling in towards a closed leaf.
So, we see that the absence of closed leaves really does simplify
matters, perhaps to a bizarre extent, when we consider the picture
in $\hyp^2$.
\subsection{... and sometimes even when there \emph{are} closed (isolated) leaves...}
We now have a nice picture of the lift of a lamination to $\hyp^2$.
There are principal regions, which look rather nice when there are
no closed leaves. If there are no closed leaves, we either have
finite sided ideal polygons, or compact cores and crowns, both of
which are quite simple phenomenon. In fact, this picture can be
useful even when there are no closed leaves: given a lamination $L$,
take out the (necessarily finitely many) closed leaves, look at the
principal regions, and put the closed leaves back in again. However
it's not that simple: our closed leaf may be isolated; or it may be
isolated only on one side, i.e. a non-isolated boundary leaf; or it
may be isolated on neither side, i.e. a non-boundary leaf. Only when
the closed leaf is isolated is the picture easy to see.
What does this amount to? Let us consider the picture in $\hyp^2$.
If our closed leaf is isolated, then inserting it into the picture
simply subdivides some existing principal regions. So this amounts
to inserting a geodesic into the interior of finite sided ideal
polygons and compact-cores-with-crowns. But the closed leaves can't
cross the non-closed leaves --- a lamination consists of
\emph{disjoint} geodesics! So each closed leaf either lies inside a
finite sided ideal polygon, or lies inside a
compact-core-with-crowns.
If an isolated closed leaf lies inside a finite sided ideal polygon,
it's easy to see it must just be a diagonal. Actually, as it turns
out this picture can't happen. The reason why is the argument of the
previous lemma. The deck translation $g$ corresponding to our closed
leaf $\gamma$ must carry some sides of the polygon closer to
$\gamma$, contradicting the fact that we have a principal region.
(And, for that matter, showing that $\gamma$ is not isolated.
Because diagonals in ideal polygons are isolated.)
If an isolated closed leaf lies in the interior of a
compact-core-with-crowns, it must lie inside the compact core. If an
endpoint in $\hyp^2$ lifts to a crown, it can't possibly project to
a closed leaf, since it has infinite distance going out to a point
of a crown. So its endpoints must lie in the compact core; and as
the core is convex, so must the entire leaf.
This case includes our simplest example: when $L$ is a finite
collection of simple closed geodesics. Then we remove the closed
geodesics, obtaining the empty lamination; we take a principal
region, which is the entire surface, for which there are no crowns
and the entire surface is the compact core; and then we re-insert
the closed leaves, which certainly lie within the compact core!
Maybe not such an interesting example.
Things are more complicated when we have a non-isolated closed
curve. The problem is, once it is removed, we no longer have a
lamination; the set is no longer closed. So all our previous
arguments for laminations without closed leaves no longer apply.
\subsection{... and take up all the space...}
We have already said that geodesics can't be too close; that a
lamination is nowhere dense in the surface (proposition
\ref{nowhere_dense}); and that boundary leaves are dense in the
lamination (lemma \ref{dense}). It would seem, then, that they
should be small. But we have not said anything about the \emph{area}
of the lamination. It should be measure zero, surely! Thankfully it
is.
\begin{lem}
Let $S$ be a closed orientable hyperbolic surface and $L$ a
lamination on $S$. The area of $L$ is zero; the area of $S-L$ is the
full area of $S$.
\end{lem}
This is a vector-field and Gauss--Bonnet bash. Construct some vector
fields on the principal regions, with some singularities, in a
canonical way, and use that the Euler characterisic.
\subsection{... while the rest is rather weird.}
Now for perhaps the most confusing part of the picture of all. We've
got an excellent picture of principal regions, at least when there
are no closed leaves. We have principal regions all over $\hyp^2$.
You might have got the impression that they in fact
\emph{tessellate} $\hyp^2$. Sadly, if you have this impression, it's
not quite correct. It will be correct only if every leaf is
isolated. We actually have the plane split into principal regions,
with strange ``twilight zones" in between of non-boundary leaves
--- except that there is no zone as such. This sounds weird. What does this mean?
Well, certainly if we have a boundary leaf which is not isolated,
beyond it will be non-boundary leaves. But leaves are nowhere dense,
and moreover, boundary leaves are dense in the lamination. So
there's no region of non-boundary leaves or anything; in fact, there
must be a boundary leaf, and hence a principal region, arbitrarily
close to any non-boundary leaf. On the other hand, if we have a
non-boundary leaf, there are leaves which come arbitrarily close to
it; and then it is not difficult to show that there are leaves which
come arbitrarily close to any point of the leaf (to be close at a
particular point, it suffices to be very very \emph{very} close at
another point!).
Reconciling these apparently-contradictory-but-not statements gives
a picture of the non-boundary leaves. Well, not so contradictory at
all, come to think of it. The non-boundary leaves can't just
disappear from the picture in $\hyp^2$ --- and they can't bound any
principal regions! It follows that they have to be doing something
like this.
\section{Derivations and their properties}
\subsection{Stability, Perfection and Their Consequences}
We now look a bit more closely at derivations. Turns out a derived
lamination $L'$ has interesting stability properties. As we will
see, \emph{the minimal sublaminations of $L'$ are precisely its
components}. Let's see what this amounts to.
Well, it implies some pretty strong and amazing stuff. Because $L''$
is a sublamination of $L'$, hence a union of minimal sublaminations
--- so $L''$ just consists of a subset of the components of $L'$.
Some of the components of $L'$, upon being derived, cease to exist;
others of them persist --- and since they are minimal, their
derivation must be themselves. So the components of $L'$ either
disappear under derivation, or are perfect; and $L''$ consists
precisely of the perfect components. It follows that $L''$ is
perfect and $L''' = L''$. So once we prove the above statement, we
have proved theorem \ref{three_derivations}.
It also implies theorem \ref{two_derivations}: if $L$ has no closed
leaves then $L'' = L'$. For when we consider the components of $L'$,
the ones that disappear are sublaminations whose derivation is empty
--- they are sublaminations consisting entirely of isolated leaves.
Any lamination consisting entirely of isolated leaves actually
consists entirely of closed leaves; an infinite leaf will spiral
towards something. But if we rule out closed leaves, then none of
these occur, so $L'$ is perfect.
There is another corollary as well:
\begin{thm}
The following are equivalent:
\begin{enumerate}
\item
every leaf of $L$ is dense in $L$;
\item
$L$ is connected and $L' = L$.
\end{enumerate}
\end{thm}
If every leaf is dense, then the lamination is certainly connected,
and there are no isolated leaves, so $L' = L$. On the other hand, if
$L$ is connected and $L' = L$, then $L' = L$ is a union of perfect
components. Since it is connected, it is one perfect component. So
the closure of any leaf is that component, hence the whole
lamination.
\subsection{How the proof goes.}
Now, let's see why it might be true that the minimal sublaminations
of $L'$ are precisely its components. A good example to bear in mind
is a lamination consisting of 3 leaves: a closed leaf, and two
non-closed leaves spiralling towards it from either side.
A component of a lamination is always a sublamination; but not
necessarily a minimal one. There may be extra leaves in there. In
our example there is only one component, and the whole lamination is
certainly a sublamination, but it's not minimal. For instance, just
take the closed leaf. On the other hand, if you take a minimal
sublamination, it is the closure of any leaf in it; but this minimal
sublamination may not be a component. In our example, the closed
leaf is a minimal sublamination, but certainly not a component.
In our example, however, deriving removes both the spiralling
leaves, so we are left only with one leaf, and clearly the minimal
sublaminations are the components now.
Note what this means: upon taking $L'$, everything near closed
leaves gets removed --- leaving closed leaves isolated. This makes
sense --- closed leaves will now become isolated components of $L'$,
and will be among the components of $L'$ that disappear under
further derivation. And in $L''$, there are no closed leaves, and no
leaves spiralling towards closed leaves --- which is good, for such
things are not perfect.
In fact, in our example there is a general phenomenon occurring: any
leaf close to a closed leaf is isolated.
\begin{lem}
Let $S$ be a closed oriented hyperbolic surface and let $L$ be a
lamination on $S$. Let $C$ be a closed leaf of $L$. Then any
other leaf sufficiently close to $C$ is isolated. That is, there
exists a neighbourhood $N$ of $C$ such that any leaf other than
$C$ intersecting $N$ is isolated; equivalently, $L' \cap N
\subseteq C$.
\end{lem}
To see why, again we look to $\hyp^2$. The closed leaf $C$ lifts to
a line $\tilde{C}$ with a deck transformation $g$ translating along
it. Think about what a leaf close to $C$ looks like. We might expect
it to share an endpoint at infinity with $\tilde{C}$. This is right.
If it shares no endpoint at infinity, and is sufficiently close,
then translating it by $g$ will see it intersecting its translate.
This is bad, because there is no self-intersection in leaves. So it
does share an endpoint at infinity with $\tilde{C}$. But we know
that there are only finitely many leaves of $\tilde{L}$ intersecting
at a point at infinity, so $C$ is isolated.
Now for the main theorem.
\begin{thm}
\label{union_of_components} Let $S$ be a closed oriented hyperbolic
surface and $L$ a lamination on $S$. If $L_1$ is a sublamination of
$L'$ then $L_1$ is a union of components of $L'$. Equivalently, the
minimal sublaminations of $L'$ are precisely its components.
\end{thm}
\begin{Proof}
The idea is simply to analyse what's in $L_1 \subset L'$ and where
they lie in the picture in $\hyp^2$. In particular, $L_1$, being a
subset of $L'$, contains no isolated leaves of $L$: it contains only
non-isolated leaves of $L$. Amongst these there are
\begin{enumerate}
\item
closed leaves; and
\item
non-closed leaves --- these we call $L_2$.
\end{enumerate}
Now $L_1$ will consist of closed leaves, as well as $L_2$. We have
showed that anything near a closed leaf is isolated, so the leaves
in category (i) will be isolated in $L'$, hence components of $L'$.
We only need to show that the leaves in category (ii), namely $L_2$,
also form a union of connected components of $L'$.
Now $L_2$ is a lamination, being a sublamination of $L'$ where
closed leaves in $L'$ are isolated. The point is to show that if we
take any other leaf in $L'$, it lies far away from $L_2$; for then
$L_2$ is a union of connected components of $L'$. The way we do this
is to consider the principal regions of $L_2$; anything in $L'$ but
not $L_2$, namely anything in $L' - L_2$, we will show lies in the
compact cores of the principal regions of $L_2$. Thus they are far
from $L_2$, which are all near boundary leaves and the boundaries of
the principal regions.
We consider the possibilities for a principal region $U$ of $L_2$.
If $\tilde{U}$ is a finite-sided ideal polygon, and $\gamma$ is a
leaf of $L' - L_2$, then $\tilde{\gamma}$ is a diagonal, so is
isolated. In fact, it is also in $L$ of course ($L' \subset L$!),
and must have been isolated there, a contradiction since then it
wouldn't be in $L'$!
If $\tilde{U}$ consists of a compact core and crown regions, then
$\tilde{\gamma}$ either lies in a compact core or has an end going
to an end of a crown region. The first case is what we want. In the
second case, the leaf is isolated (and was in $L$), a contradiction
again to being in $L'$.
\end{Proof}
In fact we can do something which is maybe slightly stronger and
definitely much more convoluted! (This is the statement Casson and
Bleiler prove, which is perhaps not the most friendly approach from
the point of view of understanding.) We have shown that any
sublamination of $L'$ consists of components of $L'$, but in fact,
\emph{any sublamination of $L$}, when intersected with $L'$,
consists of components of $L'$.
\begin{cor}
\label{grr} Let $S$ be a closed oriented hyperbolic surface and $L$
a lamination on $S$. If $L_1$ is a sublamination of $L$, then $L_1
\cap L'$ is a union of components of $L'$.
\end{cor}
For $L_1 \cap L'$ is certainly a sublamination of $L'$: the leaves
of $L_1$ which intersect $L'$ are certainly leaves of $L'$; and the
intersection of two closed sets is closed. The corollary now follows
immediately.
\section{Surface Automorphisms}
The most amazing thing about surface automorphisms is their
classification: periodic, reducible, or pseudo-Anosov. In essence,
reducible ones are annoying because they really belong not on our
surface, but on a decomposition of the surface into smaller pieces.
Periodic ones also just get in the way; it's pretty unusual to have
a periodic automorphism. What's amazing is that \emph{everything
else} is pseudo-Anosov, which means having some very interesting
dynamical properties (defined in terms of measured foliations). It's
not clear at all why ``most" nontrivial automorphisms should have
such an interesting dynamical description.
These definitions are topological; they do not involve hyperbolic
geometry, or any other type of geometry for that matter. It is by
introducing geometry that we get so much interesting structure. The
definitions are up to homotopy, and we'll be homotoping stuff around
quite freely.
\begin{defn}
Let $S$ be a closed orientable surface. Let $h: S \To S$ be an
automorphism of $S$. The automorphism $h$ is called:
\begin{enumerate}
\item
\emph{periodic} if there exists a positive integer $n$ such that
$h^n$ is homotopic to the identity;
\item
\emph{reducible} if $h$ is homotopic to an automorphism which
leaves invariant an essential closed 1-submanifold of $S$, i.e.
leaves invariant a finite collection of essential closed curves.
\item
\emph{pseudo-Anosov} if there exist transverse singular
foliations $\F^s, \F^u$ (called \emph{stable} and
\emph{unstable} foliations) equipped with transverse measures
$\mu^s, \mu^u$ such that for some $\lambda > 1$,
\[
h(\F^s, \mu^s) = (\F^s, \lambda \mu^s), \quad h(\F^u, \mu^u)
= (\F^u, \lambda^{-1} \mu^u)
\]
\end{enumerate}
\end{defn}
The definition of pseudo-Anosov means that the leaves of the stable
foliation attract, and the leaves of the unstable foliation repel.
The idea is that reducible automorphisms really belong on the
surface obtained by cutting along the invariant 1-submanifold. And
periodic automorphisms cannot possibly have the pseudo-Anosov
property.
The above-mentioned amazing fact is contained in the following
theorem.
\begin{thm}
\label{amazing} Every non-periodic irreducible automorphism of a
closed oriented hyperbolic surface is isotopic to a pseudo-Anosov
automorphism.
\end{thm}
We'll get to this eventually.
\subsection{Telling them apart}
We can tell something about reducibility or periodicity from looking
at the induced map of a surface automorphism on its homology. As
always, let $S$ be a closed oriented surface --- for now it's just
topological, not hyperbolic, not geometric. Then the automorphism
$h: S \To S$ induces $h_*: H_1(S) \To H_1(S)$. The first homology
over $\Z$ is just a free abelian group (of rank $2g$ where $g$ is
the genus of $S$). So $h_*$ can be written as a ($2g \times 2g$)
matrix $A$ with integer entries and which is invertible. We write
$\chi_h(t)$ for the (degree $2g$) characteristic polynomial of this
matrix.
\begin{lem}
If $h$ is periodic then all the zeroes of $\chi_h(t)$ are roots of
unity.
\end{lem}
This is clear; since $h_*^n = 1$, we have $A^n = 1$; so the minimal
polynomial of $A$ is a factor of $t^n - 1$, hence has all zeroes
being roots of unity; hence also the characteristic polynomial.
\begin{lem}
If $h$ is reducible then either $\chi_h(t)$ has a root of unity as a
zero, or is reducible, or is a polynomial in $t^n$ for some positive
integer $n>1$.
\end{lem}
This is just a little linear algebra, and requires different
considerations depending on how the invariant 1-submanifold
separates the surface. If you can get $h_*(C) = C$ for some
(homology class of a) closed curve $C$, you'll get a root of unity
in the characteristic polynomial. But $h_*$ may permute the
components of the invariant 1-submanifold, and then we have to think
about the matrix: the characteristic polynomial then may be
reducible; if it is not reducible, we find it is of the form $\det(B
- t^n I)$ for some $n>1$, and hence a polynomial in $t^n$.
\subsection{Surface automorphisms meet laminations: The invariant lamination}
So far we have only discussed surface automorphisms in the abstract.
No mention of what they have to do with laminations! The definition
of pseudo-anosov has to do with \emph{foliations}, which may look a
little like laminations, but not really; they're very different, in
fact.
The connection is: an automorphism which is not periodic --- (which
we secretly know will be reducible or pseudo-Anosov) --- has an
invariant \emph{lamination}.
\begin{thm}
\label{invariant_lamination} Let $S$ be a closed oriented hyperbolic
surface and let $h$ be a non-periodic automorphism of $S$. Then
there exists a lamination $L$ on $S$ such that $h(L) = L$.
\end{thm}
In saying that $h(L) = L$, we are employing the standard procedure
of taking geodesics to geodesics by a straightening process. (We are
homotoping rather blithely, then.) This is done by lifting $h$ to
the universal cover $\hyp^2$, where it acts on $\hyp^2$ and extends
continuously to the circle at infinity $S^1_\infty$. (I have not
proved this, but it's a standard fact, and it's proved in Casson and
Bleiler.) Then straightening out geodesics by noting where their
endpoints map to at infinity.
The proof is a bit tricky. It uses a clever trick to obtain an
infinite sequence of distinct geodesics; and then it uses the
compactness of the space $\L$ of laminations; and then some more!
\begin{Proof}
We first show that there exists a simple closed geodesic $C$ such
that for all $n \geq 1$, $h^n(C) \neq C$. Suppose to the contrary
that there is not; then for all simple closed geodesics $C$, there
is a sufficiently high power of $h$ which fixes it. Taking enough
geodesics and taking a multiple of all the sufficiently high powers
--- which is alright since $h$ is not periodic --- we fix so much
of the surface that $h$ must in fact be periodic. This is a
contradiction. So there is such a $C$.
Now we look at $h^n(C)$. This is a sequence of simple closed
geodesics in $S$, hence a sequence of laminations. But $\L$, the
space of laminations, is compact --- so this sequence has a
convergent subsequence $h^{n_i}(C)$, which converges to a lamination
$K$.
Seems like we might be done. Well we would be, if $h(K) = K$. But
sadly, there is no reason why this should true. (We would have been
fine, if our sequence $h^n(C)$ were convergent --- but we only have
a subsequence.) We're going to try to get an invariant lamination
out of this.
It is therefore natural to consider the laminations $h^n(K)$. If
they were disjoint, we could just take their union, (and close it,)
and then we would have an invariant lamination. In fact, even if
they were not disjoint, but contained some overlaps of the same
geodesics, this would be good enough: we still just take union and
closure. But they may not be disjoint! So we ask: how much might
they intersect? Now, \emph{transverse} interections are what matter;
degenerate intersections of geodesics mean that both laminations
contain the same geodesic, which as we just noticed is no problem at
all.
We therefore need to consider $h^r(K) \cap h^s(K)$. Well we can
cancel some $h$'s here so it is sufficient to consider $K \cap
h^r(K)$, and transverse intersections in it. This is a limit of
$h^{n_i}(C) \cap h^{n_i + r}(C)$, which has the same number of
transverse intersections as $C \cap h^r(C)$. When we take a limit,
the number of transverse intersections might change
--- but it \emph{can only decrease}. For take a small neighbourhood
of each transverse intersection point in the limit; these also
contain transverse intersections of the converging laminations ---
and possibly more of them.
Therefore, we let $N_r$ be the number of transverse intersections of
$C \cap h^r(C)$. Then $K \cap h^r(K)$ has $\leq N_r$ transverse
intersections. In particular, this number \emph{is finite}, and
consists of discrete points. So every leaf where a transverse
isolation occurs \emph{is isolated}. It follows that $K'$ and
$h^r(K')$ have no transverse intersection points; and hence, any
$h^r(K')$ and $h^s(K')$ have no transverse intersections.
So we now carry out our strategy, take the union $\cup^\infty
h^r(K')$ and its closure $\overline{\cup^\infty}(K')$. This is the
lamination we want.
\end{Proof}
\subsection{What the invariant lamination means}
We have looked at several features of laminations. Some of these
correspond to features of the automorphism.
As a first example, suppose that $L$ has a closed leaf. Since $h$
fixes $L$, it takes a closed leaf to another closed leaf. There can
be only finitely many closed leaves (because they are disjoint
simple closed geodesics), and hence after possibly cycling around we
have a finite collection of disjoint closed leaves invariant under
$L$; so $L$ is reducible.
\begin{lem}
Let $S$ be a closed oriented hyperbolic surface and $L$ a lamination
on $S$ which is invariant under the automorphism $h$. If $L$ has a
closed leaf then $h$ is reducible. Equivalently, if $h$ is
irreducible then $L$ has no closed leaves.
\end{lem}
This is the real proof that closed leaves are annoying. Closed
leaves are annoying because reducible automorphisms are annoying;
and closed leaves imply reducible automorphisms.
As another example, consider the principal regions. If
there are no closed leaves, then these come in two types: finite
sided ideal polygons; and surfaces with compact cores with crowns
attached. If the lamination is fixed under an automorphism $h$,
however, then these regions must be carried to themselves. In
particular:
\begin{enumerate}
\item
boundary leaves map to boundary leaves
\item
principal regions map to principal regions; and in particular
\item
finite ideal polygon principal regions map to finite ideal
polygon principal regions, and
\item
compact cores and crowns map to compact cores and crowns.
\item
In particular, boundaries of the compact cores map to boundaries
of compact cores.
\end{enumerate}
However, boundaries of compact cores are simple closed curves in our
surface. So they give us a 1-submanifold invariant under $h$.
\begin{lem}
Let $S$ be a closed oriented hyperbolic surface and $L$ a lamination
on $S$. Suppose $L$ is invariant under an automorphism $h: S \To S$.
If $L$ has no closed leaves, and there exists a principal region of
$L$ which has a compact core, then $h$ is reducible. Equivalently,
(and better!), if $h$ is irreducible then (as we previously saw,
there are no closed leaves and) each principal region of $L$ is a
finite sided ideal polygon.
\end{lem}
Now, this picture says even more. If we only have finite sided ideal
polygons for principal regions, then \emph{the lamination is
connected.} Just go from one principal region to another along
boundary leaves --- which meet at infinity (and hence are
arbitrarily close, so in the same connected component). Boundary
leaves are dense in the lamination, as we know, so the lamination is
connected.
\begin{lem}
We still suppose $L$ is invariant under $h$ as above. If $h$ is
irreducible, then $L$ is connected.
\end{lem}
We can still say more. Being connected as a lamination has even more
relevance if you are of the form $L'$, i.e. you are a derived
lamination. If $h(L) = L$, then certainly $h(L')=L'$. If $h$ is
irreducible then $L$ has no closed leaves, so certainly not $L'$. So
what does $L'$ look like? As we know, its components are precisely
its minimal sublaminations, and there are no closed leaves. If $L'$
is connected then: it's just one component, perfect, and has no
proper sublaminations. So $L'' = L'$ and the closure of any leaf of
$L'$ is all $L'$. Said another way, every leaf of $L'$ is dense in
$L'$.
Even more than this: take a leaf $\gamma$ of $L$; as $h$ is
irreducible, $\gamma$ is not closed. Then its closure $\bar{\gamma}$
is a sublamination of $L$, and by corollary \ref{grr}, $\bar{\gamma}
\cap L'$ is a union of components of $L$. (If you don't like
annoying corollary \ref{grr}, you can quite easily still use the
more approachable theorem \ref{union_of_components}.) It's nonempty,
since otherwise $\bar{\gamma}$ would consist entirely of isolated
leaves of $L$, impossible as $\gamma$ is not closed. Since $L'$ is
connected, $\bar{\gamma} \cap L' = L'$, so $\gamma$ is dense in $L$.
So not only is every leaf of $L'$ dense in $L'$; better, every leaf
of $L$ is dense in $L'$.
\begin{lem}
We still let $L$ be invariant under $h$. If $h$ is irreducible, then
$L'$ is connected and perfect and every leaf of $L$ is dense in
$L'$.
\end{lem}
We are not done yet. Let us consider $L'$ further; in particular,
let us consider its principal regions. We know that $h(L') = L'$, so
from the argument we applied above to $L$ we see that each principal
region of $L'$ is a finite sided ideal polygon. But now, how do
these join up? Suppose one edge of such a polygon in $\hyp^2$ joins
directly to an edge of another polygon; then the edge along which
they were joined is isolated (in $L'$). But $L'$ contains no
isolated leaves; in fact, it is connected and perfect. What must
happen is that non-boundary leaves interfere on the other side of
the edge, which are nowhere dense but come arbitrarily close to our
edge; and then there is a whole mess of boundary leaves among them
(since boundary leaves are dense in the lamination). In any case it
follows that the vertices at infinity of the polygon principal
region have degree precisely 2. If they had larger degree, the other
edges would interfere with nearby non-boundary leaves.
\begin{lem}
Still let $L$ be invariant under $h$. If $h$ is irreducible, then
every principal region of $L'$ is a finite sided ideal polygon for
which every ideal vertex is the endpoint of precisely two leaves of
$L'$.
\end{lem}
To summarise:
\begin{prop}
Let $S$ be an oriented closed hyperbolic surface. Let $h$ be an
irreducible automorphism of $S$ and let $L$ be a lamination
invariant under $S$. Then:
\begin{enumerate}
\item
$L$ has no closed leaves;
\item
every principal region of $L$ is a finite sided ideal polygon,
and every principal region of $L'$ is a finite sided ideal
polygon for which every ideal vertex is the endpoint of precisely
two leaves of $L'$;
\item
$L$ is connected;
\item
$L'$ is connected and perfect;
\item
every leaf of $L$ is dense in $L'$.
\end{enumerate}
\end{prop}
That's quite a lot to say about a lamination, from the simple fact
of irreducibility --- and every automorphism is irreducible once you
cut up the surface enough.
\subsection{Attraction and Repulsion}
In nature, as in life, some things attract, and other things repel.
Pseudo-anosov automorphisms do both: the leaves of the stable
foliation repel, while the leaves of the unstable foliation attract.
We want to see why an irreducible and non-periodic automorphism has
these attracting and repelling properties.
Let's consider the stable foliation. What we hope to see is a
singular foliation, leaves of which attract. The leaves attract each
other, but to conserve area (it's a finite area surface!), each
individual leaf should be elongated, in some sense.
As it turns out, the stable and unstable foliation of a
pseudo-anosov automorphism will arise from an invariant lamination.
How do you go from a lamination to a foliation? This will be
demonstrated shortly, in a dazzling burst of analysis that uses a
Cantor function. (Yes, as in the Cantor set and the devil's
staircase.)
So then, in a pseudo-anosov automorphism and an invariant lamination
we should perhaps expect to see some fixed leaves; and nearby ones
are attracted to them. (Such a fixed leaf will not be closed, since
otherwise it would be a reducible automorphism, by the previous
section.) This doesn't seem to make much sense, since we have shown
that leaves cannot be too close --- for instance, they are nowhere
dense; furthermore, boundary leaves are dense, and there are only
finitely many of them --- and boundary leaves are all that show up
in the hyperbolic plane! What we will see in $\hyp^2$, however, is
boundary leaves shifting around to get ``closer" to our fixed leaf,
even though they are still actually quite far away.
In the universal cover $\hyp^2$, such a leaf $\gamma$ becomes to a
geodesic $\tilde{\gamma}$; even if it's a ``singular" leaf which
ends at a singularity, its continuation is a geodesic. On either
side of $\gamma$, other leaves should be attracted. So upon
iteration of our iteration $h$, they should move closer and closer,
and in fact converge to $\tilde{\gamma}$. Since we know that each
vertex at infinity has only finitely many geodesics ending there,
this means that the vertices at infinity converge to the endpoints
of $\tilde{\gamma}$.
So, we require that under iteration of $h$, the circle at infinity
moves in such a way that every point converges to an endpoint of
$\tilde{\gamma}$. Well, this is not quite right. For one thing, it
is impossible since $h$ acts continuously on the circle at infinity;
there must be intervening repelling fixed points. For another, this
only applies if the geodesic $\tilde{\gamma}$ is ``innermost" in
some sense.
How could we define ``innermost" here? Consider the interval on the
circle at infinity $S^1_\infty$ cut of by $\tilde{\gamma}$. What we
\emph{can't} have is a situation, say, where this interval is cut in
half by two geodesics, lifts of leaves, which form an ideal triangle
with $\tilde{\gamma}$. If this happens, the automorphism (being an
automorphism of the interval, and leaving $\tilde{L}$ invariant)
must fix those geodesics
--- so that $\tilde{\gamma}$ is not ``innermost". When $\tilde{\gamma}$
is ``innermost", we call the interval it cuts off on $S_\infty^1$
\emph{stable}.
Try to find a definition that makes all this rigorous. If you
can, it's probably equivalent to the following definition.
\begin{defn}
Let $S$ be a closed oriented hyperbolic surface and $L$ a lamination
on $S$. Let $\tilde{L}$ be the lift of $L$ to the universal cover
$\tilde{S} \cong \hyp^2$. A \emph{stable interval for $L$} is a
closed interval $I \subset S^1_\infty$ such that for any two points
$P, Q$ in the interior of $I$, there is a leaf of $\tilde{L}$ which
separates $P$ and $Q$ from $I$.
\end{defn}
It follows that any stable interval for $L$ is a (lift of a) leaf of
$L$ --- just take $P$ and $Q$ arbitrarily close to the endpoints of
$I$. It also follows that there are leaves of the lamination which
have endpoints at infinity arbitrarily close to the endpoints of
$I$.
We should obtain in this situation a point of $I$ which is a point
at infinity fixed under $h$, and which is the unique repelling fixed
point of $h$ in $I$. The endpoints of $I$ will be only attracting
fixed points; the interior fixed point will attract
everything else.
To prove our amazing theorem \ref{amazing}, we will show: that this
definition of stable interval, in our situation, implies something
nice about attracting and repelling fixed points; that when $h$ is
irreducible and non-periodic, it has nicely behaved fixed points at
infinity, alternately repelling and attracting; it can be taken to
fix a lamination which involves stable intervals; and that this
implies the pseudo-anosov property.
\subsection{The Proof I: How good is our invariant lamination?}
We now embark on the first part of the proof of our amazing theorem
\ref{amazing}. We will consider an irreducible and non-periodic $h$,
and the invariant lamination $L$ we obtained earlier in theorem
\ref{invariant_lamination}. Note this required $h$ to be
non-periodic.
Where did $L$ come from? Recall we took a curve $C$ for which
$h^n(C) \neq C$ --- we showed one exists, but any would do. Then we
considered $h^{n}(C)$ and took a convergent subsequence, converging
to a lamination $K$ --- such a convergent subsequence exists, but
any would do. Then we considered $h^n(K)$, showed that these all
contain no transverse intersections, so that $h^n(K')$ were all
disjoint; and took their union for $L$.
It is worth thinking about how $C$, $K$ and $L$ interact. First let
us consider $K$ and $L$. Can they intersect? They certainly can, in
fact all of $K'$ lies in $L = \overline{\cup^\infty h^n(K')}$. But
what about transverse intersection? Well, any leaf of $K$ which has
a transverse intersection with $L$ has a transverse intersection
with some $h^n(K')$; and hence has infinitely many points of
transverse intersection with $h^n(K)$, a contradiction to what we
found in our proof of theorem \ref{invariant_lamination}. So $K$ and
$L$ have no transverse intersection.
We know quite a lot about $L'$, from section 9.3. In particular, we
know that every principal region is a finite sided ideal polygon,
and that arbitrarily close to each side, there are leaves on the
outside. We can ask where our special curve $C$ lies in this
picture. Clearly $C$ can't intersect $L'$ tangentially; this would
mean it coincides with a leaf of $L'$, but $L'$ contains no closed
leaves. Can it avoid $L'$ altogether? If it doesn't intersect $L$,
then it lies in one principal region; and hence it must be a
diagonal in the ideal polygon. But since $C$ is a closed curve, this
would mean that leaves get arbitrarily close to $C$; and then deck
transformatoins corresponding to $C$ would contradict our picture of
our finite sided ideal polygon. So $C$ intersects $L'$ transversely
somewhere.
This is interesting: $C$ intersects $L'$ transversely, but $K$,
which is the limit of $h^{n_i}(C)$ for some sequence $n_i$, does not
intersect $L'$ (in fact, not even $L$) transversely. What has
happened? In $\hyp^2$, $C$ crosses leaves of $L'$; but upon
iterating with $h$ this has straightened out into a leaf of $L'$. We
will exploit this interesting property in regard to stable
intervals.
Now let's suppose we have a stable interval $I$ on $S^1_\infty$ cut
off by a geodesic $\tilde{\gamma}$. As we have seen, $\tilde{\gamma}
\in L$. It follows from the definition of stable interval that there
are other leaves arbitrarily close to $\tilde{\gamma}$ on one side,
so $\gamma$ is not isolated, and $\gamma$ is in $L'$.
We want to see what happens if $h$ fixes $\gamma$, and hence fixes
$I$.
Well, we consider $C$, which intersects $L'$. We lift to $\hyp^2$
and take a component $\tilde{C}$ which intersects $\tilde{\gamma}$,
with endpoints $A,B$. Without loss of generality, $A$ lies in the
interior of $I$, and $B$ lies outside $I$. Upon iterating $h$, we
know that $h^{n_i}(C) \To K$, so $h^{n_i}(\tilde{C})$ converges to a
geodesic $\tilde{C}_\infty$, with endpoints $A_\infty, B_\infty$.
Clearly $\tilde{C}_\infty$ is a lift of a leaf of $K$, which does
not intersect $L'$ transversely. There are not many ways this can
happen. Since $B$ lies outside $I$, so does $B_\infty$. And since
$A$ is in the interior of $I$, $A_\infty \in I$. The only way to
avoid transverse intersection with $I$ is to have $A_\infty \in
\partial I$. So we have shown that $h^{n_i}(A) \To \partial I$. It
follows that $A_\infty$ is an attracting fixed point of $h|_I$.
Are there any other fixed points? Since $h$ fixes $L'$, and since
boundary leaves are dense, we can see that the open subinterval $U =
(A, A_\infty)$ contains no fixed points; $A$ moves towards
$A_\infty$, and so does everything else in between. What about the
other end of the interval? Since one end attracts, and by the
stable-interval property there are leaves arbitrarily close by, the
other end attracts too. What about in the interior? Well if we a
leaf close to $\tilde{\gamma}$, and iterate under $h^{-1}$, it will
converge either to a geodesic or a single point at infinity. If it
converges to a geodesic, however, then we have ourselves a nested
stable interval, which has attracting endpoints under $h$; but we
just showed it attracts points under $h^{-1}$. This is a
contradiction unless the geodesic is degenerate, consisting of a
single point $Z$ and all in the interior of $I$ attract to $Z$ under
$h^{-1}$.
Our invariant lamination, then, is good. How good? Very good. Stable
intervals which are invariant under $h$ actually are stable: their
endpoints are attracting fixed points, and there is a unique
repelling fixed point in the interior.
\begin{lem}
Let $S$ be a closed oriented hyperbolic surface, and $h$ an
irreducible non-periodic automorphism of $S$. Let $L$ be the
lamination invariant under $h$ constructed in theorem
\ref{invariant_lamination}. If any stable interval $I$ is fixed by
$h$ then the two endpoints of $I$ are attracting fixed points and
the only other fixed point of $I$ is a repelling fixed point in the
interior of $I$.
\end{lem}
Why did we need non-periodic? We needed this for the proof of the
invariant lamination. Why did we need irreducible? To use nice
properties of the principal regions of $L'$.
\subsection{The Proof II: Alternating attraction and repulsion}
We have shown in our situation of a non-periodic irreducible $h$,
stable intervals actually have the simple dynamical property we hope
they have. Now we will show that such a property of alternate
attraction and repulsion applies globally.
\begin{lem}
Let $S$ be a closed orientable surface and $h$ an irreducible
non-periodic automorphism of $S$. Then $h$ has finitely many fixed
points on $S^1_\infty$, alternately attracting and repelling.
\end{lem}
\begin{Proof}
We use theorem \ref{invariant_lamination} to get a lamination $L$
such that $h(L) = L$. We now know a lot about $L$ and $L'$; and we
know that stable intervals invariant under $h$ have endpoints which
are attracting and unique repelling points in their interior.
We consider three cases, which obviously exhaust all the
possibilities.
\begin{enumerate}
\item
$h$ fixes the endpoints at infinity of a boundary leaf of $L'$;
\item
$h$ fixed the endpoints at infinity of a non-boundary leaf of
$L'$;
\item
$h$ does not fix the endpoints of any leaf of $L'$.
\end{enumerate}
In the first case, the boundary leaf $\gamma$ in question is part of
the boundary of a the lift $\tilde{U}$ of a principal region $U$.
Here $\tilde{U}$ is a finite sided ideal polygon; and as discussed
previously, each ideal vertex is the endpoint of precisely two
leaves. It follows that $h$ fixes the entire polygon. On the outside
of each edge, there are arbitrarily close leaves, which come
arbitrarily close to any point of $U$. It follows that each edge of
the polygon cuts off an interval on the circle at infinity which is
a stable interval. In this stable interval there is a unique
repelling fixed point. So we have alternate repelling and attracting
fixed points.
In the second case, a similar argument applies, though the picture
is a bit more complicated to see. Nearby to our fixed non-boundary
leaf there are again arbitrarily close leaves, and hence that it
cuts off a stable interval, on both sides. So the endpoints are
attracting fixed points; and there is a unique repelling fixed point
on either side.
In the final case, $h$ may or may not fix any points at infinity. If
it has no fixed points at infinity, there is nothing to prove; so
suppose $h$ fixes a point $x$ on the circle at infinity. For any
lift $\tilde{\gamma}$ of a leaf of $L'$, we can consider the open
interval $U(\tilde{\gamma})$ on the circle at infinity which has the
same endpoints as $\tilde{\gamma}$ and avoids $x$. All the sets
$U(\tilde{\gamma})$ are either nested or disjoint. It follows that
under $h$, either $x$ is an attracting fixed point and there is
precisely one other repelling fixed point, or vice versa.
\end{Proof}
In fact, we can say more. Note that $L'$ is invariant under $h$; and
is perfect; and connected. If $h$ has more than one attracting fixed
point at infinity, then the geodesics joining consecutive attracting
fixed points are in $L'$. But this is not just true for $h$; it is
true for any positive power of $h$ as well: if $h^m$ has an
attracting fixed point then it's an attracting fixed point of $h$
also. Conversely, (well, almost conversely,) every boundary leaf of
$L'$ connects two endpoints at infinity fixed under some positive
power of $h$: since there are only finitely many boundary leaves,
there is a power of $h$ which fixes that leaf and its orientation.
And this property actually is sufficient to define $L'$: the
boundary leaves are precisely those which have endpoints which are
fixed under some positive power of $h$; and boundary leaves are
dense. So $L'$ is in this sense unique.
\begin{lem}
Make the same assumptions as above. Any lift of a strictly positive
power of $h$ has finitely many fixed points at infinity, alternately
attracting and repelling. There is a unique perfect lamination $L^s$
which contains the geodesics joining consecutive attracting fixed
points of any positive power of $h$.
\end{lem}
\subsection{The Proof III: The End}
We are now almost done. Given a non-periodic irreducible
automorphism $h$ of $S$, we can obtain a stable lamination $L^s$,
and an unstable lamination $L^u$ (stable for $h^{-1}$).
These two laminations intersect transversely: we can see this from
the alternating attracting and repelling fixed points of $h$. Their
intersection points are mapped to intersection points by $h$. So we
get rectangles bounded by arcs of $L^s$ and $L^u$, which are mapped
to other rectangles under $h$. This allows us to straighten out $h$.
We call points on $S$ equivalent if they are in the same rectangle:
that is, they are either in the (closure of the) same component of
$L^s - L^u$; or in the (closure of the) same component of $L^u -
L^s$; or in the (closure of the) same component of $S - L^s - L^u$.
The quotient by this equivalence relation is just like the Cantor
function, and gives us a homeomorphic surface on which the
lamination has become a foliation. Using this idea, we can finish
the proof of the amazing theorem \ref{amazing}.
\begin{thebibliography}{99}
\bibitem{CB}
Andrew J. Casson and Steven A. Bleiler, Automorphisms of Surfaces
after Nielsen and Thurston
\end{thebibliography}
\end{document}