\documentclass{article}
\usepackage{graphicx}
%\input epsf.sty
%\usepackage[dvips]{graphicx}
\usepackage{latexsym}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{verbatim}
% THEOREMS -------------------------------------------------------
%From thesis...
%\documentclass{book}
%\usepackage{graphicx}
%\input epsf.sty
%\usepackage[dvips]{graphicx}
%\usepackage{latexsym}
%\usepackage{amsfonts}
%\usepackage{amssymb}
%\usepackage{verbatim}
%\usepackage{afterpage}
%\usepackage{amsmath}
%\usepackage{cite}
%\usepackage{titling}
%\usepackage{appendix}
\usepackage[standard, thmmarks]{ntheorem}
%\usepackage{smartref}
%\addtoreflist{figure}
%\usepackage[margin=3cm]{geometry}
%\usepackage{fancyhdr}
%End from thesis...
\newtheorem{prob}{Problem}
\newtheorem{ax}{Axiom}[section]
\newtheorem{thm}{Theorem}[section]
\newtheorem{cor}[thm]{Corollary}
\newtheorem{lem}[thm]{Lemma}
\newtheorem{prop}[thm]{Proposition}
%\theoremstyle{definition}
\newtheorem{defn}[thm]{Definition}
%\theoremstyle{remark}
\newtheorem{rem}[thm]{Remark}
%\numberwithin{equation}{section}
% MATH -----------------------------------------------------------
\newcommand{\norm}[1]{\left\Vert#1\right\Vert}
\newcommand{\abs}[1]{\left\vert#1\right\vert}
\newcommand{\set}[1]{\left\{#1\right\}}
\newcommand{\R}{\mathbb R}
\newcommand{\RP}{\mathbb R P}
\newcommand{\CP}{\mathbb C P}
\newcommand{\Z}{\mathbb Z}
\newcommand{\C}{\mathbb C}
\newcommand{\Q}{\mathbb Q}
\newcommand{\F}{\mathcal F}
\newcommand{\hyp}{\mathbb H}
\newcommand{\M}{\mathcal M}
\newcommand{\Lob}{\mathcal L}
\newcommand{\Vect}{\text{Vect\,}}
\newcommand{\Div}{\text{Div\,}}
\newcommand{\im}{\text{Im\,}}
\newcommand{\St}{\text{St\,}}
\newcommand{\Cl}{\text{Cl\,}}
\newcommand{\Li}{\text{Li\,}}
\newcommand{\sgn}{\text{sgn\,}}
\newcommand{\vol}{\text{vol\,}}
\newcommand{\Conv}{\text{Conv\,}}
\newcommand{\Fol}{\text{Fol\,}}
\newcommand{\eps}{\varepsilon}
\newcommand{\To}{\longrightarrow}
\newcommand{\BX}{\mathbf{B}(X)}
\newcommand{\A}{\mathcal{A}}
\newcommand{\N}{\mathbb N}
\newcommand{\diagram}{\vspace*{5cm}}
\renewcommand{\theenumi}{(\roman{enumi})}
\renewcommand{\labelenumi}{\theenumi}
%\newenvironment{proof}{\noindent \emph{Proof. }}{ \begin{flushright}$\square$\end{flushright} }
% ----------------------------------------------------------------
\begin{document}
\title{Notes on Giroux's 1991 paper, ``Convexit\'{e} en topologie de contact"}
\author{Daniel Mathews}%
\maketitle
% ----------------------------------------------------------------
\tableofcontents
\section{Introduction}
This paper of Giroux is absolutely seminal in the study of contact
geometry. Convexity at first glance may not seem so crucial or
natural, but it is! Giroux showed why, and along the way proved a
number of basic results that are standard today.
Much of Giroux's paper deals with higher-dimensional contact
manifolds. I will stick only to 3-manifolds. Also, I am more
interested in convex \emph{surfaces} than the other notions of
convexity discussed by Giroux: namely, those relating to convex
contact structures.
Let $(M, \xi)$ be a closed oriented contact 3-manifold throughout.
Let $\alpha$ be a contact form for $\xi$.
\section{So what is a convex surface?}
\begin{defn}
A \emph{convex surface} $S$ in $(M, \xi)$ is an embedded surface
(possibly with boundary) for which there exists a transverse contact
vector field $X$, i.e. a vector field transverse to $S$ for which
the flow of $X$ preserves $\xi$.
\end{defn}
So, the crucial thing appears to be a contact vector field. Are
there many contact vector fields on $\xi$? It turns out there are
very many --- as many as there are sections of the line bundle
$TM/\xi$ on $M$. Giroux refers to Arnold for this, but it's not too
difficult to prove. So, it seems that it should be fairly generic
that this should be transverse to $S$, though maybe you might think
there are topological obstructions. It turns out that there are no
obstructions, and a generic surface \emph{is} convex.
This is special to 3 dimensions.
A generic surface is convex. This is one of Giroux's greatest
achievements. Maybe it wasn't that hard to prove once
he set his mind to it, but discovering this was a great
breakthrough.
\section{Why?}
Indeed. Why convex surfaces.
By this one might mean: what is the motivation for this definition?
Well, it comes from pseudoconvex embeddings in complex
geometry: these often give you contact structures. They are all
about, relevant to our context, Morse functions and gradient vector
fields keeping things invariant. ``Convex" in the complex context
says something about a Morse function ``pointing out" of the
manifold, for instance as a gradient vector field for a Morse
function points out of a level set. So this is where the invariant
vector field and the transversality come from.
But by the question one might mean: Why do we bother with these
things?
And the most basic layperson's reason could be: ``Because humans are
so bad at visualising 3 dimensions!"
It's true. A field of planes moving about in 3-dimensional space is
potentially something you can visualise. It's not an unacceptably
high number of dimensions. Planes are not that complicated. If we
could see them better, we could probably prove a lot more directly.
Why are they useful? We will soon see that the contact structure
near a surface is determined by its characteristic foliation; and a
foliation on a surface is much easier to keep track of than a whole
contact structure. Some lines on a surface are much easier to
visualise than planes which rotate all over the place! At the price
of only seeing the contact structure in the neighbourhood of a
surface, we gain the pleasure of not having to tax our 3-dimensional
visualisation abilities. This was known before Giroux's paper, it
seems; but Giroux's paper is the earliest location I know of where
there is a written proof of this result.
So: the contact structure near a surface is determined by its
characteristic foliation. But this has nothing to do with convexity,
yet. However a generic surface is convex. The crucial blow is struck
by the following result: the characteristic foliation on a convex
surface is more or less determined by its \emph{dividing set}. We
will define what a dividing set is in due course: it is a certain
finite set of curves on a surface. What the ``more or less" means
will become clear shortly. But note the obvious point: as easy as a
characteristic foliation is to understand, a finite set of curves is
much simpler again.
For anyone who has ever struggled to draw a picture even of the
standard contact structure and figure out which way all those planes
were wiggling, convex surfaces, then, offer a simplification of
contact structures that is truly awesome.
\section{What a convex surface looks like I}
Before we go proving all these major results, we would like to get a
nice picture of a convex surface $S$ in $(M, \xi)$. Note that,
having a transverse contact vector field, $S$ is automatically
oriented.
\subsection{First example: a contactization}
An example: a \emph{contactization} of a symplectic manifold. Yes,
symplectic and contact are \emph{that} closely related --- you can
symplectize a contact manifold, and contactize a symplectic
manifold. In both cases, you ``-ize" it by crossing it with $\R$,
and defining an appropriate contact/symplectic form on the product.
Actually, you can't contactize \emph{any} symplectic manifold. It
has to be \emph{exact}. Well, how else were you going to get a
contact 1-form canonically out of a symplectic 2-form? So take an
exact symplectic manifold $(W, \omega)$ where $\omega = d\beta$. Let
$W$ have dimension $2n$ Then on $W \times \R$ we want to write a
contact $1$-form. What about $\beta$? The simplest possibility,
perhaps, but that's not going to work; for $\beta$ doesn't change in
the $\R$ direction. So, you might try the next simplest possibility,
say $\beta + dt$, where $t$ is the coordinate in the $\R$ direction.
Is that a contact form? Yes:
\[
(\beta + dt) \wedge \left( d(\beta + dt) \right)^n = \beta \wedge (d\beta)^n +
dt \wedge (d\beta)^n = dt \wedge (d\beta)^n = dt \wedge
\omega^n \neq 0.
\]
Why does $\beta \wedge (d\beta)^n = 0$? Because it's a $(2n+1)$-form
but is only nontrivial in the $W$ direction, which is
$2n$-dimensional. And why is $dt \wedge \omega^n \neq 0$? From the
definition of a symplectic form, $\omega^n$ is a volume form on $W$;
and $dt$ is a volume form on $\R$; so their product is a volume form
on $W \times \R$.
Consider now $M = W \times \R$ as a 3-manifold (so $W$ is a
2-manifold); however the discussion all still works in higher
dimensions, it we define everything properly.
Our contact form is very simple: $\beta + dt$. In particular it is
invariant under translations in the $\R$ direction. That is, it is
invariant under the flow of the vector field $X =
\partial/\partial t$. So $X = \partial/\partial t$ is a contact vector
field. Any horizontal surface $W \times \{t\}$ is therefore a convex
surface. In fact, any slice of $W \times \R$ which is ``never
vertical" is going to give us a convex surface. More
precisely, take any function $f: W \To \R$ and consider its graph,
which is a surface in $W \times \R$. Being the graph of a function
is what me mean by ``never vertical"; this will also be a convex
surface.
\subsection{Second (not just an) example: vertically invariant
contact structures}
But in fact \emph{every} convex structure looks a bit (not exactly!)
like this, and this is the key to understanding what's going on. Any
convex surface $S$ certainly has a neighbourhood diffeomorphic to $S
\times \R$; and we can choose the $\R$ coordinate so that our
transverse contact vector field is $X = \partial/\partial t$; then
in these coordinates, the contact structure is invariant under the
flow of $X$, i.e. invariant under translations in the $\R$
direction. So if we can understand the situation of $S \times \R$
with $X = \partial/\partial t$ as a contact vector field, we will
understand what any convex surface looks like.
\subsubsection{The form of the form}
So, we have $S \times \R$, and want to consider contact structures
which are invariant under translations in the $\R$ direction. What
does the 1-form $\alpha$ look like? If we write $x,y$ for some local
coordinates on $S$ and $t$ for the coordinate on $\R$, we have
\[
\alpha = \alpha_x \; dx + \alpha_y \; dy + \alpha_t \; dt
\]
where $\alpha_x, \alpha_y, \alpha_t$ are functions $S \times \R \To
\R$. However if the contact structure is to be invariant under
translations in the $t$ direction, we must have $L_X \alpha$ being a
multiple of $\alpha$; actually, by rescaling $\alpha$, we can choose
$\alpha$ to satisfy $L_X \alpha = \alpha$. That is, $\alpha$ is to
be invariant under translations in the $\R$ direction. So $\alpha_x,
\alpha_y, \alpha_t$ are all functions on $S$ alone. From now on we
write $\beta$ for the local expression $\alpha_x \; dx + \alpha_y \;
dy$ and have a $1$-form on $S$; we write $u$ for $\alpha_t$ and $u$
is a function on $S$. So we have
\[
\alpha = \beta + u \; dt.
\]
This is the general form of a vertically invariant 1-form on $S
\times \R$.
But it's not the general form of a vertically invariant
\emph{contact} 1-form on $S \times \R$. For that we need to do a
computation:
\begin{align*}
\alpha \wedge d\alpha &= (\beta + u \; dt) \wedge (d\beta + du
\wedge dt) \\
&= \beta \wedge d\beta + \beta \wedge du \wedge dt
+ u \; dt \wedge d\beta \\
&= \beta \wedge du \wedge dt + u \; dt
\wedge d\beta \\
&= (\beta \wedge du + u \; d\beta) \wedge dt
\end{align*}
Again $\beta \wedge d\beta = 0$ since this is a $3$-form on $S$. So
the contact-ness of $\alpha$ depends on the form
\[
\theta = \beta \wedge du + u \; d\beta.
\]
This is a $2$-form on $S$: if it is nondegenerate everywhere, then
wedging it with $dt$ will give us an everywhere nondegenerate
$3$-form on $S \times \R$; if it is degenerate anywhere, then
wedging it with $dt$ will give us a degenerate $3$-form on $S \times
\R$. So $\alpha$ is contact if and only if $\theta$ is
nondegenerate, i.e. an area form on $S$.
Clearly for any such $\alpha$, provided $\theta$ is nondegenerate,
any surface $S \times \{t\}$ is a convex surface in $S \times \R$
with contact vector field $X = \partial/\partial t$. In fact, again,
for any function $S \To \R$, the graph of the function inside $S
\times \R$ is a convex surface in $(S \times \R, \alpha)$.
\subsubsection{It's mostly contactizations}
Note that the expression for $\alpha$, namely $\beta + u \; dt$, is
actually quite close to the contact form on a contactization, as
given above. In fact, we usually do have mostly a contactization!
The $u$ is the only difference. If we had $u \neq 0$, then we could
divide through by $u$ and obtain a different form
--- of the contactization type --- giving the same contact
structure. So: whenever $u$ is nonzero, we really have a
contactization. The region of $S$ where $u=0$ is generically a set
of curves which we will call $\Gamma$; away from $\Gamma$, on $(S
\backslash \Gamma) \times \R$, we can take the contact form as
$\beta/u + dt$, so we have the contactization of the exact
symplectic manifold $(S \backslash \Gamma, d(\beta/u))$.
A priori, of course, $\Gamma$ could be a nasty set, but generically
it will be a set of curves. In fact more is true: if $\alpha$ is a
contact form, so that $\theta$ is nondegenerate, then $\Gamma$
\emph{must be} a set of curves. For whenever $u = 0$, to have
$\theta \neq 0$ means $\beta \wedge du + u \; d\beta = \beta \wedge
du \neq 0$. Hence $du \neq 0$, and so the tangent space to $\Gamma$
is precisely the one-dimensional kernel of $du$; so $\Gamma$ is a
1-manifold.
Actually, note that we now have \emph{two} area/symplectic forms on
$S$. From considering the contact-ness of $\beta + u \; dt$ we have
the form $\theta = \beta \wedge du + u \; d\beta$. And from
considering the exact symplectic manifold $(S \backslash \Gamma)$
$1$-form we have $d(\beta/u)$. These are both area forms, so differ
by a function which is nowhere zero. It turns out that function is
$u^2$, which is certainly positive away from $\Gamma$. In
particular, $\theta = u^2 d(\beta/u)$:
\[
d \left( \frac{\beta}{u} \right) = \frac{d\beta}{u} + d \left(
\frac{1}{u} \right) \wedge \beta = \frac{1}{u} \; d\beta - \frac{1}{u^2}
du \wedge \beta = \frac{1}{u^2} \left( u \; d\beta + \beta
\wedge du \right) = \frac{1}{u^2} \; \theta.
\]
\subsubsection{Where do the contact planes go?}
Let's consider the surface $S \times \{0\}$ (or $S$ times any
number) in more detail. What does the contact structure look like
here? The contact plane is the kernel of $\alpha = \beta + u \; dt$.
The plane is usually transverse to $S \times \{0\}$; the contact
plane is only tangent at isolated points of $S \times \{0\}$. At
these points, $\alpha$ is of the form $u \; dt$; these are the
points where $\beta = 0$. (It's possible to have $\beta = 0$ but
still have $\theta \neq 0$.)
The plane can sometimes be vertical. At such points, the contact
plane contains $\partial/\partial t$, so $u = 0$, and $\alpha$ is
simply of the form $\beta$; the points where $\xi$ is vertical are
the points where $u=0$. If you like, the value of $u$ tells us how
non-vertical the plane is, well, not really, but it may be useful to
think this way.
\subsubsection{Where does the characteristic foliation go?}
We will obtain a characteristic foliation on $S \times \{0\}$. It is
the intersection of $\ker \alpha$ with the horizontal surface; but
since the $u \; dt$ term has nothing to say about horizontal
vectors, the characteristic foliation is completely determined by
$\beta$. In fact, the characteristic foliation is simply given by
$\ker \beta$.
We can even describe a vector $Y$ directing the characteristic
foliation on $S \times \{0\}$. Well, we could just take $Y$ to be an
arbitrary vector in $\ker \beta$, but that's not so canonical.
Rather, we have a $1$-form on $S$ and an area/symplectic $2$-form
$\theta$. Recall that in symplectic manifolds, $1$-forms have dual
vectors and vice versa; and on a $2$-dimensional surface, the dual
vector to a $1$-form lies in its kernel. So, we can take $Y$ to be
the vector dual to the $1$-form $\beta$ with respect to the
area/symplectic form $\theta$ on $S$. That is: we define the vector
field $Y$ on $S$ by
\[
\beta = i(Y) \; \theta.
\]
Then $Y$ directs the characteristic foliation on $S \times \{0\}$.
Actually, there was another potential, highly canonical, choice for
a vector field directing the characteristic foliation; at least,
away from $\Gamma$, that is, where $u=0$. Because recall that $(S
\backslash \Gamma) \times \R$ is the contactization of the exact
symplectic surface $(S \backslash \Gamma, d(\beta/u))$. And the
characteristic foliation can be taken as given not just by the
kernel of $\beta$, but by the kernel of the highly canonical
Liouville 1-form $\beta/u$. So by the same reasoning as the previous
paragraph, we can define a vector field $Z$ directing the
characteristic foliation on $S \times \{0\}$ by being dual to the
Liouville form $\beta/u$ with respect to $d(\beta/u)$:
\[
\frac{\beta}{u} = i(Z) \; d \left( \frac{\beta}{u} \right).
\]
Are these closely related? One would hope so! Well, of course $Y$
and $Z$ both direct the same characteristic foliation, so they must
be scalar multiples of each other. In fact, that multiple is just
the function $u$: $Z = uY$. This follows from a computation:
plugging both $Y$ and $Z$ into the same area form $d(\beta/u)$ we
can see what the factor is.
\begin{align*}
i(Z) d \left( \frac{\beta}{u} \right) &= \frac{\beta}{u} \\
i(Y) d \left( \frac{\beta}{u} \right) &= \frac{1}{u^2} i(Y)
\theta = \frac{1}{u^2} \beta = \frac{1}{u} \cdot
\frac{\beta}{u}.
\end{align*}
So $Y,Z$ point in the same direction along the characteristic
foliation when $u>0$; and in the opposite direction when $u<0$.
Now, since $Y$ is defined on the whole surface and is a smooth
vector field, $Y$ should proceed smoothly across the set $\Gamma$,
where $u=0$. From our proceeding comment, as we cross $\Gamma$, $u$
changes sign. Thus $Z$ abruptly changes direction every time you
pass through $\Gamma$. Actually, not \emph{so} abruptly, since $Z =
uY$; so $Z$ approaches $0$ near $\Gamma$, and $Z$ extends
continuously to all of $S$, with singularities along $\Gamma$.
So, which way do these vector fields point as we cross $\Gamma$?
Well, $Z$ isn't defined on $\Gamma$, but $Y$ is. Recall that being
on $\Gamma$ means $u=0$. And we want to know the change in $u$ in
the direction of $Y$, that is, $du(Y) = i(Y) \; du$. Well, from the
equation (the only one we really have!)
\[
\beta = i(Y) \; \theta = i(Y) \left( \beta \wedge du + u d\beta
\right),
\]
noting that here $u=0$ and $i(Y) \beta = 0$ (because $Y$ directs the
characteristic foliation, which is the kernel of $\beta$), we have
\[
\beta = - \beta i(Y) \; du, \quad \text{hence} \quad i(Y) \;
du = -1.
\]
We conclude that $Y$ points from positive towards negative $u$.
Since $Z$ agrees with $Y$ when $u>0$, $Z$ points out of positive $u$
pieces of $S$. And since $Z$ disagrees with $Y$ when $u<0$, $Z$
points out of negative $u$ pieces of $S$ also. So $Z$ points out of
each piece of $S \backslash \Gamma$, towards $\Gamma$.
Actually, we could have seen this more quickly and more slickly.
Think about the effect of our two vector fields on their respective
area forms. As you flow along them, do they dilate or compress area?
Well, flowing along $Y$ produces a messy result on $\theta$. But
because $S \backslash \Gamma$ is exact and has a Liouville form, the
effect of flowing $Z$ on the area form $d(\beta/u)$ is very simple:
\[
L_Z \; d \left( \frac{\beta}{u} \right) = d i(Z) d \left(
\frac{\beta}{u} \right) = d \left( \frac{\beta}{u} \right).
\]
Here all we used was $\beta/u = i(Z) d(\beta/u)$. So $Z$ dilates
the area form $d(\beta/u)$. If you expand an area form, you
certainly can't be coming \emph{inwards} through the entire
boundary!
\subsection{Characterising vertically invariant contact
structures}
\label{characterising}
Let's summarise what we find on $S \times \R$ with a vertically
invariant contact structure.
\begin{enumerate}
\item
The contact form has the expression $\beta + u \; dt$ where
$\beta$ is a $1$-form on $S$ and $u$ is a function on $S$.
\item
The $2$-form $\theta = \beta \wedge du + u \; d\beta$ must be an
area/symplectic form on $S$.
\item
The contact plane is vertical where $u=0$, and horizontal where
$\beta = 0$.
\item
Away from $\Gamma = \{u=0\}$, the manifold is actually a
contactization. Precisely: $( (S \backslash \Gamma) \times \R,
\beta/u + dt)$ is the contactization of $(S \backslash
\Gamma, d(\beta/u))$.
\end{enumerate}
The characteristic foliation on $S \times \{0\}$ is given by any of:
\begin{enumerate}
\item
The kernel of $\beta$. (Or, away from $\Gamma$, it's the kernel
of $\beta/u$.)
\item
It's directed by $Y$, which is dual to $\beta$ with respect to
the area/symplectic form $\theta$. Across $\Gamma$, $Y$ points
in the direction of decreasing $u$, i.e. from positive to
negative $u$ regions.
\item
Away from $\Gamma$, it's directed by $Z$, which is dual to the
Liouville form $\beta/u$ with respect to the exact symplectic
form $d(\beta/u)$. The flow of $Z$ expands the area form
$d(\beta/u)$; consequently, $Z$ always points outwards across
$\Gamma$, and changes direction as you cross $\Gamma$ along a
leaf. However since $Z \To 0$ as we approach $\Gamma$, $Z$
extends continuously to a singular vector field on all of $S$.
\end{enumerate}
This is more than enough to characterise what's going on here. The
question is: if you're given a (singular) foliation $\F$ on $S$, is
it the characteristic foliation on $S \times \{0\}$ for some
vertically invariant contact structure on $S \times \R$? Turns out
that the set $\Gamma$, the dilating an area form property, and the
change of direction along $\Gamma$, is enough. We can make this
precise: this is a 3-dimensional version of Giroux's theorem I.3.2;
Giroux's version is stronger and works in higher dimensions.
\begin{prop}
\label{I.3.2} Let $S$ be a closed surface and let $\F$ be a
dimension-1 singular foliation on $S$. The following are equivalent:
\begin{enumerate}
\item
There exists on $S \times \R$ a vertically invariant contact
structure which induces $\F$ as the characteristic foliation on $S
\times \{0\}$.
\item
There exists on $S$ a 1-manifold $\Gamma$ (i.e.
a finite set of disjoint closed curves) transverse to $\F$ (in
particular, avoiding the singularities of $\F$ such that
\begin{enumerate}
\item
the complement $S'$ of an open tubular neighbourhood of
$\Gamma$, with the fibres in $\F$, is an exact symplectic
manifold for which a Liouville vector field directs $\F$ and
exits transversely through the boundary; and
\item
the involution of the double cover $\partial S' \To \Gamma$
obtained by following leaves of $\F$ reverses the
orientation on the leaves. (That is, the Liouville vector
field changes direction on leaves passing through $\Gamma$.)
\end{enumerate}
\end{enumerate}
\end{prop}
\begin{Proof}
That (i) implies (ii) we have already seen: the 1-manifold $\Gamma$
where $u = 0$ divides $S$ into pieces where we have Liouville vector
fields; and the Liouville vector fields change direction on leaves
passing through $\Gamma$.
In the other direction, a contactization $S \times \R$ of an exact
symplectic manifold $(S, d\beta$) has a vertically invariant contact
structure with contact form $\beta + dt$ with the Liouville vector
field $Z$ (defined via $\beta = i(Z) d\beta$) directing the
characteristic foliation, which is the kernel of $\beta$. So we have
a vertically invariant contact structure $\beta + dt$ on $S' \times
\R$. All we need to do is glue the pieces together, knowing that the
Liouville vector field changes direction as we pass through the
boundary $\Gamma$.
This is a silly little fiddle, but the idea is clear: you can glue
it all together. You have to fiddle because the vector field goes to
zero on $\Gamma$; you need to rescale somehow to get it to work.
So, we consider a tubular neighbourhood of $\Gamma$ fibred by leaves
of $\F$. Consider it as $\Gamma \times (-1-\epsilon, 1 + \epsilon)$,
where at the $\pm 1$ points it joins with $S'$. We know that $Z$
points outwards near $\partial S'$, so we can choose a coordinate
$s$ on $S' \cap (\Gamma \times (-1-\epsilon, 1+\epsilon)) = \Gamma
\times (-1-\epsilon, -1] \cup \Gamma \times [1, 1+\epsilon)$ such
that there $Z = -s(\partial/\partial s)$. We need to extend our
contact form $\beta + dt$ over $\Gamma \times (-1-\epsilon,
1+\epsilon) \times \R$, remaining vertically invariant.
What does $\beta$ look like on $\Gamma \times [1,1+\epsilon)$? We
should be able to work this out, since we have chosen coordinates in
the region. We defined $Z$ by $\beta = i(Z) d\beta$. And $i(Z) \beta
= 0$, either from the previous equation or since $Z$ directs $\F$,
which is the kernel of $\beta$. So $L_Z \beta = d i(Z) \beta + i(Z)
d\beta = d0+\beta = \beta$, and flowing along $Z =
-s(\partial/\partial s)$ dilates $\beta$. This gives us a
differential equation for what happens to $\beta$ as you increase
$s$. Note that flowing along $Z$, in our coordinates $\beta$ always
has $\partial/\partial s$ in its kernel, so $\beta(s) = f(s)
\beta(1)$. Now $L_Z \beta = \beta$ becomes
\[
-s \frac{df}{ds} = f \quad \text{which implies} \quad f(s) =
\frac{f(1)}{s}.
\]
So $\beta(s) = \beta(1)/s$. Write $\gamma$ for $\beta(1)$, and we
can consider $\gamma$ as a $1$-form on all of $\Gamma \times
(-1-\epsilon, 1+\epsilon)$. For $s \in (-1-\epsilon,-1] \cup
[1,1+\epsilon)$ we have $\beta = \gamma/s$ and our original contact
form was $\beta + dt = \gamma/s + dt$. But this gives the same
contact structure as $\gamma + s \; dt$, and $\gamma+s \; dt$ gives
us a vertically invariant contact structure on all of $\Gamma \times
(-1-\epsilon, 1+\epsilon)$. We just need to check it's contact for
$s \in (-1,1)$, which follows upon inspecting the expression
$(\gamma + s \; dt) \wedge d(\gamma + s \; dt)$ and noting it's
contact for $s = \pm 1$.
A silly little fiddle indeed.
\end{Proof}
\section{The power of characteristic foliations}
We now know that if we have a convex surface, we can choose
coordinates so a neighbourhood looks like $S \times \R$, with the
contact vector field $X = \partial/\partial t$, so that the contact
form is $\beta + u \; dt$, with $\beta$ a $1$-form and $u$ a
function on $S$, and with the characteristic foliation $\F$ directed
by a Liouville vector field away from $\Gamma = \{u = 0\}$, which
changes direction over $\Gamma$.
On the other hand, suppose you have all this data on the surface
$S$: a $\Gamma$, an $\F$, an $X$, Liouville forms, Liouville vector
fields and all the rest of it. Does it follow that the surface is
convex?
We don't know yet: to conclude that the surface is convex, we need
to know that the contact structure near $S$ looks like our standard
picture. We \emph{could} conclude that, if we knew that the
characteristic foliation determines the contact structure nearby.
In other words, to go any further we need to harness the power of
the characteristic foliation. We need the following proposition
(Giroux's proposition II.1.2(b)).
\begin{prop}
\label{II.1.2(b)} Let $\F$ be a singular foliation on a closed
surface $S$. Fix an orientation on $\wedge^2 TS$, and we are
interested only in positive contact structures (i.e. with $\alpha
\wedge d\alpha > 0$). Two germs of contact structures which induce
the same characteristic foliation $\F$ on $S$ are conjugate by a
germ of a diffeomorphism which is isotopic to the identity through
diffeomorphisms preserving $\F$. (Consequently, the two germs of
contact structures are isomorphic!)
\end{prop}
So, we commence a thorough study of what characteristic foliations
look like, from the ground up.
\subsection{What singularities can occur in characteristic foliations?}
A first question is: characteristic foliations can be singular. How
bad can those singularities be? We have to worry about the types of
singularities in the foliation $\F$.
Nobody else in the history of contact geometry (at least, the recent
history) seems to have worried about this issue. Everyone else just
says you can assume, generically, that there are only nice
singularities. Not Giroux. Well, he is proving the result that
everyone will use, so we all owe him a debt of gratitude.
To be sure, we are assuming \emph{something}, namely that our
foliation is a \emph{singular foliation}. We are assuming that $\F$
is only degenerate at singularities which are isolated points. You
can easily find characteristic surfaces where the singularities are
worse (e.g. take a surface tangent to the contact structure along a
whole curve). Beyond this, we are making no assumptions on the types
of singularities.
A good way to study a singularity is by looking at the
\emph{linearisation} there. We have a singular foliation $\F$ on $S$
and a vector field $Y$ directing it. At a singularity $x$ of $\F$ we
have $Y=0$. Now we have a family of diffeomorphisms $\varphi_t$, the
flow of $Y$. Clearly $\varphi_t(x) = x$; $x$ isn't going anywhere!
However around $x$, we can take note of what happens. For a tangent
vector in $T_x S$, we can exponentiate it, see what happens to it
under $\varphi_t$, and (for small enough $t$) return to the tangent
space. As $t \to 0$ we obtain the linearisation of $\varphi_t$,
which is a linear map $T_x S \To T_x S$. The trace of this linear
transformation tells us what $\varphi_t$ is doing to areas as $t \to
0$; this trace is the divergence of $Y$ at $x$.
\begin{defn}
A singularity $x$ of a vector field $Y$ is \emph{isochore} if the
divergence of $Y$ at $x$ is zero.
\end{defn}
Perhaps there's an English translation for ``isochore", I don't
know. I'm just copying the word from the French.
Note this includes hyperbolic fixed points where the attracting and
repelling eigenvalues are equally strong --- so there is no effect
on area. Note also that a degenerate singularity can be okay, and
not isochore --- a degenerate singularity has non-invertible
linearisation, but the trace may still be nonzero.
It turns out that isochore singularities can't occur in
characteristic foliations; but this is all. Happily, a singular
foliation without isochore singularities can be realised as a
characteristic foliation. Thank goodness! This may seem somewhat
strange, since it implies that some fairly degenerate singularities
can occur in characeristic foliations; but that's not a problem, as
we'll see, we can construct the contact structure explicitly nearby
anyway.
This is Giroux's proposition II.1.2(a)
\begin{prop}
\label{II.1.2(a)} Let $\F$ be a singular foliation on a surface $S$.
We fix an orientation on the manifold $\wedge^2 TS$ and we are
interested only in germs of contact structures along $S$ which give
this orientation.
$\F$ is the characteristic foliation induced on $S$ by a germ of
contact structures if and only if $\F$ is without isochore
singularities.
\end{prop}
So this theorem answers two questions. It answers our original
question: what singularities can occur in characteristic foliations?
But it also answers a more general question: which singular
foliations are characteristic foliations? This is a very important
answer to know.
To prove this, we'll need to understand how contact structures work
in neighbourhood $S \times \R$ --- but, more generally than before,
we no longer require the structure to be vertically invariant.
\subsection{Contact structures on $S \times \R$ and germs}
Let us consider what a general contact $1$-form $\alpha$
on $S \times \R$ looks like. No matter what the $t$ coordinate, at
any point in $S \times \{t\}$, $\alpha$ can be written in the form
$\beta_t + u_t \; dt$ where $\beta_t$ is a $1$-form and $u_t$ a
function on $S \times \{t\}$. When is this a contact form? We do the
usual computation:
\begin{align*}
(\beta_t + u_t \; dt) \wedge d \left( \beta_t + u_t \; dt
\right) &=
(\beta_t + u_t \; dt) \wedge \left( d \beta_t + \frac{\partial
\beta_t}{\partial t} \wedge dt + du_t \wedge dt \right) \\
&= \beta_t \wedge \frac{\partial \beta_t}{\partial t} \wedge dt
+ \beta_t \wedge du_t \wedge dt + u_t \; dt \wedge d\beta_t \\
&= dt \wedge \left( u_t \; d \beta_t + \beta_t \wedge \left( du_t
- \frac{\partial \beta_t}{\partial t} \right) \right)
\end{align*}
Note that when we write $d\beta_t$ we mean taking the differential,
of a form on $S$; so the differential as a form on $S \times \R$ is
$d \beta_t + (\partial \beta_t / \partial t) dt$. So $\beta_t \wedge
d\beta_t = 0$ as a $3$-form on $S$. The $2$-form in brackets is a
$2$-form on $S$; so we have a contact form if and only if this is
nowhere degenerate. The condition is:
\[
u_t \; d\beta_t +\beta_t \wedge \left( du_t -
\frac{\partial \beta_t}{\partial t} \right) \neq 0
\]
In our situation, we are given the foliation on $S \times \{0\}$, so
we are given $\beta_0$. To define the contact structure nearby we
need nearby $\beta_t$ and $u_t$ and $\partial \beta_t/\partial t$.
However, to define the \emph{germ} of a contact structure near $S
\times \{0\}$ we need less: we only need, in addition to our
$\beta_0$, the function $u_0$ and the partial derivative $(\partial
\beta_t/\partial t)|_{t=0}$. These need to satisfy
\[
u_0 \; d\beta_0 + \beta_0 \wedge \left( du_0 - \frac{\partial
\beta_t}{\partial t}|_{t=0} \right) \neq 0
\]
on $S \times \{0\}$. A pair $(u_0, (\partial \beta_t/\partial
t)|_{t=0})$ satisfying this condition is all we need.
One important observation is that he set of all pairs $(u_0,
(\partial \beta_t/\partial t)|_{t=0})$ satisfying our condition is
basically \emph{convex}: no, not a convex surface in a contact
manifold, it's a convex set in a vector space, in the old-fashioned
sense. If we require our pair $(u_0, (\partial \beta_t/\partial
t)|_{t=0})$ to satisfy the condition above, with a particular sign
(so say $>0$ rather than $\neq 0$), the space of all such pairs is
convex.
For if we have two such pairs $(u_0, (\partial \beta_t/\partial
t)|_{t=0})$ and $(u'_0, (\partial \beta_t/\partial t)|'_{t=0})$,
with
\begin{align*}
u_0 \; d\beta_0 + \beta_0 \wedge \left( du_0 - \frac{\partial
\beta_t}{\partial t}|_{t=0} \right) &>0 \\
u'_0 \; d\beta_0 + \beta_0 \wedge \left( du'_0 - \frac{\partial
\beta_t}{\partial t}|'_{t=0} \right) &>0
\end{align*}
then
\[
\left[ (1-s)u_0 + s u'_0 \right] \; d\beta_0 + \beta_t \wedge
\left( d \left[ (1-s) u_0 + s u'_0 \right] - \left[ (1-s)
\frac{\partial \beta_t}{\partial t}|_{t=0} + s \frac{\partial
\beta_t}{\partial t}|'_{t=0} \right] \right) >0
\]
also.
End disgusting computation. A disgusting computation indeed, but
this is the essential reason why a characteristic foliation
determine the germ of contact structure. It means that the ``germ
data" $(u_0, (\partial \beta_t/\partial t)|_{t=0})$ forms a
contractible space; and with a bit of Mosering, it means that all
the relevant germs are isomorphic.
\subsection{A characterisation of isochore singularities}
So far we've defined an isochore singularity as one satisfying a
condition about linearisations and divergences. There's a slightly
more approachable way of describing them.
As usual, let our foliation $\F$ be defined by a $1$-form $\beta$.
But now we will take a dual. So let $\omega$ be an area/symplectic
form on $S$ and take the vector field $X$ dual to $\beta$ with
respect to $\omega$. So $\beta = i(X) \omega$ and $X$ directs $\F$.
The isochore condition can be written using the divergence with
respect to the area form $\omega$, i.e. $L_X \omega$. To be isochore
is to have $L_X \omega = 0$ at a singularity. We can expand this and
obtain
\[
0 = L_X \omega = i(X) d \omega + d i(X) \omega = 0 + d \beta.
\]
So at a singularity, i.e. when $\beta = X = 0$, we have $d\beta =
0$. The above computation shows that the converse is true also.
\begin{lem}
A singularity $x$ of a foliation $\F$ on a surface $S$ defined by a
1-form $\beta$ (so that $\beta = 0$ at $x$) is isochore if and only
if $d\beta = 0$ there.
\end{lem}
\subsection{Which foliations are characteristic foliations?}
We can now answer the question: precisely those without isochore
singularities. And we can prove it. We can prove proposition
\ref{II.1.2(a)}.
To see why, suppose we have a characteristic foliation $\F$ on $S$;
we will show there are no isochore singularities. Let $\alpha$ be
the contact form, and $\beta$ the induced form on $S$. At a
singularity $x$ of $\F$, $\alpha$ is tangent to $S$, so for any
vector $V$ tangent to $S$, $\alpha(V) = 0$. Since $\alpha \wedge
d\alpha = 0$, $d\alpha$ is nondegenerate on $T_x S$; hence so is its
restriction $d\beta$. So $d\beta \neq 0$, and by the above
characterisation, $x$ is not isochore.
In the other direction, given a singular foliation $\F$ without
isochore singularities, we want to construct a germ of a contact
structure on $S \times \R$. As discussed in the previous section,
giving the foliation is equivalent to giving a $1$-form $\beta_0$ on
$S \times \{0\}$, defined up to multiplication by nonzero function.
To define the germ of the contact structure near $S \times \{0\}$ we
need to add the information of $u_0$ and $(\partial \beta_t/\partial
t)|_{t=0}$, satisfying
\[
u_0 \; d\beta_0 + \beta_0 \wedge \left( du_0 - \frac{\partial
\beta_t}{\partial t}|_{t=0} \right) \neq 0.
\]
So, we just need to find such $u_0$ and $(\partial \beta_t/\partial
t)|_{t=0}$. The isochore condition, we have discovered, means that
when $\beta_0 = 0$, then $d\beta_0 \neq 0$.
The condition we must satisfy basically says that a certain $2$-form
on $S$ must be an area form. So let's take an area form $\omega$ and
compare everything to it; any $2$-form will be some function times
$\omega$. So, in particular, $d\beta_0 = f \omega$ for some function
$f$. Then the first term is $u_0 d\beta_0 = u_0 f \omega$. We want
to choose $u_0$ to get things positive, say; so one sneaky thing to
do would be to take $u_0 = f$. Then $u_0 d\beta_0 = u_0^2 \omega
\geq 0$. This is never negative; and in fact, from the isochore
condition we know that $u_0 \neq 0$ at singularities, so $u_0^2 > 0$
there.
What about the other term? Well, note that whenever $\beta_0$ is
nonzero, there is always something you can wedge it with to get
$\omega$. Let's call this $1$-form $\gamma$, so $\beta \wedge \gamma
= \omega$. This can't work at singularities; but there does exist
$\gamma$ such that $\beta \wedge \gamma = g \omega$ where $g \geq
0$. It would be very good to have the term in brackets being
$\gamma$, namely $du_0 - (\partial \beta_t/\partial t)|_{t=0} =
\gamma$. We can certainly do this: we have already chosen $u_0$, and
we can define $(\partial \beta_t/\partial t)|_{t=0}$ by this
equation. Then we have
\[
u_0 \; d\beta_0 + \beta_0 \wedge \left( du_0 - \frac{\partial
\beta_t}{\partial t}|_{t=0} \right)
= u_0^2 \omega + \beta \wedge \gamma = (u_0^2 + g) \omega.
\]
Now we know $u_0^2 + g > 0$. At the singularities of $\beta$, thanks
to the non-isochore condition we know $u_0 > 0$; and away from the
singularities of $\beta$ from our definition of $\gamma$ we have
$g>0$. So this is a positive form and we are done.
This proves the proposition. Well, you might want to worry a little
about orientations, but any problem with orientations are solvable
by taking a double (or quadruple!) cover.
\subsection{The characteristic foliation determines the germ of the
contact structure... up to isotopy...}
Now we have seen that, from the characteristic foliation, you can
construct the germ of a contact structure. A characteristic
foliation, we know, has no isochore singularities; and from a
foliation without isochore singularities we can determine the germ
of a contact structure by choosing $(u_0, (d\beta_t/dt)|_{t=0})$ as
we just did above.
What we want to know, is that the germ of the contact structure is
uniquely determined. This will prove proposition \ref{II.1.2(b)}.
Well, as we mentioned previously, given the foliation $\F$ and hence
$\beta_0$, the space of all germ data $(u_0, (d\beta_t/dt)|_{t=0})$
for contact structures forms a contractible space, in fact, convex
(in the vector space sense!).
That is, given any two germs of contact structures $((u_0^0,
(d\beta_t/dt)|_{t=0}^0), (u_0^1, d\beta_t/dt)|_{t=0}^1)$ with the
same $\F$, and hence the same $\beta_0$, we can linearly homotope
their germ data one to the other:
\[
\left( u_0^s, \frac{\partial \beta_t}{\partial t}|_{t=0}^s
\right) = \left( (1-s) u_0^0 + s u_0^1, (1-s) \frac{\partial
\beta_t}{\partial t}|_{t=0}^0 + s \frac{\partial
\beta_t}{\partial }|_{t=0}^1 \right).
\]
These give a germ of a contact form for all $s \in [0,1]$.
So, to the proof! Take two contact structures $\xi^0, \xi^1$ with
forms $\alpha^0, \alpha^1$, inducing the same foliation $\F$ on $S$.
We write $\alpha^0 = \beta_t^0 + u_t^0 \; dt$, $\alpha^1 = \beta_t^1
+ u_t^1 \; dt$. The fact that they both give the characteristic
foliation means that we can take $\beta_0^0 = \beta_0^1$. The two
pairs of germ data are above. We let $\alpha^s = (1-s) \alpha^0 + s
\alpha^1$. Then the germ data is also linearly interpolated, and is
exactly as given above, and from the convexity argument the contact
property is always satisfied; so $\alpha^s$ gives a (germ of a)
contact structure near $S$.
This proves that any two contact structures inducing the
characteristic foliation $\F$ on $S$ have germs which are isotopic
(through germs of contact structures). But we want more.
\subsection{... and up to isomorphism}
Given our $\xi^1, \xi^2$ as above, we not only want their germs to
be isotopic. We want them to be related by a ``germ of a
diffeomorphism which is isotopic to the identity through
diffeomorphisms preserving $\F$". That is, we want a one-parameter
family of diffeomorphisms
--- or rather, germs of such diffeomorphisms near $S$ --- starting
at the identity, ending at a (germ of a) diffeomorphism which takes
one contact structure to the other, and which always preserves the
surface $S$ and the foliation $\F$. That is, this one-parameter
family of (germs of) diffeomorphisms must look like flows along
leaves of $\F$. And indeed this is precisely what we do: we will
flow along leaves of $\F$, realising the family of (germs of)
contact forms $\alpha^s$.
We are already very close. We have shown that simply linearly
interpolating the (germs of) contact forms $\alpha^0, \alpha^1$ to
obtain $\alpha^s$, we have a family of (germs) of contact forms
already. They just need to be realised. If we can find a 1-parameter
family of diffeomorphisms $\varphi_s$ near $S$, which fix $S$ and
flow along leaves of $\F$, and have $\varphi_s^* \alpha^s$
proportional to $\alpha^0$, we are done. (They don't have to be
equal, just proportional, to define the contact structure.)
We will use Moser's method. How does Moser's method work? Instead of
looking at the diffeomorphisms, you look at the flow; you look at
vectors; and using the properties of the contact form and whatever
else, you work out what they must be. A good way of writing the
condition we want is:
\[
\varphi_s^* \alpha^s \wedge \frac{\partial}{\partial s} \left(
\varphi_s^* \alpha^s \right) = 0.
\]
This only refers to one particular point in time $s$, which is
useful. If $X^s$ is the vector field with flow $\varphi_s$, this
translates to
\[
L(X^s) \alpha^s = - \frac{\partial \alpha^s}{\partial s}.
\]
Expanding this out gives
\[
d i(X^s) \alpha^s + i(X^s) d\alpha^s = - \frac{\partial
\alpha^s}{\partial s}.
\]
Given the situation, it seems natural to require $i(X^s)\alpha^s =
0$. (In particular, $X^s \in \xi^s$.) Then we must have $i(X^s)
d\alpha^s = - \partial \alpha^s/\partial s$. But since $X^s \in
\xi^s$, and for any contact form $d\alpha^s|_{\xi^s}$ is
nondegenerate, we can find such an $X^s$; and we have our
one-parameter family of (germs of) diffeomorphisms!
One more thing to check: we wanted $X^s \in \F$, so that this was a
flow along leaves of $\F$. Since all $\alpha^s$ have the same part
$\beta_0$ which describes the foliation, for any vector $V$ pointing
along a leaf, $(\partial \alpha_s/\partial s)(V) = 0$. From the
equation $i(X^s) d\alpha^s = - \partial \alpha^s/\partial s$, we
then have $d\alpha^s(X^s, V) = 0$. But both $X^s,V$ lie in $\xi^s$,
so $d\alpha^s$ is nondegenerate on them; hence they are scalar
multiples. So indeed $\varphi_s$ is a flow along leaves.
And that is the power of the characteristic foliation.
\section{Singularities and their orientations}
Recall that the singularities which can occur in characteristic
foliations are the non-isochore singularities. We have avoided the
issue of orientations up to now; unfortunately now we have to deal
with it.
Note that $M$ is naturally oriented by the contact structure $\xi$
(for any contact form $\alpha$ we get an orientation from $\alpha
\wedge d\alpha$). If the orientation on $S$ is reversed, this
amounts to reversing the $\R$ coordinate on an $S \times \R$
neighbourhood, so $\alpha = \beta + u \; dt$ changes to $\beta - u
\; dt$. In effect we replace $u$ with $-u$ while $\beta$ is
unaffected. It's then easy to see that we must replace $\theta = u
\; d\beta + \beta \wedge du$ with $-\theta$; recalling that $\theta$
is an area form on $S$, we see that the orientation is reversed as
it should be. Then $\beta = i(X) \theta$ implies that we must
replace $X$ with $-X$; so the direction on the foliation is
reversed. Basically, orienting the surface (and hence foliation)
amounts to choosing a sign for $u$.
According to the choice of orientation, an elliptic singularity
becomes a source or a sink. At a hyperbolic singularity $x$, we have
two eigenvalues $\lambda_1, \lambda_2$ associated with the two
separatrices, which are the eigenvalues of the linearisation of the
flow at $x$. Here $\lambda>0$ means that the foliation points
outward and $\lambda<0$ inward; for a hyperbolic singularity
obviously there is one eigenvalue of each sign. The non-isochore
condition means that $\lambda_1 + \lambda_2 \neq 0$. Reversing the
orientation of $S$ and $\F$ will replace the eigenvalues $\lambda_1,
\lambda_2$ with the ``same eigenvalues, but in the opposite
direction" --- namely, they become $-\lambda_1, -\lambda_2$.
Now, at a singularity $x$ we have $\beta = 0$. Recall that at a
singularity $u \neq 0$. (If $u=0$ and $\beta = 0$ at $x$ then
$\theta = 0$ there too.) So $x$ is not on $\Gamma$; and
choosing an orientation at $x$ amounts to choosing the sign of $u$.
Supposing that $u$ is given in advance, of course, we have an
orientation; let us call this the \emph{pre-orientation}. But we can
also choose a standard orientation at the singularity $x$. Recall
that away from $\Gamma$, $\F$ is dilating for some area form. Well,
with one direction $\F$ will be dilating, and with the opposite
direction it will be contracting. We can define the \emph{positive
orientation} of $\F$ at $x$ to be the one that makes it expanding.
\begin{defn}
Let $x$ be a non-isochore singularity of a singular foliation $\F$.
The \emph{positive orientation} of $\F$ (and hence $S$) at $x$ is a
direction of $\F$ near $x$ by a vector field for which the
divergence with respect to some area form at $x$ is positive.
\end{defn}
So, at an elliptic singularity, with the positive orientation it is
a source; with the negative orientation it is a sink. At a
hyperbolic singularity, with eigenvalues $\lambda_1 < 0 <
\lambda_2$, we have $\lambda_1 + \lambda_2 > 0$, which means that
the ``out direction", the ``unstable direction", dominates, and
$|\lambda_2| > |\lambda_1|$.
Since an orientation at a singularity $x$ amounts to choosing a sign
for $u$, we can ask: if $x$ is positively oriented, what is the sign
of $u$? It seems like it should be positive: and indeed it is. Let
us see why.
Taking our area form $\theta$ on $S$ and the vector field $Y$
directing $\F$ given by $\beta = i(Y) \theta$, the divergence is
given by
\[
\Div_\theta Y = \frac{L_Y \theta}{\theta} = \frac{ i(Y) \; d\theta
+ d \; i(Y) \theta }{\theta} = \frac{d\beta}{\theta}.
\]
So to say that the singularity $x$ is oriented positively amounts to
$u$ being chosen so that $d\beta$ and $\theta$ have the same sign.
At the singularity, we have $\beta=0$ so $\theta = u \; d\beta +
\beta \wedge du = u \; d\beta$. Thus, the positive orientation at
$x$ puts $u>0$.
Using this, we can express the orientation condition in another way.
We give a more geometric interpretation to the area form $\theta = u
\; d\beta + \beta \wedge du$. In the first term, the $d\beta$, as we
have seen, is essentially the divergence of $Y$. In the second term,
since $\beta$ is dual to $Y$, the wedge product is essentially
$du(Y)$. Indeed, expanding out $i(Y)(\theta \wedge du) = 0$ (this is
a 3-form on $S$) we obtain $i(Y) \theta \wedge du = - (i(Y) du)
\theta$ so $\beta \wedge du = - (i(Y) du) \theta$. This gives
\[
\theta = u \; d\beta + \beta \wedge du = u \theta \; \Div_\theta
Y - \left( i(Y) du \right) \theta = \left( u \Div_\theta Y - i(Y) du
\right) \theta
\]
That $\theta$ is nondegenerate amounts to $u \Div_\theta Y - i(Y) \;
du \neq 0$. The positive choice of orientation (and hence choice of
sign of $u$ and $\theta$) gives us, at the singularity $x$, that
$u>0$, and since at $x$ we have $Y=0$, we obtain $u \Div_\theta Y -
i(Y) du > 0$. Since this expression is never allowed to equal $0$,
it must remain positive over the whole of $S$.
We record all these conclusions.
\begin{lem}
Let $S$ be a convex surface in a contact 3-manifold $M$. Let $\F$ be
the characteristic foliation on $S$ and $x$ a singularity. Take a
neighbourhood $S \times \R$ of $S = S \times \{0\}$ with contact
form $\alpha = \beta + u \; dt$, where $\beta$ is a 1-form and $u$ a
function on $S$, so that $\theta = u \; d\beta + \beta \wedge du$ is
an area form on $S$. The positive orientation of $\F$ and $S$ at $x$
gives:
\begin{enumerate}
\item
$d\beta$ and $\theta$ differ by a positive factor at $x$;
\item
$u>0$ at $x$;
\item
$u \; \Div_\theta Y - i(Y) du > 0$ on all of $S$.
\end{enumerate}
\end{lem}
If we start with a pre-determined $u$, how do the pre-determined
orientation and the positive orientation at a singularity differ?
Clearly they agree at singularities where $u>0$, and they disagree
at singularities where $u<0$. So all sources have positive
orientation; all sinks have negative orientation;
``outward-dominated" hyperbolic singularities have positive
orientation; ``inward-dominated" hyperbolic singularities have
negative orientation. If we cut $S$ along the dividing set $\Gamma =
\{u=0\}$, into the pieces $S_+ = \{u>0\}$ and $S_- = \{u<0\}$, then
the singularities with positive orientation lie in $S_+$ and the
singularities with negative orientation lie in $S_-$.
\section{What a convex surface looks like II}
Let us now return to a question we had before: we have a surface $S$
in $(M, \xi)$ with a characteristic foliation $\F$. Is $S$ convex?
\subsection{It has expanding vector fields directing the
characteristic foliation, and dividing sets...}
We know from previously that if $S$ is convex, then we can choose
coordinates so that $S$ is $S \times \{0\}$ in $S \times \R$, that
$\alpha = \beta + u \; dt$, and away from $\Gamma = \{u=0\}$ we have
Liouville vector fields and all the rest of it. But \emph{now} we
can go in the other direction too. If we have a characteristic
foliation $\F$ with this $\Gamma$ and dilating Liouville vector
fields, then we have a nice contact structure. On the other hand
though, because of the nature of the nice foliation, by proposition
\ref{I.3.2} there is a vertically invariant contact structure on $S
\times \R$ inducing the foliation $\F$. By the power of the
characteristic foliation, namely proposition \ref{II.1.2(b)}, these
the two germs of these contact structures are equivalent. So they
are locally isomorphic; and there is a contact vector field on $S$
which is transverse, because there was for the vertically invariant
version. So $S$ is convex. Thus we obtain Giroux's proposition
II.2.1.
\begin{prop}
Let $(M, \xi)$ be a contact 3-manifold, $S$ an embedded closed
orientable surface and $\F$ its characeristic foliation. The
following are equivalent:
\begin{enumerate}
\item
$S$ is convex.
\item
There exists on $S$ a curve $\Gamma$ transverse to $\F$,
possibly disconnected, which decomposes $S$ into subsurfaces
where $\F$ can be directed by a dilating vector field for a
certain area form, and exiting through the boundary.
\end{enumerate}
\end{prop}
\subsection{...if it's closed, it can't have a characteristic foliation defined by a
closed form...}
Suppose we have, as usual, a convex surface $S$, with an $S \times
\R$ neighbourhood and contact form $\beta + u \; dt$. Recall that
the characteristic foliation is defined on $S$ by the 1-form
$\beta$; and recall that the condition for $\beta, u$ to define a
contact structure is that the 2-form $\theta = u \; d\beta + \beta
\wedge du$ on $S \times \{0\}$ is nondegenerate.
If $\beta$ is closed, then $\theta = \beta \wedge du$ must be
nondegenerate. But on a closed surface, the function $u$ must have
critical points, where $du = 0$, and so this requirement of
nondegeneracy is impossible to satisfy.
This is something of a strange condition: if $\beta$ is closed, then
taking a nonzero function $f$ on $S$, $f\beta$ defines the same
foliation as $\beta$ but $f\beta$ may not be closed. But the same
argument should still work upon doing some Leibnitz rule and
rearranging the terms.
\begin{lem}
Let $S$ be a closed surface in a contact manifold $M$ with
characteristic foliation $\F$ defined by a 1-form $\beta$ on $S$. If
$\beta$ is closed (as a form on $S$) then $S$ is not convex.
\end{lem}
\subsection{...or a closed leaf with return map tangent to the
identity...}
Nor can there be a closed leaf $F$ in the characteristic foliation
$\F$ of a certain type. Every closed leaf $F$ defines a return map,
by taking a transverse interval which is sufficiently small; we then
get a return map $f: [-1,1] \To [-1,1]$, say, where the point $0$ on
the interval corresponds to the closed leaf. Then we have $f(0) =
0$. It turns out that $f'(0) = 0$ can't happen; this contradicts the
convexity of the surface.
Why is this? Along the leaves of the foliation we have $\beta = 0$.
The return map is determined by $d\beta|_F$; if we choose a product
neighbourhood of $F$ and a good coordinate system, we can easily see
that we can take $\beta$ such that $d\beta|_F = 0$ identically. But
then along $F$, $\theta = u \; d\beta + \beta \wedge du = \beta
\wedge du$... and $u$ must have a maximum on the leaf $F$, at which
$\beta \wedge du = 0$, contradicting the requirement $\theta \neq
0$.
\begin{lem}
Let $S$ be a surface in a contact manifold $M$ with characteristic
foliation $\F$. If $\F$ contains a closed leaf $F$ with return map
tangent to the identity along $F$, then $S$ is not convex.
\end{lem}
\subsection{... the dividing set avoids certain separatrices...}
If $S$ is convex, we can also consider where the dividing set
$\Gamma$ is in relation to our foliation $\F$. Of course we know
about expanding area forms and so on, but we can pin down more: it
turns out the dividing set can't go in certain places.
The idea is the following: taking a product neighbourhood and nice
coordinate system $S \times \R$ we may take $\alpha = \beta + u \;
dt$ as we have previously. The dividing set $\Gamma$ is defined by
$u=0$. The characteristic foliation is defined by $\beta$, so the
singular points of the foliation are where $\beta = 0$. If we know
something about how $u$ interacts with the characteristic foliation,
we can say more. We will find that on certain leaves of the
foliation --- namely, leaves connecting hyperbolic singularities in
a particular way --- $u$ cannot equal zero along them; and so
$\Gamma$ cannot intersect them.
Take a hyperbolic singularity $x$. Considering $u$ as given a
priori, and defining a pre-orientation, $S$ has either a positive or
negative orientation at $x$ according to the sign of $u$. Suppose
$S$ has a positive orientation at $x$. As discussed previously, we
may positively orient $S$ and $\F$ at $x$. This means that
$u>0$ at $x$; and it means that $u \; \Div_\theta Y - i(Y) du >
0$ on all of $S$.
Now, there is one crucial point, which we discovered back in section
\ref{characterising}. (Recall that there we characterised vertically
invariant contact structures: but now we know that the germ of the
contact structure on any convex surface can be described in this
way.) The crucial point is: across $\Gamma$, $Y$ points in the
direction of decreasing $u$. So from a positively oriented
hyperbolic singularity, where $u>0$, $\Gamma$ cannot intersect the
stable (inwards) separatrix: the stable separatrix is oriented
towards $x$, and $u$ would have to decrease as we cross $\Gamma$,
where $u=0$, towards $x$; but $u>0$ at $x$; this is a contradiction.
Similarly, from a negatively oriented hyperbolic singularity $x$,
$\Gamma$ cannot intersect the unstable (outwards) separatrix: for
$u<0$ at $x$, and at an intersection with $\Gamma$ on the unstable
separatrix we would have $u=0$ and increasing towards $x$, another
contradiction.
Now, if such the stable separatrix from a positively oriented
hyperbolic fixed point \emph{intersects} the unstable separatrix
from a negatively oriented hyperbolic fixed point, then we are in
trouble! For $u$ is positive at one fixed point and negative at the
other, hence must be zero somewhere in between; but for the above
reasons, this is impossible. This means that there cannot be such a
leaf connecting the two hyperbolic fixed points; and it means that
the two separatrices cannot intersect at a separate singularity.
Let us record these conclusions.
\begin{lem}
Let $S$ be a convex surface in a contact $3$-manifold $M$. Let the
contact form be $\beta + u \; dt$ as before and let $x$ be a
hyperbolic singularity of the characteristic foliation of $S$.
\begin{enumerate}
\item
If $x$ is positively (resp. negatively) oriented then $\Gamma$ does not intersect
the stable (resp. unstable) manifold/separatrix of $x$.
\item
The stable separatrix of a positively oriented
hyperbolic singularity and the unstable separatrix of a negatively oriented
hyperbolic singularity do not meet.
\end{enumerate}
\end{lem}
\subsection{...it's sufficient that the characteristic foliation is
almost Morse-Smale...}
Now we return to the situation where we have a surface $S$ in a
contact manifold, and want to know whether it is convex, by looking
at its characteristic foliation. We already have a sufficient
condition in terms of expanding area forms and dividing sets, but we
can do even better.
In the previous section we noted that in the characteristic
foliation of a convex surface, then certain separatrices do not
meet. We can actually formulate a chain of possible properties of
our foliation, where each clearly implies the next.
\begin{enumerate}
\item(Morse--Smale)
For any two separatrices of hyperbolic fixed points (possibly the same
point) they do not intersect. (Except possibly in emanating from the
same singularity.)
\item(Almost-Morse--Smale)
Positively orient all hyperbolic fixed points and
taking their stable separatrices. For any two such separatrices,
they do not meet. (Except possibly in emanating from the same
singularity.)
\item
With respect to a given pre-orientation, we consider each
hyperbolic fixed point to be positively or negatively oriented.
For any given pair of hyperbolic fixed points, one positive and
one negative, the stable separatrix of the positively oriented hyperbolic
singularity and the unstable separatrix of the negatively oriented
hyperbolic singularity do not meet.
\end{enumerate}
If the foliation is otherwise nice, then the first property is the
Morse--Smale property; the second property is called by Giroux the
almost-Morse--Smale property. And the third is not called anything
in particular, being such a mess, but is a property of
characteristic foliations on convex surfaces, by the previous
section. They are all strict implications and not equivalences.
(It's easy to find an almost Morse--Smale foliation which is not
Morse--Smale. And to find a foliation satisfying property (iii) but
not (ii), take two positively oriented hyperbolic fixed points whose
stable separatrices both connect to an elliptic fixed point, i.e. a
source.) So convex implies property (iii). And we will shortly see
that property (ii) implies convex. So in fact (i) (Morse--Smale)
implies (ii) (almost Morse--Smale) implies convex implies (iii). In
fact having (iii) but not (ii) implies we have connected elliptic
and hyperbolic fixed points of the same orientation and can apply
the elimination lemma, which we will come to shortly. But let us not
get ahead of ourselves!
To be more precise about these definitions:
\begin{defn}
A singular foliation $\F$ on a closed surface $S$ is called
\emph{Morse--Smale} if it satisfies the following conditions:
\begin{enumerate}
\item
the singularities and closed leaves of $\F$ are hyperbolic (in
the dynamical systems sense!);
\item
the limit set of each half-leaf is a singularity or a closed
leaf;
\item
For any two separatrices of hyperbolic fixed points (possibly the same
point) they do not intersect. (Except possibly in emanating from the
same singularity.) As Giroux puts it, there are ``no connections between
saddles".
\end{enumerate}
\end{defn}
\begin{defn}
A singular foliation $\F$ on a closed surface $S$ is called
\emph{almost-Morse--Smale} if it satisfies the first two conditions
above, and instead of (iii), satisfies the following: when we orient
$\F$ positively near hyperbolic fixed points, the associated stable
manifolds do not meet each other.
\end{defn}
Now let us see why almost Morse--Smale implies convex. Given a
closed surface $S$ with characeristic foliation $\F$, we explicitly
find regions on which a vector field directing $\F$ dilates (or
contracts) an area form. Around each elliptic point we take a small
disc. Around each closed leaf we take a small annulus. At each
hyperbolic singularity, we orient it positively and consider the
stable separatrix; we take a small band around it. We take the union
of these annuli, discs and bands, which forms a (possibly
disconnected) surface $S_0$. Note that from previous discussion that
$S_0$ will not intersect $\Gamma$. In fact the idea is that $S_0$ is
homeomorphic to $S \backslash \Gamma$. Note that the orientations on
components of $S_0$ may not agree with a pre-orientation on $S$; but
on each individual component of $S_0$, the positive orientations on
singularities agree. (This doesn't (and can't!) rely on any
convex-specific assumptions --- it's clear from the topology.)
By our construction, we can find an area form $\omega$ on $S_0$ and
a vector field $Y$ directing $\F$ such that $\Div_\omega Y > 0$.
(We can do this because, positively orienting everything, the
foliation points out of the discs and bands; and we can easily do
the same around closed leaves.) We are then basically done. For what
does the rest of the surface look like? It has a foliation without
any singularities or closed leaves, and it is part of the connected
oriented closed surface $S$. Hence it must be a union of annuli, and
the foliation must be the standard product foliation of intervals
from one boundary component to the other. Each one of these annuli
will contain a curve of $\Gamma$ in its core. Taking $\Gamma$ as
such, making some components of $S_0$ positive and others negative,
and gluing together, we have all the requirements of a
characteristic foliation required to be convex. Now we have Giroux's
proposition II.2.6.
\begin{prop}
Let $S$ be an orientable closed surface embedded in a contact
3-manifold. If the characteristic foliation of $S$ is almost
Morse--Smale, then $S$ is convex.
\end{prop}
Giroux quotes a theorem of M. Peixoto that a vector field on a
closed manifold is $C^\infty$-generically Morse--Smale. Since a
$C^\infty$ perturbation of a surface $C^\infty$-perturbs the
characteristic foliation, it follows that convex surfaces are
generic.
\begin{prop}
Let $S$ be a closed orientable surface in a contact 3-manifold. If
$\xi$ is transversely orientable, then the characteristic foliation
of $S$ is directed by a vector field that we can make Morse--Smale
by a $C^\infty$-small isotopy of $S$ in $V$; and hence generically
$S$ is convex.
\end{prop}
\section{What Giroux says about eliminating singularities}
Giroux has what I believe is the first proof of ``the elimination
lemma", allowing you to remove elliptic and hyperbolic fixed points
which are connected by a leaf and which have the same sign. Giroux's
(1991) version is significantly weaker than the version quoted by
Eliashberg (1992) in ``Contact 3-manifolds 20 years since J.
Martinet's work". Eliashberg there says the improvement is due to
Fuchs (but cites no paper of Fuchs), and refers to another (1991)
Eliashberg paper, ``Legendrian and transversal knots in tight
contact manifolds", for a proof.
We can compare the two versions of the statement.
\begin{prop}[Giroux]
Let $S$ be a closed orientable surface embedded in a contact
3-manifold with a Mose--Smale characteristic foliation $\F$.
(Hence $S$ is convex; let $\Gamma$ be the dividing set.)
Let $x_0$ be an elliptic and $x_1$ a hyperbolic
fixed point such that when we positively orient the foliation
near $x_1$, one or both of the stable separatrices comes from
$x_0$. Then:
\begin{enumerate}
\item
There exists in $S$ an annulus $A$ disjoint from $\Gamma$
and satisfying the following:
\begin{itemize}
\item
the only singularities of $\F$ on $A$ are $x_0$ and
$x_1$;
\item
$\F|_A$ has no closed leaf;
\item
$\F$ is transverse to the boundary of $A$/
\end{itemize}
\item
There exists an $S \times \R$ neighbourhood of $S = S \times \{0\}$ and a
function $k: A \To (-\infty, 0]$ with support in the
interior of $A$ such that the characteristic foliation of
the graph of $k$ (which is an arbitrarily small perturbation of
$S$) has no singularities.
\end{enumerate}
\end{prop}
On the other hand, the stronger version, cited by Eliashberg, is as
follows.
\begin{prop}[Giroux, Fuchs]
Let $S$ be a surface with (possibly empty) Legendrian boundary and
characteristic foliation $\F$ in a contact 3-manifold $(M, \xi)$. Let $C$ be a trajectory of $\F$
whose closure contains an elliptic point $x_0$ and a hyperbolic
point $x_1$ with the same orientation. Let $U$ be a
neighbourhood of $C$ in $M$ which contains no other singular
points of $\F$. Then there exists a $C^0$-small isotopy of $S$
in $M$ which is supported in $U$, fixed at $C$ and such that the
new surface $S'$ has no singular points of the characteristic
foliation $\F$ inside $U$. If $x_0$ and $x_1$ belong to the
Legendrian boundary of $S$ then one can kill them leaving
$\partial S$ fixed.
\end{prop}
I won't detail Giroux's proof, because it's probably better to refer
to the later paper. Finding the annulus in part (i) is easy, one
just has to fiddle a little to deal with the various possibilities.
Then finding the function $k$ one finds a function which depends
purely on the ``radial" coordinate of the annulus, supported in its
interior and which is sufficiently large that its ``Hamiltonian"
vector field $Y_k$ --- defined by $i(Y_k) \omega = dk$ for some
area form $\omega$ --- is very large away from the boundary of the
annulus. Hence considering $Y+Y_k$ is nonzero, and the dual form
$\beta +dk$ has no singularities nearby; and $\beta+dk$ describes
the characteristic foliation on the perturbed surface which is the
graph of $k$.
Giroux also shows how to simplify the characteristic foliation to
reduce an overtwisted disc to a standard form; ; and how to
eliminate elliptic singularities --- much as Eliashberg does in his
``20 years on" paper, in greater generality. So I won't detail it
here.
\section{The dividing set is all powerful}
So far we have seen that a convex surface is closely related to the
dividing set $\Gamma$. A convex surface certainly has a dividing
set. And a surface's convexity can be detected by finding a dividing
set $\Gamma$, on the complement of which the characteristic
foliation is directed by a vector field which dilates an area form.
If $S$ has an almost Morse--Smale foliation, then we showed that $S$
is convex by constructing regions on which $S$ dilated area forms,
and the remainder was just a tubular neighbourhood of $\Gamma$; so
$\Gamma$ was quite canonical, topologically.
But the dividing set is even more powerful than this. Suppose you
have a closed orientable surface $S$ in a contact 3-manifold $M$,
which is convex. So you have $\F$ and $\Gamma$. But suppose you only
have the information of $\Gamma$; you don't know what the
characteristic foliation is. Is $\F$ determined? Well obviously not:
there are potentially many foliations dilating area forms away from
$\Gamma$; if you take one, you can perturb it. But, it turns out
that all possible characteristic foliations $\F$ compatible with
$\Gamma$ can be achieved by perturbing $S$ a little. In effect,
$\Gamma$ captures precisely the important information about $\F$;
once you have $\Gamma$, provided you don't mind wiggling the surface
around a tiny bit, you can have whatever (compatible) $\F$ you like.
This is quite amazing. Humans have a hard time visualizing contact
planes fluttering around, even on a surface, let alone in space. But
it turns out, as we have seen, that the (germ of a) contact
structure near a surface is determined by its characteristic
foliation. And now, the characteristic foliation, up to a bit of
surface-jiggling, is determined by the dividing set. So we need not
visualize fluttering planes; we need not even visualize foliations
and trajectories; we only need visualize a few curves on a surface.
Let us make this precise. When we say that $\F$ is ``compatible"
with $\Gamma$, we mean something about area forms and stuff. What we
actually mean is:
\begin{defn}
Let $S$ be a closed orientable surface and $\Gamma$ a 1-manifold in
$S$. Let $S_\Gamma$ denote the compact surface with boundary
obtained by cutting $S$ along $\Gamma$. A singular foliation $\F$ on
$S$ is \emph{adapted} to $\Gamma$ on $S$ if $\F$ is directed by a
vector field which dilates an area form on $S_\Gamma$, and which
exits transversely through the boundary $\partial S_\Gamma$.
\end{defn}
Given $\Gamma$, we are going to wiggle the convex surface $S$. This
is means an isotopy $\delta_s: S \To M$ for $s \in [0,1]$ where
$\delta_0$ is the identity. Thinking of an $S \times \R$
neighbourhood where the transverse contact vector field is vertical,
we see that $S$ will remain convex as long as it is the graph of a
function; and the dividing set will always be the intersection of
$\Gamma \times \R$ with the jiggled surface. We have a
characteristic foliation to begin with, $\F_0$ say, and a foliation
we would like to get as a characteristic foliation, $\F_1$; both are
adapted to $\Gamma$.
So, taking an area form such as $\theta$ on $S$, we see that our
contact form $\alpha$ can be written $\beta + u \; dt$. Away from
$\Gamma$, we can write our contact form as $\beta + dt$; it's a
contactization. We can direct $\F_0$ by $Y_0$, defined by $i(Y_0)
\theta = \beta$. We can direct $\F_1$ by $Y_1$, where $Y_1$ dilates
a (possibly different) area form on $S_\Gamma$. But $\Div_{\pm e^g
\theta} Y = e^{-g} \Div_\theta(e^g Y)$, so by adjusting $Y_1$ by a
nonzero function, we can arrange that $Y_1$ dilates $\theta$ also.
The idea is simply to set $Y_s = (1-s) Y_0 + s Y_1$, hence $\beta_s
= i(Y_s) \theta$, and once we obtain a $u_s$ we will have a family
of contact forms $\alpha_s$ and we can Moser away to get our
isotopy. Note the vector fields $Y_s$ can certainly involve new
singularities; these are the singularities moving around, being
created and destroyed on the shifting foliations $\F_s$
There are some things to think about, though. Near $\Gamma$ we have
to fiddle a little. We take vector fields there to agree with $\pm
Y_0, \pm Y_1$ on the boundary of a tubular neighbourhood, and glue
together and linearly interpolate. So we can extend $Y_s$ over all
of $S$. We must also take a bunch of functions $u_s$ (they only
matter near $\Gamma$, though, since elsewhere it's a contactization)
such that the contact condition is satisfied: $u_s \Div_\omega (Y_s)
- Y_s \cdot u_s > 0$. This requires a fiddle also. But then we have
a family of contact structures $\xi_s$, and can Moser to obtain an
isotopy realising them. Since our $S \times \R$ neighbourhood can be
arbitrarily small, we obtain the following result.
\begin{prop}
Let $S$ be a convex closed orientable surface in a contact
3-manifold $M$, with transverse contact vector field $X$ and
dividing set $\Gamma$. Let $\F$ be a singular foliation on $S$
adapted to $\Gamma$. Then there exists an isotopy $\delta_s: S \To
V$ for $s \in [0,1]$ such that:
\begin{enumerate}
\item
$\delta_0$ is the identity;
\item
each surface $\delta_s S$ is transverse to $X$, hence convex;
\item
the dividing set of $\delta_s S$ (with respect to $X$) is $\delta_s
\Gamma$.
\end{enumerate}
\end{prop}
So indeed; by a small perturbation, we can get whatever
characteristic foliation we like. This is the power of convex
surfaces. Note, in particular, that it implies all the elimination
lemmas and results about standard forms of characteristic
foliations, immediately. It subsumes them all.
\begin{thebibliography}{99}
\bibitem{Giroux91}
Emmanuel Giroux, ``Convexit\'{e} en topologie de contact", Comment.
Math. Helvetici, 66 (1991) 637-677.
\end{thebibliography}
\end{document}