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\begin{document}
\title{Notes on Eliashberg's 1992 paper, ``Contact 3-manifolds twenty years
since J. Martinet's work}
\author{Daniel Mathews}%
\maketitle
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\tableofcontents
\section{Introduction}
These are the most relevant points (to me!) arising in this paper. I
have written them informally and without proper proof. It's no
substitute for reading the real paper! I just like to convey some of
the gist, without doing any real work.
\section{Significance of the paper}
It's one of the seminal papers spurring the resurgence of contact
geometry, along with Eliashberg's classification of overtwisted
contact structures and Giroux's work on convex surfaces.
Until this paper, the definition of overtwisted was ``contains a
standard overtwisted disk": that is, a disk with characterstic
foliation which has precisely one elliptic point, in the interior,
and precisely one limit cycle on the boundary. After this paper, an
equivalent definition was ``contains a surface with a limit cycle in
its characteristic foliation", or eqivalently, ``contains a surface
with a closed loop without singular points in its singular
foliation". Eliashberg proved they are equivalent, by manipulating
characteristic foliations. One direction is easy; the other
direction is his theorem.
\begin{thm}
\label{thm1}
Let $(M, \xi)$ be a closed oriented contact 3-manifold. Suppose
there is an embedded disc $D$ in $M$ with a limit cycle in its characteristic foliation. Then
$M$ contains a standard overtwisted disc.
\end{thm}
Clearly, this is a result about simplifying the characteristic
foliation: if we have a limit cycle, then we have a simple-looking
limit cycle. The next result is also about simplifying the
characteristic foliation: it turns out to amount to removing
elliptic points wherever possible. The result is (ostensibly!) about
the Euler class $e(\xi)$ of the contact structure on a closed
oriented surface $S$ in $M$.
\begin{thm}
\label{thm2}
Let $(M, \xi)$ be a tight contact manifold.
If $S=S^2$ then $e(\xi)[S] = 0$. Otherwise
\[
\chi(S) \leq e(\xi)[F] \leq - \chi(S).
\]
\end{thm}
This is clearly a finiteness result about the possible homology
classes of contact structures. In fact we can say more (relying also
on later results).
\begin{thm}
\label{thm3}
Let $M$ be a closed manifold. Only finitely many homotopy
classes of plane fields can contain tight contact structures.
\end{thm}
This is in stark contrast to the result for overtwisted contact
structures --- there is precisely one (up to isotopy) in each
homotopy class of plane fields.
The other major theorem is the classification of tight contact
structures on $S^3$. (Since the classification of overtwisted
contact structures was already known, and he just clarified the
distinction between tight and overtwisted, this is a complete
classification of \emph{all} contact structures on $S^3$.)
\begin{thm}
\label{thm4}
A tight contact structure on $S^3$ is isotopic to the standard
contact structure $\xi_0$.
\end{thm}
In fact we can prove a stronger result. (To see it's stronger,
consider $S^3$ as a union of two balls.)
\begin{thm}
\label{thm5}
Two tight contact structures on $B^3$ which coincide at the
boundary are isotopic relative to $\partial B^3$.
\end{thm}
The corollaries are many and varied. They are not all complete
pushovers, though.
\begin{thm}[Cerf's theorem]
\label{thm6}
Any diffeomorphism of $S^3$ extends to $B^4$.
\end{thm}
\begin{thm}
\label{thm7}
All tight contact structures on $\R^3$ are isomorphic.
\end{thm}
\section{Clearing up overtwisted disks}
The first task is to clear up all these previous difficulties with
overtwisted discs, and prove theorem \ref{thm1}. The idea is to take
a disc with a limit cycle; and then manipulate the foliation,
perturbing the disc, until it contains a standard overtwisted disc.
Eliashberg in this paper gives new techniques for manipulating the
foliation; they are now all completely standard.
In general, if you have a surface in a contact manifold, you have a
characteristic foliation on it. If you move the surface around, you
alter the foliation. And if you want to alter the foliation in
certain ways, you can sometimes move the surface around to achieve
it. The detail of how this works is crucial and is in general not
too difficult.
\subsection{$e_\pm, h_\pm, d_\pm, \chi, c$}
We begin by defining some variables. We have a contact $3$-manifold
$(M, \xi)$ and a surface in it $S$. The surface $S$ may or may not
have boundary. As usual $\chi(S)$ denotes the Euler characteristic
of the surface. We let:
\begin{enumerate}
\item
$e_\pm$ denote the number of positive/negative interior elliptic
singular points of $S_\xi$;
\item
$h_\pm$ denote the number of positive/negative interior
hyperbolic singular points of $S_\xi$;
\item
$d_\pm = e_\pm - h_\pm$.
\end{enumerate}
We also let $c(S)$ denote the ``Euler class of $\xi$ on $S$". This
makes sense if $S$ is a closed surface, and denotes the obstruction
to finding a vector field tangent to $S$ and $\xi$. If $S$ has
boundary, then we take a vector field tangent to $S$ and $\xi$ along
$\partial S$, and consider the obstruction to extending this to a
vector field tangent to $S$ and $\xi$ all over $S$.
From Poincare-Hopf, the Euler characteristic is given by the sums of
indices of singular points in a vector field on $S$, so
\[
\chi(S) = e_+ + e_- - h_+ - h_- = d_+ + d_-.
\]
On the other hand, the obstruction to extending a vector field along
$S_\xi$ takes into account the signs of singularities. A positive
elliptic point is a $+1$ obstruction, but a negative elliptic point
is a $-1$ obstruction. (It's easy to see they can cancel each other
out.) Similarly for hyperbolic singularities. We obtain
\[
c(S) = e_+ - e_- - h_+ + h_- = d_+ - d_-.
\]
Thus we find the following proposition, which Eliashberg in this
1992 paper attributed to Harlamov--Eliashberg in 1982 or Eliashberg
1991.
\begin{prop}
\[ d_\pm = \frac{1}{2} \left( \chi(S) \pm c(S) \right). \]
\end{prop}
\subsection{Manipulating Legendrian boundaries}
This proposition is stated as ``easy to see": maybe for some!
\begin{prop}
Let $S$ be a surface bounded by a piecewise smooth legendrian curve
$\Gamma$. Then $S$ can be deformed to an embedded surface
$\tilde{S}$ bounded by $\partial \tilde{S} = \tilde{\Gamma}$ that
preserves the Thurston-Bennequin invariant, and such that the
characteristic foliations $S_\xi, \tilde{S}_\xi$ are homeomorphic.
Hyperbolic corners of $\Gamma$ disappear, elliptic corners become
smooth elliptic points, and all interior singular points remain the
same.
\end{prop}
Maybe it's easier to see this after reading Giroux's paper or
something...
In any case, the Thurston--Bennequin invariant can be found simply
by looking at the characteristic foliation on the Legendrian
boundary of a surface. If the singular points are all the same sign,
then the contact structure never gets to turn around very far, so
must turn zero in total.
\begin{prop}
Let $S$ be a surface in a contact manifold $(M, \xi)$ bounded by
a Legendrian curve $\gamma$. If all singular points on $\gamma$ are of the
same sign then $tb(\gamma|S) = 0$.
\end{prop}
\subsection{The elimination lemma}
It turns out that it's possible to kill some singular points in
pairs. Namely, if there exist an elliptic point and an hyperbolic
point of the same sign, connected by a trajectory $\Gamma$ of the
characteristic foliation, then we can cancel them both. The change
only occurs in a neighbourhood of the single trajectory. It's
achieved by a $C^0$ small perturbation of our surface $S$ in $M$,
but the perturbation is fixed on the trajectory $\Gamma$ itself.
This works equally well, whether the points are in the interior or
boundary of $S$. In fact, if the trajectory is along the boundary,
we can say a little more: the perturbation can be taken to fix the
boundary (not just the trajectory from one singular point to the
other, but little chunks on either side in the support of the
perturbation, also).
Eliashberg attributes this proposition to Fuchs, improving a
slightly weaker result of Giroux.
\begin{lem}[Elimination Lemma]
Let $M, \xi, S$ be as above. Let $\Gamma$ be a trajectory of
$S_\xi$ between an elliptic point $p$ and a hyperbolic point $q$ of the same
sign. Let $U$ be a neighbourhood of $\Gamma$ in $M$ which
contains no other singular points of $S_\xi$.
Then there exists an isotopy of $S$ in $M$ such that:
\begin{enumerate}
\item
it is $C^0$ small;
\item
it is supported in $U$;
\item
it is fixed at $\Gamma$;
\item
the new surface $S'$ has no singular points in its
characteristic foliation inside $U$.
\item
If $p,q$ are both on the (Legendrian) boundary of $S$ then
we can keep $\partial S$ fixed.
\end{enumerate}
\end{lem}
We can also reverse the process (a \emph{creation} lemma!) and
create elliptic and hyperbolic singular points along a trajectory
--- having the same, but pre-specified sign, determined by the
relative orientation of $\xi$ and $S$ along a trajectory.
\begin{lem}[Creation lemma!]
Let $M,\xi,S$ be as before. Near a nonsingular point of $S_\xi$,
one can create a pair of singular points, one elliptic and one
hyperbolic, having the same (pre-specified) sign.
\end{lem}
Why are these true? Well at least the creation of singular points is
believable, simply by drawing a picture. We can take a standard
neighbourhood of a legendrian curve in the standard contact $\R^3$,
say along the $x$-axis. Perturbing a little horizontal strip by
bending it a little can give us the singularities we want. And by a
Darboux-type argument, the elimination of such singularities doesn't
sound too far fetched either.
\subsection{Manipulating the characteristic foliation I: simplifying
inside a limit cycle}
Now we can give the proof of theorem \ref{thm1}. Recall we have a
limit cycle in a disc $D$: we have to find a limit cycle in a disc
with only one elliptic point inside. The idea is to use the tricks
we have just discussed to simplify the characteristic foliation
inside a limit cycle.
WLOG we can assume the limit cycle is innermost on $D$, and we can
assume it is attractive. We also assume the characteristic foliation
is \emph{generic}: so all singularities are standard and there are
no connections between hyperbolic points.
What are the possibilities for the singularities inside $D$? By
Poincare--Hopf, we know that $e - h = \chi(D) = 1$. So if there is
only $1$ elliptic point, there are no hyperbolic points, and we are
done. If there are more elliptic points, then there are more
hyperbolics, also; in particular, there exists a hyperbolic point
$q$. We will proceed by removing hyperbolic points, so that the
situation is eventually reduced to the case of a single elliptic
point, which is a standard overtwisted disc.
Where can the separatrices of $q$ go? The incoming ones can only
come from positive elliptic points. (Not other hyperbolic points,
since $D_\xi$ is generic; and not from the attracting limit cycle
either!) So if $q$ is positive, we can apply the elimination lemma
and cancel $q$, and simplify the situation; by induction then we are
done.
We may assume then that $q$ is (indeed all hyperbolic points are)
negative. Now where do the outgoing separatrices form $q$ go? They
can go to negative elliptic points or to the limit cycle. (Not to
other hyperbolic points, by genericity!) If anything goes to a
negative elliptic point, we can apply the elimination lemma and
cancel $q$. So we may assume the outgoing separatrices go to the
limit cycle.
But in this case there is a sneaky trick: we use the creation lemma
to create a pair of singularities, so that the separatrices from $q$
go to the hyperbolic singularity. (This is possible, if you think
about it.) We can even do it so that the two separatrices now form a
smooth closed loop $\gamma$. And if we take the created
singularities to be negative, then all singularities along $\gamma$
are negative; so its Thurston-Bennequin number is $0$. Now we can
use the Legendrian perturbation lemma to smooth it into a limit
cycle. This gives us a limit cycle with fewer hyperbolic points
inside, and we are done.
\section{Manipulating the characteristic foliation II: removing
elliptic points}
We will shortly see that, by use of the tricks previously discussed,
we can almost remove any elliptic singularity from a characteristic
foliation. In particular we can prove:
\begin{prop}
Any elliptic point of $S$ can be destroyed via a perturbation of
$S$, cancelling it with a hyperbolic point, unless $S$ is a
sphere and $e_+ = e_- = 1$, $h_+ = h_- = 0$.
\end{prop}
Theorem \ref{thm2} follows immediately. For if $S$ is a sphere then
we can cancel until we are left with $e_+ = e_- = 1$, $h_+ = h_- =
0$, so that $d_+ = d_- = 1$ and $c(S) = e(\xi)[S] = d_+ - d_- = 0$.
If $S$ is not a sphere then we can cancel until $e_+ = e_- = 0$ so
that $d_+, d_- \leq 0$. Then $c(S) = d_+ - d_-$ and we have bounds:
\[
\chi(S) = d_+ + d_- \leq c(S) = d_+ - d_- \leq - d_+ - d_- = - \chi(S)
\]
so we are done. Theorem \ref{thm2} is really just about cancelling
elliptic points.
\subsection{Basins and Legendrian polygons}
To remove elliptic points, we are going to look at their
\emph{basins}. Given a subset $V$ of a surface $S$ in $M$, the
\emph{basin} $B(V)$ is the set of points of $S$ which can be reached
from $V$ along (oriented!) trajectories of $S_\xi$.
What does the basin of a set look like? Well, if you start at a
point and follow the trajectory, you can end up at a negative fixed
point, or a hyperbolic fixed point, or a limit cycle. Put limit
cycles to one side for a minute. Then the basin looks like a
``Legendrian polygon", with boundary consisting of Legendrian
curves.
We can formalise the idea of ``Legendrian polygons".
Such a polygon will arise where you have a surface (not necessarily
a polygon! The ``polygonal" part of the polygon is solely its
boundary!) with a piecewise smooth boundary; and you immerse it into
a surface $S$ in a contact manifold $(M, \xi)$ so that the boundary
runs along the characteristic foliation, with vertices mapping to
singularities.
Note that this is a bit ambiguous: a smooth edge of $S$ could run
through a singularity of $S_\xi$. Do we still count it as a vertex?
In a strange convention, we \emph{do} if the singularity is
\emph{elliptic}. We can have hyperbolic singularities along smooth
edges, and they may not count as vertices; if so, they are called
\emph{pseudovertices}.
A Legendrian polygon is required to be ``almost" injective: it can
overlap on its vertices or edges, but not in its interior. So, for
instance, a Legendrian polygon which is a disc with two vertices and
two edges could fold up into a ball. (The two edges glued together.)
Basins (still putting limit cycles to one side) have this structure
as Legendrian polygons; vertices can overlap; singularities can be
approached from different directions; trajectories may join up along
a edge; but the basin cannot overlap with itself in its interior.
It turns out, even if there are limit cycles present, then the
polygon can still be defined. Suppose there are trajectories from
$V$ approaching a limit cycle $C$. Note that if we were happy to say
``a limit cycle is a vertex", we could define an extended version of
Legendrian polygon. But we won't do that; we'll perturb to avoid the
entire situation! Suppose two adjacent boundary edges of our
proposed polygon approach the limit cycle $C$. We perturb $S$ near
$C$ to create a hyperbolic and elliptic fixed point; so that the
edges approach the elliptic point. (The hyperbolic and elliptic
point we create are negative.) In this way, by a series of small
perturbations of $S$, we can assume that $B(V)$ is in fact a
Legendrian polygon.
\subsection{Killing elliptic points}
So, to the proof of theorem \ref{thm2}. Let $E$ be an elliptic point
on $S$; we want to kill it without creating any more elliptic
points. WLOG we can assume $E$ is positive. And WLOG we may also
perturb $S$ so that it is generic; in particular, we can asssume
there are no separatrix connections between hyperbolic points.
If $\overline{B(E)}$ is a sphere, then certainly $S$ must be a
sphere! The polygon must fold up so that its boundary maps to one
point; and there are precisely two singularities. One is positive
elliptic; the other is negative elliptic. So $e_+ = e_- = 1$, $h_+ =
h_- = 0$ as desired.
We now assume $\overline{B(E)}$ is not a sphere. Well then, what
does its boundary look like? It may involve sinks, hyperbolic points
and limit cycles; it may be complicated. But we can say immediately
that if there is any positive hyperbolic point involved, we can
cancel it by the elimination lemma, and reduce the number of
elliptic points; by induction, we are done.
So we may assume that $\overline{B(E)}$ involves no positive
hyperbolic points. But it may contain limit cycles. So we perturb
$S$ near limit cycles, if necessary, so that the basin $B(E)$ is
nice and its closure $\overline{B(E)}$ is a Legendrian polygon in
$S$. As discussed above, this involves creating negative elliptic
and negative hyperbolic. (No matter that we have created extra
elliptic points, since we will now find a contradiction!)
The boundary of $\overline{B(E)}$ then consists of piecewise
Legendrian curves, with singularities on them, negative elliptic
(sinks!) and negative hyperbolic. By genericity, there are no
separatrices connecting hyperbolics. This only makes sense if the
elliptic and hyperbolic points alternate along the boundary. So the
picture is very standard. And by the elimination lemma, we can
cancel all these singular points in pairs! This then reduces
$\overline{B(E)}$ to something bounded by a limit cycle; and with
only one singularity inside; so it's a standard overtwisted disc,
contradicting tightness.
\section{Extending a characteristic foliation over a 3-ball}
We now enter upon the proof of theorem \ref{thm5}, that a tight
contact structure on $B^3$ is determined up to isotopy by its
restriction to the boundary $S$: that is, determined by the
characteristic foliation $S_\xi$.
\subsection{Strategy of the proof.}
Given a contact structure $\xi$ on $B$, with characteristic
foliation $S_\xi$ on the boundary, we want to find a standard
contact structure $\zeta$, which only depends on the foliation $\F =
S_\xi$, such that $\xi$ is isotopic to $\zeta$.
How are we going to find $\zeta$, starting only from a foliation
$\F$ on $S$?
We're going to use all manner of tricks about foliations and
pseudoconvex embeddings. As it turns out, given an embedding
$\alpha: B \To \R^3$ with certain properties and a map $\gamma:
\alpha(B) \To \R$ (also with certain properties!), we can obtain a
contact structure on $B$! How do we get this? We consider our $\R^3$
as lying inside $\C^2$, where the coordinates on $\C^2$ are $(z_1 =
x_1 + i y_1, z_2 = x_2 + i y_2)$ and the coordinates on $\R^3$ are
$(x_1, y_1, x_2)$. Then from $\alpha$ (the \emph{embedding}) and
$\gamma$ (the \emph{function}) we obtain the \emph{graph}
\[
\Gamma_\gamma = \left\{ y_2 = \gamma(u) : u \in \alpha(B)
\right\} \subset \C^2.
\]
If $\alpha$ and $\gamma$ satisfy certain requirements, it turns out
that the complex tangencies on $\Gamma_\gamma$, a 3-manifold in
$\C^2$, give a contact structure on $\alpha(B)$, and hence on $B$.
For instance, one of these requirements is that $\gamma = 0$ on $S$
and is positive on the interior of $B$; so that in $\Gamma_\gamma$,
$S$ correpsonds to $y_2 = 0$.
It will turn out that from a foliation $\F$ on $S$, we will find a
function $\varphi: S \To \R$ which is said to \emph{tame} $\F$.
There are many choices of such $\varphi$, but it turns out they are
homotopic. The pair $(\varphi, \F)$ form a \emph{function--foliation
pair} and will lie in a space called $FF$. From $(\varphi, \F)$ we
can obtain an embedding $\alpha: B \To \R^3$ satisfying the required
properties mentioned above, in fact so that $\varphi$ is the
$y_1$-coordinate of the restriction of $\alpha$ to $S$. The pair
$(\alpha, \F)$ is an \emph{embedding--foliation pair} and lies in a
space called $EF$. The choice of $\alpha$ is unique up to homotopy;
it turns out $EF \To FF$ is a Serre fibration with contractible
fibre. From $(\alpha, \F)$ we want to obtain a function $\gamma:
\alpha(B) \To \R$ which has the properties desired of $\gamma$
above. The foliation of $\F$ determines the 1-jet of $\gamma$ on the
boundary $\alpha(S)$, and so we have a pair $(\alpha,
\gamma_1^\partial)$ where $\alpha$ is an embedding and
$\gamma_1^\partial$ is a 1-jet of a function $\alpha(S) \To \R$;
such a pair lies in the space $\Conv_1^\partial$, which is
homeomorphic to $EF$. From $(\alpha, \gamma_1^\partial)$ the 1-jet
gives us a germ $\gamma^\partial$ of a function $\alpha(S) \To \R$;
the pair $(\alpha, \gamma^\partial)$ lies in a space
$\Conv^\partial$. The choice of $\gamma_1$ is unique up to homotopy,
and indeed the map $\Conv^\partial \To \Conv_1^\partial$ is a Serre
fibration with contractible fibre. Then from $\gamma_1$ we obtain a
full function $\gamma: \alpha(B) \To \R$ of the desired type. The
pair $(\alpha, \gamma)$ lies in a space $\Conv$ and the space of
such $\gamma$ is unique up to homotopy; again we have a Serre
fibration $\Conv \To \Conv^\partial$ which has contractible fibre.
Our pair $(\alpha, \gamma)$ now determines a graph $\Gamma_\gamma$
and a contact structure $\zeta$ on $B$, which is unique up to
homotopy; hence only depends on the original foliation $\F$.
That's quite a lot of construction! We have a whole lot of maps,
which are all Serre fibrations with contractible fibres (or better).
\begin{align*}
\Conv & \To & \Conv^\partial & \To & \Conv_1^\partial & \To & EF
& \To & FF & \To & \Fol\\
(\alpha, \gamma) && (\alpha, \gamma^\partial) && (\alpha,
\gamma_1^\partial) && (\alpha, \F) && (\varphi, \F) && \F
\end{align*}
Having obtained our $\zeta$, we then need to show it's isotopic to
the original contact structure $\xi$. We do this by taking a family
of spheres $S_t$, $t \in [0,1]$, which fill $B$ (okay, $S_0$ is a
point). We can apply the procedure above to obtain a $\zeta_t$ for
each $t$, and since $\zeta_t$ has the same characteristic foliation
on $S_t$ as $\xi$, we can replace the foliation $\xi$ with $\zeta_t$
on expanding balls. (Okay, we can't do this at $t=0$, but by Darboux
both $\xi$ and $\zeta$ can be assumed standard there.) As long as
$\zeta_t$ varies smoothly, we will have our isotopy. The upshot is:
provided we can do the whole intricate construction of the previous
paragraph not only once, but \emph{in families}, we are done. It's
lucky that we mentioned everything is a Serre fibration, because
that means homotopies lift and this is all possible!
And that is the gist of the proof.
(I won't go into further detail on all the Serre fibrations, since
I'm not that interested, I don't know much of the background about
pseudoconvex embeddings, and Eliashberg doesn't go into that
background.)
Where is the tightness used? It's used in constructing the taming
function. We couldn't just start with any foliation $\F$: it had to
be a characeristic foliation of a tight contact structure. Thus the
taming function is crucial. I \emph{will} go into this part! (Well,
not homotopies of families of them, but why they exist.)
\subsection{Taming functions}
We start with a two-sphere $S$ in $M$ and a foliation $\F$ on $S$.
Since $\F$ is eventually going to lie in a family, it might not be
quite so generic. We allow simple singularities or birth--death
singularities. Let $X$ be a vector field generating $\F$. We will
define a function $\varphi: S \To \R$ to \emph{tame} $\F$ if:
\begin{enumerate}
\item
$X$ is gradient-like for $\varphi$ (recall this means that there
exists a Riemannian metric so that it behaves like a gradient,
well not quite, with an inequality, more or less, critical
points are the same);
\item
the positive/negative elliptic points of $\F$ are the local
minima/maxima of $\varphi$ (so that with the increasing flow of $\varphi$,
positive elliptic points are indeed sources and vice versa);
\item
if we pass through a critical value with a
negative/positive hyperbolic singularity
in the increasing direction of $\varphi$, the
number of components of the level set increases/decreases.
\end{enumerate}
For a sphere $S$ embedded in a tight $(M, \xi)$, it turns out there
always exists a function taming $S_\xi$; and any two such taming
functions are homotopic through taming functions; and given a
nicely-behaved family of contact structures (with only simple
singularities or birth-death singularities on the characteristic
foliation of $S$), we can obtain a smooth family of taming
functions; and so any two such smooth families of taming functions
are homotopic through families of taming functions.
Let's see why.
\subsection{Tricks with Legendrian boundaries}
Let's recall some tricks with Legendrian boundaries of surfaces in
contact manifolds.
First, suppose we have a surface $S$ bounded by a piecewise smooth
legendrian curve $\Gamma$. If there are no singularities on
$\Gamma$, we have a limit cycle. If not, then there may be
singularities of any type: elliptic or hyperbolic, positive or
negative. Generically recall there are no connections between
hyperbolic singularities. We can often perturb $S$, sometimes even
fixing $\Gamma$, to simplify the situation.
\begin{enumerate}
\item
If the Thurston-Bennequin invariant is $0$, then we can $C^0$ perturb
$S$ near $\Gamma$ (but hold $\Gamma$ fixed!) to remove the
singularities. Thus we obtain a limit cycle.
\item
If all the singularities on $\Gamma$ have the same sign, then
the Thurston-Bennequin invariant is $0$, so by the same
reasoning we can perturb $S$, fixing $\Gamma$, to obtain a limit
cycle.
\item
Any hyperbolic singularities can be made to disappear by a
perturbation of $S$ (moving $\Gamma$ along with it). So if there
are only hyperbolic singularities, we can remove them all. But
on the other hand, connections between hyperbolic singularities
are extremely rare and non-generic, so this shouldn't happen in
the first place.
\end{enumerate}
\subsection{Manipulating the characteristic foliation III: removing
hyperbolic points, Legendrian polygons}
Recall, given a characteristic foliation $\F = S_\xi$ on a sphere in
a tight contact manifold, we want to construct a taming function.
How are we going to do it? We are going to start from the bottom and
work our way up! The minima will be the sources in $\F$, and we will
then construct $\varphi$ so that as $C$ increases, the subsets
$\varphi \leq C$ engulf one other singularity at a time. How will we
make sure we can do this? We will look at the basins of whatever we
have already engulfed! These, as we know, are Legendrian polygons.
So we will now perform some more tricks in this regard.
\begin{lem}
Let $D$ be an embedded disc in a tight contact manifold $(M,\xi)$, such
that $\partial D$ is transverse to $\xi$ and the characteristic
foliation exits through $\partial D$. Then by a $C^0$ small isotopy of $D$,
fixed near $\partial D$, we can kill all
positive hyperbolic points of $D_\xi$.
\end{lem}
Just think of the incoming separatrices to a positive hyperbolic
point: where do they come from? Not a limit cycle, because it's
tight; not a hyperbolic point if $D$ is generic; hence from a
positive elliptic point. Then we can cancel with the elimination
lemma.
\begin{lem}
Suppose we have a Legendrian polygon $Q$ in a surface $S$
in a tight $(M, \xi)$,
where $Q$ is an immersed disc,
possibly with vertices identified (but not edges). Then
$\partial Q$ contains at least one positive and at least
one negative singular point.
\end{lem}
Suppose otherwise. If there are no singularities on the boundary we
have a limit cycle and are done. If there are singularities, we can
perturb $S$ to make the boundary smooth; and if all the
singularities have the same sign, we can perform the legendrian
boundary trick described above and perturb $Q$, fixed on the
boundary, to obtain a limit cycle bounding a disc, a contradiction.
\begin{lem}
Suppose we have a Legendrian polygon which is a (closure of a) basin
$\overline{B(V)}$ in a sphere $S$ in a tight $(M, \xi)$,
and which is injective except possibly
for identified vertices. Then the boundary $\Gamma$ has at least one positive
hyperbolic point.
\end{lem}
The polygon must be an immersed disc, and by the previous lemma it
must contain at least one positive singular point; it can't be a
positive elliptic (source!) since it's a basin, so must be positive
hyperbolic.
\begin{lem}
Suppose we have a Legendrian polygon which is a (closure of a)
basin $\overline{B(V)}$ in a sphere in a tight $(M, \xi)$, where
$d_+(V) = 1$. Then any identified pseudovertices (recall these
must be hyperbolic) are negative.
\end{lem}
The region $V$ is an embedded disc as in the first lemma above; so
we can cancel positive hyperbolic points in $V$. Hence we may assume
$V$ has $e_+ = 1$, $h_+ = 0$. So the basin $\overline{B(V)}$ has a
simple form. Any identified hyperbolic pseudovertices will form a
smooth loop when we join them to our positive hyperbolic point in
$V$ (after we perturb the Legendrian curve). This loop only has
positive singularities, so we can perturb to get a limit cycle,
which bounds a disc since we're on a sphere.
In particular, in our case of a sphere $S$ in a tight $(M, \xi)$: if
we are looking at basins $B(V)$ then what can happen?
\begin{enumerate}
\item
The boundary $\Gamma$ of the basin could be null; then we have
the whole of $S^2$; then extending $\varphi$ is easy.
\item
The polygon $\overline{B(V)}$ could be injective except possibly
for identified vertices. Then from above we have at least one
positive hyperbolic point on $\Gamma$. We'll engulf this for the
next extension of $\Gamma$ and add it to $V$. Note this joins
together two components of $\varphi \leq C$, so if both these
regions had $d_+ = 1$, the new combined region has $d_+ = 1 + 1
- 1 = 1$ also.
\item
The polygon $\overline{B(V)}$ could have identified
pseudovertices. But we can assume $d_+(V) = 1$, and so by the above
lemma the identified pseudovertices are negative hyperbolic. We
engulf this to extend $\varphi$, and $d_+ = 1$ still.
\end{enumerate}
This is how we construct $\varphi$; and that is all I am going to
say about the proof of theorem \ref{thm5} and its corollary theorem
\ref{thm4}.
\section{Finiteness result}
So we have proved theorems \ref{thm2} and \ref{thm5}. Now theorem
\ref{thm3} is not too difficult.
Given $M$, we want to show that there are only finitely many
homotopy classes of plane fields which can occur as tight contact
structures. For any such plane field $\xi$ and any 2-dimensional
homology class $S$, we have $e(\xi)[S] = 0$ (if $S$ is a sphere) or
$e(\xi)[S] \leq -\chi(S)$ (otherwise). How many 2-dimensional
cohomology classes on $M$ are there with satisfying these
conditions? Only finitely many.
On the other hand, consider homotopy classes of 2-plane fields on
$M$. Consider in the method of obstruction theory, skeleton by
skeleton. Since $S^2$ is connected and $\pi_1(S^2) = 1$, we can
extend over $1$-skeleta in only one way. To extend over $2$-skeleta
is possible since $\pi_1(S^2) = 1$ but we have many choices since
$\pi_2(S^2) \cong \Z$. For the plane field to be extendable over $M$
we require that in each $3$-cell the boundary obstructions sum (with
signs) to $0$; so the homotopy classes of plane fields on the
$2$-skeleton of $M$ which extend over $M$ is in bijective
correspondence with $H^2(M)$. From theorem \ref{thm2}, there are
only finitely many such cohomology classes available for $e(\xi)$.
Now suppose we are given $\xi$ on the 2-skeleton. Can $\xi$ be
extended over $M$ as a tight contact structure? By theorem
\ref{thm5}, and in at most one way, up to isotopy, on each 3-cell.
Hence, being given the homotopy class of $\xi$ on the $2$-skeleton
determines the homotopy class of $\xi$ as a 2-plane field. So there
are only finitely many possibilities.
\section{Cerf}
The idea of proving Cerf's theorem is now easy. If we have an
orientation preserving diffeomorphism $f: S^3 \To S^3$, we want to
extend it to $B^4$. The idea is that $f$ is isotopic to a
contactomorphism --- this is easy to show, since the space of tight
contact structures on $S^3$ is connected.
Then one needs to show that any contactomorhism of $S^3$ with the
standard tight contact structure extends to $B^4$. This requires
some holomorphic curve methods which refer back to Gromov.
\section{Tight contact structures on $\R^3$}
We can prove theorem \ref{thm7}, that all tight contact structures
on $\R^3$ are isomorphic. The idea is to exhaust $\R^3$ by spheres.
Consider a sphere $S$ in $\R^3$. For a tight contact structure $\xi$
we know now that $e(\xi)[S] = 0$, and clearly $\chi(S) = 2$; so $d_+
= d_- = 1$. Using the elimination lemma, then, we can perturb our
sphere a little to have $e_+ = e_- = 1$, $h_+ = h_- = 0$. And then
the characteristic foliation is standard, just like the unit sphere
in the standard $\R^3$! If we perturb a little more, the
characteristic foliation is \emph{diffeomorphic} to the standard
foliation.
So, we can exhaust $\R^3$ by a nested sequence of such spheres.
Given a tight contact structure $\xi$ on $\R^3$, we will show it
isomorphic to the standard $\xi_0$. So take nested spheres $\R^3 =
\cup^\infty V_k$ for $\xi$ and $\R^3 = \cup^\infty B_k$ for $\xi_0$.
For each $k$, $(B_k, \xi_0)$ and $(V_k, \xi)$ are contactomorphic to
the standard unit ball, hence to each other, by a map $h_k$. Since
the space of embeddings of the standard ball to itself is connected,
we can take each $h_k$ to agree with $h_{k-1}$ on $B_{k-1} \cong
V_{k-1}$. Then taking their union gives us a contactomorphism of
$\R^3$.
\begin{thebibliography}{99}
\bibitem{El92}
Yakov Eliashberg, Contact 3-manifolds twenty years since J.
Martinet's work, Annales de l'institut Fourier, 42, 1--2 (1992),
165--192.
\end{thebibliography}
\end{document}