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symplectomorphism, hamiltonian flow, primitive function,  
inclusion
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\noindent{\LARGE \bf Interpolation of an exact symplectomorphism
\\ by a Hamiltonian flow.}

\begin{quote}
\noindent {\bf A. Haro} \\
{\small
\noindent
 Departament de Matem\`atica Aplicada i An\`alisi, Facultat de Matem\`atiques,
 Universitat de Barcelona.
 Gran Via de les Corts Catalanes 585, 08007 Barcelona, Spain. \\
\noindent E-mail: {\tt haro@cerber.mat.ub.es}

\noindent{\bf Abstract.}
%{\em Exact symplectomorphisms} appear naturally in analytical mechanics,
%because the time-$t$ flow of a Hamiltonian vector field defined on a
%cotangent bundle (the {\em phase space}) is an example of them.
%
Let ${\cal O}$ be the zero-section of the cotangent bundle $T^*\M$ of
a real analytic manifold $\cal M$. Let
$F:(T^*\M,{\cal O})\to (T^*\M,{\cal O})$ be a real analytic local
diffeomorphism preserving the canonical symplectic form $\o= \dif\a$ of
$T^* M$ ($\a$ denotes the Liouville form).
Suppose that $F^*\a-a$ is an exact form $\dif S$. Then:
\begin{itemize}
\item We can reconstruct $F$ from $S$ and $f= F_{|{\cal O}}$.
\item $F$ can be included into a Hamiltonian flow, provided $f$ is
included into a flow.
\end{itemize}
The proofs are constructive.
They are related with a derivation
on the Lie algebra of functions (endowed with the Poisson bracket).
}
\end{quote}

\section{Introduction}
\label{sec:sp}

As the time-$1$ flow of a Hamiltonian vector field on an exact symplectic
manifold is exact symplectic, a natural question arises:
\begin{quote}
Given an exact symplectomorphism, is it the time-$1$ flow of a time-dependent
Hamiltonian vector field?
\end{quote}
In such a case, we shall say that our symplectomorphism is {\em homologous to
the identity}. Once
we have interpolated our exact symplectomorphism by a time-de\-pen\-dent
Hamiltonian flow, next question is:
\begin{quote}
Can we get our Hamiltonian be $1$-periodic in time?
\end{quote}

This subject has been studied for many authors, and it has many variants.
It is a particular case of the more general problem of {\it inclusion
of a map into a flow}.
Moser \cite{AI} already dealt with this problem
when he proved the analyticity of the Birkhoff normal form around a
hyperbolic fixed point of an area preserving map.
Douady \cite{UDD} solved the
problem in the smooth symplectic case provided our map is given by a
generating function and Conley and Zehnder \cite{BLFPT} solved it
for smooth diffeomorphism of a torus which leaves the center of mass fixed.
On the other side, Douady \cite{UDD}, Kuksin \cite{OI1} and
Kuksin and P\"oschel \cite{OI2} solved the problem in analytic set up for
maps which are close to integrable ones.

Our exact symplectic manifold is
$T^*\M$,
endowed with the canonical symplectic form given by the differential of
the Liouville form $\a= y\ \dif x$.
$z:\M\to T^* \M$ is the canonical inclusion
and ${\cal O}= z(M)$ is the zero-section of $T^* \M$.
$q:T^* \M\to\M$ is the standard projection.
In fact, we shall
work in a tubular neighbourhood of $\cal O$, $\N$.
We want to include
a symplectomorphism $F:\N\to T^* M$ into a Hamiltonian flow,
and, hence, F has to be exact. That is, there exists a function $S:\N\to\nr$
such that $F^*\a-\a= \dif S$. This function is a {\em primimitive
function} of $F$.
Moreover, we shall suppose that the zero
section is invariant.

This reduction is not so restrictive, due to a Weinstein's theorem
\cite{LSM}. It states that we can send via a
symplectomorphism a certain neighbourhood
of any Lagrangian manifold onto a neighborhood of the zero-section of its
cotangent bundle.  Moreover, using a generalized Poincar\'e's lemma, he also
proved that if our Lagrangian manifold is exact then the symplectomorphism
is also exact (between two different manifolds, of course).
See also \cite{sticky} for analytic versions of this theorem.
Lagrangian manifolds are important in Hamiltonian dynamics.
For instance, KAM tori are Lagrangian, and also the stable and unstable
manifolds of a hyperbolic fixed point.

The theorem that we shall prove along this paper is

\vspace{.25cm}
\noindent{\bf Theorem\ }{\em
Let $\M$ be a real analytic manifold, and
$\N= T^*\M$ a tubular neighbourhood of its zero-section.
Let $F:\N\to T^* M$ be a real analytic exact symplectomorphism,
such that the zero-section is invariant, and
$f= q\comp F\comp z:\M\to\M$ is its dynamics.
Suppose that $f$ is interpolated
by the flow $f_t= f_{t,0}$ of
a real analytic time-dependent vector field $X_t\in{\cal X}(\M)$:
$f= f_1$. Then,
\begin{quote}
   $F$ is (analytically) homologous to the identity (at least in a
   tubular neighbourhood of the zero-section).
\end{quote}
}

So then, we deal with
the first problem. In order to get periodicity in time
we can apply, in some cases, a theorem
by Pronin and Treschev \cite{IAMAF}.
We remark that they studied the analytic case and applied a constructive
method to obtain the
Hamiltonian, a kind of averaging method, and they began from a non-periodic
Hamiltonian. That is to say, the began from a homologous to the
identity symplectomorphism, which is that we shall obtain. Moreover,
they worked on a compact symplectic manifold, and we can apply their theorem
in a relatively compact tubular neighbourhood of the zero-section, provided
it is compact (say, a torus).

We also work in analytic set up, and our proof is also constructive. We remark
that the only information that we need to obtain the Hamiltonian is
the primitive function of our symplectomorphism and the vector field
that interpolates the dynamics on the zero-section. If this dynamics
is not included into a flow, then we conclude that our
symplectomorphism is homologous to the corresponding lift.


\section{Definitions}
\label{sec:def}
Though we shall work in the analytic category, the definitions can be
done in the differentiable category.
Let $\M$ be a real analytic manifold.

\subsection{Exact symplectomorphisms}
\label{sec:sp.esm}
The {\em Liouville form} of the cotangent bundle {$T^*\M$} of a manifold
$\M$ is the Pfaffian form defined on each `point' $\rho_x\in T^*\M$ and
for any $X_{\rho_x}\in T_{\rho_x} T^*\M$ by
\beqn
  \a_{\rho_x}(X_{\rho_x}) = \rho_x (q_*(\rho_x) X_{\rho_x}).
\eeqn
Moreover, it is the unique Pfaffian form on $T^*\M$ which satisfies
$ \rho^* \a= \rho $
for any $1$-form $\rho$ on $\M$.
Then, $\o= \dif\a$ is
the {\it canonical symplectic structure} on $T^*\M$ (and $\a$ is
an action form for $\o$).
In cotangent coordinates $T^*\M$,
$(x,y)= (x_1,\dots,x_d,y_1,\dots,y_d)\in {\cal U}\times\nr^d$, these forms
are $\a= y\ dx= \sum_{i=1}^d y_i\ dx_i ,
\ \omega= dy\wedge dx= \sum_{i= 1}^d dy_i \wedge dx_i$.

Let $F:\N\to T^*\M$ be a diffeomorphism from a tubular neighbourhood
of the zero-section, $\N$, onto its image.
We shall refer to the first-order partial
differential equation on $\N$
\beqn
      F^*\a-\a & = & \dif S
\eeqn
as the {\em exactness equation} of $F$. If it is solvable,
then we say that $S$ is a {\em primitive function} of $F$, and hence
$F$ preserves
the symplectic form. We say that $F$  is an {\it exact symplectomorphism}.

\Rem{
If $\phi:\M\to\M$ is a diffeomorphism on the base $\M$ then its {\em lift}
$\hat\phi\deq (\phi^{-1})^*:T^*\M\to T^*\M$
preserves not only the symplectic form $\o=\dif\a$ but also
the Liouville form. All the {\it actionmorphisms} on the whole
cotangent bundle can be obtained in this way (see, for instance,
\cite{SGAM}). In fact,
it is enough to be defined in a tubular neighbourhood of the zero-section.

In the literature, the primitive function is often called generating
function. As we see, this function does not generate the
symplectomorphism.
This is the reason we have followed the nomenclature used in
\cite{DSIII}.
Anyway, the primitive function and the different types of generating
functions are very close.}

\subsection{Hamiltonian flow.}
\label{sec:sp.hvf}

It is well known that to a function $H:\N\to\nr$ we associate
a {\em Hamiltonian vector field} $X_H$, which is uniquely determined
by $\pint{X_H}{\o}= -\dif H$.
The {\em Poisson bracket} between two functions $K$,$H$ is defined by
$\PPoi{K}{H}= \o(X_K,X_H)= -\dif K(X_H) = \dif H(X_K)$.
These operations are natural with respect to pull back
by symplectomorphisms, that is, for any symplectomorphism $F$:
$F^* X_H \deq (F^{-1})_* X_H \comp F = X_{H\comp F}$,
$\PPoi{K}{H}\comp F= \PPoi{K\comp F}{H\comp F}$.

In symplectic coordinates $(x,y)$, we write
\[
   X_H= {\left( \frac{\partial H}{\partial y} \ \
      -\frac{\partial H}{\partial x} \right)}^\top \ ,\
   \PPoi{K}{H}=
        \frac{\partial K}{\partial y}\cdot \frac{\partial H}{\partial x} -
        \frac{\partial K}{\partial x}\cdot \frac{\partial H}{\partial y},
\]
where $\cdot$ is the inner product.

Let $H_t$ be a time-dependent Hamiltonian function
(where the subscript $t$ means the dependence on time),
and $X_{H_t}$ be the corresponding time-dependent Hamiltonian vector field.
It is well known that the time-t flow from $t_0$, $\varphi_{t,t_0}$,
is an exact symplectomorphism (for the sake of simplicity,
we shall suppose completeness,that is, the flow is defined for all the
values of $t_0$ and $t$). In fact,
$\varphi_{t,t_0}^* \a - \a = \dif S_{t,t_0}$,
where
\beqn
S_{t,t_0} = \int_{t_0}^t \L(H_s)\comp\varphi_{s,t_0}\ ds
\eeqn
and $\L(H)= \a(X_H) - H$.


\subsection{The Liouville derivative}
\label{sec:sp.dl}
On one hand, recall that the space of functions ${\cal F}(\N)$
endowed with the Poisson bracket is a Lie algebra, and
the relation between the Lie bracket and the
Poisson bracket is given by $X_{\PPoi{K}{H}}= \PLie{X_K}{X_H}$.
On the other hand, we have defined an operator in this space,
given by $\L(H)= \a(X_H)-H$.

\begin{Proposition}
The linear operator $\L$ associated to $\a$
is a derivation in the Lie algebra ${\cal F}(\N)$.
Moreover, it is natural with respect to pull back by actionmorphisms.
\end{Proposition}
\bproof
The proof of the
product rule is:
\beqn
 \L(\PPoi{H_1}{H_2})\!
  &=&\!  \a(\PLie{X_{H_1}}{X_{H_2}}) - \PPoi{H_1}{H_2}
  =  \dif(\a(X_{H_2})) X_{H_1} - \Lie{X_{H_1}}{\a}\ X_{H_2} -
       \PPoi{H_1}{H_2} \\
  &=&\! \PPoi{H_1}{\a(X_{H_2})} - \dif(\L(H_1)) X_{H_2} - \PPoi{H_1}{H_2}
  = \PPoi{H_1}{\L(H_2)} + \PPoi{\L(H_1)}{H_2}.
\eeqn
If $L^*\a=\a$ and $H\in{\cal F}(\N)$:
$\a(X_H)\comp L= L^*(\pint{X_H}{\a})= \pint{L^* X_H}{L^*\a}=
\pint{X_{H\comp L}}{\a}= \a(X_{H\comp L})$.
\eproof

In fact, to any action form of any exact symplectic form we can associate
a derivation in the Lie algebra of functions. In our case, the derivation
associated to the Liouville form is named the {\em Liouville derivative}.
Moreover, $\L(H)$ is also known as the {\em
elementary action} associated to  the Hamiltonian $H$,
because it is used in order to
define a variational principle for its orbits
(see, for instance, \cite{MMMC,SGAM}). It is very close
to the {\em Legendre transformation}.

In cotangent coordinates $(x,y)\in {\cal U}\times\nr^d$ the
Liouville derivative is written
\beqn
        \L(H)(x,y) &=&  y \cdot \nabla_y H (x,y) - H(x,y).
\eeqn
$\L$ is a vertical operator, because the value of
$\L(H)$ on a fiber only depends on the value of $H$ on such fiber.

\section{Determination of an exact symplectomorphism.}
\label{sec:d}
As we have seen, an exact symplectomorphism $F:\N\to T^*\M$ is
not determined by its primitive function $S$.
In general, we can obtain all the exact
symplectomorphisms
with such primitive function composing $F$ on the left with
actionmorphisms $L$:
\[
 (L\comp F)^*\a-\a= F^* L^* \a - \a= F^* \a - \a= \dif S.
\]

So then, in order to answer the question
\begin{quote}
%\paragraph{The determination problem}
What additional information do we need in order to determine an exact
symplectomorphism from its primitive function?
\end{quote}
we have to look for the actionmorphisms of our exact symplectic manifold,
because
\begin{quote}
   An exact symplectomorphism is determined by its primitive function save an
actionmorphism.
\end{quote}

An actionmorphism defined in a tubular neighbourhood of the zero-section
does not move the zero-section (see \refsec{cpf}), and it is the lift of
the diffeomorphism induced in the zero-section. In order to see
this, it is enough to proof that the only
actionmorphism $I:\N\to T^*\M$ that is the identity
on the zero-section is the identity. We write the proof for completeness.

\begin{Proposition}
Let $I:\N\to T^*\M$ be an actionmorphism that fixes all the points
of the zero-section. Then, $I$ is the identity map.
\end{Proposition}
\bproof Take cotangent coordinates $(x,y)\in {\cal U}\times\nr^d$ around each
point of the zero-section $\{y=0\}$. Hence,
$I$ is given by
\[
       \bar x= f(x,y)\ ,\ \bar y= g(x,y),
\]
with $f(x,0)= x, g(x,0)= 0$.
The condition $I^*\alpha= \alpha$ is
\[
      0= g(x,y)^\top A(x,y) - y^\top\ ,\
      0= g(x,y)^\top B(x,y),
\]
where the matrices
\[
        A(x,y)= \frac{\partial f}{\partial x}(x,y)\ ,\
        B(x,y)= \frac{\partial f}{\partial y}(x,y),
\]
satisfy $A B^\top = B A^\top$. Note that $g(x,0)= 0$, because
the matrix $(A\ B)$ has maximum rank.
Hence, we have
\beqn
      0 &=& (g(x,y)^\top A(x,y) - y^\top) B(x,y)^\top =
          -y^\top B(x,y)^\top,
\eeqn
and then each $x$-component $f_i$ is a homogeneous function of degree $0$
in the $y$-variables.
By regularity on the zero-section, these functions $f_i$ are constant with
respect to the $y$-variables: $f(x,y)= f(x,0)= x$.
Finally, $g(x,y)= y$, because $A$ is the identity matrix.
\eproof

\section{Proof of the theorem.}
\label{sec:ies.sup}

\subsection{The homotopy method.}

Let $H:\N\times\nr\to\nr$ be a time-dependent Hamiltonian function,
$X_{H_t}$ be the corresponding vector field and $\varphi_t= \varphi_{t,0}$ be
the corresponding flow from $t_0= 0$.
We would like
\beqn
  \varphi_1^*\a-\a = \dif S.
\eeqn
In fact, we impose $\forall t\ \varphi_t^*\a-\a = t\ \dif S$
(this is the idea of a {\it homotopy method}). That is to say,
we want that $S_{t,0}= t\ S$ (with the notation of \refsec{sp.hvf}).
Hence, if we derive with respect to the time the previous homotopy formula we
shall see that it is enough to impose $S = \L(H_t)\comp\varphi_t$.
Therefore, if $H_0$ satisfies $S =  \L(H_0)$ and
\beqn
0=  \frac{d}{dt}(\L(H_t)\comp\varphi_t)=
   \dif (\L(H_t))(\varphi_t) \frac{\partial\varphi_t}{\partial t} +
         \frac{\partial}{\partial t}(\L(H_t))\comp \varphi_t
   = \{H_t, \L(H_t)\}\comp \varphi_t +
     \frac{\partial}{\partial t}(\L(H_t))\comp \varphi_t,
\eeqn
then $H_t$ is a time-dependent Hamiltonian whose time-$1$ flow is an exact
symplectomorphism whose primitive function is $S$. We obtain the next
{\em homotopy problem}.

\begin{Proposition}
Let $S:\N\to\nr$ be a function. Then, the time-$t$ flow
$\varphi_t= \varphi_{t,0}$ of
a time-dependent Hamiltonian $H_t:\N\to\nr$ that satisfies  $S= \L(H_0)$
and
\beqn
     \frac{d}{dt}(\L(H_t))&=&  -\PPoi{H_t}{\L(H_t)},
\eeqn
is exact symplectic and its primitive function is $t\ S$.
\end{Proposition}

\subsection{Generating solutions from a particular one.}
\label{sec:ies.sup.gs}

if we  find a solution of the
homotopy problem, we only can assure that
the primitive function of its time-$1$ flow is $S$,
but this does not determine $F$.
This is related to the existence of
functions whose $\L$-derivative vanish (the `constants').
In fact, we can obtain many families of exact symplectomorphisms satisfying
the same homotopy problem.

\begin{Proposition}
Let $S:\N\to\nr$ be a function,
$\bar H_t$ be a solution of the corresponding
homotopy problem, and
$\varphi_t= \varphi_{t,0}$ be the flow of the corresponding Hamiltonian
vector field.
Let $L_t= \hat f_t$ be a family of lifts generated (on the zero-section)
by the flow $f_t= f_{t,0}$
of a vector field $X_t\in{\cal X}(\M)$ on the zero-section.
Consider the new family of exact symplectomorphisms $\psi_t=
L_t\comp\varphi_t$, which is also connected to the identity. Then:
\begin{itemize}
\item $\psi_t= \psi_{t,0}$ is the flow of the Hamiltonian
$H_t= h_t + \bar H_t\comp L_t^{-1}$, where $h_t$ is the Hamiltonian lift
of $X_t$;
\item $H_t$ is also a solution of the homotopy problem.
\end{itemize}
\end{Proposition}
\bproof The Hamiltonian lift of $X_t$ is defined by
$h_t(\rho_x)= \rho_x(X_{t}(x))$ (note that $h_t\in\ker(\L)$).
It is well known that its flow is just the lift of the flow on
the zero section: $L_t= \hat f_t$. Then
\beqn
\frac{d\psi_t}{dt}
&=&
\frac{d}{dt}(L_t\comp\varphi_t)
=
\frac{\partial L_t}{\partial t}\comp \varphi_t +
(L_t)_*(\varphi_t)\ \frac{d\varphi_t}{dt}
=
X_{h_t}\comp L_t\comp\varphi_t+ ((L_t)_*\ X_{\bar H_t}) \comp \varphi_t \\
&=&
X_{h_t}\comp \psi_t + (L_t)_*\ X_{\bar H_t}\comp L_t^{-1}\comp \psi_t
=
X_{h_t}\comp \psi_t + X_{\bar H_t\comp L_t^{-1}}\comp\psi_t
=
X_{h_t + H_t\comp L_t^{-1}}\comp\psi_t.
\eeqn
and the first point is proved (cf. \cite{LCE}).
The second one follows from
\beqn
\frac{d}{dt}(\L(H_t)) &=&
\frac{\partial}{\partial t}(\L(\bar H_t))\comp L_t^{-1} +
\dif(\L(\bar H_t))(L_t^{-1})\ \frac{\partial L_t^{-1}}{\partial t} \\ \\
&=&
-\PPoi{\bar H_t}{\L(\bar H_t)}\comp L_t^{-1}
-\dif(\L(\bar H_t))(L_t^{-1})\ (L_t^{-1})_*\ X_{h_t} \\
&=&
-\PPoi{\bar H_t\comp L_t^{-1}}{\L(\bar H_t\comp L_t^{-1})}
-\PPoi{h_t}{\L(\bar H_t\comp L_t^{-1})} \\
&=&
-\PPoi{H_t}{\L(H_t)}
\eeqn
and
$\L(H_0)= \L(h_0 + \bar H_0\comp L_0^{-1}) = \L(\bar H_0)= S$.
\eproof


\subsection{A condition on the primitive function.}
\label{sec:cpf}

Since the zero-section is fixed,
we have some restrictions on the primitive function.

\begin{Proposition}
Let $F:\N\to T^*\M$ be an exact symplectomorphism and $S:\N\to\nr$ be
its primitive function. Then,
\[
    F \mbox{ fixes the zero-section } \Leftrightarrow
     \forall x\in\M\ \dif S(0_x)= 0.
\]
\end{Proposition}
\bproof Let $x\in\M$ be any point on the zero-section. Since $\alpha(0_x)= 0$
and
\[
   (F^*\alpha)(0_x) = \alpha(F(0_x))\comp F_*(0_x)=
   F(0_x)\comp q_*(F(0_x))\comp F_*(0_x)= F(0_x)\comp (q\comp F)_*(0_x),
\]
then
\[
   \dif S(0_x) = F(0_x) \comp (q\comp F)_*(0_x).
\]
Finally, as $(q\comp F)_*$ is an epimorphism in all points, we reach to
the result.
\eproof

\subsection{A splitting lemma}

\begin{Lemma}
The space of functions ${\cal F} \deq {\cal F}(\N)$ splits as
\beqn
     \cal F &=& \ker\L \oplus \L(\cal F).
\eeqn
Moreover, $\L({\cal F})$ is the subspace of functions whose
vertical derivatives vanish on the zero-section and
$\ker\L$ is the subspace of fiberwise homogeneous functions
of degree $1$.
\end{Lemma}
\bproof
We have to solve the equation $\L(H)= S$.
Since $\L$ is a vertical operator, we can restrict our attention to each
fiber, where is easy to work.
On each fiber (fixed $x\in\M$) we have a linear operator
transforming $y$-valued functions. Its properties are inherited by $\L$.

Hence,
let ${\cal U}\subset\nr^d$ be an open star-shaped neighbourhood of the origin
in $\nr^d$, with coordinates $y= (y_1,\dots,y_d)$,
and $S:{\cal U}\to\nr$ be a function. We have to
solve the p.d.e.
\beqn
   y\cdot \nabla_y H (y) - H(y) = S(y).
\eeqn
If $\nabla_y S(0) = 0$, its solutions are
\beqn
   H(y) = a\cdot y + \int_0^1 \frac{1}{t^2} (S(ty)-S(0))\ dt, 
\eeqn
where $a\in\nr^d$.
\eproof

\subsection{An iterative method.}
\label{sec:ies.sup.fep}

We can consider the homotopy problem as a family of evolution problems.
We can specify a particular one,
thanks to the previous lemma.
It says that $\L_\mid \deq \L_{\mid \L({\cal F})}$
is an isomorphism in $\L(\cal F)$.
If we look for a solution
$\bar H_t\in\L({\cal F})$,
we define $S_t= \L(H_t)\in\L({\cal F})$, and then we look for $S_t$
as a solution of the {\em evolution problem}
\[
        \left\{\begin{array}{l}
           \disp\frac{d S_t}{dt}= -\PPoi{\L_\mid^{-1}(S_t)}{S_t}, \\ \\
           S_0= S.
        \end{array} \right.
\]
If we solve this problem, then $\bar H_t= \L_\mid^{-1}(S_t)$
is a solution of the original one.
Of course, it is necessary that $\L({\cal F})$
be invariant under these operations to work inside this subspace.
In fact, not only the vertical
derivatives of $S_0$ vanish on the zero-section, but also {\em all} of
them (recall that $\dif S\comp z= 0$). Next lemma
can be easily proved using cotangent coordinates.

\begin{Lemma}
Let $S,T\in{\cal F}(\N)$ be two functions
such that $\dif S\comp z= 0$ and $\dif T\comp z= 0$.
Then,
\[
 \dif(\L_\mid^{-1}(S))\comp z= 0\ ,\ \dif\{S,T\}\comp z= 0.
\]
\end{Lemma}

To obtain the solution we use expansions in powers of $t$.
If $S_t= \sum_{k\geq 0} S_k t^k$ is the expansion of $S_t$ (where $S_0= S$),
then we can compute all the terms by the recurrence
\[
   S_{k+1}= \frac{-1}{k+1} \sum_{u+v=k} \PPoi{\L_|^{-1}(S_u)}{S_v}.
\]
Hence, all the terms of the expansion verify $\dif S_k\comp z= 0$.
In fact, as the
function $S_0= S$ has $y$-order $2$, the $y$-orders of the $S_k$
increase: the $y$-order of $S_k$ is $k\!+\!2$.
This is the key point in order to prove the convergence of the
expansions.

\subsection{Proof of the convergence of the expansions.}
\label{sec:ies.ce}

We want the series be analytic
in the `spatial' variables, at least until a time $t>1$ in a
neighbourhood of the zero-section. Now, the analysis is local, and we shall
prove it using majorant estimates.

Recall that for any two functions $f(z)$, $g(z)$ ($z=(z_1,\dots,z_m)$)
analytic at $z=0$:
\[
   f(z)= \sum_n f_n z^n,\ g(z)= \sum_n g_n z^n
\]
(using multi-index notation), we say that $g$ is a majorant for $f$
($f \ll g$) iff $\forall n\ |f_n|\leq g_n$.

\begin{Lemma}
The relation $\ll$ satisfies the following properties:
\begin{enumerate}
\item $\disp f_1\ll g_1,\ f_2\ll g_2\ \Rightarrow\
      f_1+f_2\ll g_1+g_2,\ f_1 g_1\ll g_1 g_2$;

\item $\disp f\ll g \ \Rightarrow \ \frac{\partial f}{\partial z_i}\ll
                          \frac{\partial g}{\partial z_i}\ (i= 1\div m)$;

\item $\forall t\in[a,b]\ \disp f_t \ll g_t\ \Rightarrow\
       \int_a^b f_t(z) dt \ll \int_a^b g_t(z) dt$.
\end{enumerate}

Let $w_b$ be the product $w_b(z)= \prod_{i=1}^m (b-z_i)$, where $b>0$. Hence:
\begin{enumerate}
\setcounter{enumi}{3}
\item $\forall i,k=1\div m$
\[
      1\ll \frac{b}{b-z_i}\ ,\ \frac{z_i}{w_b}\ll \frac{b}{w_b}\ ,
      \ \frac{1}{(b-z_1)\dots (b-z_k)}\ll\frac{b^{m-k}}{w_b};
\]
\item $\disp \forall z\ |\ \norm{z}_\infty <b,\ |f(z)|<c, \ \Rightarrow\
f\ll \frac{cb^m}{w_b}$.
\end{enumerate}
\end{Lemma}

We complete the previous rather standard lemma with a more specific one.

\begin{Lemma}
Take $m= 2d, z=(x_1,\dots,x_d,y_1,\dots,y_d), w(z)=
\prod_{i=1}^d (b-x_i)(b-y_i)$. Then:
\begin{enumerate}
\setcounter{enumi}{5}
\item $\disp f\ll\Prod{u}{i} w^{-k}, g\ll\Prod{v}{j} w^{-l}
\ (u>0,v>0) \Rightarrow $
\beqn
   \{f,g\} &{\ll}& b^{2d-1}
   \left( \begin{array}{c}
      \disp c_{uvkl} \Prod{u}{i} \cdot \sum_{s=1}^v \PROD{v}{j}
      \\ + \\
      \disp c_{uvlk} \Prod{v}{j} \cdot \sum_{s=1}^u \PROD{u}{i}
   \end{array} \right) w^{-(1+k+l)},
\eeqn
where $c_{uvkl}= k+\frac{2dkl}{u+v}$;

\item $\disp f\ll\Prod{u}{i} w^{-k} \ (u>1) \Rightarrow
 \L_|^{-1}(f) \ll \frac{1}{u-1}\Prod{u}{i} w^{-k}$.
\end{enumerate}
\end{Lemma}
\bproof
\begin{enumerate}
\setcounter{enumi}{5}
\item
Since
\beqn \disp
      \sum_{j=1}^d \frac{\partial f}{\partial x_j}
               \frac{\partial g}{\partial y_j}
         &{\ll}&
            \sum_{j=1}^d
            \left(
             \Prod{u}{i} \frac{kw^{-k}}{b-x_j}
                  \left(
            \frac{\partial}{\partial y_j}\left(\Prod{v}{j}\right) w^{-l} +
            \Prod{v}{j} \frac{l w^{-l}}{b-y_j}
                  \right)
            \right)
\eeqn
and
\beqn
\disp \sum_{j=1}^d \frac{\partial}{\partial y_j}\left(\Prod{v}{j}\right)
   = \disp \sum_{s=1}^v \PROD{v}{j},
\eeqn
then
\beqn
 \disp \{f,g\} &{\ll}& b^{2d-1}
        \left(
        \left(\begin{array}{c}
         \disp l \Prod{v}{j} \cdot \sum_{s=1}^u \PROD{u}{i} \\
      +                                                    \\
         \disp k \Prod{u}{i} \cdot \sum_{s=1}^v \PROD{v}{j}
             \end{array}
        \right)
        +
        \frac{2dkl}{b} \Prod{u}{i} \Prod{v}{j}\right) w^{-(1+k+l)}
    \\ \\
 &{\ll}& b^{2d-1}
   \left( \begin{array}{c}
      \disp c_{uvkl} \Prod{u}{i} \cdot \sum_{s=1}^v \PROD{v}{j}
      \\ + \\
      \disp c_{uvlk} \Prod{v}{j} \cdot \sum_{s=1}^u \PROD{u}{i}
   \end{array} \right) w^{-(1+k+l)}.
\eeqn
\item Finally,
\beqn
   \disp \L_|^{-1}(f) = \int_0^1 t^{-2} f(x,ty) dt
   {\ll} \int_0^1 t^{-2} \prod_{s=1}^u y_{i_s} t^u w^{-k}(x,y) dt
   = \frac{1}{u-1} \prod_{s=1}^u y_{i_s} w^{-k}.
\eeqn
\end{enumerate}
\eproof

Finally, we are going to prove the convergence of the expansions until
a large enough time.

\begin{Proposition}
Let $S:\N\to\nr$ be a real analytic function, with $\dif S\comp z= 0$.
Then,
there exists a tubular neighbourhood of the zero-section where the
solution of the evolution problem $S_t$ is defined until a time $t>1$.
\end{Proposition}
\bproof
We can use cotangent coordinates $(x,y)$ in a neighbourhood of
each point of the zero-section.
It is sufficient to prove that we can get a small neighbourhood of zero
where the series $\sum_{k\geq 0} S_k(x,y) t^k$ is defined for $t<T$ and
$T\geq 1$.

$S(x,0)$ is constant, and we suppose that this constant is $0$.
Moreover, since $\forall x\ \frac{\partial S}{\partial y}(x,0)= 0$
we can write
\beqn
        S(x,y)  =  \sum_{i,j} y_i y_j s_{ij}(x,y),
\eeqn
where de functions $s_{ij}$ are real analytic (and $s_{ij}= s_{ji}$).
Given a small enough radius $b>0$, let $c$ be the maximum of the sup-norms
of the functions $s_{ij}$ on $\{\norm{(x,y)}_\infty< b\}$.
So then, $\forall i,j=1\div d\ s_{ij}  {\ll}  c b^{2d} w^{-1}$,
where $w(x,y)= \prod_{i=1}^{d} (b-x_i)(b-y_i)$, and
\beqn
        S_0 {\ll} c b^{2d} \sum_{i_1,i_2} y_{i_1} y_{i_2}\ w^{-1},
\eeqn

Suppose that $\forall u\leq n$
\beqn
     S_u {\ll} \gamma_u \sum_{i_1,\dots,i_{u+2}} \Prod{u+2}{i}\ w^{-(2u+1)},
\eeqn
where $\gamma_0= c b^{2d}, \gamma_1,\dots,\gamma_n$ are constants.
We want to estimate $S_{n+1}$. So then, applying the majorant
estimates of the previous lemma, we obtain
\beqn
   \disp
   S_{n+1} &=& \frac{-1}{n+1} \sum_{u+v= n} \{\L_|^{-1}(S_u),S_v\}
   \\ \\
          &{\ll}&
   \frac{b^{2d-1}}{n+1} \disp
   \sum_{\scriptsize \begin{array}{c}  u + v= n \\
                                       i_1, \dots, i_{u+2}
                                     \\ j_1, \dots, j_{v+2}
        \end{array} }
        \frac{\gamma_u \gamma_v}{u+1}
           \left(\begin{array}{c}
        \disp {\hat c}_{uv} \Prod{u+2}{i}\cdot \sum_{s=1}^{v+2} \PROD{v+2}{j}
              \\ + \\
        \disp {\hat c}_{vu} \Prod{v+2}{j}\cdot \sum_{s=1}^{u+2} \PROD{u+2}{i}
           \end{array}\right)
           w^{-(2n+3)},
\eeqn
where ${\hat c}_{uv}= c_{(u+2),(v+2),(2u+1),(2v+1)}=
(2u+1) + \frac{2d(2u+1)(2v+1)}{u+v+4}$.
Since
\beqn
  \disp
  \sum_{\scriptsize \begin{array}{l} i_1,\dots,i_{u+2}
        \\ j_1,\dots,j_{v+2} \end{array}}
  \left(\Prod{u+2}{i}\cdot \sum_{s=1}^{v+2} \PROD{v+2}{j} \right)
  &=& d(v+2) \sum_{k_1,\dots,k_{n+3}} \Prod{n+3}{k},
\eeqn
we reach to
\beqn
    S_{n+1} {\ll}
    \gamma_{n+1} \sum_{k_1,\dots,k_{n+3}} \Prod{n+3}{k}\ w^{-(2n+3)},
\eeqn
where
\[
\disp
    \frac{d b^{2d-1}}{n+1}
     \sum_{u+v= n} \frac{\gamma_u \gamma_v}{u+1}
        ({\hat c}_{uv} (v\!+\!2) + {\hat c}_{vu} (u\!+\!2))
    \leq
     4d(1+2d)b^{2d-1} \sum_{u+v= n} \gamma_u \gamma_v \deq \gamma_{n+1}.
\]

Hence, we have majored the sequence of $S_n$ by
\beqn
    S_n {\ll} \gamma_n \sum_{i_1,\dots,i_{n+2}} \Prod{n+2}{i}\ w^{-(2n+1)},
\eeqn
where the sequence $\{\gamma_n\}$ is defined by
\[
     \left\{
        \begin{array}{l}
        \gamma_0= cb^{2d},  \\
        \displaystyle
        \gamma_{n+1}= K \sum_{u+v= n}  \gamma_u \gamma_v,
        \end{array}
     \right.
\]
where $K= 4d(1+2d)b^{2d-1}$. They are the coefficients of the Taylor
series of the function
\[
        f(t) = \frac{1-\sqrt{1-4K\gamma_0 t}}{2Kt},
\]
and then $\lim_n \frac{\gamma_n}{\gamma_{n+1}}= \frac{1}{4 K \gamma_0}$.

Finally, let $\rho\in [0,1[$ be a ratio we shall choose later.  If
$\norm{x}_{\infty}\leq\rho b$ and $\norm{y}_{\infty}\leq\rho b$, then
\beqn
        |S_n(x,y)| {\leq} \gamma_n (d\rho b)^{n+2} (b(1-\rho))^{-2d(2n+1)}
                    \deq \beta_n.
\eeqn
Therefore, we have bounded
all the terms of the expansion in a domain of $x,y$:
\beqn
        \sum_{n\geq 0}|S_n(x,y)| t^n {\leq} \sum_{n\geq 0} \beta_n t^n.
\eeqn
Then, as
\beqn
    \lim_n \frac{\beta_{n}}{\beta_{n+1}}
    = \frac{b^{4d-1}(1-\rho)^{4d}}{4K\gamma_0 d\rho},
\eeqn
the convergence radius of $\sum_{n\geq 0} \beta_n t^n$ is greater than $1$
provided that $\rho$ is small enough.
\eproof

\subsection{End of proof of the Theorem.}
\label{sec:ies.sip}

Let $S_t$ be the solution of the evolution problem, that belongs
to $\L(\cal F)$. Then $\bar H_t= \L_\mid^{-1}(S_t)$ is a Hamiltonian
whose flow
$\varphi_t= \varphi_{t,0}$ has primitive function $t\ S$.
Since $\dif S_t\comp z= 0$ then $\dif\bar H_t\comp z= 0$, and all the points
of the zero-section are fixed. Hence, $\varphi_1$ is {\em the} exact
symplectomorphism whose primitive function is $S$ and fixes all the points
of the zero-section.

Finally, if the dynamics {\em on} the
zero-section can be interpolated by a flow, then the dynamics {\em around}
the zero-section can be interpolated by a Hamiltonian flow.
It is enough to lift the dynamics on the zero-section and apply
\refsec{ies.sup.gs}, \refsec{ies.sup.fep}.


\section*{Conclusion}

As we have seen, the primitive function of an exact symplectomorphism gives
us some information about it, but not all the information. In order to
obtain all the information we need, from a geometrical point of view,
where and how our symplectomorphism sends an exact Lagrangian manifold.

The primitive function can be useful
from a methodological point of view, because it let us obtain {\em all} the
symplectic dynamics around exact Lagrangian manifolds. Dynamics
which are not necessarily generated by generating functions. See the
appendix to make effective computations of these dynamics.

As a corollary of our results, the dynamics of a
symplectomorphism around an invariant torus
whose dynamics is conjugated to an ergodic translation is homologous to the
identity, and the Hamiltonian can be chosen periodic in time
\cite{IAMAF}.

\appendix
\section*{Appendix \\ Effective computation.}
Assume that our manifold $\M$ is $\nr^d$ and
$S$ is a function on $T^*\nr\simeq\nr^d\times\nr^d$ that expands
\[
        S(x,y)= \sum_n s_n(x) y^n,
\]
where the $s_n$ are $x$-functions (we use multi-indices
$n= (n_1,\dots,n_d)\in\nn^d$).
We look for the only symplectomorphism $F=(f,g)$ that fixes all the
points of the zero-section $\{y=0\}$ and whose primitive function is just $S$.
We expand $f$ and $g$ by
\[
    \disp f(x,y)= \sum_n f_n(x) y^n \ ,\
    \disp g(x,y)= \sum_n g_n(x) y^n,
\]
where
$f_n= (f_n^1,\dots,f_n^d)^\top,\ g_n= (g_n^1,\dots,g_n^d)^\top$
are vector functions,
being $f_0= x$ and $g_0= 0$.

We equate the terms of the same $y$-order in the exactness equation
\cite{Thesis} for details).  Order zero gives 
that $s_0$ is constant and each $s_{e_i}$ vanishes:
$\Dif S (x, 0) = 0$ (we already knew this). Then, we obtain 
the next recurrence for 
the $x$-functions $f_n$ and $g_n$ 
(where $\sum_i$ means
$\sum_{i=1}^d$ , $u,v\in\nn^d$ are multi-indices, $|n|= n_1+\dots+n_d$ and
terms with `wrong' multi-indices are taken zero):
\begin{itemize}
\item (Step $1$)
$\forall i, j=1\div d\ \
    g_{e_i}^j= \delta_{ij}\ ,\ f_{e_i}^j= (1+\delta_{ij})\ s_{e_i+e_j}$.
\item (Step $k$)
$\forall |n|= k,\ \forall j= 1\div d\ \
   g_n^j= G_n^j\ ,\
   f_n^j= \frac{n_j+1}{k} \sum_i F_{n+e_j-e_i}^i - F_n^j, $
where
\beqn
   G_n^j & = & \frac{\partial s_n}{\partial x_j}(x) -
     \sum_i \sum_{\scriptsize \begin{array}{c} u+v= n \\ u\neq 0,n\end{array}}
               \frac{\partial f_u^i}{\partial x_j}(x) g_v^i(x), \\ \\
   F_n^j & = & (n_j+1) s_{n+e_j}(x) -
    \sum_i \sum_{\scriptsize \begin{array}{c} u+v= n \\ |v|> 1 \end{array}}
         (u_j+1) f_{u+e_j}^i(x) g_v^i(x).
\eeqn
\end{itemize}


\irem{If $\M= \nt^d$ the functions $s_n$, $f_n$ and $g_n$
1-periodic in all their variables (save $f_0(x)= x$) and, hence, can be 
expanded in Fourier series.
An example is that the dynamics on the 
zero-section $\nt^d\times \{0\}$
is an ergodic shift $\phi(x)= x + \omega$, where $\omega\in\nr\setminus\nq$.}

\lrem{
This method does not uses the implicit function theorem,
as when one tries to obtain a symplectomorphism from a generating function.
Moreover, if the primitive function is real analytic, the expansions of $f$
and $g$ also converge.
}


\section*{Acknowledgements}
I specially thank C. Sim\'o for his suggestions, helpful discussions and
encouragement.

The research has been
supported by DGICYT grant PB 94--0215 (Spain). Partial support of the
EC grant ER\-BCHRXCT\-940460, and the catalan grant CIRIT 1996S0GR-00105
also is acknowledged.

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\end{document}
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