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\begin{document}
\title[Connecting orbits]{Connecting orbits between static
classes for generic Lagrangian systems}

\author[G. Contreras]{Gonzalo Contreras}\thanks{Gonzalo Contreras was
partially supported by CNPq-Brasil and CONACYT-M\'exico.}
\address{CIMAT \\
         A.P. 402, 3600 \\
         Guanajuato. Gto. \\
         M\'exico.}
\email{gonzalo@@cimat.mx}


\author[G. P. Paternain]{Gabriel P. Paternain}
\address{Centro de Matem\'atica\\
Facultad de Ciencias\\
Igu\'a 4225\\
11400 Montevideo\\
Uruguay}
\email {gabriel@@cmat.edu.uy}
\curraddr{CIMAT \\
         A.P. 402, 3600 \\
         Guanajuato. Gto. \\
         M\'exico.}
\email{paternain@@cimat.mx}

\subjclass{58F05, 58F11}



\date{1999}
%\maketitle
\begin{abstract} Let $L$ be a $C^{\infty}$ convex superlinear Lagrangian
on a closed manifold $M$. We show that if the number
of static classes is finite, then
there exist chains of semistatic orbits that connect
 any two given static classes.
Using this property we show that if there is only
one static class, then
the homoclinic orbits to the set of static orbits
 generate over $\re$ the relative homology of the pair
$(M,U)$, where $U$ is a sufficiently small connected 
neighborhood of the set of static orbits in $M$.

We show that generically in the sense of Ma\~n\'e \cite{Ma} the set
of semistatic orbits coincides with the support of a uniquely minimizing
 measure, therefore generically, the homoclinic orbits to the support
 of the minimizing measure
 generate over $\re$ the relative homology of the pair
$(M,U)$, where $U$ is a sufficiently small connected
neighborhood of the projection
of the support of the measure to $M$.
 
This last result was obtained -with a different proof- by S. Bolotin in \cite{Bo1,Bo2} assuming the existence of a $C^{1+Lip}$ function
$f:M\to \re$ such that $L+c-df\geq 0$, where $c$ is the critical value
of $L$.


Finally we obtain two consequences.
The first one says that if $M$ is a closed
 manifold with first Betti
number $\geq 2$ then there exists a generic set ${\cal O}\subset C^{\infty}(M,\re)$
such that if $\psi\in{\cal O}$ the Lagrangian $L+\psi$ has a unique
minimizing measure and this measure
is uniquely ergodic. When this measure is supported on a periodic orbit, this
orbit is hyperbolic and the stable and unstable manifolds have transverse homoclinic intersections. 
The second consequence says that if $M$ is a closed manifold with
first Betti number different from zero and if
$L$ is a symmetric Lagrangian, then there exists a generic
set ${\cal O}\subset C^{\infty}(M,\re)$
such that if $\psi\in{\cal O}$, then $L+\psi$ has a unique
minimizing measure and this measure
is supported on a hyperbolic fixed point
whose stable and unstable manifolds have transverse
homoclinic intersections. 

\end{abstract}

\maketitle

\section{Introduction}

Let $M$ be a closed connected smooth manifold and let $L:TM\to \re$
be a smooth convex superlinear Lagrangian. This means that $L$ restricted
to each $T_{x}M$ has positive definite Hessian and that for some Riemannian
metric we have that
\[\lim_{|v|\to \infty}\frac{L(x,v)}{|v|}=\infty,\]
 uniformly on $x\in M.$
Since $M$ is compact, the extremals of $L$ give rise to a complete flow
$f_{t}:TM\to TM$ called the Euler-Lagrange flow of the Lagrangian.
The extremals are solutions of the Euler-Lagrange equation
which in local coordinates is given by

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{d}{dt}\frac{\partial L}{\partial v}=\frac{\partial L}{\partial x}.
\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mbox{(E-L)}$$
The {\it energy} $E:TM\rightarrow \re$ is defined by
\[E(x,v)=\frac{\partial L}{\partial v}(x,v).v-L(x,v).\]
Since $L$ is autonomous, $E$ is a first integral of the flow $f_{t}$.

Recall that the action of the Lagrangian $L$ on
 an absolutely continuous curve $\ga:[a,b]\rightarrow M$ is defined by
\[A_{L}(\ga)=\int_{a}^{b}L(\ga(t),\dot{\ga}(t))\,dt.\]
Given two points, $x_{1}$ and $x_{2}$ in $M$ and $T>0$
 denote by ${\cal C}_{T}(x_{1},x_{2})$ the set of absolutely continuous curves $\ga:[0,T]\rightarrow M$, with $\ga(0)=x_{1}$ and $\ga(T)=x_{2}$.
For each $k\in \re$ we define the {\it action potential} $\Phi_{k}:M\times M\rightarrow \re$ by
\[\Phi_{k}(x_{1},x_{2})=\inf\{A_{L+k}(\ga):\;\ga\in \cup_{T>0}{\cal C}_{T}(x_{1},x_{2})\}.\]

The {\it critical value of $L$}, which was introduced
 by Ma\~n\'e in \cite{Ma}, is the real number $c(L)$ defined as
the infimum of $k\in\re$ such that for some $x\in M$, $\Phi_{k}(x,x)>-\infty$.
Since $L$ is convex and superlinear and $M$ is compact such a number exists
 and it has various important properties that we review in Section 2.
We briefly mention a few of them since we shall need them below.
For any $k\geq c(L)$, the action potential $\Phi_{k}$ is a Lipschitz
function that satisfies a triangle inequality. In general the action potential
is not symmetric but if we define $d_{k}:M\times M\to\re$ by setting
\[d_{k}(x,y)=\Phi_{k}(x,y)+\Phi_{k}(y,x),\]
then $d_{k}$ is a distance function for all $k>c(L)$ and a pseudo-distance
for $k=c(L)$.

Since $d_{k}\geq 0$, for every absolutely continuous curve $\ga:[a,b]\to M$
and all $k\geq c(L)$ we have
\begin{equation}
A_{L+k}(\ga)\geq \Phi_{k}(\ga(a),\ga(b))\geq -\Phi_{k}(\ga(b),\ga(a)).\label{se}
\end{equation}
Set $c=c(L)$. We say that an absolutely continuous curve
 $\ga:[a,b]\to M$ is {\it semistatic} if
\[A_{L+c}(\ga|_{[t_{0},t_{1}]})=\Phi_{c}(\ga(t_{0}),\ga(t_{1})),\]
for all $a<t_{0}\leq t_{1}<b$; and that it is {\it static}
if 
\[A_{L+c}(\ga|_{[t_{0},t_{1}]})=-\Phi_{c}(\ga(t_{1}),\ga(t_{0})),\]
for all $a<t_{0}\leq t_{1}<b$.
Clearly, by (\ref{se}) a static curve is semistatic.
One could also say that $\ga|_{[a,b]}$ is static
if it is semistatic and for all $a<t_{0}\leq t_{1}<b$,
 $d_{c}(\ga(t_0),\ga(t_1))=0$.
Semistatic curves are solutions of the Euler-Lagrange equation because
of their minimizing properties. Also it is not hard
to check that semistatic curves have energy precisely $c$ \cite{Ma, CDI}.
The notions of semistatic and static curves are closely related to Mather's notions
of {\it c-minimal trajectories} and {\it regular c-minimal trajectories} respectively, see \cite{M'}.

Given a vector $v\in TM$ we shall denote by $x_v:\re\to M$ the
solution of the Euler-Lagrange equation with $\dot{x}(0)=v$.

The set of vectors $v$ in $TM$ that give rise to static
 curves $x_v:\re\to M$ is a closed invariant set
 that we shall denote by $\hS:=\hS(L)$.
Similarly, 
the set of vectors $v$ in $TM$ that give rise to semistatic
 curves $x_v:\re\to M$ is a closed invariant set that
 we shall denote by $\Sigma:=\Sigma(L)$. The set $\hS$ is chain recurrent
 and the set $\Sigma$ is chain transitive \cite[Theorem V]{Ma,CDI}.
As we mentioned before $\hS\subset \Sigma$.
We need to recall (cf. Section 3, Theorem \ref{GraphProperty}) the following important {\it Lipschitz graph property} which was shown in \cite{Ma, CDI} and \cite[Theorem 6.1]{M'} that generalizes the celebrated
Lipschitz Graph Theorem of Mather \cite{M}: the set $\hS$ is a Lipschitz graph, that is, if $\pi:TM\to M$
denotes the canonical projection then the map $\pi|_{\hS}:\hS\to \pi(\hS)$ is bijective with Lipschitz inverse.
Using the graph property we can define an equivalence relation
in $\hS$ by saying that two vectors $v$ and $w$ in $\hS$ are equivalent iff $d_{c}(\pi(v),\pi(w))=0$.
The equivalence relation breaks $\hS$ into classes that we shall
call {\it static classes}.
 Let $\boldsymbol{\La}$ be the set of static classes.
Define a reflexive partial order $\preceq$ in $\boldsymbol{\La}$ by
 \begin{itemize}
 \item[(a)] $\preceq$ is reflexive.
 \item[(b)] $\preceq$ is transitive.
 \item[(c)] If there is $v\in\Sigma$ with the $\alpha$-limit set $\a(v)\subseteq\La_i$ and $\om$-limit set
            $\om(v)\subseteq\La_j$, then $\La_i\preceq\La_j$.
 \end{itemize} 


\vspace{.5cm}

\noindent {\bf Theorem A.} {\it Suppose that the number
of static classes is finite.
Then given $\La_i$ and $\La_j$ in $\boldsymbol{\La}$,
we have that $\La_i\preceq\La_j$.
  }

\vspace{.5cm}

\begin{figure}

\special{psfile=homfig2.eps hoffset=70 voffset=-170
 hscale=70 vscale=50 angle=0}

\vspace{7cm} 

\caption{Connecting orbits between static classes. The three closed
curves represent the static classes and the other curves represent
semistatic orbits connecting them.}

\label{fig1}

\end{figure}
 

Theorem A could be restated by saying that if
 the cardinality of $\boldsymbol{\La}$ is finite, then given two static
classes $\La_i$ and $\La_j$ there exist classes $\La_i=\La_1,\dots,\La_n=\La_j$
and semistatic vectors $v_1,\dots,v_{n-1}\in \Sigma$ such that
for all $1\leq k\leq n-1$ we have that $\alpha(v_{k})\subseteq \La_k$
and $\omega(v_k)\subseteq\La_{k+1}$.
In other words, between two static classes there exists a chain of static classes connected by heteroclinic semistatic orbits
(cf. Figure \ref{fig1}).

Let us assume now that $\hS$ contains only {\it one} static class.
We shall see in Section 3 (cf. Proposition \ref{ConnectedClass})
 that the static
classes are always connected, thus if we assume
 that there is only
 one static class, $\hS$ must be connected.

Given $\varepsilon>0$, let $U_{\varepsilon}$ be the $\varepsilon$-neighbourhood of $\pi(\hS)$. Since $\hS$ is connected,
the open set $U_{\varepsilon}$ is connected for $\varepsilon$ sufficiently small. 
 Let $H_{1}(M,U_{\varepsilon},\re)$ denote the first
 relative singular homology group of the pair $(M,U_{\varepsilon})$ with real coefficients. 

We shall say that an orbit of $L$ is {\it homoclinic} to a closed invariant set $K\subset TM$ if
its $\alpha$ and $\omega$-limit sets are contained in $K$.


Observe that to each homoclinic orbit $x:\re\to M$
 to the set of static orbits $\hS$ we can associate
a homology class in $H_{1}(M,U_{\varepsilon},\re)$. 
Indeed, since there exists $t_{0}>0$ such that
for all $t$ with $|t|\geq t_{0}$, $x(t)\in U_{\varepsilon}$, the class of
$x|_{[-t_{0},t_{0}]}$ defines an element in $H_{1}(M,U_{\varepsilon},\re)$.
Let us denote by ${\cal H}$ the subset of $H_{1}(M,U_{\varepsilon},\re)$ given by all the classes corresponding to homoclinic orbits to $\hS$.

In Section 4 we shall show the following result.

\vspace{.5cm}

\noindent {\bf Theorem B.} {\it Suppose that $\hS$ contains
only one static class.
 Then for any $\varepsilon$ sufficiently small
the set ${\cal H}$ generates over $\re$ the relative homology
$H_{1}(M,U_{\varepsilon},\re)$.
In particular, there exist at least ${\rm \dim}\,H_{1}(M,U_{\varepsilon},\re)$
 homoclinic orbits to the set of static orbits $\hS$.
  }

\vspace{.5cm}
 





In \cite{Ma1}, Ma\~n\'e introduced the concept of {\it generic property
of a Lagrangian }. A property $P$ is said to be {\it generic} for the
Lagrangian $L$ if there exists a generic set ${\cal O}$ (in the Baire sense) of the set $C^{\infty}(M,\re)$ of all $C^{\infty}$ functions from $M$ to $\re$
such that if $\psi\in{\cal O}$ the Lagrangian $L+\psi$ has the property $P$.
One of Ma\~n\'e's objectives in \cite{Ma1} was to show that Mather's
theory of minimizing measures becomes much more accurate and stronger
if one searches for generic properties.

Our next result describes a generic property of Lagrangians
on closed manifolds.
Let ${\cal M}(L)$ be the set of probabilities on the Borel $\sigma$-algebra of $TM$
 that have compact support and are invariant under the flow $f_{t}$.
We shall say that a measure $\mu\in {\cal M}(L)$ is {\it minimizing}
if
\[\int L\,d\mu=-c.\]
We shall denote by ${\cal M}^{0}(L)$ the set of minimizing measures. We say that a measure $\mu$ is {\it uniquely minimizing}
if the set ${\cal M}^{0}(L)$ contains $\mu$ only. 
It was shown in \cite{Ma,CDI} that a measure $\mu$ is minimizing if and only
if the support of $\mu$ is contained in $\hS$. Mather has shown in
\cite{M'} that if $\mu$ is a minimizing measure then its support
is contained in the set of Mather's regular c-minimal curves.

The following important generic property was proved in \cite{CDII} and
\cite{Ma1}.  Given a
Lagrangian $L$ there exists a generic set ${\cal O}\subset C^{\infty}(M,\re)$
such that if $\psi\in{\cal O}$ the Lagrangian $L+\psi$ has a unique
minimizing measure in ${\cal M}^{0}(L+\psi)$ and this measure
is uniquely ergodic. When this measure is supported on a periodic orbit, this
orbit is hyperbolic and {\bf if} the stable and unstable manifolds intersect, they must do it transversally. 
It is conjectured in \cite{Ma} that the unique
minimizing measure in ${\cal M}^{0}(L+\psi)$ is always supported on a periodic orbit.

We will prove in Section 5:

\vspace{.5cm}

\noindent {\bf Theorem C.} {\it  Let
    $$
    \cG_2 := \{\, \psi\in C^\infty(M,\re)\,\vert\,
    \M^{0}(L+\psi)=\{\mu\} \text{ and } \supp(\mu)=\hS(L+\psi)=\Sigma(L+\psi)\,\}.$$
Then,
\begin{itemize}
    \item[(a)] $\cG_2$ is generic in $C^\infty(M,\re)$.
    \item[(b)] If $\psi_0\in\cG_2$, then 
               $\lim_{\psi\to\psi_0}d_{H}\big(\hS(L+\psi),\,\hS(L+\psi_0)\big)=0$
               where $d_{H}$ is the Hausdorff metric between compact subsets of $TM$.
    \item[(c)] If $\psi\in C^\infty(M,\re)$, $\mu_\psi\in\M^{0}(L+\psi)$
               and $\psi_0\in\cG_2$, then
               $$
               \lim_{\psi\to\psi_0}d_{H}\big(\supp(\mu_\psi)\;,\supp(\mu_{\psi_0})\big)=0.
               $$
               \end{itemize}}


\vspace{.5cm}

Note that since $\mu$ is also uniquely ergodic, the set ${\rm supp}(\mu)$
must be a static class.
Therefore, generically, the set of static orbits contains only one
static class and it
coincides with the support of the uniquely minimizing measure.

Let us denote by $U_{\varepsilon}$ the $\varepsilon$-neighborhood
of the set ${\rm supp}(\mu)$.
From Theorems B and C we obtain right away the following
generic property.

\vspace{.5cm}

\noindent {\bf Corollary 1.} {\it Given a
Lagrangian $L$ there exists a generic set ${\cal O}\subset C^{\infty}(M,\re)$
such that if $\psi\in{\cal O}$ the Lagrangian $L+\psi$ has a unique
minimizing measure $\mu$ in ${\cal M}^{0}(L+\psi)$ and this measure
is uniquely ergodic.
For any $\varepsilon$ sufficiently small
the set ${\cal H}$ of homoclinic orbits to ${\rm supp}(\mu)$ generates over $\re$ the relative homology
$H_{1}(M,U_{\varepsilon},\re)$.
In particular, there exist at least ${\rm \dim}\,H_{1}(M,U_{\varepsilon},\re)$
 homoclinic orbits to ${\rm supp}(\mu)$. 
  }

\vspace{.5cm}

S. Bolotin has shown in \cite[Theorem 3.4]{Bo1} and
\cite[Theorem 4.3]{Bo2} (cf. also \cite{Bo3})
 that if there exists a $C^{1+Lip}$ function
$f:M\to \re$ such that
\[L+c-df\geq 0,\]
then the set ${\cal H}$ of homoclinic orbits to ${\rm supp}(\mu)$ generates over $\re^+$ the relative homology
$H_{1}(M,U_{\varepsilon},\re)$.
In particular, he gets at least $2\,{\rm \dim}\,H_{1}(M,U_{\varepsilon},\re)$
 homoclinic orbits to ${\rm supp}(\mu)$, twice as much as we do in Corollary 1.
However we do not know if his condition
is generic.

Bolotin uses methods different from ours. To prove Theorem B
we consider finite coverings $M_{0}$ of $M$ whose group of deck transformations
is given by the quotient
of $H_{1}(M,U_{\varepsilon},\ze)/(\mbox{\rm torsion})$
 by a finite index 
subgroup.
Using that the lifted Lagrangian $L_{0}$ has the same critical value as $L$,
we conclude that the number of static classes of
$L_0$ must be finite.
Hence we can apply Theorem A to
 $L_{0}$ to deduce that the group generated by the homoclinic orbits to the set
of static orbits of $L$ coincides with $H_{1}(M,U_{\varepsilon},\re)$.

We note that the homoclinic orbits that we obtain in Theorem
B and Corollary 1 have energy $c$ but they are not semistatic
orbits of $L$ (cf. Theorem C). However, they are semistatic for
lifts of $L$ to suitable finite covers.


Using Corollary 1, we shall show in Section 6:

\vspace{.5cm}

\noindent {\bf Corollary 2.} {\it Let $M$ be a closed manifold with first Betti
number $\geq 2$.
 Given a
Lagrangian $L$ there exists a generic set ${\cal O}\subset C^{\infty}(M,\re)$
such that if $\psi\in{\cal O}$ the Lagrangian $L+\psi$ has a unique
minimizing measure in ${\cal M}^{0}(L+\psi)$ and this measure
is uniquely ergodic. When this measure is supported on a periodic orbit, this
orbit is hyperbolic and the stable and unstable manifolds have transverse homoclinic intersections. 
  }

\vspace{.5cm}

We say that a Lagrangian $L$ is {\it symmetric} if for all $(x,v)\in TM$,
$L(x,v)=L(x,-v)$. Note that if $L$ is symmetric and $\psi\in C^{\infty}(M,\re)$
then $L+\psi$ is also symmetric.

\vspace{.5cm}

\noindent {\bf Corollary 3.} {\it Let $M$ be a closed manifold with
first Betti number different from zero.
Given a symmetric Lagrangian $L$ there exists a generic
 set ${\cal O}\subset C^{\infty}(M,\re)$
such that if $\psi\in{\cal O}$, then $L+\psi$ has a unique
minimizing measure in ${\cal M}^{0}(L+\psi)$ and this measure
is supported on a hyperbolic fixed point
whose stable and unstable manifolds have transverse
homoclinic intersections. 
  }

\vspace{.5cm}

In \cite{Fa3} Albert Fathi has obtained independently
results which have a considerable overlap with Theorem B.
He defines a set ${\cal C}_{0}$ by
\[{\cal C}_{0}=\bigcup_{p}\,dp(\Sigma_{0}),\]
where the union is taken over all finite and
abelian Galois covers $p:M_{0}\to M$ and $\Sigma_{0}$ is the set
of semistatic orbits of the lift of $L$ to $M_{0}$.
He shows that the connected invariant set ${\cal C}_{0}$
is contained in $W^{s}(\hS)\cap W^{u}(\hS)$ and that for any
connected open set $V$ containing ${\cal C}_{0}$ one has
$H_{1}(TM,V,\ze)=0$. As a corollary, he also obtains
the existence of at least ${\rm \dim}\,H_{1}(M,U_{\varepsilon},\re)$
homoclinic orbits to the set of static orbits $\hS$ and without
assuming that $\hS$ contains only one static class.

At this point it seems useful to note
that there are various terminologies in the literature
for several of the concepts that we use here.
Fathi refers in \cite{Fa1,Fa2,Fa3,Fa4} to the closure of the union
of the support of minimizing measures as the {\it Aubry-Mather set}. 
What we call here the static and semistatic sets, Fathi calls
the {\it Peierls set} and the {\it Ma\~n\'e set} respectively.
As we mentioned before, semistatic and static curves are closely
 related to Mather's notions
of {\it c-minimal trajectories} and {\it regular c-minimal trajectories} respectively.
The terminology we follow in this paper is that of Ma\~n\'e in \cite{Ma}.


\vspace{.5cm}

{\it Acknowledgments:} The authors thank the IMPA for hospitality while this work was in progress.
G.P. Paternain thanks the ICTP for hospitality while this work was in progress.
Finally, we wish to thank the referee for pointing out a mistake in a previous
version of this manuscript.


\section{Critical values, static and semistatic curves}

Let $M$ be a closed connected manifold and
$L:TM\to\re$ a convex superlinear Lagrangian.

The action of the Lagrangian $L$ on
 an absolutely continuous curve $\ga:[a,b]\rightarrow M$ is defined by
\[A_{L}(\ga)=\int_{a}^{b}L(\ga(t),\dot{\ga}(t))\,dt.\]
Given two points, $x$ and $y$ in $M$ and $T>0$
 denote by ${\cal C}_{T}(x,y)$ the set of absolutely continuous curves $\ga:[0,T]\rightarrow M$, with $\ga(0)=x$ and $\ga(T)=y$.
For each $k\in \re$ we define the {\it action potential} $\Phi_{k}:M\times M\rightarrow \re$ by
\[\Phi_{k}(x,y)=\inf\{A_{L+k}(\ga):\;\ga\in\cup_{T>0}\, {\cal C}_T(x,y)\}.\]

\begin{Theorem}[Basic properties of the critical value \cite{Ma,CDI}]
There exists $c(L)\in \re$ such that
\begin{enumerate}
\item if $k<c(L)$, then $\Phi_{k}(x_{1},x_{2})=-\infty$, for all $x_{1}$ and $x_{2}$ in $M$;
\item if $k\geq c(L)$, then $\Phi_{k}(x_{1},x_{2})>-\infty$ for all $x_{1}$ and $x_{2}$ in $M$ and $\Phi_{k}$ is a Lipschitz function;
\item if $k\geq c(L)$, then
\[\p(x_{1},x_{3})\leq \p(x_{1},x_{2})+\p(x_{2},x_{3}),\]
for all $x_{1}$, $x_{2}$ and $x_{3}$ in $M$ and
\[\p(x_{1},x_{2})+\p(x_{2},x_{1})\geq 0,\]
for all $x_{1}$ and $x_{2}$ in $M$;
\item if $k>c(L)$, then for $x_{1}\neq x_{2}$ we have
\[\p(x_{1},x_{2})+\p(x_{2},x_{1})> 0.\]
\end{enumerate}               \label{basic}
\end{Theorem}

Observe that in general the action potential $\p$ is {\bf not} symmetric, however defining $d_{k}:M\times M\rightarrow \re$  by
\[d_{k}(x,y)=\p(x,y)+\p(y,x),\]
Theorem \ref{basic} says that $d_{k}$ is a metric for $k>c(L)$ and a pseudometric for $k=c(L)$.
The number $c(L)$ is called the {\it critical value} of $L$.

It is important for our purposes to indicate that the theorem above also hold for
coverings of $M$, i.e. suppose $\widehat{M}$ is a covering of $M$ with
 covering projection $p$. 
Take the lift of the Lagrangian $L$ to $\widehat{M}$ which is given by 
\[\widehat{L}(x,v)=L(p(x),dp(v)).\]
Then we define for each $k\in \re$ the action potential just as above
and the results hold for $\widehat{L}$. Thus we have a critical value for $\widehat{L}$.


Using the theorem it is straightforward to check that if
$M_{1}$ and $M_{2}$ are coverings of $M$ such that $M_{1}$
covers $M_{2}$, then
\begin{equation}
c(L_{1})\leq c(L_{2}),  \label{cubre}
\end{equation}
where $L_{1}$ and $L_{2}$ denote the lifts of the Lagrangian $L$ to $M_{1}$
and $M_{2}$ respectively.
Also we have the following lemma.

\begin{Lemma}
If $M_{1}$ is a {\it finite} covering of $M_{2}$ then
$c(L_{1})= c(L_{2}). $  \label{igual}
\end{Lemma}

\PROOF We know that $c(L_{1})\leq c(L_{2})$. Suppose that
the strict inequality holds and let $k$ be such that
$c(L_{1})<k<c(L_{2})$. Hence there exists a closed curve
$\ga$ in $M_{2}$ with negative $(L_{2}+k)$-action. Since $M_{1}$
is a finite covering of $M_{2}$ some iterate of $\ga$ lifts to
a closed curve in $M_{1}$ with negative $(L_{1}+k)$-action
which contradicts $c(L_{1})<k$.

\qed

Note that for every absolutely continuous curve $\ga:[a,b]\to M$
and all $k\geq c(L)$ Theorem \ref{basic} implies that:
\begin{equation}
A_{L+k}(\ga)\geq \Phi_{k}(\ga(a),\ga(b))\geq -\Phi_{k}(\ga(b),\ga(a)).\label{se'}
\end{equation}
Set $c=c(L)$. We say that an absolutely continuous curve
 $\ga:[a,b]\to M$ is {\it semistatic} if
\[A_{L+c}(\ga|_{[t_{0},t_{1}]})=\Phi_{c}(\ga(t_{0}),\ga(t_{1})),\]
for all $a<t_{0}\leq t_{1}<b$; and that is {\it static}
if 
\[A_{L+c}(\ga|_{[t_{0},t_{1}]})=-\Phi_{c}(\ga(t_{1}),\ga(t_{0})),\]
for all $a<t_{0}\leq t_{1}<b$.
Clearly, by (\ref{se'}) a static curve is semistatic.
One could also say that $\ga|_{[a,b]}$ is static
if it is semistatic and for all $a<t_{0}\leq t_{1}<b$,
 $d_{c}(\ga(t_0),\ga(t_1))=0$.
Semistatic curves are solutions of the Euler-Lagrange equation because
of their minimizing properties. Also it is not hard
to check that semistatic curves have energy precisely $c$ \cite{Ma, CDI}.

Given a vector $v\in TM$ we shall denote by $x_v:\re\to M$ the
solution of the Euler-Lagrange equation with $\dot{x}(0)=v$.

The set of vectors $v$ in $TM$ that give rise to static curves $x_v:\re\to M$ is an invariant set that we shall denote by $\hS:=\hS(L)$.
Similarly, 
the set of vectors $v$ in $TM$ that give rise to semistatic curves $x_v:\re\to M$ is an invariant set that we shall denote by $\Sigma:=\Sigma(L)$.
As we mentioned before $\hS\subset \Sigma$.
The continuity properties of $A_{L+c}$ and $\Phi_{c}$ imply that
$\Sigma$ and $\hS$ are closed sets.

\begin{Lemma}\label{lifts}
Let $p:M_{1}\to M_{2}$ be a covering such that $c(L_{1})=c(L_{2})$.
Then any lift of a semistatic curve of $L_{2}$ is a semistatic curve
of $L_{1}$. Also the projection of a static curve of $L_{1}$ is
a static curve of $L_{2}$. If in addition, $p$ is a finite covering,
then any lift of a static curve of $L_{2}$ is a static curve
of $L_{1}$.
\end{Lemma}

\PROOF Observe first that for any $k\in\re$ we have that
\[\Phi_{k}^{1}(x,y)\geq \Phi_{k}^{2}(px,py),\]
for all $x$ and $y$ in $M_{1}$. Hence if we write $c=c(L_{1})=c(L_{2})$
we have
\begin{equation}
\Phi_{c}^{1}(x,y)\geq \Phi_{c}^{2}(px,py),\label{eins}
\end{equation}
for all $x$ and $y$ in $M_{1}$.

Suppose now that $x_{2}:\re\to M_{2}$ is a semistatic curve of $L_{2}$
and let $x_{1}:\re\to M_{1}$ be any lift of $x_{2}$ to $M_{1}$.
Using (\ref{eins}) and the fact that $x_{2}$ is semistatic we
have for $s\leq t$,

\begin{align*}
\Phi_{c}^{1}(x_{1}(s),x_{1}(t)) 
    & \leq A_{L_{1}+c}(x_{1}|_{[s,t]})=A_{L_{2}+c}(x_{2}|_{[s,t]})\\
    & =\Phi_{c}^{2}(x_{2}(s),x_{2}(t))\leq \Phi_{c}^{1}(x_{1}(s),x_{1}(t)).
\end{align*}
Hence $x_{1}$ is semistatic for $L_{1}$.

Suppose now that $x_{1}:\re\to M_{1}$ is a static curve of $L_{1}$
and let $x_{2}:\re\to M_{2}$ be $p\circ x_{1}$.
Using (\ref{eins}) and the fact that $x_{1}$ is static we
have for $s\leq t$,
\begin{align*}
-\Phi_{c}^{1}(x_{1}(t),x_{1}(s))&=\Phi_{c}^{1}(x_{1}(s),x_{1}(t)) 
  = A_{L_{1}+c}(x_{1}|_{[s,t]})=A_{L_{2}+c}(x_{2}|_{[s,t]})\\
    & \geq \Phi_{c}^{2}(x_{2}(s),x_{2}(t)) \geq -\Phi_{c}^{2}(x_{2}(t),x_{2}(s)) \geq -\Phi_{c}^{1}(x_{1}(t),x_{1}(s)).
\end{align*}
Hence $x_{2}$ is static for $L_{2}$.

Suppose now that $p$ is a {\it finite} covering and let
$x_{2}:\re\to M_{2}$ be a static curve of $L_{2}$. Let
$x_{1}:\re\to M_{1}$ be any lift of $x_{2}$ to $M_{1}$. Since $x_{2}$
is static, given $s\leq t$ and $\e>0$, there exists a curve
$\alpha:[0,T]\to M_{2}$ with $\alpha(0)=x_{2}(t)$, $\alpha(T)=x_{2}(s)$
such that 
\[A_{L_{2}+c}(x_{2}|_{[s,t]})+A_{L_{2}+c}(\alpha)\leq \e.\]
Since $p$ is a finite covering, there exists a positive integer $n$,
bounded from above by the number of sheets of the covering,
such that the $n$-th iterate
of $x_{2}|_{[s,t]}*\alpha$ lifts to $M_{1}$ as a closed curve.
Hence, there exists a curve $\beta$ joining $x_{1}(t)$ to $x_{1}(s)$
such that
\[A_{L_{1}+c}(x_{1}|_{[s,t]})+A_{L_{1}+c}(\beta)\leq n\e,\]
and thus $x_{1}$ is static for $L_{1}$.

\qed

 \section{Proof of Theorem A}
  We shall endow $M$ with a Riemannian metric and we consider in $TM$
  the associated Sasaki metric. Let $d_{M}$ and $d_{TM}$ be the corresponding    distance functions of these Riemannian metrics.
  Given $v\in TM$ denote by $\a(v)$ and $\om(v)$ its $\a$ and
  $\om$-limits respectively.
  We recall the following:
  
  \begin{Lemma}[\cite{CDI}]
   If $v\in \Sigma$ is semistatic, then $\a(v)\subset\widehat{\Sigma}$
  and $\om(v)\subset \widehat{\Sigma}$. Moreover, $\a(v)$ and $\om(v)$
  are each included in a static class.\label{wlim}
  \end{Lemma}
 
 
 Set
 $$\Sigma^\e :=\{w\in TM \,\vert\, x_w:[0,\e)\to M \text{ or }
             x_w:(-\e,0]\to M \text{ is semistatic }\}.$$
 
 
 \begin{Theorem}[Graph Property, see \cite{Ma,CDI} and \cite{M'}]
  For all $p\in \pi(\hS)$ there exists a unique $\xi(p)\in T_pM$
  such that $(p,\xi(p))\in\Sigma^\e$, in particular
  $(p,\xi(p))\in\widehat{\Sigma}$ and $\widehat{\Sigma}=
  \text{graph}(\xi)$. Moreover, there exist positive constants
 $\eta$ and $K$ such that if $(p,v)\in \hS$, $(q,w)\in \Sigma^{\e}$
and $d_{M}(p,q)<\eta$ then 
\[d_{TM}((p,v),(q,w))<K\,d_{M}(p,q).\] 
In particular, the map $\xi:\pi(\hS)\to \Sigma$  
  is Lipschitz.  \label{GraphProperty}
 \end{Theorem}
  
 
 Using the Graph Property we can define an equivalence relation on $\widehat\Sigma$ by
  $$
  u,v\in \widehat{\Sigma}, \quad u\equiv v \iff d_c(\pi(u),\pi(v)) =0.
  $$
  The equivalence classes are called {\it static classes}.
  Let $\boldsymbol{\La}$ be the set of static classes. Define a
  reflexive partial order $\preceq$ in $\boldsymbol{\La}$ by
  \begin{itemize}
  \item[(a)] $\preceq$ is reflexive.
  \item[(b)] $\preceq$ is transitive.
  \item[(c)] If there is $v\in\Sigma$ with $\a(v)\subseteq\La_i$ and
             $\om(v)\subseteq\La_j$, then $\La_i\preceq\La_j$.
  \end{itemize}            
  
 
 Let us begin with the proof of the theorem.
 We shall prove in Proposition \ref{ConnectedClass} below that
 the static classes are connected. Hence if we assume
 that there are only finitely many of them, 
  the connected components of $\hS$ are
  finite and must coincide with the static classes.
  For $\e>0$, let $\hS(\e)$ be the $\e$-neighborhood of
  $\hS$, i.e.
  $$
  \hS(\e) :=\{ \, v\in TM \, \vert \,
  d_{TM}(v,\hS)<\e\,\}.
  $$
  Fix $\e>0$ small enough such that $\e<\eta$ where $\eta$ is the positive
  constant given by Theorem \ref{GraphProperty} and such that
  the connected components of
  $\hS(\e)$ are the $\e$-neighborhoods of the static classes.
  Thus for $0<\de<\e$, $\hS(\de)=\sum_{i=1}^{N(\e)}\La_i(\de)$,
  where $\La_i(\de)$ are disjoint open sets containing exactly one
  static class and the number of components $N(\e)$ is fixed for all
  $0<\de<\e$.
  
  Now suppose that the theorem is false. This means that there exists
  $\La\in\boldsymbol{\La}$ such that the following two sets are not
  empty:
  $$
  {\Bbb A}:=\bigcup_{\{\,\La_j\in\boldsymbol{\La}\,\vert\,
  \La\preceq\La_j\,\}}\La_j
  \quad , \quad
  {\Bbb B} :=\bigcup_{ \{\,\La_j\in\boldsymbol{\La}\,\vert\,
  \La\not\preceq\La_j\,\}}\La_j.
  $$
  
  Given $v\in\Sigma$ with $\a(v)\subseteq{\Bbb A}$ and $0<\de<\e$,
  define inductively $s_k(v)$, $t_k(v)$, $T_k(v)$ as follows. Let
  \begin{align*}
  s_1(v) &:= \inf \{\, s\in\re\,\vert\, f_s(v)\notin{\Bbb A}(\e)\,\}
  \in\re\cup\{+\infty\}.\\
  \intertext{If $s_k(v)<+\infty$, $k\geq 1$, define }
  t_k(v) &:= \sup\{\, t<s_k(v)\,\vert\, f_t(v)\in{\Bbb A}(\de)\,\},  \\ 
  T_k(v) &:= \inf\{\,t>s_k(v)\,\vert\,f_t(v)\in{\Bbb A}(\de)\,\}.   \\          
  \intertext{Observe that $s_k(v)<+\infty$ implies that
  $T_k(v)<+\infty$ because by the definition of ${\Bbb B}$ and the 
  transitivity of $\preceq$ we have that $\om(v)\subseteq {\Bbb A}$.
  Define}
  A_k=A_k(\de) &:= \sup \{\,\vert\,T_k(v)-t_k(v)\,\vert\;:\;
                 v\in\Sigma,\;\a(v)\subseteq{\Bbb A},\,
		 s_k(v)<+\infty\,\},
  \end{align*}
  if $s_k(v)=+\infty$ for all  $v\in\Sigma$ with $\a(v)\subseteq{\Bbb
  A}$, write $A_\ell(\de)\equiv 0$ for all $\ell\ge k$. Now set:
  \[s_{k+1}(v):= \inf\{\,s>T_k(v)\,\vert\,f_t(v)\notin{\Bbb A}(\e)\,\}.\]
  Observe that $s_k(v)$, $t_k(v)$ and $T_k(v)$ are invariant under $f_{t}$.
  
  \bigskip
  
  We split the rest of the proof of Theorem A into
  the following claims:
 
  \medskip
  
  \noindent{\bf Claim 1.} 
  {\it $A_k(\de)<+\infty$ for all $k=1,2,\ldots$ and all $0<\de<\e$.}
  
  \medskip
  
  Define
 % $$
  %{\Bbb M} := \{\, f_t(v)\,\vert\,
 % t\in\re,\;v\in\Sigma,\;\a(v)\subseteq{\Bbb A}\,\}.
 % $$
 $${\Bbb M} := \{\,v\,\vert\,\;v\in\Sigma,\;\a(v)\subseteq{\Bbb A}\,\}.
  $$
  
 \medskip
 
  \noindent{\bf Claim 2.}
  {\it\begin{itemize}
  \item[(a)] $\ov{\Bbb M}\cap{\Bbb B}\ne \nulo$.
  \item[(b)] $\limsup_{k}A_k(\de)=\sup_k A_k(\de)=+\infty$.
  \end{itemize}}
  
  \medskip
  
  \noindent{\bf Claim 3.}
  {\it There exist sequences $v_n\in\Sigma$, $0<s_n<t_n$ such that
  $v_n\to u_1\in{\Bbb A}$, $f_{s_n}(v_n)\to u_2\notin{\Bbb A}(\e)$,
  $f_{t_n}(v_n)\to u_3\in{\Bbb A}$ and $d_c(\pi u_1, \pi u_3) = 0$.} 
  
  \medskip
  
  We now use Claim~3 to complete the proof of Theorem A. If
  $u_1\in\La_j\subseteq{\Bbb A}$, we shall prove that
  $u_2\in\La_j\setminus{\Bbb A}(\e)$, obtaining a contradiction
  and thus proving Theorem A. It is enough to show
  that $d_c(\pi u_1,\pi u_2)=0$. Indeed
  \begin{align*}
  d_c(\pi u_1,\pi u_2) 
     &= \Phi_c(\pi u_1,\pi u_2) + \Phi_c(\pi u_2,\pi u_1) \\
     &\le \Phi_c(\pi u_1,\pi u_2) + \Phi_c(\pi u_2, \pi u_3)
          + \Phi_c(\pi u_3, \pi u_1) \\
     &= \lim_n\bigl[\Phi_c(\pi v_n,\pi f_{s_n}(v_n))
          + \Phi_c(\pi f_{s_n}(v_n),\pi f_{t_n}(v_n))\bigr]
          + \Phi_c(\pi u_3,\pi u_1) \\
     &=\lim_n \Phi_c(\pi v_n, \pi f_{t_n}(v_n))
          + \Phi_c(\pi u_3,\pi u_1) \\
     &= d_c(\pi u_1, \pi u_3) = 0,
   \end{align*}
   where the fourth equality holds because $v_n$ is a semistatic
   vector.
   
   \qed           
   
   
   \medskip
  
   We need the following
 
 \begin{Lemma}\cite[Corollary 1.4]{CDI}\label{bounded}
   
   There exists $A>0$ such that if $p,q\in M$ and $x\in {\cal C}_{T}(p,q)$
   satisfy
   \begin{itemize}
   \item[(a)] $A_L(x)=\min\{\,A_L(y)\,\vert\, y\in {\cal C}_{T}(p,q)\,\}$;
   \item[(b)] $A_{L+c}(x) < \Phi_c(p,q) + d_M(p,q)$,
   \end{itemize}
   then  $| \dot{x}(t)| < A$ for all $t\in[0,T]$.
   \end{Lemma} 
   
   \bigskip
   
   \noindent{\em Proof of Claim 1.}
    Suppose that $A_i<+\infty$ for $i=1,\ldots,k-1$ and $A_k=+\infty$.
   The case $k=1$ is similar. Then there exists $v_n\in\Sigma$, with
   $\a(v_n)\subset{\Bbb A}$ and $T_k(v_n)-t_k(v_n)\to +\infty$. We can
   assume that $t_k(v_n)=0$ and that $v_n$ converges ($\Sigma$ is compact). Let
   $u= \lim_n v_n\in\partial{\Bbb A}(\de)$. Then for all $n$ we have
   \begin{equation}
   m\{\, t<0\,\vert\,f_t(v_n)\notin {\Bbb
   A}(\e)\,\}\le\sum_{i=1}^{k-1}A_i\,,\label{medida}
   \end{equation}
   where $m$ is the Lebesgue measure on $\re$. We claim that
   $\a(u)\subset{\Bbb A}$. To prove the claim it suffices to show
   that there is a sequence $r_{m}\to -\infty$ such that $f_{r_{m}}(u)\in
   \overline{{\Bbb A}(\e)}$. (Recall that $\alpha(u)$ must be contained in a
   unique static class by Lemma \ref{wlim}.) Suppose that such a sequence does    not exist. This means that there exists $R\leq 0$ such that for all
   $t\leq R$, $f_{t}(u)\not\in\overline{{\Bbb A}(\e)}$. Since $v_{n}\to u$, there   exists $n$ sufficiently large for which
   $f_{t}(v_{n})\not\in\overline{{\Bbb A}(\e)}$ for all $t\in[R-\sum_{i=1}^{k-1}A   _i-2,R-1]$. This contradicts (\ref{medida}).

   Since $f_t(v_n)\notin{\Bbb A}(\e)$
   for $0<t<T_k(v_n)$ and $T_k(v_n)\to+\infty$, then
   $f_t(u)\notin{\Bbb A}(\e)$ for all $t>0$ and hence
   $\om(u)\subseteq{\Bbb B}$. But then the orbit of $u$ contradicts the
   definition of ${\Bbb B}$.
   
   \qed
   
    
   \bigskip
   
   {\noindent \em Proof of Claim 2.}
   
   \smallskip
   
   (a) Let $p\in\pi{\Bbb A}$, $q\in\pi{\Bbb B}$. For $n>0$, let
   $x_{v_n}:[a_n,b_n]\to M$ be a solution of (E-L) such that
   $x_{v_n}(a_n)=p$, $x_{v_n}(b_n)=q$ and
   $$
   A_{L+c}(x_{v_n})\le \Phi_c(p,q) + \tfrac 1n.
   $$
   This implies that
   \begin{equation}\label{E:m.1}
   A_{L+c}\left(x_{v_n}\vert_{[s,t]}\right)\le
   \Phi_c\bigl(x_{v_n}(s),x_{v_n}(t)\bigr)+\tfrac 1n
   \end{equation}
   for all $a_n\le s\le t\le b_n$. We can assume that
   $$
   \inf\{\, s>a_n\,\vert\,x_{v_n}(s)\in\pi\,{\Bbb B}(\de)\,\}=0,
   $$
   and that the sequence $v_n$ converges (cf. Lemma~\ref{bounded}). 
   Let $u=\lim_n v_n\in\pi^{-1}(\partial\,\pi\,{\Bbb B}(\de))$. 
   Taking limits in
   \eqref{E:m.1} we obtain that $x_u\vert_{[s,t]}$ is
   semistatic for all $\lim\inf_n a_n \le s\le t \le \lim\sup_n b_n$.
   
   Any limit point $w$ of $\dot{x}_{v_n}(a_n)=f_{a_n}(v_n)$ satisfies
   $\pi(w)=p\in\pi{\Bbb A}$, and by the Graph
   Property (Theorem~\ref{GraphProperty}), $w\in{\Bbb A}$. Similarly, any limit
   point of $f_{b_n}(v_n)$ is in ${\Bbb B}$. Since ${\Bbb A}\cup {\Bbb
   B}$ is invariant and $u\notin{\Bbb A}\cup{\Bbb B}$, then $\lim_n
   a_n=-\infty$, $\lim_n b_n=+\infty$. Hence $u\in\Sigma$. Since
   $f_t(v_n)\notin{\Bbb B}(\de)$ for all $a_n\le t<0$ and $a_n\to
   -\infty$, then $f_t(u)\notin{\Bbb B}(\de)$ for all $t<0$. Hence
   $\a(u)\subseteq{\Bbb A}$ and thus $u\in {\Bbb M}$. Since 
   $u\in\pi^{-1}(\partial\,\pi\,{\Bbb B}(\de))$ there exists $z\in{\Bbb B}$
   such that $d_{M}(\pi(u),\pi(z))\leq \de$. Since $z\in \hS$ and $u\in\Sigma$    by Theorem \ref{GraphProperty} we have
   \[d_{TM}((\pi(u),u),(\pi(z),z))\leq K\,\de.\]
   Thus $u\in{\Bbb M}\cap {\Bbb B}(K\,\de)$. Letting $\de\to 0$, we obtain that
   $\ov{\Bbb M}\cap{\Bbb B}\ne\nulo$.
   
                  
   \medskip               
                  
   (b) By Claim~1 it is enough to show that $\sup_k A_k(\de)=+\infty$.
   If $\sup_k A_k(\de)<T$, then ${\Bbb M}\subseteq{\Bbb M}(\de,T)$,
   where ${\Bbb M}(\de,T)$ is the compact set given by:
   $$
   {\Bbb M}(\de,T)=\{\, v\in\Sigma\,\vert\, f_{[-T,T]}(v)\cap
   \ov{{\Bbb A}(\de)}\ne \nulo\,\}=f_{[-T,T]}\left(\overline{{\Bbb A}(\de)}\cap   \Sigma\right).
   $$
   Note that ${\Bbb M}(\de,T)\cap{\Bbb
   B}=\nulo$, because ${\Bbb B}$ is invariant and ${\Bbb
   B}\cap{\Bbb A}(\de)=\nulo$. On the other hand, since ${\Bbb M}\subseteq
   {\Bbb M}(\de,T)$ we have
   $\overline{{\Bbb M}}\cap {\Bbb B}\subseteq{\Bbb M}(\de,T)
   \cap{\Bbb B}=\nulo$. 
   This contradicts item (a).
   
   \qed
   
   \medskip
   
   \noindent{\em Proof of Claim~3.}
   Given $0<\de<\e$, by Claim~2(b) there exists $k>N(\e)$ such that
   $A_k(\de)>0$. Hence there is $v=v_\de\in\Sigma$ with
   $\a(v)\subset{\Bbb A}$, such that the orbit of $v$ leaves ${\Bbb
   A}(\e)$ and returns to ${\Bbb A}(\de)$ at least $k$ times. Since
   $k>N(\e)$ there is one component $\La_j(\de)\subseteq{\Bbb A}(\de)$
   with two of these returns, i.e. there exist
   $\tau_1(\de)<s(\de)<\tau_2(\de)$ with $f_{\tau_1}(v)\in\La_j(\de)$,
   $f_s(v)\notin{\Bbb A}(\e)$ and $f_{\tau_2}(v)\in\La_j(\de)$. 
   We can choose $v_\de$ so that $\tau_1(\de)=0$. Now,
   there exists a sequence $\de_n\downarrow 0$
   such that the repeated component
   $\La_j\subset\La_j(\de_n)$ is always the same. Let
   $s_n:=s(\de_n)$, $t_n:=\tau_2(\de_n)$, $v_{n}:=v_{\de_{n}}$
   and choose a subsequence 
   such that $v_{n}$, $f_{s_n}(v_n)$ and
   $f_{t_n}(v_n)$ converge. Let $u_1=\lim_n v_n\in \cap_n
   \La_j(\de_n)=\La_j$, $u_3=\lim_n f_{t_n}(v_n)\in\La_j$ and
   $u_2=\lim_n f_{s_n}(v_n)\notin{\Bbb A}(\e)$. Since
   $u_1,u_3\in\La_j$, then $d_c(\pi u_1,\pi u_3)=0$.
   
   \qed
   
   
   \bigskip
 
  \begin{Proposition}\label{ConnectedClass}
  Every static class is connected.
  \end{Proposition}
   
   \PROOF   
  Let $\La$ be a static class and suppose that it is not connected. 
    Let $U_1,U_2$ be disjoint open sets such that $\La\subseteq U_1\cup U_2$
    and $\La\cap U_i\ne\nulo$, $i=1,2$. Let $p_i\in\pi(U_i\cap\La)$,
    $i=1,2$.
  Since $U_{1}$ and $U_{2}$ are disjoint sets we can take
  a solution
  $x_{v_n}:[a_n,b_n]\to M$, $a_n<0<b_n$ of (E-L)
  such that $x_{v_n}(0)\notin \pi(U_1\cup U_2)$,
    $x_{v_n}(a_n)=p_1$, $x_{v_n}(b_n)=p_2$ and
 \begin{equation}
    A_{L+c}(x_{v_n})\le\Phi_c(p_1,p_2)+\tfrac 1n.  \label{zwei}
 \end{equation}
    Let $u$ be a limit point of $v_n$,
    then $x_u:\re\to M$ is semistatic (see the proof of Claim~2 item (a)). 
 Then, for $a_n\le s\le t\le
    b_n$,
 $$d_c(p_1,p_2) 
      \leq\Phi_c(p_1,x_{v_n}(s))+\Phi_c(x_{v_n}(s),x_{v_n}(t))
           +\Phi_c(x_{v_n}(t),p_2)+\Phi_c(p_2,p_1),$$
 therefore
    \begin{align*}
 d_c(p_1,p_2) 
      &\leq\liminf_n\left[\Phi_c(p_1,x_{v_n}(s))
             +\Phi_c(x_{v_n}(s),x_{v_n}(t))+\Phi_c(x_{v_n}(t),p_2)\right]
             +\Phi_c(p_2,p_1)  \\
      &\leq \liminf_{n} A_{L+c}(x_{v_{n}})+\Phi_{c}(p_{2},p_{1})\\
      &\leq d_{c}(p_{1},p_{2})=0,
 \end{align*}
 where in the last inequality we used (\ref{zwei}).
 Hence
 $$\Phi_c(p_1,x_{u}(s))+\Phi_c(x_{u}(s),x_{u}(t))
           +\Phi_c(x_{u}(t),p_2)+\Phi_c(p_2,p_1)=0.$$
    Combining the last equation with the triangle inequality
 we obtain
    $$
    d_c(x_u(s),x_u(t))\le \Phi_c(x_u(s),x_u(t))
          +\left[\Phi_c(x_u(t),p_2)+\Phi_c(p_2,p_1)
          +\Phi_c(p_1,x_u(s))\right]
          =0.
    $$
  So that $u\in\widehat{\Sigma}$. Moreover, for $s=0,t=1$:
  $$
  d_c(x_u(0),p_1)\le\Phi_c(p_1,x_u(0))
        +\left[\Phi_c(x_u(0),x_u(1))+\Phi_c(x_u(1),p_2)
        +\Phi_c(p_2,p_1)\right]=0.
  $$
  Hence $x_{u}(0)\in\pi(\La)$. On the other hand $x_{u}(0)\notin \pi(U_{1}\cup U_{2})$.
  This contradicts the fact that $\La\subseteq U_1\cup U_2$.
  
  \qed      
                
          




\section{Proof of Theorem B}



Let $U\df U_{\varepsilon}$ denote the $\varepsilon$-neighborhood of $\pi(\hS(L))$,
where $\hS(L)$ is the set of static vectors of $L$.
Since we are assuming that $\hS(L)$ contains only one static class, the set $U$ is also connected
for small $\varepsilon$. Let $i:U\to M$ be the inclusion map. 
The vector space $H_{1}(M,U,\re)$ is isomorphic to the quotient
of $H_{1}(M,\re)$ by $i_{*}(H_{1}(U,\re))$.

Let $G$ be the quotient of $H_{1}(M,U,\ze)$ by its torsion part. Since $G$ 
is free we can write $G=\ze\oplus\overset{k}{\ldots}\oplus\ze$
where $k={\rm \dim}\,H_{1}(M,U,\re)$. The group $G$ can be seen as a lattice
in $H_{1}(M,U,\re)$.
Let $J$ be a finite index subgroup of $G$. 
There is a surjective homomorphism $j:G\to G/J$ given by the projection.

If we take the Hurewicz map
\[\pi_{1}(M)\mapsto H_{1}(M,\ze),\]
and we compose it with the surjective homomorphisms $H_{1}(M,\ze)\mapsto H_{1}(M,U,\ze)$, $H_{1}(M,U,\ze)\mapsto G$ and
$j:G\to G/J$, we obtain a surjective homomorphism
\[\pi_{1}(M)\mapsto G/J,\]
whose kernel will be the fundamental group of a finite Galois covering
$M_{0}$ of $M$ with covering projection map $p:M_{0}\to M$
and group of deck transformations given by the finite abelian group $G/J$.

Observe that $G/J$ acts transitively
and freely on the set of connected components of $p^{-1}(U)$ which
coincides with the set of connected components of $p^{-1}(\pi(\hS(L)))$.
Therefore we have

\begin{Lemma}There is a one to one correspondence between elements
in $G/J$ and connected components of $p^{-1}(\pi(\hS(L)))$.  \label{components}
\end{Lemma}

Observe that to each homoclinic orbit $x:\re\to M$
 to $\hS(L)$ we can associate
a homology class in $G$. Indeed, since there exists $t_{0}>0$ such that
for all $t$ with $|t|\geq t_{0}$, $x(t)\in U$, the class of
$x|_{[-t_{0},t_{0}]}$ defines an element in $H_{1}(M,U,\ze)$ and hence in $G$.
Let us denote by ${\cal H}$ the subset of $G$ given by all the classes
corresponding to homoclinic orbits to $\hS(L)$.


\begin{figure}

\special{psfile=bump1.eps hoffset=70 hscale=50 vscale=54 angle=-90}

\vspace{7cm} 

\caption{Creating homoclinic connections with finite coverings and
Theorem A.}

\label{fig2}

\end{figure}

\begin{Lemma} For any $J$ as above, the image of $\langle{\cal H}\rangle$
under $j$ is precisely $G/J$.  \label{j}
\end{Lemma}

\PROOF Let $L_{0}$ denote the lift of the Lagrangian $L$ to $M_{0}$.
Observe first that by Lemma \ref{igual}, $c(L)=c(L_{0})$ and therefore
 by Lemma \ref{lifts} we have
\begin{equation} 
\pi_{0}(\hS(L_{0}))= p^{-1}(\pi(\hS(L))),  \label{Igualdad}
\end{equation}
where $\pi_{0}:TM_{0}\to M_{0}$ is the canonical projection
of the tangent bundle $TM_{0}$ to $M_{0}$.

Let us prove now that $L_{0}$ satisfies the hypothesis of Theorem A, that
is, the number of static
classes of $L_{0}$ is finite. In fact, we shall show that the projection
to $M_{0}$ of a static
class of $L_{0}$ coincides with a connected component of
$p^{-1}(\pi(\hS(L)))$.
Using (\ref{Igualdad}) and Proposition \ref{ConnectedClass} we 
see that the projection of a static class of $L_0$ to $M_0$
must be contained in a single connected component of $p^{-1}(\pi(\hS(L)))$.
Hence, it suffices to show that if $x$ and $y$ belong to a
connected component of $p^{-1}(\pi(\hS(L)))$
then $d^{0}_{c}(x,y)=0$. Since we are assuming that $\hS(L)$ contains
only one static class we have that $d_{c}(px,py)=0$.
Since $p:M_{0}\to M$ is a finite covering there are lifts $x_{1}$ of $px$
and $y_{1}$ of $py$ such that $d_{c}^{0}(x_{1},y_{1})=0$.
Since static classes are connected $x_{1}$ and $y_{1}$ must belong to
the same connected component of $p^{-1}(\pi(\hS(L)))$ and thus
there is a covering transformation taking $x_{1}$ into $x$ and $y_{1}$
into $y$ which implies that $d^{0}_{c}(x,y)=0$ as desired.

Now Theorem A and (\ref{Igualdad}) imply that every covering transformation
in $G/J$ can be written as the composition of covering transformations
that arise from elements in ${\cal H}$, that is, 
$j(\langle{\cal H}\rangle)=G/J$.


\qed



We shall need the following algebraic lemma.

\begin{Lemma}Let $G=\ze\oplus\overset{k}{\ldots}\oplus\ze$.
Given a finite index subgroup $J\subset G$ let us denote by
$j:G\to G/J$ the projection homomorphism.

Let $A$ be a subgroup of $G$. If $A$ has the property
that for all $J$ as above $j(A)=G/J$, then
$A=G$.  \label{fato}
\end{Lemma}

\PROOF The hypothesis readily implies that
\begin{equation}
A/A\cap J\;\mbox{ is isomorphic to}\;G/J.          \label{isom}
\end{equation}
\begin{itemize}
\item If the rank of $A$ is strictly less than the rank of $G$,
one can easily construct a subgroup $J\subset G$ with finite index
such that $A\subseteq J$ and $G/J\neq\{0\}$.
But this contradicts (\ref{isom}) because $A/A\cap J=\{0\}$.
\item If the rank of $A$ equals the rank of $G$, then $A$ has finite
index in $G$ and by (\ref{isom}) $G/A=\{0\}$ and thus $G=A$.
\end{itemize}

\qed 




Observe now that any set ${\cal H}$ of a free abelian group $G$ of rank $k$ such that the group generated by ${\cal H}$ is $G$ must have at least
$k$ elements.
Therefore if we combine Lemma \ref{j} and Lemma \ref{fato} with
$\langle{\cal H}\rangle=A$ we deduce
that the set ${\cal H}$ of classes corresponding to homoclinic orbits
generates $G$ and
must have at least $k$ elements thus concluding the proof
of Theorem B.
\qed

\section{Proof of Theorem C}

To prove the theorem we shall show first several
lemmas. We will use the following notation:

 \begin{itemize}
  \item $\M^{0}(L)=$ minimizing measures of $L$;
  \item $\S(L)=$ semistatic vectors of $L$;
  \item $\hS(L)=$ static vectors of $L$;
  \item $\La(L)=$ closure of $\bigcup_{\mu\in \M^{0}(L)}\supp(\mu)$.
  \end{itemize} 

Recall that we always have $\La(L)\subseteq \hS(L)\subseteq \S(L)$. 

\begin{Lemma}The function 
$C^\infty(M,\re)\ni\psi\mapsto c(L+\psi)$ is continuous.
\label{uno}
\end{Lemma}

\PROOF Suppose that $\psi_{n}\to\psi$ and let $c_{n}:=c(L+\psi_{n})$
and $c:=c(L+\psi)$. We will prove that $c_n\to c$.

Fix $\e>0$. Since $c-\e<c$, by the definition of critical value
there exists a closed curve $\ga:[0,T]\to M$ such that
\[A_{L+\psi+c-\e}(\ga)<0,\]
hence for all $n$ sufficiently large
\[A_{L+\psi_{n}+c-\e}(\ga)<0,\]
therefore for all $n$ sufficiently large
\[c-\e<c_n,\]
and thus
\[c-\e\leq \liminf_n c_n.\]
Since $\e$ was arbitrary we have
\[c\leq \liminf_n c_n.\]

Let us show now that $\limsup_n c_n\leq c$. Suppose that
$c<\limsup_n c_n$. Take $\e$ such that
\begin{equation}
c<c+\e<\limsup_n c_n.
\label{ch}
\end{equation}
Since $\psi_{n}\to\psi$, there exists $n_0$ such that
for all $n\geq n_0$,
\begin{equation}
-\e\leq\psi_{n}-\psi\leq\e.
\label{ch1}
\end{equation}
By (\ref{ch}), there exists $m\geq n_0$ such that
\[c<c+\e<c_m.\]
By the definition of critical value
there exists a closed curve $\ga:[0,T]\to M$ such that
\[A_{L+\psi_{m}+c+\e}(\ga)<0,\]
and hence using (\ref{ch1}) we have
\[A_{L+\psi+c}(\ga)\leq A_{L+\psi_{m}+c+\e}(\ga)<0,\]
which yields a contradiction to the definition of the
critical value $c$.


\qed

This proof also shows that $L\mapsto c(L)$ is continuous
if we endow the set of Lagrangians $L$ with the
topology of uniform convergence on compact subsets of $TM$.


\begin{Lemma}$\lim_{\psi\to 0}\S(L+\psi)\subset
\S(L)$, where $\lim_{\psi\to 0}\S(L+\psi)$ is the set of accumulation
points of sequences $v_n\in\S(L+\psi_n)\subset TM$ with
$\psi_n\to 0$.
\label{dos}
\end{Lemma}


\PROOF  Let $\psi_n\to 0$ and $v_n\in \S(L+\psi_n)$ with $v_n\to v$.
     Let $T>0$ and write $x_{v_n}(t):=\pi_\circ f^n_t(v_n)$,
     $x_v(t):=\pi_\circ f_t(v)$, $x_n=x_{v_n}(-T)$, $y_n=y_{v_n}(T)$,
     $x=x_v(-T)$, $y=x_v(T)$, $c_n=c(L+\psi_n)$ and $c=c(L)$,
     where $f^n_t$ and $f_t$  are the Euler-Lagrange flows
     of $L+\psi_n$ and $L$ respectively. Then
     \begin{equation}\label{cvs.1}
     \Phi_c(x,y)\le A_{L+c}\bigl(x_v\vert_{[-T,T]}\bigr)
     =\lim_n A_{L+\psi_n+c_n}\bigl(x_{v_n}\vert_{[-T,T]}\bigr)
     =\lim_n\Phi^n_{c_n}(x_n,y_n)\,,
     \end{equation}
     where $\Phi^n$ and $\Phi$ are the action potentials of $L+\psi_n$
     and of $L$ respectively. Write $\Delta:=\lim_n\Phi^n_{c_n}(x_n,y_n)$.
     We shall prove that $\Delta=\Phi_c(x,y)$, then~\eqref{cvs.1}
     becomes an equality and hence $x_v\vert_{[-T,T]}$ is semistatic.
     Since $T>0$ is arbitrary, then $v\in\S(L)$.
     
     Suppose that $\Phi_c(x,y)<\Delta - \e$, then there exists
     a curve $\eta:[0,S]\to M$ with $\eta(0)=x$, $\eta(S)=y$ and
     $A_{L+c}(\eta)<\Delta-\e$. Then
     \begin{equation}\label{cvs.2}
     \Phi^n_{c_n}(x_n,y_n)\le A_{L+\psi_n+c_n}(\eta)
     +\Phi^n_{c_n}(x,x_n)+\Phi^n_{c_n}(y,y_n).
     \end{equation}
     Fix a Riemannian metric on $M$.
     Using a speed~1 geodesic from $z_1\in M$ to $z_2\in M$, we get
     that
     $$
     \Phi^n_{c_n}(z_1,z_2)\le \bigl(\max_{(x,v)\in TM:\;|v|=1}
     |L(x,v)| +\max_{x\in M}
     |\psi_{n}(x)|  +c_n
     \bigr)\; d_M(z_1,z_2).
     $$
     Hence there exist $K>0$ such that for all $n$ sufficiently large we have
     $\Phi^n_{c_n}(z_1,z_2)\le K\; d_M(z_1,z_2)$.
     Letting $n\to\infty$ on equation~\eqref{cvs.2}, we get that
     $\lim_n\Phi^n_{c_n}(x_n,y_n)\le\Delta-\e$. This contradicts
     the definition of $\Delta$.
     

\qed

\begin{Lemma} If $\M^{0}(L)=\{\mu\}$ then $\hS(L)=\S(L)$.
\label{tres}
\end{Lemma}

\PROOF We show first that $\supp(\mu)$ is inside a static class.
    Since $\M^{0}(L)=\{\mu\}$, then $\mu$ is ergodic.
    In particular, $\mu$-almost every point has a dense orbit
    on $\supp(\mu)$. Let $v\in\supp(\mu)$ be such that it has
    a dense orbit on $\supp(\mu)$. Let $u,w\in\supp(\mu)$ and let
    $0<r_n<s_n<t_n$ be such that
    $\lim_n f_{r_n}(v)=u=\lim_n f_{t_n}(v)$ and
    $\lim_n f_{s_n}(v)=w$.
    Then
    \begin{align*}
    d_c\bigl(\pi f_{r_n}(v),\pi f_{t_n}(v)\bigr)
     &\le A_{L+c}\bigl(\pi f_{[r_n,s_n]}(v)\bigr)
         + A_{L+c}\bigl(\pi f_{[s_n,t_n]}(v)\bigr)
     \\
     &= A_{L+c}\bigl(\pi f_{r_n}(v),\pi f_{t_n}(v)\bigr)
     \\
     &=\Phi_c\bigl(\pi f_{r_n}(v),\pi f_{t_n}(v)\bigr),
    \end{align*}
    and
    \begin{align*}
    d_c(\pi u, \pi w)
       &=\lim_n d_c\bigl(\pi f_{r_n}(v),\pi f_{t_n}(v)\bigr)
       \\
       &\le \lim_n \Phi_c\bigl(\pi f_{r_n}(v),\pi f_{t_n}(v)\bigr)
       = \Phi_c(\pi u, \pi u) = 0,
    \end{align*}
 and hence $\supp(\mu)$ is inside a static class.

 Now let $v\in\S(L)$. For $S,T>0$ consider the probabilities
  $\nu_{ST}$ defined by
  $$
  \int g \; d\nu_{ST} = \tfrac{1}{S+T} \int_{-S}^T g(f_s(v))\;ds\;,
  $$
  for any $g:TM\to\re$ continuous. Since the $\om$~and~$\a$-limits
  of $v$ are in $\hS(L)$ and any weak limit of $\{\nu_{ST}\}_{S,T>0}$
  is invariant, then by Theorem IV in \cite{Ma,CDI},
  any weak* limit of $\nu_{ST}$ is minimizing and hence 
  it is $\mu$.
  
  Then $\a$-limit$(v)\subset\supp(\mu)$ and $\om$-limit$(v)\subset\supp(\mu)$.
  Let $u\in\a$-limit$(v)$, $w\in\om$-limit$(v)$ and
  $S_n,\,T_n\to+\infty$ such that $\lim_{n}f_{-S_n}(v)=u$ and
  $\lim_{n}f_{T_n}(v)=w$. For $s,t>0$ define
  \begin{equation}\label{cvs.3}
  \de(s,t)=A_{L+c}\bigl(\pi f_{[-s,t]}(v)\bigr)
  +\Phi_c(\pi f_t(v),f_{-s}(v)\bigr).
  \end{equation}
  Then the triangle inequality for $\Phi_c$ implies that
  $\de(-s,t)$ is increasing on $s>0$ and $t>0$. Also, since $v$ is
  semistatic, $\de(-s,t)=d_c\bigl(\pi f_{-s}(v),\pi f_t(v)\bigr)\ge 0$.
  But then, since $\supp(\mu)$ is inside a static class,
  $$
  \lim_n\de(-S_n,T_n)=d_c(\pi u, \pi w) = 0.
  $$
  Hence $\de(-s,t)\equiv 0$ for all $s,t>0$, and thus
  equation~\eqref{cvs.3} implies that $v\in\hS(L)$.

\qed

\begin{Lemma} Let
    $$
    \cG_2 := \{\, \psi\in C^\infty(M,\re)\,\vert\,
    \M^{0}(L+\psi)=\{\mu\} \text{ and } \hS(L+\psi)=\supp(\mu)\,\}.
    $$
    Then
    \begin{itemize}
    \item[(a)] $\cG_2$ is dense in $C^\infty(M,\re)$.
    \item[(b)] If $\psi_0\in\cG_2$, then 
               $\lim_{\psi\to\psi_0}d_{H}\big(\hS(L+\psi),\,\hS(L+\psi_0)\big)=0$
               where $d_{H}$ is the Hausdorff metric between compact subsets of $TM$.
    \item[(c)] If $\psi\in C^\infty(M,\re)$, $\mu_\psi\in\M^{0}(L+\psi)$
               and $\psi_0\in\cG_2$, then
               $$
               \lim_{\psi\to\psi_0}d_{H}\big(\supp(\mu_\psi)\;,\supp(\mu_{\psi_0})\big)=0
               $$.
               \end{itemize}     
\label{cuatro}
\end{Lemma}

\PROOF  Let us prove (a).
 By Theorem C in \cite{Ma1}, the set
     $$
     \cG_1:=\{\,\psi\in C^\infty(M,\re)\,\vert\,\#\M^{0}(L+\psi)=1\,\}
     $$
     is generic in $C^\infty(M,\re)$. We shall see that if
     $\psi_0\in\cG_1$, $\M^{0}(L+\psi_0)=\{\mu\}$ and $\psi_1\in
     C^\infty(M,\re)$ is such that $\psi_1\ge 0$ and
     $\{x:\;\psi_1(x)=0\}=\pi(\supp(\mu))$, then $\hS(L+\psi_0+\psi_1)=\supp(\mu)$
     and $\M^{0}(L+\psi_0+\psi_1)=\{\mu\}$. This implies that $\cG_2$
     is dense in $C^\infty(M,\re)$.
     
     Observe that $\mu\in\M^{0}(L+\psi_0+\psi_1)$ and hence
     $\supp(\mu)\subseteq\hS(L+\psi_0+\psi_1)$. If $v_0\in TM$ and
     $\pi(v_0)\notin\pi(\supp(\mu))$, we shall see that 
     $v_0\notin\hS(L+\psi_0+\psi_1)$. Then the Graph Property (cf.
     Theorem \ref{GraphProperty}) implies
     that $\hS(L+\psi_0+\psi_1)=\supp(\mu)$. Indeed, if
     $v_1=f_1^{L+\psi_0+\psi_1}(v_0)$, then
     \begin{align*}
     A&_{L+\psi_0+\psi_1+c}\left(\pi\,f_{[0,1]}^{L+\psi_0+\psi_1}(v_0)\right)
     +\Phi_c^{L+\psi_0+\psi_1}(\pi\,v_1\,,\,\pi\,v_0)
     \\
     &\ge
     \int_0^1\psi_1\left(\pi\,f_s^{L+\psi_0+\psi_1}(v_0)\right)\,ds
     +\Phi_c^{L+\psi_0}(\pi\,v_0\,,\,\pi\,v_1)
     +\Phi_c^{L+\psi_0}(\pi\,v_1\,,\,\pi\,v_0)
     \\
     &> d_c^{L+\psi_0}(\pi\,v_0,\pi\,v_1)\geq 0.
     \end{align*}
     Hence $v_0$ is not static.
     
     \medskip
     
     We now prove (b) and (c).
     From Lemmas \ref{dos} and \ref{tres}, if $\psi_0\in\cG_2$, then
     $\lim_{\psi\to\psi_0}\hS(L+\psi)\subseteq\hS(L+\psi_{0})=
     \supp(\mu_{\psi_0})$.
     On the other hand the continuity of the critical value ensures
     that the weak* limit of minimizing measures of $L+\psi$
     is minimizing for $L+\psi_0$ and hence 
     $\lim_{\psi\to\psi_0}\hS(L+\psi)\supseteq\hS(L+\psi_{0})$
     and thus $\lim_{\psi\to\psi_0}\hS(L+\psi)=\supp(\mu_{\psi_0})$.

     This implies that for any neighborhood $U$ of
     $\supp(\mu_{\psi_0})$ there is a neighborhood $\cV$ of
     $\psi_0$ such that $\hS(L+\psi)\subseteq U$ for all 
     $\psi\in\cV$. Let $d$ be the distance function of some Riemannian
     metric on $TM$.
     Using neighborhoods 
     $$U_\e:=\{\,z\in TM\,\vert\,d(z\,,\,\supp(\mu_{\psi_0}))<\e\,\},$$
     one gets that
     \begin{equation}\label{cvs.4}
     \lim_{\psi\to\psi_0}\sup_{z\in
     \La(L+\psi)}
     d\bigl(z\,,\,\supp(\mu_{\psi_0})\bigr)
     \le \lim_{\psi\to\psi_0}\;\sup_{z\in\hS(L+\psi)}\;
     d\bigl(z\,,\,\S(L+\psi_0)\bigr)=0.
     \end{equation}
     
     \medskip
     
     Given $\e>0$, let
     $\{z_1,\ldots,z_N\}\subset\supp(\mu_{\psi_0})$ be such that 
     $\supp(\mu_{\psi_0})\subset\bigcup_{i=1}^N B(z,\e_i)$, where
     $B(z,\e):=\{\,w\in TM\,\vert\, d(z,w)<\e\,\}$ and let
     $g_i:TM\to [0,1]$ be a non-constant positive continuous function
     with $\supp(g_i)\subseteq B(z_i,\e)$. Then $\int
     g_i\,d\mu_{\psi_0}>0$. The continuity of $c(L)$ implies that if
     $\psi\to\psi_0$ and $\mu_\psi\in\M^{0}(L+\psi)$ then
     $\mu_\psi\to\mu_{\psi_0}$ weakly*. Hence there is a neighborhood
     $\cV$ of $\psi_0$ such that if $\psi\in\cV$ and
     $\mu_\psi\in\M^{0}(L+\psi)$, then $\int g_i\, d\mu_\psi>0$ for all
     $i=1,\ldots,N$. Hence
     $$
     \lim_{\psi\to\psi_0}\sup_{z\in\supp(\mu_{\psi_0})}
     d\left(z\,,\La(L+\psi)\right)\le\e.
     $$
     Since this holds for any $\e>0$, then
     \begin{equation}\label{cvs.5}
     \lim_{\psi\to\psi_0}\sup_{z\in\hS(L+\psi_0)}
     d\big(z\,,\hS(L+\psi)\big)
     \le\lim_{\psi\to\psi_0}\sup_{z\in\,\supp(\mu_{\psi_0})}
     d\big(z\,,
     \La(L+\psi)\big)=0.
     \end{equation}
     From~\eqref{cvs.4} and~\eqref{cvs.5} we get that
     $$
     \lim_{\psi\to\psi_0} d_H\big(\hS(L+\psi),\hS(L+\psi_0)\big)
     =\lim_{\psi\to\psi_0}d_H\big(
     \La(L+\psi)\,,\,\supp(\mu_{\psi_0})\big)=0.
     $$
     
\qed


 To complete the proof of Theorem C we now show
that $\cG_2$ is generic.

We claim that the set 
     $$
     \cU(\e):=\{\,\psi\in C^\infty(M,\re)\,\vert\, 
     d_H\big(\hS(L+\psi)\;,
     \La(L+\psi)\big)<\e\,\}
     $$
     contains a neighborhood of $\cG_2$.
 
     This follows from parts (b) and (c) in Lemma \ref{cuatro} and
     the triangle inequality for
     the Hausdorff distance, i.e. using that
     $\hS(L+\psi_0)=\supp(\mu_{\psi_0})$ for $\psi_0\in\cG_2$, we have
     that 
     \begin{align*}
     d_H&\big(\hS(L+\psi),
     \La(L+\psi)\big)
     \le 
     \\
     &\le d_H\big(\hS(L+\psi),\hS(L+\psi_0)\big)
     +d_H\big(\supp(\mu_{\psi_0}),
     \La(L+\psi)\big).
     \end{align*}
     
     \medskip
     
     Since $\cG_2$ is dense,
     the set $\cU(\e)$ contains a open and dense set. Then
     $$
     \bigcap_{n>0}\cU(\tfrac 1n) =\big\{\,
     \psi\in C^\infty(M,\re)\,\vert\,
     \hS(L+\psi)=
     \La(L+\psi)\;\big\}
     $$
     is generic. Since $\cG_2=\cG_1\cap\big[\underset{n>0}\cap\cU(\tfrac
     1n)\,\big]$ and $\cG_1$ is generic, then $\cG_2$ is generic.
     
 
 
   \qed 
 









\section{Proof of Corollaries 2 and 3}

We need the following easy lemma.

\begin{Lemma}Let $M$ be a closed manifold with first Betti
number $b_{1}(M,\re)\geq 2$. Then if $A\subset M$ is a closed
submanifold diffeomorphic to $S^{1}$ and $U_{\varepsilon}$ denotes
the $\varepsilon$ neighborhood of $A$, we have that $H_{1}(M,U_{\varepsilon},\re)$
is non zero for all $\varepsilon$ sufficiently small.  \label{2}
\end{Lemma}

\PROOF Since $A$ is diffeomorphic to a circle, the singular homology 
of the pair $(M,U_{\varepsilon})$ coincides with the singular homology
of the pair $(M,A)$ and therefore
the vector space $H_{1}(M,U_{\varepsilon},\re)$ must have dimension $\geq b_{1}(M,\re)-1\geq 1$.

\qed

We recall the following generic property proved in \cite{CDII} and
\cite{Ma1} that we already mentioned in the introduction. 

\begin{Theorem} Given a
Lagrangian $L$ there exists a generic set ${\cal O}\subset C^{\infty}(M,\re)$
such that if $\psi\in{\cal O}$ the Lagrangian $L+\psi$ has a unique
minimizing measure in ${\cal M}^{0}(L+\psi)$ and this measure
is uniquely ergodic. When this measure is supported on a periodic orbit, this
orbit is hyperbolic and {\bf if} the stable and unstable manifolds intersect, they must do it transversally.   \label{teo}
\end{Theorem}

It is conjectured in \cite{Ma} that the unique
minimizing measure in ${\cal M}^{0}(L+\psi)$ is always supported on a periodic orbit.

Observe now that if we combine Corollary 1, Lemma \ref{2} and Theorem \ref{teo} we obtain Corollary 2.

\qed

To prove Corollary 3 we need the following lemma.
A proof can be found in \cite[Proposition 8]{Fa3}.

\begin{Lemma}If $L$ is a symmetric Lagrangian, then
\[c(L)=-\inf_{x\in M}L(x,0),\]
and
\[\La(L)=\hS(L)=\{(x,0):\;L(x,0)=-c(L)\}.\]
Moreover, the ergodic minimizing measures are the Dirac measures
concentrated on the fixed points $(x,0)$ of the
Euler-Lagrange flow with $L(x,0)=-c(L)$.
\label{3}
\end{Lemma}

Finally observe that if we combine Corollary 1, Lemma \ref{3} and Theorem \ref{teo} we obtain Corollary 3.





\qed





 
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