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\begin{document}

\title[Lorentzian Manifolds with no Null Conjugate Points]{Lorentzian Manifolds
with no\\ Null Conjugate Points}

\author[M. Guti\'{e}rrez, F. J. Palomo, A. Romero]{MANUEL GUTI\'{E}RREZ, FRANCISCO
J. PALOMO \\
Departamento de \'Algebra, Geometr\'{\i}a y Topolog\'{\i}a,
\addressbreak Universidad de M\'alaga, 29071- M\'alaga, Spain.
\addressbreak E-mail: {\rm M.G.:} {\ttfamily mgl@agt.cie.uma.es},
{\rm F.J.P.:} {\ttfamily fjpalomo@eresmas.com} \\ {\em A{\small ND}} {\rm ALFONSO ROMERO}\\
Departamento de Geometr\'{\i}a y Topolog\'{\i}a,
\addressbreak Universidad de Granada, 18071-Granada, Spain.
\addressbreak E-mail: {\ttfamily aromero@ugr.es}}

\maketitle

\begin{abstract}
An integral inequality for a compact Lorentzian manifold which
admits a timelike conformal vector field and has no conjugate
points along its null geodesics is given. Moreover, equality holds
if and only if the manifold has constant sectional curvature. This
inequality can be improved if the timelike vector field is assumed
to be Killing. As an indirect application of our technique, it is
proved that a Lorentzian torus with no conjugate points along its
timelike geodesics and admitting a timelike Killing vector field
must be flat.

MSC2000: Primary 53C50; Secondary 53C22.

Keywords: Lorentzian manifolds, timelike conformal vector fields,
null geodesics, conjugate points.

\end{abstract}


\section{Introduction}

Conjugate points on geodesics of a Lorentzian manifold have been
sistematically studied \cite[Chaps. 10, 11]{BeeEhrEas96},
\cite[Chap. 10]{One83}. Of course, their definition in Lorentzian
geometry is formally the same as the Riemannian case, but there
are here several geometric behaviours which are not possible or
have no sense for a Riemannian manifold. In fact, on a Lorentzian
manifold the geodesics are divided into three classes, spacelike,
null and timelike, according to their causal character, and
consequently conjugate points are classified into spacelike, null
and timelike points (last two cases, i.e. causal conjugate points,
have been extensively considered because their Physical meaning).
Let us remark that there are also different approaches and
techniques to deal with timelike or null conjugate points. With
respect to this class of conjugate points, let us recall that no
null geodesic in a $2-$dimensional Lorentzian manifold has
conjugate points \cite[Lemma 10.45]{BeeEhrEas96}. On the other
hand, there is a simple but surprising fact which asserts that
there are no null conjugate points (i.e conjugate points on null
geodesics) in a Lorentzian manifold of constant sectional
curvature. The converse obviously is not true; in fact, it is easy
to construct a Lorentzian manifold (even geodesically complete)
with no null conjugate points and which does not have constant
sectional curvature. The following question arises then in a
natural way:

\vspace{2mm} \noindent \textit{Under what assumption a Lorentzian manifold with no
conjugate points along its null geodesics has constant sectional curvature?}
\vspace{2mm}

The main aim of this paper is to find a satisfactory answer to
this question. In order to face it, we will introduce an integral
inequality on a remarkable family of Lorentzian manifolds such
that no null geodesic contains a pair of mutually conjugate
points.

Our efforts to deal with this problem have been inspired from
several important contributions in Riemannian geometry. Following
the philosophy to extend properties of Riemannian manifolds of
nonpositive seccional curvature to Riemannian manifolds with no
conjugate points, E. Hopf \cite{Hopf48} proved in 1948 that

\vspace{2mm} \noindent \textit{The total scalar curvature of a closed surface with
no conjugate points is nonpositive and vanishes only if the surface is flat.}
\vspace{2mm}

Therefore, this result combined with the Gauss-Bonnet theorem
gives the following relevant fact

\vspace{2mm} \noindent \textit{A Riemannian torus with no conjugate points must be
flat.} \vspace{2mm}

Note that the same conclusion trivially holds if the Riemannian metric on the torus
is assumed to have nonpositive Gauss curvature. The Hopf theorem was extended by
L.W. Green \cite{Gre58} to any dimension to obtain

\vspace{2mm} \noindent \textit{If a compact Riemannian manifold}
$(B,g)$ \textit{has no conjugate points then its scalar curvature}
$S$ \textit{satisfies}
$$\int_{B}S\, d\mu_{g} \leq 0,$$
\textit{and equality holds if and only if} \textit{the metric is
flat.} \vspace{2mm}

More recently, F. Guimaraes \cite{Gui92} has generalized Green's
theorem to complete Riemannian manifolds under the assumption that
the Ricci curvature has an integrable positive or negative part on
the unit tangent bundle.

Let us recall now that the no conjugancy assumption on a fixed
geodesic in a Riemannian manifold or a causal geodesic in a
Lorentzian manifold has been used by Ehrlich and Kim in \cite[Th.
3.1]{EhrKim94} to obtain a generalization of the Hawking-Penrose
conjugancy theorem (see \cite[Th. 12.18]{BeeEhrEas96}, for
instance) of Singularity theory.

In this paper we will consider two assumptions on the Lorentzian
manifolds to give an answer to the previous question: compacteness
and the existence of certain symmetry of the metric tensor defined
by a timelike conformal vector field. The first requirement is due
to technical reasons (we will use integration on the manifold) and
that we set our approach into global geometry. Of course, a
compact Lorentzian manifold is not always geodesically complete
but, as it was proved in \cite{RomSan95} a compact Lorentzian
manifold which admits a timelike conformal vector field must be
geodesically complete.

Let us remark that the study of conformal vector fields (with
certain causal character) on Lorentzian manifolds is a topic of
special importance. It has been developed mainly under assumptions
of interest in Physics. On the other hand, the Bochner technique
on compact Lorentzian manifolds admitting a timelike conformal
(and more general) vector field  \cite {RomSan96},
\cite{RomSan98}, \cite{Romero} has shown to be fruitfull to get
several obstructions and classifications results. The particular
case of timelike Killing vector fields was previously considered
as a useful tool to get several kinds of classification results in
Lorentzian Geometry \cite {KulRay85}, \cite{Kam93}, \cite{San97},
\cite{AliRomSan97}, \cite{San97b} and references therein.

The main results of this paper are Theorem \ref{Teorema1} and
Corollary \ref{cornuevo} and can be summarized as follows:

\vspace{2mm} \noindent {\it If $(M,g)$ is an $n(\geq 3)-$dimensional compact
Lorentzian manifold with no null conjugate points and which admits a timelike
conformal vector field $K$, then}
$$\int_{M}[n\mathrm{Ric}(U,U)+S]h^{n}d\mu_{g}\leq 0$$ {\it and equality holds if and
only if $(M,g)$ has constant sectional curvature.}\vspace{2mm}

Here $\textrm{Ric}$ and $S$ are, respectively the Ricci tensor and
the scalar curvature of $(M,g)$, $h$ is the function
$1/\sqrt{-g(K,K)}$ and $U$ is the unit timelike vector field $hK$.

If the conformal vector field $K$ is assumed to be Killing this
result can be improved, Theorem \ref{corollary1}, to get

\vspace{2mm} \noindent {\it If $(M,g)$ is an $n(\geq
3)-$dimensional compact Lorentzian manifold with no null conjugate
points and which admits a timelike Killing vector field $K$, then}
$$\int_{M}Sh^{n}d\mu _{g}\leq 0$$ {\it and equality holds if and only
if $(M,g)$ is isometric to a flat Lorentzian $n-$torus up to a (finite) covering. In
particular, in this case $K$ is parallel, the first Betti number of $M$ is not zero
and the Levi-Civita connection of $g$ is Riemannian.} \vspace{2mm}

This result widely extends a theorem of Kamishima. In fact, he
proved in \cite[Th. A]{Kam93} that if a compact Lorentzian
manifold with constant sectional curvature $k \in \R$ admits a
timelike Killing vector field, then $k \leq 0$. Moreover, if
$k=0$, then the manifold is affinelly diffeomorphic to a
Riemannian manifold with non zero first Betti number. It should be
noted that  Kamishima uses a very different technique in
\cite{Kam93}, strongly depending on the Lie group machinery of
Lorentzian space forms, than the present one.


Moreover, Theorem \ref{corollary1} permits to reprove in Corollary
\ref{Green} the Hopf-Green inequality in Riemannian Geometry
\cite{Gre58} previously mentioned.

Although we deal here with null conjugate points on $n\geq
3-$dimensional Lorentzian manifolds, by the use of a trick we are
able to prove Corollary \ref{chuli} that

\vspace{2mm} \noindent {\it A compact Lorentzian surface admitting a timelike
Killing vector field with no conjugate point along its timelike geodesic must be
flat.} \vspace{2mm}

This result resembles Hopf theorem \cite{Hopf48}, but in our case
the no conjugancy hypothesis involves only timelike geodesics. It
should be noted that a Lorentzian torus which admits a timelike
Killing vector field is conformally flat \cite{San97}. Theorem
\ref{chuli} gives an answer to the natural question to decide when
a Lorentzian torus with a timelike Killing vector field must be
flat. Finally, in the last section we illustrate our results on
remarkable examples of compact Lorentzian manifolds.

\section{Preliminaries}

Let $(M,g)$ be an $n(\geq 2)-$dimensional Lorentzian manifold;
that is a (connected) smooth manifold $M$ endowed with a
non-degenerate metric $g$ with signature ($-,+,...,+).$ We shall
write $\nabla $ for the Levi-Civita connection of $g$, $R$ for its
Riemannian curvature tensor (our convention on the curvature
tensor is $R(X,Y)Z=\nabla _{X}\nabla _{Y}Z-\nabla _{Y}\nabla
_{X}Z-\nabla _{[X,Y]}Z$), $\mathrm{Ric}$ for its Ricci tensor, $S$
for its scalar curvature and $d\mu _{g}$ for the canonical measure
associated with $g$.

As usual, the causal character of a tangent vector $v\in T_{p}M$
is timelike (resp. null, spacelike) if $g(v,v)<0$ (resp.
$g(v,v)=0$ and $v\neq 0,$ $g(v,v)>0$ or $v=0)$. If $v\in T_{p}M$
then, $\gamma _{v}$ will denote the unique (maximal) geodesic
which satisfies $\gamma _{v}^{\prime }(0)=v$. The causal character
of $\gamma _{v}^{\prime }(t)$ is independent of $t$ and it is
called the causal character of the geodesic. A null geodesic
$\gamma_{v}$ of $(M,g)$ is then a geodesic such that $v$ is a null
vector. From now on, we will write  $P_{a}^{b}$ for the parallel
translation along $\gamma _{v} $ from $\gamma _{v}(a)$ to $\gamma
_{v}(b)$.

Unless we indicate otherwise, $(M,g)$ will denote an $n(\geq
3)-$dimensional Lorentzian manifold which admits a timelike vector
field $K$. We put $h=1/ \sqrt{-g(K,K)}$ and write $U=hK$ for the
normalized vector field obtained from $K$.

Recall \cite{Har85}, \cite{Koc85} that the \textit{null
congruence} on $M$ associated to $K$ is the subset of the tangent
bundle $TM$ given by
$$
C_{K}M=\{ v\in TM:g(v,v)=0\textrm{ and }g(v,K_{\pi v})=1\} ,
$$
where $\pi :TM\longrightarrow M$ is the natural projection. This
subset is an orientable embedded submanifold of $TM$ with
dimension $2(n-1)$, and can be naturally endowed with the
Lorentzian metric $\widehat{g}$ obtained by restriction of the
Sasaki metric $\bf{g}$ of $TM$ (note that $\bf{g}$ is
non-degenerate with signature $(-,-,+,...,+)$). Moreover, now
$\pi:(C_{K}M, \widehat{g})\rightarrow (M,g)$ is a semi-Riemannian
submersion with spacelike fibers.

Let us suppose that the vector field $K$ is conformal (i.e.
$\mathfrak{L}_{K}g=\rho g$ with $\rho \in C^{\infty}(M))$. Under
that assumption we have that $C_{K}M$ is invariant by any (local)
flow of the geodesic vector field. Furthermore, if $M$ is also
assumed to be compact, then it must be geodesically complete
\cite{RomSan95}, and moreover, in this case, we have

\begin{equation}\label{Liouville}
\int_{C_{K}M}(f\circ \Phi_{t})\,
d\mu_{\widehat{g}}=\int_{C_{K}M}f\ d\mu_{\widehat{g}},
\end{equation}
for all  $f\in C^{0}(M)$ and $t\in \R$ where
$\Phi_{t}(v)=\gamma_{v}^{\prime}(t)$ is the geodesic flow. (See
details on those properties of $C_{K}M$ in \cite{Gut01}).

As a key tool to get our main results in this paper we will use
\cite[Th. 3.2]{Gut01} which reads as follows: {\it Let $(M,g)$ be
an $n(\geq 3)$-dimensional compact Lorentzian manifold that admits
a timelike conformal vector field $K$. If there is $a>0$ such that
every null geodesic $\gamma_{v}$, with $v\in C_{K}M$, has no
conjugate point of $\gamma_{v}(0)$ in $[0,a)$, then

\begin{equation}\label{vieja}
\mathrm{Vol}(C_{K}M)\geq
\frac{a^{2}}{\pi^{2}(n-2)}\int_{C_{K}M}\widetilde{\mathrm{Ric}}\
d\mu_{\widehat{g}},
\end{equation}
where $\widetilde{\mathrm{Ric}}$ denotes the quadratic form
associated with the Ricci tensor.} Moreover, equality in
(\ref{vieja}) has been characterized by using null sectional
curvature.

In order to fix the notation to be used later, we recall the
well-known notion of warped product (see \cite[Def. 7.33]{One83},
for instance). Let $(B,g_{B})$ and $(F,g_{F})$ be semi-Riemannian
manifolds. We write $\pi_B$ and $\pi_F$ for the projections of $
B\times F$ onto $B$ and $F$, respectively. Let $f>0$ a smooth
function on $B$. The warped product $B\times _{f}F$ is the product
manifold $B\times F$ furnished with the metric tensor
$g_{f}={\pi_B}^{*}(g_{B})+(f\circ \pi_B)^{2}{\pi_F}^{*}(g_{F})$.



\section{Null Geodesics with no Conjugate Points}

It is well-known that if $\gamma _{v}:[0,a]\longrightarrow M$ is a
null geodesic such that there is no conjugate point of $\gamma
_{v}(0)$ in $[0,a)$, then the \textit{Hessian form} $H_{\gamma
_{v}}^{\perp }$ is positive semidefinite, i.e. it satisfies

\begin{equation}
H_{\gamma _{v}}^{\perp }(V,V)=\int_{0}^{a}\Big[ g\Big(\frac{\nabla
V}{dt},\frac{ \nabla V}{dt}\Big)-g(R(V,\gamma _{v}^{\prime
})\gamma _{v}^{\prime },V)\Big] dt\geq 0,  \label{Hessiano}
\end{equation}
for every piecewise smooth vector field $V$ along $\gamma _{v}$
such that $V(0)=0$, $V(a)=0$ and $g(\gamma _{v}^{\prime },V)=0,$
\cite[p. 290-1]{One83}.

The following construction, similar to the one made in \cite
[Chap. 10]{BeeEhrEas96}, emerges in a natural way in our setting.
For every null tangent vector $v\in T_{p}M$, we consider the
quotient vector space $\mathcal{N}(v)=\left\langle v \right\rangle
^{\perp }\diagup \langle v\rangle $, where $\langle v \rangle$
denotes the vector subspace of $T_pM$ spanned by $v$. If $x\in
\langle v\rangle ^{\perp }$, then $ [x]$ will denote the element
of $\mathcal{N}(v)$ defined by its equivalence class. Next, we
will project on $\mathcal {N}(v)$ several geometric objects to be
used later. Given $[x]$, $[y] \in \mathcal{N}(v)$, if we put
$\overline{g}([ x] , [y] )=g(x,y)$, then $\overline{g}$ is a
(well-defined) positive definite scalar product on
$\mathcal{N}(v)$. The curvature operator may also be projected on
$\mathcal{N}(v)$ as $\overline{R}([x],v)v=[R(x,v)v]$.  Finally,
since $P_{0}^{t}(v)=\gamma_{v}^{\prime}(t)$ we can define
$\overline{P}_{0}^{t}[x]=[P_{0}^{t}(x)]\in
\mathcal{N}(\gamma_{v}^{\prime}(t))$.

To face our end, $\gamma_{v}(t)$ is considered to be defined for
all $t \in \R$. Let $\mathfrak{X}(\gamma_{v})$ be the
$C^{\infty}(\R)-$module of vector fields along $\gamma_{v}$ and
$\mathfrak{X}^{\perp}(\gamma_{v})=\{V\in
\mathfrak{X}(\gamma_{v}):g(V,\gamma_{v}^{\prime})=0\}$, which is
clearly a submodule of $\mathfrak{X}(\gamma_{v})$. If $[V]\in
\mathfrak{X}^{\perp}(\gamma_{v})/\langle \gamma_{v}^{\prime}
\rangle$, where $\langle \gamma_{v}^{\prime} \rangle$ denotes the
submodule of $\mathfrak{X}^{\perp}(\gamma_{v})$ spanned by
$\gamma_{v}^{\prime}$, then we can define $[V](t)$ to be
$[V(t)]\in \mathcal{N}(\gamma_{v}^{\prime}(t))$. Now, it is
natural to put $\frac{\overline{\nabla}[V]}{dt}=\Big[\frac {\nabla
V}{dt}\Big]$. Finally, if $[V](0)=[0], [V](a)=[0]$, we set
$\overline{H}_{\gamma_{v}}^{\perp}([V],[V])=H_{\gamma_{v}}^{\perp}(V,V)$
where $V\in [V]$ with $V(0)=0$ and $V(a)=0$.


A class $[V]$ is said to be a {\it Jacobi class} along
$\gamma_{v}$ if
$$\frac{\overline{\nabla}^{2}[V]}{dt^{2}}+\overline{R}([V],\gamma_{v}^{\prime})
\gamma_{v}^{\prime}=[0].$$ It is not difficult to show that
$\gamma_{v}(0)$ and $\gamma_{v}(a)$, with $a \not= 0$, are
conjugate along $\gamma_{v}$ if and only if there exists a Jacobi
class $[V](\not=[0])$ along $\gamma_{v}$ which satisfies
$[V](0)=[0]$ and $[V](a)=[0]$. In fact, any representant $V \in
\mathfrak{X}^{\perp}(\gamma_{v})$ of such a Jacobi class must
satisfy
$$\frac{\nabla^{2}V}{dt^{2}}+R(V,\gamma_{v}^{\prime})\gamma_{v}^{\prime}=
f\gamma_{v}^{\prime},$$
where $V(0)=\tau_{1}v$, $V(a)=\tau_{2}\gamma_{v}^{\prime}(a)$,
with $\tau_{1}, \tau_{2}\in \R$, and $f$ is a smooth function. In
this case, if we choose a function $F$ such that
$\frac{d^{2}F}{dt^{2}}=f$, then
$$J(t)=V(t)+\Big[\frac{(a-t)(F(0)-\tau_{1})+t(F(a)-\tau_{2})}{a}-F(t)\Big]
\gamma _{v}^{\prime }(t)$$
is a nonzero Jacobi vector field along $\gamma_{v}$, with $J(0)=0$
and $J(a)=0$.


Several properties of Jacobi classes below are similar to the
well-known ones of the Jacobi vector fields; for instance
\cite[Lemmas 10.15, 10.16]{One83} give respectively:


\begin{lemma} \label{nuevo}

$(1)$ For every Jacobi classes $[V],[W]$ along $\gamma_{v}$, with
$[V](c)=[W](c)=[0]$ for some $c\in\R$, we have
\begin{equation}\label{losJ1}
\overline{g}\Big(\frac{\overline{\nabla}[V]}{dt},[W]\Big)=\overline{g}\Big([V],
\frac{\overline{\nabla}[W]}{dt}\Big).
\end{equation}

$(2)$ If $[J_{1}],...,[J_{k}]$ are Jacobi classes along
$\gamma_{v}$ such that $[J_{i}]$ and $[J_{j}]$ satisfy previous
equation $(\ref{losJ1})$ for all $i,j$ and we put
$[X]=\sum_{j=1}^{k}\lambda_{j}[J_{j}]\in
\mathfrak{X}^{\perp}(\gamma_{v})/\langle \gamma_{v}^{\prime}
\rangle$, with $[X](0)=[0]$ and $[X](a)=[0]$, then

\begin{equation}\label{losJ2}
H_{\gamma_{v}\mid_{[0,a]}}^{\perp}([X],[X])=\int_{0}^{a}\overline{g}
\Big(\sum_{i=1}^{k}\frac{d\lambda_{i}}{dt}[J_{i}],\sum_{j=1}^{k}\frac{d\lambda_{j}}
{dt}[J_{j}]\Big)dt.
\end{equation}
\end{lemma}
$\blacksquare$

Let $\mathrm{End}(\mathcal{N}(v))$ be the space of all $\R-$linear
operators of $\mathcal{N}(v)$ and $\mathcal{R}:  \R  \rightarrow
\mathrm{End}(\mathcal{N}(v))$ the smooth path in
$\mathrm{End}(\mathcal{N}(v))$ given by
$\mathcal{R}(t)=\overline{P}_{t}^{0}\circ \overline{R}(\ \ ,\gamma
_{v}^{\prime }(t)) \gamma _{v}^{\prime }(t)\circ
\overline{P}_{0}^{t}$. From the classical symmetries of the
curvature tensor $R$ we get that $\mathcal{R}(t)$ is self-adjoint
with respect to $\overline{g}$. One associates to $\mathcal{R}$
the {\it matrix Jacobi equation}

\begin{equation}
\frac{d^{2}\mathcal{X}}{dt^{2}}+\mathcal{R}(t)\circ \mathcal{X}=0,
\label{Jacobiequa}
\end{equation}
where $\mathcal{X}:\R \rightarrow \mathcal{N}(v)$ is smooth. One
easily checks that $[J]$ is a Jacobi class along $\gamma_{v}$ if
and only if $\mathcal{X}(t)=\overline{P}_{t}^{0}[J(t)]$ is a
solution to (\ref{Jacobiequa}).

In all what follows $ \mathcal{A}$ will represent the solution to
(\ref{Jacobiequa}) determined from the initial conditions
$\mathcal{A}(0)=0$ and $\frac{d \mathcal{A}}{dt}\mid_{0}=Id$,
where $Id$ denotes the identity transformation. For every $x\in
\langle v \rangle ^{\perp}$, let us denote $J_{x}$ the unique
Jacobi vector field along $\gamma_{v}$ such that $J_{x}(0)=0$ and
$\frac{\nabla J_{x}}{dt}\mid_{0}=x$. It is not difficult to show
that $\mathcal{A}(t)[x]=\overline{P}_{t}^{0}[J_{x}(t)]$.

\begin{lemma}
Let $\gamma_{v}$ be a null geodesic, then there is no conjugate
point of $\gamma_{v}(0)$ along $\gamma_{v}$ if and only if
$\mathrm{Ker}\, \mathcal{A}(t)=0$ for all $t\not=0$.
\end{lemma}
\noindent {\it Proof.} Let $J$ be a nonzero Jacobi field along
$\gamma_{v}$ such that $J(0)=0$ and $J(a)=0$ for $a\not=0$. If
$\frac{\nabla J}{dt}\mid_{0}=x$, then $0\not= [x]\in
\mathrm{Ker}\, \mathcal{A}(a)$. If $0\not=[x] \in \mathrm{Ker}\,
\mathcal{A}(a)$, then $[Z]=[J_{x}]$ is a Jacobi class and
satisfies $[Z](0)=[0]$, $[Z](a)=[0]$ but $[Z]\not= [0]$.
$\blacksquare$



If $\gamma _{v}$ has no pair of mutually conjugate points, then
for any $c\not= 0$, let $\mathcal{B}_{c}$ be the solution to
(\ref{Jacobiequa}) determined from the boundary data
$\mathcal{B}_{c}(0)=Id$, and $\mathcal{B} _{c}(c)=0$. Clearly, it
satisfies $\mathcal{B}_{c}(t)[x]
=\overline{P}_{t}^{0}[I^{c}_{x}(t)] $ where $I^{c}_{x}$ is the
unique Jacobi vector field along $\gamma _{v}$ such that
$I^{c}_{x}(0)=x$ and $I^{c}_{x}(c)=0$. A similar argument as in
\cite{Gre58} or \cite[Chap.5]{Cha93}, permits assert that there
exists $\lim_{c\rightarrow +\infty } \frac{dB_c}{dt}\mid_0=
\frac{dB}{dt}\mid_0$ and taking $B$ as the solution to
(\ref{Jacobiequa}) determined from the initial data $B(0)=Id$ and
$ \frac{dB}{dt}\mid_0$, the continuous dependence of solutions on
the initial data gives that $\lim_{c\rightarrow +\infty }
B_{c}(t)=B(t)$ and $\lim_{c\rightarrow +\infty }
\frac{dB_c}{dt}\mid_t= \frac{dB}{dt}\mid_t$, with $B(t)$ non
singular for all $t\geq 0$.


\begin{corollary}\label{Corollary2}
If $\gamma_{v}$ has no a pair of mutually conjugate points, then
for every $[x]\in \mathcal{N}(v)$, $[x]\not=[0]$, there exists a
Jacobi class $[I_{x}]$ such that $[I_{x}]=\lim_{c\rightarrow
+\infty }[I_{x}^{c}]$ and $\frac{\overline{\nabla}
[I_{x}]}{dt}=\lim_{c\rightarrow +\infty }\frac{\overline {\nabla}
[I^{c}_{x}]}{dt}$. Moreover $[I_{x}](t)\not= 0$ for all $t\geq 0$.
\end{corollary}
\noindent {\it Proof.} It is enought to take
$[I_{x}](t)=\overline{P}_{0}^{t}( \mathcal{B}(t)[x])$.
$\blacksquare$

\begin{remark}
{\rm It should be pointed out that the previous construction is
analogous to the one made in \cite[Chap. 12]{BeeEhrEas96}.
However, the ingredients of equation (\ref{Jacobiequa}) differs
from the used ones in the above mentioned cite.}
\end{remark}

\section{Integral Inequalities and their Applications}

\begin{theorem}\label{Teorema1}
Let $(M,g)$ be an $n(\geq 3)$-dimensional compact Lorentzian
manifold admitting a timelike conformal vector field $K$. If
$(M,g)$ has no conjugate points along its null geodesics, then

\begin{equation}\label{desigualdad2'}
\int_{C_{K}M}\widetilde{\mathrm{Ric}}\ d\mu _{\widehat{g}}\leq 0.
\end{equation}
Moreover, equality holds if and only if $(M,g)$ has constant
sectional curvature $k\leq 0$.
\end{theorem}
\noindent {\it Proof.} The inequality (\ref{desigualdad2'})
follows taking into account that if $(M,g)$ has no conjugate
points along its null geodesics, then previous inequality
(\ref{vieja}) is valid for any positive real number $a$.

Now, we will show, in the equality case, that
$\widetilde{\mathrm{Ric}}=0$ on null tangent vectors. For every
$v\in C_{K}M$, let $\{E_{1},...,E_{n-2}\}$ be a parallel frame
along $\gamma _{v}$ such that
$\{E_{1}(0)=e_{1},...,E_{n-2}(0)=e_{n-2}\}$ give us an orthonormal
basis for $\langle v \rangle^{\perp} \cap K_{\pi v}^{\perp}$ with
$\widetilde{\mathrm{Ric}}(\gamma_{v}^{\prime})=\sum_{j=1}^{n-2}g(R(E_{j},
\gamma_{v}^{\prime})\gamma_{v}^{\prime},E_{j})$.
It is easily seem that $\{[E_{1}](t),...,[E_{n-2}](t)\}$ is a
basis of $\mathcal{N}(\gamma_{v}^{\prime}(t))$ for all $t$. We put
$X_{i}^{c}(t)=\cos ( \frac{\pi t}{2c}) E_{i}(t)$ for every integer
$c\geq 1$, and take $I^{c}_{i}$ the unique Jacobi vector field
along $\gamma_{v}$ such that $I^{c}_{i}(0)=e_{i}$ and
$I^{c}_{i}(c)=0$.

It follows that $\{[I^{c}_{1}](t),...,[I^{c}_{n-2}](t)\} $ is a
basis of $\mathcal{N}(\gamma_{v}^{\prime}(t))$, for each $ t\not=
c$, since there are no conjugate points along $\gamma _{v}$.
Therefore, there are smooth functions $\lambda _{ij}^{c}$ such
that
\begin{equation} \label{200}
[ X_{i}^{c}] =\sum_{j=1}^{n-2}\lambda _{ij}^{c}[I_{j}^{c}],
\end{equation}
on $\R-\{ c\}$, but taking into account that
$X_{i}^{c}(c)=I_{i}^{c}(c)=0$ it is not hard to show that every
function $\lambda _{ij}^{c}$ has a smooth extension to $ \R$. From
Corollary \ref{Corollary2}, there exist Jacobi classes $[I_{j}]
=\lim_{c\rightarrow +\infty }[I_{j}^{c}]$, and it is clear that
$\lim_{c\rightarrow \infty}[X_{i}^{c}]=[E_{i}]$. Hence, there are
smooth functions $\lambda_{ij}=\lim_{c\rightarrow
\infty}\lambda_{ij}^{c}$ such that



\begin{equation}\label{combinacion1}
[E_{i}]=\sum_{j=1}^{n-2}\lambda_{ij}[I_{j}].
\end{equation}
Taking into account that $\lim_{c\rightarrow +\infty}
\frac{\overline{\nabla} [ X_{i}^{c}]}{dt}=0$ and $
\lim_{c\rightarrow +\infty }\frac{\overline{\nabla}
[I_{j}^{c}]}{dt}=\frac{\overline{\nabla} [I_{j}]}{dt}$, a
straightforward computation from (\ref{200}) and
(\ref{combinacion1}) gives $\lim_{c\rightarrow +\infty}
\frac{d\lambda_{ij}^{c}}{dt}=\frac{d\lambda_{ij}}{dt}$.


On the other hand, we get
$$
\sum_{i=1}^{n-2}\overline{H}_{\gamma _{v}\mid _{[
-c,c]}}^{\perp}([X_{i}^{c}],[X_{i}^{c}])=
$$

$$
\int_{-c}^{c}\Big[ \frac{(n-2)\pi ^{2}}{4c^{2}}\sin ^{2}\Big(
\frac{\pi t}{2c}\Big) -\cos ^{2}\Big( \frac{\pi
t}{2c}\Big)\widetilde{\mathrm{Ric}}(\gamma _{v}^{\prime })\Big]
dt\geq 0.
$$

Next, let $\{f_{c}\}$ be the following sequence of nonnegative
continuous functions
$$
f_{c}:C_{K}M\rightarrow \R,\ \ \ \
v\mapsto\sum_{i=1}^{n-2}\overline{H}_{\gamma _{v}\mid _{[ -c,c]
}}^{\perp }([X_{i}^{c}],[X_{i}^{c}]).
$$
If equality holds in (\ref{desigualdad2'}) and we take into
account (\ref{Liouville}) and $\int_{-c}^{c}\sin^{2}( \frac{\pi
t}{2c}) dt=c$, then we have
$$
\int_{C_{K}M}f_{c}\ d\mu _{\widehat{g}}=\frac{(n-2)\pi
^{2}}{4c}\mathrm{Vol} (C_{K}M,\widehat{g}).
$$
Threrefore, the Fatou lemma implies that $\int_{C_{K}M}\lim \inf
f_{c}\ d\mu _{\widehat{g}}=0$, and so $ \lim \inf f_{c}(v)=0$ for
almost all $v\in C_{K}M$. Assume now we are considering such a
$v\in C_{K}M$. Since $0\leq \overline{H}_{\gamma _{v}\mid _{[
-c,c] }}^{\perp }([X_{i}^{c}],[X_{i}^{c}])\leq f_{c}(v)$ holds, we
have

\begin{equation}\label{Fatou2}
\lim \inf \overline{H}_{\gamma _{v}\mid _{[ -c,c]}}^{\perp
}([X_{i}^{c}],[X_{i}^{c}])=0,
\end{equation}
for every $i=1,...,n-2$.

From Lemma \ref{nuevo} we obtain
\begin{equation}\label{Hessiano1}
\overline{H}_{\gamma _{v}\mid _{[ -c,c] }}^{\perp
}([X_{i}^{c}],[X_{i}^{c}])=\int_{-c}^{c}\overline{g}\Big(
\sum_{j=1}^{n-2}\frac{ d\lambda
_{i,j}^{c}}{dt}[I_{j}^{c}],\sum_{j=1}^{n-2}\frac{d\lambda
_{i,j}^{c}}{dt}[I_{j}^{c}]\Big) dt,
\end{equation}
and so, from (\ref{Fatou2}), by using the Fatou lemma again, we
get
\begin{equation}\label{Fatou3}
\overline{g}\Big( \sum_{j=1}^{n-2}\frac{d\lambda
_{ij}}{dt}[I_{j}], \sum_{j=1}^{n-2}\frac{d\lambda
_{ij}}{dt}[I_{j}]\Big)=0,\ \  \textrm{ for almost all }t\geq 0.
\end{equation}
But $\overline{g}$ is positive definite, and therefore previous
(\ref{Fatou3}) implies $\sum_{j=1}^{n-2}\frac{d\lambda
_{ij}}{dt}[I_{j}]=[0]$ on $[0,+\infty)$. Since $[I_{j}](t)\not=
[0]$ for all $j=1,...,n-2$ and $t\geq 0$, we infer from
(\ref{combinacion1}) that $\{[I_{1}](t),...,[I_{n-2}](t)\}$ is a
basis of $\mathcal{N}(\gamma_{v}^{\prime}(t))$ which implies that
all $\lambda_{ij}$ are constant on $[0,+\infty )$. Therefore,
every $[E_i]$ is a Jacobi class on $[0,+\infty )$. We achieve that
$\widetilde{\mathrm{ Ric}}(v)=0$ for almost every $v\in C_{K}M$.
An standard continuity argument gives $\widetilde{\mathrm{Ric}}=0$
on null tangent vectors of $(M,g)$.

By using \cite{EhrKim94}, we deduce that $\overline{R}(\
,\gamma_{v}^{\prime})\gamma_{v}^{\prime}=[0]$ for every null
vector $v$ and so \cite[Prop. 8.28]{One83} combined with the Schur
theorem gives that $(M,g)$ has constant sectional curvature $k$.

Since $(M,g)$ must be geodesic complete from a result of
completeness by Klingler \cite{Kli96} (this fact can be also
deduced from \cite{RomSan95}) it should be globally isometric to a
quotient of the De Sitter space. Nevertheless, it is classical
\cite{CalMar62} that no discrete subgroup of the isometry group of
the De Sitter space which acts properly and discontinouslly
produces a compact quotient of the De Sitter space. In brief,
there exists no compact Lorentzian manifold of positive constant
curvature. Therefore, we conclude $k\leq 0$ as required.
$\blacksquare$

\begin{remark}
\rm{Compare Theorem \ref{Teorema1} with \cite{EhrKim94} where it
is assumed that there is no conjugate points along a fixed null
geodesic, and the conclusion only concerns to that geodesic.}
\end{remark}

Taking into account \cite[Prop. 2.3 and Lemma 3.4]{Gut01}, Theorem
\ref{Teorema1} can be rewriten as follows

\begin{corollary}\label{cornuevo} Let $(M,g)$ be an $n(\geq 3)$-dimensional
compact Lorentzian manifold admitting a timelike conformal vector field $K$. If
$(M,g)$ has no conjugate points along its null geodesics, then
\begin{equation}\label{fornueva}
\int_{M}[ n \, \widetilde{\mathrm{Ric}}(U)+S] \, h^{n}d\mu
_{g}\leq 0.
\end{equation}
Moreover, equality holds if and only if $(M,g)$ has constant
sectional curvature $k\leq 0$.
\end{corollary}

\begin{counterexample}
{\rm  Corollary \ref{cornuevo} (or Theorem \ref{Teorema1}) does
not remain true for compact Lorentzian surfaces. In fact, let
$(M,g)$ be a compact Lorentzian surface admitting a timelike
conformal vector field. Now, we recall that there are no null
conjugate points on two dimensional Lorentzian manifolds, and
that, clearly, $\widetilde{\mathrm{Ric}}=0$ holds on  null tangent
vectors in this case. Thus, if Corollary \ref{cornuevo} were
applicable, then $(M,g)$ should have constant Gauss curvature. But
now the Gauss-Bonnet theorem for Lorentzian surfaces (see
\cite{Birman-Nomizu}, for instance) can be claimed to get that
$(M,g)$ must be flat. However, it is known that there exist
nonflat Lorentzian tori which admit a timelike Killing vector
field \cite{San97}.}
\end{counterexample}

If the vector field $K$ is assumed, more restrictibly, to be
Killing, then \cite[Cor. 3.10]{Gut01} can be improved as follows

\begin{theorem}\label{corollary1}
Let $(M,g)$ be an $n\geq 3$-dimensional compact Lorentzian
manifold with no conjugate points along its null geodesics and
admitting a timelike Killing vector field $K$, then
\begin{equation}
\int_{M}S h^{n}d\mu _{g}\leq 0.  \label{desigualdad3}
\end{equation}
Equality holds if and only if $(M,g)$ is isometric to a flat
Lorentzian $n-$torus up to a (finite) covering. In particular, in
this case $U$ is parallel, the first Betti number of $M$ is not
zero and the Levi-Civita connection of $g$ is Riemannian.
\end{theorem}
\noindent {\it Proof.} The inequality is clear from Corollary
\ref{cornuevo} taking into account \cite[Lemma 3.7]{Gut01}. If
equality holds in (\ref{desigualdad3}), then
$\int_{M}h^{n}\widetilde{\mathrm{Ric}}(U)\, d\mu _{g}=0$. On the
other hand, Corollary \ref{cornuevo} says that $(M,g)$ has
constant sectional curvature $k \leq 0$; in fact, $k=0$  from our
assumption. Let us consider the Riemannian metric $g_{R}$ defined
by $g_{R}(X,Y)=g(X,Y)+2g(X,U)g(Y,U)$, for every $X,Y\in
\mathfrak{X}(M)$. Making use of \cite[Lemma 3.7]{Gut01}, the
vector field $U$ must be parallel. Hence, the Levi-Civita
connection associated to $g_R$ on $M$ agrees with the Levi-Civita
connection of $g$ (in particular, the $1-$form $\omega$, given by
$\omega (X)=g(X,U)$, is closed and clearly is not exact).
Therefore, $(M,g_{R})$ is a compact flat Riemannian manifold and
then, from \cite[Cor. 3.46]{Wol79}, there is a (finite) Riemannian
covering $\rho : (\T^{n},g_{R}^{0})\rightarrow (M,g_{R})$ by a
flat Riemannian torus. It is easily seen that $\rho :
(\T^{n},g^{0}) \rightarrow (M,g)$ is a Lorentzian covering, where
$g_{R}^{0}$ and $g^{0}$ are related in a similar way as previous
$g_{R}$ and $g$. $\blacksquare$

\begin{remark}
{\rm Kamishima proved in \cite[Th. A]{Kam93} that if a compact
Lorentzian manifold with constant sectional curvature $k \in \R$
admits a timelike Killing vector field, then it is geodesically
complete and $k \leq 0$. Moreover, if $k=0$, then the manifold is
affinelly diffeomorphic to a Riemannian manifold with non zero
first Betti number.  So that Corollary \ref{corollary1} is a wide
extension to this result. It should be noted that  Kamishima uses
a very different technique in \cite{Kam93} than the present one;
in fact his tools are strongly depending on the Lie group
machinery of Lorentzian space forms (see \cite{RomSan96},
\cite{RomSan98} for another extensions of the Kamishima result).}


\end{remark}

Theorem \ref{corollary1} contains, as a very particular case, the
classical Green result in Riemannian geometry (\cite{Gre58},
\cite[Th. 5.11]{Cha93}) mentioned in the introduction.

\begin{corollary}\label{Green}
Let $(B,g)$ be a compact Riemannian manifold of dimension $n\geq
2$ and scalar curvature $S$. If $(B,g)$ has no conjugate points
then $$\int_{M}S\,d\mu _{g}\leq 0.$$ Moreover, equality holds if
and only if $(B,g)$ is flat.
\end{corollary}

\noindent {\it Proof.} Consider the Lorentzian manifold
$(M,g_{L})=(\S^{1}\times B,-g_{can}+g)$, where $\S^1$ is the unit
circle and $g_{can}$ denote its canonical metric. We may take the
vector field $K$ as the lift to $\S^{1}\times B$ of the vector
field $z\longmapsto iz$ on $\S^{1}\subset \C$. Now, we have only
to specialize Theorem \ref{corollary1} to this Lorentzian
manifold. $\blacksquare$




\begin{remark} \rm{ Theorem \ref{corollary1} cannot be extended to the case in which $K$ is
assumed to be conformal. In fact, let $(\T^{2},g_{0})$ be a flat
Riemannian torus and let us consider $(\S^{1}, -g_{can})$ as
previously. For every $f\in C^{\infty}(\S^{1})$, $f>0$, the
Lorentzian manifold $\S^{1}\times_{f}\T^{2}$ has no conjugate
points along its null geodesics \cite[Th. 5.4]{FloSan02}. If $U$
is the vector field on $\S^{1}\times F$ given by the lift of the
vector field $z\longmapsto iz$ on $\S ^{1}\subset \C$, then $K=fU$
is a timelike conformal vector field (see \cite{AliRomSan97-1},
for instance).

Making use of the well-known formulae for the curvature of a
warped product metric \cite[Chap. 7]{One83}, we can write
$$
\int_{\S^{1}\times
\T^{2}}Sh^{3}d\mu_{g_{f}}=\textrm{area}(\T^{2})\int_{\S^{1}}\left(4f\Delta
f+2\parallel \textrm{grad} f \parallel ^{2}\right)/f^{3} \,
d\mu_{g_{can}},
$$
where $\Delta $, $\textrm{grad}$ and $\parallel\ \ \parallel$ are,
respectively, the Laplacian, the gradient and the norm of
$(\S^{1},g_{can})$. Now, the classical Green divergence theorem
permits us to write
$$
0=\int_{\S^{1}}\Delta
\Big(\frac{1}{f}\Big)d\mu_{g_{can}}=\int_{\S^{1}}\left(-\frac{\Delta
f}{f^{2}}+\frac{2\parallel \textrm{grad} f \parallel
^{2}}{f^{3}}\right)\, d\mu_{g_{can}}.
$$
Hence, if $f$ is not constant, then $\int_{\S^{1}\times
\T^{2}}Sh^{3}d\mu_{g_{f}}>0$.}
\end{remark}

After showing Theorem \ref{corollary1}, it might be natural to ask
if the function $h^n$ which appears in the integral inequality
(\ref{desigualdad3}) could be omitted. At this time we don't know
the answer, but we think that this is not possible.


As it was noted above, Corollary \ref{corollary1} cannot be used
to study Lorentain surfaces. On the other hand, it does not give
consequences from assumptions on timelike conjugate points.
However, by means of a trick, the following result can be
obtained.

\begin{corollary}\label{chuli}
Let $(M,g)$ be a compact Lorentzian surface admitting a timelike
Killing vector field $K$. If $(M,g)$ has no conjugate points along
its timelike geodesics, then $(M,g)$ must be flat.
\end{corollary}
\noindent{\it Proof.} Because $M$ is $2-$dimensional, no of its
null geodesics has conjugate points. Using this fact and the form
of the geodesics in a semi-Riemannian product, we conclude that
the Lorentzian manifold $( M\times \S^{1}, g+g_{can})$ has no
conjugate points along its null geodesics. Moreover, $K$ naturally
induces a timelike Killing vector field on this semi-Riemannian
product. From Theorem \ref{corollary1} we obtain

$$
\int_{M}G h^{3}d\mu_{g} \leq 0,
$$
where $G=(1/2)S$ is the Gauss curvature of $M$. Moreover equality
holds if and only if $(M,g)$ is flat.  On the other hand, the
integrand in (\ref{fornueva}) is $[3
\widetilde{\mathrm{Ric}}(U)+S] h^{3}=[-3G+2G]h^{3}=-Gh^{3}$. So,
Corollary \ref{cornuevo} can be also claimed to get the opposite
inequality to the previous one. $\blacksquare$

\section{Examples}

$1.$ {\it Compact standard static} Lorentzian manifolds are warped
products $B\times _{f}F$ with $(B,g_{B})$ a compact Riemannian
manifold, $\dim B=n \geq 2$, and $(F,g_{F})=(\S^{1},-g_{can})$. If
$K$ is the timelike vector field on $B\times \S^{1}$ given by the
lift of the vector field $z\longmapsto iz$ on $\S ^{1}\subset \C$,
then $h=1/\sqrt{-g_f(K,K)}=1/({f\circ \pi_B})$. Moreover, from
\cite[Lemma 12.37]{One83} $K$ is Killing.

Assume $B\times_{f}\S^{1}$ has no conjugate points along its null
geodesics. Using \cite[Chap. 7]{One83},  previous Corollary
\ref{cornuevo} gives
$$
\int_{B}\Big[ \frac{(n-1)\Delta
f}{f^{n+1}}+\frac{S^{B}}{f^{n}}\Big] d\mu _{g_{B}} \leq 0,
$$
where $\Delta $ and $S^{B}$ are the Laplacian and the scalar
curvature of $(B,g_{B})$, respectively. Taking now into account
$\Delta \Big(\frac{-1}{nf^{n}}\Big)= \frac{\Delta
f}{f^{n+1}}-\frac{n+1}{f^{n+2}}\|\mathrm{grad}f\|^{2}$, the
classical Green divergence theorem permits us to write
$$
\int_{B}\frac{\Delta f}{f^{n+1}}d\mu
_{g_{B}}=(n+1)\int_{B}\frac{\|\mathrm{grad}f
\|^{2}}{f^{n+2}}d\mu_{g_{B}} \geq 0.
$$

Therefore, if $B\times _{f}\S^{1}$ has no conjugate points along
its null geodesics, then $$\int_{B}\frac{S^{B}}{f^{n}}d\mu
_{g_{B}} \leq 0,$$ and equality holds if and only if $f$ is
constant and $(B,g_{B})$ is flat.

$2.$ {\it Generalized Robertson-Walker Compact Spacetimes} are
warped products $B\times_{f} F$ with $(B,g_{B})=(\S^{1},-g_{can})$
and $(F,g_{F})$ a compact Riemannian manifold with $\mathrm{dim}
F=n\geq 2$. If $U$ is the vector field on $\S^{1}\times F$ given
by the lift of the vector field $z\longmapsto iz$ on $\S
^{1}\subset \C$, then $K=fU$ is a timelike conformal vector field,
\cite{AliRomSan97-1} and $h=1/({f\circ \pi_{\S ^1}})$. Assume
$\S^{1}\times_{f}F$ has no conjugate points along its null
geodesics (this happen if the fiber $(F,g_{F})$ has no conjugate
points \cite[Theor. 5.4]{FloSan02}). A straightforward
computation, by using the curvature formulae of a warped product
\cite[Chap. 7]{One83}, permits to write (\ref{fornueva}) as
follows

\begin{equation}\label{ejemplo2}
n(n-1)\mathrm{Vol}(F)\int_{\S^{1}} \Big[
\frac{(f^{\prime})^{2}}{f^{3}}-\frac {f^{\prime
\prime}}{f^{2}}\Big] d\mu _{g_{can}}+\int_{F}S^{F}d\mu_{g_{F}}
\int_{\S^{1}}\frac{1}{f^{3}}d\mu_{g_{can}} \leq 0,
\end{equation}
where $f^{\prime}$ denotes $U(f)$. This inequality can be also
obtained from an inderect method. In fact, we have $\Delta
\Big(\frac{-1}{f}\Big)=-\frac{f^{\prime \prime}}
{f^{2}}+\frac{2(f^{\prime })^{2}}{f^{3}}$ ($\Delta$ denotes here
the Laplacian operator for $-g_{can}$; i.e. $\Delta=-\Delta^0$,
where $\Delta^0=\frac{d^2}{d\theta^2}$ is the usual Laplacian of
$\S ^1$). Now, with the help of the classical Green divergence
theorem, (\ref{ejemplo2}) is rewriten as follows
$$
\int_{F}S^{F}d\mu_{g_{F}}\int_{\S^{1}}\frac{1}{f^{3}}d\mu_{g_{can}}
\leq n(n-1) \mathrm{Vol}(F)\int_{\S^{1}}
\frac{(f^{\prime})^{2}}{f^{3}}d\mu_{g_{can}},
$$
which is of course true if it is noted that $\int_FS^Fd\mu_g \leq
0$ because of the Green theorem. In the equality case,
$\S^{1}\times_{f}F$ has constant sectional curvature $k \leq 0$.
Therefore $\widetilde{\mathrm{Ric}}(U)=-nk=-\frac{nf^{\prime
\prime}}{f}$. On the other hand, $\Delta
\left(\mathrm{ln}f\right)=-\frac{f^{\prime
\prime}}{f}+\frac{(f^{\prime})^{2}}{f^{2}}$, and so we use again
the divergence theorem to get that $f$ is constant (i.e.
$\S^{1}\times_{f}F$ is, in fact, a semi-Riemannian product) and
$k=0$. Therefore the fiber $(F,g_{F})$ must be flat
(alternatively, this can be also achieved observing that equality
in (\ref{ejemplo2}) gives $\int_FS^Fd\mu_g = 0$, and now the Green
theorem can be claimed again).


\begin{acknowledgments}
The first author has been partially supported by MCYT-FEDER Grant
BFM2001-1825, and the third by MCYT-FEDER Grant
BFM2001-2871-C04-01.
\end{acknowledgments}



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\end{document}
