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% Author Package file for use with AMS-LaTeX 1.2
\controldates{4-DEC-2000,4-DEC-2000,4-DEC-2000,4-DEC-2000}
 
\documentclass{era-l}

\issueinfo{6}{13}{}{2000}
\dateposted{December 7, 2000}
\pagespan{98}{104}
\PII{S 1079-6762(00)00086-X}
%\def\copyrightyear{2000}
\copyrightinfo{2000}{American Mathematical Society}

\newtheorem{Th}{Theorem}
\newtheorem{Lem}{Lemma}
\newtheorem{Cor}{Corollary}

\theoremstyle{definition}
\newtheorem{Def}{Definition}

\theoremstyle{remark}
\newtheorem{Rem}{Remark}
\newcommand{\eqdef}{\stackrel{{\rm def}}{=}}
\newcommand{\si}{\sigma}
\newcommand{\om}{\omega}
\newcommand{\bg}{\bar g}
\newcommand{\dif}{\mbox{\rm d}}

\begin{document}
  
\title[Ergodic metrics are determined by geodesics]{Metric with ergodic
geodesic flow is completely determined  by unparameterized geodesics}
\author{Vladimir S. Matveev}
\address{Isaac Newton Institute,   
 Cambridge  CB3  0EH, 
   UK}
\email{v.matveev@newton.cam.ac.uk}



\author{Petar J. Topalov}
\address{Department of Differential Equations,
Institute of Mathematics and Informatics, BAS,
Acad. G. Bonchev Street, Bl. 8, 
Sofia 1113,  
Bulgaria}
\email{topalov@math.bas.bg}

\date{June 16, 2000}
\revdate{October 1, 2000}
\commby{Dmitri Burago}
\subjclass[2000]{Primary  53C20; Secondary 37J35,  37C40, 53A20, 53C22, 
53B10}
\keywords{Projectively  equivalent metrics, ergodic geodesic flows}

\begin{abstract}
Let $g$ be a Riemannian metric with 
 ergodic geodesic flow. Then 
if some metric $\bar g$ has the same 
 geodesics (regarded as unparameterized curves) with $g$, then 
 the metrics are homothetic. If two metrics on a closed surface 
of genus greater than one  have the same geodesics, then they are 
 homothetic. 
 \end{abstract}

\maketitle


Let $g$,  $\bar g$ be two $C^2$-smooth 
 Riemannian metrics on a manifold $M^n$ of dimension $n\ge 2$. 
They  are {\em projectively  equivalent} 
 if they have the same geodesics 
regarded as unparameterized curves. 
For a given metric $g$, there always exist trivial examples of 
projectively
 equivalent metrics: for any positive constant 
$C$, the metric $Cg$ is   projectively equivalent to $g$. 



 
 Consider the fiberwise-linear mapping $G:TM^n\to TM^n$ given by   
 $G= g^{i\alpha}\bar g_{\alpha j}$
  and the fiberwise-linear mapping $A:TM^n\to TM^n$ given by 
\[
A\eqdef (\det(G))^{\frac{1}{n+1}}G^{-1}. 
\]
Consider the characteristic polynomial  
$
\det(A- \lambda \  \mbox{Id})=c_0\lambda^n+c_1\lambda^{n-1} + \dots + c_n
$ and 
 the mappings
$
 S_0, S_1,  \dots ,S_{n-1}:TM^n\to TM^n 
 $ 
given by  
\[
S_{n-m}\eqdef\sum_{i=0}^{m-1}c_i A^{m-i-1}
.\] 
Here $m$ lies in the set $\{1,2,\dots,n\}$  so that $k = n-m $ lies 
in the set $\{0,1,\dots,n-1\}$.
Consider the functions  
$
{I}_0,  {I}_1, \dots ,{I}_{n-1}:T^*M^n\to R
$
given by the general formula 
\[
I_k(x,p)\eqdef  g^{\alpha j}(S_k)_\alpha^i p_ip_j.
\]

In invariant terms, if we identify $TM^n$ with 
 $T^*M^n$ 
via the metric $g$, the functions $I_k$ are given by  
$I_k(x,\xi) = g(S_k \ \xi, \xi)$. 

\begin{Th}[\cite{MT}]\hspace{-2mm}
\label{classical}
If $g$ and $\bar g$   
are  projectively equivalent,
then   
the functions 
$I_k$  are commutative 
 integrals for 
  the geodesic flow of the metric $g$.
\end{Th} 

If the geodesic flow of a metric is ergodic, any smooth 
integral  is a function of the Hamiltonian and therefore only trivial 
examples of projective 
 equivalence are 
possible. Basically we need not ergodicity but transitivity 
(= the existence  of an orbit of the geodesic flow which goes arbitrary 
close to any point of the tangent bundle). 
By definition,   almost any  orbit of an ergodic geodesic flow is 
transitive. 


\begin{Cor}\label{main}
Let the geodesic flow of $g$ be ergodic.  
If some  metric $\bar g$  
is   projectively equivalent  to $g$, then 
 $\bar  g = C g$ for some constant $C$. 
\end{Cor}


We will give a simple 
 self-contained proof of Corollary~\ref{main} which does 
 not use Theorem~\ref{classical}.   
Actually, in the proof of Corollary~\ref{main}, we will demonstrate 
that the 
function $I_0$  is an integral for the geodesic flow of $g$. 
 It is easy to see that the function $I_0$ is given by 
 \begin{equation} \label{integral}
I_0(x,\xi)= 
-  \left(\frac{\det(g)}{\det(\bar g)}\right)^{\frac{2}{n+1}}\bar 
g(\xi,\xi). 
\end{equation}
More precisely, by 
Hamilton-Cayley theorem we have 
$c_0A^{n}+ \dots +c_{n}\equiv 0$.  Then 
\begin{eqnarray*}
S_0 & = & c_0A^{n-1}+ c_1A^{n-2}+ \dots+ c_{n-1}  \\ 
    & = & A^{-1}\left( c_0A^n + \dots +c_{n-1}A+ c_n\right) - c_n A^{-1}
  = -\det(A) A^{-1} \\ &=& -  \left(\frac{\det(g)}{\det(\bar 
g)}\right)^{\frac{2}{n+1}}G. 
\end{eqnarray*}


The geodesic flow of a metric  with negative  sectional curvature on 
a closed manifold is ergodic; see the appendix by Brin in 
the book \cite{Bal}.

\begin{Cor}{\label{manifolds}} 
Let g be a
Riemannian metric on  $M^n$. Suppose that $M^n$ is closed, connected,
  and 
 the sectional curvature of $g$ is negative. 
Then  
if some  metric $\bar g$  
is   projectively equivalent  to $g$, then 
$\bar  g = C g$ for some constant $C$.  
\end{Cor}

For closed surfaces, 
the integral of the curvature form over the surface 
is equal to the Euler characteristic. 
Then a metric with negative curvature can exist only on the surfaces 
of negative Euler characteristic.  
For  such 
surfaces, we do not 
need the curvature restriction in Corollary~\ref{manifolds}. 

\begin{Cor}{\label{Kol}} 
Let $M^2$ be a closed connected  surface of negative Euler 
characteristic, and let $g$ be a Riemannian metric on $M^2$. 
Then another Riemannian metric 
$ \bar g$ on $M^2$ is projectively equivalent to $g$ if and only if 
 $g=C  \bar g$, where $C$ is a positive constant.  
\end{Cor}

This corollary gives 
us a partial  answer to the following classical 
question: for closed surfaces,  
what local  structures  
 constructed by a given metric, do not determine the metric
 locally,   but  
determine the metric globally? Corollary~\ref{main} shows  that, 
 for surfaces of negative Euler characteristic,  
{\it the projective class} of a metric  (= the 
set of all metrics projectively equivalent to the  metric)
  uniquely defines  the metric modulo multiplication by a constant. 

The projective class  of a  
metric is a locally defined 
 object. It can be given in terms of the so-called projective connection 
(see for example  Chapter 4 of Kobayashi   \cite{Kobayashi}) or 
in terms of the Thomas vector field (see Thomas \cite{Thomas}). 

Generally speaking, 
 the projective class of a  
 metric   defines the metric neither locally nor on closed 
surfaces  with 
 Euler  characteristic
  greater than or equal to zero. 
 For the sphere, the first 
nontrivial examples of projectively equivalent metrics
 are due to Beltrami \cite{Beltrami}; 
for the torus such examples follow easily from  Dini 
\cite{Dini}. For   the projective plane and for  the Klein 
bottle  such examples can be easily 
 constructed  using Beltrami's examples on the  sphere 
and Dini's examples on the torus. 

Corollary \ref{Kol} almost follows from Kolokol'tsov \cite{Kol} or 
Kiyohara \cite{Kio}. More  precisely, the integral $I_0$ is 
quadratic in velocities so that the existence of a projectively 
equivalent metric 
implies the existence of an integral quadratic in velocities. 

   In  \cite{Kol} it was shown 
that the  geodesic flow of  a metric on a surface of 
negative Euler 
characteristic does not admit an integral which is
\begin{enumerate}
\item quadratic in velocities and 
\item functionally independent of the Hamiltonian almost everywhere. 
\end{enumerate}
The second condition does not allow us to apply directly the result 
of \cite{Kol} to our situation. 

If  a quadratic in velocities  integral
for the  geodesic flow of a metric on a connected surface 
 is functionally independent of the 
 Hamiltonian 
 at least at one point, then the 
integral  is functionally independent of the 
 Hamiltonian at almost every point. 
 This fact was shown in \cite{Kio} provided 
that the integral is not a linear combination
 of the Hamiltonian and the square of some linear in velocities 
integral; the latter 
 condition  does not 
allow us to apply directly the result 
of \cite{Kio} to our situation.

In our paper we give a simple self-contained proof of 
Corollary~\ref{Kol}. 
The main construction  is based on the idea of  
 Birkhoff (see Chapter 2 of \cite{Birkhoff}) and has been used in  
 \cite{Kol}. 

\begin{proof}[Proof of Corollary~\ref{main}]
 Let us  show  how  the existence of $\bar g$ projectively
 equivalent to $g$
 allows one to construct an integral for the geodesic flow of $g$. 
The construction is  essentially due to  \cite{TensInv} (see also 
\cite{MT});
it is slightly more general and allows one 
to construct an integral by an  
 orbital  diffeomorphism between  
 two Hamiltonian systems. 

  Let  $v$ and $\bar v$ be Hamiltonian systems
 on  
 symplectic manifolds $(N^{2n},\omega)$ and $(\bar N^{2n},\bar\omega)$
 with Hamiltonians $H$ and
$\bar H$,  respectively. 
Consider the 
isoenergy surfaces  (which are not surfaces but 
submanifolds of codimension $1$)
\[
Q^{2n-1}\eqdef\left\{p\in N^{2n}: H(x)=h\right\}, \ \ 
\bar Q^{2n-1}\eqdef\left\{p\in \bar  N^{2n}: \bar H(x)=\bar h\right\},
\]
where $h$ and $\bar h$ are regular values of the functions 
$H$, $\bar H$,  respectively. 
A diffeomorphism $\phi : Q^{2n-1}\longrightarrow \bar Q^{2n-1}$ 
 is   {\em
orbital} 
 if it 
 takes the orbits (regarded as unparameterized curves)
 of the system $v$ to
the orbits of the system $\bar v$.

 Given an orbital diffeomorphism, we can invariantly construct an 
integral for the system $v$. 
 Denote by $\sigma$, $ \bar \sigma$
the restrictions of
 the forms $\omega, \bar \omega$  to $Q^{2n-1}$, $\bar Q^{2n-1}$,  
respectively.    
Consider the pull-back  $\phi^*\bar \sigma$, which is a skew-symmetric 
two-form on $Q^{2n-1}$.
By definition of the vector field $v$, for any vector field 
$u$ we have $\omega(v,u) = dH(u)$. Then   the kernel of the form
 $\sigma$ 
 coincides (in the space ${T}_qQ^{2n-1}$ at
 each point $q\in Q^{2n-1}$) 
with the linear  span of the  vector  $v$. Since the mapping $\phi$ 
takes 
orbits to orbits, the differential $d\phi$ takes the vector field $v$ to 
a vector field which is proportional to $\bar v$.  
Therefore  
the kernel of the form 
$\phi^{*}\bar\sigma$ also coincides  (in the space ${T}_qQ^{2n-1}$ at
 each point $q\in Q^{2n-1}$) 
with the linear  span of
the  vector  $v$. 
 Therefore the forms $\sigma, \phi^*\bar\sigma$
  induce two
nondegenerate
 two-forms  on the quotient bundle 
${ T}Q^{2n-1}/\langle v\rangle$. 
Recall that the  quotient bundle 
${T}Q^{2n-1}/\langle v\rangle$ is the  linear   bundle over $Q^{2n-1}$; 
the vectors  of this bundle are equivalence classes of the vectors of 
$TQ^{2n-1}$; for any $q\in Q^{2n-1}$, 
two vectors $u_1, u_2 \in T_qQ^{2n-1}$ belong to  the 
same class $U\in {T}_qQ^{2n-1}/\langle v\rangle$ if and only if 
the vector $u_1 - u_2$ is proportional to $v$.  We will denote 
the corresponding forms on ${ T}Q^{2n-1}/\langle v\rangle$ also by 
$\sigma, \phi^*\bar\sigma$. Consider the fiberwise-linear mapping 
$\Sigma: { T}Q^{2n-1}/\langle v\rangle \to { T}Q^{2n-1}/\langle 
v\rangle$ uniquely determined by the condition 
$
\sigma(\Sigma(U_1),U_2) = \phi^*\bar \sigma(U_1,U_2)$  
for all  $U_1, U_2\in {T}Q^{2n-1}/\langle v\rangle $.
Then the determinant $\det(\Sigma)$,
 which is a function on $Q^{2n-1}$, is an integral of the system $v$.  
More precisely,  by Cartan's formula,  the  Lie derivative 
$L_v\phi^*\bar \sigma$ 
of the form $\phi^*\bar \sigma$ on $Q^{2n-1}$ 
 along the vector field $v$ satisfies  
\[
L_v\phi^{*}\bar\sigma=\dif\left[{\imath}_v\phi^{*}\bar\sigma\right]+
{\imath}_v
\dif\left[\phi^{*}\bar\sigma\right]. 
\]
On the right-hand  side  both terms vanish: the term 
$\dif\left[{\imath}_v\phi^{*}\bar\sigma\right]$ vanishes since $v$ lies
 in the kernel of the form $\phi^{*}\bar\sigma$, and the  term 
${\imath}_v
\dif\left[\phi^{*}\bar\sigma\right]$ vanishes since the form $\bar 
\omega$
 is closed and therefore the form $\phi^{*}\bar\sigma$ is also closed.
Hence  the  form $\phi^*\bar \sigma$ on $Q^{2n-1}$  is preserved by the 
flow of $v$. It is known that the form $\omega $ and the Hamiltonian $H$ 
are also preserved by  the 
flow of $v$. Therefore the form $\sigma$ and the vector field $v$
 are preserved by  the 
flow of $v$; hence the    forms    $\sigma, \phi^*\bar\sigma$ on ${ 
T}Q/\langle v\rangle$ are also preserved by the 
flow of $v$ and finally the  fiberwise-linear mapping $\Sigma$ and all
 its invariants, in particular the determinant,  are preserved by  the 
flow of $v$. 



  The  geodesic flows of projectively equivalent metrics $g$, $\bar g$ 
on $M^n$
are 
orbitally equivalent systems on the symplectic  manifolds   
$N^{2n}=\bar N^{2n}=TM^n$ with  symplectic   forms 
\[
\omega \eqdef\dif[g_{ij}\xi^jdx^i], \qquad  \bar\omega \eqdef\dif[\bar 
g_{ij}\xi^jdx^i]
\]
and Hamiltonians 
$H,\bar H: TM^n\to R$ given by 
  $H(x,\xi)\eqdef \frac{1}{2}g(\xi, \xi)$,  
$
\bar H(x,\xi)\eqdef \frac{1}{2}\bar g(\xi, \xi). 
$ 
Here $(x,\xi)\in TM^n$  assuming 
that $x= ( x^1,\dots,x^n)\in M^n$ 
and $\xi = (\xi^1,\dots,\xi^n)\in T_xM^n$.   
 The orbital diffeomorphism 
$\phi:TM^n\to TM^n$  is essentially the re-parameterization of 
geodesics, from
the natural parameter of $g$ to the natural parameter of $\bar g$:  
\[
\phi(x,\xi)\eqdef \left(x,\frac{\|\xi\|_g}{\|\xi\|_{\bar g}}\xi\right) 
=\left(x,\frac{\sqrt{H}}{\sqrt{\bar H}}\xi\right).
\]
The  diffeomorphism 
$\phi$  
obviously takes the isoenergy surface $Q^{2n-1}\eqdef \{ g(\xi, \xi) =1 
\}$
to the isoenergy surface $\bar Q^{2n-1}\eqdef \{ \bar g(\xi, \xi) =1 
\}$. 
Let us calculate the integral $\det(\Sigma)$  for these systems.



Consider the fiberwise-linear mapping  
$\Omega: T(TM^n)\to  T(TM^n)$ uniquely determined 
 by the condition
$
\omega(\Omega(u_1),u_2) = \phi^*\bar \omega(u_1,u_2)$
for any vectors $u_1, u_2\in T(TM^n)$.

Let us first calculate  the determinant of $\Omega$.
 The forms $\omega, \phi^* \bar \omega$ are  given by 
\begin{align}
\label{omega1}\omega &=   d\left[\frac{\partial H}{\partial 
\xi^i}dx^i\right]   =  
  g_{ij}d\xi^j\wedge dx^i     +   
\frac{\partial^2 H}{\partial \xi^i \partial x^j } dx^j\wedge dx^i,\\
\phi^* \bar \omega & = 
d\left[\frac{\sqrt{H}}{\sqrt{\bar H}}\frac{\partial \bar H}{\partial 
\xi^i}dx^i\right] 
 =  \frac{\sqrt{H}}{\sqrt{\bar H}}
  \left(\bar g_{ij} + \frac{1}{2}\frac{\partial \bar H}{\partial 
\xi^i}\left(
 \frac{1}{H}\frac{\partial  H}{\partial \xi^j} -\frac{1}{\bar 
H}\frac{\partial \bar  H}{\partial \xi^j}\right)\right) 
 d\xi^j \wedge dx^i \label{omega2}\\ &\quad  +  
\frac{\partial^2 \left(\frac{\sqrt{H}}{\sqrt{\bar H}}\bar 
H\right)}{\partial \xi^i \partial x^j } dx^j\wedge dx^i.\notag 
\end{align}
We see that the formulae~\eqref{omega1}, \eqref{omega2} contain no element 
of 
type $d\xi^i\wedge d\xi^j$. Then in the coordinates 
$(x^1,\dots,x^n,\xi^1,\dots,\xi^n)$ the forms $\omega, \phi^*\bar\omega$ 
are given by  
$(2n\times2n)$-matrices  with  zero  lower right-hand side 
 $(n\times{n})$-blocks. Therefore, 
 the determinant of $\Omega$ is equal, up to the sign
 $(-1)^{n}$, to the 
 square of the ratio $\left(\frac{\det(a)}{\det(g)}\right)$, where 
$a$ is the two-form given by 
\[
a_{ij}\eqdef   \frac{\sqrt{H}}{\sqrt{\bar H}}
  \left(\bar g_{ij} + \frac{1}{2}\frac{\partial \bar H}{\partial 
\xi^i}\left(
 \frac{1}{H}\frac{\partial  H}{\partial \xi^j} -\frac{1}{\bar 
H}\frac{\partial \bar  H}{\partial \xi^j}\right)\right).
\] 
In coordinates, the matrix of the two-form $a$ is given by 
\[
\frac{\sqrt{H}}{\sqrt{\bar H}}{\bar{\bf G}}({\bf 1} + {\bf P}{\bf B}),
\] 
where $\bar{\bf G}$ denotes the matrix of the metric $\bar g$; 
$\bf 1$ denotes the identity matrix;   the matrices $\bf P,B$ are given 
by 
 \[
{\bf  P}_{ij} = \frac{1}{2}\xi_i\xi_j , \  \ {\bf B}_{ij} = 
\left(\frac{g_{ij}}{H}-\frac{\bar g_{ij}}{\bar H}\right).
\]
Using that
 \[
\sum_{i=1}^n\frac{1}{H}\frac{\partial H}{\partial \xi_i}\xi_i =
\sum_{i=1}^n\frac{1}{\bar H}\frac{\partial \bar H}{\partial \xi_i}\xi_i 
= 2
\] 
we have  ${\bf PBP} =0 $ and therefore 
$
({\bf 1} + {\bf P}{\bf B})^m  =  ({\bf 1} + m {\bf P}{\bf B}). 
$ 
Hence 
\begin{equation*}\label{growth}
\left(\det({\bf 1}+ {\bf P}{\bf B})\right)^m  =  
\det({\bf 1} + m {\bf P}{\bf B}). \tag4
\end{equation*}
The left-hand side of (\ref{growth}) depends exponentially  on $m$, 
whereas the right-hand side of (\ref{growth}) depends polynomially  on $m$;
 it can happen  only if $\det({\bf 1} + {\bf P}{\bf B}) = 1$.
 Thus the determinant of the matrix of the form $a$ is 
equal to 
$\left(\frac{\sqrt{H}}{\sqrt{\bar H}}\right)^{n}\det(\bar{\bf G})$
 and the determinant of the fiberwise-linear mapping 
$\Omega$ is equal (up to the sign $(-1)^n$) 
 to 
$\left(\frac{\det(\bar g)}{\det(g)}\right)^2\left(\frac{H}{\bar 
H}\right)^{n}$. 


Now let us prove that the determinants of $\Sigma$ and $\Omega$ are 
connected by  \[\det(\Sigma)= \frac{{H}}{{\bar H}}\det(\Omega).\] 

First of all, at each point $q\in Q^{2n-1}$, 
the Hamiltonian vector field $v$ is an eigenvector 
of $\Omega$ with the eigenvalue $\frac{\sqrt{\bar H}}{\sqrt{H}}$.
 Indeed,  it is easy to understand  
 that 
$d\phi(v) = \frac{\sqrt{\bar H}}{\sqrt{H}} \bar v$: the natural 
projection 
$\pi:TM^n\to M^n$ takes the vector $v$ to the vector of length one in 
the 
metric $g$ and takes the vector $\bar v$ to the vector of length one in 
the 
metric $\bar g$. 
Then 
for any  $u\in T_qQ^{2n-1}$ we have 
\begin{eqnarray*}
\omega(\Omega(v),u) & = & \phi^*\bar \omega(v,u)
    = \bar\omega(d\phi(v),d\phi(u)) \\
 &=& \frac{\sqrt{\bar H}}{\sqrt{H}}\bar\omega(v,d\phi(u)) = 
\frac{\sqrt{\bar H}}{\sqrt{H}}d\bar H(d\phi(u)).
\end{eqnarray*}
Since the mapping $\phi$ takes the $H$ to $\bar H$, then 
$\frac{\sqrt{\bar H}}{\sqrt{H}}d\bar H(d\phi(u))
 = 
\frac{\sqrt{\bar H}}{\sqrt{H}}dH(u)$ and therefore
 $\omega(\Omega(v),u)= \frac{\sqrt{\bar H}}{\sqrt{H}}\omega(v,u)$. 
Thus $\Omega(v) =\frac{\sqrt{\bar H}}{\sqrt{H}} v$. 

Secondly, for any point $q\in Q^{2n-1}$, 
the  subspace $T_qQ^{2n-1}$ of the tangent space 
$  T_q(TM^n)$ is invariant under $\Omega$. Indeed, the   necessary and 
sufficient condition for  $u\in T_q(TM^n)$ to  lie in 
 $T_qQ^{2n-1}$  is  $\omega(u,v) =0$. Using that  
\[
\omega(\Omega(u),v) = \omega(u, \Omega(v)) = 
\frac{\sqrt{\bar H}}{\sqrt{ H}}\omega(u,v)
\]
we obtain  that  
 if   $u\in T_qQ^{2n-1}$ then   $\Omega(u) \in T_qQ^{2n-1}$. 





Finally, since $v$ 
is an eigenvector of
 $\Omega$ and since 
$T_qQ^{2n-1}$ is invariant under $\Omega$, any eigenvalue of $\Sigma$ is 
an eigenvalue of 
$\Omega$. The number of eigenvalues  of  $\Sigma$ (with multiplicities) 
is 
$2n-2$; the number of eigenvalues  of  $\Omega$ (with multiplicities) is 
$2n$; the two eigenvalues  left are evidently the eigenvalue 
$\frac{\sqrt{\bar H}}{\sqrt{H}}$  of the vector 
  $v$  and the eigenvalue 
which corresponds to the eigenvector which does  not lie in 
$T_qQ^{2n-1}$. 
 Using that the    
forms $\omega, \bar \omega ,  \sigma, \bar \sigma$ are skew-symmetric 
and therefore 
all eigenvalues of $\Omega$, $\Sigma$ have even multiplicities, we have 
that 
the eigenvalue of the eigenvector which does  not lie in $T_qQ^{2n-1}$ 
is 
 also $\frac{\sqrt{\bar H}}{\sqrt{H}}$. 
Using that  the determinant is the product of all eigenvalues, we have 
$\frac{\bar H}{H} \det(\Sigma) =  \det(\Omega)$ and therefore the 
function  
$\left(\frac{\det(\bar g)}{\det(g)}\right)\left(\frac{H}{\bar 
H}\right)^{n+1}$ 
is an integral for  the geodesic flow of $g$. 
Using that
 the Hamiltonian $H$ is also an integral 
of the geodesic flow of $g$, we conclude that the function 
(\ref{integral}) 
is an integral for the geodesic flow of $g$. 

\begin{Rem}  
Other characteristic invariants of the fiberwise-linear 
mapping $\Sigma$ give us the other integrals $I_k$.  
\end{Rem}

Now suppose that the geodesic flow of the metric $g$  is  transitive, 
in particular ergodic. Then  
any smooth integral must be a function of the Hamiltonian and 
 the 
differentials of the functions $H$ and $I_0$ must be linear dependent 
at each point of $TM^n$. The
 partial derivatives  $\frac{\partial H}{\partial \xi}$, 
$\frac{\partial I_0}{\partial \xi}$ are equal to 
%\[
%\begin{array}{ccrccccl}
%\frac{\partial H}{\partial \xi}&  = & (&  g_{1j}\xi^j, & 
%g_{2j}\xi^j, & \dots, &  g_{nj}\xi^j&) \\
%\frac{\partial I_0}{\partial \xi}&  = & (& -  
%2 \left(\frac{det(g)}{det(\bar g)}\right)^{\frac{2}{n+1}} \bar 
%g_{1j}\xi^j, & 
%- 2\left(\frac{det(g)}{det(\bar g)}\right)^{\frac{2}{n+1}} \bar 
%g_{2j}\xi^j, & %\dots, & - 2 \left(\frac{det(g)}{det(\bar 
%g)}\right)^{\frac{2}{n+1}} \bar %g_{nj}\xi^j&).  
%\end{array}
%\]
\begin{eqnarray*}
\frac{\partial H}{\partial \xi}\!\!\!\!\!&  = & \!\!\!\!\!( \:g_{1j}\xi^j, 
\:g_{2j}\xi^j,\dots, \; g_{nj}\xi^j), \\
\frac{\partial I_0}{\partial \xi}\!\!\!\!\! &  = &\! \!\!\!\!\!\bigg( \!\!\!-  
2 \left(\frac{\det(g)}{\det(\bar g)}\right)^{\frac{2}{n+1}} \bar 
g_{1j}\xi^j,  
- 2\left(\frac{\det(g)}{\det(\bar g)}\right)^{\frac{2}{n+1}} \bar 
g_{2j}\xi^j,  \dots,  - 2 \left(\frac{\det(g)}{\det(\bar 
g)}\right)^{\frac{2}{n+1}} \bar g_{nj}\xi^j\bigg).  
\end{eqnarray*}
Then the metrics must be at least 
conformally equivalent: 
$\bar g =   f(x) g$. Let us prove that the conformal coefficient $f$ is 
 constant. The integral $I_0$ for the pair of metrics 
$g$, $f(x) g$ is equal to
 $-\big(\frac{1}{f(x)^n}\big)^{\frac{2}{n+1}}f(x)g(\xi,\xi)
 = -\big(\frac{1}{f(x)}\big)^{\frac{n-1}{n+1}}g(\xi,\xi)$. Using 
that the
 Hamiltonian 
$\frac{1}{2}g(\xi,\xi)$ is an integral for the geodesic flow of $g$,  we 
obtain  
that  the function  $\big(\frac{1}{f(x)}\big)^{\frac{n-1}{n+1}}$
is 
an integral for the geodesic flow of $g$. Joining any two points of 
$M^n$ by 
a geodesic,  we see that $\big(\frac{1}{f(x)}\big)^{\frac{n-1}{n+1}}$
is a constant.   Thus $\bar g = Cg$.  
Corollary~\ref{main} is proved. 
\renewcommand{\qed}{}\end{proof}

\begin{proof}[Proof of Corollary~\ref{Kol}]
Let $g, \bar g$ be projectively 
equivalent metrics  on a closed 
  surface $M^2$ of negative Euler characteristic.
Without loss of generality we can assume that the surface $M^2$ 
is orientable. 
Consider the complex structure on $M^2$ corresponding to the metric $g$. 
Let $z$ be a complex coordinate in an open domain $U\subset M^2$, that is 
the metric $g$  reads $\lambda(z) dzd\bar z$, where 
$\lambda $ is a real-valued 
 function. Consider the complex 
momentum $p$. In complex  variables the Hamiltonian $H:{T}^*M^2\to R$
for  the geodesic flow of the metric $g$ reads $2 \frac{p\bar 
p}{\lambda(z)}$. 
In complex coordinates, let the integral $F$ be  given by 
$
  F=A(z)p^2+B(z)p\bar p+\bar A(z)\bar p^2. 
$
Then 
$\frac{1}{A(z)}dzdz$ is a meromorphic two-form without zeros.  More 
precisely, 
it is obviously a two-form:   if we change 
holomorphically the coordinate system, then the 
coefficient $A$ changes as the reciprocal of the coefficient of a  
two-form.  
By definition, $\frac{1}{A}$ has no zero. We show that in each
 coordinate chart the function $A$ is holomorphic.  
 Since $F$ is an integral of the Hamiltonian system with Hamiltonian
$H$, the  Poisson bracket $\{H, F\}$ equals zero. We have 
\begin{equation}\label{zero}
0= \{H, F\}=
 \frac{\partial H}{\partial p}\frac{\partial F}{\partial z}
-\frac{\partial H}{\partial z}\frac{\partial F}{\partial p}
+\frac{\partial H}{\partial \bar p}\frac{\partial F}{\partial \bar z}
-\frac{\partial H}{\partial \bar z}\frac{\partial F}{\partial \bar p}.
\end{equation}
On the right-hand 
 side  of (\ref{zero}) each term is 
a polynomial of third degree in  momenta. Then the bracket is also a 
polynomial of third degree in momenta. In order for  the 
polynomial  to equal zero, 
all coefficients must be zero, 
and in particular the coefficient of  $p^3$. 
Thus $\frac{1}{\lambda}\frac{\partial A}{\partial {\bar z}}$ equals 
zero, 
and $A$ is holomorphic. Finally, 
  $\frac{1}{A(z)}dzdz$ is a meromorphic two-form without zeros. 

By Abel's theorem, 
 for a meromorphic nontrivial   two-form on a closed Riemann
 surface, the  number 
of poles $P$   minus the number of zeros $Z$ 
is equal to  twice the   Euler characteristic. Then either the Euler 
characteristic of $M^2$ is negative or  $A\equiv 0$. In the 
second case the metrics
 are conformally equivalent; in the proof of Theorem~{\ref{main}} we
 already showed that if two metrics are projectively equivalent
 and conformally equivalent, then the conformal coefficient is
 constant so that $\bar g = Cg$. Corollary~\ref{Kol} is proved. 
\renewcommand{\qed}{}\end{proof}

The authors are grateful to  Professors Ursula  Hamenst\"adt, 
 Werner Ballmann,  Gerhard  Knieper, 
Anton Petrunin, 
Udo Simon, and Gudlaugur  Thorbergsson  
 for useful discussions.  The first author thanks   
European Post-Doctoral Institute  for 
financial support and
  Warwick University  and Isaac Newton Institute 
for hospitality. 
The second author was partially
supported by MESC grant MM-810/98 and by CNRS.  




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