Projectively equivalent Finsler metrics on surfaces of negative Euler characteristic

TL;DR

This paper characterizes when two real-analytic Finsler metrics on surfaces with negative Euler characteristic share the same unparametrized geodesics, showing they differ only by a scaling and a closed 1-form.

Contribution

It provides a complete characterization of projectively equivalent Finsler metrics on such surfaces, extending understanding of their geometric structure.

Findings

Two metrics have the same unparametrized geodesics iff they differ by a scaling and a closed 1-form.

The result applies specifically to surfaces with negative Euler characteristic.

The proof involves properties of real-analytic Finsler metrics and geodesic equivalence.

Abstract

We proof that on a surface of negative Euler characteristic, two real-analytic Finsler metrics have the same unparametrized oriented geodesics, if and only if they differ by a scaling constant and addition of a closed 1-form.

Equations

$$h{|}_{(x,\xi )}=\frac{\mathrm{tr}h{|}_{(x,\xi )}}{({\xi }^1{)}^2+({\xi }^2{)}^2}\left(\begin{array}{cc}({\xi }^2{)}^2& -{\xi }^1{\xi }^2\\ -{\xi }^1{\xi }^2& ({\xi }^1{)}^2\end{array}\right).$$

$$F(x,\xi):=\tfrac{1}{4}\int_{\begin{subarray}{c}\eta\in T_{x}S^{2}\\ |\eta|=1\end{subarray}}\lambda\big{(}x\times\eta\big{)}|\nu_{\xi}|\qquad\text{with }\nu_{\xi}(\cdot)=\langle\xi,\cdot\rangle,$$

$$E_i(\frac{1}{2}{F}^2,c)=F{F}_{x^i}-\frac{d}{dt}(F{F}_{{\xi }^i})=F\cdot E_i(F,c)-\frac{dF}{dt}{F}_{{\xi }^i},$$

$${\check{F}}_{x^i}-{\check{F}}_{{\xi }^i{x}^j}{\xi }^j+2G^j{\check{F}}_{{\xi }^i{\xi }^j}=0.$$

$${\beta }_{x^i}-{\beta }_{{\xi }^i{x}^j}{\xi }^j+2G^j{\beta }_{{\xi }^i{\xi }^j}=0,$$

$$F_{x^i}-F_{{\xi }^i{x}^{\ell }}{\xi }^{\ell }+2G^{\ell }{h}_{i\ell }=0,$$

$$S(h_{ii})-2G_i^{\ell }{h}_{i\ell }=0.$$

$$S(\mathrm{tr}h)=2G_1^{\ell }{h}_{1\ell }+2G_2^{\ell }{h}_{2\ell }=\underset{c(x,\xi ):=}{\underbrace{2\frac{G_1^1({\xi }^2{)}^2-(G_2^1+G_1^2){\xi }^1{\xi }^2+G_2^2({\xi }^1{)}^2}{({\xi }^1{)}^2+({\xi }^2{)}^2}}}\mathrm{tr}h.$$

$$c\mathrm{tr}\hat{h}=S(\mathrm{tr}\hat{h})=S(I\mathrm{tr}\check{h})=S(I)\mathrm{tr}\check{h}+cI\mathrm{tr}\check{h}=S(I)\mathrm{tr}\check{h}+c\mathrm{tr}\hat{h},$$

$$d^t(x,y)=\underset{0\le \tau \le t}{\max }d(S^{\tau }(x),S^{\tau }(y)).$$

$$h_{\mathrm{top}}=\underset{ϵ\to 0}{\lim }\underset{t\to \infty }{\mathrm{limsup}}\frac{1}{t}\log (H_{ϵ}^t).$$

$$\exists s_{0},d_{0},\mu_{0},k>0\leavevmode\nobreak\ \leavevmode\nobreak\ \forall n\in\mathbb{N}:\leavevmode\nobreak\ \leavevmode\nobreak\ \tilde{\mu}\big{(}B_{s_{0}n+d_{0}}(\tilde{x})\big{)}\geq\mu_{0}e^{kn}.$$

$$s_{0}:=\max_{i}\Big{(}\operatorname{length}_{F}(a_{i})\Big{)}\qquad d_{0}:=\max_{i}\Big{(}\operatorname{diam}_{d}(U_{i})\Big{)}\qquad\mu_{0}:=\min_{i}\Big{(}\mu(U_{i})\Big{)}.$$

$$\tilde{d}(\tilde{x},\Gamma(a)\tilde{y})\leq\inf_{\begin{subarray}{c}c\text{ curve in }U_{i}\\ \text{from $y$ to $x$ }\end{subarray}}\Big{(}\operatorname{length}(\tilde{a})+\operatorname{length}(\tilde{c}^{-1})\Big{)}\leq s_{0}|a|+d_{0},$$

$$\tilde{\mu}(B_{s_{0}n+d_{0}})\geq\tilde{\mu}\Big{(}\bigcup_{|a|\leq n}\Gamma(a)V_{ij}\Big{)}\geq\mu_{0}\cdot\#B_{n}\geq\mu_{0}e^{kn}.$$

$$c_0\cdot \#Q_{r_i}\ge \tilde{\mu }(U_{r_i})\ge e^{\frac{k}{2}{r}_i}.$$

$$\tilde{\rho }({\gamma }_1(r),{\gamma }_2(r))$$

$$\displaystyle\geq 2\epsilon-\frac{1+\lambda}{2}\Big{(}\tilde{d}(\gamma_{1}(r),q_{1})+\tilde{d}(\gamma_{2}(r),q_{1})\Big{)}$$

$$\ge 2ϵ-\frac{1+\lambda }{2}\cdot 2\delta$$

$$h_{\mathrm{top}}\ge \underset{r_i\to \infty }{\mathrm{limsup}}\frac{1}{r_i}\log (\frac{1}{c_0}{e}^{\frac{k}{2}{r}_i})\ge \frac{k}{2}>0.$$

Full text

1. Introduction

Two Finsler metrics on the same manifold are called projectively equivalent, if they have the same unparametrized, oriented geodesics. It is trivial that any two Finsler metrics $\hat{F},\check{F}:TM\to\mathbb{R}$ related by $\hat{F}=\lambda\check{F}+\beta$, where $\lambda>0$ is a constant and $\beta$ is a closed 1-form on $M$, are projectively equivalent. In this paper we proof that, under the assumption of real-analicity, on a closed surface of negative Euler characteristic any projective equivalent metrics must be related in this way:

Theorem 1**.**

On a surface $\mathcal{S}$ of negative Euler characteristic, two real-analytic Finsler metrics $\hat{F},\check{F}$ are projectively equivalent, if and only if $\hat{F}=\lambda\check{F}+\beta$ for some $\lambda>0$ and a closed 1-form $\beta$.

The corresponding result for Riemannian metrics has been obtained in [7, Corollary 3] (see also [8, 9, 10]), where the assumption of real-analicity is not necessary.

The outline of the proof is the following. Recall that the geodesic spray $S$ of a Finsler metric $F$ can be obtained as the Hamiltonian vector field of the Hamiltonian system on $T\mathcal{S}\backslash 0$ with symplectic form $\omega=d\theta$, where $\theta=(\tfrac{1}{2}F^{2})_{\xi^{i}}dx^{i}$ is the Hilbert 1-form, and Hamiltonian $\tfrac{1}{2}F^{2}$. A Hamiltonian system on $T\mathcal{S}\backslash 0$ is called integrable, if there exists a function $I:T\mathcal{S}\backslash 0\to\mathbb{R}$ constant along the integral curves of the Hamiltonian system and such that the differentials of the Hamiltonian and $I$ are linearly independent on an open and dense subset. A function $I$ constant along the integral curves of $S$ is called an integral.

As we are in dimension 2, in all local coordinates the fiber-Hessians of two metrics $\hat{F}$ and $\check{F}$ given by $(\hat{h}_{ij})=(\hat{F}_{\xi^{i}\xi^{j}})$ and $(\check{h}_{ij})=(\check{F}_{\xi^{i}\xi^{j}})$ must be proportional at every vector $(x,\xi)\in T\mathcal{S}\backslash 0$ and the factor of proportionality $I:T\mathcal{S}\backslash 0\to\mathbb{R}$ is given by $I(x,\xi)=\frac{\operatorname{tr}\hat{h}}{\operatorname{tr}\check{h}}|_{(x,\xi)}$. Indeed by 1-homogeneity of a metric $F$ we have $h_{ij}|_{(x,\xi)}\xi^{j}=0$ and it follows that

[TABLE]

When the metrics $\hat{F},\check{F}$ are projectively equivalent, we show:

1. (a)

The factor of proportionality $I$ is independent of the choice of local coordinates and an integral for the geodesic flow of both metrics.

This follows from an investigation of the so called Rapscak conditions for projective metrization similar to [3]. The proof is given in section 2. Alternatively, one can obtain the integral by a general construction for trajectory equivalent Hamiltonian systems, similarly to [6].

In order to show that the integral $I$ must be constant on $T\mathcal{S}\backslash 0$, we combine two classical results from the theory of integrable systems:

1. (b)

The topological entropy of the geodesic flow of a metric on a compact manifold, whose fundamental group is of exponential growth, is positive. 2. (c)

If a 4-dimensional Hamiltonian system is integrable by a real-analytic integral independent of the Hamiltonian, then its topological entropy vanishes.

We recall the definition of topological entropy of a Hamiltonian system and exponential growth of a group in section 3. Proposition b was proven for the geodesic flow of Riemannian metrics and it seems to be commonly accepted that it generalizes straight-forwardly to the Finsler case. Nonetheless, in section 3 we give a proof based on the classical proofs for the Riemannian case from [4, 5]. A proof of c can be found in [11]. A similar argument to show that an integral must be constant was used in [12].

The propositions a, b and c imply Theorem 1:

Proof of Theorem 1.

Let $\hat{F},\check{F}$ be projectively equivalent real-analytic Finsler metrics on a surface $\mathcal{S}$ of negative Euler characteristic. Then the fundamental group of $\mathcal{S}$ is of exponential growth. By b, the geodesic flow of $\hat{F}$ has positive entropy and thus by c the differential of any real-analytic integral must be proportional to the differential of $\tfrac{1}{2}\hat{F}^{2}$ at least on a set $A\subseteq T\mathcal{S}\backslash 0$ admitting an accumulation point. By a the function $I$ is such an integral and by homogeneity $V(\tfrac{1}{2}\hat{F}^{2})|_{(x,\xi)}=\hat{F}^{2}(x,\xi)\not=0$ and $V(I)|_{(x,\xi)}=0$, where $V=\xi^{i}\partial_{\xi^{i}}$ is the vertical vector field. Hence the differential of $I$ must vanish on $A$ and by analicity everywhere. Thus $I$ must be a constant $\lambda$ on $T\mathcal{S}\backslash 0$.

This implies that $\operatorname{tr}(\hat{h})=\lambda\operatorname{tr}(\check{h})$. As the Hessians have only one independent component - see equation (1), it follows that $\hat{h}_{ij}=\lambda\check{h}_{ij}$ and thus $\hat{F}=\lambda\check{F}+\beta$ for some 1-form $\beta$ on $\mathcal{S}$. But as $\hat{F},\check{F}$ are projectively equivalent, so are $\lambda\hat{F},\check{F}$, which differ by the 1-form $\beta$, which then must be closed (see Example 2). ∎

The assumption of real-analicity is necessary: On any closed surface there are (non real-analytic) projectively equivalent metrics that are not related by scaling and addition of a closed 1-form:

Example 1**.**

Let $\hat{F}$ be the round metric on $S^{2}$. We claim that there is a smooth metric $\check{F}$ on $S^{2}$ projectively equivalent to $\hat{F}$, such that

• •

$\hat{F}$* and $\check{F}$ coincide over an open, non-empty set $V\subseteq S^{2}$,*

• •

but are not related by $\hat{F}=\lambda F+\beta$ for any $\lambda>0$ and any 1-form $\beta$.

Then by attaching orientable or non-orientable handles to $S^{2}$ (Figure 1(a)) in the set $V$, we obtain projectively equivalent metrics on any closed surface, that are not related by scaling and addition of a closed 1-form.

The metric $\check{F}$ can be constructed by the method from [1, 2]: Suppose the space of unparametrized geodesics of a reversible metric on a surface forms a smooth manifold endowed with a positive measure. Define the distance $d:\mathcal{S}\times\mathcal{S}\to\mathbb{R}$ of two points on the surface as the measure of curves intersecting the unique shortest geodesic segment conencting the points (Figure 1(b)). Then the unparametrized geodesics of the original metric are shortest for the constructed distance function. Then the function $F(x,\xi):=\frac{d}{dt}|_{t=0}d(x,c(t))$, where $c$ is any curve such that $c(0)=x$ and $\dot{c}(0)=\xi$, is a Finsler metric projectively equivalent to the original metric.

Applying this procedure to the round sphere with a density function $\lambda:S^{2}\to\mathbb{R}_{>0}$ satisfying $\lambda(-x)=\lambda(x)$ and identifying an oriented great circle by its normal (Figure 1(c)), one obtains the following family of Finsler metrics

[TABLE]

all projectively equivalent to the round metric, where $x\times\eta$ denotes the cross-product in $\mathbb{R}^{3}$ and $\langle\cdot,\cdot\rangle$ the Euclidean inner product.

For $\hat{\lambda}\equiv 1$, we obtain the round metric $\hat{F}$. Let $U=\{(x_{1},x_{2},x_{3})\in S^{2}\mid|x_{3}|\leq\tfrac{1}{\sqrt{2}}\}$, denote by $V$ its complement and choose any $\check{\lambda}:S^{2}\to\mathbb{R}_{>0}$ such that $\check{\lambda}|_{U}\equiv 1$, but $\check{\lambda}|_{V}>1$. The obtained metric $\check{F}$ coincides with $\hat{F}$ over $V$, because for $(x,\xi)\in TV$ with $|\xi|=1$, the cross product $x\times\xi$ is in $U$ (Figure 1(c)).

2. Projectively equivalent Finsler metrics and proof of a

Let $M$ be a smooth manifold, $TM\backslash 0$ the tangent bundle with the origins removed and $(x,\xi)$ local coordinates on $TM$.

Definition 1**.**

1. (1)

A Finsler metric is a function $TM\to\mathbb{R}_{\geq 0}$, such that

• •

$F(x,\lambda\xi)=\lambda F(x,\xi)$* for all $\lambda>0$.*

• •

$F|_{TM\backslash 0}$* is smooth and the matrix g_{ij}|_{(x,\xi)}:=\Big{(}\tfrac{1}{2}\frac{\partial^{2}F^{2}}{\partial{\xi^{i}}\partial{\xi^{j}}}\Big{)}_{ij} is positive definite for all $(x,\xi)\in TM\backslash 0$.* 2. (2)

*The geodesics of $F$ are defined as the solutions to the Euler-Lagrange equation $E_{i}(L,c):=L_{x^{i}}-\frac{d}{dt}(L_{\xi^{i}})=0$ for the Lagrangian $L=\tfrac{1}{2}F^{2}$. * 3. (3)

The geodesic spray of $F$ is the globally defined vector field $S$ on $TM\backslash 0$, whose integral curves projected to $M$ are exactly the geodesics. 4. (4)

Two Finsler metrics $\hat{F},\check{F}$ are projectively equivalent, if any geodesic of $\hat{F}$ is a geodesic of $\check{F}$ after an orientation preserving reparametrization.

The Euler-Lagrange equation for a Finsler metric $F$ and $\tfrac{1}{2}F^{2}$ are related by

[TABLE]

so that the solutions to $E_{i}(F,c)=0$ are all orientation preserving reparametrizations of geodesics of $F$.

Two metrics $\hat{F},\check{F}$ are projectively equivalent, if and only if every geodesic of $\hat{F}$ is a solution of the Euler-Lagrange equation $E_{i}(\check{F},c)=0$. Let $S=\xi^{i}\partial_{x^{i}}-2G^{i}\partial_{\xi^{i}}$ be the geodesic spray of $\hat{F}$. Then $\hat{F},\check{F}$ are projectively equivalent, if and only if

[TABLE]

Example 2**.**

Suppose two Finsler metrics are related by $\check{F}=\lambda\hat{F}+\beta$, where $\lambda>0$ and $\beta$ is a 1-form on $M$. Then $\hat{F}$ and $\check{F}$ are projectively equivalent, if and only if $\beta$ is closed.

Indeed, let $\beta=\beta_{k}dx^{k}$. Using that $\hat{F}$ is projectively equivalent to itself equation (2) is equivalent to

[TABLE]

which is satisfied if and only if $(\beta_{j})_{x^{i}}-(\beta_{i})_{x^{j}}\equiv 0$ for all $i,j$, that is if $\beta$ is closed.

Let us now proof proposition a.

Proof of a.

Let $\hat{F}$ and $\check{F}$ be projectively equivalent Finsler metrics on a surfaces $\mathcal{S}$. We shall show that $I:T\mathcal{S}\backslash 0\to\mathbb{R}$ defined in local coordinates by I(x,\xi):=\frac{\operatorname{tr}\hat{h}}{\operatorname{tr}\check{h}}\big{|}_{(x,\xi)}=\frac{\hat{F}_{\xi^{1}\xi^{1}}+\hat{F}_{\xi^{2}\xi^{2}}}{\check{F}_{\xi^{1}\xi^{1}}+\check{F}_{\xi^{2}\xi^{2}}}\big{|}_{(x,\xi)} is well-defined and constant along the integral curves of the geodesic sprays of $\hat{F}$ and $\check{F}$.

The Hessians of $F$ and $\tfrac{1}{2}F^{2}$ are related by $g_{ij}=Fh_{ij}+F_{\xi^{i}}F_{\xi^{j}}$ and it follows by the positive definiteness of $g_{ij}$ and homogeneity, that $h_{ij}|_{(x,\xi)}\nu^{i}\nu^{j}=0$ if and only if $\nu$ is a multiple of $\xi$. Thus $\det h=0$ and $\operatorname{tr}h\not=0$, as otherwise $h$ would vanish.

Let $S=\xi^{i}\partial_{x^{i}}-2G^{i}\partial_{\xi^{i}}$ be the geodesic spray of $\hat{F}$. By (2) for $F\in\{\hat{F},\check{F}\}$ we have

[TABLE]

and thus by differentiating by $\xi^{i}$ and changing sign

[TABLE]

Adding the two equations and using (1) gives

[TABLE]

As this is a linear ODE along the integral curves of $S$, any two solutions must be a constant multiple of each other along the integral curves. Let $\operatorname{tr}\hat{h}=I(x,\xi)\operatorname{tr}\check{h}$. Then

[TABLE]

thus $S(I)=0$ as claimed.

Now let us show that the function $I$ is well-defined. The value $I(x,\xi)$ is defined such that $\hat{F}_{\xi^{i}\xi^{j}}|_{(x,\xi)}=I(x,\xi)\check{F}_{\xi^{i}\xi^{j}}|_{(x,\xi)}$. Let $\overline{x}^{i}(x)$ be a change of coordinates. Then $\overline{\xi}^{i}(x,\xi)=\frac{\partial\overline{x}^{i}}{\partial x^{j}}\xi^{j}$ and $F_{\overline{\xi}^{i}\overline{\xi}^{j}}=F_{\xi^{k}\xi^{\ell}}\frac{\partial\xi^{k}}{\partial\overline{\xi}^{i}}\frac{\partial\xi^{\ell}}{\partial\overline{\xi}^{j}}=F_{\xi^{k}\xi^{\ell}}\frac{\partial x^{k}}{\partial\overline{x}^{i}}\frac{\partial x^{\ell}}{\partial\overline{x}^{j}}$. It follows that $\hat{F}_{\overline{\xi}^{i}\overline{\xi}^{j}}|_{(\overline{x},\overline{\xi})}=I(\overline{x},\overline{\xi})\check{F}_{\overline{\xi}^{i}\overline{\xi}^{j}}|_{(\overline{x},\overline{\xi})}$. Thus $I:T\mathcal{S}\backslash 0\to\mathbb{R}$ is defined independent of the choice of coordinates. ∎

3. Topological entropy and proof of b

Definition 2**.**

Let $G$ be a group generated by finite set $S\subseteq G$. The group is of exponential growth, if for some $k>0$ it holds $\#B_{n}\geq C^{kn}$ for all $n\in\mathbb{N}$, where $\#B_{n}$ denotes the number of elements in $G$ that can be written as a product of at most $n$ elements from $S$ and inverses of those.

It can be shown that this definition does not depend on the choice of the finite generator $S$.

Definition 3**.**

Let $(X,d)$ be a compact metric space with symmetric distance function $d$ and $S^{t}:\mathbb{R}\times X\to X$ a flow. Define the family of distance functions

[TABLE]

For $\epsilon,t>0$ let $H^{t}_{\epsilon}$ be the maximal cardinality of $\epsilon$-separated sets in the metric space $(X,d^{t})$. The topological entropy of the flow $S^{t}$ is defined as

[TABLE]

The topological entropy always exists and is a number in $[0,\infty]$. It does not depend on the choice of the metric $d$, but only on the induced topology.

Let us now proof proposition b: The geodesic flow of any Finsler metric on a compact manifold, whose fundamental group is of exponential growth, has positive topological entropy. To deal with the irreversibility of the metrics, we shall use the reversibility number $\lambda_{F}:=\sup\limits_{\xi\in TM}\frac{F(-\xi)}{F(\xi)}\geq 1$, which is finite if $M$ is compact as its unit sphere bundle is compact.

Furthermore, let $d_{F}:M\times M\to\mathbb{R}$ the (possibly not symmetric) distance function of $F$, where $d_{F}(x,y)$ is defined as the infimum of the $F$-length of all curves from $x$ to $y$. Then the reversibility number of the $d_{F}$ given by $\lambda_{d_{F}}:=\sup\limits_{x,y\in M}\frac{d(y,x)}{d(x,y)}$ is at most $\lambda_{F}$.

Proof of b.

Let $(\widetilde{M},p:\tilde{M}\to M)$ be the universal cover of $M$. Let $d\mu$ be a volume form on $M$ invariant under $F$-isometries (e.g. the Holmes-Thompson or Busemann volume) and $d:M\times M\to\mathbb{R}$ the distance function induced by $F$. Let $\tilde{F}(\tilde{\xi}):=F(d\pi(\tilde{\xi}))$ be the lift of the Finsler metric to the universal cover $\widetilde{M}$, so that $p$ is a local isometry. Denote by $d\tilde{\mu}$ the corresponding volume form of $\tilde{F}$ and by $\tilde{d}$ the induced distance function on $\widetilde{M}$.

Firstly, we show that the volume of closed forward balls $B_{r}(\tilde{x})=\{\tilde{y}\in\widetilde{M}\mid\tilde{d}(\tilde{x},\tilde{y})\leq r\}$ in $\widetilde{M}$ grows exponentially with their radius as a consequence of the exponential growth of the fundamental group, that is for any $\tilde{x}\in\widetilde{M}$

[TABLE]

By compactness and definition of the universal covering, there is a finite family of open connected, simply connected subsets $U_{i}\subseteq M$, that covers $M$ and such that $p^{-1}(U_{i})$ is the union of open, disjoint subsets $V_{ij}\subseteq\widetilde{M}$, such that $p:(V_{ij},\tilde{F})\to(U_{i},F)$ is an isometry.

Fix $\tilde{x}\in\widetilde{M}$ and set $x:=p(\tilde{x})$. Let $S=\{a_{1},..,a_{\ell}\}$ be a set of closed, smooth curves through $x$ generating the fundamental group $\pi(M,x)$ and assume that $S$ is closed under inversion. Let $\#B_{n}$ the number of elements that can we written as a product of at most $n$ elements from $S$. By assumption, there is $k>0$, such that $\#B_{n}\geq e^{kn}$ for all $n\in\mathbb{N}$. Set

[TABLE]

All three are positive. Let $\tilde{x}\in V_{ij}$ and note that $\tilde{\mu}(V_{ij})=\mu(U_{i})\geq\mu_{0}$. For $a\in\pi(M,x)$, let $|a|$ be the smallest number of elements from $S$ whose product gives $a$. Consider the covering transformation $\Gamma(a):V_{ij}\to\widetilde{M}$, that maps a $\tilde{y}\in V_{ij}$ to the endpoint of the unique lift of the curve $cac^{-1}$ starting at $\tilde{y}$, where $c$ is any curve from $y:=p(\tilde{y})$ to $x$ inside $U_{i}$. Then

[TABLE]

where $\tilde{a}$ and $\tilde{c}^{-1}$ are the unique lifts of $a$ and $c^{-1}$ to $\widetilde{M}$ starting at $\tilde{x}$ and $\Gamma(a)\tilde{x}$ respectively, and have the same length as $a$ and $c^{-1}$ because $p$ is a local isometry. Thus $\Gamma(a)V_{ij}\subseteq B_{s_{0}|a|+d_{0}}(\tilde{x})$. Furthermore for different $a$ the sets $\Gamma(a)V_{ij}$ are disjoint and we have p\big{(}\Gamma(a)V_{ij}\big{)}=U_{i}, hence $\tilde{\mu}(\Gamma(a)V_{ij})=\mu(U_{i})\geq\mu_{0}$. It follows that

[TABLE]

Secondly, let $\rho$ and $\tilde{\rho}$ be the symmetrizations of $d$ and $\tilde{d}$ on $M$ and $\widetilde{M}$ respectively, that is $\rho(y_{1},y_{2}):=\frac{d(y_{1},y_{2})+d(y_{2},y_{1})}{2}$. and $\tilde{\rho}(\tilde{y}_{1},\tilde{y}_{2}):=\frac{\tilde{d}(\tilde{y}_{1},\tilde{y}_{2})+\tilde{d}(\tilde{y}_{2},\tilde{y}_{1})}{2}$. Choose $\epsilon>0$ such that any $\rho$-ball of radius $2\epsilon$ in $M$ is contained in a set $U_{i}$, for example quarter of the Lebesgue number of the covering $U_{i}$ for the distance $\rho$. Then any $\tilde{\rho}$-ball of radius $2\epsilon$ in $\widetilde{M}$ is contained in one of the sets $V_{ij}$. In particular is the $\tilde{\mu}$-measure of $\tilde{\rho}$-balls of radius $2\epsilon$ bounded from above by a finite number $c_{0}>0$.

For fixed $\tilde{x}\in\widetilde{M}$, we show existence of a sequence $r_{i}\to\infty$, such that for each $r_{i}$ there are at least $\tfrac{1}{c_{0}}e^{\frac{k}{2}r_{i}}$ unit speed geodesics $\tilde{\gamma}_{j}^{r_{i}}$ of length $r_{i}$ starting from $\tilde{x}$, whose endpoints are $\epsilon$-seperated for the symmetrized distance $\tilde{\rho}$, where $k>0$ is as in the first part.

Let $\delta>0$ and consider the $\tilde{d}$-annuli $U_{r}:=B_{r+\delta}(\tilde{x})\backslash B_{r}(\tilde{x})$. There is a sequence $r_{i}\to\infty$ such that $\tilde{\mu}(U_{r_{i}})\geq e^{\tfrac{k}{2}r_{i}}$. Indeed, suppose the inequality is violated for all but finitely many members of the sequence $r_{i}=i\delta$. Then $\tilde{\mu}(B_{n\delta}(\tilde{x}))=\sum_{i=0}^{n-1}\tilde{\mu}(U_{i\delta})\leq\frac{2}{k\delta}e^{\tfrac{k}{2}\delta n}+C$, where $C$ is a constant independent of $n$. This contradicts (3).

Let $Q_{r_{i}}$ be a maximal $2\epsilon$-separated set for $\tilde{\rho}$ in $U_{r_{i}}$. Then the $\tilde{\rho}$-balls of radius $2\epsilon$ with centers $\tilde{q}\in Q_{r_{i}}$ must cover $U_{r_{i}}$ and hence

[TABLE]

As $(M,F)$ is forward complete, so is $(\widetilde{M},\tilde{F})$ and for each $\tilde{q}\in Q_{r_{i}}$ we may choose a unit speed geodesic from $\tilde{x}$ to $\tilde{q}$ of length between $r_{i}$ and $r_{i}+\delta$. Let $\tilde{\gamma}_{1},\tilde{\gamma}_{2}$ be two such geodesics ending at $\tilde{q}_{1},\tilde{q}_{2}$. Then using $\tilde{\rho}(\tilde{y}_{1},\tilde{y}_{2})\leq\frac{1+\lambda}{2}\tilde{d}(\tilde{y}_{1},\tilde{y}_{2})$, where $\lambda$ is the reversibility number of $F$, we have

[TABLE]

Thus choosing $\delta=\frac{\epsilon}{1+\lambda}$ gives the desired sequence $r_{i}$ and geodesics $\gamma^{r_{i}}_{j}$.

Finally, let $\hat{\rho}$ be any symmetric distance on $TM\backslash 0$, such that \hat{\rho}\big{(}\xi,\nu\big{)}\geq\rho\big{(}\pi(\xi),\pi(\nu)\big{)}, where $\pi:TM\to M$ is the bundle projection. By definition, the topological entropy of the geodesic flow is $h_{\operatorname{top}}=\lim\limits_{\epsilon\to 0}\limsup\limits_{t\to\infty}\tfrac{1}{t}\log(H^{t}_{\epsilon})$, where $H^{t}_{\epsilon}$ is the maximal cardinality of an $\epsilon$-separated set with respect to the distance $\hat{\rho}^{t}$ on $TM\backslash 0$ defined by \hat{\rho}^{t}(\xi,\nu)=\max\limits_{0\leq\tau\leq t}\hat{\rho}\big{(}S^{\tau}\xi,S^{\tau}\nu\big{)}, where $S^{\tau}$ is the geodesic flow of $F$.

Let $r_{i}$ and $\gamma_{j}^{r_{i}}$ as before. Then the starting vectors of the to $M$ projected geodesics $\gamma^{r_{i}}_{j}:=p(\tilde{\gamma}^{r_{i}}_{j})$ are $\epsilon$-separated with respect to $\hat{\rho}^{r_{i}}$. Indeed, let $\gamma_{1},\gamma_{2}$ be two such geodesics and $t\in(0,r_{i}]$ the smallest value, such that \tilde{\rho}\big{(}\tilde{\gamma}_{1}(t),\tilde{\gamma}_{2}(t)\big{)}=\epsilon. Then $\tilde{\gamma}_{1}(t)$ and $\tilde{\gamma}_{2}(t)$ lie in the same $V_{ij}$ and as $p$ is a local isometry also for the symmetrized distances, we have \rho\big{(}\gamma_{1}(t),\gamma_{2}(t)\big{)}=\epsilon. Because $H^{t}_{\epsilon}$ is monotonously increasing as $\epsilon\to 0$, it follows that

[TABLE]

∎