# Anosov geodesic flows on surfaces

Hao-Tong Yan
The Ohio State University

##  1\. Introduction

This paper is an exposition of the major results of P. Eberlein’s paper, _When
is a geodesic flow of Anosov type? I_ [Ebe73], in the special case when the
manifold $M$ is a surface. We follow Eberlein’s coverage closely, adding
details when helpful, and taking advantage of simplifications given by the
dimension two case. The objective is to give readers a more tractable
introduction to the important topic of Anosov geodesic flows, which highlights
key concepts and arguments without worrying about generalizations to arbitrary
dimension. As such, this paper is suited for a beginning graduate or advanced
undergraduate student who is familiar with basic Riemannian geometry, such as
that covered by the first half of [dC92].

The main arc of this paper is divided into three parts. Section 2 highlights
some basic concepts and results from Riemannian geometry. For more in-depth
coverage, refer to [dC92]. Section 3 details useful intermediary results when
the manifold $M$ has no conjugate points. The work in these two sections
culminate in the statement and proof of the main result in Section 4, which
defines an Anosov flow and specifies conditions on Jacobi fields and curvature
that are equivalent to the geodesic flow being of Anosov type.

The author would like to thank Andrey Gogolev (The Ohio State University) for
advising this project and Jingyin Huang (The Ohio State University) for
providing helpful feedback.

##  2\. Preliminaries

We start with basic definitions and results. While our main theorem assumes
$M$ is a complete $C^{\infty}$ Riemannian manifold of dimension two, the
results in this section are dimension agnostic; hence for now we may assume
$M$ is a complete $C^{\infty}$ Riemannian manifold of dimension $n\geqslant 2$
with Riemannian metric $\left\langle,\right\rangle$. In all cases, $M$ is
equipped with the Levi-Civita connection. Let $TM$ and $SM$ denote the full
tangent bundle and unit tangent bundle of $M$, respectively, and let $\pi$ be
the projection map onto $M$ in either case. For any vector $v\in TM$, let
$\gamma_{v}$ be the unique geodesic in $M$ such that
$\gamma_{v}^{\prime}(0)=v$, which is guaranteed by completeness of $M$.

A _complete flow_ $t\to\varphi^{t}:N\to N$ on a $C^{\infty}$ manifold $N$ is a
homeomorphism of $(\mathbb{R},+)$ into the group of homeomorphisms of $N$. We
say the flow is $C^{k}$ differentiable, $0\leqslant k\leqslant\infty$, if the
map $(t,n)\to\varphi^{t}(n):\mathbb{R}\times N\to N$ is $C^{k}$
differentiable.

######  Definition 2.1.

Let $t\in\mathbb{R}$ and let $v\in TM$. We define the _geodesic flow_
$g^{t}:TM\to TM$ on $TM$ as

$g^{t}v=\gamma_{v}^{\prime}(t)\;,$

which is the velocity of $\gamma_{v}$ at time $t$.

The geodesic flow is a complete $C^{\infty}$ flow on $TM$, and also in $SM$,
since $g^{t}$ leaves $SM$ invariant for all $t\in\mathbb{R}$. Let $Y$ denote
the vector field in both $TM$ and $SM$ induced by the geodesic flow. In other
words, for any real-valued function $f:TM\to\mathbb{R}$ (or
$f:SM\to\mathbb{R}$), we have

$Y(v)f=\left.\frac{d}{dt}f(g^{t}v)\right\rvert_{t=0}\;.$

Working with this definition, we have for any $s\in\mathbb{R}$,

$\left(dg^{s}Y(v)\right)f=\left.\frac{d}{dt}f(g^{s}g^{t}v)\right\rvert_{t=0}=\left.\frac{d}{dt}f(g^{t}g^{s}v)\right\rvert_{t=0}=Y(g^{s}v)f\;,$

so $dg^{s}Y(v)=Y(g^{s}v)$.

######  Definition 2.2.

A vector field $J$ along a geodesic $\gamma$ of $M$ is called a _Jacobi field_
if

$J^{\prime\prime}(t)+R(\gamma^{\prime}(t),J(t))\gamma^{\prime}(t)=0\;,$

where the ticks over $J$ denote covariant differentiation along $\gamma$ and
$R$ is the curvature tensor.

By passing to local coordinate systems along points of $\gamma$, we see that
this equation yields a system of $n$ linear ODEs of the second order, so $J$
is uniquely determined by the values $J(0)$ and $J^{\prime}(0)$. Therefore, we
have $2n$ linearly independent Jacobi fields along $\gamma$; let $J(\gamma)$
denote their span. Let $J_{0}(\gamma)\subseteq J(\gamma)$ be the
$(2n-2)$-dimensional subspace of _perpendicular_ Jacobi fields $J$ along
$\gamma$, that is, where $\left\langle J(t),\gamma^{\prime}(t)\right\rangle=0$
for all $t\in\mathbb{R}$. The _tangential_ Jacobi fields $J$ along $\gamma$
can be expressed in the form $J(t)=(\alpha+\beta t)\gamma^{\prime}(t)$, where
$\alpha,\beta\in\mathbb{R}$.

######  Definition 2.3.

For $a\neq b$, we say that $\gamma(a)$ and $\gamma(b)$ are _conjugate_ if
there exists a nonzero Jacobi field $J$ along $\gamma$ such that
$J(a)=J(b)=0$. We say that $M$ has _no conjugate points_ if there exists no
pair of conjugate points along any geodesic in $M$, or in other words, for any
geodesic $\gamma$ in $M$ and any nonzero Jacobi field $J$ along $\gamma$,
$J(t)=0$ for at most one number $t\in\mathbb{R}$.

Let $J$ be a Jacobi field along a geodesic $\gamma$. Recall that if $a\neq b$
and $\gamma(a)$ and $\gamma(b)$ are not conjugate, then $J$ is uniquely
determined by $J(a)$ and $J(b)$. Further, if $J$ is perpendicular to $\gamma$
at any two nonconjugate points, then it is perpendicular everywhere.

Let $N\subseteq M$ be a proper $C^{\infty}$ Riemannian submanifold of $M$, and
$\gamma$ be a unit speed geodesic of $M$ such that $\gamma^{\prime}(0)$ is
perpendicular to $T_{\gamma(0)}N$. A Jacobi field $J$ along $\gamma$ is an
_$N$ -Jacobi field_ if $J$ is perpendicular to $\gamma$, $J(0)\in
T_{\gamma(0)}N$, and
$J^{\prime}(0)+S_{\gamma^{\prime}(0)}(J(0))\in(T_{\gamma(0)}N)^{\perp}$, where
$S_{\gamma^{\prime}(0)}$ is the linear operator given by the second
fundamental form on $N$. Equivalently, $J$ is an $N$-Jacobi field if it can be
written as

$J(s)=\frac{\partial r}{\partial t}(s,0)\;,$

where $r:(-\infty,\infty)\times(-\varepsilon,\varepsilon)\to M$ is a
$C^{\infty}$ variation of the form

$r(s,t)=\exp_{\pi Z(t)}(sZ(t))=(\pi\circ g^{s})(Z(t))\;,$

where $Z(t)$ is a $C^{\infty}$ curve in the unit normal bundle of $N$ such
that $Z(0)=\gamma^{\prime}(0)$. Note that $\partial r/\partial t$ denotes
shorthand for $dr(\partial/\partial t)$.

If $N$ is a point, then a Jacobi field $J$ is an $N$-Jacobi field if and only
if $J(0)=0$ and $J$ is perpendicular to $\gamma$. If $N$ is a geodesic
$\sigma$ and $J(0)\neq 0$, then a Jacobi field $J$ is an $N$-Jacobi field if
and only if $J(0)$ is tangent to $\sigma$, $J$ is perpendicular to $\gamma$,
and $\left\langle J(0),J^{\prime}(0)\right\rangle=0$.

######  Definition 2.4.

For an arbitrary proper submanifold $N\subseteq M$ and a perpendicular unit
speed geodesic $\gamma$, the point $\gamma(a)$, $a\neq 0$, is a _focal point_
of $N$ along $\gamma$ if there exists a nontrivial $N$-Jacobi field $J$ along
$\gamma$ such that $J(a)=0$. $M$ is said to have _no focal points_ if no
geodesic $N=\sigma$ has focal points along any unit speed geodesic
perpendicular to $\sigma$.

######  Proposition 2.5.

$M$ has no focal points if and only if the following condition is satisfied:
For any unit speed geodesic $\gamma$ in $M$ and any Jacobi field $J$ on
$\gamma$ such that $J(0)=0$ and $J^{\prime}(0)\neq 0$, we have

$\frac{d}{dt}\left\langle J(t),J(t)\right\rangle>0$

for all $t>0$.

###### Proof.

We follow the proof from [O’S74]. Since $J(0)=0$, we can write
$J(t)=J_{1}(t)+\alpha t\gamma^{\prime}(t)$, where $J_{1}$ is a perpendicular
Jacobi field with $J_{1}(0)=0$ and $\alpha\in\mathbb{R}$. Then

$\frac{d}{dt}\left\langle J(t),J(t)\right\rangle=\frac{d}{dt}\left\langle
J_{1}(t),J_{1}(t)\right\rangle+2\alpha^{2}t\;,$

so it suffices to show the result for nonzero perpendicular Jacobi fields
vanishing at $t=0$.

Let $J$ be such a Jacobi field, suppose $M$ has no focal points, and suppose
there exists $t_{0}>0$ such that $\frac{d}{dt}\left\langle
J(t),J(t)\right\rangle\rvert_{t=t_{0}}=0$. Then $\left\langle
J^{\prime}(t_{0}),J(t_{0})\right\rangle=0$ and by picking the smallest $t_{0}$
with this property, we can ensure that $J(t_{0})\neq 0$. Let $\sigma$ be the
geodesic with $\sigma^{\prime}(0)=J(t_{0})$, and let $L=J(t_{0}-t)$, so $L$ is
Jacobi field along the geodesic $\gamma(t_{0}-t)$. Thus $\left\langle
L^{\prime}(0),L(0)\right\rangle=0$, and by the above discussion on focal
points, $L$ is an $N$-Jacobi field with $N=\sigma$. But $L(t_{0})=0$, so
$\gamma(t_{0})$ is a focal point of $\sigma$ along $\gamma(t_{0}-t)$, a
contradiction. Since $\frac{d}{dt}\left\langle J(t),J(t)\right\rangle>0$ for
sufficiently small $t>0$, this contradiction shows that
$\frac{d}{dt}\left\langle J(t),J(t)\right\rangle>0$ for all $t>0$.

Conversely, suppose $M$ has focal points. Then there exist geodesics $\gamma$
and $\sigma$ with $\gamma(0)$ in $\sigma$ and $\gamma^{\prime}(0)$
perpendicular to $\sigma$ and a non-trivial $N$-Jacobi field $J$ with
$N=\sigma$ such that $J(t_{0})=0$ for some $t_{0}>0$. Then $L(t)=J(t_{0}-t)$
is a non-trivial perpendicular Jacobi field with $L(0)=0$. But
$\frac{d}{dt}\left\langle L,L\right\rangle\rvert_{t=t_{0}}=-2\left\langle
J^{\prime}(0),J(0)\right\rangle=0$ since $J$ is an $N$-Jacobi field. ∎

This immediately yields the following result, which will be useful later.

######  Corollary 2.6.

Suppose $M$ has no focal points and let $\gamma$ be a geodesic. Then for
$t\geqslant s\geqslant 0$ and any Jacobi field $J$ along $\gamma$ with
$J(0)=0$ and $J^{\prime}(0)\neq 0$, we have

$\left\lVert J(t)\right\rVert\geqslant\left\lVert J(s)\right\rVert\;.$

We want to define a natural isomorphism between $T_{v}(TM)$ and
$J(\gamma_{v})$ for any complete $C^{\infty}$ Riemannian manifold $M$ and any
$v\in TM$. For each $v\in TM$, $d\pi:T_{v}(TM)\to T_{\pi v}M$ is linear and
the kernel of $d\pi$ is the $n$-dimensional _vertical subspace_ of
$T_{v}(TM)$. We define a _connection map_ $K:T(TM)\to TM$ such that for each
$v\in TM$, $K:T_{v}(TM)\to T_{\pi v}M$ is linear. The kernel of $K$ is the
$n$-dimensional _horizontal subspace_ of $T_{v}(TM)$, and the intersection of
the horizontal and vertical subspaces is the zero vector.

Given a vector $\xi\in T_{v}(TM)$, let $Z:(-\varepsilon,\varepsilon)\to TM$ be
a $C^{\infty}$ curve with initial velocity $\xi$, and let $\alpha=\pi\circ
Z:(-\varepsilon,\varepsilon)\to M$. We define $K(\xi)=Z^{\prime}(0)\in T_{\pi
v}M$, where $Z^{\prime}(0)$ is the covariant derivative of $Z$ along $\alpha$
evaluated at $t=0$. By passing to a local coordinate system about $v$, we can
show that $K(\xi)$ does not depend on the chosen curve $Z$. Note that for
$\xi\in T_{v}(SM)$, we can choose the curve $Z$ to lie in $SM$.

We define a natural metric on $TM$ called the _Sasaki metric_ , which has the
property that the horizontal and vertical subspaces of $T_{v}(TM)$ are
orthogonal. Given vectors $\xi,\eta\in T_{v}(TM)$, we define

$\left\langle\xi,\eta\right\rangle_{v}=\left\langle
d\pi\xi,d\pi\eta\right\rangle_{\pi v}+\left\langle
K\xi,K\eta\right\rangle_{\pi v}\;.$

We can also define this metric on $SM$ in the same way, since $d\pi|_{T(SM)}$
and $K|_{T(SM)}$ both map $T(SM)$ to $SM$. From now on, this will be the
metric we use whenever dealing with $TM$ or $SM$.

######  Proposition 2.7.

Let $p:N\to M$ be a surjective local isometry of complete Riemannian
manifolds. Then

  1. (1)

$P=dp:TN\to TM$ is a surjective local isometry carrying $SN$ onto $SM$,

  2. (2)

$\left\lVert dg^{t}\xi\right\rVert=\left\lVert dg^{t}dP(\xi)\right\rVert$ for
any $t\in\mathbb{R}$ and any $\xi\in T(TN)$, where $g^{t}$ denotes the
geodesic flow on both $TN$ and $TM$, and

  3. (3)

$dPY(v)=Y(Pv)$ for any $v\in TN$, where $Y$ denotes the vector field defined
by $g^{t}$ in both $TN$ and $TM$.

###### Proof.

Define the projection maps $\pi_{1}:TN\to N$, $\pi_{2}:TM\to M$, and the
connection maps $K_{1}:T(TN)\to TN$ and $K_{2}:T(TM)\to TM$. We have

  1. (i)

$p\circ\pi_{1}=\pi_{2}\circ P$

  2. (ii)

$P\circ d\pi_{1}=d\pi_{2}\circ dp$

  3. (iii)

$P\circ K_{1}=K_{2}\circ dP$, and

  4. (iv)

$P\circ g^{t}=g^{t}\circ P$ for any $t\in\mathbb{R}$,

which can be verified from the definitions and by passing to local
coordinates. Relations (ii) and (iii) show that $P$ is a local isometry,
surjectivity of $p$ guarantees surjectivity of $P$, and the fact that
$P(SN)=SM$ follows from the fact that $p$ is a local isometry. Assertion (2)
follows from (iv) and (1), since $\left\lVert
dg^{t}dP(\xi)\right\rVert=\left\lVert dP\circ
dg^{t}(\xi)\right\rVert=\left\lVert dg^{t}\xi\right\rVert$. Assertion (3)
follows from (iv) and working with the explicit definition of $Y(v)$ defined
after Definition 2.1. ∎

######  Definition 2.8.

For any $v\in TM$ and any $\xi\in T_{v}(TM)$, let $J_{\xi}$ be the unique
Jacobi field on $\gamma_{v}$ such that $J_{\xi}(0)=d\pi\xi$ and
$J_{\xi}^{\prime}(0)=K\xi$.

We recall Lemma 3.4 from [dC92], which is a symmetry result used to prove
Gauss’s Lemma.

######  Lemma 2.9.

Let $A\subseteq\mathbb{R}^{2}$ be a (possibly unbounded) open rectangle. Let
$r(u,v):A\to M$ be differentiable. Then

$\frac{D}{\partial t}\frac{\partial r}{\partial s}(s,t)=\frac{D}{\partial
s}\frac{\partial r}{\partial t}(s,t)\;,$

where $D/\partial t$ denotes covariant differentiation with respect to $t$
along the curve $t\to r(s,t)$ for fixed $s$, and $D/\partial s$ denotes
covariant differentiation with respect to $s$ along the curve $s\to r(s,t)$
for fixed $t$.

Let $v\in SM$ and let $r(s,t)=\exp_{\pi Z(t)}(sZ(t))=(\pi\circ g^{s})(Z(t))$
be the variation of $\gamma_{v}$ defined in the discussion of focal points
above, with the difference that now $Z$ is any curve in $TM$ with initial
velocity $\xi\in T_{v}(TM)$. Then for any $s\in\mathbb{R}$, we claim that

$J_{\xi}(s)=\frac{\partial r}{\partial t}(s,0)\;.$

It suffices to show that $J_{\xi}(0)=\frac{\partial r}{\partial t}(0,0)$ and
$J_{\xi}^{\prime}(0)=\frac{D}{\partial s}\frac{\partial r}{\partial t}(0,0)$.
By direct computation, we have $\frac{\partial r}{\partial t}(s,0)=d\pi\circ
dg^{s}\xi$, so $\frac{\partial r}{\partial t}(0,0)=d\pi\xi=J_{\xi}(0)$. Next,
notice that $\frac{\partial r}{\partial
s}(s,t)=\gamma_{Z(t)}^{\prime}(s)=g^{s}(Z(t))$, so the preceding lemma gives
us that

$\frac{D}{\partial s}\frac{\partial r}{\partial t}(s,t)=\frac{D}{\partial
t}\frac{\partial r}{\partial s}(s,t)=\frac{D}{\partial t}g^{s}(Z(t))\;,$

which gives us

$\frac{D}{\partial s}\frac{\partial r}{\partial
t}(0,0)=Z^{\prime}(0)=K\xi=J_{\xi}^{\prime}(0)\;,$

as desired.

If $\xi\in T_{v}(SM)$, we may choose the curve $Z$ to lie in $SM$; the
$u$-parameter curves of the variation are then unit speed geodesics of $M$. We
claim that the map $s\to\left\langle
J_{\xi}(s),\gamma_{v}^{\prime}(s)\right\rangle$ is a constant function if and
only if $\xi\in T_{v}(SM)$. Indeed, consider

$\frac{d}{ds}\left\langle
J_{\xi}(s),\gamma_{v}^{\prime}(s)\right\rangle=\left\langle
J_{\xi}^{\prime}(s),\gamma_{v}^{\prime}(s)\right\rangle+\left\langle
J_{\xi}(s),\frac{D\gamma_{v}^{\prime}}{ds}(s)\right\rangle=\left\langle
J_{\xi}^{\prime}(s),\gamma_{v}^{\prime}(s)\right\rangle$

since $\gamma_{v}$ is a geodesic. We have

$\displaystyle\left\langle
J_{\xi}^{\prime}(s),\gamma_{v}^{\prime}(s)\right\rangle$
$\displaystyle=\left\langle\frac{D}{\partial s}\frac{\partial r}{\partial
t}(s,0),\frac{\partial r}{\partial s}(s,0)\right\rangle$
$\displaystyle=\left\langle\frac{D}{\partial t}\frac{\partial r}{\partial
s}(s,0),\frac{\partial r}{\partial s}(s,0)\right\rangle$
$\displaystyle=\frac{1}{2}\frac{\partial}{\partial
t}\left.\left\langle\frac{\partial r}{\partial s}(s,t),\frac{\partial
r}{\partial s}(s,t)\right\rangle\right\rvert_{t=0}$
$\displaystyle=\frac{1}{2}\frac{\partial}{\partial
t}\left.\left\langle\gamma_{Z(t)}^{\prime}(s),\gamma_{Z(t)}^{\prime}(s)\right\rangle\right\rvert_{t=0}$
$\displaystyle=\frac{1}{2}\frac{d}{dt}\left.(\left\lVert
Z(t)\right\rVert^{2})\right\rvert_{t=0}\;,$

which is zero if and only if we can choose $Z(t)$ to be a curve in $SM$, which
is true if and only if $\xi\in T_{v}(SM)$. The following result is now
straightforward to verify. For (5) we also need the result that $KY(v)=0$ and
$d\pi Y(v)=v$ for all $v\in TM$, which follows from the definitions.

######  Proposition 2.10.

Let $v\in TM$. Then

  1. (1)

$\xi\to J_{\xi}$ is a linear isomorphism of $T_{v}(TM)$ onto $J(\gamma_{v})$,

  2. (2)

$J_{\xi}(t)=d\pi\circ dg^{t}(\xi)$ and $J_{\xi}^{\prime}(t)=K\circ
dg^{t}(\xi)$ for $t\in\mathbb{R}$,

  3. (3)

$\xi\in T_{v}(TM)$ lies in $T_{v}(SM)$ for $v\in SM$ if and only if
$\left\langle K\circ dg^{t}(\xi),g^{t}v\right\rangle=\left\langle
J_{\xi}^{\prime}(t),\gamma_{v}^{\prime}(t)\right\rangle=0$ for all
$t\in\mathbb{R}$, if and only if $t\to\left\langle
J_{\xi}(t),\gamma_{v}^{\prime}(t)\right\rangle$ is a constant function,

  4. (4)

$\left\langle\xi,Y(v)\right\rangle=0$ for $v\in SM$ and $\xi\in T_{v}(SM)$ if
and only if $\left\langle J_{\xi}(t),\gamma_{v}^{\prime}(t)\right\rangle=0$
for all $t\in\mathbb{R}$, where $Y$ is the flow vector field, and

  5. (5)

$\left\langle dg^{t}\xi,Y(dg^{t}v)\right\rangle=0$ for all $t\in\mathbb{R}$,
if $v\in SM$ and $\xi\in T_{v}(SM)$ satisfy
$\left\langle\xi,Y(v)\right\rangle=0$.

######  Remark 2.11.

It follows from (2) above that for any $t\in\mathbb{R}$, any $v\in TM$ and any
$\xi\in T_{v}(TM)$ that $\left\lVert dg^{t}\xi\right\rVert^{2}=\left\lVert
J_{\xi}(t)\right\rVert^{2}+\left\lVert J_{\xi}^{\prime}(t)\right\rVert^{2}$.

######  Definition 2.12.

Let $\gamma_{n}$ be a sequence of geodesics in $M$, and let $J_{n}$ be a
sequence of Jacobi fields such that $J_{n}$ is defined on $\gamma_{n}$ for
every $n$. If $v_{n}=\gamma_{n}^{\prime}(0)$, choose $\xi_{n}\in
T_{v_{n}}(TM)$ so that $J_{n}=J_{\xi_{n}}$. We say that the Jacobi fields
$J_{n}$ converge to a Jacobi field $J$ on a geodesic $\gamma$ if
$\xi_{n}\to\xi$ in $T(TM)$, where $v=\gamma^{\prime}(0)$, $\xi\in T_{v}(TM)$,
and $J=J_{\xi}$.

Note that $\xi_{n}\to\xi$ in $T(TM)$ if and only if $K\xi_{n}\to K\xi$ and
$d\pi\xi_{n}\to d\pi\xi$. Therefore $J_{n}$ converges to $J$ if and only if
$\gamma_{n}^{\prime}(0)\to\gamma^{\prime}(0)$, $J_{n}(0)\to J(0)$, and
$J_{n}^{\prime}(0)\to J^{\prime}(0)$. If $J_{n}\to J$ and
$(u_{n})\subseteq\mathbb{R}$ is a sequence converging to a finite number $u$,
then by (2) of Proposition 2.10, $J_{n}(u_{n})\to J(u)$ and
$J_{n}^{\prime}(u_{n})\to J^{\prime}(u)$.

######  Definition 2.13.

$M$ is said to be _compactly homogenous_ if there exists a compact set
$B\subseteq M$ such that $M$ is the union of all translates of $B$ by the
isometries of $M$.

If $M$ is homogenous or a Galois Riemannian covering of a compact Riemannian
manifold, then $M$ is compactly homogenous. Since isometries preserve
sectional curvature, all values of the sectional curvature of $M$ (or Gaussian
curvature if $M$ is a surface) are taken at points of $B$, and therefore the
sectional curvature of $M$ is uniformly bounded above and below.

######  Proposition 2.14.

Let $M$ be compactly homogenous. For each integer $n>0$, let $J_{n}$ be a
Jacobi field on a geodesic $\gamma_{n}$ with initial velocity $v_{n}$. If each
of the sequences $\left\lVert v_{n}\right\rVert$, $\left\lVert
J_{n}(0)\right\rVert$, and $\left\lVert J_{n}^{\prime}(0)\right\rVert$ is
uniformly bounded above, then we can find a sequence $\varphi_{n}$ of
isometries of $M$ and a Jacobi field $L$ on a geodesic $\sigma$, such that by
passing to a subsequence, $L_{n}=d\varphi_{n}J_{n}\to L$ in the sense of
Definition 2.12.

###### Proof.

Let $p_{n}=\pi v_{n}$ and choose a sequence $\varphi_{n}$ of isometries of $M$
so that the sequence $q_{n}=\varphi_{n}p_{n}$ is contained in a compact subset
of $M$. $L_{n}=d\varphi_{n}J_{n}$ is a Jacobi field on
$\sigma_{n}=\varphi_{n}\circ\gamma_{n}$, and
$L_{n}^{\prime}(0)=d\varphi_{n}J_{n}^{\prime}(0)$. If
$w_{n}=\sigma_{n}^{\prime}(0)=d\varphi_{n}v_{n}$, then the sequences
$\left\lVert w_{n}\right\rVert$, $\left\lVert L_{n}(0)\right\rVert$, and
$\left\lVert L_{n}^{\prime}(0)\right\rVert$ are uniformly bounded above, since
$d\varphi_{n}$ is an isometry. Passing to a subsequence, let $q_{n}\to q$ in
$M$ and let $U$ be a local coordinate system about $q$. For each $n$, the
three vectors $w_{n}$, $L_{n}(0)$, and $L_{n}^{\prime}(0)$ lie in
$T_{q_{n}}M$, and relative to the induced coordinate system $\pi^{-1}(U)$ in
$TM$, it is easy to see that the coordinates of each of the sequences $w_{n}$,
$L_{n}(0)$, and $L_{n}^{\prime}(0)$ are uniformly bounded in absolute value,
so they are contained in a compact subset of $TM$. Passing to a further
subsequence, let $w_{n}\to w$, $L_{n}(0)\to u$, and $L_{n}^{\prime}(0)\to v$,
where $u$, $v$, and $w$ are vectors in $T_{q}M$. Let $\sigma=\gamma_{w}$, and
let $L$ be the Jacobi field on $\sigma$ such that $L(0)=u$ and
$L^{\prime}(0)=v$. Then $L_{n}\to L$ by the discussion following Definition
2.12. ∎

If $\varphi$ is an isometry of $M$, then $T_{\varphi}=d\varphi$ is an isometry
of $TM$, which leaves $SM$ invariant. The following result is equivalent to
that just proved.

######  Proposition 2.15.

Let $M$ be compactly homogenous. Let $(v_{n})\subseteq TM$ and
$(\xi_{n})\subseteq T(TM)$ be sequences such that $\xi_{n}\in T_{v_{n}}(TM)$
for every $n$, and that the sequences $\left\lVert v_{n}\right\rVert$ and
$\left\lVert\xi_{n}\right\rVert$ are each uniformly bounded above. Then there
exists a sequence $\varphi_{n}$ of isometries of $M$ such that by passing to a
subsequence,$\xi_{n}^{*}=dT_{\varphi_{n}}\xi_{n}$ converges to a vector
$\xi^{*}\in T(TM)$.

##  3\. Surfaces without conjugate points

From here on, assume $M$ is a _surface_. In this section we continue building
our framework in order to prove the main result in §4, and we focus on the
case where $M$ has no conjugate points. In many cases, we consider the unit
tangent bundle $SM$ instead of $TM$. Assume all geodesics are unit speed. The
main goal of this section is to define the $1$-dimensional subspaces
$X_{s}(v)$ and $X_{u}(v)$ of $T_{v}(SM)$, which, as we will eventually show,
coincide precisely with the subspaces $X_{s}^{*}(v)$ and $X_{u}^{*}(v)$ in the
definition of an Anosov flow when certain conditions are met. We will also
state and prove several propositions relevant to these subspaces that will be
useful in the final section.

We start by defining a useful method of working with the tangent space along a
geodesic. Let $\gamma$ be a geodesic, and let $E_{1}(s)$, $E_{2}(s)$ be a
parallel orthonormal frame along $\gamma$ with $E_{2}(s)=\gamma^{\prime}(s)$;
indeed, $E_{1}(s)$ can be defined by parallel transport. Then a perpendicular
vector field $J$ along $\gamma$ can be written as $J(s)=y(s)E_{1}(s)$, and we
can identify $J$ with the “curve“ $s\longmapsto y(s)$ in $\mathbb{R}$. Note
that $J^{\prime}(s)=y^{\prime}(s)E_{1}(s)$. We call such a frame an adapted
frame field.

Using an adapted frame field along $\gamma$, consider the following Jacobi
equation in local coordinates

$f^{\prime\prime}(s)+\kappa(s)f(s)=0\;,$

where $\kappa(s)$ is the Gaussian curvature. Then the solutions $f(s)$ to this
equation correspond precisely to the perpendicular Jacobi fields
$J(s)=f(s)E_{1}(s)$ along $\gamma$.

Next, recall the following useful definition from differential equations.

######  Definition 3.1.

Let $f(s)$ and $g(s)$ be differentiable. Then the Wronskian $W(f,g)(s)$ is
given by

$W(f,g)(s)=f^{\prime}(s)g(s)-f(s)g^{\prime}(s)\;.$

We have that for any two solutions $x,y$ of (J), the Wronskian is constant,
for

$\displaystyle W(f,g)^{\prime}(s)$
$\displaystyle=f^{\prime\prime}(s)g(s)-f(s)g^{\prime\prime}(s)$
$\displaystyle=-\kappa(s)f(s)g(s)+\kappa(s)f(s)g(s)$ $\displaystyle=0\;.$

Now we define some Jacobi fields that will be helpful soon. Let $a(s)$ be the
unique solution to (J) such that

$a(0)=0\text{ and }a^{\prime}(0)=1\;.$

Since $M$ has no conjugate points, $a(s)\neq 0$ for all $s\neq 0$. In
particular, $a(s)<0$ for $s<0$ and $a(s)>0$ for $s>0$ by the above conditions.
Next, for $t>0$, let $d_{t}(s)$ be the unique solution to (J) such that

$d_{t}(0)=1\text{ and }d_{t}(t)=0\;.$

In fact, for $s>0$, $d_{t}(s)$ is given by

$d_{t}(s)=a(s)\int_{s}^{t}a(u)^{-2}du\;,$

where the superscript denotes exponentiation, not function composition and
inversion. To see this, let $b_{t}(s)$ be the integral expression above for
$s>0$. Then direct computation shows that $b_{t}(s)$ satisfies (J) and that
$b_{t}(t)=0$ and $b_{t}^{\prime}(t)=-1/a(t)$. Thus for $s>0$, $b_{t}(s)$ is
the unique solution of (J) such that $b_{t}(t)=0$ and
$b_{t}^{\prime}(t)=-1/a(t)$. Extend $b_{t}(s)$ to all $s\in\mathbb{R}$ using
the existence and uniqueness theorem on (J). It is left to show that
$b_{t}(0)=1$, for then again by the existence and uniqueness theorem, we shall
have $b_{t}(s)\equiv d_{t}(s)$ for all $s\in\mathbb{R}$. Indeed, we have
$W(a,b_{t})(t)=-a(t)b_{t}^{\prime}(t)=1$, and since the Wronskian is constant,

$1=W(a,b_{t})(0)=b_{t}(0)\;,$

as desired.

Next, for $0<r<t$ and all $s>0$, we have

$d_{t}(s)-d_{r}(s)=a(s)\int_{r}^{t}a(u)^{-2}du\;.$

Notice that since $a(s)>0$ for $s>0$, $d_{t}(s)$ is strictly increasing in $t$
for fixed $s>0$. By direct computation,

$d_{t}^{\prime}(0)-d_{r}^{\prime}(0)=\int_{r}^{t}a(u)^{-2}du\;,$

which is positive and strictly increasing in $t$. We are now ready for the
following result, which will be useful for defining $X_{s}(v)$ and $X_{u}(v)$.

######  Proposition 3.2.

With the definitions from above, we have

  1. (1)

$\lim_{t\to\infty}d_{t}^{\prime}(0)$ exists.

  2. (2)

Let $d(s)$ be the solution of (J) such that

$d(0)=1\text{ and }d^{\prime}(0)=\lim_{t\to\infty}d_{t}^{\prime}(0)\;.$

Then $d(s)=\lim_{t\to\infty}d_{t}(s)$.

  3. (3)

$d(s)\neq 0$ for all $s$.

###### Proof.

This proof is an adaptation of Lemma 1 in [Gre58]. For (1), it suffices to
show that $\lim_{t\to\infty}d_{t}^{\prime}(0)-d_{r}^{\prime}(0)$ exists for
fixed $r$ with $0<r<t$. Given $q>0$, we have by direct computation that the
solution $d_{-q}$ of (J) for which $d_{-q}(0)=1$ and $d_{-q}(-q)=0$ can be
given by

$d_{-q}(s)=a(s)n_{t,-q}+d_{t}(s)$

for any $t>0$, where

$n_{t,-q}=-a(-q)^{-1}d_{t}(-q)\;.$

Notice that the choice of $t$ does not matter by uniqueness of $d_{-q}$. Next,
we claim that $n_{t,-q}$ is positive. To see this, we have $a(s)<0$ when
$s<0$, as previously mentioned. Next, since $d_{t}(0)=1$ and $d_{t}(t)=0$, the
fact that $M$ has no conjugate points implies $d_{t}(s)>0$ for $s<t$, so
$d_{t}(-q)>0$. By direct computation, we have

$d_{-1}^{\prime}(0)-d_{t}^{\prime}(0)=n_{t,-1}>0$

for any $t>0$. Thus $d_{t}^{\prime}(0)-d_{r}^{\prime}(0)$ is increasing in $t$
and bounded above by $d_{-1}^{\prime}(0)-d_{r}^{\prime}(0)$, and hence the
desired limit exists. The fact that $d(s)=\lim_{t\to\infty}d_{t}(s)$ follows
from the fact that the solutions of (J) depend continuously on the initial
conditions.

Finally, we show that $d(s)\neq 0$ for all $s$. Assume toward contradiction
that $d(t_{0})=0$ for some $t_{0}\in\mathbb{R}$. Since $d(0)=1$ by definition,
we have that $t_{0}\neq 0$, and by uniqueness of the solutions of (J), we have
$d(s)\equiv d_{t_{0}}(s)$. Since $d_{t}(s)$ is strictly increasing in $t$ for
$t>0$ for fixed $s>0$, we have $d_{t_{0}}(s)>d_{t}(s)$ for $t>0$ and $s>0$,
which means $t_{0}$ cannot be positive. Thus $t_{0}=-q_{0}$ for some
$q_{0}>0$, so $d(s)\equiv d_{-q_{0}}(s)$. Notice that since
$d_{-q}^{\prime}(0)=n_{t,-q}+d_{t}^{\prime}(0)$, we have $d_{-q}^{\prime}(0)$
is continuous in $q$ for $q>0$. Let $U$ be a positive interval containing
$q_{0}$. By the existence and uniqueness theorem, we have that the map
$q\longmapsto d_{-q}^{\prime}(0)$ sends $U$ homeomorphically onto an interval
containing $d_{-q}^{\prime}(0)$. Since
$\lim_{t\to\infty}d_{t}^{\prime}(0)=d^{\prime}(0)=d_{-q_{0}}^{\prime}(0)$,
this implies that $d_{t}^{\prime}(0)=d_{-q}^{\prime}(0)$ for some $t>0$ and
$q>0$, a contradiction by the existence and uniqueness theorem. Thus $d(s)$ is
nonvanishing. ∎

######  Remark 3.3.

For $s>0$, $d(s)$ can be given by

$d(s)=a(s)\int_{s}^{\infty}a(u)^{-2}du\;.$

######  Corollary 3.4.

Again with the definitions from above, we have

  1. (1)

$\lim_{t\to-\infty}d_{t}^{\prime}(0)$ exists.

  2. (2)

Let $\overline{d}(s)$ be the solution of (J) such that

$\overline{d}(0)=1\text{ and
}\overline{d}^{\prime}(0)=\lim_{t\to-\infty}d_{t}^{\prime}(0)\;.$

Then $\overline{d}(s)=\lim_{t\to-\infty}d_{t}(s)$.

  3. (3)

$\overline{d}(s)\neq 0$ for all $s$.

###### Proof.

Apply the results from the preceding proposition to the reverse geodesic
$\gamma(-s)$. ∎

Next, for any number $t\neq 0$, we define a linear map
$\xi\to\xi_{t}:T_{v}(TM)\to T_{v}(TM)$ for every $v\in TM$. Given a vector
$v\in TM$ and a vector $\xi\in T_{v}(TM)$, let $J(s)$ be the unique Jacobi
field along $\gamma_{v}$ such that

$J(0)=d\pi(\xi)\text{ and }J(t)=0\;.$

Let $\xi_{t}\in T_{v}(TM)$ be the vector that corresponds to $J$ under our
isomorphism, i.e., we have $J=J_{\xi_{t}}$. Thus $\xi_{t}$ is the unique
vector such that

$d\pi(\xi_{t})=J_{\xi_{t}}(0)=d\pi(\xi)\text{ and }d\pi\circ
dg^{t}(\xi_{t})=J_{\xi_{t}}(t)=0\;.$

The kernel of this map is $\ker d\pi$, which is vertical subspace of
$T_{v}(TM)$.

For $v\in SM$ it is not true in general that this map leaves $T_{v}(SM)$
invariant; in fact, for $\xi\in T_{v}(SM)$ we have $\xi_{t}\in T_{v}(SM)$ if
and only if $\left\langle\xi,Y(v)\right\rangle=0$. Indeed, if $\xi_{t}\in
T_{v}(SM)$, then $s\to\left\langle
J_{\xi_{t}}(s),\gamma_{v}^{\prime}(s)\right\rangle$ is a constant function by
Proposition 2.10 (3), but since $J_{\xi_{t}}(t)=0$, we have $\left\langle
J_{\xi_{t}}(s),\gamma_{v}^{\prime}(s)\right\rangle\equiv 0$, and by
Proposition 2.10 (5), $\left\langle\xi,Y(v)\right\rangle=0$. Conversely,
suppose $\left\langle\xi,Y(v)\right\rangle=0$. Since $KY(v)=0$ and $d\pi
Y(v)=v$, we have

$0=\left\langle\xi,Y(v)\right\rangle=\left\langle
d\pi(\xi),v\right\rangle=\left\langle
d\pi(\xi_{t}),v\right\rangle=\left\langle
J_{\xi_{t}}(0),\gamma_{v}^{\prime}(0)\right\rangle\;.$

Since $J_{\xi_{t}}(t)=0$, we also have that it is perpendicular at $t$, and
hence perpendicular everywhere. Thus again by Proposition 2.10 (3),
$\xi_{t}\in T_{v}(SM)$.

######  Definition 3.5.

Let $v\in SM$. We define

  1. (i)

$X_{s}(v)=\left\\{\xi\in T_{v}(SM)\text{ such that
}\left\langle\xi,Y(v)\right\rangle=0\text{ and }\xi_{t}\to\xi\text{ as
}t\to\infty\right\\}$, and

  2. (ii)

$X_{u}(v)=\left\\{\xi\in T_{v}(SM)\text{ such that
}\left\langle\xi,Y(v)\right\rangle=0\text{ and }\xi_{t}\to\xi\text{ as
}t\to-\infty\right\\}$.

These are called the _stable_ and _unstable subspaces_ determined by $v$.

######  Definition 3.6.

Let $\gamma$ be a unit speed geodesic in $M$ with initial velocity $v$. We
define $J_{s}(\gamma)$ and $J_{u}(\gamma)$ to be the images in $J(\gamma)$ of
$X_{s}(v)$ and $X_{u}(v)$, respectively, under our isomorphism. These are
called the _stable_ and _unstable subspaces_ of perpendicular Jacobi fields
along $\gamma$.

######  Remark 3.7.

(1) Let $\xi\in X_{s}(v)$. Since $J_{\xi_{t}}$ is perpendicular, we have that
relative to an adapted frame field $E_{1},E_{2}$ along $\gamma$ with
$E_{2}(s)=\gamma^{\prime}(s)$, $J_{\xi_{t}}$ can be given by

$J_{\xi_{t}}(s)=cd_{t}(s)E_{1}(s)$

for an appropriately chosen $c\in\mathbb{R}$ so that
$d\pi\xi=cd_{t}(0)E_{1}(0)=cE_{1}(0)$. Notice that $c$ does not depend on $t$.
Taking $t\to\infty$ shows that

$J_{\xi}(s)=cd(s)E_{1}(s)\;.$

For $\xi\in X_{u}(v)$, we have by similar reasoning that
$J_{\xi}(s)=c\overline{d}(s)E_{1}(s)$. Therefore, we can express
$J_{s}(\gamma)$ and $J_{u}(\gamma)$ as

  1. (i)

$J_{s}(\gamma)=\left\\{cd(s)E_{1}(s):c\in\mathbb{R}\right\\}$ and

  2. (ii)

$J_{u}(\gamma)=\left\\{c\overline{d}(s)E_{1}(s):c\in\mathbb{R}\right\\}$.

Clearly these are 1-dimensional vector subspaces of $J(\gamma)$, which means
$X_{s}(v)$ and $X_{u}(v)$ are1-dimensional vector subspaces of $T_{v}(SM)$.

(2) Let $\gamma$ be a unit speed geodesic in $M$. If $\kappa\equiv 0$, then
$J_{s}(\gamma)=J_{u}(\gamma)=$ the vector space of all perpendicular parallel
vector fields on $\gamma$. It will follow by Theorem 4.2 that for a compact
surface $M$ without conjugate points the geodesic flow on $SM$ is of Anosov
type if and only if $J_{s}(\gamma)\cap J_{u}(\gamma)=\left\\{0\right\\}$ for
every unit speed geodesic $\gamma$ in $M$.

######  Proposition 3.8.

If $p:N\to M$ is a surjective local isometry of complete Riemannian manifolds
and $M$ has no conjugate points, then $N$ has no conjugate points, and

$dPX_{s}(v)=X_{s}(Pv)\text{ and }dPX_{u}(v)=X_{u}(Pv)$

for every $v\in SN$, where $P=dp:TN\to TM$.

###### Proof.

$N$ has no conjugate points since for any Jacobi field $J$ on a geodesic
$\gamma$ in $N$, $PJ$ is a Jacobi field on the geodesic $p\circ\gamma$ in $M$.
If $\pi_{1}:TN\to N$ and $\pi_{2}:TM\to M$ are the projection maps, then
$p\circ\pi_{1}=\pi_{2}\circ P$. We also have $P\circ g^{t}=g^{t}\circ P$ for
every $t\in\mathbb{R}$, where $g^{t}$ is the geodesic flow on both $TN$ and
$TM$. From these two relations it follows that $(dP\xi)_{t}=dP(\xi_{t})$ for
any $t\neq 0$, any $v\in TN$, and any $\xi\in T_{v}(TN)$. If
$\left\langle\xi,Y(v)\right\rangle=0$, then $\left\langle
dP\xi,Y(Pv)\right\rangle=0$ by Proposition 2.7. Therefore
$dPX_{s}(v)=X_{s}(Pv)$ and $dPX_{u}(v)=X_{u}(Pv)$. ∎

######  Proposition 3.9.

We have

  1. (1)

If $S:SM\to SM$ is the map which takes a vector $v$ to $-v$, then
$X_{u}(-v)=dSX_{s}(v)$ and $X_{s}(-v)=dSX_{u}(v)$, and

  2. (2)

For any $t\in\mathbb{R}$ and any $v\in SM$, $dg^{t}X_{s}(v)=X_{s}(g^{t}v)$ and
$dg^{t}X_{u}(v)=X_{u}(g^{t}v)$.

###### Proof of (1).

Let $\xi\in T_{v}(TM)$ and $t\neq 0$. It suffices to show that
$dS(\xi_{t})=(dS\xi)_{-t}$. By definition of $(dS\xi)_{-t}$, it suffices to
show that $d\pi\circ dS(\xi_{t})=d\pi\circ dS(\xi)$ and $d\pi\circ
dg^{-t}\circ dS(\xi_{t})=0$. The first condition follows from the fact that
$\pi\circ S=\pi$. The second condition also requires the fact that $S\circ
g^{a}=g^{-a}\circ S$ for any $a\in\mathbb{R}$, which we have by

$S\circ
g^{a}(v)=S(\gamma_{v}^{\prime}(a))=-\gamma_{v}^{\prime}(a)=\gamma_{-v}^{\prime}(-a)=g^{-a}(-v)=g^{-a}\circ
S(v)\;.$

∎

Before we prove (2), we need the following lemma.

######  Lemma 3.10.

Let $v\in SM$ be given. Then there exist numbers $b=b(v)>0$ and
$t_{0}=t_{0}(v)>0$ such that if $\xi\in T_{v}(TM)$ satisfies $d\pi\circ
dg^{t}(\xi)=0$ for some $t\geqslant t_{0}$, then $\left\lVert
K\xi\right\rVert\leqslant b\left\lVert d\pi\xi\right\rVert$.

###### Proof.

We may assume that $d\pi\xi\neq 0$, or otherwise the Jacobi field
$J_{\xi}\equiv 0$ since it vanishes at $0$ and $t\geqslant t_{0}>0$, and
$K\xi=J_{\xi}^{\prime}(0)=0$. First, we consider the case when $J_{\xi}$ is
perpendicular to $\gamma_{v}$. Relative to an adapted frame field along
$\gamma_{v}$, if $d\pi\circ dg^{t}(\xi)=0$ for some $t>0$, we have
$J_{\xi}=cd_{t}(s)E_{1}(s)$, with $c$ chosen so that
$cd_{t}(0)E_{1}(0)=cE_{1}(0)=d\pi\xi$. Thus we have

$\left\lVert K\xi\right\rVert=\left\lVert
J_{\xi}^{\prime}(0)\right\rVert=\left\lvert
d_{t}^{\prime}(0)\right\rvert\left\lvert c\right\rvert=\left\lvert
d_{t}^{\prime}(0)\right\rvert\left\lVert d\pi\xi\right\rVert\;.$

Since $d_{t}^{\prime}(0)\to d^{\prime}(0)$ as $t\to\infty$, we can choose
$t_{0}=t_{0}(v)>1$ so that $\left\lvert d_{t}^{\prime}(0)\right\rvert\leqslant
1+\left\lvert d^{\prime}(0)\right\rvert$ for $t\geqslant t_{0}$. Setting
$b=1+\left\lvert d^{\prime}(0)\right\rvert$, if $d\pi\circ dg^{t}(\xi)=0$ for
some $t\geqslant t_{0}$, then $\left\lVert K\xi\right\rVert\leqslant
b\left\lVert d\pi\xi\right\rVert$.

If $J$ is an arbitrary Jacobi field on $\gamma$, we may write
$J(s)=J_{1}(s)+J_{2}(s)$, where $J_{1}$ is a perpendicular Jacobi field and
$J_{2}$ is a tangential Jacobi field of the form $J_{2}(s)=(\alpha
s+\beta)\gamma^{\prime}(s)$ for suitable constants $\alpha$ and $\beta$. If
$J(t)=0$ for some $t\geqslant t_{0}$, then $J_{1}(t)=J_{2}(t)=0$, and hence
$\alpha=-\beta/t$. Since $t\geqslant t_{0}>1$, we have $\left\lVert
J_{2}^{\prime}(0)\right\rVert=\left\lvert\alpha\right\rvert\leqslant\left\lvert\beta\right\rvert=\left\lVert
J_{2}(0)\right\rVert$ and $\left\lVert J_{1}^{\prime}(0)\right\rVert\leqslant
b\left\lVert J_{1}(0)\right\rVert$. Since $b\geqslant 1$, we have

$\displaystyle\left\lVert K\xi\right\rVert^{2}=\left\lVert
J^{\prime}(0)\right\rVert^{2}$ $\displaystyle=\left\lVert
J_{1}^{\prime}(0)\right\rVert^{2}+\left\lVert
J_{2}^{\prime}(0)\right\rVert^{2}$ $\displaystyle\leqslant b^{2}\left\lVert
J_{1}(0)\right\rVert^{2}+b^{2}\left\lVert J_{2}(0)\right\rVert^{2}$
$\displaystyle=b^{2}\left\lVert J(0)\right\rVert^{2}\;,$

and the result follows. ∎

###### Proof of (2).

Let $v\in SM$ and let $\xi\in T_{v}(SM)$ be given, and fix a number
$a\in\mathbb{R}$. For any $t\neq 0$, we have by the triangle inequality that
$\left\lVert dg^{a}\xi-(dg^{a}\xi)_{t}\right\rVert\leqslant\left\lVert
dg^{a}(\xi-\xi_{t+a})\right\rVert+\left\lVert
dg^{a}(\xi_{t+a})-(dg^{a}\xi)_{t}\right\rVert$. If we let
$\psi_{t}=dg^{a}(\xi_{t+a})-(dg^{a}\xi)_{t}$, then $d\pi\circ
dg^{t}\psi_{t}=0$ and $d\pi\psi_{t}=d\pi\circ dg^{a}(\xi_{t+a}-\xi)$.
If$t\geqslant t_{0}=t_{0}(g^{a}v)>0$, then by the previous lemma, $\left\lVert
K\psi_{t}\right\rVert\leqslant b\left\lVert d\pi\psi_{t}\right\rVert$, where
$b=b(g^{a}v)>0$ is independent of $t$ and $\psi_{t}$. Therefore
$\left\lVert\psi_{t}\right\rVert^{2}=\left\lVert
K\psi_{t}\right\rVert^{2}+\left\lVert
d\pi\psi_{t}\right\rVert^{2}\leqslant(1+b^{2})\left\lVert d\pi\circ
dg^{a}(\xi_{t+a}-\xi)\right\rVert^{2}\leqslant(1+b^{2})\left\lVert
dg^{a}(\xi_{t+a}-\xi)\right\rVert^{2}$, and

$\left\lVert
dg^{a}\xi-(dg^{a}\xi)_{t}\right\rVert\leqslant[1+(1+b^{2})^{1/2}]\left\lVert
dg^{a}(\xi_{t+a}-\xi)\right\rVert$

for $t\geqslant t_{0}$. If $\left\langle\xi,Y(v)\right\rangle=0$, then
$\left\langle dg^{a}\xi,Y(g^{a}v)\right\rangle=0$ by Proposition 2.10 (5).
Since $a\in\mathbb{R}$ is fixed, $\xi\in X_{s}(v)$ if and only if
$dg^{a}\xi\in X_{s}(g^{a}v)$. This fact and (1) of this Proposition imply the
similar invariance relation for $X_{u}$. ∎

######  Proposition 3.11.

Assume the Gaussian curvature $\kappa$ of $M$ is bounded from below, so
$\kappa>-k^{2}$ for some $k>0$, and let $J$ be a perpendicular Jacobi field on
a unit speed geodesic $\gamma$ in $M$ such that $J(0)=0$. Then $\left\lVert
J^{\prime}(s)\right\rVert\leqslant k\coth(ks)\left\lVert J(s)\right\rVert$ for
every $s>0$.

To show this, we first state and prove two lemmas, the first of which will
also be important later.

######  Lemma 3.12.

Let $r(s)$ be a real-valued function defined for all $s\in\mathbb{R}$ such
that $r(s)\geqslant-k^{2}$ for all $s$ for some $k>0$. If $u(s)$ is a solution
of the Ricatti equation

$u^{\prime}(s)+u(s)^{2}+r(s)=0\;,$

defined everywhere, then $|u(s)|\leqslant k$.

###### Proof.

This proof is an adaptation of Lemma 2.1 in [Gre54]. Let $b>k$ and suppose
there exists $s_{0}$ with $u(s_{0})>b$. Then there is a solution $v(s)$ of

$v^{\prime}(s)=b^{2}-v(s)^{2}$

for which $v(s_{0})=u(s_{0})$. For a suitably chosen constant $d$, we have
$v(s)=b\coth(bs-d)$. Subtracting the above by (R), we have

$v^{\prime}(s)-u^{\prime}(s)=b^{2}+r(s)-v(s)^{2}+u(s)^{2}\;.$

For $s=s_{0}$, we have $v^{\prime}(s_{0})-u^{\prime}(s_{0})=b^{2}+r(s)>0$, so

$v(s)\leqslant u(s)$

in at least a left-sided neighborhood $s_{0}-\varepsilon\leqslant s\leqslant
s_{0}$ of $s_{0}$. This, however, implies that the inequality holds for
$s\leqslant s_{0}$, since for any $s_{1}$ where $v(s_{1})=u(s_{1})$, we have
that $v^{\prime}(s_{1})-u^{\prime}(s_{1})>0$, and so $v(s)\leqslant u(s)$ for
a left-sided neighborhood of $s_{1}$. Now if we observe the behavior of
$v(s)$, we notice that $\lim_{s\to d/b^{+}}v(s)=\infty$, so $\lim_{s\to
d/b^{+}}u(s)=\infty$, but this contradicts the fact that $u(s)$ is defined
everywhere. Since $b$ can be taken arbitrarily close to $k$, we have
$u(s)\leqslant k$.

For the lower bound, suppose there exists $s_{0}$ with $u(s_{0})<-b$. We then
consider $w(s)-u(s)$, where $w(s)=b\coth(bs-c)$ with $c$ chosen such that
$w(s_{0})=u(s_{0})$. Using arguments similar to the above, we have
$w^{\prime}(s_{0})-u^{\prime}(s_{0})>0$, so $w(s)\geqslant u(s)$ for a right-
sided neighborhood $s_{0}\leqslant s\leqslant s_{0}+\varepsilon$ of $s_{0}$,
and hence for all $s\geqslant s_{0}$. Since $\lim_{s\to c/b^{-}}v(s)=-\infty$,
we also have $\lim_{s\to c/b^{-}}u(s)=-\infty$, which again contradicts the
fact that $u(s)$ is defined everywhere. Thus $u(s)\geqslant-k$. ∎

######  Lemma 3.13.

If $u(s)$ is a solution of the Ricatti equation (R) defined for all $s>0$,
then $\left\lvert u(s)\right\rvert\leqslant k\coth(ks)$ for all $s>0$.

###### Proof.

The first part of this proof is an adaptation of Lemma 3 in [Gre58]. For fixed
$s_{0}>0$, choose a number $d>0$ such that $u(s_{0})<k\coth(ks_{0}-d)$. We
show that $u(s)<k\coth(ks-d)$ for $s\geqslant s_{0}$. We have
$v(s)=k\coth(ks-d)$ is a solution to the equation

$v^{\prime}(s)+v(s)^{2}-k^{2}=0\;,$

defined for $s>d/k>0$. If $f(s)=u(s)-v(s)$, then $f(s_{0})<0$. If $f(s)=0$ for
some $s>s_{0}$, let $s_{1}$ be the first such value. Differentiating at
$s_{1}$, we obtain

$f^{\prime}(s_{1})=v(s_{1})^{2}-u(s_{1})^{2}-r(s_{1})-k^{2}=-r(s_{1})-k^{2}<0\;.$

This means that $f(s)>0$ in some left-sided neighborhood
$s_{1}-\varepsilon<s<s_{1}$ of $s_{1}$, but since $s_{0}<s_{1}$ and
$f(s_{0})<0$, this contradicts the fact that $s_{1}$ is the first point after
$s_{0}$ for which $f(s)=0$. Thus $u(s)<k\coth(ks-d)$ for all $s\geqslant
s_{0}$.

Next, for each integer $n>0$, choose a number $d_{n}>0$ such that
$u(1/n)<k\coth(k/n-d_{n})$. If $s>0$ is given, then $s>1/n$ for sufficiently
large $n$, and the above argument shows that $u(s)<k\coth(ks-d_{n})$. Since
$d_{n}<k/n$, $d_{n}\to 0$ and we have $u(s)\leqslant k\coth(ks)$. Finally, the
argument in the last paragraph of the proof of Lemma 3.12 works if $u(s)$ is
defined only for $s>0$; we assume $s_{0}>0$ and contradict the fact that
$u(s)$ is defined for $s>0$, which then gives us that $u(s)\geqslant-k$ for
any $s>0$, completing the proof. ∎

###### Proof of Proposition 3.11.

Let $\gamma$ be a unit speed geodesic in $M$. Relative to an adapted frame
field along $\gamma$, we consider the solution $a(s)$ to equation (J) defined
earlier in this section. Let $J$ be a perpendicular Jacobi field on $\gamma$
such that $J(0)=0$. We may assume $J^{\prime}(0)\neq 0$, since otherwise there
is nothing to prove, and by linearity it suffices to consider the case where
$\left\lVert J^{\prime}(0)\right\rVert=1$. Since $a(s)$ is the unique solution
to (J) with $a(0)=0$ and $a^{\prime}(0)=1$, we have that $J(s)=\pm
a(s)E_{1}(s)$. If $u(s)=a^{\prime}(s)a(s)^{-1}$, then $u(s)$ is defined for
$s>0$ and is a solution to the Ricatti equation

$u^{\prime}(s)+u(s)^{2}+\kappa(s)=0\;.$

Applying the lemma we just proved, we have

$\left\lVert J^{\prime}(s)\right\rVert=\left\lvert
a^{\prime}(s)\right\rvert=\left\lvert u(s)a(s)\right\rvert=\left\lvert
u(s)\right\rvert\left\lvert a(s)\right\rvert=\left\lvert
u(s)\right\rvert\left\lVert J(s)\right\rVert\leqslant k\coth(ks)\left\lVert
J(s)\right\rVert\;.$

∎

######  Proposition 3.14.

Assume the Gaussian curvature $\kappa$ of $M$ is bounded from below, so
$\kappa>-k^{2}$ for some $k>0$. Let $\gamma$ be a unit speed geodesic in $M$.
For any $R>0$ we can find a number $T=T(R,\gamma)>0$ such that $\left\lVert
J(s)\right\rVert\geqslant R\left\lVert J^{\prime}(0)\right\rVert$ for
$s\geqslant T$, where $J$ is any perpendicular Jacobi field on $\gamma$ such
that $J(0)=0$.

###### Proof.

It suffices to consider the case where $\left\lVert
J^{\prime}(0)\right\rVert=1$. As before, we have $J(s)=\pm a(s)E_{1}(s)$
relative to an adapted frame field $E_{1},E_{2}$ along $\gamma$ with
$E_{2}(s)=\gamma^{\prime}(s)$. Thus we want to show that $\left\lvert
a(s)\right\rvert\geqslant R$ for $s\geqslant T$.

For any $s>0$, let
$m(s)=d^{\prime}(0)-d_{s}^{\prime}(0)=\displaystyle\int_{s}^{\infty}a(u)^{-2}du$,
and let $u(s)=a^{\prime}(s)a(s)^{-1}$ and$v(s)=d^{\prime}(s)d(s)^{-1}$. Then
$u(s)$ and $v(s)$ are solutions defined for $s>0$ of the Ricatti equation
$u^{\prime}(s)+u(s)^{2}+\kappa(s)=0$. Since $d(s)=a(s)m(s)$, we have
$d^{\prime}(s)=a^{\prime}(s)m(s)+a(s)m^{\prime}(s)=a^{\prime}(s)m(s)-a(s)^{-1}$,
and consequently

$u(s)-v(s)=a(s)^{-2}m(s)^{-1}\;.$

Choosing $s_{0}>0$ such that $k\coth ks\leqslant 2k$ for $s\geqslant s_{0}$,
we have by Lemma 3.13 that

$\left\lvert a(s)^{-2}m(s)^{-1}\right\rvert\leqslant\left\lvert
u(s)\right\rvert+\left\lvert v(s)\right\rvert\leqslant 4k$

for $s\geqslant s_{0}$, or in other words, that

$\left\lvert a(s)\right\rvert=1/\left\lvert
a(s)^{-1}\right\rvert\geqslant(4k\left\lvert m(s)\right\rvert)^{-1/2}$

for $s\geqslant s_{0}$. Let $R>0$ be given. Since $m(s)\to 0$ as $s\to\infty$,
we may choose $T>s_{0}>0$ such that $s\geqslant T$ implies $\left\lvert
m(s)\right\rvert\leqslant 1/(4kR^{2})$, hence giving us $\left\lvert
a(s)\right\rvert\geqslant R$ for $s\geqslant T$. ∎

######  Proposition 3.15.

Assume that $\kappa>-k^{2}$ for some $k>0$. Then for any $v\in SM$ and any
$\xi\in X_{s}(v)$ or $X_{u}(v)$ we have $\left\lVert K\xi\right\rVert\leqslant
k\left\lVert d\pi\xi\right\rVert$.

###### Proof.

First let $\xi\in X_{s}(v)$. Relative to an adapted frame field along
$\gamma$, we have $\xi$ corresponds to $J_{\xi}=cd(s)E_{1}(s)$ for some
$c\in\mathbb{R}$. Further, for $t\neq 0$, $\xi_{t}$ corresponds to
$J_{\xi_{t}}(s)=cd_{t}(s)E_{1}(s)$. By reversing our geodesic and shifting by
$t$, i.e., by considering the geodesic $\gamma(t-s)$ instead of $\gamma(s)$,
we have the corresponding Jacobi field
$L_{\xi_{t}}(s)=J_{\xi_{t}}(t-s)=d_{t}(t-s)E_{1}(t-s)$. In particular, we have
$L_{\xi_{t}}(0)=J_{\xi_{t}}(t)=0$, so by Proposition 3.11, we have

$\left\lVert L_{\xi_{t}}^{\prime}(s)\right\rVert\leqslant
k\coth(ks)\left\lVert L_{\xi_{t}}(s)\right\rVert$

for $s>0$. Setting $s=t$, we have

$\left\lVert J_{\xi_{t}}^{\prime}(0)\right\rVert=\left\lVert
L_{\xi_{t}}^{\prime}(t)\right\rVert\leqslant k\coth(kt)\left\lVert
L_{\xi_{t}}(t)\right\rVert=k\coth(kt)\left\lVert
J_{\xi_{t}}(0)\right\rVert\;.$

By our isomorphism, we have

$\left\lVert K\xi_{t}\right\rVert\leqslant k\coth(kt)\left\lVert
d\pi(\xi_{t})\right\rVert=k\coth(kt)\left\lVert d\pi\xi\right\rVert\;.$

Taking $t\to\infty$ yields the desired result. A similar approach shows the
case where $\xi\in X_{u}(v)$. ∎

######  Proposition 3.16.

Assume that $\kappa>-k^{2}$ for some $k>0$. Let $v\in SM$ and let $\xi\in
T_{v}(SM)$ be such that $\left\langle\xi,Y(v)\right\rangle=0$ and $\left\lVert
d\pi\circ dg^{t}\xi\right\rVert$ is bounded above for all $t\geqslant 0$
(resp. for all $t\leqslant 0$). Then $\xi\in X_{s}(v)$ (resp. $\xi\in
X_{u}(v)$).

###### Proof.

Suppose $\left\lVert d\pi\circ dg^{t}\xi\right\rVert\leqslant A$ for some
$A>0$. For any $t>0$, $\xi_{t}\in T_{v}(SM)$ by Proposition 2.10 (3), since
$J_{\xi_{t}}$ is perpendicular at 0 and $t$ and hence everywhere. Thus
$\xi-\xi_{t}\in T_{v}(SM)$. By linearity,
$J_{\xi-\xi_{t}}(0)=J_{\xi}(0)-J_{\xi_{t}}(0)=0$ and
$J_{\xi-\xi_{t}}(t)=J_{\xi}(t)$, which is perpendicular by assumption, so
$J_{\xi-\xi_{t}}$ is also perpendicular everywhere. By Proposition 3.14, for
each integer $n>0$ we have

$\left\lVert\xi-\xi_{t}\right\rVert=\left\lVert
K(\xi-\xi_{t})\right\rVert=\left\lVert
J_{\xi-\xi_{t}}^{\prime}(0)\right\rVert\leqslant\left\lVert
J_{\xi-\xi_{t}}(s)\right\rVert/n$

for any $s\geqslant T(n,\gamma)$. Thus for $t\geqslant T(n,\gamma)$,

$\left\lVert\xi-\xi_{t}\right\rVert\leqslant\left\lVert
J_{\xi-\xi_{t}}(t)\right\rVert/n=\left\lVert
J_{\xi}(t)\right\rVert/n=\left\lVert d\pi\circ
dg^{t}\xi\right\rVert/n\leqslant A/n\;,$

so $\xi_{t}\to\xi$ as $t\to\infty$, as desired. The proof for the other case
works similarly. ∎

######  Proposition 3.17.

Assume that $\kappa>-k^{2}$ for some $k>0$. Suppose that there exist constants
$A>0$ and $s_{0}\geqslant 0$ so that for any perpendicular Jacobi vector field
$J$ such that $J(0)=0$ and for any numbers $t\geqslant s\geqslant s_{0}$ we
have $\left\lVert J(t)\right\rVert\geqslant A\left\lVert J(s)\right\rVert$.
Then for any $v\in SM$ and any $\xi\in T_{v}(SM)$, $\xi\in X_{s}(v)$ (resp.
$\xi\in X_{u}(v)$) if and only if $\left\langle\xi,Y(v)\right\rangle=0$ and
$\left\lVert d\pi\circ dg^{t}\xi\right\rVert$ is bounded above for all
$t\geqslant 0$ (resp. for all $t\leqslant 0$). In either case,
$(1/A)\left\lVert d\pi\xi\right\rVert$ is an upper bound.

###### Proof.

The reverse direction follows immediately from the preceding result. For the
forward direction, let $v\in SM$ and $\xi\in X_{s}(v)$, and fix a number
$u\geqslant 0$. Let $t\geqslant u+s_{0}$ and consider the Jacobi field
$L_{\xi_{t}}(s)=J_{\xi_{t}}(t-s)$. Since $L_{\xi_{t}}(0)=0$, we have

$\left\lVert d\pi\circ dg^{u}(\xi_{t})\right\rVert=\left\lVert
L_{\xi_{t}}(t-u)\right\rVert\leqslant(1/A)\left\lVert
L_{\xi_{t}}(t)\right\rVert=(1/A)\left\lVert
d\pi(\xi_{t})\right\rVert=(1/A)\left\lVert d\pi\xi\right\rVert\;.$

Since $\xi_{t}\to\xi$ as $t\to\infty$, we have

$\left\lVert d\pi\circ dg^{u}\xi\right\rVert\leqslant(1/A)\left\lVert
d\pi\xi\right\rVert$

by continuity. The proof for the case where $\xi\in X_{u}(v)$ is similar. ∎

######  Remark 3.18.

This result shows that assuming the hypothesis of Proposition 3.17 and given
any unit speed geodesic $\gamma$ in $M$, a perpendicular Jacobi field $J$ lies
in $J_{s}(\gamma)$ (resp. $J_{u}(\gamma)$) if and only if $\left\lVert
J(t)\right\rVert\leqslant(1/A)\left\lVert J(0)\right\rVert$ for $t\geqslant 0$
(resp. $t\leqslant 0$). We have from Corollary 2.6 that the hypothesis is
satisfied for $A=1$ and $s_{0}=0$.

######  Corollary 3.19.

Let $M$ satisfy the hypothesis of Proposition 3.17 and let
$B=[(1+k^{2})/A^{2}]^{1/2}$. Then for any $v\in SM$ and any $\xi\in
T_{v}(SM)$, $\xi\in X_{s}(v)$ (resp. $X_{u}(v)$) if and only if
$\left\langle\xi,Y(v)\right\rangle=0$ and $\left\lVert dg^{t}\xi\right\rVert$
is bounded above for all $t\geqslant 0$ (resp. $t\leqslant 0$). In either
case, $B\left\lVert\xi\right\rVert$ is an upper bound.

###### Proof.

If $\left\lVert dg^{t}\xi\right\rVert$ is bounded above for all $t\geqslant 0$
(resp. $t\leqslant 0$), then $\left\lVert d\pi\circ
dg^{t}\xi\right\rVert\leqslant\left\lVert dg^{t}\xi\right\rVert$ is bounded
above, and $\xi\in X_{s}(v)$ (resp. $X_{u}(v)$) by the previous proposition.
Conversely, if $\xi\in X_{s}(v)$ (resp. $X_{u}(v)$), then by Propositions 3.9
(2), 3.15, and 3.17 we have for $t\geqslant 0$ (resp. $t\leqslant 0$),

$\left\lVert dg^{t}\xi\right\rVert^{2}=\left\lVert d\pi\circ
dg^{t}\xi\right\rVert^{2}+\left\lVert K\circ
dg^{t}\xi\right\rVert^{2}\leqslant(1+k^{2})\left\lVert d\pi\circ
dg^{t}\xi\right\rVert^{2}\leqslant((1+k^{2})/A^{2})\left\lVert
d\pi\xi\right\rVert^{2}\leqslant B^{2}\left\lVert\xi\right\rVert^{2}\;.$

∎

Let $J_{0}^{*}$ be the set of all perpendicular Jacobi fields on unit speed
geodesics of $M$.

######  Proposition 3.20.

Let the universal cover $H$ of $M$ be compactly homogenous. For each
$s\geqslant 0$, let $g(s)=\inf\left\\{\left\lVert J(s)\right\rVert:J\in
J_{0}^{*},J(0)=0,\text{ and }\left\lVert
J^{\prime}(0)\right\rVert=1\right\\}$. Then $g(s)>0$ for each $s>0$, and $g$
is semi-continuous on $(0,\infty)$.

###### Proof.

Since $g$ is the infimum of a set of continuous functions, we have for any
$\alpha>0$, the set $\left\\{s>0:g(s)<\alpha\right\\}$ is open in
$(0,\infty)$. Thus $g$ is semi-continuous. If $g^{*}(s)$ is the function
defined in the same way relative to Jacobi fields in $H$, then $g^{*}(s)=g(s)$
since $p:H\to M$ is a surjective local isometry, and the Jacobi fields along
geodesics in $M$ are the projections of Jacobi fields along geodesics in $H$.
Let $L_{0}^{*}$ be the set of all perpendicular Jacobi fields on unit speed
geodesics in $H$. To prove $g(s)=g^{*}(s)$ is positive for $s>0$, it suffices
to show for fixed $s>0$ that $g^{*}(s)=\left\lVert L(s)\right\rVert$ for some
$L\in L_{0}^{*}$ such that $L(0)=0$ and $\left\lVert
L^{\prime}(0)\right\rVert=1$.

Let $s>0$ be fixed, and let $(L_{n})\subseteq L_{0}^{*}$ be a sequence of
Jacobi fields with $L_{n}(0)=0$ and $\left\lVert
L_{n}^{\prime}(0)\right\rVert=1$ such that $L_{n}(s)\to g^{*}(s)$ as
$n\to\infty$. This corresponds to a sequence $(\xi_{n})\subseteq T(SH)$ with
$d\pi\xi_{n}=0$ and $\left\lVert\xi_{n}\right\rVert=1$ with $\left\lVert
d\pi\circ dg^{s}\xi_{n}\right\rVert\to g^{*}(s)$ as $n\to\infty$. By
Proposition 2.15, there exist a sequence $\varphi_{n}$ of isometries of $H$
and a vector $\xi^{*}\in T(SH)$ such that
$\xi_{n}^{*}=dT_{\varphi_{n}}\xi_{n}\to\xi^{*}$ by passing to a subsequence,
where $T_{\varphi_{n}}=d\varphi_{n}:TH\to TH$. Since $T_{\varphi_{n}}$ is an
isometry, we have by the relations shown in the proof of Lemma 2.7 that
$d\pi\xi_{n}^{*}=0$, $\left\lVert\xi_{n}^{*}\right\rVert=1$, and

$d\pi\circ dg^{s}\xi_{n}^{*}=d\pi\circ dg^{s}\circ
dT_{\varphi_{n}}(\xi_{n})=T_{\varphi_{n}}\circ d\pi\circ dg^{s}(\xi_{n})\;.$

Therefore,

$\left\lVert d\pi\circ dg^{s}\xi_{n}^{*}\right\rVert=\left\lVert d\pi\circ
dg^{s}\xi_{n}\right\rVert\to g^{*}(s)\;.$

If $L$ is the Jacobi field corresponding to $\xi^{*}$, then
$g^{*}(s)=\left\lVert L(s)\right\rVert$. ∎

##  4\. Anosov equivalences

We are finally able to state and prove the main result of this paper. First we
need the definition of an Anosov flow.

######  Definition 4.1.

Let $\varphi^{t}$ be a complete $C^{\infty}$ flow on a (complete) $C^{\infty}$
Riemannian manifold $N$ of dimension $n\geqslant 3$. Then the flow is said to
be _Anosov_ or of _Anosov type_ if the following conditions are satisfied:

  1. (1)

The vector field $Y=\left.\frac{\partial\varphi^{t}}{\partial
t}\right\rvert_{t=0}$ defined by the flow never vanishes on $N$.

  2. (2)

For each $n\in N$ the tangent space $T_{n}N$ splits in to a direct sum

$T_{n}N=X_{s}^{*}(n)\oplus X_{u}^{*}(n)\oplus Z(n)$

($\dim X_{s}^{*}=k>0,\dim X_{u}^{*}=l>0,\dim Z=1$), where $Z(n)$ is generated
by $Y(n)$, i.e., $Z(n)=\left\\{cY(n):c\in\mathbb{R}\right\\}$, and there exist
positive number $a$, $b$, and $c$ such that

    1. (i)

for any $\xi\in X_{s}^{*}(n)$,

$\left\lVert d\varphi^{t}\xi\right\rVert\leqslant a\left\lVert\xi\right\rVert
e^{-ct}\text{ for }t\geqslant 0\text{ and }\left\lVert
d\varphi^{t}\xi\right\rVert\geqslant b\left\lVert\xi\right\rVert e^{-ct}\text{
for }t\leqslant 0\;,\text{ and}$

    2. (ii)

for any $\eta\in X_{u}^{*}(n)$,

$\left\lVert d\varphi^{t}\eta\right\rVert\leqslant
a\left\lVert\eta\right\rVert e^{ct}\text{ for }t\leqslant 0\text{ and
}\left\lVert d\varphi^{t}\eta\right\rVert\geqslant
b\left\lVert\eta\right\rVert e^{ct}\text{ for }t\geqslant 0\;.$

We now state the main theorem. For the rest of this section we assume that $M$
is a complete, $C^{\infty}$ Riemannian manifold of dimension two without
conjugate points such that the universal cover $H$ is compactly homogenous.
Let $g^{t}$ denote the geodesic flow on $TM$, $SM$, $TH$, and $SH$. Since $M$
has the same values for the Gaussian curvature as $H$, we have that $\kappa$
is bounded, so there exists a number $k>0$ such that $\kappa>-k^{2}$ in $M$.

######  Theorem 4.2.

Let $M$ be a surface without conjugate points such that universal cover $H$ of
$M$ is compactly homogenous. Then the following properties are equivalent:

  1. (1)

The geodesic flow on $SM$ is of Anosov type.

  2. (2)

For every $v\in SM$, $X_{s}(v)\cap X_{u}(v)=\left\\{0\right\\}$.

  3. (3)

For every $v\in SM$, $T_{v}(SM)=X_{s}(v)\oplus X_{u}(v)\oplus Z(v)$, where
$Z(v)$ is the one-dimensional subspace generated by $Y(v)$.

  4. (4)

There exists no nonzero perpendicular Jacobi field $J$ on a unit speed
geodesic $\gamma$ of $M$ such that $\left\lVert J(t)\right\rVert$ is bounded
for all $t\in\mathbb{R}$.

If, further, $M$ has no focal points, the above conditions are also equivalent
to the following:

  1. (5)

There exists no nonzero perpendicular parallel Jacobi field $J$ on a unit
speed geodesic $\gamma$ of $M$.

  2. (6)

Every geodesic of $M$ passes through a point of negative Gaussian curvature.

We shall prove this theorem in multiple steps, which will require intermediary
results.

######  Proposition 4.3.

Assume that $M$ admits no nonzero perpendicular Jacobi field $J$ on a unit
speed geodesic $\gamma$ such that $\left\lVert J(t)\right\rVert$ is bounded
above for all $t\in\mathbb{R}$. Then there exists a constant $A>0$ such that
if $J$ is a nonzero perpendicular Jacobi field on a unit speed geodesic
$\gamma$ such that $J(0)=0$, then $\left\lVert J(t)\right\rVert\geqslant
A\left\lVert J(s)\right\rVert$ for any numbers $t\geqslant s\geqslant 1$.

###### Proof.

It suffices to prove this result for the universal cover $H$ of $M$. Indeed,
the covering map is a surjective local isometry, so Jacobi fields along a
geodesic in $M$ are projections of Jacobi fields along geodesics in $H$.

Assume toward contradiction that the proposition is false. Then there exist
nonzero perpendicular Jacobi fields $J_{n}$ on unit speed geodesics
$\gamma_{n}$ in $H$, along with sequences $1\leqslant s_{n}\leqslant t_{n}$
such that for each integer $n>0$, we have $J_{n}(0)=0$ and $\left\lVert
J_{n}(t_{n})\right\rVert\leqslant(1/n)\left\lVert J_{n}(s_{n})\right\rVert$.
By linearity, we may assume $\left\lVert J_{n}^{\prime}(0)\right\rVert=1$.

By the extreme value theorem, we can choose a sequence
$(u_{n})\subseteq\mathbb{R}$ such that $0\leqslant u_{n}\leqslant t_{n}$ and
$\left\lVert J_{n}(s)\right\rVert\leqslant\left\lVert
J_{n}(u_{n})\right\rVert$ for every $0\leqslant s\leqslant t_{n}$. We claim
that the $u_{n}$ are lower bounded by some $\delta>0$. For if this were not
the case, then $u_{n}\to 0$ by passing to a subsequence. By Proposition 2.14,
there exists a sequence $\varphi_{n}$ of isometries of $H$ and a Jacobi field
$L$ on a unit speed geodesic $\gamma$ in $H$ such that
$L_{n}=d\varphi_{n}J_{n}\to L$ by passing to a subsequence. By continuity,

$\lim_{n\to\infty}\left\lVert
J_{n}(u_{n})\right\rVert=\lim_{n\to\infty}\left\lVert
L_{n}(u_{n})\right\rVert=\left\lVert L(0)\right\rVert=0\;.$

On the other hand, since $t_{n}\geqslant 1$ for all $n$, we have by
Proposition 3.20 that

$\left\lVert J_{n}(u_{n})\right\rVert\geqslant\left\lVert
J_{n}(1)\right\rVert\geqslant g(1)>0\;,$

a contradiction.

Next, for each $n$, let $\sigma_{n}(t)=\gamma_{n}(u_{n}+t)$ and
$L_{n}(t)=J_{n}(u_{n}+t)/\left\lVert J_{n}(u_{n})\right\rVert$. Then $L_{n}$
is a Jacobi field along $\sigma_{n}$ with $\left\lVert
L_{n}(0)\right\rVert=1$, $\left\lVert L_{n}(s)\right\rVert\leqslant 1$ for
$-u_{n}\leqslant s\leqslant t_{n}-u_{n}$, $L_{n}(-u_{n})=0$, and $\left\lVert
L_{n}(t_{n}-u_{n})\right\rVert\leqslant 1/n$. Since $H$ is compactly
homogenous, $\kappa>-k^{2}$ for some $k>0$, and by Proposition 3.11,
$\left\lVert L_{n}^{\prime}(0)\right\rVert=\left\lVert
J_{n}^{\prime}(u_{n})\right\rVert/\left\lVert
J_{n}(u_{n})\right\rVert\leqslant k\coth(ku_{n})\leqslant k\coth(k\delta)$. By
Proposition 2.14, there exists a sequence $(\varphi_{n})$ of isometries of $H$
and a Jacobi field $L^{*}$ on a unit speed geodesic $\gamma$ such that
$L_{n}^{*}=d\varphi_{n}L_{n}\to L^{*}$ by passing to a subsequence. Since
$\left\lVert L^{*}(0)\right\rVert=1$ by continuity, we have $L^{*}\not\equiv
0$.

By passing to a subsequence, we now have four cases for
$\lim_{n\to\infty}t_{n}-u_{n}$ and $\lim_{n\to\infty}u_{n}$, namely each limit
can either be finite and nonnegative or be positive infinity. We shall derive
contradictions in each of these cases.

First, suppose $t_{n}-u_{n}\to l$ and $u_{n}\to u$ for $l\geqslant 0$ and
$u\geqslant\delta>0$. Continuity yields $L^{*}(-u)=0$ and $L^{*}(l)=0$. But
$t\geqslant 0$ and $-u\leqslant-\delta$, and since $H$ has no conjugate
points, we have $L^{*}\equiv 0$, a contradiction.

Next, suppose $u_{n}\to\infty$ and $t_{n}-u_{n}\to l$ for $l\geqslant 0$. By
continuity, $L^{*}(l)=0$ and $\left\lVert L^{*}(u)\right\rVert\leqslant 1$ for
all $u\leqslant l$. By Proposition 3.14 applied to $L^{*}(l-t)$, however,
$\left\lVert L^{*}(t)\right\rVert\to\infty$ as $t\to-\infty$, a contradiction.
The case when $t_{n}-u_{n}\to\infty$ and $u_{n}\to u$ for $u\geqslant\delta>0$
is proved similarly.

Finally, when both sequences approach infinity, then $\left\lVert
L^{*}(t)\right\rVert\leqslant 1$ for all $t\in\mathbb{R}$ by continuity, but
this contradicts the assumption of the proposition. ∎

###### Proof of (2) $\iff$ (3) $\iff$ (4).

For any $v\in SM$, we have $T_{v}(SM)=Z(v)^{\perp}\oplus Z(v)$, where $Z(v)$
is the 1-dimensional subspace generated by $Y(v)$. Since $SM$ is
3-dimensional, $Z(v)^{\perp}$ is 2-dimensional. Since $X_{s}(v)$ and
$X_{u}(v)$ are 1-dimensional subspaces of $Z(v)^{\perp}$, we immediately have
(2) $\iff$ (3).

Next, suppose that $M$ admits a nonzero perpendicular Jacobi field $J$ on a
unit speed geodesic $\gamma$ such that $\left\lVert J(t)\right\rVert$ is
bounded everywhere. If $v=\gamma^{\prime}(0)$ and $\xi\in T_{v}(SM)$
corresponds to $J$, then Proposition 3.16 shows that $\xi\in X_{s}(v)\cap
X_{u}(v)$, and thus (2) $\Longrightarrow$ (4). The reverse direction follows
from Propositions 3.17 and 4.3. ∎

######  Proposition 4.4.

Assume that $M$ admits no nonzero perpendicular Jacobi field $J$ on a unit
speed geodesic $\gamma$ such that $\left\lVert J(t)\right\rVert$ is bounded
above for all $t\in\mathbb{R}$. Then for any number $\varepsilon>0$, there
exists a number $T>0$ such that

  1. (1)

for any $v\in SM$ and any $\xi\in X_{s}(v)$, $\left\lVert
dg^{t}\xi\right\rVert\leqslant\varepsilon\left\lVert\xi\right\rVert$ for
$t\geqslant T$, and

  2. (2)

for any $v\in SM$ and any $\eta\in X_{u}(v)$, $\left\lVert
dg^{t}\eta\right\rVert\leqslant\varepsilon\left\lVert\eta\right\rVert$ for
$t\leqslant-T$.

###### Proof.

It suffices by Propositions 2.7 and 3.8 to prove these assertions for the
universal cover $H$ of $M$. We start with the first assertion, and assume
toward contradiction that there exists an $\varepsilon>0$ for which there are
sequences $(t_{n})\subseteq\mathbb{R}$, $(v_{n})\subseteq SH$, and
$(\xi_{n})\subseteq T(SH)$ such that $\xi_{n}\in X_{s}(v_{n})$,
$t_{n}\to\infty$, and $\left\lVert
dg^{t_{n}}\xi_{n}\right\rVert>\varepsilon\left\lVert\xi_{n}\right\rVert$ for
every integer $n$. By linearity, we may assume
$\left\lVert\xi_{n}\right\rVert=1$. If $\psi_{n}=dg^{t_{n}}\xi_{n}$, then
$\psi_{n}\in A_{s}=\bigcup_{v\in SH}X_{s}(v)$ and
$\left\lVert\psi_{n}\right\rVert>\varepsilon$ for every $n$. Corollary 3.19
and Proposition 4.3 then imply that there exists a number $B>0$ such that for
every $n$, $\left\lVert dg^{t}\psi_{n}\right\rVert\leqslant B$ for
$-t_{n}\leqslant t<\infty$. By Proposition 2.15, there exist a sequence
$\varphi_{n}$ of isometries of $H$ and a vector $\psi^{*}\in T(SH)$ such that
$\psi_{n}^{*}=dT_{\varphi_{n}}\psi_{n}\to\psi^{*}$ by passing to a
subsequence, where $T_{\varphi_{n}}=d\varphi_{n}$. By Proposition 3.8, we have
$\psi_{n}^{*}\in A_{s}$ for every $n$. Since $g^{t}\circ
d\varphi_{n}=d\varphi_{n}\circ g^{t}$ for all $t\in\mathbb{R}$, we have

$\displaystyle\left\lVert dg^{t}\psi_{n}^{*}\right\rVert=\left\lVert
dg^{t}\circ dT_{\varphi_{n}}\psi_{n}\right\rVert=\left\lVert
dT_{\varphi_{n}}\circ dg^{t}\psi_{n}\right\rVert=\left\lVert
dg^{t}\psi_{n}\right\rVert\leqslant B\;$

for $-t_{n}\leqslant t<\infty$, where the last equality follows because
$\varphi_{n}$ is an isometry. The vector $\psi^{*}\neq 0$ since
$\left\lVert\psi_{n}^{*}\right\rVert=\left\lVert\psi_{n}\right\rVert>\varepsilon$
for every $n$, and by continuity $\left\lVert
dg^{t}\psi^{*}\right\rVert\leqslant B$ for all $t\in\mathbb{R}$. If
$\psi^{*}\in T_{v^{*}}(SH)$ and $\psi_{n}^{*}\in T_{v_{n}^{*}}(SH)$, then
$\left\langle\psi^{*},Y(v^{*})\right\rangle=0$ since
$\left\langle\psi_{n}^{*},Y(v_{n}^{*})\right\rangle=0$ for every $n$. Thus
$J_{\psi^{*}}$ is a nonzero perpendicular Jacobi field on $\gamma_{v^{*}}$ in
$H$ whose magnitude is uniformly bounded by $B$ for all $t\in\mathbb{R}$. Then
$J(t)=dpJ_{\psi^{*}}(t)$ is a nonzero perpendicular Jacobi field on the
geodesic $p\circ\gamma_{v^{*}}$ in $M$ such that $\left\lVert
J(t)\right\rVert=\left\lVert J_{\psi^{*}(t)}\right\rVert\leqslant B$ for all
$t\in\mathbb{R}$, contradicting the assumption in the proposition.

For the second assertion, we simply note that $\eta\in X_{u}(v)$ if and only
if $\xi=dS(\eta)\in X_{s}(-v)$, where $S:SM\to SM$ maps $v\longmapsto-v$.
Since $S$ is an isometry of $SM$ satisfying the relation $S\circ
g^{t}=g^{-t}\circ S$ for all $t\in\mathbb{R}$, we have that $\left\lVert
dg^{-t}\eta\right\rVert=\left\lVert dS\circ dg^{t}\xi\right\rVert=\left\lVert
dg^{t}\xi\right\rVert\leqslant\varepsilon\left\lVert\xi\right\rVert=\varepsilon\left\lVert\eta\right\rVert$
for all $t\geqslant T$. ∎

######  Lemma 4.5.

Let $M$ be a complete manifold without conjugate points, which satisfies the
hypothesis of Proposition 3.17. For each $s\geqslant 0$, let
$\varphi(s)=\sup\left\\{\left\lVert dg^{s}\xi\right\rVert:\xi\in
A_{s},\left\lVert\xi\right\rVert=1\right\\}$. Then

  1. (1)

there exists a constant $B>0$ such that $0\leqslant\varphi(s)\leqslant B$ for
all $s\geqslant 0$, and

  2. (2)

$\varphi(s+t)\leqslant\varphi(s)\cdot\varphi(t)$ for all $s\geqslant 0$ and
$t\geqslant 0$.

Furthermore, if the universal cover $H$ is compactly homogenous and $M$ admits
no nonzero perpendicular Jacobi field $J$ on a unit speed geodesic $\gamma$
such that $\left\lVert J(t)\right\rVert$ is bounded above for all
$t\in\mathbb{R}$, then

  1. (3)

$\varphi(s)\to 0$ as $s\to\infty$.

###### Proof.

First, if $B>0$ is the constant in Corollary 3.19, then
$0\leqslant\varphi(s)\leqslant B$ for $s\geqslant 0$.

Next, for any $\xi\in A_{s}$ and any $s\geqslant 0$, $\left\lVert
dg^{s}\xi\right\rVert\leqslant\varphi(s)\left\lVert\xi\right\rVert$ by
definition. Let $\xi\in A_{s}$, $\left\lVert\xi\right\rVert=1$ be given, and
let $s\geqslant 0$, $t\geqslant 0$ be arbitrary. Then

$\left\lVert dg^{s+t}\xi\right\rVert=\left\lVert dg^{s}\circ
dg^{t}\xi\right\rVert\leqslant\varphi(s)\left\lVert
dg^{t}\xi\right\rVert\leqslant\varphi(s)\cdot\varphi(t)\;,$

since $A_{s}$ is invariant under $dg^{t}$. Taking the above over all $\xi\in
A_{s}$ yields the desired result.

The final property follows immediately from Proposition 4.4. ∎

######  Lemma 4.6.

Let $\varphi:(0,\infty)\to(0,\infty)$ be a function satisfying all three
properties of the previous lemma. Then there exist numbers $a>0$ and $c>0$
such that $\varphi(s)\leqslant ae^{-cs}$ for $s\geqslant 0$.

###### Proof.

Notice that the second property implies that if $s\geqslant 0$ is any number
and $n>0$ is any integer, then $\varphi(ns)\leqslant\varphi(s)^{n}$. The third
property tells us we can choose $s_{0}>0$ so that $\varphi(s)<1/2$ for
$s\geqslant s_{0}$. Thus it suffices to show that $\varphi(s)\leqslant
e^{-cs}$ for $s\geqslant s_{0}$, where $c=\frac{1}{2}(\log 2)/s_{0}$. Given
$s\geqslant s_{0}$, we may choose an integer $n\geqslant 1$ such that
$s_{0}\leqslant s/n\leqslant 2s_{0}$. Let $s^{*}=s/n$. Then

$\frac{\log\varphi(s)}{s}=\frac{\log\varphi(ns^{*})}{ns^{*}}\leqslant\frac{\log\varphi(s^{*})^{n}}{ns^{*}}=\frac{\log\varphi(s^{*})}{s^{*}}\leqslant-\frac{\log
2}{2s_{0}}=-c\;.$

∎

###### Proof of (1) $\iff$ (4)..

We start with the reverse direction. Proposition 4.3, Lemma 4.5, and Lemma 4.6
together imply that for every $v\in SM$ and every $\xi\in X_{s}(v)$, we have

$\left\lVert dg^{t}\xi\right\rVert\leqslant a\left\lVert\xi\right\rVert
e^{-ct}$

for $t\geqslant 0$, where $a>0$ and $c>0$ do not depend on $\xi$ or $v$. By
replacing $\xi$ with $dg^{-t}\xi$, we have that

$\left\lVert dg^{-t}\xi\right\rVert\geqslant(1/a)\left\lVert\xi\right\rVert
e^{ct}$

for $t\geqslant 0$. Now let $\eta\in X_{u}(v)$, so $\eta=dS(\xi)$ for $\xi\in
X_{s}(-v)$. Since the first inequality is independent of $v$, we have

$\left\lVert dg^{t}\eta\right\rVert=\left\lVert dg^{t}\circ
dS(\xi)\right\rVert=\left\lVert dS\circ dg^{-t}\xi\right\rVert=\left\lVert
dg^{-t}\xi\right\rVert\geqslant(1/a)\left\lVert\xi\right\rVert
e^{ct}=(1/a)\left\lVert\eta\right\rVert e^{ct}$

for $t\geqslant 0$. Since $(1/a)\left\lVert\eta\right\rVert
e^{ct}>a\left\lVert\eta\right\rVert e^{-ct}$ for sufficiently large $t$, we
have $\eta\notin X_{s}(v)$ for any $\eta\in X_{u}(v)$, so $X_{s}(v)\cap
X_{u}(v)=\left\\{0\right\\}$, and by using (2) $\Longrightarrow$ (3) we have
that

$T_{v}(SM)=X_{s}(v)\oplus X_{u}(v)\oplus Z(v)\;.$

Setting $X_{s}^{*}(v)=X_{s}(v)$ and $X_{u}^{*}(v)=X_{u}(v)$, the conditions
for an Anosov flow are satisfied.

Next, we show the forward direction. Suppose toward contradiction that there
exists a nonzero perpendicular Jacobi field $J$ on a unit speed geodesic such
that $\left\lVert J(t)\right\rVert\leqslant c$ for some $c>0$ and all
$t\in\mathbb{R}$. If $v=\gamma^{\prime}(0)$, then let $\xi\in T_{v}(SM)$ be
such that $J=J_{\xi}$. Choose $k>0$ such that $\kappa>-k^{2}$. Since $J$ is
perpendicular, $\left\langle\xi,Y(v)\right\rangle=0$, and $\xi\in X_{s}(v)\cap
X_{u}(v)$ by Proposition 3.16. For any $t\in\mathbb{R}$, we have $dg^{t}\xi\in
X_{s}(g^{t}v)\cap X_{u}(g^{t}v)$ and

$\left\lVert K\circ dg^{t}\xi\right\rVert\leqslant k\left\lVert d\pi\circ
dg^{t}\xi\right\rVert\leqslant kc$

by Proposition 3.15. Thus by definition of the Sasaki metric, $\left\lVert
dg^{t}\xi\right\rVert\leqslant c(1+k^{2})^{1/2}$ for all $t$. By definition of
an Anosov flow, we have $\xi=\xi_{1}+\xi_{2}+\xi_{3}$, where $\xi_{1}\in
X_{s}^{*}(v)$, $\xi_{2}\in X_{u}^{*}(v)$, and $\xi_{3}=aY(v)$ for some
$a\in\mathbb{R}$. Notice that $\left\lVert
dg^{t}\xi_{3}\right\rVert=\left\lVert\xi_{3}\right\rVert$ for all $t$, since
$dg^{t}Y(w)=Y(g^{t}w)$ and $\left\lVert Y(w)\right\rVert=1$ for $w\in SM$. By
the properties of the spaces $X_{s}^{*}(v)$ and $X_{u}^{*}(v)$, it follows
that if $\xi_{1}\neq 0$, then $\left\lVert dg^{t}\xi\right\rVert\to\infty$ as
$t\to-\infty$, while if $\xi_{2}\neq 0$, then $\left\lVert
dg^{t}\xi\right\rVert\to\infty$ as $t\to\infty$. Thus $\xi_{1}=\xi_{2}=0$,
which contradicts the assumption that $\xi$ is nonzero and orthogonal to
$Y(v)$. ∎

For the rest of the proof, we assume that $M$ has no focal points.

###### Proof of (4) $\iff$ (5)..

We start with the forward direction and show the contrapositive. Suppose there
exists a nonzero perpendicular parallel Jacobi field $J$ on a unit speed
geodesic $\gamma$. Since $J^{\prime}(t)=0$, we have $\left\lVert
J(t)\right\rVert>0$ is constant over all $t$. Then $J$ is a nonzero
perpendicular Jacobi field whose magnitude is uniformly bounded for all $t$.

Before we show the reverse direction, we need to make some observations.
First, we claim that if a Jacobi field $J$ is in $J_{s}(\gamma)$ (resp.
$J_{u}(\gamma)$), then $\left\lVert J(t)\right\rVert$ is nonincreasing (resp.
nondecreasing) in $t$. For let $J\in J_{s}(\gamma)$ and $a<b$ be given. By
Remark 3.18, the condition of Proposition 3.17 is satisfied relative to the
constants $A=1$ and $s_{0}=0$. Thus we have $\left\lVert
J(t)\right\rVert\leqslant\left\lVert J(0)\right\rVert$ for $t\geqslant 0$ by
Proposition 3.17. For each $u\in\mathbb{R}$, let $\gamma^{*}(u)=\gamma(a+u)$
and $J^{*}(u)=J(a+u)$, effectively shifting the parameter of our geodesic and
Jacobi field by $a$. Since $\left\lVert
J^{*}(u)\right\rVert\leqslant\left\lVert J(0)\right\rVert$ for $u\geqslant-a$,
Proposition 3.16 shows that $J^{*}\in J_{s}(\gamma^{*})$. Then again using
Proposition 3.17, this time on $J^{*}$, we have that $\left\lVert
J^{*}(t)\right\rVert\leqslant\left\lVert J^{*}(0)\right\rVert$ for $t\geqslant
0$. This yields

$\left\lVert J(b)\right\rVert=\left\lVert
J^{*}(b-a)\right\rVert\leqslant\left\lVert J^{*}(0)\right\rVert=\left\lVert
J(a)\right\rVert\;,$

as desired. A similar argument shows the result for $J\in J_{u}(\gamma)$.

We now prove the reverse direction by showing the contrapositive. Suppose that
$J$ is a nonzero perpendicular Jacobi field on a unit speed geodesic $\gamma$
such that $\left\lVert J(t)\right\rVert$ is uniformly bounded for all
$t\in\mathbb{R}$. It follows from Proposition 3.16 that $J\in
J_{s}(\gamma)\cap J_{u}(\gamma)$. The claim above shows that $\left\lVert
J(t)\right\rVert$ is both nonincreasing and nondecreasing in $t$, and hence
constant. Since $J$ perpendicular, relative an adapted frame field
$E_{1},E_{2}$ along $\gamma$ with $E_{2}(s)=\gamma^{\prime}(s)$, we have

$J(t)=f(t)E_{1}(t)$

for some real-valued function $f$, which then implies $f(t)$ is constant.
Since $J^{\prime}(t)=f^{\prime}(t)E_{1}(t)\equiv 0$, we conclude $J$ is
parallel. ∎

###### Proof of (1) $\iff$ (6).

For the forward direction, we show the contrapositive. Let $\gamma$ be a unit
speed geodesic such that $\kappa(t)\geqslant 0$ along $\gamma$ for all $t$.
Relative to an adapted frame field along $\gamma$, it suffices to show that
$E_{1}$ is a Jacobi field, for then $E_{1}$ is a nonzero perpendicular
parallel Jacobi field along $\gamma$, which by (1) $\Longrightarrow$ (5)
implies the geodesic flow on $SM$ is not Anosov.

We first show that $\kappa(t)\equiv 0$ on $\gamma$. Recall that
$u(s)=d^{\prime}(s)d^{-1}(s)$ is a solution of the Ricatti equation

$u^{\prime}(s)+u(s)^{2}+\kappa(s)=0$

along $\gamma$. Since the Gaussian curvature is nonnegative, given any
$\varepsilon>0$, $\kappa(s)>-\varepsilon^{2}$ for all $s\in\mathbb{R}$. Since
$u(s)$ is defined everywhere, we have $\left\lvert
u(s)\right\rvert\leqslant\varepsilon$ by Lemma 3.12, so $u(s)\equiv 0$, which
then gives us $\kappa(s)\equiv 0$ using the Ricatti equation. This in turn
implies that the curvature tensor is identically zero along $\gamma$, so the
Jacobi equation along $\gamma$ simplifies to

$J^{\prime\prime}(s)=0\;.$

Since $E_{1}(s)$ is parallel, $E_{1}^{\prime\prime}(s)\equiv 0$, so $E_{1}$ is
a Jacobi field, as desired.

For the reverse direction, suppose every geodesic in $M$ passes through a
point of negative Gaussian curvature. Assume toward contradiction that the
geodesic flow on $SM$ is not of Anosov type, so (5) $\Longrightarrow$ (1)
implies there exists a nonzero perpendicular parallel Jacobi field $J$ along
$\gamma$. Relative to an adapted frame field along $\gamma$ and using the fact
that $J\in J_{s}(\gamma)$, we have by Remark 3.7 that $J(s)=cd(s)E_{1}(s)$ for
some $c\neq 0$. Without loss of generality, we may assume $c=1$. Since $J$ is
parallel, $d(s)$ is constant, so $d^{\prime}(s)\equiv 0$ along $\gamma$.
Plugging $u(s)=d^{\prime}(s)d^{-1}(s)\equiv 0$ into the Ricatti equation shows
that $\kappa(s)\equiv 0$, a contradiction. ∎

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