**Tema 1: The geodesic flow, the tangent bundle $TTM$, Jacobi fields**


**The geodesic flow**

Let $M$ be a closed (compact and boundaryless) Riemannian manifold. We represent
the covariant derivation by $\nabla$. The tangent bundle of $M$, denoted by $TM$,
is a manifold with dimension $2n$. An element $v\in TM$ represents a speed
of a curve in $M$. Writing $v\in (TM)_p$ means that $p$ is the basepoint of $v$, e.g.
the curve $\gamma(t)$ for which $\gamma'(0)=v$ satisfies $\gamma(0)=p$.

Among the curves in $M$, there is a class of them which is highly important in the geometric
viewpoint: the geodesics are the curves in $M$ that minimize locally the distance between two points.
In other words, let $\gamma$ be a curve parametrized by arc length, then we say that
$\gamma$ is a geodesic if the distance in $M$ between $\gamma(t)$ and $\gamma(t')$ is exactly
$|t-t'|$ whenever $t\approx t'$. This property has an infinitesimal analogue, given by the equality
$$
\gamma''=0,
$$
where $\gamma''$ represents the covariant derivative $\nabla_{\gamma'}\gamma'$. Therefore
geodesics are the curves with no acceleration.

In local coordinates, if $\gamma(t)=(x_1(t),\ldots,x_n(t))$, then $\gamma$ is a geodesic iff
$$
\frac{d^2x_k}{dt^2}+\sum_{i,j}\Gamma_{ij}^k\frac{dx_i}{dt}\frac{dx_j}{dt}=0,\ \text{ for }k=1,2,\ldots,n.
$$
This is a system of second order differential equation, hence for all initial conditions
$\gamma(0),\gamma'(0)$ it has a unique solution.
 

The //geodesic flow// on $M$ is $g=\{g^t\}:TM\to TM$ with $g^t(v)=\gamma_v'(t)$, where
$\gamma_v(t)$ is the unique geodesic such that $\gamma_v'(0)=v$.


In other words, $g^t$ sends a speed vector $v$
to the speed vector at time $t$ of the evolution of the geodesic in the direction of $v$.
If $\gamma$ is a geodesic, then the norm of $\gamma'$ is constant. Therefore, if we want to analyze 
the geodesic flow, we can restrict ourselves to unit vectors. Letting
$N:=T^1M$, the unit tangent bundle of $M$ (which is a manifold with dimension $2n-1$),
we also call the restriction $g=\{g^t\}:N\to N$ the geodesic flow on $M$.



**Geometry of the tangent bundle $TTM$**

In order to understand properties of $g$ is Anosov, we need to understand
the derivative $Dg$. This derivative is a flow on $TN$, the tangent bundle of $N$. This latter tangent 
bundle is a fiber bundle with fibers of dimension $2n-1$. More generally, since
$g$ is defined in the whole tangle bundle $TM$, $Dg$ is defined in $TTM$, which is a fiber bundle 
with fibers of dimension $2n$. The fiber bundle $TTM$ has a special geometry on its own,
as we will now see. 


__Horizontal and vertical vectors__

To understand $TTM$, we need to consider curves in $TM$ and their velocities.
Recall that $TM$ is a fiber bundle whose fibers are vertical subspaces isomorphic to $\mathbb R^n$.
We use this property introduce vertical and horizontal vectors on $TTM$. More generally,
for $\xi\in (TTM)_v$, $v\in TM$, we will define a decomposition of $\xi$ into its horizontal and vertical components.

Since $TM$ is fibered by vertical subspaces, it is easy to define vertical vectors in $TTM$
(and horizontal components).
Consider the natural projection $\pi:TM\to M$, $\pi(v)=\gamma_v(0)$, the basis point of $v$.
Then $d\pi$ measures how much curves in $ TM$ move on $M$ (with respect to their base points).

//Vertical vectors:// $\xi\in TTM$ is called a //vertical vector// if $\xi\in {\rm ker}(d\pi)$.
Denote the vertical vectors at $(TTM)_v$ by $V(v)$.

Since $d\pi$ is surjective, each $V(v)$ has dimension $n$ and so
the space of vertical vectors is a subbundle of $TTM$ of dimension $n$.
By definition, vertical vectors are those $\xi=X'(0)$ tangent to curves $X(t)$ whose basepoint does not move infinitesimally at $t=0$. As we will see, $d\pi(\xi)$ will represent the horizontal component of $\xi$,
so that $\xi$ is vertical iff its horizontal component is zero. 

Since the above definition is intrinsic, we now show how to generate vertical vectors in a simple way.
Letting $p=\pi(v)$, we will lift $(TM)_p$ to vertical vectors in $(TTM)_v$. Given $w\in (TM)_p$,
consider $X(t)=v+tw$, which is a curve on $TM$. It satisfies $X(0)=v$ and $\pi\circ X=p$, so
$w=X'(0)\in V(v)$. This defines an isomorphism $(TM)_p\to V(v)$.

Defining horizontal vectors is a little bit more complicated. We want them to be those that
represent no displacement inside the fibers of $TM$. If $TM$ was a product, this would be easy.
The tool to trivialize $TM$ is by parallel transport, hence we will define horizontal vectors
as those obtained from parallel transport. This is measured by covariant derivatives, as follows.
Let $X(t)$ be a curve on $TM$ with $X(0)=v\in TM$. If $c=\pi\circ X$, then $X$ is a vector field
over $c$. The variation of $X$ along $c$ at $t=0$ is $\nabla_{\dot{c}}X(0)$, so we arrive at the following
definition.

//Connection map//: The //connection map// is the $\kappa:TTM\to TM$ defined by
$\kappa(\xi)=\nabla_{\dot{c}}X(0)$ for some (any) $X:(-\varepsilon,\varepsilon)\to TM$
s.t. $X'(0)=\xi$, where $c=\pi\circ X$.


//Horizontal vectors:// $\xi\in TTM$ is called a //horizontal vector// if $\xi\in{\rm ker}(\kappa)$.
Denote the horizontal vectors at $(TTM)_v$ by $H(v)$.

This amounts to saying that the displacement inside the fibers of $TM$ is zero.
The space of horizontal vectors is also a subbundle of $TTM$ of dimension $n$, because
$\kappa$ is surjective (exercise).
We will see below that $\kappa(\xi)$ represents the vertical component of $\xi$, so that
$\xi$ is horizontal iff its vertical component is zero.

Now we show how to lift $(TM)_p$ to horizontal vectors in $(TTM)_v$.
Given $w\in (TM)_p$, let $c$ be a curve on $M$ with $c'(0)=w$, and let
$X(t)$ be the parallel transport of $v$ along $c$. Then $X'(0)$ is a horizontal 
vector. This defines an isomorphism $L_v:(TM)_p\to H(v)$, see the next lemma.
In other words, we consider all possible parallel transports of $v$.
It is worth understanding the above definitions when $M=\mathbb R^n$, in which case
vertical/horizontal vectors are really vertical/horizontal.


The map $L_v$ is related to $d\pi$ and $\kappa$.

**Lemma 1.** The following are true.

(1) $L_v:(TM)_p\to H(v)$ is an isomorphism.

(2) $d\pi\circ L_v={\rm Id}_{(TM)_p}$, hence $V(v)\cap H(v)=\{0\}$.

(3) The restrictions $d\pi\restriction_{H(v)}$ and $\kappa\restriction_{V(v)}$
are isomorphisms onto $(TM)_p$. In particular, $(TTM)_v=H(v)\oplus V(v)$.



**Proof.** (1) By definition, ${\rm Im}(L_v)\subset H(v)$. It is clear that $L_v$ is injective.
Since both $(TM)_p$ and $H(v)$ have dimension $n$, the result follows.


(2) Let $w\in (TM)_p$, let $c$ be a curve on $M$ with $c'(0)=w$, and let $X(t)$ be the parallel
transport of $v$ along $c$. We have $L_v(w)=X'(0)$, hence 
$$
d\pi\circ L_v(w)=\tfrac{d}{dt}|_{t=0}(\pi\circ X)=c'(0)=w.
$$
Now assume that $\xi\in V(v)\cap H(v)$. By (1), $\xi=L_v(w)$.
Since $\xi\in V(v)$, we have $d\pi(\xi)=0$ and so
$$
w=(d\pi\circ L_v)(w)=d\pi(\xi)=0\ \Rightarrow \ \xi=0.
$$


(3) Since we have injective linear maps from $(TM)_p$ to both $V(v),H(v)$, we have that
$V(v),H(v)$ have dimension at least $n$. Therefore, it is enough to prove injectivity.
If $\xi\in H(v)$ with $d\pi(\xi)=0$, then $\xi\in V(v)$ and so $\xi\in V(v)\cap H(v)=\{0\}$.
Similarly, if $\xi\in V(v)$ with $\kappa(\xi)=0$ then $\xi\in H(v)$ and so $\xi\in V(v)\cap H(v)=\{0\}$.


Now the following proposition is direct.

**Proposition 2.** For each $v\in TM$, the map $j_v:(TTM)_v\to (TM)_p\times (TM)_p$,
$$
j_v(\xi)=(d\pi(\xi),\kappa(\xi)),
$$
is an isomorphism. Also, $j_v$ maps $V(v)$ to $\{0\}\times (TM)_p$ and $H(v)$
to $(TM)_p\times\{0\}$.


We call $\xi_h=d\pi(\xi)$ the //horizontal component// of $\xi$, and
$\xi_v=\kappa(\xi)$ the //vertical component// of $\xi$.

In this coordinates, it is easy to identify the generator of the geodesic flow.
If $v\in TM$, then the generator at $v$ is $\xi=X'(0)$ for $X(t)=g^t(v)$.
The curve $\gamma=\pi\circ X$ is the geodesic with $\gamma'(0)=v$, hence
$$
j_v(\xi)=(D\pi(\xi),\nabla_{\gamma'}X)=(\gamma'(0),\nabla_{\gamma'}\gamma'(0))=(v,0).
$$


__The tangent bundle $TN$__

Now we identify the tangent bundle of $N=T^1M$, using the identification  
$(TTM)_v\cong (TM)_p\times (TM)_p$.

//Tangent bundle of $N$//: The //tangent bundle// $TN$ is the subbundle of $TTM$ whose fibers are
$$
(TN)_v=\{(v_1,v_2)\in (TM)_p\times (TM)_p:v_2\perp v\}.
$$

This is easy to see: if $X(t)$ be a curve on $N$, then
$$
\langle X,X\rangle=1\ \Rightarrow\ 2\langle \nabla_{\dot{c}}X,X\rangle=0\ \Rightarrow\ \kappa(X'(0))=\nabla_{\dot{c}}X(0)\in v^\perp.
$$

Hence $(TN)_v \subset \{(v_1,v_2)\in (TM)_p\times (TM)_p:v_2\perp v\}$. Since both subspaces
have dimension $2n-1$, we conclude the equality. 

__Sasaki metric__

The symmetry obtained in the representation of $TTM$ allows to introduce
a product Riemannian metric on $TTM$.
Let $\langle \cdot,\cdot\rangle$ be the metric on $M$.


//Sasaki metric//: The //Sasaki metric// is the metric on $TTM$ given in the above coordinates by
$$
\langle \xi,\eta\rangle_{\rm Sas}=\langle \xi_h,\eta_h\rangle+\langle \xi_v,\eta_v\rangle.
$$


__Symplectic form__

The symmetry of $TTM$ also allows to define a symplectic form on $TM$. Recall the definition below.

//Symplectic form//: Given a manifold $M$, a //symplectic form// on $M$ is a 2--form
$\omega$ s.t.:
(1) $\omega$ is closed, i.e. $d\omega=0$.
(2) $\omega$ is non-degenerate, i.e. if $Y\in (TM)_p\mapsto \omega(X,Y)$ is the null
map then $X=0$.
The pair $(M,\omega)$ is called a {\em symplectic manifold}.


//Symplectic form on $TM$//: Define the //symplectic form// $\omega$ on $TM$ by
$$
\omega(\xi,\eta)=\langle \xi_v,\eta_h\rangle-\langle \xi_h,\eta_v\rangle.
$$

Below, we will show that $\omega$ is invariant under the geodesic flow.


**Jacobi fields**


Jacobi vector fields are fields obtained as variations of geodesics. If one thinks for a second,
that is exactly what we need to understand $Dg^t(\xi)$: we consider a curve $X(s)$ on $N$ such that
$X'(0)=\xi$, then we consider the composition function $s\mapsto (g^t\circ X)(s)$, and finally differentiate 
this function at $s=0$.
For each $s$, the curve $t\mapsto g^t\circ X(s)$ is a geodesic, and the derivative with respect to $s$ is exactly
the measurement of variation of the geodesics emanating from $X$. Let us be more formal now.
Let $c$ be a curve on $M$.


//Jacobi field//: A vector field $J$ along $c$ is called a //Jacobi (vector) field// if there is
a map $\alpha:[-1,1]^2\to M$ such that
(1) For each $s\in[-1,1]$, the curve $\alpha(s,\cdot)$ is a geodesic.
(2) $J(t)=\tfrac{\partial\alpha}{\partial s}(0,t)$.


__Jacobi equation__

Let us denote by $J'(t)$ the covariant
derivative $\nabla_{\dot{c}}J$. Then $J'$ is another vector field along $c$. Similarly,
let $J''$ be the covariant derivative $\nabla_{\dot{c}}J'$, another vector field along $c$.
Let us understand what defines a Jacobi field.
If $R$ denotes the curvature tensor on $M$, then the property (1) above implies that $J$ satisfies the 
//Jacobi equation//
\begin{equation}\label{equation-jacobi}
J''+R(\gamma',J)\gamma'=0.
\end{equation}
The above equation is a second order ordinary differential equation (ODE),
thus it is defined by the values of $J(0)$ and $J'(0)$.
Therefore the space of Jacobi fields along a geodesic is a linear space of dimension $2n$.
We write this result below.

**Proposition 3.** Let $\gamma$ be a geodesic with $\gamma(0)=p$. For any $v,w\in (TM)_p$,
there is a unique Jacobi field $J$ along $\gamma$ s.t. $J(0)=v$ and $J'(0)=w$.



Let us understand this equation on local coordinartes. Fix an orthonormal basis of parallel vector 
fields $e_1=\gamma',e_2,\ldots,e_m$. Any Jacobi field can be written as $J=\sum_{k=1}^n y_ke_k$ (if $J$ is
perpendicular then $y_1=0$). By the Jacobi equation, we find out that $y_1,\ldots,y_m$ satisfy
$$
\sum y_k''e_k+\sum y_kR(e_1,e_k)e_1=0.
$$
If we denote $\mathcal R_{kj}:=\langle R(e_1,e_k)e_1,e_j\rangle$, then the above equation implies that
\begin{equation}\label{equation-jacobi-2}
y_k''+\sum_j y_j\mathcal R_{kj}=0,\ \ k=1,\ldots,m.
\end{equation}
Letting $Y=[y_1,\ldots,y_m]^T$, the above equalities become the vectorial equation
$Y''+\mathcal RY=0$ on $\mathbb R^m$.
As one usually does when dealing with ODEs, we look at the matrix form of the system of equations above and
solve it. Indeed, note that if
$J^1,\ldots,J^n$ are linearly independent solutions for the system of equations,
then the $n\times n$ invertible matrix $\mathcal J=[J^1,\ldots,J^n]$ satisfies
$$
\mathcal J''+\mathcal R\mathcal J=0.
$$


__Horizontal/vertical representation $\times$ Jacobi fields__


By Proposition 3, we can represent $(TTM)_v$ and $(TN)_v$ in terms of Jacobi fields:
\begin{align*}
(TTM)_v&=\{(J(0),J'(0)):J\text{ is Jacobi field along }\gamma_v\}.\\
(TN)_v&=\{(J(0),J'(0)):J\text{ is Jacobi field along }\gamma_v\text{ s.t. }J'(0)\perp v\}.
\end{align*}
With the representation, the action of $Dg^t$ is very simple.

**Claim:** If $\xi=(J(0),J'(0))\in (TTM)_v$ then $D_vg^\tau(\xi)=(J(\tau),J'(\tau))$.

The claim is almost a tautology, after a short thinking. Let $\alpha(s,t)$ be a variation by geodesics
on $M$ s.t. $\xi=(J(0),J'(0))$. Then $\beta(s,t):=g^\tau\circ\alpha(s,t)=\alpha(s,t+\tau)$ is a
variation by geodesics s.t.
$\tfrac{\partial\beta}{\partial t}(0,0)=g^t(v)$ and
$(\tfrac{\partial\beta}{\partial s}(0,0),\tfrac{\partial^2\beta}{\partial t\partial s}(0,0))=(J(\tau),J'(\tau))$.


Using this representation, it is easy to prove the following invariance.

**Lemma 4.** The symplectic form $\omega$ is invariant under $Dg$.


**Proof.** Let $J_1,J_2$ be two Jacobi fields. Since $Dg^t(J_i(0),J_i'(0))=(J_i(t),J_i'(t))$,
we have
\begin{align*}
&\omega(Dg^t(J_1(0),J_1'(0)),Dg^t(J_2(0),J_2'(0)))=\omega((J_1(t),J_1'(t)),(J_2(t),J_2'(t)))\\
&=\left(\langle J_1',J_2\rangle -\langle J_1,J_2'\rangle\right)(t)
\end{align*}
and so the lemma is equivalent to showing that
$f(t)=\left(\langle J_1',J_2\rangle -\langle J_1,J_2'\rangle\right)(t)$ is constant. We have
$$
f'=\langle J_1'', J_2\rangle -\langle J_1,J_2''\rangle=-\langle R(\gamma',J_1)\gamma',J_2\rangle
-\langle J_1,R(\gamma',J_2)\gamma'\rangle=0,
$$
by the symmetry of $R$.


To understand the dynamical properties of $g$, we need to consider
$Dg$ restricted to the orthogonal complement of the generator of the geodesic flow. Recalling that
this later one is $(v,0)$, we have the following.

//Orthogonal complement on $TN$//: Given $v\in N$, the //orthogonal complement// of
$(v,0)$ in the Sasaki metric is
$$
{\rm Perp}_v=\{(v_1,v_2)\in (TM)_p\times (TM)_p:v_1,v_2\perp v\}.
$$

This is direct. 

**Lemma 5.** $\{{\rm Perp}_v\}_{v\in N}$ is invariant under the geodesic flow.


**Proof.** Let $(J(0),J'(0))$ Jacobi field along $\gamma$ s.t. $J(0),J'(0)\perp \gamma'(0)$.
Since 
$$
\langle J',\gamma'\rangle '=\langle J'',\gamma'\rangle+\langle J',\gamma''\rangle
=\langle R(\gamma',J)\gamma',\gamma'\rangle=0,
$$
it follows that $J'\perp\gamma'$.
Hence $\langle J,\gamma'\rangle'=\langle J',\gamma'\rangle=0$, which 
also implies that $J\perp \gamma'$.













~~DISCUSSIONS~~