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--- ebsd2021:tema1 [2021/06/28 16:10] – external edit 127.0.0.1
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-Tema1-Minicurso Knieper
+**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~~