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--- ebsd2021:tema3 [2021/06/28 16:09] – external edit 127.0.0.1
+++ ebsd2021:tema3 [2021/08/31 22:38] (current) – escola
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-Tema 3
+**Tema 3: Jacobi equation in Hamiltonian dynamics**
+Recall that the Jacobi equation is defined by the equation $J''+R(\gamma',J)
+\gamma'=0$, which is a second order differential equation (ODE) in the manifold. Using an orthonormal frame of parallel vector fields, this equation can be rewritten as the ODE in $\mathbb R^n$, see the first post.
+The benefit of this latter representation is that the underlying space is fixed. Finally,
+the ODE also has a matrix form, see also the first post. In this section, we will
+consider the generalization of this point of view and study it from the perpective
+of Hamiltonian vector fields.
+Let $(E,\langle\cdot,\rangle)$ be a vector space with inner product, and define
+\begin{align*}
+{\rm End}(E)&=\{A:E\to E:A \text{ is linear}\}\\
+{\rm Sym}(E)&=\{A:E\to E:A\text{ is linear and symmetric}\}.
+\end{align*}
+Also, let $(E\times E,\omega)$ be the canonical symplectic vector space, as defined in the second post.
+//Curvature operator:// A //curvature operator// on $E$ is a $C^\infty$ map
+$R:\mathbb R\to {\rm Sym}(E)$.
+From now on, fix a curvature operator $R$.
+//Jacobi equation, Jacobi field, and Jacobi tensor:// The //Jacobi equation// is
+$$
+\ddot{J}(t)+R(t)J(t)=0
+$$
+A solution $J:\mathbb R\to E$ of the Jacobi equation is called a //Jacobi field//. The Jacobi equation has the matrix version
+$$
+\ddot{B}(t)+R(t)B(t)=0,
+$$
+and a solution $B:\mathbb R\to {\rm End}(E)$ of it is called a //Jacobi tensor//.
+We can interpret Jacobi solutions as Hamiltonian vector fields of a //time-dependent//
+Hamiltonian function.
+//Time-dependent Hamiltonian function:// For each $t\in\mathbb R$, define the function
+$H_t:E\times E\to\mathbb R$ by
+$$
+H_t(x,y)=\tfrac{1}{2}\left[\langle y,y\rangle+\langle R(t)x,x\rangle\right].
+$$
+We have
+\begin{align*}
+\tfrac{\partial H_t}{\partial x}(x,y)\cdot(v,w)&=\tfrac{1}{2}\left[\langle R(t)v,x\rangle+\langle R(t)x,v\rangle\right]
+= \langle R(t)x,v\rangle\\
+\tfrac{\partial H_t}{\partial y}(x,y)\cdot(v,w)&=\tfrac{1}{2}\left[\langle w,y\rangle+\langle y,w\rangle\right]
+= \langle y,w\rangle,
+\end{align*}
+and so the Hamilton equations are
+$$
+\left\{
+\begin{array}{l}
+\dot{X}=\tfrac{\partial H_t}{\partial y}(X,Y)=Y\\
+\\
+\dot{Y}=-\tfrac{\partial H_t}{\partial x}(X,Y)=-R(t)X
+\end{array}
+\right.,
+$$
+whose solutions are exactly the Jacobi solutions $(X,Y)=(J,J')$.
+To study how to generate these solutions, note that if $(A,\dot{A}),(B,\dot{B})$ are Jacobi tensors then
+\begin{align*}
+&\omega((A(t)x,\dot{A}(t)x),(B(t)y,\dot{B}(t)y))=\langle \dot{A}(t)x,B(t)y\rangle - \langle A(t)x,\dot{B}(t)y\rangle\\
+&\langle B(t)^T\dot{A}(t)x,y\rangle - \langle \dot{B}(t)^TA(t)x,y\rangle=
+\left\langle \left[B(t)^T\dot{A}(t)-\dot{B}(t)^TA(t)\right]x,y\right\rangle.
+\end{align*}
+This motivates the following definition.
+//Wronskian:// Given two Jacobi tensors $A,B$, the //Wronskian// of $A,B$ is the function
+$\Omega(A,B):\mathbb R\to {\rm End}(E)$ defined by
+$$
+\Omega(A,B)(t)=B(t)^T\dot{A}(t)-\dot{B}(t)^TA(t).
+$$
+Hence $\omega((A(t)x,\dot{A}(t)x),(B(t)y,\dot{B}(t)y))=\langle \Omega(A,B)(t)x,y\rangle$.
+The Wronskian has important invariance properties.
+**Lemma 1.** If $A,B$ are Jacobi tensors then $\Omega(A,B)$ is constant.
+**Proof.**
+Differentiating, we have
+\begin{align*}
+&\frac{d}{dt}\Omega(A,B)=\frac{d}{dt}\left[B(t)^T\dot{A}(t)-\dot{B}(t)^TA(t)\right]\\
+&=\left[\dot{B}(t)^T\dot{A}(t)+B(t)^T\ddot{A}(t)\right]-\left[\ddot{B}(t)^TA(t)+\dot{B}(t)^T\dot{A}(t)\right]\\
+&=B(t)^T\ddot{A}(t)-\ddot{B}(t)^TA(t)=B(t)^T[-R(t)A(t)]-[-R(t)B(t)]^TA(t)\\
+&=-B(t)^TR(t)A(t)+B(t)^TR(t)A(t)=0,
+\end{align*}
+where in the last passage we used that $R(t)$ is symmetric.
+It is also useful to see that $\Omega(A,B)=0$ iff $\dot{B}B^{-1}=(\dot{A}A^{-1})^T$ (regardless
+$A,B$ are Jacobi tensors). The proof is by direct calculation.
+In particular, $\Omega(A,A)=0$ iff $\dot{A}A^{-1}\in{\rm Sym}(E)$.
+**Lemma 2.** Given $A,B\in{\rm End}(E)$, let
+$$
+L=\{(Ax,Bx):x\in E\}.
+$$
+Then $L$ is isotropic iff $A^TB\in{\rm Sym}(E)$. If additionally the map $x\mapsto (Ax,Bx)$ is injective,
+then $L$ is Lagrangian. In particular, the graph
+$$
+L=\{(x,Ax):x\in E\}
+$$
+is Lagrangian iff $A\in{\rm Sym}(E)$.
+**Proof.**
+We have
+\begin{align*}
+\omega((Ax,Bx),(Ay,By))=\langle Bx,Ay\rangle-\langle Ax,By\rangle=\langle (A^TB-B^TA)x,y\rangle,
+\end{align*}
+hence $L$ is isotropic iff $A^TB=B^TA$. The other claims are direct.
+**Corollary.**
+If $A$ is a Jacobi tensor, then
+$$
+L_t=\{(A(t)x,\dot{A}(t)x):x\in E\}
+$$
+defines a family of isotropic subspaces iff $\Omega(A,A)=0$.
+Hence, $L_a$ is isotropic for some $a\in\mathbb R$ iff $L_t$ is isotropic for all $t$.
+In particular, $\{L_t\}$ is a family of Lagrangians iff $\Omega(A,A)=0$ and ${\rm dim}(L_t)={\rm dim}(E)$
+for every $t$.
+**Proof.**
+By Lemma 1, $\Omega(A,A)$ is constant. We have
+$\Omega(A,A)(t)=0$ iff $A(t)^TA(t)\in{\rm Sym}(E)$ iff $L_t$ is isotropic.
+Here is a simple example: if $A$ is a Jacobi tensor with $A(a)=0$, then
+$\Omega(A,A)(a)=0$ and so $\{L_t\}$ is a family of isotropic subspaces. If additionally $\dot{A}(a)={\rm Id}$,
+then $L_a$ is a Lagrangian.
+Coming back to the Hamilton equations, they define for $a,t\in\mathbb R$ a symplectic map (that
+preserves the symplectic form $\omega$) $\psi_{a,t}:E\times E\to E\times E$
+given by
+$$
+\psi_{a,t}(J(a),\dot{J}(a))=(J(t),\dot{J}(t)).
+$$
+Note that $\psi_{a,a}={\rm Id}$. Nowadays the family $\{\psi_{a,t}\}$ is called a //Hamiltonian isotopy//.
+In the terminology of the last corollary, $L_t=\psi_{a,t}L_a$ and so $\{L_t\}$ is a family of deformations
+of $L_a$ induced by the Hamiltonian isotopy.
+//Lagrangian deformation:// $\{L_t\}$ as above is called a //Lagrangian deformation// if each
+$L_t$ is a Lagrangian subspace.
+As we will see next, Lagrangian deformations are related to the absence of conjugate points.
+**No conjugate points**
+Conjugate points are the infinitesimal notion of positions where geodesics focus together.
+If we expect that chaos is related to divergence of geodesics, then chaotic geodesic flows
+have no conjugate points.
+//No conjugate points:// We say that the Jacobi equation $\ddot{J}(t)+R(t)J(t)=0$ has
+//no conjugate points// on $[a,b]$ if every Jacobi field $J$ s.t. $J(a)=0$ and $\dot{J}(a)\neq 0$
+satisfies $J(t)\neq 0$ for all $t\in (a,b]$.
+Alternatively, there are no conjugate points on $[a,b]$ iff the Jacobi tensor
+$A$ with $A(a)=0$ and $\dot{A}={\rm Id}$ is non-singular for all $t\in(a,b]$.
+In this case, $L_t=\{(A(t)x,\dot{A}(t)x):x\in E\}$ is a Lagrangian graph for all $t\in(a,b]$
+(but not for $t=a$). The next result provides a result in the other direction.
+**Proposition.**
+Assume that $L_t=\psi_{a,t}L_a$ is a Lagrangian graph for all $t\in[a,b]$. Then
+the Jacobi equation has no conjugate points on $[a,b]$.
+Note that the condition of the proposition is slightly stronger than the one obtained
+in the preceding paragraph.
+**Proof.** By assumption, $L_t=\{(B(t),\dot{B}(t)):t\in E\}$ where $B(t)$ is non-singular for $t\in[a,b]$ and satisfies the
+Jacobi equation. Let $A(t)$ be the solution of the Jacobi equation with $A(a)=0$ and $\dot{A}(a)={\rm Id}$.
+We wish to prove that $A(t)$ is non-singular for all $t\in (a,b]$. The idea is that, being the generator of Lagrangian
+graphs, $B$ encodes all the information of Jacobi fields and hence $A$ is uniquely
+determined by $B$. To see that, we show that the derivative of $B^{-1}A$ only depends on $B$, and hence
+$A(t)$ has an explicit formula in terms of $B$. In the sequel, we will omit the parameter $t$ when possible.
+Recall that:
+  * $L$ is isotropic implies that $\Omega(B,B)=0$ and so $\dot{B}B^{-1}\in{\rm Sym}(E)$.
+  * $\Omega(A,B)=\Omega(A,B)(a)=B(a)^T$, hence
+$$
+B^{-1}\dot{A}=B^{-1}(B^{-1})^TB(a)^T+B^{-1}\dot{B}B^{-1}A.
+$$
+Therefore,
+\begin{align*}
+\dot{\widehat{B^{-1}A}}=-B^{-1}\dot{B}B^{-1}A+B^{-1}\dot{A}=B^{-1}(B^{-1})^TB(a)^T
+\end{align*}
+and so
+$$
+A(t)=B(t)\int_a^t B(s)^{-1}[B(s)^{-1}]^TB(a)^Tds
+$$
+is non-singular for all $t\in(a,b]$.
 ~~DISCUSSIONS~~