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 $t\in(a,b]$.