Tema 2: Geodesic flows in Hamiltonian dynamics

Symplectic form and symplectic manifold

We start recalling some known facts on linear algebra. Let $V$ be a vector space.

Symplectic bilinear map: We say that a map $\omega:V\times V\to\mathbb R$ is symplectic if it is:

• bilinear: $\omega$ is linear on each coordinate.
• antisymmetric: $\omega(x,y)=-\omega(y,x)$ for all $x,y\in V$.
• non-degenerate: the map $x\in V\mapsto \omega_x(\cdot)=\omega(x,\cdot)\in V^*$ is an isomorphism, i.e. $\omega(x,y)=0$ for all $y\in V$ iff $x=0$.

In this case, we call $(V,\omega)$ a symplectic vector space.

Symplectic bilinear forms have canonical representations: if $\omega$ is symplectic, then there is a basis $\{x_1,\ldots,x_n,y_1,\ldots,y_n\}$ s.t. $$ \left\{ \begin{array}{l} \omega(x_i,x_j)=\omega(y_i,y_j)=0\\ \omega(x_i,y_j)=\delta_{i,j} \end{array}\right. ,\ \ \forall i,j=1,\ldots,n. $$ In particular, $V$ has even dimension, equal to $2n$. In this basis, $\omega$ takes the form $$ \omega(x,y)=x^T \left[ \begin{array}{ccc|ccc} &&&&&\hspace{.2cm}\\ &0&&&{\rm Id}&\\ &&&\hspace{.2cm}&&\\ \hline &&&&&\\ &{\rm -Id}&&&0&\\ &&&&&\\ \end{array}\right] y. $$

Here is the prototype of a symplectic vector space: given $(E,\langle\cdot,\cdot\rangle)$ a vector space with inner product, let $V=E\times E$ and define $\omega:V\times V\to \mathbb R$ by $$ \omega((x_1,y_1),(x_2,y_2))= \langle y_1,x_2\rangle - \langle x_1,y_2\rangle. $$ Then $(V,\omega)$ is a symplectic vector space. The literature usually considers $E=\mathbb R^n$ for this example, in which case the symplectic bilinear map is defined in $\mathbb R^{2n}$.

With respect to the symplectic structure, there are some special subspaces.

Isotropic and Lagrangian subspaces: Let $(V,\omega)$ be a symplectic vector space. Given a subspace $Y\subset V$, define $$ Y^\omega=\{x\in V:\omega(x,y)=0\text{ for all }y\in Y\}. $$ It is easy to see that ${\rm dim}(Y)+{\rm dim}(Y^\omega)=2n$. We call $Y$:

• isotropic if $Y\subset Y^\omega$, i.e. if $\omega(x,y)=0$ for all $x,y\in Y$. Clearly ${\rm dim}(Y)\leq n$.
• Lagrangian if it is isotropic and ${\rm dim}(Y)=n$. Alternatively, $Y$ is Lagrangian iff $Y=Y^\omega$.

The next definition introduces the symplectic structure on manifolds. Let $M$ be a $m$-dimensional differentiable manifold, and recall the definition of a symplectic form (which is the choice of symplectic bilinear maps on each tangent space of $M$).

Symplectic manifold: We call $(M,\omega)$ a symplectic manifold if $M$ is a differentiable manifold and $\omega$ is a symplectic form on $M$.

In particular, $m=2n$ is even. The simplest example of a symplectic manifold is $(\mathbb R^{2n},\omega)$ as defined above: if $(x_1,\ldots,x_n,$ $p_1,\ldots,p_n)$ denotes the coordinates of $\mathbb R^{2n}$, then $$ \omega=dx_1\wedge dp_1+\cdots+dx_n\wedge dp_n. $$

Compare this with Riemannian manifolds. A Riemannian manifold is a differentiable manifold equipped with a Riemannian metric. The difference between metrics and two-forms is that the first is symmetric, while the other is antisymmetric. This simple difference makes the two contexts extremely different.

$TM$ is a symplectic manifold

We already know that $(TM,\omega)$ is a symplectic manifold. Let us complete this discussion showing another definition of $\omega$.

1–form $\Theta$ on $TM$: Define the 1–form $\Theta$ on $TM$ taking $\Theta_v:TTM_v\to\mathbb R$ by $$ \Theta_v(x,y)=\langle x,v\rangle. $$

Above, $TTM_v\cong TM_p\times TM_p$ via the horizontal and vertical coordinates.

Lemma. We have $d\Theta=\omega$. In particular, $\omega$ is closed.

Hamiltonian vector fields

Let $(M,\omega)$ be a symplectic manifold, and let $H:M\to\mathbb R$ be a $C^1$ function. The derivative $dH$ is a 1–form.

Hamiltonian vector field: The Hamiltonian vector field of $H$ is the unique vector field $X_H:M\to TM$ defined by $$ \omega_p(\cdot,X_H(p))=dH_p(\cdot). $$ The flow generated by $X_H$ is called the Hamiltonian flow of $H$.

In other words, the contraction 1–form $\omega(\cdot,X_H)$ equals the derivative $dH$. Existence and uniqueness follow from the non-degeneracy of $\omega$.

Let us express the Hamiltonian vector field in the prototypical example of $(\mathbb R^{2n},\omega)$. Let $H:\mathbb R^{2n}\to\mathbb R$ be a $C^1$ function. We will show that $X_H$ satisfies the Hamilton equations. For $(v,w)\in\mathbb R^{2n}$, we have $$ dH(v,w)=\sum_{i=1}^n\tfrac{\partial H}{\partial x_i}v_i+\sum_{i=1}^n\tfrac{\partial H}{\partial p_i}w_i. $$ On the other expression, if $X_H=(X,Y)=(X_1,\ldots,X_n,Y_1,\ldots,Y_n)$ then $$ \omega((v,w),X_H)=\sum_{i=1}^n w_iX_i-\sum_{i=1}^n v_iY_i. $$ Comparing the last two equations, we conclude that $X_i=\tfrac{\partial H}{\partial p_i}$ and $Y_i=-\tfrac{\partial H}{\partial x_i}$ for $i=1,\ldots,n$, and so the Hamilton equations $$ \left\{ \begin{array}{l} X=\tfrac{\partial H}{\partial p}\\ \\ Y =-\tfrac{\partial H}{\partial x} \end{array} \right. $$ characterize $X_H$.

The Halmiltonian vector field has some important invariance properties. Let $\phi=\{\phi^t\}$ be flow generated by $X_H$.

Theorem. The following holds:

(1) $H\circ\phi^t=H$ for all $t\in\mathbb R$, hence $H$ is constant along flow lines.

(2) $\omega$ is preserved by $\phi$, i.e. $\omega(d\phi^t v,d\phi^t w)=\omega(v,w)$ for all $v,w\in TM$ with same basepoint and all $t\in\mathbb R$.

(3) $\phi$ preserves the volume form $\omega^n$.

The geodesic flow is a Hamiltonian flow

We already know that $(TM,\omega)$ is a symplectic manifold. We claim that the geodesic flow $g:TM\to TM$ is the Hamiltonian flow of the function $G:TM\to\mathbb R$ defined by $$ G(v)=\tfrac{1}{2}\langle v,v\rangle. $$ To prove this, write $X_G=(X,Y)$. For $\xi=(x,y)\in TTM$, we have $$ \omega(\xi,X_G)=\langle y,X\rangle-\langle x,Y\rangle. $$ If $x,y\in TM_v$, let $Z:(-\varepsilon,\varepsilon)\to TM$ s.t. $Z(0)=v$ and $Z'(0)=(\dot{c}(0),\nabla_{c}Z(0))=\xi$ where $c=\pi\circ Z$. Then $$ dG_v(\xi)=\tfrac{d}{dt}|_{t=0}(G\circ Z)=\tfrac{d}{dt}|_{t=0}\tfrac{1}{2}\langle Z(t),Z(t)\rangle=\langle Z,\nabla_{\dot{c}}Z\rangle(0)= \langle v,y\rangle=\langle y,v\rangle $$ and so $(X,Y)=(v,0)$. This later vector is the generator of the geodesic flow, and so the assertion is proved.