Hamiltonian flows are symplectic

Question

I want to show, in coordinates $(x,\xi)\in T^*\mathbb{R}$, that the Hamiltonian flow $\Phi_t = \exp(t H_p)$ is symplectic for each $t$. Here, $H_p$ is the Hamiltonian vector field determined by the smooth function $p(x,\xi)$. We know that $\begin{cases} \dot{x}(t) = \partial_\xi p \\ \dot{\xi}(t) = -\partial_x p\end{cases}$. So, in order to prove that $\Phi_t$ is a symplectomorphism, we must show $d_{x,\xi}(\xi - t\partial_x p)\wedge d_{x,\xi}(x + t\partial_\xi p) = d\xi\wedge dx$. But this holds, it seems, (if and) only if $p_{xx}p_{\xi\xi} - (p_{x\xi})^2 = 0$ (i.e., only if the Hessian matrix of $p$ is singular). But aren't all Hamilton flows symplectic?

Answer 1

Let $M$ be a simplectic manifold - that is a manifold with a smoothly varying antisymmetric $2$-form $w$ on its tangent bundles, satisfying:

1. $w$ is non-degenerate everywhere,
2. $dw=0$.

The first condition above gives us an isomorphism ${}^*\colon T^*_xM\cong T_xM$. So for any smooth function $H\colon M\to \mathbb{R}$, we can map the cotangent field $\nabla H$ to a tangent vector field $(\nabla H)^*$. We can then consider the resulting flow $f: M\times \mathbb{R}\to M$, which at time $t=0$ is the identity, and for any fixed point has time derivative $(\nabla H)^*$.

In index notation:

3. $f(x,0)=x$,
4. $\frac{\partial f^j(x,t)}{\partial t}=w^{ji}\frac{\partial H}{\partial f^i}$.

Now at any time $t$, we have a map $f_t: M \to M$ and this map is indeed a simplectomorphism. Indeed this follows from condition 2 on $w$ above, which we have so far not used.

That is we defined our flow using the simplectic form $w$, and the property $dw=0$ is exactly what we need to deduce that the resulting flow preserves $w$.

Let us write $dw(u,v,z)$ out in index notation, so it is ready for use when we need it:

5. $dw(u,v,z)=\left(\frac{\partial w_{ij}}{\partial f^k}+\frac{\partial w_{jk}}{\partial f^i}+\frac{\partial w_{ki}}{\partial f^j}\right)u^iv^jz^k$.

To show that $f_t$ is a simplectomorphism, we must show that $w$ applied to a pair of vectors carried by the flow has $0$ derivative over time:

$$\frac{\partial }{\partial t}w\left( \frac{\partial f}{\partial x^a}, \frac{\partial f}{\partial x^b} \right)=0.$$

We proceed:

$$\frac{\partial }{\partial t}w\left( \frac{\partial f}{\partial x^a}, \frac{\partial f}{\partial x^b} \right)= \frac{\partial }{\partial t}\left(w_{ij} \frac{\partial f^i}{\partial x^a} \frac{\partial f^j}{\partial x^b} \right) $$

(by Leibniz rule)

$$ =\frac{\partial w_{ij}}{\partial t} \frac{\partial f^i}{\partial x^a} \frac{\partial f^j}{\partial x^b} + w_{ij} \frac{\partial^2 f^i}{\partial x^a{\partial t}} \frac{\partial f^j}{\partial x^b} +w_{ij} \frac{\partial f^i}{\partial x^a} \frac{\partial^2 f^j}{\partial x^b{\partial t}} $$

(by the chain rule)

$$ =\frac{\partial w_{ij}}{\partial f^k}\frac{\partial f^k}{\partial t} \frac{\partial f^i}{\partial x^a} \frac{\partial f^j}{\partial x^b} + w_{ij} \frac{\partial^2 f^i}{\partial x^a{\partial t}} \frac{\partial f^j}{\partial x^b} +w_{ij} \frac{\partial f^i}{\partial x^a} \frac{\partial^2 f^j}{\partial x^b{\partial t}} $$

(applying (4) to all three terms)

$$ =\frac{\partial w_{ij}}{\partial f^k}w^{ks}\frac{\partial H}{\partial f^s} \frac{\partial f^i}{\partial x^a} \frac{\partial f^j}{\partial x^b} + w_{ij} \frac{\partial}{\partial x^a} \left( w^{is}\frac{\partial H}{\partial f^s}\right) \frac{\partial f^j}{\partial x^b} +w_{ij} \frac{\partial f^i}{\partial x^a} \frac{\partial}{\partial x^b} \left( w^{js}\frac{\partial H}{\partial f^s}\right) $$

(by the chain rule) $$ =\frac{\partial w_{ij}}{\partial f^k}w^{ks}\frac{\partial H}{\partial f^s} \frac{\partial f^i}{\partial x^a} \frac{\partial f^j}{\partial x^b} + w_{ij} \frac{\partial}{\partial f^l} \left( w^{is}\frac{\partial H}{\partial f^s}\right)\frac{\partial f^l}{\partial x^a} \frac{\partial f^j}{\partial x^b} +w_{ij} \frac{\partial}{\partial f^l} \left( w^{js}\frac{\partial H}{\partial f^s}\right)\frac{\partial f^i}{\partial x^a} \frac{\partial f^l}{\partial x^b} $$

(by Leibniz' rule)

\begin{eqnarray*} =\frac{\partial w_{ij}}{\partial f^k}w^{ks}\frac{\partial H}{\partial f^s} \frac{\partial f^i}{\partial x^a} \frac{\partial f^j}{\partial x^b} \\+ w_{ij} \frac{\partial w^{is}}{\partial f^l} \frac{\partial H}{\partial f^s}\frac{\partial f^l}{\partial x^a} \frac{\partial f^j}{\partial x^b} \\+w_{ij} \frac{\partial w^{js}}{\partial f^l} \frac{\partial H}{\partial f^s}\frac{\partial f^i}{\partial x^a} \frac{\partial f^l}{\partial x^b} \\- \frac{\partial^2 H}{\partial f^l\partial f^j}\frac{\partial f^l}{\partial x^a} \frac{\partial f^j}{\partial x^b} \\+ \frac{\partial^2 H}{\partial f^l\partial f^i}\frac{\partial f^i}{\partial x^a} \frac{\partial f^l}{\partial x^b} \end{eqnarray*}

(cancelling the last two terms and using

$\frac{\partial w_{ji}}{\partial f_l}w^{is}+w_{ji}\frac{\partial w^{is}}{\partial f_l}=\frac{\partial (w_{ji}w^{is})}{\partial f_l}=\frac{\partial \delta_j^s}{\partial f_l}=0$)

\begin{eqnarray*} =\frac{\partial w_{ij}}{\partial f^k}w^{ks}\frac{\partial H}{\partial f^s} \frac{\partial f^i}{\partial x^a} \frac{\partial f^j}{\partial x^b} \\+ \frac{\partial w_{ji}}{\partial f^l}w^{is} \frac{\partial H}{\partial f^s}\frac{\partial f^l}{\partial x^a} \frac{\partial f^j}{\partial x^b} \\- \frac{\partial w_{ij}}{\partial f^l} w^{js} \frac{\partial H}{\partial f^s}\frac{\partial f^i}{\partial x^a} \frac{\partial f^l}{\partial x^b} \end{eqnarray*}

(relabelling indices)

$$ =\left(\frac{\partial w_{ij}}{\partial f^k}+\frac{\partial w_{jk}}{\partial f^i}+\frac{\partial w_{ki}}{\partial f^j}\right) w^{ks}\frac{\partial H}{\partial f^s} \frac{\partial f^i}{\partial x^a} \frac{\partial f^j}{\partial x^b} $$

(using (5))

$$=dw\left(\frac{\partial f}{\partial x^a}, \frac{\partial f}{\partial x^b}, \frac{\partial f}{\partial t} \right).$$

So to summarise: $$\frac{\partial }{\partial t}w\left( \frac{\partial f}{\partial x^a}, \frac{\partial f}{\partial x^b} \right) =dw\left(\frac{\partial f}{\partial x^a}, \frac{\partial f}{\partial x^b}, \frac{\partial f}{\partial t} \right), $$ from which it follows immediately that condition 2 implies that $w$ is preserved by $f_t$.