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

Being a flow of symplectomorphism means:

$\Phi_t^*\omega =\omega$

In case you flow by a time independent vector field $X$ the infinitesimal version of this is:

$\mathcal{L}_X\omega = \frac{d}{d t}\Phi_t ^*\omega |_{t=0}=0$

As mentioned in the comments to check this the most convenient way is to use Cartan's formula. However, if you really intend to do it in coordinates that is also perfectly fine, just a little more to write. We obtain:

$\Phi_t ^*\omega= (\partial_x x(t) \partial_{\xi}\xi(t) -\partial_{\xi}x(t)\partial_x \xi(t))dx\wedge d\xi$

Now takeing the time derivative yields, we obtain using the product rule, the fact that $\Phi_0=id$ and the Hamiltonian equations:

$\frac{d}{d t}\Phi_t ^*\omega |_{t=0}=\partial_{x}\dot x(0)+\partial_{\xi} \dot \xi (0)=0$

Answer 2

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 $f_t$.