# Hamiltonian flows are symplectic

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?

• I'm not sure what you mean by $d_{x,\xi}(\xi - t\partial_x p)\wedge d_{x,\xi}(x + t\partial_\xi p)$. Anyway Hamiltonian flows being symplectic follows from a 1-line calculation from Cartan's formula ($L_X=i_Xd+di_X$) and the fact that a flow $\omega$ along a vector field $X$ preserves a form if the Lie derivative $L_X\omega$ vanishes. Apr 11 '17 at 17:33
• In general given a vector field $X$, it is generally impossible (as far as I know) to get an explicit formula for the flow as it is generally constructed using the contraction mapping principle in a really essential way. Apr 11 '17 at 17:52

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:

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