Let $H:M \to \mathbb{R}$ be a smooth function on a smooth manifold $M$ and consider the pullback $\pi^* H$ of this function to the cotangent bundle $\pi:T^*M \to M$. There is a standard symplectic form on the cotangent bundle $\omega = d \lambda$. Here, $\lambda$ at a point $(x,\phi) \in T^* M$ acts on $Y \in T_{(x,\phi)} T^*M$ in the following way: $\lambda(Y) = \phi(d\pi(Y))$.
Now, $dH$ is a 1-form and can be thought of as a section of $T^*M$. In fact, the image of $dH$ is a Lagrangian submanifold, call it $L$. I would like to know what the flow of the Hamiltonian vector field $X_{\pi^* H}$ does to $L$. As a reminder, $\omega$ is nondegenerate so we can define $X_{\pi^* H}$ by $\omega(X_{\pi^* H},-) = \pi^* dH(-)$.
Suppose $Y$ is a tangent to a fiber $F$ of the bundle. Then $\omega(X_{\pi^*H},Y) = \pi^*dH(Y) = 0$ since $d\pi(Y)=0$. This is true for any $Y$ tangent to the fiber and hence, $X_{\pi^* H}$ is $\omega$-orthogonal to $TF$. But $F$ is a Lagrangian as well and hence $TF=TF^\omega$. Hence, $X_{\pi^*H} \in TF$. Therefore, the flow of $X_{\pi^*H}$ preserves the fibers of the bundle.
But I suspect we can say even more: the time-1 flow $\phi$ should map $L$ to the zero section. I'm basing this off of a toy example: $H: \mathbb{R} \to \mathbb{R}, H(x) = x^2$. Then $X_{\pi^* H} = -2x \, \partial_y$ on $\mathbb{R}^2$ and one notices that $dH = 2x\, dx$. If I haven't made mistakes, the graph of $dH$ is mapped to the zero section.
But then, as I analyze $X_{\pi^*H}$, I seem to be getting another conclusion: $X_{\pi^*H}$ vanishes on $L$! I know I must be making a mistake somewhere but I can't seem to locate it in the following computation.
$\omega(X_{\pi^*H},Y) = d\lambda(X_{\pi^*H},Y) = X_{\pi^*H} \cdot \lambda(Y) - Y \cdot \lambda(X_{\pi^*H}) - \lambda([X_{\pi^*H},Y])$. Suppose we're at a point $(x,dH_x)$. The second term has $dH\circ d\pi(X_{\pi^*H}) = \omega(X_{\pi^*H},X_{\pi^*H}) = 0$.
The third term is $[X_{\pi^*H},Y] \cdot \pi^*H = X_{\pi^*H} \cdot Y \cdot \pi^*H - Y \cdot X_{\pi^*H} \cdot \pi^*H$. The last term in this line vanishes for the same reason as above.
Hence, overall, we have $\omega(X_{\pi^*H},Y) = X_{\pi^*H} \cdot \pi^*dH(Y) - X_{\pi^*H} \cdot Y \cdot \pi^*H = 0$ (since $X\cdot f = df(X)$). Since $Y$ is arbitrary and $\omega$ is nondegenerate, this says that $X_{\pi^* H}$ vanishes on $L$. But that contradicts the toy example I wrote on above; so something I've written is wrong.
