Let $TQ^*$ be equipped with its standard symplectic structure and let $X_H$ be a Hamiltonian vector field which is tangent to the fibers of $\pi: TQ^* \to Q.$ I need to show that $$H=h \circ \pi = \pi ^* (h)$$ where $h$ is a smooth function on $Q$.
I am self-reading a book and encountered this as an exercise but could not solve it. Please offer some help!
Let $(x^1, \cdots, x^n)$ be local coordinate of $U \subset Q$, so a local coordinate on $\pi^{-1}(U) \subset T^*Q$ is given by $(x^1\cdots x^n , y^1, \cdots y^n)$ and the symplectic form is given by
$$\omega = \sum_{i=1}^n dx^i \wedge dy^i . $$
If $X_H$ is the Hamiltonian vector field defined by a function $H : T^*Q \to \mathbb R$, then
$$dH = -\iota _{X_H} \omega . $$
Given that $X_H$ is tangential to the fiber, locally we have
$$X_H = \sum_{i=1}^n a^i(x, y)\frac{\partial}{\partial y^i} \Rightarrow dH = \sum_{i=1}^n a^i (x, y) dx^i $$
For each $x\in U$, let $I_x: T^*_x Q\to T^*Q$ be the inclusion. Then
$$d(H|_{T_x^*Q}) = d (I_x^*H) = I_x^* dH= 0 $$
thus $H$ is constant in $T_x^*Q$. So $H = h\circ \pi$ for some function $h: Q \to \mathbb R$. Let $i : Q \to T^*Q$ be the inclusion to the zero section, then
$$h = h\circ \pi \circ i = H \circ i.$$
Hence $h$ is smooth.