I want to proof the following:
Let $Y: Q \rightarrow TQ$ be a global vector field with corresponding flow $\psi_t$. Now let $X:T^{*}Q \rightarrow T(T^{*}Q)$ be the vector field generated by the corresponding flow
$\Psi_t (\alpha):= \alpha \circ d(\varphi_t^{-1})$.
Then $X$ is the Hamiltonian vector field for the function $H:T^{*}Q \rightarrow \mathbb{R}$ defined as $H(x, \alpha)=\alpha(Y(x))$.
What I have so far:
I want to show that $\omega_{can}(X,.):= d \lambda_{can}(X,.)=-dH$, where $\lambda_{can}$ is the Tautological 1-form on $T^{*}Q$.
On the right hand side, let $x \in Q, \alpha \in T^{*}_xQ$, then
$dH_{x, \alpha} : T_{\alpha}(T_x^{*}Q) \rightarrow T_{H(x, \alpha)} \mathbb{R}, v \mapsto (f \mapsto v(f \circ H)) (= f \mapsto v(f(\alpha(Y(x)))))$ for each $f \in C^{\infty} (\mathbb{R})$. Now I look at the left hand side:
$d \lambda_{can}(X(x, \alpha),v)= \sum\limits_j d p_j(X(x, \alpha)) d q_j(v)$ where $p_j, q_j$ are local coordinates on $T^{*}Q$. Now $d p_j(X(x, \alpha))=d p_j(\dfrac{d}{dt}( \alpha \circ d(\varphi_t^{-1})))$.
From here, I don't know how to proceed. Can anyone give me a hint?