For a configuration manifold $M$ we expect there to be a bracket on $\mathcal C^\infty (T^*M)$ endowing the space with a Lie algebra structure. Here we assume that $T^*M$ is also equipped with a non-degenerate symplectic 2-form $\omega_H$ and that the bracket can be written $\omega_H(X_f,X_g)$ for two Hamiltonian vector fields $X_f,X_g\in \chi(T^*M)$. We note that there exists a 1-form $\theta_H$ on $T^*M$ such that $\omega_H=-d\theta$ and that $\theta_H$ can be pulled back to $TM$ to give $\theta_L$. In charts we may write,

\begin{equation}\theta_L=\frac{\partial L}{\partial \dot q^i}dq^i\end{equation}

Hence we expect the exterior derivative to give an induced 2-form on $TM$.

\begin{equation}\omega_L=\frac{\partial ^2}{\partial q\partial v}dq\wedge dq +\frac{\partial ^2L}{\partial v\partial v}dq\wedge dv \end{equation}

At this point I am interested in evaluating $\omega _L(X_f,X_g)$ in order to see what expression is generated. My attempt is to assume that $X_f$ and $X_g$ are now Lagrangian vector fields on $TM$ and hence can be written,

\begin{equation}X_f=\dot q\frac{\partial f}{\partial q}+\dot v\frac{\partial f}{\partial v}\end{equation}

This approach however does not successfully lead anywhere (could be my shoddy maths skills to blame however, so I would be extremely interested to see if it did in fact lead somewhere?). If we cannot develop a bracket on $TM$ does this mean that $f,g\in\mathcal C^\infty (TM)$ do not have a Lie algebra structure, and at what point would they inherit such a structure during their transition to the cotangent bundle?

Thank you for your time and I hope this question is appropriate for this forum (?) apologies if not!