How does a Legendre transformation $H\to L$ work in non-canonical coordinates? Let $H(z)$ be a Hamiltonian and $\omega_{ij}$ the symplectic form on the phase-space and $\omega^{ij}$ its inverse $\omega_{ij} \omega^{jk} = \delta^k_i$. We know that the Hamilton's equations are then given as
$$\dot{z}^i = \omega^{ij} \partial_j H$$
In canonical coordinates $z\to p_i,q^j$ we just have
$$\omega^{ij} =  \begin{pmatrix}
  0 & -\mathbf{1} \\
  \mathbf{1} & 0 
 \end{pmatrix}$$
and thus the usual coordinate form of Hamilton's equations and the Legendre transformation
$$L = p_i \frac{\partial H}{\partial p_i} - H(p,q)$$
However, there exist systems (one example would be a Hamiltonian for spinning tops) where $\omega^{ij}$ cannot be globally put into the canonical form given above. How does one then execute a Legendre transform?  
In other words: Is there a closed general formula for a Lagrangian $L$ in terms of $H$, general phase-space coordinates $z$, and the symplectic form $\omega_{ij}$? 

To add some context: What I want in fact is to write the action on the phase space
$$S[z(t)] = \int p_i \dot{q}^i - H(p,q) \mathrm{d}t \,,$$
where $\dot{q}^i$ is not given in terms of phase-space variables. This is then useful in the variational approach to symplectic structure as discussed e.g. by Marsden et al. (1986).
 A: *

*Given a $2n$-dimensional symplectic manifold $(M,\omega)$, $$\mathrm{d}\omega~=~0 \tag{1}$$ with globally defined Hamiltonian function $H: M \to \mathbb{R}$. 

*Note that there is no unique notion of position and momentum variables, even locally. So the inverse Legendre transformation from the Hamiltonian to the Lagrangian formalism is not a unique or well-defined notion. But there is no need to perform an inverse Legendre transformation: We can still construct a Hamiltonian action, as shown in my Phys.SE answer here. Here we will just repeat the main action formula (4).

*Locally in a contractible open coordinate neighborhood $U\subseteq M$ there exists a symplectic potential 1-form $$\vartheta ~=~\sum_{I=1}^{2n}\vartheta_I~\mathrm{d}z^I  ~\in~ \Gamma(T^{\ast}M|_U),\tag{2}$$ such that 
$$ \omega|_U ~=~\mathrm{d}\vartheta .\tag{3}$$

*Given a path $\gamma \subset U$. Define the local Hamiltonian action 
$$S_U[\gamma]~:=~\int_{\gamma} \left( \vartheta - H ~\mathrm{d}t\right)
~=~\int_{t_i}^{t_f}\! \mathrm{d}t\left( \sum_{I=1}^{2n}\vartheta_I~\dot{z}^I -H\right). \tag{4}$$
