Finding an explicit formula for a Hamiltonian vector field I've been looking at this question: Existence of vector field given a smooth function
That is: Given a symplectic manifold $M$ of dimension $2n$, with a symplectic form $\omega \in \Omega^2(M)$, do we have for all smooth function $f\in C^\infty(M)$ a vector field $U_f$ on $M$ such that $df=i_{U_f}\omega$?
Alex M. gives an answer there using the musical isomorphisms, which assumes some previous knowledge with pseudo-Riemannian manifolds. I was wondering if one can give an explicit expression to the obtained vector field, as I am not so strong on the subject of Riemannian metrics, and cannot really decipher in terms of tangent vector what the expression should be. 
I apologize if this is a trivial question, and would appreciate any help gaining insight as to how one can obtain an explicit formula for a Hamiltonian vector field?
 A: My attempt at an answer using the help given by user3257842:
Let $\{ e_1(p),....,e_{2n}(p) \}$ be a basis for $T_p(M)$, orthonormal with respect to the Riemannian metric $\rho$. Define the representing matrix of $\omega_p$ to be $B(p)\in \mathbb{R}_{2n \times 2n}$, by:  $ B_{i,j}(p):= \omega_p \Big( e_i, e_j \Big) $
We know that for two tangent vectors $\eta_{(1)}$ and $\eta_{(2)}$ we have:
$ \omega(\eta_{(1)},\eta_{(2)}) = \eta_{(1)}^t \cdot B(p)\cdot \eta_{(2)}  $, 
 with coordinate vectors with respect to the above orthonormal basis.
For all $\eta\in T_p(M)$ we have that:
   $ dH_p(\eta)= \rho( \nabla H(p), \eta ) $
And by our choice of basis, we have that:
   $ dH_p(\eta)= \Big( \nabla H(p) \Big) ^t \cdot \eta $, 
   with the coordinate vectors with respect to our orthonormal basis.
Let us define $X_h(p)$ by:
   $ X_H(p):=\Big( \nabla H(p) \Big) ^t\cdot \Big( B(p) \Big)^{-1} $
By this definition, we can see that:
$ \Big( X_H\Big)^t B(p)= \Big( \nabla H(p) \Big) ^t\cdot \Big( B(p) \Big)^{-1} \cdot B(p)=\Big( \nabla H(p) \Big)^t $
And therefore:
$ \omega(X_H(p), \eta)= dH(\eta) \quad \text{for all} \quad \eta \in T_p(M) $
This will be our Hamiltonian vector field. 
   Let us notice, that since $\omega$ is invertible and smooth, if we denote it's inverse $\omega^{-1}$, then it is also smooth. i.e, the entries of $\Big( B(p) \Big)^{-1}$ are also smooth as a function of $p$. And we obtain that $X_H$ is a result of a composition of smooth maps, and is therefore smooth.
