Let $M$ be a smooth manifold over $\mathbb{R}$ and $\omega$ a symplectic form on $M$, i.e. a regular, non-degenerate, closed 2-form. Given a vector field on $M$, $X \in \mathcal{V}(M)$, we define the contraction of $\omega$ with respect to $X$ as $$i_X\omega:=\omega(X,-).$$
The problem I am addressing is the following one
For any $f \in \mathcal{O}(M)$ there exists a vector field $X$ such that $i_X\omega = df$
I have two ideas to address this problem. First, fixed $f$ I write explicitly the $1$-forms $df$ and $i_X\omega$ locally and then I set the equality. This gives me sufficient local conditions to determine $X$.
The second idea (which, if correct, is equivalent to the first one) is the following. Let me write the symplectic form as a skew-symmetric non-degenerate bilinear map $$\omega : \mathcal{V}(M) \times \mathcal{V}(M) \rightarrow \mathcal{O}(M).$$
Then, if we restrict ourselves in a local chart, we have a basis for $\mathcal{V}(M)$ and there exists a matrix $M \in M_{ n \times n}(\mathcal{O}(M))$ such that it expresses $\omega$ with respect to the fixed basis of $\mathcal{V}(M)$. For the property of $\omega$ I would say that $M$ has an inverse matrix in $M_n(\mathcal{O}(M))$. This is sufficient to conclude. In fact, in coordinates, by the identity $i_X\omega = df$, we would have that
$$\begin{bmatrix}
X_1, & \dots & X_n \end{bmatrix}=\begin{bmatrix} \partial_{x_1}f, & \dots & ,\partial_{x_n}f \end{bmatrix} \cdot M^{-1}$$
which define $X$ locally. Now remain to prove that these "local definitions" can be put together to define a global vector fields. I would prove this last step using just the definition (if is there any shortcut I will be happy to know that).
A third way to prove it could be the following one.
Since $\omega$ is a symplectic form, it defines a smooth bundle isomorphism
$$TM \rightarrow T^*M, \ T_pM \ni (p,\nu) \mapsto \omega_p(\nu,-) \in T_p^*M$$
(This can be seen as a consequence of the Bundle homomorphism characterization Lemma)
So, the thesis is straightforward.
I would like to know if this is a correct way to proceed.
Thank you
Best Answer
Your approaches are not quite correct since you seem to assume that you are given $i_Xw=df$. The whole point is to prove that such a vector field $X$ exists. This is why you can't just assume that you are given $i_Xw=df$ and solve for $X$, you are not guaranteed a priori that there exists an object $X$ that satisfies this identity.
Your second approach is close to the correct one (although, again, at the end you assume the identity). the approach is as follows. As you observed, at every point $p \in M$, $w$ has an invertible matrix, hence define the matrix $P^{ij}$ such that $$P^{ij}w_{jk}=\delta^{i}_{k}$$
Then define the rank-2 contravariant tensor (the so called Poisson-bivector) to be:
$$P=P^{ij} \dfrac{\partial}{\partial x^i} \wedge \dfrac{\partial}{\partial x^j}$$ You can prove this is a legitimate tensor by checking how $P^{ij}$ transform under a change of coordinates using its definition; it's a straightforward exercise but requires a bit computation so I'll leave that to you. Now define the two maps:
$$F:\Omega(M) \to V(M)$$ by $$F(\alpha) = P(\alpha, -)$$ and a second map $$G: V(M) \to \Omega(M)$$ $$G(X)=w(X,-)$$
Given how we defined $P$ it is easy to show that $G \circ F = Id : \Omega(M) \to \Omega(M)$ is in fact the identity map.
Hence for each $f \in C^{\infty}(M)$ define $$X=P(df,-)$$
Note that $$i_Xw(-)=w(X,-)= G \circ F (df) = Id(df)=df$$