# Symplectic manifold $M$, unique vector field $X_f$ on $M$ so that $df(Y)= \omega(X_f, Y)$ for all vector fields $Y$?

Let $M$ be a manifold. A symplectic form is a closed $2$-form $\omega$ suh that $X(p) \to \omega(X,\cdot)(p)$ is an isomorphism from $T_pM$ to $T_p^*M$ for each $p \in M$. A manifold together with a symplectic form is called a symplectic manifold.

Suppose $M$ is a symplectic manifold, and let $f$ be a smooth function on $M$. Is there a unique vector field $X_f$ on $M$ so that $df(Y)= \omega(X_f, Y)$ for all vector fields $Y$?

Existence is proved exactly by the formula $$\textrm d f(Y)= \omega(X_f, Y)$$; in other words, $$\omega (X_f, \cdot) = \textrm d f$$, or equivalently $$X_f ^\flat = \textrm d f$$, so $$X_f = (\textrm d f) ^\sharp$$, where $$\flat$$ and $$\sharp$$ are the musical isomorphisms.
Uniqueness is proved using the same formula: assume there exist $$Y_f$$ with the same properties. Then $$\omega (X_f - Y_f, Y) = \omega (X_f, Y) - \omega (Y_f, Y) = (\textrm d f) (Y) - (\textrm d f) (Y) = 0 \ \forall Y$$, and since $$\omega$$ is not degenerate it follows that $$X_f = Y_f$$.
$$X_f$$ is called the Hamiltonian vector field generated by $$f$$, or the symplectic gradient of $$f$$.
Yes, this is called the Hamiltonian vector field of $f$. If $\phi_p$ is your fiberwise isomorphism $T_p(M)\to T_P^*(M)$, you just define $(X_f)_p=\phi_p^{-1}(df_p)$.