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

Question

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 $Y$?

Answer 1

Yes, there is.

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 $f$.

Answer 2

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 $(X_f)_p=\phi_p^{-1}(df_p)$.