When is the gradient of a Hamiltonian function a Liouville vector field?

Question

Let $(M, \omega)$ be a symplectic manifold, $H$ a Hamiltonian function on $M$, $Y = H^{-1}(c)$ for a regular value $c$, and $J$ a compatible almost complex structure.

If $X_H$ is the Hamiltonian vector field associated to $H$, then $-JX_H$ is the gradient of $H$ with respect to the metric associated to $\omega$ and $J$.

Under what conditions will $Y$ be of contact type with a (local) Liouville vector field given near $Y$ by $-JX_H$?

Under what conditions will there exist any contact structure on $Y$ such that $X_H$ is the Reeb vector field?

Note: I'm fine with answers "up to scaling" - i.e. multiplying the metric, the symplectic form, or any of the vector fields by a nonvanishing smooth function.

Answer 1

If you allow us to multiply the vector fields by functions, there is no condition. More precisely: The hypersurface $Y$ is of contact-type if and only if there exists a nowhere vanishing function $f \colon Y \to \mathbb{R}$ such that $f X_H$ is a Reeb vector field.

Indeed, suppose $Y$ is of contact-type. Then there exists a local Liouville vector field $Y$ so $\alpha := (i_Y\omega)|_Y$ is a contact form on $Y$. Notice that the Reeb vector field has $d\alpha(R, \cdot) = 0$, which means $\omega|_Y(R, \cdot) =0$. Notice that $\omega(X_H, \cdot)|_{Y} = 0$, so by the non-degeneracy, $X_H$ and $R$ are pointwise multiples of each other. Since $Y$ is a regular level, $X_H$ is nowhere vanishing and so the claim follows.