Let $(M^n, g)$ be a closed, connected Riemannian manifold, and let $T^*M$ be its cotangent bundle. Let $\tilde{g}$ be a Riemannian metric on $T^*M$. Let $\omega$ be the canonical symplectic 2-form on $T^*M$.
There are 3 isomorphisms associated to $(T^*M, \tilde{g})$: (1) the "musical isomorphism" given by the Riemannian metric $\tilde{g}$ from $\Gamma(T^*M)$ to $\Omega^1(T^*M)$, which associates a vector field on $T^*M$ (that is to say, a smooth map $\xi: T^*M \to TT^*M$) a 1-form on $T^*M$ $\xi^{\flat}$ with $\xi^{\flat}(p,v^{\flat})(p,v^{\flat},w_1,w_2) = \tilde{g}(\xi(p,v^{\flat}),(p,v^{\flat}, w_1, w_2))$, with inverse $\theta^{\#}$; (2) an isomorphism I will call the "poetic isomorphism" given by the canonical symplectic form $\omega$ from $\Gamma(T^*M)$ to $\Omega^1(T^*M)$, which associates to a vector field on $T^*M$ a 1-form on $T^*M$ $\xi^{\rho}$ with $\xi^{\rho}(p,v^{\flat})(p,v^{\flat},w_1,w_2)$ $= \omega(\xi(p,v^{\flat}),(p,v^{\flat},w_1,w_2))$, with inverse $\theta^{P}$; and (3) the Hodge star operator from $\Omega^k(T^*M)$ to $\Omega^{n-k}(T^*M)$.
Let $H$ be a Hamiltonian function on $T^*M$; let $\text{grad}(H) = dH^{\#}$ be the gradient vector field associated to $dH$ under the inverse of the musical isomorphism, $X_H = dH^{\rho}$ be the vector field associated to $dH$ under the poetic isomorphism, and $\chi_H = X_H^{\flat}$ the 1-form associated to $X_H$ under the musical isomorphism.
Clearly, $dH$ is closed; that is, $ddH = 0$. This post appears to indicate that $\chi_H$ is coclosed, that is, $^*d^*\chi_H = \delta\chi_H = 0$ (using the Hodge star isomorphism on $T^*M$).
Is this the case? That is to say, is it the case that the only vector fields on $T^*M$ that are both gradient and Hamiltonian are the musical isomorphism images of the harmonic 1-forms on $T^*M$?
Thanks in advance for any assistance you can provide.