Let $M$ be an $n$-dimensional Riemannian orientable manifold. Denote the space of vector fields on $M$ by $V(M)$ and of $k$-forms on $M$ by $\Omega(M)$.
Let $\xi = \delta \beta$ be a co-exact 1-form for some 2-form $\beta$. Modulo a sign, my definition of the co-differential is $$\delta = \star d \star : \Omega^k \to \Omega^{k-1}$$ where $\star: \Omega^k \to \Omega^{n-k}$ is the Hodge operator.
The isomorphism between vector fields and 1-form induced by the Riemannian metric on $M$ is denoted by $$\flat = V(M) \to \Omega(M)$$ $$\sharp = (\flat)^{-1}: \Omega(M) \to V(M)$$
Consider the 2-form $\omega = d\xi$ and note that this is a pre-symplectic form (closed 2-form, possibily degenerate). With respect to this pre-symplectic structure, a vector field $X$ on $M$ is Hamiltonian iff its $\omega$-dual 1-form is exact:
$$ X \text{ Hamiltonian } \iff \omega(X, \cdot) \equiv \iota_X\omega \text{ is exact} $$
Assume $M$ to be simply connected, so that a 1-form is closed if and only if it is exact.
Question Is it true that the vector field $\xi^{\sharp}$ is Hamiltonian?
My attempts. Some equivalent conditions for $\xi^{\sharp}$ to be Hamiltonian are
- $\iota_{\xi^{\sharp}}\omega$ is an exact 1-form (by definition of being Hamiltonian)
- $\iota_{\xi^{\sharp}}\omega$ is a closed 1-form (since $M$ is simply connected)
- $\mathcal{L}_{\xi^{\sharp}} \omega = 0$ by Cartan's magic formula
Points 1. and 2. are clearly equivalent; as for 2. $\iff$ 3. denote $X = \xi^{\sharp}$ and by Cartan'f formula $$\mathcal{L}_X\omega = \iota_X(d\omega) + d \, \iota_X(\omega) = d \, \iota_X(\omega) $$ since $\omega$ is closed, so $\iota_X \omega$ is closed iff $\mathcal{L}_X \omega = 0$
$n=2$
In general, for an $n$-oriented Riemannian manifold with Riemannian volume form $\nu_g$, it holds true that $$\iota_X\nu_g = \star X^\flat \quad (n-1)-\text{form}$$ for any vector field $X$ (see e.g. Abate M, Tovena F (2011), Geometria Differenziale, p. 371).
If $n = 2$ the presyplectic form $\omega = d\xi$ must be multiple of the Riemannian volume form, i.e. $\omega = f \nu_g$ for some $f \in C^{\infty}(M)$. So $$ \begin{split} d\left(\iota_{\xi^{\sharp}}\omega \right) & = d \left( \iota_{\xi^{\sharp}}f\nu_g \right) = d\left( f \, \iota_{\xi^{\sharp}}\nu_g \right) = d \left(f \, \star \! \xi \right) \\ & = df \wedge \star \xi + f \, d (\star \xi) \\ \end{split} $$ replacing now $\xi = \delta\beta = \star d \star \beta$
$$ \begin{split} d\left(\iota_{\xi^{\sharp}}\omega \right) & = d \left( \iota_{\xi^{\sharp}}f\nu_g \right) = d\left( f \, \iota_{\xi^{\sharp}}\nu_g \right) = d \left(f \, \star \! \xi \right) \\ & = df \wedge \star \xi + f \, d (\star \xi) \\ & = df \wedge (\star d \star d \beta) + fd(\star \star d \star d \beta) \end{split} $$ so the second term vanishes since $\star \star$ is either $+1$ or $-1$ and $d^2 = 0$, but I don't know what to do with the first one.