Characteristics of Metric Dual to Killing Vector Field For a Killing vector field on a compact manifold, can we say that the metric dual one-form is closed, co-closed or both without appealing to additional information?
 A: $X$ is a Killing field on a Riemannian manifold $(M,g)$ iff
$$g(D_Y X, Z) + g(Y, D_Z X) = 0$$
for any vector fields $Y$ and $Z$. On the other hand, if $\xi = g(X,-)$ is the $g$-dual of $X$, then
$$ (d\xi)(Y,Z) = D_Y(g(X,Z)) - D_Z(g(X,Y)) - g(X,[Y,Z]) = g(D_YX,Z) - g(D_ZX,Y),$$
using that the Levi-Civita connection is torsion-free and compatible with $g$. We thus see that $d\xi = 0$, that is to say $\xi$ is closed, iff
$$ g(D_YX,Z) - g(D_ZX,Y) = 0.$$
The two equations are not equivalent.
Edit 1: here is an easy counterexample. Let $g = \frac{dx^2 + dy^2}{y^2}$. Then $X = \partial_x$ is a Killing field of $g$. Its $g$-dual is
$$\xi = \frac{dx}{y^2}$$
which is not closed. It is also not co-closed, since its Hodge star is simply
$$\frac{dy}{y^2}$$
which is not closed.
Edit 2: I just realized that you asked about compact manifolds. My example in edit 1 is non-compact, in fact it is the hyperbolic plane. Here is another example, which is compact.
Let $M = S^2$ with spherical coordinates $\phi$ (the longitude) and $\theta$ (the co-latitude). Its round metric $g$ is
$$g = \sin^2(\theta) d\phi^2 + d\theta^2.$$
Then $\partial_\phi$ is a Killing field of $M$. Its metric dual is
$$\xi = \sin^2(\theta) d\phi$$
which is not closed. However, in this example, $\xi$ is actually co-closed.
Edit 3: I should probably mention Bochner's result. First note that if $(M,g)$ is compact oriented, then a $1$-form is harmonic iff it is closed and co-closed. Bochner's theorem says that if, moreover, $(M,g)$ has nonnegative Ricci, then any harmonic $1$-form is actually parallel.
