# Vanishing product of a closed and coclosed form on a Riemannian manifold

For a (compact) Riemannian manifold $$(M,g)$$, can it happen that for a non-zero form $$\text{d}^*\omega$$, and a smooth function $$f$$ such that $$\text{d}f \neq 0$$, we can have $$\text{d}f \wedge \text{d}^*\omega = 0?$$ Note that $$*$$ denotes the codifferential with respect to $$g$$.

Yes, this can happen: Take the flat torus $$T=\mathbb R^2/\mathbb Z^2$$, $$f\colon T\to \mathbb R; x\mapsto \sin(2\pi x),$$ $$g\colon T\to \mathbb R; y\mapsto \sin(2\pi y),$$ and $$\omega=g \text{vol}= g dx\wedge dy.$$ Then, $$df\wedge d^* g=\pm \cos(2\pi x)\cos(2\pi y) dx\wedge dx=0,$$ where the actual sign does not matter.