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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 $g$.

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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.

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