Interior contraction of the Riemannian volume form (${\iota }_X{\omega }_g={\sum }_i(-1{)}^{i-1}{f}_i{\theta }_i$)

In a positive Riemannian orthonormal frame $X={\sum }_i{f}_i{E}_i$ with dual coframe ${\omega }_i$, the Riemannian volume form is ${\omega }_1\wedge \cdots \wedge {\omega }_n$. Inserting $E_i$ in its first slot omits ${\omega }_i$ and produces sign $(-1{)}^{i-1}$, giving the displayed interior product formula.

At a point in a normal orthonormal frame, torsion-freeness makes the frame brackets vanish there, hence the dual one-forms have zero exterior derivative there. Differentiating the interior product formula then leaves only ${\sum }_i{E}_i(f_i){\omega }_g$, the divergence of a Riemannian vector field. Since the point is arbitrary, the identity holds globally.

On a compact oriented Riemannian manifold without boundary, the Generalized Stokes theorem applied to $f{\iota }_X{\omega }_g$ gives the displayed identity. The product expansion is $d(f{\iota }_X{\omega }_g)=[X(f)+f\mathrm{div}X]{\omega }_g$. With boundary, the omitted boundary integral is ${\int }_{\partial M}f{\iota }_X{\omega }_g$.