Let $(M,g)$ be a compact Riemannian manifold with boundary, and suppose $N$ is the outward unit normal vector field along $\partial M$. I am trying to prove the following integration by parts formula: $$ \int_M \langle \nabla F, G \rangle \, dV_g = \int_{\partial M} \left\langle F \otimes N^\flat, G \right\rangle \, dV_{\widehat g} - \int_M \left\langle F, \mathrm{div}\,G\right\rangle \,dV_g $$ where $F$ is a covariant $k$-tensor field, $G$ is a covariant $(k+1)$-tensor field, and $\mathrm{div}\,G = \mathrm{tr}_g(\nabla G) = \mathrm{tr}\left(\left(\nabla G\right)^\sharp\right)$.
I'm trying to emulate the proof of the more traditional integration by parts formula $$ \int_M \left\langle \mathrm{grad}\,u, X \right\rangle \,dV_g = \int_{\partial M} u \left\langle X, N \right\rangle \,dV_{\widehat g} - \int_M u \,\mathrm{div}\,X \, dV_g $$ which is proven in the following steps:
- Prove $\iota_{\partial M}^*\left(X ^\big\lrcorner dV_g\right) = \langle X, N \rangle \, dV_{\widehat g}$, where $\iota_{\partial M} : \partial M \hookrightarrow M$ is inclusion and $X^\big\lrcorner dV_g$ denotes interior multiplication;
- Use this result and Stokes' theorem to prove the divergence theorem: $$ \int_M(\mathrm{div}\,X)\,dV_g = \int_{\partial M} \langle X, N \rangle \, dV_{\widehat g} $$ where the divergence is defined in this case as $(\mathrm{div}\,X)\,dV_g = d\left(X^\big\lrcorner dV_g\right)$;
- Prove the product rule for the divergence operator: $\mathrm{div}(uX) = u\,\mathrm{div}\,X + \langle \mathrm{grad}\,u, X \rangle$
But I'm not sure what the parallel steps would be. And after taking a closer look, I'm not sure I'm going about this the right way at all, since Stokes' theorem requires the integrand to be a differential form, and I don't see at all how to express the divergence of an arbitrary covariant tensor field as an alternating covariant tensor field. Help on how to begin?
EDIT: See comments below for an outline of a solution.