Why is negative divergence an adjoint of gradient?

Question

In my notes, I have $\langle F, \nabla f\rangle_{L^2(\mathcal{TX})} = \langle \nabla^* F, f\rangle_{L^2(\mathcal{X})} = \langle -\operatorname{div} F, f\rangle_{L^2(\mathcal{X})}$, where $f$ is a scalar field, $F$ is a vector field, $\mathcal X$ is a manifold and $T\mathcal X$ is a tangent plane. My question is why is negative divergence an adjoint of gradient?

Answer 1

In general, for a function $f$ and a vector field $F$, we have the following easily verified formula, see for example this wikipedia page:

$$\nabla \cdot (fF) = \langle \nabla f, F \rangle + f \,\nabla \cdot F; \tag 1$$

then if $\Omega \subset \mathcal X$ is an open set of finite measure with a sufficiently nice boundary $\partial \Omega$,

$$ \int_\Omega \nabla \cdot (fF) \, dV = \int_\Omega (\langle \nabla f, F \rangle + f \, \nabla \cdot F) \, dV = \int_\Omega \langle \nabla f, F \rangle \, dV + \int_\Omega f \, \nabla \cdot F \, dV; \tag 2$$

by the divergence theorem,

$$ \int_\Omega \nabla \cdot (fF) \, dV = \int_{\partial \Omega} (fF)\cdot \vec n \, dS, \tag 3$$

where $\vec n$ is an outward pointing unit vector field on $\partial \Omega$; using (3) in (2) yields

$$ \int_{\partial \Omega} (fF)\cdot \vec n \, dS = \int_\Omega \langle \nabla f, F \rangle \, dV + \int_\Omega f \, \nabla \cdot F \, dV; \tag 4$$

if we now make an additional assumption such as $\Omega$ is without boundary, i.e. $\partial \Omega = \emptyset$ or that $f$ or $F$ vanish on $\partial \Omega$, we have

$$ \int_{\partial \Omega} (fF)\cdot \vec n \, dS = 0, \tag 5$$

and then (4) immediately becomes

$$ \int_\Omega \langle \nabla f, F \rangle \, dV = -\int_\Omega f \, \nabla \cdot F \, dV = \int_\Omega (-\nabla \cdot F) f \, dV. \tag 6$$

Note: Though the above argument uses a slightly different notation than that of our OP Aha, it establishes the desired result with the caveat that some assumptions on the behavior of $f$ and $F$ on $\partial \Omega$ must be made. In fact, $\nabla$ and $\nabla \cdot$ require such an assumption if they are to be adjoints of one another, as indicated by (4). We are essentially performing integration by parts on $\bar \Omega = \Omega \cup \partial \Omega \subset \mathcal X$. End of Note.

Answer 2

That is because of the identity $ \operatorname{div}(f F) =\langle \operatorname{grad}f, F \rangle +f \operatorname{div} F$, so in a (compact , orientable) (semi-)Riemannian $n$-manifold without boundary you have $$\int_{M}\langle \operatorname{grad}f, F \rangle\, dV= \int_M \operatorname{div}(f F)\, dV - \int_M f \operatorname{div} F \, dV = - \int_M f \operatorname{div} F \, dV $$ because Stokes imples that $\int_M \operatorname{div}(f F)\, dV = \int_M d \big( \iota_{f F}\, dV \big)= \int_{\partial M} \iota_{fF} \, dV = 0$