# Why is negative divergence an adjoint of gradient?

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?

• See my edits to this question for proper MathJax usage. Commented Nov 7, 2017 at 17:06
• \begin{align} F(x,y,z) = {} & (F_1(x,y,z), F_2(x,y,z), F_3(x,y,z)) \\ \\ \operatorname{div} F(x,y,z) = {} & \frac{\partial F_1}{\partial x} + \frac{\partial F_1}{\partial y} + \frac{\partial F_1}{\partial z} \\ \\ \nabla f(x,y,z) = {} & \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}, \right) \end{align} Commented Nov 7, 2017 at 17:53
• \begin{align} & F_1 \frac{\partial f}{\partial x} + F_2 \frac{\partial f}{\partial y} + F_3\frac{\partial f}{\partial z} \\ \\ \overset{\Large\text{?}}= {} & -f\frac{\partial F_1}{\partial x} -f\frac{\partial F_2}{\partial y} - f\frac{\partial F_3}{\partial z} \end{align} Commented Nov 7, 2017 at 17:53
• \begin{align} & \left( F_1 \frac{\partial f}{\partial x} + f\frac{\partial F_1}{\partial x} \right) + \left( F_2 \frac{\partial f}{\partial y} + f\frac{\partial F_2}{\partial y} \right) + \left( F_3 \frac{\partial f}{\partial z} + f\frac{\partial F_3}{\partial z} \right) \\ \\ = {} & \frac \partial {\partial x} (F_1 f) + \frac \partial {\partial y} (F_2 f) + \frac \partial {\partial z} (F_3 f) \qquad \text{by the product rule} \end{align} Commented Nov 7, 2017 at 17:54
• @MichaelHardy Thanks for edit. Commented Nov 8, 2017 at 5:24

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.

• @Michael Hardy: I always learn something useful from your edits; unfortunately I haven't yet put much of it into regular practice. Thanks. Commented Nov 7, 2017 at 18:01
• I took the liberty of changing $f dV$ to $f\,dV$ and $\vec n dS$ to $\vec n \, dS,$ etc. I also put a bit of space between $f$ and $\nabla\cdot F,$ for the same reason. Commented Nov 7, 2017 at 18:01
• @MichaelHardy: yes it looks better now. Also I didn't know one could use double dollar signs to avoid the "\displaystyle" command. Commented Nov 7, 2017 at 18:03
• I'm glad you found it helpful. Commented Nov 7, 2017 at 18:23
• Can I know why 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 $\partial \Omega$"? Commented Nov 8, 2017 at 5:20

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 $\int_M \operatorname{div}(f F)\, dV = \int_M d \big( \iota_{f F}\, dV \big)= \int_{\partial M} \iota_{fF} \, dV = 0$

• What does iota represent? Commented Nov 8, 2017 at 5:39
• The contraction of a form by a vector field. if $\omega$ is a $k$-form $\iota_X \omega$ is a $k-1$-form such that $(\iota_X\omega)(Y_1,\dots,Y_{k-1}) = \omega(X,Y_1,\dots,Y_{k-1})$. It holds that $d \iota_X dV = \operatorname{div}_X dV$ where $dV$ is the Riemannian volume form. Commented Nov 8, 2017 at 9:01
• ( to be precise, actually it is sufficient to be a volume form, it can be also not a Riemannian one) Commented Nov 8, 2017 at 9:29