coordinate free proof that $\text{div}(\nabla f \times \nabla g) = 0$

Question

Let $V$ be a Euclidean $3$-dimensional space. Does there exist a coordinate-free proof that for any two $C^1$-functions $f, g: \mathbb{R}^3 \to \mathbb{R}$ we have $$\text{div}(\nabla f \times \nabla g) = 0?$$

Answer 1

I'm going to use $\nabla \cdot F$ instead of $\text{div}(F)$.

$\nabla \cdot (A \times B) = B \cdot (\nabla \times A) - A \cdot (\nabla \times B)$ is a well known vector calculus identity for $A, B$ vector fields.

If $A = \nabla f$ and $B = \nabla g$ then $\nabla \cdot (\nabla f \times \nabla g) = \nabla g \cdot (\nabla \times \nabla f) - \nabla f \cdot (\nabla \times \nabla g)$

It is also well known that $\nabla \times \nabla f = 0$ (also known as curl of gradient of function is $0$) for any $f: \mathbb{R}^3 \rightarrow \mathbb{R}$. So then

$\nabla \cdot (\nabla f \times \nabla g) = \nabla g \times 0 - \nabla f \times 0 = 0$.

Answer 2

Yes, there does. By definition of $\text{div}$, it suffices to show that $d(\lrcorner\nabla f \times \nabla g) = 0$. We want to use the fact that the cross-product $v \times w$ of two vectors can be defined by the formula $$v \times w = \mathcal{D}^{-1} \star (\mathcal{D}(v) \wedge \mathcal{D}(w)),$$ where $\star$ denotes the Hodge star operator. With this in mind, since we know that $\mathcal{D}(\nabla f) = df$ and similarly $\mathcal{D}(\nabla g) = dg$, this is the same as$$d\lrcorner\mathcal{D}^{-1} \star (df \wedge dg) = d \star \mathcal{D}\mathcal{D}^{-1} \star (df \wedge dg) = d(df \wedge dg) = d^2(f\,dg) = 0.$$In the above, we have used that $\lrcorner = \star\mathcal{D}$, $\star^2 = \text{Id}$, and $d^2 = 0$.

Answer 3

Assuming the functions involved are $C^2) as Robert Lewis pointed out, we have

$$\begin{aligned} \operatorname{div}(\nabla f \times \nabla g) & = \nabla \cdot (\nabla f \times \nabla g) \\ & = \nabla \cdot ( (\nabla f)_c \times \nabla g) + \nabla \cdot (\nabla f \times (\nabla g)_c) \\ & = - (\nabla f)_c \cdot (\nabla \times \nabla g) + (\nabla g)_c \cdot (\nabla \times \nabla f) \\ & = - \nabla f \cdot (\nabla \times \nabla g) + \nabla g \cdot (\nabla \times \nabla f) \\ & = 0, \end{aligned}$$

since $\nabla \times \nabla$ is zero. The subscript $c$ denotes a quantity held fixed when applying the operation. This means that in the product $\nabla \cdot ((\nabla f)_c \times \nabla g)$ we treat $ (\nabla f)_c$ as a constant vector and the result is $\nabla \times \nabla g$.

Answer 4

There is such a proof, assuming at least that $f$ and $g$ are $C^2$: $$\text{div}(\nabla f \times \nabla g) = \text{div}(\text{curl}(f\nabla g) - f\text{curl}(\nabla g)) = \text{div}\, \text{curl}(f\nabla g) = 0.$$

Here we used that the divergence of a curl is zero and the curl of a gradient is zero.

Answer 5

This is essentially the same as the other solutions here (esp. Kevin Dong's), but it exploits the efficiency of abstract index notation, and makes very clear what essential features we need for this identity to hold.

Given a metric $h_{ab}$ on a $3$-manifold with volume form $\epsilon_{abc}$, we can write the cross product as $$(X \times Y)^a := h^{ad} \epsilon_{bcd} X^b Y^c.$$ Now, $(\nabla f)^b = h^{bc} f_{, e},$ and the cross product of two gradients takes a particularly simple form: $$(\nabla f \times \nabla g)^a := h^{ad} \epsilon_{bcd} h^{be} f_{, e} h^{ci} g_{, i} = \epsilon^{eia} f_{, e} g_{, i}. \qquad (\ast)$$ In particular, notice that this expression doesn't depend on the metric directly, only the volume form $\epsilon$ and the covariant derivative $\nabla$.

Taking the divergence of this quantity, which coincides with $\text{tr} \circ \nabla$, gives $$\text{div} (\nabla f \times \nabla g) = (\nabla f \times \nabla g)^a{}_{, a} = (\epsilon^{eia} f_{, e} g_{, i})_{, a} = \epsilon^{eia}{}_{, a} f_{,e} g_{,i} + \epsilon^{eia} f_{,e a} g_{,i} + \epsilon^{eia} f_{,e} g_{,i a}.$$ Now, the volume form $\epsilon^{eia}$ is parallel, so the first term on the right vanishes. Next, $f_{, ea}$ is a Hessian of a function, and so is symmetric in the $e$ and $a$ indices; but these indices are contracted with the $e$ and $a$ indices of the totally skew volume form $\epsilon^{eia}$, and so this term is zero; similarly, so is the last term, which gives $$\text{div} (\nabla f \times \nabla g) = 0$$ as desired.

Note that if we we take the right-hand side of $(\ast)$ to be the definition of $\nabla f \times \nabla g$, then all we need for this identity to hold is a volume form $\epsilon_{abc}$ and a torsion-free connection that preserves it, that is, for which it is parallel. (Torsion-freeness here is necessary for the Hessian to be symmetric.)

Answer 6

This is basically Kevin Dong's proof, but using geometric calculus instead of differential forms notation.

Write the expression as

$$\nabla \cdot (\nabla f \times \nabla g) = \nabla \cdot [\epsilon^{-1} (\nabla f \wedge \nabla g)]$$

where $\epsilon$ is the right-handed pseudoscalar. It can be pulled out as follows:

$$\nabla \cdot [\epsilon^{-1} (\nabla f \wedge \nabla g)] = [\nabla \wedge (\nabla f \wedge \nabla g)] \epsilon^{-1}$$

That evaluates to

$$[(\nabla \wedge \nabla f) \wedge \nabla g - \nabla f \wedge (\nabla \wedge \nabla g)] \epsilon^{-1}$$

But for any scalar field $h$, $\nabla \wedge \nabla h = 0$ by the equality of mixed partial derivatives, so this is identically zero.