TITLE: coordinate free proof that $\text{div}(\nabla f \times \nabla g) = 0$
QUESTION [8 upvotes]: 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 $$\text{div}(\nabla f \times \nabla g) = 0?$$

REPLY [2 votes]: 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.