How to prove that div (fv) =grad f * v + div v f without coordinates?

Question

Here f is a scalar and v a vector field.

If there is no standard proof can this be proved in geometric algebra without a basis?

Answer 1

Gradient and divergence rely on a metric. If $M$ is a Riemannian manifold, the metric induces isomorphisms between vector fields and one-forms. If $V$ is a vector field, let $V^\sharp$ be the corresponding one-form, and if $\alpha$ is a one-form, let $\alpha^\flat$ be the corresponding vector field. They are defined by $V^\sharp(X) = (V,X)$ and $(\alpha^\flat,X) = \alpha(X)$ for any vector field $X$.

The metric produces a volume form $\omega$, and Hodge star operator $\star$, which takes $k$-forms to $(n-k)$-forms. Then we can define: $$\begin{aligned} \operatorname{grad} f &= (df)^\flat \\ \operatorname{div} V &= \mathop\star d \mathop\star V^\sharp \end{aligned}$$

Using these definitions, the product rule that you are seeking comes from the fact that $d$ is a derivation. But the fact that all of this relies on on a metric makes it no more general than the version which uses a basis.

Answer 2

You could try checking 'Multivector Calculus' by David Hestenes; J. Math. Anal. and Appl., Vol. 24, No. 2, 313–325.

Hestenes gives a coordinate-free definition of multivector derivative in terms of directed integration as equation 4.1. This is the directed integral of the function over the surface of a neighbourhood divided by its volume, as the volume tends to zero. The vector derivative is a special case of this. When applied to a scalar field it gives grad, when applied to a vector field it gives scalar (div) and bivector (curl) parts (equation 4.4). You need to pick out just the scalar part. The Leibniz rule for a product is shown as equation 4.14 and proved with 4.15.

You will probably need to fill in some gaps in the argument, but hopefully this should help you figure out where to start looking.