Basis-Independent Definition of Gradient of Multivector Fields

Question

The definition of the gradient operator on multivector fields in geometric calculus seems to be $$\nabla = \sum_i e_i\partial_i$$ where $\{e_i\}$ is an orthonormal basis.

That's useful, but it's inconvenient to have to derive the equivalent expressions for the gradient in a non-Cartesian basis.

Is there a basis independent definition of the gradient, similar to the one from vector calculus $$\nabla f = \lim_{\Delta V \to 0} \frac 1{\Delta V}\iint_{S} \hat n f\ dS $$ for multivector fields?

Answer 1

According to Hestenes and Sobczyk's book, "From Clifford Algebra to Geometric Calculus", the derivative of a multivector-valued function $f = f(x)$ defined on the manifold $\mathcal{M}$ ''can be defined by any of the equivalent limits

\begin{equation} \partial f(x) = \lim_{\mathcal{R}\rightarrow 0}\,\frac{1}{\mathcal{R}} \int dS f = \lim_{\left|\mathcal{R}\right|\rightarrow 0}\,\frac{I^{-1}(x)}{\left|\mathcal{R(x)}\right|} \int dS f \end{equation}

where $[\dots]$" (and there follows a full description of all the elements in this definition).

That statement appears on page 252, equation 2.1.

Is that statement an answer to your question ?

Answer 2

My book Vector and Geometric Calculus, Theorem 5.14, gives an expression for the gradient in a non-Cartesian basis.