Prove that $\partial x = \partial \cdot x = n$ (Geometric algebra)

Question

I’m trying to understand equation (2-1.34) on page 51 of Hestenes and Sobczyk’s “Clifford Algebra to Geometric Calculus”.

$\partial x = \partial \cdot x = n \tag{1.34}$

According to the book, this follows from

$\partial \wedge x = 0, \tag{1.33}$

$\partial_x = P(\partial_x) = \sum_k a^ka_k\cdot \partial_x, \tag {1.5}$

$a\cdot\partial_x x = P(a) = \partial_x(x\cdot a), \tag{1.18}$

because

$\partial \cdot x = \sum_k a^k\cdot(a_k\cdot\partial x) = \sum_k a^k \cdot a_k =n. \tag{*}$

Here $x$ is a vector from an $n$-dimensional vector space $x\in \mathcal A_n$ with orthonormal basis $a_1,…, a_n$. As $\partial x = \partial \wedge x + \partial \cdot x$, the first equation of $(1.34)$ follows from $(1.33)$.

My problem is understanding the first equation on the left of $(\ast)$.

Answer 1

The equality $$\partial\cdot x = \sum_ka^k(a_k\cdot\partial x)$$ follows directly from $\partial = \sum_ka^k(a_k\cdot\partial)$ and the fact that $a_k\cdot\partial$ is scalar-like.

Answer 2

This is only a partial answer.

For the second equality notice that $a_k \cdot \partial x$ is by (1.2) the directional derivative of the identity function in the $a_k$ direction, so the operator is $a_k \cdot \partial$ and the function is $x$.

Thus in coordinates as if $F(x)=x$ is the identity $F(x+\tau a_t)=x+\tau\, a_t$ you get $$a_k \cdot \partial x=\lim_{\tau \to 0}\frac{x+\tau\,a_t}{\tau}=a_t.$$