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

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 $$(\ast)$$.

• Like I said in my previous comment to you, you are seriously misunderstanding how to read these expressions. $a_k\cdot\partial x = (a_k\cdot\partial)x$ by convention. Even worse though is that you don't seem to be employing any convention and instead are treating the inner and geometric products as if they are associative with each other, something like $$(a\cdot b)c = a\cdot(bc).$$ This is completely false. Feb 26 at 17:17
• @NicholasTodoroff, thank you. I saw my mistake. Feb 26 at 17:22

## 2 Answers

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.

• Treating $a_k\cdot \partial$ as a scalar I can obtain $\partial \cdot x = \sum a^k\cdot a_k$ directly by writing $\partial \cdot x = \left(\sum a^k(a_k\cdot \partial)\right)\cdot x = \sum(a_k\cdot \dot\partial)a^k\cdot \dot x = \sum a^k \cdot a_k$. However, I don’t obtain the intermediate expression either $\sum a^k\cdot(a_k\cdot x)$ (as you wrote) or $\sum a^k\cdot(a_k\cdot \partial x)$ (as it appears in the book I quoted). Feb 26 at 18:14
• @Rodrigo I made a typo, the last expression is the correct one except for the dot. As for how to obtain the expression, if you understand that $a_k\cdot\partial$ is scalar like, then that's it; there is nothing else to say: $$\partial\cdot x = \left(\sum_ka^k(a_k\cdot\partial)\right)\cdot x = \sum_k[a^k(a_k\cdot\partial)]\cdot x = \sum_ka^k\cdot[(a_k\cdot\partial)x].$$ Feb 26 at 18:59
• @Rodrigo Apologies for my comment about your use of the dot being wrong, I was confused for some reason. Your use of the dot is correct here. Feb 26 at 19:03

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 $$a_k \cdot \partial x=\lim_{\tau \to 0}\frac{x+\tau\,a_t}{\tau}=a_t.$$