Show that $\partial x^2 = 2P(x)$ and $\partial \wedge x = 0$ (Geometric algebra)I’m trying to show equations (2-1.32) and (2-1.33) on page 51 of 20 of Hestenes and Sobczyk’s “Clifford Algebra to Geometric Calculus”.
$\partial |x|^2 = \partial x^2 = 2P(x) = 2x, \tag{1.32}$ $\partial \wedge x = 0, \tag{1.33}$
In trying to obtain $(1.32)$, I used the definition $(1.2)$
$a\cdot\partial F(x) \equiv \left.\frac{\partial F(x+\tau a)}{\partial \tau}\right\vert_{\tau = 0} = \lim_{\tau\rightarrow 0} \frac{F(x+\tau a)-F(x)}{\tau},\tag{1.2}$
along with equation $(1.5)$
$\partial_x = P(\partial_x) = \sum_k a^ka_k\cdot \partial_x \tag{1.5}$
and the product rule $(1.24a)$
$\partial(FG) = \dot \partial\dot F G + \dot \partial F\dot G, \tag{1.24a}$
where the overdot designates the quantities to be differentiated (with the remaining ones to be treated as constants), to rewrite $\partial x^2$ as follows:
$$\begin{aligned}\partial x^2 &= \partial (xx) = \sum_k a^ka_k\cdot\dot\partial \dot x x + \sum_k a^ka_k\cdot\dot\partial x \dot x \\ \end{aligned}$$
$$ = \sum_k a^ka_k x + \sum_k a^k\lim_{\tau\rightarrow 0}\frac{x(x+a_k\tau)-xx}{\tau} $$
$$ = P(x) + \sum_k a^kxa_k $$
But then, why is $\sum_k a^kxa_k = P(x)$?
Finally, how do I obtain $(1.33)$ from $(1.32)$ and the so-called integrability condition
$\partial_x\wedge \partial_x = 0 \tag{1.29}$
?
$\sum_ka^ka_kx$ is not the same thing as $\sum_ka^ka_k\cdot x$. Instead, $$ \partial x^2 = \sum_k(a^ka_kx + a^kxa_k) = \sum_ka^k(a_kx + xa_k) = 2\sum_ka^ka_k\cdot x = 2P(x) = 2x. $$ Better yet, if you know that $b\cdot\partial x = \partial b\cdot x = P(b)$ for constant $b$ then $$ \partial x^2 = \partial(x\cdot x) = \dot\partial(\dot x\cdot x) + \dot\partial(x\cdot\dot x) = 2\dot\partial x\cdot\dot x = 2P(x) = 2x. $$
For $\partial\wedge x$ just directly apply the formula for $\partial x^2$: $$ \partial\wedge x = \partial\wedge\left(\frac12\partial x^2\right) = \frac12(\partial\wedge\partial)x^2 = 0. $$ We can rebracket like this because $x^2$ is a scalar.