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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

$\partial_x\wedge \partial_x = 0 \tag{1.29}$

?

Rodrigo
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1 Answers1

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$\sum_ka^ka_kx$ 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.

Rodrigo
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Nicholas Todoroff
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  • Can I argue that $b\cdot \partial x = \partial b \cdot x$ simply because I already have that $b\cdot \partial x = \partial x\cdot b$ and that the inner product of two $1$-vectors commutes? – Rodrigo Feb 26 '24 at 14:12
  • @Rodrigo No, absolutely not. You are misunderstanding the bracketing. The usual convention in geometric algebra is for the geometric product to have the least precence, i.e. you always do it last. With parentheses my equation reads $$(b\cdot\partial)x = \partial(b\cdot x) = P(b).$$ The reason this is true is because, more generally, $$(b\cdot\partial)L(x) = L(P(b))$$ for any linear $L$, and $$\partial(b\cdot L(x)) = P(\bar L(b))$$ where $\bar L$ is the adjoint. The expressions $(b\cdot\partial)x = \partial(b\cdot x)$ are only equal because $x \mapsto x$ is the identity function. – Nicholas Todoroff Feb 26 '24 at 17:09
  • I don’t see how I misunderstood the bracketing here. I meant $(b\cdot \partial)x = \partial (x\cdot b) = \partial (b\cdot x)$ because $x\cdot b = b\cdot x$. – Rodrigo Feb 26 '24 at 17:27
  • @Rodrigo You said "I already have that $b\cdot\partial x = \partial x\cdot b$". How? It looks to me like you're saying $b\cdot(\partial x) = (\partial x)\cdot b$ which is true but irrelevant (in fact, both expressions are $0$). If you are really just saying that you already know $(b\cdot\partial)x = \partial(x\cdot b)$, then yes $x\cdot b = b\cdot x$ for any two vectors. – Nicholas Todoroff Feb 26 '24 at 17:39
  • Yes, I meant I already knew $\partial (x \cdot b) = \sum a^k a_k\cdot \partial (x\cdot b) = \sum a^k a_k\cdot b = P(b) = b\cdot P(\partial x) = b\cdot P(\partial) x = b\cdot \partial x$. Nevertheless, thank you for posting $\partial (b\cdot L(x)) = P(\overline L(b))$. – Rodrigo Feb 26 '24 at 17:54
  • How do you obtain $\partial (b\cdot L(x)) = P(\overline L (b))$? – Rodrigo Feb 26 '24 at 18:20
  • @Rodrigo Expand $\partial = \sum_ka^k(a_k\cdot\partial)$, use the defining equation of the adjoint $b\cdot L(x) = \bar L(b)\cdot x$, and use the more fundamental relation $(a_k\cdot\partial)x = a_k$. – Nicholas Todoroff Feb 26 '24 at 19:06
  • Isn’t the defining equation of the adjoint $\overline L(b) = \partial L \cdot b$? – Rodrigo Feb 26 '24 at 19:31
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    @Rodrigo Perhaps that is how Hestenes and Sobczyk do it, but $b\cdot L(a) = \bar L(b)\cdot a$ for all vectors $b, a$ is the more standard definition of the "adjoint of $L$ under the symmetric bilinear form $\cdot$". They note this relation in Eq. (2.19b). You can derive it from their definition by dotting by $a$: $a\cdot(\partial L(x)\cdot b) = a\cdot\partial L(x)\cdot b = L(a)\cdot b$. – Nicholas Todoroff Feb 26 '24 at 19:40
  • I see, so is the following a correct proof of $\partial(b\cdot L(x)) = P(\overline L(b))$? $\partial(b\cdot L(x)) = \sum a^k (a_k \cdot \partial)(\overline L(b)\cdot x) = \sum a^k (\overline L(b) \cdot a_k) = \sum a^k a_k \cdot \overline L(b)$, where the last equation follows because both $\overline L(b)$ and $a_k$ are vectors so that their inner product commutes? – Rodrigo Feb 26 '24 at 20:21
  • @Rodrigo Yes, that is correct. – Nicholas Todoroff Feb 26 '24 at 21:09
  • Let us continue this discussion in chat. – Nicholas Todoroff Feb 26 '24 at 21:19