Help with proof of Curl Double Product identity using Geometric Algebra. Most things seem to fall in place, but having a few issues.

Question

So I'm pretty new to GA/Clifford Algebras, but it's been fairly interesting so far. I figured I'd try to prove some basic vector calculus identities with it, just to help me get my bearings. I decided first to try the "curl double product" identity, generally written as $$\nabla\times\nabla\times F = \nabla (\nabla\cdot F) - \nabla^2F$$

I remember this being hell when trying to prove it with standard dot/cross product rules in Griffiths. I first though to rearrange a bit, since the gradient is redefined/re-contextualized in GA. $$\nabla (\nabla\cdot F) = \nabla\times\nabla\times F + \nabla^2F$$

Now, the gradient is treated as a geometric product in GA/GC, so equivalently we can write $$\nabla (\nabla\cdot F) = \nabla\cdot(\nabla\cdot F) + \nabla\wedge(\nabla\cdot F)$$ I'll leave the left term alone for now, as it is the cause of my troubles. $$\nabla\wedge(\nabla\cdot F) = (\nabla\wedge\nabla)\cdot F \text{ (I think)}$$ $$ = -(\nabla\times\nabla)^*\cdot F$$ Now, since $\nabla\times\nabla$ is a vector, its dual should be a bivector in $\mathbb{G}^3$. $F$ is a vector/vector field. Then, if you multiply the two terms, and move all of the terms of $F$ to the left side, it will take 2 swaps for every element of F. Thus, doing so will leave the sign unchanged $$ = -F\cdot(\nabla\times\nabla)^*$$ $$ = -(F\wedge(\nabla\times\nabla))^*$$ $$ = -F\times(\nabla\times\nabla)$$ $$=\nabla\times\nabla\times F$$ So we're left with $$\nabla(\nabla\cdot F) = \nabla\cdot(\nabla\cdot F) + \nabla\times\nabla\times F$$ And this is where I'm stumped. Everything seems to almost work out. But if I'm not mistaken, the divergence term doesn't collapse to the Laplacian? What am I missing here?

I have considered the identity $\textbf{A}\cdot \textbf{B} = \textbf{AB}$ if $\textbf{A}\subseteq\textbf{B}$. However, as far as I am aware, this only holds for blades no?

Answer 1

You write:

$$\boldsymbol{\nabla} (\boldsymbol{\nabla}\cdot \mathbf{F}) = \boldsymbol{\nabla}\cdot(\boldsymbol{\nabla}\cdot \mathbf{F}) + \boldsymbol{\nabla}\wedge(\boldsymbol{\nabla}\cdot \mathbf{F}),$$

but it appears that you are trying to use the vector identity

$$\mathbf{a} \mathbf{b} = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \wedge \mathbf{b},$$

(true for vectors $ \mathbf{a}, \mathbf{b} $), but not neccessarily for multivectors. In your case $ \boldsymbol{\nabla} \cdot \mathbf{F} $ is a scalar, not a vector, so there is not really any meaningful reduction that you can make of $ \boldsymbol{\nabla} (\boldsymbol{\nabla}\cdot \mathbf{F}) $.

The essense of the identity you are trying to find is given by the following manipulation

$$\begin{aligned}\mathbf{a} \times \left( { \mathbf{a} \times \mathbf{b} } \right)&=-\left( { I \mathbf{a} } \right) \wedge \left( { \mathbf{a} \times \mathbf{b} } \right) \\ &={\left\langle{{-\left( { I \mathbf{a} } \right) \wedge \left( { \mathbf{a} \times \mathbf{b} } \right) }}\right\rangle}_{1} \\ &=-{\left\langle{{I \mathbf{a} \left( { \mathbf{a} \times \mathbf{b} } \right) }}\right\rangle}_{1} \\ &=-{\left\langle{{I \mathbf{a} \left( { -I \mathbf{a} \wedge \mathbf{b} } \right) }}\right\rangle}_{1} \\ &=-{\left\langle{{\mathbf{a} \left( { \mathbf{a} \wedge \mathbf{b} } \right) }}\right\rangle}_{1} \\ &=-{\left\langle{{\mathbf{a} \cdot \left( { \mathbf{a} \wedge \mathbf{b} } \right) +\mathbf{a} \wedge \left( { \mathbf{a} \wedge \mathbf{b} } \right) }}\right\rangle}_{1} \\ &=-\mathbf{a} \cdot \left( { \mathbf{a} \wedge \mathbf{b} } \right) \\ &=-\mathbf{a}^2 \mathbf{b} + \mathbf{a} \left( { \mathbf{a} \cdot \mathbf{b} } \right).\end{aligned}$$

Notes:

• In this case, it's also true that $ \mathbf{a} B = \mathbf{a} \cdot B + \mathbf{a} \wedge B $ for bivector $ B $ and vector $\mathbf{a} $. Here the dot and wedge products have the following meanings:

$$\begin{aligned}\mathbf{a} \cdot B &= {\left\langle{{ \mathbf{a} B }}\right\rangle}_{1} \\ \mathbf{a} \wedge B &= {\left\langle{{ \mathbf{a} B }}\right\rangle}_{3}.\end{aligned}$$

• Also observe that we are free to insert a no-op grade-one selection around any vector without changing the meaning. That allows us to discard the trivector term of the vector-bivector product.

• We also used the distribution identity

$$\mathbf{x} \cdot \left( { \mathbf{y} \wedge \mathbf{z} } \right) = \left( { \mathbf{x} \cdot \mathbf{y} } \right) \mathbf{z} - \left( { \mathbf{x} \cdot \mathbf{z} } \right) \mathbf{y}.$$

If we summarize what we've done, essentially, we've first observed that

$$\mathbf{a} \times \left( { \mathbf{b} \times \mathbf{c} } \right)=-\mathbf{a} \cdot \left( { \mathbf{b} \wedge \mathbf{c} } \right),$$

for any vectors $ \mathbf{a}, \mathbf{b}, \mathbf{c} $. In particular

$$\mathbf{a} \times \left( { \mathbf{a} \times \mathbf{F} } \right)=-\mathbf{a} \cdot \left( { \mathbf{a} \wedge \mathbf{F} } \right) = -\left( { \mathbf{a} \cdot \mathbf{a} } \right) \mathbf{F} + \mathbf{a} \left( { \mathbf{a} \cdot \mathbf{F} } \right),$$

This also holds for the case where $ \mathbf{a} = \boldsymbol{\nabla} $, provided we take care to keep the gradients on the left (or use notational sugar to indicate the scope of their action.)