$\let\oldcdot\cdot \renewcommand{\cdot}{\kern-.2em\oldcdot\kern-.2em} \let\oldwedge\wedge \renewcommand{\wedge}{\kern-.2em\oldwedge\kern-.2em} \renewcommand{\vec}[1]{\mathbf{#1}} \newcommand{\e}[1]{(-1)^{#1}} \newcommand{\eh}{(-1)^{r(r-1)/2}}$

I’ve been reading Hestenes’ brilliant New Foundations for Classical Mechanics, and I would love to be able to reach for Geometric Algebra as naturally as complex numbers, but sadly, the exercises in NFCM paint a different picture, so I thought I’d work through some of them here. And with the added yak shaving delights of mathjax in hugo, who could resist?

I use the following precedence rules so I can omit parentheses:

$(\vec{A}\wedge\vec{B})\vec{C} = \vec{A}\wedge\vec{B}\vec{C}, (\vec{A}\cdot\vec{B})\vec{C}=\vec{A}\cdot\vec{B}\vec{C}$ and $\vec{A}\cdot(\vec{B}\wedge\vec{C}) = \vec{A}\cdot\vec{B}\wedge\vec{C}$

Please let me know of any errors and sloppiness @rfistman.

1.1

$a)$

$(\vec{a} \wedge \vec{b}) \cdot (\vec{c} \wedge \vec{d}) = \vec{a} \cdot ( \vec{b} \cdot \vec{c} \wedge \vec{d}) $ by 1.18

$ = \vec{a} \cdot (\vec{b}\cdot\vec{c}\vec{d} - \vec{b}\cdot\vec{d}\vec{c})$ by 1.14

$ = \vec{a}\cdot\vec{d}\vec{b}\cdot\vec{c}-\vec{a}\cdot\vec{c}\vec{b}\cdot\vec{d}$

$b)$

$\vec{a}\cdot(\vec{b}\wedge\vec{c}\wedge\vec{d})$

$ = \vec{a}\cdot\vec{b}\vec{c}\wedge\vec{d} - \vec{b}\wedge(\vec{a}\cdot \vec{c}\wedge\vec{d})$ reducing the 3-blade to 2-blades with 1.15

$ = \vec{a}\cdot\vec{b}\vec{c}\wedge\vec{d} - \vec{b}\wedge(\vec{a}\cdot\vec{c}\vec{d}-\vec{a}\cdot\vec{d}\vec{c})$ by 1.14

$ = \vec{a}\cdot\vec{b}\vec{c}\wedge\vec{d}-\vec{a}\cdot\vec{c}\vec{b}\wedge\vec{d}+\vec{a}\cdot\vec{d}\vec{b}\wedge\vec{c}$

$c)$

$(\vec{u}\wedge\vec{v})\cdot(\vec{a}\wedge\vec{b}\wedge\vec{c})$

$ = \vec{u}\cdot(\vec{v}\cdot\vec{a}\wedge\vec{b}\wedge\vec{c})$ by 1.18

$= \vec{u}\cdot(\vec{v}\cdot\vec{a}\vec{b}\wedge\vec{c}-\vec{v}\cdot\vec{b}\vec{a}\wedge\vec{c}+\vec{v}\cdot\vec{c}\vec{a}\wedge\vec{b})$ by exercise 1.1b

$ = \vec{v}\cdot\vec{a} \vec{u}\cdot\vec{b}\wedge\vec{c}-\vec{v}\cdot\vec{b}\vec{u}\cdot\vec{a}\wedge\vec{c}+\vec{v}\cdot\vec{c}\vec{u}\cdot\vec{a}\wedge\vec{b}$

1.2 TODO

1.3

I’m ashamed to admit I peeked, as these kinds of GA equations still stop me in my tracks:

$\alpha\vec{x} + \vec{a}(\vec{x}\cdot\vec{b}) = \vec{c}$

dotting both sides with $\vec{b}$ gives

$\alpha\vec{x}\cdot\vec{b} + \vec{a}\cdot\vec{b}\vec{x}\cdot\vec{b} = \vec{x}\cdot\vec{b}(\alpha +\vec{a}\cdot\vec{b}) = \vec{c}\cdot\vec{b}$

which, when $\alpha + \vec{a}\cdot\vec{b} \neq 0$, lets you solve for $\vec{x}\cdot\vec{b}$:

$$\vec{x}\cdot\vec{b} = \frac{\vec{c}\cdot\vec{b}}{(\alpha+\vec{a}\cdot\vec{b})}$$

and eliminate the pesky inner product, so you can solve for $\vec{x}$:

$$\vec{x} = \frac{\vec{c}}{\alpha} - \frac{\vec{a}\vec{c}\cdot\vec{b}}{\alpha(\alpha+\vec{a}\cdot\vec{b})}$$

1.4

Solve for $\vec{x}$ in $\alpha\vec{x} + \vec{x}\cdot\vec{B} = \vec{a}$, where $\vec{B}$ is a 2-blade.

Taking the outer product of both sides with $\vec{B}$ gets us

$\alpha\vec{x}\wedge\vec{B} + (\vec{x}\cdot\vec{B})\wedge\vec{B} = \vec{a}\wedge\vec{B}$

But since $\vec{B}$ is a 2-blade

$(\vec{x}\cdot\vec{B})\wedge\vec{B}=(\vec{x}\cdot\vec{c}\wedge\vec{d})\wedge\vec{c}\wedge\vec{d}=(\vec{x}\cdot\vec{c}\vec{d}-\vec{x}\cdot\vec{d}\vec{c})\wedge(\vec{c}\wedge\vec{d}) = \vec{x}\cdot\vec{c}\vec{d}\wedge\vec{c}\wedge\vec{d} - \vec{x}\cdot\vec{d}\vec{c}\wedge\vec{c}\wedge\vec{d} = 0$

leaving us with

$$\vec{x}\wedge\vec{B} = \frac{\vec{a}\wedge\vec{B}}{\alpha}$$

This looks like it wants to be a geometric product, so add it to the original equation to get

$$\alpha\vec{x} + \vec{x}\cdot\vec{B} + \vec{x}\wedge\vec{B} = \alpha\vec{x} + \vec{x}\vec{B} = \vec{x}(\alpha+\vec{B}) = \vec{a} + \frac{\vec{a}\wedge\vec{B}}{\alpha}$$

providing that $\alpha+\vec{B} \neq 0$, we can right multiply by $(\alpha + \vec{B})^{-1}$ to get

$$\vec{x} = (\vec{a}+\frac{\vec{a}\wedge\vec{B}}{\alpha})(\alpha+\vec{B})^{-1}$$

1.5

Solve the simultaneous equations $\vec{c}\cdot\vec{a} \neq 0, \vec{a}\wedge\vec{x} = \vec{B}, \vec{c}\cdot\vec{x} = \alpha$ for $\vec{x}$.

$$ \begin{align} \vec{c}\cdot\vec{B} &= \vec{c}\cdot\vec{a}\wedge\vec{x} \
&= \vec{c}\cdot\vec{a}\vec{x}-\vec{c}\cdot\vec{x}\vec{a} \
&= \vec{c}\cdot\vec{a}\vec{x}-\alpha\vec{a} \
\text{or } \vec{x} &= \frac{1}{\vec{c}\cdot\vec{a}}(\alpha\vec{a} + \vec{c}\cdot\vec{B}) \end{align} $$

1.6

Exercise 1.1a gives $(\vec{a}\wedge\vec{b})\cdot(\vec{b}\wedge\vec{c}) = \vec{a}\cdot\vec{c}\vec{b}^2 - \vec{a}\cdot\vec{b}\vec{b}\cdot\vec{c}$ which rearranges to $\vec{b}^2 \vec{a}\cdot\vec{c} = \vec{a}\cdot\vec{b}\vec{b}\cdot\vec{c} + (\vec{a}\wedge\vec{b})\cdot(\vec{b}\wedge\vec{c})$

and

$ \vec{a}\wedge\vec{b}\vec{c}\wedge\vec{b} = -\vec{a}\wedge\vec{b}\vec{b}\wedge\vec{c} = -(\vec{a}\vec{b}-\vec{a}\cdot\vec{b})(\vec{b}\vec{c}-\vec{b}\cdot\vec{c}) $

$ = -\vec{b}^2\vec{a}\vec{c} + \vec{b}\cdot\vec{c}\vec{a}\vec{b} + \vec{a}\cdot\vec{b}\vec{b}\vec{c} - \vec{a}\cdot\vec{b}\vec{b}\cdot\vec{c} $

reading off the bivector components gives

$ \langle\vec{a}\wedge\vec{b}\vec{c}\wedge\vec{b}\rangle_2 = -\vec{b}^2\vec{a}\wedge\vec{c} + \vec{b}\cdot\vec{c}\vec{a}\wedge\vec{b} + \vec{b}\cdot\vec{a}\vec{b}\wedge\vec{c} $

$ = \vec{b}\cdot(\vec{a}\wedge\vec{b}\wedge\vec{c}) $ by exercise 1.1b

$ = (\vec{a}\wedge\vec{b}\wedge\vec{c})\cdot\vec{b} $ by 1.5

1.7

This question asks you to reduce two rank zero expressions to “inner products of vectors”. The results make you sorry you asked and also make you question what “coordinate free” really means and what you hoped to gain from it.

i)

Applying 1.15 twice gives

$(\vec{a}\wedge\vec{b}\wedge\vec{c})\cdot(\vec{u}\wedge\vec{v}\wedge\vec{w}) = (\vec{a}\wedge\vec{b})\cdot(\vec{c}\cdot\vec{u}\wedge\vec{v}\wedge\vec{w}) = \vec{a}\cdot(\vec{b}\cdot(\vec{c}\cdot\vec{u}\wedge\vec{v}\wedge\vec{w}))$

on to an expression of inner products of vectors and bivectors with exercise 1.1b

$=\vec{a}\cdot(\vec{b}\cdot(\vec{c}\cdot\vec{u}\vec{v}\wedge\vec{w} - \vec{c}\cdot\vec{v}\vec{u}\wedge\vec{w} + \vec{c}\cdot\vec{w}\vec{u}\wedge\vec{v}))$

$=\vec{a}\cdot(\vec{c}\cdot\vec{u}\vec{b}\cdot\vec{v}\wedge\vec{w} - \vec{c}\cdot\vec{v}\vec{b}\cdot\vec{u}\wedge\vec{w} + \vec{c}\cdot\vec{w}\vec{b}\cdot\vec{u}\wedge\vec{v})$

applying the GA vector triple product gets you an inner product of vectors, so I guess we’ve finished:

$=\vec{a}\cdot(\vec{c}\cdot\vec{u}(\vec{b}\cdot\vec{v}\vec{w} -\vec{b}\cdot\vec{w}\vec{v}) - \vec{c}\cdot\vec{v}(\vec{b}\cdot\vec{u}\vec{w}-\vec{b}\cdot\vec{w}\vec{u}) +\vec{c}\cdot\vec{w}(\vec{b}\cdot\vec{u}\vec{v}-\vec{b}\cdot\vec{v}\vec{u}))$

The expression can be rearranged further, but all the interesting or symmetric arrangements are just vector inner product versions of the more compact 2-blade expansions.

ii)

$\vec{a}\vec{b}\vec{u}\vec{v} = (\vec{a}\cdot\vec{b}+\vec{a}\wedge\vec{b})\vec{u}\vec{v} = (\vec{a}\cdot\vec{b}\vec{u}+\vec{a}\wedge\vec{b}\vec{u})\vec{v}$

$=(\vec{a}\cdot\vec{b}\vec{u} + \vec{a}\wedge\vec{b}\cdot\vec{u}+\vec{a}\wedge\vec{b}\wedge\vec{u})\vec{v} = (\vec{a}\cdot\vec{b}\vec{u} + \vec{a}\wedge\vec{b}\cdot\vec{u} + \vec{a}\wedge\vec{b}\wedge\vec{u})\vec{v}$

$= \vec{a}\cdot\vec{b}\vec{u}\vec{v} + (\vec{a}\wedge\vec{b}\cdot\vec{u})\vec{v} + \vec{a}\wedge\vec{b}\wedge\vec{u}\vec{v} = \vec{a}\cdot\vec{b}\vec{u}\cdot\vec{v} + \vec{a}\cdot\vec{b}\vec{u}\wedge\vec{v} + (\vec{a}\cdot\vec{b}\vec{a}-\vec{u}\cdot\vec{a}\vec{b})\vec{v} + \vec{a}\wedge\vec{b}\wedge\vec{u}\cdot\vec{v}+\vec{a}\wedge\vec{b}\wedge\vec{u}\wedge\vec{v}$

this gives a mixture of rank 0, 2 and 4 multivectors

$= (\vec{a}\cdot\vec{b}\vec{u}\cdot\vec{v} - \vec{u}\cdot\vec{a}\vec{b}\cdot\vec{v} + \vec{u}\cdot\vec{b}\vec{a}\cdot\vec{v}) + (\vec{a}\cdot\vec{b}\vec{u}\wedge\vec{v} -\vec{u}\cdot\vec{b}\vec{a}\wedge\vec{v} +\vec{u}\cdot\vec{b}\vec{a}\wedge\vec{v} + \vec{a}\wedge\vec{b}\wedge\vec{u}\cdot\vec{v}) + \vec{a}\wedge\vec{b}\wedge\vec{u}\wedge\vec{v}$

reading off the rank zero components gives

$\langle\vec{a}\vec{b}\vec{u}\vec{v}\rangle_0 = \vec{a}\cdot\vec{b}\vec{u}\cdot\vec{v}-\vec{a}\cdot\vec{u}\vec{b}\cdot\vec{v}+\vec{a}\cdot\vec{v}\vec{b}\cdot\vec{u}$

1.10 TODO Prove the general reduction formula

$$ \begin{align} \vec{a}\vec{C}_r &= \vec{a}\cdot\vec{C}_r + \vec{a}\wedge\vec{C}_r \
\vec{C}_r\vec{a} &= \vec{C}_r\cdot\vec{a} + \vec{C}_r\wedge\vec{a} \
&= -\e{r}\vec{a}\cdot\vec{C}_r + \e{r}\vec{a}\wedge\vec{C}_r \
&= \e{r}(\vec{a}\wedge\vec{C}_r-\vec{a}\cdot\vec{C}_r) \
\e{r}\vec{C}_r\vec{a} &= \vec{a}\wedge\vec{C}_r-\vec{a}\cdot\vec{C}_r \
\e{r}\vec{C}_r\vec{a} + 2\vec{a}\cdot\vec{C}_r &= \vec{a}\vec{C}_r \
\vec{a}\vec{b}\vec{C}_r &= \vec{a}(\e{r}\vec{C}_r\vec{b} + 2\vec{b}\cdot\vec{C}_r) \
\vec{a}\vec{b}\vec{C}_r &= (-\vec{b}\vec{a} + 2\vec{a}\cdot\vec{b})\vec{C}_r \
&= -\vec{b}\vec{a}\vec{C}_r + 2\vec{a}\cdot\vec{b}\vec{C}_r \
&= -\vec{b}(\e{r}\vec{C}_r\vec{a} + 2\vec{a}\cdot\vec{C}_r) + 2\vec{a}\cdot\vec{b}\vec{C}_r \
&= - \e{r}\vec{b}\vec{C}_r\vec{a} + 2\vec{a}\cdot\vec{b}\vec{C}_r - 2\vec{b}\vec{a}\cdot\vec{C}_r \
\frac{1}{2}(\vec{a}(\vec{b}\vec{C}_r) + \e{r}(\vec{b}\vec{C}_r)\vec{a}) &= \vec{a}\cdot\vec{b}\vec{C}_r - \vec{b}\vec{a}\cdot\vec{C}_r \
&= \frac{1}{2}(\vec{a}(\vec{b}\cdot\vec{C}_r + \vec{b}\wedge\vec{C}_r) + \e{r}(\vec{b}\cdot\vec{C}_r + \vec{b}\wedge\vec{C}_r)\vec{a}) \
&= \frac{1}{2}(\vec{a}(\vec{b}\cdot\vec{C}_r) + \e{r}(\vec{b}\cdot\vec{C}_r)\vec{a} + \vec{a}(\vec{b}\wedge\vec{C}_r) + \e{r}(\vec{b}\wedge\vec{C}_r)\vec{a}) \
&= \frac{1}{2}(\vec{a}(\vec{b}\cdot\vec{C}_r) - \e{r-1}(\vec{b}\cdot\vec{C}_r)\vec{a} + \vec{a}(\vec{b}\wedge\vec{C}_r) - \e{r+1}(\vec{b}\wedge\vec{C}_r)\vec{a}) \
&= \vec{a}\cdot(\vec{b}\cdot\vec{C}_r) + \vec{a}\cdot(\vec{b}\wedge\vec{C}_r) = \vec{a}\cdot(\vec{b}\vec{C}_r) \end{align} $$

1.11 Jacobi Identity

The hint says you can go via both 1.14 and 1.18

1.14 only:

$\vec{a}\cdot\vec{b}\wedge\vec{c} + \vec{b}\cdot\vec{c}\wedge\vec{a} + \vec{c}\cdot\vec{a}\wedge\vec{b}$

$= \vec{a}\cdot\vec{b}\vec{c}-\vec{a}\cdot\vec{c}\vec{b} + \vec{b}\cdot\vec{c}\vec{a}-\vec{b}\cdot\vec{a}\vec{c} + \vec{c}\cdot\vec{a}\vec{b} - \vec{c}\cdot\vec{b}\vec{a}$ by 1.14

$= (\vec{a}\cdot\vec{b}-\vec{b}\cdot\vec{a})\vec{c} + (\vec{c}\cdot\vec{a}-\vec{a}\cdot\vec{c})\vec{b} + (\vec{b}\cdot\vec{c}-\vec{c}\cdot\vec{b})\vec{a}$

$ = 0$

via 1.8 and 1.14 because I was left with some extra terms to mop up:

$\vec{a}\cdot\vec{b}\wedge\vec{c} + \vec{b}\cdot\vec{c}\wedge\vec{a} + \vec{c}\cdot\vec{a}\wedge\vec{b}$

$ = \vec{a}\cdot\vec{b}\vec{c} - \vec{b}\cdot\vec{c}\vec{a} + \vec{b}\cdot\vec{c}\wedge\vec{a}$ by 1.8

$ = \vec{a}\cdot\vec{b}\vec{c} - \vec{b}\cdot\vec{c}\vec{a} + \vec{c}\cdot\vec{a}\vec{b}-\vec{c}\cdot\vec{b}\vec{a}$

$ = 0 $

1.13

i) Show $\langle\vec{A}^\sim\rangle_r = \langle\vec{A}\rangle_r^\sim = \eh\langle\vec{A}\rangle_r$

Multivectors can be written as a sum of $r$-vector/$r$-grade components (this is Axiom 7.1 in Chapter 1): $\vec{A} = \sum\limits_r\langle\vec{A}\rangle^\sim_r$, and the reverse is additive, by definition (1.24b):

$\vec{A}^\sim = \sum\limits_r \langle\vec{A}\rangle^\sim_r$

Taking the r-grade of both sides, using additivity and assuming idempotency of the grade operator (seems reasonable?), demonstrates the first equality: $\langle\vec{A}^\sim\rangle_r = \langle\vec{A}\rangle^\sim_r$.

Even though $\langle\vec{A}\rangle_r$ is an $r$-vector, I am going to assume it is an $r$-blade. This shouldn’t be a problem, as the reverse is additive and each of the $r$-vector’s $r$-blades all have the same grade ($r$). zzz rvector blade components???

So write $r$-blade $\langle\vec{A}\rangle_r$ as an outer product of vectors $\vec{a}_1\wedge\dots\wedge\vec{a}_r$. zzz it is shown (where?) that to reverse an $r$-blade, you reverse this product. We can do this by swapping neighbouring pairs of vectors. Starting from either side, the first vector requires $r-1$ swaps, the next $r-2$ and so on. Each swap changes the sign of the product, and there are $\sum\limits_{n=1}^{r-1} n = r(r-1)/2$ swaps,

giving $\langle\vec{A}\rangle^\sim_r = \eh\langle\vec{A}\rangle_r$.

ii) By i)

$$\begin{align} \langle(\vec{A}\vec{B})^\sim\rangle_r &= \eh\langle\vec{A}\vec{B}\rangle_r \
&=\langle\vec{B}^\sim\vec{A}^\sim\rangle_r \end{align}$$

so $\langle\vec{A}\vec{B}\rangle_r = \eh\langle\vec{B}^\sim\vec{A}^\sim\rangle_r$

1.14

Show $\langle\vec{A}\vec{B}\rangle_0 = \langle\vec{B}\vec{A}\rangle_0$

The zero grade reverse is defined as being $\langle\vec{A}^\sim\rangle_0 \equiv \langle\vec{A}\rangle_0$ (1.24c),

so $\langle\vec{A}\vec{B}\rangle_0 = \langle(\vec{A}\vec{B})^\sim\rangle_0 = \langle\vec{B}\vec{A}\rangle_0$.

Here GA loses me a little as some of its definitions seem to do too much heavy lifting. 1.24c looks like something you should be able to prove.

Identities from the book

1.5 Inner Product Definition

$ \vec{a}\cdot\vec{A}_r \equiv \frac{1}{2}(\vec{a}\vec{A}_r-(-1)^r\vec{A}_r \vec{a}) = (-1)^{r+1}\vec{A}_r\cdot\vec{a} $

1.6 Outer Product Definition

$ \vec{a}\wedge\vec{A}_r \equiv \frac{1}{2}(\vec{a}\vec{A}_r+(-1)^r\vec{A}_r \vec{a}) = (-1)^r\vec{A}_r\wedge\vec{a} $

1.8

A nice identity derived from the associativity of multiplication and equating grades:

$ \vec{a}(\vec{b}\vec{c}) = (\vec{a}\vec{b})\vec{c} \implies \vec{a}\cdot\vec{b}\vec{c} + \vec{a}\cdot\vec{b}\wedge\vec{c} = \vec{a}\cdot\vec{b}\vec{c} + \vec{a}\wedge\vec{b}\cdot\vec{c}$ which I rearranged to

$ \vec{a}\cdot\vec{b}\wedge\vec{c} +\vec{c}\cdot\vec{a}\wedge\vec{b} = \vec{a}\cdot\vec{b}\vec{c}-\vec{b}\cdot\vec{c}\vec{a}$

1.14

A useful and memorable identity. It’s the Geometric Algebra version of the Vector Triple Product, but unlike the vector triple product it works in any number of dimensions:

$ \vec{a}\cdot\vec{b}\wedge\vec{c} = \vec{a}\cdot\vec{b}\vec{c}-\vec{b}\cdot\vec{c}\vec{a} $

1.15

A general reduction formula, the higher rank version of 1.14

$\vec{a}\cdot(\vec{b}\wedge\vec{C}_r) = \vec{a}\cdot\vec{b}\vec{C}_r - \vec{b}\wedge(\vec{a}\cdot\vec{C}_r)$

1.18

$(\vec{A}_r \wedge \vec{b})\cdot\vec{C}_s = \vec{A}_r \cdot (\vec{b}\cdot\vec{C}_s)$ for $0 < r < s$.

Next: chapter 3 of NFCM