$\renewcommand{\vec}[1]{\mathbf{#1}} \let\oldtimes\times \renewcommand{\times}{\kern-.2em\oldtimes\kern-.2em} \let\oldcdot\cdot \renewcommand{\cdot}{\kern-.2em\oldcdot\kern-.2em} \let\oldwedge\wedge \renewcommand{\wedge}{\kern-.2em\oldwedge\kern-.2em} \newcommand{\et}{e^{\vec{i}\theta}} \newcommand{\net}{e^{-\vec{i}\theta}} \newcommand{\eti}[1]{e^{#1\vec{i}\theta}} \newcommand{\gt}{\mathcal{G}_3} \newcommand{\hf}{\frac{1}{2}} \newcommand{\rn}{\rangle^\hf_0} $
TODO: projection/rejection
This chapter introduces rotations in the $\vec{i}$-plane where $\vec{i} = \vec{a}\wedge\vec{b}$ is a unit 2-blade. Unit blades have an orthonormal factorization, so for our 2-blade
$$\vec{i}^2 = \vec{\sigma_1\sigma_2\sigma_1\sigma_2} = -\vec{\sigma_1\sigma_1\sigma_2\sigma_2} = -1$$
The cosine and the sine of the angle between two unit vectors $\vec{a}$ and $\vec{b}$ are (temporarily) defined as the scalar and bivector components of $\vec{ab}$
$$ \vec{a}\cdot\vec{b} = \cos(\theta) \qquad\text{(4.8a)}$$ $$ \vec{a}\wedge\vec{b} = \vec{i}\sin(\theta) \qquad\text{(4.8b)}$$
These equations are used as the definition (also temporary) of a tantalizingly useful looking “2-blade exponential”:
$$ e^{\vec{i}\theta} = \cos(\theta) + \vec{i}\sin(\theta) \qquad\text{(4.10)}$$
This acts upon vectors in the same way that that the complex exponential acts on complex numbers, however in this case the vectors can be n-dimensional and the $\vec{i}$-plane itself can be variable. Handy.
Because we have only just defined $\cos$ and $\sin$ here, for these exercises we have to re-derive their properties.
So if $\vec{ab} = e^{\vec{i}\theta}$ rotates $\vec{b}$ to $\vec{a}$ through an angle $\theta$, then $\vec{ba}$ rotates $\vec{a}$ back to $\vec{b}$, through an angle of $-\theta$, so we can deduce
$$ e^{\vec{i}\theta}e^{-\vec{i}\theta} = 1$$
We can also deduce
$$ \vec{ab} = \vec{a}\cdot\vec{b} + \vec{i}\vec{a}\wedge\vec{b} = \cos\theta + \vec{i}\sin\theta$$ $$ \vec{ba} = \vec{a}\cdot\vec{b} - \vec{i}\vec{a}\wedge\vec{b} = \cos(-\theta) + \vec{i}\sin(-\theta)$$
which gives the evenness of the cosine: $\cos(-\theta) = \cos\theta$ and the oddness of sine: $\sin(-\theta) = -\sin\theta$.
From this we get the useful result $$ 1 = \vec{abba} = (\cos\theta + \vec{i}\sin{\theta})(\cos\theta-\vec{i}\sin\theta) = \cos^2\theta + \sin^2\theta$$
or
$$\cos^2\theta + \sin^2\theta = 1$$
4.1
Show $$e^{\vec{i}\pi} = -1 \qquad\mbox{(4.14b)} $$ from $$e^{\vec{i}\pi/2} = \vec{i} \qquad \text{(4.14a)}$$ by using de Moivre’s theorem $$(e^{\vec{i}\theta})^n = e^{\vec{i}n\theta} \qquad\text{(4.16)}$$
Squaring $e^{\vec{i}\pi/2}$
$$(e^{\vec{i}\pi/2})^2 = \vec{i}^2 = -1$$ but by de Moivre’s theorem $$(e^{\vec{i}\pi/2})^2 = e^{\vec{i}\pi}$$
so $$e^{\vec{i}\pi} = -1$$
4.2
Remembering not to appeal directly to the properties of sines and cosines (and trying to avoid circular reasoning?), prove and interpret the identities $e^{\vec{i}(\theta + 2\pi n)} = e^{\vec{i}\theta}$, $e^{\vec{i}\theta}e^{-\vec{i}\theta} = 1$
i)
Rotating $\vec{b}$ to $\vec{a}$ through the angle $\theta$ defines the plane $\vec{i}$. If you first rotate $\vec{b}$ one full revolution in this plane, through an angle of $2\pi$, back to $\vec{b}$ then nothing has changed and you have
$$ e^{\vec{i}2\pi}\vec{b} = \vec{b} $$
If you repeat the process $n$ times and use the angle addition formula, then
$$ e^{\vec{i}2\pi} \dots e^{\vec{i}2\pi}\vec{b} = e^{\vec{i}2\pi n}\vec{b} = \vec{b} $$
or, canceling $\vec{b}$,
$$ e^{\vec{i}2\pi n} = 1 $$
and so
$$ e^{\vec{i}(\theta+2\pi n)} = e^{\vec{i}\theta}e^{\vec{i}2\pi n} = e^{\vec{i}\theta}$$
This says that rotational angles are only unique up to an integer multiple of $2\pi$.
ii)
If rotating $\vec{b}$ to $\vec{a}$ defines an angle $\theta$, then rotating $\vec{a}$ back to $\vec{b}$ defines the same angle, but in the opposite direction, $-\theta$, and takes you back to where you started, or symbolically.
$$ e^{-\vec{i}\theta}e^{\vec{i}\theta} = 1$$
Reversing the expression gives the result.
This says that rotating in a plane by $\theta$ radians forward, then back by the same amount, returns you to your original position.
4.3
Derive the $\sin$ and $\cos$ angle addition formulas from $e^{\vec{i}\theta}e^{\vec{i}\phi} = e^{\vec{i}(\theta+\phi)}$
$$\begin{align}
e^{\vec{i}(\theta+\phi)} &= \cos(\theta+\phi) + \vec{i}\sin(\theta+\phi)\
= e^{\vec{i}\theta}e^{\vec{i}\phi} &= (\cos\theta + \vec{i}\sin\theta)(\cos\phi + \vec{i}\sin\phi)\
&= \cos\theta\cos\phi-\sin\theta\sin\phi + \vec{i}(\cos\theta\sin\phi+\sin\theta\cos\phi)
\end{align}$$
Equating scalar and bivector parts gives
$$ \cos(\theta+\phi) = \cos\theta\cos\phi-\sin\theta\sin\phi $$
$$ \sin(\theta+\phi) = \cos\theta\sin\phi+\sin\theta\cos\phi $$
4.4
Prove $$ \cos\theta = \frac{e^{\vec{i}\theta}+e^{-\vec{i}\theta}}{2}$$ $$ \sin\theta = \frac{e^{\vec{i}\theta}-e^{-\vec{i}\theta}}{2\vec{i}}$$ $$ \vec{i}\tan\theta = \frac{e^{\vec{i}2\theta}-1}{e^{\vec{i}2\theta}+1}$$
Using the evenness and oddness of cosine and sine
$$e^{-\vec{i}\theta} = \cos\theta - \vec{i}\sin\theta = (e^{\vec{i}\theta})^\sim$$
and the first two results are recognised as the formulas for extracting scalar and bivector parts using reversion.
From these two results, and giving the ratio $\frac{\sin\theta}{\cos\theta}$ the name $\tan\theta$,
$$ \vec{i}\tan\theta = \frac{2\vec{i}\sin\theta}{2\cos\theta} $$ $$ = \frac{\et-\net}{\et+\net} = \frac{(\et-\net)\et}{(\et+\net)\et} $$ $$ = \frac{e^{\vec{i}2\theta}-1}{e^{\vec{i}2\theta}+1} $$
4.5
Prove that $\vec{ab} = e^{\vec{i}\theta}$ and $\vec{a}^2 = \vec{b}^2 = 1$ imply
a)
$(\vec{ab})^\sim = \vec{ba} = e^{-\vec{i}\theta}$
By exercise 4.2 $e^{\vec{i}\theta}e^{-\vec{i}\theta} = 1$, or $\vec{ab}e^{-\vec{i}\theta} = 1$.
Multiplying on both sides by $\vec{ba}$ gives $\vec{ba} = e^{-\vec{i}\theta}$
b) and c)
These two proofs can be obtained as a “two for one”.
Here we use the addition formula to write $\vec{ab}$ as a product of two half angle rotations. $$\vec{ab} = e^{\vec{i}\theta} = e^{\frac{\vec{i}\theta}{2}} e^{\frac{\vec{i}\theta}{2}} = (\cos\frac{1}{2}\theta + \vec{i}\sin\frac{1}{2}\theta)(\cos\frac{1}{2}\theta + \vec{i}\sin\frac{1}{2}\theta)$$ $$ = \cos^2\frac{1}{2}\theta - \sin^2\frac{1}{2}\theta + 2\vec{i}\cos\frac{1}{2}\theta\sin\frac{1}{2}\theta$$
Reversing $\vec{ab}$ negates the bivector component, so
$$\vec{ab} + \vec{ba} = 2\cos^2\frac{1}{2}\theta - 2\sin^2\frac{1}{2}\theta$$
and we can line up our “two for one”:
$$(\vec{a}+\vec{b})^2 + \vec{i}(\vec{a}-\vec{b})^2 = \vec{a}^2 + \vec{b}^2 + \vec{ab}+\vec{ba} + \vec{i}(\vec{a}^2+\vec{b}^2 - \vec{ab}-\vec{ba}) $$
$$ = 2 + 2\cos^2\frac{1}{2}\theta - 2\sin^2\frac{1}{2}\theta + \vec{i}(2 - 2\cos^2\frac{1}{2}\theta + 2\sin^2\frac{1}{2}\theta) $$
From exercise 4.4, it follows that $\cos^2\theta + \sin^2\theta = 1$, which allows us to eliminate a $\sin$ and $\cos$:
$$ = 4\cos^2\frac{1}{2}\theta + \vec{i}(4\sin^2\frac{1}{2}\theta) $$
Equating the bivector parts gives
b): $(\vec{a}-\vec{b})^2 = 4\sin^2\frac{1}{2}\theta$,
and scalar parts,
c): $(\vec{a}+\vec{b})^2 = 4\cos^2\frac{1}{2}\theta$
4.7
Solve the equation $1 + \et + \eti{2} + \eti{3} + \eti{4} + \eti{5} = 0$ by interpreting the terms as operators on a vector and identifying the geometrical figure generated.
$$1\vec{v} + \et\vec{v} + \eti{2}\vec{v} + \eti{3}\vec{v} + \eti{4}\vec{v} + \eti{5}\vec{v} = 0$$
This can be read as series of vectors placed head to tail, each successive vector rotated an extra $\theta$ radians from the previous, with the last vector pointing at the tail of the first. There are 6 vectors of equal length, and 6 equal angles, so this is a hexagon.
Alternatively, you can factor this into
$$ (1 + \et(1+\et(1+\et(1+\et(1+\et(1+\et))))))\vec{v} = \vec{T}^5\vec{v} = 0 $$
where $\vec{T} = 1+\et$ is “copy and then turn by $\theta$” operator, which generates a 5+1-gon.
OR, notice that, multiplying $1 + \et + \eti{2} + \eti{3} + \eti{4} + \eti{5}$ by $\et$ gives
$$ 1 + \et + \eti{2} + \eti{3} + \eti{4} + \eti{5} = \et + \eti{2} + \eti{3} + \eti{4} + \eti{5} + \eti{6}$$
or $\eti{6} = 1$, which has solutions $\theta = \frac{2\pi n }{6}$. $n = 0$ can’t be a solution, because $6 \ne 0$. $n = 3$ is a solution, but it’s degenerate, being a flat line. $n = 1, 2, 4, 5$ gives hexagons of clockwise and counterclockwise orientations.
4.8
Prove the following identities, and identify the trigonometric identities to which they reduce when $\vec{a}\wedge\vec{b}\wedge\vec{c} = 0$.
This condition is the same as saying that we’re working in $\gt$ (3-space), where the familiar (if flawed!) vector cross product is dual to the outer product $\vec{a}\wedge\vec{b} = i\vec{a}\times\vec{b}$ .
a)
$$ (\vec{a}\cdot\vec{b})^2-(\vec{a}\wedge\vec{b})^2 = \vec{a}^2\vec{b}^2 $$
Since $\vec{b}^2$ is a scalar, it can commute inside the product $\vec{a}^2=\vec{aa}$: $$ \vec{a}^2\vec{b}^2 = \vec{aa}\vec{b}^2 = \vec{a}\vec{b}^2\vec{a} = \vec{abba}$$ $$ = (\vec{a}\cdot\vec{b} + \vec{a}\wedge\vec{b})(\vec{b}\cdot\vec{a}+\vec{b}\wedge\vec{a})$$ $$ = (\vec{a}\cdot\vec{b} + \vec{a}\wedge\vec{b})(\vec{a}\cdot\vec{b}-\vec{a}\wedge\vec{b})$$ $$ = (\vec{a}\cdot\vec{b})^2 - (\vec{a}\wedge\vec{b})^2$$
TODO: 3D TODO: could this be the Pythagorean theorem?
b)
$$ \vec{b}\cdot(\vec{a}\wedge\vec{b}\wedge\vec{c}) = \vec{b}\cdot\vec{a}\vec{b}\wedge\vec{c} - \vec{b}^2\vec{a}\wedge\vec{c} + \vec{b}\cdot\vec{c}\vec{a}\wedge\vec{b} $$
This result is proved in chapter 2. When $\vec{a}\wedge\vec{b}\wedge\vec{c} = 0$, it reduces to
$$ \vec{b}\cdot\vec{a}\vec{b}\wedge\vec{c} + \vec{b}^2\vec{c}\wedge\vec{a} + \vec{b}\cdot\vec{c}\vec{a}\wedge\vec{b} = 0 $$
or, switching to cross products and cancelling the pseudoscalar factor $i$
$$ \vec{b}\cdot\vec{a}\vec{b}\times\vec{c} + \vec{b}^2\vec{c}\times\vec{a} + \vec{b}\cdot\vec{c}\vec{a}\times\vec{b} = 0 $$
TODO: what?
c)
$$ (\vec{a}\wedge\vec{b})\cdot(\vec{b}\wedge\vec{c}) = \vec{b}^2\vec{a}\cdot\vec{c}-\vec{a}\cdot\vec{b}\vec{b}\cdot\vec{c}$$
This result is proved in chapter 2.
In $\gt$, the outer product of vectors can be written as cross products, and using the general inner product definition:
$$ (i\vec{a}\times\vec{b})\cdot(i\vec{b}\times\vec{c}) = \langle(i\vec{a}\times\vec{b})(i\vec{b}\times\vec{c})\rangle_0 $$
in odd dimensions, like $\gt$, $i$ commutes through everything, so
$$ = \langle i^2(\vec{a}\times\vec{b})(\vec{b}\times\vec{c})\rangle_0 $$ $$ = -\langle(\vec{a}\times\vec{b})(\vec{b}\times\vec{c})\rangle_0 $$ $$ = -(\vec{a}\times\vec{b})\cdot(\vec{b}\times\vec{c}) $$
or
$$(\vec{a}\times\vec{b})\cdot(\vec{b}\times\vec{c}) = \vec{a}\cdot\vec{b}\vec{b}\cdot\vec{c} - \vec{b}^2\vec{a}\cdot\vec{c}$$
This appears to be a special case of the Scalar Quadruple Product. Is that a trigonometric identity?
I was hoping to do this without using the general inner product, by showing $(\vec{a}\wedge\vec{b})\cdot(\vec{b}\wedge\vec{c}) = (\vec{a}\wedge\vec{b})(\vec{b}\wedge\vec{c})$ in $\gt$, but all attempts resulted in a mess.
d)
e)
Formulas from the book
General Inner Product Definition
$$\vec{A}_r\cdot\vec{B}_s \equiv \langle\vec{A}_r\vec{B}_s\rangle_{\lvert r-s\rvert} \qquad\text{(1.17a)}$$
The Angle Addition Formula
$$e^{\vec{i}\theta}e^{\vec{i}\phi} = e^{\vec{i}(\theta+\phi)} \qquad\text{(4.15)}$$