
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.

## Formulas from the book

### General Inner Product Definition

$$\vec{A}_r\cdot\vec{B}_s \equiv \langle\vec{A}_r\vec{B}_s\rangle_{\lvert r-s\rvert} \qquad\text{(1.17a)}$$

$$e^{\vec{i}\theta}e^{\vec{i}\phi} = e^{\vec{i}(\theta+\phi)} \qquad\text{(4.15)}$$