## Definition

The existence of complex number presented a question for mathematicians, if a complex number exists in a 2D complex plane, could there be a 3D equivalent?

Sir William Rowan Hamilton among many other mathematicians of the 18th and 19th century had been searching for the answer, Hamilton conjectured that a 3D complex number could be represented by the triple $$a + bi + cj$$ where $$i$$ and $$j$$ are imaginary quantities and square to $$-1$$, when he was developing the algebra for this triplet the product of them raised a problem when expanded

\begin{align*} z_1 &= a_1 + b_1i + c_1j \\ z_2 &= a_2 + b_2i + c_2j \\ z_1z_2 &= (a_1 + b_1i + c_1j)(a_2 + b_2i + c_2j) \\ &= (a_1b_1 - b_1b_2 - c_1c_2) + (a_1b_2 + b_1a_2)i + (a_1c_2 + c_1a_2)j \\ & \quad + b_1c_2ij + c_1b_2ji \end{align*}

The quantities $$ij$$ and $$ji$$ represented a problem for Hamilton, even if $$ij = -ji$$ we are still left with $$(b_1c_2 - c_1b_2)ij$$

On October 16th, 1843, while he was walking with his wife along the Royal Canal in Ireland he saw the solution as a quadruple instead of a triple, instead of using two imaginary terms, three imaginary terms provided the necessary quantities to resolve products like $$ij$$

Hamilton defined a quaternion $$q$$ as

$q = s + ai + bj + ck \quad s,a,b,c \in \mathbb{R} \\ i^2 = j^2 = k^2 = ijk = -1 \\ ij = k \quad jk = i \quad ki = j \\ ji = -k \quad kj = -1 \quad ik = -j$

If a complex number $$i$$ is capable of rotating points on the plane by $$\deg{90}$$ then perhaps a triple rotates points in space by $$\deg{90}$$, in the end the triplet was replaced by a quaternion

## Notation

There are three ways of annotating a quaternion $$q$$

\begin{align} q &= s + xi + yj + zk \\ q &= s + \mbold{v} \\ q &= [s, \mbold{v}] \\ & \text{where s,x,y,z \in \mathbb{R}, \mbold{v} \in \mathbb{R}^3} \nonumber \\ & \text{and i^2 = j^2 = k^2 = ijk = -1} \nonumber \end{align}

### Real quaternion

A real quaternion has a zero vector term

$q = [s, \mbold{0}]$

### Pure quaternion

A pure quaternion is a quaternion having a zero scalar term

$q = [0, \mbold{v}]$

### Quaternion conjugate

Given

$q = [s, \mbold{v}]$

The quaternion conjugate is defined as

$q^* = [s, - \mbold{v}]$

### Quaternion norm

The norm of a quaternion $$q = [s, \mbold{v}]$$ is defined as the square root of the product of itself and its conjugate (the multiplication operation is defined later)

\begin{align*} \norm{q} &= \sqrt{qq^*} \\ &= \sqrt{s^2 + x^2 + y^2 + z^2} \end{align*}

Also note that

$\norm{q}^2 = qq^*$

Norm facts

• $$\norm{qq^{*}} = \norm{q}\norm{q^{*}}$$
• $$\norm{q^{*}} = \norm{q}$$

### Unit quaternion

A unit quaternion is a quaternion of norm one given by

\begin{align} q &= [s, \lambda \unit{n}] \quad s,\lambda \in \mathbb{R}, \unit{n} \in \mathbb{R}^3 \label{unit-norm-quaternion}\\ \left | \unit{n} \right | &= 1 \nonumber \\ s^2 + \lambda^2 &= 1 \nonumber \end{align}

Note: dividing a non-zero quaternion by its norm produces a unit norm quaternion

## Operations

### Quaternion Product

Given two quaternions

\begin{align*} q_a = [s_a, \mbold{a}] \quad \quad \mbold{a} = x_a i + y_a j + z_a k \\ q_b = [s_b, \mbold{b}] \quad \quad \mbold{b} = x_b i + y_b j + z_b k \end{align*}

The product $$q_aq_b$$ is computed as follows

\begin{align} q_aq_b &= (s_a + x_a i + y_a j + z_a k)(s_b + x_b i + y_b j + z_b k) \nonumber \\ &= (s_as_b - x_ax_b - y_ay_b - z_az_b) \nonumber \\ & \quad + (s_ax_b + s_bx_a + y_az_b - y_bz_a)i \nonumber \\ & \quad + (s_ay_b + s_by_a + z_ax_b - z_bx_a)j \nonumber \\ & \quad + (s_az_b + s_bz_a + x_ay_b - x_by_a)k \label{quaternion-product} \end{align}

Replacing the imaginaries by the ordered pairs (which are themselves quaternion units)

$i = [0, \mbold{i}] \quad j = [0, \mbold{j}] \quad k = [0, \mbold{k}] \quad 1 = [1, \mbold{0}]$

And substituting them in \eqref{quaternion-product}

\begin{align*} q_aq_b &= (s_as_b - x_ax_b - y_ay_b - z_az_b)[1, \mbold{0}] \\ & \quad + (s_ax_b + s_bx_a + y_az_b - y_bz_a)[0, \mbold{i}] \\ & \quad + (s_ay_b + s_by_a + z_ax_b - z_bx_a)[0, \mbold{j}] \\ & \quad + (s_az_b + s_bz_a + x_ay_b - x_by_a)[0, \mbold{k}] \end{align*}

By doing some aggrupations

\begin{align*} q_aq_b &= [s_as_b - x_ax_b - y_ay_b - z_az_b, \\ & \quad s_a(x_b \mbold{i} + y_b \mbold{j} + z_b \mbold{k}) + s_b(x_a \mbold{i} + y_a \mbold{j} + z_a \mbold{k}) \\ & \quad + (y_az_b - y_bz_a) \mbold{i} + (z_ax_b - z_bx_a) \mbold{j} + (x_ay_b - x_by_a) \mbold{k}] \\ &= [s_as_b - \mbold{a} \cdot \mbold{b}, s_a\mbold{b} + s_b\mbold{a} + \mbold{a} \times \mbold{b}] \end{align*}

Now let's compute the product $$q_bq_a$$

$q_bq_a = [s_bs_a - \mbold{b} \cdot \mbold{a}, s_b\mbold{a} + s_a\mbold{b} + \mbold{b} \times \mbold{a}]$

Note that the scalar quantity of both products is the same however the vector quantity varies (the cross product sign is changed) therefore

$q_aq_b \not = q_bq_a$

This is an important fact to note since for complex number the product commutes however for quaternions it doesn't

#### Product of a scalar and a quaternion

Let $$k$$ be a scalar represented as a quaternion as $$q_k = [k, \mathbf{0}]$$ and $$q = [s, \mathbf{v}]$$

Their product is

\begin{align*} q_kq &= [k, \mathbf{0}][s, \mathbf{v}] \\ &= [ks, k\mathbf{v}] \end{align*}

Note that this product is commutative

#### Product of a quaternion with itself (square of a quaternion)

\begin{align*} q &= [s, \mbold{v}] \\ q^2 &= [s, \mbold{v}] [s, \mbold{v}] \\ &= [s^2 - \mbold{v} \cdot \mbold{v}, 2s\mbold{v} + \mbold{v} \times \mbold{v}] \\ &= [s^2 - \norm{v}^2, 2s\mbold{v}] \\ &= [s^2 - (x^2 + y^2 + z^2), 2s(x\mbold{i} + y\mbold{j} + z\mbold{k})] \end{align*}

#### Product of a quaternion and its conjugate

Let $$q = [s, \mathbf{v}]$$

\begin{align*} qq^* &= [s, \mathbf{v}][s, -\mathbf{v}] \\ &= [s^2 + \mathbf{v} \cdot \mathbf{v}, -s \mathbf{v} + s\mathbf{v} - \mathbf{v} \times \mathbf{v}] \\ &= [s^2 + \mathbf{v} \cdot \mathbf{v}, \mathbf{0}] \\ &= s^2 + x^2 + y^2 + z^2 \end{align*}

Note that this product commutes i.e. $$qq^{*} = q^{*}q$$

#### Product of unit quaternions

Given

$q_a = [s_a, \mbold{a}] \\ q_b = [s_b, \mbold{b}]$

Where $$\norm{q_a} = \norm{q_b} = 1$$, the product is another unit-norm quaternion

$q_c = [s_c, \mbold{c}]$

Where $$\norm{q_c} = 1$$

#### Product of pure quaternions

Let

$q_a = [0, \mbold{a}] \\ q_b = [0, \mbold{b}]$

The product $$q_aq_b$$ is defined as

\begin{align*} q_aq_b &= [-\mbold{a} \cdot \mbold{b}, \mbold{a} \times \mbold{b}] \end{align*}

Note that the resulting quaternion is no longer a pure quaternion as some information has propagated into the real part via the dot product

#### Product of a pure quaternion with itself (square of a pure quaternion)

\begin{align*} q &= [0, \mbold{v}] \\ q^2 &= [0, \mbold{v}] [0, \mbold{v}] \\ &= [-\mbold{v} \cdot \mbold{v}, \mbold{v} \times \mbold{v}] \\ &= [-(x^2 + y^2 + z^2), \mbold{0}] \\ &= -\norm{v}^2 \end{align*}

If $$q$$ is a unit norm pure quaternion then

$q^2 = -1$

#### Product of a pure quaternion with its conjugate

\begin{align*} q^*q = qq^* &= [0, \mathbf{v}][0, -\mathbf{v}] \\ &= [\mathbf{v} \cdot \mathbf{v}, -\mbold{v \times v}] \\ &= [\mathbf{v} \cdot \mathbf{v}, \mbold{0}] \\ &= \norm{v}^2 \end{align*}

### Inverse of a quaternion

By definition, the inverse $$q^{-1}$$ of $$q$$ is

$qq^{-1} = [1, \mbold{0}]$

To isolate $$q^{-1}$$ let's pre multiply both sides by $$q^*$$

\begin{align*} q^*qq^{-1} &= q^* \\ \norm{q}^2q^{-1} &= q^* \\ q^{-1} &= \frac{q^*}{\norm{q}^2} \end{align*}

### Quaternion units

Given the vector $$\mbold{v}$$

$\mbold{v} = v \hat{\mbold{v}} \quad \text{where v = |\mbold{v}|, and |\hat{\mbold{v}}| = 1}$

Combining this with the definition of a pure quaternion

\begin{align*} q &= [0, \mbold{v}] \\ &= [0, v \hat{\mbold{v}}] \\ &= v[0, \hat{\mbold{v}}] \end{align*}

It's convenient to identify the unit quaternion as $$\hat{q}$$ (where $$v = 1$$)

$\hat{q} = [0, \hat{\mbold{v}}]$

Let's check if the quaternion unit $$\mbold{i}$$ squares to the ordered pair $$[-1, \mbold{0}]$$

\begin{align*} i^2 &= [0, \mbold{i}][0, \mbold{i}] \\ &= [0 \cdot 0 - \mbold{i} \cdot \mbold{i}, 0 \cdot \mbold{i} + 0 \cdot \mbold{i} - \mbold{i} \times \mbold{i}] \\ &= [-|\mbold{i}|^2, \mbold{0}] \quad \text{\mbold{i} \times \mbold{i} = 0} \\ & = [-1, \mbold{0}] \end{align*}

### Misc operations

#### Taking the scalar part of a quaternion

To isolate the scalar part of $$q$$ we could add $$q^*$$ to it

$2 S(q) = q + q^*$