Quaternions

Definition

The existence of complex numbers 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 centuries, had been searching for the answer. Hamilton conjectured that a 3D complex number could be represented by the triple $a + b i + c j$ , 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{aligned} z_{1} & = a_{1} + b_{1} i + c_{1} j \\ z_{2} & = a_{2} + b_{2} i + c_{2} j \\ z_{1} z_{2} & = (a_{1} + b_{1} i + c_{1} j) (a_{2} + b_{2} i + c_{2} j) \\ = (a_{1} a_{2} - b_{1} b_{2} - c_{1} c_{2}) + (a_{1} b_{2} + b_{1} a_{2}) i + (a_{1} c_{2} + c_{1} a_{2}) j \\ + b_{1} c_{2} i j + c_{1} b_{2} j i \end{aligned}

The quantities $i j$ and $j i$ represented a problem for Hamilton. Even if $i j = - j i$ , we are still left with $(b_{1} c_{2} - c_{1} b_{2}) i j$ .

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 $i j$ .

Hamilton defined a quaternion $q$ as:

\begin{array}{r} q = s + a i + b j + c k, s, a, b, c \in R \\ i^{2} = j^{2} = k^{2} = i j k = - 1 \\ i j = k, j k = i, k i = j \\ j i = - k, k j = - i, i k = - j \end{array}

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

Notation

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

\begin{aligned} (1) & q & = s + x i + y j + z k \\ (2) & q & = s + v \\ (3) & q & = [s, v] \\ (4) & where s, x, y, z \in R, v \in R^{3} \\ and i^{2} = j^{2} = k^{2} = i j k = - 1 \end{aligned}

Real Quaternion

A real quaternion has a zero vector term:

q = [s, 0]

Pure Quaternion

A pure quaternion is a quaternion having a zero scalar term:

q = [0, v]

Quaternion Conjugate

Given:

q = [s, v]

The quaternion conjugate is defined as:

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

Quaternion Norm

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

\begin{aligned} | q | & = \sqrt{q q^{*}} \\ = \sqrt{s^{2} + x^{2} + y^{2} + z^{2}} \end{aligned}

Also note that:

| q |^{2} = q q^{*}

Norm facts:

$| q q^{*} | = | q | | q^{*} |$
$| q^{*} | = | q |$

Unit Quaternion

A unit quaternion is a quaternion of norm one given by:

\begin{aligned} (5) & q & = [s, λ \hat{n}] s, λ \in R, \hat{n} \in R^{3} \\ | \hat{n} | & = 1 \\ s^{2} + λ^{2} & = 1 \end{aligned}

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

Operations

Quaternion Product

Given two quaternions:

\begin{array}{r} q_{a} = [s_{a}, a], a = x_{a} i + y_{a} j + z_{a} k \\ q_{b} = [s_{b}, b], b = x_{b} i + y_{b} j + z_{b} k \end{array}

The product $q_{a} q_{b}$ is computed as follows:

\begin{aligned} q_{a} q_{b} & = (s_{a} + x_{a} i + y_{a} j + z_{a} k) (s_{b} + x_{b} i + y_{b} j + z_{b} k) \\ = (s_{a} s_{b} - x_{a} x_{b} - y_{a} y_{b} - z_{a} z_{b}) \\ + (s_{a} x_{b} + s_{b} x_{a} + y_{a} z_{b} - y_{b} z_{a}) i \\ + (s_{a} y_{b} + s_{b} y_{a} + z_{a} x_{b} - z_{b} x_{a}) j \\ (6) & + (s_{a} z_{b} + s_{b} z_{a} + x_{a} y_{b} - x_{b} y_{a}) k \end{aligned}

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

i = [0, i], j = [0, j], k = [0, k], 1 = [1, 0]

And substituting them in $(6)$ :

\begin{aligned} q_{a} q_{b} & = (s_{a} s_{b} - x_{a} x_{b} - y_{a} y_{b} - z_{a} z_{b}) [1, 0] \\ + (s_{a} x_{b} + s_{b} x_{a} + y_{a} z_{b} - y_{b} z_{a}) [0, i] \\ + (s_{a} y_{b} + s_{b} y_{a} + z_{a} x_{b} - z_{b} x_{a}) [0, j] \\ + (s_{a} z_{b} + s_{b} z_{a} + x_{a} y_{b} - x_{b} y_{a}) [0, k] \end{aligned}

By doing some groupings:

\begin{aligned} q_{a} q_{b} & = [s_{a} s_{b} - x_{a} x_{b} - y_{a} y_{b} - z_{a} z_{b}, \\ s_{a} (x_{b} i + y_{b} j + z_{b} k) + s_{b} (x_{a} i + y_{a} j + z_{a} k) \\ + (y_{a} z_{b} - y_{b} z_{a}) i + (z_{a} x_{b} - z_{b} x_{a}) j + (x_{a} y_{b} - x_{b} y_{a}) k] \\ = [s_{a} s_{b} - a \cdot b, s_{a} b + s_{b} a + a \times b] \end{aligned}

Now let’s compute the product $q_{b} q_{a}$ :

q_{b} q_{a} = [s_{b} s_{a} - b \cdot a, s_{b} a + s_{a} b + b \times 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_{a} q_{b} \neq q_{b} q_{a}

This is an important fact to note since for complex numbers 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 $q_{k} = [k, 0]$ and $q = [s, v]$ .

Their product is:

\begin{aligned} q_{k} q & = [k, 0] [s, v] \\ = [k s, k v] \end{aligned}

Note that this product is commutative.

Product of a Quaternion with Itself (Square of a Quaternion)

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

Product of a Quaternion and Its Conjugate

Let $q = [s, v]$ .

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

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

Product of Unit Quaternions

Given:

q_{a} = [s_{a}, a] q_{b} = [s_{b}, b]

Where $| q_{a} | = | q_{b} | = 1$ , the product is another unit-norm quaternion:

q_{c} = [s_{c}, c]

Where $| q_{c} | = 1$ .

Product of Pure Quaternions

Let:

q_{a} = [0, a] q_{b} = [0, b]

The product $q_{a} q_{b}$ is defined as:

\begin{aligned} q_{a} q_{b} & = [- a \cdot b, a \times b] \end{aligned}

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{aligned} q & = [0, v] \\ q^{2} & = [0, v] [0, v] \\ = [- v \cdot v, v \times v] \\ = [- (x^{2} + y^{2} + z^{2}), 0] \\ = - | v |^{2} \end{aligned}

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

q^{2} = - 1

Product of a Pure Quaternion with Its Conjugate

\begin{aligned} q^{*} q = q q^{*} & = [0, v] [0, - v] \\ = [v \cdot v, - v \times v] \\ = [v \cdot v, 0] \\ = | v |^{2} \end{aligned}

Inverse of a Quaternion

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

q q^{- 1} = [1, 0]

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

\begin{aligned} q^{*} q q^{- 1} & = q^{*} \\ | q |^{2} q^{- 1} & = q^{*} \\ q^{- 1} & = \frac{q^{*}}{| q |^{2}} \end{aligned}

Quaternion Units

Given the vector $v$ :

v = v \hat{v}, where v = | v |, and | \hat{v} | = 1

Combining this with the definition of a pure quaternion:

\begin{aligned} q & = [0, v] \\ = [0, v \hat{v}] \\ = v [0, \hat{v}] \end{aligned}

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

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

Let’s check if the quaternion unit $i$ squares to the ordered pair $[- 1, 0]$ :

\begin{aligned} i^{2} & = [0, i] [0, i] \\ = [0 \cdot 0 - i \cdot i, 0 \cdot i + 0 \cdot i - i \times i] \\ = [- | i |^{2}, 0] since i \times i = 0 \\ = [- 1, 0] \end{aligned}

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^{*}