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 $a + bi + cj$, 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:

$$ \begin{align*} z_1 &= a_1 + b_1i + c_1j \\ z_2 &= a_2 + b_2i + c_2j \\ z_1z_2 &= (a_1 + b_1i + c_1j)(a_2 + b_2i + c_2j) \\ &= (a_1a_2 - b_1b_2 - c_1c_2) + (a_1b_2 + b_1a_2)i + (a_1c_2 + c_1a_2)j \\ & \quad + b_1c_2ij + c_1b_2ji \end{align*} $$

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:

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

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 $q$:

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

Real Quaternion

A real quaternion has a zero vector term:

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

Pure Quaternion

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

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

Quaternion Conjugate

Given:

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

The quaternion conjugate is defined as:

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

Quaternion Norm

The norm of a quaternion $q = [s, \mathbf{v}]$ 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:

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

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):

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

And substituting them in \eqref{quaternion-product}:

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

By doing some groupings:

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

Now let’s compute the product $q_bq_a$:

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

Note that the scalar quantity of both products is the same; however, the vector quantity varies (the cross product sign is changed). Therefore:

$$ q_aq_b \neq q_bq_a $$

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 $q_k = [k, \mathbf{0}]$ 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)

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

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:

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

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

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

Where $\norm{q_c} = 1$.

Product of Pure Quaternions

Let:

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

The product $q_aq_b$ is defined as:

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

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)

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

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

$$ q^2 = -1 $$

Product of a Pure Quaternion with Its Conjugate

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

Inverse of a Quaternion

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

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

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 $\mathbf{v}$:

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

Combining this with the definition of a pure quaternion:

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

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

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

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

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

Misc Operations

Taking the Scalar Part of a Quaternion

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

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