Quaternion Algebra
Basic Definition
\[q = s + ai + bj + ck \quad s, a, b, c \in \mathbb{R}\] \[i^2 = j^2 = k^2 = ijk = -1\] \[\begin{aligned} ij &= k \quad jk = i \quad ki = j \\ ji &= -k \quad kj = -i \quad ik = -j \end{aligned}\]Note that $ij != ji$
Different Representation
For a quaternion, there are three representation
\[\begin{aligned} q &= s + xi + yj + zk \\ q &= s + \textbf{v} \\ q &= [s, \textbf{v}] \end{aligned}\]Where $s, x, y, z \in \mathbb{R}$, and $\textbf{v} \in \mathbb{R}^3$
Math Operations
Addition and Subtraction
\[\begin{aligned} q_a &= [s_a, \textbf{a}] \\ q_b &= [s_b, \textbf{b}] \\ q_a \pm q_b &= [s_a \pm s_b, \textbf{a} \pm \textbf{b}] \end{aligned}\]Quaternion Product
The basic quaternion multiplication is:
\[[s_a, \textbf{a}][s_b, \textbf{b}] = [s_as_b - a \cdot b, s_a \textbf{b} + s_b \textbf{a} + \textbf{a} \times \textbf{b}]\]The regular way to prove this is to expand two quaternions separately and perform the multiplication. But an easier way is to use an additive form of quaternion which I may mention later so I won’t prove it here.
Multiply a Quaternion by a Scalar
\[\begin{aligned} q &= [s, \textbf{v}] \\ \lambda q &= \lambda [s, \textbf{v}] \\ \lambda q &= [\lambda s, \lambda \textbf{v}] \end{aligned}\]Special Quaternions
Real Quaternion
A real quaternion has zero vector term.
\[q = [s, 0]\]Pure Quaternion
A pure quaternion has zero scalar term.
\[q = xi + yj + zk\]Pure Quaternion Product
\[\begin{aligned} q_aq_b &= [0, \textbf{a}][0, \textbf{b}] \\ &= [- \textbf{a} \cdot \textbf{b}, \textbf{a} \times \textbf{b}] \end{aligned}\]Proof:
\[\begin{aligned} q_aq_b &= [0, \textbf{a}][0, \textbf{b}] \\ &= (0 + a_xi + a_yj + a_zk)(0 + b_xi + b_yj + b_zk) \\ &= (0 - a_xb_x - a_yb_y - a_zb_z) + (a_yb_z - a_zb_y)i + \\ & \qquad (a_xb_z - a_zb_x)j + (a_xb_y - a_yb_x)k \end{aligned}\]Note that in the red part, we first use $a_yjb_zk = a_yb_zi$, and then use $a_zkb_yj = -a_zb_yi$. This gives $(a_yb_z - a_zb_y)i$. The same process also applies to other terms.
Recall that for two vector \(\textbf{a} = <a_x, a_y, a_z>, \textbf{b} = <b_x, b_y, b_z>\)
The cross product of these two vectors is:
\[\begin{aligned} \textbf{a} \times \textbf{b} &= \begin{vmatrix} i & j & k \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix} \\ &= \begin{vmatrix} a_y & a_z \\ b_y & b_z \end{vmatrix} i + \begin{vmatrix} a_x & a_z \\ b_x & b_z \end{vmatrix} j + \begin{vmatrix} a_x & a_y \\ b_x & b_y \end{vmatrix} k \\ &= (a_yb_z - a_zb_y)i + (a_xb_z - a_zb_x)j + (a_xb_y - a_yb_x)k \end{aligned}\]And the dot product of these two vectors is:
\[\textbf{a} \cdot \textbf{b} = a_xb_x + a_yb_y + a_zb_z\]Thus
\[\begin{aligned} q_aq_b &= 0 - \textbf{a} \cdot \textbf{b} + \textbf{a} \times \textbf{b} \\ &= [-\textbf{a} \cdot \textbf{b}, \textbf{a} \times \textbf{b}] \end{aligned}\]Unit Quaternion
A unit quaternion comprises a zero scalar and a unit vector.
\[q = v\hat{\textbf{v}}, \text{where } v = \vert \textbf{v} \vert \text{ and } \vert \hat{\textbf{v}} \vert = 1\]Additive Form of Quaternion
Two quaternions can be represented by two of their components added together.
\[\begin{aligned} q_a &= [s_a, 0] + [0, \textbf{a}] \\ q_b &= [s_b, 0] + [0, \textbf{b}] \\ q_aq_b &= ([s_a, 0] + [0, \textbf{a}])([s_b, 0] + [0, \textbf{b}])\\ &= [s_a, 0][s_b, 0] + [s_a,0][0, \textbf{b}] + [0, \textbf{a}][s_b, 0] + [0, \textbf{a}][0, \textbf{b}] \end{aligned}\]Here, note that $[s_a, 0][s_b, 0]$ is two scalar multiplied together, so $[s_a, 0][s_b, 0] = s_as_b$.
$[s_a,0][0, \textbf{b}]$ is multiplying a quaternion by a scalar, so $[s_a,0][0, \textbf{b}] = [0, s_a \textbf{b}]$. Same rule also applies to $[0, \textbf{a}][s_b, 0]$.
Finally, $[0, \textbf{a}][0, \textbf{b}] = [-\textbf{a} \cdot \textbf{b}, \textbf{a} \times \textbf{b}]$ can be known based on pure quaternion product we deducted above.
Thus,
\[\begin{aligned} q_a &= [s_a, 0] + [0, \textbf{a}] \\ q_b &= [s_b, 0] + [0, \textbf{b}] \\ q_aq_b &= ([s_a, 0] + [0, \textbf{a}])([s_b, 0] + [0, \textbf{b}])\\ &= [s_a, 0][s_b, 0] + [s_a,0][0, \textbf{b}] + [0, \textbf{a}][s_b, 0] + [0, \textbf{a}][0, \textbf{b}] \\ &= [s_as_b, 0] + [0, s_a\textbf{b}] + [0, s_b\textbf{a}] + [-\textbf{a} \cdot \textbf{b}, \textbf{a} \times \textbf{b}] \\ &= [s_as_b - \textbf{a} \cdot \textbf{b}, s_a\textbf{b} + s_b\textbf{a} + \textbf{a} \times \textbf{b}] \end{aligned}\]Binary Form of Quaternion
\[\begin{aligned} a &= [s, \textbf{v}] \\ &= [s, 0] + [0, \textbf{v}] \\ &= [s, 0] + v[0, \hat{\textbf{v}}] \\ &= s + v\hat{q} \end{aligned}\]Here, $\hat{q}$ is the unit quaternion $[0, \hat{\textbf{v}}]$. Unlike $i$ in complex number representation, $\hat{q}$ is not a constant.
Conjugate
The idea of conjugate comes from the binary form and conjugate of complex number. The conjugate of a quaternion $q = [s, \textbf{v}]$ is $q^\ast = [s, -\textbf{v}]$
Conjugate Properties
A quaternion multiply with its conjugate will eliminate its vector part:
\[\begin{aligned} qq^\ast &= [s, \textbf{v}][s, -\textbf{v}] \\ &= [s^2 - (-\textbf{v}) \cdot \textbf{v}, s\textbf{v} - s\textbf{v} + \textbf{v} \times (-\textbf{v})] \\ &= [s^2 + v^2, 0] \end{aligned}\]\[qq^\ast = q^\ast q\]
\[\begin{aligned} (q_aq_b)^\ast &= q_b^\ast q_a^\ast \\ q_a &= [s_a, \textbf{a}] \\ q_b &= [s_b, \textbf{b}] \\ q_aq_b &= [s_a, \textbf{a}][s_b, \textbf{b}] \\ &= [s_as_b - \textbf{a} \cdot \textbf{b}, s_a\textbf{b} + s_b\textbf{a} + \textbf{a} \times \textbf{b}] \\ (q_aq_b)^\ast &= [s_as_b - \textbf{a} \cdot \textbf{b}, -s_a\textbf{b} - s_b\textbf{a} - \textbf{a} \times \textbf{b}] \\ q_a^\ast &= [s_a, -\textbf{a}] \\ q_b^\ast &= [s_b, -\textbf{b}] \\ q_b^\ast q_a^\ast &= [s_bs_a - \textbf{b} \cdot \textbf{a}, -s_b\textbf{a} - s_a\textbf{b} + (-\textbf{b}) \times (-\textbf{a})] \\ &= [s_bs_a - \textbf{b} \cdot \textbf{a}, -s_b\textbf{a} - s_a\textbf{b} + \textbf{b} \times \textbf{a}] \\ &= [s_as_b - \textbf{a} \cdot \textbf{b}, -s_a\textbf{b} - s_b\textbf{a} - \textbf{a} \times \textbf{b}] \end{aligned}\]