## Quaternions as rotations

Let \(p\) be a 3d point represented as a quaternion using its homogeneous coordinates, \(p = [w, \mathbf{v}]\) and let \(q\) be any non-zero quaternion then

Theorem: The product \(qpq^{-1}\) takes \(p = [w, \mathbf{v}]\) to \(p' = [w, \mathbf{v'}]\)

Before proving this theorem let's make the following observation, we can express \(q\) as a multiplication of a scalar quaternion \(s\) and a unit quaternion \(\mathbf{U}q\), \(q = s\mathbf{U}q\), then \(qpq^{-1}=s\mathbf{U}qp(s\mathbf{U}q)^{-1}=s\mathbf{U}qp\mathbf{U}q^{-1}s^{-1}\), because the scalar multiplication is commutative \(\mathbf{U}qp\mathbf{U}q^{-1}ss^{-1}=\mathbf{U}qp\mathbf{U}q^{-1}\) so the product doesn't change irrespective of whether $q$ is a unit quaternion or not, finally notice that \(\mathbf{U}q^{-1} = \mathbf{U}q^{*}\) so we can write the action as \(qpq^*\) **note that from now on, \(q\) is assumed to be a unit quaternion without loss of generality**

Next, let's prove that the scalar part \(qpq^{*}\) is the same as the scalar of \(p\) (we can use the formula to find the scalar component of a quaternion)

\[ \begin{align*} 2S(qpq^*) &= qpq^* + (qpq^*)^* \\ &= qpq^* + qp^*q^* \\ &= q(p + p^*)q^* \\ &= q2S(p)q^* \\ &= 2qS(p)q^* \\ &= 2[s_q, \mathbf{v_q}][s_p, \mathbf{0}][s_q, -\mathbf{v_q}] \\ &= 2[s_ps_q, s_p\mathbf{v_q}][s_q, -\mathbf{v_q}] \\ &= 2[s_ps_q^2 - s_p (\mathbf{v_q} \cdot -\mathbf{v_q}), -s_ps_q\mathbf{v_q} + s_ps_q\mathbf{v_q} + s_p\mathbf{v_q \times v_q}] \\ &= 2[s_ps_q^2 + s_p\norm{v}^2, \mathbf{0}] \\ &= 2[s_ps_q^2 + s_p(1 - s_q^2), \mathbf{0}] \quad \text{because of the definition of a unit quaternion} \\ &= 2[s_p, \mathbf{0}] \\ &= 2S(p) \end{align*} \]

Therefore the scalar part of \(p\) remains constants in the operation i.e. if \(p = [w, \mathbf{v}]\) then \(p' = qpq^{*} = [w, \mathbf{v'}]\), and because multiplication preserves norms then \(\norm{p} = \norm{p'}\) and also \(\norm{v} = \norm{v'}\) \(\blacksquare\)

Theorem: if \(\norm{q} = 1\) then \(q = [\cos{\theta}, \unit{v} \sin{\theta}]\) acts to rotate around unit axis \(\unit{v}\) by \(2 \theta\)

Let

\[ v_0 = [0, \mathbf{v_0}] \quad \norm{v_0} = \norm{\mathbf{v_0}} = 1 \\ v_1 = [0, \mathbf{v_1}] \quad \norm{v_1} = \norm{\mathbf{v_1}} = 1 \]

Be two pure quaternions (which can be represented in 3d space), and an arbitrary quaternion \(q\) which has the form

\[ \begin{align} q &= v_1v_0^* \label{q} \\ &= [0, \mathbf{v_1}][0, -\mathbf{v_0}] \nonumber \\ &= [\mathbf{v_0 \cdot v_1}, \mathbf{v_0 \times v_1}] \label{q3d} \end{align} \]

Let \(\theta\) be the angle between \(\mathbf{v_0}\) and \(\mathbf{v_1}\) then \(\mathbf{v_0 \cdot v_1} = \cos{\theta}\), also let \(\mathbf{v_0 \times v_1} = \sin{\theta} \; \unit{v}\), then \eqref{q} becomes

\[ \begin{equation} \label{q2} q = [\cos{\theta}, \sin{\theta} \unit{v}] \end{equation} \]

Let's prove first that the product \(v_2 = qv_0q^{*}\) lies in the same plane as \(\mathbf{v_0}\) and \(\mathbf{v_1}\), we do so by proving first that the product \(v_2v_1^{*}\) has the same components (dot and cross products) as \(v_1v_0^{*}\)

\[ \begin{align*} v_2v_1^* &= (qv_0q^*) v_1^* \\ &= (q v_0 (v_1v_0^*)^*) v_1^* \\ &= (q v_0 v_0 v_1^*) v_1^* \\ &= q (v_0v_0)(v_1^*v_1^*) \\ &= q (-1)(-1) \quad \text{since they're unit quaternions they square to $-1$} \\ &= v_1v_0^* \end{align*} \]

Then if \(v_2v_1^* = v_1v_0^*\) that means that \(v_2=qv_0q^*\) lies in the same plane as \(v_0\) and \(v_1\), also \(v_2\) forms an angle of \(\theta\) with \(v_1\), furthermore \(\mathbf{v_1} \times \mathbf{v_2} = \unit{v} \sin{\theta}\), finally if the angle between \(v_0\) and \(v_1\) is \(\theta\) then the angle between \(v_0\) and \(v_2\) is \(2\theta\) which confirms what's seen on the image above

Furthermore the same can be said of \(q\) acting on \(v_1\), let \(v_3 = qv_1q^{*}\) then

\[ \begin{align*} v_3v_2^* &= (qv_1q^*)(qv_0q^*)^* \\ &= (q(qv_0)q^*)(qv_0q^*)^* \quad \text{by finding $v_1$ from \eqref{q}} \\ &= q (qv_0q^*)(qv_0q^*)^* \\ &= q \\ &= v_1v_0^* \end{align*} \]

Now any vector \(p\) can be represented in terms of the base \(v_0\), \(v_1\) and \(\unit{v}\) e.g. \(p = s_1\mathbf{v_0} + s_1\mathbf{v_1} + s_2\unit{v}\), we've seen what \(q\) does to \(v_0\) and \(v_1\) so let's see what it does to \(\unit{v}\)

Before computing \(q\unit{v}q^{*}\) see that

\[ \begin{align*} q\unit{v} &= [\cos{\theta}, \sin{\theta} \unit{v}][0, \unit{v}] \\ &= [\ldots, \ldots - \sin{\theta} (\unit{v} \times \unit{v})] \\ &= [\ldots, \ldots - \mathbf{0}] \end{align*} \]

So \(q\unit{v}\) is a commutative operation because the cross product is the only term that makes the quaternion operation non-commutable and in \(q\unit{v}\) that therm is zero therefore \(q\unit{v}q^{*} = \unit{v}qq^{*} = \unit{v}\) which means that \(q\) does not modify \(\unit{v}\)

Thus the action of \(q\) on any vector \(p\) is a rotation around \(\unit{v}\) by \(2\theta\) \(\blacksquare\)

## Quaternion rotation facts

Let \(q_1\) be a quaternion which rotates the pure quaternion \(p_1\) to \(p_2\) and also let \(q_2\) be a quaternion which rotates the vector \(p_2\) to \(p_3\) then \(p_3\) will have the form

\[ \begin{align*} p_3 &= q_2p_2q_2^* \\ &= q_2(q_1p_1q_1^*)q_2^* \\ &= (q_2q_1)p_1(q_1^*q_2^*) \\ &= (q_2q_1)p_1(q_2q_1)^* \end{align*} \]

Therefore the combination of rotation \(q_1\) followed by \(q_2\) is given by \(q = q_2q_1\)

When the rotations \(q_1, q_2, \ldots, q_n\) are applied to the pure quaternion \(p\) the result is equal to \(qpq^*\) where \(q = q_n q_{n-1} \ldots q_2 q_1\)