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 $qp{q}^{-1}$ takes $p=[w,\mathbf{v}]$ to $p^{\prime }=[w,{\mathbf{v}}^{\prime }]$

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 $qp{q}^{-1}=s\mathbf{U}qp(s\mathbf{U}q{)}^{-1}=s\mathbf{U}qp\mathbf{U}{q}^{-1}{s}^{-1}$, because the scalar multiplication is commutative $\mathbf{U}qp\mathbf{U}{q}^{-1}s{s}^{-1}=\mathbf{U}qp\mathbf{U}{q}^{-1}$ so the product doesn’t change irrespective of whether $q$ is a unit quaternion or not, finally notice that $\mathbf{U}{q}^{-1}=\mathbf{U}{q}^{\ast }$ so we can write the action as $qp{q}^{\ast }$ 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 $qp{q}^{\ast }$ is the same as the scalar of $p$ (we can use the formula to find the scalar component of a quaternion)

Therefore the scalar part of $p$ remains constants in the operation i.e. if $p=[w,\mathbf{v}]$ then $p^{\prime }=qp{q}^{\ast }=[w,{\mathbf{v}}^{\prime }]$, and because multiplication preserves norms then $|p|=|p^{\prime }|$ and also $|v|=|v^{\prime }|$ $◼$

Theorem: if $|q|=1$ then $q=[\cos \theta ,\hat{\mathbf{v}}\sin \theta ]$ acts to rotate around unit axis $\hat{\mathbf{v}}$ by $2\theta$

Let

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

Let $\theta$ be the angle between ${\mathbf{v}}_{\mathbf{0}}$ and ${\mathbf{v}}_{\mathbf{1}}$ then ${\mathbf{v}}_{\mathbf{0}}\cdot {\mathbf{v}}_{\mathbf{1}}=\cos \theta$, also let ${\mathbf{v}}_{\mathbf{0}}\times {\mathbf{v}}_{\mathbf{1}}=\sin \theta \hat{\mathbf{v}}$, then $\text{(1)}$ becomes

Let’s prove first that the product $v_2=q{v}_0{q}^{\ast }$ lies in the same plane as ${\mathbf{v}}_{\mathbf{0}}$ and ${\mathbf{v}}_{\mathbf{1}}$, we do so by proving first that the product $v_2{v}_1^{\ast }$ has the same components (dot and cross products) as $v_1{v}_0^{\ast }$

Then if $v_2{v}_1^{\ast }=v_1{v}_0^{\ast }$ that means that $v_2=q{v}_0{q}^{\ast }$ lies in the same plane as $v_0$ and $v_1$, also $v_2$ forms an angle of $\theta$ with $v_1$, furthermore ${\mathbf{v}}_{\mathbf{1}}\times {\mathbf{v}}_{\mathbf{2}}=\hat{\mathbf{v}}\sin \theta$, 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 $v_3=q{v}_1{q}^{\ast }$ then

Now any vector $p$ can be represented in terms of the base $v_0$, $v_1$ and $\hat{\mathbf{v}}$ e.g. $p=s_1{\mathbf{v}}_{\mathbf{0}}+s_1{\mathbf{v}}_{\mathbf{1}}+s_2\hat{\mathbf{v}}$, we’ve seen what $q$ does to $v_0$ and $v_1$ so let’s see what it does to $\hat{\mathbf{v}}$

Before computing $q\hat{\mathbf{v}}{q}^{\ast }$ see that

So $q\hat{\mathbf{v}}$ is a commutative operation because the cross product is the only term that makes the quaternion operation non-commutable and in $q\hat{\mathbf{v}}$ that therm is zero therefore $q\hat{\mathbf{v}}{q}^{\ast }=\hat{\mathbf{v}}q{q}^{\ast }=\hat{\mathbf{v}}$ which means that $q$ does not modify $\hat{\mathbf{v}}$

Thus the action of $q$ on any vector $p$ is a rotation around $\hat{\mathbf{v}}$ by $2\theta$ $◼$

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

Therefore the combination of rotation $q_1$ followed by $q_2$ is given by $q=q_2{q}_1$

When the rotations $q_1,q_2,\dots ,q_n$ are applied to the pure quaternion $p$ the result is equal to $qp{q}^{\ast }$ where $q=q_n{q}_{n-1}\dots q_2{q}_1$