# Quaternion rotation intuition

Say the quaternions real and imaginary part are written as $(q_1, \vec q)$. One useful multiplication property is $qr=(q_1r_1 - \langle\vec q, \vec r\rangle, q_1\vec r + r_1\vec q + \vec q \times \vec r)$. I am interested in why for unit quaternion $q=\left(cos\left(\frac{\phi}{2}\right), \vec v \sin\left(\frac{\phi}{2}\right)\right), |v|=1$ the formula $x'=qxq^{-1}=qxq^*$ is a rotation of the point $x$ (represented as imaginary quaternion) by an angle $\phi$ around rotation axis $\vec v$

I know a proof similar to the wikipedia proof, it justs plugs in $q$ and $q^*$, uses the multiplication property, then simplifies and the key step is to recognize that what comes out is a rotation formula .

I am looking for something more intuitive. Like for complex numbers it is easy to see that multiplying $z=re^{i\phi}$ and $e^{i\theta}$ gives you a rotation with $re^{i(\phi+\theta)}$

In this post one has for imaginary quaternion $\mathbf v$ that $e^\mathbf{v}= \cos|v|+ \mathbf{v}\;\dfrac{\sin |v|}{|v|}$ so it looks somewhat similar to $q$ from above, but I don't see how it helps me.

• Is it really completely obvious that multiplication by $e^{i\theta}$ is a rotation by $\theta$? I suspect it only seems so because one uses it so much that it becomes almost second nature. Most of us don't get nearly that much exercise with quaternions. Aug 10 '15 at 2:46

Fun fact: the square roots of $-1$ in $\Bbb H$ are precisely the purely imaginary unit quaternions. For any such quaternion $\bf u$, the subspace $\Bbb R\oplus\Bbb R{\bf u}\subset \Bbb H$ is a real subalgebra isomorphic to $\Bbb C$. Indeed all of the numbers $a+b{\bf u}$ would just act among themselves like complex numbers $a+bi$. Let's denote this subspace by $A({\bf u})$, and its orthogonal complement by $B({\bf u})$. (Use the standard Euclidean inner product on $\Bbb H$ inducing the standard Euclidean norm $\|\cdot\|$.)
Write $B({\bf u})=\langle {\bf v},{\bf w}\rangle$ where $\{{\bf u},{\bf v},{\bf w}\}$ is an ordered orthonormal basis for the purely imaginary quaternions with the same orientation as $\{{\bf i},{\bf j},{\bf k}\}$. Then in particular, $\{1,{\bf u},{\bf v},{\bf w}\}$ has the same multiplication table as $\{1,{\bf i},{\bf j},{\bf k}\}$ (this follows from the rule ${\bf ab}=-{\bf a}\cdot{\bf b}+{\bf a}\times{\bf b}$).
Denote ${\bf q}=\exp(\theta{\bf u})=\cos\theta+{\bf u}\sin\theta$ and let $L_{\bf q}({\bf x})={\bf q}{\bf x}$ and $R_{\bf q}({\bf x})={\bf xq}$. Observe $L_{\bf q}$ and $R_{\bf q}$ are both counterclockwise rotations in the $\langle 1,{\bf u}\rangle$-plane $A({\bf u})$ by an angle $\theta$, and $L_{\bf q}$ is a clockwise rotation whereas $R_{\bf q}$ is a counterclockwise rotation in the $\langle {\bf v},{\bf w}\rangle$-plane $B({\bf u})$. Composing the two maps would double up their effect in the first plane and cancel their effects in the second; if instead we compose the maps $L_{\bf q}$ and $R_{\bf q^{-1}}$ then the effects in the first plane will cancel and the effects in the second plane will double up.
Therefore, ${\bf x}\mapsto {\bf qxq}^{-1}$ where ${\bf q}=\exp(\theta{\bf u})$ will rotate purely imaginary quaternions around the axis $\Bbb R\bf u$ by the angle $2\theta$ according to the right-hand rule.