Geometric significance of map, quaternions.

Question

Let $u, v, w \in \mathbb{R}^3$ be a triple of vectors which form an orthonormal basis in $\mathbb{R}^3$ (with the standard orientation). Identify $u, v, w$ with quaternions in the $\mathbb{R}$-linear span of the elements $\{i, j, k\}$ (call such a span $\mathbb{V}$) and, given an angle $0 \le \theta \le \pi$, put $q := \cos \theta + u \sin \theta \in \mathbb{H}$. My question is, what is the geometric significance of the linear map$$\text{Ad}\,q: \mathbb{V} \to \mathbb{V},\,x \mapsto qxq^{-1}?$$Thanks in advance.

Answer 1

Use the standard $\Bbb R$-basis $\{1,{\bf i},{\bf j},{\bf k}\}$ for $\Bbb H$ and equip it with the Euclidean inner product. The pure imaginary quaternions are then the orthogonal complement of the reals, $\Bbb R^\perp$. Because $\Bbb H$ is normed and satisfies $\|{\bf ab}\|=\|{\bf a}\|\|{\bf b}\|$ for all ${\bf a},{\bf b}\in\Bbb H$, multiplication by a unit quaternion ($\|{\bf q}\|=1$) is an isometry of $\Bbb H$. (Both left and right multiplication, which are distinct maps.) As such, conjugation by the quaternion $\bf q$ is also an isometry.

Every unit quaternion ${\bf q}$ is expressible as ${\bf q}=\cos\theta+{\bf u}\sin\theta$. Denote $\varphi_{\bf q}({\bf x})={\bf qxq}^{-1}$. One easily observes $\bf u$ is a fixed point of $\varphi_{\bf q}$, and as $\varphi_{\bf q}$ is $\Bbb R$-linear it fixes the line $\Bbb R{\bf u}$ pointwise. Since it's an isometry, it must be a rotation about the axis $\Bbb R{\bf u}$, but by what angle? Pick a basis $\{{\bf u},{\bf v}\}$ for $\Bbb R{\bf u}$'s orthogonal complement in $\Bbb R^\perp$, so that the ordered orthonormal basis $\{{\bf u},{\bf v},{\bf w}\}$ has the same orientation as $\{{\bf i},{\bf j},{\bf k}\}$. In particular, it has the same multiplication table, a fact that follows by applying the formula ${\bf ab}=-{\bf a}\cdot{\bf b}+{\bf a}\times{\bf b}$ for pure imaginary quaternions. Compute

$$\begin{array}{ll} \varphi_{\bf q}({\bf v}) & =(\cos\theta+{\bf u}\sin\theta){\bf v}(\cos\theta-{\bf u}\sin\theta) \\ & = ({\bf v}\cos\theta+{\bf w}\sin\theta)(\cos\theta-{\bf u}\sin\theta) \\ & = {\bf v}(\cos^2\theta-\sin^2\theta)+{\bf w}(2\cos\theta\sin\theta) \\ & = {\bf v}\cos(2\theta)+{\bf w}\sin(2\theta) \end{array} $$

by the double-angle formulas. Therefore, $\varphi_{\bf q}$ is a rotation in $\Bbb R^\perp$ around $\bf u$ by an angle of $2\theta$ according to the right-hand rule.