Vector space of quaternions with vanishing real part, description of map in geometric terms.

Question

This is relevant to the question here.

Let $\mathbb{H}$ be the quaternions and $V = \mathbb{R}^3 \subset \mathbb{H}$ the vector space of quaternions with vanishing real part. In the linked question, it was asked whether the set $S = \{x \in \mathbb{H}: x^2 = -1\}$ is finite or infinite.

Question. For each $x \in S$, how do we describe in geometric terms the map$$T_x: V \to V, \text{ }v \mapsto -xvx^{-1},$$viewed as a linear map of the Euclidean vector space $\mathbb{R}^3$?

Answer 1

It's just a rotation composed with a reflection. (Or just a reflection, depending on your terminology: see the last paragraph.)

$v\mapsto xvx^{-1}$ is a rotation (even though $x$ isn't a unit vector.) After all, $xvx^{-1}=\frac{x}{\|x\|}v(x^{-1}\|x\|)=\frac{x}{\|x\|}v(\frac{x}{\|x\|})^{-1}$ and $\frac{x}{\|x\|}$ is a unit vector.

The transformation $v\mapsto -v$ can be viewed in multiple ways as a rotation and a reflection. You could say that it's a rotation by $\pi$ around the $z$ axis followed by a reflection through the $x,y$ plane. (Or a rotation by $\pi$ around the $y$ axis and a reflection through the $x,z$ plane or...)

However you do it, you can combine those two initial rotations into one and then you have a single rotation followed by a reflection.

Actually depending on your terminology, you might simply call it a "reflection." Some authors use the term "reflection" to mean "orthogonal transformation with determinant $-1$". Here I was thinking of "reflection" only as a transformation of $\mathbb R^3$ with eigenvalues $\{1,1,-1\}$.

Answer 2

$$\mathbb{H} = \text{Re}(\mathbb{H}) \, \oplus \, \text{Im}(\mathbb{H}) = \mathbb{R} \, \oplus \, \text{Im}(\mathbb{H}) = \mathbb{R} \, \oplus \, \mathbb{R}^3 $$ because $$\text{Im}(\mathbb{H}) = \mathbb{R}^3$$ If $x \in \mathbb{H}$ then $$x = r + w$$ where $r \in \mathbb{R}$ and $w \in \text{Im}(\mathbb{H})$. Then if $v \in \text{Im}(\mathbb{H})$ then $v^2 = - |v|^2$ is equivalent to the dot product. Therefore $$- |x\, v \, x^{-1}|^2 = (x\, v \, x^{-1})^2 = (x\, v \, x^{-1}) (x\, v \, x^{-1}) = x \, v^2 \, x^{-1} = - |v|^2 \, x x^{-1} = -|v|^2$$ so this transformation is orthogonal, so a rotation. $$x = r + w = |r+w| \left(\frac{r}{|r+w|} + \frac{w}{|r+w|}\right) = |r+w| \left(\frac{r}{|r+w|} + \frac{|w|}{|r+w|} \, \frac{w}{|w|}\right) $$ $$= |r+w| \left(\big(\cos{\theta}\big) + \big(\sin{\theta} \big)\,\, \frac{w}{|w|}\right) = |x| \Big(\big(\cos{\theta}\big) + \big(\sin{\theta} \big)\,\, \hat{w}\Big) $$ where $\theta$ is angle determined by $\cos\theta = \frac{r}{|r+w|}$ and $\hat{w} = \frac{1}{|w|} \, w \in \text{Im}(\mathbb{H})$ is a unit imaginary quaternion. The map $$v \mapsto x\, v\, x^{-1} = \Big(\big(\cos{\theta}\big) + \big(\sin{\theta} \big)\,\, \hat{w}\Big) \, v \, \Big(\big(\cos{\theta}\big) + \big(\sin{\theta} \big)\,\, \hat{w}\Big)^{-1} = $$ $$ = \Big(\big(\cos{\theta}\big) + \big(\sin{\theta} \big)\,\, \hat{w}\Big) \, v \, \Big(\big(\cos{\theta}\big) - \big(\sin{\theta} \big)\,\, \hat{w}\Big) $$ represents a rotation around the unit vector $\hat{w}$ with angle $\theta$.