1.5 Stillwell's Naive Lie Theory rotation by conjugation theorem. Order of multiplication

Question

In section 1.5 of Naive Lie Theory by John Stillwell the rotation by quaternion $t = \cos\left(\theta\right) + u \cdot \sin\left(\theta\right)$ is defined as conjugation by $t$: $q \mapsto t^{-1} q t$ and is said to rotate the set of pure imaginary quaternions $\mathbb{R}i + \mathbb{R}j + \mathbb{R}k$ by angle $2\theta$.

I understand the proof that this is indeed the rotation, what I'm struggling with is the order of the multiplication which leads to clockwise rotation on the plane orthogonal to $u$, which differs from the convention for complex numbers earlier in the book and in general that rotation is clockwise if you look in the direction of the axis around which one rotates. E.x. $q = 0 + 1i + 0j + 0k$ would be mapped by $t = \frac{\sqrt{2}}{2} + 0i + 0j + \frac{\sqrt{2}}{2}k$ to $q_{rot} = 0 + 0i - 1j + 0k$. With a map $q \mapsto t q t^{-1}$ the result is as expected: $q_{rot} = 0 + 0i + 1j + 0k$

Is it just taking different convention and not stating it or is there something I'm missing?

Answer 1

rotation is clockwise if you look in the direction of the axis around which one rotates

This indeed is the convention. In other words, you get a counterclockwise (aka anticlockwise) if you look from the direction in which the axis of rotation points. This is consistent with the right hand rule, i.e. $\mathbf{k}\times\mathbf{i}=\mathbf{j}$ rotates $\mathbf{i}$ counterclockwise (looking down from the $+\mathbf{k}$-axis) to $\mathbf{j}$.

And indeed, if $t=\exp(\theta\mathbf{u})$, then $t\mathbf{x}t^{-1}$ is the counterclockwise rotation of $\mathbf{x}$ around $\mathbf{u}$ by $2\theta$. That means $t^{-1}\mathbf{x}t$ would be the clockwise rotation (looking from the $+\mathbf{k}$-axis "downwards"), or a counterclockwise if looking up from below (i.e. in the $+\mathbf{k}$-direction).

IIRC when I read Stillwell the unprovoked use of right actions instead of left actions did irk me.