0
$\begingroup$

I have two unit quaternions $q = e^{i2\pi{53}/60}$ and $q'= e^{i2\pi{17}/60}$, and I want to find an inner-automorphism of $S^3$ that maps $q$ to $q'$. I know that rotations in quaternions can be represented by conjugation, and it seems that the above mapping is a rotation by an angle of $2\pi{3}/5$, but I am unable to find a global inner-automorphism that maps $q$ to $q'$.

I would appreciate help in this matter very much.

$\endgroup$

1 Answer 1

1
$\begingroup$

Inner automorphisms do not act transitively on $\mathrm{Sp}(1)$ (the group of unit quaternions, also known as $S^3$). Indeed, inner automorphisms do not act transitively on any nontrivial group, since the identity is its own conjugacy class. In particular, conjugating a quaternion by another cannot change its real part. However, any two unit quaternions with the same real part are conjugate.

In particular, if $x=\exp(\theta\mathbf{u})=\cos(\theta)+\sin(\theta)\mathbf{u}$ where $\mathbf{u}$ is a square root of negative one, or equivalently a pure imaginary unit quaternion, and $\mathbf{v}$ is a pure imaginary quaternion, then the conjugation $x\mathbf{u}x^{-1}$ is just $\mathbf{u}$ rotated around the oriented axis in the $\mathbf{u}$ direction by an angle of $2\theta$.

Write $\theta=2\pi(53/60)$. Observe your quaternions are $q=\exp(\theta\mathbf{i})$ and $q'=\exp(-\theta\mathbf{i})$. Within the space of imaginary quaternions, $\mathrm{Im}(\mathbb{H})\cong\mathbb{R}^3=\mathrm{span}\{\mathbf{i},\mathbf{j},\mathbf{k}\}$, to rotate $\mathbf{i}$ to $-\mathbf{i}$ we rotate by an angle of $\pi$ around any axis orthogonal to $\mathbf{i}$, so in particular, write $\mathbf{v}=\cos(\phi)\mathbf{j}+\sin(\phi)\mathbf{k}$ for any angle $\phi$, then define the unit quaternion $x=\exp(\frac{\pi}{2}\mathbf{v})$, which is actually equal to $\mathbf{v}$, in which case we have $\mathbf{v}q\mathbf{v}^{-1}=q'$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .