Quaternion representation of rotations How would one show that 1/3 turns correspond to the eight antipodal pairs among the 16 quaternions :
$$
\pm \frac{1}{2} \pm \frac{i}{2} \pm \frac{j}{2} \pm \frac{k}{2}
$$
knowing that the rotation about axis u through angle theta corresponds to the quaternion pair $\pm q$ where,
$$
q = \cos(\theta/2) + u \sin(\theta/2)
$$
 A: Let's start with the original equation, and then take a look at what it means for $\cos(\theta/2)+u\sin(\theta/2)= \frac{1}{2} \pm \frac{i}{2} \pm \frac{j}{2} \pm \frac{k}{2}$.
First note that $\pm \frac{i}{2} \pm \frac{j}{2} \pm \frac{k}{2}=\frac{\sqrt{3}}{2}u$ where $u=\pm\frac{i}{\sqrt{3}}\pm\frac{j}{\sqrt{3}}\pm\frac{k}{\sqrt{3}}$. It doesn't matter how you pick the signs: it's always a unit vector.
So you are now looking at $\cos(\theta/2)+u\sin(\theta/2)= \frac{1}{2} + \frac{\sqrt{3}}{2}u$. From $\cos(\theta/2)=\frac12$, you learn that $\theta/2\in\{-\pi/3,\pi/3\}$. Given also $\sin(\theta/2)=\frac{\sqrt{3}}{2}$, we deduce $\theta/2=\pi/3$. This attests that the quaternion rotates by $2\pi/3$, one third of a full turn. Given that you can freely pick the three sets of signs in $u$, that accounts for $8$ quaternions.
The same story holds for $-\frac12$ in front, but these quaternions are precisely $-1$ times the other set of quaternions we discovered. Each of the two sets contains one half of an antipodal pair.
It should also be noted that this does not find all $1/3$ turn quaternions, it just finds the ones that rotate about the eight axes $\pm i\pm j\pm k$.
A: I think you can write down the rotation matrix explicitly by fixing a basis. Then using the same basis, try to find a quarterion such that its action's corresponding matrix is the same. This should not be difficult since 1/3 rotation has special properties available. 
