Generators of the group $S^3$ of unit quaternions

Question

Let $S^3$ be the group of unit quaternions, and let $A$ be the set of all purely imaginary unit quaternions (so $A$ is homeomorphic to $S^2$). How can we show that $S^3$ can be generated by the elements of the form $\alpha\beta$ where $\alpha,\beta \in A$? (This is asserted in p.9 of Morgan's book on Seiberg-Witten equations but I can't see why.)

Answer 1

Suppose $q\in S^3$. If $q = \pm 1$, then $q = (\mp i) i$ so has the form you want. Thus, we may assume $q$ has non-zero imaginary part. I'll write $q = r + p$ with $r\in \mathbb{R}$ and $p\neq 0$ purely imaginary.

Then, since $1 = |q| = r^2 + |p|^2$, we may assume $r = \cos\theta$ and $|r| = \sin\theta$ for some real angle $\theta$. Thus, we may write $q = \cos\theta + \frac{p}{|p|}\sin\theta.$

So, we are looking for two purely imaginary quaternions $\alpha,\beta$ with $\alpha\beta = \cos\theta + \frac{p}{|p|}\sin\theta$. From the formula in the comments, this amounts to finding two unit vectors in $\mathbb{R}^3$ (which I'll still call $\alpha$ and $\beta$) with $\alpha\cdot \beta = -\cos\theta$ and $|\alpha\times\beta| = \sin\theta$ and with $\alpha\times\beta$ pointing in the same direction as $p$.

But this is easy: if we pick $\alpha,\beta$ both in the plane perpendicular to $p$, then $\pm(\alpha\times \beta)$ points in the same direction as $p$, we now we just adjust the angle between $\alpha$ and $\beta$ to be $\pi-\theta$.