The geometry of $\operatorname{PSO}(4)$ and the quaternions

Question: Given a twist of the projective space, how do I find unit quaternions that represent it?

Backgroud and what do I mean: Following Conway & Smith's On Quaternions and Octonions, every pair of unitary quaternions $l,r$ gives a map $[l,r]:x \mapsto \bar{l}xr$, and a map $\ast[l,r]:x \mapsto \bar{l} \bar{x} r$, and every element of $\operatorname{GO}(4)$ can be seen as one of these two mappings (and $\operatorname{SO}(4)$ consists of those of the first type). When working projectively, the four expressions $[\pm l, \pm r]$ are all the same projective map.

On the other hand, the elements of $\operatorname{PSO}(4)$ can be thought, in a geometric way, as twists, that is, the product of two rotations on polar lines with different angles (in particular a rotation happens when one of the two angles is zero).

I would like to see, geometrically, what is the twist obtained by the product of two twists that I know (in terms of their lines). Using quaternions that would be straightforward, if I knew, given a (geometric) twist (ie. its two polar lines and the angles) how can we find the unit quaternions $l,r$ that represent it?

• I don't know much about this, but it seems to me like you could just figure out the matrix of $[l,r]$ and work backwards? – Aaron Mazel-Gee Feb 21 '11 at 18:32
• When “one of the two angles is zero”, then versors $l,r$ have the same angle (up to 180°). Conversely, when one of versors is ±1, the “twist” is an isoclinic rotation (a.k.a. Clifford translation). I’d explain more but the question is apparently faded. – Incnis Mrsi Nov 8 '14 at 20:21