One of the main theorems about the classification of Mobius transformations states that pure rotations of the Riemann sphere (without translation and dilatation) correspond to unitary Mobius transformation; that is, Mobius transformations of the form:
$$f(z) = \frac{\alpha z +\beta}{-\bar{\beta}z+\bar{\alpha}} = \frac{(a+bi) z +(c+di)}{-(c-di)z+(a-bi)}$$
where $|\alpha|^2+|\beta|^2 = a^2+b^2+c^2+d^2 = 1$.
I was wondering how to correspond a unit quaternion to a given unitary Mobius transformation, as both can be identified with rotation of the Riemann sphere. To calculate the corresponding quaternion, i outlined the following procedure:
- First, find the fixed points $z_1,z_2$ of the given unitary Mobius transformation by solving the quadratic equation: $z=f(z)\implies -\bar{\beta}z^2+(\bar{\alpha}-\alpha)z-\beta=0$. This step actually enables to find the axis of rotation of the Riemann sphere, which correspond to the vector part of the desired quaternion $q$. The reason for this is that the fixed points of $f(z)$ are exactly the images of the intersection of the axis with the Riemann sphere (the poles of the rotation), under the stereographic projection.
- Secondly, as explained in the first step, one needs to find the pre-images of $z_1,z_2$ under the stereograpic projection. This can be done by the equation: $z_1 = cot(\frac{1}{2}\theta)e^{i\phi}\implies |z_1|=cot(\frac{1}{2}\theta), arg(z_2) = \phi$, where $\theta,\phi$ are the zenith angle and the azimute of the pre-image.This completes the calculation of the axis direction (which we denote by $\eta_1$), and the vector part of $q$.
- Now the problem is to find the angle of rotation $\gamma$, which correspond to the real part of the $q$. To do this, apply $f(z)$ to the simplest complex number $z = 1$ to get $f(1) = \frac{\alpha +\beta}{-\bar{\beta}+\bar{\alpha}}$. Now apply the inverse stereographic projection (which we denote $S^{-1}$) to find the point on the sphere that is projected to $f(1)$. We now have three known vectors: $\eta_1, \eta_2 = (1,0,0), \eta_3 = S^{-1}(f(1))$.
- As a last step, calculate the length of difference vector: $l = |\eta_3-\eta_2|$. Since $l$ is a chord in a circle whose radius is $r = \sqrt{1-(\eta_1 \cdot \eta_2)^2}$ and central angle $\gamma$, one can solve for $\gamma$ by the equation $l = 2rsin(\gamma/2)$. The desired quaternion is: $q = cos(\gamma/2)+(\eta_{1x}i+\eta_{1y}j+\eta_{1z}k)sin(\gamma/2)$.
Although this procedure might appear practical, the algebra i got is so complicated that i didn't succeed in finding a closed form expression for $q$ in terms of $a,b,c,d$. Therefore, my questions are:
- Are there any known results on the correspondence between unitary Mobius transformation and quaternions?
- Is there a conceptually more transparent, and simpler algebraical, way of deriving $q$? if the answer is yes, what is the resulting formula?