One of the main theorems about the classification of Möbius transformations states that pure rotations of the Riemann sphere (without translation and dilatation) correspond to unitary Möbius transformation; that is, Möbius 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 $\lvert\alpha\rvert^2+\lvert\beta\rvert^2 = a^2+b^2+c^2+d^2 = 1$.

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: $z_1 = \cot(\frac{1}{2}\theta)e^{i\phi}\implies \lvert z_1\rvert=\cot(\frac{1}{2}\theta), arg(z_2) = \phi$, where $\theta$, $\phi$ are the zenith angle and the azimuth 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: $l = \lvert\eta_3-\eta_2\rvert$. 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 $l = 2r\sin(\gamma/2)$. 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 $a$, $b$, $c$, and $d$. Therefore, my questions are:

- Are there any known results on the correspondence between unitary Möbius transformation and quaternions?
- Is there a conceptually more transparent, and simpler algebraic, way of deriving $q$? If the answer is yes, what is the resulting formula?

**Remark:**

I've already asked this question on Math StackExchange and didn't get any responses except a comment that the angle of rotation is easily computed by the derivative of the Mobius transformation at one of the fixed points $f'(z_1)$.

I've already asked this question on Math StackExchange and didn't get any responses- I think the norm is that you should wait a few days, and not crosspost the same day. $\endgroup$ – Kimball Aug 10 at 14:56