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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 $f'(z_1)$.

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  • $\begingroup$ See L. Ahlfors, Mobius transformations in several dimensions, 1981, Ch. II. General case, to this end. $\endgroup$ – user64494 Aug 9 at 14:49
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    $\begingroup$ @user64494's reference doesn't seem obviously available online, but there's a snippet view on Google Books: Ahlfors - Möbius transformations in several dimensions (MSN). $\endgroup$ – LSpice Aug 9 at 15:19
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    $\begingroup$ @LSpice: Thank you for the link. It's kind of you. $\endgroup$ – user64494 Aug 9 at 15:27
  • $\begingroup$ try this: researchgate.net/publication/… $\endgroup$ – Mirco A. Mannucci Aug 9 at 17:42
  • $\begingroup$ 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
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After getting some very useful comments by user10354138 from math stackexchange, i succeeded in finding a solution, which is remarkably simple. According to his comments, the angle of rotation $\gamma$ can be computed by taking the derivative of Mobius transformation $f(z)$ at one of the fixed points of the transformation; that is: $e^{i\gamma} = f'(z_1)$. First, we need to find the fixed point:

  • the resulting quadratic equation is: $-(c-di)z^2-2biz-(c+di)=0\implies z_{1,2} = \frac{-d + ci}{c^2+d^2} (-b\pm \sqrt{1-a^2})$.

The calculation of the derivative yields:

$ e^{i\gamma}=f'(z_1)= \frac{2a+\sqrt{(2bi)^2-4(c^2+d^2)}}{2a-\sqrt{(2bi)^2-4(c^2+d^2)}} = \frac{2a+2i\sqrt{1-a^2}}{2a-2i\sqrt{1-a^2}} = \frac{a+i\sqrt{1-a^2}}{a-i\sqrt{1-a^2}}$.

According to several trigonometric identities, this means that $cos(\gamma/2) = a$. So the real part of the desired quaternion $q$ is $a$. Now, the norm of $z_1$ is:

$cot(\theta/2)=Norm(z_1) = \frac{-b+\sqrt{1-a^2}}{\sqrt{c^2+d^2}}\implies cos\theta = \frac{cot^2(\theta/2)-1}{cot^2(\theta/2)+1} = -\frac{b}{\sqrt{1-a^2}}$. Since the representation of $q$ is:

$$q = cos(\gamma/2)+(\eta_{1x}i+\eta_{1y}j+\eta_{1z}k)sin(\gamma/2) = a+(\eta_{1x}i+\eta_{1y}j+\eta_{1z}k)\sqrt{1-a^2}$$, one gets that the $k$-component of $q$ is $-b$. In addition, since $\frac{\eta_{1y}}{\eta_{1x}} = -\frac{d}{c}$, and $\eta_{1x}^2+\eta_{1y}^2+\eta_{1z}^2 = 1$, one gets the final result that the desired quaternion is:

$$q = a + ci - dj -bk$$.

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  • $\begingroup$ L. Ahlfors writes in "Mobius transformations in several dimension" that the quaternion technics works only for $M(H^3)$, where $H$ is the upper half-space and is not particularly useful. $\endgroup$ – user64494 Aug 10 at 11:47
  • $\begingroup$ This question was not intended to demonstrate the usefulness of quaternions, but just to identify which rotation of the Riemann sphere (which can be described by unit quaternion) corresponds to a given unitary Mobius transformation. Amazingly, the quaternions coefficients are just $a, c, -d, - b$. $\endgroup$ – user2554 Aug 10 at 12:22

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