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How can we go about finding a Moebius map that fixes the set $\{z_1=x+iy,\,\,\, z_2={1\over iy-x}\}$ for some $x,y\in \mathbb R$ that does not correspond to rotation about any arbitrary axis of the Riemann sphere?

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I suggest choosing a third point $z_3$. You know you want your map to satisfy $(w,z_1,z_2,w_3)=(z,z_1,z_2,z_3)$ for some $w_3$ distinct from $z_1,z_2$. What does a rotation about an axis of the sphere look like? How can $w_3$ be chosen to avoid this? – Cameron Buie Apr 27 '12 at 20:04
More precisely, you may want to consider this. Let $H$ be the unique affine space in $\mathbb{R}^3$ with $z_3$'s corresponding spherical point on it, normal to the axis through the spherical points corresponding to $z_1,z_2$. Let $C$ be the intersection of the plane $H$ with the Riemann sphere. Where would a rotation about the axis map $C$? $z_3$ would be mapped into the corresponding line or circle in the complex plane by such a rotation. How then must we choose $w_3$? – Cameron Buie Apr 27 '12 at 20:46
@CameronBuie: Thank you! – userabc Apr 27 '12 at 22:09
Glad to help. I'll go ahead and put it as an actual answer, since it worked out for you. – Cameron Buie Apr 27 '12 at 22:48
up vote 0 down vote accepted

Choose a third point $z_3$ distinct from $z_1,z_2$. The map has to satisfy $(w,z_1,z_2,w_3)=(z,z_1,z_2,z_3)$ for some $w_3$ distinct from $z_1,z_2$, so we need only determine how to appropriately choose $w_3$.

Let $Z_k$ be the point on the sphere corresponding to $z_k$ ($k=1,2,3$). The line through $Z_1,Z_2$ is an axis of the sphere, and so there is a unique plane $H$ in $\mathbb{R}^3$ normal to this axis and containing $Z_3$. If $C$ is the intersection of the sphere $H$, it isn't difficult to see that $C$ would be mapped to itself by any rotation about the axis, so $w_3$ must be chosen not to lie on the line or circle in the complex plane corresponding to $C$.

Having done so, it remains only to show that the given transformation isn't a rotation about any other axis of the sphere, either, but this is not difficult, as $z_1,z_2$ should be the only points fixed by the transformation.

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