Past exam of the metric spaces and complex analysis course of the University of Oxford / 2014 / 6

The line through the north pole and $p=(a,b,c)$ isIts intersection with the equatorial complex plane has third coordinate zero, giving $t=1/(1-c)$ when $p$ is not the north pole. The first two coordinates are therefore $a/(1-c)$ and $b/(1-c)$, henceThe north pole maps to the point at infinity. This is stereographic projection onto the Riemann sphere.

For $z=x+iy$ and $r^2=|z{|}^2$, solving the sphere and projection equations givesFor $z\ne 0$, the reciprocal has real and imaginary parts $x/r^2$ and $-y/r^2$, and squared modulus $1/r^2$. Substituting these into the inverse formula producesThusThe reciprocal exchanges zero and infinity, corresponding to exchanging the south and north poles, so the formula holds there too. On the sphere this is a rotation through $\pi$ about the first coordinate axis, not the antipodal map.

When all three points are finite, useThe distinctness assumptions ensure that it is a nonconstant Möbius transformation, and direct substitution gives the three required values. If one of the specified points is infinity, the limiting formulas areThese cover every case because the points are distinct.

For uniqueness, if $g$ and $\tilde{g}$ have the same three images, the Möbius map $\tilde{g}\circ g^{-1}$ fixes $0,1,\infty$. Fixing zero and infinity makes it $z\mapsto cz$ with $c\ne 0$; fixing one gives $c=1$. Therefore the normalized map is unique. This is the sharp three-transitivity of Möbius transformations.

Let $G$ be the subgroup preserving the three-point set. Its action gives a group homomorphism $\Phi :G\to S_3$. If $\Phi (h)$ is the identity permutation, then $h$ fixes all three points, so part (i) forces $h$ to be the identity Möbius map. Thus the kernel is trivial and $\Phi$ is injective.

For any permutation $\sigma$, part (i) gives a normalization $g_{\sigma }$ sending the ordered triple $(z_{\sigma (1)},z_{\sigma (2)},z_{\sigma (3)})$ to $(0,1,\infty )$. If $g$ normalizes the original triple, then $h_{\sigma }=g_{\sigma }^{-1}\circ g$ sends $z_j$ to $z_{\sigma (j)}$. Hence every permutation is attained, so $\Phi$ is surjective.This is the setwise stabilizer; its six maps need not fix each point individually.

Choose a Möbius map $T$ with $T(z_1)=0$ and $T(z_2)=\infty$, for example by normalizing the ordered triple $(z_1,z_3,z_2)$. Conjugation $h\mapsto T\circ h\circ T^{-1}$ takes the requested subgroup to the subgroup fixing zero and infinity individually.

Such a transformation must be $z\mapsto cz$ with $c\in {\mathbb{C}}^{\ast }$: the zero condition removes the constant numerator term and the infinity condition removes the nonconstant denominator term. Conversely every such dilation is Möbius and fixes the two points. Composition multiplies its nonzero parameter.This identifies the Möbius pointwise stabilizer of two points. Allowing an interchange of the two fixed points would give a larger group, but those interchanges are excluded by the stated equalities.