Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 1 / 3F (Geometry of Group Actions)

For the unit sphere, project from its north pole $N=(0,0,1)$ onto the equatorial plane, identified with the complex plane. The line through $N$ and $(X,Y,Z)$ meets the plane at the stereographic projectionThe north pole is omitted from the plane chart; adding its image infinity gives the Riemann sphere.

For a point whose image and antipodal image are finite, stereographic projection of $(-X,-Y,-Z)$ gives $v=-(X+iY)/(1+Z)$. Thereforeusing $X^2+Y^2+Z^2=1$. Conversely $u\overline{v}=-1$ implies $v=-1/\overline{u}$. Substituting this into the inverse stereographic projection changes the signs of all three coordinates, proving the converse. Thus finite chart images correspond to antipodes exactly when $u\overline{v}=-1$. On the extended chart the exceptional antipodal pair is $0,\infty$, for which the displayed ordinary complex product is not defined.