Ahlfors -complex analysis p.20
Consider a stereographic projection between the 2-sphere and $\overline{\mathbb{C}}$ (i.e. one-point compactification of $\mathbb{C}$)
Let $z,w$ be complex numbers.
Let $d(z,w)$ denote the 2-norm distance between stereographic projections of $z,w$.
That is, $d(z,w)\triangleq \frac{2|z-w|}{\sqrt{(1+|z|^2)(1+|w|^2)}}$.
Moreover, define $d(\infty,z)=d(z,\infty)=\frac{2}{\sqrt{1+|z|^2}}$, for all complex $z$.
What is this metric $d$ called?
And does this metric induce topology on $\overline{\mathbb{C}}$ identical to that by one-point compactification?
No apparent name, and it is not the same...the key line is on page 20, maybe I can figure out how to do a quote box. Nope, it keeps splitting lines within formulas, then putting in > characters.
"If the points on the sphere are denoted by $(x_1,x_2,x_3),(x_1', x_2', x_3'),$ we have first $$(x_1 - x_1')^2 +(x_2 - x_2')^2 + (x_3 - x_3')^2 =2-2(x_1 x_1' +x_2 x_2' +x_3 x_3') " $$
so he is taking the ambient distance in $\mathbb R^3$ rather than the submanifold distance on the surface of the sphere. You can see the difference through the lack of any mention of $\pi.$ If you do this recipe with the more complicated distance along the surface of the sphere by great circles, the longest distance possible would be $\pi$ rather than $2.$ Just look at the distance between $0$ and $\infty$ given by the final formula before the exercises.
Same topology, though.
EEDDIITTT, 12:42 pm
Given a complex number $z = x + i y,$ its stereographic projection back to the unit sphere is $$ \left( \frac{2x}{1 + x^2 + y^2}, \frac{2y}{1 + x^2 + y^2}, \frac{x^2 + y^2 - 1}{1 + x^2 + y^2} \right). $$ You can check that the sum of the squares of these three expresssions is $1,$ and this lies along a straight line between $(x,y,0)$ and $(0,0,1).$
Next, given two unit vectors, we know the angle between them using the dot product. This is the same as the great-circle distance along the surface of the sphere. So, given vectors/ponts $a,b$ on the standard unit sphere, the great circle distance between them is $$ \arccos a \cdot b $$
So, given $z = x + i y$ and $z_1 = x_1 + i y_1$ in the complex plane, the distance induced from great circles is $$ d_\bigcirc(z,z_1) = \arccos \left( \frac{4 x x_1 + 4 y y_1 + (x^2 + y^2 - 1)(x_1^2 + y_1^2 - 1)}{(1 + x^2 + y^2)(1 + x_1^2 + y_1^2)} \right) $$ and $$ d_\bigcirc(z,\infty) = \arccos \left( \frac{ (x^2 + y^2 - 1)}{(1 + x^2 + y^2)} \right) = \arccos \left( \; 1 \; - \; \frac{ 2}{1 + x^2 + y^2} \; \right) $$