How to prove that the stereographic projection of the $S^2$ is conformal

Question

I read on Wikipedia that the stereographic projection of the $S^2$ sphere is conformal. However, there was no proof provided. How can one show it?

Answer 1

In an infinitesimal neighbourhood of a point $p$ of $S^2$, the stereographic projection amounts to the reflection at the plane bisecting the angle between the tangent plane to $p$ and the projection plane.

Answer 2

We can stereographically project from any sphere $S$ to any plane $\Pi$ from a pole, so long as $\Pi$ is parallel to (but distinct from) the pole's tangent space. Moving $\Pi$ (along the direction of $S$'s surface normal at $p$) simply has the effect of a homothety on the stereographic projection, which is conformal, and reflecting the plane $\Pi$ across the pole's tangent space has the effect of a central inversion on the projection, which is also conformal, so without loss of generality $\Pi$ is tangent to $S$ at the poles antipode, or equivalently $\Pi$ is the tangent space of the pole's antipode.

Also, in $n$ dimensions, any two intersecting lines are contained in a $3$-dimensional vector subspace, so if we can show stereographic projection is conformal for $n=3$ then it is true for all $n>3$ too.

Stereographic projection between a sphere minus a pole and the pole's antipode's tangent plane is actually the restriction of an involution (self-inverse function) on all of $\mathbb{R}^3$. Namely, it is inversion across the larger sphere centered at the pole and in contact with pole's antipode (geometry exercise). Spherical inversion is the 3D version of circular inversion in the plane. The fact that lines invert into circles through the origin carries over into 3D.

Given two lines $L_1,L_2$ intersecting at a point $P$ (tangent to two curves, say), they invert to two circles $C_1,C_2$ at $Q$. The tangent lines $T_1,T_2$ of the two circles at $Q$ are the same angle apart as the circles' tangent lines $U_1,U_2$ at the circle center $O$, but the lines $U_1,U_2$ are parallel to the lines $L_1,L_2$ by symmetry. Thus, we can say $\angle T_1T_2=\angle U_1U_2=\angle L_1L_2$. This means inversion preserves the angles between tangent lines.

To see why circular inversion inverts circles into lines into circles through the origin, notice in the picture below similar triangles $\triangle AOB\sim\triangle B'OA'$ (geometry exercise):

There's also a boring way to show inversion is conformal by showing its differential is an orthogonal transformation between tangent spaces. That is, show its Jacobian is an orthogonal matrix. The formula for inversion across a sphere centered at $\mathbf{a}$ of radius $r$ is

$$ f(\mathbf{x})=\frac{\mathbf{x}-\mathbf{a}}{\|\mathbf{x}-\mathbf{a}\|^2}r+\mathbf{a}. $$

Without loss of generality, $r=1$ and $\mathbf{a}=\mathbf{0}$.