Stereographic projection is conformal

Statement

Setup

Further information: Stereographic projection

Consider $\mathbb {R} ^{3}$, three-dimensional Euclidean space, with coordinates $x,y,z$. Denote:

$S^{2}:=\{(x,y,z)\in \mathbb {R} ^{3}\mid x^{2}+y^{2}+z^{2}=1\}$

Identify $\mathbb {C}$ with the $xy$-plane under the map:

$(x,y,0)\mapsto x+iy$

Let $N=(0,0,1)$ denote the north pole in $S^{2}$, and define the stereographic projection as a bijective map:

$S^{2}\setminus \{N\}\to \mathbb {C}$

which sends a point $P\in S^{2}\setminus \{N\}$ to the unique point in $\mathbb {C}$ that is collinear with $N$ and $P$.

Actual statement

Pick any point $P\in S^{2}\setminus \{N\}$. Then there is a natural induced map from the tangent space to $P$ in $S^{2}$ to the tangent space to its image, in $\mathbb {C}$. This map is conformal, i.e. it preserves angles.

In other words, if we make two smooth curves in $S^{2}$ that intersect at $P$, the angle of intersection between those curves at $P$ equals the angle of intersection of their images under stereographic projection.

Alternative formulation

Consider $S^{2}\setminus \{N\}$ and $\mathbb {C}$ as Riemannian manifolds, with the former getting the induced structure from its embedding in $\mathbb {R} ^{3}$.

Then, the stereographic projection is a conformal map of Riemannian manifolds.

Alternatively, the Riemannian metric on $\mathbb {C}$ obtained using that on $S^{2}\setminus \{N\}$, is conformally equivalent to the standard metric.

Facts used

• Formula for inverse stereographic projection: If $x+iy\in \mathbb {C}$, and $\Phi$ denotes inverse stereographic projection, we have:

$\Phi (x,y)=\left({\frac {2x}{x^{2}+y^{2}+1}},{\frac {2y}{x^{2}+y^{2}+1}},{\frac {x^{2}+y^{2}-1}{x^{2}+y^{2}+1}}\right)$

Proof

Computational proof

Consider a point $z_{0}=x_{0}+iy_{0}$ and consider two straight lines through $z_{0}$, parallel to complex numbers $x_{1}+iy_{1}$ and $x_{2}+iy_{2}$ respectively. We need to show that the angle between their images under stereographic projection, equals the angle between the lines themselves.

First, let us compute the tangent vectors to the images of these lines, in $S^{2}$.