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

If $\gamma :(-\varepsilon ,\varepsilon )\to \mathbb {C}$ is a smooth curve in $\mathbb {C}$ with $\gamma (0)=z_{0}$, then using the formula for inverse stereographic we obtain the desired result.