Stereographic projection is conformal

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Statement

Setup

Further information: Stereographic projection

Consider R3, three-dimensional Euclidean space, with coordinates x,y,z. Denote:

S2:={(x,y,z)R3x2+y2+z2=1}

Identify C with the xy-plane under the map:

(x,y,0)x+iy

Let N=(0,0,1) denote the north pole in S2, and define the stereographic projection as a bijective map:

S2{N}C

which sends a point PS2{N} to the unique point in C that is collinear with N and P.

Actual statement

Pick any point PS2{N}. Then there is a natural induced map from the tangent space to P in S2 to the tangent space to its image, in C. This map is conformal, i.e. it preserves angles.

In other words, if we make two smooth curves in S2 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 S2{N} and C as Riemannian manifolds, with the former getting the induced structure from its embedding in R3.

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

Alternatively, the Riemannian metric on C obtained using that on S2{N}, is conformally equivalent to the standard metric.

Facts used

  • Formula for inverse stereographic projection: If x+iyC, and Φ denotes inverse stereographic projection, we have:

Φ(x,y)=(2xx2+y2+1,2yx2+y2+1,x2+y21x2+y2+1)

Proof

Geometric proof

Computational proof

Consider a point z0=x0+iy0 and consider two straight lines through z0, parallel to complex numbers x1+iy1 and x2+iy2 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 S2.