Ngô Quốc Anh

October 3, 2017

Stereographic projection not at the North pole and an example of conformal transformation on S^n

Filed under: Riemannian geometry — Tags: , — Ngô Quốc Anh @ 12:09

Denote \pi_P : \mathbb S^n \to \mathbb R^n the stereographic projection performed with P as the north pole to the equatorial plane of \mathbb S^n. Clearly when P is the north pole N, i.e. N = (0,...,0,1), then \pi_N is the usual stereographic projection.

Clearly, for arbitrary x \in \mathbb S^n, the image of x is

\displaystyle \pi_P : x \mapsto y = P+\frac{x-P}{1-x \cdot P}.

For the inverse map, it is not hard to see that

\displaystyle \pi_P^{-1} : y \mapsto x =\frac{|y|^2-1}{|y|^2+1}P+\frac 2{|y|^2+1}y.

Derivation of \pi_P and \pi_P^{-1} are easy, for interested reader, I refer to an answer in . Let us now define the usual conformal transformation \varphi_{P,t} : \mathbb S^n \to \mathbb S^n given by

\displaystyle \varphi_{P,t} : x \mapsto \pi_P^{-1} \big( t \pi_P ( x) \big)

and try to compute \varphi_{P,t} precisely. Clearly,

\displaystyle t \pi_P ( x) = t\Big( P+\frac{x-P}{1-x \cdot P} \Big) = t\frac{x-(x \cdot P) P}{1-x \cdot P}.

Therefore, via the formula |z-w|^2=|z|^2-2z \cdot w+|w|^2, we obtain

\displaystyle \big|t \pi_P ( x)\big|^2 = t^2\frac{|x|^2- 2 (x \cdot P) P \cdot x +| (x \cdot P) P|^2 }{(1-x \cdot P)^2}= t^2\frac{1+x \cdot P}{1-x \cdot P}

since |P|=|x|=1. Consequently,

\displaystyle \pi_P^{-1} \big( t \pi_P ( x) \big) = \frac{t^2 \dfrac{1+x \cdot P}{1-x \cdot P}-1}{t^2 \dfrac{1+x \cdot P}{1-x \cdot P}+1}P+\frac 2{t^2 \dfrac{1+x \cdot P}{1-x \cdot P}+1}t \frac{x - (x \cdot P)P}{1-x \cdot P}.

Hence, if we denote

\displaystyle D =t^2 (1+x\cdot P) + (1-x \cdot P),

then

\displaystyle \pi_P^{-1} \big( t \pi_P ( x) \big) = D^{-1} \big( [t^2 (1+x\cdot P) - (1-x \cdot P)]P + 2t [x - (x \cdot P)P] \big).

This formula is well-known and does appear quite often in may research papers, for example, in Lemma 5.1 on page 374 in a paper due to Yan Yan Li.

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