# Limit of Distance on Stereographically Projected Sphere for $K\to 0$

Let $$S_K^n$$ be the stereographically projected sphere (on the Euclidean plane $$E^n$$). Where $$n$$ is the dimensionality of the sphere and $$K$$ is its sectional cruvature. Then the distance function for $$x,y\in S_K^n$$ is: $$d_K(x,y)= \frac{1}{\sqrt{K}} \arccos\left( \frac{ 4\langle x,y\rangle + (||Kx||_2^2-1)(||Ky||_2^2-1) }{ (||Kx||_2^2+1)(||Ky||_2^2+1) } \right)$$ Prove or disprove that in the limit we have: $$\lim_{K\to 0}d_K(x,y) =c||x-y||_2, \tag{*}$$ where $$c$$ is a constant factor (most likely $$2$$).

This would mean that in the limit we recover Euclidean-like geometry.

Note that the distance function above results from the stereographic projection of the sphere through the north pole: $$S_K^n\setminus\{\text{north pole}\}\to E^n$$.

I'm not sure that this can be shown. However, for the Poincaré disk (if $$d_K$$ was the distance on the Poincaré disk and $$x,y$$ would be on the disk) one can show that the upper relationship $$(*)$$ holds with $$c=2$$.

One might have to add that $$x$$ and $$y$$ lie in the southern hemisphere for such a relationship to hold.

• As far as I can tell, taking the limit of $d_K$ using l'Hospital several times, combined with the fact that $2<x,y> = ||x||^2 + ||y||^2 - ||x-y||^2$ should yield something similar to the desired limit, potentially with a constant factor, and potentially with the desired distance squared. The computation is fairly messy and I might have potentially made a mistake somewhere. Commented Jul 16, 2019 at 13:24
• You should give some additional explanations. I guess you consider the stereographic projection $s : S^2 \setminus \{ north pole \} \to \mathbb R^2$. Then $d_K(x,y)$ for $x,y \in \mathbb R^2$ is defined as $\lVert s^{-1}(x) - s^{-1}(y) \rVert$ with Euclidean norm in $\mathbb R^3$? What is the role of the parameter $K$? Commented Jul 22, 2019 at 14:52
• @PaulFrost Yes, i consider the stereographic projection in $n$ dimensions. $K$ is the sectional curvature of the sphere. I've added everything to the question now. Commented Jul 22, 2019 at 14:57

The formula you have written for $$d_K(x,y)$$ does not posses a finite limit as $$K\to 0$$ for most choices of $$x$$ and $$y$$. Indeed, $$\cos\Bigl[\sqrt{K}\cdot d_K(x,y)\Bigr]=\frac{ 4\langle x,y\rangle + (||Kx||_2^2-1)(||Ky||_2^2-1) }{ (||Kx||_2^2+1)(||Ky||_2^2+1) } .$$ Since $$\cos$$ is continuous and $$\lim_{K\to 0}\frac{ 4\langle x,y\rangle + (||Kx||_2^2-1)(||Ky||_2^2-1) }{ (||Kx||_2^2+1)(||Ky||_2^2+1) }=1+4\langle x,y\rangle,$$ it follows that $$\sqrt{K}\cdot d_K(x,y)$$ converges to a non-zero constant as $$K\to 0$$, for suitable $$x,y$$. But since $$\frac{1}{\sqrt{K}}\to\infty$$ as $$K\to 0$$, this means that $$d_k(x,y)=\frac{1}{\sqrt{K}}\cdot \sqrt{K}\cdot d_K(x,y)$$ also tends to $$\infty$$ as $$K\to 0$$.
Thus, there does not exist any finite $$c$$ with the property that $$\lim_{K\to 0}d_K(x,y) =c||x-y||_2,$$ since the left side is infinite for most $$x,y$$ whereas the right side is always finite.