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.

  • $\begingroup$ 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. $\endgroup$ Jul 16, 2019 at 13:24
  • $\begingroup$ 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 $K$? $\endgroup$
    – Paul Frost
    Jul 22, 2019 at 14:52
  • $\begingroup$ @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. $\endgroup$
    – ndrizza
    Jul 22, 2019 at 14:57

1 Answer 1


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.


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