Geodesic segments on the unit sphere

Question

Let $\mathbb{S}^2$ be the unit sphere and $d$ be the geodesic distance. For any three points $A,B,C\in \mathbb{S}^2$ and $0<\lambda<1$, let $A_{\lambda}$ and $B_{\lambda}$ be points on the geodesic paths $[A,C]$ and $[B,C]$ such that $d(A_{\lambda},C)=\lambda d(A,C)$ and $d(B_{\lambda},C)=\lambda d(B,C)$. Assume that $d(C,A)\le \pi/2, d(C,B)\le \pi/2$. It is geometrically clear that $d(A_{\lambda},B_{\lambda})\ge \lambda d(A,B)$. But proving it mathematically seems troublesome. Does anyone have a good proof or a direct reference?

Answer 1

For $\mathbb{S}^2$, since you assme $d(C,A)$ and $d(C,B) \leq \pi/2$, you can take the geodesic normal coordinate system centered at $C$ which covers at least the "northern hemisphere". Use $\exp$ to denote the exponential map at $C$.

By the definition of geodesic distance, $d(A,B) \leq L(\gamma)$ where $\gamma$ is any curve connecting $A,B$, and $L$ its length.

Now let $$\gamma = \exp \circ (\lambda^{-1}\times \exp^{-1}(\eta_\lambda))$$ where $\eta_\lambda$ is the geodesic connecting $A_\lambda$ and $B_\lambda$. By the definition of the exponential map, $\gamma$ connects $A,B$. So it suffices to show that

$$ L(\gamma) \leq \lambda^{-1} d(A_\lambda,B_\lambda) $$

This however follows from the fact that (by explicit computation) that the standard metric on the sphere, when pulled back by the exponential map, is "decreasing" in the radial direction. (More precisely, if $v$ be a vector in $\mathbb{R}^2 = T_p(T_C\mathbb{S}^2)$, let $w_p = d(\exp_C(p))\circ v$, we have that as $p$ moves radially outward $|w_p|$ is nonincreasing.) Hence by "rescaling" $\gamma_\lambda$ to $\gamma$ in the tangent space you have that the length increases less than the linear scaling factor. This concludes the proof.