Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
If you are not afraid to use high powered machinery, this probably follows from one of the comparison theorems in Riemannian geometry, using that $\mathbb{S}^2$ has non-negative curvature. But you are right that there probably is a simpler proof in this special case. –  Willie Wong Nov 15 '11 at 13:49
    
It would have made sense to link to this related question of yours. –  joriki Nov 15 '11 at 14:05

1 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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.