## Antipodal points after Vîlcu

Posted in Math by pgadey on 2013/03/26

I’ve been thinking a lot about convex bodies in ${{\mathbb R}^3}$ lately. This post is going to be a write up of a useful lemma in the paper: Vîlcu, Constin, On Two Conjectures of Steinhaus, Geom. Dedicata 79 (2000), 267-275.

Let ${S}$ be a centrally symmetric convex body in ${{\mathbb R}^3}$. Let ${d(x,y)}$ denote thes intrinsic metric of ${S}$ and ${D = \sup_{x,y} d(x,y)}$ its intrinsic diameter. For a point ${x \in S}$ we write ${\bar{x}}$ for its image under the central symmetry.

Lemma 1 (Vîlcu) If ${d(x,y) = D}$ then ${y = \bar{x}}$.

This lemma says that if a pair realizes the inner diameter of a centrally symmetric convex body, the pair has to be centrally symmetric. This aligns well with our intuition about the sphere and cube, for example.

Proof: Let ${d(x,y) = D}$ and suppose, for contradiction, that ${y \neq \bar{x}}$. Pick some length minimizing geodesic ${\gamma_{x\bar{x}}}$ connecting ${x}$ to ${\bar{x}}$. Let ${\Gamma = \gamma_{x\bar{x}} \cup \bar{\gamma}_{x\bar{x}}}$ denote the concatenation of the paths ${\gamma_{x\bar{x}}}$ and ${\bar{\gamma}_{x\bar{x}}}$.

We check that ${\Gamma}$ is self-intersection free. Certainly ${\gamma_{x\bar{x}}}$ is self-intersection free, because it is a minimizing geodesic. Suppose that ${p \in \gamma_{x\bar{x}} \cap \bar{\gamma}_{x\bar{x}}}$. Then we have two minimizing geodesics from ${x}$ to ${\bar{x}}$ intesecting at a point on their interior. Hence they must coincide, a contradiction. Thus ${\Gamma}$ is self-intersection free and hence ${\Gamma}$ seperates ${S}$ into two open regions. Let ${S \setminus \Gamma = S_1 \cup S_2}$. We know ${S_2 = \bar{S}_1}$ since the central symmetry has to swap the components.

Suppose, for contradiction, that ${y \in \Gamma}$. Then ${d(x,y) \leq d(x,\bar{x})}$. If ${d(x,y) = d(x,\bar{x})}$ then ${y = \bar{x}}$, contradicting our hypothesis on ${y}$. If ${d(x,y) < d(x, \bar{x})}$ then we contradict the maximality of the diameter. Without loss of generality, take ${y \in S_1}$. We have ${\bar{y} \in \bar{S}_1}$. Take a minimizing geodesic ${\gamma_{y\bar{y}}}$ joining ${y}$ to ${\bar{y}}$. We have that ${\gamma_{y\bar{y}} \cap \Gamma \neq \emptyset}$ by the Jordan curve theorem. Take ${z \in \gamma_{y\bar{y}} \cap \Gamma}$. Suppose ${z \in \gamma_{x\bar{x}}}$.

The triangle inequality gives us:

$\displaystyle \begin{array}{rcl} d(x,y) & \leq & d(x,z) + d(z,y)\\ d(\bar{x},\bar{y}) & \leq & d(\bar{x},z) + d(z,\bar{y}) \end{array}$

Thus:

$\displaystyle d(x,y) + d(\bar{x},\bar{y}) \leq d(x,z) + d(z,\bar{x}) + d(y, z) + d(z, \bar{y}) = d(x,\bar{x}) + d(y,\bar{y})$

The maximality of the diameter and the equality ${d(x,y) = d(\bar{x},\bar{y})}$ then give:

$\displaystyle d(x,y) + d(\bar{x}, \bar{y}) = d(x,\bar{x}) + d(y,\bar{y})$

Suppose that ${d(x,y) < d(x,z) + d(z,y)}$ and ${d(\bar{x}, \bar{y}) < d(\bar{x},z) + d(z,\bar{y})}$. We then contradict the equality above. Thus, one of the two strict inequalities is an equality. Suppose without loss of generality that ${d(x,y) = d(x,z) + d(z,y)}$. Let ${\gamma_{xy}}$ denote a minimizing geodesic segment connecting ${x}$ to ${y}$.

We then have that ${\gamma_{x\bar{x}} \subset \gamma_{xy}}$, otherwise we’ll have two geodesics diverging from one another at a point. We then form ${\gamma'_{xy}}$ by removing ${\gamma_{x\bar{x}}}$ from ${\gamma_{xy}}$ and replacing it with ${\bar{\gamma}_{x\bar{x}}}$. We then have that ${\gamma_{xy}}$ and ${\gamma'_{xy}}$ diverge at a point, a contradiction. $\Box$

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