Parker Glynn-Adey

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})

antipode1antipode2

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

Tagged with: , ,

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: