Notes of Richard Schwartz’ third Cambridge lecture 17-05-2017

Iterated barycentric subdivisions and steerable semi-groups

In two dimensions, there are many different affinely natural procedures on simplices: the barycentric subdivision (defined by coning and induction on dimension), yielding 6 triangles; the truncation of corners, yielding 4 triangles. The second is better behaved for numerics (triangles do not get thin, they stay similar to each other, in fact only 2 shapes are encountered).

Question. What shapes of simplices arise in an iterated subdivision scheme?

Diaconis and McMullen show that almost every triangle produced gets thin.

1. The space of shapes

I am aiming at a different question. Shapes form a topological space. Let {X^n} be the space of labelled {n}-simplices mod scaling. Say we normalize so that volume stays equal to 1. Then {X^n} is a principal homogeneous space of {Sl(n,{\mathbb R})}. An affinely natural subdivision process amounts to a subdivision of the standard simplex.

Question. Does the iteration scheme produce a dense set of shapes?

The answer is positive in 2 dimensions (Barany-Beardon-Carne 1990). I gave a positive answer in 3 and 4 dimensions, I will explain my solution. My guess is that answer should be negative for {n} large enough.

The question is equivalent to the density of the semi-group generated by the matrices that map the standard simplex to each of the labelled tiles.

Lemma: the subgroup {S} generated by the subdivision is either discrete or dense.

The proof is not very enlightening. It follows that the subgroup {\langle S\rangle} generated by the subdivision is dense.

Lemma: If {S} contains a bounded infinite walk, then the semi-group {S} is dense in {\langle S\rangle}. Indeed, a long bounded walk contains segments {g_1\cdots g_k} which are close to 1, hence an expression of inverses as elements of the semi-group {S: g_1^{-1}=g_2 \dots g_n} up to small error.

1.1. {n=2}

If you label right, then all generators are infinite order elliptics.

1.2. {n=3}

A computer search revals some infinite order elliptic elements. They represent a positive fraction of words of length 3.

Experiments suggest existence of an infinite walk up to {n=10}.

2. Proof

Look at horoballs in {X}. Fix an origin {o\in X}.

Definition 1 A horoball is special if it contains the origin {o} in its boundary.

A set {W\subset X} is a strong wheel if every special horoball contains a point of {W} in its interior.

A semi-group {S} is steerable if the orbit {So} contains a steering wheel.

Lemma: If {S} is steerable, then {S} has a bounded infinite walk.

The idea is that if a walk goes far away from {o}, use a special horoball containing it in its boundary, pick a point there, start again.

Therefore, the point is to exhibit a steering wheel.

Lemma: Consider the Hadamard map {(X,o)\rightarrow H^N}, {N=\frac{n(n+1)}{2}-1} (geodesic polar coordinates) preserving min sectional curvature. This is distance non-decreasing. This allows to carry the problem to 9-dimensional hyperbolic space.

The wheel is a finite set of 144 points on the boundary of some ball {B} centered at {o}. To certify that they form a steering wheel, apply the following criterion: each of them points towards the center of a special horoball. If the convex hull of these centers contains {B}, then the horoballs cover {\partial B}, hence the wheel.

3. Questions

What makes you think the answer should be negative in high dimensions? Because {n!} is too small a number of horospheres to cover a sphere of dimension {N=\frac{n(n+1)}{2}-1}. Experiments indicate that this phenomenon could start at {n=9}.

The semi-group would be uniformly discrete?


About metric2011

metric2011 is a program of Centre Emile Borel, an activity of Institut Henri Poincaré, 11 rue Pierre et Marie Curie, 75005 Paris, France. See
This entry was posted in Course and tagged . Bookmark the permalink.

Leave a Reply

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

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

Google+ photo

You are commenting using your Google+ 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 )


Connecting to %s