Notes of Robert Young’s lecture

Filling functions and non-positive curvature

Continuation of Leuzinger’s talk. I will bring support to his conjecture and explain why it is so hard. And then describe some preliminary results.

1. {Sol_3} and {Sol_5}

{Sol_{2n-1}} is the semi-direct product of {\mathbb{R}^n} by {\mathbb{R}^{n-1}} acting by diagonal matrices of determinant 1.

{Sol_3} has exponential Dehn function whereas {Sol_5} has quadratic Dehn function.

{Sol_3} is a horosphere in {H^2 \times H^2}. In {H^2}, small disks filling a given closed curve are essentially unique. If I insist that the disk avoids a ball of radius {R}, this will have an exponentially large cost. Idem if one must avoid a horoball (note that the intersection of flat with a horoball is a diamond shaped compact set).

{Sol_5} is a horosphere in {H^2 \times H^2 \times H^2}. The intersection of a flat with a horoball is a octahedron. Now, a curve in the complement can be filled with a disk going around the octahedron, this may yield a quadratic 1-filling.

In general, we expect {k}-filling in {Sol_{2n-1}} to be easy for {k<n-1} but hard for {k=n-1}.

2. Non-uniform lattices

Drutu: {\mathbb{Q}}-rank 1 lattices have {\delta^k(L) < L^2+\epsilon}.

Young: {Sl(n,{\mathbb Z})} has quadratic Dehn function if {n>4}.

Bestvina-Eskin-Wortman: {S}-arithmetic lattices with {|S|>2} have polynomial Dehn functions.

Theorem 1 In {Sol_{2n-1}}, for {k<n-1},

\displaystyle  \begin{array}{rcl}  \delta^k(L) \sim L^{k+1}. \end{array}

2.1. I explain it for {n=3}

First show that a curve is covered by several flats.

Lemma 2 (Simple edges) Any two points are connected by a curve contained in a single flat. (avoid the octahedron).

Lemma 3 (Simple triangles) A curve in a finite union of flats can be filled with a disk in a larger finite union of flats.

Then, an arbitrary curve can be broken into triangles. Adding up areas yields {L^2}. That’s all for {Sol_5}.

2.2. In higher dimensions

It is harder to break down a sphere into simple spheres. Diameter could be arbitrarily large. In fact, Lipschitz spheres split nicely but to handle general cycles, we use Whitney decompositions (Lang and Schlichtenmaier).

We have a very general method which turns Lipschitz extension theorems into filling inequalities. The general setting is a metric space {X} with finite Assouad-Nagata dimension, and a subset {Z} in {X}. We assume a filling inequality in {X} and a Lipschitz extension property for {Z} (Lipschitz spheres extend to the disk with a controlled inscrease in Lipschitz constant). Finite Assouad-Nagata dimension allows to construct a covering of {X\setminus Z} by balls with radii comparable to distance to {Z} and bounded multiplicity (Whitney decomposition). Let {N} be the nerve of this covering. Given a cycle in {Z}, it spans a chain {C} in {X}. Approximate {C} by a cycle {C'} in a skeleton of {N} (à la Federer-Fleming). To project {C'} to {Z}, first project arbitrarily vertices of {C'} to nearest points on {Z}. Since edges in {N} have length proportional to their distance to {Z}, this projection is again Lipschitz. Apply Lipschitz extension property to get extension to 1-skeleton, and so on.

The Whitney decomposition trick was already used in the proof of the quadratic Dehn function for {Sl(5,{\mathbb Z})}.

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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 http://www.math.ens.fr/metric2011/
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