## 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 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,

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

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 ${Sl(5,{\mathbb Z})}$.