## Notes of Tullia Dymarz’s lecture

Envelopes of certain solvable groups

Problem: Let ${\Gamma}$ be cocompact lattice in ${H}$ locally compact. Given ${\Gamma}$, what can we say about ${H}$ (up to compact groups) ?

Mosher-Sageev-White: If ${\Gamma}$ is free, ${H}$ must act on a tree.

Bader-Furman-Sauer: If ${\Gamma}$ is a lattice in a group ${H_0}$ in a wide class, ${H}$ must be ${H_0}$.

I focus on abelian by cyclic groups. Such groups admit presentations of the form ${}$ for some integer matrix ${M}$ with non vanishing determnant.

Theorem 1 1. If det(M)=1 and no unit eigenvalue,

or

2. all eigenvalues are ${>1}$ in absolute value,

then ${H}$ is a subgroup of ${Isom(X_M'^r)}$ for some rational ${r}$, where ${M'}$ is the absolute Jordan form of ${M}$.

Example 1 ${M=\begin{pmatrix} 2& 1\\ 1 & 1 \end{pmatrix}}$. Then ${\Gamma}$ is a lattice in ${Sol}$.

Example 2 ${M=(m)}$. Then ${\Gamma = BS(1,m)}$.

Outline : next,

– I describe ${X_{M}}$ and its isometry group (focussing on case 2).

– I outline the proof.

– I mention case 1 and case where ${M}$ has eigenvalues ${>1}$ and ${<1}$.

1. ${X_M}$

It is made of two pieces, a tree and a negatively curved Lie group. Let ${d=det(M)}$. Let ${T}$ be the regular tree with degree ${d+1}$, oriented so that there are 1 incoming and ${d}$ outgoing edges at each vertex. This defines a height function ${h_T}$. Let ${G=\mathbb{R}^n \times\mathbb{R}}$ be the semi-direct product where ${\mathbb{R}}$ acts on ${\mathbb{R}^n}$ by the matrices ${M^t}$. This comes with a height function ${h_G}$ as well. Let

$\displaystyle \begin{array}{rcl} X_M'=\{(x,y) \in G \times T \,;\, h_G(x)+h_T(y)=0\}, \end{array}$

with the path metric.

Example 3 : For ${BS(1,2)}$, ${X_M}$ is homeomorphic to ${T \times\mathbb{R}}$, in which ${\mathbb{R}}$ factors are horocycles in hyperbolic planes.

${X_M}$ is a hyperbolic metric space, whose ideal boundary is the one point compactification of ${\mathbb{Q}_d \times\mathbb{R}^n}$ where ${\mathbb{Q}_d}$ is the field of ${d}$-adic numbers, and ${\mathbb{R}^n}$ has the Rickmann rug metric (product of snowflaked lines). Every isometry of ${X_M}$ acts on boundary by a pair of similarities ${(f,g)}$, ${f}$ a similarity of ${\mathbb{Q}_d}$, ${g}$ a similarity of ${\mathbb{R}^n}$, with reciprocal scaling factors.

2. Proof sketch

If ${\Gamma}$ is cocompact in ${H}$, then ${H}$ becomes (up to compact groups) a group of uniform quasiisometries of ${X_M'}$.

Theorem 2 (Farb-Mosher)

$\displaystyle \begin{array}{rcl} \mathrm{QI}(X_M')=\mathrm{Bilip}(\mathbb{Q}_d) \times \mathrm{Bilip}(\mathbb{R}^n). \end{array}$

Theorem 3 (Mosher-Sageev-Whyte) Every uniform subgroup of ${\mathrm{Bilip}(\mathbb{Q}_d)}$ can be conjugated in ${\mathrm{Simil}(\mathbb{Q}_p)}$ for some ${p}$ depending on ${d}$.

Theorem 4 Every uniform subgroup of ${\mathrm{Bilip}(\mathbb{R}^n)}$ can be almost conjugated in ${\mathrm{Simil}(\mathbb{R}^n)}$. Furthermore, if ${M'}$ is scalar, I can remove the almost.

Thus ${H}$ can be conjugated into ${\mathrm{Simil}(\mathbb{Q}_d) \times \mathrm{Simil}(\mathbb{R}^n)}$. Using uniformity, one then clears up the scaling factors and conclude that ${H}$ is contained in Isom${(X_{M'^r})}$ for some rational ${r}$.

Note if ${M'}$ is not diagonal, a uniform biLipschitz cyclic group of ${\mathbb{R}^n}$ is has a triangular form involving non diagonal translations ; I call this almost-similarity.

3. Case 1: ${det(M)=1}$

If ${det(M)=1}$, ${X_{M'}}$ is a solvable Lie group. Its quasi-isometry group is understood again (Eskin-Fisher-Whyte).

However, the analysis of biLipschitz maps in the non diagonalisable case does not work any more.