** Envelopes of certain solvable groups **

**Problem**: Let be cocompact lattice in locally compact. Given , what can we say about (up to compact groups) ?

Mosher-Sageev-White: If is free, must act on a tree.

Bader-Furman-Sauer: If is a lattice in a group in a wide class, must be .

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

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

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or

2. all eigenvalues are in absolute value,

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then is a subgroup of for some rational , where is the absolute Jordan form of . *

**Example 1** * . Then is a lattice in . *

**Example 2** * . Then . *

Outline : next,

– I describe and its isometry group (focussing on case 2).

– I outline the proof.

– I mention case 1 and case where has eigenvalues and .

**1. **

It is made of two pieces, a tree and a negatively curved Lie group. Let . Let be the regular tree with degree , oriented so that there are 1 incoming and outgoing edges at each vertex. This defines a height function . Let be the semi-direct product where acts on by the matrices . This comes with a height function as well. Let

with the path metric.

**Example 3** * : For , is homeomorphic to , in which factors are horocycles in hyperbolic planes. *

is a hyperbolic metric space, whose ideal boundary is the one point compactification of where is the field of -adic numbers, and has the Rickmann rug metric (product of snowflaked lines). Every isometry of acts on boundary by a pair of similarities , a similarity of , a similarity of , with reciprocal scaling factors.

**2. Proof sketch **

If is cocompact in , then becomes (up to compact groups) a group of uniform quasiisometries of .

**Theorem 2 (Farb-Mosher)** *
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** *

**Theorem 3 (Mosher-Sageev-Whyte)** * Every uniform subgroup of can be conjugated in for some depending on . *

**Theorem 4** * Every uniform subgroup of can be almost conjugated in . Furthermore, if is scalar, I can remove the almost. *

Thus can be conjugated into . Using uniformity, one then clears up the scaling factors and conclude that is contained in Isom for some rational .

Note if is not diagonal, a uniform biLipschitz cyclic group of is has a triangular form involving non diagonal translations ; I call this almost-similarity.

**3. Case 1: **

If , 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.

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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/