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 {<a,b_1,...,b_n|[b_i,b_j], ab_ia^{-1}=b_1^{m_i1}... b_n^{m_in}>} 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.

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/
This entry was posted in Workshop lecture and tagged . Bookmark the permalink.

Leave a Reply

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

WordPress.com Logo

You are commenting using your WordPress.com 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 )

Google+ photo

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

Connecting to %s