From the notes Coarse differentiation and the geometry of polycyclic groups, I found a theorem

$\Gamma$ is polycyclic iff $\Gamma$ is a lattice in a solvable unimodular lie group $G$ - Mostow

Then I started looking up definitions:

  • What is an example of $\Gamma$ and $G$ ?

It is known that polycyclic groups are amenable. How does one write down the averaging procedure?

My knowledge of infinite group is very limited:

  • Wikipedia says amenable groups have a $G$-invariant averaging procedure
    • for finite $G$ this leads to the character theory of finite groups
    • for $\mathbb{Z}$ the averaging procedure is pretty clear, such as $\frac{1}{N} \sum_{i=1}^N$
    • Wikipedia lists solvable groups as example of amenable groups
    • In search of an example of an infinite solvable group, I found that polycyclic and supersolvable groups are solvable.

These definitions are too abstract for me to follow-- I would like a "concrete" example of a polycyclic group.

It may not be so obvious abelian groups are amenable after all...

I am interested since I read a blog about amenable groups and I needed an example to try out.

  • $\begingroup$ 1) A polycyclic group is is a solvable unimodular Lie group, and is a lattice in itself. Do you mean connected or with finitely many components? 2) I can't guess the meaning of the sentence including "I will just say". 3) $SL_n(\mathbf{Z})$ is not polycyclic. $\endgroup$ – YCor Jun 21 '14 at 18:54
  • $\begingroup$ @YCor I can't find definitions or example of the notions of the theorem. I would like to see an explicit example of a "solvable unimodular lie group" $G$ and a lattice $\Gamma$ in that group. $\endgroup$ – cactus314 Jun 21 '14 at 19:01
  • $\begingroup$ @YCor, re: your point (1): is $\mathbb{Z}/2\mathbb{Z}$ (which is certainly polycyclic) a Lie group? The definition I'm familiar with would say no, but maybe I'm missing something. $\endgroup$ – Noah S Jun 21 '14 at 19:26
  • $\begingroup$ @johnmangual, definitions of "polycyclic," "lattice," and "solvable unimodular Lie group" are easily found online; as for examples, Corollary 1 of the paper math.lsu.edu/~pdani/conferences/goa2010/SpeakerNotes/… states that any locally compact topological group which admits a lattice is unimodular, so that should help you get lots of examples. $\endgroup$ – Noah S Jun 21 '14 at 19:31
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    $\begingroup$ Is $U(n)$ solvable? $\endgroup$ – Noah S Jun 21 '14 at 20:32

Here is an example of a solvable unimodular (connected) Lie group $G$, and a lattice $\Gamma$ in that group:

  • $G$ is the semidirect product $\mathbb{R}^2 \rtimes \mathbb{R}$ with respect to the action of $\mathbb{R}$ on $\mathbb{R}^2$ defined by $$t \mapsto \pmatrix{e^t & 0 \\ 0 & e^{-t}} $$ This Lie group is isomorphic (as a Lie group) to any semidirect product with respect to an action defined by $$t \mapsto M^t $$ for which the generating matrix $M \in SL(2,\mathbb{R})$ has distinct real eigenvalues $\lambda>1$ and $\lambda^{-1} \in (0,1)$. For example one could use $$M = \pmatrix{2 & 1 \\ 1 & 1} $$ Of course to define $M^t$ one first picks a matrix $P$ in the Lie algebra $sl(2,\mathbb{R})$ such that $M=\exp(P)$, and then $M^t = \exp(tP)$.
  • $\Gamma$ is the semidirect product $\mathbb{Z}^2 \rtimes \mathbb{Z}$ with respect to the action of $\mathbb{Z}$ on $\mathbb{Z}^2$ defined using the above matrix $M$.

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