Lattices in amenable groups

I talk about lattices a lot on this blog, but almost always in the context of symmetric spaces because I find the theory so interesting. However, upon reflection there are certain natural things I have never done. First, I realise that I never actually say anything about what a lattice is in any of the posts, so I record this here. Second, it is a common theme in geometric group theory that amenable groups should be tractable somehow, but I have never really blogged about them. I'll do that in this post, but I still don't know what the best way to introduce amenable groups is, so I'll leave it to the reader to explore. The basics of the discrete side are covered in these lecture notes. Some of the definitions of amenability carry over to the non-discrete setting and many of the basic properties do as well. We will be following Gelander's lectures.

What's a lattice?

I think that the best examples of lattices are the fundamental groups of hyperbolic surfaces and closed and cusped hyperbolic 3-manifolds, living as a discrete subgroup inside of the isometry group of the universal cover. They act on the universal cover, and the quotient has finite volume. Sometimes it is compact, but in the cusped case it is not. There are much wilder groups and lattices out there, so unfortunately one must mumble some more words to be precise, but I think this conveys the feeling. Anyway, here are the words.

Let $G$ be a locally compact group equipped with a left Haar measure $\mu$, i.e. a Borel regular measure which is finite on compact sets, positive on open sets and invariant under left multiplications (i.e. $\mu(gA)=\mu(A)$ for every $g\in G$ and measurable $A\subset G$) --- by Haar's theorem such $\mu$ exists and is unique up to normalization. The group $G$ is called unimodular if $\mu$ is also right invariant, or equivalently if it is symmetric in the sense that $\mu (A)=\mu (A^{-1})$ for every measurable set $A$. Note that $G$ is compact iff $\mu(G)<\infty$. 

For example, discrete groups, abelian groups, compact groups and perfect groups (i.e. groups that are equal to their commutator subgroup) are unimodular. Below are two examples of groups failing to be unimodular that I found on MSE.  

Let $G$ be a Lie group. $G$ acts on itself by conjugation, fixing the identity, and thus $G$ acts on $T_e(G)$, the tangent space at the identity. Write $Ad(g)$ for the action of $g \in G$ on $T_e(G)$. The group is unimodular if and only if $\det(g)= \pm 1$ for all $g \in G$. On a Lie group, a measure is basically a volume form (up to some sign). (Normally we pull back differential forms, not push them forward. In what follows, if $h$ is an automorphism of a manifold $X$, $h_{\ast}$ means $(h^{-1})^{\ast}$;) If we choose a volume form $\omega$ on the tangent space to the identity (known as $\mathfrak{g}$), then we get a volume form on $T_g(G)$ by $(Lg)_{\ast} \omega$ where $Lg$ is left multiplication by $g$. The form on $G$ which is $(Lg)_{\ast} \omega$ on $g$ is left invariant,  so it must be the Haar measure. (Again, up to sign nuisances due to  measures versus volume forms.) Simultaneously, we get a right invariant volume form by $(Rg)_{\ast} \omega$. So we want to know whether $(Lg)_{\ast} \omega = (Rg)_{\ast} \omega$. Equivalent, we want $(Rg^{-1})_{\ast} (Lg)_{\ast} \omega = \omega$. The left hand side is precisely the action of conjugation by $g$ on volume forms at the identity. Explicitly, $(R g^{-1})_{\ast} (Lg)_{\ast} \omega = \det Ad(g) \omega$. Unwinding the issues about signs, we actually want $\det Ad(g) = \pm 1$.

A closed subgroup $H\le G$ is said to be co-finite if the quotient space $G/H$ admits a non-trivial finite $G$ invariant Borel regular measure. Like in Haar's theorem, it can be shown that a $G$ invariant measure on $G/H$, if exists, is unique up to scaling. 
A lattice in $G$ is a co-finite discrete subgroup. A discrete subgroup $\Gamma\le G$ is a lattice iff it admits a finite measure fundamental domain, i.e. a measurable set $\Omega$ of finite measure which forms a set of right coset representatives for $\Gamma$ in $G$ --- we will always normalize the measures so that  $\mathrm{vol}(G/\Gamma)=\mu(\Omega)$. Any two fundamental domains have the same measure, so a computation shows that if $G$ admits a lattice then it is unimodular. 

Non-unimodular examples 

Here are two examples of groups which aren't uni-modular

 Example: Let $G$ be the group $\left( \begin{smallmatrix} u & v \\ 0 & 1 \end{smallmatrix} \right)$, where $u$ is a nonzero real and $v$ is any real. Then
$$ \begin{pmatrix} u & v \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1+x & y \\ 0 & 1 \end{pmatrix} \begin{pmatrix} u & v \\ 0 & 1 \end{pmatrix}^{-1}=\begin{pmatrix} 1+x & -vx + uy \\ 0 & 1 \end{pmatrix}.$$
So $G$ acts on the tangent space by
$$ \begin{pmatrix} u & v \\ 0 & 1 \end{pmatrix} : \begin{pmatrix} x \\ y \end{pmatrix} \mapsto \begin{pmatrix} x  \\ -vx + uy \end{pmatrix}$$
which is described by the matrix
$$\begin{pmatrix} 1 & 0 \\ -v & u \end{pmatrix}.$$
Since the determinant is $u \neq 1$, the group is not unimodular.

Here is a way to make the above more concrete and geometric. Let $S$ be the square
$$\left\{ \begin{pmatrix} 1+x & y \\ 0 & 1 \end{pmatrix} : |x|, |y| \leq 1/2 \right\}$$
and let $p$ be an odd number. Then
$$\bigcup_{i=-(p-1)/2}^{(p-1)/2} \ \begin{pmatrix} \frac{1}{p} & \frac{i}{p} \\ 0 & 1 \end{pmatrix}\cdot  S \cdot  \begin{pmatrix} \vphantom{\frac{1}{p}} p & 0 \\ 0 & 1 \end{pmatrix} \ = \ \bigcup_{i=-(p-1)/2}^{(p-1)/2} \left\{ \left( x, \frac{y+i}{p} \right) : (x,y) \in S \right\} = S.$$ 


So, if we had a left and right invariant measure on $G$, we would have $\mu(S) = p \mu(S)$ and thus $\mu(S) = 0$ or $\infty$. But $S$ is compact and contains an open set, so it should have positive finite measure.

Example (automorphisms of locally finite graphs): Let $T$ be the infinite $r$-regular for $r \geq 3$ tree. Let $(v_0, v_1, v_2, \ldots)$ be some infinite non-backtracking path through it. Let $K$ be the group of all symmetries $g$ of $T$ so that $g(v_i)=v_i$. Let $G$ be the group of all symmetries $g$ of $T$ for which there exist $N \geq 0$ and $d \in \mathbb{Z}$ so that $g(v_i) = v_{i+d}$ for $i \geq N$, topologized in the usual way.

The next three items then show that $G$ isn't unimodular. Here are some proof sketches:

$K$ is compact: Let $U(i,d)$ be the set of all vertices of $T$ at distance $\leq d$ from $v_i$. Then $T = \bigcup_{i \geq 0, d \geq 0} U(i,d)$ and $K$ takes each $U(i,d)$ to itself, so $K$ embeds into the product of symmetric group $\prod_{i,d} S(U(i,d))$. With a little thought you can see that $K$ is closed in the product topology, so $K$ is compact. $\square$

$K$ is an open neighborhood of the identity in $G$: Define a continuous homomorphism $G \to \mathbb{Z}$ by $g \mapsto d$ where $g(v_i) = v_{i+d}$ for $d$ sufficiently large. Let $G_0$ be the kernel; since $\mathbb{Z}$ is discrete, $G_0$ is an open neighborhood of the identity in $G$. Now let $L$ be the set of all vertices $x$ of $T$ so that $x$ is at distance $i$ from $v_i$ for some $i$. We have a map $\phi: G_0 \to L$ sending $g$ to $g(v_0)$, and this is continuous for the discrete topology on $L$. So $\phi^{-1}(v_0)$ is an open neighborhood of the identity in $G_0$, and we claim $\phi^{-1}(v_0)=K$. By definition, if $g \in K$, then $g(v_0)=v_0$. Conversely, if $g \in G_0$ and $g(v_0)=v_0$, then $g$ fixes $(v_0, v_i, v_{i+1}, v_{i+2}, \ldots, )$ for some $i$. But then it also fixes all points on the path from $v_0$ to $v_i$, so $g$ is in $K$. $\square$

$G$ is not unimodular: Suppose there were a left and right invariant measure $\mu$. Since $K$ is open and compact, we would have $0 < \mu(K) < \infty$. Now, let $g \in G$ be some map with $g(v_i) = v_{i+1}$ for all $i$. If $\mu$ were left and right invariant, we would have $\mu(g K g^{-1}) = \mu(K)$. But it is easy to see that $g K g^{-1}$ is the set of symmetries of the tree which fix $(v_1, v_2, \ldots, )$. So $K$, being those symmetries which also fix $v_0$, is an index $r-1$ subgroup of $g K g^{-1}$, so $\mu(g K g^{-1}) = (r-1) \mu(K)$. 

I'm a little unsure what the takeaway from these examples should be. To my extremely naive eyes, it seems like what goes wrong is that the stretching and squashing is allowed to be arbitrarily extreme, which makes sense (in retrospect, at least). I suppose that even for non-uniform lattices the fundamental domain is quite nice (I have in mind a cusped hyperbolic manifold).

Soluble Lie groups

We first discuss soluble groups before saying a bit about general amenable groups. For soluble Lie groups Mostow proved the following classical result:

Mostow's theorem: Let $G$ be a connected soluble Lie group. Then every co-finite subgroup of $G$ is uniform.

While the first proofs of this were rather involved, there now exist much simpler proofs that don't involve any Lie theory. Let's see what machinery we need to give one of these simpler proofs.

Propositon 1: Let $G$ be a nilpotent locally compact group and $H \leq G$ a closed subgroup. Then $H$ has finite covolume if and only if $H$ is cocompact.

The ``if" part of Proposition 1 holds in the much greater generality of amenable groups. Indeed, if $G$ is amenable and $H\le G$ is a closed uniform subgroup, the compact $G$-space $G/H$ (as any compact $G$-space) admits a $G$ invariant probability measure. The ``only if" direction however, is more involved for non-nilpotent groups. In the proof of the proposition we will use a couple of facts:

Proposition 2: Every nilpotent locally compact group is unimodular.

Sketch: A locally compact group $G$ is unimodular if and only if $G/Z(G)$ is unimodular where $Z(G)$ is the center of $G$. Combined with all abelian groups being unimodular, an induction on nilpotency class shows the result.

Proposition 3: For any locally compact groups $H\le F\le G$, $H$ is co-finite in $G$ iff it is co-finite in $F$ and $F$ is co-finite in $G$.

I think this is pretty plausible so I won't subject the reader to the proof.

Proof of Proposition 1: It remains to prove the ``only if" direction. Let $H \leq  G$ be a co-finite subgroup,  let $Z = Z(G)$ be the centre of $G$ and let $F = \overline{Z\cdot H}$. By induction on the nilpotency degree, we infer that $F/Z$ is uniform in $G/Z$ and, hence, that $F$ is uniform in $G$. It is thus sufficient to prove that $H$ is uniform in $F$. The latter fact is clear since $H$ is normal in $F$ by definition. $\blacksquare$

We need a couple more results in the proof of Mostow's theorem
Proposition 4: (a): If $G$ is a connected Lie group and $\Gamma$ is a finitely generated (or more generally compactly generated) abelian co-finite lattice, then $\Gamma$ is uniform. 
(b): A connected soluble Lie group is Noetherian in the sense that every closed subgroup is compactly generated. 

The general case of (b)  can be deduced from the abelian case using induction on the solvability degree.

Proof of Mostow's Theorem: We shall prove the result for every compactly generated soluble Lie group (note that a connected Lie group is compactly generated).
Let $G$ be a compactly generated soluble Lie group and $H\le G$ a co-finite subgroup. Up to replacing $G$ and $H$ by finite index subgroups we may assume that the commutator $G'$ is nilpotent. (Indeed, by the Ado--Iwasawa theorem $G$ admits an almost faithful complex linear representation $\rho:G\to\mathrm{GL}_d(\mathbb{C})$, and after replacing $G$ by a finite index subgroup, the Zariski closure of the image is connected, in which case, by Lie's theorem the commutator is nilpotent.)   
We may argue by induction on the nilpotency degree of $G'$, where the base case when $G'$ is trivial follows from Proposition 1. Let $Z$ be the centre of $G'$. By induction $(G/Z)/(\overline{HZ}/Z)$ is compact, hence, we are left to show that $H$ is uniform in $E:=\overline{HZ}$, which is again compactly generated by Proposition 4(b). Clearly $F=H\cap E'$ is normal in $E$. Dividing by $F$ we are left to prove that $H/F$ is cocompact in $E/F$. Since $H/F$ is abelian, the result follows from Proposition 4(a). $\blacksquare$

Generalisations

This is, however, not the end of the story by any stretch. The analogue of Mostow's theorem holds whenever $G$ is a linear group over a local field. This property is important enough that it has been given a name. 

Definition: A locally compact amenable group $G$ has property $(M)$ if every closed subgroup of finite covolume in $G$ is cocompact.

Mostow's theorem that we discussed above can be restated as connected soluble Lie groups have property $(M)$. Bader-Caprace-Gelander-Mozes show that amenable linear locally compact groups and compactly generated nilpotent-by-nilpotent groups $G$ have property $(M)$ and point to connections with ergodic theory and spectral gap in their paper. I leave this to the interested reader to delve into. Instead, I would like to close by presenting an example of a metabelian locally compact group with a non-uniform lattice from their paper.

Let $p_n$ be a sequence of primes such that $\prod_{n=1}^\infty\frac{p_n}{p_n-1}<\infty$. Let $G$ be the compact by discrete metabelian group \[G=(\prod_{n=1}^\infty \mathbb{F}_{p_n}^*)\ltimes (\bigoplus_{n=1}^\infty\mathbb{F}_{p_n})\] and let $\Gamma$ be the set of sequences $(a_n,a_n-1)$ where $a_n\in\mathbb{F}_{p_n}$ and $a_n=1$ for all but finitely many $n$'s. This forms a discrete subgroup, and we briefly indicate how to show that $\Gamma$ is a non-uniform lattice in $G$.
 For $m\in\mathbb{N}$ let \[G_m:=(\prod_{n=1}^\infty \mathbb{F}_{p_n}^*)\ltimes (\bigoplus_{n=1}^m\mathbb{F}_{p_n})~~~\text{and}~~~ \Gamma_m=\Gamma\cap G_m.\] One computes that $[G_{m}:G_{m-1}]=p_m$ and $[\Gamma_{m}:\Gamma_{m-1}]=p_m-1$.

Hence, if we normalise the Haar measure on $G$ so that its compact subgroup $\prod_{n=1}^\infty \mathbb{F}_{p_n}^*$ has measure $1$ then \[\mathrm{vol} (G_m/\Gamma_m)=\frac{\mathrm{vol}(G_m)}{|\Gamma_m|}=\prod_{n=1}^m\frac{p_n}{p_n-1}.\]
 $\mathrm{vol} (G/\Gamma)=\prod_{n=1}^\infty\frac{p_n}{p_n-1}<\infty$ and hence $\Gamma$ is a lattice.
Finally, $G$ is a direct limit of the open compact subgroups $G_n$. The above sequence of covolumes doesn't stabilise, so none of the $G_N$ contains a fundamental domain for $\Gamma$ in $G$, hence $\Gamma$ is nonuniform.

This example shows that a locally compact group $G$ may fail to have $(M)$ even if it has a cocompact closed normal subgroup with property $(M)$, and understanding exhaustively the lattices in a cocompact normal subgroup of a locally compact group $G$ is not sufficient to understand all lattices in $G$. This points to subtleties when trying to understand property $(M)$, and gives context to show why the result on nilpotent-by-nilpotent groups is non-trivial.