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Let $G_n$ be the Lie group consisting of $n \times n$ upper triangular matrices of determinant $1$ with real entries. In other words, $$G_n = \{\text{$\left(\begin{matrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n} \\ 0 & a_{22} & a_{23} & \cdots & a_{2n} \\ 0 & 0 & a_{33} & \cdots & a_{3n} \\ \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 & 0 & \cdots & a_{nn} \end{matrix}\right)$ $|$ $a_{ij} \in \mathbb{R}$ and $a_{11} \cdots a_{nn} = 1$}\}.$$ Does $G_n$ contain a lattice (i.e. a discrete subgroup of finite covolume)? The obvious thing to try is to take the subgroup consisting of matrices with integer entries, but that does not work (though it would work if we were working with strictly upper triangular matrices).

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If a locally compact group $G$ contains a lattice, then $G$ has to be unimodular and you $G_n$ is not. – Misha Dec 5 '12 at 4:41
@Misha : Can you give me a reference for the fact that locally compact groups that contain lattices have to be unimodular? – Edward Cooper Dec 5 '12 at 4:50
See e.g. Corollary 1 on page 3 of Venkataramana's notes on lattices:… – Misha Dec 5 '12 at 5:08
To Misha's remarks I'd add the comment that Example 3 (on page 2 of the lecture notes he links from Tata) illustrates for $n=2$ why this type of solvable group isn't unimodular. I'm not sure how to sort out those solvable Lie groups which are unimodular, but an old theorem of Mostow shows that any lattice must then be arithmetic. (Venkatarama gives a few of the standard references.) – Jim Humphreys Dec 5 '12 at 17:05
The proof that the group of upper triangular matrices $U_n$ is not unimodular could be found for instance in Bump's book "Automorphic forms and representations", page 426. – Misha Dec 5 '12 at 22:25
up vote 14 down vote accepted

@Edward, here is a short proof which I wrote awhile ago for my notes on group theory.

Lemma. If a 2nd countable locally compact group $G$ contains a lattice $\Gamma$ then $G$ is unimodular.

Proof. For arbitrary $g \in G$ consider the push-forward $\nu=R_g(\mu)$ of the (left) Haar measure $\mu$ on $G$; here $R_g$ is the right multiplication by $g$: $$ \nu(E)= \mu(Eg). $$ Then $\nu$ is also a left Haar measure on $G$. By the uniqueness of Haar measure, $\nu= c\mu$ for some constant $c > 0$. The lattice $\Gamma \le G$ has a fundamental domain $D\subset G$, i.e., a measurable subset of $G$ such that $$ \bigcup_{\gamma\in \Gamma} \gamma D= G, \quad \mu(\gamma D\cap D)=0, \quad \forall \gamma\in \Gamma \setminus 1. $$ (Proof of existence of such domain uses 2nd countable assumption.) In particular, $0<\mu(D)<\infty$, since $\Gamma$ is a lattice. Then $Dg$ is again a fundamental domain for $\Gamma$ and, thus, $\mu(D)=\mu(Dg)$. Hence, $\mu(D) = \mu(Dg) = c\nu(D)$. It follows that $c=1$. Thus, $\mu$ is also a right Haar measure. qed

Another proof could be found in Chapter I of Raghunathan's book "Discrete subgroups of Lie groups".

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A reference is also in Deitmar-Echterhoff "Principles of Harmonic Analysis" Chapter 9 somewhere, not sure though if they refer only to uniform lattices. Nice proof +1. – Marc Palm Dec 5 '12 at 9:37
Thanks for this! – Edward Cooper Dec 5 '12 at 16:21
It's not clear why two fundamental domains have the same measure. – Guntram Dec 5 '12 at 18:21
@Guntram: Because the measure of a fundamental domain is the same as the measure of the quotient $\Gamma\backslash G$. – Misha Dec 5 '12 at 20:47
If $G$ is locally compact and abelian, and $D$ is a lattice in $G$, do you know if one can prove the existence of a measurable FD without the second countability assumption?.. Should I ask this as a question, or is it OK if I ask here? – Nick S Feb 15 '14 at 20:48

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