Bianchi classification of solvable Lie groups and cocompact subgroups Recall that abelian is contained in nilpotent is contained in solvable.
There is a unique 3 dimensional connected simply connected abelian Lie group $ \mathbb{R}^3 $.
And there is a unique connected simply connected nonabelian nilpotent 3 dimensional Lie group, the Heisenberg group
$$
Nil=\{ \begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix} : a,b,c \in \mathbb{R} \} 
$$
However there are lots of (simply connected, connected, 3 dimensional) solvable non nilpotent groups.
The two solvable (non nilpotent) groups I can think of are, first,
$$
{SE}_2= \{
\begin{bmatrix}
a & b & x \\
-b & a & y \\
0 & 0 & 1 
\end{bmatrix} : a^2+b^2=1 \} 
$$
which is the connected component of the identity in the Euclidean group $ E_2= \mathbb{R}^2 \rtimes O_2 $. And, second, the isometry group of the Minkowski plane $ E_{1,1}= \mathbb{R}^2 \rtimes O_{1,1} $
$$
E_{1,1}= \{
\begin{bmatrix}
a & b & x \\
-b & a & y \\
0 & 0 & 1 
\end{bmatrix} : a^2-b^2=1 \} 
$$
which is isomorphic to
$$
 \{
\begin{bmatrix}
a & 0 & x \\
0 & b & y \\
0 & 0 & 1 
\end{bmatrix} : ab=1 \} 
$$
For solvable non nilpotent Lie groups the Bianchi classification https://en.wikipedia.org/wiki/Bianchi_classification lists 6 distinct types $ 3,4,5,6,6_0,7_0 $ as well as an infinite family of of distinct groups, type $ 7 $.
Which of these groups $ 3,4,5,6,6_0,7_0,7 $ admit cocompact discrete subgroups? A group admitting a cocompact discrete subgroup must be unimodular (see for example https://arxiv.org/pdf/0903.2926.pdf) and of these solvable groups only $ 6_0 $ (corresponding to $ E_{1,1} $) and $ 7_0 $ (corresponding to $ E_2 $) are unimodular and thus possibly contain cocompact discrete subgroups.
For both unimodular groups $ E_2 $ and $ E_{1,1} $ do there exist cocompact discrete subgroups?
If not why not? If so what is an example of a cocompact discrete subgroup?
And which of these groups is the Sol geometry for 3 manifolds based on?

Great answer from Lee Mosher (I already accepted it). Just wanted to provide some extra information with some examples from this source https://arxiv.org/abs/0903.2926 for the sake of completeness.
As Lee Mosher notes, type $ 7_0 $ is $ E_2 $ which is contained in $ E_3 $ so these 3d compact manifolds have Euclidean geometry, just like the 3 torus. However topologically they can be quite different from the 3 torus. For example take
$$
{SE}_2= \{
\begin{bmatrix}
a & b & x \\
-b & a & y \\
0 & 0 & 1 
\end{bmatrix} : a^2+b^2=1 \} 
$$
and mod out by
$$
\mathbb{Z}^2 \cong
\{
\begin{bmatrix}
1 & 0 & n \\
0 & 1 & m \\
0 & 0 & 1 
\end{bmatrix} : n,m \in \mathbb{Z} \}
$$
the resulting compact homogeneous Euclidean 3 manifold has fundamental group $ \mathbb{Z}^2 \rtimes \mathbb{Z} $ where the semi direct product is with respect to
$$ 
n \mapsto  \begin{bmatrix} -1 & 0 \\ 0 & -1  \end{bmatrix}^n
$$
and the abelianization is $ H_1 \cong \mathbb{Z} $ ( thus this manifold is certainly not the 3 torus).
Type $ 6_0 $, the group on which the Sol geometry is based, is the isometry group of the Minkowski plane $ E_{1,1}= \mathbb{R}^2 \rtimes O_{1,1} $ with unimodular subgroup
$$
SE_{1,1}= \{
\begin{bmatrix}
a & b & x \\
b & a & y \\
0 & 0 & 1 
\end{bmatrix} : a^2-b^2=1 \} 
$$
which is isomorphic to
$$
\{
\begin{bmatrix}
a & 0 & x \\
0 & b & y \\
0 & 0 & 1 
\end{bmatrix} : ab=1 
\} 
$$
which contains the following (cocompact) lattice
$$
H=
\{
\begin{bmatrix}
\beta^k & 0 & n+m \beta ​\\
0 & \beta^{-k} & n+m \beta^{-1} \\
0 & 0 & 1 
\end{bmatrix} 
: k,n,m \in \mathbb{Z} \}
$$
where $ \beta $ is the root of $ x^2+3x+1 $ (or $ x^2+dx+1 $ for any integer $ d $ such that the roots of the polynomial are real and not integers ( $ |d| \geq 3 $ )). The fact that this forms a group follows from the fact that the unimodular companion matrix for the polynomial $ x^2+dx+1 $ satisfies the following matrix equation
$$
\begin{bmatrix} \beta & 0 \\ 0 & \beta^{-1} \end{bmatrix}
\begin{bmatrix} 1 & \beta  \\ 1 & \beta^{-1} ​\end{bmatrix}
=\begin{bmatrix} 1 & \beta  \\ 1 & \beta^{-1} ​\end{bmatrix}
\begin{bmatrix} 0 & -1  \\ 1 & -d ​\end{bmatrix}
 $$
Note that in the reference they construct a lattice using $ d=-3 $ and $ \beta= \frac{3+\sqrt{5}}{2} $. The compact coset manifold $ SE_{1,1}/H $ has fundamental group $ \mathbb{Z}^2 \rtimes \mathbb{Z} $ but the semidirect product is with respect to the map
$$
n \mapsto \begin{bmatrix} 0 & -1  \\ 1 & -d ​\end{bmatrix}^n
$$
and the abelianization is $ H_1 \cong \mathbb{Z} $.
