Bianchi classification of solvable Lie groups and cocompact subgroups

Question

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?

Answer 1

The group $6_0$ is the isometry group of solv geometry, which is the geometry of $\mathbb R^3$ equipped with the Riemannian metric $$e^{2t} dx^2 + e^{-2t} dy^2 + dt^2 $$ The lattices in $6_0$ correspond to those compact Riemannian 3-manifolds whose universal cover is isometric to solv geometry. These 3-manifolds are the mapping tori of Anosov homeomorphisms of the torus $T^2$, described as follows. Start with a matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL_2(\mathbb Z)$ with $\left|\text{trace}(A)\right| = \left|a+d\right| > 2$, and so $A$ has an independent pair of eigenvectors with eigenvalues $\lambda, \frac{1}{\lambda}$ with $\left|\lambda\right| > 1$. The matrix $A$ preserves the lattice $\mathbb Z \oplus \mathbb Z$ and so its action on $\mathbb R^2$ descends to an Anosov diffeomorphism $f : T^2 \to T^2$. Using the mapping torus construction one obtains the desired 3-manifold $M = T^2 \times [0,1] / (x,1) \sim (f(x),0)$. The fact that the eigenvalues of $M$ form a multiplicative inverse pair allows you to diagonalize $A$ to the form $\begin{pmatrix} \lambda & 0 \\ 0 & 1/\lambda\end{pmatrix}$ and using that to construct a deck transformation invariant solv-geometry metric on the universal cover of $M$.

The group $7_0$ is actually a subgroup of the full isometry group of Euclidean geometry $E^3$. The lattices in that full isometry group are the famous 3-dimensional crystallographic groups, also known as the space groups, and their number is known (look up that link to see). When you add the constraint of being a torsion free subgroup of the Lie subgroup $7_0$ (torsion free is needed for the quotient of $\mathbb R^3$ to be a 3-manifold), then the number of possibilities is cut wayyyy down. These 3-manifolds can in fact be described almost exactly as above, i.e. as certain mapping toruses of $T^2$ diffeomorphisms, with the difference that one requires $\left|\text{trace}(A)\right| = \left|a+d\right| < 1$. The 3-manifolds that you obtain are exactly the mapping toruses of (orientation preserving) finite order diffeomorphisms $f : T^2 \to T^2$. Up to isometry I count 5 of these: the identity (whose mapping torus is $T^3$); an order 2 rotation with 4 fixed points, and an order 4 rotation with 2 fixed points and 2 points of order 2 (picture rotations of a square with opposite sides glued); an order 3 rotation with 3 fixed points, and an order 6 rotation with 1 fixed point, 2 points of order 2, and 3 points of order 3 (picture rotations of a regular hexagon with opposite sides glued).

Answer 2

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} $.