**Does every three dimensional compact solvmanifold admit either Euclidean, nil, or sol geometry?**


definitions/motivation/background:


A solvmanifold is a manifold $ M $ admitting a transitive action by a solvable Lie group $ G $. In other words
$$
M \cong G/H
$$
where $ G $ is solvable and $ H $ is a closed subgroup of $ G $.


In dimension $ 2 $ there is strange coincidence where $ M $ is a solvmanifold if and only if it admits a transitive action by the Euclidean group $ E\_2 $ if and only if it is a flat manifold. This coincidence arises because $ SO\_2 $ is abelian so special Euclidean group $ SE\_2 $ is solvable.


Plenty of solvmanifolds are flat i.e. Euclidean. Some examples in dimension $ 3 $ include
the three torus $ T^3 $ (since abelian groups are solvable) as well as more exotic flat manifold like the one constructed from
$$
G=SE\_2= \{
\begin{bmatrix}
a & b & x \\
-b & a & y \\
0 & 0 & 1 
\end{bmatrix} : a,b,x,y \in \mathbb{R}, a^2+b^2=1 \} 
$$
and
$$
H=
\{
\begin{bmatrix}
-1 & 0 & n \\
0 & -1 & m \\
0 & 0 & 1 
\end{bmatrix} : n,m \in \mathbb{Z} \}
$$
the resulting compact flat 3 dimensional solvmanifold $ G/H $ 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). Indeed five of the six possible flat compact orientable 3-manifolds ( the sixth is addressed in the EDIT) can be constructed this way where $ H $ is the semidirect product of a lattice in $ \mathbb{Z}^2 $ with a finite (orders $ 1,2,3,4,6 $ respectively, this corresponds to the monodromy) cyclic subgroup of $ SL\_2(\mathbb{Z}) $ preserving that lattice.


Another geometry solvmanifolds admit is nil geometry (since every niloptent group is solvable), for example
$$
G= \{ 
\begin{bmatrix} 
1 & x & y \\
0 & 1 & z \\
0 & 0 & 1 
\end{bmatrix}
: x,y,z \in \mathbb{R} \}
\, , \,
H= \{ 
\begin{bmatrix} 
1 & k & n \\
0 & 1 & m \\
0 & 0 & 1 \\
\end{bmatrix}
: k,n,m \in \mathbb{Z} \}
$$
yielding $ G/H $ the Heisenberg nilmanifold.


And, finally, some solvmanifolds have sol geometry. For example, the quotient $ G/H $ for
$$
G=
\{
\begin{bmatrix}
a & 0 & x \\
0 & b & y \\
0 & 0 & 1 
\end{bmatrix} : ab=1 
\} 
$$
and
$$
H=
\{
\begin{bmatrix}
\beta^k & 0 & n+m \beta ​\\
0 & \beta^{-k} & n+m \beta^{-1} \\
0 & 0 & 1 
\end{bmatrix} 
: k,n,m \in \mathbb{Z} \}
$$
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 discrete subset of matrices forms a subgroup 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 <https://arxiv.org/abs/0903.2926> they construct the lattice $ H $ using $ d=-3 $ and $ \beta= \frac{3+\sqrt{5}}{2} $. This compact three dimensional solvmanifold with sol geometry has fundamental group $ \mathbb{Z}^2 \rtimes \mathbb{Z} $. Here 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} $.


EDIT: there was originally a second part to this question asking if every flat compact 3 manifold is a solvmanifold. The answer is no. The Hantzsche-Wendt Manifold, mentioned in the comment by HJRW, is a flat compact 3-manifold that does not admit a transitive action by any Lie group so in particular there is no transitive action by a solvable Lie group. See


<https://math.stackexchange.com/questions/4374758/transitive-lie-group-action-on-the-hantzsche-wendt-manifold>



3 dimensional solvmanifolds and Thurston geometries
Yes.


First observe that every compact 3d solvmanifold admits a Thurston geometry. To see why: Let $ M $ be a compact solvmanifold. Then by theorem of Auslander $ M $ is finitely covered by a compact special solvmanifold $ M' $ (a compact special solvmanifold is the quotient of a simply connected solvable Lie group $ G $ by a cocompact lattice). So $ G $ counts as a model geometry (with trivial stabilizer) for $ M' $. Since $ M' $ covers $ M $ then $ M $ also has model geometry $ G $.


A solvmanifold is aspherical and has solvable fundamental group. Since it is aspherical the geometry cannot be $ S^3 $ or $ S^2 \times E^1 $. Since it has solvable fundamental group it certainly has virtually solvable fundamental group and thus the geometry cannot be $ \tilde{SL\_2} $ or $ H^2 \times E^1 $ or $ H^3 $. Thus every three dimensional compact solvmanifold admits either $ E^3 $ Nil or Solv geometry.


