Is every Nil manifold a nilmanifold?

Question

First of all this is an absolutely superlative answer it almost brought me to tears: https://math.stackexchange.com/a/3791368/758507

I am fairly certain that all compact 3 dimensional Nil manifolds (manifolds admitting Nil geometry) are nilmanifolds (admit a transitive action by a nilpotent Lie group). Every compact 3d nilmanifold (besides the torus which is abelian) is of the form $N_r:=H(3, \mathbb{R})/\Gamma_r$ where

$$H(3, \mathbb{R}) = \left\{\begin{bmatrix} 1 & x & z\\ 0 & 1 & y\\ 0 & 0 & 1\end{bmatrix} : x, y, z \in \mathbb{R}\right\}$$

is the three dimensional Heisenberg group, and

$$\Gamma_r = \left\{\begin{bmatrix} 1 & a & \frac{c}{r}\\ 0 & 1 & b\\ 0 & 0 & 1\end{bmatrix} : a, b, c \in \mathbb{Z}\right\}$$

where $ r $ is a nonzero integer. For a reference see theorem 5.4 bottom of page 25 in https://arxiv.org/abs/0903.2926 . I just want to confirm that these nilmanifolds $ N_r$ exhaust all compact 3d Nil manifolds. In particular I believe $ N_r $ is just the mapping torus of $ T^2 $ associated to $$ \begin{bmatrix} 1 & r \\ 0 & 1 \end{bmatrix} \in SL_2(\mathbb{Z}) $$ where again $ r $ is a nonzero integer. At this point it is worth mentioning that $ N_{\pm r} $ are diffeomorphic and moreover the mapping tori associated to the four matrices $$ \begin{bmatrix} 1 & \pm r \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ \pm r & 1 \end{bmatrix} $$ are all diffeomorphic.

If the $ N_r $ are all the compact Nil manifolds then by guess is correct. If there are other compact Nil manifolds then my guess is wrong. Again my question is:

Is every compact Nil manifold a nilmanifold?

Answer 1

No not every compact Nil manifold is a nilmanifold.

There are exactly two types of compact Nil manifolds. This can be checked from classifications based on Seifert fibrations.

The first type are circle bundles (in fact principal circle bundles) over the torus $ T^2 $. These are classified by their Euler number/ first Chern number $ r \neq 0 $. They are precisely the mapping tori of the maps (this map is just a Dehn twist repeated $ r $ times) $$ \begin{bmatrix} 1 & r \\ 0 & 1 \end{bmatrix} \in SL_2(\mathbb{Z}) $$ These are the nilmanifolds $ N_r \cong H(3,\mathbb{R})/\Gamma_r $ described in the question. These fully exhaust all compact 3d nilmanifolds, again see the question.

It turns out that this is only half the compact Nil manifolds. The other half are exactly the mapping tori of
$$ \begin{bmatrix} -1 & r \\ 0 & -1 \end{bmatrix} \in SL_2(\mathbb{Z}) $$ again $ r \neq 0 $. Since we have already exhausted all the compact 3d nilmanifolds above it is enough to show that this second list of mapping tori is distinct from the first list. One way to see that they are distinct is by appealing to the fact that every Nil manifold has a unique Seifert fibration and this second list corresponds to $$ \{b; (o_1, 0); (2, 1), (2, 1), (2, 1), (2, 1) \} $$ with $ b=-2 $ corresponding to the Euclidean mapping torus $ r =0 $ (the unit tangent bundle of the Klein bottle) and all the rest being distinct Nil manifolds. In particular, this second list is distinct from the first list which just has Siefert fibration type of actual circle bundles over the torus $ T^2 $.

For a proof without Seifert fibrations see Are these mapping tori different?

There are no other finite volume Nil manifolds since every finite volume Nil manifold is compact and orientable and for all geometries except $ H^3 $ and Sol the Seifert fiber spaces include all compact orientable examples.