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Edit: Thoughts updated (22/3/2021).

I've come across with the following problem.

Let $G=\mathbb{R}^s \ltimes_\varphi \mathbb{R}^k$ where $\varphi:\mathbb{R}^s\to \mathrm{Aut}(\mathbb{R}^k)=\mathsf{GL}(k,\mathbb{R})$ is an homomorphism such that $\mathbb{R}^k=N$ the nilradical of $G$ (i.e. $N$ is the maximal connected nilpotent normal subgroup of $G$ so that $\mathfrak{n}$ is the maximal nilpotent ideal of $\mathfrak{g}$).

If we take $\Gamma_1$ a lattice (discrete and cocompact subgroup) in $\mathbb{R}^s$ and $\Gamma_2$ a lattice in $\mathbb{R}^k$ such that $\varphi(\Gamma_1)$ preserves $\Gamma_2$ then $\Gamma=\Gamma_1\ltimes_\varphi \Gamma_2$ is a lattice of $G$. Let us call $\Gamma$ a splittable lattice.

Question: Are all the lattices in $G=\mathbb{R}^s\ltimes_\varphi\mathbb{R}^k$ splittable lattices?

Thoughts: By Mostow's theorem on solvmanifolds, we have that $\Gamma\cap\mathbb{R^k}$ is lattice in $\mathbb{R}^k$ and $\Gamma/\Gamma\cap\mathbb{R}^k$ is a lattice in $G/\mathbb{R}^k\cong\mathbb{R}^s$.

Then we have the following exact short sequences:

$0\to N\to G \to G/N\to 0$ and

$0\to \Gamma\cap N \to \Gamma \to \Gamma/\Gamma\cap N\to 0$.

The above sequence splits and we have the inclusion of the groups below in the groups above. But I don't know if it possible to conclude that the sequence below also splits. If it helps, the group $\Gamma$ is a polycyclic group.

Another important hypothesis is that the group $\Gamma$ is isomorphic to a Bieberbach group (so it has a finite index abelian subgroup $\Lambda$, according to First Bieberbach's Theorem)

Thanks!

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    $\begingroup$ Note that $\Gamma/(\Gamma \cap N)$ is a free abelian group of rank $s$, so the only question is whether there exist $s$ commuting elements of $\Gamma$ lifting a set of generators (e.g. for $s \leq 1$ this is automatic). $\endgroup$
    – R. van Dobben de Bruyn
    Commented Mar 2, 2021 at 2:07
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    $\begingroup$ I don't understand your claim that Gamma has a finite index abelian subgroup. This is does not seem true at all since lattices are Zariski dense. $\endgroup$
    – Nicolast
    Commented Mar 22, 2021 at 20:59
  • $\begingroup$ Every (group isomorphic to a) Bieberbach group of $\mathrm{Iso}({R^k})$ has a maximal abelian normal subgroup of finite index (this is the First Bieberbach Theorem) $\endgroup$
    – Alejandro Tolcachier
    Commented Mar 23, 2021 at 12:29
  • $\begingroup$ Why is Gamma isomorphic to a Bieberbach group? Typically such groups, which are torsion-free and virtually abelian, occur inside semidirect products of the form ${\mathbb R}^d \rtimes O(d)$, and they don't have to be solvable $\endgroup$
    – Yemon Choi
    Commented Mar 23, 2021 at 13:04
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    $\begingroup$ Sorry, I will try to clarify a little. The action $\varphi$ is (in my case) multiplying by skew-symmetric matrices. For example Taking $s=1$ and $k=2$, $\varphi(t)=\begin{pmatrix} \cos(t) & -\sin(t) \\ \sin(t) & \cos(t) \end{pmatrix}$, and a lattice $\Gamma$ is $\frac{2\pi}{3} \mathbb{Z}\ltimes_{\varphi} P\mathbb{Z}^2$ where $P^{-1} \varphi(\frac{2\pi}{3}) P$ is an integer matrix. The subgroup $\Lambda=2\pi \mathbb{Z}\ltimes_{\varphi} P\mathbb{Z}^2$ is abelian and $\Gamma/\Lambda=\mathbb{Z}_3$ $\endgroup$
    – Alejandro Tolcachier
    Commented Mar 23, 2021 at 16:30

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