# When does a subgroup of $\operatorname{GL}(n, \mathbb Q)$ have a bounded fundamental domain on $\mathbb R^n$?

$$\DeclareMathOperator\GL{GL}$$Let $$G \subset M_{n\times n~}(\mathbb Z)$$ be a finitely generated subgroup of $$\GL(n,\mathbb Q)$$ (i.e. $$g\in G$$ is an invertible matrix with entries in $$\mathbb Z$$). Then $$G$$ acts on $$\mathbb R^n = \mathbb Z^n \otimes_{\mathbb Z} \mathbb R$$ through $$\GL(n,\mathbb Q)$$.

Suppose that there is a rational affine subspace $$V \subset \mathbb R^n$$ (by this, I mean that there is a sub-lattice $$L \subset \mathbb Z^n$$ and $$a \in \mathbb Z^n$$ such that $$V = a + (L \otimes_{\mathbb Z} \mathbb R)$$), and $$V$$ is invariant under the action of $$G$$ (i.e. for any $$v\in V, g\in G$$, we have $$g\cdot v \in V$$). Moreover, there exists $$v \in L$$ (in fact, we can take $$v=a$$) such that $$G \cdot v = L.$$

Question: is there a bounded subset $$P \subset V$$ such that $$\bigcup_{g \in G}\ g\cdot P = V \quad ?$$

Any suggestion on relevant questions/references is very welcome! Particularly, I don't know which field studies such problems ….

Edit:

Example. Consider $$(0,1)+L:=(\mathbb Z,1) \subset \mathbb Z^2$$, and $$G=\{\begin{pmatrix} 1&k\\0&1\end{pmatrix}\mid k\in\mathbb Z\}.$$ For $$v=(0,1)$$, we have $$G \cdot v =(0,1)+L$$. In this case, we can take $$P$$ to be the interval from $$(0,1)$$ to $$(1,1)$$.

• Usually a fundamental domain is required to satisfy stricter conditions than your $P$. You only need $P$ bounded? Jan 9, 2021 at 3:14
• Yes, for me "$P$ bounded" is enough. I don't know how to characterize such set in one word in the title, so I coin the name "fundamental domain". Jan 9, 2021 at 3:22
• The hypothesis cannot be satisfied by any subgroup of ${\rm GL}_n({\bf Z})$, finitely generated or not, because the action of ${\rm GL}_n({\bf Z})$ preserves the subgroups $M {\bf Z}^n$ of ${\bf Z}^n$, so for example cannot take a nonzero vector $v$ to $2v$, let alone to the zero vector. Did you mean to require that $G \cdot v$ consist of all primitive integer vectors? Jan 9, 2021 at 3:41
• You are right! I should say "$\mathbb Z$-linear combinations of the elements in $G \cdot v$ is equal to $\mathbb Z^n$" Jan 9, 2021 at 4:46
• Then the answer is negative, you can take for instance the standard linear representation of the permutation group on $n$ symbols. Jan 9, 2021 at 5:26

After reading the comments, I think the underlying algebraic question is: " if $$G=U$$ is a unipotent linear algebraic group defined over $$\mathbb{Q}$$, then is the arithmetic unipotent subgroup $$U(\mathbb{Z})$$ cocompact in $$U(\mathbb{R})$$? "
E.g. the integral Heisenberg group $$H(\mathbb{Z})$$ is cocompact in $$H(\mathbb{R})$$.
If the OP does not study the full arithmetic unipotent $$U(\mathbb{Z})$$, but restricts to a subgroup $$G \subset U(\mathbb{Z})$$, then the answer to the above question is No unless $$G$$ is finite index in $$U(\mathbb{Z})$$. Indeed it's evident that discrete subgroups $$L'$$ of $$\mathbb{Z}^n$$ are cocompact in $$\mathbb{R}^n$$ if and only if $$[L':\mathbb{Z}^n]<+\infty$$. 