$\DeclareMathOperator\GL{GL}$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 ….


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 $(1,1)$.

  • $\begingroup$ Usually a fundamental domain is required to satisfy stricter conditions than your $P$. You only need $P$ bounded? $\endgroup$
    – LSpice
    Commented Jan 9, 2021 at 3:14
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    $\begingroup$ 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". $\endgroup$
    – Li Yutong
    Commented Jan 9, 2021 at 3:22
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    $\begingroup$ 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? $\endgroup$ Commented Jan 9, 2021 at 3:41
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    $\begingroup$ You are right! I should say "$\mathbb Z$-linear combinations of the elements in $G \cdot v$ is equal to $\mathbb Z^n$" $\endgroup$
    – Li Yutong
    Commented Jan 9, 2021 at 4:46
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    $\begingroup$ Then the answer is negative, you can take for instance the standard linear representation of the permutation group on $n$ symbols. $\endgroup$ Commented Jan 9, 2021 at 5:26

1 Answer 1


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 $U(\mathbb{R})$? "

The answer to the above question is Yes, as well known to students in geometry of numbers, reduction theory of quadratic forms, and arithmetic groups. For example, a proof is contained in Borel/Harish-Chandra, S 6.10. (See image below).

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 $[L':\mathbb{Z}^n]<+\infty$.

enter image description here

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    $\begingroup$ It is unclear to me that this is the intended question. But the question was rewritten so many times (and will be probably rewritten again) that it's hard to tell. $\endgroup$ Commented Jan 10, 2021 at 18:00

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