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Li Yutong
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$\DeclareMathOperator\GL{GL}$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 Q^n$$a \in \mathbb Z^n$ such that $V = a + (L \otimes_{\mathbb Z} \mathbb R)$), and $V$ is invariant under the action of $G$ (i.e. for any $v\in V, g\in G$, we have $g\cdot v \in V$). Moreover, there exists $v \in L$ (in fact, we can take $v=a$) such that $$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 $L:=(\mathbb Z,1) \subset \mathbb Z^2$$(0,1)+L:=(\mathbb Z,1) \subset \mathbb Z^2$, and $$G=\{\begin{pmatrix} 1&k\\0&1\end{pmatrix}\mid k\in\mathbb Z\}.$$ Then forFor $v=(0,1)^t$$v=(0,1)$, we have $G \cdot v =L$$G \cdot v =(0,1)+L$. In this case, we can take $P$ to be the interval from $(0,1)^t$$(0,1)$ to $(1,1)^t$$(1,1)$.

$\DeclareMathOperator\GL{GL}$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 $a \in \mathbb Q^n$ 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 $v \in L$ 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 $L:=(\mathbb Z,1) \subset \mathbb Z^2$, and $$G=\{\begin{pmatrix} 1&k\\0&1\end{pmatrix}\mid k\in\mathbb Z\}.$$ Then for $v=(0,1)^t$, $G \cdot v =L$. In this case, we can take $P$ to be the interval from $(0,1)^t$ to $(1,1)^t$.

$\DeclareMathOperator\GL{GL}$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 $(1,1)$.

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Li Yutong
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$\DeclareMathOperator\GL{GL}$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 $a \in \mathbb Q^n$ 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 $v \in L$ 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 $L:=(\mathbb Z,1) \subset \mathbb Z^2$, and $$G=\{\begin{pmatrix} 1&k\\0&1\end{pmatrix}\mid k\in\mathbb Z\}.$$ Then for $v=(0,1)^t$, $G \cdot v =L$. In this case, we can take $P$ to be the interval from $(0,1)^t$ to $(1,1)^t$.

$\DeclareMathOperator\GL{GL}$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 $a \in \mathbb Q^n$ 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 $v \in L$ 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 ….

$\DeclareMathOperator\GL{GL}$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 $a \in \mathbb Q^n$ 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 $v \in L$ 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 $L:=(\mathbb Z,1) \subset \mathbb Z^2$, and $$G=\{\begin{pmatrix} 1&k\\0&1\end{pmatrix}\mid k\in\mathbb Z\}.$$ Then for $v=(0,1)^t$, $G \cdot v =L$. In this case, we can take $P$ to be the interval from $(0,1)^t$ to $(1,1)^t$.

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

$\DeclareMathOperator\GL{GL}$Let $G$$G \subset M_{n\times n~}(\mathbb Z)$ be a finitely generated subgroup of $\GL(n,\mathbb Z)$$\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 Z)$$\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 $a \in \mathbb Q^n$ 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 $v \in L$ 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 ….

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

$\DeclareMathOperator\GL{GL}$Let $G$ be a finitely generated subgroup of $\GL(n,\mathbb Z)$. Then $G$ acts on $\mathbb R^n = \mathbb Z^n \otimes_{\mathbb Z} \mathbb R$ through $\GL(n,\mathbb Z)$.

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 $a \in \mathbb Q^n$ 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 $v \in L$ 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 ….

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 $a \in \mathbb Q^n$ 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 $v \in L$ 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 ….

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