# co.combinatorics – When does a subgroup of \$operatorname{GL}(n, mathbb Q)\$ have a bounded basic province on \$mathbb R^n\$? Answer

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co.combinatorics – When does a subgroup of \$operatorname{GL}(n, mathbb Q)\$ have a bounded basic province on \$mathbb R^n\$?

$$DeclareMathOperatorGL{GL}$$Let $$G subset M_{ntimes n~}(mathbb Z)$$ breathe a finitely generated subgroup of $$GL(n,mathbb Q)$$ (i.e. $$gin G$$ is an invertible matrix with entries in $$mathbb Z$$). Then $$G$$ acts on $$mathbb R^n = mathbb Z^n otimes_{mathbb Z} mathbb R$$ via $$GL(n,mathbb Q)$$.

Suppose that there’s a rational affine subspace $$V subset mathbb R^n$$ (by this, I denote that there’s 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 underneath the motion of $$G$$ (i.e. for any $$vin V, gin G$$, now we have $$gcdot v in V$$). Moreover, there exists $$v in L$$ (in truth, we are able to take $$v=a$$) such that
$$G cdot v = L.$$

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

Any suggestion on pertinent questions/references may be very welcome! Particularly, I do not know which bailiwick research such issues ….

Edit:

Example. Consider $$(0,1)+L:=(mathbb Z,1) subset mathbb Z^2$$, and $$G={commence{pmatrix} 1&okay&1end{pmatrix}mid kinmathbb Z}.$$ For $$v=(0,1)$$, now we have $$G cdot v =(0,1)+L$$. In this illustration, we are able to take $$P$$ to breathe the interval from $$(0,1)$$ to $$(1,1)$$.

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