# Tag Info

To illustrate how such a question is of arithmetic nature (and what a complete answer should look like), here is a partial answer in a specific case. Namely: I specify to $M=\begin{pmatrix}2 & 1 \\... 1 vote ### Finite index subgroup of$\mathbb{Z}^2$that is invariant under a non-singular matrix For A fixed$M$, such finite index subgroups may be categorized as follows. Let$W\in GL(2,\mathbb{Z})$and$b$a positive integer dividing the lower left element of the matrix$W^{-1} M W$. Then the ... 3 votes ### Finite index subgroup of$\mathbb{Z}^2$that is invariant under a non-singular matrix If$v$is not scalar, the ring$\mathbb Z[M]$generated by$M$has rank$2$over$\mathbb Z$. It is therefore either an order in a number field, a finite-index subring of$\mathbb Z^2$, or$\mathbb Z[\...
Note that $\sqrt2\Lambda$ is even, and $\mathrm{vol}(\sqrt2\Lambda)^2=\det(A)$, where $A$ is the Gram matrix of $\sqrt2\Lambda$. Thus $\mathrm{vol}(\Lambda)=2^{-n/2}\sqrt{\det(A)}$, so the smallest ...
After scaling your lattice by $\sqrt{2}$, the Gram matrix has integral entries, so the absolute value of its determinant, being a nonzero integer, is at least $1$. This gives a lower bound of \$2^{-n/2}...