Let $\Lambda\subset \mathbb{R}^4$ be a lattice. We identify $\mathbb{R}^4$ with the space $M\_2(\mathbb{R})$ of $2\times 2$ matrices over $\mathbb R$. It then is is clear that the set
$$
\det(\Lambda)=\big\{\det(\lambda):\lambda\in\Lambda\big\}
$$
is not necessarily discrete in $\mathbb R$.
But now we additionally insist that the group
$$
\Gamma=\big\{\gamma\in\mathrm{SL}\_2(\mathbb{R}): \gamma\Lambda=\Lambda\big\}
$$
be a cocompact lattice in $\mathrm{SL}\_2(\mathbb{R})$.


Under this condition, is the set $\det(\Lambda)$ discrete in $\mathbb R$?



Are the determinants of a lattice discrete?
This seems to be related to issues of Diophantine approximation. Fix such $\Lambda$ and $\Gamma$. Fix $\lambda \in \Lambda$ so that $\det(\lambda) \neq 0$. After rescaling $\Lambda$, we may assume that $\det(\lambda) = 1$. Fix some $\gamma \in \Gamma$ other than the identity. Conjugating $\Gamma$ (and transforming $\Lambda$ via the conjugating matrix) we may assume that $\gamma$ is of the following form.
\begin{pmatrix}
\alpha & 0 \\
0 & 1/\alpha
\end{pmatrix}


Suppose that $p$ and $q$ are integers.
Then $p\lambda$ and $q\lambda$ lie in $\Lambda$, thus so does $p\gamma\lambda$ and thus so does $p\gamma\lambda - q\lambda = (p\gamma - q I) \lambda$. Here $I$ is the two-by-two identity matrix.


We now compute:
$$
\det(p\gamma\lambda - q\lambda) = \det(p\gamma - qI) \det(\lambda)
 = p^2 + q^2 - pq(\alpha + 1/\alpha)
$$
So if $\alpha$ has unusually good Diophantine approximations then this quadratic form should "approximate zero". In this case, $\det(\Lambda)$ is indiscrete.


