Fix the standard $4$-dimensional linear symplectic space $(\mathbb{R}^4, \omega)$. Let $e_1, e_2, f_1, f_2$ denote the standard symplectic basis, and $J$ the standard almost-complex structure such that $\omega(x,Jy)$ coincides with the euclidean inner product $\langle x ,y \rangle$ for every $x,y \in \mathbb{R}^4$.
Suppose $A \in Sp(4,\mathbb{R})$ satisfies the following properties: (i) for every nonzero $x \in \mathbb{Z}^4, ||Ax|| \geq 1$, (ii) the areas of the parallelepipeds spanned by $Ae_1, Ae_2$ and $Af_1, Af_2$ coincide and this common area $m(A)$ is $\geq 1$, (iii) if $x,y \in \mathbb{Z}^4$ span a lagrangian subspace distinct from either the $\mathbb{R}$ -spans of $e_1, e_2$ and $f_1, f_2$, then the area of the 2-dimensional lagrangian parallelepiped spanned by $Ax, Ay$ is $\geq m(A)$.
Now a question: can the areas of the lagrangian parallelepipeds spanned by $Ae_1, Af_2$ and $Ae_2, Af_1$ $both$ be $> m(A)$?
A remark: the vectors $Ae_1, Ae_2, Af_1, Af_2$ span a 4-dimensional parallepiped of volume equal to 1. There are 6 standard 2-dimensional 'faces' to this parallelepiped corresponding to the 6 distinct pairs ${Ae_1 \wedge Ae_2, \ldots, Af_1 \wedge Af_2}$. Since $A$ is symplectic we know further that the `symplectic' 2-dimensional parallelepipeds spanned by $Ae_1, Af_1$ and $Ae_2, Af_2$ have area equal to 1. Now what appears amazing is that we should not have a relation expressing the total 4-dimensional volume of this parallelepiped as a homogeneous quadratic polynomial in the areas of these 6 standard 'faces'.
