Let's fix the standard symplectic structure $(\mathbb{R}^{2g}, \omega, J)$. A (marked) symplectic lattice then has the form $A\mathbb{Z}^{2g}$ for $A \in Sp_{2g}\mathbb{R}$. We say a vector subspace $W$ is rational in $\mathbb{Z}^{2g}$ if $W \cap \mathbb{Z}^{2g}$ is a cocompact lattice in $W$, ie. the flat torus $W/W \cap \mathbb{Z}^{2g}$ has finite volume. Likewise we say $W$ is rational in $A \mathbb{Z}^{2g}$ if $A' W$ is rational in $\mathbb{Z}^{2g}$ (where the prime $'$ denotes inverse). 

The following problem arises in computations with symplectic lattices: Suppose $W$ is a rational lagrangian subspace in the symplectic lattice $\Lambda$, and suppose we have a basis $w_1, w_2, \ldots$ for $W$ in $\Lambda$, (ie. each $w_i$ is a lattice vector in $\Lambda$). Now for any $A \in Sp_{2g}\mathbb{R}$ how do we compute a basis for $A'W$ in $A\Lambda$?

Some remarks: 

(i) The rationality of $A'W $ in $A \Lambda$ is rather particular to our situation. It is guaranteed by $A$ being symplectic, $W$ being lagrangian, the identity $W^o=JW^\perp=JW$, and the following fact from geometry of numbers: if $W$ is a rational subspace in a lattice $\Lambda$, then $W^o$ is rational in the dual lattice $\Lambda^*$ (here 'dual lattice' is meant in the Lekkerkerker or Conway/Sloane sense). 

(ii) We might not even have $A'W$ be lagrangian in $A\Lambda$, but that's acceptable for us. 

(iii) This question arises from trying to understand how the volume of a rational lagrangian subspace grows under linear symplectomorphisms. I'd be happy to discuss further.