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 $A\Lambda$?

Note: the rationality of $A'W$ in $A \Lambda$ is guaranteed by $A$ being symplectic and $W$ being lagrangian (and generally false otherwise). It essentially follows from the following two facts: it is generally true that $W^o$ (euclidean orthogonal complement) will be rational in the dual lattice $\Lambda^*$ (("dual" in the standard Lekkerkerker geometry-of-numbers sense)); however in the symplectic case we have $W^o=JW^\perp=JW$ (where $\perp$ denotes $\omega$-orthogonal complement) and moreover $\Lambda^*=J\Lambda$. Finally, one should observe that $A'W$ might not be lagrangian in $A\Lambda$ (but that's acceptable in our case).  


This question arises from considering configurations of rational lagrangian subspaces in a symplectic lattice and trying to see how their volumes grow under linear symplectomorphisms. I'd be very pleased to discuss the matter further.