Isotropic subspaces in a symplectic vectorspace over $GF(q)$ Let $V$ be a symplectic vectorspace of dimension $2n$, and $r\mid n$. Is this statement true?"There is an isotropic spread of $r$ dimensional subspaces in $V$". By an isotropis subspace I mean a subspace $U$, that $U\subset U^{\perp}$.
any comment is appreciated.
 A: Can you construct it by restriction of scalars? Namely, as Nick Gill says, it is enough to consider the case $r=n$ (i.e., Lagrangian subspaces). Secondly, let us fix a non-zero functional $\phi:\mathbb F_{q^n}\to\mathbb F_q$ and a symplectic space $(V,\omega)$ of dimension $2$ over $\mathbb F_{q^n}$. Then $(V,\phi\circ\omega)$ is a symplectic space of dimension $2n$ over $\mathbb F_q$. Any $\mathbb F_{q^n}$-line in $V$ is going to be Lagrangian (over $\mathbb F_{q^n}$, and therefore also over $\mathbb F_q$), and such lines form a spread. 
P.S. I assume here that you are looking at $V$ over the finite field $\mathbb F_q$: this is mentioned in the title, but not in the body of the question. 
A: What you can show is that, if $W$ is a maximal isotropic subspace of $V$ (and so the dimension of $W$ is $n$), then there exists another maximal isotropic subspace $U$ of $V$ such that $U\oplus W=V$. Now, the number $r$ dividing $n$, you can construct your spread taking $r$-dimensional subspaces of $U$ and $W$.
