Given a symplectic vector space $V$, and a subspace $ \ U \subset V$, if $ \ U^\bot \not\subset \ U$, ($U \ $ is not co-isotropic), is it necessarily the case that $U \subseteq U^\bot \ $ or $ \ U \cap U^\bot = {0}$?
Another way of phrasing it would be, is containment, equality and trivial intersection the only possibilities for a subspace and its (symplectic) complement?
(For the technical definitions: are the cases isotropic, co-isotropic, Lagrangian and symplectic the complete list of possibilities for a subspace and its complement?)
Let $(V,\Omega)$ be a symplectic vector space of $\text{dim} V=2n$, with basis $e_1,...,e_n,f_1,...,f_n$, let us consider the four subspaces.
Isotropic: $\mathscr{L}\{e_1,e_2\}$
Symplectic: $\mathscr{L}\{e_1,f_1\}$
Lagrangian: $\mathscr{L}\{e_1,..,e_n\}$
Co-isotropic: $\mathscr{L}\{e_1,e_2,...,e_n,f_1\}$
By definition, a Lagrangian subspace is both isotropic and co-isotropic. Also if $Y\nsubseteq Y^\Omega$, then either $Y^\Omega\subseteq Y$ or $Y\cap Y^\Omega=\{0\}$. So these are the only possibilities. Another way to check this is to ask, can you come up with a subspace which doesn't have any of the above four subspace properties ?