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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?)

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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 ?

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  • $\begingroup$ What about $W =\mathscr L \{e_1,e_2,f_1\}$? $W^\bot = \mathscr L \{e_2,e_3,\ldots, e_n, f_3,\ldots, f_n\}$, so it is not one of the above. $\endgroup$ – mlainz Jan 16 at 14:25

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