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Let $(V,\omega)$ be a vector space over $K$ together with symplectic form $\omega : V \times V \rightarrow K$. Let $U \subseteq V$ be a Lagrangian subspace (in other words, $U = U^{\perp}$). I want to know whether there must exist another Lagrangian subspace $W$ such that $U \oplus W = V.$

So far I have only been able to guarantee that an isotropic space $W$ exists such that $U \oplus W = V$; in other words, $W \subseteq W^{\perp}.$ I used Zorn's lemma to construct $W$.

If $V$ is finite-dimensional, then $\mathrm{dim} W = \mathrm{dim}(V) / 2 = \mathrm{dim}(W^{\perp})$ immediately implies that $W = W^{\perp}$.

Does anyone know what happens when $V$ is infinite-dimensional?

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  • $\begingroup$ Wow, out for a week and nothing! $\endgroup$ – user253055 Jul 12 '15 at 22:03
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It is not true in general, for infinite dimensional spaces, even when they are strong symplectic. There is a counterexample briefly mentioned by Kalton and Swanson of a subspace of a reflexive Banach space, that is maximally isotropic (i.e. Lagrangian), but has no Lagrangian complement.

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  • $\begingroup$ Note that the result of Kalton and Swanson concerns closed subspaces in a symplectic vector space induced by a particular topology, while the question concerns vector subspaces not necessarily closed with respect to some topology. $\endgroup$ – Blazej Dec 19 '18 at 10:14

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