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Let $(V,\omega)$ be a symplectic vector space but assume that $\omega$ is degenerate. Let $A$ be a Lagrangian subspace. When $\omega$ is non-degenerate, it is known that $dim \ A= \frac{1}{2} dim \ V$. Can one say anything in the degenerate case? Could it be true that $dim \ A \leq \frac{1}{2} dim \ V$? The trick is that the decomposition $V=A \oplus A^\perp$ doesn't hold when the form is degenerate.

Edit: Summarizing the answer of Moya, it seems that the equality $dim \ A \leq \frac{1}{2} dim \ V$ still holds when the form is non-degenerate. Here is the proof. The map $P:V/A^{\perp}\to A^{\ast}$ by $P(v+A^{\perp})=\omega(-,v)$ is always surjective (why is this true? for $\omega$ non-degenerate this is trivial, one can pick an orthonormal basis, for the degenerate case, it doesn't seem clear) Therefore we obtain
$ dim(A)=dim(A^*) \leq dim(V)-dim(A^\perp), $
with equality when the form is non-degenerate. In particular, when $A$ is Lagrangian this gives the desired result.

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If $A$ is Lagrangian, then necessarily $dim\, A=dim\, V/2$. This is because $A=A^{\perp}$ and, while the formula $V=A\oplus A^{\perp}$ is not always true, it is always true that $dim\, V=dim\, A+dim\, A^{\perp}$ for symplectic forms.

Edit: The statement is not in general true for a degenerate form. However, I have seen no reference for a symplectic vector space wherein the form is nondegenerate. If you relax the conditions to a pseudo-symplectic vector space, the statement is not true.

The reason is because we can define a map $P:V/A^{\perp}\to A^{\ast}$ by $P(v+A^{\perp})=\omega(-,v)$. This is an isomorphism if and only if $\omega$ is nondegenerate. With this, we get that $dim V-dim A^{\perp}=dim A$, leading to the result.

Edit 2: The map is emphatically not injective in the case of a degenerate map. Actually, I'm not convinced the map is surjective either. I actually don't see a meaningful statement being made in the nondegenerate case honestly.

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  • $\begingroup$ Thank you! Can you please expand on the statement $dim V = dim A + dim A^\perp$ for a degenerate form? $\endgroup$ Commented Feb 27, 2015 at 8:02
  • $\begingroup$ I've added a description for this case. $\endgroup$
    – Moya
    Commented Feb 27, 2015 at 8:24
  • $\begingroup$ Okay, using your answer, I edited my question to try and prove the inequality dim(A)≤\frac{1}{2}dim(V) in the pseudo-symplectic case. However, an argument is still missing. $\endgroup$ Commented Feb 27, 2015 at 10:41
  • $\begingroup$ I've added another edit. I actually do not see a reason for the space to be surjective. I'll think about it a bit more. $\endgroup$
    – Moya
    Commented Feb 27, 2015 at 21:00

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