Let $W$ be the maximal symplectic subspace of a presymplectic vector space $(V,\omega)$. Then $W^\omega=\text{Rad}(\omega)$.

Question

Let $(V,\omega)$ be a presymplectic vector space and let $$\text{Rad}(\omega)=\{v\in V\colon\omega(v,v')=0\,\,\forall v'\in V\}.$$

Let $(W,\omega|_W)$ be a maximal symplectic subspace, i.e. $W$ is not contained in a bigger symplectic quotiënt. Then I want to show that the space $$ W^\omega:=\{v\in V\colon\omega(v,w)=0\,\,\forall w\in W\}=\text{Rad}(\omega). $$ For sure, we have that $\text{Rad}(\omega)\subset W^\omega$. For the other inclusion, I think I need to exploit the maximality of $W$. Suppose the other inclusion is false, i.e. there exists a $v\in W^\omega$ such that $v\notin\text{Rad}(\omega)$, so there exists $v'\in V\setminus W$ such that $\omega(v,v')\neq 0$. But this means that the space $\text{Span}(W\cup v')$ is symplectic. Since $W$ is already maximal, we have that $v'\in W$, so we arrive at a contradiction. Is this correct?

Answer 1

Assume $u\in W^\omega\setminus {\rm rad}(\omega)$. Then

• as $\omega|_W$ is nondegenerate, $u\notin W$.
• there is a $v\in V$ such that $\omega(u,v)\ne 0$. Now this can't be of the form $\lambda u+w$ with $w\in W$ as $\omega(u,\, \lambda u+w)=0$.

It means we could extend $W$ by these two vectors $u,v$. We have to check that this extension is indeed symplectic, and then it contradicts maximality of $W$.