Let $(V,\omega)$ be an infinite dimensional symplectic vector space with a symplectic form $\omega.$ Is the symplectic complement $W^\omega$ of a symplectic subspace $W\subset V$ is still symplectic?
I came out with this question when I was trying to work on the problem which statement does not assume $V$ is finite dimensional. I can prove the finite dimensional case, and here is my proof.
Taking a symplectic basis $\{e_1,\cdots,e_n,f_1,\cdots,f_n\}$ for $V\cong\mathbb R^{2n}$, let's say that $W=\text{span}_{\mathbb R}\{e_1,\cdots,e_k,f_1,\cdots,f_k\}$. Assume that $W^\omega$ is not symplectic, then, by definition, there exists $v\in W^\omega$ such that $\omega(v,e_m)=\omega(v,f_m)=0$ for all $m=k+1,\cdots,n.$ Also by definition of symplectic complement, we have $\omega(v,e_m)=\omega(v,f_m)=0$ for all $m=1,\cdots,k.$ Thereforth $\omega(v,e_m)=\omega(v,f_m)=0$ for all $m=1,\cdots,n,$ i.e. $\omega(v,w)=0$ for all $w\in V$, deducing that $\omega$ is degenerated in $V,$ a contradiction.
It seems that such an idea does not work on the infinite dimensional case, since I have no idea how to find a symplectic basis in an infinite dimensional space.
Could anyone help please?
