# Topological vector space completion with respect to a symplectic form?

Suppose we have an infinite-dimensional vector space $$V$$ with a symplectic form $$\omega:V\times V\to\mathbb R$$. It can be given a weak topology that makes $$\omega$$ continuous.

Does it make sense to complete $$V$$ with respect to this topology? (Should something stronger be used?)

Will $$\omega$$ still be non-degenerate on the completion? (Meaning: if $$\forall a,\omega(a,b)=0$$, then $$b=0$$.) ...Obviously, yes. In fact, even if $$\omega$$ was originally degenerate on $$V$$, the process of completion would "quotient out" the degenerate subspace, as any sequence in it converges to $$0$$ according to $$\omega$$.

Is $$\omega$$ necessarily "strongly symplectic", meaning $$a\mapsto\omega(a,\cdot)$$ is a bijection between $$V$$ and its topological dual $$V^*$$? If not, can we make it so by changing the topology on $$V$$ (thus changing $$V^*$$)?

In general, I'm wondering what the conditions and implications are for a symplectic space to be complete and self-dual, similar to a Hilbert space (but without a norm).

• This question generalizes to arbitrary linear functions from $V$ to $V^*$. It turns out that, assuming $V$ has a countable basis (so it's isomorphic to some subspace of the sequence space $\mathbb R^{\mathbb N}$), it's always isomorphic to a Hilbert space with the standard weak topology. I don't know if I'll get around to composing an answer. – mr_e_man yesterday