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).

  • $\begingroup$ 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. $\endgroup$ – mr_e_man yesterday

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