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Let $E$ be a Frechet space and $\mathcal{F}$ be a non-degenerate bounded skew symmetric bilinear map $\mathcal{F}: E\times E\to \mathbb R$ on $E$. We can identify $TE$ with $E\times E$, with this identification, $\mathcal{F}$ defines 2 form $\omega$ on $E$, by the following: $$\omega_x: T_x E\times T_xE\to \mathbb R$$ $$\omega_x(v, w):= F(\tilde{v}, \tilde{w})$$ where $\tilde{v}$ and $\tilde{w}$ are identification of $v,w\in T_x E$ with $E$.

I feel this definition is well defined for Frechet manifold and also the form defined in this way is closed. Is this true.

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  • $\begingroup$ for the safe side, assume $E$ is a nuclear Frechet space where $TE\simeq E\times E$. $\endgroup$
    – Jonujohn
    Dec 4, 2013 at 19:28
  • $\begingroup$ I'd like just to mention that in infinite dimensions there are (at least) two notions of a symplectic structure: a weak symplectic structure, when $\omega$ induces only an injective map $TE \to T^{*}E$ and a strong symplectic structure when $\omega$ induces an isomorphism $TE \to T^{*}E$. A non-degenerate bilinear map as above will give you only the weak version. $\endgroup$ Dec 5, 2013 at 15:02

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Yes, it is. See Section 48 of

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