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Let $(\omega_t)_{t\in [0,1]}$ be a path of cohomologous symplectic forms on $X$. The standard Moser's argument shows that there exists a family of diffeomorphisms $(\psi_t)_{t\in [0,1]}$ of $X$ with $\psi_0=id_X$ and $\psi_t^*\omega_t=\omega_0$.

Questions: If $(\omega_t)_{t\in S^1}$ is a loop of symplectic forms, is it possible (or under what assumptions it is possible) to find $\psi_t$ such that $\psi_1= id $ as well?

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    $\begingroup$ If I understand correctly, you are asking about the fundamental group of the group of symplectic diffeomorphisms of a symplectic manifold. Dusa McDuff has a survey article on the topology of groups of symplectic diffeomorphisms: arxiv.org/pdf/math/0404340.pdf $\endgroup$
    – Jason Starr
    Commented Aug 1, 2016 at 14:48
  • $\begingroup$ Thanks, I should have read a version of this before; I will take a look again. I think what am I asking is of different nature. $\psi_t$ depends on a choice of a path of $1$-forms $\sigma_t$ such that $\frac{d}{dt} \omega_t = d\sigma_t$. The question is if there is a choice such that $\psi_1=\psi_0$? $\endgroup$
    – Mohammad Farajzadeh-Tehrani
    Commented Aug 1, 2016 at 18:41

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Answer: no.

Example: take the square of a four-dimensional Dehn twist, call it $\phi_1$. This is isotopic to the identity as a diffeomorphism, so let's fix such an isotopy $(\phi_t)$. Consider $\omega_t = \phi_t^*\omega$. If what you said could be done, we could find a symplectic isotopy from $\phi_1$ to the identity, but it's known that this is usually impossible.

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