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In Lectures on Symplectic Geometry by AC da Silva, we have the following:

$\textbf{Moser Theorem}$: Suppose that a manifold $M$ is compact, the 2-forms $\omega_0$ and $\omega_1$ on $M$ are such that $[\omega_0] = [\omega_1]$ and the 2-form $\omega_t$ where $\omega_t = (1-t)\omega_0 + t \omega_1$ is symplectic for each $t \in [0,1]$. Then $\exists$ an isotopy $\rho: M \times \mathbb{R} \rightarrow M$ such that $\rho_t^{*} \omega_t = \omega_0$ $\forall$ $t \in [0,1]$.

Why would we need $[\omega_0] = [\omega_1]$? In the proof, I get that it is used directly to say there exists a 1-form $\mu$ such that $\omega_0-\omega_1 = d\mu$ but, it still isn't intuitively clear why this condition on the cohomology class is so essential to the Moser Trick. Thank you all in advance for any pointers.

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  • $\begingroup$ Recall that homotopic smooth applications induce identical morphisms in cohomology. $\endgroup$
    – Aphelli
    Jul 12, 2019 at 12:32
  • $\begingroup$ See my edited answer to your previous question. $\endgroup$ Jul 12, 2019 at 21:44

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The condition $[\omega_0]=[\omega_1]$ is necessary for the existence of an isotopy $\rho_t$ such that $$\rho_t^*\omega_t=\omega_0.$$ The standard way to see this is to use the fact that isotopies preserve cohomology classes, and this is how Da Silva's justifies it.

A more direct (computational) way to do it can be obtained by reading Da Silva's proof of the Moser Theorem itself: assume that there is such an isotopy, and consider $\partial_t(\rho_t^*\omega_t)$. On one hand this is $0$, since $\omega_0$ does not depend on $t$. On the other hand, by Proposition $6.4$ in da Silva's Lectures on Symplectic Geometry we have $$\partial_t(\rho_t^*\omega_t)=\rho_t^*\left(\mathcal{L}_{X_t}\omega_t+\dot{\omega_t}\right)$$ where $X_t$ is the time-dependent vector field generated by $\rho_t$. Comparing the two expressions for $t=0$ we find $$0=\mathcal{L}_{X_t}\omega_t+\dot{\omega_t}=\mathcal{L}_{X_t}\omega_t-\omega_0+\omega_1.$$ But since $\omega_0$ is symplectic we can use Cartan's Formula to compute $$\mathcal{L}_{X_0}\omega_0=\mathrm{d}\left(X_0\lrcorner\omega_0\right)+X_0\lrcorner\mathrm{d}\omega_0=\mathrm{d}\left(X_0\lrcorner\omega_0\right).$$ Putting all this together we see that the difference between $\omega_0$ and $\omega_1$ is an exact $2$-form, $$\omega_0-\omega_1=\mathrm{d}\left(X_0\lrcorner\omega_0\right)$$ so they live in the same cohomology class.

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Maybe you should read the paragraph just before, I can't say more than her I think. If you require that there is a symplectomotphism homotopic to the identity then the equality of cohomology class is a necessary condition, and often one ask himself if it is a sufficient condition: it is not except with further hypothesis.

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