# Cohomology classes in the Moser Theorem

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.

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

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.