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I have a problem solving the following exercise (Ex. 22-15) from John M. Lee‘s „Introduction to smooth manifolds“.

The problem is the following: Use the same technique as in the proof of the Darboux theorem to prove the following theorem: If $M$ is an oriented compact smooth $n$ manifold ($n\geq 1$) and if $\omega_0,\omega_1$ are smooth orientation forms with $\int_M\omega_0=\int_M\omega_1,$ then there exists a diffeomorphism $F:M\to M$ with $F^*\omega_1=\omega_0.$

For more context I‘ll quickly give an overview of the argument in the proof of the Darboux theorem: Given two symplectic forms $\omega_0,\omega_1$ one could locally (in a coordinate ball) write \begin{equation} \omega_t:=(1-t)\omega_0+t\omega_1:=\omega_0+t\eta=\omega_0+td\alpha \end{equation}(because $\eta$ is defined in a coordinate ball and closed and therefore exact). Then one defines a time dependent VF in a open set where all $\omega_t$‘s are non-degenerate by \begin{equation} V_t\lrcorner \omega_t=-\alpha_t \end{equation} and then calculates \begin{equation} \frac{d}{dt}|_{t=t_0}\theta_{t,0}^*=\theta_{t_0,0}^*(d(V_t\lrcorner \omega_t)+\frac{d}{dt}|_{t=t_0}\omega_t)=0, \end{equation}where $\theta_{t,0}$ is one of the time dependent flows of $V_t.$ This implies that $\theta_{1,0}^*\omega_1=\omega_0.$

My problem when attacking the problem is now the following: As the problem is a global problem, I can‘t just write $\eta=\omega_1-\omega_0$ as the differential of a form $\alpha.$ If this would be possible, then the rest would be exactly the same because \begin{equation} \Gamma (TM)\ni X \mapsto X\lrcorner \omega \in \Omega^{n-1}(M) \end{equation} is a smooth bundle isomorphism for each orientation form and because each $\omega_t$ is an orientation form for $0\leq t\leq 1.$

I would be glad for hints or solutions how to surpass my problem or for other approaches.

Thanks in advance.

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1 Answer 1

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Indeed, $\omega_1-\omega_2$ is exact since $\int_M(\omega_1-\omega_2)=0$.

If $\omega$ is an $n$-form on a compact $n$-manifold $M$ without boundary, then $\omega $ is exact if and only if $\int\limits_{M}\omega=0$

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  • $\begingroup$ This helped. Thanks! $\endgroup$ Aug 23, 2018 at 13:21

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