Let $\Phi_t,\Psi_s$ be the <local flows> of $X,Y$. The derivative of the pullback of $Y$ by $\Phi_t$ is $\Phi_t^*[X,Y]$. Therefore $[X,Y]=0$ makes $Y$ invariant under the $X$ flow. For fixed $t$, the two curves $\Phi_t(\Psi_s(p))$ and $\Psi_s(\Phi_t(p))$ then solve the same initial-value problem for $Y$. Uniqueness gives commuting flows wherever both compositions are defined. Conversely, commutation implies invariance of $Y$ under $\Phi_t$, and differentiation at $t=0$ gives $\boxed{[X,Y]=0}$. This proves <vanishing Lie bracket is equivalent to commuting local flows>.

For the contraction $\eta$, commute each <Lie derivative> past the other interior products using $[\mathcal L_{X_i},\iota_{X_j}]=\iota_{[X_i,X_j]}=0$. Each $\mathcal L_{X_i}\omega$ is zero by volume preservation. <Cartan's magic formula> $d\iota_X=\mathcal L_X-\iota_Xd$, applied successively, now gives
$$
d\eta=\sum_{i=1}^n(-1)^{i-1}\iota_{X_1}\cdots\widehat{\iota_{X_i}}\cdots\iota_{X_n}\mathcal L_{X_i}\omega
+(-1)^n\iota_{X_1}\cdots\iota_{X_n}d\omega=0.
$$
The last term vanishes since $\omega$ has top degree. Hence $\boxed{\eta\text{ is closed}}$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 115 / 1 / 1 / 3 / Solution

For smooth <vector fields> $X,Y$, the <local flows> commute wherever both compositions are defined exactly when $[X,Y]=0$. The derivative of the pullback of $Y$ by the $X$ flow is the pullback of $[X,Y]$. A zero bracket therefore makes $Y$ invariant under that flow; uniqueness of solutions to its ordinary differential equation gives commutation. Conversely commutation implies invariance, whose derivative gives the zero bracket. Neither field need have a complete flow.

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