 Solution

Take the usual convention that the smooth isotopy starts at ${\rho }_0=\mathrm{id}$. The exterior derivative commutes with a pullback of a differential form, soConsequentlyThe right-hand condition says precisely that ${\rho }_t$ is a symplectic isotopy.

Here an exact symplectic manifold is necessarily noncompact in positive dimension: if $\dim M=2n$ and $\omega =-d\lambda$, then ${\omega }^n=-d(\lambda \wedge {\omega }^{n-1})$, contradicting the positive symplectic volume and the Generalized Stokes theorem on a compact smooth manifold without boundary. Thus the paper's introductory word “closed” must here be read as “without boundary”, as its parenthesis suggests, rather than as including compactness.