I have been going through 'Lectures on Symplectic geometry' by AC da Silva and in the proof of the Moser Theorem (notable for the 'Moser Trick'), there is an application of the Cartan Magic Formula that eventually leads to the equation $$i_{v_t}\omega_t + \mu=0$$ where $\mu$ is a 1-form and it says we can 'solve for the vector field $v_t$' but I'm not sure how to proceed. Any insight would be much appreciated. It would also be great to know the reason for requiring that the 2-forms $\omega_0$ and $\omega_1$ be in the same cohomology class. I see it being used to say that $\omega_1-\omega_0 = d\mu$ for some 1-form $\mu$ but some intuition would be most welcome.

  • $\begingroup$ Since any symplectic form is nondegenerate, you can build a tangent-cotangent bundle isomorphism as in the Riemannian case. Now solving for your vector fields basically amounts in applying this isomorphism to a one form. As a side remark: there should be no exterior differential in front of $\mu$. $\endgroup$ Jul 11, 2019 at 14:04

1 Answer 1


Let $(M,\omega)$ be a symplectic manifold. Then define $\tilde{\omega} \colon \mathfrak{X}(M) \to \Omega^1(M)$ by $\tilde{\omega}(X)(Y) := \omega(X,Y)$, called the tangent-cotangent bundle isomorphism. One can show that this is a smooth isomorphism (a bundle isomorphism if you see it as a map $TM \to T^*M$). Moreover, for any one form $\eta \in \Omega^1(M)$ we can solve $i_X\omega = \eta$ by setting $X := \tilde{\omega}^{-1}(\eta)$. Indeed, we compute

$$(i_X\omega)(Y) =\omega(X,Y) = \tilde{\omega}(X)(Y) = \eta(Y)$$

for every $Y \in \mathfrak{X}(M)$. For a proof of the tangent-cotangent bundle isomorphism, see theorem 2.40 in my notes on classical mechanics.

To answer your second question, in order to apply the moser trick you need the family $\omega_t$ to satisfy $$\frac{d}{dt} \omega_t = d\eta_t$$ for a smooth family of one forms $\eta_t$ (see for example Symplectic Topology by Dusa McDuff and Dietmar Salamon). Coinciding cohomology classes for example is a sufficient condition if you consider a linear interpolation.

  • 1
    $\begingroup$ Thank you for the resource. I'm looking forward to read it. $\endgroup$ Jul 12, 2019 at 4:10
  • $\begingroup$ @TheGeekGreek This has been immensely helpful. Thank you for taking the time to explain! $\endgroup$ Jul 13, 2019 at 2:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .