Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Reading a paper I had the need to complete a proof, and come up with a certain argument(see below). My question is: could I reduce it to a special case of some theorem? I ask this question in order to give a correct reference, instead of my trivial ad hoc argument.

I had to prove that:
Given a smooth action $\Psi$ of $\mathbb{T}^k$ on a symplectic manifold $(M,\omega)$, if $\omega$ is exact and there exists a smooth map $\pi:M\to P$ constant on the orbit of $\Psi$ and such that $\zeta_X\lrcorner\omega\in\pi^*(\Omega^1(P)),\forall X\in\textrm{Lie}(\mathbb{T}^k)$,( being $\zeta$ the action of $\textrm{Lie}(\mathbb{T}^k)$ on $M$ induced by $\Psi$), then the $\Psi$ is an hamiltonian action w.r.t. $\omega$.

For completeness, I sketch also the trivial proof:
Let $\eta$ be a primitive of $\omega$, and $\mu_i\in C^{\infty}(M,\mathbb{R})$ be defined by $\mu_i=\int_0^1(\textrm{Fl}_t^{\zeta_{e_i}})^*(\zeta_{e_i}\lrcorner\eta)dt$.
We content that $d\mu_i=\zeta_{e_i}\lrcorner\omega$, for any $i=1,\ldots,k$.
By H.Cartan's formula and the theorem on Lie derivative we get $d\mu_i=\int_0^1\frac{d}{dt}((\textrm{Fl}_t^{\zeta_{e_i}})^*\eta)dt-\int_0^1(\textrm{Fl}_t^{\zeta_{e_i}})^*(\zeta_{e_i}\lrcorner\omega)dt$.
The first integral is identically zero for periodicity.
The second one is $\zeta_{e_i}\lrcorner\omega$, because its integrand is a constant function of $t$, i.e. $\mathcal{L}(\zeta_{e_i}).(\zeta_{e_i}\lrcorner\omega)=0$ by the hypothesis.

share|cite|improve this question
This is a old question of yours, but you might want to look at this post:… – R.S. May 17 '14 at 1:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.