Take the 2-minute tour ×
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.

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|improve this question
add comment

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.