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I posted this question here on math.stackexchange.com. I have not had an answer, and I thought it could be more appropriate here.
(Please, If you judge this my opinion is wrong, then I will delete this question)

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

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. $\omega$.
Let $\eta$ be a primitive of $\omega$, and $\mu_i\in C^{\infty}(M,\mathbb{R})$ be defined by $\mu_i=\int_0^1(\textrm{Fl}t^{{\zeta{{e_i}}}})^*(\zeta_{e_i}\lrcorner\eta)dt$ .$\mu_i=\int_{0}^{1} \big(\textrm{Fl}_{t}^{\zeta_{e_i}}\big)^\ast (\zeta_{e_i}\lrcorner\eta)\,dt.$$ We content contend that d\mu_i=\zeta_{e_i}\lrcorner\omega, for any i=1,\ldots,k. By H.Cartan's formula and the theorem on Lie derivative we get d\mu_i=\int_0^1\frac{d}{dt}((\textrm{Fl}t^{\zeta{e_i}})^*\eta)dt-\int_0^1(\textrm{Fl}t^{\zeta{e_i}})^*(\zeta_{e_i}\lrcorner\omega)dt.d\mu_i=\int_0^1\frac{d}{dt}\bigg(\big(\textrm{Fl}_t^{\zeta_{e_i}}\big)^\ast\eta\bigg)\,dt-\int_0^1\big(\textrm{Fl}_t^{\zeta_{e_i}}\big)^\ast(\zeta_{e_i}\lrcorner\omega)\,dt.$$ 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. 3 added 4 characters in body; added 1 characters in body I posted this question here on math.stackexchange.com. I have not had an answer, and I thought it could be more appropriate here. (Please, If you judge this my opinion is wrong, then I will delete this question) Reading a paper I had the need to complete a proof, and came up with a certain argument(see below). My question is: at your knowledge, could I reduce it to a special case of some other theorem? I ask this question in order to give a correct reference, instead of my trivial ad hoc argument, in the case the answer is positive. 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$\mu_i=\int_0^1(\textrm{Fl}t^{\zetat^{{\zeta{e_i}})^*(\zeta_{e_i}\lrcorner\eta)dt$.{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 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.