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The following question - proposal came to my mind about 4 years ago but I did not find any solution to this question and did not find any answer via e-personal comunication with some researchers so I present these questions here hope I can find some ideas or collaborator for work. Your comment or message to my personal email are very appreciated.

Let we have a $G$-principal bundle $\pi: M\to X$ where $G$ is a Lie group with Lie algebra $\mathfrak{g}$. Moreover assume that $M$ is at the same time a symplectice manifold and the action of $G$ on $M$ is a symplectic action. Let $\phi: M\to \mathfrak{g}^ *$ be the momentum map whose differential is $d \phi: TM\to M\times \mathfrak{g}^*$. We equip $M$ with an invariant metric then we get the corresponding LC connection. This connection gives us a connection form $\omega: TM\to \mathfrak{g}$ which can be considered as a bundle morphism $\omega: TM\to M\times \mathfrak{g}$. We consider the adjoint operator $\omega^*:M\times \mathfrak{g}^* \to T^*M $ of the connection form. The composition $\omega^* \circ d\phi: TM\to T^*M$ defines a $2$ linear map $\alpha:TM\times TM \to \mathbb{R}$. So we are interested to study this $2$ linear map. If it is not necessarily an anti symmetric $2$ linear map, we denote again by $\alpha$ its antisymmetric component. Is $\alpha$ a closed $2 form$? What is its cohomological class? Is it a trivial class? Is it cohomologue to the original symplectic $2$ form of $M$? To apply this idea to the tangent frame bundle of a manifold it is first necessary to ask that for what type of manifolds $M$ the total space of the tangent frame bundle $GL(M)$ admits a symplectic structure? When the base space is a symplectic manifold then it is guaranteed that the total space has a symplectic structure, but what about the general case? After we determine the manifolds whose total space of the Frame bundle is a symplectic manifold then we study the above mentioned $2$-form $\alpha$ which is a byproduct of combination of principal and symplectic structures.

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    $\begingroup$ If I'm not mistaken, then $\alpha(X, Y) = \sigma(X^v, Y)$, where $\sigma$ is the symplectic form and $X^v$ denotes the vertical part of $X$ wrt to the connection. So unless you assume some compatibility of the connection with the symplectic form, I don't see how to answer any of your questions. $\endgroup$ Commented Jul 24, 2023 at 9:13
  • $\begingroup$ @TobiasDiez Thank you very much for your comment. I think about some compatibilities. $\endgroup$ Commented Jul 26, 2023 at 4:41
  • $\begingroup$ @TobiasDiez what kind of compatibility do you siggest? $\endgroup$ Commented Jul 26, 2023 at 4:49
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    $\begingroup$ For example, you might want to require your metric to be compatible with the symplectic form so that the LC is a symplectic connection, $\nabla \sigma = 0$. $\endgroup$ Commented Jul 26, 2023 at 9:35
  • $\begingroup$ @TobiasDiez Yes I see ,thank you $\endgroup$ Commented Jul 27, 2023 at 8:33

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