Consider a 1-dimensional symplectic group action of $G$ on $(M,\omega)$. An infinitesimal generator $\xi \in \mathfrak{g}$ gives rise to the fundamental vector field $\xi_M$ on $M$, which could be a Hamiltonian vector field, i.e., there exists a smooth function $A$ such that $\xi_M=\{-,A\}$; or it might not be. In the affirmative case it is called a 1-parameter Hamiltonian group action.
A Hamiltonian group action is a group action that is Hamiltonian in the sense that it is generated by Hamiltonian vector fields.
In other words, a symplectic group action of $G$ on a symplectic manifold $(M,\omega)$ is Hamiltonian if there exists a momentum mapping, i.e. a map
$$ J: M \to \mathfrak{g}^*, $$where $\mathfrak{g}^*$ is the dual of the Lie algebra of $G$, such that for all $\xi\in\mathfrak{g}$,
$$ d\langle J,\xi\rangle = \omega(\xi_M,\cdot) $$where $\xi_M$ is the vector field generated by the action of $\xi$.
If we consider a 1-parameter subgroup generated by $\xi$, then the infinitesimal generator is $X_{\xi_M}$. Thus, if the group action is Hamiltonian, we have
$$ d\langle J,\xi\rangle = \omega(\xi_M,\cdot) $$and so $\xi_M=X_{\langle J,\xi\rangle}$ so the infinitesimal generators or the group action are Hamiltonian vector fields
In other words, the momentum mapping provides a method for constructing the function $A$ associated to the infinitesimal generator $\xi_M=X_A$ of a Hamiltonian group action.
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Author of the notes: Antonio J. Pan-Collantes
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