Equivalence of "different" versions of Noether's theorem

Question

1. There is a (perhaps?) less-well-known version of Noether's theorem in John Lee's Introduction to Smooth Manifolds, i.e. Theorem 22.22, which roughly states that if a vector field $V$ such that $${\cal L}_V \omega={\cal L}_{V}X_H=0,\tag{1}$$ then $$V=X_f\tag{2}$$ for some function $f$ and this $f$ is a conserved quantity along $X_H$.

   Note: Here $X_f$ denotes the Hamiltonian vector field associated to the function $f$.

2. I know this should be related to the usual / Lagrangian version of the Noether's theorem for physicists: if an infinitesimal (quasi-)symmetry $$\delta L = \frac {dF}{dt}\tag{3}$$ for some function $F$, then $$\frac {\partial L}{\partial \dot{q}}\delta q - F\tag{4}$$ is constant along the motion. However, I don't know what the exact relationship is, and are they equivalent from each other? Is the "Hamiltonian" version a generalisation of the "Lagrangian" version?

   Edit: More specifically, I want to know what is the $\delta q$ and $F$ in the correspondence? I am assuming that it should be $$\begin{align}\delta q & = V^q \\ F & = pV^q-f,\end{align}\tag{5}$$ where $V^q$ is the $q$-component of the vector field $V$, and $f$ is the function whose Hamiltonian vector field is $V$. Is this correct?

Answer 1

1. Since a symplectic vector field $V$ is locally a Hamiltonian vector field $X_Q$, then OP's Hamiltonian Noether's theorem given in terms of vector fields $V$ and $X_H$ can locally be recast $$0~=~{\cal L}_VX_H~=~[V,X_H]~=~[X_Q,X_H]~=~X_{\{Q,H\}} \tag{A}$$ as the somewhat trivial fact that a charge $Q\equiv f$ that Poisson commutes with the Hamiltonian $H$ is an constant of motion on-shell: $$ \{Q,H\}~=~0 \qquad \Rightarrow \qquad \frac{dQ}{dt}~\approx~0. \tag{B}$$

2. Concerning an Hamiltonian action formulation of Noether's theorem, Yes, OP's identifications are essentially correct. For details, see e.g. my related Phys.SE answer here.