For the classical (=field theoretical) Noether theorem there exist a well known approach using symplectic geometry. Recall, that in classical terminology the conserved current (a $1+3$-vector time+spatial keeping physical framework) $j^{\mu}$ is related to the conserved charge through $$Q= \int_V j^0 dV.$$
In symplectic geometry the role of conserved charge $Q$ is replaced by so-called momentum map $\Phi: P \to \mathfrak{g}^* $ where in physical context $P$ is the phase manifold and $\mathfrak{g}^*$ the dual of the Lie algebra of the symmetry group $G$, which we consider to be a Lie group acting on $P$. Now the symplectic version of Noether lemma says, that $\Phi$ can be related to the conserved charge $Q$ in classical sense, by the fact that if $\zeta \in \mathfrak{g}$ is any generator of the Lie group of symmetries, then - reinterpreting $\Phi$ as a map $\phi: P \times \mathfrak{g} \to \mathbb{R}$ the restriction of $\Phi(\zeta,-)$ to the Hamilton integral curves - in english: the "physical" paths on the phase space; ie solutions of Hamilton equation - is constant, so can interpreted as "the" conserved charge.
Question: What is the counterpart for "conserved current" $j^{\mu}$ in this symplectic formulation of Noether theorem?