Symplectic comoment maps and Hamiltonian observable actions

Let $\mathfrak{g}$ be a real Lie algebra and let $(M,\omega )$ be a symplectic manifold. Put $A=C^{\mathrm{\infty }}(M,\mathbb{R})$, with pointwise multiplication and the symplectic Poisson bracket $\{F,G\}=X_G(F)=-X_F(G)$ in the library convention (Smooth functions form a Poisson algebra, Poisson bracket on a symplectic manifold). Thus $A$ is a commutative associative real algebra; its bracket is real-bilinear, skew, satisfies Jacobi, and obeys the Leibniz rule in each argument. These are the real Poisson-algebra properties used below. A comoment map in this page's usage is a real-linear map $J:\mathfrak{g}\to A$ satisfying

$J([\xi ,\eta ])=\{J(\xi ),J(\eta )\}(\xi ,\eta \in \mathfrak{g})\mathrm{.}$

For $\xi \in \mathfrak{g}$, define its observable action on a smooth real-valued function $F$ by

${\rho }_J(\xi )F:=\{J(\xi ),F\}\mathrm{.}$

Bracket bilinearity makes each ${\rho }_J(\xi )$ a real-linear operator on $A$.

Here an observable means an element of $A$. For a real vector space $A$, an operator $D\in {\mathrm{End}}_{\mathbb{R}}(A)$ is a Poisson derivation when it satisfies both operation rules

$D(FG)=D(F)G+FD(G),D\{F,G\}=\{D(F),G\}+\{F,D(G)\}(F,G\in A)\mathrm{.}$

Thus a Poisson derivation preserves pointwise multiplication and the Poisson bracket by the Leibniz rule. The comoment condition says that $J$ is a Lie algebra homomorphism to the Poisson Lie algebra of functions. This definition does not require a Lie-group action, select a moment map, or assert integration of a Lie-algebra action to a group.