Poisson actions defined in terms of coactions.

Question

If $(M,\{ \cdot,\cdot \}_{M})$ and $(M',\{ \cdot,\cdot \}_{M'})$ are two Poisson manifolds, then a smooth mapping $\varphi: M \to M'$ is called a Poisson map if it respects the Poisson structures, namely, if for all $x \in M$ and smooth functions $f,g \in {C^{\infty}}(M')$ , we have: $$ {\{ f,g \}_{M'}}(\varphi(x)) = {\{ f \circ \varphi,g \circ \varphi \}_{M}}(x). $$ Suppose that $G$ is a reductive group and $G$ acts on $M$. Then we have a map \begin{align} \psi: G \times M \to M \end{align} The action $\psi$ is Poisson if if for all $(g, x) \in G \times M$ and smooth functions $f,f' \in {C^{\infty}}(M)$ , we have: $$ {\{ f, f' \}_{M}}(\psi(g, x)) = {\{ f \circ \psi, f' \circ \psi \}_{G \times M}}(g, x). \quad (1) $$ Can the Poisson action $\psi$ defined in terms of coactions? The action $\psi$ has a corresponding coaction: $\delta: {C^{\infty}}(M) \to {C^{\infty}}(G) \otimes {C^{\infty}}(M)$. We denote $\delta(f) = f_{(-1)} \otimes f_{(0)}$, where we use the Sweedler notation. I think that (1) is equivalent to the following equation. For $f, f' \in {C^{\infty}}(M)$, $$ \delta(\{ f, f' \}_{M}) = \{\delta(f), \delta(f') \}_{G \times M}. \quad (2) $$ Is (1) equivalent to (2)? Are there some reference about this? Thank you very much.

Answer 1

By the definition of the coaction, we have $\delta(f)(g, x) = f_{(-1)}(g)f_{(0)}(x) = f(\psi(g,x))$. Therefore $\delta(f) = f \circ \psi$ and $\delta(f') = f' \circ \psi$. Therefore the right hand side of (1) corresponds to the right hand side of (2).

The left hand side of (1) corresponds to the left hand side of (2) since $(\delta(\{ f, f' \}_{M}))(g, x) = (\{ f, f' \}_{M})(\psi(g, x))$.