Poisson actions

Let now $\phi\: G\times M\to M$ be a Lie group action. Let us fix the following notations

$$\phi(g, x)=g\cdot x$$

$$\forall\; g\in G,\quad \phi_g\: M\to M,\quad x\mapsto g\cdot x$$

$$\forall\; x\in M,\quad \phi_x\: G\to M,\quad g\mapsto g\cdot x$$

Remark that

$$\phi_{g\cdot x}=\phi_x R_g,\quad \phi_g\phi_x = \phi_x L_g.$$

For $f\in\Coo(M)$ let $\th_f\: M\to \gerg^*$ be defined by

$$\th_f(x) = d_g f(g\cdot x)\big\vert _{g=e}$$

If we have only the infinitesimal action we can define equivalently

$$\scalar{\th_f}{Y} = \rho(Y)f$$

We can give a slightly different look on conditions (3)-(4).

$$\th\:\Om^1(M)\to \Coo(M; \gerg^*)\in \gX(M)\tensor \Coo(M;\gerg^*)$$

Recall the Poisson coboundary introduced in ().

$$d_{\Pi}\:\gX^p(M)\to \gX^{p+1}(M),\quad d_{\Pi}(P)=[\Pi, P]$$

$$(d_{\Pi}X)(df, dg)=(L_X\Pi)(df, dg)=X\poiss{f}{g}-\poiss{Xf}{g} -\poiss{f}{Xg}$$

Therefore LHS of (3) can be rewritten as

$$(d_{\Pi}\th)(df, dg)$$

and phrased with suitable conventions as

$$(d_{\Pi}\th - \half[\th, \th])(df, dg)=0$$

i.e. $\th$ satisfies a Maurer-Cartan type of equation.