Let $G$ be a Lie group, $e$ its unite element and let $\pi$ be an bivector filed, do we have that the following are equivalent:
$\pi$ is left-invariant Poisson structure.i.e, for all $g\in G$: $$\pi(L^*_g\alpha,L^*_g\beta) =\pi(\alpha,\beta) \circ L_g$$ where $\alpha, \beta \in \Omega^1(G)$.
for all $g\in G$: $$ \pi_{\#,g} = T_eL_g\circ \pi_{\#,e}\circ(T_eL_g)^*$$ more details, ($T_gG \stackrel{(T_eL_g)^*}{\longrightarrow} T_e^*G \stackrel{\pi_{\#,e}}{\longrightarrow} T_e^*G \stackrel{T_eL_g}{\longrightarrow} T_g^*G$ ),then the map $\pi_{\#,g} : T^*_gG \longrightarrow T_gG$
$\forall$ $g\in G$ $\forall \; \varphi\; \psi \in \mathcal{C}^{\infty}(G)$ $$ \{ \varphi \circ L_g, \psi \circ L_g\} = \{\varphi,\psi\}\circ L_g $$
if G is connected, for any $u\in \mathcal{G}$(lie algebra), $$L_u \pi =0 $$
where $L_g :G \longrightarrow G $ is the left translation on $G$.