TITLE: Is the Poisson tensor associated to a left invariant symplectic form, also left invariant?
QUESTION [6 upvotes]: Given a left invariant symplectic form $\omega$ on a Lie group $G$, the Poisson tensor associated to $\omega$ is given by
$$\pi(df,dg)=\omega(X_f,X_g)$$
where $X_f$ is the hamiltonian vector field associated to the function $f$ ; $i_{X_f}\omega=-df$.
I would like to check if the Poisson tensor $\pi$ is also left invariant or not?
Thanks for you help.

REPLY [4 votes]: Yes, it is also left-invariant.  This should be clear intuitively since $\pi$ is just $\omega$ under the isomorphism $T^*M \to TM$ setup by $\omega$.  To see this rigorously let $h\in G$ and let $L_h$ denote left multiplication by $h$.  Then
$$
(L_h^* \pi) (df, dg) = \pi(L_{h^{-1}}^* df, L_{h^{-1}}^* dg) = \pi( d(f\circ L_{h^{-1}}), d(g\circ L_{h^{-1}})) = \omega(X_{f\circ L_{h^{-1}}}, X_{f\circ L_{h^{-1}}}).
$$
But $X_{f\circ L_{h^{-1}}}$ is characterized by $\omega(X_{f\circ L_{h^{-1}}},V) = d(f\circ L_{h^{-1}})(V) = (L_{h^{-1}}^* df)(V) = df({L_{h^{-1}}}_* V) = \omega(X_f, {L_{h^{-1}}}_* V) = \omega({L_h}_* X_f, V)$, where this last part follows from the left invariance of $\omega$.  Therefore $X_{f\circ L_{h^{-1}}} = {L_h}_* X_f$.  Putting this into the above equation we see
$$
(L_h^* \pi)(df,dg) = \omega({L_h}_* X_f, {L_h}_* X_g) = \omega(X_f,X_g) = \pi(df,dg).
$$