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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.

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up vote 2 down vote accepted

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 $$ (L_h^* \pi)(df,dg) = \omega({L_h}_* X_f, {L_h}_* X_g) = \omega(X_f,X_g) = \pi(df,dg). $$

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Many thanks, Eric for the nice answer! –  amine Aug 9 '11 at 17:47
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