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Given two Poisson tensors $\pi_0$ and $\pi$ (on a Lie group $G$) ; if the two tensors are compatible, i.e. $$[\pi_0,\pi]=0$$ what are the relations between their hamiltonians ?

If we denote by $X_f$ and $H_f$ the hamiltonians of $f\in C^\infty(G)$, with respect to $\pi_0$ and $\pi$, what can we say about the lie bracket $[X_f,H_f]$ ?

All what I managed to do is to use the graded Jacobi identity, for the Schouten bracket, to show that $$\mathscr{L}_{H_f}\pi_0=\mathscr{L}_{X_f}\pi$$

Any suggestions?

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2 Answers 2

The graded Jacobi identity for the Schouten bracket implies that for any multivector $P$ you have $[\pi_0,P]=[\pi_1,P]$ (up to sign), and therefore the corresponding Poisson coboundary operators $d_{\pi_0}$, $d_{\pi_1}$ are graded commuting. I guess that any interesting information on your kind of question should follow from the relation between the single Poisson cohomologies and the total complex with differential the sum of the two differentials.

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Thanks Nicola, for your answer! What about the relation between $[\pi_0,f]=-X_f$ and $[\pi,f]=-H_f$ (in the case of $0$-multivector $P=f$)? –  amine Aug 11 '11 at 19:46
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Definitively, for any multivector $P$, we have $$[\pi_0,[\pi,P]]=-[\pi,[\pi_0,P]].$$ Thus, for the coboundary operators, we have $$\delta_{\pi_0}\circ\delta_\pi=-\delta_\pi\circ\delta_{\pi_0}.$$ In particular, $$\mathscr{L}_{H_f}\pi_0=-\mathscr{L}_{X_f}\pi.$$ If $X_f$, respectively $H_f$, denote the hamiltonian with respect to $\pi_0$, respectively $\pi$ then: $$[H_f,X_g]=X_{H_f(g)}-[\mathscr{L}_{H_f}\pi_0,g]=H_{X_g(f)}+[\mathscr{L}_{X_g}\pi,f].$$ One can explore the above identities to try to find relations between the hamiltonians!

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In the last identity when $f=g$ you have $X_{H_f(g)}=0$ and I guess that's the answer to your question. –  Nicola Ciccoli Aug 12 '11 at 14:14
Yes, Nicola! I am interested in the case where $\pi_O$ is the Poisson tensor associated to a left invariant symplectic form and $\pi$ is a multiplicatif Poisson tensor. If $\pi_0$ and $\pi$ are compatible, how to characterize the dual group? –  amine Aug 12 '11 at 22:19
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