Hamiltonians of compatible Poisson tensors - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T06:45:51Z http://mathoverflow.net/feeds/question/72584 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/72584/hamiltonians-of-compatible-poisson-tensors Hamiltonians of compatible Poisson tensors amine 2011-08-10T13:07:28Z 2011-08-12T11:08:01Z <p>Hi!</p> <p>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 ?</p> <p>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 $[X_f,H_f]$ ?</p> <p>All what I managed to do is to use the graded Jacobi identity, for the Schouten bracket, to show that $$\mathscr{L}_{H_f}\pi_0=\mathscr{L}_{X_f}\pi$$</p> <p>Any suggestions?</p> http://mathoverflow.net/questions/72584/hamiltonians-of-compatible-poisson-tensors/72705#72705 Answer by Nicola Ciccoli for Hamiltonians of compatible Poisson tensors Nicola Ciccoli 2011-08-11T18:22:01Z 2011-08-11T18:22:01Z <p>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.</p> http://mathoverflow.net/questions/72584/hamiltonians-of-compatible-poisson-tensors/72759#72759 Answer by amine for Hamiltonians of compatible Poisson tensors amine 2011-08-12T11:08:01Z 2011-08-12T11:08:01Z <p>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!</p>