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Given a graded manifold with symplectic form $\omega$ of degree $n$, I have seen two expressions for the corresponding Poisson bracket of degree $-n$. Cattaneo-Fiorenza-Longoni, http://www.math.uzh.ch/fileadmin/math/preprints/15-05.pdf in section 2.7, give $$\lbrace f,g \rbrace=\iota_{X_{f}}\iota_{X_{g}} \omega,$$ while Cattaneo-Schatz, arXiv:1011.3401 in example 4.9, give $$\lbrace f,g \rbrace=(-1)^{|f|+1}X_{f}(g).$$

It seems to me that it is the latter that gives the proper graded anticommutativity so that the Poisson bracket is a Lie bracket of degree $-n$.

Is one of the two definitions mistaken or is it a matter of differing conventions?

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If one adopts the convention that the Hamiltonian vector field $X_h$ is defined by the equation $\iota_{X_h}\omega=dh$ (as both Cattaneo-Fiorenza-Longoni and Cattaneo-Schaetz do), and that $\iota_X(df)=X(f)$, then yes, there is a missing $(-1)^{|f|+1}$ factor in front of the formula in Cattaneo-Fiorenza-Longoni. But I can imagine other graded symplectic geometry sign conventions in which $\{f,g\}=\iota_{X_f}\iota_{X_g}\omega$ is the correct formula. For instance, in ordinary symplectic geometry, it is customary to define Hamiltonian vector fields by the equation $\iota_{X_h}\omega+dh=0$ rather than by $\iota_{X_h}\omega=dh$, and with this convetion one has $\iota_{X_f}\iota_{X_g}\omega=-X_f(g)$ with no need of additional sign corrections.

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Thank you. Is there a reason for the customary definition of Hamiltonian vector field? Does this have something to do with the interpretation of $h$ as energy, which should be positive? –  dhagbert Apr 22 '13 at 13:30
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Indeed that could be a reason, but I have to admit not be expert at all on motivations from classical mechanics. From a purely mathematical point of view, using the defining equation $\iota_{X_h}\omega+dh=0$ one has $X_{\{f,g\}}=[X_f,X_g]$ and so mapping a smooth function to its Hamiltonian vector field is a Lie algebra homomorphism from the Poisson algebra of the symplectic manifold to the Lie algebra of vector fields, rather than a Lie algebra antihomomorphism as is the case using $\iota_{X_h}\omega=dh$. –  domenico fiorenza Apr 22 '13 at 18:11
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This comes from stating Hamilton's equations in the standard form for which rate-of-change-of-position-and-momentum is on one side of the equation and derivatives-of-the-Hamiltonian is on the other. Signs are inserted to make this come out right. Every since Newton physicists usually state their equations of motion this way (rate of change = source of rate of change) -- unless they are Richard Feynman and are joking around, then they may bring each and everything to the left and have a zero on the right. –  Urs Schreiber Apr 22 '13 at 18:50
    
@domenico. I think your comment about the Lie algebra homomorphism is worth adding to your answer. –  dhagbert Apr 23 '13 at 10:02
    
@Urs. I remember hearing about Feynman's joke, but I don't remember it well enough to find it funny. Could you refresh my memory? –  dhagbert Apr 23 '13 at 10:04
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