0
$\begingroup$

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?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

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.

Improve this answer
$\endgroup$
9
  • $\begingroup$ 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? $\endgroup$
    – dhagbert
    Apr 22, 2013 at 13:30
  • 1
    $\begingroup$ 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 $\iota_{X_h}\omega=dh$. $\endgroup$
    – domenico fiorenza
    Apr 22, 2013 at 18:11
  • 1
    $\begingroup$ 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. $\endgroup$
    – Urs Schreiber
    Apr 22, 2013 at 18:50
  • $\begingroup$ @domenico. I think your comment about the Lie algebra homomorphism is worth adding to your answer. $\endgroup$
    – dhagbert
    Apr 23, 2013 at 10:02
  • $\begingroup$ @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? $\endgroup$
    – dhagbert
    Apr 23, 2013 at 10:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.