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Non-degenerate Poisson bracket and even-dimensional manifold

  1. Oct 17, 2004 #1
    From this reference:

    http://www.amazon.com/exec/obidos/t...f=sr_1_1/103-6229518-2985440?v=glance&s=books titled From Classical to Quantum Mechanics,

    I quote the following: ( [tex]\xi^i [/tex] are coordinate functions)

    Let M be a manifold of dimension n. If we consider a non-degenerate Poisson bracket, i.e. such that

    [tex]\{\xi^i,\xi^j\} \equiv \omega^i^j[/tex]

    is an inversible matrix, we may define the inverse [tex]\omega_i_j[/tex] by requiring

    [tex]\omega_i_j \omega^j^k = \delta_i^k[/tex]

    We define a tensorial quantity

    [tex]\omega \equiv \frac{1}{2}\;\omega_i_j \; d\xi^i \wedge d\xi^j[/tex]

    which turns out to be a non-degenerate 2-form.
    This implies that the dimension of the manifold M is necessarily even.

    My questions are the following:

    I dont understand the two statement that I have put in red above.
    What is a non-degenerate 2-form?
    Why does this one above 'turns out' to be non-degenerate?
    Why does that imply that M is even?
    Additional comments would be welcome. Like concerning the meaning of [tex]\omega [/tex] above.

    In addition, I guess the point here by a shorter way: I think that all odd-dimensional antisymmetric matrices are singular. Is there a link with the language used above?

    Warm thanks in advance,

  2. jcsd
  3. Oct 21, 2004 #2

    matt grime

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    A non-degenerate two form is essentially a non-degenerate symplectic form on the tangent bundle at all points. This implies that the tangent bundle has even dimension. Which is what the things in red are saying.

    given any two-form, it is not necessarily non-degenerate, just as any symplectic form is not necessarily non-degenerate.

    'it turns out' means 'in this case with these hypotheses we can prove it is'
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