Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Tensor analog of Poincare Lemma

  1. Apr 9, 2012 #1
    Hi everyone,

    I know that there is a result that corresponding to a closed p-form α, I can find a p-1 form β,
    such that dβ = α.

    I wanted to ask, what the tensorial analog of this would be. I mean would it be right to say that on a manifold with a lorentzian metric,

    If I have a vector field A, such that

    [itex]\nabla_{\mu} A^{\mu} = 0[/itex]

    then there exists an anti-symmetric tensor B such that,

    [itex]\nabla_{\nu} B^{\mu \nu} = A^{\mu} [/itex]

    where [itex]\nabla [/itex] is the covariant derivative comaptible with the metric
    (So that covariant derivative of the metric is zero)

    Also, if possible, please tell me where I can find the proof of the correct statement.

    Thanks in advance.
  2. jcsd
  3. Apr 12, 2012 #2
    I think what you are looking for is a Poincare Lemma for the co-differential, if I get this right. So the problem is not so much about using vector fields or forms, when you have a metric available (as you say) to switch between them.

    Essentially the divergence [itex]\mathrm{div}(A)[/itex] of a vector field [itex]A[/itex] is the co-differential [itex]\delta(A^{\flat})[/itex] of its corresponding one-form [itex]A^{\flat}[/itex]. As the co-differential is also nil-potent as is the exterior derivative, there is of course also a Poincare Lemma for the co-differential. Something like: Every co-closed form is co-exact.

    But note that these statement generally only hold locally on manifolds with non-vanishing first Betti number.
  4. Apr 12, 2012 #3

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    On any Riemannian manifold one has the Hodge decomposition, which states that any n-form [itex]\varphi_n[/itex] can be written

    [tex]\varphi_n = d \alpha_{n-1} + \delta \beta_{n+1} + \gamma_n[/tex]
    for some [itex]\alpha_{n-1}, \beta_{n+1}, \gamma_n[/itex] (the subscript indicates the degree of the form), where [itex]\delta[/itex] is the co-differential, and [itex]\gamma_n[/itex] is harmonic (i.e. [itex]d \gamma_n = \delta \gamma_n = 0[/itex]).

    If we assume that [itex]\varphi_n[/itex] is co-closed, then

    [tex]\delta \varphi_n = 0 = \delta d \alpha_{n-1} + 0 + 0[/tex]
    hence we must have [itex]\alpha_{n-1} = 0[/itex]. If the n-th Betti number is zero, then we must also have [itex]\gamma_n = 0[/itex], and in this case [itex]\varphi_n[/itex] can be written

    [tex]\varphi_n = \delta \beta_{n+1}[/tex]
    This statement is always true locally.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook