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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

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    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.
     
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