Tensor analog of Poincare Lemma

  • #1

Main Question or Discussion Point

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.
 

Answers and Replies

  • #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.
 
  • #3
Ben Niehoff
Science Advisor
Gold Member
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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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