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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.