# Tensor analog of Poincare Lemma

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

$\nabla_{\mu} A^{\mu} = 0$

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

$\nabla_{\nu} B^{\mu \nu} = A^{\mu}$

where $\nabla$ 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.

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 $\mathrm{div}(A)$ of a vector field $A$ is the co-differential $\delta(A^{\flat})$ of its corresponding one-form $A^{\flat}$. 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.

Ben Niehoff
Gold Member
On any Riemannian manifold one has the Hodge decomposition, which states that any n-form $\varphi_n$ can be written

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

If we assume that $\varphi_n$ is co-closed, then

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

$$\varphi_n = \delta \beta_{n+1}$$
This statement is always true locally.