If $ \partial_a F_{bc} + \partial_b F_{ca} + \partial_c F_{ab} = 0 $ then $ F_{ab} $ is the curl of a 4-vector

Question

Any skew-symmetric tensor $ F_{\alpha\beta} $ which is the curl of a 4-vector $A_\mu$, that is each tensor having the form $ F_{\alpha\beta} = \partial_\alpha A_\beta - \partial_\beta A_\alpha $, will satisfy the relation $$ \partial_\alpha F_{\beta\gamma} + \partial_\beta F_{\gamma\alpha} + \partial_\gamma F_{\alpha\beta} = 0. $$

That is very easy to see. But I have some heuristic reasons (related to the electromagnetic tensor) to think that the converse is also true.

Question: How can you prove that if $ \partial_\alpha F_{\beta\gamma} + \partial_\beta F_{\gamma\alpha} + \partial_\gamma F_{\alpha\beta} = 0 $ then $F_{\alpha\beta}$ is necessarily the curl of a 4-vector $A_\mu$?

Answer 1

Hint: The implication $$\mathrm{d}F=0\quad\Rightarrow\quad\exists A \text{ locally}:~F=\mathrm{d}A$$ is a special case of Poincare lemma. A proof can be found in any good textbook on differential forms. Note that there can in principle be topological obstructions that hinter a globally defined 1-form $A$.

Answer 2

If the coordinate domain on which $F_{\mu\nu}$ is defined contains the origin and is star-shaped with respect to it (all points of the domain can be connected to the origin with a straight line), define $$ A_\nu(x)=\int_0^1\mathrm dt\ tF_{\kappa\nu}(tx)x^\kappa. $$

This $A_\nu$ will satisfy $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ provided the integrability condition $$ \partial_\mu F_{\nu\kappa}+\partial_\nu F_{\kappa\mu}+\partial_\kappa F_{\mu\nu}=0 $$ holds.