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

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 $$A_\mu$$?

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 $$A$$.
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.