 Solution

The preceding condition is symmetric in $b,d$, so ${\ell }_{[b;d]}=0$. On a patch where the potential representation supplied in the question holds, write ${\ell }_a={\partial }_a\varphi$. DefineThen $k_{a;b}=e^{\varphi }({\ell }_{a;b}+{\ell }_a{\ell }_b)$. Since $e^{\varphi }\ne 0$, the condition on $\ell$ is exactly $\boxed{k_{a;b}=0}$, so $k$ is a parallel covector field. Closed covectors have this scalar-potential representation locally by the Poincare lemma; a global potential additionally requires exactness, for example on a simply connected domain.