= 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 $\ell_a=\partial_a\phi$. Define
$$
\boxed{k_a=e^\phi\ell_a=\partial_a(e^\phi).}
$$
Then $k_{a;b}=e^\phi(\ell_{a;b}+\ell_a\ell_b)$. Since $e^\phi\ne0$, the condition on $\ell$ is exactly \b[$\boxed{k_{a;b}=0}$], so $k$ is a <parallel covector field>. Closed covectors have this scalar-potential representation locally by the <Poincaré lemma>; a global potential additionally requires exactness, for example on a simply connected domain.
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