Scalar potential and differential forms

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Given a Riemannian manifold (M^3, g), for a vector field X over M we can define the generalized curl \nabla \times X = \star (dX) where dX is the exterior derivative of (the one-form corresponding to) X and \star is the Hodge star operator with regards to the Riemannian volume form. So for example the vector calculus identity \nabla \times \nabla f = 0 in this language becomes \star d(df)=0 which follows from d^2 = 0. Similarly we can define the divergence \nabla \cdot X = \star d( \star X) and again the identity “divergence of curl is zero” translates to the equality \star \ d \star (\star dX) =0, which again holds because here we have \star \star = \mathrm{Id}.

In the physical setting, we may have a vector field E over M and a Killing field \eta (i.e. a field with \mathcal{L}_{\eta} g = 0, which will be a Lorentz transformation), and we ask whether there exists a scalar potential for it, meaning that a function f \in C^\infty(M) with \iota_{\eta}(\star E) = df. And what physicists do to test for existence of such an f is to take the curl of the vector field \iota_{\eta}(\star E) and check if it’s zero. But we now see how such a check is justifiable, because \iota_{\eta}(\star E) being curl-free corresponds to it being closed, and in \mathbb{R}^4 spacetime closed forms are exact, i.e. potential fields will exist.

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