The inhomogeneous Maxwell equation ${*}\mathrm d{*}F=J^\flat$ is only true for the signature $(+---)$?

Question

In short, we would expect that considering the Riemannian manifold $(M,g)$ and the Riemannian manifold $(M,-g)$ leads to the same Maxwell equations, i.e. we would expect that $\newcommand{\imult}{\mathbin{\lrcorner}}$ $\newcommand{\hodge}{{*}}$ $\newcommand{\dif}{\mathrm{d}}$ \begin{equation} \begin{cases}\dif F=0\\\hodge\dif\hodge F=J^\flat\end{cases} \end{equation} is true for both metrics. However I think that this is not the case and I would like to know whether my reasoning is correct:

On the one hand, equations like $$F=\dif(A^\flat)$$ and $$f^\flat=qU\imult F$$ show that $F$ does depend on the signature: If $F$ is the EM tensor w.r.t. to one signature, then $-F$ is the EM tensor w.r.t. the other signature. However, then \begin{equation}\tag{1} \hodge\dif\hodge F=J^\flat \end{equation} is not invariant under a change of the signature: $(1)$ is equivalent to \begin{equation} \dif\hodge F=*J^\flat \end{equation} and we see that the RHS is independent of the signature, but the LHS does depend on the signature because $F$ does (the $*$ on $\Omega^2$ and $\dif$ do not depend on the signature).

In summary, the equality can only hold for one signature - as I said, I think that $(+---)$ is the "correct" one. Am I right?

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$^1$ I think that $\dif\hodge F=-{*}J^\flat$ for the other signature.

Answer 1

Yes, if you construct the theory as being based on vector fields $A$ and then defining $F = \mathrm{d}(A^\flat)$, then there is a dependence on the metric signature here.

However, an alternative formalization is to construct the theory with the connection form $A$ being the fundamental quantity, and likewise a current form $J$. If we want to write everything in terms of forms anyway, there is little point to starting with vector fields. It is then the vector fields $A^\sharp, J^\sharp$ that are signature dependent, not $F$ or the equations of motion.

You just need to pick a convention and stick with it, neither of these approaches is more "correct" than the other.