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The problem is:

For any closed 2-form in the Minkowski space $\mathbb{R}^{3,1}$ satisfiying $dF=0$ and $\delta F \ne 0$ (with $\delta$ denoting the codifferential), does there exist a Lorentz manifold $(M,g)$ and a corrsponding diffeomorpism $f:M\to\mathbb{R}^{3,1}$ such that the pullback of $F$ on $M$, $f^{*}F$, satisfies $d(f^{*}F)=0$ and $\delta(f^{*}F)=0$?

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    $\begingroup$ Well, the condition on $d(f^*F)$ is automatic since exterior derivative commutes with pullbacks. $\endgroup$
    – Jeanne Clelland
    Commented Mar 17, 2018 at 20:01

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Your question can be equivalently phrased as:

Given a closed two-form $F$ on $\mathbb{R}^4$, is it always possible to find a Lorentzian metric on $\mathbb{R}^4$ such that $\delta F = 0$.

Then the answer is, in general, no.

First, consider the case if you require $(\mathbb{R}^4,g)$ to be globally hyperbolic.

Counterexample:

Let $A$ be non-closed, compactly supported one form on $\mathbb{R}^4$. Its exterior derivative $F = dA$ is closed, compactly supported, and not identically zero.

Suppose $g$ is globally hyperbolic. Then by the compact support of $F$ there exists a Cauchy hypersurface $\Sigma$ of $(\mathbb{R}^4, g)$ such that $F, \nabla F$ restricts to zero along $\Sigma$. By the uniqueness of solutions to hyperbolic PDEs, since $dF = \delta F = 0$, this implies $F \equiv 0$, contradicting the original assumption. Q.E.D.

The same argument can be localized as follows:

  1. Start with the same compactly supported (in space-time) non-trivial $F$.
  2. Suppose, for contradiction, that there exists a metric $g$ that does the job.
  3. Find a point $x$ on $\partial~\mathrm{supp}~ F$ such that the boundary is space-like with respect to $g$.
  4. Let $\Sigma$ be a space-like hypersurface through $x$ such that $\mathrm{supp}~ F$ is, in a causal convex neighborhood of $x$, to the future of $\Sigma$.
  5. Within the convex neighborhood we can find a globally hyperbolic neighborhood of $x$ for which $\Sigma$ is Cauchy. Running the argument above we see that $F$ must vanish identically in this neighborhood, which means that $x$ cannot in fact be on the boundary of $F$'s support. A contradiction.
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  • $\begingroup$ Thank you very much for give this counterexample！I wonder what if I weaken the condition of the example from compact support to vanishing in infinity would it possible to make some modification to make the proccess work as well？ $\endgroup$
    – jacktang1996
    Commented Mar 19, 2018 at 13:05
  • $\begingroup$ @jacktang1996: you will need to complement vanishing at infinity with certain growth/decay rates on your assumption of the new metric $g$. For example, you can start with an arbitrary solution to the Maxwell equations on $\mathbb{R}^{1,3}$, and define a diffeomorphism of $\mathbb{R}^{1,3}$ to itself so that it stretches as you go toward infinity (for example, imagine the mapping on $\mathbb{R}^4$ that looks like $x \mapsto e^{|x|} x$ for $x$ outside the ball of radius 2 and the identity within). Compact support avoids this issue as 0 is smaller than any positive number. $\endgroup$
    – Willie Wong
    Commented Mar 19, 2018 at 15:36

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