A problem about closed 2-forms on Minkowski space 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 $\delta(f^{*}F)=0$?
 A: 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:


*

*Start with the same compactly supported (in space-time) non-trivial $F$. 

*Suppose, for contradiction, that there exists a metric $g$ that does the job. 

*Find a point $x$ on $\partial~\mathrm{supp}~ F$ such that the boundary is space-like with respect to $g$. 

*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 $\Sigma$. 

*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. 

