This question is motivated by the physical description of magnetic monopoles. I will give the motivation, but you can also jump to the last section.

Let us recall Maxwell’s equations: Given a semi-riemannian 4-manifold and a 3-form $j$. We describe the field-strength differential form $F$ as a solution of the equations

$\mathrm{d}F=0$

$\mathrm{d}\star F=j$ (where $\star$ denotes the Hodge star).

If the second de-Rham-cohomology vanishes (for example in Minkowski space), $F$ is exact and we can write it as $F=\mathrm{d}A$, where $A$ denotes a 1-form.

Now let us consider monopoles: We use two 3-forms $j_m$ (magnetic current) and $j_e$ (electric current) and consider the equations

$\mathrm{d}F=j_m$

$\mathrm{d}\star F=j_e$.

Essentially, it is described in this paper, but the author Frédéric Moulin (a physicist) uses coordinates. Now he assumes that (in Minkowski space) $F$ can be decomposed using two potentials — into an exact (in the image of the derivative) and a coexact (in the image of the coderivative) form: $F=\mathrm{d}A-\star\mathrm{d}C$. Is there a mathematical justification for this assumption (maybe it is just very pragmatic)?

### The actual question

Given a 2-form $F$ on 4-dimensional Minkowski space (more generally: semi-riemannian manifolds)—are there any known conditions such that $F$ decomposes into an exact and a coexact form: $F=\mathrm{d}A+\star\mathrm{d}C$)?

For *compact riemannian* manifolds there is the well-known Hodge decomposition: There is always a decomposition into an exact, a coexact and a harmonic form. In the non-compact case you might be able to get rid of the harmonic form by only considering “rapidly decaying” forms (Wikipedia suggests that, but I do not have a good reference, in euclidean space there is the Helmholtz decomposition, and non-trivial (smooth) harmonic 1-forms do not vanish at infinity).

That is why I also ask: Are there “rapidly decaying” harmonic 2-forms in Minkowski space? Any references where I could see what is known about harmonic forms and Hodge theory in the semi-riemannian case are also welcome.