I began studying differential forms to gain a deeper understanding of the Maxwell's equations. Let's say we have a $2$-form representing the electric flux density (the $D$-field). We can take the exterior derivative of it (which in vector field language will be divergence), to get a $3$-form. The resulting distribution can be interpreted as a scalar field of electric charge density.
It is clearly possible to go back from the $3$-form to a $2$-form by taking the superposition (Coulomb) integral of the charge distribution at each point in space. I understand that this isn't exactly the inverse of the exterior derivative because the original $D$-field could've had a component with zero divergence that got annihilated by the derivative. Nevertheless, the Coulomb's integral seems to be acting as some kind of inverse of an exterior derivative.
What kind of operator is the Coulomb's integral exactly? I don't know much about differential geometry, but I saw the term interior derivative thrown around, and that it takes a $p$-form and turns it into a $(p-1)$-form. However, I'm not that far into differential geometry, and I'm not able to decipher whether this is the thing I'm looking for.
Bonus points if someone could shed light on how the Biot-savart integral, in a similar fashon, seems to be the inverse of curl, in the language of forms.
The interior product you found can be used in the definition, but you need to integrate. Basically, with suitable convergence hypotheses, the $3$-form $\omega=f(x,y,z)\,dz\wedge dx\wedge dy$ on $\Bbb R^3$ will map to $\mathscr I\omega = \left(\int_{-\infty}^\infty f(x,y,z)\,dz\right)dx\wedge dy$. The fundamental proposition is that $\omega = d(\mathscr I\omega) + \mathscr I(d\omega)$.
This makes sense in the general setting of fiber bundles and is called integration over the fiber.