electromagnetism - Does existence of magnetic monopole break covariant form of Maxwell’s equations for potentials?

Absence of magnetic charges is reflected in one of Maxwell's fundamental equations: $$\operatorname{div} \vec B = 0. \tag1$$ This equation allows us to introducte concept of vector potential: $$\vec B = \operatorname{rot} \vec A.$$ Using this concept, it is possible to express Maxwell's equations in a graceful symmetric form: $$\nabla^2 \vec A - \frac{1}{c^2}\frac{\partial^2 \vec A}{\partial t^2} = - \frac{\vec j}{\epsilon_0 c^2}, \tag2$$ $$ \nabla^2 \phi -\frac{1}{c^2}\frac{\partial^2 \phi}{\partial t^2} = - \frac{\rho}{\epsilon_0}. \tag3 $$

Noticing, that vector $\vec A$ and scalar $\phi$ potentials, as well as electric current density $\vec j$ and charge density $\rho$, form a 4-vector in Minkovsky space-time. Therefore, Maxwell's equations can be expressed in a covariant form, using d'Alembertian: $$\nabla_{\mu}\nabla^{\mu} A_{\nu} = \frac{j_{\nu}}{\epsilon_0}. \tag4$$

If magnetic monopols exist, Maxwell's equation $(1)$ will look as: $$\operatorname{div} \vec B = \mu_0 c \rho_{\mathrm{magnet}}.$$

As the divergence of $\vec{B}$ isn't equal to zero, it impossible to introduce concept of vector potential. Thus, the equation in the form of $(4)$ will not be possible to express.

Answer

Another option, besides modifying the potential $A_\mu = (A_i, \phi)$ in some way, is to introduce another 4-potential $C_\mu = (C_i, \psi)$. Then the electric and magnetic field are given by $$E = - \nabla \times C - \frac{\partial A}{\partial t} - \nabla \phi$$ $$B = \nabla \times A - \frac{\partial C}{\partial t} - \nabla \psi$$

More on this 2-potential approach can be found here: http://arxiv.org/abs/math-ph/0203043