I have read at many places that the pure Maxwell theory (without any matter) is self-dual. This is the general form for Maxwell Lagrangian density: $$\mathcal{L} = - \frac{1}{4} F_{\mu\nu} F^{\mu\nu},$$ where $F_{\mu\nu} \equiv \partial_\mu A_\nu - \partial_\nu A_\mu.$
There are two ways to see the self-duality that I know of. Either we can write it in terms of electric and magnetic fields, $$ \mathcal{L} = \frac{1}{2}\left(\mathbf{E}^2 - \mathbf{B}^2\right);$$ under the duality transformation $\mathbf{E}\rightarrow\mathbf{B}$ and $\mathbf{B} \rightarrow-\mathbf{E}$, the form of the Lagrangian remains the same (up to a negative sign), and so the equations of motion also remain the same. Alternatively, we can also write (in the language of differential forms): $$\mathcal{L} = -\frac{1}{2} F \wedge *F,$$ where $*F$ is the Hodge dual of $F$. This is invariant under the duality transformation $ F \rightarrow *F \ ,\ *F \rightarrow **F = -F $ which again lands us with the same form of Lagrangian (with a negative sign).
My actual problem concerns the duality in presence of monopoles. But I think my confusion can be answered at the level of pure Maxwell theory itself. As described in this answer, we can introduce an extra gauge field (call it $A^m$, and call the electric gauge field $A^e$) that couples with magnetic monopoles. I am assuming that under the self-duality, the $A^m$ and $A^e$ fields will be swapped (up to a negative sign). But then I don't see how the kinetic term will remain invariant. So perhaps I can word my question like this:
1) How can I express the Maxwell Lagrangian in terms of $A^e$ and $A^m$, so that the self-duality is obvious?
Let me also phrase this question in slightly different way:
2) Where is the $A^m$ field hidden in Maxwell Lagrangian? (Is it hidden as a constraint, for example?)
My understanding might be completely incorrect about this. If you have any comment or reference about it, that will be appreciated.