Why is the electromagnetic duality an S-duality? One of the examples that Wikipedia gives of S-duality is the EM duality. Namely that
$$
\begin{align}
\mathbf{E} &\rightarrow\mathbf{B} \\
\mathbf{B} &\rightarrow -\frac{1}{c^2}\mathbf{E} \quad \text{ or }\quad (E, B) \rightarrow (E \cos \theta - B \sin \theta, B \cos \theta + E \sin \theta)
\end{align}
$$
or if you prefer
$$
F^{\mu \nu} \mapsto{ }^{\star} F^{\mu \nu} \quad{ }^{\star} F^{\mu \nu} \mapsto-F^{\mu \nu}.
$$
As far as I understand, an S-duality maps an strongly interactive theory to a weakly interactive one, i.e. generally, $g\mapsto 1/g$. Where can I see this?
I am assuming one would see this at the Hamiltonian/action level but I am a bit lost on how this would work out...
 A: You can see it in lots of ways. Here is one of them: Consider the first-order Lagrangian
for the Maxwell theory
$$L=-\frac{1}{4g^2}F^{\mu\nu}F_{\mu\nu}+\partial_{\mu}\widehat{A}_{\nu}(\star F)^{\mu\nu}.$$
The fields $F^{\mu\nu}$,$\widehat{A}_\mu$ are assumed to be independent and $F^{\mu\nu}$ is not the field strength of a 1-form at this level. Varying this Lagrangian w.r.t. $\widehat{A}_\mu$ will give you the Bianchi identity on $F$, namely
$$\partial_{[\mu}F_{\nu\rho]}=0\,,$$
which can be locally solved as $F_{\mu\nu}=\partial_{[\mu}A_{\nu]}$ for another 1-form $A_\mu$.
Plugging this solution back into $L$ will give you the standard second order Maxwell Lagrangian for the $A$ field.
Alternatively, one can vary $L$ w.r.t. $F$ and obtain the so-called duality relation (up to an irrelevant overall factor that can be absorbed into the field $\widehat{A}$)
$$F_{\mu\nu}\propto 2g^2\star(\partial_{[\mu}\widehat{A}_{\nu]}):=g^2(\star\widehat{F})_{\mu\nu}.$$
Using this relation, you can see that the first order Lagrangian $L$ becomes the second order Maxwell theory (this time for the dual field $\widehat{A}$) but with the reverse coupling constant, i.e.
$$L\propto -\frac{g^2}{4}\widehat{F}^{\mu\nu}\widehat{F}_{\mu\nu}.$$
Thus, you can see that the coupling gets inverted. This justifies why EM duality is a strong-weak one.
