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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...

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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.

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    $\begingroup$ Thank you for your answer! Could you perhaps explain a bit more explicitly where the duality comes in to play? namely, is $\hat{F}^{\mu \nu} =^{\star} F^{\mu \nu}$ or something analogous? $\endgroup$ Feb 3, 2021 at 15:06
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    $\begingroup$ Dear @FriendlyLagrangian, unfortunately there was a tedious typo in my first reply and it is related to your comment. I edited my answer and now everything should be clear. $\endgroup$
    – Ozz
    Feb 3, 2021 at 15:27
  • $\begingroup$ So if I understood you correctly, $F^{\mu \nu} \mapsto *F^{\mu \nu}$ and $*F^{\mu \nu} \mapsto -F^{\mu \nu}$ is a symmetry (I agree). Further more, we can define an object $\widehat{F}_{\mu \nu} \propto -\frac{1}{g^2}*F_{\mu \nu}$ and its dual (using $*^2=-1$). With $\widehat{F}_{\mu \nu} $ we arrive at a second order Maxwell theory but with inverse coupling constant $g$. What is the claim? $\widehat{F}^{\mu \nu} \neq * F^{\mu \nu}$ $\endgroup$ Feb 3, 2021 at 17:29
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    $\begingroup$ Dear @FriendlyLagrangian, yes that is exactly the point. Another way to think about this: Let me switch to standard differential notation for brevity and suppose that we have the duality relation $F=g^2\star\widehat{F}$. If you act on both sides with the differential $d$, you will get $0=d\star d\widehat{F}$ (the Maxwell equation of motion for $\widehat{A}$), simply because $F$ satisfies the Bianchi identity $dF=0$. Now, let's act with $d\star$ on the duality relation. The result should convince you that, under EM duality, one has an ''exchange'' of Bianchi identities with equations of motion. $\endgroup$
    – Ozz
    Feb 3, 2021 at 18:47
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    $\begingroup$ Regarding the citation, my answer relies on the first-order approach. This is a very basic approach and I cannot really cite one particular reference. I guess you can find the same approach (even for free theories involving different gauge fields in arbitrary number of spacetime dimensions) in any textbook or relevant article :) $\endgroup$
    – Ozz
    Feb 3, 2021 at 18:52

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