The Noether Current for Electric-Magnetic Duality
I will briefly review, in covariant notation, a result from the following paper:
Duality Transformations of Abelian and Nonabelian Gauge Fields
S. Deser and C. Teitelboim
Phys. Rev. D 13 (1976),
1592-1597.
Noether Currents
Consider an action governing the dynamics of some set of fields $ \phi ^ {a} $ of the form: \begin{equation} S = \int _ {\text{M}} ^ {} \mathcal{L}\left( \phi ^{a}, \partial \phi ^ {a} \right) \ . \end{equation} An arbitrary variation of this action will be of the form \begin{equation} \delta S = \int _ {\text{M}} ^ {} E _{a} \delta \phi ^{a} + \text{d}\Theta \ , \end{equation} where $ E _ {a} $ are the equations of motion governing the field $ \phi ^ {a} $ and $ \Theta $ is the symplectic potential. For a symmetry transformation, the Lagrangian can at most change by a total derivative, so: \begin{equation} \delta S = \int _ {\text{M}} ^ {} \text{d} H \ . \end{equation} On equating the above two and working on-shell, i.e. where $ E _ {a} = 0 $, we find that the Noether current $ J = \Theta - H $ is conserved, i.e. $ \text{d} J = 0 $.
A small comment on the degree of the forms above: $ J $, $ \Theta $ and $ H $ are all $ (d-1) $-forms. One can also consider their Hodge duals (each denoted by the corresponding lowercase symbols) in which case the Noether current would be the $ 1 $-form $ j = \theta - h $ and the conservation law would look like $ \text{d} \star j = 0 $.
Duality Rotations
Let’s work out an example: the Noether current corresponding to electric-magnetic duality rotations in source-free electromagnetism. Recall that Maxwell’s equations take the form \begin{equation} \text{d} \star F = 0 \quad \text{and} \quad \text{d} F = 0 \ , \end{equation} and that electric-magnetic duality rotates these two equations into each other: \begin{equation} \delta F = \star F \quad \text{and} \quad \delta \star F = -F \ . \end{equation} Bianchi’s identity $ \text{d} F = 0 $ implies there exists some $ F = \text{d} A $. It is therefore natural to ask: how do electric-magnetic duality rotations act on $ A $?
The answer here is complicated by the following fact: let’s say there were a $ \delta A $ such that $ \text{d} \left( \delta A \right) = \star F $. Then, by acting on both sides of this equation with an exterior derivative, we’d find \begin{equation} d ^ {2} \left( \delta A \right) = 0 = \text{d} \star F \ . \end{equation} Note that the first equality is identically true, whereas the second is only true on-shell. For this reason, it is necessary to extend electric-magnetic duality rotations off-shell in an appropriate manner. This is done by considering a more general transformation \begin{equation} \delta A = Z \quad \text{such that} \quad \text{d} Z = \star F + G \ , \end{equation} where $ G $ is zero on-shell. In particular, $ \delta F = \text{d} Z $, a fact we’ll use shortly.
Free Maxwell theory is governed by the action \begin{equation} S = \int _ {\text{M}} ^ {} -\frac{1}{2} F \wedge \star F \ , \end{equation} an arbitrary variation of which yields \begin{equation} \delta S = \int _ {\text{M}} ^ {} - \delta A \wedge \text{d} \star F - \text{d} \left( \delta A \wedge \star F \right) \ . \end{equation} Since we’re interested in the Noether current, we can safely ignore the equation of motion term. Plugging in the appropriately off-shell-extended $ \delta A $ for electric-magnetic duality rotations yields \begin{equation} \delta S = \int _ {\text{M}} ^ {} - \text{d} Z \wedge \star F \ , \end{equation} which we can write as \begin{equation} \delta S = \int _ {\text{M}} ^ {} - Z \wedge \text{d} \star F - \text{d} \left( Z \wedge \star F \right) \ . \end{equation} This allows us to identify $ H _ {\text{D}} = -Z \wedge \star F $.
We now use the following algebraic fact: \begin{equation} -\text{d} Z \wedge \star F = \frac{1}{2} G \wedge G + \frac{1}{2} \text{d} \left[ A \wedge F - Z \wedge \text{d} Z \right] \ . \end{equation} Assuming that $ G \wedge G = 0 $, we read off \begin{equation} \Theta _ {\text{D}} = \frac{1}{2} \left[ A \wedge F - Z \wedge \text{d} Z \right] \ , \end{equation} and so the duality current $ J _ {\text{D}} = \Theta _ {\text{D}} - H _ {\text{D}} $ is \begin{equation} J _ {\text{D}} = \frac{1}{2} \left[ A \wedge F + 2 Z \wedge \star F - Z \wedge \text{d} Z \right] \ , \end{equation} and $ \text{d} J _ {\text{D}} = 0 $. Using the relation $ \text{d} Z = \star F + G $, we can also write this as \begin{equation} J _ {\text{D}} = \frac{1}{2} \left[ A \wedge F + Z \wedge \star F - Z \wedge G \right] \ . \end{equation} so it is clear that on-shell (when $ G = 0 $) the current is simply \begin{equation} J _ {\text{D}} \Big\vert _ {\text{on-shell}} = \frac{1}{2} \left[ A \wedge F + Z \wedge \star F \right] \ . \end{equation}
To the best of my knowledge, the corresponding charge was first computed in:
An Invariance Property of the Free Electromagnetic Field
M. G. Calkin
Am. J. Phys. 33 (1965)
958–960.
The charge $ Q _ {\text{D}} $ is proportional to the difference in the number of left- and right-circularly polarised photons.