In 3D $\mathcal{N}=2$ supersymmetric field theory with abelian gauge fields, the gauge field $A_{\mu}$ is often dualised to a real scalar $\gamma$. Does a Chern-Simons term prevent this dual description? Since, as I understand, the construction uses the dual field strength $\star F$, which is closed by Maxwell's equations thus locally exact: $ \star F = d \gamma$. This is the scalar field dual to the gauge field $A_{\mu}$. The Chern-Simons action $F \wedge A$ does however not depend on the dual field strength or is symmetric under $F \rightarrow \star F$ and therefore the dual description is unattainable. Is this correct?

And if I do have such a dual description, does this always complete to the supersymmetric duality: linear multiplet $\rightarrow$ chiral fields? Since the $\sigma, A^{\mu} \rightarrow \sigma + i\gamma$ is the lowest component of that transformation ($\sigma$ is the scalar field of the vector multiplet).


Indeed, the explicit occurrence of an $A$ in the action through the Chern-Simons term means that you cannot dualize at the level of the action.

Note that the Chern-Simons term can be physically interpreted as the coupling of $A$ to a "Chern-Simons current" (this interpretation is prominent in the CS explanation of the quantum Hall effect), and the presence of sources always breaks Yang-Mills electro-magnetic duality.

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  • $\begingroup$ Thanks! But what do you mean by dualising at the level of the action? Is there still a dual theory describing a scalar on some other level? $\endgroup$ – Graphite Aug 12 '18 at 15:26

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