In 3D ${\cal N}=2$ SUSY, the linear multiplet contains a global current. How is this related to the gauge field?

Question

I am reading the paper on 3d $\mathcal{N}=2$ supersymmetry by O. Aharony et al. (https://arxiv.org/abs/hep-th/9703110) and I am a bit confused about linear multiplets in section 2.3. A linear multiplet is defined as $\Sigma = \epsilon^{\alpha\beta} \bar{D}_{\alpha}D_{\beta} V$ where $V$ is the vector multiplet corresponding to a $U(1)$ gauge symmetry. Then they say the linear multiplet can be used to describe global currents generating a $U(1)_{J}$ global symmetry. What kind of current is this and what does it has to do with the $U(1)$ gauge symmetry of the vector multiplet $V$ $\Sigma$ is built out of? My confusion is that the gauge symmetry of $V$ is a local $U(1)$ symmetry, not a global one. Or do they just mean the current corresponding to global gauge transformations as a subset of the local transformations?

Answer 1

A linear multiplet is defined as $\Sigma=\epsilon^{\dot{\alpha}\beta}\bar{D}_{\dot{\alpha}}D_{\beta}V$ where $V$ is the vector multiplet corresponding to a U(1) gauge symmetry.

You have no "dotted indices" for the representations of the Lorentz group in $d=3$ (since the (real) Lorentz algebra is $\mathfrak{sl}(2)$ or $\mathfrak{su}(2)$ for the Minkowskian/Euclidean spacetime respectively). Moreover the definition is valid for any gauge symmetry vector potential, not only $\mathsf{U}(1)$.

What kind of current is this and what does it has to do with the U(1) gauge symmetry of the vector multiplet V Σ is built out of?

When you reduce the $d=4,\mathcal{N}=1$ theory to $d=3,\mathcal{N}=2$ (the crucial point is that you have $4$ generators of SUSY in both theories) the vector superpotential decomposes as \begin{equation} V=-i\theta\bar{\theta}\sigma -\theta\gamma^i\bar{\theta} A_i+i\theta^2\bar{\theta}\bar{\lambda}-\bar{\theta}^2\theta\lambda +\frac{1}{2}\theta^2\bar{\theta}^2 D,\end{equation} where $\sigma$ is now a (Lorentz) scalar field which corresponds to the component of the $d=4$ vector field in the reduced direction. The scalar $\sigma$ can take a VEV which breaks the gauge group to its maximal torus $\mathsf{U}(1)^{\operatorname{rank}\mathfrak{g}}$ (topologically this is $(\mathbb{S}^1)^r$, i.e. an $r$-torus), thus to each Abelian factor you can associate a stress-energy field $F^i$. For simplicity, let's take the case of only one Abelian factor, and call $F_{ij}$ the corresponding field strength, which is a Lie algebra valued $2$-form, then in $d=3$ the Hodge star will map it to a Lie algebra valued $1$-form, which is closed (Maxwell's eqns.), hence locally exact, we can introduce a scalar field $\gamma$ and write (locally) \begin{equation} \phi:= \sigma+i\gamma,\quad \star_3F=d\gamma, \end{equation} $\gamma$ is known as the dualised scalar photon. Now the crucial point is that its charge is quantised, i.e. \begin{equation} q=\frac{1}{2\pi}\int F\in\mathbb{Z}, \end{equation} therefore the $\gamma$ field is compact, which is to say, $\phi$ and $\phi+2\pi i$ are to be identified, or equivalently the action is invariant w.r.t. a shift of $\gamma$ by $2\pi\mathbb{Z}$: we call this symmetry $\mathsf{U}(1)_{\scriptscriptstyle\rm J}$ or topological symmetry. As you can see from $d\star_3F=0$, $J\sim\star_3F$ is the generator of the $\mathsf{U}(1)_{\scriptscriptstyle\rm J}$ global symmetry, with conserved charge $\sim q$ (the precise proportionality factors depend on your definitions). Defining \begin{equation} \Sigma=-\frac{i}{2}\epsilon^{{\alpha}\beta}\bar{D}_{{\alpha}}D_{\beta}V, \end{equation} then \begin{equation} 2\pi J=\sigma+\dots+\frac{1}{2}\theta\gamma^i\bar{\theta}(\star_3F)_i. \end{equation}

Try looking at appendices A and B of this article https://arxiv.org/abs/1406.6684 for a more careful explanation.