 Solution

Under the Abelian supergauge transformation $V\mapsto V+i(\Omega -{\Omega }^{†})$, the real and imaginary components of $\omega$, together with $\psi$ and $F$, shift $C,\chi$ and $M$. They can be chosen to setleaving Wess-Zumino gaugeThe remaining imaginary scalar gauge parameter acts as the ordinary transformation $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\alpha$.

The chiral field-strength superfieldis chiral and gauge invariant in the Abelian theory. Henceis supersymmetric. Its components are the Maxwell kinetic term, the gaugino kinetic term, and the auxiliary term $D^2/(2e^2)$.

For the component check, vary $F_{\mu \nu }$ using $\delta A_{\mu }=ϵ{\sigma }_{\mu }\bar{\lambda }+\lambda {\sigma }_{\mu }\overline{ϵ}$ and vary the gaugino term using $\delta \lambda =F_{\rho \sigma }{\sigma }^{\rho \sigma }ϵ$ and its conjugate. After integration by parts, the terms proportional to ${\partial }_{\mu }{F}^{\mu \nu }$ cancel between the two variations. The remaining terms reduce, by the stated sigma-matrix identity, to a contraction of ${\partial }_{[\mu }{F}_{\nu \rho ]}$, which vanishes by the Bianchi identity. Thus $\delta \mathcal{L}$ is a total divergence and the action is invariant. The auxiliary field would make this supersymmetry close off shell; setting $D=0$ gives the displayed on-shell transformations.