Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 307 / 3 / Solution

Put $\theta ={\theta }^{+}$, $\bar{\theta }={\bar{\theta }}^{+}$ and $y^{+}=x^{+}+i\theta \bar{\theta }$. The general two-dimensional N=(0,2) supersymmetry chiral superfield isIn ordinary coordinates this is $\Phi =\varphi +\sqrt{2}\theta {\psi }_{+}+i\theta \bar{\theta }{\partial }_{+}\varphi$. A supersymmetric kinetic action iswhose component form, up to light-cone conventions and total derivatives, is

A Fermi superfield satisfying ${\bar{D}}_{+}{\Lambda }_{-}=f(\Phi )$ has expansionHere $G$ is a complex bosonic auxiliary field. With the normalization in the question, the component action isup to equivalent sign conventions for the fermions.

The chiral integral in $S_3$ is supersymmetric only if its integrand is chiral. Applying ${\bar{D}}_{+}$ gives the necessary and sufficient conditionwith every $J^a$ holomorphic and with gauge charges chosen so that each product ${\Lambda }_{-a}{J}^a$ is gauge invariant. Its component expansion couples $G_a$ linearly to $J^a$ and supplies the corresponding Yukawa term ${\lambda }_{-a}{\psi }_{+}^i{\partial }_i{J}^a$.

Eliminating each $G_a$ by its algebraic field equation gives the nonnegative scalar potential. In the normalization displayed above and in the question,If the conventional definitions ${\bar{D}}_{+}{\Lambda }_{-}^a=\sqrt{2}{E}_a$ and $S_J=-\int d\theta {\Lambda }_{-a}{J}^a/\sqrt{2}+h.c.$ are used instead, the same result is written $V={\sum }_a(|E_a{|}^2+|J^a{|}^2)$; the two forms differ only by the normalization of $f$ and $J$.