unconstrained superfield. Defining ${\bf P}=(-\frac{1}{4}\bar{\nabla}^{2}+2R)$ equation is replaced by ($\partial_{S}W\equiv W_{S}$), $$-3{\bf P}(Ue^{-K/3})+\frac{k}{4}{\cal W}^{2}+W_{S}=0.\label{eq:WSconstraint}$$ Thus we need to replace $S$ in equation by $$S=W_{S}^{-1}(3{\bf P}(Ue^{-K/3})-\frac{k}{4}{\cal W}^{2}).\label{eq:SUtilde}$$ Note that in the above $W_{S}^{-1}$ is the functional inverse of $W_{S}$.

Substituting into (with $W(\phi)\rightarrow W(\phi)+\Delta W(S)$ we have for the chiral superspace terms $$\int d^{8}z\frac{{\bf E}}{2R}[W(\phi)+(\Delta W(S)-S\partial_{S}\Delta W)]+h.c.\label{eq:ASUtilde}$$ where $S$ is now given as a function of $U$ by . Clearly gauge kinetic terms are hidden in this expression, but the coupling functions are essentially power series in ${\bf P}(Ue^{-K/3})$ which can always be rewritten as a chiral projection of some redefined unconstrained real field $\Omega$ i.e. as ${\bf P}\Omega$. For instance in the simplest case where $\Delta W=S^{2}/2$ the expression becomes $$\begin{aligned}
\int d^{8}z\frac{{\bf E}}{2R}[W(\phi)-\frac{1}{2}(3{\bf P}(Ue^{-K/3})-\frac{k}{4}{\cal W}^{2})^{2}]+h.c. & \sim\\
\int d^{8}z\frac{{\bf E}}{2R}[W(\phi)-(3{\bf P}(Ue^{-K/3})\frac{k}{4}{\cal W}^{2}] & + & h.c.\end{aligned}$$ so that the gauge coupling function is $3{\bf P}(Ue^{-K/3})$. As can be easily seen this is the case whatever the functional form $\Delta W$ takes, since this relates to the coefficient of the leading term in the expansion in ${\cal W}^{2}$. This is obviously of the same form as the original chiral field representation in terms of $S$ for the gauge coupling function, 