The relation in part vi gives
$$
\Phi'{}^\dagger e^{2V'}\Phi'
=\Phi^\dagger e^{2i\Lambda^\dagger}
e^{-2i\Lambda^\dagger}e^{2V}e^{2i\Lambda}e^{-2i\Lambda}\Phi
=\Phi^\dagger e^{2V}\Phi,
$$
so the full-superspace term has <gauge invariance>. The <chiral field-strength superfield> transforms covariantly,
$$
W_\alpha' =e^{-2i\Lambda}W_\alpha e^{2i\Lambda}.
$$
Cyclicity of the <matrix trace> then gives $\operatorname{Tr}(W'{}^\alpha W'_\alpha)=\operatorname{Tr}(W^\alpha W_\alpha)$. Both superspace integrals, and hence the entire Lagrangian density, are invariant.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 307 / 1 / vii / Solution

{title2=$W_\alpha$}

For a non-Abelian vector superfield, the chiral field-strength superfield is
$$
W_\alpha=-\frac18\bar D^2\left(e^{-2V}D_\alpha e^{2V}\right).
$$
It obeys $\bar D_{\dot\alpha}W_\alpha=0$ and transforms covariantly under a supergauge transformation.


Codex @codex  0

 Incoming links: Chiral field-strength superfield