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Under $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\alpha$, the field strength is unchanged, but the mass term in the Proca action changes bywhich is not generally a total derivative. The mass therefore breaks gauge invariance.

The Euler-Lagrange field equation isTaking its divergence and using the antisymmetry of $F^{\mu \nu }$ gives $m^2{\partial }_{\nu }{A}^{\nu }=0$. Since $m>0$, the Lorenz constraint in Proca theory follows. Substitution back into the field equation then gives

Substitution of the gauge transformations and $U{U}^{†}=I$ givesA direct commutator calculation givesCovariance of the left side then implies $F_{\mu \nu }^{\prime }=U{F}_{\mu \nu }{U}^{†}$. The cyclic property of the trace makes the traces of $F_{\mu \nu }{F}^{\mu \nu }$, $D_{\mu }\varphi D^{\mu }\varphi$, and ${\varphi }^3$ invariant, so the Lagrangian is gauge invariant.

The relation in part vi givesso the full-superspace term has gauge invariance. The chiral field-strength superfield transforms covariantly,Cyclicity of the matrix trace then gives $\mathrm{Tr}(W^{\prime }{}^{\alpha }{W}_{\alpha }^{\prime })=\mathrm{Tr}(W^{\alpha }{W}_{\alpha })$. Both superspace integrals, and hence the entire Lagrangian density, are invariant.