Definition: Invariant and prepared actions

Write ${S^{[1]}}_{\mathrm{inv}}={S^{[2]}}_{\mathrm{EH}}+{S^{[3]}}_{\mathrm{EH}}+{S^{\mathrm{sc}}}_{\mathrm{inv}}+S_{\mathrm{af}}$, $V_{\sigma }=\kappa \int ({\sigma }_{\epsilon }-1){\ell }_3[h]$, and $V_g$ for the switched-scalar deformation. Then ${S^{[1]}}_D={S^{[1]}}_{\mathrm{inv}}+V_{\sigma }+V_g$. Equivalently,

Setting ${\sigma }_{\epsilon }\equiv 1$ removes only $V_{\sigma }$. The switched theory $S_D-V_{\sigma }=S_{\mathrm{inv}}+V_g$ still carries the breaking $B_g$. It is not $S_{\mathrm{inv}}$.