Proposition: Nonzero breaking on the unrestricted space

Let $\varphi ={\varphi }_m$, with ${\varphi }_m=\sin (mt)/m$ for $m>0$ and ${\varphi }_0=t$, let $\gamma =0$, and let $c=\vartheta \zeta (t,\mathbf{x}){\partial }_t$ with $\vartheta$ an odd generator and $0\ne \zeta \in {C_c}^{\infty }(O)$, $\zeta \ge 0$. Then

In particular $(S_{\mathrm{ext}},S_{\mathrm{ext}})\ne 0$ on $\mathfrak{F}$.

Proof. At $\gamma =0$: $A\gamma =0$, and $J_3$, $C_2$, $M_2$, ${{\mathcal{D}}_{\sigma }}_{\epsilon }$, ${Z_{\sigma }}_{\epsilon }$, ${{\mathfrak{R}}_{\sigma }}_{\epsilon }$ are quadratic in $\gamma$ and vanish; ${\mathfrak{L}}_q={\mathfrak{L}}_{\eta }$; $P_m{\varphi }_m=0$. Hence ${\mathcal{K}}_{\nu }=({\partial }_{\mu }g){T_0}^{\mu }{}_{\nu }[{\varphi }_m]$. For ${\varphi }_m$, ${\partial }_t{\varphi }_m=\cos (mt)$, $(\partial \varphi {)}^2-m^2{\varphi }^2=\cos (2mt)$, so ${T_0}^{00}={\cos }^2(mt)-\frac{1}{2}\cos (2mt)=\frac{1}{2}$ and ${T_0}^{0i}=0$. Thus $c^{\nu }{\mathcal{K}}_{\nu }=c^0({\partial }_{\mu }g){T_0}^{\mu 0}=\frac{1}{2}\vartheta \zeta {\partial }_{t}g$, and (eq:witnessA1) follows from Lemma lem:fall. The Grassmann coefficient is a strictly negative real number.