Proposition: The defect class is nonzero

Let $m=0$. For every nonzero $\chi \in {C_c}^{\infty }(U)$ there exist $f,g,\delta$ satisfying (eq:separation) with the given $U$ such that, for $\xi =\chi {\partial }_1\in {\mathfrak{g}}_U$,

Explicitly, the smooth solutions

give ${\mathcal{L}}_{\xi }h=\chi {{\delta }^2}_{\mu }{{\delta }^2}_{\nu }$ and

Consequently the ${\kappa }^2$ identity of Theorem thm:ward fails on the frozen field content for every nonempty plateau: shrinking ${\mathfrak{g}}_U$ does not remove the defect.

Proof. $\square h=0$, so $P_{\mathrm{gr}}h=-\frac{1}{4}\mathcal{P}\square h=0$; $\square (t{x}^1)=0$. Directly, $({\mathcal{L}}_{\xi }h{)}_{\mu \nu }={\xi }^1{\partial }_1{h}_{\mu \nu }+h_{1\nu }{\partial }_{\mu }\chi +h_{\mu 1}{\partial }_{\nu }\chi =\chi {{\delta }^2}_{\mu }{{\delta }^2}_{\nu }$, since $h_{1\cdot }=0$.

Put $k=\chi {\delta }^2\otimes {\delta }^2$. On $\mathrm{supp}k\subset U$ both $g$ and ${\sigma }_{\epsilon }$ equal $1$. Then ${\mathrm{tr}}_{\eta }k=-\chi$, $k^{22}=\chi$, and (eq:density) gives the diagonal $a^{00}=-\frac{1}{2}\chi$, $a^{11}=\frac{1}{2}\chi$, $a^{22}=-\frac{1}{2}\chi$, $a^{33}=\frac{1}{2}\chi$. Let $w^{\mu }={\partial }_{\nu }{a}^{\nu \mu }$, so $w^0=-\frac{1}{2}{\partial }_t\chi$, $w^1=\frac{1}{2}{\partial }_1\chi$, and ${{\mathsf{X}}_1}^{\mu }[k]=-{{\mathsf{E}}_0}^{\mathrm{R}}{w}^{\mu }$. X-A's first-order relational scalar is

by the symmetry of ${\mathsf{L}}_{1,g}$ and $⟨{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}f,v⟩=⟨f,{{\mathsf{E}}_0}^{\mathrm{R}}v⟩$. With ${\partial }_t{\varphi }_0=x^1$, ${\partial }_1{\varphi }_0=t$: ${\partial }_{\mu }(a^{\mu \nu }{\partial }_{\nu }{\varphi }_0)=x^1{w}^0+t{w}^1$ and ${{\mathsf{X}}_1}^{\mu }{\partial }_{\mu }{\varphi }_0=-x^1{{\mathsf{E}}_0}^{\mathrm{R}}{w}^0-t{{\mathsf{E}}_0}^{\mathrm{R}}{w}^1$. The commutator identities ${{\mathsf{E}}_0}^{\mathrm{R}}(x^{1}v)=x^1{{\mathsf{E}}_0}^{\mathrm{R}}v+2{{\mathsf{E}}_0}^{\mathrm{R}}{\partial }_1{{\mathsf{E}}_0}^{\mathrm{R}}v$ and ${{\mathsf{E}}_0}^{\mathrm{R}}(tv)=t{{\mathsf{E}}_0}^{\mathrm{R}}v-2{{\mathsf{E}}_0}^{\mathrm{R}}{\partial }_t{{\mathsf{E}}_0}^{\mathrm{R}}v$, valid for $v$ with past-compact support because $\square (x^{1}u)=x^1\square u-2{\partial }_{1}u$ and $\square (tu)=t\square u+2{\partial }_{t}u$, give

which is (eq:witness).

Let $\Phi :={\partial }_1{\partial }_t({{\mathsf{E}}_0}^{\mathrm{R}}{)}^2\chi$, smooth, supported in $J^{+}(\mathrm{supp}\chi )\subset \{t>2\delta \}$. If $\Phi$ vanished on $\{t<T\}$, then ${\square }^2\Phi ={\partial }_1{\partial }_t\chi$ would vanish there, hence on $\mathrm{supp}\chi$, and a compactly supported $\chi$ with ${\partial }_1{\partial }_t\chi =0$ is zero. So $\Phi \ne 0$ on an open set $O\subset \{2\delta <t<T\}$. Choose $f\ge 0$ supported in $O$ with $\int f\Phi \ne 0$, then $g$ equal to $1$ on a neighbourhood of $\overline{U}$, supported in $\{2\delta <t<T\}$, and shrink $\delta$ if needed. All of (eq:separation) holds. By Lemma lem:ideal(ii), ${\hat{A}}_1[{\mathcal{L}}_{\xi }\hat{h}]\notin \mathfrak{J}$.