Theorem: Existence of a nonzero supported defect

Suppose $m=0$. Inside any nonempty regular observation region $\mathcal{O}$, one can choose admissible $f,g,U,\xi$, fix them before evaluation, and find smooth free fields $u,r$ such that

Consequently ${\hat{A}}_1[{\mathcal{L}}_{\xi }\mathbf{r}]\notin \mathcal{J}$ for that fixed instance.

The assertion is existential in the observation and switch data. It does not say that every previously fixed smearing detects a defect.

Proof. Choose nonempty $V⋐\mathcal{O}$, with its closure away from $\Sigma$, and a nonzero real ${\chi }_0\in {C_c}^{\infty }(V)$. Choose a plateau $U$ containing $\overline{V}$ within the regular reserve and a compact switch equal to one near $\overline{U}$. Take

Both fields are allowed nondecaying free configurations: $\square r=0$ and $\square u=0$. They are not required to satisfy the additional strong condition $C[r]=0$.

Because $r_{1\mu }=0$,

On its support, $g=\sigma =1$. Set $w_{\ast }^{\mu }={\partial }_{\nu }{a_1}^{\nu \mu }[l]$. Then

The first scalar response is

For this $u,l$, the term with the Hessian of $u$ vanishes, and the source is $x^1{w}_{\ast }^0+t{w}_{\ast }^1$.

The retarded commutator formulas

follow by applying $\square$ and zero-past uniqueness. They give

Let $\Psi ={\partial }_1{\partial }_t({{\mathsf{E}}_0}^{\mathrm{R}}{)}^2{\chi }_0$. It is smooth, and ${\square }^2\Psi ={\partial }_1{\partial }_t{\chi }_0$. If $\Psi$ vanished throughout $V$, this derivative would vanish there, hence everywhere since ${\chi }_0$ is supported in $V$. A compactly supported function with ${\partial }_1{\partial }_t{\chi }_0=0$ is zero: ${\partial }_t{\chi }_0$ is independent of $x^1$ and compactly supported, and then ${\chi }_0$ is independent of $t$ and compactly supported. This is a contradiction.

There is therefore a sign-definite nonzero patch of $\Psi$ in $V$. Choose a nonnegative observation test $f$ supported there. Then

Choose $\delta$ and then $\epsilon <\delta$ to satisfy (eq:separation). All data are now fixed. Lemma lem:ideal proves nonmembership in $\mathcal{J}$.