Theorem: Fixed-preparation residual and plateau intertwining

For every permitted contact prescription,

If $\xi$ is supported in the plateau $t>\epsilon$, the right side is zero. Thus multiplication by the fixed ramp intertwines that plateau-supported linear action, not the unrestricted action.

For every nonconstant ramp (eq:ramp) and every $T>\epsilon$, there exist real compact $\xi$ in the preparation layer and a real compact tensor test $j$ supported in $t>T$ such that the residual is nonzero. For that detecting pair its value is unchanged by all contacts (eq:contacts).

The residual concerns this prepared external source; it is not the companion's density-weighted cubic source.

Proof. By (eq:productrule), the change in the source is $K({\sigma }_{\epsilon }\xi )-B_{\epsilon }(\xi )$. The first term is annihilated by Theorem thm:ward, proving (eq:residual). On the plateau $\partial {\sigma }_{\epsilon }=0$ on $\mathrm{supp}\xi$, so $B_{\epsilon }(\xi )=0$.

It remains to prove a late nonzero detector. Choose a real nonzero $\chi \in {C_c}^{\infty }((0,\epsilon ))$ such that $b(t)={\sigma }_{\epsilon }^{\prime }(t)\chi (t)$ is nonzero. Take

where $\psi$ is real, compact, and chosen with $\hat{\psi }(0,q,0)\ne 0$ for some $q\ne 0$. Then $B_{\epsilon }(\xi )$ has only its $01$ and $10$ components, equal to $b(t)\psi (\mathbf{x})$.

Fix spatial momentum $\mathbf{p}=(0,q,0)$. For timelike $p=(\omega ,0,q,0)$, the transverse trace of this tensor is zero because $p^1=0$. Its spin-two projection is nonzero when its scalar Fourier factor is nonzero and $q\ne 0$. The entire function $\hat{b}(\omega )$ is not identically zero; it therefore does not vanish on every interval above ${\omega }_{\ast }=\sqrt{q^2+a}$. Consequently the vector-valued function

has a nonzero discontinuity across an open portion of $\omega >{\omega }_{\ast }$, by (eq:absorptive). Local real contacts have no such discontinuity.

Suppose $u={\Pi }^{\mathrm{R}}\ast B_{\epsilon }(\xi )$ had compact support. Its Fourier transform would then be entire, including its restriction to this complex $\omega$ line. On $|\omega |<{\omega }_{\ast }$, it agrees with the analytic spectral expression. The identity theorem on

would make that entire function agree with both boundary values of the spectral expression, forcing its discontinuity to vanish. This contradicts the nonzero spin-two discontinuity. Thus $u$ is not compactly supported.

Retarded support gives $\mathrm{supp}u\subset J^{+}(\mathrm{supp}{B}_{\epsilon }(\xi ))$. If its support were bounded above in time, this causal containment and the compact source support would make it compact in spacetime. Hence its support reaches times greater than every $T$. There is a compact real tensor test $j$ supported in $t>T$ with $⟨j,u⟩\ne 0$. The reality of $u$ permits a real detector.

A contact changes $u$ by a finite differential operator applied to $B_{\epsilon }(\xi )$, supported in the preparation layer. Such a change is invisible to this late $j$. The nonzero residual is therefore contact independent for the constructed detector.