Theorem: Plateau-switch obstruction

There exist

and a real compact-data free scalar ${\varphi }_0$, such that

1. $f\ge 0$ and $\int f{\mathrm{d}}^{4}x=1$;

2. $g=1$ on a neighborhood of $\mathrm{supp}f$;

3. $gK\xi =0$ everywhere;

4. both $0$ and $K\xi$ satisfy the vacuum linearized Einstein equation;

5. $${{\mathcal{A}}_{f,g}}^{(1)}(K\xi ,{\varphi }_0)-{{\mathcal{A}}_{f,g}}^{(1)}(0,{\varphi }_0)\le -\frac{1}{2}.$$

Thus the candidate (eq:A1g) does not descend through (eq:free-representative-identification), even when the switch is constant near the observation region.

Proof. Choose $t_{\ast }\in (0,T)$. At that time prescribe compact smooth scalar data

where $\chi =1$ near $x=0$. The Klein–Gordon solution has compact smooth Cauchy data also at $t=0$, by finite propagation. At $x_{\ast }=(t_{\ast },0)$,

Choose a relatively compact neighborhood $U⋐(0,T)\times {\mathbb{R}}^3$ on which ${\partial }_1{\varphi }_0\ge 1/2$. Choose a nonnegative unit-integral test $f$ supported in $U$.

Choose a compact smooth switch $g$ equal to one near $\mathrm{supp}f$. Next choose ${\chi }_{\xi }\in {C_c}^{\infty }((0,T)\times {\mathbb{R}}^3)$ equal to one on a neighborhood of $\mathrm{supp}g$, and set

All derivatives of $\xi$ vanish near $\mathrm{supp}g$. Therefore $gK\xi =0$, and hence

The zero metric has zero response as well.

On the other hand, $w[K\xi ]=\xi$, and $\xi ={\partial }_1$ on $\mathrm{supp}f$. Thus

Finally, $G^{(1)}K=0$ proves the vacuum statement.