Lemma: Compact tensor-sector witness

There are a real smooth solution $U=(h,0)$ and a real compact test $f=(j,0)$ such that

The solution has uniformly compact spatial support on compact time intervals.

Proof. First choose a nonzero compact smooth symmetric tensor $q_{ij}$ on ${\mathbb{R}}^3$ satisfying

An explicit local construction is as follows. Put

choose a real constant symmetric matrix $S$ and a compact real smooth $\chi$, and set

The Fourier symbol of $D$ is $|\mathbf{k}{|}^2{\delta }_{ij}-k_i{k}_j$. It makes the displayed $q$ divergence-free and trace-free. Taking $S=\mathrm{diag}(1,-1,0)$ and a suitable nonzero $\chi$ makes $q\ne 0$.

Fix $t_0\in I$, and solve

The scalar normally hyperbolic Cauchy theorem applies componentwise. Divergence and trace satisfy the same equation with zero data and hence remain zero.

Set $h_{ij}=a^2{\gamma }_{ij}$, $h_{00}=h_{0i}=0$, and $\phi =0$. Substitution in (eq:curvature)–(eq:scalar) gives $\mathcal{H}U=0$ precisely when (eq:tensorequation) holds. The scalar variation vanishes because the background spatial Hessian of $\bar{\varphi }$ is proportional to ${\delta }_{ij}$. The gauge condition vanishes by the same trace and divergence identities.

Equivalently, this substitution reduces the quadratic action to

whose Euler equation is (eq:tensorequation). The background Euler identities remove the terms without derivatives. This provides a direct check of the normalization.

Choose a nonnegative time bump $\rho$ supported sufficiently close to $t_0$, and set

Its spacetime covariant divergence vanishes: the spatial components use ${\partial }_i{q}_{ij}=0$, and the temporal component uses ${\delta }^{ij}{q}_{ij}=0$. Therefore $K^{†}f=0$. Also

is strictly positive for a sufficiently small support of $\rho$. Finite propagation proves the support claim.