Lemma: Sharp charge separation for every nonzero test

For every nonzero $f\in \mathcal{T}$, there is a real smooth scalar configuration $\varphi$, which can be chosen compactly supported in $M_{+}$, such that

No directional descendant is needed.

Proof. The metric derivative is

Choose $x_0=(t_0,{\mathbf{x}}_0)$ with $f(x_0)\ne 0$. Select $\mathbf{z}$ with

Since $t_0>0$, a neighborhood $U⋐M_{+}$ of $x_0$ can be chosen away from the spatial point $\mathbf{z}$. Let

on that neighborhood. Choose $\chi \in {C_c}^{\infty }(U;\mathbb{R})$, nonnegative and positive at $x_0$, and form the complex scalar

Extended by zero outside $U$, it is compact smooth.

On the cone $t=r(\mathbf{x})$,

Using (eq:advancedtrace) and (eq:Bderivative),

The denominator is bounded away from zero on the relevant support. Positivity holds on an open spatial patch near ${\mathbf{x}}_0$.

Write ${\varphi }_{\mathbb{C}}={\varphi }_1+i{\varphi }_2$ with real compact smooth ${\varphi }_1,{\varphi }_2$. The trace is complex linear in the scalar configuration. Since (eq:positivewitness) is nonzero, at least one of the traces for ${\varphi }_1,{\varphi }_2$ is nonzero. Choose that real configuration. The metric configuration may be zero, since (eq:Bderivative) is independent of $h$.