Lemma: Restricted sharp-generator separation

Let $f\in \mathcal{T}$. Suppose

for every smooth symmetric metric direction $k$ with $k_{00}=0$ and every smooth scalar $\phi$. Then $f=0$. It is enough to use compact metric directions supported in $M_{+}$.

Proof. For $\chi \in {C_c}^{\infty }(M_{+})$, let $H$ have its only nonzero component

Then $k_{00}=0$. Because $C$ has constant coefficients,

The vector $CH$ is compactly supported away from $\Sigma$, so has zero Cauchy data there. Cauchy uniqueness gives

Directly from (eq:C),

Choose $\phi =t$. This is an allowed smooth configuration. Alternatively, use a compact smooth scalar agreeing with $t$ near $\mathrm{supp}f$. The hypothesis becomes

for every $\chi$. Hence ${\partial }_{t}f=0$. A smooth function independent of $t$ on $M_{+}$ and compactly supported there must vanish. Thus $f=0$.

Real $\chi$ already suffice for complex $f$, by applying the distributional identity to its real and imaginary parts.