Lemma: Grading and punctured identities

For $k=4,5,6$, define

Then

On the punctured domain,

All three graded subspaces are nonzero.

Proof. Each element of ${\mathcal{H}}_k$ is homogeneous of degree $-k$. If $u_4+u_5+u_6=0$, dilation gives $a^{-4}{u}_4+a^{-5}{u}_5+a^{-6}{u}_6=0$ for every $a>0$. Three distinct dilations give an invertible Vandermonde system, proving directness.

The products satisfy the distributional product rule where their wavefront criterion holds, giving (eq:Leibniz1)–(eq:Leibniz2). By Lemma lem:families, $\square \mathcal{S}=\mathrm{i}\delta$; hence $\square \mathcal{S}=0$ on the punctured domain, which proves (eq:trace).

The massless time-ordered square is nonzero and has scaling degree four; this is also stated in FR (Section 7, following Theorem 7.2). Thus ${\mathcal{H}}_4=\mathrm{span}\{{\mathcal{S}}^2\}\ne 0$. If ${\mathcal{H}}_5=0$, all first derivatives of ${\mathcal{S}}^2$ vanish by (eq:Leibniz1). On the connected punctured space it is then constant, incompatible with nonzero homogeneity of degree $-4$. If ${\mathcal{H}}_6=0$, equation (eq:Leibniz2) similarly makes every $\mathcal{S}{\partial }_{\alpha }\mathcal{S}$ constant, and its degree $-5$ makes it zero. This contradicts ${\mathcal{H}}_5\ne 0$.