The cubic tree is not individually closed

The witness of Section sec:split also evaluates (eq:dC). For $\gamma =t\eta$, $A\gamma =0$. In the causal cylinder,

Consequently (eq:duvac) gives

For this tensor,

in the cylinder. With the massless scalar used there,

It follows that

Thus $s_0\mathcal{T}=0$ does not imply closure of this cubic Einstein response. Equation (eq:cubicwitness) is a computed variation, not a failed split cancellation: the latter already canceled $-I$ against $+I$ in Theorem thm:cancel.

A complete interacting second-order identity would have to include the other gravitational, scalar, ghost, gauge-fixing, and antifield terms with their proper response maps. We do not claim that identity from (eq:dJ). In particular, membership of ${\mathcal{B}}_f$ and $\mathcal{T}$ in ${\mathcal{E}}_H$ is not a quantum retarded-product theorem for the unswitched cubic Einstein source.