Decoupling (proof, exact in d )

• H1. $sh$ is longitudinal (above), so $X_{hh}$ depends on $G$ only through $p_m{G}^{(mn)(rs)}$. Because $p\in$ QdS, $p\cdot \tilde{g}=0$, so $\tilde{g}$ on the pair $(mn)$ drops out.

• H2/H3. $p_m(G_{D01}+G_{D03}{)}^{(mn)(rs)}=0$ identically in $d$. This holds for the 4D part and for the evanescent part separately: each of the six structures of $p_{m}G$ (for example ${\tilde{g}}_{rs}{p}_n$: $-\frac{i{p}^4}{8(d-1)}$ from D01 and $+\frac{i{p}^4}{8(d-1)}$ from D03) cancels exactly. Hence $X_{D01+D03}=0$ for all six external assignments (Q4S, QdS, mixed $\hat{}/\stackrel{~}{}$). With $D02=D12=0$, the hh slot receives nothing from the evaluated graphs.

• H4 (controls). $X_{D01}$ and $X_{D03}$ are each nonzero, and so are their evanescent parts alone. The cancellation is therefore a property of the SUM (the ghost loop cancels the longitudinal part of the bubble), not of each graph. The map is not trivially zero.

• Order. At one loop the ch projection is linear in ${\Gamma }_1$, so no product of evanescent poles occurs. The cancellation is exact in $d$ and not merely $O(ϵ)$.

Mutations, each rejected (summed slot becomes nonzero): wrong relative sign of D03; D03 doubled (symmetry factor); D03 with its evanescent part dropped (naive 4D); the trace operator ${\eta }_{mn}{p}_{\nu }$ in place of $sh$.

Verdict: DECOUPLED (proof), as the sum D01 + D03 (+ D02 = D12 = 0), and not graph by graph. No evanescent structure survives in the slot. No evanescent $h\cdot h$ counterterm is added, so the counterterm quotient is unchanged and no contract file changes for this. The summed evanescent poles of the hh Green function are nonzero (eight structures, recorded in the certificate) but transverse. Renormalizing the hh Green function itself would need evanescent $h\cdot h$ counterterms; that matters only outside this projection and is not claimed. D11, in the same row, is open.