Proposition: Unique fully contracted selected triangle

The fully contracted order-${\hslash }^3$ graph for

has one scalar edge between the first and third vertices and one metric edge from each sharp vertex to the middle vertex. The two assignments to the middle slots give the factor $2$ in (eq:T). No $ww$-edge occurs.

Proof. Each sharp vertex has one scalar and one metric slot, and the middle vertex has two metric slots. Since mixed scalar–metric covariances vanish, the scalar slots must contract with each other. The remaining sharp metric slots must contract with the two middle metric slots, in two possible assignments.

An edge joining the two sharp metric slots would leave the middle slots without partners, and its covariance is also zero by (eq:tensor). The inherited tensor identity selects vector index $0$ at both sharp vertices. Thus the scalar edge gives $D_{\epsilon }$, and the mixed edges give the two $K_{\epsilon }$ factors. The minus sign in the first $B_f$ and the minus sign in the third $B_{f^{\prime }}$ multiply to $+1$, as on the VII-D convention sheet VIID (sc:spectral, prop:positive).