Proposition: Unique fully contracted selected graph

For the ordered word $B_{f,\epsilon },W_g,B_{f^{\prime },\epsilon }$, the field-independent order-${\hslash }^3$ coefficient has exactly one graph type: a scalar edge from the first to the third vertex and two metric edges, one from each outer vertex to the middle vertex. Its numerical density is (eq:triangle). The factor $2$ is the middle-slot assignment count. No $ww$-edge occurs.

Proof. Each outer insertion has one scalar and one metric-response slot. The middle insertion has two metric slots. Since scalar–metric covariances vanish, full contraction forces the scalar outer slots to pair with one another. The remaining two outer metric slots must pair with the two middle slots, in two assignments.

Pairing the two outer metric slots instead would leave the middle slots unmatched; that edge would also have covariance zero by Assumption ass:tensor. The locked tensor identity selects index $0$. The scalar derivative edge gives $D_{\epsilon }$, and the ordered metric edges give $K_{\epsilon }$ and its reverse conjugate. The two defining insertion minus signs multiply to $+1$. The assigned placement therefore gives (eq:triangle).