Manuscript II@2 · II.tex · proposition · 50d64def

Proposition: Exact original pivots

Complete matrix encoding and the finite certificate › Original pivot columns · lines 1504–1538 · proof included or linked

II@2#prop:pivots, II@2#eq:pivots

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(N_{S})_{:, ℓ_{a}} = 080, (N_{S})_{:, ℓ_{b}} = 008 .

Proof. At $β_{+}$ , the traced factor of $O_{a}$ must receive the longitudinal leg; all physical tensors are trace free. The other species-one factor receives $A$ . The two species-two factors receive $B, B$ in two orders. The required contractions are

tr_{η} G = - 2, k_{1} \cdot k_{2} = - 2, A_{μν} B^{μ ρ} B^{ν}_{ρ} = - 1.

Each assignment contributes

- (k_{1} \cdot k_{2}) (tr_{η} G) A_{μν} B^{μ ρ} B^{ν}_{ρ} = - (- 2) (- 2) (- 1) = 4.

Thus $n_{a} (β_{+}) = 8$ .

For $O_{b}$ , its traced species-two factor receives a physical $B$ , so the response is zero. Species exchange gives the other nonzero entry. Blindness gives both first entries.

Displayed formulas in this passage

II@2#eq:pivots · line 1506
$(N_{S})_{:, ℓ_{a}} = 080, (N_{S})_{:, ℓ_{b}} = 008 .$
unlabeled · line 1522
$tr_{η} G = - 2, k_{1} \cdot k_{2} = - 2, A_{μν} B^{μ ρ} B^{ν}_{ρ} = - 1.$
unlabeled · line 1527
$- (k_{1} \cdot k_{2}) (tr_{η} G) A_{μν} B^{μ ρ} B^{ν}_{ρ} = - (- 2) (- 2) (- 1) = 4.$

Proposition: Exact original pivots

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