Manuscript II-B@1 · II-B-v1.tex · lemma · bfd2eb3c

Lemma: Two original columns

Complete contacts and the exact three-row certificate › Original pivots, rank, fit, and kernels · lines 1081–1106 · proof included or linked

II-B@1#lem:pivots

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

Proof. At $β_{+}$ , only the longitudinal tensor can occupy the trace in $O_{a}$ . The other species-one factor receives $A$ ; the species-two factors receive $B, B$ in two orders. Since

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

each order contributes

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

The total is $8$ . The trace of a physical species-two tensor makes $O_{b}$ vanish there. Species exchange gives the other row, and blindness gives the first entries.

Displayed formulas in this passage

unlabeled · line 1083
$(N_{S})_{:, ℓ_{a}} = 080, (N_{S})_{:, ℓ_{b}} = 008 .$
unlabeled · line 1094
$tr_{η} G = - 2, k_{1} \cdot k_{2} = - 2, A_{μν} B^{μ ρ} B^{ν}_{ρ} = - 1,$
unlabeled · line 1099
$- (k_{1} \cdot k_{2}) (tr_{η} G) A_{μν} B^{μ ρ} B^{ν}_{ρ} = 4.$

Lemma: Two original columns

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