Why it is a question
A structured or translational score is a norm of F_{h⊴r} x_h − F_{t⊴r} x_t, plus a translation, which is a coboundary on the edge (THEORY-299). A bilinear score x_hᵀ M_r x_t is not visibly of that form. It is a pairing, not a difference. LIT-232 says such models “do not fit” (App. A.1), and NOTE-843 rates that moderate: stated, not proved impossible. The paper suggests sheaves valued in other categories might take them in (App. E). Until 2026-10-19 THEORY-299’s promote_when asked for “a proof, or a counterexample”. At the owner’s direction, from the promotion audit’s batch 013 (the audit’s decision file for THEORY-299, ADR-047), THEORY-299 was promoted without it, because its title does not claim it, and the request moved here.
What would answer it
A proof that no sheaf of vector spaces on the knowledge graph, with embeddings as 0-cochains, has a coboundary whose norm (or a fixed monotone function of it) reproduces a bilinear score, or a construction that does. A construction must say what it changes: the category of the stalks, the cochain degree, or the scoring function. Benchmark results cannot settle it, because they bear on how well the models work, not on what they are.