A Proof of the Below–Krummeck–Richter-Gebert Conjecture on Cyclic Determinant Quotients

For points A_1, …, A_n, A'_1, …, A'_n in K², where K is a field, write [XY] for the 2 × 2 determinant and let α(π) be the cyclic quotient ∏_k [A_{π_k} A_{π_{k+1}}] / ∏_k [A'_{π_k} A'_{π_{k+1}}], with indices read cyclically. Below, Krummeck and Richter-Gebert conjectured in 2003 that Σ sgn(π) α(π) = 0, where the sum runs over the permutations π of {1, …, n} with π_1 = 1. Richter-Gebert posed the problem again at Oberwolfach in 2009. It had been proved for n = 5 and checked by computer algebra for n = 6, and it follows from a symmetry argument when n ≡ 0, 3 (mod 4). We prove the identity for every n ≥ 3 and every field, assuming only that [A'_iA'_j] ≠ 0 for i ≠ j. Equivalently, the associated bracket polynomial vanishes identically. The proof writes the numerator as the trace of a product of traceless 2 × 2 matrices. The reciprocals of the denominators are Parke–Taylor factors. They satisfy a shuffle identity, known from the Kleiss–Kuijf relations of gauge theory, and hence are the coefficients of a Lie polynomial. Mapping this Lie polynomial into 2 × 2 matrices over an exterior algebra, where the commutators of the generators are central, gives zero. The same argument shows that Σ_σ sgn(σ) M_{σ_2} ⋯ M_{σ_n} / ([A'_1 A'_{σ_2}] ⋯ [A'_{σ_n} A'_1]) = 0 for all traceless 2 × 2 matrices M_2, …, M_n when n ≥ 4, and it recovers the vanishing of the same sum without the matrices. This is an unrefereed note. Unrefereed preprint released for independent mathematical scrutiny. Publication on Zenodo does not constitute peer review. AI-assisted tools supported research, computation, proof development, and manuscript preparation. The author remains responsible for all claims and the final text. Corpus identifier: OWR-3385-008 (ulamai/UnsolvedMath; Oberwolfach Report 2/2009, Problem 6 (J. Richter-Gebert): the Below–Krummeck–Richter-Gebert alternating-sum conjecture for cyclic determinant quotients).