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.