Collinear Triple Fixed Point Census
Abstract
Nonidentity translations fix precisely the cosets of an admissible three-cycle.
For every positive modulus n, let G(n) be ZMod(n) times ZMod(n). Let T(n) consist of unordered finite subsets S of G(n) with exactly three elements, with both coordinate projections injective on S, and with the following determinant identity for every choice of points in S:
Write F(n,t) for the members of T(n) fixed by translation by t. Addition to a set denotes its pointwise translate; its cardinality counts each unordered set once. The identity translation is excluded only from the zero-count and quartet assertions. No prime-modulus hypothesis is used.
The displayed floor of n squared divided by three denotes natural Euclidean division, with any remainder discarded.
Theorem 1.1 (Three-cycle classification, exact fibers, and census).
Proof. Machine-checked in Lean as D5/S3/Factorization/CollinearTripleFixedPointCensus.fixed_point_census (✓ std3). ∎
Source. Repository-derived.
Commentary.
If three times t is zero and both coordinates of t are nonzero, each coordinate has additive order three. The set C of its first three multiples consequently has three points and both coordinate maps are injective. Differences of multiples of the same vector have zero determinant, so C belongs to T(n). Translation by t cyclically permutes C.
If t fixes S, choose p in S. Invariance places all three points of p+C in S; equality follows from their common cardinality three. Conversely, every translate of C is fixed. A translation stabilizes C exactly when its vector belongs to C: necessity follows by translating zero, and sufficiency follows from the cyclic permutation. This gives the displayed exact fibers and a stabilizer of cardinality three. Orbit-stabilizer then gives n squared divided by three fixed triples.
For any nonzero fixing vector, summing the three points forces three times the vector to vanish. A zero first or second coordinate would contradict the corresponding injectivity, because a nonzero translation cannot fix an individual point. For n=3m, the standard representative of a coordinate annihilated by three is zero, m, or 2m. Keeping both coordinates nonzero gives exactly the four displayed pairs. Every other nonzero translation fixes no member of T(n).