Translation-invariant example with an asymmetric fiber matrix. This refutes the tempting stronger claim that abelian translation invariance plus Eq677 forces F to be symmetric. It is idempotent and satisfies Eq255, so it is not a counterexample to the main implication.
Carrier F11 x H, where H=F3[T]/(T^4-T^3+T^2-T+1) has 81 elements. Let k be multiplication by T and Q={1,3,4,5,9} modulo 11. Define (i,s)*(j,t)=(6j-5i,h_(j-i)(s,t)), with h_0(s,t)=s+k(t-s), h_d(s,t)=t for d in Q, and h_d(s,t)=-s-t otherwise.
The diagonal fiber satisfies Eq677 because k^4-k^3+k^2-k+I=0. For distinct base coordinates, the four differences encountered in Eq677 are d,5d,2d,2d; their QR pattern is AABB or BBAA. Both patterns of A(s,t)=t and B(s,t)=-s-t give the required identity.
All three fiber operations are translation invariant: for B this uses characteristic 3. The complete magma therefore has a regular C11 x C3^4 group of translation automorphisms.
Let B[i,j]=1 when i-j is a nonzero quadratic residue mod 11. Its fiber matrix is F=J_891+B tensor (81 I_81-J_81). In the construction's numbering, F[81,0]=81 but F[0,81]=1. Nevertheless F is normal, as expected for convolution on an abelian group; BB^T=B^TB=3I_11+2J_11. The rank is 881.
Verified all 793881 Eq677 pairs, Eq255, all 891 idempotents, the complete fiber matrix formula, and translation invariance under five generators. A separate scalar verifier checked the exported table, the asymmetry witness, and normality independently.
dwrensha · 2026-09-05 02:25:51