C01.1 · Derivation

The multiplication tensor

The table in the stdout block below is generated from the product, not entered as an expected answer. Each basis-pair result is checked to be one signed basis vector and compared with the separately implemented quaternion doubling. The output reports **64 signed-unit product comparisons**, with **24 negative products**. These are actual counts for this test, not the old production record’s unexplained sign-count assertion. They are not 64 independent experiments or separate peer reviews.

The norm, null cone and zero divisors

The run gives

\[ N(x)=s^2+E_1^2+E_2^2+E_3^2-F_1^2-F_2^2-F_3^2-t^2, \]

with diagonal Gram matrix \(\operatorname{diag}(1,1,1,1,-1,-1,-1,-1)\) and inertia \((4,4,0)\), ordered as positive, negative, zero. These are outputs of the symbolic determinant and Hessian calculation, not a count of the signs of the multiplication table’s diagonal.

Indeed the computed table has \(e_i^2=-1\), while \(f_i^2=l^2=+1\). Since the imaginary basis elements change sign under conjugation, their norm signs are the **opposite** of their square signs. Confusing these two counts is an error in the supplied cold verification of the old table.

The null cone is the quadratic locus \(N(x)=0\). The output exhibits nonzero zero divisors:

\[ (1+l)(1-l)=0,\qquad N(1+l)=N(1-l)=0. \]

Their sum has norm four, so the entire null cone is not a vector subspace. The span of \(f_1,f_2,f_3,l\) is negative definite under the computed norm and is not the null span requested by P1. A different subspace really is totally null: in Zorn coordinates impose \(b=0\) and \(v=0\), leaving \(a,u\) free. The run substitutes those constraints and obtains zero norm with four free coordinates. This is a null linear subspace **inside** the cone, not the whole cone. A nilpotent example is also computed: \((e_1+f_1)^2=0\).

For nonzero norm, the checked conjugation identities supply the two-sided inverse \(\bar X/N(X)\). For nonzero null \(X\), \(\bar X\) is a nonzero annihilating partner. Consequently the algebra is a split composition algebra, not a division algebra. No “division-algebra axioms up to alternativity” conclusion is used.

Alternativity beyond a basis spot check

For the associator \([x,y,z]=(xy)z-x(yz)\), checking only \([b_i,b_i,b_j]=0\) would not be a proof for arbitrary linear combinations. The run therefore checks both polarized identities

\[ [b_i,b_j,b_k]+[b_j,b_i,b_k]=0,\qquad [b_i,b_j,b_k]+[b_i,b_k,b_j]=0 \]

over every basis triple. It reports **1,024 vector equalities**, or **8,192 scalar coefficients**, all zero. Trilinearity then extends these antisymmetries to arbitrary real vectors; setting the exchanged arguments equal gives left and right alternativity. This explains exactly why this finite coefficient check suffices for the polynomial identities. The separate **128** repeated-basis checks are redundant controls, not the general proof by themselves.

The algebra is not associative: the computed witness is

\[ [e_1,e_2,f_1]=-2f_2\ne0. \]

The run also checks the stated deterministic sample of integer pairs. This sample is an additional implementation control; the general claims above rely on coefficientwise identities, not on random sampling.

Negative control: the supplied March table

The program extracts the table from the actual attached production JSON, without correcting its entries, and extends it bilinearly. It finds **30 failed left-alternative basis tests out of 64**, and **30 failed right-alternative tests out of 64**. One witness is

\[ [e_1,e_1,e_2]_{\rm March}=-2e_2\ne0. \]

This is stronger than merely observing that the original generator did not show its work: the supplied table fails the algebraic identity it asserted. It is also not a harmless basis relabeling, because a change of basis cannot turn a failed polynomial identity into a valid one.

Under the conjugation fixing the scalar basis element and negating the other basis elements, the old table yields the candidate form with inertia \((5,3,0)\), not the intended neutral form. The code checks that its product with the conjugate really gives that scalar quadratic form. Composition already fails on a basis pair: the candidate norm of \(e_1e_2\) is minus one, whereas the product of the separate norms is plus one. These are computations from the supplied table, not claims that the old verifier established them. The original verifier’s signature confirmation must not be copied into the new article.