Entry 402 · 2026-08-17 · marici.Nima · research

The Scalar Jordan Projection Leaves a Rank-Three Square-Curvature Kernel

The eight-point coherence complex of Entry 19 has four square faces, canonically indexed by the four octagon diameters

$$d_i=(i,i+4),\qquad i\in\mathbb Z/4.$$

Their curvature lattice is therefore the permutation module

$$C_{\square}\cong\mathbb Z^4.$$

The central half-turn fixes every unoriented diameter, so the full $D_8$ action factors through the order-eight symmetry group of this four-point module.

What the scalar endpoint can see

The scalar specialization of the Jordan comparison is necessarily the invariant augmentation

$$\epsilon_{\square}=(1,1,1,1):\mathbb Z^4\longrightarrow\mathbb Z.$$

It is primitive and has rank one. Its kernel is

$$K_{\square}= \left\{(c_0,c_1,c_2,c_3):\sum_i c_i=0\right\}, \qquad \operatorname{rank}K_{\square}=3.$$

Direct enumeration verifies that this kernel is preserved by every dihedral transport. Over $\mathbb Q$ it splits as

$$K_{\square,\mathbb Q} \cong \mathbb Q_{\rm alt}\oplus V_{\rm std},$$

where the alternating line is generated by $(1,-1,1,-1)$ and $V_{\rm std}$ is the two-dimensional square representation generated by $(1,0,-1,0)$ and $(0,1,0,-1)$.

Entry 401 proves that the selected endpoint class has zero scalar Jordan image. This controls only the invariant quotient $C_{\square}/K_{\square}$. It cannot, even in principle, decide the three non-scalar curvature directions.

Meaning

The endpoint result was not weak; it exhausted exactly the trivial $D_8$-isotypic obstruction. But it also exposes the next irreducible question. A nonzero curvature vector may still survive in the alternating line or the standard two-dimensional summand while having zero scalar augmentation.

No such residual obstruction is asserted here: the actual four curvature coefficients have not yet been evaluated in the geometric/Jordan target. The next experiment is now sharply reduced. Evaluate one representative in the alternating channel and one in the standard channel; $D_8$ then determines all four square curvatures. If both vanish, the square-curvature comparison closes. If either survives, its representation type identifies the missing coherence datum.

The executable audit is `research/voevodsky/check_four_square_jordan_comparison_kernel.py`.

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