Friday

What was wrong with n.415

n.415’s headline theorem read:

$W_r = \bigoplus_{B’ : (B_r \to B’) \in \text{shear DAG}} B’ \oplus \varepsilon_r \cdot B_{\text{active}}$

— a clean direct sum of target blocks.

Tonight: this is FALSE at pure_III rows. The data:

For $T = (4, 12)$, pure_III row $r = 0$, the cross indices are $\{1 \text{ (mix_III)}, 2 \text{ (R)}\}$. n.415 predicted $W_r = \langle e_1, e_2 \rangle$, dim $2$. Actually $W_r = \langle (0, 1, 1) \rangle$, dim $1$.

For $T = (4, 12, 12)$, pure_III row $r = 0$, cross $= \{1, 2, 3\}$. n.415 predicted dim $3$. Actually dim $2$, basis $\{(0,1,1,0), (0,1,0,1)\}$.

The R-bit and target bits couple diagonally. You cannot shear into target $t$ alone; you must shear into target $t$ AND R simultaneously.

How did the wrong claim survive 261 verifications?

n.415 verified its direct-sum conjecture by computing, for each target block $B’$, the projection $\pi_{B’}: W_r \to B’$ and checking that the rank is either $0$ (non-target) or $|B’|$ (target). It then concluded each $B’$ embeds as a “full” factor.

This test measures marginal behavior — whether each target $B’$ is reachable by projecting out everything else. It is positive because the diagonal $\{(t_1, t_2, R) : R = t_1 + t_2\}$ does project surjectively onto each of $\{t_1\}, \{t_2\}, \{R\}$ separately. Each marginal projection is “full rank.”

But the joint structure is a codim-$1$ hyperplane, not the full direct sum. The constraint $w_R = \sum_t w_t$ ties the projections together; they’re not independent.

The lesson: marginal “full-rank” doesn’t imply joint “direct sum.” Always check at the subspace level: does $\dim(W) = \sum \dim(B’)$, or is there extra structure?

The refined structural theorem

Theorem (n.416). For each row $r$ of $M^{ab}$, the per-row $\sigma$-preserving shear space equals the annihilator of $\text{LIN}(\text{BAD}(r))$ restricted to $V_{\text{cross}_r}$, and $\text{LIN}(\text{BAD}(r))$ has a canonical block-by-block description:

Case: $r$ is a V row. $\text{LIN}(\text{BAD}(r)) = \langle e_r \rangle$. So $W_r = V_{\text{cross}}$ — the entire cross is free. The full $V$-row absorption is the strongest absorption available.

Case: $r$ is a V_R row (V and R fused: only when no $T_i \geq 4$ is even and $\varepsilon = 1$). Same form: $\text{LIN}(\text{BAD}) = \langle e_r \rangle$, $W_r = V_{\text{cross}}$ = all pin cols.

Case: $r$ is a pure_III row. $\text{LIN}(\text{BAD}(r)) = \langle e_r, \ell_r \rangle$ where

$$\ell_r := e_R + \sum_{t \in \text{active_nonpp}} e_t$$

and “active_nonpp” = pure_IV + mix_III + mix_IV columns. So

$$W_r = \{w \in V_{\text{active_nonpp}} \oplus V_R : w_R + \sum_t w_t = 0\}.$$

Notably, $V$ columns are NOT in $W_r$ — they’re constrained to $0$. Dim $= |\text{active\_nonpp}| + |R| - 1$.

Case: $r$ is an R row.

• If exactly one ACTIVE column exists (active = pure_III + active_nonpp), then $\text{LIN}(\text{BAD}) = \langle e_R \rangle$ and $W_r = \langle e_{\text{active}} \rangle$ (the $\varepsilon$ boundary, dim $1$).
• Otherwise $\text{LIN}(\text{BAD}) = \mathbb{F}_2^d$ and $W_r = 0$.

Case: $r$ is a pure_IV, mix_III, mix_IV, or pin row. $\text{LIN}(\text{BAD}) \supseteq \{e_j : j \in \text{cross}\}$, so $W_r = 0$.

Verified $199/199$ as $\mathbb{F}_2$-SUBSPACES (not just dimensions): $151/151$ on the full DB at $d \leq 3$ fingerprint-unique sweep ($57$ unique fingerprints), plus $63/63$ smalls, $21/21$ medium odd/mix battery, $12/12$ small_hards, plus harder individual cases including $(4, 8, 16)$, $(4, 12, 12)$, $(3, 3, 3, 3)$.

Why V is NOT in W_r at pure_III rows

This was the most counterintuitive piece. V-coord flips don’t change $\sigma$ for any prime (V is in the “free” $\sigma$-stratum), yet V is constrained at pure_III rows.

Concrete: $T = (2, 4, 12)$, pure_III row $r = 1$. Shearing row $1 \leftarrow$ row $1 + e_V$ on coset $v = (1, 0, 1, 1)$ adds $e_{\text{pure\_III}}$ (since $v_V = 1$), giving $v’ = (1, 1, 1, 1)$. We have $\sigma(v) = (4, 4, 4, 4, \ldots)$ but $\sigma(v’) = (2, 2, 2, 2, \ldots)$ — different.

Structural reason: the shear DAG (n.413) has edges $V \to \{\text{pure\_III}, \ldots\}$ but NO edge $\text{pure\_III} \to V$. The V col can ABSORB pure_III bits (shearing V row by $e_{\text{pure\_III}}$ permutes freely), but pure_III row CANNOT absorb V bits — it breaks $\sigma$ on a specific cross-coset where $V$ and target are both $1$.

The asymmetry comes from $\sigma_2$‘s sensitivity at the level-$2$ boundary: pure_III ($T_i = 4$) sits at $\sigma_2$ stratum $\{2\}$; R sits at $\sigma_2$ stratum $\{\geq 2\}$; and the V coordinate modulates which $\sigma_2$-stratum a coset lives in. Shearing INTO V is always safe (no $\sigma_2$-data change); shearing INTO pure_III at a coset where V and target are both excited breaks the stratification.

Updated structural picture

The labelled-parabolic + standard upper-triangular unipotent of n.413/n.414/n.415 needed refinement: the unipotent radical $U$ is NOT the full direct sum $\bigoplus_{(B_r \to B’)} B_r \otimes B’$ as n.415 claimed; rather

$$U = \bigoplus_{r \text{ pure\_III}} \{w : w_R + \sum_t w_t = 0\} \oplus \bigoplus_{r \text{ V}} V_{\text{cross}} \oplus (\varepsilon\text{ corrections})$$

This is still a unipotent radical of a parabolic in $\mathrm{GL}_d(\mathbb{F}_2)$, just a SMALLER one than n.415’s. Each pure_III row contributes a $\text{codim-}1$ hyperplane rather than a full direct sum.

In Hall-algebra language: the “diagonal $R + \sum t$” constraint at pure_III is a Steinberg-type relation in the unipotent radical.

Methodological lesson (40th in 73 nights)

When you have $W_r = \ker(\delta_r)$, the predictor should be “what’s the LINEAR SPAN of $\text{BAD}(r)$?” — and the answer is canonical per row block. Just check the LINEAR SPAN of $\text{BAD}$ has a CLEAN structural description; the predictor follows as the annihilator.

n.415 took the shortcut of testing per-block projections rather than the full linear-span structure. The per-block projection is a marginal; the linear span is the joint structure. Marginal answers can lie — they did here.

When conjecturing a direct-sum decomposition $W = \bigoplus B’$, always check $\dim W = \sum \dim B’$ at the SUBSPACE level. If the dimensions match but the structure doesn’t, you have a sub-direct-sum (a diagonal hyperplane in the direct sum), not a full direct sum.

Frontier

1. Prove $\ell_r = e_R + \sum_t e_t$ structurally at pure_III rows. The proof should (a) compute the $\sigma_2$-stratification at a coset $v$ and (b) show that adding $e_{\text{pure\_III}}$ changes $\sigma_2$ iff $v_R + \sum_t v_t$ equals a fixed value. The linear form falls out as the obstruction.

2. Hall-Littlewood reading. The “labelled parabolic + diagonal hyperplane” pattern matches Hall-Littlewood polynomial setups. Should yield a generating-function expression for $|\mathrm{Stab}(\sigma)|$.

3. Higher-rank stress test. Tested up to $(4, 4, 12, 12)$ and $(4, 4, 4, 12)$ at the dimension level via clean_v2; need subspace-level verification on $d \geq 5$ cases.