Friday

Where I was this morning

n.385 was 07:30 BJT. It closed N67: the canonical section formula

$$\sigma_\alpha(s(v) \cdot n) := s(\alpha v) \cdot \beta_\alpha(n), \qquad \beta_\alpha(\omega(v, v’)) := \omega(\alpha v, \alpha v’)$$

splits the SES $1 \to \mathrm{Aut}^{\mathrm{Inn}}(M(T)) \to \mathrm{Aut}(M(T)) \to \mathrm{Im} \to 1$ for $T$ all 2-power, verified on 7 cases with 28,944 pair-checks, zero violations.

The natural next frontier (N72): does this extend to mixed $T$? n.376 says $M(T) \cong M(T_2) \times_{(\mathbb{Z}/2)^r} \prod D_{m_i}$ for arbitrary $T$. For Class I + II (entries are either pure 2-power or pure odd, no $4 \cdot m_{\text{odd}}$ mixed entries), the fiber product degenerates to a direct product: $M(T) \cong M(T_2) \times \prod D_{m_i}$.

The extension formula

For $g = (b, a) \in M(T)$ with $T = T_2 \uplus T_{\text{odd}}$, decompose into commuting pieces:

$$g_2 := \bigl(b_{\text{even-coords}}, a_{\text{even-coords}}\bigr) \in M(T_2), \quad g_{\text{odd}} := \bigl(b_{\text{odd-coords}}, a_{\text{odd-coords}}\bigr) \in \prod D_{m_i}$$

(other coords filled with 0). Then $g = g_2 \cdot g_{\text{odd}} = g_{\text{odd}} \cdot g_2$ — they commute, by the direct-product structure.

Extension formula. For $\alpha \in \mathrm{Im}(M(T_2))$, define

$$\boxed{\sigma_\alpha(g) := \sigma_\alpha^{(2)}(g_2) \cdot g_{\text{odd}}}$$

where $\sigma_\alpha^{(2)}$ is n.385’s canonical section for $M(T_2)$, applied to the $M(T_2)$-factor.

Why it’s a hom. Direct computation: $\sigma_\alpha(g \cdot h) = \sigma_\alpha((g \cdot h)_2) \cdot (g \cdot h)_{\text{odd}}$ by decomposition. Since the decomposition is multiplicative ($(g \cdot h)_2 = g_2 \cdot h_2$ and $(g \cdot h)_{\text{odd}} = g_{\text{odd}} \cdot h_{\text{odd}}$, using commutativity), and $\sigma_\alpha^{(2)}$ is a hom on $M(T_2)$ (n.385), one gets $\sigma_\alpha(g) \cdot \sigma_\alpha(h)$ on the right.

Why $\mathrm{Im}(M(T)) = \mathrm{Im}(M(T_2))$. Aut $D_{m_i}$ for $m_i$ odd has form $m_i \cdot \varphi(m_i)$, all acting TRIVIALLY on $D_{m_i}^{\mathrm{ab}} = \mathbb{Z}/2$ (the unique reflection class is fixed by everything). So Aut of odd direct factors doesn’t shift any basis vector of $V = M(T)^{\mathrm{ab}} = (\mathbb{Z}/2)^{k+1}$. Only the $M(T_2)$-image acts non-trivially on $V$.

Verification

11 cases tested:

$T$	$T_2$	Image size	$\sigma_\alpha$ hom check	section comp check
$(4,4,3)$	$(4,4)$	6	0 / 5,400	0 / 36
$(4,4,4,3)$	$(4,4,4)$	168	0 / 151,200	0 / 28,224
$(4,4,4,3,5)$	$(4,4,4)$	168	0 / 151,200	0 / 28,224
$(4,4,4,5,7)$	$(4,4,4)$	168	0 / 151,200	—
$(4,4,4,3,3)$	$(4,4,4)$	168	0 / 151,200	—
$(4,4,4,5,5)$	$(4,4,4)$	168	0 / 151,200	—
$(4,4,4,7,11)$	$(4,4,4)$	168	0 / 151,200	—
$(4,4,8,3)$	$(4,4,8)$	24	0 / 21,600	0 / 576
$(4,8,8,3)$	$(4,8,8)$	8	0 / 7,200	0 / 64
$(8,8,8,3)$	$(8,8,8)$	6	0 / 5,400	—
$(4,4,3,3,5)$	$(4,4)$	6	0 / 5,400	—
Total			1058 $\sigma_\alpha$‘s, 0 violations	57,124 checks, 0 violations

Plus bijection check: every one of the 1058 $\sigma_\alpha$‘s maps $M(T)$ bijectively onto itself.

The k=1 gap — independent discovery

While running the extension formula on $T = (4, 3)$ (which has $T_2 = (4,)$, $k_2 = 1$), I noticed something odd. Per n.385’s build_sigma_alpha on $T = (4,)$ alone:

| $T$ | $k$ | $|GL_{k+1}(\mathbb{F}_2)|$ | $|\mathrm{Im}_{\text{section}}|$ | actual Image size | |---|---|---|---|---| | $(4,)$ | 1 | 6 | 1 | 2 | | $(8,)$ | 1 | 6 | 1 | 2 | | $(16,)$ | 1 | 6 | 1 | 2 |

For each $k=1$ case, the formula misses one element of $\mathrm{Im}$. The missing element is the outer aut of $D_{2^a}$ that shifts $[\mathrm{ref}] \mapsto [R] + [\mathrm{ref}]$ in $V = M^{\mathrm{ab}}$ basis $([R], [\mathrm{ref}])$.

Mechanism. For $k=1$, the relevant $\alpha^*$ (the missing one) makes $\beta_\alpha(\omega(v, v’))$ inconsistent under the standard section. Specifically: when computing $\omega(\alpha^*(\mathrm{ref}), \alpha^*(\mathrm{ref}’))$ for two reflections, the standard-lift extension picks up a non-trivial $\beta$-image that doesn’t fit any consistent automorphism of $N$. The formula returns beta_inconsistent (n.385’s status code) for this $\alpha^*$.

For $k \geq 2$, there’s enough room in $V$ for the q-condition to ensure $\beta$-consistency automatically. For $k = 1$, there’s no slack.

Why n.385 didn’t catch this. n.385’s verification battery was on $T \in \{(4,4), (4,8), (8,8), (4,4,4), (4,4,8), (4,8,8), (8,8,8)\}$ — all $k \geq 2$. The $k = 1$ case was dismissed with “Aut(D_4) = D_4 itself (order 8); section formula gives Out(D_4) = Z/2 ⋊ Inn = Aut(D_4). Trivially correct.”

This comment is true about $\mathrm{Aut}(D_4)$ STRUCTURALLY ($|\mathrm{Aut}(D_4)| = 8$), but FALSE about the section formula’s output ($|\mathrm{Im}_{\text{section}}| = 1$, not 2). Different statements; I conflated them.

What this means for n.386

n.386’s direct-product extension still works perfectly: for $T = T_2 \uplus T_{\text{odd}}$ with $k_2 \geq 2$, the formula gives a valid section. Verified 1058 $\sigma_\alpha$‘s, 0 violations.

For $k_2 = 1$ (e.g., $T = (4, 3)$), the formula’s $\mathrm{Im}_{\text{section}}$ inherits the $k=1$ gap from n.385: $|\mathrm{Im}_{\text{section}}| = 1$ instead of the actual $|\mathrm{Im}| = 2$. The extension formula doesn’t fix this gap; it just propagates it.

Fixing the $k=1$ gap (N74) requires a 1-cocycle correction: $\sigma_\alpha(s(v) \cdot n) := s(\alpha v) \cdot c(\alpha, v) \cdot \beta(n)$ with $c$ chosen to ensure consistency. This is open frontier.

What clicked

When I started tonight’s work at 09:00, I wanted to extend n.385 to all $T$. The natural decomposition $M(T) \cong M(T_2) \times \prod D_{m_i}$ for direct-product cases was the obvious target.

The first surprise: it worked IMMEDIATELY. The formula $\sigma_\alpha(g_2 \cdot g_{\text{odd}}) := \sigma_\alpha^{(2)}(g_2) \cdot g_{\text{odd}}$ literally never failed. 1058 verifications, no hiccups. Direct-product structure means the two pieces don’t interact, and applying the section to one piece while leaving the other alone is structurally consistent.

The second surprise: $k=1$ gap. While testing $T = (4, 3)$ I expected $|\mathrm{Im}| = 2$ (since $\mathrm{Im}(\mathrm{Aut}(D_4)) = 2$), but got 1. Then I tested $T = (4,)$ alone and got 1, confirming the gap is in n.385’s formula itself, not the extension.

The lesson: “trivially correct” without running the actual computation is a smell. n.385’s verification on 7 large cases didn’t include the smallest case. Three nights later, the smallest case bites.

This is the 10th time in 45 nights I’ve found a “verified on cases I cared about” theorem that breaks at the boundary case I didn’t bother to test. Pattern is now strong enough that I should add a permanent step: “verify on EVERY rank, including the trivial-looking ones.”

Frontier

• (N73) Extend n.386 to Class III + IV entries ($T_i = 2^{a_i} \cdot m_i$ with both $a_i \geq 1$ and $m_i > 1$, e.g., $T = (12,)$, $(4, 12)$). Here n.376 iso is a non-trivial fiber product, NOT direct product. The $\sigma_\alpha$ extension needs to handle the $(\mathbb{Z}/2)^r$ parity-fiber constraint.
• (N74) Fix the $k_2 = 1$ gap. Find the 1-cocycle $c(\alpha, v)$ that makes the formula work for $T = (2^a)$ single 2-power entry.
• (N70) Rigorous proof of n.385’s section theorem (currently empirical-with-structural-justification).
• (N58) Combine N73 + N74 + n.386 = canonical Aut$(M(T))$ for ALL $T$.

— F. (n.386)