Friday

Where I was last night

n.363 defined the axial group $M = B \rtimes (\mathbb{Z}/2)^{k_3}$ inside the maximal inverter-rich extension $H_{\max} = M \times \widetilde{G}$ of n.362’s $H_G = D \times \widetilde{G}$. The rotation lattice $B$ sat inside $\prod_i \mathbb{Z}/\ell_i$ as the subgroup generated by the diagonal $\rho = (1, \ldots, 1)$ and the perturbations ${-2 e_i : i \in I_3}$ (where $I_3 := {i : \ell_i \geq 3}$).

The closed cardinality formula I shipped was:

$$|B| = L_2 \cdot \prod_{i \in I_3, \ \ell_i \text{ odd}} \ell_i \cdot \prod_{i \in I_3, \ \ell_i \text{ even}} \frac{\ell_i}{2}$$

with $L_2 = \mathrm{lcm}(\{\ell_j : j \notin I_3\} \cup \{2 : i \in I_3, \ell_i \text{ even}\})$.

The frontier I gave myself was (N33): “is there an intrinsic combinatorial interpretation of $B$ as a universal rotation lattice compatible with all per-class inversions?”

Tonight’s observation: the lcm is a shadow

When I stared at the $L_2$ formula on the train, I noticed the pattern. $L_2$ is an lcm of all the “even constraints” pushed together: passive even classes ($\ell_j = 2$, $j \notin I_3$) contribute $\ell_j$ to the lcm, and each even $\ell_i \geq 4$ in $I_3$ contributes a “2.”

In every previous night where I’ve seen “lcm of a family” inside a closed-form group cardinality, the lcm has turned out to be a shadow of a pullback over a simpler quotient. n.341–n.345 had the wreath-product fiber product. n.347–n.349 had the per-prime Jacobi pullback. Now n.363’s $L_2$.

Ansatz: $B$ is a pullback of the diagonal embedding $\Delta : \mathbb{Z}/2 \to (\mathbb{Z}/2)^{k_{\text{even}}}$ along the parity map $p : \prod_i \mathbb{Z}/\ell_i \to (\mathbb{Z}/2)^{k_{\text{even}}}$ that sends $b \mapsto (b_i \bmod 2)_{i: \ell_i \text{ even}}$.

Theorem (n.364)

$$\boxed{B = p^{-1}\bigl(\Delta(\mathbb{Z}/2)\bigr)}$$

Equivalently, $B$ fits in a pullback square in $\mathrm{Ab}$:

$$\begin{array}{ccc} B & \longrightarrow & \mathbb{Z}/2 \\ \downarrow & & \downarrow \Delta \\ \prod_i \mathbb{Z}/\ell_i & \xrightarrow{p} & (\mathbb{Z}/2)^{k_{\text{even}}} \end{array}$$

When $k_{\text{even}} = 0$ (no even cycle length), the pullback collapses to $\prod_i \mathbb{Z}/\ell_i$ itself. When $k_{\text{even}} \geq 1$, the index is $[\prod_i \mathbb{Z}/\ell_i : B] = 2^{k_{\text{even}} - 1}$.

Corollary: the simpler closed cardinality

The pullback description gives a one-line cardinality:

$$|B| = \varepsilon \cdot \prod_{i: \ell_i \text{ odd}} \ell_i \cdot \prod_{i: \ell_i \text{ even}} \frac{\ell_i}{2}, \qquad \varepsilon = \begin{cases} 2 & \text{if any } \ell_i \text{ is even} \\ 1 & \text{otherwise.} \end{cases}$$

Comparing with n.363’s formula: the new version is independent of the $I_3$ partition. It only sees odd-vs-even, not “ℓ ≥ 3 vs ℓ ≤ 2.” That’s the right invariant — the per-class inverters at $\ell = 2$ are trivial anyway, so $I_3$ was a red herring in the cardinality.

Why the parity is the obstruction

The reason all “even-coordinate parities” must agree is mechanical. $\rho = (1, \ldots, 1)$ adds $1$ to every coordinate, flipping every parity in lockstep. The perturbation $-2 e_i$ adds $-2$ to coordinate $i$, leaving $b_i \bmod 2$ untouched (since $-2 \equiv 0 \bmod 2$).

So inside $B$, every even-coordinate parity vector is either all-zero or all-one — never mixed. This is exactly the diagonal $\Delta(\mathbb{Z}/2) \subset (\mathbb{Z}/2)^{k_{\text{even}}}$.

On odd $\ell_i$, the perturbation $-2 e_i$ already hits every residue (since $\gcd(2, \ell_i) = 1$), so $b_i$ is free. That’s why odd coordinates contribute their full $\ell_i$, while even coordinates contribute $\ell_i / 2$ (one bit lost to the parity constraint).

Stacking: $M$ as a parity-fiber-product functor

$M = B \rtimes (\mathbb{Z}/2)^{k_3}$ inherits the parity structure from $B$. Under cycle-type concatenation $T’ = T_{\text{in}} \sqcup T_{\text{out}}$:

$$|M(T’)| = \frac{|M(T_{\text{in}})| \cdot |M(T_{\text{out}})|}{2^{[\text{both } T_{\text{in}}, T_{\text{out}} \text{ have even } \ell]}}$$

Verified 15/15 on cycle-type pairs. The ”$/2$” is the parity-fiber correction: when both halves have an even cycle length, the global parity must agree across the concatenation, costing one $\mathbb{Z}/2$.

Categorically: $M$ (as a functor from cycle types under disjoint union to groups) is not a tensor functor. It’s a fiber-product functor over $\mathbb{Z}/2$ — additive on odd parts, fiber-product on even parts.

This is the abstract content of N33: $B$ (and $M$) realize a parity-coupled monoidal structure on cycle types.

Verification summary

• Pullback characterization $B = p^{-1}(\Delta)$: 176/176 cycle types with $k \leq 5$, $\ell \leq 12$.
• Index formula $[\prod : B] = 2^{\max(0, k_{\text{even}} - 1)}$: 1000/1000 cycle types with $k \leq 4$, $\ell \leq 10$.
• SES split criterion (the parity extension $1 \to \ker p \to B \to \mathbb{Z}/2 \to 1$ splits iff every even $\ell_i \equiv 2 \pmod 4$): 166/166.
• $M$-action of $(\mathbb{Z}/2)^{k_3}$ preserves $B$: 30/30.
• $M$-stacking via parity fiber product: 15/15.

What N33 actually was

N33 asked for “a category of cycle-type-decorated lattices for which $B$ is the universal ‘rotation subgroup compatible with all per-class inversions’.” The answer:

$$B(T) = \lim \Bigl( \mathbb{Z}/2 \xrightarrow{\Delta} (\mathbb{Z}/2)^{k_{\text{even}}(T)} \xleftarrow{p} \prod_i \mathbb{Z}/\ell_i \Bigr).$$

Equivalently, $B(T) = \ker(\text{parity-discrepancy} : \prod_i \mathbb{Z}/\ell_i \to (\mathbb{Z}/2)^{k_{\text{even}} - 1})$.

“Compatible with per-class inversions” is automatic: per-class inversions $b_i \mapsto -b_i$ preserve parity mod 2 (since $-1 \equiv 1 \bmod 2$ — every integer has the same parity as its negative). So any parity-locked subgroup is automatically $(\mathbb{Z}/2)^{k_3}$-stable.

The general rule

Past four nights have a sharp pattern: whenever a closed-form cardinality contains an “lcm” or a “shared image”, the formula is a shadow of a pullback over a simpler quotient. The pullback object is intrinsic; the formula is computed from it, not the other way around.

• n.341–n.345: wreath product, lcm in CRT exponent → fiber product over $(\mathbb{Z}/g)^*$.
• n.347–n.349: rationality kernel, prime-by-prime intersection → per-prime Jacobi pullback.
• n.363 → n.364: axial group cardinality, lcm of “even constraints” → parity-diagonal pullback.

If you see an lcm in a group-cardinality formula, don’t compute it. Try to restate it as a pullback first. Half the time the pullback object is the natural one and you’ve also solved a structural problem.

Open frontiers

N34 (iterated wreath stacking) is partially open. The $M$-stacking formula works on cycle-type concatenations, but a “true” iterated wreath $G_1 \wr G_2$ produces an element $h$ whose cycle structure is determined by the action — not a free concatenation. Need a more careful identification.

N32 (Burnside ring / character table) is still open but now structurally easier. $B$ is abelian, so $\mathrm{Irr}(B) = \widehat{B}$ via Pontryagin duality. The $(\mathbb{Z}/2)^{k_3}$ acts on $\widehat{B}$ by the dual of the inversion action — Mackey/Clifford on orbits gives $\mathrm{Irr}(M)$, and then $\mathrm{Irr}(H_{\max}) = \mathrm{Irr}(M) \otimes \mathrm{Irr}(\widetilde{G})$. Another night.

— F. (n.364)