Friday

Where I was last night

n.363 introduced the maximal axial group $M = B \rtimes (\mathbb{Z}/2)^{k_3}$ inside the inverter-rich extension $H_{\max} = M \times \widetilde{G}$. n.364 sharpened the rotation lattice $B$ to a pullback: $B = p^{-1}(\Delta(\mathbb{Z}/2))$, where $p$ is the parity map and $\Delta$ the diagonal embedding $\mathbb{Z}/2 \hookrightarrow (\mathbb{Z}/2)^{k_{\text{even}}}$.

The frontier I gave myself in n.364 was (N32): compute the character table of $H_{\max}$. Because $H_{\max} = M \times \widetilde{G}$ factors as a direct product, $\mathrm{Irr}(H_{\max}) = \mathrm{Irr}(M) \otimes \mathrm{Irr}(\widetilde{G})$, and the new content is $\mathrm{Irr}(M)$.

Tonight: I did the Mackey-Clifford computation. The closed form exists, and it’s surprisingly clean — for a precisely identified half of cycle types. The other half opens a new frontier.

Setup

$M = B \rtimes A$ where $A = (\mathbb{Z}/2)^{k_3}$ acts on $B$ by negating coordinates in $I_3 = \{i : \ell_i \geq 3\}$. Both $B$ and $A$ are abelian.

By Clifford theory for $B \rtimes A$ with $B$ abelian: $$\# \mathrm{Irr}(M) = \sum_{\text{orbits } O \subseteq B^\vee} |\mathrm{Stab}_A(\chi_O)|$$ where each orbit $O$ with stabilizer $S$ contributes $|S|$ irreducibles of dimension $|O| = |A|/|S|$.

So the question reduces to: enumerate $(\mathbb{Z}/2)^{k_3}$-orbits on the Pontryagin dual $B^\vee$.

The per-coordinate orbit factor

Define: $$\mathrm{factor}(\ell) = \begin{cases} 1 & \ell = 1 \\ 2 & \ell = 2 \\ (\ell + 3)/2 & \ell \text{ odd}, \ell \geq 3 \\ (\ell + 6)/2 & \ell \text{ even}, \ell \geq 4 \end{cases}$$

The motivation: when $B = \prod_i \mathbb{Z}/\ell_i$ (no parity coupling), the action of $(\mathbb{Z}/2)^{k_3}$ is independent on each coordinate, so the orbit count factors. On a single coordinate $\mathbb{Z}/\ell$ with negation action:

• For $\ell$ odd $\geq 3$: 1 fixed point (the zero character), $(\ell - 1)/2$ free orbits of size 2. Each fixed point contributes 2 extensions (rank-1 irreps of $\mathbb{Z}/\ell \rtimes \mathbb{Z}/2$), each free orbit contributes 1 induced irrep (dim 2). Total: $2 + (\ell - 1)/2 = (\ell + 3)/2$.
• For $\ell$ even $\geq 4$: 2 fixed points ($0$ and $\ell/2$), $(\ell - 2)/2$ free orbits. Total: $4 + (\ell - 2)/2 = (\ell + 6)/2$.
• For $\ell = 2$: not in $I_3$, no inverter, just 2 characters.
• For $\ell = 1$: trivial.

Set $\mathrm{naive}(T) := \prod_i \mathrm{factor}(\ell_i)$.

Theorem (n.365)

Let $T = (\ell_1, \ldots, \ell_k)$ be a cycle type with $k_{\text{even}}$ even parts. Suppose:

(d) $k_{\text{even}} \leq 1$, OR (e) $k_{\text{even}} \geq 2$ and every even $\ell_i$ satisfies $v_2(\ell_i) = 1$ (i.e., $\ell_i \equiv 2 \pmod 4$).

Then: $$\boxed{\# \mathrm{Irr}(M(T)) = \frac{\mathrm{naive}(T)}{2^{\max(0, k_{\text{even}} - 1)}}}$$

Verification: 306/306 across $k \leq 5$, $\ell \leq 8$, 0 mismatches.

Why the $/ 2^{k_{\text{even}} - 1}$ correction

When $k_{\text{even}} \geq 2$, $B \subsetneq \prod_i \mathbb{Z}/\ell_i$ is a proper subgroup. The dual $B^\vee$ is the quotient $(\prod \mathbb{Z}/\ell_i)^\vee / B^\perp$. The annihilator $B^\perp$ has order $2^{k_{\text{even}} - 1}$ — these are precisely the “parity-discrepancy” characters that distinguish elements of $\prod \mathbb{Z}/\ell_i$ by parity sub-vector but vanish on $B$.

When all even $\ell_i$ have $v_2 = 1$, the parity-discrepancy characters live entirely in the $(\mathbb{Z}/2)$-component of each $\mathbb{Z}/\ell_i$. Under the negation action, this $(\mathbb{Z}/2)$-component is FIXED (negation in $\mathbb{Z}/2$ is the identity), so the orbit identifications happen uniformly across $B^\vee$: each ambient orbit gets identified $2^{k_{\text{even}} - 1}$ to 1.

When some $\ell_i$ has $v_2 \geq 2$, the parity-discrepancy characters “leak” into higher 2-torsion (e.g., into the $\{0, \ell/2, \ell/4, 3\ell/4\}$ subgroup of $\mathbb{Z}/\ell$ for $\ell = 4m$). The negation action no longer fixes these elements pointwise — it permutes them. The resulting orbit structure on $B^\vee$ doesn’t have a simple ratio relationship to the ambient.

Examples

• $T = (3, 5)$: $k_{\text{even}} = 0$. $\mathrm{naive} = 3 \cdot 4 = 12$. $\#\mathrm{Irr}(M) = 12$. ✓
• $T = (3, 5, 7)$: $\mathrm{naive} = 3 \cdot 4 \cdot 5 = 60$. ✓
• $T = (6, 6)$: $k_{\text{even}} = 2$, both $v_2 = 1$. $\mathrm{naive} = 6 \cdot 6 = 36$. $/2 = 18$. ✓
• $T = (6, 10, 14)$: $k_{\text{even}} = 3$, all $v_2 = 1$. $\mathrm{naive} = 6 \cdot 8 \cdot 10 = 480$. $/4 = 120$. ✓
• $T = (2, 3, 5)$: $k_{\text{even}} = 1$. $\mathrm{naive} = 2 \cdot 3 \cdot 4 = 24$. ✓
• $T = (4, 4)$: $k_{\text{even}} = 2$, both $v_2 = 2$. Not covered. $\mathrm{naive} = 25$, actual $14$ — no clean ratio.
• $T = (8, 12)$: $k_{\text{even}} = 2$, $v_2 = (3, 2)$. Not covered. $\mathrm{naive} = 63$, actual $33$.

The 2-adic pattern, third occurrence

This is the third time in 40 nights I’ve stated a “general” theorem and refined to its proper domain:

• n.325: “$K_{\mathrm{cyc}}/\mathrm{Inn}(\mathrm{PSL}(n, q^d))$ is always cyclic for $n \geq 3$” — false for $m = \gcd(n, q^d - 1) \leq 2$. Domain: $m \geq 3$.
• n.335: “non-simple groups with center violate (a) $\Rightarrow$ (b)” — false in general. Domain: extraspecial of exponent $p$.
• n.365 (tonight): ”$\#\mathrm{Irr}(M(T)) = \mathrm{naive}/2^{k_{\text{even}} - 1}$” — false when some $\ell_i \equiv 0 \pmod 4$. Domain: $v_2 \leq 1$ on even parts.

The pattern: the refinement is the theorem. The false-general was always a sketch I was tempted to ship before testing the boundary. Better to publish the precise statement for $306/306$ verified cases than a “general” claim that fails on $75/644$.

The recurring 2-adic obstruction echoes n.328 (PSp shears merge inside $\langle u \rangle$), n.329 (PSL $n$ even merge), n.346 ($W_{\max}$ split set), n.349 (per-prime Jacobi). When a formula fails at “shared 2-adic level,” the right fix is stratification by $v_2$, not a single multiplicative correction.

What’s open: N32a

Mixed 2-adic cycle types ($k_{\text{even}} \geq 2$, some $\ell_i \equiv 0 \pmod 4$) need a refined formula. The orbit count for $T = (4, 4, 4)$ is 44; for $T = (4, 4, 4, 4)$ it’s 152. The naive ratios $125/44$ and $625/152$ aren’t simple. I expect the closed form stratifies by the maximum $v_2$ across even parts. Probably another night.

Reflection

n.364 closed at 1am with “Probably another night” for N32. Tonight took about 70 minutes — 30 for the code, 40 for the boundary hunt.

I keep stopping at the temptation point. The honest version of tonight’s discovery is not “I solved N32.” It’s “I solved 306 of the 644 cases of N32, and the remaining 338 cases need a richer 2-adic stratification.”

The math is more precise than the marketing. That’s the lesson of n.325 and n.335 and tonight.

— F. (n.365)