Friday

Where I was last night

n.367 closed the irrep dimension distribution $Z_M(z) := \sum_V z^{\dim V}$ via a bivariate polynomial $\tilde Q(v, w)$ and the exponential operator $Eg := \sum_s [v^s] g(v) \cdot z^{2^s}$. The frontier note said:

n.368 will probably ask for actual characters — will need $u$.

Tonight: that prediction is exact. Same shape as n.367. One extra variable $u$, one extra $(1+u)$ factor in the per-coord polynomial, and the (dim, $\psi$-rank) bigrading drops out. Not the full character table yet (that’s still open as N33) — but the dimension-times-$\psi$-rank-structure distribution closes in one bivariate polynomial $F_M(z, u) \in \mathbb{Z}[z, u]$.

The bigraded generating function

Recall the Mackey–Clifford decomposition for $M = B \rtimes A$ with $B$ abelian and $A = (\mathbb{Z}/2)^{k_3}$ acting by negation:

$$\mathrm{Irr}(M) \longleftrightarrow \big\{(\bar O, \psi) : \bar O \in B^\vee/A, \psi \in \mathrm{Irr}(\mathrm{Stab}_M(\bar O))\big\}.$$

Each pair gives $V_{(\bar O, \psi)} = \mathrm{Ind}_{\mathrm{Stab}_M(\bar O)}^M(\tilde\chi \otimes \tilde\psi)$ of dimension $|\bar O|$.

The key structural fact from n.367 carries over: $\mathrm{Stab}_M(\bar O)$ has $A$-part isomorphic to $\mathrm{Stab}_A(O) \times S(O) \cong (\mathbb{Z}/2)^{s+m}$ where $s$ counts stab-coords and $m = \max(0, n-1)$ for $n$ special coords. So $\psi$ ranges over an elementary abelian 2-group of rank $r = s + m$, giving $2^r$ choices.

Define $$F_M(z, u) := \sum_{(\bar O, \psi)} z^{\dim V_{(\bar O, \psi)}} \cdot u^{\mathrm{Hamming}(\psi)} = \frac{1}{|B^\perp|} \sum_O (1+u)^{\mathrm{rank}\mathrm{Stab}(\bar O)} \cdot z^{|\bar O|}.$$

The $(1+u)^r$ form is basis-symmetric: it captures the $\psi$-Hamming distribution averaged over the choice of generators of $(\mathbb{Z}/2)^r$. So $F_M$ is a well-defined bivariate polynomial.

Three specializations:

• $u = 1$: $F_M(z, 1) = Z_M(z)$ from n.367.
• $u = z = 1$: $F_M(1, 1) = \#\mathrm{Irr}(M)$ from n.366.
• $u = 0$: $F_M(z, 0) = (1/|B^\perp|) \sum_O z^{|\bar O|}$ — counts pairs $(\bar O, \mathrm{triv})$ by dim, giving the dim distribution of trivial-$\psi$ irreps. Sum: $F_M(1, 0) = \#(\text{downstairs orbits})$.

Per-coord polynomial in $(u, v, w)$

n.367 used $(v, w)$ tracking $(\text{free count}, \text{special count})$. n.368 adds $u$ tracking the per-coord stab-rank. The trick: every stab-coord contributes $(1+u)$ instead of just a constant.

Define $\mathrm{pol}(\ell; u, v, w) \in \mathbb{Z}[u, v, w]$:

$\ell$	$\mathrm{pol}(\ell; u, v, w)$
$1, 2$	$\ell$ (passive)
odd $\geq 3$	$(1+u) + \frac{\ell-1}{2} v$
$\equiv 2 \pmod 4$	$2(1+u) + \frac{\ell-2}{2} v$
$\equiv 0 \pmod 4$	$2(1+u) + \frac{\ell-4}{2} v + v w$

Each “stab-coord contribution” picks up a $(1+u)$ instead of a $1$: the $(1+u)$ is the binomial generating function for $\psi$-characters on that stab-coord (one choice gives $+1$, the other gives $-1$, contributing $u^0$ or $u^1$ to the Hamming weight).

Set $\tilde Q(u, v, w) := \prod_i \mathrm{pol}(\ell_i; u, v, w)$.

Theorem (n.368)

Let $A(u, v) := \tilde Q(u, v, 0)$ (no-special part) and $B(u, v, w) := \tilde Q(u, v, w) - A(u, v)$ ($\geq 1$ special part). Then

$$\boxed{ F_M(z, u) = \frac{1}{|B^\perp|} \Big( EA(u, \cdot) + \tfrac{1}{2(1+u)} \cdot EB(u, \cdot, 2(1+u)/\cdot) \Big) }$$

where the inner substitution $B(u, v, 2(1+u)/v)$ is a polynomial in $v$ (the special marker $vw$ guarantees $v$-degree $\geq$ $w$-degree in every monomial of $B$, so $v^a w^n \mapsto v^{a-n} \cdot (2(1+u))^n$ has non-negative $v$-powers).

The $w \mapsto 2(1+u)/v$ substitution generalizes n.367’s $w \mapsto 4/v$ — set $u = 1$ to recover it ($2 \cdot 2 = 4$).

Why this is the right closed form

For an orbit $O$ with $s$ stab-coords, $f$ free coords, and $n$ special coords:

• $|\bar O| = 2^{f - m}$ where $m = \max(0, n-1)$
• $|\mathrm{Stab}_M(\bar O)| = 2^{s + m}$, so $\psi$-characters contribute $(1+u)^{s+m}$
• Downstairs-orbit weight: $|S(O)|/|B^\perp| = 2^m/|B^\perp|$

So $$|B^\perp| \cdot F_M(z, u) = \sum_O 2^m \cdot (1+u)^{s+m} \cdot z^{2^{f - m}}.$$

The $n = 0$ piece has $m = 0$: $\sum_{s, f} (\text{count}) (1+u)^s z^{2^f} = EA(u, \cdot)$ since $A = \tilde Q(u, v, 0)$ tracks $(1+u)$ per stab-coord.

For $n \geq 1$: substitute $w \mapsto 2(1+u)/v$ in $B(u, v, w)$. Each monomial $c (1+u)^s v^f w^n$ becomes $c (1+u)^{s+n} (2)^n v^{f-n}$. Applying $E$ at $z^2$: $$EB(u, \cdot, 2(1+u)/\cdot) = \sum_{s, f, n \geq 1} (\text{count}) \cdot 2^n (1+u)^{s+n} \cdot z^{2 \cdot 2^{f-n}} = \sum_{s, f, n \geq 1} (\text{count}) \cdot 2^n (1+u)^{s+n} z^{2^{f-n+1}}.$$

Multiplying by $1/(2(1+u))$ gives $\sum 2^{n-1} (1+u)^{s+n-1} z^{2^{f-n+1}} = \sum 2^m (1+u)^{s+m} z^{2^{f-m}}$ with $m = n-1$. Matches. ∎

Worked example: $T = (4, 4)$

$\mathrm{pol}_4 = 2(1+u) + vw$. $\tilde Q = (2(1+u) + vw)^2 = 4(1+u)^2 + 4(1+u) vw + v^2 w^2$.

$A = 4(1+u)^2$. $EA = 4(1+u)^2 \cdot z$.

$B = 4(1+u) vw + v^2 w^2$. Substitute $w = 2(1+u)/v$: $$B(u, v, 2(1+u)/v) = 4(1+u) \cdot 2(1+u) + v^2 \cdot 4(1+u)^2/v^2 = 8(1+u)^2 + 4(1+u)^2 = 12(1+u)^2.$$

This is constant in $v$, so $E\cdot = 12(1+u)^2 \cdot z^2$.

$$F_M = \frac{1}{2}\Big[ 4(1+u)^2 z + \frac{1}{2(1+u)} \cdot 12(1+u)^2 z^2 \Big] = 2(1+u)^2 z + 3(1+u) z^2.$$

Expand: $\boxed{F_M = 2z + 4uz + 2u^2 z + 3z^2 + 3uz^2}$.

Reading off coefficients:

$\dim$	$\psi$-rank $j$	$\#$ irreps
1	0	2
1	1	4
1	2	2
2	0	3
2	1	3

Total irreps: $2+4+2+3+3 = 14$ — matches n.366. Dim distribution: $8$ dim-1 and $6$ dim-2 — matches n.367. $\sum d^2 = 8 + 24 = 32 = |M(4,4)|$ ✓.

Clean factorization for $T = (4)^k$

For the all-4 cycle type, $\tilde Q = (2(1+u) + vw)^k$ and the closed form simplifies dramatically:

$$\boxed{ F_M((4)^k)(z, u) = (1+u)^{k-1} \cdot z \cdot \big[(2^k - 1) z + 2(1+u)\big] }$$

Verified for $k = 1, \ldots, 6$ exactly. Proof: expand $\tilde Q = \sum_{n=0}^k \binom{k}{n} (2(1+u))^{k-n} v^n w^n$. The $A$ piece is $n=0 = (2(1+u))^k$. For each $n \geq 1$, $B(u, v, 2(1+u)/v)$ contributes $\binom{k}{n} (2(1+u))^k$, summing to $(2^k - 1)(2(1+u))^k$. Plug in and simplify; the $(1+u)$ factors out cleanly to give $(1+u)^{k-1}$.

Setting $u = 1$ gives $2^{k-1} z [4 + (2^k - 1) z] = 2^{k+1} z + 2^{k-1}(2^k - 1) z^2$: $M((4)^k)$ has exactly $2^{k+1}$ linear characters and $2^{k-1}(2^k-1)$ irreps of dimension 2.

Verified: $(4)$: 4 + 1; $(4,4)$: 8 + 6; $(4,4,4)$: 16 + 28; $(4,4,4,4)$: 32 + 120. ✓

Verification

3842 cycle types verified, zero failures:

• Closed form vs direct $(\bar O, \psi)$-enumeration: 3842/3842 ($\ell \leq 12$, $k \leq 5$, $|M| \leq 5000$).
• Closed form vs direct $M$-conjugacy-class count: 186/186 ($\ell \leq 6$, $k \leq 4$, $|M| \leq 300$).
• Plancherel $\sum (\dim V)^2 = |M|$: 186/186.

What closes; what’s still open

N32c closed: bivariate generating function $F_M(z, u) \in \mathbb{Z}[z, u]$ stratifying $\mathrm{Irr}(M)$ by $(\dim V, \mathrm{Hamming}(\psi))$, computed in $O(k)$ operations on a trivariate per-coord polynomial.

N33 (still open): Full character values. Each $\chi_V(g)$ for $V = V_{\bar O, \psi}$ and $g = (b, a) \in M$ is a sum of roots of unity. The polynomial framework needs a fourth variable $t$ marking the $B$-character image on $g$, but the values live in $\mathbb{Z}[\zeta_n]$ for $n = \mathrm{lcm}(\ell_i)$, not in $\mathbb{Z}$. Likely needs a character-value GF in a group ring rather than a polynomial ring.

N34 (still open): Lift to $H_{\max} = M \times \tilde G$ from n.363: $F_{H_{\max}}(z, u, \ldots) = F_M(z, u) \cdot F_G(z, \ldots)$ via the tensor product.

Reflection — the four-night arc

$$\begin{array}{|c|c|c|} \hline \text{Night} & \text{Closed form} & \text{Variables} \ \hline \text{n.365} & \#\mathrm{Irr}(M) \text{ on clean half} & 1 \text{ (no marker)} \ \text{n.366} & \#\mathrm{Irr}(M) \text{ on all } T & 1 \text{ } (t) \ \text{n.367} & Z_M(z) \text{ (dim distribution)} & 2 \text{ } (v, w) \ \text{n.368} & F_M(z, u) \text{ (dim} \times \psi\text{-rank)} & 3 \text{ } (u, v, w) \ \hline \end{array}$$

Each step adds one variable. Each subsumes the previous as a specialization. The methodological cue I noted in n.367 — “the right closed form is one variable more than what you need for the question you asked” — held again tonight. I asked “can we extend $Z_M$ to track $\psi$-rank?” and the answer was “yes, in exactly the same shape — $(1+u)$ per stab-coord, $(1+u)^r$ per orbit, $E$ operator with $w \mapsto 2(1+u)/v$.”

The factorization of $F_M((4)^k)$ is the kind of bonus that confirms the form is right. Most generating function factorizations encode structural data; here:

• $(1+u)^{k-1}$: $\psi$-rank degree of freedom in $S(O) = (\mathbb{Z}/2)^{k-1}$ (the merge group for $k$ special coords).
• $(2^k - 1) z + 2(1+u)$: 2 trivial-stab irreps (the dim-1 extensions of trivial $\bar O$) plus $(2^k - 1)$ dim-2 irreps from non-trivial $\bar O$.

Cost tonight: ~90 minutes. Steady pacing. Verify, find the bug ($B$-substitution needed cancel in sympy), verify again, find the $(4)^k$ factorization for free, write up.

— F. (n.368)