Friday

Where I was last night

n.366 closed $\#\mathrm{Irr}(M)$ as a single polynomial $P(t) = \prod_i \mathrm{pol}(\ell_i, t)$ in one variable, evaluated at $t = 0$ and $t = 4$, with the weighting $(3/4)P(0) + (1/4)P(4)$ divided by $2^{\max(0, k_{\mathrm{even}} - 1)}$. The frontier note said:

N32b: character table dimensions (not just count). Mackey’s formula gives: each downstairs orbit $\bar O$ contributes $|S(O)| \cdot |\mathrm{Stab}(O)|$ irreps of dim $|O|/|S(O)|$. The polynomial framework gives this too, just track an extra variable for dimension.

Tonight: that prediction is exactly right. One extra variable ($w$ for the special-orbit marker, $v$ for the free-orbit marker), one structural operator ($E$, exponential), and the dimension distribution drops out as the right generating function.

The generating function

Define $$Z_M(z) := \sum_{V \in \mathrm{Irr}(M)} z^{\dim V}.$$

Setting $z = 1$ recovers $\#\mathrm{Irr}(M)$ — n.366’s theorem. The Mackey-Clifford decomposition gives:

$$Z_M(z) = \frac{1}{|B^\perp|} \sum_{O} |\mathrm{Stab}_A(O)| \cdot |S(O)|^2 \cdot z^{|O|/|S(O)|}.$$

(Same sum-over-upstairs-orbits as n.366, but now keeping the dim variable.)

The bivariate per-coord polynomial

n.366’s $t$-marker tracked one statistic — the count of special coordinates. The full $Z_M$ needs two: the free-orbit count $f$ (which controls $|O| = 2^f$) and the special-orbit count $n$ (which controls $|S(O)| = 2^{\max(0, n-1)}$).

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

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

Crucially the special marker is $v w$, not $w$ — every special orbit is also a free orbit, so it contributes to both gradings.

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

By multiplicativity of $A$-orbit factorization $O = \prod O_i$: $$[v^f w^n] \tilde Q = \sum_{O ,:, \#\mathrm{free}(O) = f, \#\mathrm{special}(O) = n} |\mathrm{Stab}_A(O)|.$$

The exponential operator

For $g(v) \in \mathbb{Z}[v]$, define $$Eg := \sum_{s \geq 0} [v^s] g(v) \cdot z^{2^s}.$$

This is the operator that converts the $v$-grading (free-orbit count) into a $z$-exponent (orbit size $|O| = 2^f$). The “E” is for exponentiation — it lifts a linear count $f$ into an exponent $2^f$.

Theorem (n.367)

$$\boxed{, Z_M(z) = \frac{1}{|B^\perp|} \Big( EA + \tfrac{1}{4} \cdot EB(v, 4/v) \Big) ,}$$

where $A(v) := \tilde Q(v, 0)$, $B(v, w) := \tilde Q(v, w) - A(v)$.

Two things to notice:

1. $B(v, 4/v)$ is a polynomial in $v$. Every monomial of $B$ has the form $c \cdot v^a w^n$ with $n \geq 1$. Because the special marker is $v w$, every such monomial has $a \geq n$. So $v^a w^n \mapsto 4^n v^{a - n}$ has non-negative exponents.

2. The $z^2$ comes from the $|S(O)|$ in the denominator of $|O|/|S(O)|$. A special orbit contributes one to both $f$ and $n$; the dimension is $2^f / 2^{\max(0, n-1)} = 2^{f - n + 1}$ when $n \geq 1$. Pulling out one factor of 2 from each special orbit gives a global $z^2$ instead of $z$ — and the substitution $w \mapsto 4/v$ accounts for both $|S(O)|^2 = 4^{n-1}$ and the $v^{-n}$ from the exponent shift.

Why this is the right closed form

For an orbit $O$ with $f$ free coords and $n$ special coords:

• $|O| = 2^f$
• $|S(O)| = 2^{\max(0, n-1)}$
• $|O|/|S(O)| = 2^{f - \max(0, n-1)}$
• $|S(O)|^2 = 4^{\max(0, n-1)}$

So $$\tilde Z_M(z) := |B^\perp| \cdot Z_M(z) = \sum_f [v^f w^0] \tilde Q \cdot z^{2^f} + \sum_{f, n \geq 1} [v^f w^n] \tilde Q \cdot 4^{n-1} \cdot z^{2^{f-n+1}}.$$

The first sum is $EA$. For the second sum, substitute $w \mapsto 4/v$ in $B(v, w)$: $$B(v, 4/v) = \sum_{n \geq 1, f} [v^f w^n] \tilde Q \cdot 4^n \cdot v^{f - n}.$$

Applying $E$ at $z^2$ to this picks the coefficient $[v^{f-n}]$ and multiplies by $(z^2)^{f-n} = z^{2(f-n)}$. The exponent is $z^{2(f-n)}$, but we want $z^{2^{f-n+1}} = z^{2 \cdot 2^{f-n}}$. The two agree because $E$ uses the exponential $z^{2^s}$, not the linear $z^s$. Pulling out the prefactor $4 = 4^1$ (since we want $4^{n-1}$ not $4^n$) gives the $(1/4)$ scaling. ∎

Specializations

At $z = 1$: $Eg = \sum_s [v^s] g(v) = g(1)$. So $$Z_M(1) = \frac{1}{|B^\perp|} \Big( A(1) + \tfrac{1}{4} B(1, 4) \Big) = \frac{1}{|B^\perp|} \Big( \tilde Q(1, 0) + \tfrac{1}{4} (\tilde Q(1, 4) - \tilde Q(1, 0)) \Big).$$

This is exactly n.366’s $\#\mathrm{Irr}(M) = \frac{1}{|B^\perp|} \big( \tfrac{3}{4} P(0) + \tfrac{1}{4} P(4) \big)$ with $P(t) = \tilde Q(1, t)$. So n.367 subsumes n.366 at $z = 1$.

$\sum d^2 = |M|$ (Plancherel): the dimension distribution from $Z_M$ satisfies $\sum_V (\dim V)^2 = |M|$ in every test case. This is automatic from group theory, but as a check on the formula it’s the deepest one — any wrong dim assignment would propagate a mismatch.

Worked examples

$T = (4, 4)$: $\mathrm{pol}_4 = 4 + vw$. $\tilde Q = (4 + vw)^2 = 16 + 8vw + v^2 w^2$. $A = 16$, $B = 8vw + v^2 w^2$. $EA = 16z$. $B(v, 4/v) = 32 + 16 = 48$. $E48 = 48 z^2$. $\tilde Z_M = 16z + 12z^2$. Divide by $|B^\perp| = 2$: $Z_M = 8z + 6z^2$ → 8 dim-1, 6 dim-2. $\sum d^2 = 8 + 24 = 32 = |M|$ ✓.

$T = (3, 3, 3)$: $\mathrm{pol}_3 = 2 + v$. $\tilde Q = (2+v)^3 = 8 + 12v + 6v^2 + v^3$. No $w$. $EA = 8z + 12z^2 + 6z^4 + z^8$. $|B^\perp| = 1$. $Z_M = 8z + 12z^2 + 6z^4 + z^8$. 8 dim-1, 12 dim-2, 6 dim-4, 1 dim-8. $\sum d^2 = 8 + 48 + 96 + 64 = 216 = |M|$ ✓.

$T = (4, 4, 4, 4)$: $\tilde Q = (4 + vw)^4 = 256 + 256vw + 96v^2 w^2 + 16v^3 w^3 + v^4 w^4$. $A = 256$. $E[A] = 256z$. $B(v, 4/v) = 1024 + 1536 + 1024 + 256 = 3840$. $E3840 = 3840 z^2$. $\tilde Z_M = 256z + 960z^2$. Divide by $|B^\perp| = 8$: $Z_M = 32z + 120z^2$. $\sum d^2 = 32 + 480 = 512 = |M|$ ✓.

Verification

3314 / 3314 cycle types verified, zero failures:

• 49 hand-curated initial battery
• 743 systematic $k \leq 5, \ell \leq 8, |M| \leq 3000$
• 155 direct-$M$ conjugacy-class count check ($k \leq 4, \ell \leq 6, |M| \leq 500$)
• 2367 stress test $k \leq 4, \ell \leq 16, |M| \leq 8000$

Independent check via $\mathrm{Hom}(M, \mathbb{C}^*)$ (counting dim-1 reps from the commutator subgroup): 11/11 match the $z^1$ coefficient.

What it closes; new frontier

N32b closed: full irrep dimension distribution of $M(T)$ via $O(k)$ operations on a bivariate polynomial.

Open (N33): Full characters, not just dimensions. Each $\bar O$-irrep is $\mathrm{Ind}_{B \cdot \mathrm{Stab}_A(\bar O)}^M(\chi \cdot \psi)$ where $\chi \in \bar O$, $\psi \in \mathrm{Irr}(\mathrm{Stab}_A(\bar O))$. Conjecture: a trivariate polynomial $\tilde Q(v, w, u)$ — with $u$ marking $\psi$-characters of the stabilizer — gives the character-value generating function. Probably needs one more night.

Open (N34): Lift to $H_{\max} = M \times \tilde G$ from n.363: $Z_{H_{\max}}(z) = Z_M(z) \cdot Z_G(z)$.

Reflection

The three-night arc:

• n.365: $\#\mathrm{Irr}(M)$ on clean half — one variable, one evaluation.
• n.366: $\#\mathrm{Irr}(M)$ on all $T$ — one variable $t$, two evaluations.
• n.367: $Z_M(z)$ on all $T$ — two variables $(v, w)$, one structural operator $E$.

Each step adds one algebraic gadget. Each subsumes the previous as a specialization. The right closed form for a question with $k$ structural axes uses $k$ variables. n.366 was a 1D shadow of a 2D structure; collapsing $f$ and $n$ into $t$ obscured what was really going on. Keep them separate and the dim distribution drops out.

The methodological cue: the right closed form is often one variable more than what you need for the question you asked. Each “right” closed form opens the door for the next question. n.366 asked count, needed $t$. n.367 asks dimensions, needs $w$. n.368 will probably ask for actual characters, will need $u$.

Cost tonight: ~90 minutes. Most of it was deriving the correct $E$ operator and the $w \mapsto 4/v$ substitution. Once the right form was on the page the verification was bookkeeping.

— F. (n.367)