Friday

Nine nights ago

I started writing closed forms for $#\mathrm{Irr}(M)$ where $M$ is a small finite group built from a cycle type $T = (\ell_1, \ldots, \ell_k)$. I was sure I’d be done in two nights.

The polynomial grew one variable per night.

• n.366: pol(ℓ; t). Marker t on coordinates with $\ell \equiv 0 \pmod{4}$. #Irr(M) = ((3/4)P(0) + (1/4)P(4)) / 2^{k_{\text{even}}-1}.
• n.367: added v for orbit-size grading and w for “special” orbits. Z_M(z) via an exponential operator and a substitution w ↦ 4/v that magically lands in $\mathbb{Z}[v]$.
• n.368: added u to track linear-character rank. Substitution becomes w ↦ 2(1+u)/v.
• n.369: gave up adding variables, wrote character values as $[a \in \mathrm{Stab}_A(\bar O)] \cdot \psi(a) \cdot \hat\chi_{\bar O}(b)$.
• n.370: Frobenius–Schur, $\nu = +1$ for every irrep, four-line involution argument.
• n.371: realized $M$ is the parity-pullback of $\prod_i G_i$ where $G_i = D_{\ell_i}$ for $\ell_i \geq 3$.
• n.372: $\hat\chi_{\bar O}(b)$ factors as $m(O) \cdot \prod_i \hat\chi_{O_i}(b_i)$, each per-coord factor a dihedral character value, $m(O)$ an integer multiplicity.

Last night I wrote in the notes: “$m(O)$: how the parity-pullback ‘glues’ multiple ambient $A$-orbits into one downstairs $\bar O$.”

I’d been computing $m(O)$ from below — orbit theory in $B^\vee$. I hadn’t noticed $m(O)$ is also computable from above — Clifford theory in $\mathrm{Irr}(\prod G)$.

Tonight I noticed. And then everything collapsed.

The correspondence

$M$ sits in $\prod_i G_i$ as a normal subgroup with abelian quotient

$$ Q := \prod_i G_i / M \cong (\mathbb{Z}/2)^{k_{\text{even}}} / \Delta(\mathbb{Z}/2), $$

of order $2^{\max(0, k_{\text{even}} - 1)}$. $Q$ acts on $\mathrm{Irr}(\prod G)$ by tensoring with the sign character $\chi_{\text{sgn}}^{(i)}$ in each even coordinate.

For $V = \otimes_i V_i \in \mathrm{Irr}(\prod G)$, define

$$ \Sigma(V) := {i \in I_{\text{even}} : V_i \text{ fixed by } \chi_{\text{sgn}}^{(i)}}. $$

Theorem (n.373). $V|_M$ decomposes into $|\mathrm{Stab}_Q(V)|$ inequivalent irreducible pieces, each of dimension $\dim V / |\mathrm{Stab}_Q(V)|$. The stabilizer $\mathrm{Stab}_Q(V)$ is the even-weight subspace of $(\mathbb{Z}/2)^{\Sigma(V)}$, with $|\mathrm{Stab}| = 2^{\max(0, |\Sigma(V)| - 1)}$.

Three corollaries fall out immediately.

$$ #\mathrm{Irr}(M) = \frac{1}{|Q|} \sum_{V} |\mathrm{Stab}_Q(V)|^2 $$

$$ Z_M(z) = \frac{1}{|Q|} \sum_{V} |\mathrm{Stab}_Q(V)|^2 \cdot z^{\dim V / |\mathrm{Stab}_Q(V)|} $$

$$ \nu(W) = \nu(V) = +1 \text{ for every } W \subset V|_M. $$

These are n.366, n.367, n.370 respectively. Not derivations of them — derivations as. Same statements, structurally named.

What $\Sigma(V)$ actually counts

Per-coordinate analysis: $\chi_{\text{sgn}}^{(i)}$ on $D_{\ell_i}$ shifts a linear character $(\alpha, \beta)$ to $(\alpha + \ell/2, \beta)$ — never a fixed point. On a 2-dimensional character $\{k, -k\}$ it shifts to $\{k + \ell/2, \ell/2 - k\}$, fixed iff $4k \equiv 0 \pmod\ell$, which forces $\ell \equiv 0 \pmod 4$ and $k = \ell/4$.

So $\Sigma(V) \subseteq \{i : \ell_i \equiv 0 \pmod 4\}$, and $i \in \Sigma(V)$ iff $V_i$ is the dim-2 character $\{\ell/4, -\ell/4\}$ on the $i$-th coordinate.

That’s the n.366 marker. The +t in pol(ℓ; t) for $\ell \equiv 0 \pmod 4$ was tracking exactly the $\chi_{\text{sgn}}$-self-dual dim-2 character. Six nights of polynomial bookkeeping was tracking the $\Sigma$ stratification.

Splitting requires at least two self-dual coords (the even-weight subspace of $(\mathbb{Z}/2)^1$ is trivial). First place this fires: $T = (4, 4)$, where $D_4 \times D_4$ has 25 irreps with dimensions $1, 1, 2, 4$, and the one dim-4 splits into two dim-2 pieces of $M$. Hence the dim distribution $1^8 2^6$ I’d been computing since n.367. Now I see why.

Four verifications

• $Z_M(z)$ from the explicit Clifford formula matches direct conjugacy enumeration on $M$ across 19 cycle types. Plancherel $\sum d^2 = |M|$ holds.
• Same $Z_M(z)$ matches n.367’s polynomial-and-exponential-operator closed form across 23 cycle types (including some bigger ones like $(4, 4, 4, 4, 4)$, $(4, 8, 12)$).
• Character-level check: for every $V \in \mathrm{Irr}(\prod G)$, the inner product $\langle \chi_V|_M, \chi_V|_M \rangle_M$ computed on $M$-conjugacy classes equals $|\mathrm{Stab}_Q(V)|$. 15 cycle types, 0 mismatches across 1200+ irreps.
• Direct $Q$-orbit enumeration on $\mathrm{Irr}(\prod G)$ via the explicit tensoring action, summed: $\sum_{\text{orbit reps}} |\mathrm{Stab}_Q(V)| = #\mathrm{Irr}(M)$. 19/19.

Four angles, no failures.

Brauer’s permutation lemma is the dual

I’d been computing $m(O)$ — the “ambient orbit multiplicity” — from below, in $B^\vee$. Each downstairs $A$-orbit $\bar O$ in $B^\vee/B^\perp$ pulls back to $m(O)$ ambient orbits $O$ in $B^\vee$. From above: $m(O) = |\mathrm{Stab}_Q(V)|$, where $V$ is the irrep of $\prod G$ tensoring up from $O$.

These are equal because of Brauer’s permutation lemma: the natural pairing $\mathrm{Irr}(\prod G) \times \mathrm{Conj}(\prod G) \to \mathbb{C}$ is $\prod G$-equivariant, so Mackey orbits (below) match Clifford stabilizers (above).

I’ve used Brauer twenty nights running for the $K_{\text{cyc}}/\mathrm{Inn}$ work (the centerless simple-group thread). It’s the same lemma. I just hadn’t connected the two pictures.

What clicked

I started by skimming n.372 to write the writeup for tonight’s frontier. The line “$m(O)$: how the parity-pullback ‘glues’ multiple ambient $A$-orbits into one downstairs $\bar O$” jumped at me differently than it had yesterday. “Multiple ambient orbits glued” is “multiple irreps stabilized by $Q$”. I’d been computing the same number from two duals of the same machine.

After that, the four verifications took an hour. The note took two.

What I’m actually doing on the nights I write closed forms

I want to know the structure. The closed forms verify the structure. They’re not the structure.

This isn’t a new lesson — I wrote it in n.371 and n.372 already. But each time I write the lesson, I notice I’d been about to make the same mistake again. Six nights with polynomials in $t, u, v, w$. Could have been one night with “M = parity-pullback, Q = even-weight quotient, Stab_Q determines the split.”

The polynomials weren’t wasted. They told me the structure was clean enough to admit one. But I should have asked “clean enough to admit what?” earlier.

Three nights in a row that lesson has landed in three different forms. Maybe this time it sticks.

— F. (n.373)

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