§7 · Agent 7 · Audit: criterion and bridge · 2026-10-18T19:26:59Z

Audit 7.8 · Audit of A3-7: with disconnected stabilizers, (IG)$_n$ for all $n$ gives $\mathcal M_I=\mathrm{Irr}\,\mathcal P_I$

proved

The twisted-character basis (Lang for twisted classes, Gallagher, Clifford, Fourier inversion on $\mathbb Z/d$) is correct; brute-force counts on six pairs $(A,\sigma)$ confirm #twisted classes = #$\sigma$-stable classes = #$\sigma$-stable irreducibles; non-$\mathrm{Fr}$-stable simple objects are correctly declared invisible to (IG)$_1$.

Audit 7.8.

Holds. Keep (L1) of A3-1, finitely many orbits per level, (D1)–(D3) with simple objects $\mathrm{IC}(P,\xi)$, no connectedness or rationality of stabilizers/orbits. Then: (a) (IG)$_1$ puts every $\mathrm{Fr}$-stable simple object in $\mathcal M_I$; (b) (IG)$_n$ for all $n\ge1$ gives $\mathcal M_I=\mathrm{Irr}\,\mathcal P_I$ and $\mathcal C_I^{\mathrm{Iwa}}=\mathcal C_I$ over $\overline{\mathbb F}_q$; (c) a non-$\mathrm{Fr}$-stable simple object contributes $0$ to every trace over $\mathbb F_q$, so (IG)$_1$ alone says nothing about it. The key input, for a finite group $A$ with automorphism $\sigma$ of order $d$ and $\Gamma=A\rtimes\mathbb Z/d$: the functions $a\mapsto\tilde\xi(a\sigma)$, $\xi\in\mathrm{Irr}(A)^\sigma$, $\tilde\xi$ an extension to $\Gamma$, form an orthonormal basis of the $\sigma$-twisted class functions on $A$.

Computation (consistency check of the dimension count). For $(A,\sigma)$ = ($\mathbb Z/3$, inversion), ($V_4$, order-3 automorphism), ($D_4$, an inner automorphism), ($S_3$, identity), ($A_4$, conjugation by a transposition), ($Q_8$, $i\mapsto j\mapsto k$), the number of $\sigma$-twisted classes $\{a\sim ba\sigma(b)^{-1}\}$ is $1,1,5,3,2,2$, equal in each case to the number of $\sigma$-stable conjugacy classes, which by Brauer's permutation lemma is the number of $\sigma$-stable irreducible characters.

  1. Step 1 (points). $H=I/I_{N,M}$ connected, $o$ $\mathrm{Fr}$-stable, $y\in o(\mathbb F_q)$ (Lang); $H(\mathbb F_q)$-orbits on $o(\mathbb F_q)$ $\leftrightarrow H^1(\mathbb F_q,S_y)=H^1(\mathbb F_q,A)$, $A=\pi_0S_y$, i.e. twisted classes. Holds.
  2. Step 2 (sheaves). Equivariant irreducible local systems on $o$ are $\mathcal L_\xi$, $\xi\in\mathrm{Irr}(A)$, with $\mathrm{Fr}^*\mathcal L_\xi\simeq\mathcal L_{\xi\circ\sigma}$ (up to the convention $\sigma$ vs $\sigma^{-1}$). For $\sigma$-stable $\xi$, an extension $\tilde\xi$ to $\Gamma$ exists because $\Gamma/A$ is cyclic (Isaacs, Cor. 11.22); Frobenius on the stalk at a point of class $a$ is $\tilde\xi(a\sigma)$. Holds.
  3. Step 3 (basis). (i) $\Gamma$-classes in the coset $A\sigma$ are $A$-twisted classes: $b^{-1}(b\sigma)b=\sigma(b)\sigma$ realizes conjugation by $\sigma$ through $A$. $\mathrm{Irr}(\Gamma)$ spans class functions on $\Gamma$. By Clifford theory with cyclic $\Gamma/A$, $\chi\in\mathrm{Irr}(\Gamma)$ is either $\tilde\xi\lambda$ ($\lambda\in\mathrm{Irr}(\Gamma/A)$, Gallagher, Isaacs Cor. 6.17), or induced from the normal subgroup $A\rtimes\langle\sigma^e\rangle$, $e>1$, which misses $A\sigma$, so $\chi|{A\sigma}=0$. Hence the $\tilde\xi|{A\sigma}$ span. (ii) $\langle\tilde\xi,\tilde\eta\lambda\rangle_\Gamma=\frac1d\sum_{j}\overline{\lambda(\sigma^j)}s_j$ with $s_j=|A|^{-1}\sum_a\tilde\xi(a\sigma^j)\overline{\tilde\eta(a\sigma^j)}$; the left side is $\delta_{\xi\eta}\delta_{\lambda1}$ for all $\lambda$, so Fourier inversion on $\mathbb Z/d$ gives $s_j=\delta_{\xi\eta}$ for every $j$, in particular $s_1$. Holds. The brute-force counts above (script scripts/twisted_classes.py) agree with the resulting equality of dimensions.
  4. Step 4 (basis of $C_I^{(N,M)}$). Block-triangular in the indicators of $I(\mathcal O)$-orbits (surjectivity $I(\mathcal O)\to H(\mathbb F_q)$ needs $I_{N,M}$ connected, kept from (L1c)); diagonal blocks are the invertible twisted character tables of step 3 times $(-1)^{\delta}$; non-$\mathrm{Fr}$-stable orbits have no rational points; IC's with non-$\sigma$-stable $\xi$ are not $\mathrm{Fr}$-stable and contribute $0$ (block permutation, Audit 7.7 step 2). The chain of A3-6 then gives (a). Holds.
  5. Step 5. Each simple object of a level is $\mathrm{Fr}^n$-stable for some $n$ (finite permutation); $\mathcal C_I$ and $\mathcal M_I$ are defined over $\overline{\mathbb F}q$ and unchanged when $q$ is replaced by $q^n$; apply (a) over $\mathbb F{q^n}((t))$. Holds. (c) is the block-permutation vanishing. This is the treatment of checklist item 4 at function level: functions see only $\mathrm{Fr}$-stable simple objects, and only through traces.

Builds on

  • Proposition 3.7 · Disconnected stabilizers: (IG)$_n$ for all $n$ gives $\mathcal M_I=\mathrm{Irr}$

Built on by

  • Summary 7.11 · Verdict of the audit: (I$_I$) over $\overline{\mathbb F}_q$ and $\mathbb C$, (II$_I$) over $\overline{\mathbb F}_q$ only, converses as stated
Original record, as the agent wrote it at 19:26 UTC

Audit of A3-7: with disconnected stabilizers, (IG)$_n$ for all $n$ gives $\mathcal M_I=\mathrm{Irr}\,\mathcal P_I$

  1. Step 1 (points). $H=I/I_{N,M}$ connected, $o$ $\mathrm{Fr}$-stable, $y\in o(\mathbb F_q)$ (Lang); $H(\mathbb F_q)$-orbits on $o(\mathbb F_q)$ $\leftrightarrow H^1(\mathbb F_q,S_y)=H^1(\mathbb F_q,A)$, $A=\pi_0S_y$, i.e. twisted classes. Holds.
  2. Step 2 (sheaves). Equivariant irreducible local systems on $o$ are $\mathcal L_\xi$, $\xi\in\mathrm{Irr}(A)$, with $\mathrm{Fr}^*\mathcal L_\xi\simeq\mathcal L_{\xi\circ\sigma}$ (up to the convention $\sigma$ vs $\sigma^{-1}$). For $\sigma$-stable $\xi$, an extension $\tilde\xi$ to $\Gamma$ exists because $\Gamma/A$ is cyclic (Isaacs, Cor. 11.22); Frobenius on the stalk at a point of class $a$ is $\tilde\xi(a\sigma)$. Holds.
  3. Step 3 (basis). (i) $\Gamma$-classes in the coset $A\sigma$ are $A$-twisted classes: $b^{-1}(b\sigma)b=\sigma(b)\sigma$ realizes conjugation by $\sigma$ through $A$. $\mathrm{Irr}(\Gamma)$ spans class functions on $\Gamma$. By Clifford theory with cyclic $\Gamma/A$, $\chi\in\mathrm{Irr}(\Gamma)$ is either $\tilde\xi\lambda$ ($\lambda\in\mathrm{Irr}(\Gamma/A)$, Gallagher, Isaacs Cor. 6.17), or induced from the normal subgroup $A\rtimes\langle\sigma^e\rangle$, $e>1$, which misses $A\sigma$, so $\chi|{A\sigma}=0$. Hence the $\tilde\xi|{A\sigma}$ span. (ii) $\langle\tilde\xi,\tilde\eta\lambda\rangle_\Gamma=\frac1d\sum_{j}\overline{\lambda(\sigma^j)}s_j$ with $s_j=|A|^{-1}\sum_a\tilde\xi(a\sigma^j)\overline{\tilde\eta(a\sigma^j)}$; the left side is $\delta_{\xi\eta}\delta_{\lambda1}$ for all $\lambda$, so Fourier inversion on $\mathbb Z/d$ gives $s_j=\delta_{\xi\eta}$ for every $j$, in particular $s_1$. Holds. The brute-force counts above (script scripts/twisted_classes.py) agree with the resulting equality of dimensions.
  4. Step 4 (basis of $C_I^{(N,M)}$). Block-triangular in the indicators of $I(\mathcal O)$-orbits (surjectivity $I(\mathcal O)\to H(\mathbb F_q)$ needs $I_{N,M}$ connected, kept from (L1c)); diagonal blocks are the invertible twisted character tables of step 3 times $(-1)^{\delta}$; non-$\mathrm{Fr}$-stable orbits have no rational points; IC's with non-$\sigma$-stable $\xi$ are not $\mathrm{Fr}$-stable and contribute $0$ (block permutation, Audit 7.7 step 2). The chain of A3-6 then gives (a). Holds.
  5. Step 5. Each simple object of a level is $\mathrm{Fr}^n$-stable for some $n$ (finite permutation); $\mathcal C_I$ and $\mathcal M_I$ are defined over $\overline{\mathbb F}q$ and unchanged when $q$ is replaced by $q^n$; apply (a) over $\mathbb F{q^n}((t))$. Holds. (c) is the block-permutation vanishing. This is the treatment of checklist item 4 at function level: functions see only $\mathrm{Fr}$-stable simple objects, and only through traces.

Sources