§8 · Agent 8 · Audit: group, 𝔸ⁿ, 𝔾ₘ · 2026-10-18T19:21:32Z

Audit 8.6 · Audit of (IG) and (II$_I$) for the group (A4-6, A2-9 step 8)

proved

(IG) fails for the group over every non-archimedean field by the sign character; this is the pages' Proposition E argument, not a new one. The functions-to-sheaves implication (II_I) for X = H is correctly proved, and vacuous since (IG) fails.

Audit 8.6.

Holds. (a) For every non-archimedean \(F\) and \(W\ne1\): \(\mathcal H_I\ast\varphi_0=\operatorname{span}\{a\ast\mathbf 1_{K_H}\ast b\}\subseteq\ker\varepsilon\not\ni T_e\), so (IG) fails. (b) Over \(F=\mathbb F_{q^m}(\!(t)\!)\), \(\operatorname{span}\{\operatorname{tr}_mM_{x,y}\}=\mathcal H_{I_H}\mathbf 1_{K_H}\mathcal H_{I_H}\). (c) (II\(_I\)) for \(X=H\): if (IG) held over \(\mathbb F_q(\!(t)\!)\), then \(\mathcal M_I^\flat=\widetilde W\). With Audit 8.5 this re-derives the failure of (IG) over \(\mathbb F_q(\!(t)\!)\) without circularity.

Remark: (a) is word for word classical.html Proposition E; the function half of row 1 has no second proof in the file. The brief only excludes the pages' Propositions II and IV.1, so this is allowed.

  1. The two-sided action. With \((g_1,g_2)\cdot x=g_1xg_2^{-1}\) and \(\operatorname{vol}(I_H\times I_H)=1\), \(((f_1\otimes f_2)\ast\varphi)(x)=\iint f_1(g_1)f_2(g_2)\varphi(g_1^{-1}xg_2)=(f_1\ast\varphi\ast f_2^\vee)(x)\), \(f^\vee(g)=f(g^{-1})\); \(T_w^\vee=T_{w^{-1}}\) and \(\varepsilon(f^\vee)=\varepsilon(f)\). With \(\varphi_0=\mathbf 1_{H(\mathcal O)}=\mathbf 1_{K_H}\), \(\mathcal H_I\ast\varphi_0=\mathcal H_{I_H}\mathbf 1_{K_H}\mathcal H_{I_H}\), a two-sided ideal on which \(\varepsilon\) vanishes by Audit 8.3, while \(\varepsilon(T_e)=1\). The coefficient field does not matter: all spaces have \(\mathbb Q\)-forms spanned by the \(T_w\), and spanning is a rank condition.
  2. (b). \(\{\operatorname{tr}_m\mathrm{IC}_z\}\) is a basis (unitriangular to \(\{T_z\}\) up to sign on finite Bruhat lower sets; D3); with multiplicativity, the span of the traces of generators is the ideal.
  3. D1. \(\mathrm{Fl}_z(\mathbb F_q)=I_HzI_H/I_H\) and each \(I\)-orbit is Frobenius-stable with \(\mathbb F_q\)-points one \(I(\mathbb F_q[[t]])\)-orbit, by Lang for the connected stabilizers (Audit 8.1 step 1; the pro-algebraic limit as in Audit 8.3 step 3).
  4. (c). If \(T_\mu\in\operatorname{span}\operatorname{tr}M_{x,y}\), then \(\operatorname{tr}\mathrm{IC}_\mu\in\operatorname{span}\{\operatorname{tr}\mathrm{IC}_z:z\in\mathcal M_I^\flat\}\) because coefficients \(\sum_n(-1)^n\operatorname{tr}A_{z,n}\) vanish for \(z\notin\mathcal M^\flat_I\); independence of IC traces gives \(\mu\in\mathcal M^\flat_I\). Only “nonzero coefficient ⇒ constituent” is used, which is correct without weights. To conclude \(\mathcal C_I^{\rm Iwa}=\mathcal C_I\) one needs \(\mathcal M_I=\mathcal M_I^\flat\), i.e. the equivariant splitting (Audit 8.1(R3)). This is the Iwahori form of satake-two-levels Prop. II for this \(X\), with inputs D1–D4 checked.
  5. (d) and checklist 6. The authors are right that (a) does not imply Audit 8.5: a retract need not satisfy \(\varepsilon\circ\operatorname{tr}=0\), and \(\mathrm{IC}_e\oplus\mathrm{IC}_e[1]\) has \(\varepsilon\)-trace \(0\). The sheaf statement is genuinely stronger and needs the weights of Audit 8.4.
  6. Fields. (a) over any non-archimedean \(F\) (complex coefficients); (b), (c) over \(\mathbb F_{q^m}(\!(t)\!)\) with \(\ell\)-adic sheaves over \(\overline{\mathbb F}_q\).

Builds on

  • Corollary 4.6 · (IG) fails by the same character; (II$_I$) holds for the group

Built on by

  • Summary 8.13 · Verdict of the audit of Claim III: group, $\mathbb A^n$, $\mathbb G_m$
Original record, as the agent wrote it at 19:21 UTC

Audit of (IG) and (II$_I$) for the group (A4-6, A2-9 step 8)

  1. The two-sided action. With \((g_1,g_2)\cdot x=g_1xg_2^{-1}\) and \(\operatorname{vol}(I_H\times I_H)=1\), \(((f_1\otimes f_2)\ast\varphi)(x)=\iint f_1(g_1)f_2(g_2)\varphi(g_1^{-1}xg_2)=(f_1\ast\varphi\ast f_2^\vee)(x)\), \(f^\vee(g)=f(g^{-1})\); \(T_w^\vee=T_{w^{-1}}\) and \(\varepsilon(f^\vee)=\varepsilon(f)\). With \(\varphi_0=\mathbf 1_{H(\mathcal O)}=\mathbf 1_{K_H}\), \(\mathcal H_I\ast\varphi_0=\mathcal H_{I_H}\mathbf 1_{K_H}\mathcal H_{I_H}\), a two-sided ideal on which \(\varepsilon\) vanishes by Audit 8.3, while \(\varepsilon(T_e)=1\). The coefficient field does not matter: all spaces have \(\mathbb Q\)-forms spanned by the \(T_w\), and spanning is a rank condition.
  2. (b). \(\{\operatorname{tr}_m\mathrm{IC}_z\}\) is a basis (unitriangular to \(\{T_z\}\) up to sign on finite Bruhat lower sets; D3); with multiplicativity, the span of the traces of generators is the ideal.
  3. D1. \(\mathrm{Fl}_z(\mathbb F_q)=I_HzI_H/I_H\) and each \(I\)-orbit is Frobenius-stable with \(\mathbb F_q\)-points one \(I(\mathbb F_q[[t]])\)-orbit, by Lang for the connected stabilizers (Audit 8.1 step 1; the pro-algebraic limit as in Audit 8.3 step 3).
  4. (c). If \(T_\mu\in\operatorname{span}\operatorname{tr}M_{x,y}\), then \(\operatorname{tr}\mathrm{IC}_\mu\in\operatorname{span}\{\operatorname{tr}\mathrm{IC}_z:z\in\mathcal M_I^\flat\}\) because coefficients \(\sum_n(-1)^n\operatorname{tr}A_{z,n}\) vanish for \(z\notin\mathcal M^\flat_I\); independence of IC traces gives \(\mu\in\mathcal M^\flat_I\). Only “nonzero coefficient ⇒ constituent” is used, which is correct without weights. To conclude \(\mathcal C_I^{\rm Iwa}=\mathcal C_I\) one needs \(\mathcal M_I=\mathcal M_I^\flat\), i.e. the equivariant splitting (Audit 8.1(R3)). This is the Iwahori form of satake-two-levels Prop. II for this \(X\), with inputs D1–D4 checked.
  5. (d) and checklist 6. The authors are right that (a) does not imply Audit 8.5: a retract need not satisfy \(\varepsilon\circ\operatorname{tr}=0\), and \(\mathrm{IC}_e\oplus\mathrm{IC}_e[1]\) has \(\varepsilon\)-trace \(0\). The sheaf statement is genuinely stronger and needs the weights of Audit 8.4.
  6. Fields. (a) over any non-archimedean \(F\) (complex coefficients); (b), (c) over \(\mathbb F_{q^m}(\!(t)\!)\) with \(\ell\)-adic sheaves over \(\overline{\mathbb F}_q\).

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