§4 · Agent 4 · Audit: counterexample and mechanism · 2026-10-17T22:11:43Z

Audit 4.4 · Audit: the orbit obstruction and the injective leading-term lemma

proved

Both parts of Lemma 2.2 are re-derived. (a) Hecke translates of \(\varphi_0\) stay in \(G(F)X(\mathcal O)=G(F)x_0\). (b) With an injective leading-term map, the Hecke span meets each lower set exactly in the span of the corresponding translates. The first remark of Lemma 2.2 rests on Remark 2.12, which is outside this audit.

Audit 4.4.

(a) Let \(F=\mathbb F_q(\!(t)\!)\), \(X=G/H\) with \(H\) smooth and connected. Then \(X(\mathcal O)=K\cdot x_0\), \(\operatorname{supp}(f\ast\varphi_0)\subseteq G(F)x_0\) for all \(f\in\mathcal H_K\), and (SG) implies \(X(F)=G(F)x_0\).

(b) Let \(\{f_j\}_{j\in J}\) span \(\mathcal H_K\) (or at least have \(f_j\ast\varphi_0\) spanning \(\mathcal H_K\ast\varphi_0\)), \(I\) the \(K\)-orbits with a partial order with finite lower sets, and \(\psi:J\to I\) injective with \(\operatorname{supp}(f_j\ast\varphi_0)\subseteq\{\nu\le\psi(j)\}\) and \(f_j\ast\varphi_0\ne0\) on \(O_{\psi(j)}\). Then \(\mathcal H_K\ast\varphi_0\cap C(L)=\operatorname{span}\{f_j\ast\varphi_0:\psi(j)\in L\}\) for every lower set \(L\), and \(\mathbf 1_{O_\mu}\in\mathcal H_K\ast\varphi_0\Rightarrow\mu\in\psi(J)\).

Outcome: holds. The remark "(b) never yields a failure under (R) and (C)" depends on Remark 2.12 (not audited here). The remark that the top value comes from the coset \(t^\lambda K\) needs \(t^\lambda x_0\in O_{\varphi(\lambda)}\), which holds when \(\tau(t^\lambda x_0)=t^{\lambda-\theta\lambda}\) with \(\varphi(\lambda)\) dominant.

  1. (a), support. \((f\ast\varphi_0)(x)=\sum_{gK}f(g)\varphi_0(g^{-1}x)\ne0\) forces \(g^{-1}x\in X(\mathcal O)\).
  2. (a), integral points. \(G\to X\) is smooth and surjective, so by Hensel's lemma every \(x\in X(k[\![t]\!])\) lifts: \(L^+X(k)=L^+G(k)x_0\) with stabilizer \(L^+H\). The \(K\)-orbits on \(L^+X(\mathbb F_q)=X(\mathcal O)\) are indexed by \(\ker\big(H^1(\mathbb F_q,L^+H)\to H^1(\mathbb F_q,L^+G)\big)\), trivial because \(L^+H\) is connected when \(H\) is (Lang). So \(X(\mathcal O)=Kx_0\).
  3. (a), orbits. The \(G(F)\)-orbits on \(X(F)\) are indexed by \(\ker\big(H^1(F,H)\to H^1(F,G)\big)\). A \(K\)-orbit outside \(G(F)x_0\) is compact open (the orbit map is smooth), so its indicator lies in \(C_K\) but, by steps 1–2, not in \(\mathcal H_K\ast\varphi_0\).
  4. (b), the top term. Let \(f=\sum_jc_jf_j\ast\varphi_0\in C(L)\) be a finite sum, \(M=\{\psi(j):c_j\ne0\}\), and \(m\in M\) maximal. If \(c_j\ne0\) and \(f_j\ast\varphi_0\ne0\) on \(O_m\), then \(m\le\psi(j)\in M\), so \(\psi(j)=m\), and \(j\) is unique by injectivity. Hence \(f|_{O_m}=c_{j_0}(f_{j_0}\ast\varphi_0)|_{O_m}\ne0\) and \(m\in L\). Every element of the finite set \(M\) lies below a maximal one and \(L\) is a lower set, so \(M\subseteq L\). The reverse inclusion is the support bound. Independence of the \(f_j\ast\varphi_0\) and the one-orbit statement (take \(f=\mathbf 1_{O_\mu}\), \(L=\{\nu\le\mu\}\), so the maximal \(m\) must equal \(\mu\)) follow in the same way. "(SG) \(\iff\psi(J)=I\)" then follows with Proposition B.
  5. Injectivity is essential. In the split group case \(\varphi(a,b)=a-b\) is far from injective on \(\Lambda_T^+\), and (SG) holds although no injective \(\psi\) on all of \(\Lambda_T^+\) exists; cancellations of common top terms are what (b) excludes.
  6. Hypotheses as used in Theorem 2.6. \(I=\Lambda_H^+\) with the dominance order has finite lower sets, also when \(H\) has a central torus (\(c'\le c\) forces equal central components); \(J=\Lambda_H^+\), \(f_a=T_a\) is a basis of \(\mathcal H_K\); \(\psi(a)=2a\) is injective.

Builds on

  • Lemma 2.2 · Lemma: orbit and leading-term obstructions to \(\mathcal H_K\ast\varphi_0=C_K\)

Built on by

  • Summary 4.12 · Verdict of the audit: the unitary twist is a valid counterexample; the repairs are minor

Sources