§4 · Agent 4 · Audit: counterexample and mechanism · 2026-10-17T23:01:23Z

Audit 4.19 · Audit: under (R), (C), injective leading terms force dominant lifts

proved

The key claim of Remark 2.12 item 4 is re-derived. If \(\varphi\) is injective on \(\Lambda_T^+\) modulo cocharacters acting trivially on \(X\), then \((T^\theta)^\circ\) acts trivially, hence is central. Then \(Z_G(S)=T\), \(\varphi\) is onto \(\Lambda_S\) because \(\pi_0(T^\theta)=1\), and every dominant label has a dominant preimage. So (SG) holds and the injective-leading-term criterion of Lemma 2.2(b) never yields a failure inside Proposition II's setting.

Audit 4.19.

Assume (R), (C), \(p\) odd, and that no simple factor of \(G\) acts trivially on \(X\). (1) If \(\varphi|_{\Lambda_T^+}\) is injective, then \(X_*(T)^\theta=0\), \(T=S\), and \(Z_{G^\theta}(S)=T[2]\) is disconnected; so under (C), \(\varphi\) is never injective on \(\Lambda_T^+\) (rank \(\ge1\)). (2) If \(\varphi|_{\Lambda_T^+}\) is injective modulo the cocharacters acting trivially on \(X\), then \(Z'=(T^\theta)^\circ\) is central in \(G\), no root vanishes on \(S\), \(Z_G(S)=T=SZ'\), \(\varphi:\Lambda_T\to\Lambda_S\) is onto, and every \(\mu\in\Lambda_S^+\) has a dominant lift. So \(U=\varnothing\) and (SG) holds (Audit 4.18(e)). (3) Items 1–3 of Remark 2.12 hold (the forcing criterion is Audit 4.4(b) with \(\psi=\varphi|_{(\Lambda_T^+)^\sigma}\); for the unitary twist \(\ker\varphi\cap X_*(T)^\sigma=0\)).

Outcome: holds with minor defects. (i) "Divide \(G\) by any simple factor that acts trivially" is a reduction sketch (Hecke operators of such a factor act by their degrees), not carried out. (ii) Item 5's cyclicity formula \([M,\beta]=[N,\mathrm{Fr}^*(i)\circ\beta\circ p]\) is correct in the convention \(\beta:M\to\mathrm{Fr}^*M\), opposite to the convention \(\beta:\mathrm{Fr}^*M\to M\) of Theorem 1.3. (iii) "which Frobenius twisting cannot produce" is heuristic: under (R), Frobenius on labels is trivial, but Frobenius eigenvalues on multiplicity spaces can still cancel (Proposition 1.8(g)). Item 5 is labeled heuristic and uses no external categorical-trace theorem, as stated.

  1. (1). For \(0\ne\kappa\in\ker\varphi=X_*(T)^\theta\) and \(\xi\) regular dominant, \(N\xi+\kappa\) and \(N\xi\) are dominant for \(N\gg0\) with the same image. Injectivity forces \(X_*(T)^\theta=0\), so \(\theta=-1\) on \(X_*(T)\), \(T\) is \(\theta\)-split and equals \(S\), \(Z_G(S)=T\), and \(Z_{G^\theta}(S)=T^\theta=T[2]\cong\mu_2^{\operatorname{rk}G}\), disconnected (\(p\) odd).
  2. (2), \(Z'\) acts trivially and is central. The same pair \(N\xi+\kappa\), \(N\xi\) shows that every \(\kappa\in X_*(T)^\theta\) acts trivially on \(X\), so the torus \(Z'\) generated by these cocharacters lies in the kernel \(N\) of the action. With no simple factor acting trivially, \(N^\circ\) (normal, connected) is a central torus, so \(Z'\subseteq Z(G)\).
  3. (2), \(Z_G(S)=T\). \(T=SZ'\) (Audit 4.18 step 2). A root vanishing on \(S\) would vanish on the central \(Z'\) too, hence on \(T\); impossible. So \(Z_G(S)=T\) and \(Z_{G^\theta}(S)=T^\theta\).
  4. (2), surjectivity of \(\varphi\). \(f:T\to S\), \(t\mapsto t\theta(t)^{-1}\), lands in \(S\) (connected image of \(\theta\)-split elements), is onto (\(f(s)=s^2\)), has kernel \(T^\theta\), and \(X_*(f)=\varphi\). From \(1\to T^\theta\to T\to S\to1\): \(0\to X^*(S)\to X^*(T)\to X^*(T^\theta)\to0\), and dualizing gives \(\operatorname{coker}X_*(f)\cong\operatorname{Ext}^1(X^*(T^\theta),\mathbb Z)\), of order \(|\pi_0(T^\theta)|\). Under (C), \(T^\theta=Z_{G^\theta}(S)\) is connected, so \(\varphi\) is onto. Checks: \(\mathrm{SL}_2/\mathrm{SO}_2\): \(\varphi=\times2\), \(\pi_0=\mathbb Z/2\); \(\mathrm{GL}_2/T\): \(\varphi\) onto, \(T^\theta=\) centre.
  5. (2), dominance. A preimage of \(\mu\) has the form \(\lambda=\tfrac12\mu+\zeta\) with \(\zeta\in X_*(Z')_{\mathbb Q}\), since \(\varphi\) is \(2\cdot\)(projection to \(\Lambda_S\otimes\mathbb Q\)). Roots vanish on the central \(Z'\), so \(\langle\alpha,\lambda\rangle=\tfrac12\langle\alpha,\mu\rangle\ge0\): \(\lambda\) is dominant. Hence \(U=\varnothing\), and (SG) holds by Audit 4.18(e) (= Corollary C; the run's Lemma 4.8).
  6. Conclusion. Inside (R) and (C), either \(\varphi\) is not injective modulo trivial cocharacters (the criterion does not apply), or every label lifts dominantly ((SG) holds). A counterexample there needs an uncovered label \(\mu\) with \(\mathrm{IC}_\mu\) a summand but \(\mathbf 1_{\mathcal O_\mu}\notin\mathcal H_K\ast\varphi_0\) (Audit 4.18(e)). This confirms the scope statement of Summary 4.12.

Builds on

  • Remark 2.12 · Remark: why the trace misses thick generation, and what (R) removes

Built on by

  • Summary 4.20 · Verdict of the second audit: the mechanism claims hold, with minor repairs
  • Theorem 6.3 · Verdict: sheaf generation over \(\overline{\mathbb F}_q\) does not imply spherical generation over \(\mathbb F_q(\!(t)\!)\)

Sources