§7 · Agent 7 · Adversarial audit · 2026-10-17T04:00:32Z

Audit 7.27 · Audit of the open-cell lemma: support bound, monotonicity and the black-cone step hold

proved audit: minor defect

Minor defect (second audit). Correct audit of Lemma 8.4. The closing 'citation fix' (that \(\tau^{-1}(L^+G)=L^+X\) is in the proof of Chen–Yi's Lemma 40, not in the lemma) refers to a citation that Lemma 8.4 does not make; it belongs to the source list of Computation 8.6, whose text already says 'proof of Lemma 40'. A stale cross-reference only.

Each step of the open-cell lemma is rechecked: the support bound is a pointwise bound on the translate, \(\varphi\) is monotone, the black-cone argument excludes smaller cells with the same leading term, and the fibres are closed. No defect found.

Audit 7.27.

Lemma 8.4 holds: both fibres lie in \(\Gr^{\lambda_k}\), and their stalks are computed by proper base change without any IC stalk of the singular \(\Gr^{\le\lambda_k}\).

  1. Support bound. If \(gL^+G\in\Gr^{\lambda''}\) and \(g^{-1}x\in L^+X\), then \(x\in L^+Gt^{\lambda''}L^+X\). Proposition III, step 2 is a pointwise bound on this set: \(\tau(L^+Gt^{\lambda''}L^+X)\subseteq L^+Gt^{\lambda''}L^+Gt^{-\theta\lambda''}L^+G\), the Cartan convolution bound gives \(\nu\le\lambda''+(-\theta\lambda'')^+\), and the projection \(p\) gives \(\nu\le\varphi(\lambda'')\) (integrality is automatic since \(\Lambda_T=Q^\vee\)). Monotonicity: \(\varphi(\alpha_j^\vee)\in\{0,e_1,e_6\}\), so \(\lambda''\le\lambda_k\) implies \(\varphi(\lambda'')\le\mu_k'\). Hence \(\varphi(\lambda'')\in\{\mu_k,\mu_k'\}\).
  2. No dominant lift. \(3k\omega_1^\vee\) with \(k\) odd has no dominant lift: \(b=0\) forces \(c_2=\dots=c_6=0\), and then \(a=2c_1\) is even (family page, section 5).
  3. Black cone. \(\kappa=\lambda_k-\lambda''\) lies in \(\ker\varphi\cap Q^+\), the cone of the black coroots. Dominance of \(\lambda''\) and \(\langle\alpha_j,\lambda_k\rangle=0\) give \(\langle\alpha_j,\kappa\rangle\le0\) for every black \(j\), contradicting \((\kappa,\kappa)=\sum_jc_j(\alpha_j^\vee,\kappa)>0\). A box search (\(c_j\le6\), \(k=1,3,5\)) finds no dominant \(\lambda''\).
  4. Stalks. \(F_x\) is closed in \(\Gr^{\le\lambda_k}\) (\(L^+X\) is closed, and \(\tau\) is a closed immersion for \(p\ne2\)) and lies in the smooth cell, where \(V_{\lambda_k}=\underline k[D_k]\) and the twisted product restricts to \(\underline k[D_k]\). Proper base change gives \(R\Gamma(F_x)[D_k]\); no IC stalk of the singular \(\Gr^{\le\lambda_k}\) enters.

Citation fix (cosmetic): \(\tau^{-1}(L^+G)=L^+X\) is a parenthetical remark in Chen–Yi's proof of Lemma 40, not the lemma itself; it holds because \(\tau\) is a closed immersion.

Severity: none.

Verified by computation: e6_fibre_A7.py

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  • Lemma 8.4 · Both convolution fibres lie in the open cell \(\Gr^{\lambda_k}\)

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Original record, as the agent wrote it at 04:00 UTC

A8-4: both fibres lie in the open cell; support bound, monotonicity and step 3 hold

Step 1 (holds). \(gL^+G\in\mathrm{Gr}^{\lambda''}\), \(g^{-1}x\in L^+X\Rightarrow x\in L^+G\,t^{\lambda''}L^+X\). criterion.html Step 2 is a pointwise bound on this set: \(\tau(L^+Gt^{\lambda''}L^+X)\subseteq L^+Gt^{\lambda''}L^+Gt^{-\theta\lambda''}L^+G\), the Cartan convolution bound gives \(\nu\le\lambda''+(-\theta\lambda'')^+\), and \(p\) gives \(\nu\le\varphi(\lambda'')\) (integrality automatic, \(\Lambda_T=Q\)). Monotonicity: \(\varphi(\alpha_j^\vee)\in\{0,e_1,e_6\}\), so \(\lambda''\le\lambda_k\Rightarrow\varphi(\lambda'')\le\mu_k'\). Hence \(\varphi(\lambda'')\in\{\mu_k,\mu_k'\}\).

Step 2 (holds). \(3k\omega_1^\vee\), \(k\) odd, has no dominant lift (families/e6-f4.html §5: \(b=0\) forces \(c_2=\dots=c_6=0\), \(a=2c_1\) even).

Step 3 (holds). \(\kappa=\lambda_k-\lambda''\in\ker\varphi\cap Q^+\) = black cone; dominance of \(\lambda''\) and \(\langle\alpha_j,\lambda_k\rangle=0\) give \(\langle\alpha_j,\kappa\rangle\le0\) for all black \(j\), contradicting \((\kappa,\kappa)=\sum c_j(\alpha_j^\vee,\kappa)>0\). My box check (\(c_j\le6\), \(k=1,3,5\)) finds no dominant \(\lambda''\).

Step 4 (holds). \(F_x\) is closed in \(\mathrm{Gr}^{\le\lambda_k}\) (\(L^+X\) closed; \(\tau\) a closed immersion for \(p\ne2\)) and lies in the smooth cell, where \(V_{\lambda_k}=\underline k[D_k]\) and the twisted product restricts to \(\underline k[D_k]\); proper base change gives \(R\Gamma(F_x)[D_k]\). No IC stalk of the singular \(\mathrm{Gr}^{\le\lambda_k}\) enters. Citation fix (cosmetic): \(\tau^{-1}(L^+G)=L^+X\) is a parenthetical in Chen–Yi's proof of Lemma 40, not the lemma; it holds because \(\tau\) is a closed immersion. Severity: none.

Sources