§5 · Agent 5 · Audit: transfer, hypotheses · 2026-10-18T04:00:39Z
Audit 5.2 · Audit of A4-4: \(\Lambda_S=(1-\theta)X_*(T)\) and the labels \(x_\kappa\)
proved audit: correct
The \(E_6\) lattice computations of Computation 4.4 are redone independently in exact arithmetic and agree. The labels \(x_\kappa=t^\mu x_0\) are rational and independent of the lift \(\mu\).
Audit 5.2.
Outcome: holds. In coroot coordinates, \(e_1=(1-\theta)\alpha_1^\vee=(2,1,2,2,1,0)=2\omega_1^\vee-\omega_6^\vee\) and \(e_6=(0,1,1,2,2,2)=-\omega_1^\vee+2\omega_6^\vee\). \((1-\theta)\alpha_j^\vee=0\) for \(j=2,\dots,5\), \(\operatorname{rank}X_*(T)^\theta=4\) and \(3\omega_1^\vee=2e_1+e_6\). Moreover \(X_*(S)=\mathbb Ze_1\oplus\mathbb Ze_6\), and \(x_\kappa\) does not depend on \(\mu\). Notation warning (checklist item 8). Computation 4.4's \(e_i\in\Lambda_S\subseteq X_*(T)\) and Audit 3.1's \(e_i\in\Lambda_X\) (the image of \(\alpha_i^\vee\)) correspond under \(\Lambda_S\cong\Lambda_X\), \(\varphi(\mu)\mapsto\bar\mu\). Pairings with \(\mathcal X(X)\) differ by the factor \(2\): \(\langle e_1^{\mathrm{A4}},\gamma_1\rangle=4\), \(\langle e_1^{\mathrm{A3}},\gamma_1\rangle=2\). This is consistent, but the two must not be mixed.
What was computed (script a5_e6_checks.py, exact integer and rational arithmetic).
- The \(72\) roots of \(E_6\) (Bourbaki labels) are generated from the simple roots by simple reflections; the Cartan matrix is symmetric, so roots and coroots share coordinates.
- \(W(D_4)=\langle s_2,\dots,s_5\rangle\) is enumerated (\(192\) elements). Exactly one element negates \(\alpha_2,\dots,\alpha_5\); it has length \(12\). This is \(w_L\), and \(\theta=-w_L\) on \(X_*(T)=Q^\vee\).
- \(\langle\alpha_j,e_1\rangle_j=(2,0,0,0,0,-1)\) and \(\langle\alpha_j,e_6\rangle_j=(-1,0,0,0,0,2)\), giving the coweight expressions. \(\omega_1^\vee=(\tfrac43,1,\tfrac53,2,\tfrac43,\tfrac23)\), so \(\omega_1^\vee\notin Q^\vee\), and \(3\omega_1^\vee=(4,3,5,6,4,2)=2e_1+e_6\).
- \(\operatorname{rank}(\theta-1)=2\), so \(\ker(\theta-1)=X_*(T)^\theta\) has rank \(4\).
Proofs re-derived.
- \(\theta\mu=-\mu\iff w_L\mu=\mu\iff\mu\perp\{\alpha_2,\dots,\alpha_5\}\). So \(X_*(S)=Q^\vee\cap(\mathbb Z\omega_1^\vee\oplus\mathbb Z\omega_6^\vee)\). This is the kernel of the surjection \(\mathbb Z\omega_1^\vee\oplus\mathbb Z\omega_6^\vee\to P^\vee/Q^\vee\cong\mathbb Z/3\), so it has index \(3\). \(\mathbb Ze_1\oplus\mathbb Ze_6\) lies in it with index \(\det\begin{pmatrix}2&-1\\-1&2\end{pmatrix}=3\). Hence equality, and \(\varphi(X_*(T))=\mathbb Ze_1+\mathbb Ze_6\) because \(\varphi\) kills \(\alpha_2^\vee,\dots,\alpha_5^\vee\).
- If \(\varphi(\mu)=\varphi(\mu')\), then \(\mu-\mu'\in X_*(T)^\theta=X_*(T_8)\), a saturated sublattice. The corresponding subtorus \(T_8\) is split over \(\Lambda\) and fixed pointwise by \(\theta\), so \(t^{\mu-\mu'}\in H(E)\) and \(t^\mu x_0=t^{\mu'}x_0\). Also \(\tau(t^\mu x_0)=t^{\mu-\theta\mu}=t^{\varphi(\mu)}\). No \(\sqrt\varpi\) is needed (cf. Proposition 4.9, where it is).
- Related check used later. On \(X^*(T)=P\), \(1-\theta^*=1+w_L\) sends \(\omega_1,\dots,\omega_6\) to \((2,0),(1,1),(2,1),(2,2),(1,2),(0,2)\) in the basis \(\omega_1,\omega_6\). Its elementary divisors are \((1,1)\), so \((1-\theta^*)P=\mathbb Z\omega_1\oplus\mathbb Z\omega_6\) and \(X^*(T^\theta)\) is torsion-free of rank \(4\). This confirms Proposition 3.2 step 3(a) and Proposition 3.7 step 1.
Builds on
- Computation 4.4 · Computation: \(\Lambda_S=(1-\theta)X_*(T)\) and \((T^\theta)^\circ\) is maximal in \(F_4\)
Built on by
- Summary 5.13 · Verdict of audit A5: transfer route A4-2…A4-9 and char-p hypotheses A3-1…A3-10
Sources
- Computation 4.4 (claim audited)
- script a5_e6_checks.py (parts 1–2)
- spherical-generation/families/e6-f4.html (E6/F4 data)