Audit 5.11 · Audit of A3-11, A3-2: no n-dependence; retracts reach the trace, Hecke does not

Re-derivation.

1. A3-11(a). The counts of A3-7 are polynomials in \(q\) whose derivation is uniform over \(\mathbb F_{q^n}\) (verified in Computation 5.7). At relevant pairs the leading term is \(c_{\nu\lambda}q^{f}\); otherwise the degree is at most \(f-1\), because \(r_\lambda(\nu)\) is even. Lang–Weil then gives exactly \(c_{\nu\lambda}\) top components, all fixed by Frobenius. Step 1 of A3-11 (towers of rational fibrations) is plausible, but it is not needed and is not proved in detail.
2. A3-11(b), (c). Frobenius acts on \(H^{2f}_c\) by the scalar \(q^f\), so no \(\operatorname{tr}A=0\) cancellation occurs. Audit 5.5 is uniform in \(F\). So the obstruction is the fixed matrix \(c\).
3. A3-2(a). By D1 the Weil structure is block-triangular for the perverse filtration. Its \((\nu,i)\) diagonal block is \(\Phi^{\mathrm{can}}_\nu\otimes A_{\nu,i}\), because \(\operatorname{End}\mathrm{IC}_\nu=\overline{\mathbb Q}_\ell\) and \(\operatorname{Hom}(\mathrm{IC}_\nu,\mathrm{IC}_{\nu'})=0\). Compressing to one summand gives the formula. This is correct but formal: \(\operatorname{tr}\mathrm{IC}_\nu\) is itself a twisted trace with \(e\) a primitive idempotent.
4. A3-11 remark. A multiplicity space with \(\operatorname{tr}A=0\) at \(n=1\) forces a zero entry, and possibly a zero row, at \(n=1\), and so a failure of \((\mathrm{SG})_1\). At even \(n\) the row is nonzero, but that alone does not decide \((\mathrm{SG})_n\) (columns may still fail to span). "Disappears at even \(n\)" is therefore unjustified. This is a heuristic remark and is not used.
5. Retracts versus Hecke operators. Thick closure adds retracts; on traces this adds the classes \(\operatorname{tr}(\Phi\circ\mathrm{Fr}^*e)\) for idempotents \(e\in\operatorname{End}(V_\lambda\ast\mathrm{IC}_0)\). The Hecke algebra only produces \(e=\mathrm{id}\). For this \(X\), Audit 5.8 gives \(V_{((1,1),0)}\ast\mathrm{IC}_0\simeq\mathrm{IC}_{(2,0,0)}\oplus\mathrm{IC}_{(1,0,1)}\oplus\mathrm{IC}_{(0,0,2)}\): restricting the adjoint representation of \(\mathrm{SO}_5\) gives \(V_2\boxtimes V_0\boxtimes V_0+V_0\boxtimes V_2\boxtimes V_0+V_1\boxtimes V_1\boxtimes V_0\). So the endomorphism algebra is \(\overline{\mathbb Q}_\ell^{\,3}\), and the idempotent onto \(\mathrm{IC}_{(0,0,2)}\) is not in the image of \(\operatorname{End}(V_\lambda)=\overline{\mathbb Q}_\ell\). The fusion functional of Audit 5.9 shows that no combination of the \(e=\mathrm{id}\) traces reaches \(\operatorname{tr}\mathrm{IC}_{(0,0,2)}\).