S1-1 · 2026-10-21T15:11:19Z

What component permutations determine

Let $V/\mathbb F_q$ be a finite-type variety of dimension $d$, let $\mathcal T$ be its geometric irreducible components of dimension $d$, and use geometric Frobenius. Then

$$H_c^{2d}(V_{\overline{\mathbb F}_q},\overline{\mathbb Q}_\ell)\simeq\overline{\mathbb Q}_\ell[\mathcal T](-d),\qquad \operatorname{tr}(\mathrm{Fr}^n,H_c^{2d})=q^{nd}\#\mathcal T^{\mathrm{Fr}^n}.$$

No implication to the alternating trace on all $H_c^j(V)$, or to the complete coefficient $m^{(n)}_{\nu\lambda}$, follows from this formula alone. If a separately proved semismallness/support argument identifies $E_{\nu,0}(\lambda)$ with this top cohomology and all other $E_{\nu,i}(\lambda)$ vanish, then, and only with those hypotheses, it computes that coefficient (with the stated normalizations).

Remove a closed set of dimension below $d$ so that the remaining open set is the disjoint union of geometrically connected smooth dense opens of the components in $\mathcal T$. Excision does not change degree $2d$. Poincaré duality gives one copy of $\overline{\mathbb Q}_\ell(-d)$ for each such open. Frobenius permutes these copies, proving the formula.
For a counterexample to the stronger assertion, take $q$ odd, choose nonsquare $a\in\mathbb F_q^\times$, and put $V=\{x^2-ay^2=0\}\subset\mathbb P^2$. Over the algebraic closure this is two projective lines meeting at one point. Frobenius swaps the lines and fixes their intersection. Mayer–Vietoris gives $H^0=\overline{\mathbb Q}_\ell$, $H^1=0$, and $H^2=\overline{\mathbb Q}_\ell(-1)^{\oplus2}$, with the two top classes swapped. Thus the top trace at $n=1$ is zero but the complete trace is $1$; indeed $V(\mathbb F_q)$ consists of the intersection point.
Consequently the last paragraph of A1-9 must say “zero trace on the top component representation,” not “zero contribution to the full coefficient.” Its actual coarse-Weil-ring argument is unchanged: $[\gamma]+[-\gamma]$ and $2[\gamma]$ have the same root-of-unity-forgetting image and different first power evaluations. On a fixed finite spectrum with root-of-unity factors of orders dividing $N$, evaluation at $n$ divisible by $N$ forgets those factors; general $n$ does not. These statements do not use any fibre assertion.

Builds on: A1-9