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I am looking for examples $(X,\eta)$ where the integral hard Lefschetz is an isomorphism:

$X/k$ is a smooth projective variety of dimension $d$ over an algebraic closure of a finite field and $\eta \in H^2(X,\mathbf{Z}_\ell(1))$ the class of an ample hypersurface section. Consider all cohomology groups modulo torsion. By the (rational) hard Lefschetz, $(\cup\eta)^{d-1}: H^1(X,F) \to H^{2d-1}(X,F(d-1))$, with $F$ an $\ell$-adic sheaf, is injective with finite cokernel. For $F$ take $\mathbf{Z}_\ell$ or $T_\ell\mathscr{A}$ with $\mathscr{A}/X$ an Abelian scheme.

For example, I this holds if $X = \mathbf{P}^d_k$ and $F = \mathbf{Z}_\ell$. (OK, this is clear since $H^1(\mathbf{P}^d_k,\mathbf{Z}_\ell) = 0$.)

If it holds for $F = \mathbf{Z}_\ell$, it holds for $F = T_\ell\mathscr{A}$ if $\mathscr{A} = A \times_k X$ is constant, so I am interested in cases where $\mathscr{A}/X$ is not constant.

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    $\begingroup$ For an abelian variety $A$ with a principal polarization with class $\theta \in H^2(A,\mathbb{Z}_{\ell}(1))$, multiplication by $\theta ^i$ induces an isomorphism $H^{n-i}(A,\mathbb{Z}_{\ell})\buildrel\sim\over\longrightarrow H^{n+i}(A,\mathbb{Z}_{\ell}(i))$ for all $i$. $\endgroup$
    – abx
    Apr 2, 2014 at 10:07
  • $\begingroup$ Thank you! Can you give me a reference for this result? $\endgroup$
    – user19475
    Apr 2, 2014 at 10:11
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    $\begingroup$ I am afraid I don't know a precise reference. All you need to know is that $H^*(A,\mathbb{Z}_{\ell})$ is the exterior algebra of $H^1$, and that the class $\theta $ can be written $e_1\wedge e'_1+\ldots +e_g\wedge e'_g$ in a symplectic basis -- that follows easily from the definition of a principal polarization. $\endgroup$
    – abx
    Apr 2, 2014 at 10:31
  • $\begingroup$ @abx Sorry for pinging you on an old question, but I'm somewhat confused by this computation: by Riemann-Roch $\chi(L_{\theta})=\theta^g/g!$ where $L_{\theta}$ is the associated line bundle so in particular $\theta^g$ is divisible by $g!$ (equal to $g!$ if $\theta$ is a principal) and does not induce an integral isomorphism $H^0\to H^{2g}$ if $l\leq g$. Similarly, in terms of the basis, raising the expression you wrote into $g$th power gives $g! e_1\wedge e_1'\wedge\ldots e_g\wedge e'_{g}$ which does not generate $H^{2g}$ if $g!$ is not invertible in the coefficient ring. $\endgroup$
    – SashaP
    Jun 1, 2023 at 18:21

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