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Let $U$ be an affine smooth surface over an algebraic closure of a finite field. Let $\mathscr{A}/U$ be an Abelian scheme and $\ell \neq \mathrm{char}(k)$ be prime.

Are there vanishing results for the étale cohomology with compact supports $H^2_c(U,\mathscr{A}[\ell^n]) = H^2(U,\mathscr{A}[\ell^n]^\vee(2))$ (the equality by Poincaré duality)? (My question is if there are $U$ such that this holds true, not if it is true for all $U$.)

Edit: $H^2(U,\mu_{\ell^n})$ is related to $\mathrm{Pic}(U)$ via the cycle class map.

Edit 2: What I want: Let $S/\bar{\mathrm{F}}_q$ be a smooth projective surface. Is there a (smooth?) projective ample hypersurface section $C \hookrightarrow S$ with affine complement $U \hookrightarrow S$ such that $H^2_c(U,\mathscr{A}[\ell^n]) = H^2(U,\mathscr{A}[\ell^n]^\vee(2)) = 0$?

My idea: Excise generators of $\mathrm{Pic}(S)/\ell^n$ and $\mathrm{Br}(S)[\ell^n]$ in order for $H^2(U,\mathscr{A}[\ell^n]^\vee(2))$ sitting in $$0 \to \mathrm{Pic}(U)/\ell^n \to H^2(U,\mu_{\ell^n}^{2g}) \to \mathrm{Br}(U)[\ell^n] \to 0$$ to vanish (the quotient of the Picard group $\mathrm{Pic}(S)/\ell^n$ is finitely generated; one could also assume $\mathrm{Br}(S)[\ell^n]$ to be finite). Can someone tell me if this works out correctly?

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    $\begingroup$ This seems unlikely. E.g. if $\mathcal{A}$ is a relative elliptic curve with full $\ell$-torsion, then $\mathcal{A}[\ell]$ is isomorphic to $(\mu_\ell)^2$ (ignoring twists since the field is algebraically closed) and $H^2(U,\mu_\ell)$ very much depends on the geometry of $U$. $\endgroup$
    – Martin Bright
    Commented Jun 14, 2014 at 9:41
  • $\begingroup$ My question was if there are $U$ such that this holds true, not if it is true for all $U$. $\endgroup$
    – user19475
    Commented Jun 14, 2014 at 10:03
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    $\begingroup$ Certainly there are some. Take $U$ to be $\mathbb{A}^2$ and $\mathscr{A}$ a constant Abelian variety over $U$. Then you have $\mathrm{Pic}(U)=0$ and $\mathrm{Br}(U)=0$, so $H^2(U,\mu_\ell)$ vanishes. Perhaps you could clarify what you mean by "vanishing results". $\endgroup$
    – Martin Bright
    Commented Jun 14, 2014 at 13:34
  • $\begingroup$ Thank you. What I mean is: I am looking for cases of $U$ and $\mathscr{A}/U$ such that $H^2(U,\mathscr{A}[\ell^n]^\vee(2)) = 0$. $\endgroup$
    – user19475
    Commented Jun 14, 2014 at 13:39

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