Let $S/\mathbf{F}_q$ be a $d$-dimensional smooth projective variety and $A/S$ be an Abelian scheme. Is there an easy description of $H^0(S, A(\ell)(d-1))$?` ($A(\ell)$ = union of $A_{\ell^n}$)

Maybe you have misstated the question - as it stands the answer is always zero (because the same is true for a single fibre).
– Tony SchollOct 6 '10 at 15:31

Right. It should be $A(\ell)$ rather than $T_\ell{A}$.
– user6960Oct 6 '10 at 15:42

1

Assume $S$ is connected and let $s\in S(\mathbb{F}_q)$. Then your $H^0$ injects into $H^0(k(s), A_\bar{s}(\ell)(d-1))$, which is the Pontryagin dual of the coinvariants $T/(F_s - q^d)$, $T=$ Tate module of abelian var. dual to $A_s$. So by Riemann hypothesis for eigenvalues of Frobenius, that group is finite, of order something like $P_s(q^d)$ where $P_s=$ char.poly of Frob (I may have twists/duals not exactly correct). So order of your $H^0$ divides the HCF of these integers for closed points $s$. Is that the sort of thing you are looking for?
– Tony SchollOct 6 '10 at 17:32

`$s\in S(\mathbb{F}_q)$`

. Then your $H^0$ injects into $H^0(k(s), A_\bar{s}(\ell)(d-1))$, which is the Pontryagin dual of the coinvariants $T/(F_s - q^d)$, $T=$ Tate module of abelian var. dual to $A_s$. So by Riemann hypothesis for eigenvalues of Frobenius, that group is finite, of order something like $P_s(q^d)$ where $P_s=$ char.poly of Frob (I may have twists/duals not exactly correct). So order of your $H^0$ divides the HCF of these integers for closed points $s$. Is that the sort of thing you are looking for? – Tony Scholl Oct 6 '10 at 17:32