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Let $A$ be an abelian variety of dimension $g$ over a number field $K$, and $\ell$ be a rational prime. Suppose that the Galois action on the $\ell$-adic cohomology $H^k(A, \mathbb{Z}_\ell) \otimes_{\mathbb{Z}_\ell} \overline{\mathbb{Q}_\ell}$ has a one-dimensional Jordan-Holder quotient. Is there a (conjectural or known) classification of the possible characters $\psi : Gal(K^{ab} / K) \to \overline{\mathbb{Q}_\ell}^*$ which can give the action on this quotient (in terms of $K$, $g$, and $\ell$)?

(When $g = 1$, there is a simple explicit description: For $H^0$ and $H^2$, it must always be the trivial and cyclotomic character respectively. For $H^1$, there can only be a one-dimensional Jordan-Holder quotient if the elliptic curve has CM, in which case there is a well-known explicit description. This follows from Serre's open image theorem, since any reducible subgroup of $GL_2(\mathbb{Z}_\ell)$ has infinite index.)

Even if there is not a good description in full generality, are there any interesting classes of examples (besides those examples coming from abelian varieties of CM-type) which admit an explicit description?

EDIT: I would like to know not just whether a particular $\psi$ can occur for some abelian variety over some number field (which is answered below by Aşağı Güzdək), but rather for which pairs $(K, g)$ it can occur for an abelian variety of dimension $g$ defined over the number field $K$.

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Any character $\psi$ arising in this way will be de Rham. The only such characters arise (up to twist: EDIT by finite order characters) from Grössencharacters from some CM field. By purity, one may determine the infinity type of $\psi$. By the Tate conjecture (proved by Faltings), any such quotient which arises in $H^1$ of an abelian variety $A$ is induced from a morphism $A \rightarrow X$ where $X$ is (the Weil restriction of) a CM abelian variety.

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    $\begingroup$ This is a great start, but it's not a full answer, right? (or did I misunderstand?). For example say $K$ is an imaginary quadratic extension of the rationals, $p$ splits in $K$, and $\psi$ is an algebraic grossencharacter with Hodge-Tate weight 0 at one place above $p$ and 2 at the other. Do we expect some unramified twist of $\psi$ to show up in the cohomology of some abelian variety? Maybe there a "Weil number" obstruction to this (eigenvalues at unramified primes had better be Weil numbers), and, if there is, then what if we impose that all these eigenvalues are Weil numbers of the same wt? $\endgroup$
    – Kevin Buzzard
    Commented Jan 29, 2011 at 12:35
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    $\begingroup$ Yes, this example occurs --- in $H^2$ of the product of the elliptic curve with CM over $K$ with itself. And I'm pretty sure it can occur for abelian varieties over $K$ of dimension $g$ if and only if $g \geq 2$ and the Hilbert class field of $K$ is contained in a quadratic extension of $K'$. $\endgroup$
    – Eric Larson
    Commented Jan 29, 2011 at 17:08
  • $\begingroup$ Kevin, the infinity type of any algebraic Hecke character is a $\mathbf{Z}$-linear combination of CM-types. This implies that (possibly after a tate twist) we can find CM-Hecke characters $\chi_1, \ldots, \chi_n$ such that $\psi$ and $\chi:= \prod \chi_i$ have the same infinity type, and hence $\xi := \psi \chi^{-1}$ will be of finite order. The characters $\chi_i$ arise in $H^1(A_i)$ for some CM abelian varieties $A_i$, and thus $\psi$ occurs in some Weil restriction (to account for $\xi$) of $H^n$ of $\prod A_i$. $\endgroup$
    – user631
    Commented Jan 29, 2011 at 18:36
  • $\begingroup$ Moreover, the Tate conjecture for Abelian varieties implies that one doesn't have to worry about the Tate twist: the $\psi$ which arise will exactly be those whose infinity type is an effective linear combination of CM types. $\endgroup$
    – user631
    Commented Jan 29, 2011 at 18:39
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    $\begingroup$ I didn't say "simple". I merely meant to suggest that I had answered your original question and wasn't really interested in the edited version. What's an example of a geometrically simple abelian variety $A$ with trivial endomorphisms with a non-trivial Grossencharacter occurring in the cohomology of $A$? $\endgroup$
    – user631
    Commented Jan 30, 2011 at 3:38

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