Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $k$ be a number field. Recall that Faltings proved the famous Tate conjecture, which states that for any abelian variety $X$ over $k$ and any prime $\ell$, the natural map $$\mathrm{End}(X) \otimes \mathbb{Q}_\ell \to \mathrm{End}(V_\ell(X))^{\mathrm{Gal}(\bar k/k)},\qquad (*)$$ is an isomorphism, where $V_\ell(X)$ denotes the usual Tate module of $X$ tensored with $\mathbb{Q}_\ell$.

If I understand it correctly however, this is but one part of the Tate conjecture, which in general states that the cycle class map $$A^r(X) \otimes \mathbb{Q}_\ell \to H^{2r}(\bar X, \mathbb{Q}_\ell)^{\mathrm{Gal}(\bar k/k)},$$ is an isomorphism for all $r$. The "Tate conjecture" as I state it in $(*)$ is but the case $r=1$ of this conjecture.

Is the full Tate conjecture still open for abelian varieties? Are there any interesting special cases where it is known?

share|improve this question
3  
Most cases where the full Tate conjecture is known is when the Galois invariants are generated by classes of divisors (which is known not to be true in general). This is proved by showing that the image of the Galois group acting on $H^2$ is as big as possible, e.g., if the endomorphism ring is $\mathbb{Z}$ then one would like to show that the Zariski closure of the image of Galois is the symplectic group. However, these types of results require restrictions on the dimension. (The known results are due to Ribet, Serre, Chi, Pink and probably others I don't know about.) –  ulrich May 1 at 5:12

1 Answer 1

up vote 12 down vote accepted

Let me put it this way, Tate's conjecture for abelian varieties is known to imply the Hodge conjecture for abelian varieties, and the last is very much open for this class. For the implication, see the article by Deligne and Milne on "Hodge cycles on abelian varieties" (you can get a copy off of Milne's website).

There are lot's of interesting cases known for Tate. People like Zarhin, who seems to be around and has done a lot of this, can give more precise statements.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.