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

I would like to know, what is known on algebraic cycles of dimension 2 modulo algebraic or rational equivalence on the square of a generic abelian surface.

First, let $A$ be a generic abelian surface (generic abelian variety of dimension 2) over $\mathbb{C}$. Then the group of cycles of dimension 1 (divisors) up to rational equivalence is known, it its the Picard group of $A$, see e.g. the answers to this question and Fulton, Intersection Theory, Chapter 19. Still I have a stupid question: Can one "write down" all the (positive?) divisors on $A$?

The real question concerns the square $A^2=A\times_{\mathbb{C}} A$ of a generic abelian surface $A$. This is an abelian variety of dimension 4. I am interested in algebraic cycles of dimension 2 on $A^2$. I think I know the group of algebraic cycles of dimension 2 on $A^2$ modulo homological equivalence and modulo torsion, it is $\mathbb{Z}^6$ (because the space of invariants of $\mathrm{Sp}_{4,\mathbb{Q}}$ in $\wedge^4(\mathbb{Q}^4\oplus\mathbb{Q}^4)$ is of dimension 6). What is known about the group of algebraic cycles of dimension 2 on $A^2$ modulo rational or algebraic equivalence? In particular, what is known about the Griffiths group? Again a stupid question: Can one "write down" all the cycles of dimension 2 on $A^2$ (in some sense)?

share|improve this question
    
Regarding the first question (divisors on A): why isn't the theta dunction concrete enough for generating all of them ? –  David Lehavi Apr 7 '10 at 7:03
    
@David Lehavi: Please give a reference! –  Mikhail Borovoi Apr 7 '10 at 9:05
add comment

4 Answers 4

Here is an easy $5$ dimensional space of cycles: Inside $A \times A$, consider the subvarieties $\{ (a,b) : a=mb \}$, for $m=0$, $1$, $2$, $3$, $4$. I will show that these are linearly independent over $\mathbb{Q}$.

By Kunneth and Poincare, $$H^4(A \times A, \mathbb{Q}) \cong \bigoplus_{i=0}^4 H^{i}(A, \mathbb{Q}) \otimes H^{4-i}(A, \mathbb{Q}) \cong \bigoplus_{i=0}^4 \mathrm{End}(H^{i}(A, \mathbb{Q})).$$

The graph of multiplication by $m$, in this presentation, has class $$(\mathrm{Id}, m \mathrm{Id}, m^2 \mathrm{Id}, m^3 \mathrm{Id}, m^4 \mathrm{Id})$$

Since the Vandermonde matrix $$\begin{pmatrix} 0^0 & 0^1 & 0^2 & 0^3 & 0^4 \\ 1^0 & 1^1 & 1^2 & 1^3 & 1^4 \\ 2^0 & 2^1 & 2^2 & 2^3 & 2^4 \\ 3^0 & 3^1 & 3^2 & 3^3 & 3^4 \\ 4^0 & 4^1 & 4^2 & 4^3 & 4^4 \end{pmatrix}$$ has nonzero determinant, the $5$ classes I listed are linearly independent over $\mathbb{Q}$.

share|improve this answer
    
Thanks! But what I really want is to see somehow all the cycles! –  Mikhail Borovoi Apr 6 '10 at 18:27
2  
After careful reading the answer I noticed that the dimension of the space of algebraic cycles in $H^4(A\times A)$ is 6, and not 5, as I erroneously wrote previously. Indeed, the space $H^2(A,\mathbb{Q})$ is a reducible representation of Sp$_4$, say, $V\oplus W$, where $V$ and $W$ are non-equivalent irreducible representations. It follows that End$(H^2(A,\mathbb{Q}))$ has a 2-dimensional space of invariants generated by Id$_V$ and Id$_W$. Thus we obtain a 6-dimensional space of Hodge cycles. They are linear combinations of intersections of divisors, hence they are algebraic. –  Mikhail Borovoi Apr 7 '10 at 19:43
    
Oh, good point. Another way to see this is that, if $\Theta$ is a $\Theta$ divisor in $A$, then $\Theta \times \Theta$ is in $\mathrm{End}(H^2(A))$ and is not a multiple of the identity. –  David Speyer Apr 8 '10 at 1:42
    
How do you see that $\Theta\times\Theta$ in End$(H^2(A))$ is not a multiple of the identity? –  Mikhail Borovoi Apr 8 '10 at 6:31
add comment

As far as I know, there is no smooth projective variety over $\mathbb{C}$ of dimension $n>2$ with all possible Hodge numbers nonzero (i.e. $h^{p,q} \neq 0$ for all $p+q = n$) for which the Griffiths group of codimension $r$ cycles is known to be zero for any $1<r<n$.

For codimension $2$ cycles the Abel-Jacobi map is expected to detect the Griffiths group, however the computations in Nori: Algebraic cycles and Hodge theoretic connectivity, p. 372, suggest that for the self product of the generic abelian surface the Abel-Jacobi map on the Griffiths group might well be nonzero.

share|improve this answer
add comment

Not an answer per se, but you might be interested in http://arxiv.org/abs/1003.3183, where similar questions are investigated.

share|improve this answer
add comment

Probably you know this, but I might point out that Nori [Proc. Indian Acad, 1989] proved that the Griffiths group of a generic abelian 3-fold is infinitely generated. It may be worth looking at, even though I have some doubts about whether his method would give anything useful in your case.

share|improve this answer
    
Thanks! Yes, I know the paper of Nori of 1989 and the paper of Fakhruddin of 1996. –  Mikhail Borovoi Apr 8 '10 at 6:24
add comment

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.