## Algebraic cycles of dimension 2 on the square of a generic abelian surface

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)?

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 Regarding the first question (divisors on A): why isn't the theta dunction concrete enough for generating all of them ? – David Lehavi Apr 7 2010 at 7:03 @David Lehavi: Please give a reference! – Mikhail Borovoi Apr 7 2010 at 9:05

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 $\mathbb{Q}$.

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Thanks! But what I really want is to see somehow all the cycles! – Mikhail Borovoi Apr 6 2010 at 18:27
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 2010 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 2010 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 2010 at 6:31

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.

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

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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.

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 Thanks! Yes, I know the paper of Nori of 1989 and the paper of Fakhruddin of 1996. – Mikhail Borovoi Apr 8 2010 at 6:24