Let $A$ be a $g$-dimensional, complex abelian variety, let $H$ be an ample divisor, let $D\in Pic^0(A)$, and let $0\leq k\leq g$.

Question 1: Does $D^k\neq 0\in CH^k(A,\mathbb{Q})$ imply that $H^{g-k}\cdot D^k\neq 0\in CH^g(A,\mathbb{Q})$?

(As Jason points out, it is necessary to work with $\mathbb{Q}$-coefficients.)

Question 2: Assuming the answer to Question 1 is "yes", is there a simple proof of this fact?

Primarily, we are interested in abelian varieties over $\mathbb{C}$. (As ulrich points out, Kunnemann answered Question 1 affirmatively for abelian varieties over finite fields.)

  • $\begingroup$ That is not true. Consider what happens when you replace $H$ by a positive integer multiple $nH$. Then $H^{g-k}\cdot D^k$ is replaced by $n^{g-k}(H^{g-k}\cdot D^k)$. Thus, if $D$ is a $n$-torsion divisor class, then $(nH)^{g-1}\cdot D$ is zero. $\endgroup$ May 24, 2018 at 15:13
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    $\begingroup$ This is indeed true with rational coefficients. More general results are proven in the paper "A Lefschetz decomposition for Chow motives of abelian schemes" by Klaus Kunnemann. $\endgroup$
    – naf
    May 24, 2018 at 19:39
  • $\begingroup$ Thank you both. I will read through the Kunnerman paper. $\endgroup$ May 25, 2018 at 2:48
  • $\begingroup$ I want to remark that I approved the suggested edit because I am interested in the case over $\mathbb{C}$ $\endgroup$ May 30, 2018 at 20:34
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    $\begingroup$ Proposition 5.5 in this paper gives what you want. However I would not say that the proof is simple... $\endgroup$
    – abx
    May 31, 2018 at 8:53

1 Answer 1


Let me just explain the statement of the result which alluded to by abx and proved by Beauville. Let $A$ be an abelian variety over $\mathbb{C}$, and let $\operatorname{CH}^*(A)$ be the Chow ring of cycles with rational coefficients modulo rational equivalence.

Beauville considers eigenspaces for the action of pulling back by multiplication by $n$ of $\operatorname{CH}^*(A)$ defined as follows

$$\operatorname{CH}^p_s(A):= \{x\in \operatorname{CH}^p(A) | n^*_Ax = n^{2p-s}x \text{ for all }n\in \mathbb{Z}\}.$$

The Chow ring of $A$ completely decomposes into these eigenspaces. (I believe that conjecturally this decomposition is a splitting of the Bloch-Beilinson filtration, but I couldn't define the latter filtration for you.)

The first observation that can be made is if $D \in \operatorname{CH}^1(A)$ is topologically trivial (i.e. if $D$ is in $\operatorname{Pic}^0(A)$), then $D\in \operatorname{CH}^1_1(A)$. This is because pulling back by multiplication by $n$ induces multiplication by $n$ on $\operatorname{Pic}^0(A)$.

Second, if we choose $H$ to be a symmetric ample divisor (i.e. choose $H$ to be invariant under $\pm 1$), then $H\in \operatorname{CH}^1_0(A)$, this should follow from the theorem of the square.

Now we state Beauville's result.

Proposition 5.5. (Beauville) If $H$ is a symmetric ample divisor on $A$, then the multiplication map $$ -\times H^{g-p}\colon \operatorname{CH}^p_s(A) \rightarrow \operatorname{CH}^q_s(A) $$ is injective for $p+q\le g+s$ and surjective for $p+q \ge g+s$. In particular it is bijective for $p+q=g+s$.

In our situation, we have $D^k \in CH^k_k(A)$. Thus the map of interest is $$ -\times H^{g-k} \colon \operatorname{CH}^k_k(A) \rightarrow \operatorname{CH}^g_k(A). $$ Thus we have $p+q=g+s$ in the proposition and therefore the map is an isomorphism, which answers the question.

If $H$ is not symmetric, then one can write $H=H^{sym}+\epsilon$ where $H^{sym}$ is a symmetric ample divisor and $\epsilon\in \operatorname{Pic}^0(A)$. Then the proposition applied to $(H^{sym})^k$ answers the question after the observation that all the terms in the expansion of $(H^{sym}+\epsilon)^{g-k}$ are in distinct eigenspaces in $\operatorname{CH}^{g-k}(A)$.


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