Let $X$ be a smooth projective variety defined over a number field $F$ and consider the Abel-Jacobi map $\mathrm{AJ}_k:\mathrm{CH}_0^k(X_{\overline{\mathbb{Q}}})\rightarrow \mathrm{Jac}^{2k-1}(X)$, where the LHS is the subgroup of the Chow group consisting of cohomologically trivial cycles and the RHS is the intermediate Jacobian considered as complex tori. In general, it is a conjecture that if the AJ image of a cycle is zero, then the cycle is zero, i.e., rationally equivalent to zero.

My question is about a simple situation: suppose $\mathrm{H}^1(X,\mathbb{C})$ vanishes so that $\mathrm{AJ}_1$ is the trivial map, would the whole of $\mathrm{CH}_0^1(X_{\overline{\mathbb{Q}}})$ Or $\mathrm{CH}_0^1(X_{\overline{\mathbb{Q}}})\otimes_{\mathbb{Z}} \mathbb{Q}$ be trivial?

  • $\begingroup$ Please edit: are you looking at the case $k=1$? $\endgroup$
    – abx
    Oct 31, 2013 at 8:38
  • 2
    $\begingroup$ the Chow group $CH^1(X)$ of codimension 1 cycles can be identified with the Picard group $Pic(X)$. The complex analytic space associated with the latter can be written as the quotient of $H^1(X^{an},\mathcal{O}_X)=H^0(X^{an},\Omega^1_X)^*$ by the lattice $H^1(X^{an},\mathbb{Z})$. Indeed if $H^1(X^{an},\mathbb{C})=H^1(X^{an},\mathcal{O}_X)\oplus H^0(X^{an},\Omega^1_X)$ vanishes you get what you expect. $\endgroup$
    – Nicola
    Oct 31, 2013 at 9:52

1 Answer 1


Indeed $\mathrm{CH}_0^1(X_{\overline{\mathbb{Q}}})=\mathrm{Pic}^0(X_{\overline{\mathbb{Q}}})=0 $. The point is that since $X_{\overline{\mathbb{Q}}}$ has a rational point, the group $\mathrm{Pic}^0(X_{\overline{\mathbb{Q}}})$ is isomorphic to the group of $\overline{\mathbb{Q}}$-points of the Jacobian, which is trivial.


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