I have been read many papers,But I don"t know a integral hodge class of the calabi-Yau 3-fold is algebraic or non-algebraic?Hope give some help and nice reference. Calabi-Yau 3-fold is a Kahler 3-fold with trival canonical bundle.Is it a open question?

  • $\begingroup$ I think we don't even know for abelian varieties... $\endgroup$
    – diverietti
    Oct 24, 2012 at 16:27
  • $\begingroup$ The integral Hodge conjecture is known to be false for Calabi-Yau 3-folds. See Kollar's "Classification of Irregular Varieties" or Voisin's "On integral Hodge classes on uniruled or Calabi-Yau threefolds." $\endgroup$
    – Matt
    Oct 24, 2012 at 17:03
  • $\begingroup$ As I was indicated by Totaro, the case of abelian threefolds has been answered positively by Craig Grabowski in his thesis "On the integral Hodge conjecture for $3$-folds". $\endgroup$ Aug 26, 2016 at 14:38

1 Answer 1


Even though there are several examples showing that the Integral Hodge conjecture (IHC) fails in general, there are some positive results as long as the canonical bundle is not too positive:

Theorem (Voisin) Let $X$ be a smooth projective threefold over $\mathbb C$ which is either uniruled or strongly Calabi-Yau ($K_X\simeq O_X$ and $b_1(X) = 0$). Then the IHC is true for X, i.e., $H_2(X, \mathbb Z)$ is generated by algebraic curves.

There is also the following extension to the theorem above proved by Totaro:

Theorem (Totaro) The IHC holds for Calbi-Yau 3-folds holds without the assumption that $b_1(X) = 0$. In particular, IHC holds for 1-cycles on abelian 3-folds.

See Voisin's paper On integral Hodge classes on uniruled or Calabi-Yau threefolds and these notes.

This raises the question when $H_2(X,\mathbb Z)$ is generated by rational curves for simply connected Calabi-Yaus (as is the case for K3 surfaces). This is of course a difficult question since it is not even known whether a simply connected CY 3-fold always contains a rational curve.

  • 1
    $\begingroup$ Where can we find the proof of Totaro's theorem? $\endgroup$
    – HYL
    Feb 8, 2014 at 17:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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