MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
I think we don't even know for abelian varieties... – diverietti Oct 24 '12 at 16:27
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." – Matt Oct 24 '12 at 17:03

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.

share|cite|improve this answer
Where can we find the proof of Totaro's theorem? – HYL Feb 8 '14 at 17:02

Your Answer


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.