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

$\begingroup$ I think we don't even know for abelian varieties... $\endgroup$– diveriettiOct 24, 2012 at 16:27

$\begingroup$ The integral Hodge conjecture is known to be false for CalabiYau 3folds. See Kollar's "Classification of Irregular Varieties" or Voisin's "On integral Hodge classes on uniruled or CalabiYau threefolds." $\endgroup$– MattOct 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". $\endgroup$– Olivier BenoistAug 26, 2016 at 14:38
1 Answer
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 CalabiYau ($K_X\simeq O_X$ 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 CalbiYau 3folds holds without the assumption that $b_1(X) = 0$. In particular, IHC holds for 1cycles on abelian 3folds.
See Voisin's paper On integral Hodge classes on uniruled or CalabiYau threefolds and these notes.
This raises the question when $H_2(X,\mathbb Z)$ is generated by rational curves for simply connected CalabiYaus (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 3fold always contains a rational curve.

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