Question

Are there examples of non-Kahler complex manifolds with holomorphically trivial canonical bundle?

Answer 1

Yes, you might look at the following paper by J. Fine and D. Panov: http://arxiv.org/abs/0905.3237

Answer 2

This is covered in Andrei Halanay's answer, but it's worth mentioning the simplest examples, which are primary Kodaira surfaces. For the simplest of these:

Take C^2 and quotient by the group generated by these a_k:

a_1 : z -> z + 1

a_2 : z -> z + i

a_3 : w -> w + z + 1

a_4 : w -> w - iz + i

(I think this is it.)

The quotient group is nonabelian. Here z is the fiber and w the base.

Answer 3

There are Hopf surfaces which give an example. Topologically they are given by $S^3\times S^1$ and can be realized as a complex manifold by the quotient $\mathbb{C}^2-(0,0)/\mathbb{Z}$ where $n\in \mathbb{Z}$ acts by $(x,y)\mapsto (\lambda^n x,\beta^n y)$ for some fixed non-zero complex numbers $\lambda$ and $\beta$. The Picard group of these guys is a $\mathbb{C}^*$ (a line bundle is determined by its monodromy around the $S^1$ factor). The monodromy of the canonical line bundle on the above Hopf surface is $\lambda\cdot \beta$ and so if we take $\beta=\lambda^{-1}$, it will be trivial. Indeed, the section $dx\wedge dy$ of the canonical line bundle on $\mathbb{C}^2-(0,0)$ will then descend to the quotient.

Answer 4

Any non-trivial principal elliptic bundle $\pi:X \to B$ over a Calabi-Yau basis is non-Kaehler but has trivial canonical bundle (because $\mathcal{K}_X \simeq \pi^*\mathcal{K}_B$).

Answer 5

There is a reasonably extensive literature on non-Kahler Calabi-Yau threefolds. They are of interest in string theory; see for example http://xxx.lanl.gov/abs/hep-th/0301161, as well as http://xxx.lanl.gov/abs/0809.4748 for an analogue of Calabi's conjecture in this context.

Answer 6

One can ask for non-Kähler compact manifolds with holomorphically trivial tangent bundle, and still get many examples. By a result of Wang ( Proc. AMS 5 ), these are quotients of a complex Lie group $G$ by a discrete subgroup $\Gamma$.

If the quotient is compact Kähler then $G$ must be abelian. Indeed, every vector subspace of the Lie algebra of $G$ gives rise to a holomorphic differential form on $G/\Gamma$. If $G$ is not abelian then one can choose a subspace not closed under the Lie bracket. The corresponding differential form is clearly not closed, what cannot happen in compact Kähler manifolds.

For a thorough study of examples of this kind see this book by Winkelmann.