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

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7 Answers 7

up vote 16 down vote accepted

Yes, you might look at the following paper by J. Fine and D. Panov:

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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.

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Let me add that it's obvious that dz dw is preserved in the quotienting (CY). Also, b_1 = 3, meaning it's non-Kahler. Here is a link to Kodaira's paper: – Eric Zaslow Nov 19 '10 at 12:59

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.

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Nice example. I was aware that the standard Hopf surface has non-trival canonical bundle, but I did not know that the are other Hopf surfaces with trival canonical bundle. The paper I cited above constructs simply connected examples, though. – Andrei Moroianu Nov 18 '10 at 20:53
These manifolds are not Hopf surfaces. They are not compact. – jvp Nov 19 '10 at 1:55
oops! I think jvp is correct, my example is not compact. The map from my surface to $\mathbb{C}$ given by $(x,y)\mapsto xy$ is surjective. Wikipedia on Hopf surfaces tells me that in order for the above class of surfaces to be a Hopf surface, I need $0<|\alpha|\leq|\beta|<1$ and so their canonicals will all have non-trivial canonical bundle. Mea Culpa. – Jim Bryan Nov 19 '10 at 5:29
Ah, thanks for the clarification jvp & Jim. – Dave Anderson Nov 19 '10 at 5:56

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

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Nice example! How do you see that it is non-Kahler? Do you have a reference? – JME Jul 30 '11 at 10:53
This is an old result of Andre Blanchard dating back in the '50s. A very thorough discussion may be found in a paper of Thomas Hofer:… – Andrei Halanay Aug 1 '11 at 19:32

There is a reasonably extensive literature on non-Kahler Calabi-Yau threefolds. They are of interest in string theory; see for example, as well as for an analogue of Calabi's conjecture in this context.

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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.

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More examples are given by nilmanifolds.

Barberis, Dotti and Verbitsky proved in Theorem 2.7 in that nilmanifolds endowed with invariant complex structures have trivial canonical bundle. See also Cavalcanti and Gualtieri's Theorem 3.1 in

On the other hand, non-tori nilmanifolds never admit a Kaehler structure, because of Benson and Gordon, , or Hasegawa, , or ...

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