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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 15 down vote accepted

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

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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: jstor.org/pss/2373157 –  Eric Zaslow Nov 19 '10 at 12:59
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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
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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
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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
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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: projecteuclid.org/DPubS/Repository/1.0/… –  Andrei Halanay Aug 1 '11 at 19:32
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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.

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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 http://www.ams.org/mathscinet-getitem?mr=2496748 that nilmanifolds endowed with invariant complex structures have trivial canonical bundle. See also Cavalcanti and Gualtieri's Theorem 3.1 in http://www.ams.org/mathscinet-getitem?mr=2131642

On the other hand, non-tori nilmanifolds never admit a Kaehler structure, because of Benson and Gordon, http://www.ams.org/mathscinet-getitem?mr=976592 , or Hasegawa, http://www.ams.org/mathscinet-getitem?mr=946638 , or ...

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