Are there examples of compact complex manifolds $X$ with $K_X$ nef, but $X$ is not Kähler? Perhaps even non-Moishezon examples?

Here, nef can be defined as follows: For any $\varepsilon>0$ there is a Hermitian metric $h_{\varepsilon}$ on $K_X$ with curvature $\Theta_{h_{\varepsilon}} \geq - \varepsilon \omega$, where $\omega$ is a positive-definite real $(1,1)$--form on $X$.


1 Answer 1


Let $X$ and $Y$ be compact complex manifolds. Note that $K_{X\times Y} \cong \pi_1^*K_X\otimes \pi_2^*K_Y$. If $Y$ has trivial canonical bundle, then $K_{X\times Y} \cong \pi_1^*K_X$. Now the pullback of a nef line bundle is again nef, see Proposition 1.8 (i) of Compact Complex Manifolds with Numerically Effective Tangent Bundles by Demailly, Peternell, and Schneider. So one can construct many examples by choosing $X$ with $K_X$ nef and $Y$ non-Kähler with $K_Y$ trivial.

Example: Let $X$ be a curve of genus $g > 1$ and $Y$ be a primary Kodaira surface. Then $X\times Y$ is a non-Kähler threefold with $K_{X\times Y}$ nef. Note that $X\times Y$ is also not Moishezon as it contains $Y$ as a complex submanifold and $Y$ is not Moishezon.

For more examples of non-Kähler manifolds with $K_Y$ trivial, see the introduction of Non-Kähler Calabi-Yau Manifolds by Tosatti. As is pointed out in Proposition 1.1, if $K_Y$ is trivial (or even just torsion), then it admits a metric $h$ with curvature $\Theta_h = 0$ and hence $K_Y$ is nef, so we don't even need to take a product with $X$ in the above construction.

  • $\begingroup$ Thank you! This is a very nice answer! Thank you for taking the time :-) $\endgroup$
    – ABBC
    Commented May 30, 2022 at 23:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.