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2 votes
0 answers
68 views

On spin structure for Kähler manifolds and square roots of $\det (TX)$

I'm stuck on the proof that for a (compact) Kähler manifold $X$ (of complex dimension $n$), a spin structure on the tangent bundle $TX$ is equivalent to a line bundle $L$ together with an isomorphism $...
Alessandro Nanto's user avatar
1 vote
0 answers
146 views

Compact complex manifolds with nef canonical bundle have nonnegative Kodaira dimension

Let $X$ be a compact Kähler manifold with nef canonical bundle. The (Kähler extension of the) abundance conjecture asserts that $K_X$ is semi-ample, and thus $K_X^{\otimes m}$ admits a section for ...
ABBC's user avatar
  • 275
5 votes
1 answer
390 views

Compact complex non-Kähler manifolds with nef canonical bundle

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 ...
ABBC's user avatar
  • 275
0 votes
1 answer
192 views

Why vanish the integer m of an ample line bundle in the Kodaira embedding theorem?

I try to understand the following version of the Kodaira embedding theorem: Let $X$ be a compact Kähler manifold. A line bundle $L$ is positiv if and only if it is ample. I have a problem with the '...
danielg's user avatar
4 votes
0 answers
179 views

How the existence of holomorphic sections depends on the choice of complex structure

In this Mathoverflow question it is asked how many invariant complex structures exist on the full flag manifold of $SU(m)$. In this question it is asked when a line bundle over a flag manifold has ...
Han Jin Ma's user avatar
9 votes
1 answer
1k views

Non-compact Kähler manifolds which admit a positive line bundle

A complex manifold which admits a positive line bundle is automatically Kähler. Furthermore, if the manifold is compact, then it is projective by the Kodaira Embedding Theorem. In particular, not ...
Michael Albanese's user avatar