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Questions about Kähler manifolds and Kähler metrics.

2 votes

Examples of surfaces with negative Kahler curvature operator

Examples of compact Kähler manifolds with non positive holomorphic bisectional curvature are given by: Closed submanifolds of complex tori. Smooth compact quotients of bounded symmetric domains. M …
diverietti's user avatar
  • 7,902
7 votes
1 answer
940 views

Riemannian holonomy of generic manifolds

It is well known, as well as absolutely intuitive, that the Riemannian holonomy of a generic Riemannian manifold is $O(n)$, the Riemannian holonomy of a generic orientable Riemannian manifold is $SO(n …
diverietti's user avatar
  • 7,902
22 votes
1 answer
1k views

Relationship between the signs of different notions of curvature in complex geometry

Let $(X,\omega)$ be a complex hermitian manifold, and call $\Theta$ its Chern curvature tensor. Out of this we can consider different notions of curvature, namely the holomorphic bisectional curvature …
diverietti's user avatar
  • 7,902
2 votes

Geometrical meaning of admissible hermitian metric on a line bundle

I don't know if this answers to your question, and very likely you already know it, but one easy fact is the following. First of all an obvious necessary condition is that at the level of cohomology …
diverietti's user avatar
  • 7,902
0 votes

A question about nef classes on compact Kähler manifolds

Just for fun, here is an answer in the purely algebraic setting. So, suppose that $X$ is irreducible projective algebraic of dimension $n$, $\alpha=c_1(\mathcal O_X(D))$ is the class of a nef diviso …
diverietti's user avatar
  • 7,902
2 votes

What is the holomorphic sectional curvature?

Maybe you already were aware of that, or maybe it really doesn't answer to your question, but I'll try anyhow... Take a look at this, Subsection 7.5 on page 39. The construction you talk about in you …
diverietti's user avatar
  • 7,902
11 votes

Why can we not always take a Kähler class to be in rational cohomology?

Since Artie Prendergast-Smith is not expanding his comment in an answer, let me do it. As I said in the comments, his comment is essentially THE answer to the OP question. But let me give some more de …
diverietti's user avatar
  • 7,902
15 votes

Specific line bundle over complex manifold implies Kähler?

As Dima said, it is much more: in fact it is projective. But let me give you some more insights on this kind of questions. I shall give you the definition of four different classes of compact comple …
diverietti's user avatar
  • 7,902
9 votes

Kähler metric on projectivised bundle

Let me recall you briefly how to obtain such a Kähler metric on the total space of the projectivized bundle. Start with any given hermitian metric $h$ on $E$ and consider on the projectivized bundle …
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  • 7,902
4 votes

Noncompact Kähler manifolds with nonzero Ricci tensor but vanishing scalar curvature

On a $n$-dimensional Kähler manifold $(X,\omega)$, the Ricci form is (minus) the curvature of the canonical bundle $K_X$ endowed with the induced metric. Thus, if $X$ has zero Ricci curvature then its …
diverietti's user avatar
  • 7,902
9 votes
1 answer
2k views

Deform a compact Kähler manifold to a non Kähler one

Could you give me an example of a compact Kähler manifold which analytically deforms to a non Kähler one? For example, there is no hope to find a complex structure on a Hopf manifold in order to make …
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