Good evening,

I am reading a paper of 1961 by Andreotti e Vesentini "Sopra un teorema di Kodaira" (unfortunately available only in Italian), that extends under some peculiar assumptions the Kodaira's embedding theorem of 1954. To begin my question, I recall two definitions that are written in the mentioned paper. Let $X$ be a complex manifold. A form $\varphi \in \Omega^{1,1}(X)$ is called a upper bound form (not necessarily closed) if the following conditions are satisfied:

  1. $\forall p \in X$ a differential form $\sigma_p$ can be chosen such that $\varphi - \sigma_p$ is positive in $X \setminus \{p\}.$
  2. There exists a Hermitian metric on the canonical bundle $K$, such that, if $\Theta_K$ is the curvature two form induced from that metric, then $\varphi - \Theta_K$ is positive.
  3. There exists a complete Hermitian metric on $X$ defined from $\varphi - \Theta_K.$

A holomorphic line bundle $E \rightarrow X$ is called uniformly positive, if it is possible to choose a upper bound form $\varphi$ on $X,$ a positive integer $\mu_0,$ and a metric on the fibers of $E$ such that $h_0 \Theta_F - \varphi$ is positive $\forall p \in X.$

My questions are the following:

  • How a manifold that admits a uniformly positive line bundle should look like?
  • Can someone give me an example of a manifold that does not admits a uniformly positive holomorphic line bundle?
  • 1
    $\begingroup$ The conditions seem difficult to read, but roughly speaking, I think the authors wish to extend the Kodaira embedding theorem to complex manifolds admitting a holomorphic line bundle satisfying some positivity condition, where the manifolds are not necessarily compact. This is useful, despite the conditions looking not so nice. I presume that open subsets of compact manifolds satisfying the usual conditions of the Kodaira theorem satisfy the authors' conditions, but please double check. I also presume that complex manifolds which are not projective probably do not satisfy these conditions. $\endgroup$ – Malkoun Sep 22 '17 at 14:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.