2
$\begingroup$

Let $M$ be a compact complex manifold, $L$ a holomorphic line bundle over $M$, and $\nabla$ a connection extending the holomorphic structure map $\overline{\partial}$ of $L$. In general can it happen that the curvature form has a $(2,0)$ component? In the case that it does not, can the $(1,1)$-curvature form be non-degenerate?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ It certainly can contain a $(2,0)$-component. In general, if $\theta$ is the connection matrix in a local frame, then the curvature is $\Theta=d\theta+\theta^2=d\theta$ (since $\theta$ is a $1$-form). Locally, the connection matrix could be any $1$-form, and if it is compatible with $\bar\partial$, it could be any $(1,0)$-form. $\endgroup$
    – Richard Lärkäng
    Commented Jun 3, 2022 at 4:22
  • 1
    $\begingroup$ Regarding your second question, it seems like the Chern connection associated to a hermitian metric on $\mathcal{O}_{\mathbb{P}^n}(1)$ would answer that. $\endgroup$
    – Richard Lärkäng
    Commented Jun 3, 2022 at 4:27

1 Answer 1

Reset to default
1
$\begingroup$

Sure, any closed 2-form $\eta$ with integer cohomology can serve as the curvature of a connection on a line bundle. This can be seen if you take a line bundle with the same Chern class and connection $\nabla$ (which is possible using the $C^\infty$ exponential sequence) and modifying the connection by taking $\nabla_1:=\nabla +\alpha$ where $\alpha$ is a 1-form such that $d\alpha= \eta-\eta_0$, where $\eta_0$ is the curvature of $\nabla$. If your form $\eta$ was of type (1,1)+(2,0), the resulting connection $\nabla_1$ induces a holomorphic structure on $L$. The (1,1)-part of the curvature can be non-degenerate, degenerate or even zero.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .