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?

$\endgroup$
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. $\endgroup$ 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. $\endgroup$ Jun 3, 2022 at 4:27

1 Answer 1

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.

$\endgroup$

Your Answer

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

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