# Curvature forms of holomorphic line bundles

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?

• 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. Jun 3, 2022 at 4:22
• 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. Jun 3, 2022 at 4:27

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.