Simple example of non-integrable holomorphic connection

Question

Let $X$ be a complex manifold with complex dimension $d$ and structure sheaf $\mathcal{O}_X$. Let $E$ be a locally free sheaf on $X$. A $holomorphic$ connection on $E$ is a morphism of sheaves $$\nabla: E \to E \text{ }\otimes_{\mathcal{O}_X} \Omega_{X}^{1} $$ satisfying the product rule $\nabla(fs) = s \otimes df + f\nabla(s)$ for all open $U \subset X$ with $f \in \mathcal{O}_X(U), s \in E(U)$ . The connection $\nabla$ is said to be $flat$ or $integrable$ if the composite $$ E \xrightarrow{\nabla} E \text{ } \otimes_{\mathcal{O}_X} \Omega_{X}^{1} \xrightarrow{\nabla_1} E \text{ } \otimes_{\mathcal{O}_X} \Omega_{X}^{2}$$ is $0$ where $\nabla_1$ is the map to 2-forms $s \otimes w \mapsto \nabla(s) \wedge w \text{ } + s \otimes dw$. What is a simple example of a triple $(X, E, \nabla)$ with $\nabla$ non-integrable? Any such example must necessarily have $d \geq 2$.

Answer 1

Any degree nonzero line bundle on a compact Riemann surface has a non-integrable holomorphic connection. The reason is that its first Chern class is nonzero.

More precisely, take a compact Riemann surface $X$, and a line bundle together with Hermitian metric $(\mathcal{L},h)$ on $X$, then the curvature form is $$\Omega=-\partial\bar{\partial}\log h$$ This curvature form $\Omega$ is exactly the composite $\nabla_1\circ\nabla$, where $\nabla$ is the Chern connection associated to $h$. (You can read more on Griffiths & Harris's Principles of Algebraic Geometry, Chapter 0, section 5.)

Moreover, the first Chern class of $\mathcal{L}$ is $c_1(\mathcal{L})=-\frac{1}{2\pi i}\Omega,$ and the degree of the line bundle is $$\deg(\mathcal{L})=\int_Xc_1(\mathcal{L}).$$ In particular, degree nonzero implies $c_1(\mathcal{L})$ (and therefore the curvature form) is nonzero.