Simple example of non-integrable holomorphic connection

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 $$d \geq 2$$. This is crossposted from SE here. From that discussion it is apparent that integrability is a strict condition, as even the existence of a holomorphic connection on a holomorphic vector bundle implies vanishing of Chern classes in the compact Kahler case, and integrability is a still stronger condition, but I have not found an explicit example.

• If you don't impose more conditions, the answer is trivial. Take $E=\mathscr{O}_X$, with the connection given by $\nabla(1)= \omega$, where $\omega$ is any non-closed holomorphic 1-form -- for instance $\omega =xdy$ on $\mathbb{C}^2$. – abx May 17 '20 at 14:01
• There are many $G$-invariant examples on tangent bundles of quotients $G/\Gamma$, where $\Gamma$ is a lattice in a complex Lie group $G$. You can use the components of the Maurer--Cartan form to provide connection 1-forms. – Ben McKay May 17 '20 at 14:20
• @abx: Is there an example for compact complex manifold? I'm thinking that when $X$ is compact Kahler, all holomorphic 1-forms are closed, so there is no such trick for nonflat connection on $\mathcal{O}_X$. – AG learner May 17 '20 at 15:57
• Yes. Take for $X$ an abelian surface, $(\alpha ,\beta )$ a basis of holomorphic 1-forms, and define $\nabla$ on $\mathscr{O}_X^2$ by $\nabla(e_1)=e_2\otimes \alpha$, $\nabla (e_2)=e_1\otimes \beta$ (here $(e_1,e_2)$ is the natural basis of $\mathscr{O}_X^2$). – abx May 17 '20 at 17:29
• My examples are compact. – Ben McKay May 17 '20 at 18:59

1 Answer

Probably this answer intersects with the previous ones. Consider the complex Heisenberg group $$H$$ of the $$3\times 3$$ complex matrices with 1 on the diagonal and 0 under the diagonal. Let $$x, y, z$$ be the other entries, $$z$$ being in the corner. The center $$Z$$ is $$\{x=y=0\}$$. One has the (trivial) line bundle $$Z\to H\to H/Z\cong C^2$$ The left-invariant $$2$$-plane field $$\nabla$$ whose value at $$I$$ is $$z=0$$ is a non-integrable holomorphic connection on this bundle. If one insists that the manifolds be compact, one just has to quotient $$H$$ on the right by the lattice $$\Gamma$$ of matrices with entries in $$Z[i]$$. Let $$E:=C/Z[i]$$. One has the bundle $$E\to (H/\Gamma)\to E^2$$ and the image of $$\nabla$$ is a non-integrable holomorphic connection on this bundle.

Another famous explicit example is the hyperplane distribution $$\nabla$$ orthogonal to the complex lines in $$C^{n+1}\setminus 0$$, which is a holomorphic connection for the tautological line bundle over $$CP^{n}$$, and not integrable for $$n\ge 2$$.