2
$\begingroup$

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.

$\endgroup$
5
  • 2
    $\begingroup$ 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 $\mathbb{C}^2$. $\endgroup$ – abx May 17 '20 at 14:01
  • 1
    $\begingroup$ 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. $\endgroup$ – Ben McKay May 17 '20 at 14:20
  • $\begingroup$ @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 $\mathcal{O}_X$. $\endgroup$ – AG learner May 17 '20 at 15:57
  • 1
    $\begingroup$ 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 $\mathscr{O}_X^2$). $\endgroup$ – abx May 17 '20 at 17:29
  • $\begingroup$ My examples are compact. $\endgroup$ – Ben McKay May 17 '20 at 18:59
3
$\begingroup$

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 $n\ge 2$.

$\endgroup$

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.