For vector bundles $(\pi: V \rightarrow M )$ over a complex manifold, there is a notion of holomorphicity that can be defined in two equivalent ways :

  • $V$ is a complex manifolds and $\pi:V \rightarrow M$ is holomorphic

  • we can find transition functions for V, $\psi_{ij}:M\rightarrow M_{n\times n}(\mathbb{C})$ that are holomorphic function of the base manifolds

(then the holomorphic coordinates that defines the complex structure of the first definition are simply the coordinates in the special trivializations of the second definition )

This allows to define a Dolbeault operator $\bar{\partial}: \Gamma(V) \rightarrow T^*_{0,1}M \otimes \Gamma(V)$ on sections of V: If $s: M \rightarrow V$ is a section of V, and $s^i$ are its coordinates in one of the trivializations given by the second definition, then the coordinates of $\bar{\partial}s$ are simply $\bar{\partial}s^i$ and it's clear that this definition does not depend on the trivialization.

Now I am interested in a generalization of this procedure to the case where the base manifold only has an almost complex structure.

Let $(M,J_M)$ be an almost complex manifold and let $(\pi: V \rightarrow M , J_L )$ be "an almost holomorphic vector bundle" over M.

I am not completely familiar with the terminology here and I am not sure if this is standard so let me be explicit again. By almost holomorphic vector bundle I mean a vector bundle $\pi: V \rightarrow M$, equiped with an almost complex structure $J_V$ such that $J_M \circ d\pi = d\pi \circ J_V$.

I think that part of my problems is that I don't know how to make sense of this definition in terms of trivialization.

Now here is my question: For holomorphic vector bundles over complex manifolds we can define a Dolbeault operator $\bar{\partial}$ that acts on sections $\bar{\partial}: \Gamma(V) \rightarrow T^*_{0,1}M \otimes \Gamma(V)$. Is there a way to generalize this construction to almost holomorphic vector bundle?

If yes this would allow to define "almost holomorphic section" as section s such that $\bar{\partial}s=0$.


1 Answer 1


A complex structure $J$ on a manifold $M$ induces an algebraic splitting of the complexified cotangent bundle, $$ TM \otimes \mathbf{C} = T^{1,0}M \oplus T^{0,1} M, $$ and one can define $\bar{\partial}$ to be the exterior derivative $d$ followed by projection to the $(0, 1)$-subbundle.

However, the condition $\bar{\partial}^{2} = 0$ is equivalent to integrability of $J$, i.e., to $J$ being a holomorphic structure. (The wikipedia link contains more details. Wells' Differential Analysis on Complex Manifolds also discusses the issue, if memory serves.) Similar remarks can be made for vector bundles over $M$.

I think you'll find the formal condition $\bar{\partial} s = 0$ "behaves as you'd like it to" primarily because of integrability. Particularly, $d = \partial + \bar{\partial}$ is equivalent to integrability.

In other words, you can make the requested definitions without problem, but they're probably not as useful as you might expect.

  • $\begingroup$ Thanks for answering but I still don't really see how to define this operator in the case I want to consider here. I think my question wasn't sufficiently clear anyway so I edited it to make it more explicit. I know that non integrable almost complex structure have less nice properties than integrable ones but I am nonetheless interested in what type of operations you can define with it. $\endgroup$
    – Ayriann
    Jul 10, 2015 at 11:48
  • $\begingroup$ @Hwang Do you have a reference for the claim "$d = \partial + \overline{\partial}$ is equivalent to integrability"? $\endgroup$
    – AmorFati
    Jan 15, 2019 at 6:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .