# Confusion surrounding the Koszul-Malgrange theorem

I recently had the need to appeal to some complex geometry in my research and have been trying to unravel the various relationships surrounding the Koszul-Malgrange theorem.

According to nlab, the theorem goes as follows.

Theorem 1. (Koszul-Malgrange theorem)

Holomorphic vector bundles over a complex manifold are equivalently complex vector bundles which are equipped with a (hermitean) holomorphic flat connection. Under this identification the Dolbeault operator $$\overline{\partial}$$ acting on the sections of the holomorphic vector bundle is identified with the holomorphic component of the covariant derivative of the given connection.

Since the original reference is in french, which I have trouble reading, I haven't been able to go though the proof of the theorem. From the bit of complex geometry I know though, I am a bit confused.

I know that, given a complex vector bundle $$E\to X$$ and a flat connection, the $$(0,1)$$-part of the connection defines a holomorphic structure, while the $$(1,0)$$-component defines a holomorphic connection on the corresponding holomorphic bundle. By Chern-Weil theory, the rational chern classes of $$E$$ vanish, as do the rational Chern classes of a holomorphic vector bundle equipped with holomorphic connection.

On the other hand, I know that in general, a holomorphic bundle may have nonvanishing rational Chern classes. In fact, over compact, complex projective manifolds, these classes generate certain Hodge classes (or possibly all according to the Hodge conjecture).

This makes me think the theorem should be a correspondence between flat bundles and holomorphic vector bundles with holomorphic connection.

Question

What is the exact relationship between smooth complex vector bundles equipped with connection and holomorphic structures? I realize this question is probably a bit elementary, but complex geometry is not my specialty. References are also welcome.

• You could try Griffith-Harris, or Ballmann's book on Kähler geometry. As an educated guess, you should read "holomorphically flat", which means something like ($\partial^2=0$ or rather) $\bar\partial^2=0$. The Chern classes still see the $(1,1)$-part of $(\partial+\bar\partial)^2$. I am only irritated by the word "Hermitean" - one can always introduce a Hermitean metric, but then "equivalently" does not make so much sense ... Apr 16 '16 at 8:27
• @SebastianGoette I see, so your guess is that "holomorphically flat" should mean something like "flat in the (0,1) direction", but not flat as a smooth connection? That makes more sense to me. I'll take a look at the reference you provided. Thanks! Apr 16 '16 at 8:45
• Apr 16 '16 at 9:49
• That's how I am used to using "holomorphically flat", maybe it's my physics background. In any case, to disambiguate, the sentence continued with the precise statement. Apr 18 '16 at 16:39
• @UrsSchreiber I took this quote from the page titled "holomorphic vector bundle". I now see that you have a page devoted to the theorem where its a bit clearer. Apr 19 '16 at 3:28

Since I was rather surprised that those authors would make such a claim, I looked up the original reference [Koszul, Malgrange, Sur certaines structures fibrées complexes.] Here is a rough translation of the relevant theorem:

Theorem 2. Let $G$ be a complex Lie group, $V$ a complex manifold, $P$ a principal bundle with group $G$ over $V$. Let $\gamma$ be a connection form on $P$. The following are equivalent

• The almost complex structure defined by $\gamma$ is a complex structure.
• ...
• The $(0,2)$ part of the curvature $R_\gamma$ is $0$.
• ...

[Footnote in orignal: ...For $R_\gamma$ to be of type $(2,0)$ it is necessary and sufficient that $\gamma$ be holomorphic...]

So that they are not saying that the curvature is zero, or that the connection is holomorphic. Clearly the version of the theorem you quoted is needlessly ambiguous, at best. Perhaps you should contact the author of that nlab page.

• I agree this is much clearer than the entry on the nlab page. I have the author's contact, so I'll let them know. Thanks! Apr 16 '16 at 13:12