References for “modern” proof of Newlander-Nirenberg Theorem

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Hi,

I'm starting to prepare a graduate topics course on Complex and Kahler manifolds for January 2011. I want to use this course as an excuse to teach the students some geometric analysis. In particular, I want to concentrate on the Hodge theorem, the Newlander-Nirenberg theorem, and the Calabi-Yau theorem.

I have many excellent references (and have lectured before) on the Hodge and CY theorems. However, for the Newlander-Nirenberg theorem, I am finding it hard to find a "modern" treatment. I recall going through the original paper in my graduate student days, but I hope that there is a more streamlined version floating around somewhere. (I want to consider the general smooth case, not the easy real-analytic version). Besides the original paper, so far I can only find these references:

J. J. Kohn, "Harmonic Integrals on Strongly Pseudo-Convex Manifolds, I and II" (Annals of Math, 1963)

and

L. Hormander, "An introduction to complex analysis in several variables" (Third Edition, 1990)

Both are easier to follow than the original paper. But my question is: are there any other proofs in the literature, preferably from books rather than papers? The standard texts on complex and Kahler geometry that I have looked at don't have this.

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Have you looked at the paper by Nijenhuis and Wolf in the Annals in the early sixties where they also prove existence of pseudoholomorphic curves. – Mohan Ramachandran Mar 2 2011 at 21:23

5 Answers

There is a proof due to Malgrange which can be found in Nirenberg's, Lectures on Linear Partial Differential Equations. I am not sure that one can call the proof modern, but it is the simplest proof that I know.

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Thanks for this one too. I will check it out. – Spiro Karigiannis Jul 7 2010 at 13:42

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Hi Spiro: I have had much the same difficulties as you, but I now know a modern proof.

At heart, the original proof is an application of the implicit function theorem. More specifically, let $U$ be a polydisk in $C^n$ consider the sequence of Banach manifolds $Diff^{k,\alpha}(U,C^n) \to AC^{k-1,\alpha}(U) \to (A^{0,2})^{k-2,\alpha}(U,TU).$

These are respectively the diffeomorphisms $U\to C^n$ of class $(k,\alpha)$, the almost complex structures on $U$ of class $(k-1,\alpha)$ and the $(0,2)$ forms on $U$ with values in the holomorphic tangent bundle, of class $(k-2,\alpha)$. The first map is the pullback of the standard complex structure, and the second is the Frobenius integrability form $\phi \mapsto \overline \partial \phi - \frac 12 [\phi\wedge \phi].$

The object is to show that the first map is locally surjective onto the inverse image of $0$ by the second. These spaces are Banach manifolds, and if you can show that the sequence of derivatives (respectively at the identity, at the standard complex structure and at 0) is split exact, the result follows from the implicit function theorem.

This sequence of derivatives is the Dolbeault sequence on $U$ (in the appropriate class), and it is split exact, though this is NOT obvious. There is an error in the original paper, or rather in the paper of Chern's that it depends on, but the result is true. The remainder of the mess in the original proof is due to the authors writing out the Picard iteration in the specific case, rather than isolating the needed result.

I am working on getting this written up with Milena Pabiniak, a graduate student here at Cornell. Write me at jhh8@cornell.edu if you are interested in seeing details.

John Hubbard

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Mister Hubbard, welcome to MO! – Victor Protsak Mar 2 2011 at 21:05
@John: Something using the Banach space implicit function theorem and a split short exact sequence definitely sounds like a modern proof to me. I would certainly love to see the details of this. I'll contact you.  Thanks! – Spiro Karigiannis Mar 4 2011 at 12:31

It's covered in Demailly's too-little-known book, Complex analytic and differential geometry, though the proof given there is apparently modelled on the references you cited.

Edit: I just noticed that the MathOnline link currently seems to be non-functional, so here's a link to Demailly's webpage.

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Thanks. Indeed, I did not know about this book. It has a ton of useful other stuff too. I'd still be interested to hear about any other references, though. – Spiro Karigiannis Jun 17 2010 at 15:39

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+1 for the mathonline plug! :) – Kevin Lin Jun 17 2010 at 17:32

This is not quite an answer to your question, but you might consult the book by Donaldson and Kronheimer "The geometry of 4-manifolds". In chapter 2 they prove an integrability theorem for holomorphic vector bundles, the point being that this can be regarded as a simpler version of the Newlander-Nirenberg theorem, and (in my view) very suitable for your course. You might also want to mention the following simple example for instructional purposes: the nilpotent Lie group H^3 x R where H^3 is the Heisenberg group has an obvious left-invariant almost-complex structure whose Nijenhius tensor vanishes. Although not a complex Lie group, it is easy to find independent local complex coordinates z_1, z_2. I suspect that there are similar classes of almost-complex examples where the integration is elementary.

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Thanks, Simon. I am indeed considering the simpler version from Donaldson-Kronheimer. My course plans may be a little bit too ambitious. – Spiro Karigiannis Jul 26 2010 at 13:40

In my continued searches for modern proofs of Newlander-Nirenberg, I found this great source: it's "Applications of Partial Differential Equations to Some Problems in Geometry", a set of lecture notes by Jerry Kazdan, which are available on his website at UPenn. The proof of Newlander-Nirenberg in here is based on Malgrange's proof, which is also in Nirenberg's book (mentioned by Morris KaLka in his answer above), but these notes by Kazdan cover a lot of basic geometric analysis theorems, so they're an excellent resource. I wish I knew about these when I was in graduate school...

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Hi Spiro, thank you for the reference. By the way, have you get some examples you can do the explicit calculation following there proofs? – Jay Nov 16 at 17:33
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