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Quite a long ago, I tried to work out explicitly the content of the Newlander-Nirenberg theorem. My aim was trying to understand wether a direct proof could work in the simplest possible case, namely that of surfaces. The result is that the most explicit statement I could get is a PDE I was not able to solve.

Assume a quasi-complex structure $J$ is given on the surface $S$; we want to prove that this is induced by a complex one (in this case there are no compatibility conditions). This can be easily transformed in the problem of local existence for a second order PDE, as follows.

We look for local charts on $S$ which are holomorphic (with respect to the quasi-complex structure on $S$). Two such charts are then automatically compatible. So the problem is local.

Fix a small open set $U \subset S$ and identify it with a neighboorhood of $0 \in \mathbb{R}^2$ via a differentiable chart. Locally we can write $J = \left(\begin{matrix}[a | b] \\ [c | d ]\end{matrix}\right)$ for some functions $a, \cdots, d$ (pretend it is a two by two matrix).

A chart is given by a complex valued function $f = u + iv$. The condition that the differential is $\mathbb{C}$-linear can be verified on a basis of the tangent space; moreover if it is true for a vector v, it remains true for Jv, which is linearly independent. Here we have used that $J^2 = -1$.

So we need only to check it for the vector $\partial_x$. Since $J \partial_x = a \partial_x + c \partial_y$, the condition says

$-v_x = a u_x + c u_y$

$u_x = a v_x + c v_y$

Hence we need to solve this system, with $f = u + i v$ non singular ($f$ will be then locally invertible). Since $a$ and $c$ do not vanish simultaneously, we can assume $c(0) \neq 0$, hence $c \neq 0$ on $U$ provided $U$ is small.

We can then solve for $u_y$ and get the equivalent system

$u_x = a v_x + c v_y$

$-u_y = \frac{1 + a^2}{c}v_x +a v_y$

Moreover the Jacobian $J_f = u_x v_y + u_y v_x = \frac{1}{c}(v_x^2 + (a v_x + c v_y)^2)$, so $f$ is nonsigular if $v$ is. By Poincaré's lemma, the system admits a local solution if and only if

$\frac{\partial}{\partial_y} \left( a v_x + c v_y \right) - \frac{\partial}{\partial_x} \left( \frac{1 + a^2}{c}v_x +a v_y \right) = 0$.

Hence we are looking for a local solution of the last equation with $(v_x(0), v_y(0)) \neq (0, 0)$.

So my question is:

Is there a simple way to prove local existence for a nonsingular solution of the last displayed equation?

I should make clear that I'm not looking for a proof of Newlander-Nirenberg; of this there are plenty. I am more interested in seeing what Newlander-Nirenberg becomes in terms of PDE in the simplest possible case, and then see that the PDE thus obtained is solvable. According to the answer of Andy, the equation which comes out is the Beltrami equation, so I will have a look at it. Still, I'm curious if any standard PDE technique can solve the equation I derived in the most stupid way above.

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I could be wrong, but I could have sworn the person's name is spelt Nirenberg. (Sorry to come across like a pedant!) –  Joel Fine Mar 9 '10 at 8:41
    
Thank you, corrected. –  Andrea Ferretti Mar 9 '10 at 12:04
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3 Answers

up vote 2 down vote accepted

I'm not sure if the following is elementary enough, but it does only use standard PDE machinery (plus some basic Riemannian geometry). It's also nice in that it suggests an approach to proving the uniformization theorem (via metrics of constant curvature).

Say you have an almost-complex structure on the unit disk. Your goal is to find a conformal isomorphism of this disk (or at least some neighborhood of the origin) with an open subset of the complex plane. To do it, first choose a metric $g$ on the disk which is compatible with the given complex structure. Let $K$ be the curvature of this metric. If you can find a flat metric $\tilde{g}$ on on the disk which is conformally equivalent to $g$ then you'll be done, since the exponential map with respect to $\tilde{g}$ will be an isometry, hence also a conformal isomorphism.

So, multiply $g$ by an arbitrary positive function $e^f$, and compute the curvature of the new metric. You'll find that it's given by the formula:

$\tilde{K} = e^{-2f}(K - \Delta f) $

where $\Delta$ is the Laplacian with respect to the metric $g$. Setting the left hand side equal to zero, you have reduced to solving the Laplace equation, which can be done locally using standard PDE techniques.

As for a more "direct" approach...

The equation you wrote down should reduce to solving the Laplace equation as well, using the notion of conjugate harmonic functions. However, solving your equation will inevitably be a bit more subtle due to your requirement that the solution have nonvanishing differential at the origin. There is a proof along the lines you're suggesting in Taylor's PDE book, chapter 5, section 11, and I think there's a similar one in Jost's "Postmodern analysis". Basically the idea is to rescale your coordinate system so that the metric is nearly flat, in which case you should have a conformal map that is close to the identity map in a high enough sobolev space, and therefore has a nonvanishing derivative at the origin.

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I don't have time to derive it from your equation (maybe someone else can?), but it turns out that the PDE you need to solve to integrate an almost-complex structure in 2 dimensions is the Beltrami equation. This takes the form $$\frac{\partial f}{\partial \overline{z}} = \mu \frac{\partial f}{\partial z},$$ where $\mu$ is a complex-valued function with $\|\mu\|_{\infty} < 1$. It turns out that this is solvable in some sense even if $\mu$ is only measurable! This latter result is due to Morrey. For $\mu$ real-analytic, the result is due to Gauss. My favorite sources for this PDE include Ahlfors's book "Lectures on Quasi-Conformal Mappings", Lehto and Virtanen's book "Quasiconformal mappings in the plane", and Hubbard's book "Teichmuller Theory". Probably Hubbard's book is the easiest to read -- in it, the desired result is Theorem 4.6.1, and Gauss's proof for $\mu$ real analytic is Proposition 2.6.2 (this proof is pretty slick, but the whole Newlander-Niremberg has a slick proof for real analytic almost complex structures using the Frobenius theorem; I seem to recall that there was a blog posting on the Secret Blogging Seminar about this). I believe that Hubbard also discusses how to use this to integrate an almost complex structure.

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It might be worth adding that on a complex manifold of any dimension, you can describe all the other almost complex structures relative to the given complex structure in terms of Beltrami differentials (which are (0,1) forms with values in the tangent bundle). Then to show that the one corresponding to $\mu$ is integrable you need to find n solutions to the n-dimensional Beltrami equation whose differentials span $T^*$ at a point. Of course, in n-dimensions you'll need the integrability condition to hold. –  Joel Fine Mar 9 '10 at 8:53
    
Thank you very much. I know that the real analytic case of Newlander-Nirenberg is settles with an application of Frobenius. I was more interested in translating explicitly N-N in terms of PDE ans see what would come out; then see directly that the relevant PDE is solvable. I will have a look at Hubbard's book then. –  Andrea Ferretti Mar 9 '10 at 12:00
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Another place to look is the article of Chern called "A simple proof of the existence of isothermal parameters on a surface" (jstor.org/pss/2032933). (I think the link needs jstor access to go past the first page.) –  Joel Fine Mar 9 '10 at 21:18
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To each 2dim manifold $M$ with almost complex structure $J$ you can associate the canonicala bundle $K,$ i.e. the bundle of complex linear 1-forms on $M.$ This bundle is equipped with a natural differential operator (the exterior derivative $d$), which can be viewed as a holomorphic structure: sections $s,t$ lying in the kernel of $d$ are related by a meromorphic function $f$ (with respect to $J$): $s=ft.$ Because of Poincare s lemma, to obtain a Riemann surface you need to know that at each point $p\in M$ you have a non-vanishing holomorphic (w.r.t. $d$) section in a neighbourhood of $p.$ A nice way to prove the existence of such a section is the following: Take any connection $\tilde\nabla$ on $K$ such that its $\bar K$ part satisfies $\frac{1}{2}(\nabla+i* \nabla)=d.$ If $\nabla$ would be flat you get parallel (->holomorphic) non-vanishing sections. Of course this cannot be done globally (on comapct surfaces), but you only need the flatness locally around $p.$ Any other connection having the same $\bar K$-part is obtained by adding sections of $\omega\in\Gamma(K).$ The curvature changes by $F^{\nabla+\omega}=F^\nabla+d\omega.$ By Serre duality (in the complex geometry sense and not in the algebraic geometry sense), the curvature can be made flat locally. In this setup, Serre duality is just the complex geometry counterpart to the Hodge theory. You might find the PDE-theory for this problem in Griffiths-Harris.

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Thank you for your reply, but maybe I should add a bit of context. I'm an algebraic geometer and I know the PDE theory necessary for Hodge theory. I'm explicitly trying to see whether the most elementary approach to N-N for surfaces, i. e. writing out the equation that needs to be solved in R^2, can be carried out. Of course there are a lot of more interesting and less elementary ways to prove the theorem, but what I'm really interested in is solving the displayed equation with standard tools of PDE theory. –  Andrea Ferretti Mar 25 '10 at 12:29
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