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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 a complex 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.

2 deleted 7 characters in body

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 a 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 we you can find a flat metric $\tilde{g}$ on on the disk which is conformally equivalent to $g$ then we'll 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 elliptic 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.

1

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 a 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 we can find a flat metric $\tilde{g}$ on on the disk which is conformally equivalent to $g$ then we'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 elliptic 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.