# Can we prove uniformization by solving the Yamabe problem directly?

One version of the uniformization theorem says that a simply connected complex manifold of (complex) dimension one is biholomorphic to either the unit disc, $\Bbb C$, or $\Bbb{CP}^1$. The proof of this goes through potential theory, ultimately using analysis of subharmonic functions to obtain a biholomorphic map to an open subset of $\Bbb{CP}^1$. Somehow this always seemed like magic to me, so I'm always looking for approaches I find more straightforwardly comprehensible.

On the other hand, because of the relationship between conformal and complex geometry in (real) dimension 2, this is equivalent to solving the Yamabe problem (which was originally stated as a generalization of uniformization): does every conformal class of metrics contain one of constant scalar curvature?

Let's quickly recall the formula for the change of scalar curvature in dimension 2 under conformal change of metric:

$$S(e^f g) = e^{-f}(S(g)+\Delta_g f).$$

(Note that here we are using the $g$-Laplacian of $f$.) Fix a metric on either $\Bbb R^2$ or $S^2$. Can we directly solve this equation for the appropriate choice of constant on the LHS, leading to a more Riemannian-flavored solution to the uniformization theorem? (If we can do this without first classifying simply connected surfaces, all the better.) I would also be satisfied with a solution coming from Ricci/Yamabe flow type ideas.

We approach the problem of uniformization of general Riemann surfaces through consideration of the curvature equation, and in particular the problem of constructing Poincaré metrics (i.e., complete metrics of constant negative curvature) by solving the equation $Δu-e^{2u}=K_0(z)$ on general open surfaces. A few other topics are discussed, including boundary behavior of the conformal factor $e^{2u}$ giving the Poincare metric when the Riemann surface has smoothly bounded compact closure, and also a curvature equation proof of Koebe's disk theorem.