Uniformization theorem for $C^k$ surfaces? Does the uniformization theorem apply for surfaces that are $C^k$ ($k<\infty$)? I'm familiar with a couple of proofs of Uniformization (using Riemann-Roch, Ricci flow). But most of these proofs assume that the surface is smooth i.e. $C^\infty$ or even analytic. I was wondering if the theorem still works if the surface is only $C^k$ for some $1\leq k<\infty$. I have tried looking for a reference but was unsuccessful.
In particular, I would like to show that a compact, orientable $C^1$ genus-$0$ surface is conformal to $S^2$.
One possible route would be to use one of the several approximation/density theorems for $C^\infty$ manifolds in the space of $C^k$ manifolds.
 A: What you need is Gauss' theorem on isothermal coordinates in the low-regularity setting.  It is known to analysts as Morrey's theorem and proofs can be found in many places, for instance:
Ahlfors, Lars V., Lectures on quasiconformal mappings, Princeton, N.J.-Toronto-New York-London: D. Van Nostrand Company. Inc. 146 p. (1966). ZBL0138.06002.
Lehto, Olli, Univalent functions and Teichmüller spaces, Graduate Texts in Mathematics, 109. New York etc.: Springer-Verlag. XII, 257 p. DM 124.00 (1987). ZBL0606.30001.
Suppose that $ds^2= Edx^2 + F dxdy + Gdy^2$ is a measurable Riemannian metric on a bounded domain $U$ in the xy-plane, which is regarded as the complex plane. The Beltrami differential of this metric is defined as
$$
\mu= \frac{E - G + 2iF}{E + G + 2\sqrt{EG - F^2}}. 
$$
Suppose, in addition, that $||\mu||_\infty<1$ in $U$, which is true, for instance, if $ds^2$ is the restriction  of a continuous Riemannian metric on a larger domain $\Omega$ such that $cl(U)$ is a compact in $\Omega$. Then the measurable Riemann mapping theorem, first proven independently by Lavrentiev and Morrey in 1930s states that there exists an essentially unique (unique, up to postcomposition with a conformal automorphism of the unit disk) quasiconformal homeomorphism $f=f_\mu: U\to \Delta$ ($\Delta$ is the unit disk), which is a conformal isomorphism between $(U, ds^2)$ and the Euclidean metric on $\Delta$. I will not define conformality here, suffices to say that it preserves angles between tangent vectors a.e. in $U$ and preserves orientation.
Quasiconformal maps belong to the Sobolev class $W^{1,2}_{loc}$. If you want higher regularity, you can use for instance
Lehto, Olli, On the differentiability of quasi-conformal mappings with prescribed complex dilatation, Ann. Acad. Sci. Fenn., Ser. A I 275, 28 p. (1960). ZBL0090.05301.
which implies that if $\mu$ is locally Lipschitz-continuous, then $f_\mu$ is a $C^1$-diffeomorphism, hence, is conformal in the traditional sense of Riemannian geometry.
For surfaces this can be used as follows. Start with, say, a $C^1$-smooth surface $S$. (You can as well assume $C^\infty$, since there is always a compatible $C^\infty$ atlas.) Let $ds^2$ be, say, a Lipchitz-continuous Riemannian metric on $S$. (A $C^1$-smooth metric will, of course, suffice.) What's important here is the regularity of the metric, not the regularity of a surface.
Now, using  Lehto's theorem, find a compatible atlas consisting of $C^1$-smooth quasiconformal maps to domains in ${\mathbb C}$ which map conformally $ds^2$ to the Euclidean metric. The transition maps will be conformal and, hence, you get a Riemann surface structure on $S$. By the Uniformization theorem for Riemann surfaces, you get a conformal isomorphism from the universal cover of $S$ to, say, the unit disk. With respect to the original smooth structure on $S$, this map will be a $C^1$-diffeomorphism.  In particular, $(S,ds^2)$ will be conformally-isomorphic to a hyperbolic metric on $S$, where the conformal isomorphism is a $C^1$-diffeomorphism.  The same proof works for other conformal types: Euclidean and spherical.
