Criterion for deciding the conformal class of a metric on a complete surface

For orientable closed Riemannian surfaces $(S,g)$, there is a constant curvature metric $\overline{g}$ on $S$ that is conformal to $g$ in the sense that $\overline{g} = e^ug$ for some smooth function $u$. Moreover, by the Gauss-Bonnet theorem the sign of the curvature of $\overline{g}$ is uniquely determined by the topology of $S$.

However, when $S$ is non-compact, it may admit different kinds of complete metrics of constant curvature. For example, on $\mathbb{R}^2$ you have many flat metrics but you also have the hyperbolic metric coming from some identification of the plane with the unit disk. On the cylinder $C = \mathbb{R}^2 \backslash \mathbb{Z}$, there is the natural flat metric coming from the Euclidean metric but there is also the hyperbolic metric arising from the quotient of the upper half-plane by a horizontal translation.

I am interested in formulating some (geometric) criterion for a metric $g$ on $\mathbb{R}^2$ or on the cylinder $C$ to be conformal to a flat metric. For example, if $(C,g)$ is asymptotic to the product metric $dx^2 + d\theta^2$ with suitable decay assumptions, I was shown how to solve the PDE for a conformal deformation to a flat metric using weighted Sobolev spaces. But for weaker notions of "asymptotic flatness" of $g$, this method wouldn't work although I feel like such a metric should be conformally flat, even for very weak notions of asymptotic flatness...

So, given a metric $g$ on the plane or on the cylinder, how can I differentiate between $g$ being conformally flat or conformally hyperbolic?

• I remember that Milnor wrote a paper: J.Milnor, On deciding whether a surface is parabolic or hyperbolic. Amer. Math. Monthly, 84:43-46, 1977. Mar 28 '16 at 9:38
• I'll turn my answer into a comment, since it didn't seem very useful: non-constant positive harmonic functions can distinguish the two settings. As Robert Bryant kindly pointed out, the cylinder admits both a flat conformal class and the "cusp" hyperbolic conformal class given by the quotient of the upper half-plane by $z\mapsto z+1$. Note that the "$z$ large region can be made exactly flat, so any criterion you use will have to take into account both ends at once. Mar 28 '16 at 9:57
• You can try to use your metric to estimate the conformal modulus (of subcylinders) of your cylinder; it is flat if and only if none of its ends admits a collar of finite conformal modulus. Similarly you can study collars of the end of R^2 (once you removed 3 points). Mar 28 '16 at 10:40

This is a classical problem which is called the Type Problem of a simply connected Riemann surface: If you have a metric on an open simply connected surface, to determine whether it is conformally equivalent to the plane or to the disk.

The general situation is the following: there are necessary and sufficient conditions, but they are usually difficult to verify for specific metrics. And there are very many separate necessary or sufficient conditions which are easier to verify, especially for some special classes of metrics. There was an intensive research on this in 1930-1950, and a few modern papers.

Some references are:

MR2019938 Benjamini, Itai; Merenkov, Sergei; Schramm, Oded, A negative answer to Nevanlinna's type question and a parabolic surface with a lot of negative curvature. Proc. Amer. Math. Soc. 132 (2004), no. 3, 641–647,

MR0954627 Doyle, Peter G. On deciding whether a surface is parabolic or hyperbolic. Geometry of random motion (Ithaca, N.Y., 1987), 41–48, Contemp. Math., 73, Amer. Math. Soc., Providence, RI, 1988.

MR0428232 Milnor, John On deciding whether a surface is parabolic or hyperbolic. Amer. Math. Monthly 84 (1977), no. 1, 43–46.

MR0279280 Nevanlinna, Rolf Analytic functions. New York-Berlin 1970 viii+373 pp.

MR0049330 Volkovyskiĭ, L. I. Investigation of the type problem for a simply connected Riemann surface. (Russian) Trudy Mat. Inst. Steklov. 34, (1950). 171 pp. This is the most comprehensive exposition. Not much was added to this theory since 1950.

A Riemann surface that is topologically a disc is conformally equivalent to the plane if and only if its boundary is a puncture. That is, there is a neighbourhood of the boundary that is conformally equivalent to the punctured disc. Equivalently, and more geometrically, for any compact set K the family of curves separating infinity from K has infinite modulus.

The equivalent version with two punctures works for the punctured plane / cylinder.

(Note that this is essentially an extended version of the comment made by retract-subpolyhedron.)