# Simply connected noncompact complete 2-dimensional submanifold of $\mathbb{R}^3$

Can a simply connected noncompact complete 2-dimensional submanifold of $\mathbb{R}^3$ (endowed with the metric induced by the flat metric of $\mathbb{R}^3$ ) be conformally equivalent to the hyperbolic plane?

From the uniformization theorem I know that it must be conformmally equivalent either to the euclidean plane or to the hyperbolic plane. But can I exclude the second case?

• To be clear, you mean an open simply-connected subset $\Omega \subset \mathbb R^2$ equipped with a complete Riemannian metric? Why is the half-plane model of the hyperbolic plane not an example? – Anthony Carapetis Dec 11 '17 at 12:43
• Sorry! I meant $\R^3$!! I will edit the question! Thanks for the comment! – Onil90 Dec 11 '17 at 12:57
• In fact the hyperboloid is asymptotic to the cone $x^2+y^2=z^2$, from which it follows that the restricted Euclidean metric on the hyperboloid is asymptotically flat, meaning that its Gaussian curvature converges uniformly to $0$ as one approaches infinity. I am pretty sure (but not entirely sure) that an asymptotically flat complete Riemannian metric on a surface diffeomorphic to the plane must be conformally Euclidean. – Lee Mosher Dec 12 '17 at 12:45
• And, by the way, the question should be rewritten to rule out counterexamples such as the simply connected, complete, compact 2-dimensional submanifold $S^2$. – Lee Mosher Dec 12 '17 at 12:46
• @LeeMosher: Asymptotically flat surfaces can be conformal to the hyperbolic plane. See my answer. – Moishe Kohan Dec 13 '17 at 15:11

This question (going back to Loewner) was studied by various people, the first example, I think, is due to Robert Ossermann

R. Osserman, A hyperbolic surface in 3-space. Proc. Amer. Math. Soc. 7 (1956), 54–58.

He constructs such surfaces as graphs of certain functions $f: {\mathbb R}^2\to {\mathbb R}$. More examples were constructed in:

H. Huber Riemannsche Flächen von hyperbolischem Typus im euklidischen Raum. Math. Ann. 139 (1959) 140–146.

R.Jenkins, On hyperbolic surfaces in three-dimensional Euclidean space. Michigan Math. J. 8 (1961) 1–5.

and, presumably, other papers.

There is a related question of the conformal type of a complete simply-connected surface of negative curvature which is not bounded away from zero. Such surface can be either of hyperbolic or of parabolic type, it depends on the rate of decay of curvature at infinity. For instance, for rotationally-symmetric metrics, i.e. metrics of the form $$dr^2+g(r)^2 d\theta^2$$ the metric is conformal to the complex plane if and only if $$\int_1^\infty g^{-1}=\infty.$$

See:

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

Milnor further proves that if (for all large $r$) the curvature $K(r)=K(r,\theta)$ satisfies $$K(r)\ge −\frac{1}{r^2 \log(r)}$$ then the metric is conformal to the complex plane, while if $g$ is unbounded and (for large $r$) $$K(r)\le −\frac{1+\epsilon}{r^2\log(r)}$$ then the metric is conformal to the open unit disk.

Milnor's criterion was extended to arbitrary metrics by Peter Doyle in

P.Doyle, On deciding whether a surface is parabolic or hyperbolic, Geometry of random motion, 41–48, Contemp. Math., 73, Amer. Math. Soc., Providence, RI, 1988.

He proved that a metric $dr^2 + g^2(r,\theta)d\theta^2$ on the plane (in the polar coordinates) is conformal to the unit disk if and only if

$$\int_{0}^{2\pi} \frac{d\theta}{\int_1^\infty \frac{dr}{g}}<\infty.$$

Edit. In view of Milnor's theorem, an explicit example of a complete surface in $E^3$ conformal to the unit disk can be obtained as follows. Let $f(r), r\ge 0$, be a nonnegative function which near zero equals $0$ and for $r\ge 1$ equals $r^\beta$, $\beta> 2$. Consider the surface of revolution $S\subset E^3$ which is the graph of the function $$(r,\theta)\mapsto z=f(r)$$ (I am using cylindrical coordinates $(r,\theta, z)$ on ${\mathbb R}^3$.) Then the surface $S$ is complete and conformal to the open unit disk.