Non-compact surfaces with non-negative Gauss curvature

Question

Is there a topological classification of non-compact complete connected 2-dimensional Riemannian manifolds with non-negative Gauss curvature?

Answer 1

Here are two arguments.

1. Suppose that $M$ is a noncompact complete connected Riemannian $n$-manifold with finitely generated fundamental group and Ricci curvature $\ge 0$. By the Bishop-Gromov inequality, volumes of metric balls in $M$ satisfy $$ Vol B(x,R)\le \omega_n R^n, $$ where $\omega_n$ is the volume of the unit ball in $\mathbb R^n$. Thus, $M$ has (sub)polynomial volume growth. From this, one concludes that the fundamental group $\Gamma$ of $M$ also has (sub)polynomial growth: $$ \# \{\gamma\in \Gamma: |\gamma|\le R\}\le C R^n $$ for some constant $C$. Here $|\gamma|$ is the word-length of $\gamma$ with respect to a fixed finite generating set of $\Gamma$. Now, by Gromov's polynomial growth theorem it follows that $\Gamma$ is virtually nilpotent (i.e. contains a nilpotent subgroup of finite index). If $M$ is 2-dimensional, it follows that $\pi_1(M)$ is abelian (actually, cyclic) and, thus, $M$ is diffeomorphic to the plane or to the annulus or to the Moebius band. Lastly, if $M$ is a surface whose fundamental group is locally cyclic (i.e. every finitely generated subgroup is cyclic), then $\pi_1(M)$ itself is cyclic. This follows from the fact that fundamental groups of noncompact surfaces are free.

2. Let us prove that if $M$ is a complete connected noncompact 2-dimensional Riemannian manifold of curvature $\ge 0$, then either $M$ is simply connected or the metric on $M$ is flat. In the latter case, $M$ is isometric to a cylinder or to a flat Moebius band.

Indeed, assume that $M$ is not simply connected. Take a nontrivial element $\gamma\in \pi_1(M)$; then $\gamma$ has infinite order (since $M$ is assumed to be noncompact, which implies that its fundamental group is free). Take a Riemannian covering $p: N\to M$ such that $p_*(\pi_1(N))=\langle \gamma\rangle < \pi_1(M)$. It suffices to prove that $N$ is flat. Without loss of generality, we may assume that $N$ is orientable (otherwise, take its orientation covering space). Thus, $N$ is diffeomorphic to $S^1\times \mathbb R$. Now, take two sequences $x_n, y_n\in N$ diverging to different ends of $N$. Specifically, we can take $x_n=(1, n), y_n=(1, -n)$, $n\in \mathbb N$. Let $x_ny_n\subset N$ denote a minimizing geodesic connecting $x_n$ to $y_n$; I will parameterize this geodesic with unit speed, $c_n: [a_n, b_n]\to x_n y_n$, $a_n< 0< b_n$, $c_n(0)\in S^1\times \{0\}$. Clearly, $\lim a_n=-\infty, \lim b_n=\infty$. Then, after passing to a subsequence, the sequence of geodesics $c_n$ converges to a complete minimizing geodesic $c: \mathbb R\to N$. Next, I will apply the Cheeger-Gromoll splitting theorem: If a complete connected Riemannian manifold of Ricci curvature $\ge 0$ contains a line (a complete minimizing geodesic), then it splits isometrically as a product of $\mathbb R$ and another Riemannian manifold. In our case, this means that $N$ is isometric to the product of a circle and the real line, i.e. is flat. Thus, the original manifold $M$ was flat as well.

Now, it follows from the classification of complete flat 2-dimensional Riemannian manifolds that every such manifold $M$ is either a cyclinder or a Moebius band, or the plane.

I am sure, there are other arguments as well, these were the first two which came to mind. The result itself is probably due to Cohn-Vossen:

Cohn-Vossen, S., Kürzeste Wege und Totalkrümmung auf Flächen., Compositio math. 2, 69-133 (1935). ZBL61.0789.01.

Remark. Milnor conjectured that a complete Riemannian manifold of Ricci curvature $\ge 0$ has finitely generated fundamental group. This was recently disproven in

Bruè, Elia; Naber, Aaron; Semola, Daniele, "Fundamental Groups and the Milnor Conjecture". arXiv:2303.15347 (2023)

Answer 2

Here is a version using a bit less group theory:

Let $S$ be a complete surface with nonnegative Gauss (equivalently, sectional) curvature. By Perelman's proof of the soul conjecture (probably there are direct ways to prove this in dimension 2), if $S$ has any point with positive curvature, $S$ is diffeomorphic to a vector space, i.e. diffeomorphic to $\mathbb{R}^2$. Otherwise, $S$ is complete, noncompact, and flat. In particular, the universal cover of $S$ is flat $\mathbb{R}^2$, with fundamental group acting freely by isometries.

The isometries of the flat plane are well known, generated for example by translations and reflections. The only isometries of the plane that act without fixed points are translations and translations composed with a reflection orthogonal to the translating direction.

One may check that having two translations that are not colinear will lead to a compact quotient, as will having two (translations +) reflections across distinct parallel lines. Finally, the distances of translation must have a greatest common divisor (in order for the quotient to be a manifold of dimension 2). This tells us that the fundamental group has a single generator, given by a translation, perhaps followed by reflection across, and $S$ is either a flat cylinder or Möbius band, accordingly.

In total, $S$ is either diffeomorphic to a nonnegatively curved metric on $\mathbb{R}^2$ or isometric to a flat cylinder or Möbius band.