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Let M be a complete oriented non compact Riemannian manifold with universal covering space (endowed with covering metric) conformally equivalent to the complex plane (with euclidean metric). Then M is locally conformally equivalent to the complex plane. Now if M is not (globally) conformally equivalent to the complex plane then M has to be conformally equivalent to a Cylinder. Why? This fact is used in the proof of the second part of theorem 3 in the article of Fischer Colbrie -Schoen ' The structure of complete stable minimal surfaces in 3-Manifolds of non negative scalar curvature' Thank you

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    $\begingroup$ A $2$-manifold that is covered by $\mathbb R^2$ is either $\mathbb R^2$, a cylinder, or a torus. Since it is non compact, it can't be a torus, and if it isn't the plane, then it must be a cylinder. $\endgroup$
    – treble
    Commented Jan 21, 2013 at 15:06

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You can answer this by classifying the subgroups of $\operatorname{Aut}(\mathbb{C})$ that act freely and properly discontinuously on $\mathbb{C}$. Alternatively, you can use uniformization and treble's comment. I'll sketch out the subgroup classification.

Conformal automorphisms of $\mathbb{C}$ are Möbius transformations that fix $\infty$, rotations and translations. Since $\pi_1M$ acts freely on $\mathbb{C}$, it must act by translations and not rotations, so identify it with a discrete subgroup $\pi$ of $\mathbb{R}^2$.

Argue that this subgroup can have no more than two generators, otherwise it wouldn't be discrete in $\mathbb{R}^2$. (If $x,y,z$ generate the group, and $z = ax + by$, then you can find elements of $\mathbb{Z}x\oplus\mathbb{Z}y$ arbitrarily close to elements of $\mathbb{Z}z$.)

So now $\pi$ is either $\{0\}$, $\mathbb{Z}x$, or $\mathbb{Z}x+\mathbb{Z}y$ (for some linearly independent $x,y\in\mathbb{R}^2-\{0\}$). You have ruled out $\{0\}$ (since $M$ is not isomorphic to $\mathbb{C}$, and $\mathbb{Z}x+\mathbb{Z}y$ (since $M$ is not compact), so you're left with $\pi = \mathbb{Z}x$. The quotient $\mathbb{C}/\pi$ is a cylinder.

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