Complete non compact riemannian manifolds and cylinders 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
 A: 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.
