# 3-manifolds bounded by a non-compact Riemann surface of genus 0

Suppose $S$ is a non-compact Riemann surface in $\mathbb R^3$ that has no boundary and has genus zero (i.e. its fundamental group is generated only by its ends at infinity). A typical example would be a boundary of a tubular neigborhood of a tree, with all leaves on the sphere at infinity.

Is it true that one of the 3-manifolds into which $S$ divides $\mathbb R^3$ is homeomorphic to $\mathbb R^3$? (The answer is clear for the example given above, but how is it in general?)

• I think you should only consider connected surfaces that are properly embedded in $\mathbb{R}^3$, because otherwise $\mathbb{R}^3\setminus S$ may be connected, and not homeomorphic to $\mathbb{R}^3$ (e.g. when $S$ is the standard sphere with some disjoint closed disks removed). Oct 4, 2013 at 13:28
• Sure, in fact I should have written `closed, non compact, no boundary'. Thanks for pointing this out. Oct 4, 2013 at 13:56

No, it's not true. Here's a counterexample. Consider three concentric infinite cylinders (properly) embedded in the obvious way in $\mathbb R^3$. Join cylinders 1 and 2 with a knotted tube, and similarly join cylinders 2 and 3 with a knotted tube. The surface is now a connected sum of three annuli, and hence (connected) genus zero. A simple homology calculation shows that neither of the two complementary regions is a 3-ball.
Oops. Since the example only depends on a homology calculation, it doesn't matter whether or not the tubes are knotted -- the knots are an unnecessary distraction. (I got distracted by the standard examples of closed surfaces in $S^3$ which don't bound handle-bodies on either side.)