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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?)

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    $\begingroup$ 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). $\endgroup$
    – Roberto Frigerio
    Oct 4, 2013 at 13:28
  • $\begingroup$ Sure, in fact I should have written `closed, non compact, no boundary'. Thanks for pointing this out. $\endgroup$
    – Pawel Goldstein
    Oct 4, 2013 at 13:56

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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.)

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  • $\begingroup$ The fundamental group of each complementary component will contain a knot group as a summand, so cannot be simply-connected. $\endgroup$
    – Ian Agol
    Oct 4, 2013 at 15:38
  • $\begingroup$ @Ian, yes, that was my original idea, but then I realized that there's a simpler homology argument and that the tubes don't need to be knotted. $\endgroup$
    – Kevin Walker
    Oct 4, 2013 at 16:10
  • $\begingroup$ That is a very nice example, thanks a lot. How about a follow-up question: What if we additionally assume that S is planar, i.e. homeomorphic to an open subset of a plane? I believe that the answer should be YES in that case... $\endgroup$
    – Pawel Goldstein
    Oct 8, 2013 at 9:56
  • $\begingroup$ If I understand you correctly, the surface in my example is planar in your sense. A connected sum of three annuli is homeomorphic to a large disk with five smaller disks removed. $\endgroup$
    – Kevin Walker
    Oct 10, 2013 at 0:25
  • $\begingroup$ Of course you are right, I feel a bit ashamed not to have realized that. Thanks a lot. $\endgroup$
    – Pawel Goldstein
    Oct 10, 2013 at 7:56

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