# How wild can an open topological 3-manifold be if it has a compact quotient?

Let $$M$$ be an open, simply connected, 3-manifold. Suppose $$M$$ admits a properly discontinuous, co-compact topological action by a finitely generated group.

Question 1: If $$M$$ is 1-ended, must it be homeomorphic with $$\mathbb{R}^3$$?

More generally:

Question 2: Is $$M$$ determined up to homeomorphism by its number of ends?

(By a classical result of Hopf, this number is 1, 2, or uncountably infinite, I think.)

More generally, I'm interested in results of the form: if a 3-manifold $$M$$ covers a compact manifold/orbifold, then it cannot be as wild as a generic open 3-manifold such as e.g. the Whitehead manifold.

Edit: Having discussed this with an expert, I believe that my questions above boil down to the following conjecture:

Conjecture: Let $$G$$ be the fundamental group of a closed 3-manifold $$M$$. Then $$G$$ is simply connected at infinity. (Equivalently, the universal cover of $$M$$ is simply connected at infinity.)

This conjecture is implicit in this paper by Funar & Otera: https://www-fourier.ujf-grenoble.fr/~funar/funote.pdf

• I think you need to require that the group action is fixed point free. To show that M is homeomorphic to R^3, the key is to show the end is simply connected. Classical contractible manifolds are known not to cover manifolds nontrivially. See David Wright's paper core.ac.uk/download/pdf/81127239.pdf May 3 at 21:37
• If M is contractible and covers a closed manifold, then M is homeomorphic to R^3. This follows from the geometrization. May 3 at 23:14
• @ShijieGu : a concrete example would be appreciated! May 5 at 7:06
• Does the fact that the conjecture answers the first question use Pardon's answer here: mathoverflow.net/questions/417973/… ?
– HJRW
May 12 at 8:59
• @HJRW yes, I think Pardon’s answer together with the orbifold theorem reduces this to the manifold case by passing to a finite sheeted manifold cover of the orbifold. May 12 at 16:42

The possible universal covers of closed 3-manifolds are $$S^3-C$$, where $$|C|=0, 1, 2$$ or $$C$$ is a tame Cantor set, corresponding to the space of ends of the fundamental group as you suspect. This follows from the geometrization theorem, known to experts but might not be written down. I gave a survey talk on this once, you can find the notes here.
If $$M$$ is a closed orientable connected 3-manifold, and $$\pi_1 M$$ is finite, then its universal cover is $$S^3$$ by the Poincaré conjecture.
If $$\pi_1M$$ is infinite and $$\pi_2 M=0$$, then the universal cover $$\tilde{M} \cong \mathbb{R}^3$$. In this case, $$M$$ has a geometric decomposition. If the decomposition is trivial, then $$M$$ is modeled on one of the six geometries homeomorphic to $$\mathbb{R}^3$$, and hence the universal cover is $$\mathbb{R}^3$$. Otherwise, $$M$$ has an essential torus, hence is a Haken manifold. Waldhausen proved that Haken manifolds have universal cover $$\mathbb{R}^3$$ - see Theorem 8.1.
If $$\pi_1 M$$ is infinite but $$\pi_2 M\neq 0$$, then either $$M$$ is modeled on the $$S^2\times \mathbb{R}$$ geometry and $$M$$ is homeomorphic to $$\mathbb{RP}^3\#\mathbb{RP}^3$$ or $$S^2\times S^1$$. In this case the number of ends of $$\pi_1 M$$ is 2. Otherwise, $$M$$ is a non-trivial connect sum by the sphere theorem and its universal cover is $$S^3-C$$ where $$C$$ is a tame Cantor set. The connect summands have universal cover either $$S^3$$, $$S^2\times \mathbb{R}$$ or $$\mathbb{R}^3$$. When you take connect sums, you remove open balls from each manifold and glue the sphere boundaries together. The universal cover is obtained by gluing the universal covers of each summand punctured along balls, either finitely many in $$S^3$$ or infinitely many in $$S^2\times\mathbb{R}$$ or $$\mathbb{R}^3$$. One can see that such manifolds are built out of thrice punctured spheres, and hence the universal cover can be decomposed into thrice punctures spheres. Such a manifold is homeomorphic to $$S^3-C$$.