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Usually, when we have a complete noncompact manifold, we can look at its universal cover so that many geometric property can be lifted to a simply-connected manifold. If we take a large compact subset of this complete noncompact manifold such that the complement has only one component with compact boundary (probably many components). What is the universal covering of this complement? Is it subset of the original universal cover, what could it look like? If we forget it as a complement and just consider a noncompact manifold with compact boundary, what can we say about its universal covering?

My question is not that specific. I am not looking for a definite answer, in stead, I would like to see some examples. I have found some questions

Universal covering of boundary It asks if the lift of the compact boundary is a universal covering of the boundary. In general it is just covering. There are some conclusions in the response: simply-connected 3-manifolds must have simply connected boundary. This is only true for compact manifolds, right?

and

Universal covering of surface

I would like to know more since I am considering noncompact and higher dimensional objects. For simplicity, we can discuss 3-dimensional ones.

Edit: One more question to discuss: If a complete manifold has n ends (number of disconnected components outside a large compact subset), what is the number of the end of its universal covering? Increasing or decreasing or possibly infinite?

1, R^3 still has one end. 2, Catenoid in R^4 has two ends, the universal covering has one end. more examples?

STUDENT
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  • I suggest you work out some examples, e.g. when your manifold is $R^2$ and you remove from it a closed disk or several disjoint closed disks. – Moishe Kohan Mar 23 '24 at 14:44
  • @MoisheKohan, I think it's still the plane. Two dimensional case seems trivial. – STUDENT Mar 25 '24 at 04:02
  • I see, you are only interested in the covering space itself, not the covering map. Then for 3-dimensional manifolds the answer is that the universal covering space of every compact 3-dimensional manifold (possibly with boundary) embeds in $S^3$. – Moishe Kohan Mar 25 '24 at 23:44
  • Yes, I am more concerned about it's shape or topology. How about noncompact ones? – STUDENT Mar 26 '24 at 09:04
  • For tame noncompact manifolds the story is the same. As for non-tame manifolds, I do not think enough is known. Why do you care about these? – Moishe Kohan Mar 26 '24 at 09:36
  • I don't know too much about "tame or nontame". What I studied is about parabolicity or nonparabolicity of noncompact complete Riemannian manifold. These definitions are actually defined for ends and they somehow describe the shape or geometry of the ends. As I asked above, if we directly consider its universal covering, we get a simply connected noncompact complete Riemannian manifold. So I am curious about the universal cover of one end. The end is just a noncompact manifold with compact boundary. What does compact boundary components become in its universal covering? – STUDENT Mar 26 '24 at 15:43

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